SCHAUM’S OUTLINE OF
THEORY AND PROBLEMS OF
ORGANIC CHEMISTRY Second Edition
HERBERT MEISLICH, Ph.D. Professor Emeritus of Chemistry City College of CUNY
HOWARD NECHAMKIN, Ed.D. Professor Emeritus of Chemistry Trenton State College and
JACOB SHAREFKIN, Ph.D. Professor Emeritus of Chemistry Brooklyn College of CUNY
SCHAUM’S OUTLINE SERIES McGRAW-HILL
New York St. Louis San Francisco Auckland Bogoti Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto
HERBERT MEISLICH holds a B.A. degree from Brooklyn College and an M.A. and Ph.D. from Columbia University. He is a professor emeritus from the City College of CUNY, where he taught Organic and General Chemistry for forty years at both the undergraduate and doctoral levels. He received the Outstanding Teacher award in 1985, and has coauthored eight textbooks, three laboratory manuals in General and Organic Chemistry and 15 papers on his research interests. HOWARD NECHAMKIN is Professor Emeritus of Chemistry at Trenton State College; for 11 years of his tenure he served as Department Chairman. His Bachelor’s degree is from Brooklyn College, his Master’s from the Polytechnic Institute of Brooklyn and his Doctorate in Science Education from New York University. He is the author or coauthor of 53 papers and 6 books in the areas of inorganic, analytical, and environmental chemistry. JACOB SHAREFKIN is Professor Emeritus of Chemistry at Brooklyn College. After receiving a B.S. from City College of New York, he was awarded an M.A. from Columbia University and a Ph.D. from New York University. His publications and research interest in Qualitative Organic Analysis and organic boron and iodine compounds have been supported by grants from the American Chemical Society, for whom he has also designed national examinations in Organic Chemistry.
Schaum’s Outline of Theory and Problems of ORGANIC CHEMISTRY Copyright 0 1991,1977 by The McGraw-Hill Companies Inc. All rights reserved. Printed in the United States of America. Except as permitted under the Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 8 9 10 1 1 12 13 14 15 16 17 18 19 20 PRS PRS 9 8 7
ISBN 0-07-041458-0 Sponsoring Editor: David Beckwith Production Supervisor: Fred Schulte Editing Supervisors: Patty Andrews, Meg Tobin, Maureen Walker
Library of Congress Cataloging-in-PublicationData
Meislich. Herbert. Schaum’s outline of theory and problems of organic chemistry / Herbert Meislich, Howard Nechamkin, and Jacob Sharefkin. - - 2nd ed. cm. - - (Schaum’s outline series) p. Includes index. ISBN 0-07-041458-0 1. Chemistry, Organic- -Problems, exercises, etc. 2. Chemistry, Organic- -Outlines, syllabi, etc. I . Nechamkin, Howard, 1 1 . Sharefkin. Jacob. 111. Title. QD257.M44 1991 547- -dc20 90-6383 CI P
McGraw -Hill A Division of The McGmwHiU Companies
gz
To Amy Nechamkin, Belle D. Sharefkin, and John B. Sharefkin
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The beginning student in Organic Chemistry is often overwhelmed by facts, concepts, and new language. Each year, textbooks of Organic Chemistry grow in quantity of subject matter and in level of sophistication. This Schaum’s Outline was undertaken to give a clear view of first-year Organic Chemistry through the careful detailed solution of illustrative problems. Such problems make up over 80% of the book, the remainder being a concise presentation of the material. Our goal is for students to learn by thinking and solving problems rather than by merely being told. This book can be used in support of a standard text, as a supplement to a good set of lecture notes, as a review for taking professional examinations, and as a vehicle for self-instruction. The second edition has been reorganized by combining chapters to emphasize the similarities of functional groups and reaction types as well as the differences. Thus, polynuclear hydrocarbons are combined with benzene and aromaticity. Nucleophilic aromatic displacement is merged with aromatic substitution. Sulfonic acids are in the same chapter with carboxylic acids and their derivatives, and carbanion condensations are in a separate new chapter. Sulfur compounds are discussed with their oxygen analogs. This edition has also been brought up to date by including solvent effects, CMR spectroscopy, an elaboration of polymer chemistry, and newer concepts of stereochemistry, among other material. Our thanks and appreciation to Mr. David Beckwith and Maureen Walker for their careful and thorough proofreading. HERBERT MEISLICH HOWARD NECHAMKIN JACOBSHAREFKIN
V
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Contents Chapter
Chapter
Chapter
Chapter
1
2
3
4
STRUCTURE AND PROPERTIES. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Carbon Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Lewis Structural Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Types of Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Functional Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Formal Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1 2 5 6 6
BONDING AND MOLECULAR STRUCTURE . . . . . . . . . . . . . . . . . . . . 2.1 Atomic Orbitals., . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Covalent Bond Formation-Molecular Orbital (MO) Method . . . . . . . . . . . 2.3 Hybridization of Atomic Orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Electronegativity and Polarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Oxidation Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Intermolecular Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Solvents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Resonance and Delocalized 7~ Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
CHEMICAL REACTIVITY AND ORGANIC REACTIONS . . . . . . . . . . 3.1 Reaction Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Carbon-Containing Intermediates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Types of Organic Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Electrophilic and Nucleophilic Reagents . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Bond-Dissociation Energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Chemical Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Rates of Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Transition-State Theory and Enthalpy Diagrams . . . . . . . . . . . . . . . . . . . . . 3.10 Bronsted Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11 Basicity (Acidity) and Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.12 Lewis Acids and Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
ALKANES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
12 13 16 19 20 20 21 22
30 30 32 33 34 35 36 37 38 40 41 43
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nomenclature of Alkanes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation of Alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties of Alkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Alkane Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49 54 55 57 61
STEREOCHEMISTRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Stereoisomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Optical Isomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
4.1 4.2 4.3 4.4 4.5
Chapter
1
vii
68 69
...
CONTENTS
Vlll
5.3 Relative and Absolute Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Molecules with More Than One Chiral Center . . . . . . . . . . . . . . . . . . . . . . 5.5 Synthesis and Optical Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 76 78
ALKENES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Nomenclature and Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Geometric (cis-tram) Isomerism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Preparation of Alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Chemical Properties of Alkenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Substitution Reactions at the Allylic Position . . . . . . . . . . . . . . . . . . . . . . . 6.6 Summary of Alkene Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
5.4
Chapter
Chapter
Chapter
6
7
8
ALKYL HALIDES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Synthesis of RX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Chemical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Summary of Alkyl Halide Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
119
ALKYNES AND DIENES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
142
Chapter
9
10
119 120 121 132
Alkynes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties of Acetylenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Alkadienes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . MO Theory and Delocalized T Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . Addition Reactions of Conjugated Dienes . . . . . . . . . . . . . . . . . . . . . . . . . . Polymerization of Dienes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cycloaddition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Alkyne Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Diene Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
142 145 148 149 152 155 156 157 157
CYCLIC HYDROCARBONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Nomenclature and Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Geometric Isomerism and Chirality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Conformations of Cycloalkanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 MO Theory of Pericyclic Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Terpenes and the Isoprene Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
166
8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9
Chapter
87 88 91 96 107 108
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Aromaticity and Huckel's Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 Antiaromaticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Polynuclear Aromatic Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
166 167 170 179 181 184 188
198 198 202 203 206
CONTENTS
10.5 Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter
11
12
13
207 209 212
.
AROMATIC SUBSTITUTION ARENES . . . . . . . . . . . . . . . . . . . . . . . . 11.1 11.2 11.3 11.4 11.5
Chapter
ix
215’ Aromatic Substitution by Electrophiles (Lewis Acids. E’ or E) . . . . . . . 215 Electrophilic Substitutions in Syntheses of Benzene Derivatives . . . . . . . . 224 Nucleophilic Aromatic Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Arenes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 Summary of Arene and Aryl Halide Chemistry . . . . . . . . . . . . . . . . . . . . . 234
SPECTROSCOPY AND STRUCTURE . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Ultraviolet and Visible Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Infrared Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 Nuclear Magnetic Resonance (Proton. PMR) . . . . . . . . . . . . . . . . . . . . . . . 12.5 ‘ T N M R ( C M R ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 Mass Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
242 242 243 244 248 257 259
Chapter
ALCOHOLS AND THIOLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A . Alcohols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Nomenclature and H-Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 Summary of Alcohol Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B. Thiols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 Summary of Thiol Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
269 269 269 270 274 279 280 280 281
Chapter
14
ETHERS. EPOXIDES. GLYCOLS. AND THIOETHERS. . . . . . . . . . . . 292 A . Ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 292 14.1 Introduction and Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 14.3 Chemical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 298 14.4 Cyclic Ethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 14.5 Summary of Ether Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B. Epoxides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 14.6 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 14.7 Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 301 14.8 Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 14.9 Summary of Epoxide Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C . Glycols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 304 14.10 Preparation of 1.2.Glycols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.11 Unique Reactions of Glycols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 14.12Summary of Glycol Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306
CONTENTS
X
D. Thioethers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.13 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.14 Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.15 Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter
15
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES . . . . . . . . 315 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9
Introduction and Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation . . . ......................................... Oxidation and Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Addition Reactions of Nucleophiles to ‘C=O . . . . . . . . . . . . . . . . . . . . . Addition of Alcohols: Acetal and Ketahormation . . . . . . . . . . . . . . . . . . Attack by Ylides; Wittig Reaction . . . . ........................ Miscellaneous Reactions ........................ Summary of Aldehyde Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Ketone Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ~
Chapter
16
17
~~
Introduction and Nomenclature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation of Carboxylic Acids . . ............................ Reactions of Carboxylic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Carboxylic Acid Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . Polyfunctional Carboxylic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....... Transacylation; Interconversion of Acid Derivatives . . . . . More Chemistry of Acid Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Carboxylic Acid Derivative Chemistry . . . . . . . . . . . . . . . . . Analytical Detection of Acids and Derivatives . . . . . . . . . . . . . . . . . . . . . Carbonic Acid Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Carbonic-Acid-Derivative Chemistry ....................... Synthetic Condensation Polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derivatives of Sulfonic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
CARBANION-ENOLATES AND ENOLS . . . . . . . . . . . . . . . . . . . . . . . . . 17.1 Acidity of H’s a to C=O; Tautomerism . . . . . . . . . . . . . . . . . . . . . . . . . .
17.2 Alkylation of Simple Carbanion-Enolates . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Alkylation of Stable Carbanion-Enolates . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4 Nucleophilic Addition to Conjugated Carbonyl Compounds: Michael 3,4-Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.5 Condensations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter
18
315 317 323 325 330 332 334 336 337
CARBOXYLIC ACIDS AND THEIR DERIVATIVES . . . . . . . . . . . . . . . 344 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9 16.10 16.11 16.12 16.13
Chapter
307 307 307 307
344 347 348 354 354 358 360 367 367 369 370 371 372
384 384 389 391 396 397
AMINES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
412
Nomenclature and Physical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reactions of Quaternary Ammonium Salts . . . . . . . . . . . . . . . . . . . . . . . . Ring Reactions of Aromatic Amines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Spectral Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reactions of Aryl Diazonium Salts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.8 Summary of Amine Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
412 414 418 424 425 426 426 429
18.1 18.2 18.3 18.4 18.5 18.6 18.7
CONTENTS
Chapter
19
PHENOLIC COMPOUNDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
440
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Analytical Detection of Phenols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Phenolic Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Phenolic Ethers and Esters . . . . . . . . . . . . . . . . . . . . . . . .
440 441 443 450 451 451
19.1 19.2 19.3 19.4 19.5 19.6
Chapter
Chapter
20
21
AROMATIC HETEROCYCLIC COMPOUNDS . . . . . . . . . . . . . . . . . . . 457 20.1 Five-Membered Aromatic Heterocycles with One Heteroatom . . . . . . . . . 457
20.2 Six.Membered. Heterocycles with One Heteroatom . . . . . . . . . . . . . . . . . 20.3 Compounds with Two Heteroatoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.4 Condensed Ring Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
463 467 468
AMINO ACIDS. PEPTIDES. PROTEINS . . . . . . . . . . . . . . . . . . . . . . . . .
474
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preparation of a-Amino Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties of Amino Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Peptides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary of Amino Acid Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
474 476 478 481 486 488
CARBOHYDRATES AND NUCLEIC ACIDS . . . . . . . . . . . . . . . . . . . . .
494
21.1 21.2 21.3 21.4 21.5 21.6
Chapter
22
xi
22.1 22.2 22.3 22.4 22.5 22.6 22.7
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chemical Properties of Monosaccharides . . . . . . . . . . . . . . . . . . . . . . . . . . Evidence for Hemiacetal Formations as Exemplified with Glucose . . . . . Stereochemistry and Structure Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Disaccharides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Polysaccharides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Nucleic Acids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
494 495 498 500 507 510 511
517
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Chapter 1 Structure and Properties 1.1 CARBON COMPOUNDS
Organic chemistry is the study of carbon (C) compounds, all of which have covalent bonds. Carbon atoms can bond to each other to form open-chain compounds, Fig. l-l(o), or cyclic (ring) compounds, Fig. 1-l(c). Both types can also have branches of C atoms, Fig. 1-l(b) and (d). Saturated compounds have C’s bonded to each other by single bonds, C-C; unsaturated compounds have C’s joined by multiple bonds. Examples with double bonds and triple bonds are shown in Fig. 1-l(e). Cyclic compounds having at least one atom in the ring other than C (a heteroatom) are called heterocyclics, Fig. 1-l(f). The heteroatoms are usually oxygen (0),nitrogen (N), or sulfur (S). H H H H H
I I
l I
H-C-C-C-C-H
l I
1 I
H H H
I I
l l
H H ‘C/ H q \C,H \ / H H
l l
H-C-C-C-H
‘
H C H
H H H H
/I
H H H
H
‘c’
H
‘c’
‘H
/ \C/H
H\
F-
\
H
H
n- Butane
lsobutane
un branched, open-chain
open -chain
Cyclopropane
Met hylcyclopropane
unbranched. cyclic
branched. cyclic
(a)
(b)
(c)
(d)
branched.
H H
I
H, ,C=C,
H
HH
H
I
H-F<-H I I H-C/ \ /C-H H C
H H-CsC-H
I
H-C-C-H
/
\ I
0:
H Ethene (Ethylene)
H
\
Cyclopentene
have double bonds
Ethyne (Acetylene)
Ethylene oxide
has a triple bond
heterocyclic
(e)
(f) Fig. 1-1
Problem 1.1 Why are there so many compounds that contain carbon?
4
Bonds between C’s are covalent and strong, so that C’s can form long chains and rings, both of which may have branches. C’s can bond to almost every element in the periodic table. Also, the number of isomers increases as the organic molecules become more complex. Problem 1.2 How do the boiling points, melting points, and solubilities of covalent organic compounds differ from those of salts? Account for the differences. 4 Covalent organic compounds have much lower boiling and melting points because the forces attracting molecules to each other are weak. The electrostatic forces attracting the oppositely charged ions in inorganic salts are very strong. The forces of attraction between the atoms within a covalent molecule are, however, very strong. Salts are generally soluble in water because water aids in the separation of the ions. Salts are not soluble in organic solvents such as ether and benzene. Most organic compounds are insoluble in water but dissolve in organic solvents. Problem 1.3 Compare the chemical reactivities of covalent organic compounds and salts.
1
4
2
STRUCTURE AND PROPERTIES
[CHAP. 1
Ionic reactions often occur instunfuneously. Reactions between covalent organic molecules proceed more slowly; they often need higher temperatures and/or catalysts. 2. Many reactions of organic compounds give mixrures of products. 3. Organic compounds are less stable with respect to heat. They usually decompose at temperatures above 700 "C. which break many of their covalent bonds. 4. Organic compounds are more susceptible to oxidarion. Hydrocarbons burn in 0, to give CO, and H,O. Inorganic compounds generally remain unchanged even after intense heating. 1.
1.2 LEWIS STRUCTURAL FORMULAS
Molecular formulas merely include the kinds of atoms and the number of each in a molecule (as C,H,, for butane). Structural formulas show the arrangement of atoms in a molecule (see Fig. 1-1). When unshared electrons are included, the latter are called Lewis (electron-dot) structures [see Fig. 1-1( f)).Covalences of the common elements-the numbers of covalent bonds they usually form-are given in Table 1-1; these help us to write Lewis structures. Multicovalent elements such as C, 0 , and N may have multiple bonds, as shown in Table 1-2. In condensed structural formulas all H's and branched groups are written immediately after the C atom to which they are attached. Thus the condensed formula for isobutane [Fig. l-l(b)) is CH,CH(CH,) 2'
Group
1
2
3
4
Lewis Symbol
H.
.Be.
.B.
.c.
Covalence
1
2
3
4
Com pou nds with H
H-H Hydrogen
Beryllium hydride
7
.N.
-0.
'F:
3
2
1
H
H-B-H HI
H-Be-H
6
5
I
Boron hydride*
H<~-H
H-N-HI H
Methane
Ammonia
H+H
H-F:
Water
Hydrogen fluoride
* Exists as B,H,.
Table 1-2. Normal Covalent Bonding Bonding for N
Bonding for C
I 7A=
=c=
a =
-NI
-N=
as in
us in
as in
us in
as in
us in
H
H
I
\~4 / : ..w=ij: HX -H "/ \H
H3-H H Methane
Ethene (Ethylene)
H
Carbon dioxide
Ethyne (Acetylene)
H-N-H HI Ammonia
HUN^: Nitrous acid
Bonding for 0 :N= as in
-4%-
G=
as in
as in
H :N*-H
H&H
+c /
'H
Hydrogen cyanide
Water
Formaldehyde
CHAP. 11
STRUCTURE AND PROPERTIES
3
Problem 1.4 (a) Are the covalences and group numbers (numbers of valence electrons) of the elements in Table 1-1 related? (b) Do all the elements in Table 1-1 attain an octet of valence electrons in their bonded 4 states? (c) Why aren’t Group 1 elements included in Table 1-l? (a) Yes. For the elements in Groups 4 through 7, covalence = 8 - (group number). (b) No. The elements in Groups 4 through 7 do attain the octet, but the elements in Groups 2 and 3 have less than an octet. (The elements in the third and higher periods, such as Si, S, and P, may achieve more than an octet of valence electrons.) (c) They form ionic rather than covalent bonds. (The heavier elements in Groups 2 and 3 also form mainly ionic bonds. In general, as one proceeds down a Group in the Periodic Table, ionic bonding is preferred.)
Most carbon-containing molecules are three-dimensional. In methane, the bonds of C make equal angles of 109.5’ with each other, and each of the four H’s is at a vertex of a regular tetrahedron whose center is occupied by the C atom. The spatial relationship is indicated as in Fig. 1-2(a) (Newman projection) or in Fig. 1-2(6) (“wedge” projection). Except for ethene, which is planar, and ethyne, which is linear, the structures in Fig. 1-1 are all three-dimensional. H”s project toward viewer H h ’ s project away from viewer
projects in back of plane of paper b projects o u t of plane of paper toward reader
Hh
d:
.
H/(
Hf
Hh
Hh (a )
( b1
Fig. 1-2
Organic compounds show a widespread occurrence of isomers, which are compounds having the same molecular formula but different structural formulas, and therefore possessing different properties. This phenomenon of isomerism is exemplified by isobutane and n-butane [Fig. 1-l(a) and (b)]. The number of isomers increases as the number of atoms in the organic molecule increases. Problem 1.5 Write structural and condensed formulas for (a) three isomers with molecular formula C,H,, and ( 6 ) two isomers with molecular formula C,H,. 4 (a) Carbon forms four covalent bonds; hydrogen forms one. The carbons can bond to each other in a chain:
H H H H H H-L-L-L-L4-H
I
I
I
I
(structural formula)
I
H H H H H CH,(CH,),CH, n-Pentane
(condensed formula)
or there can be “branches” (shown circled in Fig. 1-3) on the linear backbone (shown in a rectangle).
H
t
T
H-C-H (CH,),CHCH,CH, Isopentane
Fig. 1-3
H
4
STRUCTURE AND PROPERTIES
[CHAP. 1
( 6 ) We can have a double bond or a ring.
H
CH.C=CH,
H
”I
H
Propene (Propylene)
H H Cyclopropane
Problem 1.6 Write Lewis structures for ( a ) hydrazine, N2H,; (6) phosgene, COCl,; ( c ) nitrous acid, HNO,.
4
In general, first bond the multicovalent atoms to each other and then, to achieve their normal covalences, bond them to the univalent atoms (H, Cl, Br, I, and F). If the number of univalent atoms is insufficient for this purpose, use multiple bonds or form rings. In their bonded state, the second-period elements (C, N , 0, and F) should have eight (an octet) electrons but not more. Furthermore, the number of electrons shown in the Lewis structure should equal the sum of all the valence electrons of the individual atoms in the molecule. Each bond represents a shared pair of electrons. N needs three covalent bonds. and H needs one. Each N is bonded to the other N and to two H’s:
H H
I
t
H-N-N-H C is bonded to 0 and to each C1. To satisfy the tetravalence of C and the divalence of 0, a double bond is placed between C and 0. :O:
E
/ \
:Cl:
:Cl:
The atom with the higher covalence, in this case the N, is usually the more central atom. Therefore, bond each 0 to the N. The H is bonded to one of the 0 atoms and a double bond is placed between the N and the other 0. (Convince yourself that bonding the H to the N would not lead to a viable structure.) H&N=O:
Problem 1.7 Why is none of the following Lewis structures for COCl, correct? (a) :Cl<-==]:
(6) : C l < d l :
( c ) :Cl=&@l:
(d) :ClS-*l:
4
The total number of valence electrons that must appear in the Lewis structure is 24, from (2 X 7](2Cl’s) + 4(C) + 6 ( 0 ) . Structures ( 6 ) and ( c ) can be rejected because they each show only 22 electrons. Furthermore, in ( b ) , 0 has 4 rather than 2 bonds, and, in ( c ) one Cl has 2 bonds. In ( a ) , C and 0 do not have their normal covalences. In (d), 0 has 10 electrons, though it cannot have more than an octet. Problem 1.8 Write structural formulas for (a) C,H,, (6) C,H,.
4
In each case the C’s are bonded to each other, but there are not enough H’s to satisfy the tetravalence of C. With only two C’s, ring structures are impossible; multiple bonds must therefore occur. H H (a)
I
I
H-C=C-H
Ethene
(6) H a - - H
Ethyne
Problem 1.9 Use the Lewis-Langmuir octet rule to write Lewis electron-dot structures for: (a) HCN, (6) 4 CO,, ( c ) CCl, and (d) C,H,O.
CHAP. I ]
STRUCTURE AND PROPERTIES
5
Attach the H to the C. because C has a higher covalence than N . The normal covalences of N and C are met with a triple bond. Thus H-N: is the correct Lewis structure. The C is bonded to each 0 by double bonds to achieve the normal covalences. :o=c=o: :CI:
I
:CI-c-CI:
Each of the four Cl’s is singly bonded to the tetravalent C to give
..
I
:CI: The three multicovalent atoms can be bonded as C 4 - 0 or as C-0-C. If the six H’s are placed so that C and 0 acquire their usual covalences of 4 and 2, respectively, we get two correct Lewis structures (isomers) H H H H I t H-L-ij-&-H H--C-C-FH I I HI H H H Dimethyl ether Ethanol Problem 1.10 Determine the positive or negative charge, if any, on:
H (a)
H H
H ( b ) H--d=O:
HJ-0:
A
I
I
(c) H 4 - C .
:Cl:
H
I
( e ) CXl:
( d ) H-k-&-H
I
I
4
I
:Cl:
H
The charge on a species is numerically equal to the total number of valence electrons of the unbonded atoms, minus the total number of electrons shown (as bonds or dots) in the Lewis structure. The sum of the valence electrons (6 for 0, 4 for C, and 3 for three H‘s) is 13. The electron-dot formula shows 14 electrons. The net charge is 13 - 14 = -1 and the species is the methoxide anion, CH,O-. There is no charge on the formaldehyde molecule, because the 12 electrons in the structure equals the number of valence electrons; i.e., 6 for 0, 4 for C, and 2 for two H’s. This species is neutral, because there are 13 electrons shown in the formula and 13 valence electrons: 8 from two C’s and 5 from five H’s. There are 15 valence electrons: 6 from 0, 5 from N , and 4 from four H’s. The Lewis dot structure shows 14 electrons. It has a charge of 15 - 14 = t 1 and is the hydroxylammonium cation, [H,NOH]’. There are 25 valence electrons, 21 from three Cl’s and 4 from C. The Lewis dot formula shows 26 electrons. It has a charge of 25 - 26 = - 1 and is the trichloromethide anion, :CCI ,.
TYPES OF BONDS Covalent bonds, the mainstays of organic compounds, are formed by the sharing of pairs of electrons. Sharing can occur in two ways: (1)
A- + * B + A : B
(2)
A
+ :B+
acceptor
A : B coordinate covalent
donor
In method ( l ) , each atom brings an electron for the sharing. In method (2), the donor atom (B:) brings both electrons to the “marriage” with the acceptor atom (A); in this case the covalent bond is termed a coordinate covalent bond. Problem 1.11 Each of the following molecules and ions can be thought to arise by coordinate covalent bonding. Write an equation for the formation of each one and indicate the donor and acceptor molecule or ion. 4 (4 N H m BF, (4 (CH,),OMgCI, ( 4 Fe(CO)s
STRUCTURE AND PROPERTIES
6
acceptor
donor
(4
H'
+ :NH,
+NHS
(6)
FIB
+ :F:
-BF,
(c)
:Cl-Mgxl:
+ : & C H ,-( I
(4
Fe
+5
(All N-H
[CHAP. 1
bonds are alike,)
(All B-F
bonds are alike.)
CH ),%MgCI,
CH, : G O : -Fe(CzO:),
Notice that in each of the products there is at least one element that does not have its usual covalence-this is typical of coordinate covalent bonding. Recall that an ionic bond results from a transfer of electrons (M. + A.-, M ' + : A - ) . Although C usually forms covalent bonds, it sometimes forms an ionic bond (see Section 3.2). Other organic ions, such as CH,COO (acetate ion), have charges on heteroatoms. Problem 1.12 Show how the ionic compound Lit F- forms from atoms of Li and F.
4
These elements react to achieve a stable noble-gas electron configuration (NGEC). Li(3) has one electron more than He and loses it. F(9) has one electron less than Ne and therefore accepts the electron from Li. Lie + .F:-Li
1.4
' :F:
(or simply LiF)
FUNCTIONAL GROUPS
Hydrocarbons contain only C and hydrogen (H). H's in hydrocarbons can be replaced by other atoms or groups of atoms. These replacements, called functional groups, are the reactive sites in molecules. The C-to-C double and triple bonds are considered to be functional groups. Some common functional groups are given in Table 1-3. Compounds with the same functional group form a homologous series having similar characteristic chemical properties and often exhibiting a regular gradation in physical properties with increasing molecular weight. Problem 1.13 Methane, CH,; ethane, C,H,; and propane, C,H, are the first three members of the alkane 4 homologous series. By what structural unit does each member differ from its predecessor?
These members differ by a C and two H's; the unit is -CH,-
(a methylene group).
Problem 1.14 ( a ) Write possible Lewis structural formulas for (1) CH,O; (2) CH,O; (3) CH,O,; (4)CH,N; (5) CH,SH. (6) Indicate and name the functional group in each case. 4
The atom with the higher valence is usually the one to which most of the other atoms are bonded.
H ( 6 ) (1)
-OH
alcohol
(2)
)c=o:
aldehyde
0:
/
(3) -C-0-H car box y Iic acid
(4) -NH2
amine
(5) -SH thiol
1.5 FORMAL CHARGE
The formal charge on a covalently bonded atom equals the number of valence electrons of the unbonded atom (the Group number) minus the number of electrons assigned to the atom in its bonded state. The assigned number is one half the number of shared electrons plus the total number of unshared electrons. The sum of all formal charges in a molecule equals the charge on the species. In this outline formal charges and actual ionic charges (e.g., Na') are both indicated by the signs + and -.
CHAP. 11
7
STRUCTURE AND PROPERTIES
Problem 1.15 Determine the formal charge on each atom in the following species: (a) H,NBF,,; ( b ) C H , N H j ; and (c) SO:-.
H
I H-N'-B I H
H atoms F atoms N atom €3 atom
GROUP NUMBER
-
1
-
7 5 3
-
-
:F:
I
I
-F:
..
:F:
UNSHARED ELECTRONS 0 6 0 0
SHARED ELECTRONS
+
+ + + +
- FORMAL -
1
-
CHARGE 0
I
-
0
4
=
+1
4
-
-1
pit-{ +
The sum of all formal charges equals the charge on the species. In this case, the + I on N and the - 1 on B cancel and the species is an uncharged molecule.
H H
C atom N atom H atoms
GROUP NUMBER 4
UNSHARED ELECTRONS 0
5
0
1
+
1/2
+ + +
SHARED ELECTRONS 4
- FORMAL -
4
=
I
J =
Net charge on species =
S atom each 0 atom
GROUP NUMBER 6
UNSHARED ELECTRONS 0
+
6
6
+
+
1/2
SHARED ELECTRONS 4 1
Net charge is + 2
CHARGE 0
+1
+I
- FORMAL =
CHARGE +2
_I=
-1
=
-2
+ 4(-1)
These examples reveal that formal charges appear on an atom that does not have its usual covalence and does not have more than an octet of valence electrons. Formal charges always occur in a molecule or ion that can conceived to be formed as a result of coordinate covalent bonding. Problem 1.16 Show how (a) H,NBF, and ( b ) CH,NHf can be formed from coordinate covalent bonding. Indicate the donor and acceptor, and show the formal charges. 4 donor
acceptor
+ BF,+H,N-BF, t
(U)
H,N:
(h)
CH, NH2 + H ' -[CH,6H,]
Problem 1.17 From Problems 1 . 1 1 and 1.16 generalize about the formal charges on an atom and the role the 4 atom plays as the donor or acceptor site.
[CHAP. 1
STRUCTURE AND PROPERTIES
8
Functional Group
General Formula
I
Table 1-3 Some Common Functional Croups Example
General Name
None
I
H m H --cl
Chloride
CH,CH2CI
-Br
Bromide
CH,Br
--OH
Alcohol
CH,CH,OH
U
Ether
CH,CH,OCH,CH,
I Amine'
-NH,
RNH,
-NR',X-
R,N+X-
Quaternary ammonium
R--C=O
Aldehyde
I
- C 4
0
II
<--OH 0
II
<--OR'
CH,CH,CH ,NH,
II
R-C-OH
Methyl bromide
Ethanol
Ethyl alcohol
Decyl trimethylammonium chloride
CH,CH,CH==O
Propanaf
Propionaldehyde
2-BU tan one
Methyl ethyl ketone
Ethanoic acid
Acetic acid
Ethyl ethanoate
Ethyl acetate
I
I '
CH,CH,C=O
0
Carboxylic acid
II
CH,-C-OH 0
II
CH ,--C--OC,H,
I
I
Ethoxyethane
Decyltrimet hylammonium chloride
,
CH
0
I I
Bromomethane
CH (CH ),N( CH, ) f C1-
R Ketone
I
Chloroethane
Propylamine
H
R<=O
I
Ethyne
1- Aminopropane'
H
I
I
Ethylene
~
1
CHAP. 11
0
0
0
II
I --C-NH,
R<-NH,
0
Amide
II
CH ,--C-NH
II
R-C-4
<-Cl
0
0
II
I1
II
Et hanarnide
Acetamide
Ethanoyl chloride
Acetyl chloride
0
0
II
0
9
STRUCTURE AND PROPERTIES
Acid chloride
It
CH,--C--CI
0
0
II
I1
0
II
R4-O-C-R
Acid anhydride
CH,--C-O-C-CH,
Ethanoic anhydride
Acetic anhydride
-C=N
R N -
Nitrile
CH,(%N
Et ha n e nitrile
Acetonitrile
-NO,
R-NO,
Nitro
CH ,-N
Nitromethane
Nitromethane
Thiol
CH,-SH
Methanethiol
Methyl mercaptan
X-0-C-
0
~~
-SH
~
-
~
S
H
-~
-s-
R-S-R
Thioether (sulfide)
CH,-S
Dimet h yl thioether
Dimethyl sulfide
-s-s
R-S-S-R
Disulfide
CH,--S--S--CH,
Dimethyl disulfide
Dimethyl disulfide
Methanesulfonic acid
Methanesulfonic acid
Dimethyl sulfoxide
Dimethyl sulfoxide
0
0
II
0
II -!+OH II
R-WH
0
0
0
0
0
0
II
-s-
II
I R-S-R
Sulfonic acid
Sulfoxide
II II
CH,-S4H
I
CH,-SXH,
&
Dimethyl sulfone
' The italicized portion indicates the group. * A primary (1") amine; there are also secondary (2"), R,NH, and tertiary (3"). R,N, amines. Another name is propanamine.
10
STRUCTURE AND PROPERTIES
[CHAP. 1
The donor atom experiences an increase in positive charge [the N atom in Problem 1.16(b)] or a decrease in negative charge [F in Problem l.ll(b)]. The acceptor atom gains negative charge [the B atom in Problem I . 16(a)] or loses positive charge [the H ' in Problem 1.16(6)].
Supplementary Problems Problem 1.18 Why are the compounds of carbon covalent rather than ionic?
4
With four valence electrons, it would take too much energy for C to give up or accept four electrons. Therefore carbon shares electrons and forms covalent bonds. Problem 1.19 Classify the following as (i) branched chain, (ii) unbranched chain, (iii) cyclic, (iv) multiple bonded, or (v) heterocyclic:
CH
CH,
CH
1 ,
/\
l 3
Problem 1.20 Refer to a Periodic Chart and predict the covalencies of the following in their hydrogen 4 compounds: ( a ) 0; (b) S; (c) CI; ( d ) C; (e) Si; ( f ) P; (g) Ge; ( h ) Br; (i) N; ( j ) Se. The number of covalent bonds typically formed by an element is 8 minus the Group number. Thus: ( a ) 2; (42; (c) 1; ( 4 4; ( 4 4; ( f )3; ( g ) 4; ( h ) 1; (i) 3; ( j ) 2. Problem 1.21 Which of the following are isomers of 2-hexene, CH,CH=CHCH,CH,CH,? ( a ) CH ,CH,CH=CHCH ,CH
,
(c) CH,CH,CH,CH=CHCH,
(6) CH,=CHCH,CH,CH,CH, (d)
H
\ /
(4
CH,
I
H ,C--C--CH I
H2C CH2
I
H I
,
H C-C-CH I H
1
H , C- CH ,
4
All but (c), which is 2-hexene itself. Problem 1.22 Find the formal charge on each element of
and the net charge on the species (BF,Ar).
A tom each F B Ar
Group Number 7 3 8
:F: :&:B:F: :F:
Unshared $ # Shared Electrons + Electrons 6 1 0 4 6 1
4
# -
-
Formal Charge of Atom 0 -1 +1
0 = net charge
CHAP. I ]
11
STRUCTURE AND PROPERTIES
Problem 1.23 Replacing the OH group of carboxylic acids, RCO(OH), by another functional group gives a broad class of compounds called acid derivatives. Identify the acid derivatives in IUPAC column of Table 1-3. 4 There are four: ethyl ethanoate, ethanamide, ethanoyl chloride, and ethanoic anhydride. Problem 1.24 Write Lewis structures for the isomers of C,H,N. Name the compound and classify the 4 functional group in each case.
The three C's can be attached to the N as part of one, two, or three alkyl groups. The compound is named by naming each alkyl group attached to N, in one word whose suffix is amine. CH
I ,
I
CH,CHNH, Isopropylamine U 1" amine
CH,CH,CH,NH, Propylamine a 1" arnine
CH
H
I '
CH,CH,NCH, Ethylmethylamine a 2" arnine
CH'NCH, Trimet hylamine a 3" amine
Problem 1.25 Write Lewis structures for the nine isomers having the molecular formula C,H,O, in which C, 4 H, and 0 have their usual covalences; name the functional group(s) present in each isomer.
One cannot predict the number of isomers by mere inspection of the molecular formula. A logical method runs as follows. First write the different bonding skeletons for the multivalent atoms, in this case the three C's and the 0. There are three such skeletons: (i) C--C-C-0
(ii) C-4
(iii) C 4 - C
I
0 To attain the covalences of 4 for C and 2 for 0, eight H's are needed. Since the molecular formula has only six H's, a double bond or ring must be introduced onto the skeleton. In ( i ) the double bond can be situated three ways, between either pair of C's or between the C and 0. If the H's are then added, we get three isomers: (1) (2), and (3) In (ii) a double bond can be placed only between adjacent C's to give (4).In (iii), a double bond can be placed between a pair of C's or C and 0 giving (5) and (6) respectively. (1) H,C=CHCH20H
(2) CH,CH==CHOH
(3) CH,CH,CH,CH=O
alkene alcohols( enols)
(4) CH,OCH=CH,
an aldehyde
(5) H,C=CHCH,
an alkene ether
(6) CH'CCH,
I :OH ..
I1
:0: a ketone
an enol
In addition three ring compounds are possible (7) H , C - C H O H \/ CH, a cyclic alcohol
(8) H, - C H C H ,
7/ :0:
(9) H , C - C %
I
I
H,C--CH, heterocyclic erhers
Chapter 2 Bonding and Molecular Structure 2.1
ATOMIC ORBITALS
An atomic orbital (AO) is a region of space about the nucleus in which there is a high probability of finding an electron. An electron has a given energy as designated by ( a ) the principal energy level (quantum number) n related to the size of the orbital; ( 6 ) the sublevel s , p , d , f, or g , related to the shape of the orbital; ( c ) except for the s, each sublevel having some number of equal-energy (degenerate) orbitals differing in their spatial orientation; ( d ) the electron spin, designated or 5. . Table 2-1 shows the distribution and designation of orbitals.
Principal energy level, n
1
2
3
4
Maximum no. of electrons, 2n'
2
8
18
32
Sublevels [n in number]
1s
2s,
2p
3s, 3p, 3d
4s, 4p, 4 4
4f
Maximum electrons per sublevel
2
2,
6
2,
2,
10,
14
Designation of filled orbitals
1s2
22, 2ph
3s2, 3p", 3d"'
4s2, 4ph, 4d'", 4f"
Orbitals per sublevel
1.
1,
1,
1,
3
6,
3,
10
5
6,
3,
5,
7
The s orbital is a sphere around t h e nucleus, as shown in cross section in Fig. 2 - l ( a ) . A p orbital is t w o spherical lobes touching on opposite sides of the nucleus. The three p orbitals are labeled p x , p , , and p , because they are oriented along the x - , y - , and z-axes, respectively [Fig. 2 - 1 ( 6 ) ] .In a p orbital there is no chance of finding an electron at the nucleus-the nucleus is called a node point. Regions of an orbital separated by a node are assigned + and - signs. These signs are not associated with electrical or ionic churges. The s orbital has no node and is usually assigned a +. Three principles are used to distribute electrons in orbitals. "Aufbau" or building-up principle. Orbitals are filled in order of increasing energy: I s , 2 s , 2 p , 3 s , 3p, 4s, 3 d , 4p, 5s, 4d, 5p, 6s, 4f, 5 d , 6 p , etc. 2 . Pauli exclusion principle. No more than two electrons can occupy an orbital and then only if they have opposite spins. 3 . Hund's rule. One electron is placed in each equal-energy orbital so that the electrons have parallel spins, before pairing occurs. (Substances with unpaired electrons are paramagnetic-they are attracted to a magnetic field.) 1.
Problem 2.1
Show the distribution of electrons in the atomic orbitals of ( a ) carbon and (6) oxygen.
4
A dash represents an orbital; a horizontal space between dashes indicates an energy difference. Energy increases from left to right. ( a ) Atomic number o f C is 6.
The t w o 2p electrons are unpaired in each of two p orbitals (Hund's rule).
12
BONDING AND MOLECULAR STRUCTURE
CHAP. 2 )
13
y-axis
(a) s Orbital
PY
P I
(6 ) p Orbitals Fig. 2-1 (b) Atomic number of 0 is 8.
t.1 Is
2.2
T.1 2s
T.1
t
1
2 p , 2p,. 2p:
COVALENT BOND FORMATION-MOLECULAR ORBITAL (MO) METHOD
A covalent bond forms by overlap (fusion) of two AO’s--one from each atom. This overlap produces a new orbital, called a molecular orbital (MO), which embraces both atoms. The interaction of two AO’s can produce two kinds of MO’s. If orbitals with like signs overlap, a bonding MO results which has a high electron density between the atoms and therefore has a lower energy (greater stability) than the individual AO’s. If AO’s of unlike signs overlap, an antibonding MO* results which has a node (site of zero electron density) between the atoms and therefore has a higher energy than the individual AO’s. Asterisk indicates antibonding. Head-to-head overlap of AO’s gives a sigma (U)MO-the bonds are called a bonds, Fig. 2-2(a). The corresponding antibonding MO* is designated U ” , Fig. 2-2(b). The imaginary line joining the nuclei of the bonding atoms is the bond axis, whose length is the bond length. Two parallel p orbitals overlap side-by-side to form a pi ( m ) bond, Fig. 2-3(a), or a m* bond, Fig. 2-3(6). The bond axis lies in a nodal plane (plane of zero electronic density) perpendicular to the cross-sectional plane of the 7~ bond. Single bonds are U bonds. A double bond is one o and one m bond. A triple bond is one (T and two 7~ bonds (a rzand a ry,if the triple bond is taken along the x-axis). Although MO’s encompass the entire molecule, it is best to visualize most of them as being localized between pairs of bonding atoms. This description of bonding is called linear combination of atomic orbitals (LCAO). Problem 2.2 What type of M O results from side-to-side overlap of an s and a p orbital?
BONDING AND M O L E C U L A R S T R U C T U R E
14
and
---
I_*
-
P
S
and P
P
( a ) U Bonding
0
and
P
S
+
and P
P
-
+
U*(PP)
( 6 ) U * Antibonding Fig. 2-2
P Y
[CHAP. 2
PY
(b)
7r*
Antibonding Fig. 2-3
The overlap is depicted in Fig. 2-4. The bonding strength generated from the overlap between the + s A 0 and the + portion of the p orbital is canceled by the antibonding effect generated from overlap between the + s and the - portion of the p . The MO is nonbonding ( 0 ) ; it is no better than two isolated AO's.
Problem 2.3 List the differences between a
U
bond and a
7r
bond.
4
CHAP. 2)
15
BONDING AND MOLECULAR STRUCTURE
Qj
P
Fig. 2-4 1.
s Bond
U Bond Formed by head-to-head overlap of AO’s.
1. Formed by lateral overlap of p orbitals (or p and d orbitals). 2. Has maximum charge density in the cross-sectional plane of the orbitals. 3. N o free rotation. 4. Higher energy. 5. One or two bonds can exist between two atoms.
2. Has cylindrical charge symmetry about bond axis. 3. Has free rotation. 4. Lower energy. 5 . Only one bond can exist between two atoms.
Problem 2.4 Show the electron distribution in MO’s of ( a ) H,, ( b ) H f , (c) H i , (d) He,. Predict which are unstable. 4 Fill the lower-energy M O first with no more than two electrons. H, has a total of two electrons, therefore
t.1
U
U*
Stable (excess of two bonding electrons). H i , formed from H’ and Ha, has one electron:
t -
-
U
U*
Stable (excess of one bonding electron). Has less bonding strength than H,. HI, formed theoretically from H r and He, has three electrons: U
U*
Stable (has net bond strength of one bonding electron). The antibonding electron cancels the bonding strength of one of the bonding electrons. He, has four electrons, two from each He atom. The electron distribution is
Not stable (antibonding and bonding electrons cancel and there is no net bonding). Two He atoms are more stable than a He, molecule. Problem 2.5 Since the U MO formed from 2s AO’s has a higher energy than the AO’s, predict whether ( a ) Li,, ( b ) Be, can exist.
U*
MO formed from Is 4
The MO levels are: u , ~ u ~ ~ u ,with ,~u energy ~ , . increasing from left to right. Li, has six electrons, which fill the MO levels to give
designated ( U , , ) ~ ( ~ ; ~ ) ~ Li, ( U ,has ~ ) an ~ . excess of two electrons in bonding MO’s and therefore can exist; it is by no means the most stable form of lithium.
16
BONDING AND MOLECULAR STRUCTURE
[CHAP. 2
( h ) Be, would have eight electrons:
There are no net bonding electrons, and Be, does not exist Stabilities of molecules can be qualitatively related to the bond order, defined as (number of valence electrons in MO’s) - (number of valence electrons in MO*’s) 2 The bond order is usually equal to the number of U and T bonds between two atoms; i.e., 1 for a single bond, 2 for a double bond, 3 for a triple bond. Bond order =
Problem 2.6 The MO’s formed when the two sets of the three 2 p orbitals overlap are (the 7~ and r* pairs are degenerate). (a) Show how M O theory predicts the paramagnetism of 0,. (b) What is the bond order in O,? 4 The valence sequence of MO’s formed from overlap of the n
*
U 2 9 0 25 =,I,,
=2,p?p,
*
= = 2p,
*
2p,
=2
*
AO’s of diatomic molecules is:
zp,
0, has 12 electrons to be placed in these MO’s, giving (a) The electrons in the two, equal - energy, rr* MO*’s are unpaired; therefore, 0, is paramagnetic. ( b ) Electrons in the first two molecular orbitals cancel each other‘s effect. There are 6 electrons in the next 3 bonding orbitals and 2 electrons in the next 2 antibonding orbitals. There is a net bonding effect due to 4 electrons. The bond order is 112 of 4, or 2; the two 0 ’ s are joined by a net double bond.
2.3
HYBRIDIZATION OF ATOMIC ORBITALS
A carbon atom must provide four equal-energy orbitals in order to form four equivalent (T bonds, as in methane. CH,. It is assumed that the four equivalent orbitals are formed by blending the 2s and the three 2 p AO’s to give four new hybrid orbitals, called sp’ HO’s, Fig. 2-5. The shape of an sp3 HO is shown in Fig. 2-6. The larger lobe, the “head,” having most of the electron density, overlaps with an orbital of its bonding mate to form the bond. The smaller lobe, the “tail” is often omitted when depicting HO’s (see Fig. 2-11). However, at times the “tail” plays an important role in an organic reaction. M x
rs
t t -
2P
P)
2s
LL
2sp‘
ground-state valence shell of carbon atom
t -
t -
t -
sp’-hybridized state of carbon atom (before bonding)
Fig. 2-5
The AO’s of carbon can hybridize in ways other than sp3, as shown in Fig. 2-7. Repulsion between pairs of electrons causes these HO’s to have the maximum bond angles and geometries summarized in Table 2-2. The sp’ and s p HO‘s induce geometries about the C’s as shown in Fig. 2-8. Only U bonds, not 7~ bonds, determine molecular shapes.
17
BONDING AND MOLECULAR STRUCTURE
CHAP. 21
Fig. 2-6
Lt-
2P
---r t t t
r t r ---
SP
SP
2s
=
7 4 Hybridized states
Ground state
Table 2-2 N u mbe r of Remaining p's
Type of Bond Formed
Tetrahedral*
0
(T
120"
Trigonal planar
1
U
180"
Linear
2
U
Type
Bond Angle
Geometry
sp'
109.5"
sp2 SP
sp
SP
Fig. 2-8
Problem 2.7 The H,O molecule has a bond angle of 105". ( a ) What type of AO's does 0 use to form the two equivalent a bonds with H ? (6) Why is this bond angle less than 109.5"? 4
0 has two degenerate orbitals, the p, and p , . , with which to form two equivalent bonds t o H. However, if 0 used these AO's, the bond angle would be 90°, which is the angle between the y- and z-axes. Since the angle is actually 105". which is close to 1Oc).S0, 0 is presumed to use sp' HO's.
? & 'TJ ?J. t T ,o= 1s 2sp'
( s p ' HO's)
BONDING AND MOLECULAR STRUCTURE
18
[CHAP. 2
( h ) Unshared pairs of electrons exert a greater repulsive force than do shared pairs, which causes a contraction of bond angles. The more unshared pairs there are, the greater is the contraction. Problem 2.8
Each H-N-H
bond angle in :NH, is 107". What type of AO's does N use'? ,N = TS.
2s
1s
4
(ground state)
2p: 2p, 2p,
If the ground-state N atom were to use its three equal-energy p AO's to form three equivalent N-H bonds, each H-N-H bond angle would be 90". Since the actual bond angle is 107" rather than YO", N . like 0, uses sp' HO's Apparently. for atoms in the second period forming more than one covalent bond (Be, B. C, N . and 0).a hybrid orbital must be provided for each a bond und each unshured puir of electrons. Atoms in higher periods also often use HO's. Problem 2.9 Predict the shape of ( a ) the boron trifluoride molecule (BF,) and ( 6 ) the boron tetrafluoride 4 anion (BF,). All bonds are equivalent. ( a ) The HO's used by the central atom, in this case B, determine the shape of the molecule.
5B = TJ Is
t.l--t 2s 2p, 2p,.
(ground state)
2p,
There are three sigma bonds in BF, and no unshared pairs; therefore, three HO's are needed. Hence, B uses sp' HO's, and the shape is trigonal planar. Each F-B-F bond angle is 120".
The empty pI orbital is at right angles to the plane of the molecule. ( b ) B in BFS has four a bonds and needs four HO's. B is now in an sp' hybrid state 5 B = - tS. Is
t t -t 2sp
(sp' hybrid state)
used for bonding
The empty sp' hybrid orbital overlaps with a filled orbital of F - which holds two electrons, (coordinate covalent bonding) :F:- + BF,4 BF, The shape is tetrahedral; the bond angles are 109.5". Problem 2.10
Arrange the s, p , and the three sp-type HO's in order of decreasing energy.
4
The more s character in the orbital, the lower the energy. Therefore, the order of decreasing energy is p > sp.' > sp2 > sp
s
Problem 2.11 What effect does hybridization have on the stability of bonds?
4
Hybrid orbitals can ( a ) overlap better and ( b ) provide greater bond angles, thereby minimizing the repulsion between pairs of electrons and making for greater stability.
By use of the generalization that each unshared and o-bonded pair of electrons needs a hybrid orbital, but 7~ bonds do not, the number of hybrid orbitals (HON)needed by C or any other central atom can be obtained as HON = (number of
(J
bonds) + (number of unshared pairs of electrons)
CHAP. 21
19
BONDING AND MOLECULAR STRUCTURE
The hybridized state of the atom can then be predicted from Table 2-3. If more than four HO's are needed, d orbitals are hybridized with the s and the three p's. If five HO's are needed, as in PCI,, one d orbital is included to give trigonal-bipyramidal sp3d HO's, Fig. 2-9(a). For six HO's, as in SF,, two d orbitals are included to give octahedral sp3d2HO's, Fig. 2-9(b). The above method can also be used for the multicovalent elements in the second and, with few exceptions, in the higher periods of the periodic table. Table 2-3
Predicted Hybrid State
sp sp ' d sp'd
I
6
%3 120"
90"
(a) sp& HO's
( 6 ) s p s d 2 HO's Fig. 2-9
Problem 2.12 Use the HON method to determine the hybridized state of the underlined elements: (a) CHCl,
Number of
U
(6) H 2 C e H , Bonds
+
(c) -4
Number of Unshared Electron Pairs
4
3 2 2 1
3
2.4
(d) H e N :
0 0 0 0 1 1
=
( e ) H&+
4
HON
Hybrid State
4
SP3
3 2 2 2
SP
4
sps
SP2 SP
SP
ELECTRONEGATIVITY AND POLARITY
The relative tendency of a bonded atom in a molecule to attract electrons is expressed by the term electronegativity. The higher the electronegativity, the more effectively does the atom attract and hold electrons. A bond formed by atoms of dissimilar electronegativities is called polar. A nonpolar covalent bond exists between atoms having a very small or zero difference in electronegativity. A few relative electronegativities are F (4.0) > O(3.5) > C1, N (3.0) > Br (2.8) > S , C, I ( 2 . 5 ) > H (2.1)
20
B O N D I N G A N D M O L E C U L A R STRUCTURE
[CHAP. 2
The more electronegative element of a covalent bond is relatively negative in charge, while the less electronegatirie eienient is relutiveiy positive. The symbols 6 + and S - represent partial charges (bond polarity). These partial charges should not be confused with ionic charges. Polar bonds are indicated by ++; the head points toward the more electronegative atom. The vector sum o f all individual bond moments gives the net dipole moment of the molecule. Problem 2.73 What d o the molecular dipole moments g the shapes o f thcse molecules'?
=0
for CO2 and
,U
= 1.84
D for H,O tell you about
4
In CO,
0 is more electronegative than C, and each C-0 bond is polar as shown. A zero dipole moment indicates a symmetrical distribution o f 6 - charges about the 6 + carbon. The geometry must be linear; in this way, in div id ua I bond momen t s ca nce I : c--fct---,
o=c=o
H , O also has polar bonds. However, since there is a net dipole moment, the individual bond moments do not cancel. and the molecule must have a bent shape:
pqH H
f rmrltunt moment
2.5 OXIDATION NUMBER
The oxidation number (ON) is a value assigned to an atom based on relative electronegativities. It equals the group number minus the number of assigned electrons, when the bonding electrons are assigned to the more electronegative atom. The sum of all ( 0 N ) ' s equals the charge on the species. Problem 2.74 Determine the oxidation number of each C, (ON),., in: ( a ) CH,. (6) C H , O H , (c) CH,NH,, 4 ( d ) H,C=CH,. Use the data ( O N ) , = -3; (ON),, = 1; (ON),, = -2. All examples are molecules; therefore the sum of all ( O N ) values is 0. (U)
(ON), + 4(ON),, = 0;
(ON),
+ (4 x
1) = O ;
(ON),. = -4
( 6 ) ( O N ) , + (ON),) + 4(ON),, = 0;
(ON),. + (-2)
(c) ( O N ) , + ( O N ) , + 5(ON),,
(ON),
= 0;
+ (-3)
+ 4 =0; +5 =O;
(ON),. = -2
(ON),. = -2
( d ) Since both C atoms are equivalent,
2(ON),
2.6
+ 4(ON),,
= 0;
2(ON), + 4 = 0 ;
(ON),.= - 2
INTERMOLECULAR FORCES
(a) Dipole-dipole interaction results from the attraction of the 6 + end of one polar molecule for the 6- end of another polar molecule. (b) Hydrogen-bond. X-H and :Y may be bridged X-H---:Y if X and Y are small, highly electronegative atoms such as F, 0, and N. H-bonds also occur intramolecularly. (c) London (van der Waals) forces. Electrons of a nonpolar molecule may momentarily cause an imbalance of charge distribution in neighboring molecules, thereby inducing a temporary dipole moment. Although constantly changing, these induced dipoles result in a weak net attractive force.
CHAP. 21
BONDING AND MOLECULAR STRUCTURE
21
The greater the molecular weight of the molecule, the greater the number of electrons and the greater these forces. The order of attraction is H-bond S dipole-dipole > London forces Problem 2.15 Account for the following progressions in boiling point. (a) CH,, -161.5"C; Cl,, -34°C; 4 CH,CI, -24 "C. ( b ) CH,CH,OH, 78 "C; CH,CH,F, 46 "C; CH,CH,CH,, -42 "C.
The greater the intermolecular force, the higher the boiling point. Polarity and molecular weight must be considered. ( a ) Only CH,CI is polar, and it has the highest boiling point. CH, has a lower molecular weight (16 g/mole) than has Cl, (71 g/mole) and therefore has the lowest boiling point. (6) Only CH,CH,OH has H-bonding, which is a stronger force of intermolecular attraction than the dipole-dipole attraction of CH,CH,F. CH,CH,CH, has only London forces, the weakest attraction of all.
Problem 2.16 Account for the following progression in boiling point: CH,CI, -24 "C; CH,Br, 5 "C; CH,I, 43 "C. 4
The order of polarity is CH,CI>CH,Br>CH,I. The order of molecular weight is CH,I>CH,Br> CH,CI. The two trends oppose each other in affecting the boiling point. The order of molecular weight predominates here. Problem 2.17 The boiling points of n-pentane and its isomer neopentane are 36.2 "C and 9.5 " C , respectively. Account for this difference. (See Problem 1.4 for the structural formulas.) 4
These isomers are both nonpolar. Therefore, another factor, the shape of the molecule, influences the boiling point. The shape of n-pentane is rodlike, whereas that of neopentane is spherelike. Rods can touch along their entire length; spheres touch only at a point. The more contact between molecules, the greater the London forces. Thus, the boiling point of n-pentane is higher. Problem 2.18 Ethyl alcohol, C,H,OH, boils at 78.3"C. Its isomer, dimethyl ether, CH,OCH,, boils at -24 "C. Explain. 4
Both are polar molecules and have dipole-dipole attraction. However. in ethyl alcohol, H-bonding can exist:
H
I
H,CCH,O----H-OCH,CH, In CH,OCH, the H's are on C and so cannot H-bond.
2.7 SOLVENTS
The oppositely charged ions of salts are strongly attracted by electrostatic forces, thereby accounting for the high melting and boiling points of salts. These forces of attraction must be overcome in order for salts to dissolve in a solvent. Nonpolar solvents have a zero or very small dipole moment. Protic solvents are highly polar molecules having an H that can form an H-bond. Aprotic solvents are highly polar molecules that do not have an H that can form an H-bond. Problem 2.19 Classify the following solvents: (a) (CH,),S=O, tetrachloride; (c) C,H,, benzene;
dimethyl sulfoxide; ( b ) CCI,, carbon
22
BONDING AND MOLECULAR STRUCTURE
[CHAP. 2
F
(d) H N(CH,)2 Dimethylformamide 0
(e) CH,OH, methanol; ( f ) liquid NH,.
4
Nonpofar: (b) Because of the symmetrical tetrahedral molecular shape, the individual C-CI bond moments cancel. (c) With few exceptions, hydrocarbons are nonpolar. Protic: (e) and ( f ) . Aprotic: ( a ) and (d). The C--0and C=O groups are strongly polar and the H's attached to C do not typically form H-bonds. Problem 2.20 Mineral oil, a mixture of high-molecular-weight hydrocarbons, dissolves in n-hexane but not in 4 water or ethyl alcohol, CH,CH,OH. Explain. Attractive forces between nonpolar molecules such as mineral oil and n-hexane are very weak. Therefore, such molecules can mutually mix and solution is easy. The attractive forces between polar H,O or C,H,OH molecules are strong H-bonds. Most nonpolar molecules cannot overcome these H-bonds and therefore do not dissolve in such polar protic solvents. Problem 2.21 Explain why CH,CH,OH is much more soluble in water than is CH,(CH,),CH,OH.
4
The OH portion of an alcohol molecule tends to interact with water-it is hydrophilic. The hydrocarbon portion does not interact, rather it is repelled- it is hydrophobic. The larger the hydrophobic portion, the less soluble in water is the molecule. Problem 2.22 Explain why NaCl dissolves in water but not in hexane.
4
Water, a protic solvent, helps separate the strongly attracting ions of the solid salt by solvation. Several water molecules surround each positive ion (Na') by an ion-dipole attraction. The 0 atoms, which are the negative ends of the molecular dipole, are attracted to the cation. H,O typically forms an H-bond with the H-bond negative ion (in this case CIF). ion-dipole attraction
artracrion
61
6*
CI----H
H
6 -/ Na'---0
\6+
H
and
\a-
0
6+/
H
Problem 2.23 Compare the ways in which NaCl dissolves in water and in dimethyl sulfoxide.
4
The way in which NaCI, a typical salt, dissolves in water, a typical protic solvent, was discussed in Problem 2.22. Dimethyl sulfoxide also solvates positive ions by an ion-dipole attraction; the 0 of the C-0 group is attracted to the cation. However, since this is an aprotic solvent, there is no way for an H-bond to be formed and the negative ions are not solvated when salts dissolve in aprotic solvents. The S, the positive pole, is surrounded by the methyl groups and cannot get close enough to solvate the anion.
The bare negative ions discussed in Problem 2.23 have a greatly enhanced reactivity. The small amounts of salts that dissolve in nonpolar or weakly polar solvents exist mainly as ion-pairs or ion-clusters, where the oppositely charged ions are close to each other and move about as units. Tight ion-pairs have no solvent molecules between the ions; loose ion-pairs are separated by a small number of solvent molecules. 2.8
RESONANCE AND DELOCALIZED n ELECTRONS
Resonance theory describes species for which a single Lewis electron structure cannot be written. As an example, consider dinitrogen oxide, N,O:
23
BONDING AND MOLECULAR STRUCTURE
CHAP. 21
-
..
-
t
:N=N=O:
Calculated bond length, nm Observed bond length, nm
:N-N-Q:
+
.. -
resonance
0.120
0.115
0.110
0.147
0.112
0.119
0.112
0.119
(with formal charges)
A comparison of the calculated and observed bond lengths shows that neither structure is correct. Nevertheless, these contributing (resonance) structures tell us that the actual resonance hybrid has some double-bond character between N and 0, and some triple-bond character between N and N . This state of affairs is described by the non-Lewis structure 6-
4;
+
:N=NcO:
in which broken lines stand for the partial bonds in which there are delocalized p electrons in an extended ?T bond created from overlap of p orbitals on each atom. See also the orbital diagram, Fig. denotes resonance, not equilibrium. 2-10. The symbol
Fig. 2-10
The energy of the hybrid, E,, is always less than the calculated energy of any hypothetical contributing structure, E,. The difference between these energies is the resonance (delocalization) energy, E,:
E, = E,
-
E,,
The more nearly equal in energy the contributing structures, the greater the resonance energy and the less the hybrid looks like any of the contributing structures. When contributing structures have dissimilar energies, the hybrid looks most like the lowest-energy structure. Contributing structures ( a ) differ only in positions of electrons (atomic nuclei must have the same positions) and (6) must have the same number of paired electrons. Relative energies of contributing structures are assessed by the following rules.
1. Structures with the greatest number of covalent bonds are most stable. However, for second-period elements (C, 0, N) the octet rule must be observed. 2. With few exceptions, structures with the least number or amount of formal charges are most stable. 3. If all structures have formal charge, the most stable (lowest energy) one has - on the more electronegative atom and + on the more electropositive atom. 4. Structures with like formal charges on adjacent atoms have very high energies. 5 . Resonance structures with electron-deficient, positively charged atoms have very high energy, and are usually ignored. Problem 2.24 Write contributing structures, showing formal charges when necessary, for ( a ) ozone, 0,; ( 6 ) CO,; (c) hydrazoic acid, HN,; ( d ) isocyanic acid, HNCO. Indicate the most and least stable structures and give 4 reasons for your choices. Give the structure of the hybrid.
[CHAP. 2
BONDING AND MOLECULAR STRUCTURE
24
:o=o-o: It/ + b :
t--*
:o-O=$: 0
(equal-energy structures).
9.-
2
The hybrid is :O-0-0:.
4;
( 1 ) is most stable; it has no formal charge. (2) and (3) have equal energy and are least stable because they have formal charges. In addition, in both (2) and ( 3 ) , one 0, an electronegative element, bears a + formal charge. Since (1) is so much more stable than ( 2 ) and (3), the hybrid is :O=C=O:, which is just ( 1 ) .
(1) And (2) have about the same energy and are the most stable, since they have the least amount of formal charge. (3) has very high energy since it has + charge on adjacent atoms and, in terms of absolute value, a total formal charge of 4. (4) has a very high energy because the N bonded to H has only six electrons. The hybrid, composed of (1) and (2), is
(1) has no formal charge and is most stable. (2) is least stable since the - charge is on N rather than on the
more electronegative 0 as in (3). The hybrid is H-N=C-+: contributing structure.
(the same as ( I ) ) , the most stable
Problem 2.25 ( a ) Write contributing structures and the delocalized structure for (i) NOT and (ii) NO;. ( b ) Use p AO’s to draw a structure showing the delocalization of the p electrons in an extended T bond for (i) and 4 (ii). (c) Compare the stability of the hybrids of each.
The - is delocalized over both 0’s so that each can be assumed to have a the same bond length.
-
f charge. Each N - 4 bond has
The - charges are delocalized over three 0’s so that each has a - $ charge. (6) See Fig. 2-11. (c) We can use resonance theory to compare the stability of these two ions because they differ in only one feature-the number of 0 ’ s on each N, which is related to the oxidation numbers of the N’s. We could not, for example, compare NOT and HSO;, since they differ in more than one way; N and S are in different groups and periods of the periodic table. NO, is more stable than NOT since the charge on NOS is delocalized (dispersed) over a greater number of 0 ’ s and since NO, has a more extended T bond system. Problem 2.26 Indicate which one of each of the following pairs of resonance structures is the less stable and is an unlikely contributing structure. Give reasons in each case.
25
BONDING A N D MOLECULAR STRUCTURE
CHAP. 21
P
Y
f+l
Fig. 2-11
I
I1
(c) H,C=CH-CH,
+
.c-+
V (e)
H,c-CI: IX
-
H,C-CH-CH,
-
-
*
VI ++
H,C=~I:
4
X
I has fewer covalent bonds, more formal charge and an electron-deficient N. (6) IV has + on the more electronegative 0. (c) VI has similar - charges on adjacent C's, fewer covalent bonds, more formal charge and an electrondeficient C. ( d ) VII has fewer covalent bonds and a + on the more electronegative N , which is also electron-deficient. (e) C in X has 10 electrons; this is not possible with the elements of the second period. (a)
Supplementary Problems Problem 2.27 Distinguish between an AO, an HO, an MO and a localized MO.
4
An A 0 is a region of space in an atom in which an electron may exist. An HO is mathematically fabricated from some number of AO's to explain equivalency of bonds. An M O is a region of space about the entire molecule capable of accommodating electrons. A localized MO is a region of space between a pair of bonded atoms in which the bonding electrons are assumed to be present. Problem 2.28 Show the orbital population of electrons for unbonded N in (a) ground state, (6) sp3, (c) sp'. 4 and (d) sp hybrid states.
BONDING AND MOLECULAR STRUCTURE
26
(a)
t.l
Is
fi 2s
r r2P
t
(4 TJ.
tL t
t
- --
2sp‘
[CHAP. 2
1 2p
ts- t -t t ( d ) 7s. -2sp -’ 2SP 2P Note that since the energy difference between hybrid and p orbitals is so small, Hund’s rule prevails over the Aufbau principle. (6)
tS. t.l -t t t
Problem 2.29 ( a ) NO; is linear, ( 6 ) NO, is bent. Explain in terms of the hybrid orbitals used by N.
4
(a) NO;, :&N=6:. N has two a bonds, no unshared pairs of electrons and therefore needs two hybrid orbitals. N uses s p hybrid orbitals and the U bonds are linear. The geometry is controlled by the arrangement of the sigma bonds. ( 6 ) NO,, : b N : O F . N has two U bonds, one unshared pair of electrons and, therefore, needs three hybrid orbitals. N uses sp‘ hybrid HO’s, and the bond angle is about 120”. Problem 2.30 Draw an orbital representation for the cyanide ion. :GENT.
4
See Fig. 2-12. The C and N each have one U bond and one unshared pair of electrons, and therefore each needs two s p hybrid HO’s. On each atom one s p hybrid orbital forms a (r bond while the other has the unshared pair. Each atom has a p , A 0 and a p , AO. The two p , orbitals overlap to form a n, bond in the xy-plane; the two p , orbitals overlap to form a T , bond in the xz-plane. Thus, two T bonds at right angles to each other and a U bond exist between the C and N atoms.
SP
only
Fig. 2-12
Problem 2.31 (a) Which of the following molecules possess polar bonds: F,, HF, BrCl, CH,, CHCI,, CH,OH? ( b ) Which are polar molecules? 4 (a)
HF, BrCl, CH,, CHCI,, CH,OH.
( b ) HF, BrCl, CHCI,, CH,OH. The symmetrical individual bond moments in CH, cancel.
Problem 2.32 Considering the difference in electronegativity between 0 and S, would H’O or H,S exhibit greater (a) dipole-dipole attraction, (b) H-bonding? 4 (a)
H,O,
(6) H,O.
Problem 2.33 Nitrogen trifluoride (NF,) and ammonia (NH,) have an electron pair at the fourth corner of a tetrahedron and have similar electronegativity differences between the elements (1.0 for N and F and 0.9 for N and H). Explain the larger dipole moment of ammonia (1.46 D) as compared with that of NF, (0.24 D). 4
CHAP. 21
27
BONDING AND MOLECULAR STRUCTURE
Net dipole moment is very small; actual direction of moment is not known.
(It
.. ..
:F
yjp 1
:F..
H
Netmoment
H H
The dipoles in the three N-F bonds are toward F, see Fig. 2-13(a), and oppose and tend to cancel the effect of the unshared electron pair on N. In NH,, the moments for the three N-H bonds are toward N , see Fig. 2-13(6), and add to the effect of the electron pair. Problem 2.34 Explain why the dipole moment of CH2C1, is greater than that of CHCl,.
4
The resultant moment of the H and the encircled CI in CHCI, opposes the resultant moment of the other two Cl's, as shown in Fig. 2-14(b). The actual values for CH,Cl2 and CHCI, are 1.6 and l.OD, respectively.
Net moment ( a1
Net moment
Fig. 2-14
( b1
Problem 2.35 When 50 mL of C,H,OH and 50 mL of H,O are mixed, the volume of the solution is less than 100 mL. Explain. 4
The H-bonds between H,O and C,H,OH molecules are stronger than the H-bond between the like molecules of the individual compounds. The H,O and C,H,OH molecules are closer when mixed, and the volume shrinks. Problem 2.36 NH; salts are much more soluble in water than are the corresponding Na' salts. Explain.
Na' is solvated merely by an ion-dipole interaction. NHZ is solvated by H-bonding
H
which is a stronger attractive force. Problem 2.37 The F - of dissolved NaF is more reactive in dimethyl sulfoxide,
P
CH, CH, and in acetonitrile, CH,C=N, than in CH,OH. Explain.
4
28
[CHAP. 2
BONDING AND MOLECULAR STRUCTURE
H-bonding prevails in CH,OH (a protic solvent), CH,OH---F-, thereby decreasing the reactivity of F-. CH,SOCH, and CH,CN are aprotic solvents; their C-H H’s do not H-bond. Problem 2.38 Find the oxidation number of the C in ( a ) CH,CI, ( 6 ) CH,CI,, (c) H,CO, (d) HCOOH, and 4 (e) CO,, if (ON),., = - 1 . From Section 2.5:
Problem 2.39 Give a True or False answer to each question and justify your answer. ( a ) Since in polyatomic anions XYL,-(such as SO:- and B F I ) , the central atom X is usually less electronegative than the peripheral atom Y,it tends to acquire a positive oxidation number. (6) Oxidation numbers tend to be smaller values than formal charges. (c) A bond between dissimilar atoms always leads to nonzero oxidation numbers. ( d ) Fluorine 4 never has a positive oxidation number. True. The bonding electrons will be alloted to the more electronegative peripheral atoms, leaving the central atoms with a positive oxidation number. False. In determining formal charges an electron of each shared pair is assigned to each bonded atom, In determinating oxidation numbers pairs of electrons are involved, and more electrons are moved to or away from an atom. Hence larger oxidation numbers result. False. The oxidation numbers will be zero if the dissimilar atoms have the same electronegativity, as in PH,. True. F is the most electronegative element; in F, it has a zero oxidation number. Which of the following transformations of organic compounds are oxidations, which are Problem 2.40 reductions, and which are neither?
( a ) H,C=CH,-CH,CH,OH
( c ) CH,CHO-CH,COOH
(6) CH,CH20H-CH,CH=0
(d) H2C=CH,-CH,CH,CI
(e) HC%CH-H,C=CH, 4
To answer the question determine the average oxidation numbers (ON) of the C atoms in reactant and in product. An increase (more positive or less negative) in ON signals an oxidation; a decrease (more negative or less positive) signals a reduction; no change means neither. ( a ) and (d) are neither, because (ON),. is invariant at -2. (b) and (c) are oxidations, the respective changes being from -2 to -1 and from -1 to 0. (e) is a reduction, the change being from - 1 to -2. Problem 2.41 Irradiation with ultraviolet (uv) light permits rotation about a bonding and antibonding MO’s.
7r
bond. Explain in terms of 4
Two p AO’s overlap to form two pi MO’s, 7r (bonding) and 7r* (antibonding). The two electrons in the original p AO’s fill only the 7r MO (ground state). A photon of uv causes excitation of one electron from 7~ to 7r* (excited state).
t J. -
-
7r
7r*
(ground state)-
uv
.1 t 7 7r
(excited state)
(Initially the excited electron does not change its spin.) The bonding effects of the two electrons cancel. There is now only a sigma bond between the bonded atoms, and rotation about the bond can occur. Problem 2.42 Write the contributing resonance structures and the delocalized hybrid for ( a ) BCI,, (b) H,CN, (diazomethane). 4
CHAP. 21
29
BONDING AND MOLECULAR STRUCTURE
( a ) Boron has six electrons in its outer shell in BCI, and can accommodate eight electrons by having a B-CI bond assume some double-bond character.
Problem 2.43 Arrange the contributing structures for (a) vinyl chloride, H,C==CHCI, and (6) formic acid, HCOOH, in order of increasing importance (increasing stability) by assigning numbers starting with 1 for most important and stable. 4
H,CkHKCl:
( a)
-
H,C--CH=el:
I
c-*
H,&-CH=CI; 111
I1
I is most stable because it has no formal charge. 111 is least stable since it has an electron-deficient C. In 111, Cl uses an empty 3d orbital to accommodate a fifth pair of electrons. Fluorine could not do this. The order of stability is :O:
II
H--C-+-H
( b)
-
V
I ( 1 ) > II(2) > III(3)
:o; I +
H-c=~-H
:or c--*
+
:():
I
H-4-0-H
c--,
VII
VI
H-C-Q-H
..
VIII
V and VI have the greater number of covalent bonds and are more stable than either VII or VIII. V has no formal charge and is more stable than VI. VIII is less stable than VII since VIII's electron deficiency is on 0, which is a more electronegative atom than the electron-deficient C of VII. The order of stability is
V( 1 ) > VI(2) > VII(3) > VIII(4) Problem 2.44
What is the difference between isomers and contributing resonance structures?
4
Isomers are real compounds that differ in the arrangement of their atoms. Contributing structures have the same arrangement of atoms; they differ only in the distribution of their electrons. These imaginary structures are written to give some indication of the electronic structure of certain species for which a typical Lewis structure cannot be written. Problem 2.45 Use the HON method to determine the hybridized state of the underlined elements: (a) H - m H
( 6 ) H&=O
Number of
U
Bonds
(c) H m - -
+
( d ) &ICI:-
(e) PF,
Number of Unshared Electron Pairs
=
(f) HON
CH,QCH, Hybrid State SP
sp SP
4
sp'd' sp ' d sp
4
Chapter 3 Chemical Reactivity and Organic Reactions 3.1
REACTION MECHANISM
The way in which a reaction occurs is called a mechanism. A reaction may occur in one step or, more often, by a sequence of several steps. For example, A + B-* X + Y may proceed in two steps: (1) A - I + X
followed by
(2) B + I + Y
Substances such as I, formed in intermediate steps and consumed in later steps, are called intermediates. Sometimes the same reactants can give different products via different mechanisms. 3.2 CARBON-CONTAINING INTERMEDIATES Carbon-containing intermediates often arise from two types of bond cleavage:
Heterolytic (polar) cleavage. Both electrons go with one group, e.g. A:B-A'
+ :B-
or
A:-
+ B'
Homolytic (radical) cleavage. Each separating group takes one electron, e.g.
A:B----* A. + .B 1. Carbocations are positively charged ions containing a carbon atom having only six electrons in three bonds:
2. Carbanions are negatively charged ions containing a carbon atom with three bonds and an unshared pair of electrons: - .
I
3. Radicals (or free radicals) are species with at least one unpaired electron. This is a broad category in which carbon radicals,
I
are just one example. 4. Carbenes are neutral species having a carbon atom with two bonds and two electrons. There are two kinds: singlet
15.
-C-
in which the two electrons have opposite spins and are paired in one orbital, and triplet
t t
-C-
in which the two electrons have the same spin and are in different orbitals. 30
CHAP. 31
31
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
Problem 3.1 Determine the hybrid orbital number (HON) of the five C-containing intermediates tabulated below and give the hybrid state of the C atom. Unpaired electrons do nor require an HO and should not be counted in your determination. 4 Number of Bonds
U
+
Number of Unshared Electron Pairs
( a ) carbocation (b) carbanion (c) radical (d) singlet carbene (e) triplet carbene
=
HONl Hybrid State
sp sp .’ sp sp
SP
Recall that the two unshared electrons of the triplet carbene are not paired and, hence, are not counted; they are in different orbitals. Problem 3.2 Give three-dimensional representations for the orbitals used by the C’s of the five carbon 4 intermediates of Problem 3.1. Place all unshared electrons in the appropriate orbitals. A carbocation has three trigonal planar sp2 HO’s to form three 0 bonds, and an empty P A 0 perpendicular to the plane of the U bonds. See Fig. 3-l(a). (b) A carbanion has four tetrahedral sp3 HO’s; three form three 0 bonds and one has the unshared pair of electrons. See Fig. 3-l(b). ( c ) A radical has thesame ‘orbitals as the carbocation. The difference lies in the presence of the odd electron in the p orbital of the radical. See Fig. 3-l(c). (d) A singlet carbene has three sp2 HO’s; two form two U bonds and the third holds the unshared pair of electrons. It also has an empty p A 0 perpendicular to the plane of the sp’ HO’s. See Fig. 3-l(d). (e) A triplet carbene uses two sp HO’s to form two linear U bonds. Each of the two unhybridized p AO’s has one electron. See Fig. 3-l(e). (a)
P,
P,
Fig. 3-1
32
CHEMICAL REACTIVITY A N D O R G A N I C REACTIONS
Problem 3.3 Write formulas for the species resulting from the of the C-C bond in ethane. C,H,,, and classify these species. H ,C:CH
((1)
[CHAP. 3
homolytic cleavage. ( b ) heterolytic cleavage 4
H ,C*+ *CH,
+
tth'rnc
hleth>l radical*,
H,C:CH,-H,C' Et hanc
+ :CH,
('arbocLition ('arbanion
3.3 TYPES OF ORGANIC REACTIONS 1. Displacement (substitution). An atom o r group of atoms in a molecule or ion is replaced by another atom or group. 2. Addition. Two molecules combine t o yield a single molecule. Addition frequently occurs at a double or triple bond and sometimes at small-size rings. 3. Elimination. This is the reverse of addition. Two atoms or groups are removed from a molecule. If the atoms o r groups are taken from adjacent atoms (p-elimination), a multiple bond is formed; if they are taken from other than adjacent atoms, a ring results. Removal of two atoms or groups from the same atom (a-elimination) produces a carbene. 4. Rearrangement. Bonds in the molecule are scrambled, converting it to its isomer. 5 . Oxidation-reduction (redox), These reactions involve transfer o f electrons or change in oxidation number. A decrease in the number o f H atoms bonded to C and an increase in the number o f bonds to other atoms such as C, 0, N , CI, Br, F. and S signals oxidation. Problem 3.4 The following represents the steps in the mechanism for chlorination of methane: Initiation Step
Propagation Steps
(1)
I I
:CI:CI: + energy
__*
:CI-+ .CI: Chlorine radicals
H (2) H:C:H H
H
+ .CI: --, H:CI:+ H:C. H
Methyl radical
H
H
The propagation \teps constitute the overall reaction. ( U ) Write the equation for the overall reaction. (b) What are the intermediate\ in the overall reaction'! (c) Which reactions are homolytic? (d) Which is a displacement reaction'? ( U ) I n which reaction is addition taking place? (f')The collision of which species would lead to side 4 products'? Add Steps 2 and 3: CH, + CI,-+CH,CI + HCI. The intermediates formed and then consumed are the H,C. and radicals. Each step is homolytic. In Steps 1 and 3, Cl, cleaves; in Step 2, CH, cleaves. Step (3) involves the displacement of one .cl: of CI, by a C H , group. I n Step 2, .Cl: displaces a C H , group from an H. None. H ,C- + .CH, -+ H ,CCH (ethane) Problem 3.5
ldcntify each of the following as ( 1 ) carbocations, (2) carhanions, (3) radicals o r (4) carbenes:
4
CHAP. 31
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
33
Problem 3.6 Classify the following as substitution, addition, elimination, rearrangement, or redox reactions. (A reaction may have more than one designation.)
+ Br,-CH,BrCH,Br C,H,OH + HCI-C,H,CI + H,O CH ,CH=CHCH , + ZnCI, CH ,CHCICHCICH ,+ Zn CH,=CH,
NH:(CNO)--H,NCNH,
-
-
II
0
CH,CH,CH,CH,
-
+ 2Mn03 + O H - A 3CH,COO- + 2Mn0, + 2H,O + OH :CCI, + H,O + CI
3CH,CHO HCCl,
(CH ,),CH
BrCH,CH,CH,Br
+ Zn-,
H,C--&H,
(A means heat.)
+ ZnBr,
Addition and redox. In this reaction the two Br’s add to the two double-bonded C atoms (1.2-addition). The oxidation number (ON) for C has changed from 4 - 2(2) - 2 = -2 to 4 - 2(2) - 1 = - 1; (ON) for Br has changed from 7 - 7 = 0 to 7 - 8 = -1. Substitution of a CI for an OH. Elimination and redox. Zn removes two Cl atoms from adjoining C atoms to form a double bond and ZnCI, (a p-elimination). The organic compound is reduced and Zn is oxidized. Rearrangement (isomerization). Rearrangement (isomerization). Addition and redox. The Br’s add to two C atoms of the ring. These C’s are oxidized and the Br’s are reduced. Redox. CH,CHO is oxidized and MnO, is reduced. Elimination. An H’ and Cl- are removed from the same carbon (a-elimination). Elimination. The two Br’s are removed from nonadjacent C’s giving a ring. Redox [see (c)].
ELECTROPHILIC AND NUCLEOPHILIC REAGENTS Reactions generally occur at the reactive sites of molecules and ions. These sites fall mainly into two categories. One category has a high electron density because the site ( a ) has an unshared pair of electrons or (6) is the S- end of a polar bond or (c) has C=C n electrons. Such electron-rich sites are nucleophilic and the species possessing such sites are called nucleophiles or electron-donors. The second category ( a ) is capable of acquiring more electrons or (b) is the S + end of a polar bond. These electron-deficient sites are electrophilic and the species possessing such sites are called electrophiles or electron-acceptors. Many reactions occur by coordinate covalent bond formation between a nucleophilic and an electrophilic site.
Nu:
+ E----*
Nu:E
Problem 3.7 Classify the following species as being (1) nucleophiles or ( 2 ) electrophiles and give the reason for your classification: ( a ) HO:, (b) :C-=N:-, (c) : B r + , ( d ) BF,, (e) H 2 0 : . ( f ) AICI,, ( g ) :NH,, ( h ) H,Cf 4 (a carbanion), (i) SiF,, ( j ) Ag’, (k) H,C+ (a carbocation), ( I ) H,C: (a carbene), ( m ) ::I:. (1) ( a ) , ( b ) , (e), (g), ( h ) , and ( m ) . They all have unshared pairs of electrons. All anions are potential nucleophiles.
34
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
[CHAP. 3
(2) (d) and ( f ) are molecules whose central atoms (B and AI) have only six electrons rather than the more desirable octet; they are electron-deficient. (c), ( j ) and (k)have positive charges and therefore are electron-deficient. Most cations are potential electrophiles. The Si in (i) can acquire more than eight electrons by utilizing its d orbitals, e.g. SiF, (an electrophile) + 2:F:
-
SiFi-
Although the C in (I) has an unshared pair of electrons, (I) is electrophilic because the C has only six electrons. Problem 3.8 Why is the reaction CH,Br
+ OH- --* CH,OH + Br a nucleophilic displacement?
The :OH- has unshared electrons and is a nucleophile. Because of the polar nature of the C-Br \ii+
-C-Br /
4
bond,
6
C acts as an electrophilic site. The displacement of Br- by OH- is initiated by the nucleophile HO: Problem 3.9 Indicate whether reactant (1) or (2) is the nucleophile or electrophile in the following reactions:
H,C=CH2 (1) + Br, (2)+ BrCH,-CH,Br CH,NH; (1) + CH,COO- (2)--*CH,NH2 + CH,COOH C H , C X l : (1) + AlCl, (2)+ CH, +AICL II 0 0
ii'
CH,CH--V (1) + :SO,H- (2)+ CH,--CHSO,H
4 4 1 I
0-
4
Nucleophile
THERMODYNAMICS thermodynamics and the rate of a reaction determine whether the reaction proceeds. The thermodynamics of a system is described in terms of several important functions: (1) A E , the change in energy, equals q,, the heat transferred to or from a system at constant volume: A E = q,. ( 2 ) A H , the change in enthalpy, equals q p ,the heat transferred to or from a system at constant pressure: A H = qp. Since most organic reactions are performed at atmospheric pressure in open vessels, A H is used more often than is A E . For reactions involving only liquids or solids: A E = A H . A H of a chemical reaction is the difference in the enthalpies of the products, H , , and the reactants, H R :
AH=Hp
- HR
If the bonds in the products are more stable than the bonds in the reactants, energy is released, and A H is negative. The reaction is exothermic. (3) AS is the change in entropy. Entropy is a measure of randomness. The more the randomness, the greater is S ; the greater the order, the smaller is S. For a reaction, A S = s, - SR
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
CHAP. 31
(4)
AG = G,
-
35
G, is the change in free energy. At constant temperature, AG
= AH
- 7' AS ( T = absolute temperature)
For a reaction to be spontaneous, AG must be negative. Problem 3.10 State whether the following reactions have a positive or negative AS and give a reason for your choice.
H2 + H2C=CH2
-
H,CCH,
-
H,C-CH=CH2 CH,COO-(aq)
+ H,O'(aq)
CH,COOH
+ H20
4
Negative. Two molecules are changing into one molecule and there is more order (less randomness) in the product (S, < SR). (6) Positive. The rigid ring opens to give compound having free rotation about the C - C single bond. There is now more randomness (S, > SR). (c) Positive. The ions are solvated by more H,O molecules than is CH,COOH. When ions form molecules, many of these H,O molecules are set free and therefore have more randomness (S,> S,). (a)
Problem 3.11 Predict the most stable state of H,O (steam, liquid, or ice) in terms of ( a ) enthalpy, (6) 4 entropy, and (c) free energy. ( a ) Gas-, Liquid- Solid are exothermic processes and, therefore, ice has the least enthalpy. For this reason,
ice should be most stable. (6) Solid-, Liquid-, Gas shows increasing randomness and therefore increasing entropy. For this reason, steam should be most stable. (c) Here the trends to lowest enthalpy and highest entropy are in opposition; neither can be used independently to predict the favored state. Only G, which gives the balance between Hand S, can be used. The state with lowest G or the reaction with the most negative AG is favored. For H,O, this is the liquid state, a fact which cannot be predicted until a calculation is made using the equation G = H - TS.
3.6
BOND-DISSOCIATION ENERGIES
The bond-dissociation energy, AH, is the energy needed for the endothermic homolysis of a covalent bond A:B- A. + -B; AH is positive. Bond formation,the reverse of this reaction, is exothermic and the AH values are negative. The more positive the AH value, the stronger is the bond. The AH of reaction is the sum of all the (positive) AH values for bond cleavages plus the sum of all the (negative) A H values for bond formations. Problem 3.12 Calculate AH for the reaction CH, + Cl,-, CH,Cl + HCl. The bond-dissociation energies, in kJ/mol, are 427 for C-H, 243 for C l - C l , 339 for C - C l , and 431 for H - C l . 4
The values are shown under the bonds involved:
+ Cl--Cl+ 427 + 243 +
-
H,C-H
cleavages (endothermic)
-
H,C--Cl+ H-Cl (-339) + ( 4 3 1 )
=-loo
formations (exothermic)
The reaction is exothermic, with AH = - 100 kJ/mol.
36
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
8
[CHAP. 3
Problem 3.13 Compare the strengths of bonds between similar atoms having: ( a ) single bonds between atoms with and without unshared electron pairs; (6) triple, double, and single bonds. 4 ( a ) Bonds are weaker between atoms with unshared electron pairs because of interelectron repulsion. (6) Overlapping of p orbitals strengthens bonds, and bond energies are greatest for triple and smallest for single bonds.
3.7 CHEMICAL EQUILJBRIUM Every chemical reaction can proceed in either direction, dA + eB Z f X + gY, even if it goes in one direction to a microscopic extent. A state of equilibrium is reached when the concentrations of A , B , X , and Y no longer change even though the reverse and forward reactions are taking place. Every reversible reaction has an equilibrium expression in which K,, the equilibrium constant, is defined in terms of molar concentrations (mol/L) as indicated by the square brackets: dA + eB= f X Ke
=
+ gY
Products favored; K , is large Reactants favored; K, is small
[X]f[Y]x [A]'[B]'
K , varies only with temperature. The AG of a reaction is related to K , by AG where R is the molar gas constant (R
=
-2.303 RT log K,
= 8.314 J
mol-' K - ' ) and T is the absolute temperature (in K).
Problem 3.14 Given the reversible reaction
C,H,OH
+ CH,COOH eCH,COOC,H, + H,O
what changes could you make to increase the yield of CH,COOC2H,?
4
The equilibrium must be shifted to the right, the side of the equilibrium where CH,COOC2H, exists. This is achieved by any combination of the following: adding C,H,OH, adding CH,COOH, removing H,O, removing CH ,COOC,H,. Problem 3.15 Summarize the relationships between the signs of A H , T A S , and A G , and the magnitude of K,, and state whether a reaction proceeds to the right or to the left for the reaction equation as written. 4
See Table 3-1 Table 3-1
AH
-
TAS
=
AG
Reaction Direction
K,
-
+
-
Forward + right
>1
+
-
+
Reverse --* left
<1
-
-
Usually - if A H < -60 kJ/mol
Depends on conditions
?
+
+
Usually + if A H > +60 kJ/mol
Depends on conditions
?
CHAP. 31
CHEMICAL REACTIVITY A N D ORGANIC REACTIONS
37
Problem 3.16 At 25°C the following reactions have the indicated K , values:
CH,CH,OH Ethanol
H2C-OH
+ CH,COOH
CH,COOCH,CH,
Acetic acid
Ethyl acetate
+ H20
K , = 4.0
Water
Hg-0 a
lactone
Since the changes in bonding in these reactions are similar, both reactions have about the same A H . Use 4 thermodynamic functions to explain the large difference in the magnitude of K , . A larger K,, means a more negative AG. Since S H is about the same for both reactions, a more negative AG means that AS for this reaction is more positive. A more positive AS (greater randomness) is expected in reaction (6) because one molecule is converted into two molecules, whereas in reaction ( a ) two molecules are changed into two other molecules. When two reacting sites, such as O H and C O O H , are in the same molecule, the reaction is intramolecular. When reaction sites are in different molecules. as in ( a ) , the reaction is intermolecular. Intramolecular reactions often have a more positive AS than similar intermolecular reactions.
3.8 RATES OF REACTIONS The rate of a reaction is how fast reactants disappear or products appear. For the general reaction + gD, the rate is given by a rate equation
dA + eB-.fC
Rate = k[A].'[B]' where k is the rate constant at the given temperature, T, and [ A ] and [ B ]are molar concentrations. Exponents x and y may be integers, fractions, or zero; their sum defines the order of the reaction. The values of x and y are found experimentally, and they may differ from the stoichiometric coefficients d and e. Experimental conditions, other than concentrations, that affect rates of reactions are: Temperature. A rough rule is that the value of k doubles for every rise in temperature of 10°C. Particle size. Increasing the surface area of solids by pulverization increases the reaction rate. Catalysts and inhibitors. A catalyst is a substance that increases the rate of a reaction but is recovered unchanged at the end of the reaction. Inhibitors decrease the rate.
At a given set of conditions, the factors that determine the rate of a given reaction are: 1 . Number of collisions per unit time. The greater the chances for molecular collision, the faster the reaction. Probability of collision is related to the number of molecules of each reactant and is proportional to the molar concentrations. 2. Enthalpy of activation (activation energy, EpCt) ( A H ' ) . Reaction may take place only when colliding molecules have some enthalpy content, AH', in excess of the average. The smaller the value of AH', the more successful will be the collisions and the faster the reaction. (AH' = E,,, at constant pressure .) 3. Entropy of activation ( A S ' ) , also called the probability factor. Not all collisions between molecules possessing the requisite AH' result in reaction. Often collisions between molecules must also occur in a certain orientation, reflected by the value of AS'. The more organized or less random the required orientation of the colliding molecules, the lower the entropy of activation and the slower the reaction.
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
38
[CHAP. 3
Problem 3.17 Predict the effect on the rate of a reaction if a change in the solvent causes: ( a ) an increase in AHS and a decrease in ASS,(6) a decrease in AH' and an increase in ASS,(c) an increase in AHSand in AS*, (d) a decrease in AHS and in AS'. 4 ( a ) Decrease in rate. (6) Increase in rate. (c) The increase in AHS tends to decrease the rate but the increase in ASS tends to increase the rate. The combined effect is unpredictable. (d) The trends here are opposite to those in part (c); the effect is also unpredictable. In many cases the change in AH' is more important than the change in ASs in affecting the rate of reaction.
3.9 TRANSITION-STATE THEORY AND ENTHALPY DIAGRAMS When reactants have collided with sufficient enthalpy of activation (AHt) and with the proper orientation, they pass through a hypothetical transition state in which some bonds are breaking while others may be forming. The relationship of the transition state (TS) to the reactants (R) and products (P) is shown by the enthalpy (energy) diagram, Fig. 3-2, for a one-step exothermic reaction A + B 4 C + D. At equilibrium, formation of molecules with lower enthalpy is favored, i.e., C + D. However, this applies only if A H of reaction predominates over T AS of reaction in determining the equilibrium state. The reaction rate is actually related to the free energy of activation, AC', where AG' = AH' - T AS'.
T
Progress of Reaction (Reaction Coordinate) Fig. 3-2
In multistep reactions, each step has its own transition state. The step with the highest-enthalpy transition state is the slowest step and determines the overall reaction rate. The number of species colliding in a step is called the molecularity (of that step). If only one species breaks down, the reaction is unimolecular. If two species collide and react, the reaction is bimolecular. Rarely do three species collide (termolecular) at the same instant. The rate equation gives molecules and ions and the number of each (from the exponents) involved in the slow step and in any preceding fast step@). Intermediates do not appear in rate equations, although products occasionally do appear. Problem 3.18 Draw an enthalpy diagram for a one-step endothermic reaction. Indicate AH of reaction and AH*. 4 See Fig. 3-3. Problem 3.19 Draw a reaction-enthalpy diagram for an exothermic two-step reaction in which ( a ) the first step 4 is slow, (6) the second step is slow. Why do these reactions occur?
T
39
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
CHAP. 3)
TS
I
Progress of Reaction
> React ion Progress (a)
Fig. 3-4
Fig. 3-3
See Fig. 3-4, in which R = reactants, I = intermediates, P = products, TS, = transition state of first step, TS, = transition state of second step. Because the reactions are exothermic, H , < H , . Problem 3.20 In Problem 3.19(6), the first step is not only fast but is also reversible. Explain.
4
The A H S for I to revert to reactants R is less than that for I to form the products P. Therefore, most 1’s re-form R, so that the first step is fast and reversible. A few 1’s have enough enthalpy to go through the higher-enthalpy TS, and form the products. The A H S for P to revert to I is prohibitively high; hence the products accumulate, and the second step is at best insignificantly reversible. Problem 3.21 Catalysts generally speed up reactions by lowering A H S . Explain how this occurs in terms of 4 ground-state and transition-state enthalpies ( H , and H T s ) .
A H S can be decreased by ( a ) lowering H T s , (6) raising H , or (c) both of these. Problem 3.22 The reaction A + B + C
+ D has ( a ) rate mechanisms consistent wih these rate expressions.
or (6) rate
= k[A][B],
= k[A].
Offer possible
4
( a ) Molecules A and B must collide in a bimolecular rate-controlling step. Since the balanced chemical
equation calls for reaction of one A molecule with one B molecule, the reaction must have a single (concerted) step. ( 6 ) The rate-determining step is unimolecular and involves only an A molecule. There can be no prior fast steps. Molecule B reacts in the second step, which is fast. A possible two-step mechanism is: Step 1: Step 2:
C + I ( I = intermediate) B+1 4 D
A+
Adding the two steps gives the balanced chemical equation: A + B-, C + D. Problem 3.23 For the reaction 2A + 2B-, C + D, rate = k[Al2[B). Give a mechanism using only unimolecular or bimolecular steps. 4
One B molecule and two A molecules are needed to give the species for the slow step. The three molecules do not collide simultaneously since we are disregarding the very rare termolecular steps. There must then be some number of prior fast steps to furnish at least one intermediate needed for the slow step. The second B molecule which appears in the reaction equation must be consumed in a fast step following the slow step.
-
Mechanism 1
A+B
fast
AB + AA,B
+ B-c
AB (intermediate)
slow
fast
A,B
(intermediate)
+D
A +A
A,
-
+ B-
A,B
Mechanism 2
fast
slow
+ B-C
fast
(intermediate)
A,
A,B
(intermediate)
+D
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
40
Problem 3.24 For the reaction A + 2B-C determining step is unimolecular.
+ D,
rate
=
[CHAP. 3
k[A][B]'. Offer a mechanism in which the rate4
The slow step needs an intermediate formed from one A molecule and two B molecules. Since the rate expression involves the same kinds and numbers o f molecules as does the chemical equation, there are no fast steps following the slow step. Mechanism 2
Mechanism I ld*1
A+B-AB
B+B-B,
AB + B-AB,
B2 + A-
\IOU
AB,-C+D
A,B-C
fd\t
fast
A,B
+D
clow
Notice that often the rate expression is insufficient to allow the suggestion of an unequivocal mechanism. More experimental information is often needed.
BRONSTED ACIDS AND BASES
3.10
In the Bronsted definition, an acid donates a proton and a base accepts a proton. The strengths of acids and bases are measured by the extent to which they lose or gain protons, respectively. In these reactions acids are converted to their conjugate bases and bases to their conjugate acids. Acid-base reactions go in the direction of forming the weaker acid and the weaker base. Problem 3.25 Show the conjugate acids and bases in the reaction of H,O with gaseous ( a ) HCI, ( b ) :NH,.
4
H,O, behaving as a Bronsted base, accepts a proton from HCI, the Bronsted acid. They are converted to the conjugate acid H,O + and the conjugate base Cl -, respectively. HCI + H,O,-H,O'
+ CK
acid ba-id? base I (stronger) (stronger) (weaker) (weaker)
The conjugate acid-base pairs have the same subscript and are bracketed together. This reaction goes almost to completion because HCI is a good proton donor and hence a strong acid. H,O is amphoteric and can also act as an acid by donating a proton to :NH,. H,O is converted to its conjugate base, OH-., and :NHl to its conjugate acid. NHf. H : O H + N H , e N H i +:OHagd I
ba-di,
base,
:NH, is a poor proton acceptor (a weak base); the arrows are written to show that the equilibrium lies mainly to the left. To be called an acid, the species must be more acidic than water and be able to donate a proton to water. Some compounds, such as alcohols, are not acidic toward water, but have an H which is acidic enough to react with very strong bases or with Na. Problem 3.26 Write an equation for the reaction of ethanol with ( a ) NH;, a very strong base; ( b ) Na.
(4 (b)
CH,CH,OH
+ :NHi -CH,CH20:-
2CH,CH20H + 2Na-2CH,CH20:-
4
+ :NH, + 2Na' + H,
Relative quantitative strengths of acids and bases are given either by their ionization constants, K , and K,, o r by their pK, and p K , values as defined by:
41
C H EM I CA L RE ACT1V ITY AN D 0RG A N I C REACT10N S
CHAP. 31
PK,, = -log K,,
P K , = - 1% K',
The stronger an acid or base, the larger its ionization constant and the smaller its pK value. The strengths of bases can be evaluated from those of their conjugate acids; the strengths of acids can be evaluated from those of their conjugate bases. The strongest acids have the weakest conjugate bases, and the strongest bases have the weakest conjugate acids. This follows from the relationships K,,
= (K[,)(K,,)= 1Wl4
PK,,+ PK,, = P K ,
=
14
in which K w , the ion-product of water = (H,O'][OH-1.
3.1 1 BASICITY (ACIDITY) AND STRUCTURE
The basicity of a species depends on the reactivity of the atom with the unshared pair of electrons, this atom being the basic site for accepting the H'. The more spread-out (dispersed, delocalized) is the electron density on the basic site, the less basic is the species. The acidity of a species can be determined from the basicity of its conjugate base.
DELOCALIZATION RULES (1) For bases of binary acids (H,,X) of elements in the same Group, the larger the basic site X , the more spread-out is the charge. Compared bases must have the same charge. (2) For like-charged bases of binary acids in the same period, the more unshared pairs of electrons the basic site has, the more delocalized is the charge. (3) The more s character in the orbital with the unshared pair of electrons, the more delocalized the electronic charge. (4) Extended p - p T bonding between the basic site and an adjacent T system (resonance) delocalizes the electronic charge. ( 5 ) Extended p - d T bonding with adjacent atoms that are able to acquire more than an octet of electrons delocalizes the electronic charge. (6) Delocalization can occur via the inductive effect, whereby an electronegative atom transmits its electron-withdrawing effect through a chain of 0 bonds. Electropositive groups are electrondonating and localize more electron density on the basic site. are
With reference to (4) and ( 5 ) . some common
I
-C=O carbonyl
&N nitrile
-N=O
I
0
nitro
7~
systems that participate in extended 0
I -s=o I
7~
bonding
<<-
a-
; ( b ) :NH,
( N uses sp' HO's)
I
I
0
sulfon y I ( p-d 77 bonding)
Problem 3.27 Compare the basicities of the following pairs ( a ) RS and R O and :PH, ( P uses p AO's for its three bonds); ( c ) N H , and O H .
4
(a) S and 0 are in the same periodic group, but S is larger and its charge is more delocalized. Therefore. RS is a weaker base than R O and RSH is a stronger acid than ROH. (6) Since the bonding pairs of electrons of :PH, are in p AO's, the unshared pair of electrons is in an s AO. In :NH, the unshared pair is in an sp' HO. Consequently, the orbital of P with the unshared electrons has more s character and PH, is the weaker base. In water PH, has no basic property. ( c ) : O H - has more unshared pairs of electrons than :NH,. its charge is more delocalized. and it is the weaker base.
42
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
[CHAP. 3
Problem 3.28 Compare and account for the acidity of the underlined H in:
and
R-0--tI
R-C-0--H_
II
0 H-CH2-C=CH2
H-CH2-CH-CH,
H-CH,-C-CH,
It
I
I
4
0
Compare the stability of the conjugate bases in each case.
R-C
/Ot---*
R-C
//3
or R-C
‘0-
O\
,o \
0
the C and 0 ’ s participate in extended n bonding so that the - is distributed to each 0. In RO- the - is localized on the 0. Hence RCOO- is a weaker base than RO-, and RCOOH is a stronger acid than ROH. ( b ) The stability of the carbanions and relative acidity of these compounds is (111) > (I) > (11) Both (I) and (111) have a double bond not present in (11) that permits delocalization by extended 7r bonding. Delocalization is more effective in (111) because charge is delocalized to the electronegative 0.
Problem 3.29 In terms of delocalization explain why HCCI, is more acidic than HCF,.
4
CI,CC is less basic than F,CC . F can disperse charge only by an inductive effect [see Fig. 3-5(a)]. In addition to an inductive effect, CI uses its empty 3d orbitals to disperse charge by p-d 7~ bonding, see Fig. 3-5(b)]. F is a second-period element with n o 2d orbitals.
X
= CI.
F
Problem 3.30 Why is a sulfonic acid, such as methanesulfonic acid (CH,SO,H), a stronger acid than a selenic acid, such as methaneselenic acid (CH,SeO,H)? 4
In either conjugate base there is p-d 7r bonding between the central heteroatom and three 0 atoms which bear some of the negative charge. Thus, the conjugate bases are very weak and the acids are very strong. However, the greater the difference between the principal energy levels of the p and d AO’s, the less effective is the overlap and the less delocalization occurs. Hence, CH,SeO,, in which Se uses a 4d AO, has less
CHAP. 3)
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
-
43
delocalization of the negative charge and is a stronger base than CH,SOj, in which S uses a 3d A 0 to overlap with the 2 p AO’s of the 0 atoms, Furthermore, Se-0 is a longer bond than S - 0 . Problem 3.31 Account for the decreasing order of basicity in the following amines: CH,NH, > NH3 > NF,.
4
The F’s are very electronegative and, by their inductive effects, they delocalize electron density from the N atom. The N in NF, has less electron density than the N in NH,; NF, is a weaker base than NH,. The CH, group, on the other hand, is electron-donating and localizes more electron density on the N of CH,NH,, making CH,NH, a stronger base than NH,. Problem 3.32 Account for the decreasing order of acidity: HC-=CH > H,C=CH, > CH,CH,. We apply the HON rule to the basic site C of the conjugate bases. For H-C, the HON is 2 and the C uses sp HO’s. For H,C&H-, the HON is 3 and the C uses sp2 HO’s. For CH,CH,, the HON is 4 and the C uses sp3 HO’s. As the s character decreases, the basicities increase and the acidities decrease.
3.12 LEWIS ACIDS AND BASES A Lewis acid (electrophile) shares an electron pair furnished by a Lewis base (nucleophile) to form a covalent (coordinate) bond. The Lewis concept is especially useful in explaining the acidity of an aprotic acid (no available proton), such as BF,.
H F
F
H
H:N: H
I I
I I
F
Lewis
H F
Lewis acid
base
The three types of nucleophiles are listed in Section 3.4. Problem 3.33 Explain the observation that neither pure H,SO, nor pure HCIO, conducts an electric current 4 but a mixture of the two does. Neither pure acid is ionized. In the mixture the stronger acid, HCIO,, donates a proton to H,SO,, which acts as a base because of the unshared electron pairs on the 0’s. H:O:CIO,
+
acid,
H:O:SO,H
H’ H:O:SO,H
base
Problem 3.34 Given the following Lewis acid-base reactions: Lewis base 4H,N: 2:i%H:&
+ Lewis acid + cu2+ + SiF, + :o=c=g:
+
:O:ClO;
acidz
--, --+
__*
*
base,
Product Cu(NH,):” SiFa- .. :Q-C=(j: !)€.I
H H,C=CH,
+
H’(froma Bronsted acid)
H2C=0:
+
BF,
+
I
H,6-CH2 +
--+
H,C=O:BF,
[CHAP. 3
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
44
(a) Group the bases as follows: ( 1 ) anions, (2) molecules with an unshared pair of electrons, ( 3 ) negative end of a 7r bond dipole, and (4) available 7r electrons. ( b ) Group the acids as follows: (1) cations, (2) species with electron-deficient atoms, (3) species with an atom capable of expanding an octet, and (4) positive end of a 7r bond dipole. 4 (0)
(1)
O H - , F-
(b)
(1)
Cu”, H’
6
(2) :NH1. H,C=O (2) BF,
k
6 -
(3) H 2 C = 0 :
(4)
H,C=CH,
(3) SiF,
(4)
:O=C=Q:
6 -
4+
6-
Supplementary Problems Problem 3.35 Which of the following reactions can take place with carbocations? Give examples when reactions do occur. ( a ) acts as an acid; ( b ) reacts as an electrophile; (c) reacts as a nucleophile; ( d ) undergoes 4 rearrangement.
Carbocations may undergo ( a ) , ( b ) , and ( d ) .
CH (d)
CH,
l 3
I
CH , - C - ~ H C H , ~ CH,-C-CH,CH, H
Problem 3.36 Give three-dimensional representations for the orbitals used by the C’s in (a) H , C = C H + and ( h ) H,C=CH-. 4
I n both (a) and ( b ) the C of the CH, group uses sp2 HO’s to form U.bonds with the two H’s and the other C. In (a) the charged C uses two sp HO’s to form two U bonds, one with H and one with the other C. One p A 0 forms a TT bond with the other C, and the second unhybridized p A 0 has no electrons, see Fig. 3-6(a). In ( b ) the charged C uses three sp2 HO’s: two form U bonds with the H and the other C , and the third has the unshared pair of electrons, see Fig. 3-6(6). sp‘ HO with lone electron pair
r
P,
(b)
Fig. 3-6
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
CHAP. 3)
45
Problem 3.37 Write the formula for the carbon intermediate indicated by ?, and label as to type.
+ :N=N:
H,C-N=N
+ .Hg* + N2
(CH,),Hg-? -
H,C-N=N:
-?
(cH,),cOH
+ H + +?
+ H,O:
H--C-'C-H+Na.-?+Na'+
$H,
+ D-Br-? + Br+ Zn" + 21H,CI, + Zn:-? (CH,),C-CI + AICI,-? + AlClJ H,C=CHz
( a ) and (6) H,C., a radical.+(c) and (g) H,C:, a carbene. ( d ) and (h) (CH,),C', a carbocation. (e)
H--C-=C:,
a carbanion.
(f)H,C--CH,-D,
a carbocation.
Problem 3.38 Classify the following reactions by type.
+ OH-
H,C-CH,-Br OH I (CH,),CHOH CH2
-
C,H,
+ H,
+ HNO,
PI
H,SO,
H C O O H H,O ~ CH,=C=O
'd
+ H,O + Br-
+ H,
H,C=CHCH,
\/
H,C=CH,
H2C-CH2
(CH,),C=O
HZC-CH2 H,C-CH,Br
__*
+ :H-
-
H3C-CH,
+ :Br-
H,C--CH,
+ H,O
C,H,NO,
+ CO
+ H,O+
CH,COOH
H2C=0
+ 2Ag(NH,)S + 3 0 H - -
H,C=O
+ HCN+
HCOO-
+ 2Ag + 4NH, + 2H,O
H,C(OH)CN
4
( a ) Elimination and an intramolecular displacement; a C - 0 bond is formed in place of a C-Br bond. (6) Elimination and redox; the alcohol is oxidized to a ketone. (c) Rearrangement. ( d ) Displacement and redox; H,CCH,Br is reduced. (e) Addition and redox; H,C=CH, is reduced. (f)Substitution. ( g ) Elimination. ( h ) Addition. (i) Redox. ( j ) Addition.
Problem 3.39 Which of the following species behave as (1) a nucleophile, (2) an electrophile. (3) both, or (4) neither?
(d) AIBr,
(g)
Br'
( 1 ) NO:
( m ) CH,=CHCH f
(6) H,O:
(4
CH,W
(h)
Fe3'
(4
CH,
(4 H'
(f)
Becl,
(i)
SnCI,
(k) H,C=O (I) CH,(%N
(0)
H,C=CHCH,
(a)
:CL-
4
(1): ( a ) , (0 ( e ) , (0). (2): ( c ) , (4,(f), ( g ) , ( h ) , (i), ( j ) . (m). (3): ( k ) and ( I ) (because carbon is electrophilic; oxygen and nitrogen are nucleophilic). (4): (n). Problem 3.40 Formulate the following as a two-step reaction and label nucleophiles and electrophiles.
46
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
H,C=CH, Step 1
7
H,C=CH,
+ Br-BrP
nucleophile
Step 2
+
+ Br,
I
H,C-tH,
__*
--+
I Br
electrophile I
+ Brnucleophile
H,C-CH,
I
Br
Br electrophile,
4
I I Br Br
electrophile,
+ Br-
H,C-CH,
+H,C-CH,
[CHAP. 3
I
Br
nucleophile,
Problem 3.41 The addition of 3 mol of H, to 1 mol of benzene. C,H,.
3H2
+ C,H,
Pd(rc)
=.C,H,,
occurs at room temperature (rt); the reverse elimination reaction proceeds at 300 "C. For the addition reaction, AH and AS are both negative. Explain in terms of thermodynamic functions: (a) why A S is negative and (6) why 4 the addition doesn't proceed at room temperature without a catalyst. (a) A negative AH tends to make AG negative, but a negative AS tends to make AG positive. At room temperature, AH exceeds T AS and therefore AG is negative. At the higher temperature (300"C), T AS exceeds AH, and AG is then positive. A S is negative because four molecules become one molecule, thereby reducing the randomness of the system. (6) The addition has a very high AHS, and the rate without catalyst is extremely slow. Problem 3.42 The reaction CH, + Iz-,CH,I + HI does not occur as written because the equilibrium lies to the left. Explain in terms of the bond dissociation energies, which are 427, 151, 222, and 297 kJ/mol for C-H, 4 1-1, C-I, and H-I, respectively.
The endothermic, bond-breaking energies are +427(C-H) and + 151(1-I), for a total of +578 kJ/mol. The exothermic, bond-forming energies are -222(C--I) and -297(H--I), for a total of -519 kJ/mol. The net A H is
+ 578 - 519 = +59 kJ/mol and the reaction is endothermic. The reactants and products have similar structures, so the AS term is insignificant. Reaction does not occur because AH and AG are positive. Problem 3.43 Which of the isomers ethyl alcohol, H,CCH,OH, or dimethyl ether, H,COCH,. has the lower enthalpy? The bond dissociation energies are 356, 414, 360, and 464 kJ/mol for C - C , C-H, C - 0 and 0-H, respectively. 4
C,H,OH has 1 C - - C bond (356kJ/mol), 5 C-H bonds ( 5 x 414kJ/mol), one C - 0 bond and one 0-H bond (464kJ/mol), giving a total energy of 3250kJ/mol. CH,OCH, has 6 (6 x 414 kJ/mol) and 2 C - 4 bonds (2 x 360 kJ/mol), and the total bond energy is 3204 kJ/mol. the higher total bond energy and therefore the lower enthalpy of formation. More energy decompose C,H,OH into its elements.
(360kJ/mol) C-H bonds C,H,OH has is needed to
Problem 3.44 Consider the following sequence of steps:
(1) A-B
(2) B + C z D + E
(3) E+A---+2F
(a) Which species may be described as (i) reactant, (ii) product and (iii) intermediate? (6) Write the net chemical equation. (c) Indicate the molecularity of each step. ( d ) If the second step is rate-determining, write 4 the rate expression. (e) Draw a plausible reaction-enthalpy diagram. ( a ) (i) A, C; (ii) D, F; (iii) B, E.
CHEMICAL REACTIVITY AND ORGANIC REACTIONS
CHAP. 31
(b) (c)
47
2A + C+ D + 2F (add steps 1, 2, and 3). (1) unimolecular, (2) bimolecular, (3) bimolecular.
(d) Rate = k[C][A], since A is needed to make the intermediate, B. (e) See Fig. 3-7.
Reaction Progress Fig. 3-7
Problem 3.45 A minor step in Problem 3.44 is 2E-,G.
What is G?
4
A side product.
Problem 3.46 The rate expression for the reaction (CH,),C-Br
+ CH,COO- + Ag'
--+
CH,COOC(CH,),
+ AgBr
Rate = k[(CH,),C-Br][Ag']
is
Suggest a plausible two-step mechanism showing the reacting electrophiles and nucleophiles. The rate-determining step involves only (CH,),C-Br participate in an ensuing fast step. Step 1
Step 2
a+
6-
(CH3)$2-ur:
T
and Ag'. The acetate ion CH,COO- must
+ Ag'
nucleophilic electrophile site +
CH,COO-
nucleophile
+ C(CH,),
4
(CH,),C'
+ AgBr
CH,COOC(CH,),
electrophile
Problem 3.47 Give the conjugate acid of ( a ) CH,NH,, (6) CH,O-, (c) CH,OH, (d) :H-, (e) :CH,, ( f ) H,C=CH,. 4
(4C H , W ,
( b ) CH,OH,
(4CH,OHl, ( 4 H,,
( e ) CH,, ( f ) H,CCHI.
Problem 3.48 What are the conjugate bases, if any, for the substances in Problem 3.47?
4
( a ) CH,NH-, ( 6 ) :CH202-, (c) CH,O-, (d) none, (e) H2C:,-, ( f ) H , w H - . The bases in (6) and (e) are extremely difficult to form; from a practical point of view CH,O- and H,C:- have no conjugate bases.
Problem 3.49 Are any of the following substances amphoteric? ( a ) H,O, (6) NH,, (c) NH:, (d) Cl-, (e) HCO,, ( f ) HF. 4 ( a ) Yes, gives H,O' and OH-. (6) Yes, gives NHI and H,N-. (c) No, cannot accept H'. (d) No, cannot donate H'. (e) Yes, gives H,CO, (CO, + H,O) and CO:-. ( f ) Yes, gives H,F' and F-.
48
CHEMICAL REACTIVITY A N D ORGANIC' REACTIONS
Problem 3.50 Account f o r the fact that acetic acid, CH\COOH, is CH,OH.
;i
(CHAP. 3
stronger acid in water than in methanol. 4
The equilibrium CH,COOH
+ H,O --_L CH,COO + H,O
'
lies more to the right than does
CH,COOH
+ CH,OH
C H ,COO
+ CH,OH
This difference could result if CH,OH were a weaker base than H,O. However, this might n o t be the case. The significant difference arises from solvation of the ions. Water solvates ions better than does methanol; thus the equilibrium is shifted more toward the right to form ions that are solvated by water. Problem 3.51 Refer t o Fig. 3-8, the enthalpy diagram for the reaction A-, B . ( U ) What d o states 1 , 2, and 3 represent'? ( h ) Is the reaction exothermic or endothermic? (c.) Which is the rate-determining step, A-2 or 2 - B ? ( d ) Can substance 2 ever be isolated from the mixture? ( e ) What represents the activation enthalpy of the overall reaction A + B'? (f)Is step A 2 reversible'? 4 --j
1 and 3 are transition states; 2 is an intermediate. Since the overall product, B , is at lower energy than reactant, A , the reaction is exothermic. The rate-determining step is the one with the higher enthalpy of activation, 2 - B . Yes. The activation enthalpy needed for 2 to get through transition state 3 may be so high that 2 is stable enough to be isolated. The A H ' is represented by the difference in enthalpy between A , the reactant, and the higher transition state, 3. The A H S for 2- A is less than the A H ' for 2- B; therefore, 2 returns to A more easily than it goes on to B. The step A-2 is fast and reversible.
tA
Enthalpy
Reaction Progress
Fig. 3-8.
-
Chapter 4 Alkanes 4.1
DEFINITION
Alkanes are open-chain (acyclic) hydrocarbons constituting the homologous series with the general formula C,rH,,,+2where n is an integer. They have only single bonds and therefore are said to be saturated. Problem 4.1 ( a ) Use the superscripts I , 2, 3. etc., to indicate the different kinds of equivalent H atoms in propane, CH,CH2CH,. (b) Replace one of each kind of H by a CH, group. (c) How many isomers of butane, C4HI0,exist? 4
CH;CH;CH;
CH:CH;CH;
E
H' H2 H' H~C-C-C,H' H' H' H
-H'
CH,CH,CH,m
=
n-Butane
CH; CHSCH; (c) Two:
I
-H?
CH,CHCH,
=
'm
lsobutane
n-butane and isobutane.
Problem 4.2 ( a ) Use the superscripts 1, 2, 3, 4, etc., to indicate the different kinds of equivalent H's in (1) n-butane and (2) isobutane. (b) Replace one of each kind of H in the two butanes by a CH,. (c) Give the 4 number of isomers of pentane, C,H,,. I
(a)
(1)
CHiCHSCHSCH:
CH3 i 4 3 3 A I H7 (2) C k 3 C A C k 3 or CH(CH3)3 or HC-C-CH H' H 4 H7 1 3 CH3
I H ' H2 H' H' or HC-C-C-CH' H' H2 H2 H CH;CH;CHICH:
CH,CH,CH,CH,
+& -
Fl
@
n-Pentane
CHiCHiCHiCH;
+h CH,CHCH,CH3 - H*
Isopentane
I CH:CH*CH;
y
3
I CH3CH2CH,ICH,]
+k - H3
Isopentane
CH,
Neopentane
(c) Three: n-pentane, isopentane, and neopentane.
49
50
ALKANES
[CHAP. 4
Sigma-bonded C's can rotate about the C - 4 bond and hence a chain of singly bonded C's can be arranged in any zigzag shape (conformation). Two such arrangements, for four consecutive C's, are shown in Fig. 4-1. Since these conformations cannot be isolated, they are not isomers.
Fig. 4-1
The two extreme conformations of ethane-called eclipsed [Fig. 4-2(a)) and staggered [Fig. 4-2(b))-are shown in the "wedge" and Newman projections. With the Newman projection, we sight along the C-C bond, so that the back C is hidden by the front C. The circle aids in distinguishing the bonds on the front C (touching at the center of the circle) from those on the back C (drawn to the circumference of the circle). In the eclipsed conformation, the bonds on the back C are, for visibility, slightly offset from a truly eclipsed view. The angle between a given C-H bond on the front C and the closest C-H bond on the back C is called the dihedral(torsiona1) angle (0). The 0 values for the closest pairs of C-H bonds in the eclipsed and staggered conformations are 0" and 60°,respectively. All intermediate conformations are called skew; their 8 values lie between 0" and 60". (See Fig. 4-2.)
( a ) Eclipsed
( b ) Staggered
Fig. 4-2
Figure 4-3 traces the energies of the conformations when one CH3 of ethane is rotated 360".
A
Eclipsed
Eclipsed
,e,
12 kJ / mol
8
ua
1 (r
Staggered 1
w
Staggered I
120"
1
IW
Angle of Rotation (Dihedral Angle) Fig. 4-3
I
2w
I
3aP
+
CHAP. 41
51
ALKANES
Problem 4.3 ( a ) Are the staggered and eclipsed conformations the only ones possible for ethane? (6) Indicate the preferential conformation of ethane molecules at room temperature. ( c ) What conformational changes occur as the temperature rises? (d) Is the rotation about the (T C-C bond, as in ethane, really “free”? 4 No. There is an infinite number with energies between those of the staggered and eclipsed conformations. For simplicity we are concerned only with conformations at minimum and maximum energies. ( 6 ) The staggered form has the minimal energy and hence is the preferred conformation. (c) The concentration of eclipsed conformations increases. (d) There is an energy barrier of 12 kJ/mol (enthalpy of activation) for one staggered conformation to pass through the eclipsed conformation to give another staggered conformation. Therefore, rotation about the sigma C - C bond in ethane is somewhat restricted rather than “free.” (0)
Problem 4.4 How many distinct compounds do the following structural formulas represent? (0)
CH3-CH-CH2-CH-CH2-CH3
I
( 6 ) CH2-CH-CH2-CH-CH3
I
CH3
I
CH,
CH3
I
I
CH,
CH3
CH3 CH3
I
(c) CH3-CH-CH2-CH-CH2-CH,
I
(d)
I
CH3
CH3
CH3 (e)
I
CH3-CH-CH2
I CH3-CH2-C-C I
4
H3
H
Two. ( a ) , ( 6 ) , (c), (e), and ( f ) are conformations of the same compound. This becomes obvious when the longest chain of carbons, in this case six, is written in a linear fashion. ( d ) represents a different compound. Problem 4.5 ( a ) Which of the following compounds can exist in different conformations? (1) hydrogen peroxide, HOOH; (2) ammonia, NH,; (3) hydroxylamine, H,NOH; (4) methyl alcohol, H,COH. (6) Draw two structural formulas for each compound in ( a ) possessing conformations. 4 ( a ) A compound must have a sequence of at least three consecutive single bonds, with no
exist in different conformations. ( l ) , (3), and (4)have such a sequence. In (2),
72
bonds, in order to
H
I
H-N-H
the three single bonds are not consecutive.
/H H*. 0 *N ‘ H
’
The first-drawn structure in each case is the eclipsed conformation; the second one is staggered.
52
ALKANES
[CHAP. 4
Problem 4.6 Explain the fact that the calculated entropy for ethane is much greater than the experimentally determined value. 4 The calculated value incorrectly assumes unrestricted free rotation so that all conformations are equally probable. Since most molecules of ethane have the staggered conformation, the structural randomness is less than calculated, and the actual observed entropy is less. This discrepancy led to the concept of conformations with different energies.
I
0
60
I I20
1 180
1
I
240
300
I
m
+
Rotation, degrees Fig. 4-4
Figure 4-4 shows extreme conformations of n-butane. The two eclipsed conformations, I and 11, are least stable. The totally eclipsed structure I, having eclipsed CH,’s, has a higher energy than 11, in which CH, eclipses H. Since the other three conformations are staggered, they are at energy minima and are the stable conformations (conformers) of butane. The anti conformer, having the CH,’s farthest apart, has the lowest energy, is the most stable, and constitutes the most numerous form of butane molecules. In the two, higher-energy, staggered, gauche conformers the CH,’s are closer than they are in the more stable anti form. Problem 4.7 Give two factors that account for the resistance to rotation through the high-energy eclipsed conformation. 4 Torsional strain arises from repulsion between the bonding pairs of electrons. which is greater in the eclipsed form because the electrons are closer. Steric strain arises from the proximity and bulkiness of the bonded atoms o r group of atoms. This strain is greater in the eclipsed form because the groups are closer. The larger the atom or group, the greater is the steric strain.
Problem 4.8 How does the relative population of an eclipsed and a staggered conformation depend on the energy difference between them? 4 The greater the energy difference, the more the population of the staggered conformer exceeds that of the eclipsed.
Problem 4.9 Draw a graph of potential energy plotted against angle of rotation for conformations of (a) 4 2.3-dimethylbutane. ( b ) 2-methylbutane. Point out the factors responsible for energy differences.
CHAP. 41
53
ALKANES
Start with the conformer having a pair of CH,'s anri. Write the conformations resulting from successive rotations about the central bond of 60". ( a ) As shown in Fig. 4-5(a), structure IV has each pair of CH,'s eclipsed and has the highest energy. Structures I1 and VI have the next highest energy; they have only one pair of eclipsed CH,'s. The stable conformers at energy minima are I, 111, and V. Structure I has both pairs of CH, groups anri and has the lowest energy. Structures 111 and V have one pair of CH,'s anfi and one pair gauche. (6) As shown in Fig. 4-5(6), the conformations in decreasing order of energy are: 1. IX and XI; have eclipsing CH,'s. 2. XIII; CH, and H eclipsing. 3. X; CH,'s are all gauche. 4. VIII and XII; have a pair of anti CH,'s. ( a ) 2,3-Dimethylbutane
H
CH3
H
(XIII)
H
H
H (VIII)
(VIII)
Angle of Rotation Fig. 4-5
ALKANES
54
4.2
[CHAP. 4
NOMENCLATURE OF ALKANES
The letter n (for normal), as in n-butane, denotes an unbranched chain of C atoms. The prefix iso(i-) indicates a CH, branch on the second C from the end; e.g., isopentane is
CH,CHCH,CH3
I
CH, Alkyl groups, such as methyl (CH,) and ethyl (CH,CH,), are derived by removing one H from a1kanes. The prefixes sec- and tert- before the name of the group indicate that the H was removed from a secondary or tertiary C, respectively. A secondary C has bonds to two other C's, a tertiary to three other C's, and a primary either to three H's or to two H's and one C. The H's attached to these types of carbon atoms are also called primary, secondary and tertiary (l', 2" and 3"), respectively. A quaternary C is bonded to four other C's. The letter R is often used to represent any alkyl group. Name the alkyl groups originating from ( a ) propane, ( 6 ) n-butane, (c) isobutane.
Problem 4.10
L
CH, HCH, is isopropyl (i-Pr).
(4
CH3CH2CHZ-- is n-propyl (n-Pr);
(6)
CH3CHzCH2CH2- is n-butyl (n-Bu);
(c)
CH3-CHCHz-
I
4
I
CH,CHCH,CH, is sec -butyl (s-Bu).
I
is isobutyl (i-Bu);
is tert-butyl (t-Bu).
CH3 Problem 4.11
Use numbers I, 2, 3, and 4 to designate the l", 2", 3", and 4" C's, respectively, in
CH,CH,C(CH,),CH,CH(CH,),.Use letters a, 6, c, etc., to indicate the different kinds of 1" and 2"C's. 4
CH, Ib
Problem 4.12 Name by the IUPAC system the isomers of pentane derived in Problem 4.2. CH,CH,CH,CH,CH,
Pentane
4
(IUPAC does not use n)
The longest consecutive chain in I
CH3
21
3
4
CH,CHCH,CH, has 4 C's and therefore the IUPAC name is a substituted butane. Number the C's as shown so that the branch CH, is on the C with the lower number, in this case C2. The name is 2-methylbutane and not 3-methylbutane. Note that numbers are separated from letters by a hyphen and words are run together. The longest consecutive chain in CH3
I I
CH,-C-CH, CH3
CHAP. 41
55
ALKANES
has three C’s; the parent is propane. The IUPAC name is 2,2-dimethylpropane. Note the use of the prefix di- to show two CH, branches, and the repetition of the number 2 to show that both CH,’s are on C’. Commas separate numbers and hyphens separate numbers and words.
Problem 4.13 Name the compound in Fig. 4-6(a) by the IUPAC system.
4
H CHICH,
I l l I l l CH2CHzCH2 I l l
-C-C-C-CHI
CHIC HICH 3
(a) Fig. 4-6
The longest chain of consecutive C’s has 7 C’s [see Fig. 4 - 6 ( b ) ] , and so the compound is named as a heptane. Note that, as written, this longest chain is bent and not linear. Circle the branch alkyl groups and consecutively number the C’s in the chain so that the lower-numbered C’s hold the most branch groups. The name is 3,3.4,5-tetrame thyl-4-ethylheptane.
4.3
PREPARATION OF ALKANES
REACTIONS WITH NO CHANGE IN CARBON SKELETON 1,
Reduction of AIkyl Halides (RX, X = C1, Br, or I) (Substitution of Halogen by Hydrogen)
-
R X + Z n +H’-RH+Zn2’+ X4RX + LiAIH,4RH + LiX + AIX, (b) RX + H r or RH + X- ( H r comes from LiAIH,) R H + (n-C,H,),SnX (c) RX + (n-C,H,),SnH( d ) Via organometallic compounds (Grignard reagent). Alkyl halides react with either Mg or Li in dry ether to give organometallics having a basic carbanionic site. (a)
dry e t h e r
RX + 2Li-
R:Li’
+ LiX
then R: i i + H,O-
RH
+
-
+ LiOH
Alkyllithiurn
RX + Mg-
dry ether
R:(MgX)’
then
R(MgX) + H,O-
Grignard reagent
The net effect is replacement of X by H. 2. Hydrogenation of ‘C=C’
/
\
(alkenes) or
a(alkynes)
CH3
I
CH3-C=CH2
CH3
I
+ H2
CH3-CH-CH3
Isobutylene
CH,--C-=C-H
Isobutane Pt
+ 2H,-CH,--CH2
Propane
R H +(MgX)+(OH)-
56
ALKANES
[CHAP. 4
PRODUCTS WITH MORE CARBONS THAN THE REACTANTS
Two alkyl groups can be coupled by indirectly reacting two molecules of RX, or RX with R'X, to give R-R or R-R', respectively. The preferred method is the Corey-House synthesis, which uses the organometallic lithium dialkylcuprates, R,CuLi, as intermediates. 2 R-Li
R,CuLi
+ CuI-
+ R'-X-
[(CH,),C],CuLi
I
ether
R,CuLi
ether
R-R'
+ Br--CH,CH,CH,
H
+ LiI (Most + RCu + LiX
-
R groups are possible.)
(All R' groups except 3".)
(CH,),C--CH,CH,CH,
+ (CH3),CHCu + LiBr
I
H
2- Met h y lpe n t ane
Lithium diisopropylcuprate
Problem 4.14 Write equations to show the products obtained from the reactions: ( a ) 2-Bromo-2-methylpropane
( b ) Product of ( a ) + H,O
+ magnesium
Product of ( a ) + D 2 0
(c)
4
CH3 CH-&r(MgBr)+
I
CH,
I I
+ HOH +CH,-C-H
CH3 base,
(c)
in dry ether
+ (MgBr+)(OH-)
CH,
acidz
acid,
base
The r-butyl carbanion accepts a deuterium cation to form 2-methyl-2-deuteropropane,(CH,),CD.
Problem 4.15 Use 1-bromo-2-methylbutane and any other one- or two-carbon compounds, if needed, to synthesize the following with good yields: (a)
(4
2-methylbutane
(6) 3,6-dimethyloctane
3-methylhexane
CH,
CH,
I BrCH,CHCH,CH,-
(c)
Mr
I
BrMgCH2CHCH,CH3-
4
CH7 H20
I
CH,CHCH,CH,
BrCHzCHCH2CH, I
CH,
CHAP. 41
ALKANES
Problem 4.16 What kind of alkane can be made by the indirect coupling of two molecules of RX? The product is a symmetrical alkane, R-R,
57
4
whose left end is a duplicate of its right end.
Problem 4.17 Give the different combinations of RX and R’X that can be used to synthesize 3-methylpen4 tane. Which synthesis is “best”? From the structural formula CH,*H,~CHXH+CH, I3
CH,
we see that there are three kinds of C - C bonds, labeled 1 , 2, 3. The combinations are: for bond 1 , CH,X and XCH,CH(CH,)CH,CH,; for bond 2, CH,CH,X and XCH(CH,)CH2CH,, for bond 3, CH,X and XCH(CH,CH,),. The chosen method utilizes the simplest, and least expensive, alkyl halides. On this basis, bond 2 is the one to form on coupling.
4.4
CHEMICAL PROPERTIES OF ALKANES
Alkanes are unreactive except under vigorous conditions. 1.
Pyrolytic Cracking [heat (A) in absence of 0,; used in making gasoline]
Alkane 5mixture of smaller hydrocarbons 2. Combustion
CH,
600°C + 210, -CO,
+ 2H,O
AH of combustion = -809.2 kJ/mol
(see Problem 4.39). Problem 4.18 ( a ) Why are alkanes inert? (6) Why do the C--C rather than the C-H bonds break when alkanes are pyrolyzed? (c) Although combustion of alkanes is a strongly exothermic process, it does not occur at 4 moderate temperatures. Explain. (a) A reactive site in a molecule usually has one or more unshared pairs of electrons, a polar bond, an electron-deficient atom or an atom with an expandable octet. Alkanes have none of these. (6) The C--C bond has a lower bond energy ( A H = +347 kJ/mol) than does the C-H bond ( A H = +414 k J / mol). (c) The reaction is very slow at room temperature because of a very high AH’.
3. Halogenation
RH + X
(Reactivity of X,: F, > C1, > Br,.
2
or A
2 RX + HX
I, does not react; F, destroys the molecule)
Chlorination (and bromination) of alkanes such as methane, CH,, has a radical-chain mechanism, as follows: INITIATION STEP C I : C I A 2C1. or A
A H = +243 kJ/mol
The required enthalpy comes from ultraviolet (uv) light or heat. PROPAGATION STEPS
(i) H,C:H + C1.(ii) H,C. + C1:Cl-
H,C. + H:CI H,C:Cl + C1.
AH = -4 kJ/mol (rate-determining) AH = -96 kJ/mol
58
[CHAP. 4
ALKANES
The sum of the two propagation steps is the overall reaction,
CH,
+ Cl,--+CH,CI + HCI
AH = -100 kJ/mol
In propagation steps, the same free-radical intermediates, here CI. and H,C., are being formed and consumed. Chains terminate on those rare occasions when two free-radical intermediates form a covalent bond:
Inhibitors stop chain propagation by reacting with free-radical intermediates, e.g.
The inhibitor-here 0,-must be consumed before chlorination can occur. In more complex alkanes, the abstraction of each different kind of H atom gives a different isomeric product. Three factors determine the relative yields of the isomeric product. (1) Probability factor. This factor is based on the number of each kind of H atom in the molecule. For example, in CH,CH,CH,CH, there are six equivalent 1" H's and four equivalent 2" H's. The odds on abstracting a 1" H are thus 6 to 4, or 3 to 2. (2) Reactivity of H. The order of reactivity of H is 3" > 2" > 1". (3) Reactivity of X-. The more reactive CI. is less selective and more influenced by the probability factor. The less reactive Br. is more selective and less influenced by the probability factor. As summarized by the reactivity-selectivityprinciple: If the attacking species is more reactive, it will be less selective, and the yields will be closer to those expected from the probability factor. Problem 4.19 ( a ) List the monobromo derivatives of (i) CH,CH2CH,CH, and (ii) (CH,),CHCH,. (6) Predict the predominant isomer in each case. The order of reactivity of H for bromination is
3"(16OO) > 2"(82) > 1"( 1 )
4
H's, and there are two possible isomers for each compound: (i) CH,CH,CH,CH,Br and CH,CHBrCH,CH,; (ii) (CH,),CHCH,Br and (CH,),CBrCH,. (6) In bromination, in general, the difference in reactivity completely overshadows the probability effect in determining product yields. (i) CH,CHBrCH,CH, is formed by replacing a 2" H; (ii) (CH,),CBrCH, is formed by replacing the 3" H and predominates. ( a ) There are two kinds of
Problem 4.20
Using the bond dissociation energies for X,,
show that the initiation step for halogenation of alkanes,
x, is not rate-determining.
+2x.
CHAP. 41
ALKANES
59
The enthalpy AH' (Section 3.8) is seldom related to A H of the reaction. In this reaction, however, A H S and A H are identical. In simple homolytic dissociations of this type, the free radicals formed have the same enthalpy as does the transition state. On this basis alone, iodine, having the smallest A H and AH', should react fastest. Similarly, chlorine, with the largest A H and AH', should react slowest. But the actual order of reaction rates is F, > CI, > Br, > I, Therefore, the initiation step is not rate-determining; the rate is determined by the first propagation step, H-abstraction.
Problem 4.21 Draw the reactants, transition state, and products for
Br.
+ CH,
+ CH,+HBr
4
In the transition state, Br is losing radical character while C is becoming a radical; both atoms have partial radical character as indicated by 6 . . The C atom undergoes a change in hybridization as indicated: REACTANTS
H
\
.C-H
He* 1
+
Br.
H
-
TRANSITION STATE 1
-
C is becoming s p * (trigonal planar)
C is sp' (tetrahedral)
PRODUCTS
H
I
.C.
+
H-Br
H" H ' C is s p 2 (trigonal planar)
Problem 4.22 Bromination of methane, like chlorination, is exothermic, but it proceeds at a slower rate under the same conditions. Explain in terms of the factors that affect the rate, assuming that the rate-controlling step is
X.
HX + *CH,
+ CH,-
4
Given the same concentration of CH, and C1. or Bra, the frequency of collisions should be the same. Because of the similarity of the two reactions, A S s for each is about the same. The difference must be due to the AH*, which is less (17 kJ/mol) for C1. than for Br- (75 kJ/mol). Problem 4.23 2-Methylbutane has l", 2", and 3" H's as indicated: 3"
2"
I"
(cH, ),CHCH,CH,
( a ) Use enthalpy-reaction progress diagrams for the abstraction of each kind of hydrogen by .X.( b ) Summarize the relationships of relative (i) stabilities of transition states, (ii) A H s values, (iii) stabilities of alkyl radicals, and (iv) rates of H-abstraction. 4 ( a ) See Fig. 4-7. (6) (i) 3" > 2"> 1" since the enthalpy of the TS,, is the greatest and the enthalpy of the TS,. is the smallest. (ii) A H : . < A H i . < A H : , . (iii) 3">2">1". (iv) 3">2">1".
Problem 4.24 List and compare the differences in the properties of the transition states during chlorination 4 and bromination that account for the different reactivities for l", 2", and 3"H's.
ALKANES
60
t
[CHAP. 4
TS,.
TS,.
6.
6.
(CH,),C---H --X
+ HX
Reaction Progress Fig. 4-7
The differences may be summarized as follows:
1. Time of formation of transition state 2. Amount of breaking of C-H bond 3. Free-radical character ( 6 . ) of carbon 4. Transition state more closely resembles
Chlorination Earlier in reaction
Bromination Later in reaction
Less, H,C - - - H - - - - - C1
More, H,C- - - - H- - - Br
Less
More
Reactants
Products
These show that the greater selectivity in bromination is attributable to the greater free-radical character of carbon. With greater radical character, the differences in stability between l", 2", and 3" radicals become more important, and reactivity of H (3" > 2" > 1") also becomes more significant.
Problem 4.25 Account for the order of stability of radicals: 3" > 2" > 1".
4
The carbon atom with the odd electron is an electron-deficient site. Alkyl groups attached to such a C have an electron-releasing inductive effect which attenuates the electron deficiency, thereby stabilizing the radical.
4.
Isomerization
AICIj. HCI
CH3
I
CH3CH2CHzCH3 I CH3-CH-CH3
CHAP. 41
4.5
61
ALKANES
SUMMARY OF ALKANE CHEMISTRY
-
PREPARATION 1. C Skeleton Preserved (a) Direct Replacement of X by H
CH ,CH,CH,CH,X
PROPERTIES 1. Thermal Dehydrogenation
H2C=CH-CH=CH2
+ 2H,
2. Combustion
or
+ O,-CO,
CH,CH,CHXCH3
3. Halogenation (X = C1, Br)
(6) From Organometal
+ X,+ h
(M = Li, MgX) CH,CH,CH,CH,M* or CH ,CH,CHMCH
+ H,O
H,O --+
11
CH,CH,CHXCH, (major) + CH,CH,CH,CH,X
CH,CH2CH2CH, Butane
,
(c) Hydrogenation of Alkenes
CH,CH,CH=CH, or
CH,CH=CHCH, 4. Isomerization
2. Coupling of RX
(CH,),CH
(a) With Cuprate (Corey-House)
CH,CH,X-
1 LI
2 Cul
(CH,CH,),CuLi
( b ) With Na (Wurtz)-poor 2 CH,CH,X
+ C,H5X-
yield Na
*From RX with Li or Mg. ' Also from corresponding alkynes.
Supplementary Problems Problem 4.26 Assign numbers, ranging from (1) for LOWEST to (3) for HIGHEST. to the boiling points of the following hexane isomers: 2,2-dimethylbutane, 3-methylpentane, and n- hexane. Do not consult any tables for data. 4 Hexane has the longest chain. the greatest intermolecular attraction. and, therefore, the highest boiling point, (3). 2,2-Dimethylbutane is ( I ), since, with the most spherical shape, it has the smallest intermolecular contact and attraction. This leaves 3-methylpentane as (2).
62
ALKANES
[CHAP. 4
Problem 4.27 Write structural formulas for the five isomeric hexanes and name them by the IUPAC system.
4
The isomer with the longest chain is hexane, CH,CH,CH,CH,CH,CH,. If we use a five-carbon chain, a CH, may be placed on either C’ o r CJ to produce 2-methylpentane, or on C’ to give another isomer, 3-methylpentane.
CH3
CH3 ’
1
1
4
7
dHICH2CHCH2CHI
bH$HtHztH,6H, 2-Methylpentane
3-Methylpentane
With a four-carbon chain either a CH,CH, o r two CH,’s must be added as branches to give a total of 6 C’s. Placing CH,CH, anywhere on the chain is ruled out because it lengthens the chain. Two CH,’s are added, but only to central C’s to avoid extending the chain. If both CH,’s are introduced on the same C, the isomer is 2J-dimethylbutane. Placing one CH, on each of the central C’s gives the remaining isomer, 2,3-dimethylbutane.
CH, CH,
CH,
I
I I
CH,-C-CH,CH3
I I
CH,-C-C--CH, H
bH3
H
2,3-Dimethylbutane
2.2-Dimethylbutane
Problem 4.28 Write the structural formulas for (a) 3.4-dichloro-2,5-dimethylhexane; (b) 5-( 1.24 dimethylpropyl)-6-methyldodecane.(Complex branch groups are usually enclosed in parentheses.)
CH, CI
I
I
Cl
I
CH,
I
CH3-CH-CH-CH-CH-CH, (6) The group in parentheses is bonded to the fifth C. It is a propyl group with CH,’s on its first and second C’s (denoted I ’ and 2 ’ ) counting from the attached C.
( a ) 2,2,3-Trimethylpentane. (6) The longest chain has ten C’s, and there is an isobutyl group on the fifth C. The name is 5-(2-methylpropy1)decane o r 5-isobutyldecane.
Problem 4.30 Write the structural formulas and give the IUPAC names for all monochloro derivatives of ( a ) isopentane, (CH,),CHCH,CH,; (6) 2,2,4-trimethylpentane, (CH,),CCH,CH(CH,),. ( a ) Since there are four kinds of equivalent H’s,
4
CHAP. 41
63
ALKANES
there are four isomers: CH,
CH3
I CIC H,CC H,C H I
I
CH,-C-CH,CH,
I
1-Chloro-2-methyl-
I
C H,-C-CH2CH2CI
I
I
H CI
2-Chloro-2-methylbutane
butane
I
CH,-C-CHCH,
CI
H
CH,
CH3
I
H
2-C hloro-3-methylbutane
1 -Chloro-3-methyl-
butane
(6) There are four isomers because there are four kinds of H‘s: (CH,),CCH2CH(Ch,),. CH,
CH, H
CH,
I
CICH,-C-CH,-C-CH,
I
CH,
I I
I
l
I
l
CH,-C-C-C-CH,
CH,
l
l
CH, C1
H
1 -Chloro-2,2.4-
CH,
I
CH,-C-CH,-C-CH~
I
H
CH,
3-ChIoro-2,2,4trimethylpentane
trirnethylpentane
CH,
CH,
CH,
H
CH,
I I
CI
2-ChIoro-2,4,4trimethylpentane
I I CH,-C-CH,-C-CH,CI I I
1 -Chloro-2,4,4trirnethylpentane
Problem 4.31 Derive the structural formulas and give the IUPAC names for all dibromo derivatives of propane. 4
The two Br’s are placed first on the same C and then on different C’s. Br
Br
I
Br-CHCH,CH,
I I
BrCH,CHBrCH,
BrCH,CH,CH,Br
2.2-Dibromopropane
1,2-Dibromopropane
1,3-Dibrornopropane
CH,CCH,
Br 1 . 1 -Dibrornopropane
Problem 4.32 Give topological structural formulas for (a) propane, ( 6 ) butane, (c) isobutane, (d) 2,2dimethylpropane, (e) 2,3-dimethylbutane, ( f ) 3-ethylpentane, ( g ) l-chloro-3-methylbutane,( h ) 2,3-dichloro2-methylpentane, ( i ) 2-chloro-2,4,4-trimethylpentane. 4
In this method one writes only the C - - C bonds and all functional groups bonded to C. The approximate bond angles are used.
Problem 4.33 Synthesize ( a ) 2-methylpentane from CH,CH=CH--CH(CH,),, ( 6 ) isobutane from isobutyl chloride, (c) 2-methyl-2-deuterobutane from 2-chloro-2-methylbutane. Show all steps. 4 ( a ) The alkane and the starting compound have the same carbon skeleton.
CH3
CH,
I
CH,-CH=CH-CH-CH,
H p t __*
I
CH,-CH,-CH,--CH-CH~
ALKANES
64
[CHAP. 4
( 6 ) The alkyl chloride and alkane have the same carbon skeleton. CH,
CH,
I
I
CH3CHCH2-CI
CH3CHCH,-H
( c ) Deuterium can be bonded to C by the reaction of D,O with a Grignard reagent.
-
CH3
CH3
I dryyfhcr+ CH,CH,C-MgCI I
I CH3CH2C-CI I CH3
DZO
CH,
2-Chloro-2-methylbutane
Grignard reagent
CH3
I I
CH,CH,C-D CH3
2-Deutero-2-methylbutane
Problem 4.34 RC1 is treated with Li in ether solution to form RLi. RLi reacts with H,O to form isopentane. Using the Corey-House method, RCI is coupled to form 2,7-dimethyloctane. What is the structure of RCI? 4 To determine the structure of a compound from its reactions, the structures of the products are first considered and their formation is then deduced from the reactions. The coupling product must be a symmetrical molecule whose carbon-to-carbon bond was formed between C4 and C 5 of 2,7-dimethyloctane. The only RCI which will give this product is isopentyl chloride:
( CH3
CH,CHCH,CH,
)
CH, ,CuLi
I
+ CICH,CH,CHCH,
Lithium dusopentylcuprate
-
This alkyl halide will also yield isopentane. CH,
CH,
I
formed bond
IT
*
CH,
I
CH,CHCH,CH,-CH,CH,CHCH, 2,7-Dimethyloctane
CH,
I
CH,CHCH2CH2Li
2 CH,AHCH,CH, (Isopentane)
Problem 4.35 Give steps for the following syntheses: ( a ) propane to (CH,),CHCH(CH,),, 2-methylpentane, ( c ) "CH,CI to 'JCH,'JCH2'4CH,'4CH,.
(6) propane to 4
( a ) The symmetrical molecule is prepared by coupling an isopropyl halide. Bromination of propane is
preferred over chlorination because the ratio of isopropyl to n-propyl halide is 96%/4% in bromination and only 56%/44% in chlorination. CH3CH2CH,
Br (127 " C )
(CH,),CHBr
CH,CH,CH,CI
I
l.,!& 3
CH,CH,CH,
CH, CH,
I
CH3CH-CHCH,
CH3CHBrCH3
+ (CH,),CHCI
(sepurate the mixture)
Problem 4.36 Synthesize the following deuterated compounds: ( a ) CH,CH,D, (6) CH,DCH,D. ( a ) CH,CH,Br=
(6) H,C=CH,
Mg
+ D,-
CH,CH,MgBrPt
H2CDCH,D
CH,CH,D
4
CHAP. 41
65
ALKANES
Problem 4.37 In the dark at 150 OC, tetraethyl lead, Pb(C,H,),, catalyzes the chlorination of CH,. Explain in terms of the mechanism. 4
Pb(C,H,), readily undergoes thermal homolysis of the P b - C bond. Pb(C,H,), The CH,CH;
then generates the C1- that initiates the propagation steps. CH,CH2*+ C1:CI-
CH,CH,CI
CH, H,C* + Cl:Cl+
Problem 4.38
*Pb*+ 4CH3CH,*
+
+ C1.
1
H,C* H C 1 (propagation steps) H,CCI + CI-
"*-
+
Hydrocarbons are monochlorinated with tert-butyl hypochlorite, t-BuOC1. t-BuOCI + RH-
RCI + t-BuOH
Write the propagating steps for this reaction if the initiating step is t-BuO*+ CI.
I-BuOCl-
4
The propagating steps must give the products and also form chain-carrying free radicals. The formation of t-BuOH suggests H-abstraction from RH by t-BuO-, not by CI.. The steps are: RH + t-BuO*---* R. + t-BuOH RCI + t-BuO. R*+ t-BuOCl+
R. and t-BuO. are the chain-carrying radicals. Problem 4.39 Calculate the heat of combustion of methane at 25°C. The bond energies for C-H, C=O, and O - H are respectively 413.0, 498.3, 803.3, and 462.8 kJ/mol.
-
O=O, 4
First, write the balanced equation for the reaction. CH,
+ 20,
CO,
+ 2H,O
The energies for the bonds broken are calculated. These are endothermic processes and A H is positive. C + 4H
CH,-
2 0 , 4 4 0
AH
= 4( +413.0) =
+ 1652.0kJ/mol
AH
= 2( +498.3) =
+996.6 kJ/mol
Next, the energies are calculated for the bonds formed. Bond formation is exothermic, so the A H values are made negative. C + 20-+0=C=O 4H + 2 0 +
2 H U H
A H = 2(-803.3)
=
-1606.6 kJ/mol
A H = 4(-462.8)
= - 1851.2 kJ/mol
The enthalpy for the reaction is the sum of these values:
+ 1652.0+ 996.6 - 1606.6 - 1851.2 = -809.2
kJ/ mol (Reaction is exothermic)
Problem 4.40 ( a ) Deduce structural formulas and give IUPAC names for the nine isomers of C,H,,. (6) Why 4 is 2-ethylpentane not among the nine? (a)
seven-C chain
1. CH,CH,CH,CH,CH,CH,CH,
Heptane
CH, six-C chain
I
2. CH,CHCH,CH,CH,CH,
2-Methyl hexane
CH
1 ,
3. CH,CH,CHCH,CH,CH,
3-Methylhexane
ALKANES
66
[CHAP. 4
CH,
I
five-C chain
2,3-Dimet h ylpent ane
4. CH,CHCHCH,CH,
I
CH, CH,
CH,
I
I
2,4-Dimethylpentane
5. CH,CHCH,CHCH, CH 6.
4' 4'
CH, CH,CH,CH,
2,2-Dimet hylpentane
CH,
CH
7. CH,CH, CH,CH,
3,3-Dimet hylpen tane
CH.3
CH
:{.
8. CH,CH,CHCH,CH,
3-Ethylpentane
CH, CH,
I I
four-C chain
I
9. CH,C---CHCH,
2,2,3-Trimethylbutane
CH, (6) Because the longest chain has six C's, and it is 3-methylhexane. Problem 4.41 Singlet methylene, :CH, (Section 3.2), may be generated from diazomethane, CH2N2;the other product is N,. It can insert between C-H bonds of alkanes: :CH,
I
+ --C-H+--C
I I
Determine the selectivity and reactivity of :CH, from the yields of products from methylene insertion in pentane:
CH,CH,CA,CH,CH, I
2
2
1
+ :CH, --
HCH,CH,CH,CH,CH,
HCH,
Name Kind of CH insertion
I
Hexane, CH'
Yield
1
48%
Product Hexane 2-Methylpentane 3-Methylpentane Total
+ CH,CHCH,CH,CH, + CH,CH,CHCH,CH, I I
I
HCH,
2-Methylpentane, CH2
I
I
35%
3-Methylpentane, CH.' 17%
H's of Pentane Kind
Number
1 2 3
6 4 2 12
Proportion x 100% = Calc. % Yield 61 12 4/12 21 12
=
= =
50 33.3 16.7
Observed % Yield 48 35 17
CHAP. 41
67
ALKANES
The almost identical agreement validates the assumption that methylene is one of the most reactive and least selective species in organic chemistry.
Problem 4.42 How would the energy-conformation diagrams of ethane and propane differ?
4
They would both have the same general appearance showing minimum energy in the staggered form and maximum energy in the eclipsed form. But the difference in energy between these forms would be greater in propane than in ethane (13.8 versus 12.5 kJ/mol). The reason is that the eclipsing strain of the Me group and H is greater than that of two H’s.
Problem 4.43 1,2-Dibromoethane has a zero dipole moment, whereas ethylene glycol, C H , 0 H C H 2 0 H , has a measurable dipole moment. Explain. 4
-
1,2-Dibromoethane exists in the anti form, so that the C Br dipoles cancel and the net dipole moment is zero. When the glycol exists in the gauche form, intramolecular H-bonding occurs. Intramolecular H-bonding is a stabilizing effect which cannot occur in the anti conformer.
%: H6!Br
anti CH,BrCH,Br
H
gauche
H
CH,OHCH,OH
Chapter 5 Stereochemistry 5.1 STEREOISOMERISM
Stereoisomers (stereomers) have the same bonding order of atoms but differ in the way these atoms are arranged in space. They are classified by their symmetry properties in terms of certain symmetry elements. The two most important elements are: A symmetry plane divides a molecule into equivalent halves. It is like a mirror placed so that half the molecule is a mirror image of the other half. 2. A center (point) of symmetry is a point in the center of a molecule to which a line can be drawn from any atom such that, when extended an equal distance past the center, the line meets another atom of the same kind. 1.
A chiral stereoisomer is not superimposable on its mirror image. It does not possess a plane or center of symmetry. The nonsuperimposable mirror images are called enantiomers. A mixture of equal numbers of molecules of each enantiomer is a racemic form (racemate). The conversion of an enantiomer into a racemic form is called racemization. Resolution is the separation of a racemic form into individual enantiomers. Stereomers which are not mirror images are called diastereomers. Molecules with a plane or center of symmetry have superimposable mirror images; they are achiral. Problem 5.1 Classify as chiral or achiral: ( a ) a hand, (b) a screw, (c) an Erlenmeyer flask, (d) a clock face, (e) 4 a rectangle, ( f ) a parallelogram, ( g ) a helix, ( h ) the letter A, (i) the letter R. The achiral things are (c) and ( h ) , with planes of symmetry; (e), with a center of symmetry and two planes of symmetry; and ( f ) , with a center of symmetry. This leaves ( a ) , (b), (d), ( g ) , and (i) as chiral. [ Chiral derives from the Greek for hand.]
Problem 5.2 Indicate the planes and centers of symmetry in ( a ) methane (CH,), ( 6 ) chloroform (CHCI,), and (c) ethene (H2C==CH,). 4 In molecular structures, planes of symmetry can cut through atoms or through bonds between atoms. ( a ) CH, has no center but has two planes, as shown in Fig. 5-l(a). Each plane cuts the C and a pair of H's, while bisecting the other H-C-H bond angle. (6) CHCI, has no center and has one plane that cuts the C, H, and one of the Cl's while bisecting the CI--C--CI bond angle, as shown in Fig. 5-l(6). ( c ) In ethene all six atoms lie in one plane, which, in Fig. 5-l(c), is the plane of the paper. It has a center of symmetry between the C's. It also has three symmetry planes, one of which is the aforementioned plane of the paper. Another plane bisects the C=C bond and is perpendicular to the plane of the paper. The third plane cuts each C while bisecting each H-C-H bond.
planes
/'
t
planes /
center
(b)
Cl I
(c)
Fig. 5-1
68
I
'
I
I
plane of paper
CHAP. 51
69
STEREOCH EM I STRY
5.2 OPTICAL ISOMERISM
Plane-polarized light (light vibrating in only one plane) passed through a chiral substance emerges vibrating in a different plane. The enantiomer that rotates the plane of polarized light clockwise (to the right) as seen by an observer is dextrorotatory;the enantiomer rotating to the left is levorotatory. The symbols (+) and (-) designate rotation to the right and left, respectively. Because of this optical activity, enantiomers are called optical isomers. The racemic form ( ? ) is optically inactive since it does not rotate the plane of olarized light. The specific rotation [ a ]P*is an inherent physical property of an enantiomer which, however, varies with the solvent used, temperature (Tin "C), and wavelength of light used ( A ) . It is defined as the observed rotation per unit length of light path, per unit concentration (for a solution) or density (for a pure liquid) of the enantiomer; thus;
where
a = observed rotation, in degrees [=length of path, in decimeters (dm) c = concentration or density, in g/mL = kg/dm3
and where, by convention, the units deg.dm2/kg of [ a ] :are abbreviated to degrees ("). Problem 5.3 A 1.5-g sample of an enantiomer is dissolved in ethanol to make 50mL of solution. Find the specific rotation at 20 "C for sodium light ( A = 589.3 nm, the D line), if the solution has an observed rotation of 4 +2.79" in a 10-cm polarimeter tube. First change the data to the appropriate units: 10 cm = 1 dm and 1.5 g/S0mL = 0.03 g/mL. The specific rotation is then
[4""= a = ec
+2'790 - +93" (in ethanol) (q(0.03)
~
Problem 5.4 Calculate the observed rotation in Problem 5.3 if: ( a ) the measurement is made in a 5-cm polarimeter tube, (6) the solution is diluted from 50mL to 1S0mL and the determination is carried out in a 10-cm tube. 4
t ) = +1.39;
a is jointly proportional to f and c. ( a ) a = (93")(0.03)(
(6)a
=
93"(1)(0.01)
=
+0.93".
Problem 5.5 How could it be decided whether an observed dextrorotation of +60" is not actually a 4 levorotation of -300". Halving the concentration or the tube length would halve the number of optically active molecules and the new rotation would be +30° if the substance was dextrorotatory or -150" if levorotatory.
Many chiral organic molecules have at least one carbon atom bonded to four different atoms or groups, called ligands. Such a carbon atom is called a chiral center and is indicated as C*. A chiral center is a particular stereocenter, defined as an atom for which interchange of a pair of ligands gives a different stereomer. In the case of a chiral center, the new stereomer is the enantiomer. Enantiomers can be depicted by planar projection formulas, as shown for lactic acid, H,CCH(OH)COOH, in Fig. 5-2 [see also Fig. 1-21. In the Fischer projection, Fig. 5-2(c), the chiral C is at the intersection; horizontal groups project toward the viewer; vertical groups project away from the viewer. Problem 5.6 Draw ( a ) Newman and (6) Fischer projections for enantiomers of CH,CHIC,H,. See Fig. 5-3.
4
70
STEREOCHEMISTRY
H3c+cooH
[CHAP. 5
v HOOC+CH, OH
H0Oc(..H3
OH
in back
-:
I!
H3 HO COOH
HO
Fig. 5-2
Mirror
Mirror
(b) Fig. 5-3
Problem 5.7 Determine if the configuration of the left-hand enantiomer of Fig. 5-3(6) is changed by a rotation in the plane of the paper of (i) 90" and (ii) 180".
4
Rotations of 90" and 180" give the following Fischer projections:
The best way to see if the configuration has changed is to determine how many swaps of groups must be made to go from the initial to the final Fischer projection. An even number of swaps leaves the configuration unchanged, while an odd number results in a changed configuration. Convince yourself that to get (i) takes three swaps, while to get (ii) takes two swaps (the horizontal pairs and the vertical pairs). Problem 5.8 Write structural formulas for the monochloroisopentanes. Place an asterisk on any chiral C and indicate the 4 different groups about the C*. 4
CHAP. 51
CH3
I* CICH2-C-CH2CH3 I H
1 -Chloro-2-methyl-
butane (I) (H, CH,, CHzCH,, CICH2)
71
STEREOCHEMISTRY
CH3
CH3
CH3
I CH3-C-CH,-CH, I
I
I
* CH3-CH-CH-CH3 I
CH3-CH-CH2-CH2CI
CI
CI
2-Chloro-2-niethylbutane (11)
2-Chloro-3-methylbutane (111) (H,C1, CH,,(CH,),CH)
I -Chloro-3-methylbutane (IV)
In looking for chirality one considers the entire group, e.g. CH,CH,, attached to the C* and not just the attached atom.
5.3 RELATIVE AND ABSOLUTE CONFIGURATION
Configuration is the spatial arrangement of ligands in a stereoisomer. Enantiomers have opposite configurations. For enantiomers with a single chiral site, to pass from one configuration to the other (inversion) requires the breaking and interchanging of two bonds. A second interchange of bonds causes a return to the original configuration. Configurations may change as a result of chemical reactions. To understand the mechanism of reactions, it is necessary to assign configurations to enantiomers. For this purpose, the sign of rotation cannot be used because there is no relationship between configuration and sign of rotation. Problem 5.9 Esterification of (+)-lactic acid with methyl alcohol gives (-)-methyl lactate. Has the configuration changed?
OH
(+)-Lactic acid
=
+3.3"
OH
(-)-Methyl lactate
= -8.2"
4
No; even though the sign of rotation changes, there is no breaking of bonds to the chiral C*
The Cahn-Ingold-Prelog rules (1956) are used to designate the configuration of each chiral C in a molecule in terms of the symbols R and S. These symbols come from the Latin, R from rectus (right) and S from sinister (left). Once told that the configuration of a chiral C is R or S, a chemist can write the correct projection or Fischer structural formulas. In our statement of the three rules the numerals 1, 2, 3, and 4 are used; some chemists use letters a, b, c, and d in their place. Rule 1.
Ligands to the chiral C are assigned increasing priorities based on increasing atomic number of the atom bonded directly to the C. (Recall that the atomic number, which is the number of protons in the nucleus, may be indicated by a presubscript on the chemical symbol; e.g., *O. For isotopes, the one with higher mass has the higher priority, e.g., deuterium over hydrogen.) The priorities of the ligands will be given by numbers in parentheses, using (1) for the highest priority (the heaviest ligand) and (4) for the lowest priority (the lightest ligand). The lowest-priority ligand, (4), must project away from the viewer, behind the paper, leaving the other three ligands projecting forward. In the Fischer projection the priority-(4) ligand must be in a vertical position (if necessary, make two interchanges of ligands to achieve this configuration). Then, for the remaining three ligands, if the sequence of decreasing priority, (1) to (2) to (3), is counterclockwise, the configuration is designated S; if it is clockwise, the configuration is designated R. The rule is illustrated for l-chloro-lbromoethane in Fig. 5-4. Both configuration and sign of optical rotation are included in the complete name of a species, e.g., (S)-( +)-l-chloro-l-bromoethane.
STEREOCHEMISTRY
72
[CHAP. 5
Clockwise, R
Counterclockwise, S Fig. 5-4
Rule 2.
If the first bonded atom is the same in at least two ligands, the priority is determined by comparing the next atoms in each of these ligands. Thus, ethyl (H,CCH,-), with one C and two H’s on the first bonded C, has priority over methyl(-CH,), with three H’s on the C. For butyl groups, the order of decreasing priority (or increasing ligand number) is
Rule 3. For purposes of assigning priorities, replace:
-C=C
7
by -C-C
‘c Problem 5.10 Structures of CHClBrF are written below in seven Fischer projection formulas. Relate structures (6) through ( g ) to structure ( a ) .
If two structural formulas differ by an odd number of interchanges, they are enantiomers; if by an even number, they are identical. See Table 5-1.
STEREOCHEMISTRY
CHAP. 51
73
Table 5-1 Sequence of Group Interchanges
Number of Interchanges
Relationship to ( a )
1 (odd) 1 (odd) 2 (even) 2 (even) 2 (even) 3 (odd)
Enantiomer Enantiomer Same Same Same Enantiomer
H, Br H,F H , F; Br, Cl H, Br; CI, F F, Br; F. CI F, Br; Br, Cl; H, Cl
Problem 5.11 Put the following groups in decreasing order of priority.
(6) -CH=CH2 (e)
cf)
-C=O
I
-c=o I
(c) -C=N
(g) -CH,NH*
OH
H
( d ) -CH21
( h ) -C-NH,
(i)
It 0
-c=o I
4
CH3
In each case, the first bonded atom is a C. Therefore, the second bonded atom determines the priority. In decreasing order of priority, these are J > > ,N > 6C. The equivalencies are
(U)
-c-c\/c c
/ (6) -C-C
H
‘c
/
N
(c) -C-N
\N
/
( d ) -C-H
I‘
H (e)
/
H
-C-0
O\
/O
U) -c-0
\O
The order of decre sing priority is: ( d ) > (f)> ( h ) > (i) > ( e ) > ( c )> ( g ) > ( a ) > (b). Note that in has a greater priority than 3 0’s in ( f ) .
(t
). one
’roblem 5.12 Designate as R or S the configuration of
a) The decreasing order of priorities is C1 (1), CH,CI (2), CH(CH,), (3), and CH, (4).CH,, with the lowest priority, is projected in back of the plane of the paper and is not considered in the sequence. The sequence of decreasing priority of the other groups is counterclockwise and the configuration is S.
The compound is (S)- 1,2-dichloro-2,3-dimethylbutane.
STEREOCHEMISTRY
74
The sequence of priorities is Br ( I ) , HzC=CH-
[CHAP. 5
(2), CH3CH2- (3) and H (4).
The name is (R)-3-bromo-l-pentene. H is exchanged with NH, to put H in the vertical position. Then the other two ligands are swapped, so that, with two exchanges, the original configuration is kept. Now the other three groups can be projected forward with no change in sequence. A possible identical structure is
The sequence is clockwise or R. An alternative approach would leave H horizontal, thus generating an answer known to be wrong. If the wrong answer is R or S. the right answer is S or R. Problem 5.13 Draw and specify as R and S the enantiomers, if any, of all the monochloropentanes.
4
n-Pentane has 3 monochloro-substituted products, 1-chloro, 2-chloro, and 3-chloropentane
CICH,CH~CH,CH~CH, CH
,EHCICH,CH~CH
CH ,CH,CHCICH,CH,
Only 2-chloropentane has a chiral C , whose ligands in order of decreasing priority are C1 ( l ) , CH,CH, (2), CH, (3) and H (4).The configurations are: '1)
(1)
(R)-2-Chloropentane
(S)-2-Chloropentane
The structures of the monochloroisopentanes are given in Problem 5.8. The sequence of decreasing priority of ligands for structure 1 is CICH, > CH,CH, > CH, > H , while that for I11 is C1 > (CH,),CH > CH, > H. The con figurations a re :
Problem 5.14 Write structural formulas for the following, indicating enantiomers, if any. and their configura4 tions: ( a ) 3-methyl-3-pentanol. ( b ) 2,2-dimethyl-3-bromohexane. ( c ) 3-phenyl-3-chloro- I-propene.
The alcohol
CH~CH*- ~ - - c H ~ c H ,
I
OH
has no atom bonded to four different groups and no enantiomers; it is achiral.
CHAP. 51
75
STEREOCHEMISTRY
*
(6) (H,C),CCHBr(CH,),CH, has one chiral C (shown with an asterisk).
Relative configuration is the experimentally determined relationship between the configuration of a given chiral molecule and the arbitrarily assigned configuration of a reference chiral molecule. The D,L system (1906) uses glyceraldehyde, HOCH,C*H(OH)CHO, as the reference molecule. The (+)and (-)-rotating enantiomers were arbitrarily assigned the configurations shown below in Fischer projections, and were designated as D and L, respectively.
H+OH CHO
CHzOH
OHC HO+H HOH2C
Relative configurations about chiral C's of other compounds were then established by synthesis of these compounds from glyceraldehyde. It is easy to see that D-(+)-glyceraldehyde has the R configuration and the L-(-) isomer is S. This arbitrary assignment was shown to be correct by Bijvoet (1951), using X-ray diffraction. Consequently, we can now say that R and S denote absolute configuration, which is the actual spatial arrangement about a chiral center. Problem 5.15 D-( +)-Glyceraldehyde is oxidized to (-)-glyceric acid, HOCH,CH(OH)COOH. Give the designation of the acid.
D,L
- r0;:
Oxidation of the D-( +)-aldehyde does not affect any of the bonds to the chiral C. The acid has the same configuration even though the sign of rotation is changed. ~~
I
CHO OH CH,OH
D-(+)-Glyceraldehyde
CH,OH
D-(-)-Glyceric acid
Problem 5.16 Does configuration change in the following reactions? Designate products
H
(D,L)
and (R,S).
4 D
76
STEREOCHEMISTRY
[CHAP. 5
4
No bond to the chiral C is broken and the configuration is unchanged. Therefore, the configuration of both reactant and product is D . The product is R: S converts to R because there is a priority change. Thus, a change from R to S does not necessarily signal an inversion of configuration; it does only if the order of priority is unchanged. A bond to the chiral C is broken when I - displaces Cl . An inversion of configuration has taken place and there is a change from D to L . The product is R. This change from S to R also shows inversion, because there is no change in priorities. Inversion has occurred and there is a change from D to L . The product is R. Even though an inversion of configuration occurred, the reactant and product are both R. This is so because there is a change in the order of priority. The displaced Cl has priority (1) but the incoming CN- has priority (2).
In general. the D , L convention signals a configuration change, if any, as follows: D--* D or L-+ L means no change (retention), and D - L or L + D means change (inversion). The RS convention can also be used, but it requires working out the priority sequences. Problem 5.17 Draw three-dimensional and Fischer projections, using superscripts to show priorities, for (a) (S)-2-bromopropanaldehyde, (6) (R)-3-iodo-2-chloropropanol. 4
5.4 MOLECULES WITH MORE THAN ONE CHIRAL CENTER
With n dissimilar chiral atoms the number of stereoisomers is 2" and the number of racemic forms is 2''-', as illustrated below for 2-chloro-3-bromobutane (n = 2). The R,S configuration is shown next to the C's.
cl-7
(R) R r (R)
H
CH3 I
Br
CH, 11
CH3 111
H
(R) (S)
CH3 IV
If n = 2 and the two chiral atoms are identical in that each holds the same four different groups, there are only 3 stereoisomers, as illustrated for 2,3-dichlorobutane.
77
STEREOCHEMISTRY
CHAP. 51
Mirror
c1.F
(R)
H
(R)
c1
CH3 V
H F I (S) Cl H (s)
H
H CH3
CH3 VI
(s)
Plane
CH,
VII
VIII meso
2( S)-3(R)Dichlorobutane
2(R)-3(S)Dichlorobutane
Structures VII and VIII are identical because rotating either one 180" in the plane of the paper makes is superposable with the other one. VII possesses a symmetry plane and is achiral. Achiral stereoisomers which have chiral centers are called meso. The meso structure is a diastereomer of either of the enantiomers. The meso structure with two chiral sites always has the (RS) configuration. Problem 5.18 For the stereoisomers of 3-iodo-2-butano1, ( a ) assign R and S configurations to C-' and C7.(6) Indicate which are (i) enantiomers and (ii) diastereomers. ( c ) Will rotation about the C - C bond alter the configurations? 4
I H
I1
I11
CH,
CH3
IV I
I
I
I
I
H
OH
OH
(i) I and IV, I1 and 111. (ii) I and IV diastereomeric with I1 and 111; I1 and I11 diastereomeric with I and IV. No. Problem 5.19 Compare physical and chemical properties of ( a ) enantiomers, (6) an enantiomer and its 4 racemic form, and ( c ) diastereomers. With the exception of rotation of plane-polarized light, enantiomers have identical physical properties, e.g., boiling point, melting point, solubility. Their chemical properties are the same toward achiral reagents, solvents, and conditions. Towards chiral reagents, solvents, and catalysts, enantiomers react at different rates. The transition states produced from the chiral reactant and the individual enantiomers are diastereomeric and hence have different energies; the A H S values are different, as are the rates of reaction. Enantiomers are optically active; the racemic form is optically inactive. Other physical properties of an enantiomer and its racemic form may differ depending on the racemic form. The chemical properties are the same towards achiraf reagents, but chiral reagents react at different rates. Diastereomers have different physical properties, and have different chemical properties with both achiral and chiral reagents. The rates are different and the products may be different.
STEREOCHEMISTRY
78
[CHAP. 5
Problem 5.20 How can differences in the solubilities of diastereomers be used to resolve a racemic form into individual enantiomers? 4 The reaction of a racemic form with a chiral reagent, for example, a racemic ( 5 ) acid with a (-) base, yields two diastereomeric salts (+ )( -) and (-)( -) with different solubilities. These salts can be separated by fractional crystallization, and then each salt is treated with a strong acid (HCI) which liberates the enantiomeric organic acid. This is shown schematically:
Racemic Form (2)RCOOH
+
Chiral Base (-)B
Diastereomeric Salts (+)RCOO-(-)BH' + (-)RCOO-(-)BH'
---+
HCI
BH'CI-
sepu ru t ed
5.
+ (+)RCOOH
5. HC1 (-)RCOOH + BH'CI'
separated enant iomeric acids
The most frequently used chiral bases are the naturally occurring, optically active alkaloids, such as strychnine, brucine, and quinine. Similarly, racemic organic bases are resolved with naturally occurring, optically active, organic acids, such as tartaric acid.
5.5
SYNTHESIS AND OPTICAL ACTIVITY
1. Optically inactive reactants with achiral catalysts or solvents yield optically inactive products. With a chiral catalyst, e.g., an enzyme, any chiral product will be optically active. 2. A second chiral center generated in a chiral compound may not have an equal chance for R and S configurations; a 50 : 50 mixture of diastereomers is not usually obtained. 3. Replacement of a group or atom on a chiral center can occur with retention or inversion of configuration or with a mixture of the two (complete or partial racemization), depending on the mechanism of the reaction. Problem 5.21 ( a ) What is the stereochemistry of CH,CHC1CH2CH, isolated from the chlorination of butane? ( b ) How does the mechanism of chlorination of RH account for the formation of the product in (a)? 4 Since all reactants are achiral and optically inactive, the product is optically inactive. Since the product has a chiral center, C 2 , it must be a racemic form. See Fig. 5-5. The first propagation step produces CH,CHCH,CH,. The C of this radical uses sp2 hybridized orbitals and its three bonds are coplanar. The p orbital with the odd electron is at right angles to this plane, which is said to have two faces. Approach of a reactant to one or the other face makes contact with one or the other lobe of the p AO. In the second propagation step, the p A 0 of the sec-butyl radical can add a CI atom from CI, on either face to give rise to one or the other of a pair of enantiomers. For this reason the faces are said to be enantiotopic faces. Since the probability of attack is identical on either face, chlorination gives equal amounts of enantiomers and hence a racemic form. CI, (on either 4de of p orbital)
-
C,H,
H n - Bu tane
CI. + C,H,
HCI +
CI + CI
H
sp' sec-Butyl
free radical
(R)-2-Chlorobutane Fig. 5-5
C,H5
H (S)-2-Chlorobutane
CHAP. 51
79
STEREOCHEMISTRY
Problem 5.22 ( a ) What two products are obtained when C.’ of (R)-2-chlorobutane is chlorinated? (6) Are these diastereomers formed in equal amounts? (c) In terms of mechanism account for the fact that
*
is obtained when (R)-CICH,-CH(CH,)CH,CH, is chlorinated. CI
CI H
H
H CI (RI
4
CI CI
(RR) optically active en a n t iome r
In the products C’ retains the R configuration since none of its bonds were broken and there was no change in priority. The configuration at C3,the newly created chiral center, can be either R or S. As a result, two diastereomers are formed, the optically active RR enantiomer and the optically inactive RS meso compound. No. The numbers of molecules with S and R configurations at Cz are not equal. This is so because the presence of the C’-stereocenter causes an unequal likelihood of attack at the faces of C3.Faces which give rise to diastereomers when attacked by a fourth ligand are diastereotopic faces. Removal of the H from the chiral C leaves the achiral free radical ClCH,C(CH,)CH,CH,. Like the radical in Problem 5.21, it reacts with C1, to give a racemic form.
The C3 of 2-chlorobutane7 of the general type RCH,R’, which becomes chiral when one of its H’s is replaced by another ligand, is said to be prochiral. Problem 5.23 Answer True or False to each of the following statements and explain your choice. ( a ) There are two broad classes of stereoisomers. (6) Achiral molecules cannot possess chiral centers. (c) A reaction catalyzed by an enzyme always gives an optically active product. (d) Racemization of an enantiomer must result in the breaking of at least one bond to the chiral center. (e) An attempted resolution can distinguish a racemate 4 from a meso compound.
( a ) True: Enantiomers and diastereomers. (6) False: Meso compounds are achiral, yet they possess chiral centers. ( c ) False: The product could be achiral. (d) True: Only by breaking a bond could the configuration be changed. (e) True: A racemate can be resolved but a meso compound cannot be, because it does not consist of enantiomers. Problem 5.24 In Fig. 4-4 give the stereochemical relationship between ( a ) the two gauche conformers; (6) the 4 anti and either gauche conformer.
( a ) They are stereomers, because they have the same structural formulas but different spatial arrangements. However, since they readily interconvert by rotation about a U bond, they are not typical, isolatable, configurational stereomers; rather they are conformational stereomers. The two gauche forms are nonsuperimposable mirror images; they are conformational enantiomers. (6) They are conformational diastereomers, because they are stereomers but not mirror images.
Configurational stereomers differ from conformational stereomers in that they are interconverted only by breaking and making chemical bonds. The energy needed for such changes is of the order of 200-600 kJ/mol, which is large enough to permit their isolation, and is much larger than the energy required for interconversion of conformers.
STEREOCHEMISTRY
80
[CHAP. 5
Supplementary Problems Problem 5.25 (a) What is the necessary and sufficient condition for the existence of enantiomers? (6) What is the necessary and sufficient condition for measurement of optical activity? (c) Are all substances with chiral atoms optically active and resolvable? (d) Are enantiomers possible in molecules that do not have chiral carbon atoms'? (e) Can a prochiral carbon ever be primary or tertiary? ( f ) Can conformational enantiomers ever be resolved'? 4 ( a ) Chirality in molecules having nonsuperimposable mirror images. (6) An excess of one enantiomer and a specific rotation large enough to be measured. (c) No. Racemic forms are not optically active but are resolvable. Meso compounds are inactive and not resolvable. (d) Yes. The presence of a chiral atom is a sufficient but not necessary condition for enantiomerism. For example, properly disubstituted allenes have no plane or center of symmetry and are chiral molecules even though they have no chiral C's:
Mirror
*
a chiral allene
(nonsuperimposable enantiomers)
(e) No. Replacing one H of U 1" CH, group by an X group would leave an achiral --CH,X group. In order for a 3" CH group to be chiral when the H is replaced by X, it would already have to be chiral when bonded to the H. ( f ) Yes. There are molecules which have a large enthalpy of activation for rotating about a (+ bond because of severe steric hindrance. Examples are properly substituted biphenyls, e.g. 2,2'-dibromo-6,6'-dinitrobiphenyl, Figs. 5-6. The four bulky substituents prevent the two flat rings from being in the same plane, a requirement for free rotation.
+
Mirror
-
NO,
-
Br
Br NO,
Fig. 5-6
Problem 5.26 Select the chiral atoms in each of the following compounds: CH.
4
.J
CI
CH3
H
I
[CH,CHCH,NCH,CH,CH,] ' CI
I
OH
OH Cholesterol (0)
I
CH,CH,
0
CHAP. 5 )
81
STEREOCHEMISTRY
(a) There are eight chiral C’s: three C’s attached to a CH, group, four C’s attached to lone H’s, and one C attached to OH. ( 6 ) Since N is bonded to four different groups, it is a chiral center, as is the C bonded to the OH. ( c ) There are no chiral atoms in this molecule. The two sides of the ring -CH2CH2- joining the C’s bonded to Cl’s are the same. Hence, neither of the C’s bonded to a CI is chiral. Problem 5.27 Draw examples of (a) a meso alkane having the molecular formula C,H,, and (6) the simplest alkane with a chiral quaternary C. Name each compound. 4
(4 3.4-Dimethylhexane
(6) The chiral C in this alkane must be attached to the four simplest alkyl groups. These are CH,-, CH,CH,-, CH,CH, CH,-, and (CH,),CH-, and the compound is
CH,CH,CH,
qTH3 CHCH,
CH,CH,
2,3-Dimet hyl-3-et h ylhexane
Problem 5.28 Relative configurations of chiral atoms are sometimes established by using reactions in which there is no change in configuration because no bonds to the chiral atom are broken. Which of the following 4 reactions can be used to establish relative configurations?
(S)-CH,CHClCH,CH,
(4
+ Na’OCH,
CH,CH(OCH,)CH,CH,
+ Na’Cl-
+ CH,Br+CH,CH,
(6)
CH3
(4 (4 (4
-+
--
CH3
I + PC1, CH,CH,CClCH,CI + POCI, + HCI (CH,),C(OH)CH(CN)CH, + Na’Br(S)-(CH,),C(OH)CHBrCH, + Na’CNCH,CH,CHCH, + ;H2 (R)-CH,CH,CHCH, + Na I I 1
(R)-CH,CH,COHCH,CI
OH
-
0-Na’
(6) and (e). The others involve breaking bonds to the chiral C. Problem 5.29 Account for the disappearance of optical activity observed when (R)-2-butanol is allowed to 4 stand in aqueous H,SO, and when (S)-2-iodooctane is treated with aqueous KI solution.
Optically active compounds become inactive if they lose their chirality because the chiral center no longer has four different groups, or if they undergo racemization. In the two reactions cited, C remains chiral and it must be concluded that in both reactions racemization occurs. Problem 5.30 For the following compounds, draw projection formulas for all stereoisomers and point out their R,S specifications, optical activity (where present), and meso compounds: (a) 1,2,3,4-tetrahydroxybutane, (6) 1-chloro-2,3-dibromobu tane , (c) 2,4-diiodopentane, (d) 2,3,4-tribromohexane , (e) 2,3,4-tribromopentane .
*
*
4
(a) HOCH,CHOHCHOHCH,OH, with two similar chiral C’s, has one meso form and two optically active enantiomers.
STEREOCHEMISTRY
82
HF H3
*
OH
[CHAP. 5
‘CH,OH
‘CH,OH
H;$H
H& -H HO
OH
‘CH,OH
4CH,0H
4CH,0H
(2S.3R) meso
(2S.X)
(2R,3R)
,
1
Racemic form
*
(6) CICH,CHBrCHBrCH, has two different chiral C’s. There are four ( Z 2 ) optically active enantiomers. CH,CI
CH,CI
:+E
H+:; CH, (2S.3R)
CH,CI
CH,CI
H:* Br CH3
CH,
CH3
(2R.X)
(2S.3S)
(2R.3R)
threo enantiomers
erythro enantiomers
The two sets of diastereomers are differentiated by the prefix eryrhro for the set in which at least two identical o r similar substituents on chiral C’s are eclipsed. The other set is called threo.
(c) CH3?!HICH,6HICH, has two similar chiral C’s, C‘ and C‘, separated by a CH, group. There are two enantiomers comprising a ( 2 )pair and one meso compound.
meso
*
*
*
Brr
Racemate
( d ) With three different chiral C’s in CH,CHBrCHBrCHBrCH,CH,, there are eight (2’) enantiomers and four racemic forms.
l3;3
H t‘CH, ;;
H H
H” H4
Br
C,H, I (2S,3S,4R) Racemate
Hf; H Br
(2S,3S,4S) \
C2Hs
C,H,
111
IV
(2R,3R,4S)
(2S,3R,4R)
(2R,3S,4S)
Br3r
Racemate
(2R,3R,4R) 2
$
Racemate
Br H*
Br
Br
C2Hs
*
H
I1
v1
V
Br
C2Hs
Br
C,H,
H
Br
C2Hs
.
C,H,
VII
Vill
(2S,3R,4S)
(2R.3S.JR) Y
Race mate
J
CHAP. 51 I
1
83
STEREOCHEMISTRY
*
3
* J
5
(e) CH,CHBrCHBrCHBrCH, has two similar chiral atoms (C' and C'). There are two enantiomers in which the configuratiosn of C' and C' are the same, RR or SS. When C' and C' have different configurations, one R and one S, C7becomes a stereocenter and there are two meso forms.
Problem 5.31 The specific rotation of (R)-(-)-2-bromooctane is -36". What is the percentage composition 4 of a mixture of enantiomers of 2-bromooctane whose rotation is +18"?
Let x
= mole
fraction of R, 1 - x
= mole
x(-336')
fraction of S.
+ (1 - x)(36") = 18"
or
x=
The mixture has 25% R and 75% S; it is 50% racemic and 50% S. Problem 5.32 (S)-2-Chlorobutane yields a mixture of 71 % meso and 29% 2(S)-3( S)-2,3-dichlorobutane, while (R)-2-chlorobutane similarly gives a mixture of 71% meso and 29% 2(R)-3(R)-2,3-dichlorobutane on chlorination. 4 Account for these products by conformational analysis.
See Fig. 5-7. In the first step the radicals CH,~HCI--kH--CH, (111 through VI) are formed on an atom adjacent to a chiral C. The formation of the free radical does not change the configuration of the chiral center, which is S in one case and R in the other. A second chiral center, which may be R or S, is generated after step two. I
I
I I
I
H
I
I
/ 1
(I)
\
1
1
(VII)
(VIII)
(2R.3R)
(2R.X) (71% meso)
(2Wo enantiomer)
!
Fig. 5-7
I I
I
1
(IX)
(X)
(ZS.3R) (71% meso)
(7S.3S)
(2% enantiomer)
[CHAP. 5
STEREOCHEMISTRY
84
(R)-2-Chlorobutane ( I ) forms free radicals (111 and IV) which are conformational diastereomers with different stabilities and populations. This is also true of 1's enantiomer 11, (S)-2-chlorobutane, which gives free radicals V and VI. The more stable free-radical conformers are IV and V because their CH,'s are anti-like. The transition states for CI abstractions arising from conformations IV and V have lower AH' values than the diastereomeric transition states from the more crowded gauche-like conformations, 111 and VI. The major product is therefore the meso compound, VIII = I X . Problem 5.33 Predict the yields of stereoisomeric products, and the optical activity of the mixture of products, formed from chlorination of a racemic mixture of 2-chlorobutane to give 2,3-dichlorobutane. 4
The (S)-2-chlorobutane comprising 50% of the racemic mixture gives 35.5% of the rneso (SR) product and
14.5% of the RR enantiomer. The R enantiorner gives 35.5% rneso and 14.5% RR products. The total yield of meso product is 71% and the combination of 14.5% RR and 14.5% SS gives 29% racemic product. The total reaction mixture is optically inactive. This result confirms the generalization that optically inactive starting
materials, reagents and solvents always lead to optically inactive products.
Problem 5.34 For the following reactions give the number of stereoisomers that are isolated, their R,S configurations and their optical activities. Use Fischer projections. (4
(6)
meso- HOCH ,CHOHCHOHCH,OH ( R)-CICH ,CH( CH
,)CH2CH,
o x 1J.I I I t 1 11
CUprdlL.
0
I1
t12 NI
t)-CH,-C--CHOH--CH,-CH
~UC(
(d)
HOCH ,CHOHCHOHCOOH
CH 3CH,CH( CH ,)CH ,CH,CH( CH, )CH,CH ,
\ 1.1
(S)-CH,CH,
-EH2
CH, )CH,CH,
t i 2 NI
,CH(OH)CH(OH)CH,
CH ,CH2CH(CH ,)CH(CH, )CH,CH,
4
(a) This rneso alcohol is oxidized at either terminal CH,OH to give an optically inactive racemic form. The chiral C next to the oxidized C undergoes a change in priority order; CH,OH (3) goes to COOH (2). Therefore, if this C is R in the reactant, it becomes S in the product; if S it goes to R.
-
I l l
HO OH H&H++CH?OH H H
I I )
(1)
HO OH
HO OH + HOCH,+-COOH
HO&-~+--CH,OH
H H
H H
(7RJR) (50%)
( 2 S . 3 S ) (50% )
meso; ( 1 R . 3 S )
( b ) Replacement of CI by the isopentyl group does not change the priorities of the groups on the chiral C. There is one optically active product, whose two chiral C's have R configurations.
CH, CH,CH,+CH,CI
+
H CICH2tCH,CH,
H
CH,
(R)
(RI
-
I
HO H
\
H
CH,CH,tCH,-CH,+CH,CH, H (RI
(c) This reduction generates a second chiral center.
H
CH,
H OH
\c H ++c
HO H
H,
CHAP. 51
85
STEREOCHEMISTRY
RR and SS enantiomers are formed in equal amounts to give a racemic form. The meso and racemic forms are in unequal amounts. ( d ) Reduction of the double bond makes C3 chiral. Reduction occurs on either face of the planar 7r bond to form molecules with R and molecules with S configurations at C’. These are in unequal amounts because of the adjacent chiral C that has an S configuration. Since both chiral atoms in the product are structurally identical, the products are a meso structure (RS) and an optically active diastereomer (SS).
C H 3CH2-C
p‘
<
C H 3CH
H~INI
CH2CH.3
#
H3C CHj C H 2CH
H H
meso
H CH3
CH3CHzwCH2CH,
(9
(3R4S)
H,C €3
(3S.W
active
Problem 5.35 Designate the following compounds as erythro or threo structures.
( a ) Erythro (see Problem 5.30). (6) Erythro; it is best to examine eclipsed conformations. If either of the chiral C’s is rotated 120” to an eclipsed conformation for the two Br’s, the H’s are also eclipsed.
Br (c)
Br Threo; a 60” rotation of one of the chiral C’s eclipses the H’s but not the OH’S.
Problem 5.36 CH,CH,OH is oxidized in the presence of an enzyme to CH,CHO. When racemic CH,CHDOH is similarly oxidized, optically active CH,CHDOH is isolated after the reaction. Explain. 4
The enzyme is chiral and causes oxidation of only one enantiomer. The unreacted, optically active enantiomer is isolated. Problem 5.37 Glyceraldehyde can be converted to lactic acid by the two routes shown below. These results reveal an ambiguity in the assignment of relative D , L configuration. Explain.
CH, H+OH-H+OH COOH (R)-(-)-Lactic acid
CHO
-
-
CH,OH D-(+)-Glyceraldehyde
COOH H+OH
CH, (S)-(+)-Lactic acid
In neither route is there a change in the bonds to the chiral C. Apparently, both lactic acids should have the configuration, since the original glyceraldehyde was D. However, since the CH, and COOH groups are interchanged, the two lactic acids must be enantiomers. Indeed, one is (+) and the other is (-). This shows that for unambiguous assignment of D or L it is necessary to specify the reactions in the chemical change. Because of such ambiguity, R,S is used. The ( + ) lactic acid is S, the (-) enantiomer is R. D
86
STEREOCHEMISTRY
[CHAP. 5
Problem 5.38 Deduce the structural formula for an optically active alkene, C6H,,, which reacts with H, to form an optically inactive alkane, C,H,,. 4 The alkene has a group attached to the chiral C which must react with H, to give a group identical to one already attached, resulting in loss of chirality.
Problem 5.39 Use conformational theory to evaluate the validity of the following statement: “meso-2,34 Dibromobutane is achiral because it possesses a plane of symmetry.,, The statement is true only for the high-energy eclipsed conformation, which has an extremely low population. The more stable anti conformer has a center rather than a plane of symmetry. Including the gauche, there are an infinite number of conformations between the anti and eclipsed forms. These all exist as conformational enantiomers comprising an infinite number of conformational racemic forms. Problem 5.40 Tetra-sec-butylmethane has four optically active and one optically inactive stereoisomer. List 4 the isomers in terms of R,S designation. Each sec-butyl group has a chiral C that can be R or S. Since all four groups are equivalent, the order of writing the designation is immaterial, RRRS is the same as RRSR. The possibilities are: RRRR, RRRS, RRSS, SSRR, SSSR, and SSSS. RRRR and SSSS, and RRRS and SSSR, are enantiomeric pairs. These are the four optically active isomers. The mirror image of RRSS is SSRR. These are identical and therefore the isomer is meso. This isomer is a rare example of a compound which is achiral because it has only an improper axis of symmetry-it has no plane or center of symmetry.
Chapter 6 Alkenes 6.1 NOMENCLATURE AND STRUCTURE
Alkenes (olefins) contain the functional group
>&c( and have the general formula C,H,,. saturated cycloalkanes.
U
double bond
These unsaturated hydrocarbons are isomeric with the
H 2C-
C H JCH=C H2
Ic,H,I Propylene (Propene)
\ /cH2 CH2
Cyclopropane
In the IUPAC system the longest continuous chain of C’s containing the double bond is assigned the name of the corresponding alkane, with the suffix changed from -ane to - m e . The chain is numbered so that the position of the double bond is designated by assigning the lower possible number to the first doubly bonded C. ‘CH2
II
CH3CH=CHCH?CH?CH3 2-Hexene
I
CHj--C=C
HCH 3
2-Methyl-2-butene
*CH CHJCH2-~HCH2$H26HJ 1 4 3-Ethyl-1-hexene
A few important unsaturated groups that have trivial names are: H,C=CHH,C=CH-CH,(Allyl), and CH3CH=CH- (Propenyl).
(Vinyl),
Problem 6.1 Write structural formulas for (a) 3-bromo-2-pentene, (6) 2,4-dimethyl-3-hexene, (c) 2,4,4trimethyl-2-pentene, (d) 3-ethylcyclohexene, 4
Problem 6.2 Supply the structural formula and IUPAC name for (a) trichloroethylene, ( 6 ) sec-butylethylene, (c) sym-divinylet hylene. 4 Alkenes are also named as derivatives of ethylene. The ethylene unit is shown in a box.
87
88
ALKENES
C1
I
[CHAP. 6
H
I
Trichloroet hene
C l - w k C l CH3
I
CH,CH,CH-m] H,C=CH-\CH=CH
3-Methyl- 1-pentene
1.3,SHexatriene
I-CH=CH,
Give the IUPAC name for:
I
H2C >CH-CH2-CH=CH2 H2C
EH3(! H26H2dH2 61
CH2CH3
rl
5
2
1
CH3CHCH=C--dH2CHCH,
(d)
I
CH,
11
HC. 'CH=CH /
(a) 3-Cyclopropyl-1-propene.( b ) 4 4 1 -Methylbutyl)-l,4-hexadiene(pick longest chain with both double bonds). ( c ) 4-Ethyl-2,6-dimethyl-4-decene. ( d ) 1,3,5-Cycloheptatriene.
6.2 GEOMETRIC (CiS-franS) ISOMERISM
The C=C consists of a cr bond and a n bond. The n bond is in a plane at right angles to the plane of the single U bonds to each C (Fig. 6-1). The n bond is weaker and more reactive than the U bond. The reactivity of the n bond imparts the property of unsaturation to alkenes; alkenes therefore readily undergo addition reactions. The n bond prevents free rotation about the C S and therefore an alkene having two different substituents on each doubly bonded C has geometric isomers. For example, there are two 2-butenes:
Fig. 6-1
H3\
,CH3
dc=c\ H
CH,'s on same side; called cis-
Y
/=".
H3
FH3 H
CH,'s on opposite sides; called trans-
Geometric (cis-trans) isomers are stereoisomers because they differ only in the spatial arrangement of the groups. They are diastereomers and have different physical properties (m.p., b.p., etc.). In place of cis-tram, the letter 2 is used if the higher-priority substituents (Section 5.3) on each C are on the same side of the double bond. The letter E is used if they are on opposite sides.
CHAP. 61
89
ALKENES
Problem 6.4 Predict ( a ) the geometry of ethylene, H2C=CH,; (b) the relative C-to-C bond lengths in ethylene and ethane; (c) the relative C-H bond lengths and bond strengths in ethylene and ethane; ( d ) the 4 relative bond strengths of C--C and C==C Each C in ethylene (ethene) uses sp2 HO’s (Fig. 2-8) to form three trigonal a bonds. All five a bonds (four C-H and one C - C ) must lie in the same plane: ethylene is a planar molecule. All bond angles are approximately 120”. The C=C atoms, having four electrons between them, are closer to each other than the C - C atoms, which are separated by only two electrons. Hence, the C=C length (0.134 nm) is less than the C - C length (0.154 nm). The more s character in the hybrid orbital used by C to form a a bond, the closer the electrons are to the nucleus and the shorter is the a bond. Thus, the C-H bond length in ethylene (0.108nm) is less than the length in ethane (0.110 nm). The shorter bond is also the stronger bond. Since it takes more energy to break two bonds than one bond, the bond energy of C-C in ethylene (611 kJ/mol) is greater than that of C--C in ethane (348 kJ/mol). However, note that the bond energy of the double bond is less than twice that of the single bond. This is so because it is easier to break a n bond than a U bond. Problem 6.5 Which of the following alkenes exhibit geometric isomerism? Supply structural formulas and names for the isomers. 4 H H 743 I t ( U ) CH3CH2C==CCH2CH, (6) H,C=C(CI)CH, ( c ) C,H,C=C-CH,I
( d ) CH,CH=CH
b2H5 ( e ) CH,CH=CH-CH==CHCH,CH,
( f ) CH,CH--LH--CH=CHCH,
No geometric isomers because one double-bonded C has two C2H,’s. (b) No geometric isomers; one double-bonded C has two H’s. (c) Has geometric isomers because each double-bonded C has two different substituents: (a)
CH,CH, >=Ch
/CHJ
CH3CH, ‘C=C
H/
cis- or (Z)-l-Iodo-2-pentene
4
/H \H21
irans- or (E)I-Iodo-2-pentene
( d ) There are two geometric isomers because one of the double bonds has two different substituents. H3C\c=c /CH=CHz H/
\H
(2)-1,3-Pentadiene ( c i s )
H3C
\c=c /H
/ H
\ CH=CH,
(Ebl.3-Pentadiene (trans)
(e) Both double bonds meet the conditions for geometric isomers and there are four diastereomers of 2,4-heptadiene.
cis,cis or (2.2)
cis, trans or (2,E )
irans. cis or ( E , Z )
tram, irans or ( E . E )
Note that cis and truns and E and 2 are listed in the same order as the bonds are numbered
90
ALKENES
[CHAP. 6
( f ) There are now only three isomers because cis-tram and tram-cis geometries are identical.
*\c=c
=c=,
H ,c
H
/
/w
H,
H ,c
/H
>c=c /c=c\
\CH3
\H
H
H H
\c=c
H
H 3c /
\H
\H
H'
rrurts, rruns -2A-Hexadiene or
cis, truns-2.4-Hexadiene or (2,E )-ZP-Hexadiene
cis. cis -2.4-Hexadiene or (2,Z )-2,4-He xadiene
'
\c=c
( E , E )-2,CHexadiene
Problem 6.6 Write structural formulas for ( a ) (E)-2-methyl-3-hexene ( t r a m ) , (b) (S)-3-chloro-l-pentene, ( c ) (R) ,( 2)-2-chlor0-3-heptene (cis). 4
Problem 6.7 Supply structural formulas and systematic names for all pentene isomers including stereoisomers. 4 These are derived by first writing the carbon skeletons of all pentane isomers and then introducing one double bond. A double bond can be placed in n-pentane in two ways to give two isomers.
H2C=CHCH2CH,CH,
CH,CH=CHCH,CH, 2-Pentene
I-Pentene
2-Pentene has geometric isomers:
cis- or (2)-?-Pentene
rrans- or (E)-2-Pentene
From isopentane we get three isomeric alkenes but none has geometric isomers.
CH3
CH, H2C=CCH,CH3 I
CH,C=CHCH, I
?-Methyl- I-butene
2- Methyl-2-butene
CH,
I
CH,CHCH=CH, 3-Methyl- I -butene
N o alkene can be formed from neopentane
because a double bond would give the central C five bonds. None of the isomeric pentenes are chiral. Atoms with multiple bonds cannot be chiral centers. Problem 6.8 How do boiling points and solubilities of alkenes compare with those of corresponding alkanes?
4
Alkanes and alkenes are nonpolar compounds whose corresponding structures have almost identical molecular weights. Boiling points of alkenes are close to those of alkanes and similarly have 20" increments per
91
ALKENES
CHAP. 61
C atom. Both are soluble in nonpolar solvents and insoluble in water, except that lower-molecular-weight alkenes are slightly more water-soluble because of attraction between the 'TT bond and H,O. Problem 6.9 Show the directions of individual bond dipoles and net dipole of the molecule for ( a ) 4 1,I -dichloroethylene, (6) cis- and trans-l,2-dichloroethylene.
The individual dipoles are shown by the arrows on the bonds between C and C1. The net dipole for the molecule is represented by an arrow that bisects the angle between the two Cl's. C-H dipoles are insignificant and are disregarded.
cis
In the truns isomer the C - 4 a zero dipole moment.
trans
moments are equal but in opposite directions; they cancel and the tram isomer has
Problem 6.10 How can heats of combustion be used to compare the differences in stability of the geometric isomers of alkenes? 4
The thermodynamic stability of isomeric hydrocarbons is determined by burning them to CO, and H,O and comparing the heat evolved per mole ( - A H combustion). The more stable isomer has the smaller ( - A H ) value. Truns alkenes have the smaller values and hence are more stable than the cis isomers. This is supported by the exothermic ( A H negative) conversion of cis to tram isomers by ultraviolet light and some chemical reagents. The cis isomer has higher energy because there is greater repulsion between its alkyl groups on the same side of the double bond than between an alkyl group and an H in the trans isomer. These repulsions are greater with larger alkyl groups, which produce larger energy differences between geometric isomers. Problem 6.11 Suggest a mechanism to account for interconversion of cis and tram isomers by electromagnetic 4 radiation of wavelength less than 300nm.
See Problem 2.41. The excited molecule has the indicated major conformation. Return of the excited electron from this conformation to the ground-state 'TT bond results in formation of a mixture of geometric isomers.
excited molecule
6.3 PREPARATION OF ALKENES 1.
1,2-Eliminations
Also called p-eliminations, these constitute the principal laboratory method whereby two atoms or groups are removed from adjacent bonded C's. (a)
Dehydrohalogenation
H X I I B; + -C-C-
I
I
+ -C=C-
I
I
+
B:H
+
X-
KOH in ethanol is most often used as the source of the base, BT, which then is mainly C,H,O-.
92
ALKENES
(6) Dehydration
H OH
I I
-C-C(c)
[CHAP. 6
Dehalogenation
I I
acid
--C=C-
I
I
+
HzO
+
MgBr,(or ZnBr,)
Br Br
Mg(or Zn)
I I + -C-CI I
+
-C=C-
I
I
(d) De hydrogenation H H
I I
-C-C--
I I
Pt or Pd
-C=C-
I
I
+ H2
(mainly a special industrial process)
Cracking (Section 4.4) of petroleum hydrocarbons is the source of commercial alkenes. In dehydration and dehydrohalogenation the preferential order for removal of an H is 3”> 2” > 1” (Saytzeff rule). We can say “the poor get poorer.” This order obtains because the more R’s on the C = C group. the more stable is the alkene. The stability of alkenes in decreasing order of substitution by R is R,C=CR, > R,C=CRH > R,C=CH,, 2.
RCH=CHR > RCH=CH, > CH,=CH,
Partial Reduction of Alkynes
R
\c=c
cis-alkene
H
RC-CR an alkyne
/R
R\
,c=c
H
/H
?runs-a1kene
\R
These reactions, which each give only one of two possible stereomers, are called stereoselective. In this case they are more specifically called diastereoselective, because the stereomers are diastereomers. Problem 6.12 (a) How does the greater enthalpy of cis- vis a vis tram-2-butene affect the ratio of isomers formed during the dehydrohalogenation of 2-chlorobutane? (6) How does replacing the CH, groups of 2-chlorobutane by t-butyl groups to give CH,C(CH,),CH,CHCIC(CH,),CH, alter the distribution of the alkene geometric isomers? 4 ( a ) The transition states for the formation of the geometric isomers reflect the relative stabilities of the
isomers. The greater repulsion between the nearby CH,’s in the cis-like transition state (TS), causes this TS to have a higher enthalpy of activation ( A H ’ ) than the rrans-like TS. Consequently, the tram isomer predominates. (6) The repulsion of the bulkier r-butyl groups causes a substantial increase in the A H ’ of the cis-like TS, and the tram isomer is practically the only product.
Problem 6.13 Give the structural formulas for the alkenes formed on dehydrobromination of the following alkyl bromides and underline the principal product in each reaction: ( a ) 1-bromobutane, (6) 2-bromobutane, (c) 3-bromopentane, (d) 2-bromo-2-methylpentane, (e) 3-bromo-2-methylpentane, ( f ) 3-bromo-2,3-dimethylpen tane . 4
CHAP. 61
93
ALKENES
-
The Br is removed with an H from an adjacent C.
H2C-CHCH,CH,
( a1
$r
I, I I
-
4
H,C=CHCH,CH,
H Br H2 (c1
(only 1 adjacent H; only 1 product)
A 1
H2C-CHCHCH,
(61
H2C=CHCH2CH,
CH,CHCHCHCH,
I, I I
H Br HI
+ cis- and truns-CH,CH=CHCH, (-H2)
(-HI)
cis- and trans-CH,CH=CHCH,CH,
di-R-substituted
(adjacent H's are equivalent)
CH,
CH3
I
I
+ H2C=C-CH2CH2CH3
(d1
and CH,C=CHCH,CH,
(-H') (CH,),C-CHCHCH,
(e)
I, I I H Br H2
* (CH,),C=CHCH,CH, (-H'
)
(-H2)
+ cis- and trans-(CH,),CHCH-CHCH,
tri-R-substituted
(-H2)
H,CH~
di-R-substituted
CH,
I CH,CH-C-C(CH,), I I I H' Br H3
I
__+
cis - and trans-CH,CH=CCH(CH,), (-HI) tri-R-alkene CH,
I + C H ,C H,C=C
CH2
(C H 3,
II + C H ,C H2-C-C
(-H3)tetra-R-alkene
(
H(C H3)2
- H 2 ) di-R-alkene
Problem 6.14 ( a ) Suggest a mechanism for the dehydration of CH,CHOHCH, that proceeds through a carbocation intermediate. Assign a catalytic role to the acid and keep in mind that the 0 in ROH is a basic site like the 0 in H,O. (b) Select the slow rate-determining step and justify your choice. (c) Use transition states to 4 explain the order of reactivity of ROH: 3" > 2" > 1".
H
H (U)
Step 1
I
CH,CHCH,
I
+
H,SO,
H:~)H
:OH base
I
e CH,CHCH, + HSO; I fast
+
acid,
acid,
basez
an onium ion
Step 2
CH,C-CH,
CH,eHCH,
I
H,b'
+
H20
H lsopropyl cation
+
5CH,CH=CH,
Step 3
+
H,SO,
very Strong
acid I
base,
base I
acid
,
Instead of HSO;, a molecule of alcohol could act as the base in Step 3 to give ROH;. ( 6 ) Carbocation formation, Step 2, is the slow step, because it is a heterolysis leading t o a very high-energy carbocation possessing an electron-deficient C. ( c ) The order of reactivity of the alcohols reflects the order of stability of the incipient carbocation (3" > 2" > 1") in the TS of Step 2, the rate-determining step. See Fig. 6-2.
ALKENES
94
[CHAP. 6
REH~ - - OH^ 8+
8+
6+
R,C---OH,
+
R3
R ,CH~H,
~
0
~
2
Reaction Prope ss Fig. 6-2
Problem 6.15 Account for the fact that dehydration of: (a) CH,CH,CH,CH,OH yields mainly CH,CH==CHCH, rather than CH,CH,CH=CH,, (b) (CH,),CCHOHCH, yields mainly (CH,),C=C(CH,),.
4
( a ) The carbocation (R') formed in a reaction like Step 2 of Problem 6.14(a) is 1" and rearranges to a more stable 2" R,CH' by a hydride shift (indicated as -H:; the H migrates with its bonding pair of electrons).
H
H
I +ROH 2C H , C H 2 C + H 2 T +
I" RCH; (n-Butyl cation)
H
&
2" RICH+ (3,3-Dimethyl2-butyl cation)
+ RGH,
2" RICH+ (sec-Butyl cation)
(b) The 2" R,CH+ formed undergoes a methide shift (-:CH,)
(CH,), -CHCH3
CH,CH=CHCH,
-.CHj
+
(CH,),C
to the more stable 3" R,Ct.
9 (CH,),C=C(CH,),
+ RbH,
3" R,C+ (2,3-Dirnethyl2-butyl cation)
Carbocations are always prone to rearrangement, especially when rearrangement leads to a more stable carbocation. The alkyl group may actually begin to migrate as the leaving group (e.g., H,O) is departing-even before the carbocation is fully formed.
Problem 6.16 Supply structural formulas for the compounds formed by dehydrating the following alcohols and underline the major product. H3C
(a)
H
I I CHJ-C-C-CH, I I
HO CH3
CH3
I I
( h ) CH,-C-CH,OH
CH,
(c)
CH,CH,CHCH,CH,
I
OH
4
95
ALKENES
CHAP. 61
The principal product has more alkyl groups on the unsaturated C's. (6) Neopentyl alcohol cannot be dehydrated to an alkene without rearrangement because there is no H on the adjacent ( p ) C. The 1" neopentyl cation RCH: rearranges to a more stable 3" R,C' by a :CH, shift; this is followed by loss of an H'.
74
7H3
@I a + CH3-C-CH20H2
A
I
7H3
- CHJ
-H 0
___*
I
-H+
CH,-:-CH,-CH,
CH,
+
CH,
3" R,C+
CH,C=CHCH,
1" RCHt
CH,, /c=c\/ H H CH,CH,
CH3\/c=c / H 'H
+
7S% (runs-2- Pentene
Problem 6.17 Assign numbers from 1 for justify your choices.
LEAST
CH,CH3
to 3 for
MOST
to indicate the relative ease of dehydration and
CH,
CH,
I
( a ) CH,CHCH,CH,OH
I
+ CH,=CCH,CH,
25% cis-2-Pentene
CH,
I
CH,
I
I
( b ) CH,--C-CH,CH,
( c ) CH,CH-CH-CH,
I
I
OH
4
OH
( a ) 1, ( b )3, (c) 2. The ease of dehydration depends on the relative ease of forming an R', which depends in turn on its relative stability. This is greatest for the 3" alcohol (b) and least for the 1" alcohol ( a ) . Problem 6.18 Give structural formulas for the reactants that form 2-butene when treated with the following reagents: ( a ) heating with conc. H,SO,, (6) alcoholic KOH, (c) zinc dust and alcohol, (d) hydrogen and a 4 catalyst.
Problem 6.19 Write the structural formula and name of the principal organic compound formed in the following reactions:
CH,
I
( a ) CH,CCICH,CH3
+ alc. KOH
(6) HOCH2CH,CH,CH,0H / ( c ) Trans- H2C
--
+ BF,, heat
CH2--CH2
\
CH2 + Zn in alcohol / 'C' H B r -C H €3r
CH3
I I
( d ) CH3-C-CH3
Br
+ CH,COO-
+
96
ALKENES
[CHAP. 6
CH,
I
(a)
CH,C=CHCH,
(c)
/ c H2HZC,
( b ) H,C=CH-CH=CH,
2-Methyl-2-butene
1,3-Butadiene
CH3
CH2\ ,CH, \ / CH=CH
I
Cyclohexene
( d ) CH,-C=CH,
Isobutylene
Problem 6.20 To prepare 2-methyl-1-butene in good yield, would you employ ( a ) dehydration of 24 methylbutanol or (b) dehydrohalogenation of 2-methyl-1-chlorobutane with alcoholic KOH? Reaction (b) is preferred, because ( a ) would result in rearrangement to 2-methyl-2-butene.
6.4
CHEMICAL PROPERTIES OF ALKENES
Alkenes undergo addition reactions at the double bond. The 7r electrons of alkenes are a nucleophilic site and they react with electrophiles by three mechanisms (see Problem 3.40).
E Nu
E Intermediate R’
+ E
E
---
E- (Free radical)
E
Intermediate R.
RCH-CHR
+ E-NU
(Cyclic, one-step, rare) E
Nu
Transition State
REDUCTION TO ALKANES 1.
Addition of H2 RCH-CHR
+ H,
Pt.Pd o r NI
(heterogeneous catalysis)
RCH,CH,R
H2 can also be added under homogeneous conditions in solution by using transition-metal coordination complexes such as the rhodium compound, Rh[P( C,H,),Cl] (Wilkinson’s catalyst). The relative rates of hydrogenation H,C=CH. > RCH-CH,
> R,C=CH,,
RCH-CHR
> R,C=CHR > R,C=CR,
indicate that the rate is decreased by steric hindrance
2.
-
Reductive Hydroboration RCH-CHR
IBH
from BzH6
RCH-CHR
I
H
I
BH2
Alkylborane
CH,COOH
RCH2CH2R (homogeneous catalysis)
CHAP. 61
97
ALKENES
The compound BH, does not exist; the stable borohydride is diborane, B,H,. In syntheses B,H, is dissolved in tetrahydrofuran (THF), a cyclic ether, to give the complex THF:BH,,
in which BH, is the active reagent. Problem 6.21 Given the following heats of hydrogenation, - A H , , in kJ/mol: 1-pentene, 125.9; cis-2-pentene, 119.7; trans-2-pentene, 115.5. ( a ) Use an enthalpy diagram to derive two generalizations about the relative stabilities of alkenes. ( b ) Would the AH,, of 2-methyl-2-butene be helpful in making your generalizations? ( c ) The corresponding heats of combustion, - A H , , are: 3376, 3369, and 3365 kJImol. Are these values consistent with your generalizations in part (a)? ( d ) Would the A H < of 2-methyl-2-butene be helpful in your comparison? (e) Suggest a relative value for the A H < of 2-methyl-2-butene. 4 See Fig. 6-3. The lower AH,, , the more stable the alkene. (1) The alkene with more alkyl groups on the double bond is more stable; 2-pentene > 1-pentene. (2) The tram isomer is usually more stable than the cls. Bulky alkyl groups are anti-like in the tram isomer and eclipsed-like in the cis isomer. No. The alkenes being compared must give the same product on hydrogenation. Yes. Again the highest value indicates the least stable isomer. Yes. On combustion all four isomers give the same products, H,O and CO,. Less than 3365 kJ/mol, since this isomer is a trisubstituted alkene and the 2-pentenes are disubstituted.
1
I-Pentene
T '
cis-2-Pentene truns-2-Pentene
A H h = -125.9 I
A H h = - 119.7
Fig. 6-3
Problem 6.22 What is the stereochemistry of the catalytic addition of H, if trans- CH,CBr=CBrCH, gives ruc-CH,CHBrCHBrCH, and its cis isomer gives the meso product? 4
In hydrogenation reactions, two H atoms add stereoselectively syn (cis) to the
tram
(SS) racemate
bond of the alkene.
ALKENES
98
[CHAP. 6
meso
cls
ELECTROPHILIC POLAR ADDITION REACTIONS Table 6-1 shows the results of electrophilic addition of polar reagents to ethylene. Table 6 1
Product
Reagent
Structure
Name
Structure
CH,XCH,X
Ethylene dihalide ~~~
Ethyl halide
CH,CH,X
X:OH
Ethylene halohydrin
CH ,XCH,OH
%%SO,OH
Ethyl bisulfate
CH,CH,OSO,H
H:OH
Ethyl alcohol
CH,CH,OH
Hydrohalic acids Hypohalous acids Sulfuric acid (cold) ~
~
Water (dil. H,O')
6 + 6-
I
~
Ethyl borane
Borane
~~
Ethylene glycol
Peroxyformic acid
~
~
~
&+a-
Hg( 02CCH 3 1 2 H2O
~
~
~
~
~
CH20HCH20CH]--+HOCH,CH,OH
II
0
Mercuric acetate, H,O
~
0 Ethanol HO Hg
I
HO H
0
I
COCH, Problem 6.23 Unsymmetrical reagents like HX add to unsymmetrical alkenes such as propene according to Markovnkov's rule: the positive portion, e.g., H of HX, adds to the C that has more H's ("the rich get richer',). Explain by stability of the intermediate cation. 4
The more stable cation (3">2"> 1") has a lower A N S for the transition state and forms more rapidly (Fig. 6-4). Markovnikov additions are called regioselective since they give mainly one of several possible structural isomers. Problem 6.24 Give the structural formula of the major organic product formed from the reaction of CH,CH=CH, with: ( a ) Br,, (6) HI, (c) BrOH, (d) H,O in acid, (e) cold H,SO,, ( f ) BH, from B,H,, (g) peroxyformic acid ( H , 0 2 and HCOOH). 4
CHAP, 61
99
ALKENES
I
(CH3)2C2= C'H2 + H' Reaction Progress Fig. 6-4
The positive +
(a+)
+
part of the addendum is an electrophile (E') which forms CH,CHCH,E rather than
CH,CHECH,. The Nu: part then forms a bond with the carbocation. The E' is in a box; the Nu:- is encircled.
(6) CH3-CH-CH2
6h
2CH CHICH
A (CH3CH,CH3,B (Anti-Markovnikov orientation; with nonbulky alkyl groups, all H's of BH, add to form a trialkylborane)
Problem 6.25 Account for the anti-Markovnikov orientation in Problem 6.24( f ) .
4
The electron-deficient B of BH,, as an electrophilic site, reacts with the ?r electrons of C=C, as the nucleophilic site. In typical fashion, the bond is formed with the C having the greater number of H's, in this case the terminal C. As this bond forms, one of the H's of BH, begins to break away from the B as it forms a bond to the other doubly bonded C atom giving a four-center transition state shown in the equation. The product from this step, CH,CH,CH,BH, (n-propyl borane), reacts stepwise in a similar fashion with two more molecules of propene, eventually to give (CH,CH,CH,),B.
CH,CH=CH,
+ BH, --+
U +
CH,q+H,+ 1 1 H-4H,
CH,CH2CH2BH,
-
2 CH2=CHCH3
(CH,CH,CH,),B
This reaction is a stereoselective and regioselective syn addition.
Problem 6.26 ( a ) What principle is used to relate the mechanisms for dehydration of alcohols and hydration of alkenes? (6) What conditions favor dehydration rather than hydration reactions? 4 ( a ) The principle of microscopic reversibility states that every reaction is reversible, even if only to a
microscopic extent. Furthermore, the reverse process proceeds through the same intermediates and transition states, but in the opposite order.
RCH2CH20H
H+
RCH=CH,
+H20
ALKENES
100
[CHAP. 6
(6) Low H,O concentration and high temperature favor alkene formation by dehydration, because the volatile alkene distills out of the reaction mixture and shifts the equilibrium. Hydration of alkenes occurs at low temperature and with dilute acid which provides a high concentration of H,O as reactant. Problem 6.27 Why are dry gaseous hydrogen halides (HX) acids and not their aqueous solutions used to 4 prepare alkyl halides from alkenes?
Dry hydrogen halides are stronger acids and better electrophiles than the H 3 0 + formed in their water solutions. Furthermore, H,O is a nucleophile that can react with R' to give an alcohol. Problem 6.28 Isobutylene gas dissolves in 63% H,SO, to form a deliquescent white solid. If the H,SO, solution is diluted with H,O and heated, the organic compound obtained is a liquid boiling at 83°C. Explain.
CH3-C CH, -I
I I
+ H:OS03H
CH,-C-OSO,H
I
H O
A CH,-C-OH
I
+ H,SO,
CH3
CH3
Isobutylene
4
CH,
CH,
terl-Butyl hydrogen sulfate (reactive ester of H2S04) (white solid)
red-Butyl alcohol (b.p. 83 "C)
Problem 6.29 Arrange the following alkenes in order of increasing -reactivity on addition of hydrohalogen acids: (a) H,C=CH,, (6) (CH,),C=CH,, ( c ) CH,CH=CHCH,. 4
The relative reactivities are directly relattd to the stabilities of the intermediate R"s. Isobutylene, (b), is most reactive beca+use it forms the 3" (CH,),CCH,. The ne+xt-most reactive compound is 2-butene, (c), which forms the 2" CH,CHCH,CH,. Ethylene forms the 1" CH,CH, and is least reactive. The order of increasing reactivity is: ( a ) < (c) < ( b ) . Problem 6.30 The addition of HBr to some alkenes gives a mixture of the expected alkyl bromide and an isomer formed by rearrangement. Outline the mechanism of formation and structures of products from the 4 reaction of HBr with ( a ) 3-methyl-l-butene, (b) 3,3-dimethyl-l-butene.
No matter how formed, an R' can undergo H: or :CH, (or other alkyl) shifts to form a more stable R".
7 w
CH3
CH3
I
CH3CHBrcHCH3
2-Bromo-3-methylbutane
H2C-LHCHCH3 -% CH3-yH-C-CH3
I
CH3
I CH,CH,-:-CH,
H 2"
Br-
3"
CH3
CH,
CH,
I
H2C=-CH--C--CH,
H* I__*
I
CH3-:H-C-CH3
I
CH3
&H3 2"
lesr srahle
/
I
CH3
I 1
CH3CH2-C-CH3
Br
2-Bromo-2-methylbutane
CH,-CH-k-CH,
I
I
Br CH3 3-Bromo-2,2-dimethylbutane
k3
CH3
I
+
CH3-CH-C
I
CH, Br
I
I
-CH3 Br-_ CH,CH-C-CH,
CH3 3" more srable
I
CH, 2-Bromo-2,3-dimethylbutane
CHAP. 61
101
ALKENES
Problem 6.31 Compare and explain the relative rates of addition to alkenes (reactivities) of HCI, HBr and HI.
4
The relative reactivity depends on the ability of HX to donate an H ' (acidity) to form an R ' in the rate-controlling first step. The acidity and reactivity order is HI > HBr > HCI.
Problem 6.32 (a) What does each of the following observations tell you about the mechanism of the addition of Br, to an alkene? (i) In the presence of a C1- salt, in addition to the vic-dibromide, some vicbromochloroalkane is isolated but no dichloride is obtained. (ii) With cis-2-butene only rac-2,3-dibromobutane is formed. (iii) With tram-2-butene only rneso-2,3-dibromobutane is produced. (b) Give a mechanism 4 compat ible with these observations. (a) (i) Br, adds in two steps. If Br, added in one step, no bromochloroalkane would be formed. Furthermore, the first step must be the addition of an electrophile ( the Br ' part of Br, ) followed by addition of a
nucleophile, which could now be Br- or CI-. This explains why the products must contain at least one Br. (ii) One Br adds from above the plane of the double bond, the second Br adds from below. This is an anti (trans) addition. Since a Br' can add from above to either C, a racemic form results.
Br cis rucemic
(iii) This substantiates the anti addition.
trans
meso
The reaction is also stereospecific because different stereoisomers give stereochemically different products, e.g. cis + racemic and \runs + meso. Because of this stereospecificity , the intermediate cannot be the free carbocation CH,CHBrCHCH,. The same carbocation would arise from either cis-or trans-2-butene, and the product distribution from both reactants would be identical. ( 6 ) The open carbocation is replaced by a cyclic bridged ion having Br' partially bonded to each C (bromonium ion). In this way the stereochemical differences of the starting materials are retained in the intermediate. In the second step, the nucleophile attacks the side opposite the bridging group to yield the anti addition product.
(2S,3R)-2,3-Dibromobutane (meso )
H
H
H trans
(2R,3S)-2,3-Dibromobutane (meso)
a bromonium ion
Br, does not break up into Br' and Br-. More likely, the other as an anion (Fig. 6-5).
T
electrons attack one of the Br's. displacing the
102
ALKENES
[CHAP. 6
..
Bromonium ion + :BJ :
-
Fig. 6-5
Problem 6.33 Alkenes react with aqueous C1, or Br, to yield vic-halohydrins, -CXCOH. Give a mechanism 4 for this reaction that also explains how Brz and (CH,),C=CH2 give (CH,),C(OH)CH,Br. The reaction proceeds through a bromonium ion [Problem 6.32(6)] which reacts with the nucleophilic H,O to give
I+ I
-CBrCOH,
I
This protonated halohydrin then loses H' to the solvent, giving the halohydrin. The partial bonds between the C's and Br engender S+ charges on the C's. Since the bromonium ion of 2-methylpropene has more partial positive charge on the 3" carbon than on the 1" carbon, H,O binds to the 3°C to give the observed product. In general, X appears on the C with the greater number of H's. The addition, like that of Br,, is anti because H,O binds to the C from the side away from the side where the Br is positioned. Problem 6.34 ( U ) Describe the stereochemistry of glycol formation with peroxyformic acid (HC0,H) if cis-2-butene gives a racemic glycol and rruns-2-butene gives the meso form. (6) Give a mechanism for cis. 4 The reaction is a stereospecific anti addition similar to that of addition of Br,.
OH
racemic
I
HK..
.
I
OH racemic
Problem 6.35 Describe the stereochemistry of glycol formation with cold alkaline aqueous KMnO, if cis-2-butene gives the meso glycol and truns-2-butene gives the racemate. 4
ALKENES
CHAP. 61
103
The reaction is a stereospecific syn (cis) addition since both OH groups bond from the same side.
cis-2-Butene
(R,S)-meso-Glycol
1! __*
Cyclic intermediate from topside attack
tram-2-Butene
from underside attack
roc-Glycol
DIMERIZATION AND POLYMERIZATION Under proper conditions a carbocation (R’), formed by adding an electrophile such as H’ or BF, to an alkene, may add to the C=C bond of another alkene molecule to give a new dimeric R“; here, R’ acts as an electrophile and the 7r bond of C=C acts as a nucleophilic site. R“ may then lose an H’ to give an alkene dimer. Problem 6.36 ( a ) Suggest a mechanism for the dimerization of isobutylene, (CH,),C=CH,. (6) Why does (CH,),C’ add to the “tail” carbon rather than to the “head” carbon? (c) Why are the Bronsted acids H,SO, 4 and HF typically used as catalysts, rather than HCl, HBr, or HI?
n
Step 2 Me,C’
+ H,C=C(CH r \
CH, ) +Me,C-CHC 3 2
“tail” “head”
electrophile
Step 3 R‘+ - H’
R”
-
I
H
I
+TH H’
nucleophile
3” dimeric R’’
Me,C--CH==C(CH,),
(major, Saytzeff product) (minor, non-Saytzeff product)
- H’+ ---+ Me,C--CH,C(CH,)=CH,
( 6 ) Step 2 is a Markovnikov addition. Attack at the “tail” gives the 3” R“; attack at the “head” would give the much less stable, 1” carbocation ‘CH,C(CH,),CMe,. ( c ) The catalytic acid must have a weakly nucleophilic conjugate base to avoid addition of HX 60 the C=C. The conjugate bases of HCl, HBr, and H1 (Cl-, Br-, and I - ) are good nucleophiles that tind to R’.
104
[CHAP. 6
ALKENES
The newly formed R’’ may also add to another alkene molecule to give a trimer. The process whereby simple molecules, or monomers, are merged can continue, eventually giving high-molecularweight molecules called polymers. This reaction of alkenes is called chain-growth (addition) polymerization. The repeating unit in the polymer is called the mer. If a mixture of at least two different monomers polymerizes, there is obtained a copolymer. Problem 6.37 Write the structural formula for ( a ) the major trimeric alkene formed from (CH,),C=CH,, labeling the mer; ( b ) the dimeric alkene from CH,CH=CH,. [Indicate the dimeric R’.] 4
mer The individual combining units are boxed. (b)
(CH,),CHCH=CHCH,
; [( CH, ),CHCH,CHCH,]
S TEREOCHEMISTR Y OF POL YMERIZATION
The polymerization of propylene gives stereochemically different polypropylenes having different physical properties. H
-c-
or
H *.,
*IR
4-
The C’s of the mers are chiral, giving millions of stereoisomers, which are grouped into three classes depending on the arrangement of the branching Me (R) groups relative to the long “backbone” chain of the polymer (Fig. 6-6). CH,H CH,H CH,H CH,H CH,H CH,H
Isotactic (Me or R all on same side)
CH,H HCH, CH,H HCH, CH,H H C H , CH,H
Syndiotactic (Me or R on alternate sides)
CH,H CH,H
HCH, CH,H H C H , H C H ,
Atactic (Me or R randomly distributed)
Fk.6-6
CHAP. 6)
105
ALKENES
Problem 6.38 Suggest a mechanism for alkane addition where the key step is an intermolecular hydride (H:) transfer. 4 See Steps 1 and 2 in Problem 6.36 for formation of the dimeric R'. CH 3 CH (CH,),CCH,L
I
4'
~-@C(CH,),-+(CH,),CCH,
@+kCH,),
CH, CH 3 dimeric R' This intermolecular H: transfer forms the (CH,),C' ion which adds to another molecule of (CH,),C=CH, to continue the chain. A 3"H usually transfers to leave a 3" R'. FREE -RADICAL ADDITIONS
-
-
RCHSH,
+ HBr
RCHSH,
+ HSH
RCHSH,
+ HCCl,
RCHSH,
+ BrCCI,
RCH,CH,Br
ROOR
(anti-Markovnikov; not with HF, HCI or HI)
RCH,CH,SH
ROOR
(anti-Markovnikov)
RCHCH,[ml
-Lh ROOR
(anti-Markovnikov)
1-1
RCHmCH,
Problem 6.39 Suggest a chain-propagating free-radical mechanism for addition of HBr in which Br. attacks the alkene to form the more stable carbon radical. 4 Initiation Steps R-0-U-RRO. + HBr-
heat
2R-0
a
bond is weak)
(*
Br. + R U H
Propagation Steps For Chain Reaction CH,CHBrCH, +ICH,CH==CH,I+
Br---+ CH,CHCH2Br
( I " radical)
CH,CHCH,Br
(2" radical)
+ HBr-CH,CH,CH,Br
+ Br.
The Br. generated in the second propagation step continues the chain.
CARBENE ADDITION
1
H H
WN2
Diazomethane
CLEAVAGE REACTIONS
Ozonolysis
Ozonide
Carbonyl compounds
ALKENES
106
[CHAP. 6
Problem 6.40 Give the products formed on ozonolysis of ( a ) H,C=CHCH,CH,, ( b ) CH,CH=CHCH,, (c) (CH,),C==CHCH,CH,, (d) cyclobutene, (e) H,C=CHCH,CH=CHCH,. 4
To get the correct answers, erase the double bond and attach a =O to each of the formerly double-bonded The total numbers of C’s in the carbonyl products and in the alkene reactant must be equal.
H,C=O + O=CHCH,CH,. CH,CH=O; the alkene is symmetrical and only one carbonyl compound is formed. (CH,),C=O + O=CHCH,CH,. O=CHCH,CH,CH-V; a cycloalkene gives only a dicarbonyl compound. H , C = O + O = C H C H , C H = O + O=CHCH,. Noncyclic polyenes give a mixture of monocarbonyl compounds formed from the terminal C’s and dicarbonyl compounds from the internal doubly bonded C’S. Problem 6.41 Deduce the structures of the following alkenes. 0
II
An alkene CIOHZO o n ozonolysis yields only CH,--C--CH,CH,CH,.
0
H
I
II
An alkene C,H,, on ozonolysis gives (CH,),CC--V and C H , - C - C H , C H , . A compound C,H,, adds one mole of H, and forms on ozonolysis the dialdehyde
CH3
CH3
I
I
O=CHCHCH,CH,CHCH=O A compound C X H l 2adds two moles of H, and undergoes ozonolysis to give two moles of the dialdehyde O=CHCH2CH,CH=0. 4 The formation of only one carbonyl compound indicates that the alkene is symmetrical about the double bond. Write the structure of the ketone twice so that the C = O groups face each other. Replacement of the two 0 ’ s by a double bond gives the alkene structure.
CH3
CH3
I
CH,CH,CH,C=O
+ O=CCH2CH2CH,
CH3
I CH,-C-CH=O I
H3C CH,
I
I
CH3
CH3
I
CH,CH2CH2C=CCH2CH2CH3
t-
CHj
I I +CH3-C-C=C-CH2CH, I 1
I
+ 0-C-CH2CH3
cis or tram
CH,H
CH3
C,H,, has four fewer H’s than the corresponding alkane, CxH,,. There are two degrees of unsaturation. One of these is accounted for by a C = C because the alkene adds 1 mole of H,. The second degree of unsaturation is a ring structure. The compound is a cycloalkene whose structure is found by writing the two terminal carbonyl groups facing each other.
CH,-CH
CH=O O=CH /
‘cH-cH3 ‘C H*-cH /
’
2
01 HZO(Zn)
CH3-C
yCH=T ,CH-CHI \CH,-CH~
(d) The difference of six H’s between C,H,, and the alkane C,H,, shows three degrees of unsaturation. The 2mol of H, absorbed indicates two double bonds. The third degree of unsaturation is a ring structure. When two molecules of product are written with the pairs of C = O groups facing each other, the compound is seen to be a cyclic diene.
CHAP. 61
107
ALKENES
CH=O /
7
0-CH
I
HX\
CH=O
\
CH2
H*C
O=CH
I
~H=CH R / '&i2 H2C
o3
1.
'2.
17
HZO(Zn)
,CH2
H2C
14
5
'~H=CH
/CH2
13-Cyclooctadiene
6.5 SUBSTITUTION REACTIONS AT THE ALLYLIC POSITION
-
Allylic carbons are the ones bonded to the doubly bonded C's; the H's attached to them are called allylic H's.
+ H,C=CHCH, Br, + H,C=CHCH, C1,
high temperature
H,C=CHCH,CI
low concentration
H,C=CHCH,Br
of Br,
+ HCI + HBr
The low concentration of Br, comes from N - bromosuccinimide (NBS).
NBS
S02C12 + H2C=CHCH3
product of bromina tion
Succinimide
H2C=CHCH2C1
Sulfuryl chloride
+ HCl + SO2
These halogenations are like free-radical substitutions of alkanes (see Section 4.4). The order of reactivity of H-abstraction is allyl > 3" > 2" > 1" > vinyl Problem 6.42 Use the concepts of ( a ) resonance and ( b ) extended account for the extraordinary stability of the allyl-type radical.
7~
orbital overlap (delocalization) to
-LA-&
4
( a ) Two equivalent resonance structures can be written:
therefore the allyl-type radical has considerable resonance energy (Section 2.7) and is relatively stable. (6) The three C's in the allyl unit are sp2-hybridized and each has a p orbital lying in a common plane (Fig. 6-7). These three p orbitals overlap forming an extended 7r system, thereby delocalizing the odd electron. Such delocalization stabilizes the allyl-type free radical.
Problem 6.43 Designate the type of each set of H's in CH,CH=CHCH,CH,--CH(CH,), (e.g. 3", allylic, etc.) and show their relative reactivity toward a Br. atom, using (1) for the most reactive, (2) for the next, etc. Labeling the H's as
we have: ( a ) l", allylic (2); (6) vinylic (6); (c) 2", allylic ( 1 ) ; (d) 2" (4); ( e ) 3" (3); ( f ) 1" ( 5 ) .
4
[CHAP. 6
ALKENES
108
-7-7 0
....
AlJ
-c-c\
--U---
Fig. 6.7
6.6
SUMMARY OF ALKENE CHEMISTRY
PROPERTES
PREPARA TlON 1.
Dehydrohalogenation of RX
RCHXCH,, RCH,CH,X 2.
1. Addition Reactions
+ alc. KOH
(a) Hydrogenation
Dehydration of ROH
RCHOHCH,, RCH,CH,OH
+ H,SO,
,Homogeneous: H,
3. Dehalogenation of vic - Dihalide
4.
+ Zn A
I
Rb(PP-
RCH,CH, 2 Chemical: (BH,), CH,CO,H
heat
RCHXCH,X
-
Heterogeneous: H,/Pt
1I
///
( b ) Polar Mechanism
Dehydrogenation of Alkanes
X2+ RCHXCH,X (X = C1, Br) HX-, RCHXCH, + X,, H 2 0 RCH( OH)CH,X + H,O- H RCH(OH)CH,
RCH,CH,, heat, Pt/Pd '
+ H,SO,-+
-+
RCH(OSO,H)CH, + BH,, H,O, + NaOH- RCH,CH,OH + dil. cold KMnO,-+ RCH(OH)CH,OH + hot KMnO, RCOOH + CO, + R'C0,H- - H + R C H - C H ,
5.
--j
0 '' + RCO,H, H,O++ RCH(OH)CH,OH + HF, HCMe, 0" RCH,CH,CMe, + 0,, Zn, H,O+ RCH-0 + CH2=0 __.*
(c)
Free-Radical Mechanism
+ HBr + RCH,CH,Br
+ CHCl, + RCH,CH,CCl, + H' or BF,-,polymer
2.
Allylic Substitution Reactions
R--CH,--CH=CH,
+ X, A or uv
R--CHX-CH==CH,
109
ALKENES
CHAP. 61
Supplementary Problems Problem 6.44 Write structures for the following: 2,3-dimethyl-2-pentene (6) 4-chloro-2,4-dimeth yl-2-pen tene (c) ally1 bromide (d) 2,3-dimethylcyclohexene (e) 3-isopropyl-1- hexene ( f ) 3-isopropyl-2,6-dimethyl-3-heptene (a)
CH
I '
H,C--C----C--CH,--CH,
(U)
I
CH' CH
CH
I '
I '
(6) H,C--C=CH--C-CH, A1
Problem 6.45 (a) Give structural formulas and systematic names for all alkenes with the molecular formula C,H,, that exist as stereomers. (b) Which stereomer has the lowest heat of combustion (is the most stable), and 4 which is the least stable? (a) For a molecule to possess stereomers, it must be chiral, exhibit geometric isomerism, or both. There are five such constitutional isomers of C,H,,. Since double-bonded C's cannot be chiral, one of the remaining four C's must be chiral. This means that a sec-butyl group must be attached to the C=C group. and the enantiomers are:
H
,
H C=CH -C-CH~
CH
(R)-3-Methyl- I-pentene
(S)-3-Methyl-I-pentene
A molecule with a terminal = C H , cannot have geometric isomers. The double bond must be internal and is first placed between C' and C3to give three more sets of stereomers: H H \ / C=C H,C
/
\
H \
c=c
CH,CH,CH,
( 2 )-2-Hexene
CH,CH,CH, /
H,C
/
\
H
\
/ C=C
H
H
\
c=c
CH(CH,),
/
H
(E)-2-Hexene
(E)-3-Methyl-2-pentene
(2)-4-Methyl-2-pentene
(E)-4-Methyl-2-pentene
( 2 )-3-Meth yl-2-pen tene
110
[CHAP. 6
ALKENES
The double bond is now placed between C-i and CJ to give the isomers: CH,CH,
‘c=c
/
CH,CH H YC=C< H CH,CH,
CH,CH,
H ‘
H’
(Z ) -3-Hexene
(E)-3-Hexene
(6) The 3-methyl-2-pentenes each have three R groups on the C = C group and are more stable than one or the other isomers, which have only two R’s on C=C. The E isomer is the more stable because it has the smaller CH,’s cis to each other, whereas the 2 isomer has the larger CH,CH, group cis to a CH, group. 3-Methyl-1-pentene is the least stable, because it has the fewest R groups, only one, on the C=C group. Among the remaining isomers, the truns are more stable than the cis because the R groups are on opposite sides of the C=C and are thus more separated. Problem 6.46 Write structural formulas for the organic compounds designated by a ? and show the stereochemistry where requested. CH3
I
(U)
CH,-C-CH,CH,CH, I
?(major) + ?(minor)
+ alcoholic KOH-
Br H
\
,c=c\ /H
CH;
+
Br,
CH3
? (alkene) + ? (reagent)
(4
\ CH, / CH2, CH-C \
(Stereochemistry)
CH,
CH,
1
CH,-C-CH,-C-CH,
I
H
I
I
OH
+ CHBr, + peroxide +?
HOBr
-
’? (Stereochemistry)
CH3
CH3
cf)
+
+?
-
CH,CH,CH,CH2CH=CH, ,CH,CH,
?
-+
I
CH,CH,CHCH,OH
+ H,SO,,
major: has 3 R‘s on the C 4 .
heat --+?(major) + ?(minor)
minor: has only 2 R’s.
(6) Anti addition to a cis diastereomer gives a racemic mixture, Br H H Br CH,+CH, + CH,--$--$-CH, Br H H Br (c) The 3” alcohol is formed by acid-catalyzed hydration of CH, CH,
I
I CH,-C-CH=C-CH, I H
or
CH,
CH,
I I CH,-C-CH2-C=CH, I H
111
ALKENES
CHAP. 61
The reagent is dilute aq. H,SO,.
(4
CH,CH,CH,CH,CH,CH,CBr,.
(e)
In this polar Markovnikov addition, the positive Br adds to the C having the H. The addition is anti, so the Br will be tram to OH but cis to CH,. The product is racemic:
I
H
HO
H.
The formation of products is shown:
(f)
H CH,CH,CHCHDH CH3 I
H+ ---+
I
CH3CH2CHCH21)H CH3 I
-H20
OH
i"+ /
CH,CH,CHCH; (I")
I
;
CH3CH2C=CH2
I
- H:
CH,C H=C(CH,), _-Ht CH$H,-C(CH,), (3")
(major)
Problem 6.47 Draw an enthalpy-reaction progress diagram for addition of Br, to an alkene. See Fig. 6-8.
A
-xa d
5 C w
/---
+ Br,
Reaction Progress Fig. 6-8
Problem 6.48 Write the initiation and the propagation steps for a free-radical-catalyzed (RO-) addition of CH,CH=O to I-hexene to form methyl n-hexyl ketone,
0
II
CH,CH,CH,CH,CH,CH,-C-CH, The initiation step is
0
II
CH,-C:H
0
+ RO.
4
I + RO:H
CH,--C.
4
112
ALKENES
[CHAP. 6
and the propagation steps are
0 n-C,H,-CH=CH,
0
I
+ C-CH,
+ H-C-CH,
n-C,H,-CH-CH,-C-CH,
n-C,H,-CH-CH,-C-CH,
__*
__*
II
n-C4H9--CH2CH2-
e e
(adds to give 2" R.)
-CH,
+-
-CH,
(regenerates)
Problem 6.49 Suggest a radical mechanism to account for the interconversion of cis and tram isomers by heating with I,. 4
I, has a low bond dissociation energy (151 kJ/mol) and forms 21- on heating. 1- adds to the C==C to form a carbon radical which rotates about its sigma bond and assumes a different conformation. However, the C-I bond is also weak (235 kJ/mol) and the radical loses 1. under these conditions. The double bond is reformed and the two conformations produce a mixture of cis and tram isomers.
'
'IJ
'
+"
tram
-\
cis
Problem 6.50 Write structures for the products of the following polar addition reactions: (a) (CH,),C=CHCH, (c)
(cH,),N--cH=cH,
+ I-Cl-?
+ HSCH,+?
( b ) (CH,),C=CH,
+ HI-?
( d ) H,C=CHCF, 6 +
8-
(CH,),C@CHmCH,. I is less electronegative than Cl in I -Cl (CH 3 ) q
+ HCI-
?
4
and adds to the C with more H's.
H2
CH ,S 8 i
6-
H is less electronegative than S; H-SCH,. (CH, ),N-cH~--cH,--I
+
+
The + +c+harge on N destabilizes an adjacent + charge. (CH,),NCH,CH, is more stable than (CH,),NCHCH,. Addition is anti-Markovnikov. CICH,CH2CF,. The strong electron-attracting C+F3group destabilizes an adjacent + charge so that CH2CH,CF, is the intermediate rather than CH,CHCF,. Problem 6.51 Explain the following observations: (a) Br, and propene in C,H,OH gives not only BrCH,CHBrCH, but also BrCH,CH(OC,H,)CH,. (6) Isobutylene is more reactive than 1-butene towards peroxide-catalyzed addition of CCI,. (c) The presence of A g t salts enhances the solubility of alkenes in H,O. 4
(a) The intermediate bromonium ion reacts with both Br- and C,H,OH as nucleophiles to give the two products. (6) The more stable the intermediate free radical, the more reactive the alkene. H,C=CHCH,CH, adds -CCI, to give the less stable 2" radical CI,CCH,CHCH2CH,, whereas H,C==C(CH,), reacts to give the more stable 3" radical CI,CCH,C(CH,)2. (c) Ag' coordinates with the alkene by p-d 7r bonding to give an ion similar to a bromonium ion, but more stable:
CHAP. 61
113
ALKENES
Problem 6.52 Supply the structural formulas of the alkenes and the reagents which react to form: (a) 4 (CH,),CI, (6) CH,CHBr,, ( c ) BrCH,CHClCH,, (d) BrCH,CHOHCH,Cl.
CH
1 ,
+ HI + HBr + peroxide H,C==CHCH, + BrCl
( 6 ) H,C=CHBr + HBr (d) BrCH,--CH=CH, + HOCl
(a) CH,--C=CH,
( c ) H,C=CCI--CH,
or
or H,C=CH--CH,CI
+ HOBr
Problem 6.53 Outline the steps needed for the following syntheses in reasonable yield. Inorganic reagents and solvents may also be used. (a) 1-Chloropentane to 1,2-dichloropentane. (b) 1-Chloropentane to 2-chloropentane. ( c ) 1-Chloropentane to 1-bromopentane. (d) 1-Bromobutane to 1,2-dihydroxybutane. ( e ) Isobutyl chloride to CH, CH,
I I
I I
CH,-C-CH,-C-CH,
4
I
CH3
Syntheses are best done by working backwards, keeping in mind your starting material.
-
(a) The desired product is a vic-dichloride made by adding C1, to the appropriate alkene, which in turn is made by dehydrochlorinating the starting material. ClCH,CH,CH,CH,CH,
KOH- H,C==CHCH,CH,CH, alc
CI 7
ClCH,CHCICH,CH,CH,
(b) To get a pure product add HCI to 1-pentene as made in part ( a ) .
+ HCI+
H,C=CHCH,CH,CH,
H,CCHCICH,CH,CH,
( c ) An anti-Markovnikov addition of HBr to 1-pentene [part (a)]. H,C=CHCH,CH,CH,
+ HBr-
peroxide
BrCH,CH,CH,CH,CH,
( d ) Glycols are made by mild oxidation of alkenes. alc KMn04 BrCH, CH, CH, CH, KOH- H, M H C H , CH, RTHOCH, CHOHCH, CH, (e) The product has twice as many C's as does the starting material. The skeleton of C's in the product corresponds to that of the dimer of (CH,),C=CH,. CH3 (CH3)2CHCH2Cl
:!i+ (CH&C=CH,
"2'O4
I .(CH,),CCH=C(CH,),+ (CH,),CCH2C=CH2
1
!
(CH,),CCH,CI(CH,), Problem 6.54 Show how propene can be converted to (a) 1,s-hexadiene, (6) 1-bromopropene, ( c ) 4-methyl-lpentene. 4
I
Br CH,CH=CH,
Cl*. so0 "C
I
CH3
MgBr
+
* CICH,CH=CH,
CuBr p
,
I
CH,-CH-CH,CH=CH,
114
ALKENES
[CHAP. 6
Problem 6.55 Dehydration of 3,3-dimethyl-2-butanol, (CH,),CCHOHCH,, yields two alkenes, neither of 4 which is (CH,),CCH=CH,. What are their structures? +
The initially formed 2" (CH,),CCHCH, undergoes a :CH, shift to give H
(C€-13)2k-&CH,)2 (3" carbocation)
which forms
CH,
CH,
I
I
and H,C=C-CH-CH,
I
CH3 J
2
3
4
Problem 6.56 (a) Br, is added to (S)-H,C=CHCHBrCH,. Give Fischer projections and R,S designations for 4 the products. Are the products optically active? (6) Repeat ( a ) with HBr. ( a ) C' becomes chiral and the configuration of C-' is unchanged. There are two optically active diastereomers of 1,2,3-tribromobutane. It is best to draw formulas with H's on vertical lines.
Br H2k=EH+dH, H (S)
Br Br
__*
BrCH,
H
Br
W t H ,+ B r & - i 2 - ~ ~ H ,
H H (2R,3S) optically active
r H (2S.3S) optically active
(6) There are two diastereomers of 2,3-dibromobutane:
Problem 6.57 Polypropylene can be synthesized by the acid-catalyzed polymerization of propylene. ( a ) Show 4 the first three steps. ( b ) Indicate the repeating unit (mer).
-
+=PCH,
(U)
CH,CH=CH, -% (CH,),C+
I
H monomer
monomeric carbocation
H H
-
I l&&-&Hj (CH,),C-C--C+ I l l H H CH, dimeric carbocat ion
H
I (CH,),CH-CH2-C-CH2-C+ I
H
I
CH3 trimeric carbocation
H -CH2-C-
I
I
CH,
Problem 6.58 List the five kinds of reactions of carbocations and give an example of each. ( a ) They combine with nucleophiles.
4
CHAP. 61
ALKENES
115
( b ) As strong acids they lose a vicinal H [deprotonate), to give an alkene.
CH
I '
CH,--C--CH, +
-H4
CH
l 3
CH,--C=CH,
(c) They rearrange to give a more stable carbocation.
CH
I
CH
I '
'+
3"
1"
(d) They add to an alkene to give a carbocation with higher molecular weight.
I
C H , - C+ - C H ,
I ,
I
+ H,C=C-CH,---.+
electrophile
CH
CH
CH'
CH'
1 '
nucleophile
( e ) They may remove :H (a hydride transfer) from a tertiary position in an alkane.
CH
CH
1 ,
+ CH,-C'
Problem 6.59 From propene, prepare ( a ) CH,CHDCH,D, ( 6 ) CH'CHDCH,, (c) CH,CH,CH,D. (a)
Add D, in the presence of Pd.
(6) CH,CH==CH,
(4
+ HCl-
4
CH,CHClCH,
CH,CH=CH,
5 CH,CHMgCICH,
87Hg
CH,CHDCH,
or propene + B,D,-
(CH,CH,CH,),B-
(CH,CHDCH,),B-
C H COOD
C H COOH
product
CH,CH,CH,D
Problem 6.60 Ethylene is alkylated with isobutane in the presence of acid (HF) to give chiefly (CH,),CHCH(CH,), , not (CH,),CCH,CH,. Account for the product. 4
H,C=CH2
- HF
H,C==C(CH,),
+ q c (CH 3 l 3
H,CCH,
H'CCH,
+
+ C(CH,),
+
Problem 6.61 Give 4 simple chemical tests to distinguish an alkene from an alkane.
116
[CHAP. 6
ALKENES
A positive simple chemical test is indicated by one or more detectable events, such as a change in color, formation of a precipitate, evolution of a gas, uptake of a gas, evolution of heat.
' '+ \C=C
/
-A--&I I
5
Br,
(red)
(loss of color)
Br Br
(colorless)
(loss of color and formation of precipitate) (colorless)
\ / C-C + H,SO,(conc.) / \ \C=C
/
/
\
+ H,'*
-A-(!.?,I 1
+
+ heat
(solubility in sulfuric acid)
H OS0,H
-k-k-I I
(uptake of a gas)
H H
Alkanes give none of these tests. Problem 6.62 Give the configuration, stereochemical designation and R,S specification for the indicated tetrahydroxy products.
H H H H
I I I I I I
dil. cold
CH,-C-C=C-C-CH, OH (S)
OH
cis, meso
H H
I I
I
(RI
q
I
I
dil. cold
H OH
(U)
?+?
(6)
+?
(c)
?+?
(d)
?
(e)
?+?
cf)
?
KMno4,0H-+
I
+?
I.
H
CH,-C-C=C-C-CH, OH (9
?
KMnO,.OH-
I.
trans
(optically active)
OHH H H
I I l CH,-C-C=C-C--CH, I
H (RI
l
I
OH
cis (optically active)
(a) syn Addition of encircled OH'S:
dil. cold
KMn04. OH-
*
I
CHAP. 61
117
ALKENES
( 6 ) anti Addition of encircled OH’S:
racemic form
( c ) syn Addition; same products as part (b). (d) anti Addition; same products as part ( a ) . ( e ) syn Addition:
CH3,
w
C
, = C
H OH
H
3
~ OH
w C € H OH
?
,
One optically active stereoisomer is formed.
(f) anti Addition:
(RI (RI (RI (RI Two optically active diasteriomers. Problem 6.63 Describe ( a ) radical-induced, and ( b ) anion-induced, polymerization of alkenes. (c) What kind 4 of alkenes undergo anion-induced polymerization?
-
( a ) See Problem 6.39 for formation of a free-radical initiator, RO-, which adds according to the Markovnikov rule. a = c H c H ,
RO. + H,C--LHCH,
H,C--C*
I\
/
RO
H CH ,
H,C--C--CH,C.
I\ I\ R b H CH, H CH,
-
The polymerization can terminate when the free-radical terminal C of a long chain forms a bond with the terminal C of another long chain (combination), RC- + C R ’ RC-CR’. Termination may also occur when the terminal free-radical C’s of two long chains disproportionate, in a sort of auto-redox reaction. One C picks off an H from the C of the other chain, to give an alkane at one chain end and an alkene group at the other chain end:
RCHC. I + *CCHR’---* I RCHCH I + ,C&R’ \
I I
II
I I
I
(6) Typical anions are carbanions, R: -, generated from lithium or Grignard organometallics.
-
RCH,C:- __* RCH,C-CH2C: /\ /\ I\ X CH, H,C X H,C X
These types of polymerizations also have stereochemical consequences.
(c)
[CHAP. 6
ALKENES
118
Since alkenes do not readily undergo anionic additions, the alkene must have a functional group X (such as - - C N or O=COR') on the C=C that can stabilize the negative charge: I
Problem 6.64 Alkenes undergo oxidative cleavage with acidic KMnO, and as a result each C of the C=C ends up in a molecule in its highest oxidation state. Give the products resulting from the oxidative cleavage of ( a ) 4 H,C==CHCH,CH,, ( b ) (E)- or (2)-CH,CH=CHCH,, (c) (CH,),C=CHCH,CH,.
CH,CH=CHCH,
(6) CH3 (c)
-
H,C==CHCH,CH,
(4
I
CH,--C=CHCH,CH,CH,
KMn04
KMn04
+ HOOCCH,CH, CH,COOH + HOOCCH, CO,
(H,C== gives CO,) (RCH= gives RCOOH)
CH.3
KMnO.,
I
CH,-C=O
+ HOOCCH,CH,CH,
(R,C= gives R,C=O)
Chapter 7 Alkyl Halides 7.1 INTRODUCTION Alkyl halides have the general formula RX, where R is an alkyl or substituted alkyl group and X is any halogen atom (F, C1, Br, or I). Problem 7.1 Write structural formulas and IUPAC names for all isomers of: ( a ) C,H,,Br. Classify the isomers as to whether they are tertiary (3"), secondary (2"), or primary (1"). (6) C,H,CI,. Classify the isomers which are gem-dichlorides and uic-dichlorides. 4 Take each isomer of the parent hydrocarbon and replace one of each type of equivalent H by X. The correct IUPAC name is written to avoid duplication. ( a ) The parent hydrocarbons are the isomeric pentanes. From pentane, CH3CH,CH,CH,CH3, we get three monobromo products, shown with their classification. BrCH,CH,CH,CH,CH,
CH,CHBrCH,CH,CH,
I-Bromopentane (1")
CH,CH,CHBrCH,CH,
2-Bromopentane (2")
3-Bromopentane (2")
Classification is based on the structural features: RCH,Br is l", R,CHBr is 2", and R,CBr is 3". From isopentane, (CH,),CHCH,CH,, we get four isomers. CH3
CH3
CH3
I
I
I
BrCH,CHCH,CH,
CH,C BrCH,C H,
C H,C HC H BrC H
I-Bromo-2-methylbutane (1")
2-Bromo-2-methylbutane (3")
2-Bromo-3-methylbutane (2")
CH,
I
,
C H ,C HC H,C H,B r 1 -Bromo-3-methyl-
butane ( 1 ")
CH,Br
I
CH,CHCH,CH, is also 1-bromo-2-methylbutane;the two CH,'s on Cz are equivalent. Neopentane has 12 equivalent H's and has only one monobromo substitution product, (CH,),CCH,Br (l"), l-bromo-2,2-dimethylpropane. (b) For the dichlorobutanes the two Cl's are first placed on one C of the straight chain. These are geminal or gem-dichlorides. CH,CCI,CH,CH,
CI,CHCH,CH,CH,
2 ,2-Dichlorobutane
1 .I-Dichlorobutane
Then the Cl's are placed on different C's. The isomers with the Cl's on adjacent C's are vicinal or uic-dichlorides. CI
Cl
I
I
CICH,CHCH,CH,
ClCH ,C H,CHC H,
1.2-Dichlorobutane ( u i c )
1,3-Dichlorobutane
CICH,C H,CH,C H,CI
C H,C HClC HC IC H,
1 ,CDic hloro-
2.3-Dichlorobutane ( u i c )
butane
From isobutane we get CH,
I
C12CH-CH-CH3 1.1 -Dichloro-2-methyl-
propane (gem )
CH3
CH,
I CICH2-CCI-CH3
CICH2-CH-CH2CI
1,2-Dichloro-2-methylpropane (uic )
1,3-Dichloro-2-methylpropane
119
I
ALKYL HALIDES
120
[CHAP. 7
Problem 7.2 Give the structural formula and IUPAC name for ( a ) isobutyl chloride, (b) r-amyl bromide (amylz pentyl). 4 (a)
CH,,
,
CH3
I I
C H-CH2CI
( b ) CH3-C-CH2CH3
CH3
Br
1 -Chloro-2-methylpropane
2-Bromo-2-methylbutane
Problem 7.3 Compare and account for differences in the ( a ) dipole moment, (6) boiling point, (c) density and 4 (d) solubility in water of an alkyl halide RX and its parent alkane RH. ( a ) RX has a larger dipole moment because the C-X bond is polar. (6) RX has a higher boiling point since it has a larger molecular weight and also is more polar. (c) RX is more dense since it has a heavy X atom; the order of decreasing density is RI > RBr > RCl > RF. (d) RX, like RH, is insoluble in H,O, but RX is somewhat more soluble because some H-bonding can occur: 8+
8-
&+
8-
R-X :-- -H-OH
This effect is greatest for RF.
7.2 SYNTHESIS OF RX 1. Halogenation of alkanes with C1, or Br, (Section 4.4). 2. From alcohols (ROH) with HX or PX, (X= I, Br, Cl); SOC1, 3. Addition of HX to alkenes (Section 6.4). 4. X, ( X = Br, C1) + alkenes give vic-dihalides (Section 6.4). 5 . RX + X’ - +RX’ + X- (halogen exchange).
(major method).
Problem 7.4 Give the products of the following reactions:
(a)
+ HI(c) CH3CH,0H + PI,(P + I,)+ (e) H,C=CH, + Br2+ (g) CH,CH,CH,Br + ICH,CH,CH,I + H,O
(c)
CH,CH,I
( a ) CH,CH,CH,OH
-
+ H,PHO, (phosphorous acid)
(e) H,CBrCH,Br
( f ) CH,CH,CI(CH,),
(d)
+ NaBr + H,SO,(CH,),CHCH,OH + SOCI, +
(f)
CH,CH=C(CH,),
(b) n-C,H,OH
+ HI+ 4
+ NaHSO, + H,O (CH,),CHCH,CI + HCl(g) + SO,(g)
(6) n-C,H,Br (d)
( 8 ) CH,CH,CH,I
+ Br-
Problem 7.5 Which of the following chlorides can be made in good yield by light-catalyzed monochlorination of the corresponding hydrocarbon?
To get good yields, all the reactive H’s of the parent hydrocarbon must be equivalent. This is true for
121
ALKYL HALIDES
CHAP. 7)
Although the H's of (f),propene, are not equivalent, the three allylic H's of CH, are much more reactive than the inert vinylic H's on the double-bonded C's. The precursors for (b) and (d), which are CH,CH,CH,CH, and (CH,),CH, respectively, both have more than one type of equivalent H and would give mixtures. Problem 7.6 Prepare
(c)
(a) CH,CHBrCH,
( b ) CH,CH,CH,CH,I
CH3
CH,
I CH,-C-CH, I
I
( d ) CICH,C=C-CH,
CI
CH,CHCH,
(e)
I
I t
CI CI
CH,
from a hydrocarbon or an alcohol. Two ways:
4
CH3CH=CH, + H B r - m l a C H , C H O H C H ,
Two ways: CH,CH,CH=CH,
+ HBr-
peroxide
I-
CH3CH2CH,CH,Br-
acetone
ICH,CH,CH ,CH,
P'3
(P+Iz)
CH,CH,CH,CH,OH
HI does not undergo an anti-Markovnikov radical addition. Two ways: CH3
I
+ HCI
CH,--C=CH,
I
I
Cl
CH3
I CH,-C=C-CH, I
CH,
I
+ CI,
CH, CH,CH=CH,
CICH,-C=C-CH,
1
+ CI, -+
+ HCI
CH3 CH,CHCH,
I t
CI CI
CHEMICAL PROPERTIES Alkyl halides react mainly by heterolysis of the polar C6:X6-
bond.
NUCLEOPHILIC DISPLACEMENT or
"B
+?R X
Nu:/ + Substrate
Nucleophile
-+
--
R:Nu
or
[R:Nu]' Product
+ :x+
Leaving Group
The weaker the Bronsted basicity of X-, the better leaving group is X - and the more reactive is RX. Since the order of basicities of the halide ions X- is I- < Br- < C1- < F-, the order of reactivity of RX is RI > RBr > RCI > RF. The equilibrium of nucleophilic displacements favors the side with the weaker Bronsted base; the stronger Bronsted base displaces the weaker Bronsted base. The rate of the displacement reaction on the C of a given substrate depends on the nucleophilicity of the attacking base. Basicity and nucleophilicity differ as shown:
ALKYL HALIDES
122
basicity BTH&
w *I
B-H
nucleophilicity
[CHAP. 7
+ :x
I
+ :X -
B-C-
I
Problem 7.7 What generalizations about the relationship of basicity and nucleophilicity can be made from the following relative rates of nucleophilic displacements: ( a ) O H - + H H , O and (c)
(b) H,C:- > :OH- > :F:-
NH,+NH,
:I: - > :Br: - >:Cl: - > :F:-
( d ) CH,O- > O H - - > CH,COO-
4
Bases are better nucleophiles than their conjugate acids. In going from left to right in the Periodic Table, basicity and nucleophilicy are directly related-they both decrease. (c) In going down a Group in the Periodic Table they are inversely related, in that nucleophilicity increases and basicity decreases. (d) When the nucleophilic and basic sites are the same atom (here an 0),nucleophilicity parallels basicity. (a)
(b)
The order in Problem 7 . 7 ( c ) may occur because the valence electrons of a larger atom could be more available for bonding with the C, being further away from the nucleus and less firmly held. Alternatively, the greater ease of distortion of the valence shell (induced polarity) makes easier the approach of the larger atom to the C atom. This property is called polarizability. The larger, more polarizable species (e.g. I, Br, S, and P) exhibit enhanced nucleophilicity; they are called soft bases. The smaller, more weakly polarizable bases (e.g. N, 0, and F) have diminished nucleophilicity; they are called hard bases. Problem 7.8 Explain why the order of reactivity of Problem 7.7(c) is observed in nonpolar, weakly polar 4 aprotic, and polar protic solvents, but is reversed in polar aprotic solvents. In nonpolar and weakly polar aprotic solvents, the salts of : N u - are present as ion-pairs (or ion-clusters) in which the nearby cations diminish the reactivity of the anion. Since, with a given cation, ion-pairing is strongest with the smallest ion, F-, and weakest with the largest ion, I-, the reactivity of X- decreases as the size of the anion decreases. In polar protic solvents, hydrogen-bonding, which also lessens the reactivity of X-, is weakest with the largest ion, again making the largest ion more reactive. Polar aprotic solvents solvate only the cations, leaving free, unencumbered anions. The reactivities of all anions are enhanced, but the effect is more pronounced the smaller the anion. Hence, the order of Problem 7 . 7 ( c ) is reversed. Problem 7.9 Write equations for the reaction of RCH,X with (a)
:c
( b ) :OH-
( c ) :OR‘-
( d ) R’-
Y0
( e ) RC,
U)
H3N:
10and classify the functional group in each product.
(6)
+ RCH,X -:OH + RCH,X
(4
-:OR’+ RCH,X
(4
-R’+ RCH,X-
(0)
(4 (f1 (g )
:I-
__*
_I*
-:CN
+ RCH,X
+ :XF
RCH,OH + :X7
+ :XF RCH,NH,’ + :@ RCH,CN + :XF
__*
__*
__*
+
+ :XT
:*
RCH,OR’
RCH,R’
7 0 0 C R ’ + RCH,X :NH3 + RCH,X
RCH,I
RCH,OOCR’
Iodide Alcohol Ether Alkane (coupling) Ester Ammonium salt Nitrile (or Cyanide)
(g)
CN-
CHAP. 71
123
ALKYL HALIDES
Problem 7.10 Compare the effectiveness of acetate (CH,COO- ), phenoxide (C,H,O- ) and benzenesulfonate (C,H,SO,) anions as leaving groups if the acid strengths of their conjugate acids are given by the pK, 4 values 4.5, 10.0, and 2.6, respectively.
The best leaving group is the weakest base, C,H,SOJ; the poorest is C,H50-, which is the strongest base.
Sulfonates are excellent leaving groups-much better than the halides. O n e of the best leaving groups (10' times better than B r - ) is CF,SOi, called triflate. 1.
s N 1 and s N 2 Mechanisms
The two major mechanisms of nucleophilic displacement are outlined in Table 7-1. Table 7-1
TWO: ( 1 ) R : X ~ R +
Steps
+:x-
(2) R + + : N U H ~ ~ S ' - R N U H + = k[RX]
(1st-order)
RNu + :XR:X + :Nu-R:X + :Nu+ RNu':XR:X' + Nu- __* RNu + X: R:X' + Nu+ RNu' + X:
One: or or or
carbocation
I
= k[RX][:Nu-] s-
TS of slow step
I
Stereochemistry ~
Unimolecular
Bimolecular
Inversion and racemization
Inversion (backside attack)
3"> 2" > 1" > CH, Stability of R' RI > RBr > RCI > RF Rate increase in polar solvents
CH, > 1" > 2"> 3" Steric hindrance in R group RI > RBr > RCl > RF With Nu- there is a large rate increase in polar aprotic solvents
~~
Reactivity Structure of R Determining factor Nature of X Solvent effect on fate Effect of nucleophile
R' reacts with nucleophilic solvents rather than with :Nu(solvolysis), except when R' is relatively stable
Cat a1ysis
Lewis acid, e.g. Ag', AlCI,, ZnCI,
Competition, reaction
Elimination, rearrangement
I
Rate depends on nucleophilicity
I - > Br- > Cl-; RS:- > R0:Equilibrium lies towards weaker Bronsted base (1) Aprotic polar solvent (2) Phase-transfer
Problem 7.11 Give the 3 steps for the mechanism of the SN1 hydrolysis of
(I ) Me,C:Br: (3) Me,COH;
1 s-
:Nu---C---X (with :Nu-)
Molecularity ~
*'*.
(2d-order)
--+
Me,C'
+ :Br:-
+ H,O--tMe,COH
(2) Me,C'+ :OH,--tMe,C:OH;
+ H,O+
Elimination, especially with 3" RX in strong Bronsted base 3O
Rx, Me,CBr.
ALKYL HALIDES
124
[CHAP. 7
Problem 7.12 Give examples of the four charge-types of SN2 reactions, as shown in the first line of Table 7-1. 4 CH,CH,Br + :OH- +CH,CH20H + :Bi:
-
+ :NH,-CH,CH,NH; + :Brr (cH,),S:CH, + :OHCH, OH + (cH,), S:
CH,CH,Br +
*.-
-
a sulfonium cation
(CH,),&CH,
(leaving group)
Dimethyl sulfide
+ :N(CH,),+CH,N(CH,)f
+ (CH3)2S:
Problem 7.13 ( a ) Give an orbital representation for an s N 2 reaction with (S)-RCHDX and :Nu-,if in the transition state the C on which displacement occurs uses sp2 hybrid orbitals. ( 6 ) How does this representation explain (i) inversion, (ii) the order of reactivity 3" > 2" > l"? 4 ( a ) See Fig. 7-1. (6) (i) The reaction is initiated by the nucleophile beginning to overlap with the tail of the sp3 hybrid orbital holding X. In order for the tail to become the head, the configuration must change; inversion occurs. (ii) As H's on the attacked C are replaced by R's, the TS becomes more crowded and has a higher enthalpy. With a 3" RX, there is a higher A H S and a lower rate.
R
sp
l*
R
\
hybridized
S Configuration
sp hybridized withp orbital
TS Fig. 7-1
R
'
sp hybridized
R Configuration (Nu and X of same priority)
Problem 7.14 (a) Give a representation of an S,1 TS which assigns a role to the nucleophilic protic solvent molecules (HS:) needed to solvate the ion. (6) In view of this representation, explain why (i) the reaction is first-order; (ii) R' reacts with solvent rather than with stronger nucleophiles that may be present; (iii) catalysis by Ag' takes place; (iv) the more stable the R', the less inversion and the more racemization occurs. 4
Solvent-assisted S, 1 TS
(6) (i) Although the solvent HS: appears in the TS, solvents do not appear in the rate expression. (ii) HS: is already partially bonded, via solvation, with the incipient R'. (iii) Ag' has a stronger affinity for X- than has a solvent molecule; the dissociation of X - is accelerated. (iv) The HS: molecule solvating an unstable R' is more apt to form a bond, causing inversion. When R' is stable, the TS gives an intermediate that reacts with another HS: molecule to give a symmetrically solvated cation,
which collapses to a racemic product:
. .
CHAP. 71
ALKYL HALIDES
125
The more stable is R', the more selective it is and the more it can react with the nucleophilic anion Nu-. Problem 7.15 Give differences between SN1 and sN2 transition states.
4
1. In the SN1 TS there is considerable positive charge on C; there is much weaker bonding between the attacking group and leaving groups with C. There is little or no charge on C in the SN2TS. 2. The SN1 TS is approached by separation of the leaving group; the SN2TS by attack of :Nu- or :Nu. 3. The A H * of the SN1 TS (and the rate of the reaction) depends on the stability of the incipient R'. When R' is more stable, AH' is lower and the rate is greater. The A H * of the SN2 TS depends on the steric effects. When there are more R's on the attacked C or when the attacking :Nu- is bulkier, A H * is greater and the rate is less. Problem 7.16 How can the stability of an intermediate R' in an SN1 reaction be assessed from its enthalpy-reaction diagram? 4
The intermediate R' is a trough between two transition-state peaks. More stable R"s have deeper troughs and differ less in energy from the reactants and products. Problem 7.17 Give the solvolysis products for the reaction of (CH,),CCl with ( a ) CH,OH, (6) CH,COOH. 4
0
II
(6) (CH,),COC-CH, t-Butyl acetate.
( a ) (CH,),COCH, Methyl t-butyl ether.
Problem 7.18 ( a ) Formulate (CH,),COH + HCl+ (CH,),CCl + H,O as an s N 1 reaction. (6) Formulate the reaction CH,OH + HI+ CHJ + H,O as an s N 2 reaction. 4 Step 1
(CH,),COH base
+ HCI e(cH,),cOH,+ fast
acid,
acid I
CI-
base,
oxonium ion
+
Step 2
(CH,),COH,
Step 3
(CH,),C'
SteD 1
CH,OH
-
slow
+ C1-
(CH,),C'
+ H,O
(CH,),CCl
+
+ H I e C H , O H , + Ifast
Problem 7.19 ROH does not react with NaBr, but adding H,SO, forms RBr. Explain.
4 +
Br-, an extremely weak Bronsted base, cannot displace the strong base OH-. In acid, ROH, is first formed. Now, Br- displaces H,O, which is a very weak base and a good leaving group. Problem 7.20 Optically pure (S)-( +)-CH,CHBr-n-C,H,, has [ a ] g= +36.0". A partially racemized sample having a specific rotation of +30" is reacted with dilute NaOH to form (R)-(-)-CH,CH(OH)-n-C,H,, ([a]: = -5.97'), whose specific rotation is -10.3" when optically pure. ( a ) Write an equation for the reaction using projection formulas. (6) Calculate the percent optical purity of reactant and product. ( c ) Calculate percentages of racemization and inversion. ( d ) Calculate percentages of frontside and backside attack. (e) Draw a conclusion concerning the reactions of 2" alkyl halides. (f) What change in conditions would increase inversion? 4
126
ALKYL HALIDES
[CHAP. 7
(6) The percentage of optically active enantiomer (optical purity) is calculated by dividing the observed specific rotation by that of pure enantiomer and multiplying the quotient by 100%. The optical purities are: - 5.97" Bromide = (100%) = 83% Alcohol = -(100%) = 58% + 30" 36" - 10.3" ( c ) The percentage of inversion is calculated by dividing the percentage of optically active alcohol of opposite configuration by that of reacting bromide. The percentage of racemization is the difference between this percentage and 100%. +
58%
Percentage inversion = -(100%) = 70% 83% Percentage racemization = 100% - 70% = 30%
( d ) Inversion involves only backside attack, while racemization results from equal backside and frontside attack. The percentage of backside reaction is the sum of the inversion and one-half of the racemization; the percentage of frontside attack is the remaining half of the percentage of racemization. Percentage backside reaction = 70%
+ $ (30%) = 85%
Percentage frontside reaction = f (30%) = 15% The large percentage of inversion indicates chiefly SN2 reaction, while the smaller percentage of racemization indicates some S,1 pathway. This duality of reaction mechanism is typical of 2" alkyl halides. (f) The S,2 rate is increased by raising the concentration of the nucleophile-in this case, OH . (e)
Problem 7.21 Account for the following stereochemical results:
I 2% inversion Ho-c-c6H5 HBr 98% racemization I +
H
CH3 CH30-c-c6H5
II
+
27% inversion HBr 73% racemization
4
H fit
H,O is more nucleophilic and polar than CH,OH. It is better able to react to give HS:--- R---:SH (see Problem 7.14), leading to racemization. Problem 7.22 NH, reacts with RCH,X to form an ammonium salt, RCHzNH f X - . Show the transition state, 4 indicating the partial charges.
N gains S + as it begins to form a bond. Problem 7.23 Explain why neopentyl chloride, (CH,),CCH,CI, a 1" RCI, does not participate in typical SN2 react ions. 4
The bulky (CH,),C group sterically hinders backside attack by a nucleophile. Problem 7.24 H,C=CHCH,CI is solvolyzed faster than (CH,),CHCI. Explain.
4
127
ALKYL HALIDES
CHAP. 71
Solvolyses go by an SN1 mechanism. Relative rates of different reactants in S,1 reactions depend on the stabilities of intermediate carbocations. H,C=CHCH,CI is more reactive because
[H,C= CH--CH,]+ is more stable than (CH,),CH'. radical.)
See Problem 6.42 for a corresponding explanation of stability of an ally1
Problem 7.25 In terms of ( a ) the inductive effect and (6) steric factors, account for the decreasing stability of R': Compared to H, R has an electron-releasing inductive effect. Replacing H's on the positive C by CH,'s disperses the positive charge and thereby stabilizes R'. Steric acceleration also contributes to this order of R' stability. Some steric strain of the three Me's in Me,C-Br separated by a 109" angle ( s p , ) is relieved upon going to a 120" separation in R' with a C using sp2 hybrid orbitals. Carbocations have been prepared as long-lived species by the reaction RF + SbF, + R' + SbF,. SbF,, a covalent liquid, is called a superacid because it is a stronger Lewis acid than R'. Problem 7.26 The following are relative rates observed for the formation of alcohols from the listed alkyl halides in 80% water and 20% ethanol at 25°C. Account for the minimum value for (CH,),CHBr. i
Compound Relative rate
CH,Br
CH,CH,Br
(CH,),CHBr
(CH,),CBr
2140
171
4.99
1010
4
The first three halides react mainly by the sN2 pathway. Reactivity decreases as CH,'s replace H's on'the attacked C, because of increased steric hindrance. A change to the S,1 mechanism accounts for the sharp rise in the reactivity of (CH3),CBr. Problem 7.27 How does the S,1 mechanism for the hydrolysis of 2-bromo-3-methylbutane, a 2" alkyl bromide, to give exclusively the 3" alcohol 2-methyl-2-butano1, establish a carbocation intermediate? 4 The initial slow step is dissociation to the 2" 1,2-dimethylpropyl cation. A hydride shift yields the more stable 3" carbocation, which reacts with H,O to form the 3" alcohol. , r B r -
+
+
I CH
l 3
Br 2o
I ,
C H , - C H - C - C H , ~C H , - C H - C - C H , I + fast I I k 6H k 3"
Problem 7.28 RBr reacts with AgNO, to give RNO, and RONO. Explain. The nitrite ion,
CH
[":]-
+H
~ O
4
\b:
has two different nucleophilic sites: the N and either 0. Reaction with the unshared pair on N gives RNO,, while RONO is formed by reaction at 0. [Anions with two nucleophilic sites are called ambident anions.]
128
ALKYL HALIDES
[CHAP. 7
2. Role of the Solvent Polar solvents stabilize and lower the enthalpies of charged reactants and charged transition states. The more diffuse the charge on the species, the less effective the stabilization by the polar solvent . Problem 7.29 In terms of transition-state theory, account for the following solvent effects: (a) The rate of solvolysis of a 3" RX increases as the polarity of the protic nucleophilic solvent (:SH) increases, e.g.
H,O > HCOOH > CH,OH > CH,COOH
(6) The rate of the S,2 reaction :Nu- + RX-+RNu + :X- decreases slightly as the polarity of protic solvent increases. (c) The rate of the S,2 reaction :Nu + RX+ RNu' + :X- increases as the polarity of the solvent increases. (d) The rate of reaction in (6) is greatly increased in a polar aprotic solvent. (e) The rate of the reaction in (6) is less in nonpolar solvents than in aprotic polar solvents. 4 See Table 7-2. Table 7-2 Ground State (GS) (a)
6+
I
6-
Nu - - -R- - -X
1
RX + Nu
(c)
None in GS charge in TS
6-
HS---R---X---HS:
RX + HS: RX +Nu-
(b)
Relative Charge
TS
1
6-
Nu- - -R-- -X I
(d)
Sameas (6)
Full in GS diffuse in TS None in GS charge in TS
-~~
I I
1
Effect of Solvent Change
AHt
I I
Rate
I
A lower H of TS A lower H of GS a less lower H of TS
I I
Decreases Increases
I
Increases Decreases
A lower H of TS
Decreases
Increases
A big rise in H of GS* A rise in H of TS
Decreases
Increases
~~
* Aprotic solvents do not solvate anions.
'. ( e ) In nonpolar solvents, N u - is less reactive because it is ion-paired with its countercation, M S,2 reactions with Nu-'s are typical of reactions between water-soluble salts and organic substrates that are soluble only in nonpolar solvents. These incompatible reactants can be made to mix by adding small amounts of phase-transfer catalysts, such as quarternary ammonium salts, Q'A-. Q' has a water-soluble ionic part and nonpolar R groups that tend to be soluble in the nonpolar solvent. Hence, Q' shuttles between the two immiscible solvents while transporting Nu-, as Q'Nu-, into the nonpolar solvent; there Nu- quickly reacts with the organic substrate. 0' then moves back to the water while it transports the leaving group, X-, as Q'X-. Since the positive charge on the N of Q' is surrounded by the R groups, ion-pairing between Nu- and Q' is loose, and N u - is quite free and very reactive. Problem 7.30 Write equations for and explain the use of tetrabutyl ammonium chloride, Bu,N'Cl-, facilitate the reaction between 1-heptyl chloride and cyanide ion.
Bu,N'Cln-C,H,,CI
+ Na'CN--Bu,N'CN-
+ Bu,N'CN-+
n-C,H,,CN
+ Na'C1+ Bu,N'Cl-
to
4
(in water) (in nonpolar phase)
The phase-transfer catalyst, Bu,N'Cl , reacts with CN- to form a quaternary cyanide salt that is slightly soluble in the organic phase because of the bulky, nonpolar, butyl groups. Reaction with CN- to form the nitrile is rapid because it is not solvated or ion-paired in the organic phase; it is a free, strong nucleophile. The
CHAP. 71
129
ALKYL HALIDES
phase-transfer catalyst, regenerated in the organic phase, returns to the aqueous phase, and the chain process is propagated.
ELIMINATION REACTIONS In a 6-elimination (dehydrohalogenation) reaction a halogen and a hydrogen atom are removed from adjacent carbon atoms to form a double bond between the two C's. The reagent commonly used to remove HX is the strong base KOH in ethanol (cf. Section 6.2). H H
H
le H-C-C-H
alc.KOH
____*
I I H X
I
H
H-C-C-H
ethyl halide
I
+ HX (as KX and H 2 0 )
Ethylene
n-Propyl bromide CH,CH,CH,Br-
-
alc. KOH
sec-Butyl chloride CH,CHCICH,CH,
CH,CH=CH,
alc. KOH
Propene
CH,CH=CHCH,
2-Butene (mainly truns)
1. El and E2 Mechanisms The two major mechanisms for p-eliminations involving the removal of an H and an adjacent functional group are the E2 and El. Their features are compared and summarized in Table 7-3. A third mechanism, Elcb,is occasionally observed. Table 7-4 compares E2 and SN2. Problem 7.31 ( a ) Why do alkyl halides rarely undergo the E l reaction? (6) How can the E l mechanism be promoted? 4
Rx reacts by the E l mechanism only when the base is weak and has a very low concentration; as the base gets stronger and more concentrated, the E2 mechanism begins to prevail. On the other hand, if the base is too weak or too dilute, either the R' reacts with the nucleophilic solvent to give the S,1 product or, in nonpolar solvents, RX fails to react. ( b ) By the use of catalysts, such as Ag', which help pull away the leaving group X-. (a)
2.
H-D Isotope Effect
The C-H bond is broken at a faster rate than is the stronger C-D bond. The ratio of the rate constants, k,lk,, measures this H-D isotope effect. The observation of an isotope effect indicates that C-H bond-breaking occurs in the rate-controlling step. Problem 7.32 Why does CH,CH,I undergo loss of HI with strong base faster than CD,CH,I loses DI?
4
These are both E2 reactions in which the C-H or C-D bonds are broken in the rate-controlling step. Therefore, the H-D isotope effect accounts for the faster rate of reaction of CH,CH,I. Problem 7.33 Explain the fact that whereas 2-bromopentane undergoes dehydrohalogenation with C,H,O-K' to give mainly 2-pentene (the Sayzteff product), with Me,CO-K' it gives mainly 1-pentene (the 4 anti-Sayzteff, Hofmanrr, product).
Since Me,CO- is a bulky base, its attack is more sterically hindered at the 2"H than at the 1"H. With Me,CO -,
, ,
CH ,==CHCH,CH CH major product
~
ICY \
Ii'Br /I
indercrl
+-
-~
2
m o r e hrndcrcd
0
CH,CH--LHCH,CH, ~
minor product
ALKYL HALIDES
130
[CHAP. 7
Table 7-3 El Steps
Two: (1)
H-&-&--L
I I
E2
H-L-&+
I
I
+ L-
R' intermediate
6-
Transition states
Indicates El Kinetics
Indicates E2
First-order Rate = k[RL] Ionization determines rate Unimolecular
Second-order Rate = k[RLJ[:B-] Bimolecular
Stereochemistry
Nonstereospecific
anti Elimination (syn when anti impossible)
Reactivity order factor
3" > 2" > 1" RX Stability of R'
3" > 2" > 1" RX Stability of alkenes (Saytzeff rule)
Re arrangements
Common
None
Deuterium isotope effect
None
0bse rved
Regioselect ivi t y
Saytzeff
SN2 (See Problem 7.33)
Alkyl group
3" > 2" > 1"
3" > 2" > 1"
Loss of H
No effect
Increased acidity
Base Strength Concentration
Weak Low
Strong High
Leaving group
Weak base I - > Br- > C1- > F-
Weak base I - > Br- > C1- > F-
~~~
~
~~
~
~~
Competing reaction Favors E l
Solvent
Favors E2
~~
Catalysis
~
~
~
Phase-transfer Polar protic
Polar aprotic
Problem 7.34 Assuming that anti elimination is favored, illustrate the stereospecificity of the E2 dehydrohalogenation by predicting the products formed from (a) meso- and ( b ) either of the enantiomers of 4 2,3-dibromobutane. Use the wedge-sawhorse and Newman projections.
ALKYL HALIDES
CHAP. 7)
131
Table 7-4
Structure of R Reagent
Favors E2
Favors
3" > 2" > 1"
1" > 2" > 3"
Strong bulky Bronsted base (e.g., Me,CO-)
sN2
Strong nucleophile
Temperature
High
LOW
Low-polarity solvent
Yes
No
I > Br > C1> F
I > Br > C l > F
Favors E2
Favors SN1
3" > 2" > 1"
3" % 2" > 1"
Structure of L
Structure of R Base Strength Concentration
I
Structure of L
&
Strong High
Very weak Low
I
I>Br>Cl%F
I>Br>Cl%F
cis - or (E)-2-Bromo-2-butene
H
mesa
__* anti
--HRr
*FMe Br
111
111
Br
Me
Me
H enantiorner (RR)
tram- or (2)-2-Brorno-2-butene
Problem 7.35 Dehalogenation of vic-dihalides with active metals (Mg or Zn) is also an anti elimination. Predict the products from ( a ) meso- and (6) either enantiomer of 2,3-bromobutane. 4
[CHAP. 7
ALKYL HALIDES
132
HxMe
unri eliminnkion
Mf3 +
+
_____)
Me
tram-2-Butene
meso
< 1 'H
elimination
aH
Me.
unki
Mg + Me&H
H
+
~
Me'
enantiomer (SS)
~
MgBr,
'H
ck-2-Butene
Problem 7.36 Account for the percentages of the products, (i) (CH,),CHOC,H, and (ii) CH,CH=CH,, of the reaction of CH,CHBrCH, with 4 (b) C2H,0H+3% (ii) +97% (i) ( a ) C,H,ONa/C,H,OH-,79% (ii) + 21% (i) (a) C,H,O- is a strong base and E2 predominates. (6) C2H,0H is weakly basic but nucleophilic, and SN1 is favored.
Problem 7.37 Account for the following observations: (a) In a polar solvent such as water the SN1 and E l reactions of a 3" RX have the same rate. (b) (CH,),CI+H,O-,(CH,),COH+HI but (CH,),CI+ 4 OH- --.* ( C H , ) , M H , + H,O + I-. (a) The rate-controlling step for both E l and S,1 reactions is the same:
and therefore the rates are the same. (6) In a nucleophilic solvent in the absence of a strong base, a 3" RX undergoes an S,1 solvolysis. In the presence of a strong base (OH-) a 3" RX undergoes mainly E2 elimination.
7.4
SUMMARY OF ALKYL HALIDE CHEMISTRY PREPA RA TION
PROPERTIES
1. Direct Halogenation
1.
/,+ 2.
Replacement of OH in ROH
RCH2CH20H + NaX +
+ H2S04
/+ soc12------+ RCH2CH2X A
-
Nucleophilic Substitution + OH- +RCHzCHzOH Alcohol SHRCHzCH2SH Mercaptan + CN- +RCH2CH2CN Nitrile + OR- --+ RCHzCHzOR Ether + SR- -+ RCH2CH2SR +Thioether + NH3 --+ RCH2CH2NH3X- Amine salt + NRJ--+ RCH,CH,NR;'X- Quarternary ammonium salt ( Q 'A- )
R'COO-
+ px,
+ LiAIH4 3. Addition of HX to Alkene
RCH=CH2
+ HX
RCH=CH2
+ HBr
-
__*
-
RCH2CH200CR' Ester
RCH2CH3
H+
__*
peroxide
RCH2CH2Br
4.
Formation of Organometallics
+ Mg-
RCH,CH,MgX Grignard reagent Ornanolithium reagent
+ Li --+RCH,CH,Li
133
ALKYL HALIDES
CHAP. 71
Supplementary Problems Problem 7.38
Write the structure of the only tertiary halide having the formula C,H,,Br.
rl
In order for the halide to be tertiary, Br must be attached to a C that is attached to three other C’s; i.e., to no H’s. This gives the skeletal arrangement C
I I
C-C-C Br involving four C’s. A fifth C must be added, which must be attached to one of the C’s on the central C, giving: C
I I
C-C-C-C
Br and with the H’s. CH3
I I
CH,-C-CH,-CH Br
2-Bromo-2-methylbutane
Problem 7.39 On substitution of one H by a C1 in the isomers of C5HI2,( a ) which isomer gives only a primary 4 halide? (b) Which isomers give secondary halides? (c) Which isomer gives a tertiary halide? ( a ) 2,2-Dimethylpropane. (b) CH,CH,CH,CH,CH, gives 2-chloropentane and 3-chloropentane. 2-Methylbutane gives 2-chloro-3-methylbutane. (c) 2-Methylbutane gives 2-chloro-2-methylbutane.
Problem 7.40 Complete the following table: Substance
1. 2.
CH,CH,
Treated with
Yields
(4
CH,CH,Cl CH, HCH,
SOCI,
(6)
’iCI
CH,CHCH,Br
I
Br 4.
HBr, peroxides
CH,CH==CH,
( a ) Cl,, UV
(b) CH,CHCH,
(c)
CH,CH==CH,
(4
4
(d) CH,CH,CH,Br
AH
--
Problem 7.41 Give the organic product in the following substitution reactions. The solvent is given above the arrow. (a) HS- + CH,CH,CHBrCH,
+ (CH,),CBr HCOOH CH,CH,Br + AgCN CH,CHBrCH, + CH,NH, CH,CHBrCH, + (CH3)2S:
(6) I(c)
(d) (e)
CH OH
2 products
-
ALKYL HALIDES
[ j: 1'-
CH,CH,Br
+ :P(C,H,),
CH,CH,Br
+
(thiosuffate ion)
S-S-0
CH ,CH ,C H SHCH
,
[CHAP. 7
-
(a mercaptan).
0
II
(CH,),COCH; 3" RX undergoes S, 1 solvolysis. CH,CH,CN
i
+ CH,CH,NC;
[I+
(CH,),CHNCH,
: m N :- is an ambident anion (Problem 7.28).
Br-; an ammonium salt.
[(CH,)2CHS(CH,)2]+ Br-; a sulfonium salt. Br-; a phosphonium salt. [CH3CH2P(C,H5),]+ 0
I I
CH,CH,-S-S-0-;
S is a more nucleophilic site than is 0.
0 Problem 7.42 Account for the observation that catalytic amounts of KI enhance the rate of reaction of 4 RCH,CI with OH- to give the alcohol RCH,OH. Since I - is a better nucleophile than OH-, it reacts rapidly with RCH,CI to give RCH,I. But since I - is a much better leaving group than Cl-, RCHzI reacts faster with OH- than does RCH,CI. Only a catalytic amount of I - is needed, because the regenerated I - is recycled in the reaction. RCHzCl + OH-i
slower
RCH,OH
.p
+ C1'
Problem 7.43 Account for the following products from reaction of CH,CHOHC(CH,), with HBr: (a) CH,CHBrC( CH,), , (6) H,CF---CHC(CH, ), , ( c ) (CH,),CHCHBr(CH,), , ( d ) (CH,),CH--LH( CH,), , ( e ) (CH,),CHC=CH, 4
b3
For this 2" alcohol, oxonium-ion formation is followed by loss of H,O to give a 2" carbocation that rearranges to a 3" carbocation. Both carbocations react by two pathways: they form bonds to Br- to give alkyl bromides or they lose H' to yield alkenes. CH,CHOHC(CH,),
HBr
-HO CH,CHC(CH,), --LCH,CHC(CH,), I 2" carbocation OH-. :CH, +
J-
(CH, 1, C H b h 1 2
CH,CHBrC(CH,), -H
(4
1 H2C=CHC(CH,)2 (6)
CHAP, 71
ALKYL HALIDES
-
135
Problem 7.44 Show steps for the following conversions: (a)
BrCH,CH,CH,CH,
(6) CH,CHBrCH,CH,
CH,CHBrCH,CH,
+CH,CICHClCH,CI
(c) CH,CH,CH,
-
BrCH,CH,CH,CH, 4
To do syntheses, it is best to work backward while keeping in mind your starting material. As you do this, keepasking what is needed to make what you want. Always try to use the fewest steps. ( a ) The precursors of the possible product are the corresponding alcohol, 1-butene, and 2-butene. The alcohol is a poor choice, because it would have to be made from either of the alkenes and an extra step would be needed. 2-Butene cannot be made directly from the 1" halide, but 1-butene can.
alc. K O H
BrCH,CH,CH,CH,
CH,=CHCH,CH,
CH,CHBrCH,CH,
-
(6) To ensure getting 1-butene, the needed precursor, use a bulky base for the dehydrohalogenation of the starting material. CH,CHBrCH,CH,
Me
CO-K+
HBr
CH,==CHCH,CH,
peroxides
BrCH,CH,CH,CH,
(c) The precursor for vic dichlorides is the corresponding alkene; in this case, H,C=CHCH,CI, which is made by allylic chlorination of propene. Although free-radical chlorination of propane gives a mixture of isomeric propyl chorides, the mixture can be dehydrohalogenated to the same alkene, making this particular initial chlorination a useful reaction.
C H , C H , C H , G CICH,CH,CH,
+
CH,CHCICH,
alc.
CH,=CHCH,
2CH,=CHCH,CI-% uv
ClCH,CHClCH,CI
Problem 7.45 Indicate the products of the following reactions and point out the mechanism as SN1, SN2,El
or E2.
( a ) CH,CH,CH,Br + LiAIH, (source of :H-) (6) (CH,),CBr + C,H,OH, heat at 60°C ( c ) CH,CH==CHCl+ NaNH,
(d) BrCH,CH,Br + Mg (ether) (e) BrCH,CH,CH,Br + Mg (ether) ( f ) CH,CHBrCH, + NaOCH, in CH,OH
( a ) CH,CH,CH,; an SN2reaction, :H- of AlHa replaces Br-.
(6)
(4 (4
+ (CH3),C+]*
CH CH OH
(CHJ,COCH,CH,
3"RX
major; S,1
CH,CH=CHCl + NaNH,CH,C%CH Vinyl halides are quite inert toward sN2 reactions. BrCH,CH,Br
+ Mg-
very minor; El
+ NH, + NaCl
H,C=CH,
(E2)
+ MgBr,
This is an E2 type of p-elimination via an alkyl magnesium iodide. Mg + BrCH,CH,BrBrMg-:CH,-CH,-Br+ + Q P MgBr, (e)
CH3
+ CH,C=CH, I
+ H,C=CH,
This reaction resembles that in ( d ) and is an internal S,2 reaction.
( f ) This 2" RBr undergoes both E2 and S,2 reactions to form propylene and isopropyl methyl ether.
CH,CHBrCH,
+ &aOCH,(CH,OH)-
CH,CH=CH,
+ CH,CHCH, I
OCH,
4
136
ALKYL HALIDES
[CHAP. 7
Problem 7.46 Give structures of the organic products of the following reactions and account for their formation: C H OH
CICH,CH2CH,CH,Bra,S;t~~CN(6) CH,CHBrCH, + NaI- a c e t o n e (c) CICH,CH=CH, + NaI(a)
4
( a ) CICH,CH,CH,CH,CN.
Br- is a better leaving group than Cl-. (6) CH,CHICH,. Equilibrium is shifted to the right because NaI is soluble in acetone, while NaBr is not and precipitates. (c) ICH,CH==CH,. NaI is soluble in acetone and NaCl is insoluble.
Problem 7.47 As a rule 1" RX is the least reactive in S,1 solvolysis reactions. Account for the high reactivity 4 observed in S, 1 displacements with CICH,OCH,CH, in ethanol.
The rapid S,l reaction is attributed to the stability of a C' adjacent to -&-. The empty p orbital on the C' can overlap with the p orbital on 0, thereby delocalizing the + charge. The positive ion then reacts with the nucleophilic alcohol, giving an ether. CH3CH2-O-CH2CI
+Cl-
+
CH3CH2-0-cH,'
$
+C2H50H
+
CH3CH,-O=CH2
-H+ CH3CH2-O-CH20C2H,
Problem 7.48 Account for the following observations when (S)-CH,CH,CH,CHID is heated in acetone solution with Naf: (a) The enantiomer is racemized. (6) If radioactive *I- is present in excess, the rate of racemization is twice the rate at which the radioactive * I - is incorporated into the compound. 4 (a) Since enantiomers have identical energy, reaction proceeds in both directions until a racemic equilibrium mixture is formed.
D
*I-
+ CH3CH2CH,+I
inversion inversion
*
D I+CH,CH,CH,
H
H
(R)
6)
+ I-
( s N 2 reaction)
(6) Each radioactive *I- incorporated into the compound forms one molecule of enantiomer. Now one unreacted molecule and one molecule of its enantiomer, resulting from reaction with *I-, form a racemic modification. Since two molecules are racemized when one *I- reacts, the rate of racemization will be twice that at which *I- reacts. Problem 7.49 Indicate the effect on the rate of S,1 and s N 2 reactions of the following: ( a ) Doubling the concentration of substrate (RL) or Nu-. (6) Using a mixture of ethanol and H,O or only acetone as solvent. (c) Increasing the number of R groups on the C bonded to the leaving group, L. (d) Using a strong Nu-. 4
Doubling either [RL] or [Nu-] doubles the rate of the s N 2 reaction. For S,1 reactions the rate is doubled only by doubling [RL] and is not affected by any change in [Nu-]. (6) A mixture of ethanol and H,O has a high dielectric constant and therefore enhances the rate of S,1 reactions. This usually has little effect on S,2 reactions. Acetone has a low dielectric constant and is aprotic and favors S,2 reactions. (c) Increasing the number of R's on the reaction site enhances s N 1 reactivity through electron release and stabilization of R'. The effect is opposite in S,2 reactions because bulky R's sterically hinder formation of, and raise AHt for, the transition state. (d) Strong nucleophiles favor S,2 reactions and do not affect S,1 reactions. (a)
Problem 7.50 List the following alkyl bromides in order of decreasing reactivity in the indicated reactions,
ALKYL HALIDES
CHAP. 71
137
CH3
I
CH,CH2CH2CH2CH2Br
CH3C-CH,CH,
8,
CH,CH2CHCH2CH,
8,
s N 1 reactivity, ( b ) s N 2reactivity, (c) reactivity with alcoholic AgNO,.
4
Reactivity for the S, 1 mechanism is 3"(I) > 2"(III) > l"(11). The reverse reactivity order for SN2 reactions gives 1"(11) > 2"(111) > 3"(I). Ag' catalyzes S, 1 reactions and the reactivities are 3"(I) > 2"( 111) > 1"( 11). Problem 7.51 Potassium rert-butoxide, K'OCMe,, is used as a base in E2 reactions. (a) How does it compare in effectiveness with ethylamine, CH,CH,NH,? (6) Compare its effectiveness in the solvents tert-butyl alcohol and dimethylsulfoxide (DMSO) (c) Give the major alkene product when it reacts with (CH,),CClCH,CH,. 4
( a ) K'OCMe, is more effective because it is more basic. Its larger size also precludes SN2 reactions. (6) Its reactivity is greater in aprotic DMSO because its basic anion is not solvated. Me,COH reduces the effectiveness of Me,CO- by H-bonding. (c) Me,CO- is a bulky base and gives the anti-Saytzeff (Hofmann) product CH2=C(CH,)CH2CH,.
Problem 7.52 Indicate and explain why the following may or may not be used as synthetic reactions.
(4
H 2 0 + CH,CHCICH,
CH,CH=CH,
4
(a) A 3"RX does not undergo S,2 displacement. Instead, strong bases induce E2 elimination to give CH,==C(CH,),. (6) H,O is too weakly basic to effect E2 elimination. The E l reaction does not occur because AH' for the transition state leading to an incipient R,CH+ is very large and the rate is extremely low. Problem 7.53 Give structures of all alkenes formed and underline the major product expected from E2 elimination of: (a) 1-chloropentane, (6) 2-chloropentane. 4 (4
CH,CH,CH,CH,CH,Cl+
CH,CH,CH,CH==CH,
A 1" alkyl halide, therefore one alkene.
(6)
CH,CH2CH2CH-CH3
I c1
CH3CH2CH,CH=CH2
+ CH3CH,CH=CHCH3
A 2" alkyl halide flanked by two R's; therefore two alkenes are formed. The more substituted alkene is the major product because of its greater stability. Problem 7.54 How is conformational analysis used to explain the 6 : 1 ratio of trans- to cis-2-butene formed on dehydrochlorination of 2-chlorobutane? 4 For either enantiomer there are two conformers in which the H and Cl eliminated are anti to each other.
ALKYL HALIDES
138
[CHAP. 7
-H$Hl' $2:; H
H
trans-2-Butene
4
H
H
CHI
a
L B:---H
H
H...
--I
cis-2-Butene
Conformer I has a less crowded, lower-enthalpy transition state than conformer 11. Its A H S is less and reaction rate greater; this accounts for the greater amount of trans isomer obtained from conformer I and the smaller amount of cis isomer from conformer 11. Problem 7.55 State whether each of the following R"s is stabilized or destabilized by the attached atom or group: +
( 4 F3C-C-
(b)
I
:F3C+
H2N<--
(c)
I
( d ) H3N<-
4
I
If an electron-withdrawing group is adjacent to the positive C, it will tend to destabilize the carbocation. Electron-donating groups, on the other hand, delocalize the + charge and serve to stabilize the carbocation. Destabilized. The-strongly electron-withdrawing F's place a 6 + on the atom adjacent to C'. F
t,+ t I F
F+C+C-
(Arrows indicate withdrawn electron density.) Stabilized. Each F has an unshared pair of electrons in a p orbital which can be shifted to orbital overlap.
3-
via p-p
Stabilized. The unshared pair of electrons o n N can be contributed to C'.
Destabilized. The adjacent N has a
+
charge.
Problem 7.56 Draw structural formulas for the most important organic products of the following reactions and point out how they are formed. (a)
2,3-Dimethyl-2-bromobutane + NaNH, in decane (heat)
(b) CH,I
+ NaNOz
(c) CHCl,
CH3 H
I
CH,-C-C-CH, !lr
1
LH,
+ (CH,),CO-K' + H,C=CH,
4
CH,
+ NH;
--+
I
CH,-C=C-CH,
I
CH3
+ NH, + Br-
(E2)
139
ALKYL HALIDES
CHAP. 71
( b ) CH,NO, + some CH,ONO; SN2 reaction. (NO, is an ambident ion.) ( c ) Strong base abstracts H’. The carbanion :CCl, loses C1- to form the reactive dichlorocarbene which adds to c=c.
an a-elimina tion
Problem 7.57 Account for the inertness of the following rigid bicyclic compound to nucleophilic substitution reactions:
Cl
4
Although this is a 3” RCl, it does not undergo an S,1 reaction because the bridgehead C bonded to C1 is part of a rigid structure and therefore cannot form a planar R’. The bicyclic ring structure does not permit a backside nucleophilic attack on C, ruling out the S,2 mechanism.
Problem 7.58 Account €or the formation of CN CH,CH=CH-CH,CN
and
I
CH,-CH-CH=CH,
from the reaction with CN- of l-chloro-2-butene, CH,CH=CH-CH,CI. Formation of 1-cyano-2-butene results from SN2 reaction at the terminal C.
N-
CH3-CH=CHAttack by CN- can also occur at C” with the Cl- in an allylic rearrangement: N ~ C ~ ~ H , - k H = Cn H - C H , - C IP
__*
CH,-CH=CH-CH,-CN
+ Cl-
electrons of the double bond acting as nucleophile to displace
-
CH,-CH-CH=CH,
I CN
+ C1-
(an 5,2’
reaction)
Problem 7.59 Calculate the rate for the S,2 reaction of 0.1-M C,H,I with 0.1-M C N - if the reaction rate for 0.01 M concentrations is 5.44 x 10-9 mol/L * s. 4 The rates are proportional to the products of the concentrations, Rate
5.44 x 1 0 - ~ mol/L - s
- [O. 1][0.1] -
[O.OI][O.Ol]
Rate = 100 x 5.44 x 10-9 mol/L * s = 5.44 x lO-’ mol/L * s
Problem 7.60 Give reactions for tests that can be carried out rapidly in a test tube to differentiate the following compounds: hexane, CH,CH=CHCI, H,&CHCH,CI and CH,CH,CH,CI. 4 Hexane is readily distinguished from the other three compounds because there is a negative test for Clafter Na fusion and treatment with acidic AgNO,. The remaining three compounds are differentiated by their
ALKYL HALIDES
140
[CHAP. 7
reactivity with alcoholic AgNO, solution. CH,CH=CHCl is a vinylic chloride and does not react even on heating. H,C=CHCH,Cl is most reactive (allylic) and precipitates AgCl in the cold, while CH,CH,CH,Cl gives a precipitate of AgCl on warming with the reagent.
-
Problem 7.61 Will the following reactions be primarily displacement or elimination? ( a ) CH,CH2CH,Cl
(c)
(a) (6) (c) (d) (e)
CH,CHBrCH,
+ I+ OH-
(6) (CH,),CBr + CN- (ethanol)+ (H,O)--+ ( d ) CH,CHBrCH, + OH- (ethanol)+ (e) (CH,),CBr + H20-
4
SN2 displacement. I- is a good nucleophile, and a poor base. E2 elimination. A 3" halide and a fairly strong base. Mainly SN2 displacement. Mainly E2 elimination. A less polar solvent than that in (c) favors E2. S,1 displacement. H,O is not basic enough to remove a proton to give elimination.
Problem 7.62 Depending on the solvent, ROH reacts with SOCI, to give RCl by two pathways, each of which involves formation of a chlorosulfite ester. 0
II
ROSCl along with HC1. Use the following stereochemical results to suggest mechanisms for the two pathways: In pyridine, a 3" amine base, (R)-CH,CH(OH)CH,CH,
+(S)-CH,CHClCH,CH,
and, in ether, the same (R)-ROH-(R)-RCl.
4
Pyridine (Py) reacts with the initially formed HCl to give PyH'C1- and the free nucleophilic C1- attacks the chiral C with inversion, displacing the OSOCl- as SO, and Cl-. With no change in priority, inversion gives the (S)-RCl. Ether is too weakly basic to cause enough dissociation of HCl. In the absence of Cl-, the C1 of the - 0 S O C l attacks the chiral C from the side to which the group is attached. This internal nucleophilic substitution (S,i) reaction proceeds through an ion-pair and leads to retention of configuration.
Alkyl chlorosulfite
ion pair
(R) Problem 7.63 Discuss the validity of the following statements. Since every S,2 reaction goes with inversion, attack at a chiral C (a) with D-( +) configuration gives a chiral product with L-(-) configuration, and ( b ) with R configuration gives a chiral product with S configuration. 4 (a)
The statement is false. It is true that configurational inversion D ~ occurs, L but there is no relationship between configuration and rotation. The L enantiomer could be (+) or (-).
CHAP. 7)
ALKYL HALIDES
141
( 6 ) False. Assignment of R,S configuration depends on the assignment of priorities. It is possible for inversion to occur but the configuration to remain the same because the displacing group and the leaving group have different priorities:
Chapter 8 and Dienes 8.1
ALKYNES
NOMENCLATURE AND STRUCTURE Alkynes or acetylenes (CnH2,,-*)have a a-and are isomeric with alkadienes, which have two double bonds. In IUPAC, a --C%C-is indicated by the suffix -yne. Acetylene, C,H,, is a linear molecule in which each C uses two sp HO's to form two (T bonds with a 180" angle. The unhybridized p orbitals form two IT bonds. Problem 8.1 Name the structures below by the IUPAC system: ( a ) CH,C=CCH,
(d) HC=C-CH2CH=CH,
( 6 ) CH,C=CCH,CH,
(e)
HC=C-CH,CH,CI
(f)
CH,CH=CH-C=C-C=CH
CH3
( c ) CH,-C-C=C-&-CH, I
I
H
CH3 4
LH,
(a) 2-Butyne ( 6 ) 2-Pentyne (c) 2,2,5-Trimethyl-3-hexyne
1-Penten-4-yne C = C has priority over C%C and gets the smaller number. (e) 4-Chloro-1-butyne ( f ) 5-Hepten- 1,3-diyne (d)
Problem 8.2 Supply structural formulas and IUPAC names for all alkynes with the molecular formula (a) CSH,, ( b ) C,H,,,* 4
Insert a triple bond where possible in n-pentane, isopentane and neopentane. Placing a triple bond in an n-pentane chain gives H - €= C - - C H , C H , CH , (1-pentyne) and CH,--C-=C--CH,CH, (2-pentyne). Isopentane gives one compound, CH3
I
H-C=C-CHCH,
3-Methyl- I-butyne
because a triple bond cannot be placed on a 3°C. No alkyne is obtainable from neopentane, (CH, ),C( CH, ) z Inserting a triple bond in n-hexane gives: *
H-C=C-CH,CH,CH,CH,
CH,--C=C-CH,CH,CH,
I-Hexyne
CH,CH,--C=C-CH,CH,
2-Hexyne
3-Hexyne
Isohexane yields two alkynes, and 3-methylpentane and 2,2-dirnethylbutane one alkyne each. CH3
I
C H ,C HC H,C=CH
CH,
CH,
I
I
HC-C-CHCH,CH,
CH,CH-C=C-CH,
CH3
I
CH,--C-C=C-H
I
CH3 4-Methyl- 1 pentyne
3-Methyl-lpentyne
4-Methyl-2pentyne
3,3-Dimethyl-lbutyne
Problem 8.3 Draw models of ( a ) sp hybridized C and ( b ) C2H2to show bonds formed by orbital overlap.
4
142
CHAP. 81
143
ALKYNES AND DIENES
(a) See Fig. 8-l(a). Only one of three p orbitals of C is hybridized. The two unhybridized p orbitals ( p , and p,,) are at right angles to each other and also to the axis of the sp hybrid orbitals. (6) See Fig. 8-1(6). Sidewise overlap of the p,, and p, orbitals on each C forms the 7rv and 7r, bonds, respectively.
H
Problem 8.4 Why is the C=C
distance (0,120nm) shorter than the C=C (0.133 nm) and C--C (0.154 nm).
4
The carbon nuclei in c-=C are shielded by six electrons (from three bonds) rather than by four or two electrons as in C = C or C - C , respectively. With more shielding electrons present, the C’s of --C=Ccan get closer, thereby affording more orbital overlap and stronger bonds. Problem 8.5 Explain how the orbital picture of a accounts for ( a ) the absence of geometric isomers in CH,C-=CC,H,; (6) the acidity of an acetylenic H, e.g.
HCSCH + NH,
__+
+ NH,
HC%C
4
(pK, = 25)
(a) The sp hybridized bonds are linear, ruling out cis-tram isomers in which substituents must be on different
sides of the multiple bond. (6) We apply the principle: “The more s character in the orbital used by the C of the C-H acidic is the H.” Therefore the order of acidity of hydrocarbons is
bond, the more
=C-H > =(!‘> -€ -A-H I I SP
SP
SP
Problem 8.6 (a) Relate the observed C-H and C-4 bond lengths and bond energies given in Table 8-1 in terms of the hybrid orbitals used by the C’s involved. (6) Predict the relative C--C bond lengths in CH,CH,, CH,=CH--CH=CH,, and H a a - H .
Table 8-1
Compound (1) (2) (3) (4) (5) (6)
CH,--CH, CH,=CH, H a - H CH,--CH, CH,--CH=CH, C H , a - H
Bond -C-H =C-H =C-H C-CC& C-C=
Bond Length, nm
Bond Energy, kJ/mol
0.110 0.108 0.106 0.154 0.151 0.146
410 423 460 356 377 423
4
[CHAP. 8
ALKYNES AND DIENES
144
Bond energy increases as bond length decreases; the shorter bond length makes for greater orbital overlap and a stronger bond. (a) The hybrid nature of C is: (1) Cs,x-Hs, (2) C \ p ~ - H v(3) , C,,-H,, (4) C,px-Csp~,(5) C s p ~ - C Sand p ~ (6) C , p ~ - C cInp .going from ( I ) to (3) the C-H bond length decreases as the s character of the hybrid orbital used by C increases. The same situation prevails for the C--C bond in going from (4) to (6). Bonds to C therefore become shorter as the s character of the hybridized orbital used by C increases. Crp~-Clpx; H,C=CH-CH=CH,, ( b ) The hybrid character of the C's in the C - C bond is: for CH,
C , p ~ - C , pand ~ ; H--C+C--CkC--H, Cs,-Cr,. Bond length becomes shorter as s character increases and hence relative C - C bond lengths should decrease in the order CH,CH, > CH,=CH--CH=CH,
> H(3-=C4%CH
The observed bond lengths are, respectively, 0.154 nm, 0.149 nm and 0.138 nm.
LABORATORY METHODS OF PREPARATION 1.
Dehydrohalogenation of vic-Dihalides or gem-Dihalides
I I -C-CI I
H
H H
H X
or -
H X
x x
a gem-dihalide
a vicdihalide
liq. NH,(b.p. -33 " C ) The vinyl (alkenyl) halide requires the stronger base sodamide (NaNH,).
2.
-
Primary Alkyl Substitution in Acetylene; Acidity of s - H
R-C-C-H
+
INaoYH2
I
Na
[see Problem 8.5(b)]
R-C=CiNa'
+
an acetylide anion
4
Problem 8.7 Why is this last reaction not used with 2" or 3" RX's?
Acetylides are strongly basic (conjugate bases of very weak acids) and form alkenes with 2" and 3" RX's by E2 elimination. Na'Br-
+ CH,C-CH
+ CH,CH=CH, Propylene
4
(CH3)2CHBr
CH,C-CiNa+
(CH3)jCBr
P
CH,
I
CH,-C-CH,
+ CH,C=CH + Na'Br-
Isobutylene
Problem 8.8 Explain why CH,CHBrCH,Br does nor react with KOH to give CH,==CHCH,Br.
4
In E2 eliminations the more acidic H is removed preferably. The inductive effect of the Br's increases the acidities of the H's on the C's to which the Br's are bonded. To get this product the less acidic H (one of the CH, group) must be removed. Problem 8.9 Outline a synthesis of propyne from isopropyl or propyl bromide.
4
ALKYNES AND DIENES
CHAP. 81
I
145
The needed vic-dihalide is formed from propene, which is prepared from either of the alkyl halides. C H,C H,C H,Br n-Propyl bromide
or CH,CHBrCH,
alc . KOH P
Br
CH,-CH=CH,
NaNH
2CH,CHBrCH,Br
+
CH,CH=CHBr 2 , CH3C-cH Propenyl bromide
Propylene bromide
Propene
Isopropyl bromide
Propyne
NaNH,
Problem 8.10 Synthesize the following compounds from HC-=CH and any other organic and inorganic 4 reagents (do not repeat steps): (a) 1-pentyne, (6) 2-hexyne. NaNHZ
H a - H -
(a)
CH31
(b) Na :kC--H-
- +
H-C=C:Na-
CH3CH2CH21
- +
NaNH2
H4%C-CH,CH2CH,
CHJCHZCH21
CH,~-H-CH,-C%C:Na-CH,~-CH,CH,CH,
+ 2H2O-
Problem 8.11 Industrially, acetylene is made from calcium carbide, CaC, Formulate the reaction as a Bronsted acid-base reaction.
The carbide anion C:- is the base formed when H-H [ :C=C:I2-+ 2HOH+ base I
+ Ca(OH),.
4
loses two H”s.
H m - H
+20H-
acid
base
acid,
HC--CH
(Ca” precipitates as Ca(OH),)
8.2 CHEMICAL PROPERTIES OF ACETYLENES
ADDITION REACTIONS AT THE TRIPLE BOND
Nucleophilic TT electrons of alkynes add H2 and electrophiles in reactions similar to additions to alkenes. Alkynes can add one or two moles of reagent but are less reactive (except to H,) than alkenes. 1. Hydrogen
+ 2H2 --% CH3CH2CH2CH2CHJ
CH3-C=C-CH2CH3
(a) \
* H /Pd(Pb)
/
H /c=c\H
[CH~--CEC--C~HSI
addition
Na. liq. NHJ addition
cis-2-Pentene
2-Pentyne
tram-2-Pen tene
stereospec$c reductions
2.
-
HX (HC1, HBr, HI)-an CH,--C-=C-H
HBr
CH,-C%C-H 3.
Halogen (Br,, CI,)-an
-
anti addition for first mole
CH,-CBr=CH,
+ HBr
per oxide
HBr
CH,-CBr,-CH,
CH,-CH=CHBr
(anti-Markovnikov)
anti addition for first mole
X
I X R-C=C-H~R-C=C-H~-R-C-C-H I X
X
(Markovnikov addition)
a gem-dihalide
x x I I I t
x x
ALKYNES AND DIENES
146
[CHAP. 8
4. H 2 0 (Hydration to Carbonyl Compounds)
O H
+ H2O
CH3-C-C-H
II I I
CH3-C-CH
HIS04
(Markovnikov addition)
H
Propy ne
5.
Acetone
a vinyl alcohol (enol) (unstable)
Boron Hydride
a dialkylborane
a uinylborane
>='<
With dialkylacetylenes, the products of hydrolysis and oxidation are cis-alkenes and ketones, respectively.
-(
CHJCOOH 0°C
CH3
CH3
0 H202
NaOH
2-Butanone
a vinylborane
cis-2-Butene
II
CH3-CH2-C-CH3 a ketone
6. Dimerization CU( N H ~ ) ; C I -
2 Ha--H----------*
H2O
7.
H,C----CH--C'-=C-H Vinylacetylene
Nucleophiles
CH,C=CCH,
+ CN-,
HCN-
CH,CH=C(CN)CH,
Problem 8.12 In terms of the mechanism, explain why alkynes are less reactive than alkenes towards 4 electrophilic addition of, e.g., HX or Br,. The mechanism of electrophilic addition is similar for alkenes and alkynes. When HX adds to a triple bond the intermediate is a carbocation having a positive charge on an sp-hybridized C atom, +
-G=C-H
I
sp
This vinyl-type carbocation is less stable than its analog formed from an alkene, which has the positive charge on an sp'-hybridized C atom,
+ I
-C-C-H
I I
*-
An addendum such as Br, forms an intermediate bromonium-type ion '1
+:
Br In this ion some positive charge is dispersed to the C's, which, because of their sp-like hybrid character, are less able to bear the positive charge than the sp2 C's in the alkene's bromonium ion. Such situations cause alkynes to be less reactive than alkenes toward Br,.
147
ALKYNES AND DIENES
CHAP. 81
Problem 8.13 Alkynes differ from alkenes in adding nucleophiles such as CN-. Explain.
4
The intermediate carbanion from addition of CN- to an alkyne has the unshared electron pair on an sp2-hybridizedC. It is more stable and is formed more readily than the sp'-hybridized carbanion formed from a nucleophile and an alkene. q H-C-C-H
-CN-
-
A
HCN\ H,C=C-CN
H-C=C:CN spi
I
I
H
H
Acrylonitrile
Problem 8.14 Outline a synthesis of trans-2,3-dibromo-2-hexene from C,H, and point out the conditions for optimum yield of product. 4
tram-Dibromoalkenes can be prepared by the anti addition of 1 mol of Br, to 1 mol of an alkyne. The resulting vinyl Br's decrease the nucleophilic reactivity of the double bond. Addition of a second mole of Br, is slower. To further prevent tetrabromo formation we slowly add Br, to an excess of alkyne. The preparation of 2-hexyne from acetylene is given in Problem 8.10(6).
Br2
CH,--CH,CH,CH,
CH,,
/Br
Br /'<\ CH,CH,CH,
Problem 8.15 Dehydrohalogenation of 3-bromohexane gives a mixture of cis-2-hexene and tram-2-hexene. 4 How can this mixture be converted to pure (a) cis-2-hexene? (6) trans-2-hexene?
Relatively pure alkene geometric isomers are prepared by stereoselective reduction of alkynes.
-
( a ) Hydrogenation of 2-hexyne with Lindlar's catalyz! gives 98% cis-2-hexene.
CH,CH=CHCH,CH,CH,
NaNH2
2 CH,CH-CHCH,CH,CH, Br
cis- and 1ram-2-Hexene
Br
CH,C=CCH,CH,CH,
H /Pt(Pb)
CH3\
,c=c\
2
CH,CH,CH, /
H 2-Hexyne
H ( Z ) - or cis-2-Hexene
( b ) Reduction of 2-hexyne with Na in liquid NH, gives the trans product.
CH,C=CCH,CH,CH,
,::
CH3
+c\
H
/
H H,CH,CH,
( E ) - or rruns-2-Hexene
Problem 8.16 Outline steps for the conversion of CH,CH,CH,Br to (a) CH,CBr---CH,, (6) CH,CCl,CH,, (c) CH,CH=CHBr. 4
As usual, we think backward (the retrosynthetic approach). Each product is made from CH,C=CH, which in turn is synthesized from CH,CH=CH,. ( 6 ) CH,CCI,CH,
9
CH,CH,CH,Br-
alc. KOH
CH,CH=CH,
2CH,CHBrCH,Br-
NaNHZ
HBr
C H , C S C H T (a) C H 3 C B d H , (c) CH,CH==CHBr
~2
ALKYNES AND DIENES
148
-
[CHAP. 8
ACIDITY AND SALTS OF 1-ALKYNES [see Problem 8.5(6)]
CH,C'-=CH
+ Ag'
N"3
-H+
CH,eCAg(s)-
HN03
CH3CkCH + Ag'
Problem 8.17 Will the following compounds react? Give any products and the reason for their formation.
+ aq. Na'OH-+ + CH,OH+ CH,CH,C%C--MgI C H , m : - N a ' + NHiCH,--C%C--H
(a)
4
( a ) No. The products would be the stronger acid H,O and the stronger base CH,(%CT. (6) Yes. The products are the weaker acid CH,CH,C%C:H and the weaker base MgI(OCH,). (c) Yes. The products are the weaker acid propyne and the weaker base NH,.
Problem 8.18 Deduce the structure of a C,H, compound which forms a precipitate with Ag' and is reduced 4 to 2-methylbutane.
The precipitate shows an acetylene bond at the end of a chain with an acidic H. With --C%CH the other three carbons must be present, as a (CH3)*CH- group, because of reduction of ( C H , ) , C H e H to (CH,),CHCH2CH,.
8.3 ALKADIENES Problem 8.19 Name by the IUPAC method and classify as cumulated, conjugated, or isolated:
(U
)
H,C=CH-C
H=C HC H3
( b ) H,C=CHCH,
(c) H,C=C=CH,
(d) H,C=CH--CH=CHCH=CH,
4
( a ) 1,3-Pentadiene. Conjugated diene since it has alternating double and single bonds, i.e., -C=C-C==C-. (b) 4-Ethyl-1,4-heptadiene. Isolated diene since the double bonds are separated by at least
one sp,-hybridized C, i.e., -C=C-(CH,),,-C==C-. double bonds are on the same C, i.e., --C=C=C-. single and double bonds.
( c ) 1,2-Propadiene (allene). Cumulated diene since 2 ( d ) 1,3,5-Hexatriene. Conjugated since it has alternating
Problem 8.20 Compare the stabilities of the three types of dienes from the following heats of hydrogenation, A H , (in kJ/mol). (For comparison, A H , for 1-pentene is -126.)
H Conjugated
H,C=A-CH=CH-CH,
-230
1.3-Pentadiene Isolated
H,C=CH-CH2-CH=CH2
-252
1,4-Pentadiene Cumulated
H,C=C=C
H-C H2CH,
1,2-Pentadiene
-297
4
The calculated A H , , assuming no interaction between the double bonds, is 2(-126) = -252. The more negative the observed value of AH,, compared to -252, the less stable the diene; the less negative the observed value, the more stable the diene. Conjugated dienes are most stable and cumulated dienes are least stable; under the proper conditions allenes tend to rearrange to conjugated dienes.
CHAP. 81
149
ALKYNES AND DIENES
Problem 8.21 Predict the product of the reaction
alc. KOH
+ CH,-C-
‘
H The more stable conjugated diene
!-CH,-CH=CH,
4
__*
L-1
H H H (CH,),A-L=A--CH=CH, is formed, even though it is less substituted than the isolated diene (CH,),C=CHCH,CH=CH,, which is not formed.
Problem 8.22 Give steps for the conversion HC=CCH,CH,CH,
HC-CCH,CH,CH,
H /R(Pb)
H,C=CHCH,CH,CH,
CI
+H,C=CH-CH=CHCH,.
4
*
(allylic subsIituIion)
H,C=CHCHCICH,CH,
H,C=CHCH=CHCH,
Problem 8.23 Account for the stability of conjugated dienes by ( a ) extended w bonding, (6) resonance theory. 4 ( a ) The four p orbitals of conjugated dienes are adjacent and parallel (Fig. 8-2) and overlap to form an
extended r system involving all four C’s. This results in greater stability and decreased energy. Arrows indicate electron spin.
Fig. 8-2
(6) A conjugated diene is a resonance hybrid:
(9
Structure (i) has 11 bonds and makes a more significant contribution than the other two structures, which have only 10 bonds. Since the contributing structures are not equivalent, the resonance energy is small.
8.4 MO THEORY AND DELOCALIZED 12 SYSTEMS
Review Section 2.2. The MO theory focuses attention on the interactingp AO’s of the delocalized w systems, such as conjugated polyenes. The theory states that the number of interactingp AO’s is the same as the number of 7r molecular orbitals formed. The molecular orbitals are considered to be stationary waves and their relative energies increase as the numbers of nodal points in the corresponding waves increase. Nodes may appear at a C atom, as indicated with a 0 rather than a +
[CHAP. 8
ALKYNES AND DIENES
150
or a - sign. In a linear system with an even number of molecular orbitals, half are bonding MO’s and half are antibonding MO*’s. With an odd number of molecular orbitals, the middle-energy molecular orbital is nonbonding (MO’). The electrons in the delocalized 7r system are placed first in the bonding, then in the nonbonding (if present), and then, if necessary, into the antibonding, molecular orbitals-with no more than two electrons in each molecular orbital. Electrons in MO’s add to bonding strength, those in MO*’s diminish bonding strength; those in M0”’s have no effect. We often simplify our representations of molecular orbitals by showing only the signs of the upper lobes of the p AO’s, and not the entire orbital. Problem 8.24 Apply the MO theory to the n system of ethene.
4
Each of the doubly-bonded C’s, C=C, has a p AO. These two p AO’s provide two molecular orbitals, a lower-energy bonding MO and a higher-energy antibonding MO*. Each p A 0 has one electron, giving two electrons for placement in the 7~ molecular orbitals. Molecular orbitals receive electrons in the order of their increasing energies, with no more than two of opposite spins in any given molecular orbital. For ethene, the two p electrons, shown as t and 4, are placed in the bonding MO ( n ) ;the antibonding MO* ( n * )is devoid of electrons. Note the simplification of showing only the signs of the upper lobes of the interacting p orbitals. The stationary waves, with any nodes, are shown superimposed on the energy levels.
-
-
A+
antibonding
bonding
Problem 8.25 Apply the MO theory to 1,3-butadiene and compare the relative energies of its molecular 4 orbitals with those of ethene (Problem 8.24). Four p AO’s (see Fig. 8-2) give four molecular orbitals, as shown in Fig. 8-3. Wherever there is a switch from + to -, there is a node, as indicated by a heavy dot. Note that n, of the diene has a lower energy than n of ethene.
- -7rt Antibonding MO’s*
- -7Tf -
Ethene
1.3-Butadiene Fig. 8-3
In a linear n system the relative energies of the molecular orbitals are determined by the pairwise overlaps of adjacent p orbitals along the chain. An excess of bonding interactions, + with + or - with -, denotes a bonding MO; an excess of antibonding interactions, + with -, denotes an antibonding MO*.
CHAP. 81
ALKYNES AND DIENES
15 1
Problem 8.26 Considering the overlaps of the p orbitals of Fig. 8-3, justify the assignment of the relative 4 energies of the molecular orbitals of 1,3-butadiene and their designation as n or ?T*.
1,3-Butadiene has three adjacent-p-orbital overlaps, C'-C2, C2-Cz,and C3-C4. The energies are listed in ascending order:
3 antibonding ( + -, - +, + -); highest energy antibonding r:: 2 antibonding and 1 bonding (+ -, - -, - +); next-to-highest energy antibonding n,: 2 bonding and 1 antibonding ( + + , + - , - -); next-to-lowest energy bonding nl: 3 bonding ( + +, + +, + +); lowest energy bonding n::
Problem 8.27 Explain how the energies shown in Fig. 8-3 are consistent with the fact that a conjugated diene 4 is more stable than an isolated diene.
The energy of nl+ n2 of the conjugated diene is less than twice the energy of an ethene n bond. Two ethene n bonds correspond to an isolated diene. Problem 8.28 ( a ) Apply the MO theory to the ally1 system (cf. Problem 8.26). Indicate the relative energies of the molecular orbitals and state if they are bonding, nonbonding, or antibonding. (6) Insert the electrons for the carbocation C,Hl, the free radical C,Hs. and the carbanion C,H,, and compare the relative energies of these 4 three species. ( a ) Three p AO's give three molecular orbitals, as indicated in Fig. 8-4. Since there are an odd number of p
AO's in this linear system, the middle-energy molecular orbital is nonbonding ( n i ) .Note that the node of this MO" is at a C, indicated by a 0. An MO" can be recognized if the number of bonding pairs equals the number of antibonding pairs (see Problem 8.26) or if there is no overlap. 4
The electrons in the r; orbital do not appreciably affect the stability of the species. Therefore all three species are more stable than the corresponding alkyl systems C,H;, C,H,., and C,H;:. The extra electrons do increase the repulsive forces between electrons slightly, so the order of stability is C,HJ > C,Hs. C,H,.
'
V
lr:
antibonding (2 antibonding pairs)
n; nonbonding
(no bonding pairs)
lr,
Fig. 8-4
bonding (2 bonding pairs)
152
ALKYNES AND D I E N E S
[CHAP. 8
8.5 ADDITION REACTIONS OF CONJUGATED DIENES 1,2- AND I94-ADDITI0NS
Typical of conjugated dienes, 1,3-butadiene undergoes both 1,2- and 1,4-addition, as illustrated with HBr. H2C-CH=CHCH2 HI
14-
[ H 2 C = C H C m ]
d>H,C-CH-CH=CH :in
k
br
8r
Problem 8.29 Explain 1,4-addition in terms of the mechanism of electrophilic addition.
4
The electrophile (H') adds to form an allylic carbocation with positive charge delocalized at C2 and C4 (resonance forms I1 and 111). This cation adds the nucleophile at C' to form the 1,Z-addition product or at C4 to form the 1,4-addition product. 2
3
H2f-yH-CH=CH2 H 1
4
I
H
I 2
H2C-CH-CH=CH2
(11) + Br-
3
H2(i-CH=CH-$H2
1
Br 1,2-adduct
____)
4
H2C-CH=CH-CH2
(111)
I
H
I
Br
H 1.4-adduct
The relative rates of formation of carbocations are: 3", 2' allyl > 1" allyl, 2" > 1" > CHJ. CONTROLLING FACTORS A rate-controlled reaction is one whose major product is formed through the transition state with the lowest A H S . A thermodynamic-controlled reaction is one whose major product has the lower (more negative) AH of reaction. Reactions may shift from rate to thermodynamic control with increasing temperature, especially when the formation of the rate-controlled product is reversible. Problem 8.30 Use an enthalpy-reaction diagram to explain the following observations. Start from the allylic carbocation, the common intermediate. 1.2-Adduct
1,6Adduct H2C-CH=CH-
HBr
+ H,C=CH-CH=CH2
H
I
Br
20% H,C-CH-CH=CH,
k 8,
H
+
FH2 Br
80% H2C-CH=CH-CH2
I H
I
4
Br
The different products arise from enthalpy differences in the second step, the reaction of Br- and the allyl
R'. See Fig. 8-5. At -80°C the 1,2-adduct, the rate-controlled product, is favored because its formation has the lower AHS.1,2-Adduct formation can reverse to refurnish the intermediate allylic carbocation, R'. At 40 "C, R'
goes through the higher-energy transition state for formation of the more stable 1,4-adduct, the thermodynamiccontrolled product. The 1,4-adduct accumulates because the addition, having a greater AH', is more difficult to reverse than that for the 1,2-adduct. The 1,4-adduct has a lower enthalpy because it has more R groups on the
c=c.
CHAP. 81
ALKYNES AND DIENES
153
1 Problem 8.31 Explain why the conjugated 1,3-pentadiene reacts with one mole of Br, at a faster rate than does the isolated 1,4-pentadiene. 4 The reaction products are shown: H,C==CH-CH=CH-€H, conjugated
H,C=CH--CH,--CH=CH,
+Br2 [
+
BrCH,--CH--CH==CH--CH,
I
BrCH,--CH=CHAH--CH, an allylic
R'
1
BrCH ,--CH--CH=CH--CH,
+ Br-/
Ar
'L BrCH,--CH=CH--CH--CH,
I
Br
+ Br, --+BrCH,--&H-CH,--CH=CH,
isolated
an
R'
+ Br- ---+BrCH,--CH--CH,--CH=CH,
I
Br The intermediate carbocation formed from the conjugated diene is allylic and is more stable than the isolated carbocation from the isolated diene. Since the transition state for the rate-controlling first step leading to the lower-enthalpy allylic R' also has a lower enthalpy, AH' for this reaction is smaller and the reaction is faster. It is noteworthy that although the conjugated diene is more stable, it nevertheless reacts faster. Problem 8.32 ( a ) Which additional C's in the following R' bear some
+ charge?
(6) With which of these C's will :Nu- react to give the thermodynamic-controlled product? ( a ) C3, C', and C'. These are alternating sites. +
+
+
4
~=~-~=~-F-~=~C*~=~-~-~=F-~=~C--*~-~=~-~=F-~=~
( 6 ) :Nu- adds to equivalent C' or C7 to give the conjugated triene
I
I
I
I
I
I
I
<===C-C=C-C=C-C-Nu
I
(most stable)
ALKYNES AND DIENES
154
[CHAP. 8
Addition at C-' (or C') gives a triene with only two conjugated C=C's.
I I I I I I I
(less stable)
-C==C-C-C=C-C=C-
I
Nu
Problem 8.33 Write structural formulas for major and minor products from acid-catalyzed dehydration of H,C=CHCH,CH( OH)CH,. 4 Dehydration can occur by removal of H from either C ' or C'.
H , ~ - ~ H - - ~ H - - ~ H = ~ H--+ , H,C-CH=CH-CH=CH,
I
1
I
I
H' OH HZ
H (from -H-')
4-Hydroxy- 1-pentene (I-Penten-4-ol)
1,3-Pentadiene (major) conjugated diene
+ H,C=CH-CH-CH=CH, I
H (from -H5)
1.4-Pentadiene (minor) isolated diene
Problem 8.34 Write the structures of the intermediate R"s and the two products obtained from the reaction of H,C=C(CH,)CH=CH2 with ( a ) HBr, (6) Cl,. 4 (a)
H'+adds to C' to form the more stable allylic 3" R', rtther than to C' or C" to form the nonallylic 1" R"s H,C-CH(CH,)CH==CH, and H,C=C(CH,)CH,-CH,, respectively; or to C4 to yield the 2" allylic R' +
H,C=C( CH,)CHCH, , CH3
I
H,C=C-CH=CH,
CH3
CH3
1
a+&-
+ HBr --+ CH,-C'-CH=CH,
*--*
1
r-
I
CH,-C=CH-&H,
Er-
CH3
CH3
I + CH3-C=CH-CH2
I
CH3-C-CH=CH2
I
I
Br
Br 3-Met h yl-3- bromo1 -butene (minor)
I-Brorno-3-methyl2-butene (major)
(6) C1' also adds to C' to form a hybrid allylic R'. CH3
CH3
I
H,C=C--CH=CH,
+ CI,
__*
I
H,C-C-=CH-CH,
cI1 '
+
+ C1-
I
__*
CH3
CH3
I H,C-C-CH=CH, I I CI Cl 3-Methyl-3,4dichloro- 1-butene (minor)
+
I
2-Methyl-l,4dichloro-2-butene (major)
Problem 8.35 Consider addition of one mole of Br, to one mole of 1,3,5-hexatriene. ( a ) Deduce structures of possible products. (b) Which products will be obtained under thermodynamic-control conditions? 4
CHAP. 81
155
ALKYNES AND DIENES
First, Br' adds at C' to form an allylic-type R' having form three dibromo addition products.
5
6
- H,b-CH=CH-{H-CH=CH,2
I Br
3
I
Br
H,C-CH-CH=CH-CH=CH,
I
Br
adds to
' H,C-CH=CH-CH-CH=CH,
I I
+
+ charge on C2, C4,and C". Then Br-
-
Br4
I
(I)
Br
H,C-CH-CH=CH-CH=CH,
(11)
Qr Qr
H,C-CH=CH-CH=CH-eH,
-
I
Br
d
A fourth, very minor, product (IV) is possible from addition at C3 and H,C=CHCHBrCHBrCH=CH, . Thermodynamic-control conditions favor the most stable product. Compounds I1 and 111 are conjugated dienes and hence more stable than the isolated dienes, I and IV. 111 is more stable than I1 because its double bonds have more substituents.
Problem 8.36 Write initiation and propagation steps in radical-catalyzed addition of BrCCI, to 1,3-butadiene and show how the structure of the intermediate accounts for: ( a ) greater reactivity of conjugated dienes than alkenes, (b) orientation in addition. 4
Initiation Rad. + Br:CCI,-
Rad:Br
+ CCI,
Propagation
1,4-Add~ct (major)
1, t - A d d ~ t
(minor)
( a ) The allyl radical formed in the first propagation step is more stable and requires a lower AH' than the alkyl
free radical from alkenes. The order of free-radical stability is allyl > 3" > 2" > 1". (b) The 1,4-orientation is similar to ionic addition because of the relative stabilities of the two products.
8.6 POLYMERIZATION OF DIENES
ELECTROPHILIC CATALYSIS
I
2
3
4
5
6
?
ECH,-CH=CHCH2CH2CH=CH:H2
+
MCHFCHCHPCH~)
(E[-CH2CH=CHCHt-],+2)+ mer of polymer
156
ALKYNES AND DIENES
[CHAP. 8
NUCLEOPHILIC OR ANIONIC POLYMERIZATION
CH3 CH3 I I Nu:CH2C=CH-CH2:CH2-C=CH-CH2
CH3
n(CHrL--CH-CH~ n +2
dirneric anion
mer of polymer
The reaction is stereospecific in yielding a polymer with an all-cis configuration. Conjugated dienes undergo nucleophilic attack more easily than simple alkenes because they form more stable allyl carbanions,
Like the allyl cation, the allylic anion is stabilized by charge delocalization through extended bonding.
7~
RADICAL POLYMERIZATION
monorner
dimeric free radical
monomeric free radical
Conjugated diene polymers are modified and improved by copolymerizing them with other unsaturated compounds, such as acrylonitrile, H,C=CH-C%N.
8.7 CYCLOADDITION A useful synthetic reaction is cycloaddition of an alkene, called a dienophile, to a conjugated diene by 1,4-addition. //CH2 CH
I
CH \CH2 1,3-Butadiene
+
CH2
II
CH2 Ethylene
F H 2 \
2 0 0 ~
___,
CH
II
C?
CH,
I
,CH2 CH* Cyclohexene
Diels-Alder reaction
CHAP. 81
157
ALKYNES AND DIENES
8.8 SUMMARY OF ALKYNE CHEMISTRY
PREPARA TION
PROPERTIES
1. Industrial
1. Addition Reactions
2CH4(- 3H2)-
0 2
+ Cu(NH,),Cl--+ H,C=CH--C-=CH
H-H
H,/Pt(Pb)d RCHSH,
/+
+ Na, NH,-RCH=CH, + HX- RCX=CH,-
--
+HX
(~yn) (anti) RCX,CH,
+ HOH(Hg++,H’)+
2. Laboratory
R-CO-CH, RCX=CHX++xz RCX,--CHX,
Triple-Bond Formation
+ R;BH
De hydrohalogen at ion RCHXCH,X or RCH,CHX,
RCHdHBR;
Alkylation of Acetylene
,
HC-=CH + NaNH, --+H-Na + RMgX- HC-3CMgX H-H
RCH CH=O
-
2. Replacement of Acidic Hydrogen -
+ NaNH,+ Ag(NH,): + R’MgX+
R m N a R m A g R m M g X + R’H
8.9 SUMMARY OF DIENE CHEMISTRY
PREPARA TION
PROPERTIES
1. Dehydration of Diols
1. Hydrogen Addition
HOCH2--CH2--CH2--CH2-OH 2.
\
Dehydrogenation (industrial)
r+ 2.
Alkanes: CH,CH,CH,CH, - 2H2 Alkenes: H,C=CHCH,CH, - H, H,C=CHCH=CH,
3. Dehydrohalogenation
,
Dihalides: CH CHXCH CH ,X Ally1 halides: H,C=CHCHXCH,
2H2Butane Polar Additions + CI, -+ 1,2- + 1,4-Dichlorobutenes + 2C1, 1,2,3,4-Tetrachlorobutane + HI +3-Iodo- + 1-Iodobutenes
/
-
+ --(CH2CH4H
-
3. Free-Radical Addition
+ BrCCl, 1,2-and 1,4-Adducts + R.+-(CH2--CH=CH--CH2)npolymer
4.
Diels-Alder Reaction
CHdH, + H , C d H Z j H , C , C H , - C H ~CH2 / /
158
ALKYNES AND DIENES
[CHAP. 8
Supplementary Problems Problem 8.37 For the conjugated and isolated dienes of molecular formula C,H,,, tabulate (a) structural formula and IUPAC name, ( 6 ) possible geometric isomers, (c) ozonolysis products. 4 Table 8-2 ( a ) Formula and Name
(1)
H~c=cH--]-cH,-cH,
( b ) Geometric Is0 me rs
( c ) Ozonolysis Products
2
H $ 5 0 . O=C H-CH=O,
2
H2C=0,
O=CHCH,CH=O,
None
H2C=0,
O=CHCH,CH,CH=O,
O=C HC H2CH
1,3-Hexadiene
(2)
(CH=CHI-CH ,
HzC=CH-CH2-
O=CHCH,
1,CHexadiene
(3) H,C=CH-CH2-CH2-CH=CH2
O=CH2
1 .S-Hexadiene
CH3 (4)
CH3
I
H,C=C-v]-CH,
I
2
H2C=0,
O=C-CH=O,
None
H2C=0,
O=C-CH,CH=O,
2
H2C=0,
O=CH-C=O,
O=CHCH,
2-Methyl- I ,3-pentadiene
CH,
CH3
I
( 5 ) H,C=C-CH,-C H=C H, 2-Methyl- 1,4-pentadienc
(6) H2C=CH-
El
C=CH -CH,
I
O=CH2
CH3
I
O=CHCH,
3-Methyl- 1.3-pentadiene
CH,
CH, (7)
I
H,C=CH-CH=C--CH,
None
H2C=0,
O=CH-CH=O,
I
O=C-CH,
4-Methyl- 1.3-pentadiene
(8) C H , - ~ I - ~ ~ I - C H , 2.4- H exadie ne
3
cis, cis ; cis, trans ; trans, trans
CH,CH=O,
CH, CH,
I
t
(9) H2C=C-C=CH2
O=CH-CH=O,
O=CHCH,
CH, CH, None
H2C=0,
I
O=C-C=O,
I
O=CH2
2.3-Dimethyl- 1.3-butadiene
CH,
CH,
CH2
CH2
I
(10)
I
H,C=C-CH=CH2
None
H2C=0,
I I
O=C-CH=O,
O=CH2
2-Ethyl-l.3-butadiene
CH,
(11)
I H2C=CH-CHCH=CH2 3-Methyl- 1.4-pcntadiene
CH, None
H,C=O,
I
O=CHCH-CH=O,
O=CH,
CHAP. 81
159
ALKYNES AND DIENES
Problem 8.38 Show reagents and reactions needed to prepare the following compounds from the indicated starting compounds. (a) Acetylene to ethylidene iodide ( 1,l-diiodoethane). (b) Propyne to isopropyl bromide. ( c ) 2-Butyne to racemic 2,3-dibromobutane. (U') 2-Bromobutane to rrans-2-butene. (e) n-Propyl bromide to 4 2-hexyne. ( f ) 1-Pentene to 2-pentyne.
H-c=c-HHI.
(4 (4
CH3C'-3C-H=
-
H , C = C H I CH,CHI, ~ HBr
CH,--CH=CH,-
-
CH,CHBrCH,
Add H, first; the reaction can be stopped after 1 mol is added. H2/ PI(Pb)
CH,C=CCH,
(c)
alc K O H
( d ) * CH,CHBrCH,CH,-
meso and rac
CH,CH,CH,Br-
alc K O H
Er2
CH,CH=CH,+
H,C=CHCH,CH,CH,
-
KNH?
CH,CH=CHCH,CH, (Saytzeff product)
Na N H ,
CH,C%CCH,
-%CH,=
2CH,CHBrCH,CH,CH,
(rrans addition)
Br?
CH,CHBrCH,Br-
CH,C--CH (f)
rac-( t)-CH,CHBrCHBrCH,
cis + trans-CH,CH=CHCH,-
CH,CHBrCHBrCH, (e)
Br2
cis-CH,CH=CHCH,
truns-CH,CH=CHCH,
KNH,
n
T Na +
C,H7Br
CH ,C-=C--CH,CH,CH
alc K O H
Br?
KNt1,
2CH,CHBrCHBrCH,CH,
CH,C%CCH,CH,
(not the less stable allene, CH,CH==CH==CHCH,)
Problem 8.39 Write a structural formula for organic compounds (A) through (N): HC-=CCH,CH,CH,-
(4
CH,MgBr
(4
CH,C%CH-(C,
(4
)i
A%(NHJ
(4
CICH,
-r3
AgC%CCH,CH,CH,
(4
+ alc. KOH-(L)-(M) (A)
+
(H)
aq
dil
+ (N)
RrCCI,
peroxide
HC-=CCH,CH,CH,
CH, (C) + CH,C-=CMgBr (D)
(b)
H , O + . Hg'
CZH~I
+ BH,+W-(J)KM,o,-(K)
HCHCICH,
(B)
a gas) + (D)-(E) CH3COOH
CH3-H
HNO, CH,I
CH,CH,CkSH+Na'NH~-(F)-(G)-
(4
(A)-
CH,=cH,
4
(B)
(E) 0 -
(c)
CH,CH,C-=Na'
(4
(F)
(CH,CH=CH),B
CH,CH,C-=CCH,CH, (I)
CH,CH=CH,
(e)
(J)
II
CH,CH,--C--CH,CH,CH, CH,CHOHCH,OH
CH
CH3
I H,C=C-CH=CH,
(G)
(L) Cl,CCH, 1.2-addition product
(H)
(K)
CH
I '
(M) CI,CCH,-C=CH-CH,-Br
(N)
1,4-addition product (major)
Problem 8.40 Give possible products, with IUPAC names, of the following addition reactions: ( a ) 1,3butadiene and 1,4-pentadiene, each with 1 and 2 mol of HBr; ( b ) 2-methyl-l,3-butadiene and 1,3-pentadiene with 1 mol of HI. 4 ~~~~~~
* Tram- and cis-alkenes are made by stereospecific reductions of corresponding alkynes.
160
(a)
[CHAP. 8
ALKYNES A N D DIENES
+ HBr +
H,C=CH-CH=CH,
( 1 &Adduct)
5 CH,CH,CHBrCH,Br + CH,-CHBr-CH,-CH,Br
CH,-CH=CH-CH,Br 1 -Bromo-2-butene
1,2-Dibromobutane
+
( 1.2-Adduct)
1.3-Dibromobutane
CH,-CHBr-CH=CH,
5 CH,-CHBr-CHBr-CH,
3-Bromo-1-butene
2.3-Dibromobutane
-
+ HBr +CH,-CHBrCH,-CH=CH,
H,C=CH-CH,-CH=CH,
CH,CHBrCH,CHBrCH,
HBr
CBromo- 1-pentem
2,CDibromopentane
CH3
I
(6) HzC=C-CH=CH2
+ HI
__*
CH,
I
+ CH3-
CH,-C=CH--CHJ I-Iodo-3-methyl2-butene
ZH3
ICH=CH,
+
1CH2-
3-Iodo-3-methyl1 -butcne
ZH3
=CHCH,
I -1odo-2-methyl-
t
4-Iodo-2-pentenc (1.2- and 1 &adduct)
(a)
2%
3-Iodo-1-pcntene (1 ,tadduct)
1 -1odo-2-pentene (1.4-adduct)
to 5 for MOST to indicate the relative reactivity on HBr
(6) CH,-CH=CH-CH,
H,C=CH-CH,CH,
,
--CHCH, (2")
+ H2C=CHCHICH2CH3 + H,CICH=CHCH,CH,
CH,CHI-CH=CH-CH,
LEAST
-CHICH,
t
from less stable H2C-
+ HI +
Problem 8.41 Assign numbers from 1 for addition to the following compounds:
ZH3
3-Iodo-2-methyl1 -butene
2-butene
from more stable (CH,),CCH=CH, ( 3 " )
H,C=CH-CH=CH-CH,
+ H,C=
(c) H,C=CH-CH=CH, CH3
I
( d ) CH,-CH=CH-CH=CH,
4
Conjugated dienes form the more stable ally1 R"s and therefore are more reactive than alkenes. Alkyl groups o n the unsaturated C's increase reactivity. Relative reactivities are: ( a ) 1, (6) 2, ( c ) 3, ( d ) 4, ( e ) 5 . Problem 8.42 For the reaction of propyne with ( a ) HOBr, (6) Br, products and the mechanisms of their formation.
+ NaOH,
give the structures of the
4
H
I + H' # H--I)-Br * --+ CH,-c=yH + ?H2- -H+ H-0-Br
(a)
7 + Br-Y-H
CH,-C-C-H
H
Bromoacetone
CHAP. 81
161
ALKYNES AND DIENES
with strong bases to form a nucleophilic carbanion which displaces :BrT from Br, by (6) Propyne reacts s+ attacking Br to form 1-bromopropyne.
Problem 8.43 14CH,CH==CH2is subjected to allylic free-radical bromination. Will the reaction product be exclusively labeled H,C = CHI4CH,Br? Explain. 4
No. The product consists of an equal number of H,C = CHI4CH2Brand I4CH, = CHCH,Br molecules. H-abstraction produces a resonance hybrid of two contributing structures having both "C and I4C as equally reactive, free-radical sites that attack Br,. Br.
+ ''CH3CH=CH2
__*
HBr
+ l4CH2CH=CH2'
I
'%H,=C HC H
f
BrI4CH2CH=CH,
Brl +
, I
-
+
*
-B
r
or Br14CH2CH='4CH,
I4CH,=CHCH Br
Problem 8.44 ( a ) Write a schematic structure for the mer of the polymer from head-to-tail reaction of 2-methyl-l,3-butadiene. (b) Account for this orientation in polymerization. ( c ) Show how the structure is deduced from the product
0
II
CH~-C-CH,-CH2-CH=0
obtained from ozonolysis of the polymer. (a)
4
1'4-Addition with regular head-to-tail orientation produces a polymer with the following repeating unit (mer): CH3
I
-CH2-C=CH-CH2-(6) This orientation results from more rapid formation of the more stable intermediate free radical.
The 1" allylic site is more reactive than the 3" allylic site. Attack at the other terminal S H 2gives the less stable free radical. CH3
I
CH,=C--CH=CH, (c)
CH,
I + *R+ [CH,=C--CHCH,R
CH3
I
(2") *CH,--C=-CHCH,R] (1")
Write the ozonolysis products with the 0 ' s pointing at each other. Now erase the 0 ' s and join the C's by a double bond.
162
ALKYNES AND DIENES
CH3
CH,
CH,
I
I
[CHAP. 8
I
=C-CH,CH,-CH=C-CH,-CH,-CH=C-CH,-CH,-CH= '
piece of mer
me r
mer
::2.
H20+
piece of next mer
CH,
CH3
I
CH3
I
I + O=C-CH,CH,CH=O
+ O=C-CH,CH,CH=O
O=C-CH,CH,CH=O
Problem 8.45 ( a ) Calculate the heat of hydrogenation, A H h , of acetylene to ethylene if the AHh's to ethane are -137kJ/mol for ethylene and -314kJ/mol for acetylene. ( b ) Use these data to compare the ease of hydrogenation of acetylene to ethylene with that of ethylene to ethane. 4 Write the reaction as the algebraic sum of two other reactions whose terms cancel out to give wanted reactants, products and enthalpy. These are the hydrogenation of acetylene to ethane and the dehydrogenation of ethane to ethylene (reverse of hydrogenation of H,C=CH,). (Eq. 1) H a - H (Eq. 2)
+ 2H,
I_*
CH,--CH,+
-314 kJ/mol
H,C=CH,
+ H2+
H a - H
CH,--CH,
+ H,
+ 137 kJ/mol
H,C--LH,
- 177 kJ /mol
Equation 2 (dehydrogenation) is the reverse of hydrogenation ( A H h = -137 kJ/mol). Hence the AHh of Eq. 2 has a + value. Acetylene is less stable thermodynamically relative to ethylene than ethylene is to ethane because AH,, for acetylene+ ethylene is - 177 kJ/mol, while for ethylene +ethane it is - 137 kJ/mol. Therefore acetylene is more easily hydrogenated and the process can be stopped at the ethylene stage. In general, hydrogenation of alkynes can be stopped at the alkene stage. Problem 8.46 Deduce the structural formula of a compound of molecular formula C,H,, which adds 2 mol of H, to form 2-methylpentane, forms a carbonyl compound in aqueous H,SO,-HgSO, solution and does not react 4 with ammoniacal AgNO, solution, [Ag(NH,),]'NOJ. There are two degrees of unsaturation since the compound C,H,, lacks four H's from being an alkane. The addition of 2mol of H, excludes a cyclic compound. It may be either a diene or an alkyne, and the latter functional group is established by hydration to a carbonyl compound. The skeleton must be C
I c-c-c-c-c as established by the reduction product. The two possible alkynes with this skeleton are (CH,),CHCH,CkKH
and ( C H , ) , C H a - - C H ,
The negative test for a 1-alkyne with Ag ' establishes the second structure, 4-methyl-2-pentyne. Problem 8.47 Set up a table to show convenient chemical reactions, and the visible sign of these reactions, 4 which will differentiate n-pentane, 1-pentene and 1-pentyne. See Table 8-3. Positive and negative tests are indicated by
+ and
- signs.
Table 8-3
n- Pen tane 1-Pentene 1-Pentyne
I
Decolorization of Br, (Red) in CCI,
White Precipitate with Ag(NH,)I
+ +
+
I
CHAP. 81
163
ALKYNES AND DIENES
Problem 8.48 What is the structure of a hydrocarbon of molecular formula C,H,, if 1 mol adds 2 mol of H, 4 and also undergoes reductive ozonolysis to 2 mol of O=CHCH,CH,CH=O?
The compound C,H,, lacks 6 H’s from being an alkane, C,H,,. It has 3 degrees of unsaturation; two are in multiple bonds, as shown by the uptake of two moles of H,, and one is ascribed to a ring. The cyclic compound has either one triple or two double bonds. Since a four-carbon dicarbonyl compound is formed on ozonolysis, the hydrocarbon ( a ) cannot be a cycloalkyne because a dicarboxylic acid would be formed, (6) must have both olefin bonds as part of the ring and ( c ) must be a symmetrical cyclooctadiene with two CH, groups separating the C=C groups. The compound is 1,s-cyclooctadiene.
/CH2-cH2 H2C
I
H2C\
\
CH2
1
/CH* CH2-CH2 Cyclooctane
-
2H2
1
H2C
CH-CH /
\CH2
I
HZC,
/CH2 CH=CH
I.
o3
2.
H20.Zn+
CH=O / H2C
O=CH
\
CH2
I
I
H2C\
/CH2 CH-0
O=CH
1.5-Cyclooctadiene 1
2
3
4
5
Problem 8.49 The allene 2,3-pentadiene ( C H , C H S H C H , ) does not have a chiral C but is resolved into enantiomers. ( a ) Draw an orbital picture that accounts for the chirality [see Problem 5.25(d)]. (6) What 4 structural features must a chiral allene have?
C” is sp hybridized and forms two U bonds by sp-sp2 overlap with the orbitals of C2 and C4. The two remaining p orbitals of C-3form two 7r bonds, one with C2 and one with C4. These n bonds are at right angles to each other. The H and CH, on C2 are in a plane at right angles to the plane of the H and CH, on C4. See Fig. 8-6. Because there is no free rotation about the two 7r bonds, the two H’s and two CH,’s have a fixed spatial relationship. Whenever the two substituents on C2 are different and the two substituents on C4 are different, the molecule lacks symmetry and is chiral. Individually, the terminal C’s of the allenic system must have two different attached groups; e.g. RHC=C=CHR’(R). The groups could be other than H’s. H,C==C==CHR is not chiral. Mirror
H
‘ CH,
Fig. 8-6
Problem 8.50 Heating C,H,Br (A) with alcoholic KOH forms an alkene, C,H, (B), which reacts with bromine to give C,H,Br,(C). (C) is transformed by KNH, to a gas, C,H, (D), which forms a precipitate when passed through ammoniacal CuCl. Give the structures of compounds (A) through (D). 4
The precipitate with ammoniacal CuCl indicates that (D) is a 1-alkyne, which can only be 1-butyne. The reactions and compounds are: HaCH,CH, BrCH,CHBrCH,CH, %H,C=CHCH,CH, &BrCH,CH,CH,CH, (D) (C) (B) (A) (A) cannot be CH,CHBrCH,CH,, which would give mainly H,CCH=CHCH, and finally CH,C-=CCH,.
164
[CHAP. 8
ALKYNES AND DIENES
Problem 8.51 H , C = C H C H , W H (A) adds HBr to give H , C C H B r C H , m H . H W C H = C H , (B) adds 4 HBr to give H,C=CBrCH=CH,. Explain.
An isolated double bond is more reactive than an isolated triple bond in electrophilic additions, which accounts for the behavior of (A). If HBr had added to the double bond in (B), a butyne would have resulted. By HBr adding as it does, a more stable conjugated diene is formed. Problem 8.52 Is the fact that conjugated dienes are more stable and more reactive than isolated dienes an incongruity? 4 No. Reactivity depends on the relative A H * values. Although the ground-state enthalpy for the conjugated diene is lower than that of the isolated diene, the transition-state enthalpy for the conjugated system is lower by a greater amount (see Fig. 8-7).
A H * conjugated < AH' isolated and rateconlugated > rateisolated
t
Reaction Progress Fig. 8-7
Problem 8.53 Explain why 1,3-butadiene and 0, do not react unless irradiated by uv light to give the 1,4-adduct
c-)
4
0-0
Ordinary ground-state 0, is a diradical, :+Q:
t t
One bond could form, but the intermediate has 2 electrons with the same spin and a second bond cannot form.
0-0
t
t
&t
?i
+ H2C-CH-CH -CH,
_I*
O-O-CH2-CH=CH-CH2
t
When irradiated, 0, is excited to the singlet spin-paired state
:&&:
tl Singlet 0, reacts by a concerted mechanism to give the product.
t
CHAP. 81
165
ALKYNES AND DIENES
Problem 8.54 Is there any inconsistency between the facts that the C-H bond energy of all C-H bonds and that it is also the most acidic?
bond in acetylene has the greatest
No. Bond energy is a measure of homolytic cleavage, = C : H - - + = C cleavage, W : H + Base----,=! + H’(Base).
+ .H. Acidity is due to a heterolytic
4
Problem 8.55 (a) Apply the MO theory to the C=C-C=C-C+ carbocation, considering the signs of the upper lobes of adjacent p atomic orbitals (6) Indicate the relative energies of the molecular orbitals and state if they are bonding, nonbonding, or antibonding. (c) Show the distribution of the ?r electrons. 4 (a) Each C has a p A 0 in the same plane; we may arbitrarily assume they are p,, AO’s. These five AO’s afford five MO’s as shown in Fig. 8-8(a) using the standard notation of solid dots for nodes and zeros for nodal
C’S. (6) The relative energies are suggested in Fig. 8-8(6). Since this is a linear system with an odd number of p AO’s, the middle-energy MO is nonbonding. (c) Since there are two electrons in each of the two 7~ bonds, four electrons must be placed in the molecular orbitals. These are paired in the two bonding MO’s [Fig. 8-8(c)].
+ +.+ o + + + o - + + + + + +.-.+.--.
+.-
0 - 0
n$ (antibonding) IT: (antibonding)
n; (nonbonding) n2 (bonding)
n,
(bonding)
ri tl -
Chapter 9 Cyclic Hydrocarbons 9.1 NOMENCLATURE AND STRUCTURE
Cyclic hydrocarbons are called cycloalkanes; they are examples of alicyclic e p h a t i c cyclic) compounds. Cycloalkanes, having the general formula C,H,, , are isomeric with alkenes but, unlike alkenes, they are saturated compounds. They are named by combining the prefix cyclo- with the name of the alkane having the same number of C’s as are in the ring. Two or more substituents are listed alphabetically and are assigned the lowest possible numbers. Alicyclic compounds are usually symbolized by the appropriate-sized rings, without the C’s of the ring and with or without the attached H’s.
ji:
H.
H
..H
or
A
,.fit H.
CH,
CH3 Met hylcyclopropane
H or
0 -
H
H Cyclopentene
Polycyclic compounds have more than one ring; those with exactly two rings are bicyclic. The rings may merely be bonded to each other, as in Problem 9.l(d). Those that share a C-C bond are said to be fused, as exemplified by decalin (bicyclo[4.4.0]decane). The C’s that are common to the two rings, shown encircled in decalin below, are called bridgehead C’s.
a zero-carbon bridge
----a ;ridgehead
C’s
Decalin Unlike fused bicyclics, bridged bicyclics have one or more C’s separating the bridgehead C’s, as in Problem 9.2( f). For bicyclic compounds, the prefix bicyclo- is combined with a pair of brackets enclosing numbers separated by periods, which is followed by a name indicating the total number of atoms in the bridged rings. The bracketed numbers show how many C’s are in each bridge joining the bridgehead C’s and are cited in the order of decreasing size. Problem 9.1 Draw structural formulas for ( a ) bromocycloheptane, (6) bicyclo[3.1.O]hexane, (d) cyclobutylcyclohexane.
166
1-ethylcyclopentene, (c)
4
CHAP. 91
167
CYCLIC HYDROCARBONS
Problem 9.2 Name the following compounds:
‘(33: Me60Br 5
Me
3
2
5
CH,
CH,
Me { a
one-carbon bridge
bridgehead
Me
In ( f ) C ’ , C2, C-’, C4, C’, C6 constitute one ring; C’, C7, C4,C’, C6, another; C’,C2, C3, C4, C7, a third.
4
( a ) This fused-ring bicyclic compound has a four-carbon bridge made up of C2, C3,C4, and C”, and a two-carbon bridge of C7 and C’; the bridge between the bridgehead C’s has 0 carbon atoms. There are eight C’s in the compound. The name is bicyclo[4.2.0]octane. ( 6 ) Consider the cyclopropane ring to be a substituent on (c) The substituents, written alphabetithe longest carbon chain. The name is 1-cyclopropyl-3-methyl-1-pentene. cally, are numbered so that the c’s have the lowest possible numbers. The name is 3-bromo-l,l-dimethylcyclohexane. (d) 1,1,3-Trimethylcyclopentane. (e) 3-Nitrocyclohexene (nor 6-nitrocyclohexene). The doubly bonded C’s are numbered I and 2, to give the smaller number to the substituent. ( f ) The numbering of C’s of bicyclic compounds starts with the bridgehead C closest to a substituent. Substituents on the largest bridge get the smallest numbers. The name is 2,2,7,7-tetrame t h ylbicyclo[2.2.11heptane.
9.2 GEOMETRIC ISOMERISM AND CHlRALlTY
Review Chapter 5. The inability of atoms in rings to rotate completely about their ck-tram (geometric) isomers in cycloalkanes.
--
MeAMe
H
H
H
n
cis- 1.2-Dimeth ylcyclopropane
H
Me
CT
A
bonds leads to
__
Me truns- 1.2 - Dimethylcyclopropane
In such diagrams, either (i) the flat ring is perpendicular to the plane of the paper, with the bond(s) facing the viewer drawn heavy and with the substituents in the plane of the paper and projecting up and down, or (ii) the flat ring is in the plane of the paper, with “wedges” projecting toward the viewer and “dots” away from the viewer. Since the ring C’s are sp3-hybridized, they may be chiral centers. Therefore, substituted cycloalkanes may be geometric isomers as well as being enantiomers or meso compounds.
CYCLIC HYDROCARBONS
168
[CHAP. 9
Problem 9.3 Indicate the geometric isomers, if any, for (a) 1,1,2-trimethylcyclopropane,(b) 1,3-dirnethylcyclobutane, (c) 1,2,3-trimethyIcyclohexane,( d ) 1,2,4-trimethyIcyclopentane,( e ) 1-methyl-2-propenylcyclopen4 tane, ( f ) 2-chlorobicyclo[2.2.l)heptane(common name is 2-chloronorbornane).
&
Me (U)
Me
Me
,none
(b)
M@H
H Q
H
he tram
CiS
Me
H
Ye
Me
cis ,tram
cis,cis (all Me's up)
7
Irans,lrans
To designate cis,trans, start at C' and go around the ring increasing the numbers.
Me
Me
Me cis,cis (all Me's up)
cis ,trans
cis,tram
cis,cis
trans,lrans
tram,cis
tram,tram
The ring and double bond both show geometric isomerism; the stereomeric designation for the ring precedes that of the double bond.
ex0
endo
Relative to the smallest bridge, the Cl can be on the same side ( e m ) or on the opposite side (endo). Problem 9.4 Give the names, structural formulas and stereochemical designations of the isomers of (U) bromochlorocyclobutane, (6) dichlorocyclobutane, (c) bromochlorocyclopentane, ( d ) diiodocyclopentane, (e) dimethylcyclohexane. Indicate chiral C's. 4 (a) There is only one structure for 1-bromo-1-chlorocyclobutane:
169
CYCLIC HYDROCARBONS
CHAP. 91
With 1-bromo-2-chlorocyclobutanethere are cis and tram isomers and both substituted C's are chiral. Both geometric isomers form racemic mixtures.
*Hfl C]
Br CI
Br
H
Cl
C1
d . LS
S. R
R. R
tram
c u ,racemic
H
racemic
1 - Bromo-2-chlorocyclobutane
In I-bromo-3-chlorocyclobutane there are cis and tram isomers, but no enantiomers; C' and C7are not chiral, because a plane perpendicular to the ring bisects them and their four substituents. The sequence of atoms is identical going around the ring clockwise or counterclockwise from C' to C3. planes of
19Same
L
//
(Solid indicates an H is in the back, no dot for an H in front.)
(j=JIme :(-J
c1
/
symmetry
/
c1
cis
/
/
lrans
1 -Bromo-3-chlorocyclobutane
In these structural formulas, the other atoms on C' and C' are directly in back of those shown and are bisected by the indicated plane. Same as ( a ) except that the cis-l,2-dichlorocyclobutanehas a plane of symmetry (dashed line below) and is
meso.
c1
c1 c1
c1
qLw q y * ; p tram ,racemic
There are nine isomers because both 1,2- and 1,3-isomers have cis and tram geometric isomers, and these have enantiomers.
H
cis ,rac
I -Bromo- 1 -chlorocyclopentane
cis .rac
CI
H
tram.rac
I
1-Bromo-2-chlorocyclopentane
tram, rac
I - Bromo-3-chlorocyclopeniane
The diiodocyclopentanes are similar to the bromochloro derivative, except that both the cis-l,2- and the cis-l,3-diiodo derivatives are meso. They both have planes of symmetry. 1
1
1
cis-1.2-diiodopentane
I
cis-1.3-diiodopentane
CYCLIC HYDROCARBONS
170
[CHAP. 9
(e) There are nine isomeric dimethylcyclohexanes. .I
I
Me I
1 ,I-Dirnethyl
Me
Me
cis,meso-1,2-dirnethyl
Me
trons,rac- 1,3-Dirnethyl
cis ,meso-
Me
cis
-
Me
trans,roc- 1.2-Dirnethyl
(no chiral centers)
trans- 1.4-Dirnethyl
9.3 CONFORMATIONS OF CYCLOALKANES RING STRAIN
The relative stabilities of cycloalkanes can be determined by comparing their AH’s of combustion (Problem 4.39) on a per-CH,-unit basis. Rings have different AH’s of combustion per CH, unit because they have different amounts of ring strain. Problem 9.5 (a) Calculate AH of combustion per CH, unit for the first four cycloalkanes, given the following AH’s of combustion, in kJ/mol: cyclopropane, -2091; cyclobutane, -2744; cyclopentane, -3320; cyclohexane, -3952. (b) Write (i) the thermochemical equation for the combustion of cyclopropane and (ii) the theoretical equation for the combustion of a CH, unit of any given ring. (c) How do ring stability and ring size correlate for the first four cycloalkanes? 4
(a) Divide the given AH values by the number of CH, units in the ring (3, 4, 5, and 6, respectively), to obtain: cyclopropane, -697; cyclobutane, -686; cyclopentane, -664; cyclohexane, -659. Observe that these per-unit AH’s are in the reverse order of the total AH’s. (b)
(i)
C,H,
(ii) -CH,-+ (c)
In (ii) of (b), AH=H(CO,) ships under consideration,
+ qO,-3CO2 ~O,+CO,
+ 3H,O
AH = -2091 kJ/mol
+ H,O
+ H ( H , O ) - H( ;02)- H(unit).
AH
H(unit) = constant - AH Now, from ( a ) , AH increases (becomes less negative) with increasing size of the ring. Thus, H(unit) decreases with increasing size, which implies that H(ring) also decreases with increasing size. But a decreasing H(ring) means an increasing ring stability. In short, stability increases with ring size. Problem 9.6 Account for the ring strain in cyclopropane in terms of geometry and orbital overlap.
4
The C’s of cyclopropane form an equilateral triangle with C-C-C bond angles of @“-a significant deviation from the tetrahedral bond angle of 109.5’. This deviation from the “normal” bond angle constitutes angle strain, a major component of the ring strain of cyclopropane. In terms of orbital overlap, the strongest chemical bonds are formed by the greatest overlap of atomic orbitals. For sigma bonding, maximum overlap is achieved when the orbitals meet head-to-head along the bond axis, as in Fig. 9-l(a). This type of overlap in cyclopropane could not lead to ring closure for sp,-bybridized C’s because it would demand bond angles of 109.5”. Hence the overlap must be off the bond axis to give a bent bond, as shown in Fig. 9-1(b).
CHAP. 91
CYCLIC HYDROCARBONS
171
Fig. 9-1
In order to minimize the angle strain the C's assume more p character in the orbitals forming the ring and more s character in the external bonds, in this case the C-H bonds. Additional p character narrows the expected angle, while more s character expands the angle. The observed H-C-H bond angle of 114" confirms this suggestion. Clearly, there are deviations from pure p , s p , sp3, and sp2 hybridizations. Problem 9.7 What factor besides angle strain contributes to the ring strain of cyclopropane?
4
Because the cyclopropane molecule is a planar ring, the three pairs of H's eclipse each other as in n-butane (see Fig. 4-4)to introduce an eclipsing (torsional) strain. Problem 9.8 ( a ) Why is the A H of cis-1,2-dimethylcyclopropane greater than that of its t r a m isomer? ( b ) Which isomer is more stable? 4 ( a ) There is more eclipsing in the
stable.
cis isomer because the methyl groups are closer. ( b ) The tram is more
CONFORMATIONS OF CYCLOBUTANE AND CYCLOPENTANE Problem 9.9 Explain why the ring strain of cyclobutane is only slightly less than that of cyclopropane.
4
If the C's of the cyclobutane ring were coplanar, they would form a rigid square with internal bond angles of 90'. The deviation from 109.5' would not be as great as that for cyclopropane, and there would be less angle strain in cyclopropane. However, this is somewhat offset by the fact that the eclipsing strain involves four pairs of H's, one pair more than in cyclopropane. Actually, eclipsing strain is reduced because cyclobutane is not a rigid flat molecule. Rather, there is an equilibrium mixture of two flexible puckered conformations that rapidly flip back and forth (Fig. 9-2), thereby relieving eclipsing strain. Puckering more than offsets the slight increase in angle strain (angle is now 88"). The boxed H's in Fig. 9-2 alternate between up-and-down and outward-projecting positions relative to the ring.
Fig. 9-2
172
Problem 9.10
CYCLIC HYDROCARBONS
[CHAP. 9
Depict the flexible puckered conformation of cyclobutane in a Newman projection.
4
2 (C, is in back of c,1
Fig. 9-3
See Fig. 9-3. The circle represents the C’s of any given ring C--C bond. The other C’s of the ring bridge these two C’s, one to the front C (heavy line) and the other to the back C (ordinary line). This Newman projection formula reveals that the H’s on adjacent C’s are skewed in the puckered conformation. Problem 9.11 If cyclopentane had a planar ring, the ring would be regular pentagon with internal bond angles of 108”. As this would be very close to the tetrahedral angle of 109.5”, the ring would be free from angle strain, Nevertheless, at any given instant, one of the five C’s is puckered out of the plane, to give an envelope conformation. Account for this observation. 4
Cyclopentane with a planar ring would have five pairs of eclipsed H’s producing considerable eclipsing strain. This strain is reduced, at the expense of some increase in angle strain, when one CH, pushes out of the plane of the ring (Fig. 9-4). The puckering is not fixed but alternates around the ring, out-of-plane C
Fig. 9-4 Problem 9.12 Are the following compounds stable?
4
H
H ( a ) No. A trans-cyclohexene is too strained. The tram unit C--C=C-C cannot be bridged by two more C’s. rrans-Cycloalkenes are stable for eight-membered, ( 6 ) , and larger rings. (c) No. Cycloalkynes of fewer than which cannot be bridged eight C’s are too strained. The triple bond imposes linearity on four C’s, C-4, by two more C’s but can be bridged by four C’s, as in (d). (e) No. A bridgehead bicyclic cannot have a double bond at a bridgehead position (unless one of the rings has at least eight C’s). Such a bridgehead C and the three
CHAP. 91
173
CYCLIC HYDROCARBONS
atoms bonded to it cannot assume the flat, planar structure required of an sp2-hybridized C. This is known as Bredt’s rule. ( f ) Yes. This exists because one of the bridges has no C, and these bridgehead C’s can easily use sp*-hybridized orbitals to form triangular, planar sigma bonds. (g) Yes. Compounds (spiranes) having a single C which is a junction for two separate rings are known for all size rings. However, the rings must be at right angles. .........
CT
(h) No. Three- and six-membered rings cannot be fused trans, since there is too much strain. CONFORMATIONS OF CYCLOHEXANE
Problem 9.13 Among the cycloalkanes with up to 13 ring C’s, cyclohexane has the least ring strain. Would one 4 expect this if cyclohexane had a flat hexagonal ring?
No. A flat hexagon would have considerable ring strain. It would have six pairs of eclipsed H’s and C - C - C bond angles of 120’, which deviate by 10.5’ from the tetrahedral angle, 109.5’. Cyclohexane minimizes its ring strain by being puckered rather than flat. The two extreme conformations are the more stable chair and the less stable boat. The twist-boat conformer is less stable than the chair by about 23 kJ/mol, but is more stable than the boat. It is formed from the boat by moving one “flagpole” to the left and the other to the right. See Fig. 9-5. Problem 9.14 In terms of eclipsing interactions explain why ( a ) the chair conformation is more stable than the boat conformation and ( 6 ) the twist-boat conformation is more stable than the boat conformation. 4
H H
0.18 nm
H
H
H
H
Flagpole TH
HI
H
H I
Chair conformer most stable
Boat conformation least stable
(a)
(6) Fig. 9-5
Twist-boat conformation
(4
174
CYCLIC HYDROCARBONS
[CHAP. 9
( a ) In the boat form, Fig. 9-5(6), the following pairs of C-C bonds are eclipsed: C’-C2 with C3-CJ, and C’--Ch with C‘--C.‘. Furthermore, the H’s on C2--C3 and C’--C6 are also eclipsed. Additional strain arises from the crowding of the “flagpole” H’s on C’ and C‘, which point toward each other. This is called steric strain because the H’s tend to occupy the same space. In the chair form, Fig. 9-5(a), all C-C bonds are skew and one pair of H’s on adjacent C’s is anfi and the other pair is gauche (see the Newman projection). ( 6 ) Twisting C’ and CJ of the boat form away from each other gives the flexible twist-boat conformation in which the steric and all the eclipsed interactions are reduced [Fig. 9-5(c)].
SUBSTITUTED CYCLOHEXANES 1.
Axial and Equatorial Bonds
Six of the twelve H‘s of cyclohexane are equatorial (e); they project from the ring, forming a belt around the ring perimeter as in Fig. 9-6(a). The other 6 H’s, shown in Fig. 9-6(6), are axial (a); they are perpendicular to the plane of the ring and parallel to each other. Three of these axial H’s on alternate C’s extend up and the other three point down. Converting one chair conformer to the other, also changes the axial bonds, shown as heavy lines in Fig. 9-6(c), to equatorial bonds in Fig. 9-6(d). The equatorial bonds of Fig. 9-6(c) similarly become axial bonds in Fig. 9-6(d). Problem 9.15 Give the conformational designations of the boxed H’s in Fig. 9-2.
Axial on the left; equatorial on the right.
( a ) Equatorial Bonds
(e) Methylcyclohexane, Me(a)
(f)
Fig. 9-6
Methylcyclohexane, Me(e)
4
CYCLIC HYDROCARBONS
CHAP. 91
175
2. Monosubstituted Cyclohexanes
Replacing H by CH, gives two different chair conformations; in Fig. 9-6(e) CH, is axial, in Fig. 9-6( f ) CH, is equatorial. For methylcyclohexane, the conformer with the axial CH, is less stable and has 7.5 kJ/mol more energy. This difference in energy can be analyzed in either of two ways: 1,3-Diaxial interactions (transannular effect). In Fig. 9-6(e) the axial CH, is closer to the two axial H’s than is the equatorial CH, to the adjacent equatorial H’s in Fig. 9-6(f). The steric 1,3-diaxial interaction is 3.75 kJ/mol, and the total is 7.5 kJ/mol for strain for each CH,-H both. Gauche interaction (Fig. 9-7). An axial CH, on C’ has a gauche interaction with the C 2 - C 3 bond of the ring. One gauche interaction is also 3.75 kJ/mol; for the two the difference in energy is 7.5 kJ/mol. The equatorial CH, indicated as (CH,) is anti to the C 2 - C ” ring bond. In general, a given substituent prefers the less crowded equatorial position to the more crowded axial position.
1,l-Dimethylcyclohexane Fig. 9-7
Problem 9.16 ( U ) Draw the possible chair conformational structures for the following pairs of dimethylcyzlohexanes: (i) cis- and truns-1,2-; (ii) cis- and truns-1,3-; (iii) cis- and truns-1,4-. (6) Compare the stabilities of the more stable conformers for each pair of geometric isomers. ( c ) Determine which of the isomers of jimethylcyclohexane are chiral. 4 When using chair conformers, the better way to determine whether substituents are cis or rruns is to look at :he axial rather than the equatorial groups. If one axial bond is up and the other is down, the isomer is truns; if mth axial bonds are up (or down), the geometric isomer is cis. [U) (i) In the 1,2-isomer, since one axial bond is up and one is down, they are tram (Fig. 9-8). The equatorial bonds are also truns although this is not obvious from the structure. In the cis-1,2-isomer an H and CH, are truns to each other (Fig. 9-9).
H
H truns-1,2- (CH,’s aa); less stable
trans-l,2- (CH,’s ee); more stable Fig. 9-8
CYCLIC HYDROCARBONS
176
[CHAP. 9
CH,
q
C
H
,
H
L
H
J
T
cis-1,2-(CH,’s ea); conformational enantiomers
Fig. 9-9 (ii) In the 1,3-isomer both axial bonds are up (or down) and cis (Fig. 9-10). In the more stable conformer [Fig. 9-lO(a)] both CH,’s are equatorial. In the trans isomer, one CH, is axial and one equatorial (Fig. 9-11).
hH
H
I
H
A
-
H
I CH3 CH3 ( 6 ) cis-1,3- (CH,’s aa); less stable
( a ) ~ i s - 1 ~ 3(CH,’s ee); more stable
Fig. 9-10
=
(i d r n r r c c r l )
truns- 1,3-
y5+cH
H
(CH,’s ea)
CH,
Fig. 9-11
(iii) In the 1,4-isomer the axial bonds are in opposite directions and are tram (Fig. 9-12).
H
CH3 less stable truns- 1,4- (CH,’s aa)
more stable trans- 1,4- (CH,’s ee)
H
H
3cH I
H
CH3
cis-1,C (CH,’s ea) Fig. 9-12
CH3
CHAP. 9)
177
CYCLIC HYDROCARBONS
( 6 ) Since an (e) substituent is more stable than an (a) substituent, in each case the (CH,’s ee) isomer is more stable than the (CH,’s ea) isomer.
(i) tram > cis
(ii) cis > tram
(iii) tram > cis
(c) The best way to detect chirality in cyclic compounds is to examine the flat structures as in Problem 9.4(e); truns-1,2- and rrum-1,3- are the chiral isomers. Problem 9.17 Give your reasons for selecting the isomers of dimethylcyclohexane shown in Figs. 9-8 to 9-12 that exist as: ( a ) a pair of configurational enantiomers, each of which exists in one conformation; (6) a pair of conformational diastereomers; (c) a pair of configurational enantiomers, each of which exists as a pair of 4 conformational diastereomers; (d) a single conformation; (e) a pair of conformational enantiomers.
(a)
(6)
(c) (d) (e)
trans-l,3-Dimethylcyclohexaneis chiral and exists as two enantiomers. Each enantiomer is (ae) and has only one conformer. Both cis-l,3- and tram- 1,4-dimethylcyclohexane have conformational diastereomers, the stable (ee) and unstable (aa). Neither has configurational isomers. trum-l12-Dimethylcyclohexaneis a racemic form of a pair of configurational enantiomers. Each enantiomer has (ee) and (aa) conformational diastereomers. cis-l,4-Dimethylcyclohexanehas no chiral C’s and has only a single (ae) conformation. cis-l,2-Dimethylcyclohexanehas two (ae) conformers that are nonsuperimposable mirror images.
Problem 9.18 Write planar structures for optical isomers of cis-2-chlorocyclohexanol.Indicate chiral C’s. 4
This is not a meso compound because the two chiral C’s are different and there are two enantiomers, as shown:
Problem 9.19 Use 1,3- interactions and gauche interactions, when needed, to find the difference in energy between ( a ) cis- and trans-1,3-dimethylcyclohexane;(b) (ee) truns-l,2-and (aa) tram-1,2-dimethylcyclohexane. 4
Each CH,/H 1,3-interaction and each CHJCH, gauche interaction imparts 3.75 kJ/mol of instability to the molecule. ( a ) In the cis-1,3-isomer (Fig. 9-10) the more stable conformer has (ee) CH,’s and thus has n o 1,3-
interactions. The tram isomer has (ea) CH,’s. The axial CH, has two CHJH 1,3-interactions, accounting for 2(3.75) = 7.5 kJ/mol of instability. The cis isomer is more stable than the tram isomer by 7.5 kJ/mol. ( 6 ) See Fig. 9-13; (ee) is more stable than (aa) by 15.0 - 3.75 = 11.25 kJ/mol.
Problem 9.20 The energy difference between the cis and tram isomers of 1,1,3,5-tetramethylcyclohexanewas experimentally determined to be 15.5 kJ/mol. Compare this with the calculated value based on 1,3-diaxial interactions of 3.75 and 15.0kJ/mol for H/CH, and CH,/CH,, respectively. Consider only the more stable conformers. 4
See Fig. 9-14. Cis (ee) has 7.5 kJ/mol of strain energy from two CHJH 1,3-interactions. Tram (ea) has 22.5 kJ/mol, 15.0 kJ/mol from one CHJCH, and 7.5 kJ/mol from two CH,/H. The difference, 22.5 - 7.5 = 15.0 kJ/mol, compares favorably with the experimental value of 15.5 kJ/mol. Problem 9.21 Write the structure of the preferred conformation of ( a ) tram-1 -ethyl-3-isopropylcyclohexane, ( 6 ) cis-2-chloro-cis-4-chlorocyclohexyl chloride. 4
[CHAP. 9
CYCLIC HYDROCARBONS
178
H.
H'
no CH,/H 1,34nteractions;
1 CH,/CH, gauche interaction = 3.75 kJ/mol
axial CH,'s have no CH,/CH, gauche interactions; each axial CH, has two CHJH 1,3-interactions = 4 x 3.75 = 15.0 kJ/mol (aa) Conformation
(ee) Conformation Fig. 9-13
cis (ee)
tram (ea)
Mg. 9-14 ( a ) The rrans-1,3-isomer is (ea); the bulkier group, in this case i-propyl, is equatorial, and the smaller group,
in this case ethyl, is axial. See Fig. 9-15(a). (6) See Fig. 9-15(6).
H
c1 Fig. 9-15
Problem 9.22 You wish to determine the relative rates of reaction of an axial and an equatorial Br in an S,2 displacement. Can you compare ( a ) cis- and trans-1 -methyl-4-bromocyclohexane? (b) cis- and t r a w l -r-butyl-4bromocyclohexane? (c) cis-3,S-dimethyl-cis-1- bromocyclohexane and cis-33-dimethyl-tram-1- bromocyclohexane? 4
CHAP. 91
CYCLIC HYDROCARBONS
179
( a ) The tram substituents are (ee). The cis substituents are (ea). Although CH, is bulkier and has a greater (e)
preference than has Br, the difference in preference is small and an appreciable number of molecules exist with the Br (e) and the CH, (a). At no time are there conformers with Br only in an (a) position. These isomers, therefore, cannot be used for this purpose. (b) The bulky t-butyl group can only be (e). In practically all molecules of the cis isomer, Br is forced to be (a). All molecules of the tram isomer have an (e) Br. Because t-butyl “freezes” the conformation and prevents interconversion, these isomers can be used. ( c ) The cis-3,5-dimethyl groups are almost exclusively (ee) to avoid severe CHJCH, 1.3-interactions were they to be (aa). These cis-CH,’s freeze the conformation. When Br at C‘ is cis, it has an (e) position; when it is tram, it has an (a) position. These isomers can be used. Problem 9.23 The planar structure of cis-l,2-dimethylcyclohexane,which is meso, shows a plane of symmetry 4 [Problem 9.4(e)J. ( a ) Is the chair conformer achiral? (b) Why is this isomer optically inactive?
(a) No. ( 6 ) The two conformers formed by rapid interconversion are nonsuperimposable mirror images (Fig. 9-9). These conformational enantiomers comprise an optically inactive racemic form.
9.4
SYNTHESIS
INTRAMOLECULAR CYCLIZATION
This technique applies to many open-chain compounds, as discussed in later chapters. Pertinent here is the intramolecular cyclization of polyenes (an electrocyclic reaction). 4
2\,
1.3-Pentadiene (a diene)
1,3,5-Hexatriene (a triene)
3-Meth ylc yclobutene
I ,3-Cyclohexadiene
INTERMOLECULAR CYCLIZATION
In this method two, or occasionally more, open-chain compounds are merged into a ring. Examples are the common syntheses of cyclopropanes by the addition of carbene (CH,) or substituted carbenes to alkenes (Section 6.4). H H:C:
+ H3CCH=CH2 +
Carbene
Propylene
Me thylcyclopropane
Methylene can be transferred directly from the reagent mixture, CH212+ Zn-Cu alloy, to the alkene without being generated as an intermediate (Simmons-Smith reaction).
CH212 + CH3CH=CH2 Methylene diiodide
+ Zn12 Methylcyclopropane
(no insertion)
180
CYCLIC HYDROCARBONS
[CHAP. 9
Problem 9.24 The carbene :CCl, generated from chloroform, CHCl,, and KOH in the presence of alkenes gives substituted cyclopropanes. Write the equation for the reaction of :CCI, and propene. 4
[:CC121 + CH3CH=CH2
U
+ CH3
CI
CI
l.l-Dichloro-2methylcyclopropane
Dichlorocarbene [see Problem 7.56(c)]
Another class of intermolecular cyclizations are the Cycloaddition Reactions of Alkenes and Alkynes ( a ) [ 2 + 21. Ultraviolet-light-catalyzed dimerization of alkenes yields cyclobutanes in one step.
1.3-Butadiene
1,2-DivinylcycIobutane (one of several products)
(6) [ 2 + 4 ] (the Diels-Alder reaction; Section 8.7). A conjugated diene and an alkene form a cyclohexene. Reactive alkenes (dienophiles) have electron-attracting groups on their unsaturated C’s.
1.3-Butadiene
___*
Ethylene
1.3-Butadiene
+
/CH=CH \ CH2 /
heat
+
’
Cyclohexene
CJ H - L~ H
H:CJ
Acrylic aldehyde a dienophile
‘&,=CH+
&€I2
t
I
bat
5
CH = C H 3 /
H2C
‘
\6
\6H,-CH
‘CH=O
I /
CH2
‘CH=O 1,2,3,6-Tetrahydrobenzaldehyde
[ 2 + 2 + 2 + 21
(access to
4 H-H-
Cyclooctatetraene
Cyclohexanes may be formed by hydrogenating compounds with benzene rings, many of which are isolated from coal, as illustrated with toluene:
Toluene from cool tar
Methylcyclohexane
CHAP. 91
9.5
181
CYCLIC HYDROCARBONS
CHEMISTRY
The chemistry of cyclic hydrocarbons and their corresponding open-chain analogs is similar. Exceptions are the cyclopropanes, whose strained rings open easily, and the cyclobutanes, whose rings open with difficulty. The larger rings are stable [see Problem 9.5(c)]. Problem 9.25 Although cyclopropanes are less reactive than alkenes, they undergo similar addition reactions.
(a) Account for this by geometry and orbital overlap. (6) How does HBr addition to 1,l-dimethylcyclopropane resemble Markovnikov addition? 4
Because of the ring strain in cyclopropane (Problem 9.6), there is less orbital overlap (Fig. 9-1) and the sigma electrons are accessible to attack by electrophiles. The proton of HBr is attacked by an electron pair of a bent cyclopropane sigma bond to form a carbocation that adds Br- to give a 1,3-Markovnikov-additionproduct.
CH,
1,I-Dimethylcyclopropane
CH,
bl
2-B romo-2-met hy I butane 1,3-addition product
3" R'
Problem 9.26 Which conformation of 1,3-butadiene participates in the Diels-Alder reaction with, e.g., ethene? 4
The two conformations of 1,3-butadiene are s-cis (cisoid) and s-truns (transoid):
s-cis
s-tram
Although s-truns is the more favorable conformer, reaction occurs with s-cis because this conformation has its double bonds on the same side of the single bond connecting them; hence, the stable form of cyclohexene with a cis double bond is formed. Reaction of the s-tram conformer with ethene would give the impossibly strained tram-cyclohexene [Problem 9.12(u)]. As the s-cisconformer reacts, the equilibrium between the two conformers shifts toward the s-cis side, and in this way all the unreactive s-tram reverts to the reactive s-cis conformer. Problem 9.27 Outline a synthesis of the following alicyclic compounds from acyclic compounds. ( a)
1,l-Dimethylcyclopropane
(6) @ I = C H 2
CHJ,
5
( c ) Cyclooctane
4
CYCLIC HYDROCARBONS
182
[CHAP. 9
t{ / P d
NI(C")2
Acetylene 7 Cyclooctatetraene2Cyclooctane
(4
Problem 9.28 Starting with cyclopentanol, show the reactions and reagents needed to prepare (a) cyclopentene, (6) 3-bromocyclopentane, ( c ) 1,3-~yclopentadiene, (d) rrans-l,2-dibromocyclopentane,(e) cyclopentane. 4 Br
Cyclopentanol
C yclopentene
Problem 9.29 (a) Use the following observations to discuss stereospecificity of carbene addition. (6) Suggest mechanisms for addition of (i) singlet and (ii) triplet carbenes to alkenes. €4,c
H
'c=c
/
/CH3
\
+
JT
CH2
-+
H
H
cis-2-Butene
.'
(singlet)
cis- or trans-2-Butene
I
cis
+f
EH,
CH,
H
-+
?$ c--CH2
H
+ HK
(singlet)
trans
\
FH3
/
\
H
c-c
H
trans-2-Butene
mixture of cis- and trans-l,2-Dimethylcyclopropane
4
(triplet)
(a) Since cis reactant- cis product, and frans-, frans, the addition of singlet carbene is stereospecific. The addition of triplet carbene is nor stereospecific. ( b ) (i) The 7r electrons of the alkene and unshared pair of electrons of singlet CH, are properly paired to form simultaneously two sigma bonds. The transition state is
(ii) The electrons are not matched when triplet carbene adds. The sequence of steps is:
p
cis- or truns- CH ,-CH-C
1
/
H-CH,
\
H'k
-[
CH,-CH--TCH-CH3
i
H2Cf
t
[-I
electron 'pin inversion
CH,--CH--CH-CH,
I H,C1
f
Singlet state
Intermediate diradical
cis-
1-
trans1.2-Dimethylc yclopropane
CHAP. 91
183
CYCLIC HYDROCARBONS
Before the intermediate diradical can form the second bond, the spin of one electron must switch. This takes enough time to permit free rotation about the starred C--C bond, leading to a mixture of cis and rruns isomers.
Problem 9.30 Complete the following reactions :
( b ) ozonolysis of
(g)
,// + KMnO,
6
-
(c)
// +
/\.
(h)
H2
+ H2 KMnO,
__*
I20 YNI
’
NI
m-C3
4
CH3
Cycloalkenes behave chemically like alkenes.
C=O
(a Markovnikov addition)
(b)
u y z
( c ) CH,CH,CH,. Under these conditions the strained three-membered ring opens. (d) BrCH,CH,CH,Br. Again, the three.-membered ring opens. ( e ) CH,CH2CH,CH,. The strained four-membered ring opens, but a higher temperature is needed than in part (c). ( f )N o reaction. The five-membered ring has no ring strain. (g) NO reaction. Even the strained rings are stable toward oxidation. (h) TcooH, cyclopropanecarboxylic acid.
Problem 9.31 Cycloalkanes with more than six C’s are difficult to synthesize by intramolecular ring closures, yet they are stable. On the other hand, cyclopropanes are synthesized this way, yet they are the least stable 4 cycloalkanes. Are these facts incompatible? Explain. No. The relative ease of synthesis of cycloalkanes by intramolecular cyclization depends on both ring stability and the probability of bringing the two ends of the chain together to form a C-to-C bond, thereby closing the ring. This probability is greatest for smallest rings and decreases with increasing ring size. The interplay of ring stability and this probability factor are summarized below (numbers represent ring sizes).
Probability of ring closure Thermal stability
3 >4 > 5 > 6 > 7 > 8 > 9 6>7,5>8,9*4>3
Ease of synthesis
5 > 3 , 6 > 4 , 7, 8, 9
The high yield of cyclopropane indicates that a favorable probability factor outweighs the ring instability. For rings with more than six C’s the ring stability effect is outweighed by the highly unfavorable probability factor.
Problem 9.32 Account for the fact that intramolecular cyclizatiom to rings with more than six C‘s are effected 4 at extremely low concentrations (Ziegler method). Chains can also react intermolecularly to form longer chains. Although intramolecular reactions are ordinarily faster than intermolecular reactions, the opposite is true in the reaction of chains leading to rings with more than six C‘s. This side reaction from collisions between different chains is minimized by carrying o u t the reaction in extremely dilute solutions.
184
CYCLIC HYDROCARBONS
Very Dilute Solution
-AB
[CHAP. 9
Concentrated Solution
ACHz(CH2)n CH2B
-AB
ACHZ(CH2)n CH2-CH2(CH,)n CH2B Chain lengthening
Ring closure
9.6 MO THEORY OF PERICYCLIC REACTIONS
The formation of alicyclics by electrocyclic and cycloaddition reactions (Section 9.4) proceeds by one-step cyclic transition states having little or no ionic or free-radical character. Such pericyclic (ring closure) reactions are interpreted by the Woodward-Hoffmann rules; in the reactions, the new U bonds of the ring are formed from the “head-to-head” overlap of p orbitals of the unsaturated reactants. INTERMOLECULAR REACTIONS The rules state that: 1. Reaction occurs when the lowest unoccupied molecular orbital (LUMO) of one reactant overlaps with the highest occupied molecular orbital (HOMO) of the other reactant. If different molecules react, either can furnish the HOMO and the other the LUMO. 2. The reaction is possible only when the overlapping lobes of the p orbitals of the LUMO and the HOMO have the same sign (or shading). 3. Only the terminal p AO’s of the interacting molecular orbitals are considered, as it is their overlap that produces the two new (T bonds to close the ring. 1. Ethene Dimerization [2
+ 21 to Cyclobutane
The bracketed numbers indicate that the cycloaddition involves two species each having two n electrons. Without ultraviolet light we have the situation indicated in Fig. 9-16(a). Irradiation with uv causes a n+ n* transition (Fig. 8-3), and now the proper orbital symmetry for overlap prevails [Fig. 9-16(b)].
Fig. 9-16
CHAP. 91
2.
CYCLIC HYDROCARBONS
185
Diels-Alder Reaction [2 + 41
See Fig. 9-17.
Fig. 9-17
ELECTROCYCLIC (INTRAMOLECULAR) REACTIONS In electrocyclic reactions of conjugated polyenes, one double bond is lost and a single bond is formed between the terminal C's to give a ring. The reaction is reversible.
-
HCT~CH~
I? HC-CH2 a diene
HC-CHZ
II
HC-CH2
I
a cyclobutene
trans
heat
7
H cis,trans-2,4-Hexadiene
H
*
CH3
trans,trans isomer
cis -3,4-Dimethylcyclobutane
heat
T
CH3 trans isomer
To achieve this stereospecificity, both terminal C's rotate 90" in the same direction, called a conrotatory motion. Movement of these C's in opposite directions (one clockwise and one counterclockwise) is termed disrotatory. The Woodward-Hoffmann rule that permits the proper analysis of the stereochemistry is: The orbital symmetry of the HOMO must be considered, and rotation occurs to permit overlap of two like-signed lobes of the p orbitals to form the (T bond after rehybridization. The HOMO for the thermal reaction then requires a conrotatory motion [Fig. 9-18(a)]. Irradiation causes a disrotatory motion by exciting an electron from n2- T : , which now becomes the HOMO [Fig. 9-18(6)].
186
CYCLIC HYDROCARBONS
[CHAP. 9
Fig. 9-18
Problem 9.33 When applying the Woodward-Hoffmann rules to the Diels-Alder reaction, ( a ) would the same conclusion be drawn if the LUMO of the dienophile interacts with the HOMO of the diene? ( b )Would the 4 reaction be light-catalyzed? ( a ) Yes; see Fig. 9-19(a). (6) No; see Fig. 9-19(b).
Fig. 9-19
CHAP. 91
CYCLIC HYDROCARBONS
187
Problem 9.34 Use Woodward-Hoffmann rules to predict whether the following reaction would be expected to occur thermally or photochemically.
allyl carbanion
4
The MO energy levels of the allyl carbanion n system showing the distribution of the four v electrons (two from n double bond and two unshared) are indicated in Fig. 9-20. The 0 is used whenever a node point is at an atom. The allowed reaction occurs thermally as shown in Fig. 9-21(a). The photoreaction is forbidden [Fig. 9-21(b)] .
Fig. 9-20
Fig. 9-21
188
CYCLIC HYDROCARBONS
[CHAP. 9
9.7 TERPENES AND THE ISOPRENE RULE
The carbon skeleton C
I
c-c-c-c
of H~C=C-CH=CH~ Isoprene
is the structural unit of many naturally occurring compounds, among which are the terpenes, whose generic formula is (C,H,), . Problem 9.35 Pick out the isoprene units in the terpenes limonene, myrcene and a-phellandrene, and in vitamin A , shown below. 4
In the structures below, dashed lines separate the isoprene units.
Limonene
Myrcene
a-Phe1landrene
Supplementary Problems Problem 9.36 Draw formulas for ( a ) isopropylcyclopentane, bicyclo[4.4. llundecane, ( d ) tram-1-propyl-4-butylcyclohexane.
(6) c~-l,3-dimethylcyclooctane,(c)
4
CH,
FH,CH,CH,
CO
or CH,CH,CH,CH,
CHAP. 91
CYCLIC HYDROCARBONS
189
Problem 9.37 Name each of the following compounds and indicate which, if any, is chiral.
. 0 9 1 3
4
The rings are numbered as shown, starting at a bridgehead C and going around the larger ring first, in such a way that the first doubly-bonded C attains the lowest possible number. The name is bicyclo[4.3.0]non-7ene. The molecule is chiral; C' and C6 are chiral centers. ( 6 ) 2,3-Diethylcyclopentene.The molecule is chiral; C3 is a chiral center. (c) Bicyclo[4.4.2]dodecane. The molecule is achiral. (a)
Problem 9.38 Draw the structural formulas and give the stereochemical designation (meso, ruc, cis, tram, achiral) of all the isomers of trichlorcyclobutane. 4
@cl
$\
Cl
y$
Cl
cL2'
Cl
achiral
rac
H
H
CI
H/FH Cl
Cl
Cl
cis,cis,meso
H
CI
tram,rrans ,meso
H
c1
c1
H
tram,cis ,rac
Problem 9.39 Show steps in the synthesis of cyclohexane from phenol, C,H,OH.
OH
4
OH
Phenol
Cyclohexanol
Cyclohexene
Cyclohexane
Problem 9.40 Write structural formulas for the organic compounds designated by a ?. Indicate the stereo4 chemistry where necessary and account for the products. CH2 CHJ-CH
+ Br,
(CCI,)
-
?
+?
CYCLIC HYDROCARBONS
190
H CH3 / \c=C
/
+ CH,N,
\
H
(liquid phase) 5 ?
CH3
\ / /c=c
+
H
[CHAP. 9
‘ H
CH2N2
c
(in presence of an inert gas, argon)
Same as (d) but with added 0, /CH(CH+
CH
\iCH2-
H
+ CHCI,
CH
-
?
+
?
(CH
)
?
CO-K+
?
+?
?+?
\CH~---CH
FH2\ CH2 + Br,
C\H2 / CH3-C=CH
2-c
\
H2S04
CH2
heat
/
CCt4
’?
?
dil. aq. KMn04
*
?
‘CH,-CH~H ( a ) Two Br’s add to each of the two nonequivalent single bonds I and
9r
I1 of
a
c
b
a
to form two products, (2)-CH,Br-CH,--€HBr-CH, b
a
C
CH,Br-CH-CH,Br CH,-C H B r
by breaking (I) and by breaking (11)
LH,
by radical substitution; unlike the three-membered ring, the four-membered ring is stable. 1 ( 6 ) LH,-CH, (c) In the liquid phase we get singlet CH, which adds cis; tram-2-butene forms trans-l,2-dimethylcyclopropane:
(d) Some initially formed singlet CH, collides with inert-gas molecules and changes to triplet
which adds nonstereospecifically. cis-2-Butene yields a mixture of cis- and trans-l,2-dimethylcyclopropane:
CHAP. 91
CYCLIC HYDROCARBONS
191
0, is a diradical which combines with triplet carbenes, leaving the singlet species to react with cis-2-butene to give cis-l,2-dimethylcyclopropane. ( f ) Dichlorocarbene adds cis to C=C, but either cis or tram to the Me.
(e)
c H ando c L l ( g ) CHCIBr, loses Br-, the better leaving group, rather than CI- to give CIBrC:, which then adds so that either CI or Br can be cis to CH,.
Br CH,
HJC
Br
truns addition; 2 enantiomers ( r u c )
HO
OH
Cyclohexanol is dehydrated to cyclohexene, which forms a meso glycol by cis addition of two OH’S. Problem 9.41 Explain why (a) a carbene is formed by dehydrohalogenation of CHCl, but not from methyl, exist in chair conformations, but ethyl, or n-propyl chlorides; (b) cis-l,3-and truns-l,4-di-tert-butylcyclohexane their geometric isomers, truns-1,3-and cis-1,4-, do not. 4 ( a ) Carbene is formed from CHCI, because the three strongly electronegative Cl’s make this compound
sufficiently acidic to have its proton abstracted by a base. CH,CI has only one CI and is considerably less acidic. Carbene formation is an a-elimination of HC1 from the same C; it does not occur with ethyl or propyl chlorides because protons are more readily eliminated from the p C’s to form alkenes. (6) Both cis-1,3 and frans-1,4 compounds exist in the chair form because of the stability of their (ee) conformers. Truns-1,3-and cis-1,4-are (ea). Ad axial t-butyl group is very unstable, so that a twist-boat with a quasi-(ee) conformation (Fig. 9-22) is more stable than the chair.
Twist-boat Twisting reduces eclipsed and “flagpole” interactions
Fig. 9-22
192
[CHAP. 9
CYCLIC HYDROCARBONS
Problem 9.42 Assign structures or configurations for A through D. ( a ) Two isomers, A and B, with formula C,H,, differ in that one adds 1 mol and the other 2 mol of H,. Ozonolysis of A gives only one product, O=CH(CH,),CH=O, while the same reaction with 1 mol of B produces 2 mol of CH,=O and 1 mol of O=CH(CH,),CH--V. (6) Two stereoisomers, C and D, of 3,4-dibromocyclopentane-l, 1-dicarboxylic acid undergo decarboxylation as shown: CHBr-CH2\
I
C
CHBr-CH2/
-
/COOH \COOH
COT +
CHBr-CH2\
1
CHBr-CH2/
C
/H ‘COOH
a monocarboxylic acid
a gem-dicarboxylic acid
C gives one, while D yields two, monocarboxylic acids.
4
Both compounds have two degrees of unsaturation (see Problem 6.41). B absorbs two moles of H, and has two multiple bonds. A absorbs one mole of H, and has a ring and a double bond; it is a cycloalkene. As a cycloalkene, A can form only a single product, a dicarbonyl compound, on ozonolysis.
1 c?;H2-clHl
FHEO
CH2 \ CH2-CH2
Octane- 1 &dial
CH2 \
,CH2
CH2-CH2
Cyclooctene (A)
Cyclooctane
Since one molecule of B gives three carbonyl molecules, it must be a diene and not an alkyne.
H,C=O
+ O=HC(CH,),CH=O
+ O=CH,
2 H2C=CH(CH2),CH=CH2
2H
2CH,CH,(CH,),CH,CH,
1,7-0ctadiene (B)
1 ,dHexanedial
Octane
The Br’s of the dicarboxylic acid may be cis or trans. Decarboxylation of the cis isomer yields two isomeric products in which both Br’s are cis (E) or frans (F) with respect to COOH. The cis isomer is D. In the monocarboxylic acid G formed from C (trans isomer), one Br is cis and the other trans with respect to COOH and there is only one isomer. COOH
COOH
Q
c -‘% -
H
COOH
___* -CO2
Br
H
Br
CJ o+c) COOH
Br
(0
Br
Br
Br
Br
Br
(D) cis
trans
Problem 9.43 Outline the reactions and reagents needed to synthesize the following from any acyclic (b) compounds having up to four C’s and any needed inorganic reagents: ( a ) cis-1-methyl-2-ethylcyclopropane; 4 trans- 1,l-dichloro-2-ethyl-3-n-propylcyclopropane; (c) 4-cyanocyclohexene; (d) bromocyclobutane. ( a ) The cis-disubstituted cyclopropane is prepared by stereospecific additions of singlet carbene to cis-2-
pentene.
CH,\
CH2CH, \
,c=c
H
/
‘H
+ CH,N2
5
6-
+
N2
CHAP. 91
193
CYCLIC HYDROCARBONS
The alkene, having five C’s, is best formed from 1-butyne, a compound with four C’s.
NaNH2
H-C-CCH,CH,
CH I
Na+:C=CCH,CH,
H INi
CH,-C=CCH,CH,
2
/c=c\ H
H
CiS
addition
( b ) Add dichlorocarbene to truns-3-heptene, which is formed from 1-butyne. CH
C2HS
H /
2 5 \
+ CHCl, + (CH,),CO-
-
H~c=c\H,CH,CH,
CH,CH,C=CH
NaNH
I
CH,CH,C=C:Na+
+ C-
-
H
I
AV!H,CH~CH, Cl2
n-CjH7Br
CH,CH, Na
CH,CH,C-CCH,CH,CH,
addition tram
\
H /
H/c=c\cH,CH,CH,
(c) Cyclohexenes are best made by Diels-Alder reactions. The CN group is strongly electron-withdrawing and
when attached to the C = C engenders a good dienophile.
-
,CHZ
fH2 CH FNHC
HC
I
HC b 1 2
I ,3-Butadiene
+
II
CH*
Acrylonitrile
[
II
HC\
‘CHCN
I
(a Diels-Alder reaction)
/CHZ CH2
*m-. 1-1 Br
(electrocyclic reaction)
Cyclobutene
Problem 9.44 Write planar structures for the cyclic derivatives formed in the following reactions, and give their stereochemical labels.
3-Cyclohexenol+ dil. aq. KMnO,+
-
3-Cyclohexenol+ HC0,H and then H,O+ (+)- trans-l,2-Dibromocyclopropane + Br,
(4 (4
meso-cis-1,2-Dibromocyclopropane
1-Methylcyclohexene + HBr (peroxide)
(4 HO OH
meso
HO
rac
H
meso
+ Br,
light
light
(an anti addition)
HO OH
ra c
194
CYCLIC HYDROCARBONS
Br
H
Br
Br
H
H
H
Optically active
cis,frans Optically inactive (meso)
Br
Br
racemate
Br
[CHAP. 9
Br
Br
Br
I H
Br
I Br
all cis cis, rrans-meso no chiral c’s
(Note that in absence of chiral catalysts an optically inactive reactant gives only optically inactive products).
H,C Br cis,ruc (eryfhro)
Problem 9.45 Use cyclohexanol and any inorganic reagents to synthesize (a) frans-l,2-dibromocyclohexane, ( 6 ) c~-l,2-dibromocyclohexane, (c) trans-l,2-cyclohexanediol. 4
OH Cyclohexanol
Sr H fruns-1,2-Dibromocyclohexane
C yclohexene
(b) See Problem 9.44(e).
Br H
Br Br 1 -Bromocyclohexene
cis- 1,2-Dibromocyclohexane
HO H C yclohexene
tram- 1,2-Cyclohexanediol
Problem 9.46 For the two alkylated cyclohexanes
(a) Draw the possible chair conformations; (b) explain which is the more stable conformer.
Properly to assign (e) and (a) conformations, consider pairs of groups, e.g., 1’2-ci.s is (ea).
CHAP. 91
CYCLIC HYDROCARBONS
195
(a) As shown in Fig. 9-23(a), conformers I and I1 of compound A each have two (a) and two (e) CH,’s.
However, the two (a) CH,’s of I are on the same side of the ring and have a severe 1,3-CH3/CH, interaction. Conformer I1 is more stable because its two (a) CH,’s are on opposite sides of the ring and do not cause as much steric strain. In compound B the steric strain from two (a) CH,’s and an (e) r-butyl of conformer 111 is less than that (6) from one (a) t-butyl and two (e) CH,’s of IV, IV is less stable.
(6 ) Fig. 9-23
Problem 9.47 Account for the apparent anomalies in the following substances whose diaxial conformations are more stable than the diequatorial: ( a ) trans-l,2-Dibromocyclohexanein nonpolar solvents exists 50% in the (aa) conformation, but in polar solvents the (ee) form is favored. (b) cis-1,3-Cyclohexanediol in CCl, is shown by infrared measurement to exist as the (aa) conformer. 4 (a) The two equatorial Br’s introduce some dipole-dipole repulsion, tending to destabilize the (ee) conformer.
However, polar solvent molecules surround the Br’s and minimize the dipole interactions, and the sterically favored (ee) conformer now predominates. See Fig. 9-24(a). (6) H-bonding stabilizes the two (aa) OH’S and more than compensates for the destabilizing 1’3-interactions. See Fig. 9-24(6).
(6- close)
(6- far apart) (a 1
Fig. 9-24
196
CYCLIC HYDROCARBONS
[CHAP. 9
Problem 9.48 Decalin, CloH18,has cis and trans isomers that differ in the configurations about the two shared C's as shown below. Draw their conformational structural formulas.
0 H
cis
0 H
tram
4
For each ring the other ring can be viewed as 1,2-substituents. For the trans isomer, only the rigid (ee) conformation is possible structurally. As shown in Fig. 9-25, diaxial bonds point 180" away from each other and cannot be bridged by only four C's to complete the second ring. Cis fusion is (ea) and the bonds can be twisted to reverse the (a) and (e) positions, yielding conformation enantiomers.
Problem 9.49 Explain the following facts in terms of the structure of cyclopropane. ( a ) The H's of cyclopropane are more acidic than those of propane. (6) The Cl of chlorocyclopropane is less reactive toward 4 SN2 and SN1 displacements than the C1 in CH,CHClCH,. ( a ) The external C-H bonds of cyclopropane have more s character than those of an alkane (Fig. 9-1). The more s character in the C-H bond, the more acidic the H. (6) The C - C I bond of chlorocyclopropane also has more s character, which diminishes the reactivity of the C1. Remember that vinyl chlorides are inert in SN2and SN1 reactions. The R' formed during the S,1 reaction would have very high energy, since the C would have to use sp2 hybrid orbitals needing a bond angle of 120". The angle strain of the R' is much more severe (120"-60") than in cyclopropane itself (109"-60").
Problem 9.50 Use quantitative and qualitative tests to distinguish between ( a ) cyclohexane, cyclohexene and 4 1,3-~yclohexadiene;(6) cyclopropane and propene. ( a ) Cyclohexane does not decolorize Br, in CCl,. The uptake of H,, measured quantitatively, is 2mol for 1 mol of the diene, but 1 mol for 1 mol of the cycloalkene. (6) Cyclopropane resembles alkenes and alkynes, and differs from other cycloalkanes in decolorizing Br, slowly, adding H,, and reacting readily with H,SO,. However, it is like other cycloalkanes and differs from multiple-bonded compounds in not decolorizing aqueous KMnO, .
Problem 9.51 Draw a conclusion about the stereochemistry of the Diels-Alder reaction from Fig. 9-26.
4
The stereochemistry of the dienophile is preserved in the product, the substituted cyclohexene. This fact is consistent with a one-step concerted mechanism.
CHAP. 91
197
CYCLIC HYDROCARBONS
trans
trans
cis
cis
Fig. 9-26
Problem 9.52 ( a ) Give the structure of the major product, A , whose formula is CsH8, resulting from the dehydration of cyclobutylmethanol. On hydrogenation, A yields cyclopentane. (b) Give a mechanism for this 4 reaction. ( a ) Compound A is cyclopentene, which gives cyclopentane on hydrogenation.
+
The side of a ring migrates, thereby converting a RCH, having a strained four-membered ring to a much more stable R,CH+ with a strain-free 5-membered ring.
Chapter 10 Benzene and Polynuclear Aromatic Compounds 10.1 INTRODUCTION Benzene, C,H,, is the prototype of aromatic compounds, which are unsaturated compounds showing a low degree of reactivity. The Kekule structure (1865) for benzene,
L
H
has only one monosubstituted product (C,H,Y) since all six H's are equivalent. There are three disubstituted benzenes-the 1,2-, 1,3-, and 1,4-position isomers-designated as ortho, meta, and para, respectively. X
X
H 1.2- or orfho
H (0-)
1.3- or meta ( m - )
X
Y I .4- or par0 ( p -)
Ploblem 10.1 Benzene is a planar molecule with bond angles of 120". All six C-to-C bonds have the identical length, 0.139 nm. Is benzene the same as 1,3,5-cyclohexatriene? 4
No. The bond lengths in 1,3,5-cyclohexatrienewould alternate between 0.153 nm for the single bond and 0.132 nm for the double bond. The C-to-C bonds in benzene are intermediate between single and double bonds.
Problem 10.2 ( a ) How do the following heats of hydrogenation ( A H h , kJImo1) show that benzene is not the ordinary triene 1,3,5-~yclohexatriene?Cyclohexene, - 119.7; 1,4-~yclohexadiene,-239.3; 1,3-~yclohexadiene, -231.8; and benzene, -208.4. (6) Calculate the delocalization energy of benzene. (c) How does the delocalization energy of benzene compare to that of 1,3,5-hexatriene ( A H , = -336.8 kJ/mol)? Draw a 4 conclusion about the relative reactivities of the two compounds.
In computing the first column of Table 10-1, we assume that in the absence of any orbital interactions each double bond should contribute -119.7 kJlmol to the total A H , of the compound, since this is the A H , of an isolated C=C(in cyclohexene). Any difference between such a calculated A H , value and the observed value is the delocalization energy. Since h H h for 1,4-~yclohexadieneis 7.5 kJ/mol less than that for 1,3-~yclohexadiene, conjugation stabilizes the 1,3-isomer, [Remember that the smaller (more negative) the energy, the more stable the structure.] 1,3,5-Cyclohexatriene should behave as a typical triene and have A H , = -359.1 kJ/mol. The observed A H , for benzene is -208.4 kJ/mol. Benzene is nor 1,3,5-cyclohexatriene;in fact, the latter does not exist. (6) See Table 10-1, (a)
198
199
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
CHAP. 10)
Table 10-1 ~~~
Observed AHh, kJ/mol
Calculated AHhC
kJ/mol
I
Cyclohexene
I
I ,4-Cyclohexadiene
Cyclohexane
0 0
I
+
I ,3-Cyclohexadiene
+
'E'
- 119.7
I
I
+ 2H,
3H,
__*
0I 0
Delocalization Energy
0.0
I 2( - 119.7) = -239.4
-231.8
-7.6
3( - 119.7) = -359.1
-208.4
- 150.7
3( - 119.7) = -359.1
-336.8
-22.3
I
Benzene
H
+3H,+C,H,,
H - H H H
cis-l,3,5-Hexatriene
(c)
The delocalization energy of benzene (-150.7 kJ/mol) is much smaller than that of 1,3,5-hexatriene (-22.3 kJ/ mol). Three conjugated double bonds engender a large negative delocalization energy only when they are in a ring. Since the ground-state enthalpy of benzene is much smaller in absolute value than that of the triene, the AH' for addition of H, to benzene is much greater, and benzene reacts much slower. Benzene is less reactive than open-chain trienes towards all electrophilic addition reactions.
Problem 10.3 (a) Use the A Hh's for complete hydrogenation of cyclohexene, 1,3-~yclohexadieneand benzene, as given in Problem 10.2, to calculate AHh for the addition of 1 mole of H, to (i) 1,3-cyclohexadiene, (ii) benzene. (6) What conclusion can you draw from these values about the rate of adding 1 mol of H, to these three compounds? (The A H of a reaction step is not necessarily related to A H S of the step. However, in the cases being considered in this problem, AHreactionis directly related to AH'.) ( c ) Can cyclohexadiene and cyclohexene be isolated on controlled hydrogenation of benzene? 4
Equations are written for the reactions so that their algebraic sum gives the desired reactant, products and enthalpy. ( a ) (i) Add reactions (1) and (2):
A H = +119.7 kJ/mol Cyclohexane
Cyclohexene
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
200
(2)
0 0 0 + 2H2
1.3-Cyclohexadiene
(1)+(2)=(3)
AH,
= -231.8
AH,
=
[CHAP. 10
__*
Cyclohexane
+ H,
-112.1 kJ/mol
Cyclohexene
1.3-C yclohexadiene
Note that reaction (1) is a dehydrogenation (reverse of hydrogenation) and that its A H is positive. (ii) Add the following two reactions:
0 -0 - 2H2
Cyclohexane
1.3-Cyclohexadiene
Benzene
Cyc lohexane
0
+
Benzene
H
2
+
A H = +231.8 kJ/mol
AH,
=
+23.4 kJ/mol
1.3-C yclohexadiene
( b ) The reaction with the largest negative A H , value is the most exothermic and, in this case, also has the fastest rate. The ease of addition of 1 mol of H, is:
*
cyclohexene (- 119.7) > 1,3-cyclohexadiene(- 112.1) benzene ( +23.4) ( c ) No. When one molecule of benzene is converted to the diene, the diene is reduced all the way to cyclohexane by two more molecules of H, before more molecules of benzene react. If lmol each of benzene and H, are reacted, the product is 5 mol of cyclohexane and 3 mol of unreacted benzene. Problem 10.4 The observed heat of combustion ( A H , ) of C,H, is -3301.6kJ/mol.* Theoretical values are calculated for C,H, by adding the contributions from each bond obtained experimentally from other compounds; these are (in kJ/mol) - 492.4 for C=C, -206.3 for C--C and -225.9 for C-H. Use these data to calculate the heat of combustion for C,H, and the difference between this and the experimental value. Compare 4 the difference with that from heats of hydrogenation.
The contribution is calculated for each bond and these are totaled for the molecule. Six C-H bonds = 6( -225.9) = - 1355.4kJ /mol Three C--C bonds = 3(-206.3) = - 618.9 Three C==C bonds = 3( -492.4) = - 1477.2 TOTAL = -3451.5 (calculated A H , for C,H,) Experimental = -3301.6 DIFFERENCE = - 149.9 kJ/mol This difference is the delocalization energy of C,H,; essentially the same value is obtained from AH, (Table 10-1).
*Some books define heat of combustion as - A H , and values are given as positive numbers.
CHAP. 101
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
201
Problem 10.5 How is the structure of benzene explained by ( a ) resonance, (6) the orbital picture, (c) molecular orbital theory? 4 (a)
Benzene is a hybrid of two equal-energy (Kekule) structures differing only in the location of the double bonds:
( b ) Each C is sp2 hybridized and is U bonded to two other C’s and one H (Fig. 10-1). These U bonds comprise the skeleton of the molecule. Each C also has one electron in a p orbital at right angles to the plane of the ring. These p orbitals overlap equally with each of the two adjacent p orbitals to form a T system parallel to and above and below the plane of the ring (Fig. 10-2). The six p electrons in the T system are associated with all six C’s. They are therefore more delocalized and this accounts for the great stability and large resonance energy of aromatic rings.
Fig. 10-1
Fig. 10-2
(c) The six p AO’s discussed in part (6) interact to form six T MO’s. These are indicated in Fig. 10-3, which gives the signs of the upper lobes (cf. Fig. 8-3 for butadiene). Since benzene is cyclic, the stationary waves representing the electron clouds are cyclic and have nodal planes, shown as lines, instead of nodal points. See Problem 9.34 for the significance of a 0 sign. The six p electrons fill the three bonding MO’s, thereby accounting for the stability of C,H,.
The unusual benzene properties collectively known as aromatic character are:
1. Thermal stability. 2. Substitution rather than addition reactions with polar reagents such as HNO,, H,SO, and Br,. In these reactions the aromatic unsaturated ring is preserved.
202
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
+ +
+ +
[CHAP. 10
+ + Fig. 10-3
3. Resistance to oxidation by aq. KMnO,, HNO, and all but the most vigorous oxidants. 4. Unique nuclear magnetic resonance spectra. (See Table 12-4.) 10.2 AROMATICITY AND HUCKELS RULE
Huckel’s rule (1931) for planar species states that if the number of 7~ electrons is equal to 2 + 4n, where n equals zero or a whole number, the species is aromatic. The rule was first applied to carbon-containing monocyclics in which each C is capable of being sp2-hybridized to provide a p orbital for extended 7~ bonding; it has been extended to unsaturated heterocyclic compounds and fused-ring compounds. Note that benzene corresponds to n = 1. Problem 10.6 Account for aromaticity observed in: (a) 1,3-~yclopentadienylanion but not 1,3-cyclopentadiene; (6) 1,3,5-~ycloheptatrienylcation but not 1,3,5-~ycloheptatriene;(c) cyclopropenyl cation; (d) the 4 heterocycles pyrrole, furan and pyridine. (a)
1,3-CycIopentadiene has an sp3-hybridized C, making cyclic p orbital overlap impossible. Removal of H’ from this C leaves a carbanion whose C is now sp2-hybridized and has a p orbital capable of overlapping to give a cyclic 7r system. The four 7r electrons from 2 double bonds plus the two unshared electrons total six 7r electrons; the anion is aromatic (n = 1).
cyclopentadienyl anion ( b ) Although the triene has six p electrons in three C = C bonds, the lone sp3-hybridized C prevents cyclic overlap of p orbitals.
cycloheptatrienyl cation
CHAP. 101
(c)
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
203
Generation of a carbocation by ionization permits cyclic overlap of p orbitals on each C. With six electrons, the cation is aromatic ( n = 1). Cyclopropenyl cation has two 7r electrons and n = 0.
I
H
7r
a salt
The ions in parts ( a ) , (b), and (c) are reactive but they are much more stable than the corresponding open-chain ions. (d) Huckel’s rule is extended to heterocyclic compounds as follows:
Pyrrole (6 n electrons; 2 unshared electrons on N overlap in the n system)
Furan (6 n electrons; only 2 electrons on 0 participate in the 7r system)
Pyridine electrons; the unshared pair of electrons on N does not participate in the n overlap) (6
7r
Note that dipoles are generated in pyrrole and furan because of delocalization of electrons from the he teroatoms.
10.3 ANTlAROMATlClTY
Planar cyclic conjugated species less stable than corresponding acyclic unsaturated species are called antiaromatic. They have 4n v electrons. 1,3-Cyclobutadiene (n = l), for which one can write two equivalent contributing structures, is an extremely unstable antiaromatic molecule. This shows that the ability to write equivalent contributing structures is not sufficient to predict stability.
Problem 10.7 Cyclooctatetraene (C,H,), unlike benzene, is not aromatic; it decolorizes both dil. aq. KMnO, and Br, in CCl,. Its experimentally determined heat of combustion is -4581 kJ/mol. ( a ) Use the Hiickel rule to account for the differences in chemical properties of C,H, from those of benzene. (b) Use thermochemical data D f Problem 10.4 to calculate the resonance energy. (c) Why is this compound not antiaromatic? (d) Styrene, C,H,CH=CH,, with heat of combustion -4393 kJ/mol, is an isomer of cyclooctatetraene. Is styrene aromatic? 4
C,H, has eight rather than six p electrons. Since it is not aromatic it undergoes addition reactions. The calculated heat of combustion is: 8 C-H bonds = 8(-225.9) = -1807 kJ/mol 4 C - C bonds = 4(-206.3) = -825 4 C=C bonds = 4( -492.4) = - 1970 TOTAL -4602 kJ/mol
The difference -4602 - (-4581) = -21 kJ/mol shows small (negative) resonance energy and no aromaticity. Although the molecule has (n = 2) 7r electrons it is not antiaromatic, because it is not planar. It exists chiefly in a “tub” conformation (Fig. 10-4).
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
204
[CHAP. 10
Fig. 10-4
Styrene is aromatic; its delocalization energy is -4602 - (-4393) = -209 kJ/mol. This is attributable to the presence of the benzene ring. Styrene has a more negative energy than does benzene [- 150 kJ/rnol] because its ring is conjugated to the C=C bond, thereby extending the delocalization of the electron cloud. Problem 10.8 Deduce the structure and account for the stability of the following substances which are insoluble in nonpolar but soluble in polar solvents. ( a ) A red compound formed by reaction of 2 mol, of AgBF, with 1 mol of 1,2,3,4-tetrapheny1-3,4-dibromocyclobut-l-ene. (b) A stable compound from the reaction of 2 mol 4 of K with 1 mol of 1,3,5,7-cyclooctatetraenewith no liberation of H,.
The solubility properties suggest that these compounds are salts. The stability of the organic ions formed indicates that they conform to the Huckel rule and are aromatic.
ph-m:l r]=c
(a) Two Brf’s are abstracted by two Ag”s to form two AgBr and a tetraphenylcyclobutenyl dication.
Ph
+ 2AgBF,
Br
Br
Ph
Ph
ph)@(]
2BFi
Ph
+ 2AgBr
Ph
an aromatic cation ( n
=
0)
(6) Since K . is a strong reductant and no H, is evolved, two K’s supply two electrons to form a cyclooctatetraenyl dianion (Fig. 10-5). This planar conjugated unsaturated monocycle has 10 electrons, conforms to the Huckel rule (n = 2) and is aromatic.
2K*
+
2K’
(G)
Fig. 10-5 Problem 10.9 Sondheimer synthesized a series of interesting conjugated cyclopolyalkenes that he designated the [n]-annulenes,where n is the number of C’s in the ring.
[ 14)-Annulene
[ 18)-Annulene
Account for his observations that (a) [ 181-annulene is somewhat aromatic, [16]- and (201-annulene are not; (6) [ 181-annulene is more stable than [ 141-annulene. 4
CHAP. 101
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
205
+ 2 (n = 4) v electrons; there are 4n 7~ electrons in the nonaromatic, nonplanar [ 161- and [20]-annulenes. (6) [14]-Annulene is somewhat strained because the H’s in the center of the ring are crowded. This steric strain prevents a planar conformation, which diminishes aromaticity. ( a ) The somewhat aromatic [18]-annulene has 4n
Problem 10.10 Use the Hiickel rule to indicate whether the following planar species are aromatic or antiaromatic: H
H
H
H
H 4
( a ) Aromatic. There are 2 7r electrons from each C = C and 2 from an electron pair on S to make an aromatic sextet. (6) Antiaromatic. There are 4n ( n = 2) 7r electrons. (c) Aromatic. There are 6 v electrons. (d) Aromatic. There are 10 7r electrons and this anion conforms to the ( 4 n + 2) rule (n = 2). (e) Antiaromatic. The cation has 4n (n = 2) 7r electrons. ( f ) and (g) Antiaromatic. They have 4n ( n = 1) 7~ electrons.
Problem 10.11 The relative energies of the MO’s of conjugated cyclic polyenes can be determined by the following simple polygon rule instead of using nodal planes as in Problem 10.5(c). Inscribe a regular polygon in a circle, with one vertex at the bottom of the circle and with the total number of vertices equal to the number of MO’s. Then the height of a vertex is proportional to the energy of the associated MO. Vertices below the horizontal diameter are bonding v , those above are antibonding T * , and those on the diameter are nonbonding 7“. Apply the method to 3-, 4-, 5, 6-, 7-, and 8-carbon systems and indicate the character of the MO’s. 4 See Fig. 10-6.
Fig. 10-6
206
10.4
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
[CHAP. 10
POLYNUCLEAR AROMATIC COMPOUNDS
Most of these compounds have fused benzene rings. The prototype is naphthalene, C,,H,. a
a
a a Naphthalene
Although the Huckel 4n + 2 rule is rigorously derived for monocyclic systems, it is also applied in an approximate way to fused-ring compounds. Since two fused rings must share a pair of 7~ electrons, the aromaticity and the delocalization energy per ring is less than that of benzene itself. Decreased aromaticity of polynuclear aromatics is also revealed by the different C-C bond lengths. Problem 10.12 Draw a conclusion about the stability and aromaticity of naphthalene from the fact that the experimentally determined heat of combustion is 255 kJ/mol smaller in absolute value than that calculated from the structural formula. 4
The difference, -255 kJlmo1, is naphthalene’s resonance energy. Naphthalene is less aromatic than benzene because a per-ring resonance energy of $(-255) = -127.5 kJ/mol is smaller in absolute value than that of benzene (- 150 kJ / mol) . Problem 10.13 Deduce an orbital picture (like Fig. 10-2) for naphthalene, a planar molecule with bond angles of 120”. 4
See Fig. 10-7. The C’s use sp2 hybrid atomic orbitals to form U bonds with each other and with the H’s. The remaining p orbitals at right angles to the plane of the C’s overlap laterally to form a T electron cloud.
Fig. 10-7
Probtem 10.14 ( a ) Draw three resonance structures for naphthalene. ( b ) Which structure makes the major contribution to the structure of the hybrid in that it has the smallest energy? ( c ) There are four kinds of C-to-C bonds in naphthalene: C’-C2, C2-C3, C’-Cy, and CY-C”. Select the shortest bond and account for your choice. 4
CHAP. 101
207
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
I
I1
I11
(6) Structure I has the smallest energy because only it has two intact benzene rings. (c) The bond, in all of its positions, that most often has double-bond character in the three resonance structures is the shortest. This is true for C1-C2(also for C3-C4,C5-C6,and C7-c8).
Anthracene and phenanthrene are isomers (C,,H
'm: 8
9
1
5
10
4
lo)
having three, fused benzene rings; 4
10
6
Ant hracene
7 Phenanthrene
As the number of fused rings increases, the delocalization energy per ring continues to decrease in absolute value, and compounds become more reactive, especially toward addition. The delocalization energies per ring for anthracene and phenanthrene are - 117.2 and -126.8 kJ/mol, respectively.
10.5 NOMENCLATURE
Some common names of benzene derivatives are toluene (C,H,CH,), xylene (C,H,(CH,),), phenol (C,H,OH), aniline (C,H,NH,), benzaldehyde (C,H,CHO), benzoic acid (C,H,COOH), benzenesulfonic acid (C,H5S0,H), styrene (C6H,CH=CH2), mesitylene ( 1,3,5-(CH,),C,H,), and anisole (C,H,OCH,). Derived names combine the name of the substituent as a prefix with the word benzene. Examples are nitrobenzene (C,H,NO,), ethylbenzene (C,H,CH,CH,) and fluorobenzene (C,H,F). Some aryl (Ar-) groups are: C6H5- (phenyl), C6H5-C6H4- (biphenyl), p-CH,C6H4(p-tolyl) and (CH3),C6H3- (xylyl). Some arylalkyl groups are: C6H5CH2- (benzyl), C,H, H( b e n d ) , C , H , e (benZO), (C,H,),CH(benzhydryl), (C6H5),C- (trityl). The order of decreasing priorities of common substituents is: COOH, SO,H, CHO, CN, -0, OH, NH,, R, NO,, X. For disubstituted benzenes with a group giving the ring a common name, 0-, p-, or rn- is used to designate the position of the second group. Otherwise positions of groups are designated by the lowest combination of numbers.
7
Problem 10.15 Name the compounds: YOOH
H3C\
FH7
CH I
208
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
[CHAP. 10
( a ) p-Aminobenzoic acid. (6) m-Nitrobenzenesulfonic acid. (c) m-Isopropylphenol. (d) 2-Bromo-3-nitro-5hydroxybenzoic acid. (Named as a benzoic acid rather than a phenol because COOH has priority over OH.) (e) 3,4'-DichlorobiphenyI; the notational system in biphenyl is: m
o
0'
m'
m
o
0'
m'
Problem 10.16 Give the structural formulas for ( a ) 2,4,6-tribromoaniline, ( b ) m-toluenesulfonic acid, (c) 4 p-bromobenzalbromide, (d) di-o-tolylmethane, (e) trityl chloride.
Br
Br
CH,
H iC
Problem 10.17 Give the structures of the isomeric alkylbenzenes C,H,, that can have one, two, or three 4 isomeric monoiodo derivatives.
Six C's are needed €or the benzene ring, which leaves two C's attached to the ring either as two CH,'s or as CH,CH,. There are four C,H,, alkylbenzenes, as shown in Table 10-2. They are written with equivalent positions o n the ring designated by the letters a , b, etc. Table 10-2
Isomer
I
b
Number of monoiodo derivatives
1
11 p-Xylene
o-Xylene
rn-Xylene
2
I
3
/
Ethylbenzene
)
Problem 10.18 Write structures for (a) P-bromonaphthalene, (6) a-naphthol, (c) 1,5-dinitronaphthalene, ( d ) P-naphthoic acid. 4
NO*
CHAP. 101
209
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
Problem 10.19 Name the following compounds:
4
(a) 1-naphthalenesulfonic acid or a -naphthalenesulfonic acid, (6) 1-naphthaldehyde or a -naphthaldehyde, (c) 8-bromo-I-methoxynaphthalene.
Problem 10.20 Indicate the number of isomers of ( a ) monochloronaphthalene, (6) dichloronaphthalene. ( a ) 2 ( a - and
4
p - ) . (6) 10(1,2-; 1,3-; 1,4-; 1,5-; 1,6-; 1,7-; 1,8-; 2,3-; 2,6-; and 2,7-).
10.6 CHEMICAL REACTIONS The unusual stability of the benzene ring dominates the chemical reactions of benzene and naphthalene. Both compounds resist addition reactions which lead to destruction of the aromatic ring. Rather, they undergo substitution reactions, discussed in detail in Chapter 11, in which a group or atom replaces an H from the ring, thereby preserving the stable aromatic ring. Atoms or groups other than H may also be replaced.
REDUCTION Benzene is resistant to catalytic hydrogenation (high temperatures and high pressures of H, are needed) and to reduction with Na in alcohol. With Na in refluxing ethanol, naphthalene gains two H's giving the unconjugated 1, 4- dihydronaphthalene. In higher-boiling alcohols two more H's are gained, at C' and C4,to give the tetrahydro derivative (tetralin). Anthracene and phenanthrene each pick up H's at C9 and C f o ,and acquire no more. Problem 10.21 (a) Write equations for the reductions of (i) naphthalene, (ii) anthracene and (iii) phenanthrene. ( b ) Explain why naphthalene is reduced more easily than benzene. ( c ) Explain why anthracene and 4 phenanthrene react at the C9-C" double bond and go no further.
H
Naphthalene
\
1,2,3,4-Tetrahydronaphthalene (Tetralin)
& 0 0 =& o0
9,lO-Dihydroanthracene
Anthracene
H H
(iii) 1,4-Dihydronaphthalene
H
Na
Phenanthrene
9,lO-Dihydrophenanthrene
(6) Each ring of naphthalene is less aromatic than the ring of benzene and therefore is more reactive. (c) These reactions leave two benzene rings having a combined resonance energy of 2(-150) = -300 kJ/mol. Were attack to occur in an end ring, a naphthalene derivative having a resonance energy of -255 kJ/mol would remain. Two phenyls are less energetic (more stable) than one naphthyl. Problem 10.22 In the Birch reduction benzene is reduced with an active metal (Na or Li) in alcohol and liquid NH,(-33 "C)to a cyclohexadiene that gives only OCHCH,CHO on ozonolysis. What is the reduction product?
4
210
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
[CHAP. 10
Since the diene gives only a single product on ozonolysis, it must be symmetrical. The reduction product is 1,4-~yclohexadiene. H
I
/
C
I
\
0
+
0 %-H ‘CH
/
o=c’
1,4-Cyclohexadiene
I
H Malonic aldehyde
Problem 10.23 Typical of mechanisms for reductions with active metals in protic solvents, two electrons are transferred from the metal atoms to the substrate to give the most stable dicarbanion, which then accepts two H”s from the protic solvent molecules to give the product. (U) Give the structural formula for the dicarbanion 4 formed from C6H6 and (b) explain why it is preferentially formed.
( 6 ) The repulsion between the negative charges results in their maximum separation (paru orientation). Product formation is controlled by the stability of the intermediate dicarbanion and not by the stability of the product. In this case, the more stable product would be the conjugated 1,3-cyclohexadiene.
OXIDATION Benzene is very stable to oxidation except under very vigorous conditions. In fact, when an alkyibenzene is oxidized, the alkyl group is oxidized to a COOH group, while the benzene ring remains intact. For this reaction to proceed there must be at least one H atom on the C attached to the ring.
C,H,CHRR--+
C,H,COOH
(poor yields)
However, in naphthalene one benzene ring undergoes oxidation under mild conditions; e.g., Cr20:in H’. Under more vigorous conditions a benzene dicarboxylic acid is formed. 0
0 Naphthalene
1.4-Naphthoquinone
Phthalic
acid
Problem 10.24 Oxidation of 1-nitronaphthalene yields 3-nitrophthalic acid. However, if 1-nitronaphthalene is reduced to a-naphthylamine and if this amine is oxidized, the product is phthalic acid.
CHAP. 101
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
1-Nitronaphthalene
1
oxid.
21 1
1
a-Naphthylamine oxid
NO*
I
HOOC' 3-Nitrophthalic acid
Phthalic acid
How do these reactions establish the gross structure of naphthalene?
4
The electron-attracting -NO, stabilizes ring A of I-nitronaphthalene to oxidation, and ring B is oxidized to form 3-nitrophthalic acid. By orbital overlap, -NH2 releases electron density, making ring A more susceptible to oxidation, and a-naphthylamine is oxidized to phthalic acid. The NO, labels one ring and establishes the presence of two fused benzene rings in naphthalene. Problem 10.25 Like naphthalene, anthracene and phenanthrene are readily oxidized to a quinone. Suggest 4 the products and account for your choice.
0
0 9,lO- Anthraquinone
9,10-Phenanthraquinone
Oxidation at C' and C"' leaves two stable intact benzene rings. (See Problem 10.21(c).)
HALOGENATION Under typical polar conditions, benzene and naphthalene undergo halogen substitution and not addition. However, in the presence of uv light, benzene adds C1, to give 1,2,3,4,5,6-hexachlorocyclohexane, an insecticide. Problem 10.26 Unlike benzene and naphthalene, anthracene and phenanthrene readily add one mole of Br,. 4 Give the products and explain your choice.
Anthracene undergoes 1,4-addition and phenanthrene undergoes 1,2-addition, and in each case the 9,lO-dibromo derivative is formed. Each product has two fully aromatic benzene rings-this could not be said of any other dibromo product. H
Br
9,lo-Dibrorno-9,lO-dihydroanthracene (mixture of cis and !runs)
9,10-Dibromo-9,lO-dihydrophenanthrene (mixture of cis and tram)
[CHAP. 10
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
212
10.7 SYNTHESIS
Benzene, naphthalene, toluene, and the xylenes are naturally occurring compounds obtained from coal tar. Industrial synthetic methods, called catalytic reforming, utilize alkanes and cycloalkanes isolated from petroleum. Thus, cyclohexane is dehydrogenated (aromatization), and n-hexane(cyc1ization) and met hylcyclopentane(isomerization) are converted to benzene. Aromatization is the reverse of catalytic hydrogenation and, in the laboratory, the same catalysts-Pt, Pd, and Ni-can be used. The stability of the aromatic ring favors dehydrogenation.
acooH
Problem 10.27 Use the Diels-Alder reaction to synthesize benzoic acid, C,H,COOH.
NCH2 CH I + CH-COOH II CH
h,
CH*
-
/
CH
CH II CH
\
2\
CH-COOH I CH2
/
CH2
4
+
2H2
S and Se can be used in place of Pt, and H,S and H,Se are then the respective products.
Supplementary Problems Problem 10.28 (a) Draw two Kekule structures for 1,2-dimethyIbenzene(o-xylene). ( b ) Why are these structures not isomers? What are they? ( c ) Give the carbonyl products formed on ozonolysis. 4
( b ) These structures differ only in the position of the structures, not isomers. (c)
7~
electrons and therefore are contributing (resonance)
Although neither contributing structure exists, the isolated ozonolysis products arising from the resonance hybrid are those expected from either one: CH,C-CCH,
11 II
0 0
HC-CH
II I1
0 0
CH,C--CH
II 11
0 0
Problem 10.29 What are the necessary conditions for (a) aromaticity and (6) antiaromaticity?
4
(1) A planar cyclic molecule or ion. (2) Each atom in the ring must have a p AO. (3) These p AO’s must be parallel, so that they can overlap side-by-side. (4) The overlapping 7~ system must have (4n + 2) n electrons (Huckel). ( 6 ) In ( a ) change (4n + 2) to 4n. (a)
Problem 10.30 Design a table showing the structure, number of n electrons, energy levels of 7~ MO’s and electron distribution, and state of aromaticity of: (a) cyclopropenyl cation, ( b ) cyclopropenyl anion, (c) cyclobutadiene, (d) cyclobutadienyl dication, (e) cyclopentadienyl anion, ( f) cyclopentadienyl cation, ( g ) 4 benzene, ( h ) cycloheptatrienyl anion, (i) cyclooctatetraene, ( j ) cyclooctatetraenyl dianion.
See Table 10-3. (H’s are understood to be attached to each doubly bonded C.)
CHAP. 101
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
Table 10-3
Structure
Number of tr Electrons
7
tr
MO's
trf-
-IT;
-T)-.rrI
H
4
I I
flls*
Aromatic
Antiaromatic
+lr'
#n
Aromaticity
I
-tr'
4
tr:*
*tr; +T
Antiaromatic
I
2
Q Q
flf-
6
-T'
+tr2
IT&
Aromatic
#TI
H
-tr;
7T;-
4
v3+
+tr2
Antiaromatic
+ h l
H
Q
H
8
trt-
-tr;
-IT:
trt*
*tr:
7r3+
+7r2
*v
Nonaromatict (nonplanar)
I
10
tIf it were planar it would be antiaromatic; to avoid this, (i) is nonplanar.
213
214
BENZENE AND POLYNUCLEAR AROMATIC COMPOUNDS
Problem 10.31 Explain aromaticity and antiaromaticity in terms of the MO’s of Problem 10.30.
[CHAP. 10
4
Aromaticity is observed when all bonding MO’s are filled and nonbonding MO’s, if present, are empty or completely filled. Huckel’s rule arises from this requirement. A species is antiaromatic if it has electrons in antibonding MO’s or if it has half-filled bonding or nonbonding MO’s, provided it is planar. Problem 10.32 The deep-blue compound azulene (C,,H,) has five- and seven-membered rings fused through two adjacent C’s. It is aromatic and has a significant dipole moment of 1.0D. Explain. 4 Azulene can be written as fused cyclopentadiene and cycloheptatriene rings, neither of which alone is aromatic. However, some of its resonance structures have a fused cyclopentadienyl anion and cycloheptatrienyl cation, which accounts for its aromaticity and its dipole moment of 1.0 D.
Azulene is an example of a nonbenzenoid polynuclear aromatic compound. Problem 10.33 Name the monobromo derivatives of ( a ) anthracene, (6) phenanthrene. ( a ) There are 3 isomers: 1-bromo, 2-bromo-, and 9-bromoanthracene. (6) There are 5 isomers: 1-bromo-, 2-bromo-, 3-bromo-, 4-bromo-, and 9-bromophenanthrene.
Problem 10.3 What is the Diels-Alder addition product of anthracene and ethene?
4
Reaction occurs at the (most reactive) C9 and C’’ positions.
Problem 10.35 Which isomer of the arene C,,H,, resists vigorous oxidation to an aryl carboxylic acid? 4 For a side chain R to undergo oxidation to COOH, it must have at least one benzylic H. To resist oxidation, the benzylic C must be 3”. The arene is C,H,C(CH,),.
Chapter 11 Aromatic Substitution. Arenes 11.1 AROMATIC SUBSTITUTION BY ELECTROPHILES (LEWIS ACIDS, E' OR E)
MECHANISM SD3
Benzene
Lewis acid
c
benzenonium ion (strong Bronsted acid)
substituted benzene
or Step (1) is reminiscent of electrophilic addition to an alkene. Aromatic substitution differs in that the intermediate carbocation (a benzenonium ion) loses a cation (most often H') to give the substitution product, rather than adding a nucleophile to give the addition product. The benzenonium ion is a specific example of an arenonium ion, formed by electrophilic attack on an arene (Section 11.4). It is also called a sigma complex, because it arises by formation of a U bond between E and the ring. See Fig. 11-1 for a typical enthalpy-reaction curve for the nitration of an arene. Problem 11.1 Account for the relative stability of the benzenonium ion by charge delocalization.
(U)
resonance theory and (6) 4
contributing structures
Note that + is at C's ortho and paru to sp3 hybridized C, which is the one bonded to E'. The benzenonium ion is a type of allylic cation [see Problem 8.28(6)]. The five remaining C's using sp'-hybridized orbitals each have a p orbital capable of overlapping laterally to give a delocalized T structure, or U complex:
delocalized (hybrid) structure
The S + indicates positions where
+ charge exists.
Problem 11.2 For each electrophilic aromatic substitution in Table 11-1, give equations for formation of E' and indicate what is B - or B (several bases may be involved). In reaction (c) the electrophile is a molecule, E. 4
X, H,SO,
+ FeX,+X'(E') + FeXa(B-) (forms HX + FeX,) + + HONO,-HSO;(B-) + H,ONO,+H,O + NOl(E') 215
AROMATIC SUBSTITUTION. ARENES
216
(4
+ HSO,(B-) + SO,(E)
2H,SO,+H,O' C6H6 + so3
RX
+ AIX, --+ ROH + HF
-A=LRCOCI
-
+
/
-
H
C,Hs,
C,H,S03H
SO; (B-)
+ AIX; (B-) (forms HAIX,) R+ (E') + H,O (B) + F- (B-)
R' (E')
__*
+ H,PO,
+ AICI,
[CHAP. 11
-6-AH
+
RC=O:
L
(E')
+ H,PO; (B-)
+
t--*
RC=O: , (E+) + AlClI (B-1 .
I
oxycarbonium ion
Table 11-1
I
Reaction
FeX, (from Fe
(6) Nitration ( c ) Sulfonation
( d ) Friedel-Crafts alkylation
I
1
I
Catalyst
Reagent
+ X,)
X'
H2S04
ArNO,
+NO,
H$O4 or H,S,O,
none
ArS0,H
so3
RX, ArCH,X ROH
AlCI,
HF, H2S0,, or BF,
I
RCOCl
I R'
I
I
AICI,
Ar-R, Ar-CH,Ar Ar-R Ar-CHCH,
I
acylation
E'orE
ArCl, ArBr
H3PO4 or H F
(e) Friedel-Crafts
I
Product
R
I +
0
RC = O
Ar-4-R
Problem 11.3 How does the absence of a primary isotope effect prove experimentally that the first step in 4 aromatic electrophilic substitution is rate-determining? A C-H bond is broken faster than is a C-D bond. This rate difference (isotope effect, k H l k D )is observed only if the C-H (or C-D) bond is broken in the rate-determining step. If no difference is observed, as is the case for most aromatic electrophilic substitutions, C-H bond-breaking must occur in a fast step (in this case the second step). Therefore, the first step, involving no C-H bond-breaking, is rate-determining. This slow step requires the loss of aromaticity, the fast second step restores the aromaticity. Problem 11.4 How is E' generated, and what is the base, in the following reactions? (a) Nitration of reactive aromatics with HNO, alone. (6) Chlorination with HOCI using HCI as catalyst. ( c ) Nitrosation (introduction of 4 a NO group) of reactive aromatics with HONO in strong acid. (d) Deuteration with DCI.
(a)
HNO, + H-0-NO,
__*
" 1-
NO; + H-?-NO, unstable
i
H,O (Base) + NO, (E+) Nitronium ion
H
(61
H'
+ H-0-CI
--+
I
H-T-CI
+ H,O
(Base) + CI' (E+)
CHAP. 111
(c)
217
AROMATIC SUBSTITUTION. ARENES
H-0-N-0
+ H'
( d ) D', transferred by DCI to benzene.
base,
-
H
I+
H-0-N=O
__+
H,O (Base)
+ NO'
(E')
Nitrosonium ion
Base is CI-.
acid,
acid,
basez
Problem 11.5 Since the initial step of aromatic electrophilic substitution is identical with that of alkene addition, explain why ( a ) aromatic substitution is slower than alkene addition; ( b ) catalysts are needed for aromatic substitution; (c) the intermediate carbocation eliminates a proton instead of adding a nucleophile.
4
The intermediate benzenonium ion is less stable than benzene; hence its formation has a high AHS and the reaction is slowed. Loss of aromaticity is energetically more unfavorable than loss of T bond. (b) The catalysts are acids which polarize the reagent and make it more electrophilic. (c) The addition reaction would be endothermic and would produce the less stable cyclohexadiene. Loss of a proton, on the other hand, produces a stable aromatic ring. (a)
Problem 11.6 Sulfonation resembles nitration and halogenation in being an electrophilic substitution, but differs in being reversible and in having a moderate primary kinetic isotope effect. Illustrate with diagrams of enthalpy (If) versus reaction coordinate. 4
In nitration (and other irreversible electrophilic substitutions) the transition state (TS) for the reaction wherein
loses H' has a considerably smaller A H S than does the TS for the reaction in which NO; is lost. In sulfonation the A H S for loss of SO, from
[ir'
]
\so;
is only slightly more than that for loss of H'.
H
H Nitration
Sulfonation
L
H' + ArNO,
ArH
Reaction Fig. 11-1
+ SO,
\
H' + ArSO;
Reaction
218
AROMATIC SUBSTITUTION. ARENES
[CHAP. 11
In terms of the specific rate constants
k, is about equal to k - , . (For nitration, k , 9 k - .) Therefore, in sulfonation the intermediate can go almost equally well in either direction, and sulfonation is reversible. Furthermore, since the rate of step (2) affects the overall rate, the substitution of D for H decreases the rate because AHS for loss of D' from
+P
Ar
\SO; is greater than AH' for loss of H' from the protonated intermediate. Hence, there is a modest primary isotope effect.
Groups other than H + can be displaced during electrophilic aromatic attack. The acid-catalyzed reversal of sulfonation (desulfonation) exemplifies such a reaction; here, H + displaces SO,H as SO, and H'. ArS0,H
+H20.
5 0 % aq H,SO,,
150°C
fum. H,SO,
ArH + H,SO,
Problem 11.7 Use the principle of microscopic reversibility (Problem 6.26) to write a mechanism for 4 desulfonation. ArSO,
so;.
+ H t= = = A r c
ArH
+ SO,
H
ORIENTATION AND ACTNATION OF SUBSTITUENTS The 5 ring H's of monosubstituted benzenes, C,H5G, are not equally reactive. Introduction of E into C,H,G rarely gives the statistical distribution of 40% ortho, 40% meta, and 20% para disubstituted benzenes. The ring substituent(s) determine(s) (a) the orientation of E (meta or a mixture of ortho and paru) and (6) the reactivity of the ring toward substitution. Problem 11.8 (a) Give the delocalized structure (Problem 11.1) for the 3 benzenonium ions resulting from the common ground state for electrophilic substitution, C,H,G + E'. (b) Give resonance structures for the para-benzenonium ion when G is OH. (c) Which ions have G attached to a positively charged C? (d) If the products from this reaction are usually determined by rate control (Section 8 . 5 ) , how can the Hammond principle be used to predict the relative yields of op (i.e., the mixture of ortho and paru) as against rn (rneta) products? (e) In terms of electronic effects, what kind of G is a (i) op-director, (ii) m-director? ( f ) Classify G in 4 terms of its structure and its electronic effect. G
Q
6+
H ortho
E puru
meta
(9
219
AROMATIC SUBSTITUTION. ARENES
CHAP. 111
Q- (j- (j +Q
H
E
(major contributor because all atoms obey octet rule)
+--)
H
E
H
E
H
E
The ortho and para. This is why G is either an op- or an rn-director. Because of kinetic control, the intermediate with the lowest-enthalpy transition state (TS) is formed in the greatest amount. Since this step is endothermic, the Hammond principle says that the intermediate resembles the TS. We then evaluate the relative energies of the intermediates (op vs. m) and predict that the one with the lowest enthalpy has the lowest AHS and is formed in the greatest yield. (i) An electron-donating G can better stabilize the intermediate when it is attached directly to positively charged (op) C’s. Such G’s are op-directing. (ii) An electron-withdrawing G destabilizes the ion to a greater extent when attached directly to positively charged (op) C’s. They destabilize less when attached mefa and are thus m-directors. Electron-donating (op-directors): (i) Those that have an unshared pair of electrons on the atom bonded to the ring, which can be delocalized to the ring by extended T bonding.
Other examples are U,-X: (halogen) and -%. (ii) Those with an attached atom participating in an electron-rich 7r bond, e.g.
-LEA-
c&-
Ar-
(iii) Those without an unshared pair, which are electron-donating by induction or by hypercoqjugation (absence of bond resonance), e.g., alkyl groups.
H H-L H+
hyperconjugated structure
Electron-withdrawing (rn-directors): The attached atom has no unshared pair of electrons and has some positive charge, e.g. +
-NR3
0 -N
+//
\o-
0-
I
-S C OH
b-
X6+a+
-c+x6t Xb-
8-
08-
Ila+ -c-
a+
-C-N
6-
a+//
0
-C
\OH
X = F, Cl
Problem 11.9 Explain: ( a ) All m-directors are deactivating. (6) Most op-directing substituents make the ring more reactive than benzene itself-they are activating. (c) As exceptions, the halogens are op-directors but are deactivating. 4 ( a ) All m-directors are electron-attracting and destabilize the incipient benzenonium ion in the
TS. They therefore diminish the rate of reaction as compared to the rate of reaction of benzene. (6) Most op-directors are, on balance, electron-donating. They stabilize the incipient benzenonium ion in the TS, thereby increasing the rate of reaction as compared to the rate of reaction of benzene. For example, the ability of the --OH group to donate electrons by extended p orbital overlap (resonance) far outweighs the ability of the OH group to withdraw electrons by its inductive effect.
220
(c)
AROMATIC SUBSTITUTION. ARENES
[CHAP. 11
In the halogens, unlike the OH group, the electron-withdrawing inductive effect predominates and consequently the halogens are deactivating. The 0-,p-, and rn-benzenonium ions each have a higher A H S than does the cation from benzene itself. However, on demand, the halogens contribute electron density by extended 7r bonding.
'0%: (showing delocalization of
+ to X)
H
and thereby lower the A H S of the ortho and para intermediates but not the meta cation. Hence the halogens are op-directors, but deactivating. Problem 11.10 Compare the activating effects of the following op-directors: ( a ) -OH,
-0;
and -0C-CH,
( b ) -NH2
II
and -NH-C-CH,
I1
0
0 Explain your order.
4
( a ) The order of activation is -0- > - O H > - O C O C H , . The -0-,with a full negative charge, is best able to donate electrons, thereby giving the very stable uncharged intermediate
s+
s-
In -OCOCH, the C of the C - 0 group has + charge and makes demands on the for electron density, thereby diminishing the ability of this U to donate electrons to the benzenonium ion. (b) The order is -NH, >-NHCOCH, for the same reason that OH is a better activator than --OCOCH,.
Table 11-2 extends the results of Problem 11.10. Table 11-2
rn-Directors
op-Directors weakly deactivating
activating
-F:
-R
C6Hs-
I
-C=L-
--cl:
--er:
-I:
deactivating
..
-N=o:
..
-NRf -NO2
-CN
-SO,H
-€OH
It
0
--CCl,
11
0
- C H -CR
It
0
It
0
Problem 11.11 Account for the following. ( a ) Nitration of PhC(CH,), gives only 16% ortho product, whereas PhCH, gives 50%. ( b ) Of all the aryl halides, PhF gives the least amount of ortho product on nitration. 4 ( a ) Steric hindrance, which is more pronounced with the bulky - C ( C H , ) , group, inhibits formation of the ortho isomer, thereby increasing the yield of the para isomer. ( b ) Even though the X can donate electrons to
CHAP. 111
AROMATIC SUBSTITUTION. ARENES
22 1
stabilize the op-intermediates, the electron-withdrawing inductive effect is substantial. The inductive effect is strongest at the closer ortho position and weakest at the more distant para position. Since F has the greatest inductive effect among halogens, PhF has the smallest percentage of ortho substitution. Problem 11.12 ( a ) Draw enthalpy-reaction diagrams for the first step of electrophilic attack on benzene, toluene (meta and para) and nitrobenzene (meta and para). Assume all ground states have the same energy. (6) 4 Where would the para and meta substitution curves for C,H,CI lie on this diagram? Since CH, is an activating group, the intermediates and TS's from PhCH, have less enthalpy than those from benzene. The para intermediate has less enthalpy than the meta intermediate. The TS and intermediates for PhNO, have higher enthalpies than those for C,H,, with the meta at a lower enthalpy than the para. See Fig. 11-2. They would both lie between those for benzene and p-nitrobenzene, with the para lower than the meta.
Reaction Progress Fig. 11-2
Problem 11.13 ( a ) Explain in terms of the reactivity-selectivity principle (Section 4.4) the following yields of meta substitution observed with toluene: Br, in CH,COOH, 0.5%; HNO, in CH,COOH, 3.5%; CH,CH,Br in GaBr,, 21%. (6) In terms of kinetic vs. thermodynamic control, explain the following effect of temperature on isomer distribution in sulfonation of toluene: at O"C, 43% 0-and 53% p-; at lOO"C, 13% U - and 79% p-.
4
( a ) The most reactive electrophile is least selective and gives the most meta isomer. The order of reactivity is:
CH,CH; >NO; >Br,(Br')
(6) Sulfonation is one of the few reversible electrophilic substitutions and therefore kinetic and thermodynamic products can result. At 100 "C the thermodynamic product predominates; this is the para isomer. The ortho isomer is somewhat more favored by kinetic control at 0°C. Problem 11.14 PhNO,, but not C,H,, is used as a solvent for the Friedel-Crafts alkylation of PhBr. Explain.
4
AROMATIC SUBSTITUTION. ARENES
222
[CHAP. 11
C,H, is more reactive than PhBr and would preferentially undergo alkylation. -NO, deactivating that PhNO, does not undergo Friedel-Crafts alkylation or acylations.
is so strongly
Problem 11.15 Account for the percentages of m-orientation in the following compounds: ( a ) C,H,CH, (4.4%), C,H,CH,Cl (15.5%), C,H,CHCl, (33.8%), C,H,CCl, (64.6%); (6) C,H,N'(CH,), (loo%), C,H,CH,N'(CH,), (88%), C,H,(CH,),N'(CH,), (19%). 4 Substitution of the CH, H's by Cl's causes a change from electron-release (+CH3) to electron-attraction (+ CCI,) and rn-orientation increases. "Me, has a strong electron-attracting inductive effect and is rn-orienting. When CH, groups are placed between this N' and the ring, this inductive effect falls off rapidly, as does the rn-orientation. When two CH,'s intercede, the electron-releasing effect of the CH, bonded directly to the ring prevails, and chiefly op-orientation is observed. Problem 11.16 Predict and explain the reaction, if any, of ( a ) phenol (PhOH), (b) PhH and ( c ) benzenesul4 fonic acid with D,SO, in D,O. ( a ) D,SO, transfers D', an electrophile, to form 2,4,6-trideuterophenol. Reaction is rapid because of the activating - O H group. The metu positions are deactivated. (b) PhH reacts slowly to give hexadeuterobenzene. ( c ) The sulfonic acid does not react, because -SO,H is too deactivating.
RULES FOR PREDICTING ORIENTATION IN DISUBSTITUTED BENZENES
1. If the groups reinforce each other, the orientation can be inferred from either group. 2. If an op-director and an m-director are not reinforcing, the op-director controls the
orientation. (The incoming group goes mainly ortho to the m-director). 3. A strongly activating group, competing with a weakly activating group, controls the orientation. 4. When two weakly activating or deactivating groups or two strongly activating or deactivating groups compete, substantial amounts of both isomers are obtained; there is little preference. 5 . Very little substitution occurs in the sterically hindered position between meta substituents. 6. Very little substitution occurs ortho to a bulky op-directing group such as t-butyl. Problem 11.17 Indicate by an arrow the position(s) most likely to undergo electrophilic substitution in each of the following compounds. List the number of the above rule(s) used in making your prediction. ( a ) rn-xylene, (6) p-nitrotoluene, ( c ) rn-chloronitrobenzene, (d) p-methoxytoluene, (e) p-chlorotoluene, ( f) rn-nitrotoluene, (g) o-methylphenol (o-cresol). 4
mdjor
(Rule 1)
(Rule 5)
OCH3(moderate)
CI (weak)
CH3 (weak)
CH, (weak)
(Rule 3)
(Rule 4)
(Rule 1)
(Rule 2)
OH (strong)
(Rule 2)
(Rule 3)
CHAP. 111
223
AROMATIC SUBSTITUTION. ARENES
ELXCTROPHILIC SUBSTITUTION OF NAPHTHALENE 1. Electrophilic substitution of naphthalene occurs preferentially at a position. 2. Examples of P-substitution are: ( a ) sulfonation at high temperatures (at low temperatures, a-substitution occurs); ( b ) acylation with RCOCl and AlCl,, in C,H5N0, as solvent (in CS2 or CH,ClCH,Cl, a -substitution occurs). 3. Substitution occurs in a ring holding an activating (electron-releasing) group: ( a ) para to an a-substituent; ( b ) ortho to an a-substituent i f the para position is blocked; (c) to a ortho position if the activating group is a P-substituent. 4. A deactivating group (electron-withdrawing) directs electrophiles into the other ring, usually at a positions. Problem 11.18 Account for (a) formation of the a-isomer in nitration and halogenation of naphthalene, (6) 4 formation of a -naphthalenesulfonic acid at 80 "C and p-naphthalenesulfonic acid at 160 "C. The mechanism of electrophilic substitution is the same as that for benzene. Attack at the a position has a lower AH' because intermediate I, an allylic R' with an intact benzene ring, is more stable than intermediate I1 from P-attack.
+ #
$
11, not an allylic R' @-substitution
I, an allylic R' a-substitution
In I1 the + charge is isolated from the remaining double bond and hence there is no direct delocalization of charge to the double bond without involvement of the stable benzene ring. In both I and I1 the remaining aromatic ring has the same effect on stabilizing the + charge. Since I is more stable than 11, a-substitution predominates. a-Naphthalenesulfonic acid is the kinetic-controlled product [see part ( a ) ] . However, sulfonation is a reversible reaction and at 160 "C the thermodynamic-controlled product, P-naphthalenesulfonic acid, is formed.
Problem 11.19 Name the product and account for the orientation in the following electrophilic substitution reactions: (a)
(6) 2-Ethylnaphthalene + Cl,, Fe
1-Methylnaphthalene + Br,, Fe
(c) 2-Ethylnaphthalene + C,H,COCI (e) 2-Methoxynaphthalene + HNO,
+ AICI,
+ H,SO,
(d)
1-Methylnaphthalene + CH,COCI, AICI,(CS,)
(f)
2-Nitronaphthalene + Br,, Fe
4
(a) 1-Methyl-4-bromonaphthalene. Br substitutes in the more reactive a position of the activated ring. (6) 1-Chloro-2-ethylnaphthalene.C' (ortho and a) is activated by C,H, since C', which is also a, is meta to C,H,. (c) 1-(2-EthyInaphthyl) ethyl ketone. Same reason as in (6). (d) 4-( 1-Methylnaphthyl) methyl ketone. Same reason as in (a). (e) 1-Nitro-2-methoxynaphthalene. Same reason as in (6). ( f ) 1 -Bromo-6-nitronaphthalene and 1-bromo-7-nitronaphthalene. NO, deactivates its ring and bromination occurs at the a positions of the other ring, more at C5that bears no delocalized 6 + .
Problem 11.20 (a) Why cannot 2-naphthalenecarboxylic acid be formed by oxidation of 2-methylnaphthalene? (b) How is 2-naphthalenecarboxylic acid prepared? 4 (a) Oxidation is an electrophilic attack and CH, is activating. The CH,-substituted benzene ring is more susceptible to oxidation than is the side chain CH,. The product is a quinone:
AROMATIC SUBSTITUTION. ARENES
224
[CHAP. 11
0
0
11.2 ELECTROPHILIC SUBSTITUTIONS IN SYNTHESES OF BENZENE DERIVATIVES
ORDER OF INTRODUCING GROUPS
To d o such syntheses it is essential to introduce the substituents in the proper order, which is based on the knowledge of the orientation and activation of both ring and incoming substituents. Problem 11.21 From C,H, (PhH) or PhCH, synthesize: ( a ) p-CIC,H,N02, ( 6 ) m-ClC,H,NO,, O,NC,H,COOH, (d) m-O,NC,H,COOH.
(c) p4
In the synthesis of disubstituted benzenes, the first substituent present determines the position of the incoming second. Therefore the order of introducing substituents must be carefully planned to yield the desired isomer. ( a ) Since the two substituents are para, it is necessary to introduce the op-directing Cl first. Fe
HN03
C' 2
HzSOq
P h H d PhC1-
p-CIC,H,NO,
( 6 ) Since the substituents are rnefa, the rn-directing NOz is introduced first. HNO,
P h H x PhN02
C'2
m-CIC,H,NO,
(c) The COOH group is formed by oxidation of CH,. Since p-O,NC,H,COOH has two rn-directing groups, the NO, must be added while the op-directing CH, is still present.
PhCH,
€lNOj
p-CH,C,H,NO,
+ o-CH,C,H,NO,
The para isomer is usually easily separated from the op mixture. p-O,NC,H,CH
KMnOJ
p-O,NC,H,COOH
-
(use phase transfer catalysts, Prob. 7.30.)
( d ) Now the substituents are mefa, and NOz is introduced when the rn-directing COOH is present PhCH,
KMnO, H'
PhCOOH-
HNO, ti2SO4
rn-O,NC,H,COOH
USE OF BLOCKING GROUPS If, in most cases, electrophilic substitution of a phenyl derivative with an op-directing group yields mainly the para isomer, how are good yields of the ortho isomer obtained? Answer: First introduce an easily removed blocking group into thepara position; then introduce the ortho substituent; and, finally, remove the blocking group. Two good blocking groups are -SO,H and -C(CH,),. Problem 11.22 Show steps in the synthesis of ( a ) o-chlorotoluene and (6) 1,3-dimethyl-2-ethylbenzene. 4
CHAP. 111
AROMATIC SUBSTITUTION. ARENES
225
( a ) In this synthesis -SO,H
is the blocking group. In the second step the op-directing CH, and the m-directing SO,H reinforce each other.
( b ) In this synthesis --C(CH,), is the blocking group. In the second step, although C,H, and C(CH,), are competing op-directors, the bulkiness of the latter group inhibits attack ortho to itself.
In the reaction with HF, the electrophile H' replaces C(CH,)f, which forms (CH,),C=CH,.
11.3 NUCLEOPHILIC AROMATIC SUBSTITUTIONS
ADDITION-ELIMINATION REACTIONS Nucleophilic aromatic substitutions of H are rare. The intermediate benzenanion in aromatic nucleophilic substitution is analogous to the intermediate benzenonium ion in aromatic electrophilic substitution, except that negative charge is dispersed to the oppositions.
:Nu-
+ ArH --+
H
-/
Oxidants such as 0, and K,Fe(CN), facilitate the second step, which may be rate-controlling, by oxidizing the ejected :H-, a powerful base and a very poor leaving group, to H,O. Electron-withdrawing groups positioned at the oppositions bearing the negative charges greatly stabilize the formation of the intermediate benzenanion. Thus, groups (such as NO,, CN, and halogen) which deactivate the ring toward electrophilic attack, encourage nucleophilic attack. These groups are op-directors toward nucleophilic aromatic substitution. Problem 11.23 Account for the product in the reaction
226
AROMATIC SUBSTITUTION. ARENES
[CHAP. 11
CN- is a nucleophile. N0,’s activate the ring toward nucleophilic substitution at op-positions by withdrawing the electron density and placing charge on the 0 ’ s of NO,:
When not sterically hindered, the orrho position may be more reactive. CN- is a “thin” nucleophile and its insertion ortho to each NO, is not hindered. A good leaving group, such as halide ion ( X - ) , is more easily displaced than H - from a benzene ring by nucleophiles. Electron-attracting substituents, such as NO2 and CN, in orrho and paru positions facilitate the nucleophilic displacement of X of aryl halides. The greater the number of such ortho and puru substituents, the more rapid the reaction and the less vigorous the conditions needed.
a c l + : N u -
-
addition usually slow
elimination usually fast
Problem 11.24 Write resonance structures to account for activation in addition-elimination aromatic nucleophilic substitution from delocalization of the charge of the intermediate carbanion by the following pura 4 substituent groups: (a) -NO,, (b) - - C N , ( c ) -N=O, ( d ) CH=O.
Only the resonance structures with the negative charge on the puru C are written to show delocalization of charge from ring C to the paru substituent.
GU
X
-
N:
CHAP. 111
AROMATIC SUBSTITUTION. ARENES
227
Problem 11.25 Compare addition-elimination aromatic nucleophilic and electrophilic substitution reactions with aliphatic S,2 reactions in terms of (a) number of steps and transition states, (6) character of intermediates. 4
(a) Nucleophilic and electrophilic aromatic substitutions are two-step reactions, having a first slow and rate-determining step followed by a rapid second step. Aliphatic SN2reactions have only one step. There are 2 transition states for the aromatic and 1 for the aliphatic substitution. (b) S,2 reactions have no intermediate. In electrophilic aromatic substitution the intermediate is a carbocation, while that in nucleophilic substitution is a carbanion. Problem 11.26 Why do the typical S,2 and S, 1 mechanisms not occur in nucleophilic aromatic substitution? 4
The SN2backside attack cannot occur, because of the high electron density of the delocalized 7r cloud of the benzene ring. Furthermore, inversion at the attacked C is sterically impossible. The S,1 mechanism does not occur, because the intermediate C,HS, with a + charge on an sp-hybridized C, would have a very high energy. The ring would also have a large ring strain.
The instability of Ar' prevents ArX (X is a halogen) from reacting with Ag'. Problem 11.27 The suggestion (following Problem 11.23) that the first step is slow is based partly on the finding that ArF, of all the aryl halides, reacts at the fastest rate; whereas RF undergoes the slowest aliphatic 4 S,2 reaction. Trace the argument. If the second step (the loss of X - ) were rate-determining, ArF would react at the slowest rate, for F- is the poorest leaving group of all the X- anions. If the first step were rate-determining, the stabilization of the charge on its transition state (TS) by the X group would affect the rate. Because F is the most electronegative X, it best stabilizes the TS, causing ArF to be the most reactive aryl halide. Hence, the first step must be rate-determining.
ELIMINATION-ADDITION REACTIONS With very strong bases, such as amide ion, NH,, unactivated aryl halides undergo substitution by an elimination-addition (benzyne) mechanism.
(elimination)
bcnzyne
(addition)
Problem 11.28 How do the following observations support the benzyne mechanism? (a) Compounds lacking ortho H's, such as 2,6-dimethylchlorobenzene, do not react. ( 6 ) 2,6-Dideuterobromobenzene reacts more slowly than bromobenzene. (c) o-Bromoanisole, o-CH,OC,H,Br, reacts with NaNH,/NH, to form m-CH,OC,H,NH,. (d) Chlorobenzene with C1 bonded to 14C gives almost 50% aniline having NH, bonded to 14 C and 50% aniline with NH, bonded to an ortho C. 4 (a) With no H ortho to C1, vicinal elimination cannot occur. (6) This primary isotope effect (Problem 7.32) indicates that a bond to H is broken in the rate-determining step, which is consistent with the first step in the benzyne mechanism being rate-determining. (c) NH; need not attack the C2 from which the Br- left; it can add at C3.
228
AROMATIC SUBSTITUTION. ARENES
a benzyne
[CHAP. 11
y*
CI -H+ -cl-*
Problem 11.29 How is it that the nucleophilic substitution of nitrochlorobenzenes is by addition and 4 elimination, whereas a benzyne mechanism is found with chlorobenzene?
The transition state that leads to the intermediate anion, which is formed when the nucleophile adds, is stabilized by the electron-withdrawing nitro groups ortho-para to the leaving group. Problem 11.30 Account for the observation that NaOH reacts at 300 "C with p-bromotoluene to give rn- and 4 p-cresols, while rn-bromotoluene yields the three isomeric cresols.
The benzyne intermediate from p-bromotoluene has a triple bond between C 3 and C4; both C's are independently attacked by OH-, giving a mixture of m- and p-cresols (HOC,H,CH,). Two isomeric benzynes are formed from rn-bromotoluene, one with a CZ-to-C3triple bond and the other with a C3-to-C4triple bond. Hence, this mixture of benzynes reacts with OH- at al! three C's, giving the mixture of three isomeric cresols.
11.4 ARENES
NOMENCLATURE AND PROPERTIES Benzene derivatives with saturated or unsaturated C-containing side chains are arenes. Examples are cumene or isopropylbenzene, C,H,CH(CH,),, and styrene or phenylethene, C,H,CH=CH,. Problem 11.31 Supply systematic and, where possible, common names for:
4
(a) p-Isopropyltoluene ( p-cymene). ( b ) 1,3,5-Trimethylbenzene (mesitylene). (c) p-Methylstyrene. ( d ) 1,4-Dipheny1-2-butyne(dibenzylacetylene). (e) (2)-1,2-Diphenylethene (cis-stilbene).
CHAP. 11)
229
AROMATIC SUBSTITUTION. ARENES
Problem 11.32 Arrange the isomeric tetramethylbenzenes, prehnitene (1,2,3,4-) and durene (1,2,4,5-), in 4 order of decreasing melting point and verify this order from tables of melting points. The more symmetrical the isomer, the closer the molecules are packed in the crystal and the higher is the melting point. The order of decreasing symmetry, durene > prehnitene, corresponds to that of their respective melting points, +80 "C > -6.5 "C.
SYNTHESES Friedel-Crafts alkylations and acylations, followed by reduction of the c=O group, are most frequently used to synthesize arenes. Coupling reactions can also be employed. Ar,CuLi
+ RX-
ArR
1"
R,CuLi
or
+ ArX-
ArR
Problem 11.33 Explain the following observations about the Friedel- Crafts alkylation reaction. (a) In monoalkylating C,H6 with RX in AIX, an excess of C6H6is used. ( b ) The alkylation of PhOH and PhNH, gives poor yields. (c) Ph-Ph cannot be prepared by the reaction AICll
PhH + PhCl+> (d) At 0°C
PhH + 3CH,CI
AIC13
+ HCl
Ph-Ph
1,2,4-trimethyIbenzene
but at 100 "C one gets 1,3,5-trimethyIbenzene (mesitylene). (e) The reaction PhH
+ CH,CH,CH,Cl-
AICIl
PhCH,CH2CH,
+ HCI
gives poor yield, whereas PhH
+CH,CHClCH,z
PhCH(CH,),
+ HCl
gives very good yield.
4
( a ) The monoalkylated product, C6H,R, which is more reactive than C6H6 itself since R is an activating group, will react to give C6H,R, and some C6H,R3. To prevent polyalkylation an excess of C,H, is used to increase the chance for collision between R' and C6H6and to minimize collision between between R' and C6H,R. ( b ) OH and NH, groups react with and inactivate the catalyst. PhCl + AICI, +-B
(4
Ph +
+ AlCl;
Ph' has a very high enthalpy and doesn't form. (d) The alkylation reaction is reversible and therefore gives the kinetic-controlled product at 0°C and the thermodynamic-controlled product at 100 "C. (e) The R' intermediates, especially the 1" RCH; can undergo rearrangements. With CH,CH,CH,CI we get
CH,CH,CH;
--H\ CH,&HCH,
and the major product is PhCH(CH,),.
-
-
Problem 11.34 Prepare PhCH,CH,CH, from PhH and any open-chain compound.
-
PhH + CICH,CH==CH, or
PhH + CICOCH,CH,
AICI3
AICI3
H21 PI
PhCH,CH,CH,
Z n l H g . HCI
PhCH,CH,CH,
PhCH,CH=CH,
PhCOCH,CH,
4
(Clemmen5en reduction)
CH,CH,CH,Cl cannot be used because the intermediate 1" CH,CH,CHi rearranges to the more stable 2" (CH,),CH+, to give C,H,CH(CH,), as the major product. The latter method is for synthesizing PhCH2R.
230
AROMATIC SUBSTITUTION. ARENES
[CHAP. 11
Problem 11.35 Give the structural formula and the name for the major alkylation product: ( a ) C,H,
+ (CH,),CHCH,Cl-
( c ) C,H,
+ CH,CH,CH,CH,Cl=
AIC13
+ (CH,),CCH,OH-
(b) C,H,CH,
AIClj
(d) m-xylene + (CH,),CCl=
CH3
BF,
AIClj
4
CH3
I
I
3 CH,:CH,
CH, CH, x6H5
lsobutyl cation (1")
lsobutyl chloride
Neopentyl alcohol
Neopentyl cation (1")
1
rCH,CH,CH,CH;] ( c ) CH,CH,CH,CH,CI
-%
1-
H'
CH3CH,yHCH,
rerr-Butyl cation (3")
1
rerr- Pe nt y1 cation (3")
p-tert-Pentyltoluene
% 34% PhCH,CH,CH,CH, + 66% C,H,-CHCH,CH,
I
n-Butylbenzene
n-Butyl chloride
H 3c
fert-Butylbenzene
CH3
s-Butylbenzene
Thermodynamic product; it has less steric strain and is more stable than the kinetic-controlled isomer having a bulky t-butyl group orfho to a CH,.
Chemistry; Reactivity of the Benzylic (Ph The chemistries of the benzylic and allylic positions are very similar. Intermediate carbocations, free radicals and carbanions formed at these positions are stabilized by delocalization with the adjacent T system, the benzene ring in the case of the benzylic position. Another aspect of arene chemistry is the enhanced stability of unsaturated arenes having double bonds conjugated with the benzene ring. This property is akin to the stability of conjugated di- and polyenes. Problem 11.36 PhCH, reacts with Br, and Fe to give a mixture of three monobromo products. With Br, in light, only one compound, a fourth monobromo isomer, is isolated. What are the four products? Explain the 4 formation of the light-catalyzed product. With Fe, the products are 0-,p-, and some rn-BrC,H,CH,. In light the product is benzyl bromide, PhCH,Br. Like allylic halogenation (Section 6 . 5 ) , the latter reaction is a free-radical substitution: (1) Br2;2Br.
(2) Bra
(3) PhCH,
+ Br2-
+ PhCH, --+PhCH, + HBr PhCH,Br
+ Br.
Steps (2) and (3) are the propagating steps. Problem 11.37 Account for the following: PhCH,CH(CH,), with little or no PhCH,CBr(CH,), formed.
+ Br2&
PhCHBrCH(CH,), 4
CHAP. 111
231
AROMATIC SUBSTITUTION. ARENES
The H’s on the C attached to the ring (the benzylic H’s), although they are in this case 2”, are nevertheless more reactive toward Br- than are ordinary 3” H’s. Like a C=C group in the allylic system, the Ph group can stabilize the free radical by electron donation through extended p orbital overlap.
Problem 11.38 Irradiation of an equimolar mixture of cyclohexane, toluene and Br, in CCI, gives almost 4 exclusively benzyl bromide. A similar reaction with C1, gives mainly cyclohexyl chloride. Explain.
This is a competitive reaction that compares the reactivities of toluene and cyclohexane. Br. is less reactive and more selective than Cl., and reactivity differences of the respective H’s determine the product (benzyl > 2”). With the more reactive and less selective Cl., the statistical advantage of cyclohexane (12H’s) over toluene (three H’s) controls the product ratio. Problem 11.39 Which is more reactive to radical halogenation, PhCH, or p-xylene? Explain.
4
p-Xylene reactivity depends on the rate of formation of the benzyl-type radical. Electron-releasing groups such as CH, stabilize the transition state, producing the benzyl radical on the other CH, and thereby lowering the AH’ and increasing the reaction rate. Problem 11.40 Outline a synthesis of 2,3-dimethyl-2,3-diphenylbutane from benzene, propylene, and any needed inorganic reagents. 4
The symmetry of this hydrocarbon makes possible a self-coupling reaction with 2-bromo-2-phenylpropane. C6HS
C6HS
C6H6 + CH,CH=CH,
Br
I.
Li
CH3-L-L--CH, LH3
-
LH,
Problem 11.41 Give all possible products of the following reactions and underline the major product. ( a ) PhCH,CHOHCH(CH,),
(c) PhCH=CHCH,
+ HBr+
“ZS04
(e) P h C H - L H C H d H ,
(6) PhCH,CHBrCH(CI-I,),; (d) PhCH=CHCH,
+ Br,
(equimolar amounts)-
+ HBr-
alc
peroxide
4
PhCH,CH=C(CH,), + PhCH=CHCH(CH,),. The major product has the C=C conjugated with the benzene ring and therefore, even though it is a disubstituted alkene, it is more stable than the minor product, which is a trisubstituted nonconjugated alkene. Same as part ( a ) and for the same reason. PhCH,CHBrCH, + PhCHBrCH,CH,. H’ adds to C = C to give the more stable tenzyl-type PhCHCH,CH,. Reaction with Br- gives the major product. The benzyl-type cation PhCHR (like CH,=CHCH,’) can be stabilized by delocalizing the + to the op-positions of the ring:
PhCH,CHBrCH, + PhCHBrCH,CH,. Br. adds to give the more stable benzyl-type PhCHCHBrCH, rather than PhCHBrCHCH,. We have already discussed the stability of benzylic free radicals (Problem 11.37).
232
AROMATIC SUBSTITUTION. ARENES
[CHAP. 11
( e ) PhCHBrCHBrCH=CH, + PhCHSHCHBrCHBr + PhCHBrCH=CHCH,Br. The major product is the conjugated alkene, which is more stable than the other two products [see part ( a ) ] .
Problem 11.42 Explain the following observations. ( a ) A yellow color is obtained when Ph,COH (trityl alcohol) is reacted with concentrated H,SO,, or when Ph,CCI is treated with AICI,. On adding H,O, the color disappears and a white solid is formed. (6) Ph,CCI is prepared by the Friedel-Crafts reaction of benzene and CCI,. It does not react with more benzene to form Ph,C. (c) A deep-red solution appears when Ph,CH is added to a solution of NaNH, in liquid NH,. The color disappears on adding water. (d) A red color appears when Ph,CCI reacts with Zn in C,H,. 0, decolorizes the solution. 4
The yellow color is attributed to the stable Ph,Cf, whose rings. Ph,COH
+ H2S0,+Ph,Cf
is delocalized to the oppositions of the 3
+ H,O+ + HSO; + AlCI;
Ph,CCl
+ AlCI,+
Ph,C'
Ph,Cf
+ 2H,O--
Ph,COH
kwis
Lewis base
white solid
acid
+
+ H,O+
With AICI,, Ph,CCI forms a salt, Ph,C'AICl;, whose carbocation is too stable to react with benzene. Ph,C' may also be too sterically hindered to react further. The strong base :NHZ removes H' from Ph,CH to form the stable, deep red-purple carbanion Ph,CT, which is then decolorized on accepting H' from the feeble acid H,O.
n + :NH,
Ph,CH acid
,
Ph,Ci base,
**
__*
basez
+ H,Oacid,
H:NH, acid,
Ph,CH acid,
+ Ph,Ci base, (deep red)
+ OHbase,
The Ph,Ci is stabilized because the - can be delocalized to the oppositions of the three rings (as in the corresponding carbocation and free radicals). C1. is removed from Ph,CCI by Zn to give the colored radical Ph,C-, which decolorizes as it forms the peroxide in the presence of 0,. 2Ph,CCI
+ Zn-
2Ph,C- + ZnCl,
2Ph,C* + .c)-o*+ Ph,C:O:O:CPh, Problem 11.43 In the following hydrocarbons the alkyl H's are designated by the Greek letters a,p, y , etc. Assign each letter an Arabic number, beginning with 1 for LEAST, in order of increasing ease of abstraction by a Br..
4
See Table 1 1-3.
AROMATIC SUBSTITUTION. ARENES
CHAP. 111
233
Table 11-3
B
a
I
I
6
Y
€
(a)
1 (1")
2 (2")
4 (2", benzylic)
5 (2", dibenzylic)
3 ( I", benzylic)
(W
2 (1")
3 (2")
4 (allylic)
1 (vinylic)
5 (allylic, benzylic)
(c)
3 (OP to other CH,'s)
1 ( m o to two other CH,'s; more hindered)
2 (less sterically hindered than p ; mp to two
Problem 11.44 Use + and - signs for positive and negative tests in tabulating rapid chemical reactions that ;an be used to distinguish among the following compounds: ( a ) chlorobenzene, benzyl chloride and cyclohexyl 4 ;hloride; ( b ) ethylbenzene, styrene, and phenylacetylene. See Tables ll-4(a) and 11-4(b).
Table 11-4(a) Chlorobenzene
Benzyl chloride
Cyclohexyl chloride
Ring sulfonation is exothermic
+
+
-
Alc. AgNO, (forms AgCI, a white precipitate)
-
+ (very fast)*
+ (much slower)*
Reaction
+
*Ag' induces an S, 1 reaction; PhCH, > C,H:, .
Table ll-4(6) Reactions
Ethylbenzene
Styrene
Phenylacetylene
Br, in CCI, (is decolorized)
-
+
+
Ag(NH,),' (forms a precipitate)
-
-
+
234
[CHAP. 11
AROMATIC SUBSTITUTION. ARENES
11.5 SUMMARY OF ARENE AND ARYL HALIDE CHEMISTRY PREPARATlOU A.
PROPERTIES
ALKYLBENZENES 1. Oxidation
1. Direct Alkylation
ArH + CH,CH ArH + H,C=C ArH + CH,CH
+ CrO,CI, + KMnO, 2.
2. Coupling
ArCHC , H-,--
Ar,CuLi + RX (1") R,CuLi + ArX .-c-cC
fl -
ArA-CH, Ar-COOH
heat
Reduction +
3H2w
3. Side-Chain
+ C 1 , L ArCHCICH,
ArCH=CH2 + H, ArH + CH,COCI(AICI,)-* A r X - C H ,
-
(6) Ring (Electrophilic)
+ X,
Fe
op-C,H,C,H,X
H2S04
+ HN0,op-C,H,C,H,NO, + H,S,O,---* op-C2H,C,H,S0,H + RX- AICI, op-C,H,C,H,R
\-
+ RCOCl-C,H,C,H,COR AIClj
4.
Disproportionation
+
C6H6
B. ALKENYLBENZENES
1. Reduction
1. Dehydrogenation 2. Dehydration
ArCHOHCH, or ArCH,CH,OH
/--
+ H,
ArCH=CH2 ArCHXCH, or ArCH,CH2X + alc. KOH
-
ArCHOHCH,OH ArCOOH
x
1. Nucleophilic Displacements
+ NaOH ArOH + NaCl + NaNH, # ArNH, + HCI + NaCN--+ ArCN + NaCl
Fe HIO,/HNO, Ag'
2. Indirect Halogenation by Diazonium
ArH-
KMn0,-
pe
1. Direct Halogenation +
+ hot
+ X, ArCHXCH,X + H,O-ArCHOHCH, + HBr- r o tde ArCHBrCH, + HBrArCH,CH,Br
+ Zn
C. ARY LHALIDES
ArH xz ArH + I,
+ cold KMnO,
3. Other Addition Reactions
Dehalogenation
ArCHXCH,X
-
ArCH,CH,
2. Oxidation
\ \ // --
3. Dehydrohalogenation 4.
-
+ C,H4(C2HS)2
2. Metals
ArNH, __* Ar Nl+ X-
+ Li --+ArLi + Cu+
(see Chapter 18)
3.
CHzO
Ar-Ar
ArCH,OH
+ CuX,
Ullmann Reaction
Electrophilic Substitution
+ HN03/H2S04*
op-XC,H,NO,
AROMATIC SUBSTITUTION. ARENES
CHAP. 111
235
Supplementary Problems Problem 11.45 Write structural formulas for the principal monosubstitution products of the indicated reactions from the following monosubstituted benzenes. For each write an S or F to show whether reaction is SLOWER or FASTER than with benzene. ( a ) Monobromination, C,H,CF,. (6) Mononitration, C,H,COOCH,. (c) Monochlorination, C,H,OCH,. (d) Monosulfonation, C,H,I. (e) Mononitration, C,H,C,H,. (f)Monochlorination, C,H,CN. ( 8 ) Mononitration, C,H,NHCOCH,. (h) Monosulfonation, C,H,CH(CH,)CH,CH,. 4
Problem 11.46 Which xylene is most easily sulfonated?
4
m-Xylene is most reactive and sulfonates at C4 because its CH,’s reinforce each other [Rule 1; Problem 11.17(a)]. Problem 11.47 Write structures for the principal mononitration products of ( a ) o-cresol (o-methylphenol), (6) p-CH,CONHC,H,SO,H, (c) m-cyanotoluene (m-toluonitrile). 4
Problem 11.48 Assign numbers from 1 for monobromination of the following groups. (a)
(I) PhNH,,
(6) (I) PhCH,,
(11) PhNHlCI-,
LEAST
to 5 for
(111) PhNHCOCH,,
MOST
to designate relative reactivity to ring
(IV) PhCl, (V) PhCOCH,.
(11) PhCOOH, (111) PhH, (IV) PhBr, (V) PhNO,.
(c) (I) p-xylene, (11) p-C6H4(COOH),, (111) PhMe, (IV) p-CH,C6H,COOH, See Table 11-5. Table 11-5
(V) m-xylene.
4
236
AROMATIC SUBSTITUTION. ARENES
[CHAP. 11
Problem 11.49 Why are the negative poles of the dipole moments of C,H,OCH, and C,H,NH, in the ring 4 away from the 0 and N , even though these atoms are more electronegative than C?
Delocalization of the electron densities on 0 and N toward the ring submerges the electron-withdrawing inductive effect of these atoms. Problem 11.50 Explain why the oxidation of PhCH, to PhCOOH by KMnO, or K,Cr,O, and H' goes in 4 poor yield, whereas the same oxidation of p-O,NC,H,CH, to p-O,NC,H,COOH goes in good yield.
The oxidant is seeking electrons and is therefore an electrophile. As a side reaction, the oxidant can attack and destroy the ring. The NO2 deactivates the ring toward electrophilic attack, thereby stabilizing it toward degradative oxidation. Problem 11.51 Use PhH, PhMe, and any aliphatic or inorganic reagents to prepare the following compounds in reasonable yields: ( a ) rn-bromobenzenesulfonic acid, (6) 3 -nitro-4-bromobenzoic acid, (c) 3,4-dibromonitrobenzene , (d) 2,6-dibromo-4-nitrotoluene. 4
Br
(m-director is added first)
Nitration of p-BrC,H,CH, would have given about a 50-50 mixture of two products; 2-nitro-4bromotoluene would be unwanted. When oxidation precedes nitration, an excellent yield of the desired product is obtained.
Nitration followed by dibromination would give as the major product 2,5-dibromonitrobenzene (see Rule 2, preceding Problem 11.17).
Problem 11.52 Available organic starting materials for synthesizing the pairs of compounds below are PhH, PhMe, and PhNO,. Select the compound in each pair whose synthesis is more expensive and explain your choice. ( a ) p-C,H,(NO,), and rn-C,H,(NO,), , ( 6 ) rn-nitrotoluene and p-nitrotoluene, ( c ) 2,4,6-trinitrotoluene (TNT) and 1,3,5-trinitrobenzene7 (d) rn-C,H,CI, and p-C,H,Cl,. 4
( a ) p-C,H,(NO,),. NO, is rnetu-directing. The rnetu isomer is easily made directly by nitrating PhNO,. The paru isomer cannot be prepared directly and requires a more circuitous, expensive route. (6) rn-Nitrotoluene. Nitration of PhMe gives the pura isomer. PhNO, is too deactivated to undergo metu alkylation. (c) 1,3,5Trinitrobenzene. Two NO,'s deactivate the ring so that insertion of a third NO, is very difficult. This extreme deactivation is countered somewhat by Me. Therefore, PhMe can be more easily trinitrated than PhH. (d) rn-C,H,C12. C1 is op-directing. Problem 11.53 Give the product(s) from ( a ) mononitration and ( 6 ) dinitration of naphthalene.
4
CHAP. 111
237
AROMATIC SUBSTITUTION. ARENES
(6) Ring without NO, is deactivated less and undergoes cy -substitution:
( a ) a-Substitution; a-C,,H,NO,.
13-Dinitronaphthalene
1,&Dinitronaphthalene
Problem 11.54 Name and account for the products from the reaction of 2-ethylnaphthalene at
(U)
+ 96% H,SO,
40"C, (6) 140°C.
( a ) 2-Ethyl-1-naphthalenesulfonicacid. The activating group directs substitution to the reactive ortho position. ( 6 ) 2-Ethyl-6-naphthalenesulfonicacid. Higher-temperature p -sulfonation in the less hindered ring gives the thermodynamic product.
Problem 11.55 Supply structures for organic compounds (A) through (0).
PhBr + Mg
--
Ph-GSH
(4
Et 0
(F)
KOH
ArCH7CI
+ CH,MgX-(I)-(J)-(K)
PhCHBrCH,CH,
(A)
H C=CHCH2Br
(E)
(B) P h C H S H C H ,
(G)
N BS
(H)
LI. NH3
(C) PhCHOHCHOHCH,
(D) PhCOOH + CH,COOH ( 6 ) (E) PhMgBr
(F) PhCH,CH=CH,
(G) P h C H K H C H ,
(H) PhCH=CHCH2Br
(conjugated alkene is more stable)
Ph
(4
(I)
p-CH3C6H4
(4
(J)
PhC-CMgX (+CH,)
(L)
\
PhC=CCH,Ar Ph
/
(cis)
/'="\,
H
(M)
\ (K) ,C=C H
/
H
\cH,A~
p-CH C H CH-CH,Ph
'1
Br
+
', which is p-CH,C,H,CHCH,Ph In (M), H'+adds to give more stable R CH,C,H,CH,CHPh, because of electron release by p-CH,. p-BrC H
$9
6 4 \
(N)
/c=c\
H
/
H
CH3
( t run s )
(0)p-BrC,H,-cH,-CHCH,
8,
In (0),Br. adds to give more stable R., p-BrC,H,CHCHBrCH,, which is benzylic.
rather than p -
[CHAP. 11
AROMATIC SUBSTITUTION. ARENES
238
Problem 11.56 Assign numbers from 1 for LEAST to 3 for MOST to the Roman numerals for the indicated compounds to show their relative reactivities in the designated reactions. (a)
HBr addition to (I) P h C H S H , , (11) p - C H 3 C 6 H , C H d H 2 , (111) p-O,NC,H,CH=CH,.
(6) Dehydration of (I) p-O,NC,H,CHOHCH,, (11) p-H,NC,H,CHOHCH,,
(c)
CH3 I Dehydration of ( I ) P h - C - C H , C H , ,
CH, I (11) P h - C H - C H , O H ,
I
(111) C,H5CHOHCH,.
(111) PhCHOHCH,CH,CH,.
OH (d) Solvolysis of ( I ) C,HSCH,Cl, (11) /I-O,NC,H,CH,Cl,
(111) p-CH,OC,H,CH,Cl.
4
See Table 11-6. Table 11-6
I11
I1
I
3 (p-Me stabilizes the benzylic ) 'R
1 (p-NO, destabilizes the benzylic ) 'R
(b)
1 (p-NO, destabilizes benzylic ) ' R
3 (p-NH, stabilizes the benzylic ) ' R
2
(c)
3 (get 3" benzylic R')
I (get I", non-benzylic ) 'R
2 (get 2" benzylic ) 'R
1 (p-NO, destabilizes benzylic ) 'R
3 (p-CH,O stabilizes benzylic R')
I I* (d)
Problem 11.57 Show the syntheses of the following compounds from benzene, toluene and any inorganic reagents or aliphatic compounds having up to three C's:
CH3 (a)
p-BrC,H,CH,Cl
( b ) p-BriCbH,CH=CH,
(c) Ph-A-CH,
AH ( d ) p-0,NC,H4CH2Ph
C,H,CH,
(a)
C,H,-
(6)
C7H5CI AIC13
4
-
Er,. Fe
Ph,CHCH,
(e)
CI
p-BrC,H,CH,
uv
p-BrC,H,CH,Cl
Br2
CI2
C , H , C H 2 C H , ~p-BrC,H,CH,CH,:
p-BrC,H,CHClCH,
alc.
p-BrC,H,CH=CH,
The C = C must be introduced by a base-induced reaction. If acid is used, such as in the dehydration of an alcohol, the product would undergo polymerization.
(4 (4
PhH
+ (CH,),CHCl
5 Ph-CH(CH,),
--% Ph-C(CH,), Br I
HN03 C6H5CH3
CIZ
c:$b% Ph-C(CH,), AH
C6H6
p-02NC6H4CH3 7 p - 0 2 N C 6 H 4 C H 2 C 1 ~p-02NC6H4CH2CfjH5
CHAP. 111
239
AROMATIC SUBSTITUTION. ARENES
The deactivated C,H,NO, cannot be alkylated with PhCH,CI. PhH-
(4
C H C1 AICI:,
AIc.
PhCH,CH,
P h C H B r C H , G PhCH=CH,
Ph,CHCH,
Problem 11.58 Deduce the structural formulas of the following arenes. ( a ) (i) Compound A (C,6H16) decolorizes both Br, in CCl, and cold aqueous KMnO,. It adds an equimolar amount of H,. Oxidation with hot KMnO, gives a dicarboxylic acid, C,H,(COOH),, having only one monobromo substitution product. (ii) What structural feature is uncertain? (6) Arene B (C,oH14)has five possible monobromo derivatives (C,,,H1.7Br). Vigorous oxidation of B yields an acidic compound, C 8 H 6 0 4 ,having only one mononitro substitution product, C,H,O,NO,. 4 since it adds one H,. The other 8 degrees of unsaturation mean the presence of two benzene rings. Since oxidative cleavage gives a dicarboxylic acid, C,H,(COOH),, each benzene ring must be disubstituted. Since C,H,(COOH), has only one monobromo derivative, the COOH’s must be para to each other.
( a ) (i) Compound A has one C=C
Br
Br
(A)
\.
Br
(ii) It may be cis or tram. : 6 ) C,H,O, must be a dicarboxylic acid, C,H,(COOH),, and, as in part (a), has para COOH’s. Compound B must therefore be a p-dialkylbenzene.
@
CH,Br
/TCH3 H3C
6
B
4 6
r
F\HCH3 H3C
H3C
F\HCH3
CH H3C/ \CH2Br
monobromo substitution products
H,C
CH / \ CH3 (B)
COOH
COOH
COOH
COOH
CBr /\ H3C CH3
240
AROMATIC SUBSTITUTION. ARENES
[CHAP. 11
Problem 11.59 Outline practical laboratory syntheses from benzene or toluene and any needed inorganic reagents of: ( a ) p-chlorobenzal chloride, (6) 2,4-dinitroaniline1 ( c ) rn-chlorobenzotrichloride, (d) 2,5dibromonitrobenzene . 4
CCI,
Br
Br
Problem 11.60 Assign numbers from 1 for LEAST to 3 for MOST to show the relative reactivities of the compounds with the indicated reagents: (a) C,H,CH,CH,Br (I), C,H,CHBrCH, (11) and C,H,CH==CHBr (111) with alcoholic AgNO,; (b) CH,CH,CI ( I ) , C,H,CH,Cl (11) and C,H,Cl (111) with KCN; ( c ) M nitrochlorobenzene (I), 2,4-dinitrochlorobenzene(11) and p-nitrochlorobenzene (111) with sodium methoxide. See Table 11-7
I
I1
I11
( a ) An s N 1 reaction
2 (primary)
3 (benzylic)
1 (vinylic)
(b) An s N 2 reaction
2 (primary)
3 (benzylic)
1 (aryl)
(c) An aromatic nucleophilic displacement
1 (meta)
3 ( 0 and p )
2 (para)
Problem 11.61 Which C1 in 1,2,4-trichlorobenzene reacts with -OCH,COO- to form the herbicide “2,4-D”? 4 Give the structure of “2,4-D.” Cl’s are electron-withdrawing and activate the ring to nucleophilic attack. The Cl at C’ is displaced because it is ortho and para to the other Cl’s.
+
-OCH,COO-
-
0C H 2C00-
Qcl Cl “2.4-D”
+
CI-
CHAP. 11)
241
AROMATIC SUBSTITUTION. ARENES
Problem 11.62 Explain these observations: ( a ) p-Nitrobenzenesulfonic acid is formed from the reaction of p-nitrochlorobenzene with NaHSO, , but benzenesulfonic acid cannot be formed from chlorobenzene by this reaction. ( 6 ) 2,4,6-Trinitroanisole with NaOC,H, gives the same product as 2,4,6-trinitrophenetole with 4 NaOCH,. ( a ) Nucleophilic aromatic substitution occurs with p-nitrochlorobenzene, but not with chlorobenzene, because
NO, stabilizes the carbanion [Problem 11.24(a)]. . ? N O 1
+ Na':SO,H-
+02N
(O)S4H
+ NaCl
(6) The product is a sodium salt formed by addition of alkoxide.
Problem 11.63 After o-DC,H,F is reacted for a short time with NaNHJNH,, the recovered starting material contains some C,H,F. Under the same conditions o-DC,H,Br leaves no C,H,Br. Explain. 4
F- is a much poorer leaving group than is Br-.
&* QF
2
F
F
&D
w,
Br
Br
+ NH,
Chapter 12 Spectroscopy and Structure 12.1 INTRODUCTION Spectral properties are used to determine the structure of molecules and ions. Of special importance are ultraviolet (uv), infrared (ir), nuclear magnetic resonance (nmr), and mass spectra (ms). Free radicals are studied by electron spin resonance (esr). The various types of molecular energy, such as electronic, vibrational, and nuclear spin, are quantized. That is, only certain energy states are permitted. The molecule can be raised from its lowest energy state (ground state) to a higher energy state (excited state) by a photon (quantum of energy) of electromagnetic radiation of the correct wavelength.
ultraviolet (uv) 8 ' Far Near ultraviolet (uv)
5
?! w
8
Electronic Electronic Electronic Molecular vibration Spin (electronic
Visible Infrared (ir) Radio
100-200 nm 200-350 nm 350-800 nm 1-300 p m lm
Wavelengths ( A ) for ultraviolet spectra are expressed in nanometers (1 nm = 10-' m); for the infrared, micrometers are used ( 1 p m = 10-6 m). Frequencies (v) in the infrared are often specified by the wave number V , where V = 1/A; a common unit for V is the reciprocal centimeter (1 cm-' = 100m-l). The basic SI units for frequency and energy are the hertz (Hz) and the joule (J), respectively. Problem 12.1 ( a ) Calculate the frequencies of violet and red light if their wavelengths are 400 and 750nm, respectively. (6) Calculate and compare the energies of their photons. 4
The wavelengths are substituted into the equation v = c/A, where c = speed of light = 3.0 X 10' m / s . Thus
Red:
V =
3.0 x 10' m/s = 4 . 0 1014s--1 ~ =400THz 750 x 10-9 m
where 1 THz = 10'' Hz = 10" s-l. Violet light has the shorter wavelength and higher frequency. The frequencies from part (a) are substituted into the equation E = hv, where h = 6.624 x 10-34J - s (Planck's constant). Thus Violet:
E = (6.624 x 1 0 - J~. ~s)(7.5 x 1014s - I )
Red:
E = (6.624 x 10-34J * s)(4.0 X 1014S - ' ) = 2.7 x 10-" J
= 5.0 x
10-
J
Photons of violet light have more energy than those of red light. emblem 12.2 Express 10 micrometers ( a ) in centimeters, (6) in angstroms (1 A = 10-"' m), (c) in nanometers, (d) as a wave number. 4
(4
10 p m = (10 x 10-6 m)(
242
100 cm r ) 10-3 cm =
243
SPECTROSCOPY AND STRUCTURE
CHAP. 12)
1A 10- m = 105 A 10 p m = (10 x 10-6 m)( T) 10 p m = (10 x 10-6 m)(
109nm 7) = 104 nm
In a typical spectrophotometer, a dissolved compound is exposed to electromagnetic radiation with a continuous spread in wavelength. The radiation passing through or absorbed is recorded on a chart against the wavelength or wave number. Absorption peaks are plotted as minima in infrared, and usually as maxima in ultraviolet spectroscopy. At a given wavelength, absorption follows an exponential law of the form A=ECl
where
A = absorbance = -log,, (fraction of incident radiation transmitted) E = molar extinction coefficient, cm */mol C = concentration of solution, moI/L( = moI/cm3) l = thickness of solution presented to radiation, cm
The wavelength of maximum absorption, A,,, , and the corresponding of a compound. Units are normally omitted from specifications of E .
are identifying properties
12.2 ULTRAVIOLET AND VISIBLE SPECTROSCOPY
Ultraviolet and visible light cause an electron to be excited from a lower-energy HOMO to a high-energy LUMO. There are three kinds of electrons: those in U bonds, those in T bonds, and unshared electrons, which are designated by the letter n for nonbonding. These are illustrated in formaldehyde:
H
On absorbing energy, any of these electrons can enter excited states, which are either antibonding U * or T * . All molecules have CT and U * orbitals, but only those with n orbitals have 7r* orbitals. Only the n - , T * , n + , ~ * and , more rarely the n - , U * excitations occur in the near ultraviolet and visible regions, which are the available regions for ordinary spectrophotometers. Species which absorb in the visible region are colored, and black is observed when all visible light is absorbed. Problem 12.3 The relative energy for various electronic states (MO's) is: IT*
n I
(nonbonding)
U
List the three electronic transitions detectable by uv spectrophotometers in order of increasing A E.
4
n + n* < T + n* < n + a * Problem 12.4 List all the electronic transitions possible for (a) CH,, (6) CH,CI, (c) H,C==O. n-
( a ) U + U * .(6) u+ n* and n + n * .
U * and
n-,
U*
(there are no n or
T*
MO's). ( c ) U--+U * ,U +
4
7r*, T--+U * ,n--, U * ,
244
SPECTROSCOPY AND STRUCTURE
Problem 12.5 The uv spectrum of acetone shows two peaks of A,, = 280nm, E,,, = 100. (U) Identify the electronic transition for each. (6) Which is more intense?
[CHAP. 12
= 15
and A,,
= 190,
4
E,,,
( a ) The longer wavelength (280 nm) is associated with the smaller-energy (n+ n * ) transition. n-+ n* occurs
at 190 nm. ( 6 ) n+ n* has the larger
E,,,
and is the more intense peak.
Problem 12.6 Draw conclusions about the relationship of A,,, to the structure of the absorbing molecule from the following A,, values (in nm): ethylene (170), 1,3-butadiene (217), 2,3-dimethyl-l,3-butadiene(226), 1,3-~yclohexadiene(256), and 1,3,5-hexatriene (274). 4
1. Conjugation of n bonds causes molecules to absorb at longer wavelengths. 2. As the number of conjugated T bonds increases, A,, increases. 3. Cyclic polyenes absorb at higher wavelengths than do acyclic polyenes. 4, Substitution of alkyl groups on C=C causes a shift to longer wavelength (red shift). Problem 12.7 Account for the following variations in A,,,(nm)
of CH,X: X = C1( 173), Br(204), and I(258).
4
The transition must be n - + U * [Problem 12.4(6)]. On going from C1 to Br to I the n electrons ( a ) are found in higher principal energy levels (the principal quantum numbers are 3, 4, 5 , respectively), (6) are further away from the attractive force of the nucleus, and (c) are more easily excited. Hence absorption occurs at progressively higher A,, since less energy is required. Problem 12.8 Use MO theory (see Fig. 8-3) to explain why A,, (See Problem 12.6 for data.)
for 1,3-butadiene is higher than for ethylene. 4
For 1,3-butadiene, the excitation is from n,, the highest occupied MO (HOMO), to n:, the lowest unoccupied MO*(LUMO). The AE for this transition is less than the AE for n-+ n* for ethylene. Therefore A,, for 1,3-butadiene is greater than A,, for ethylene. Problem 12.9 Identify the two geometric isomers of stilbene, C,H,CH=CHC,H,, 294 nm and 278 nm.
from their A,,
values,
4
The higher-energy cis isomer has the shorter wavelength. Steric strain prevents full coplanarity of the cis phenyl groups, and the conjugative effect is attenuated. Problem 12.10 Each of three C,H,, isomers absorbs 2 mol of H, to form n-hexane. CO, and RCOOH are not formed on ozonolysis from any of the compounds. Deduce possible structures for the 3 compounds if their uv absorption maxima are 176, 211, and 216 nm. 4
The isomers are hexadienes. An allene or alkyne is not possible since CO, and RCOOH are not ozonolysis products. Absorption at 175 nm is attributed to the isolated 1,4-hexadiene. The conjugated cis- and truns-1,3hexadienes absorb at 21 1 and 216 nm, respectively. Problem 12.11 The complementary color pairs are: violet-yellow, blue-orange, and green-red. Given a red, 4 an orange and a yellow polyene, which is most and which is least conjugated?
The orange polyene absorbs blue, the red absorbs green and the yellow absorbs violet. The most conjugated polyene, in this case the red one, absorbs the color of longest wavelength, in this case green. Violet has the shortest wavelength and therefore the yellow polyene is the least conjugated.
12.3 INFRARED SPECTROSCOPY
Infrared radiation causes excitation of the quantized molecular vibration states. Atoms in a diatomic molecule, e.g. H-H and H<1, vibrate in only one way; they move, as though attached by a coiled
CHAP. 121
245
SPECTROSCOPY AND STRUCTURE
spring, toward and away from each other. This mode is called bond stretching. Triatomic molecules, such as CO, (O=C=O), possess two different stretching modes. In the symmetrical stretch, each 0 moves away from the C at the same time. In the antisymmetrical stretch, one 0 moves toward the C while the other 0 moves away. Molecules with more than two atoms have, in addition, continuously changing bond angles. These bending modes are indicated in Fig. 12-1. Scissoring
Rocking
IN-PLANE BENDING
+
Twisting
+ means
Wagging
coming forward
n
OUT-OF-PLANE BENDING Fig. 12-1
In a molecule, each bond such as O - H and each group of three or more atoms such as NH, and CH, absorbs ir radiation at certain frequencies to give quantized excited stretching and bending vibrational states. See Tables 12-1 and 12-2. Only vibrations that cause a change in dipole moment give rise t o an absorption band. Although absorption is affected only slightly by the molecular environment of the bond or group, it is possible to determine from small variations in the band is part of a frequencies such factors as the size of the ring containing a C=O group or whether -0 ketone, COCl, o r COOH. An observed absorption band at a specific wavelength proves the identity of a particular bond o r group of bonds in a molecule. Conversely, the absence of a certain band in the spectrum usually rules out the presence of the bond that would produce it. Between 42 and 24 THz (1400 and 800cm-') there are many peaks which are difficult to interpret. However, this range, called the fingerprint region, is useful for determining whether compounds are identical. (It is virtually impossible for two different organic compounds to have the same ir spectrum, because of the large number of peaks in the spectrum.) Problem 12.12 How do the following factors affect absorption frequencies? Use data in Tables 12-1 and 12-2. stretch, the hybrid orbitals used by C. (b) Bond strength; i.e., change in bond multiplicity. (c) Change in mass of one of the bonded atoms; e.g., 0-H versus O - D . (d) Stretching versus bending. (e) H-bonding of OH. 4 ( a ) For C-H
The more s character in the C-H
bond, the stiffer the bond and the higher the frequency:
I
I
4-H
< 4-H
I
< =C-H
Stretching frequencies parallel bond strengths. Because bond strength increases with the number of bonds between two given atoms, absorption frequencies increase with bond multiplicity:
I
<-c-
t
l l
<
I t <=c-
<
a-
SPECTROSCOPY AND STRUCTURE
246
Table 12-1.
Infrared Absorption Peaks (mostly stretching)
Intensity*
(4 (4 (m-s)
[CHAP. 12
Structure
I I
C-4
(in ethers, alcohols and esters)
SO, (in sulfonic acid derivatives)
C-H
(in alkanes) ~~~
I I
6)
c=C bond in aromatic ring (usually shows several peaks:
( 4
c=C
(4
C=O (in amides O=C-N)
(9
C=O (in carbonyl compounds and esters)
1700-1725 1770- 1820 2100-2200
1% I
(in carboxylic acids) (in acid chlorides)
22 10-2260
1
I
C-H
(of aldehyde group)
C-H
(C is part of aromatic ring)
C-H
(C is acetylenic)
(m)
C-H
(C is ethylenic)
(m-4
C-H
(in alkanes)
(m)
N-H
(in amines and amides)
320-3600
(s, b)'
0 - H (in H-bonded ROH and ArOH)
3600-3650
(s)
0-H
(w)
2500-3000 3000-3100
II--
3300-3500
I 2100
7
(4 (s)
I I
~~
*Intensities: (s) = strong, (m) = medium, (w) = weak, (b) = broad, (vb) = very broad. 'Intermolecular H-bonded peaks are sharp, not broad.
Frequencies are inversely related to the masses of the bonded atoms. Therefore, changing the lighter H to the heavier D causes a decrease in the stretching frequency. (d) Most of the stretching frequencies in Table 12-1 are higher than the bending frequencies in Table 12-2. (e) H-bonding causes a shift to lower frequencies (3600 cm-' --* 3300 cm-I). The band also becomes broader and less intense. (c)
SPECTROSCOPY AND STRUCTURE
CHAP. 121
247
Table 12-2. Bending Frequencies (cm- ') of Hydrocarbons*
Alkanes
~
CH(CHJ2 Doublet of equal intensities at 1370, 1385; also
=CH,
CH3
1430-1470
1420-1470 1375
C(CH,), Doublet at
1370 (strong) 1395 (moderate)
1170
Alkenes out-of-plane
cis
RCH-CHR
tram
Aromatic C-H out-of-plane
Monosubstituted 690-7 10
ortho 735-770
730-770
675-730 965-975
(variable)
Disubstituted meta para
690-710 810-840 750-8 10
*Most peaks are medium-to-strong or strong. Problem 12.13 Identify the peaks marked by Roman numerals in Fig. 12-2, the ir spectrum of ethyl acetate,
4
Frequency, cm-'
1
2
1
1 3
1
I
4
1 7
6
5
1
8
1
1
9
Wavelength, p m
1
I
II
10
12
( a ) Actual Spectrum
I
I 3200
I U00
I l9SO
1
I 700
1
IS00
1
1300
Frequency, cm-' (6)Simplified Spectrum Fig. 12-2
I
I loo
1
900
I
I3
I
14
1
IS
248
SPECTROSCOPY AND STRUCTURE
[CHAP. 12
The deep valleys represent transmission minima and are therefore absorption “peaks” or bands. At about B O c m - ’ , peak I is due to H--CSp3 stretching. The peak at 1700cm--’,11, is due to stretching of the
group. The two bands at 1400-15Ocm-’, 111, are again due to C-H bonds. The one at 1250cm-’, IV, is due to the C - 0 stretch. (It is extremely difficult and impractical to attempt an interpretation of each band in the spectrum .)
Problem 12.14 Which of the following vibrational modes show no ir absorption bands? ( a ) symmetrical CO, stretch, (6) antisymmetrical CO, stretch, (c) symmetrical O==C=S stretch, (d) C = C stretch in o-xylene, (e) C==C stretch in p-xylene and ( f ) C=C stretch in p-bromotoluene. 4 Those vibrations which do not result in a change in dipole moment show no band. These are ( a ) and (e), which are symmetrical about the axis of the stretched bonds. Problem 12.15 Explain the following observation. A concentrated solution of C,H,OH in CCI,, as well as one of CH,OHCH,OH, has a broad 0-H stretch near 3350cm-I. On dilution with CCI,, the spectrum of CH,OHCH,OH does not change, but that of C,H,OH shows a sharp 0-H stretch at 3600cm-‘ instead of the 4 broad band at 3350 cm- ’. H-bonding in HOCH,CH,OH is intramolecular (Section 2.6) and is not disturbed by dilution. H-bonding in C,H,OH is intermolecular dilution separates the OH’S and the H-bonds and their 3350 cm-’ bands disappear.
12.4 NUCLEAR MAGNETIC RESONANCE (PROTON, PMR)
ORIGIN OF SPECTRA Nuclei with an odd number of protons or of neutrons have permanent magnetic moments and quantized nuclear spin states. For example, an H in a molecule has two equal-energy nuclear spin states, which are assigned the quantum numbers + $ (T) and - $ ( & ) [Fig. 12-3(a)]. When a compound is placed in a magnetic field its H’s align their own fields either with or against the applied magnetic field, H,, giving rise to two separated energy states, as shown in Fig. 12-3(b). In the higher energy state, E,, the fields are aligned against each other; in the lower energy state, E,, they are aligned with each other. The difference in energy between the two states, which is directly proportional to H,, corresponds to a frequency of radiowaves. For this reason, radio-frequency photons can “flip” H nuclei from lower to higher energy states, reversing their spin in the process. When such spin-flip occurs, the nucleus (a proton, in the case of H) is said to be in resonance with the applied radiation; hence the name nuclear magnetic resonance (nmr) spectroscopy. When used for medical diagnosis this technique is called magnetic resonance imaging (mri). Applied magnetic
I
fie*d E,
,L,
T
AE (radio frequency)
Fig. 12-3
CHAP. 121
249
SPECTROSCOPY AND STRUCTURE
In practice it is easier and cheaper to fix the radio frequency and slowly to vary the magnetic field strength. When the H , value is reached enabling the proton to absorb the radio photon and to spin-flip, a signal (peak) is traced on calibrated chart paper. When the radiowaves are removed the excited nuclei quickly return to the lower-energy spin state. The same sample can then be used for obtaining repeated spectra. The sample compound is dissolved in a proton-free solvent such as CCl, or a deuterated solvent such as DCCl,. (Although D is nmr-active it does not absorb in the frequency range that H does and, therefore, does not interfere with proton spectra.) The solution is placed in a long thin tube which is spun in the magnetic field so that all molecules of the compound feel the same magnetic field strength at any given instant. Problem 12.16 Which of the following atoms do not exhibit nuclear magnetic resonance? " C , 'H, I9F, 3'P, 13C, and 32S.
I 6 0 ,
14N, "N,
4
Atoms with odd numbers of protons and/or neutrons are nmr-active. The inactive atoms are: '*C ( 6 p , 6n), 0 (Sp, 8n), and "S (16p, 16n). To detect the nmr activity of atoms other than 'H requires alteration of the nmr spectrometer. The ordinary spectrometer selects the range of radiowave frequency that excites only 'H. 16
CHEMICAL SHIFT Nmr spectroscopy is useful because not all H's change spin at the same applied magnetic field, for the energy absorbed depends on the bonding environment of the H. The magnetic field experienced by an H is not necessarily that which is applied by the magnet, because the electrons in the bond to the H and the electrons in nearby r bonds induce their own magnetic fields. This induced field, H*, partially shields the proton from the applied H,. The field "felt" by the proton, the effective field, is H, - H * . The typical nmr spectrogram, Fig. 12-4, is produced by the application of an external field Ho that very slowly decreases in time. First to appear are the signals of the most upfield (most shielded) protons; the greater the shielding, the smaller the effective field at the proton and hence the lower the signal frequency. As we move downfield, H, remains essentially constant; so the other proton signals come in, in the order of decreasing shielding (increasing effective field, increasing frequency).
e--- Signal Frequency, Hz 500
400
300
200
100
0
I
I
I
I
1
TMS
L
I
I
1
1
I
I
I
1
1
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
0
Down field less shielded
8, ppm
Fig. 12-4
Upfield more shielded
250
SPECTROSCOPY AND STRUCTURE
[CHAP. 12
The displacement of a signal from the hypothetical position of maximum shielding is called its chemical shift, notated as S (delta) and measured in parts per million (ppm). As indicated on Fig. 12-4, the zero of the 6 scale is conventionally located at the signal produced by the H’s of tetramethylsilane (TMS), (CH3)4Si. This compound serves because its H-signal is usually isolated in the extreme upfield region. Clues to the structure of an unknown compound can be obtained by comparing the chemical shifts of its spectrum to the S values in such tabulations as Table 12-3. Some generalizations about molecular structure and proton chemical shift in ‘Hnmr (pmr) are: 1. Electronegative atoms, such as N, 0, and X, lessen the shielding of H’s and cause downfield shifts. The extent of the downfield shift is directly proportional to the electronegativity of the atom and its proximity to the H. The influence of electronegativity is illustrated with the methyl halides, MeX. 6, ppm X
I
4.3
I F
3.1
2.7
2.2
c1
Br
I
Table 12-3. Proton Cbemicd Shifts Character of Underlined Proton
6, pprn
6,PPm 4-2.5
RKH,
Tertiary:
R,-CH_
I
4-3
4-3.4
I
4-43
I
4.1-3.7
I
Br-4-H
Chloride:
C1-L-a
Alcohol:
H0-L-H
Fluoride:
F-L-H
Ester (I):
R-C=O
aHto
2.2-2.0
I 1 2.62.0
t
Bromide:
II
R-CH, Secondary:
I
Character of Underlined Proton
2.7-2.0
I
Ester (11): a H to C=O
& L = O
Carboxylic acid:
!j--L--C=O
a H
2;bonyl:
I 1
Acetylenic:
3-2.2
Benzylic:
I
I
I -
I / 0-c-Jj
\
I bR
‘
bH
-c=o I -C-H
j
Aromatic:
0 F - i
12.0-4.0
Phenolic:
0 0 - B
15.0- 17.0
Enolic :
-L=L-O-H
8.5-6.0
I
3-2
alkyl 0
I
(Ar-l-J)
-4~c-H
R-0-6--Y
I
CHAP. 121
25 1
SPECTROSCOPY AND STRUCTURE
2. H’s attached to n-bonded C’s are less shielded than those in alkanes. The order of S values is:
I I
I
H - C = O > Ar-H
> H--C==C->
H a - >
H
I
3. Ar, -0, C=C, and C=C are electron-withdrawing by induction and cause a downfield shift of an H on an adjacent C, as in
I I allylic
I I benzylic-type
I I a -carbony1
-4-H
Il-C-c==c
Ar-C-H
I I propargyllic
H-C-G=C
4. For alkyl groups in similar environments, shielding increases with the number of H’s on the C. The order for 6 values is: H
4 - H >4 - H
I
Methine
I
H
>4 - H
Methylene
k
Methyl
5 . Electropositive atoms, such as Si, shield H’s; hence the reference property of TMS. 6. H’s attached to a cyclopropane ring and those situated in the T cloud of an aromatic system are strongly shielded. Some may have negative S values. 7. H’s which participate in H-bonding, e.g., OH and NH, exhibit variable S values over a wide range, depending mainly on sample concentrations. H-bonding diminishes shielding. Hence, in a non-H-bonding solvent, the OH-signal for ROH (or NH-signal for RNH,) moves downfield as the sample is concentrated, because H-bonding is enhanced. H-bonding is accompanied by exchange of H’s between the ROH molecules, resulting in broadening of the signals. Problem 12.17 Use Table 12-3 to assign approximate S values for the chemical shift of the one type of H in 4 ( U ) (CH,),C=C(CH,),, ( 6 ) (CH,),C=O, (c) benzene, ( d ) O = C H - C H = O . ( a ) 1.7 ppm, (6) 2.3 ppm, (c) 7.2 ppm, ( d ) 9.5 ppm.
Problem 12.18 Give the numbers of kinds of H’s present in ( a ) CH,CH,, (6) CH,CH,CH,, (CH,),CHCH,CH,’ ( 4 H,C----CH,, ( e ) CH,CH=CH,, (f)C,H,NO,, ( g ) C,H,CH,. (a) (6) (c) (d) (e)
One (all equivalent). Two: CH‘ICHiCHff. Four: (CH ff ),CHhCH; CH: One (all equivalent). Four:
(c)
4
.
The =CH, H’s are not equivalent since one is cis to the CH, and the other is tram. Replacement of H‘ by X gives the cis-diastereomer. Replacement of Hd gives the tram-diastereomer. (f) Three: Two ortho, two meta, and one para. (g) Theoretically there are three kinds of aromatic H’s, as in (f).Actually the ring H’s are little affected by alkyl groups and are equivalent. There are two kinds: C,HjCH!.
SPECTROSCOPY AND STRUCTURE
252
[CHAP. 12
Problem 12.19 How many kinds of equivalent H's are there in the following? ( a ) CH,CHClCH,CH,
( 6 ) p-CH,CH,-C,H,-CH,CH,
(c)
Br,CHCH,CH,CH,Br
4
( a ) Five as shown:
Hb H'
I
t
CH,"-C*-C-CH~ A1
Ad
The two H's of CH, are not equivalent, because of the presence of a chiral C. Replacing H' and Hd separately by X gives two diastereomers; H' and Hd are diasteriomeric H's. (6) Three: All four aromatic H's are equivalent, as are the six in the two CH, and the four in the CH,'s. (c) Four: Br,CH"CH,bCHzCH;Br. Problem 12.20 How many kinds of H's are there in the isomers of dimethylcyclopropane?
4
Dimethylcyclopropane has three isomers, shown with labeled H's to indicate differences and equivalencies.
Hh
CH,"
CHY
CH; I
I
I
I
I
I
CHY
Hb
I
I
I
Hb
H"
H'
1,lDimethylcyclopropane (1)
cis-1,2Dimethylcyclopropane (11)
tram-l,2Dimethylcyclopropane (111)
In 11, H' and Hd are different since H' is cis to the CH,'s and Hd is frans. In 111 the CH, H's are equivalent; they are each cis to a CH, and tram to a CH,. Problem 12.21 Label the different types of H's in ( a ) p-ethoxyacetophenone, ( 6 ) o-chlorotoluene, (c) 4 1'3-dimet h y 1benzene (M - x ylene).
RELATIVE PEAK AREAS; H-COUNTING The area under a signal graph is directly proportional to the number of equivalent H's giving the signal. For example, the compound C,H;CH,bC(CH;), has five aromatic protons ( a ) , two benzylic protons ( b ) , and nine equivalent CH, protons (c). Its nmr spectrum shows three peaks for the three different kinds of H, which appear at: ( a ) S = 7.1 ppm (aromatic H), (b) S = 2.2 ppm (benzylic H), (c) 6 = 0.9 ppm (1" H). The relative areas under the peaks are a : b : c = 5 : 2 : 9. The nmr instrument integrates the areas as follows: When no signal is present, it draws a horizontal line. When the signal is reached the line ascends and levels off when the signal ends. The relative distance from plateau to plateau gives the relative area. See Fig. 12-12 for typical nmr spectra showing integration.
CHAP. 121
SPECTROSCOPY AND STRUCTURE
253
Problem 12.22 ( a ) Suggest a structure for a compound C,H,, showing low-resolution nmr signals at 6 values 4 of 7.1, 2.2, 1.5, and 0.9 ppm. ( b ) Give the relative signal areas for the compound. ( a ) The value 7.1 ppm indicates H’s on a benzene ring. The formula shows three more C’s, which might be
attached to the ring as shown below (assuming that, since this is an alkylbenzene, all aromatic H’s are equivalent) : (1) 3 CH,’s in trimethylbenzene, (CH;),C,Hi
(2) a CH, and a CH,CH, in CH;C,H:CHfCH:
(3) a CH,CH,CH, in C,H:CH:CHZCH:
(4) a CH(CH,), in C,H:CHb(CH:),
Compounds (1) and (4) can be eliminated because they would give two and three signals, respectively, rather than the four observed signals. Although (2) has four signals, H” and H“ are different benzylic H’s and the compound should have two signals in the region 3.0-2.2ppm rather than the single observed signal. Hence (2) can be eliminated. Only (3) can give the four observed signals with the proper chemical shifts. (b) 5:2:2:3. Problem 12.23 What compound C,H,O has nmr signals at S = 7.3, 4.4, and 3.7 ppm, with relative areas 4 7 : 2.9 : 1.4, respectively?
The relative areas of the three different kinds of H become 5 : 2 : 1 on dividing by 1.4. That is, five H’s contribute to the 6 = 7.3, 2 H’s to 6 = 4.4 and 1 H to 6 = 3.7, for a total of eight H’s, which is consistent with the formula. The five H’s at 6 = 7.2 are aromatic, indicating a C,H, compound. The remaining portion of the formula comprises the CH,OH group. The H at 6 = 3.7 is part of OH. The two H’s at 6 = 4.4 are benzylic and alpha to OH. The compound is C,H,CH,OH, benzyl alcohol. Problem 12.24 The pmr (proton magnetic resonance) spectrum of CH,OCH,CH,OCH, shows chemical shifts of 3.4 and 3.2 ppm, with corresponding peak areas in the ratio 2 : 3. Are these numbers consistent with the given structure? 4
Yes. Since the CH,’s and CH,’s are each equivalent, only two signals appear. Both are shifted downfield by the 0, the CH,-signal more than the CH,-signal. Integration provides the correct ratio, 2 : 3 = 4 : 6, the actual numbers of H’s engendering the signals.
PEAK-SPLITTING; SPIN-SPIN COUPLJNG Because of spin-spin coupling, most nmr spectra do not show simple single peaks but rather groups of peaks that tend to cluster about certain S values. To see how this coupling arises we examine the molecular fragment
-LH&: present in a very large number of like molecules. The signal for Hb is shifted slightly upfield or downfield depending on whether the spin of Ha is aligned against or with the applied field. Since in about half of the molecules the H“ are spinning t and in half &, Hb gives rise to a doublet instead of a singlet. The effect is reciprocal: the two Hb’s split the signal of H“. There are four spin states of approximately equal probability for the two Hb:
t t;
t
5.9
5. t ;
5.5.
Because the middle two spin states have the same effect, the signal of H“ is split into a triplet with relative intensities 1 : 2 : 1. In the molecular fragment
SPECTROSCOPY AND STRUCTURE
254
[CHAP. 12
the three Hh’s produce a doublet, due to the effect of the single H”. H”, however, yields a quartet, due to the effects of the three Hb, which may be spinning as follows:
ttt; Intensities
ttl* tlt’ ltt;
t5.5.. l l t ’ I t ! ;
5.11
3
1
3
1
The entire quartet integrates for one H. Spin-spin coupling usually, but not always, occurs between nonequivalent H’s on adjacent atoms. In general, if n equivalent H’s are affecting the peak of H’s on an adjacent C, the peak is split into n + 1 peaks. A symmetrical multiplet is an ideal condition and not always observed in practice. Problem 12.25 In which of the following molecules does spin-spin coupling occur? If splitting is observed give the multiplicity of each kind of H. ( y 3
(a) CICH,CH,CI
H
‘C=C (d) / Br
/H
hr
I
(6) CICH,CH,I H (e)
)=c
Br
CH,CH, I
/cl
U) ‘c=c
H ‘
C/
/
H
H ‘
(g )
0 CH,CH,
4
Splitting is not observed for ( a ) or ( d ) , which each have only equivalent H’s, or for ( c ) , which has no nonequivalent H’s on adjacent C’s. The H’s of CH, in (6) are nonequivalent and each signal is split into a triplet ( n = 2; 2 + 1 = 3). In (e) the two H’s are not equivalent and each generates a doublet. The vinyl H’s in ( f ) are nonequivalent since one is cis to Cl and the other is cis to I; each gives rise to a doublet. In this case the interacting H’s are on the same C. Compound ( g ) gives a singlet for the equivalent uncoupled aromatic H’s, quartet for the H’s of the two equivalent CH, groups coupled with CH,, and a triplet for the two equivalent CH, groups coupled with CH,. Problem 12.26 Give schematic coupling patterns showing relative chemical shifts observed for the following common alkyl groups: ( a ) ethyl, --CH,CH,; ( 6 ) isopropyl -CH(CH,),; (c) 1-butyl, -C(CH,),, 4 See Fig. 12-5.
2 He : 3 H’ (a) X-CHfCH!
1 H’ : 6H’ ( b ) X 4 H ‘ (CH%
9H (c) X--C(CHih
Fig. 12-5
Problem 12.27 Why is splitting observed in 2-methylpropene but not in l-chloro-2,2-dimethyIpropane? 4 See Fig. 12-6. In (d), H“ is more downfield than Hh because Cl is more electron-withdrawing than Br. In (CH”,),C-€H:CI, H” and H b are not on adjacent C’s and are too far away from one another to couple.
In
CHS CH;’
‘c=c
/Hb \Hb
CHAP. 121
SPECTROSCOPY AND STRUCTURE
255
although H" and H bare not on adjacent C's, they are close enough to couple because of the shorter C=C bond. Problem 12.28 Sketch the nmr spectra of ( a ) 1'1-dichloroethane, (6) 1,1,2-trkhloroethane, (c) 1,1,2,2tetrachloroethane and (d) 1-bromo-2-chloroethane. In each case indicate the "staircase" curve of relative areas. 4
31 1
m
H"
'
II
1111)) I
Hb
H"
( a ) ClKH"CH,b
Hh
(6) C1tCH"CH;Cl
r-J H" (C )
C12CH"CH"Clz
Hb
( d ) CICH2"CHSBr
Fig. 12-6
Problem 12.29 F's couple H's in the same way as do other H's. Predict the splitting in the nmr spectrum of 2,2-difluoropropane. 4
In CH;CF,CH; the two F's split the Ha-signal into a 1 : 2 : 1 triplet. The F-signal, when detected by a special probe, would be a septet. Problem 12.30 Deuterium does not give a signal in the proton nmr spectrum nor does it split signals of nearby protons. Thus D's might just as well not be there. What is the difference between nmr spectra of CH,CH,CI and CH,CHDCI? 4
See Problem 12.26(a) for the spectrum of CH,CH,CI. CHfSCHhDClhas a doublet for Ha and, more downfield, a quartet for H'. Problem 12.31 The stable anti conformer of CH,CH,CI shows a nonequivalency of the CH, H's:
H* is anti to the C1 while the H,'s are gauche. Why does H* not give a signal different from the H,'s? (Instead, the three H's produce an equivalent triplet.) 4
256
SPECTROSCOPY AND STRUCTURE
[CHAP. 12
Rotation around the C - € bond is rapid. Detection by the nmr spectrometer is slower. The spectrometer therefore detects the average condition, which is the same for each H; 1/3 anti and 2 / 3 gauche. Problem 12.32 What information can you deduce from the fact that one signal in the nmr spectrum of 2,2,6,6-tetradeuterobromocyclohexanechanges to two smaller signals when the spectrum is taken at low 4 temperatures? As the ring changes its conformation from one chair form to another, the Br--C-H proton changes its position from axial to equatorial (Fig. 12-7). An axial H and an equatorial H have different chemical shifts. But at room temperature the ring "flips" too fast for the instrument to detect the difference; it senses the average condition. At low temperatures this process becomes slow enough so that the instrument can pick up the two different H,, and H,, signals. D's are used to ensure that the H under study is a singlet. Br
D
D
Fig. 12-7
COUPLING CONSTANTS
Figure 12-8 summarizes the nmr spectrum of CH,CH,Cl by using a vertical line segment for each peak. The spacing between lines within a multiplet is typically constant; furthermore, the spacing in each coupled multiplet is constant. This constant distance, which is independent of If,, is called the coupling constant, J, and is expressed in Hz. The value of the coupling constant depends on the structural relationship of the coupled H's, and becomes a valuable tool for structure proof. Some typical values are given in Table 12-4.
Fig. 12-8 Problem 12.33 A compound, CzH2BrC1,has two doublets, J
=
16 Hz. Use Table 12-4 to suggest a structure. 4
The three possibilities showing J vaues for two doublets are:
gem-vinyl H's J = 0-3 HZ
The trans isomer fits the data.
cis H's
J
= 7-12
HZ
tram H's J = 13-18Hz
257
SPECTROSCOPY AND STRUCTURE
CHAP. 121
Table 12-4
Type of H’s (free rotation of C-C)
H-C-C-H H
\
J, Hz
/
-7
13-18
trans
H
/H >C==C\
7-12
cis
/H =ch
0-3
Phenyl H’s
12.5
ortho
6-9
meta para
0- I
1-3
13C NMR (CMR)
Although I2C is not nmr-active, the I3C isotope, which has a natural abundance of about 1%, is nmr-active because it has an odd number of neutrons. Modern techniques of Fourier transform spectroscopy are capable of detecting signals of this isotope in concentrated solution. Since H and I3C absorb at different frequencies, proton signals do not appear in the I3C spectra. However, spin-spin coupling between attached H’s and a 13C (proton-coupling) is observed. The same (n + 1)-rule that is used for H’s on adjacent C’s is used to analyze the 13 C-H coupling pattern. Thus, the I’C of a I3CH, group gives rise to a quartet. To avoid this coupling, a proton-decoupled spectrum can be taken, in which each different I3C nucleus appears as a sharp singlet. Both types of spectra are useful. The decoupled spectrum permits the counting of different I3C’sin the molecule; the coupled spectrum allows the determination of the number of H’s attached to each ”C. Figure 12-9 shows ( a ) the proton-coupled and (6) the proton-decoupled spectrum of dichloroacetic acid, CHC1,COOH. Each I3C has a characteristic chemical shift 6, as listed in Table 12-5. Note that the 13Cof 1 3 C 4 is the most downfield, with aromatic 13C’ssomewhat more upfield. Integration of the peak areas affords the relative number of 13C’seach peak represents, as shown in Fig. 12-10 for p-BrC,H,COCH,. Problem 12.34 Why is spin-spin coupling between adjacent l3Cs not observed?
4
Since the natural abundance of this isotope is so low, the chance of finding two 13C’snext to each other is practically nil. However, if a compound were synthesized with only 13C’s,then coupling would be observed. Problem 12.35 How many peaks would be evidenced in the decoupled spectrum of ( a ) methylcyclohexane? (6) cyclohexene? (c) 1-methylcyclohexene? 4
33 5
(a)
Five peaks
0
11
2
(6) Three peaks
5
(c)
Seven peaks
258
[CHAP. 12
SPECTROSCOPY AND STRUCTURE
t
C2
x
.C1 U
2
U I
200
180
160
140
120 100 80 60 e--- Chemical Shift, 6 ppm ( a ) Proton-coupled
40
20
0
C2
t
x
.CI U
2
U
TMS
c (
2 200
180
160
140
120
80
100
I--- Chemical
1
I
1
I
60
40
20
0
Shift, 6 ppm
(6) Proton-decoupled
Fig. 12-9
Table 12-5
259
SPECTROSCOPY AND STRUCTURE
CHAP. 121
200
I
I
180
160
I
140
I
I
I
I
120 100 80 60 Chemical Shift, 6 ppm
1
1
1
40
20
0
Fig. 12-10
Problem 12.36 Sketch ( a ) the coupled and ( b ) the decoupled '3C-spectrumof
y . 3
The
l3C__o
1 p
2
p
3
group is electron-withdrawing and causes a downfield shift of other attached I3C atoms.
4
See Fig. 12-11.
C2
1
210
I
I
I
1
I
1
1
I
I
1
1
200
180
160
140
120
100
80
60
40
20
0
Fig. 12-11
12.6 MASS SPECTROSCOPY
When exposed to sufficient energy, a molecule may lose an electron to form a cation-radical, which then may undergo fragmentation of bonds. These processes make mass spectroscopy (ms)a useful tool for structure proof. Very small concentrations of the parent molecules (RS), in the vapor state, are ionized by a beam of energetic electrons ( e - ) ,
260
SPECTROSCOPY AND STRUCTURE
R:S+e-+2eParent molecule
+R*S'----*R*+S' Parent cation-radical
or
[CHAP. 12
R' + S .
Fragment ions
and a number of parent cation-radicals (RS') may then fragment to give other cations and neutral species. Fragment ions can undergo further bond-cleavage to give even smaller cations and neutral species. In a mass spectrogram, sharp peaks appear at the values of rnle (the mass-to-charge ratio) for the various cations. The relative heights (intensities) of the peaks represent relative abundances of the cations. Most cations have a charge of + 1 and therefore most peaks record the masses of the cations. Fragmentation tends to give the more stable cations. The most abundant cations are the most stable ones. The standard against which all peak intensities in a given mass spectrogram are measured is the most intense peak, called the base peak, which is arbitrarily assigned the value 100. If few parent molecules fragment-not a typical situation-the parent cation will furnish the base peak. Unless all the parent ions fragment, which rarely happens, the largest observed m l e value is the molecular weight of the parent (RS) molecule. This generalization overlooks the presence of naturally occurring isotopes in the parent. Thus, the chances of finding a I3C atom in an organic molecule are 1.11%; the chances of finding two are negligible. Therefore, the instrument detects a small peak at mRS+ 1, owing to the 13 C-containing parent. The chances of finding 'H in a molecule are negligible. The masses and possible structures of fragment cations, especially the more stable ones, are clues to the structure of the original molecule. However, rearrangement of cations complicates the interpretation. Mass spectra, like other spectra, are unique properties used t o identify known and unknown compounds. Problem 12.37 ( a ) What molecular formulas containing only C and H can be assigned to a cation with rnle equal to (i) 43, (ii) 65, (iii) 91? (Assume that e = + l . ) (6) What combination of C, H and N can account for an mle of (i) 43, (ii) 57? (Assume that e = + l . ) 4 Divide by 12 to get the number of C's; the remainder of the weight is due to H's. (i) C , H f , (ii) C,H;, (iii) C,H:. (b) ( i ) If one N is present, subtracting 14 leaves a mass of 29, which means 2 C's (mass of 24) are present. Therefore the formula is C,H,N+. If 2 N's are present, it is CH,Nl. (ii) CH,Nf, C2HSNI or C,H,N+
(a)
Problem 12.38 ( a ) Do parent (molecular) ions, RS', of hydrocarbons ever have odd rnle values? (6) If an RS' contains only C, H and 0, may its mle value be either odd or even? (c) If an RS' contains only C, H and N, may its mle value be either odd or even? ( d ) Why cannot an ion, rnle = 31, be C,H:? What might it be?
4
No. Hydrocarbons, and their parent ions, must have an even number of H's: CnHZn,_?, C n H z nC,H2n-2, , C n H Z n - 6etc. , Since the atomic weight of C is even (12), the mle values must be even. The presence of 0 in a formula does not change the ratio of C to H. Since the mass of 0 is even (16), the mass of RS' with C, H and 0 must be even. The presence of each N (rn = 14) requires an additional H (CnHZn+,N,C n H 2 n t l N ,C,H,,-,N). Therefore, if the number of N's is odd, an odd number of H's and an odd rnle value result. An even number of N's requires an even number of H's and an even mle value. These statements apply only to parent ions, nor to fragment ions. The largest number of H's for two C's is six (C,H,). Some possibilities are CH,O' and CH,N+. Problem 12.39 Which electron is most likely to be lost in the ionization of the following compounds? Write an 4 electronic structure for RS'. ( a ) CH,, (6) CH,CH,, ( c ) H,C=CH,, ( d ) CH,CI, (e) H , C = O . The electron in the highest-energy M O is most likely to be lost. The relative energies for electrons are: n > T > a (Problem 12.3). In a compound with no n or n electrons, the electron most likely comes from the highest-energy a bond.
CHAP. 121
H ( a ) H:C;H
(+
H
H H (b) H:C;C:H H H (c)
261
SPECTROSCOPY AND STRUCTURE
H,CLCH,
is the charge due to the missing electron)
(the C-C
bond is weaker than the C-H
(electron comes from the
7~
bond)
bond)
( d ) H3C:Cl? (an n electron is lost) (e)
H,C=O'
(an n electron is lost)
Problem 12.40 Write equations involving the electron-dot formulas for each fragmentation used to explain the following. ( a ) Isobutane, a typical branched-chain alkane, has a lower-intensity RS' peak than does n-butane, a typical unbranched alkane. (6) All 1" alcohols, RCH,CH,OH, have a prominent fragment cation at mle = 31. ( c ) All C,H,CH,R-type hydrocarbons have a prominent fragment cation at mle = 91. ( d ) Alkenes of the type H,C=CHCH,R have a prominent fragment cation at rnle = 41. ( e ) Aldehydes,
H
I
R-C=O show intense peaks at (rnle),, - 1 and mle = 29.
4
(a) Cleavage of C--C is more likely than cleavage of (the stronger) C-H. isobutane, H H
I
(H,C),C?CH,
-
I
(H,C),C+
Fragmentation of RS' for
+ CH,
gives a 2" R', which is more stable than the 1" R' from n-butane,
H H
H
I I
I + *CH,CH, IH
H,CC?CCH,
+H,CC+
I t
H H
Hence RS' of isobutane undergoes fragmentation more readily than does RS' of n-butane, and fewer RS' fragments of isobutane survive. Consequently, isobutane, typical of branched-chain alkanes, has a low-intensity RS' peak compared with n-butane.
H ,! R-&~:kH,OH
-
H
R-
I :b+*HI H'
A -€'next to an 0 is stabilized by extended generally undergoes cleavage of the bond
H
I I
C,H,: C?R
H
-
R.
*
T
+ CH,OH +
c--*
rnle
\a
H,C=()H
= 31
bonding (resonance). The RS' species of alcohols
*t 0
C E-OH
+ C,H,:CHr
__*
stable benzyl R'
. ,; a-.-
0:
H,CLCH-C;'R
a more stable aromatic cycloheptatrienyl cation, m l e = 91
+
H,C=CH-CH,
*\\
stable allylic
cation, m l e = 41
+ -R
SPECTROSCOPY AND STRUCTURE
262
[CHAP. 12
Problem 12.41 Why is less than 1mg of parent compound used for mass spectral analysis?
4
A relatively small number of molecules are taken to prevent collision and reaction between fragments. Combination of fragments might lead to ions with larger masses than RS', making it impossible to determine the molecular weight. The fragmentation pattern would also become confusing.
Problem 12.42 Suggest structures for three of the peaks (86, 43, and 42) in the mass spectrum of n-hexane.
4
The R S ' of n-hexane, CH,CH,CH,CH,CH,CH,, has a mass = 86. A split in half would give a fragment CH,CH,CH, of mass 43. The 42 fragment may be made up of three CH,'s coming from the middle of the chain: .CH,CH,CH,'. Problem 12.43 Give the structure of a compound, C,,H,,O, whose mass spectrum shows r n l e values of 15,
43, 57, 91, 105, and 148.
4
The value 15 suggests a 'CH,. Because 43 - 15 = 28, the mass of a c=O group, the value of 43 could mean an acetyl, CH,CO, group in the compound. The highest value, 148, gives the molecular weight. Cleaving an acetyl group (rnle = 43) from 148 gives 105, which is an observed peak. Next below 105 is 91, a difference of 14; this suggests a CH, attached to CH,CO. So far we have CH,COCH, adding up to 57, leaving 148; 57 = 91 to be accounted for. This peak is likely to be C,H:, whose precursor is the stable benzyl cation, C,H,CH,. The structure is CH, ---CH,---CH,-C,H,.
3 0
Problem 12.44 How could mass spectroscopy distinguish among the three deuterated forms of ethyl methyl ketone?
DCH,CH,COCH,
(1)
(2) CH,CH,COCH,D
(3) CH,CHDCOCH,
4
The expected peaks for each compound are shown in Table 12-6; each has a different combination of peaks.
b
m/e
15 16
29 30
43 44
DCH,CH,COCH, CHf DCH;
-
DCH,CH,' CH,CO'
-
CH,CH2COCH,D CHf DCH; CH,CH,'
-
DCH,CO'
CH,CHDCOCH, CHf
-
CH,CHD+ CH,CO+
-
Problem 12.45 A prominent peak in the mass spectrum of 2-methyl-2-phenylbutaneis at r n l e = 119. To what fragment cation is this peak due? Show how cleavage occurs to form this ion. 4
SPECTROSCOPY AND STRUCTURE
CHAP. 121
263
The RS is CH3 CH
' 5-1
+
NI
C-CHZCH, n
CH3
Cleavage between C" and C' gives C,H,C(CH,), , m / e = 119.
Supplementary Problems Problem 12.46 Match the type of spectrometer with the kind of information which it can provide the chemist.
1. 2. 3. 4.
Mass Infrared Ultraviolet Nuclear magnetic resonance
A. functional groups B. molecular weights C. proton environment D. conjugation
4
l B , 2 A , 3D, 4C. Problem 12.47 Why do colored organic compounds, such as @-carotene(the orange pigment isolated from carrots), have extended conjugation? 4
The more effective electron delocalization in molecules with extended conjugation narrows the energy gap between the HOMO and the LUMO. Thus, visible radiation of lower frequency (longer wavelength) is absorbed in the HOMO-, LUMO electron transition. Problem 12.48 ( a ) Account for the fact that benzene absorbs at 254nm in the uv, and phenol, C,H,OH, 4 absorbs at 280nm. ( b ) Where would one expect 1,3,5-hexatriene to absorb, relative to benzene? ( a ) The p orbital on 0, housing a pair of n electrons, overlaps with the cyclic v system of benzene, thereby
extending the electron delocalization and decreasing the energy for the HOMO +LUMO transition. Consequently, phenol absorbs at longer wavelengths. (b) The triene absorbs at a longer A,,, (275 nm). Since the cyclic 7r system of benzene has a lower energy than the linear T system of the triene, benzene absorbs radiation of shorter wavelength. Problem 12.49 Which peaks in the ir spectra distinguish cyclohexane from cyclohexene?
4
One of the C-H stretches in cyclohexene is above 3000cm-' (C,,z-H); in cyclohexane the C-H itretches are below 3000cm-' (CsP~--H). Cyclohexene has a C==C stretch at about 1650cm-'. Problem 12.50 The mass spectrum of a compound containing C, H, 0, and N gives a maximum mle of 121. [ts ir spectrum shows peaks at 700, 750, 1520, 1685, and 3100cm-', and a twin peak at 3440cm-'. What is a Feasonable structure for the compound? 4
The molecular weight is 121. Since the mass is odd, there must be an odd number of N's [Problem
12.38(c)]. The ir data indicate the following groups to be present:
1520 cm-' : aromatic ring (1450-1600 cm-' range) I 1685 cm-': C=O stretch of amide structure -CO-N (1630-1690 cm-' range) 3100 cm-I: aromatic C-H bond (3000-3100 cm-' range) 3440 cm- ' : -N-H in amine or amide (3300-3500 cm- ' range) 700, 750 cm-' : monosubstituted phenyl
264
SPECTROSCOPY AND STRUCTURE
A twin peak due to symmetric and antisymmetric N-H together we find that the compound is benzamide,
[CHAP. 12
stretches means an NH, group. By putting the pieces
0
II
C,H,-C-NH, Problem 12.51 The ir spectrum of methyl salicylate, o-HOC,H,COOCH,, has peaks at 3300, 1700, 3050, 1540, 1590, and 2990cm-'. Correlate these peaks with the following structures: ( a ) CH,, ( b ) C=O, (c) OH 4 group on the ring, and ( d ) aromatic ring. ( a ) 2990cm-'; ( b ) 1700cm-'; (c)
3300cm-'; ( d ) 3050, 1540, 1590cm-'.
Problem 12.52 Calculate E, for a compound whose maximum absorbance is A,,, = 1.2. The cell length ( 8 ) is 1.0cm and the concentration is O.O76g/L. The mass spectrum of the compound has the largest m / e value at 100. 4
The molecular weight is 100 g/mol; therefore, c = 7.6 x lO-, mol/L and
Emax =
-= Amax
CI
1.2 (7.6 x 10-,)( 1.0)
=
1600
Problem 12.53 Methanol is a good solvent for uv but not for ir determinations. Why?
4
Methanol absorbs in the uv in 183 nm, which is below 190 nm, the cutoff for most spectrophotometers, and therefore it doesn't interfere. Its ir spectrum has bands in most regions and therefore it cannot be used. Solvents such as CCI, and CS, have few interfering bands and are preferred for ir determinations. Problem 12.54 How can nmr distinguish between p-xylene and ethylbenzene?
4
p-CH;C6HtCHj has two kinds of H's, and its nmr spectrum has two singlets. C,H:CHfCHi has three kinds of H's, and its nmr spectrum has a singlet for H", a quartet for Hh, and a triplet for H'. Problem 12.55 A compound, C,H,O, contains a C=O compound is an aldehyde or a ketone?
group. How could nmr establish whether this
4
If an aldehyde, the compound is CH,CH,CHO, with three multiplet peaks and a downfield signal for
H
-c=o (6 = 9-10ppm). If a ketone, it is (CH,),C=O,
with one singlet.
Problem 12.56 Describe the expected appearance of the nmr spectrum of n-propylbenzene,
0
CH;-
CH,~-CH;
4
(All chemical shifts are given as 6 in ppm). At about 7.2 there is a signal due to the aromatic ring H's. At 2.3-3 there is a triplet for the benzylic H". If the coupling effects of H" and H' with Hb were similar, the multiplet for Hb would be a sextet ( 5 + 1). If the effects were different, Hb would be split into a triplet by an H", and each peak of the triplet would be further split into a quartet. Some of these peaks might coincide. At best, Hb is a complex multiplet at about 1.3. The H' signal is a triplet at about 0.9. The relative peak intensities will be: (ring H) : H" : Hb : H' = 5 : 2 : 2 : 3.
CHAP. 12)
SPECTROSCOPY AND STRUCTURE
265
Problem 12.57 The nmr spectrum of a compound, C,H,ClF,, has two noncoupled triplets. Triplet A is 1.5 4 times the intensity of the more downfield triplet B . Suggest a structure.
Since A is 1.5 times as intense as B and the molecular formula has five H's, A must come from CH, and B from CH,. A and B are not coupling each other; if they did, we would find a triplet and a quartet. A and B must each be coupled to two F's. B is more downfield than A , indicating a --CH,Cl group. The compound is ClCfi,CF,Cfi, See Problem 12.29 for H / F coupling. Problem 12.58 The nmr spectrum of a dichloropropane shows a quintuplet and, downfield, a triplet of about 4 twice the intensity. Is the isomer 1,l-, 1,2-, 1,3-, or 2,2-dichloropropane?
We would expect the following signals: 1,l-Dichloropropane, Cl,CH"CH,bCHi: a triplet (Hr), a complex multiplet more downfield (Hb), and a triplet still more downfield (H"). 1,2-Dichloropropane7ClCH;CH*CH;: a doublet (Hr), another doublet more downfield (H"), a complex
I
Cl multiplet most downfield (Hb). 1,3-Dichloropropane, C1CH;CHfCH;CI: a quintuplet (Hh), and, downfield, a triplet (H") C1
I
2,2-Dichloropropane, CHfl--C-CHfl: a singlet (H").
I
c1 The compound is 1,3-dichloropropane. Problem 12.59 Consider the coupled I3C nmr spectra of ( a ) 1,3,5-trimethylbenzene9( 6 ) 1,2,3-trimethylbenzene, ( c ) n-propylbenzene. (i) Write each structure and label the different kinds of C's by numerals I, 2, . . . . (ii) Show the splitting for each peak by the letters s for singlet, d for doublet, t for triplet, and q for quartet. 4
Problem 12.60 Indicate whether the following statements are true or false and give a reason in each case. ( a ) The ir spectra are identical for the enantiomers
Tir
B r T i H , and CH,
H
H
(6) The nmr spectra of the compounds in ( a ) are also identical. ( c ) The ir spectrum of 1-hexene has more peaks than the uv spectrum. (d) Compared to CH,CH2CH0, the n + 7r* for H,C=CHCHO has shifted to a shorter wavelength (blue shift). 4
266
SPECTROSCOPY AND STRUCTURE
(CHAP. 12
9
A
-
1
8
1
7
1
1
I
J
6
4
1
1
I
I
3
2
1
0
1 3
I 2
1 I
0
12
1 8
1 7
1
1
6
5
1 4
1
4 PPm
I 7
I
6
I
I 5
4 8 9
PPm
Fig. 12-12
I
3
I
2
I
I
I
0
CHAP, 121
267
SPECTROSCOPY AND STRUCTURE
(a) and ( b )
True. The compounds are enantiomers which have identical vibrational modes and proton resonances. ( c ) True. The ir spectrum has peaks for stretching and bending of all bonds, while the uv spectrum has only one peak for excitation of a ?r electron. ( d ) False. The shift is to longer wavelength (red shift), since
H,C=CHC=O
I
H is a conjugated system.
Problem 12.61 Assign the nmr spectra shown in Fig. 12-12 to the appropriate monochlorination products of 2,4-dimethylpentane (C,H,,CI) and justify your assignment. Note the integration assignments drawn in the spectra. 4 The three possible structures are:
CH3
CH3
CH3
CICH,-(!H-CH2-AH--CH3
CH3J
-CH2-CH-CH3 4 I
2IH33
I
CH3
CH3-CH--CH-&H-CH, I Cl
5
c1 1-Chloro-2,4-
2-Chloro-2.4-
dimethylpentane (I)
CH3
I
3-Chloro-2.4-
dimethylpentane (11)
dimethylpentane (111)
The best clue is the most downfield signal arising from the H's closest to CI. In spectrum ( a ) the signal with the highest 6 value is a doublet, integrating for two H's, that corresponds only to structure 1 (ClCH,-). This is confirmed by the nine H's of the 3 CH,'s that are most upfield and the four 2" and 3" H's with signals between these. In spectrum ( b ) the most downfield signal is a triplet, for one H, which arises from the
H C1 H
I l l -c-c-cI I!
grouping in 111. In addition, the most upfield signal is a doublet, integrating for 12 H's, which is produced by the H's of the 4 CH,'s split by the 3" H. This leaves I1 for spectrum ( c ) . The most downfield group of irregular signals, integrating for three H's, comes from the two 2" and one 3" H on C3 and C4, respectively. The most upfield doublet, integrating for six H's, arises from the two equivalent CH,'s on C4 split by the C4 3" H. The two CH,'s on C2 give rise (six H's) to the singlet of median S value. 0
I
I
1 2
I
3
I
4
I
J
1
6
I
7
I
8
I
9
1 10
Wavelength, pm Fig. 12-13. Infrared Spectrum
I I1
1 12
1
13
I
I4
1
IS
268
SPECTROSCOPY AND STRUCTURE
[CHAP. 12
Problem 12.62 Deduce structures for the compound whose spectral data are presented in Fig. 12-13, Table 12-7, and Fig. 12-14. Assume an 0 is present in the molecule. There was no uv absorption above 180nm. 4
1
m/e
Relative intensity , o/c of base peak
1 1 ::i 2:
Table 12-7. Mass Spectrum
2:
31
39
41
42
43
44
45
59
87
102
4
11
17
6
61
4
100
11
21
0.63
Fig. 12-14. nmr Spectrum (CDCI,) From the mass spectrum, we know the RS’ has molecular weight 102. The C, H portion of the compound has a mass of 102 - 16 (for 0)= 86. Dividing by 12 gives 6 C’s, leaving a remainder of 14 for 14 H’s. The molecular formula is C,H,,O. (Had we chosen 7 C’s we would have had C,H,O, which is an impossible formula.) The compound has no degrees of unsaturation. This fact is consistent with (but not proven by) the absence of uv absorption above 180 nm. Note also the absence of C-H stretch above 3000 cm-’ (H--C,,* or H-C,,,); the 0 must be present as C U H (an alcohol) or as C-0-C (an ether). The absence of a peak in the ir at 3300-3600cm-’ precludes the presence of an 0-H group. The strong band at about 1110cm-’ represents the C-0 stretch. The structure of the alkyl groups R U R of the ether is best revealed by the nmr spectrum. The downfield septet (see blown-up signal) and the upfield doublet integrate 1 : 6. This arrangement is typical for a CH(CH,), grouping [see Problem 12.26(b)]. Both R groups are isopropyl, since no other signals are present. The compound is (CH,),CHOCH(CH,),. The significant peaks in the m$ss spectrum that are cynsistent with our assignment are: m l e = 102 - 15 (CH,)=87, which is (CH,),CHOCHCH,; 43 is (CH,),CH; 41 would be the ally1 cation H,C----CHCH,’ formed from fragmentation of (CH,),CH. The most abundant peak, mle = 45, probably comes from rearrangement of fragment ions. It could have the formula C,H,O. The presence of this peak convinces us that 0 was indeed present in the compound. The peak cannot be from a fragment with only C and H; C,H; is impossible.
Chapter 13 Alcohols and Thiols A.
ALCOHOLS
13.1 NOMENCLATURE AND H-BONDING
ROH is an alcohol and ArOH is a phenol (Chapter 19). Some alcohols have common names, usually made up of the name of the alkyl group attached to the OH and the word "alcohol"; e.g., ethyl dcohol, C2H,0H. More generally the IUPAC method is used, in which the suffix -01 replaces the -e D f the alkane to indicate the OH. The longest chain with the OH group is used as the parent, and the C bonded to the OH is called the carbinol carbon. Problem 13.1 Give a common name for each of the following alcohols and classify them as 1". 2", or 3":
CH3
( a ) CH,CH,CH,OH
( b ) CH,CH,CHCH,
( c ) CH,CHCH,OH
I
AH CH3 (e)
I
CH,-CH-OH
I I
( d ) CH3-C--OH
CH3
CH3
CH3
cf) CH,-~-CH,OH
(g)
1
C,H,CH,OH
4
CH3 ( a ) n-propyl alcohol, 1"; (6) sec-butyl alcohol, 2"; (c) isobutyl alcohol, 1"; (d) r-butyl alcohol, 3"; (e) isopropyl alcohol, 2"; (f) neopentyl alcohol, 1"; ( g ) benzyl alcohol, 1".
Problem 13.2 Name the following alcohols by the IUPAC method.
CH,CH,CH, (U)
(6)
I
CH,CH,-C-OH I CH,CH3
H
( c ) Cl-CHCH,OH
I
c1
PH
( e ) a B OH r
OH I
H,C=CH&HCH,
( a ) 3-Ethyl-3-hexanol
(c) 2,2-Dichloroethanol
(6) 3-Buten-2-01
(d) 3-Phenyl-3-hexanol
{ote that in IUPAC the OH is given a lower number than C=C
H
(e) cis-2-Bromocyclohexanol or C1.
'roblem 13.3 Supply structures for alcohols having the IUPAC names ( a ) 2-propen-1-01, ( b ) 1-ethylcyclo4 ropanol, (c) 3-phenyl-2-butanol.
LT)
H,C=CHCH,OH Ally1 alcohol
'CH,CH,
269
ALCOHOLS AND THIOLS
270
[CHAP. 13
Problem 13.4 Explain why ( a ) propanol boils at a higher temperature than the corresponding hydrocarbon; ( 6 ) propanol, unlike propane or butane, is soluble in H,O; (c) n-hexanol is not soluble in H,O; (d) dimethyl ether (CH,OCH,) and ethyl alcohol (CH,CH20H) have the same molecular weight, yet dimethyl ether has a lower boiling point (-24°C) than ethyl alcohol (78°C). 4 ( a ) Propanol can H-bond intermolecularly. C , H , T - - - H - ( ( There is also a less important dipole-dipole interaction. H C,H, (6) Propanol can H-bond with H,O: C,H,T---H-?
H
H
(c) As the R group becomes larger, ROH resembles the hydrocarbon more closely. There is little H-bonding between H 2 0 and n-hexanol. When the ratio of C to OH is more than 4, alcohols have little solubility in water. (d) The ether CH,OCH, has no H on 0 and cannot H-bond; only the weaker dipole-dipole interaction exists.
Problem 13.5 The ir spectra of tram- and cis-1,2-cyclopentanediolshow a broad band in the region 3450-3570cm-'. On dilution with CCI,, this band of the cis isomer remains unchanged, but the band of the tram isomer shifts to a higher frequency and becomes sharper. Account for this difference in behavior. 4 The OH'S of the cis isomer participate in intramofecufarH-bonding, Fig. 13-1(u), which is not affected by dilution. In the tram isomer, the H-bonding is intermolecular, Fig. 13-1(6), and dilution breaks these bonds, causing disappearance of the broad band and its replacement by a sharp OH band at higher frequency. H
H H
cis
H- 0' Intermolecular H-bond
Intramolecular H-bond
(4
(4
Fig. 13-1
13.2 PREPARATION 1. RX + OH- --!% ROH + X- ( s N 2 or s N 1 Displacement, Table 7-1) 2. Hydration of Alkenes [see Problem 6.24(d)]
3. Hydroboration-Oxidationof Alkenes [see Problem 6.24( f )] Treatment of the alkylboranes with H,O, in OH- replaces -BRCH=CH2
+ (BH3)2
-
I
(RCHCH&-B
b
with OH.
: :RCHCH,
trialky lborant
The net addition of H - O H to alkenes is cis, anti-Markovnikov, and free from rearrangement.
27 1
ALCOHOLS AND THIOLS
CHAP. 131
4. Oxymercuration-Demercuration of Alkenes
not isolated
The net addition of H-OH
is Markovnikov and is free from rearrangement.
Problem 13.6 Give structures and IUPAC names of the alcohols formed from (CH,),CHCH=CH, reaction with ( a ) dilute H,SO,; (6) B,H,, then H,O,, OH-; ( c ) Hg(OCOCH,),, H,O, then NaBH,.
by
4
( a ) The expected product is 3-methyl-2-butano1, (CH,),CHCHOHCH,,
from a Markovnikov addition of H,O. However, the major product is likely to be 2-methyl-2-butano1, (CH,),COHCH,CH,, formed by rearrangement of the intermediate R'. +H+
(CH,),CHCH=CH,
*
(CH,),CH;HCH,
3(CH,),COHCH,CH,
(CH,),;CH,CH,
(2")
(3")
(6) Anti-Markovnikov HOH addition forms (CH, ),CHCH,CH,OH, 3-methyl-1-butanol. (c) Markovnikov HOH addition with no rearrangement gives (CH,),CHCHOHCH,, 3-methyl-2-butanol. Problem 13.7 Give the structure and IUPAC name of the product formed on hydroboration-oxidation of 1-meth ylcyclohexene. 4
H and OH add cis and therefore CH, and OH are tram.
0'"' CH,
'
_I___* b"2'
2.
H
OH-. H2O2
*.
H
trans-2-Methylcyclohexanol
H
5. Carbonyl Compounds and Grignard Reagents (Increase in Number of Carbons) Grignard reagents, RMgX and ArMgX, are reacted with aldehydes or ketones and the intermediate products hydrolyzed to alcohols.
-
H
[RJMgX
+ H-C-01
~-CH,-O(MgX)'
HO
2 \j-C H 2 -O H
(1" alcohol)
Formaldehyde
H mMgX
I
+ R'-C=O
H
H
I -
--+ R'-C-O(MgX)'
I
Hi0 __*
I
R'-C-OH
I
!E!
aldehyde
R" @MgX
I
+ R'-C=O ketone
R" --+
I
-
R'-C-O(MgX)'
I
rn
(2" alcohol)
alkoxide salts
R" "20
I
R'-C-OH
+
I
M
(3" alcohol)
ALCOHOLS AND THIOLS
272
[CHAP. 13
The boxed group in the alcohol comes from the Grignard; the remainder comes from the carbonyl compound.
R + Mg(X)(OH) electrophilic site
an alkoxide: a strong conjugate base of an alcohol
nucleoPhile
alcohol
ArMgX is best made from ArBr in ether or from ArCl in tetrahydrofuran (Arc1 is not reactive in ether). ether ArBr + MgArMgBr THF
ArCl + Mg-
ArMgCl
Problem 13.8 Give 3 combinations of RMgX and a carbonyl compound that could be used to prepare. CH,
I I
4
C,H,CH,C-CH,CH,
Off This 3” alcohol is made from RMgX and a ketone, R’COR”. The possibilities are:
0
CH, (I) m / - L - C H 2 C H 3
I
from
p ] M g C I
1I
and CH,CCH,CH,
OH
0
CH3
I
(2) C , H , C H , - - C - - ( C H ~ ~
from
1-MgBMgBr
II
and C,H,CH,CCH,
bH
P
(3) C,H,CH,-C--CH,CH,
I
0
from
MgI
II
and C,H,CH,CCH,CH,
OH
The best combination is usually the one in which the two reactants share the C content as equally as possible. In this case, (1) is best. Problem 13.9 Give four limitations of the Grignard reaction.
4
(1) The halide cannot possess a functional group with an acidic H , such as OH, COOH, NH, SH, or C-=C-H, because then the carbanion of the Grignard group would remove the acidic H and be reduced. For example: [HOCH,CH,MgBr]+ (BrMg)’ -OCH,CH,H HOCH,CH,Br + Mgunstable
(2) If the halide also has a C==O (or C=N--, B N , N 4 ,s----O, M H ) group, it reacts inter- or intramolecularly with itself. (3) The reactant cannot be a vic-dihalide, because it would undergo dehalogenation: BrCH,CH,Br + MgH,C==CH, + MgBr, (4) A ketone with two bulky R groups, e.g., <(CH,),, with an organometallic compound with a bulky R’ group.
would be too sterically hindered to react
CHAP. 131
273
ALCOHOLS AND THIOLS
-
Problem 13.10 Prepare 1-butanol from ( a ) an alkene, (b) 1-chlorobutane, ( c ) 1-chloropropane and (d) ethyl bromide. 4
CH,CH,CH=CH, HO-
1. (BH3)2
+ CH,CH,CH,CH,CI-
H20
CH,CH,CH,CH,OH
CH,CH,CH,CH,OH
+ C1-
(S,2) 1-Chloropropane has one less C than the needed 1" alcohol. The Grignard reaction is used to lengthen the chain by adding H,C==O (formaldehyde). CH,CH,CH,CI
Mg
ether7 CH,CH,CH,MgCI
1. HCH=O
CH,CH,CH,CH,OH
1-Butanol is a 1" alcohol with two C's more than CH,CH,Br. Reaction of CH,CH,MgBr with ethylene oxide followed by hydrolysis gives 1-butanol. CH,CH,Br
ez:r
-cm
ICH,CH,ICH,CH,O(MgBr) 3CH,CH,CH,CH,OH
Problem 13.11 For the following pairs of halides and carbonyl compounds, give the structure of each alcohol formed by the Grignard reaction. ( a ) Bromobenzene and acetone. ( b ) p-Chlorophenol and formaldehyde. ( c ) Isopropyl chloride and benzaldehyde. (d) Chlorocyclohexane and methyl phenyl ketone. 4
(3% ( 6 ) The weakly acidic OH in p-chlorophenol prevents formation of the Grignard reagent. H H
OH
OH
W
Methylcyclohexylphen ylcarbinol
6.
I
Addition of :H- or H, to -C=O (Fixed Number of Carbons) The :H- is best provided by sodium borohydride, NaBH,, in protic solvents such as ROH or H,O.
I
NaBH, in ethanol
I
A
Lithium aluminum hydride, LiAIH,, in anhydrous ether can also be used. problem 13.12 How do the alcohols from LiAIH, or catalytic reduction of ketones differ from those derived 'rom aldehydes? 4
ALCOHOLS AND THIOLS
274
[CHAP. 13
Ketones yield 2" alcohols while aldehydes give 1" alcohols. H Ketone:
LiAIH4 H20 +
I. 2.
R-C-R'
I
R-C-R'
H IPt
6
bH a 2" alcohol
H Aldehyde:
LiAlH, H20
I.
R-i-H
2,
I
H IPt
>R-C-HA
I
OH a 1" alcohol
Problem 13.13 ( a ) What is the expected product from catalytic hydrogenation of acetophenone, C,H,COCH,? (6) One of the products of the reaction in ( a ) is C,H,CH,CH,. Explain its formation. 4 ( a ) C,H,CHOHCH,.
( 6 ) The initial product, typical of benzylic alcohols,
R i
C6H,c-H
I
H can be further reduced with H,. This reaction is a hydrogenolysis (bond-breaking by H2).
C,H,CHOHCH,
+ H,=
C,H,CH,CH,
+ H,O
Problem 13.14 Reduction of H,C=CHCHO with NaBH, gives a product different from that of catalytic hydrogenation (H2/Ni). What are the products? 4 H,C=CHCH,OH
E
H /Ni
H,C=CHCHO 4 H,CCH,CH,OH Acrolein
selective reduction of
nonselective reduction
'c=o
/
7. Reduction of RCOOH or RCOOR with LiAIH,
The initial product is a lithium alkoxide salt, which is then hydrolyzed to a 1" alcohol.
F
C,H, - O H Propionic acid
0
II
C, H,C--OCH,
* 9
C,H,CH,O- L i ' z C , H , C H , O H a lithium alkoxide
Propanol
Methyl propionate
13.3 REACTIONS 1. The electron pairs on 0 make alcohols Lewis bases. 2. H of OH is very weakly acidic. The order of decreasing acidity is
H,O > ROH( 1") > ROH(2") > ROH(3") > R m H 9 RCH,
CHAP. 131
275
ALCOHOLS AND THIOLS
3. 1" and 2" alcohols have at least one H on the carbinol C and are oxidized to carbonyl compounds. They also lose H, in the presence of Cu (300°C) to give carbonyl compounds. 4. Formation of alkyl halides (Problem 7.4(a)-(d)). 5 . Intramolecular dehydration to alkenes [Problems 6.14 through 6.171. 6. Intermolecular dehydration 1to ethers. 2 R O H x ROR + H 2 0 H2S04
7. Ester formation. ROH
+ R'COH II
0
H2S04
R'COR
I1
+ H20
0 ester
8. Inorganic ester formation. With cold conc. H 2 S 0 4 , sulfate esters are formed. ROH + H,S04-
cold
H,O
+ ROS0,H-
ROH
ROS0,OR
an alkyl acid sulfate
a dialkyl sulfate
Similarly, alkyl phosphates are formed with H 3 P 0 4 , phosphoric acid. Problem 13.15 Supply equations for the formation from phosphoric acid of ( a ) alkyl phosphate esters by reactions with an alcohol, in which each acidic H is replaced and H,O is eliminated; (6) phosphoric anhydrides on heating to eliminate H,O. 4 / OH +ROH / O R +ROH O==P-OH-O=P-OH-O=P-OR-O=P--OR -HzO -HzO OH 'OH
'
Phosphoric acid
an alkyl dihydrogen phosphate
0
OH
'OH
II
0
II
+H3P0,
I
I
-HzO
OH
/OR
+ROH
'OR
-HzO
a dialkyl hydrogen phosphate
HO-P--O-P--OHOH
/OR
a trialkyl phosphate
0
0
II
II I
HO-PUP-O-P--OH
I
I
OH
OH
Diphosphoric acid
Phosphoric acid
0
II
OH
OH
Triphosphoric acid
Problem 13.16 Alkyl esters of di- and triphosphoric acid are important in biochemistry because they are stable in the aqueous medium of living cells and are hydrolyzed by enzymes to supply the energy needed for muscle :ontraction and other processes. ( a ) Give structural formulas for these esters. (6) Write equations for the hydrolysis reactions that are energy-liberating. 4
0
0
II II RO-P-O-P-OH I
OH
I
OH
an alkyl trihydrogen diphosphate
I
OH
I
OH
I
OH
an alkyl tetrahydrogen triphosphate
ALCOHOLS AND THIOLS
276
[CHAP. 13
( 6 ) Hydrolysis occurs at the ester(reaction I) or the anhydride bonds(reactions I1 and 111).
0
P /
0
11 II H 2 0+ R--O-P--O-PUP--OHI I I OH OH OH
\
I[
5;
RO-P4H OH I
I
OH
7 7
+HO-PUP-OH I I OH
I
OH
OH
I
OH
9. Alkyl sulfonates (esters of sulfonic acids), R S02RIAr) 0
RS-CI
8
+ HOR’ --+
1I 11
R S - O R ’ + HCI 0
a sulfonyl chloride
a sulfonate ester
Problem 13.17 How does sulfonate ester formation from sulfonyl chloride resemble nucleophilic displacements of alkyl halides? 4 The alcohol acts as a nucleophile and halide ion is displaced.
0 RS’-
11 /?
S-CI 0 II
-a 0
pyrldinc
11
R-S-ij-R’
+ (pyridine
H)’ C1-
Sulfonyl chlorides are prepared from the sulfonic acid or salt with PCl,.
Aromatic sulfonyl chlorides are also formed by ring chlorosulfonation with chlorosulfonic acid, HOS0,CI. ArH
+ HOS0,CI-
ArS0,CI
+ H,O
a sulfonyl chloride
Problem 13.18 ( a ) Why are sulfonate esters such as p-toluenesulfonyl (tosyl) esters widely used in S,2 reactions? ( 6 ) How is this reaction used for indirect displacement of --OH from ROH? 4 (a)
The p-toluenesulfonate anion, p-CH,C,H,SO;, like I -.
is a very weak base and is therefore a good leaving group,
(6) The alcohol is converted to a tosyl ester, which is then reacted with a nucleophile.
CHAP. 131
277
ALCOHOLS AND THIOLS
10. Oxidation. Alcohols with at least one H on the carbinol carbon (1" and 2") are oxidized to carbonyl compounds. R R RCH,OH1"
R C H S
\CHOH-+
and R
/
2"
>c=o
R
an aldehyde
a ketone
Aldehydes are further oxidized to carboxylic acids, RCOOH. To get aldehydes, milder reagents, such as the Jones (diluted chromic acid in acetone) or Collins reagent (a complex of CrO, with 2mol of pyridine), are used. Problem 13.19 Give the products for the reaction of isopropanol with conc. H,SO, at ( a ) O"C, (6) room 4 temperature, (c) 130 "C, (d) 180 "C. (a)
(CH,),CH&H,
+ HSO,
oxonium ion
(6) (CH,),CHOSO,H Isopropyl hydrogen sulfate ( c ) (CH,),CHOCH(CH,), Diisopropyl ether (d) CH,CH=CH, (high temperatures favor intramolecular dehydration) Problem 13.20 Write equations to show why alcohols cannot be used as solvents with Grignard reagents or with LiAlH,. 4
--
Strongly basic R i and HT react with weakly acidic alcohols.
n+ CH,CH,MgCl +
CH,OH
4CH,OH
+ LiAlH,
CH,CH, 4H,
+ (CH,O)-(MgCI)'
+ LiAI(OCH,),
Problem 13.21 Why may Na metal be used to remove the last traces of H,O from benzene but not from ethanol? 4
Ethanol is sufficiently acidic to react with Na, although not as vigorously as does H,O. 2C,H,OH
+ 2Na-+2C,HsO-Na' + H,
Problem 13.22 Give the main products of reaction of 1-propanol with ( a ) alkaline aq. KMnO, solution during 4 distillation; (6) hot Cu shavings; (c) CH,COOH, H'. (a) CH,CH,CHO.
Since aldehydes are oxidized further under these conditions, CH,CH,COOH is also obtained. Most of the aldehyde is removed before it can be oxidized. (6) CH,CH,CHO. The aldehyde can't be oxidized further. (c) CH,C--OCH,CH,CH,, an ester.
a
Problem 13.23 Explain the relative acidity of liquid l", 2", and 3" alcohols.
4
The order of decreasing acidity of alcohols, CH,OH > 1" > 2" > 3", is attributed to electron-releasing R's. These intensify the charge on the conjugate base, RO-, and destabilize this ion, making the acid weaker. Problem 13.24 Give simple chemical tests to distinguish ( a ) 1-pentanol and n-hexane; (6) n-butanol and t-butanol; ( c ) 1-butanol and 2-buten-1-01; (d) 1-hexanol and 1-bromohexane. 4
278
ALCOHOLS AND THIOLS
[CHAP. 13
( a ) Alcohols such as 1-pentanol dissolve in cold H,SO, [see Problem 13.19(a)]. Alkanes such as n-hexane are insoluble. (6) Unlike t-butanol (a 3" alcohol), n-butanol (a 1" alcohol) can be oxidized under mild conditions. The analytical reagent is chromic anhydride in H,SO,. A positive test is signaled when this orange-red solution turns a deep green because of the presence of Cr". (c) 2-Buten-1-01decolorizes Br, in CCI, solution; 1-butanol does not. (d) 1-Hexanol reduces orange-red CrO, to green Cr3+; alkyl halides such as 1-bromohexane do not. The halide on warming with AgNO,(EtOH) gives AgBr.
Problem 13.25 Write balanced ionic equations for the following redox reaction: CH,CHOHCH,
+ K,Cr,O, + H,SO,-CH,--CO--CH,
+ Cr,(SO,), + H 2 0 + K,SO,
hear
4
Write partial equations for the oxidation and the reduction. Then: (1) Balance charges by adding H' in acid solutions or OH- in basic solutions. (2) Balance the number of 0 ' s by adding H,O's to one side. (3) Balance the number of H's by adding H's to one side. The number added is the number of equivalents of oxidant or
-
OXIDATION (CH,),CHOH
Cr,O;-
(CH,),C=O
In acid balance charges with H ': (no change)
Cr,0,2-
+ 8H'
Balance 0 with H 2 0 : (no change)
Cr,Of-
+ 8H'
Balance H: (CH,),CHOH
-
(CH,),C=O
Balance equivalents:
Add:
--
REDUCTION
+ 2H
Cr,O:
2Cr3+
+X r 3 +
__*
Xr3'
+ 7H20
+ 8H' + 6H +2Cr3++ 7H20
3(CH3),CHOH-3(CH,),C=O + 6H Cr,O:- + 8 H t + 6H-2Cr3' + 7H20
3(CH,),CHOH
+ Cr,O:- + 8H'
-
3(CH3),C=0
+ 2Cr3' + 7H,O
Problem 13.26 How can the difference in reactivity of l", 2", and 3" alcohols with HCI be used to distinguish 4 among these kinds of alcohols, assuming the alcohols have six or less C's? The Lucas test uses conc. HCI and ZnCI, (to increase the acidity of the acid). 3" ROH + HCI+
soluble, by assumption
3' RCI + H,O insoluble liquid
The above reaction is immediate; a 2" ROH reacts within 5 min; a 1" ROH does not react at all at room temperature. Problem 13.27 In CCI, as solvent, the nmr spectrum of CH,OH shows two singlets. In (CH,),SO, there is a 4 doublet and a quartet. Explain in terms of the "slowness" of nmr detection. In CCI,, CH,OH H-bonds intermolecularly, leading to a rapid interchange of the H of 0 - H . The instrument senses an average situation and therefore there is no coupling between CH, and OH protons. In (CH,),SO, H-bonding is with solvent, and the H stays on the 0 of OH. Now coupling occurs. This technique can be used to distinguish among RCH20H, R2CHOH and R,COH, whose signals for H of OH are a triplet, a doublet and a singlet, respectively. Problem 13.28 H of OH has a chemical shift that varies with the extent of H-bonding. Why can we detect an 4 nmr signal from H of ROH by shaking the sample with D 2 0 and then retaking the spectrum? Through H-bonding the exchange reaction ROH + D,O+ ROD + DOH occurs. D-signals are not detected by a proton nmr spectrometer. If the signal in question disappears, it came from H of ROH. A new signal appears at about S =5ppm, due to DOH.
CHAP. 13)
279
ALCOHOLS AND THIOLS
13.4 SUMMARY OF ALCOHOL CHEMISTRY
PROP€RTES
PREPARATION 1. Hydration of Alkenes
1. Basicity
+ HX+
(a) Markovnikov RCH=CH2 + H 3 0 '
Acidity
2.
\
+ Na-
Oxymercuration-demercuration: RCHOHC
+ Hg(OAc),/H,O
RCHdH,
Hydrolysis
-P
2" (sN2): RCHXCH,
+ R'COOH/H'
(3")
+ conc. H,SO,=
( b ) Esters R'COOCH,CH,R
+ H,O'
Reduction of
- ;I:i
RCOOH + LiAlH, (b) Esters
R'COOCH,CH,R
+
, , RCH,CH,OH ~
+ LiAlH,
(c) Carbonyl Compounds Aldehydes (to 1" ROH):
ll
+ NaBH,
Formaldehyde
+ H,C=O
(b) Other Aldehydes
RLi + R'CH-0-
RCH(0H)R' (2" ROH)
(c) Ketones RMgX + R'C0R'-
(b) Ether Formation (Intermolecular)
\ 5.
+ H,SO,=
(RCH,CH,),O
Oxidation
\
-
(a) Aldehydes (from 1" ROH)
+ pyridine
*
CrO,
RCH,CHO
(b) Ketones (from 2" ROH)
Organometallic Reactions
RCH,MgX
(RCH,CH,O),SO,
(a) Alkene Formation (Intramolecular) H,SO,- heat RCH=CH,
(1")-
Ketones (to 2" ROH): RCOR' + NaBH, \-
(8)
R'COOCH,CH,R
c=O Group
(a) Carboxylic Acids
4.
__*
(c) Inorganic Esters (e.g., Sulfates)
RCH,CH,X
RCH,CHO
RCH,CH,X RCH,CH,Cl RCH,CH,Br RCH,CH,F
(b) Esterification
+ OH-
3" (sN1): R3CX + H,O-R,COH
3.
-
+ HX+ SOC1,+ PBr, + SF,
RCH=CH2 + 1. B,H,, 2. HO,
(a) Alkyl Halides, R
Alkoxide salt
(a) Formation of Alkyl Halides, RX (X = CI, Br, I
H ydroboration-Oxidation
1" (sN2):
RCH,CH,O- Na'
3. Displacements
(b) Anti-Markovnikov
2.
RCH2CH,0H; X- Onium salt
RR'R'COH (3" ROH)
+ KMnO,/H'
RCOR'
280
ALCOHOLS AND THIOLS
B.
[CHAP. 13
THIOLS
13.5 GENERAL
Also called mercaptans, these compounds, with the generic formula RSH, are sulfur analogs of alcohols, just as H2S is of H 2 0 . The 0 atom of the OH has been replaced by an S atom, and the -SH group is denominated sulfhydryl. A physical characteristic is their odor; 2-butene-1-thiol, CH,CH=CHCH,SH, and 3-methyl-lbutanethiol, CH,CH(CH,)CH,CH,SH, contribute to skunk smell. Thiols are biochemically important because they are oxidized with mild reagents to disulfides which are found in insulin and proteins.
=R-S-S-R Br2
2 R-S-H a thiol
Zn. H
a disulfide
Problem 13.29 How are thiols prepared in good yield?
4
The preparation of thiols by S,2 attack of nucleophilic HS- on an alkyl halide gives poor yields because the mercaptan loses a proton to form an anion, RS-, which reacts with a second molecule of alkyl halide to form a t hioether .
HS;+ R-X:
-:x:
-
+ RSIH-
-H+
-
RS:
t
~
R-9
--+
R-:+R
+ 1..x1
-
thioether
Dialkylation is minimized by using an excess of :SH-, and avoided by using thiourea to form an alkylisothiourea salt that is then hydrolyzed.
Thiourea
Urea
Problem 13.30 Show steps in the synthesis of ethyl ethanesulfonate, CH,CH2S0,0CH,CH,, from CH,CH,Br and any inorganic reagents. 4
The alkyl sulfonic acid is made by oxidizing the thiol, which in turn comes from the halide. excess
CH,CH,SH-
KMn04
CH,CH,SO,H A L ?
CH,CH,OH
+ CH,CH,SO,CI-
pyridine U
bust
CH,CH,SO,OCH,CH,
Problem 13.31 Why are mercaptans ( a ) more acidic ( K , = 10-l’) than alcohols ( K O= l0-l’) and (b) more 4 nucleophilic than alcohols? ( a ) There are more and stronger H-bonds in alcohols, thus producing an acid-weakening effect. Also, in the conjugate bases RS- and RO- the charge is more dispersed over the larger S, thereby making RS- the weaker base and RSH the stronger acid (Section 3.11). ( b ) The larger S is more easily polarized than the smaller 0 and therefore is more nucleophilic. For example, RSH participates more rapidly in S,2 reactions than ROH. Recall that among the halide anions, nucleophilicity also increases as size increases: F- < C1- < Br- < I - .
Problem 13.32 Offer mechanisms for
CH,CH=CH,
&CH,CH(SH)CH,
5Lc H , c H , c H , s H
Acid-catalyzed addition has an ionic mechanism (Markovnikov):
CH,CH=CH,
+ H + --+ CH,;HCH, Z CH, H‘ -
4
CHAP. 131
281
ALCOHOLS AND THIOLS
The peroxide- or light-catalyzed reaction has a free-radical mechanism (anti-Markovnikov):
H - S - H ~ HS. + . H
CH,CHSH,
+ *SH+
CH,CHCH2SH
C H , C H C H , S H x CH,CH,CH,SH
+ -SH
13.6 SUMMARY OF THIOL CHEMISTRY
PREPARA TlON
PROPERTIES
1. Alkyl Halide
1. Acidic Hydrogen
+ OH----,
RX + KSH or (H,N),C=S
.
2. Olefin
R'CH-CH, 3. Disulfide
R-S-S-R
/
Zn. H 2 S 0 4
+ RS-
( K , = 10-l')
2. Sulfur Atom as Nucleophile
+ H,S
+ R'--COCl-
+ R'--CH=O+ O H - ; R'X-
R'--COSR (thioester) R'--CH(SR,) (thioacetal) R-S-R' (thioether)
-
3. Oxidation
4. Aryl Sulfonyl Chloride
ArS0,Cl
H,O
+ Pb2+--+Pb( SR),
+ KMnO, --+R-SO,H (sulfonic acid) + I, R-S-S-R (disulfide)
ArSH
Supplementary Problems Problem 13.33 Give the IUPAC name for each of the following alcohols. Which are l", 2", and 3"? (U)
CH,CH3
I
(d) CH3C-CH2CH, AH
(e)
Cl-CHCH,CH=CHCH,OH
I
CH3-C-CH, L6H,- In-C 1
( a ) 2-Methyl- 1-butanol , 1". (6) 2-Methyl-3-phenyl-1-propanol, 1". (c) 1-Methyl-1-cyclopentanol, 3". (d) 3-Methyl-3-.pentanol, 3". (e) 5-Chloro-6-methyl-6-(3-chlorophenyl)-2-hepten-l-o1. (The longest chain with OH has seven C's and the prefix is hept-. Numbering begins at the end of the chain with OH; therefore, -1-01. The aromatic ring substituent has C1 at the 3 position counting from the point of attachment, and is put in parentheses to show that the entire ring is attached to the chain at C". C1 on the chain is at C5.)1".
Problem 13.34 Write condensed structural formulas and give IUPAC names for ( a ) vinylcarbinol, (6) 4 diphenylcarbinol, (c) dimethylethylcarbinol? (d) benzylcarbinol.
ALCOHOLS AND THIOLS
282
(a)
[CHAP. 13
H,C=CHCH:OH 2-Propen-1-01 (Ally1 alcohol)
( 6 ) (C,H,),CH--OH (c) (CH,),C--OH
Diphenylmethanol (Benzhydrol) 2-Methyl-2-butanol
LB H 2 W (d) C,H ,-CH,eH,OH 2-Phenylethanol ( /3 -Phenylethanol)
Problem 13.35 The four isomeric C,H,OH alcohols are (i) (CH,),COH
(ii) CH,CH,CH,CH,OH
(iii) (CH,),CHCH,OH
(iv) CH,CH(OH)CH,CH, Synthesize each, using a different reaction from among a reduction, an S,2 displacement, a hydration, and a 4 Grignard reaction. Synthesis of the 3" isomer, (i), has restrictions: the S,2 displacement of a 3" halide cannot be used because elimination would occur; nor is there any starting material that can be reduced to a 3" alcohol. Either of the two remaining methods can be used; arbitrarily the Grignard is chosen. CH,Mgl
+ (CH,),C=O-----+(CH,),COH after
hydrolvsrs
-
The S,2 displacement o n the corresponding RX is best for 1" alcohols such as (ii) and (iii); let us choose (ii) for this synthesis.
+ OH-
CH,CH,CH,CH,CI
-cI-
CH,CH,CH,CH,OH
The 1" alcohol, (iii), and the 2" alcohol, (iv) can be made by either of the two remaining syntheses. However, the one-step hydration with H,O to give (iv) is shorter than the two-step hydroboration-oxidation to give (iii). CH,CH,CH=CH,
+ H,O+ 2CH,CH(OH)CH,CH,
-
Finally, (iii) is made by reducing the corresponding RCH=O or RCOOH. (CH ,) ,CH ,COOH
I
LiAIHJ
2
Ht
(CH ,) ,CH ,CH ,OH
Problem 13.36 Identify the products from reaction of (CH,),C=O with: ( a ) NaBD, and H,O; ( 6 ) NaBD, and CH,OD. 4 ( a ) (CH,),CDOH; (6) (CH,),CDOD. In both reactions nucleophilic :D- first attacks and reduces the carbonyl to form a basic alkoxide salt:
I-
44 + D -
Problem 13.37 Prepare ethyl p-chlorophenylcarbinol by a Grignard reaction.
4
Prepare this 2" alcohol, p-CIC,H,CHOHCH,CH,, from RCHO and R'MgX. Since the groups on the carbinol C are different, there are two combinations possible: H (l)
1
p-C1C6H4C=0
, 1
CH,CH2MgCI H,U+
p-ClC,H,CHOHCH,CH, (ring Cl has not interfered)
(2) p-CICf34MgBr
I
CHICHZCHO H,Ot
+
p-CIC,H,CHOHCH2CH,
Br is more reactive than Cl when making a Grignard of p-CIC,H,Br.
CHAP. 131
283
ALCOHOLS AND THIOLS
Problem 13.38 Give the hydroboration-oxidation product from ( a ) cyclohexene, (b) ~&-2-phenyl-2-butene, (c) tram-2-phenyl-2-butene. 4 Addition of H,O is cis anti-Markovnikov. See Fig. 13-2. In Fig. 13-2(c) the second pair of conformations show the eclipsing of the H's with each other and of the Me's with each other. The more stable staggered conformations are not shown,
H OH Me
H
H
0
I
Me
Enantiomers
I
"'...p&+ ,
Ph staggered
eclipsed
\
7
#
Conformers
n
Q H Me 0
.**
Ph
Me
*:
Me
Ph
Me HO
'Enantiomers
Me
c o d bmer s
Fig. 13-2
Problem 13.39 Write a structural formula for a Grignard reagent and a carbonyl compound which react to give each of the following alcohols after hydrolysis: ( a ) 2-butanol, (6) benzyl alcohol, (c) 2,4-dimethyl-3pentanol, (d) t-butanol. 4 The carbinol C and all but one of its R or Ar groups come from the carbonyl compound. The substituent from the Grignard reagent is written in a box below.
284
ALCOHOLS AND THIOLS
[CHAP. 13
( a ) Two possible combinations of reactants are:
H
OH
I 7 CH,CH,C-m I
I
I
CH,CH,C=O
W p B r
A
H
OH
CH,C=O I
I
r w ] M s B r
2
H3O'
CH,-C-[-I
I I
H
I CH,] m h l p B r
I
CH,-C-CH,
II
I
CH,-C-CH,
I
OH
0
Problem 13.40 The following Grignard reagents and aldehydes or ketones are reacted and the products hydrolyzed. What alcohol is produced in each case'? ( a ) Benzaldehyde (C,H,CH=O) and C,H,MgBr. (b) Acetaldehyde and phenyl magnesium bromide. (c) Acetone and benzyl magnesium bromide. ( d ) Formaldehyde and cyclohexyl magnesium bromide. (e) Acetophenone (C,H,CCH 3 ) and ethyl magnesium bromide. 4
II
C6HTCH=0
+ C,H,MgBr
8
---
H
I
C,H,T-OH
1-Phenyl-1-propanol
C,H, H CH,CH=O
I
+ C,H,MgBr-C,H,
1-Phenyl- 1-ethano1
CH,
+ C,H,CH,MgBr-
CH,--C--CH,
II
C,H,CH,
I
-F""
l-Phenyl-2-methyl-2-propanol
OH
0 H,C=O
C,H,-C-CH,
b
+ C,H
I
,MgBr --+C6HI ,CH,OH
+ C,H,MgBr
I
__*
Cyclohexylcarbinol
2-Phenyl-2-butanol
CHAP. 131
285
ALCOHOLS AND THIOLS
Problem 13.41 Give the mechanism in each case:
CH, ( a) CH,CH,CH,CH,CH,OH
aCH,CH,CH,CH2CH2Cl
I
3 CH,CCH,CH,
( 6 ) CH,
AI CH3 ( c ) CH,-&-CH2CH3
-!% C H , - ~ ~ C H , C H ,
AH
&I
Why did rearrangement occur only in (b)?
4
( a ) The mechanism is S,2 since we are substituting C1 for H,O from 1" ROH,' (6) The mechanism is S,1.
CH,
I CH,-CH-CH-CH, AH
CH3
I CH3-C-CH-CH3 I +
CH,
I
3 CH,-C-CH,-CH,
HJ
2" R' (less stable)
CI
~
CH3
I
CH,-C-CH,-CH,
AI
3" R' (more stable)
+
( c ) S,1 mechanism. The stable 3" (CH,),CCH,CH, reacts with CI- with no rearrangement. Problem 13.42 Why does dehydration of 1-phenyl-2-propanolin acid form 1-phenyl-1-propene rather than 4 1-phenyl-2-propene?
1-Phenyl-1-propene, PhCH=CHCH,, is a more highly substituted alkene and therefore more stable than l-phenyl-2-propene, PhCH,CH=CH2. Even more important, it is more stable because the double bond is conjugated with the ring. Problem 13.43 Write the structural formulas for the alcohols formed by oxymercuration-demercuration from ( a ) 1-heptene, (6) 1-methylcyclohexene, (c) 3,3-dimethyl-l-butene. 4
The net addition of H,O is Markovnikov. CH,( CH, ),CHOHCH,
(CH,),CCHOHCH,
2-Heptanol
3,3-Dimethyl-2-butanol (no rearrangement occurs)
Problem 13.44 List the alcohols and acids that compose the inorganic esters ( a ) (CH,),COCl and (b) 4 CH,CH,ONO,. Name the esters.
Conceptually hydrolyze the 0 to the heteroatom bond while adding an H to the 0 and an OH to the heteroatom. ( a ) (CH,),COH and HOCl, t-butyl hypochlorite. (6) CH,CH,OH and HONO,, ethyl nitrate. Tert-butyl hypochlorite is used to chlorinate hydrocarbons by free-radical chain mechanisms. Problem 13.45 Starting with isopropyl alcohol as the only available organic compound, prepare 2,3-dimethyl2-butanol. 4
This 3" alcohol, (CH,),COHCH(CH,),,
is prepared from a Grignard reagent and a ketone.
ALCOHOLS A N D THIOLS
286
[CHAP. 13
CH, CH,
I
t
CH,-C4H-CH3=
C1
AMgBr -+ CH3CHCH3
CH, CH,
CH,CHCH3
I
I I
1
I
CH,C----CH-CH,
MgBr
Br
+ Mg” + Br-
OH Problem 13.46 Use formaldehyde and 2-butanol to prepare 1,2-dibrom0-2-methylbutane. The product, BrCH,CBr(CH,)CH,CH,,
4
is a vic-dibromide formed from the alkene 2-methyl-l-butene,
CH3 CH,CHL=CH,
which must be made from
CH3
I
CH,CH,CHCH,OH which in turn is prepared as follows:
CH,-CH,-CH-CH,
CH3CH2CHCH,
AH
ez*CH,CH,CHCH,
HCHO
MgBr I
Ar
CH,CH,CHCH, h,oH
2-Methyl- 1-butanol
However, dehydration of 2-methyl-1-butanol with H,SO, forms 2-methyl-2-butene by carbocation rearrangement.
CH,CH,CHCH,OH
:Hy
CH,CH,CHCH:a
LH3
CH3CH26-CH3
LH3
2CH,CH=C-CH,
LH,
LH3 2-Meth yt-2-butene
The needed alkene is prepared by dehydrohalogenation and then brominated.
Problem 13.47 Alcohols such as Ph,CHCH,OH rearrange on treatment with acid; they can be dehydrated by heating their methyl xanthates (Tschugaev reaction). The pyrolysis proceeds by a cyclic transition state. Outline the steps using Ph,CHCH,OH. 4
S
I
Ph,CHCH,OH 5 Ph,CHCH,O-Na’
Ph,CHCH,-0-C-S-Na’ a xanthate Ph,C
CH,-m,
IT
HLsf-scH3 an s-methylxanthate
The elimination is cis.
CH I
4
Ph,C=CH,
+ &C-0
+ HSCH,
CHAP. 131
287
ALCOHOLS AND THIOLS
F R I N:y*
Problem 13.48 Methyl ketones gives the haloform test: CHI,
+ RCOO-Na'
-
I
a methyl ketone
yellow precipitate
1, can also oxidize 1" and 2" alcohols to carbonyl compounds. Which butyl alcohols give a positive haloform test?
4
Alcohols with the
groups are oxidized to
cH31i-R
and give a positive test. The only butyl alcohol giving a positive test is
H
t:
CH, -CH2CH3 AH Problem 13.49 A compound, C,H,,O, is oxidized under vigorous conditions to benzoic acid. It reacts with CrO, and gives a positive iodoform test (Problem 13.48). Is this compound chiral? 4 Since benzoic acid is the product of oxidation, the compound is a monosubstituted benzene, C,H,G. Subtracting C,H, from C,H,,O gives C,H,O as the formula for a saturated side chain. A positive CrO, test means a 1" or 2" OH. Possible structures are:
CH
'-1
H H H C-C-CH H H OH
H H C,H,C-C-C H I
H
OH
I
H
H H H C,H,C-c-C-oH H H H
H
c6H5-
kH3
111
I1
YCH20H IV
Only I1 has the - C H( OH) C H, needed for a positive iodoform test. 11 is chiral.
Problem 13.50 Suggest a possible industrial preparation for ( a ) t-butyl alcohol, ( b ) ally1 alcohol, (c) glycerol (HOCH,CHOHCH,OH). 4 CH3 CH3 CH3 ( a)
I
';;rt bH2 + C H 3 - -IC = C H 2 7
CHJ-CH-CH, Isobutane
(b)
Propene
CICH2-CH=CH, Allyl chloride
AH
Isobut ylene CI
CH,-CH=CH,
ClCH2-CH=CH2
HOCH,CH=CH,
Allyl chloride
+
\
Allyl alcohol
CH2-C H-C H2
I
CI
I
I
CH3-C-CH3
I
OH C1
ALCOHOLS AND-TH!OLS
288
[CHAP. 13
Problem 13.51 Assign numbers, from 1 for the lowest to 5 for the highest, to indicate relative reactivity with HBr in forming benzyl bromides from the following benzyl alcohols: (a) p-CI--C,H,--CH,OH, (b) (C,H,),CHOH. (c) P-@,N--C,H,--CH,OH, ( d ) (C,H,);COH, ( e ) C,HsCH,OH. 4 Differences in reaction rates depend on the relative abilities of the protonated alcohols to lose H,O to form The stability of R' affects ihe A H S for forming the incipient R' in the transition state and determines the overall rate. Electron-attracting groups such as NO, and CI in tne par@ position destabilize R' by intensifying the positive charge. NO, of (c) is more effective s k e it destabilizes by both resonance and induction, while CI of (a) destabilizes only by induction. The more C,H,'s on the benzyl C, the more stable is R'.
I?'.
(0)
2
!cl
(b) 4
(4
1
(4 3
5
Problem 13.52 Supply structural formulas and stereochemical designations foi the organic compounds (A) through (H). H O ( a ) (A) + conc. H , S O , ~ C H , - C H C H , C H , ~ (B) bS03H H O
( b ) (R)-CH3CHOHCH2CH=CH2
(C) + (D)
+ KEr --+ (E) + (F) + ( G )
(c) (R)-CH3CH2-CH-CH=CH2
I OH
4
(A) is cis- or tram-CH,CH=CHCH, or H,C=CHCH,CH,; (B) is rac-CH,CHOHCH,CH,. The intermediate CH,C+HCH,CH, can be attacked from either side by HSOd to give afi optically inactive racemic hydrogen sulfate ester which is hydrolyzed to rac-2-butanol. (C) is (R,R)-CH,-T
cH 2 T C
H
3
$7
(D)is (R,S)-CH,
;
OH
CH2-C-CH,
H
H
m eso
optically active
Hydration follows Markovnikov's rule, and a new similar chiral C is formed. The original chiral C remains R, but the new chiral C may be R or S. The two diastereomers are not formed in equal amounts [see Problem 5.22(b)]. (E) is rac-CH,CH,CHBrCH=CH,; (F) is tram- and (G) cis-CH,CH2CH=CHCH,Br. The intermediate R' in this S, 1 reaction is a resonance-stabilized (charge-delocalized) allylic cation +
t
[CH,CH,CHCH=CH,
c---*
CH,CH,CH=CHCH,]
6+
or CH,CH,CH=CH-CH,
6+
+
~-
In the second step, Br- attacks either 05 the positively charged C's to give one of the three products. Since R' is flat the chiral C in (E) can be R or S; (E) is racernic. (F) is the major product because it is most stable (tram and disubstituted).
Problem 13.53 -How does the Lewis theory of acids and bases explain the functions of (a) ZnCl, in the Lucas 4 reagent? (b) ether as a solvent in the Grignard reagent? ZnC1, Lewis acid
+ 2 HCI
__*
ROH
H2ZnCl, --+ ROH;
(a stronger acid than HCI)
RCI
+ H20
R'MgX acts as a Lewis acid because Mg can coordinate with one unshared electron pair of each 0 of two ether molecules to form an addition compound,
R'
R I !! :OtMg:Q: R A R that is soluble in ether.
CHAP. 131
289
ALCOHOLS AND THIOLS
Problem 13.54 Draw Newman projections of the conformers of the following substituted ethanols and predict 4 their relative populations: ( a ) FCH,CH,OH, ( 6 ) H,NCH,CH,OH, (c) BrCH,CH,OH.
If the substituents F, H2N, and Br are designated by Z, the conformers may be generalized as anti or gauche.
gauche
anti
For ( a ) and (b) the gauche is the more stable conformer and has a greater population because of H-bonding with F and N. The anti conformer is more stable in ( c ) because there is no H-bonding with Br and dipole-dipole repulsion causes OH and Br to lie as far from each other as possible. Problem 13.55 Deduce the structure of a compound, C,H,,O, which gives the following nmr data: S = 0.8 (doublet, six H's), 6 = 1.7 (complex multiplet, one H), 6 = 3.2 (doublet, two H's) and S = 4.2 (singlet, one H; 4 disappears after shaking sample with D 2 0 ) .
The singlet at S = 4.2 which disappears after shaking with D 2 0 is for OH (Problem 13.28). The compound must be one of the four butyl alcohols. Only isobutyl alcohol, (CH,),CHCH,OH, has six equivalent H's (two CH,'s), accounting for the six-H doublet at S = 0.8, the one-H multiplet at S = 1.7, and the two-H doublet which is further downfield at S = 3.2 because of the electron-attracting 0. Problem 13.56 The attempt to remove water from ethanol by fractional distillation gives 95% ethanol, an
azeotrope that boils at a constant temperature of 78.15 "C. It has a lower boiling point than either water (100 "C)
or ethanol (78.3"C). A liquid mixture is an azeotrope if it gives a vapor of the same composition. How does 4 boiling 95% ethanol with Mg remove the remaining H 2 0 ? Mg
+ H,O
-
H,
+ Mg(OH), insoluble
The dry ethanol, called absolute, is now distilled from the insoluble Mg(OH),. Problem 13.57 Explain why the most prominent (base) peak of 1-propanol is at mle = 31, while that of allyl alcohol is at m / e = 57. 4 CH,CH,--CH,OH + cleaves mainly into +
CH,CH2- + 6 H 2 0 Hc-----,H2C=0H
( m / e = 31) rather than CH,CH,CHOH
+ H., because C - C is weaker
than C-H.
In allyl alcohol,
H
I
CH,=CHC-OH
I
H the C-H
bond cleaves to give
CH,=CHc-OH
I
H ( m l e = 57). This cation is stabilized by both the CH2=CH and the 0.
I
Problem 13.58 Inorganic acids such as H 2 S 0 , , H,PO,, and HOCl (hypochlorous acid) form esters. Write structural formulas for ( a ) dimethyl sulfate, ( 6 ) tribenzyl phosphate, (c) diphenyl hydrogen phosphate, (d) t-butyl nitrite, ( e ) lauryl hydrogen sulfate (lauryl alcohol is n-C,,H,,CH,OH), ( f ) sodium lauryl sulfate. 4
ALCOHOLS AND THIOLS
290
[CHAP. 13
Replacing the H of the OH of an acid gives an ester.
0
0
(U)
I CH,O-S-WH,
I
(6) (PhCH,O),P-O
A
(c) (PhO),P-OH
( d ) (CH,),CONO
0 (e)
I I 0
n-C, ,H,,CH,OS-OH
Cf)
n-C, ,H,,CH,OSO,O-Na'
(a detergent)
Problem 13.59 Dialkyl sulfates are good alkylating agents. ( a ) Write an equation for the reaction of dimethyl sulfate with C,H,OH in NaOH. (6) Explain why dimethyl sulfate is a good methylating agent. 4
( b ) Methyl sulfate anion (CH,OSO,) is the conjugate base of a strong acid, methyl sulfuric acid (CH,OSO,H). It is a very weak base and a good leaving group.
Problem 13.60 Explain why trialkyl phosphates are readily hydrolyzed with OH- to dialkyl phosphate salts, 4 whereas dialkyl hydrogen phosphates and alkyl dihydrogen phosphates resist alkaline hydrolysis.
n
HO-
OR
I + RO=P=O+HOR I
+
OR
dialkyl phosphate
-
HO-
1
OR
+ HO=P=O+ I
H20 +
OR
OH n~
2HO-
+ HO=P--C\---* bR
0
4
2H20 + RO-P-O
\
The hydrogen phosphates are moderately strong acids and react with bases to form anions (conjugate bases). Repulsion between negatively charged species prevents further reaction between these anions and OH-. Problem 13.61 The ir spectrum of RSH shows a weak -S-H stretching band at about 2600cm-' that does not shift significantly with concentration or nature, of solvent. Explain the difference in behavior of S - H and 0-H bonds. 4
The S-H bond is weaker than the O - H bond and therefore absorbs at a lower frequency. There is little or no H-bonding in S - H and, unlike the 0-H bond, there is little shifting of absorption frequency on dilution. Problem 13.62 Write structural formulas with stereochemical designations for the reaction products I and 11.
CHAP. 131
ALCOHOLS AND THIOLS
H
I
I = (S)-TsO--C--CH,
I
(no change in configuration)
H I1
I I
(R)-CH,-C-SH
(S,2 inversion)
29 1
Chapter 14 Ethers, Epoxides, Glycols, and Thioethers A.
ETHERS
14.1 INTRODUCTION AND NOMENCLATURE
Simple (symmetrical) ethers have the general formula R--O-R or Ar--O--Ar; mixed (unsymmetrior Ar-0-Ar' or Ar-0-R. The derived system names R and Ar as cal) ethers are R-0-R' separate words and adds the word "ether." In the IUPAC system, ethers (ROR) are named as alkoxy- (RO-) substituted alkanes. Cyclic ethers have at least one 0 in a ring. Problem 14.1 Give a derived and IUPAC name for the following ethers: (a) CH,OCH,CH,CH,CH,, (CH ,) ,CHOCH( CH ,)CH ,CH ,,(c) C,H,OCH,CH, , (d) p-NO,C,H,OCH , (e) CH,OCH ,CH,OCH, .
,
(6) 4
(a) Methyl n-butyl ether, 1-methoxybutane. (6) sec-Butyl isopropyl ether, 2-isopropoxybutane. (Select the longest chain of C's as the alkane root.) (c) Ethyl phenyl ether, ethoxybenzene (commonly called phenetole). (d) Methyl p- nitrophenyl ether, 4-nitromethoxybenzene (or p-nitroanisole). (e) 1,2-dimethoxyethane. Problem 14.2 Account for the following. (a) Ethers have significant dipole moments (-1.18 D). (6) Ethers have lower boiling points than isomeric alcohols. (c) The water solubilities of isomeric ethers and alcohols are comparable. 4 ( a ) The C - 0 - C bond angle is about 110" and the dipole moments of the two C-0 bonds do not cancel. (6) The absence of OH in ethers precludes H-bonding and therefore there is no strong intermolecular force of attraction between ether molecules as there is between alcohol molecules. The weak polarity of ethers has no appreciable effect. (c) The 0 of ethers is able to undergo H-bonding with H of H,O.
14.2 PREPARATION
SIMPLE ETHERS 1.
Intermolecular Dehydration of Alcohols (see Section 13.3)
2. 2" Alkyl Halides with Silver Oxide 2(CH,),CHCI
+ Ag,O-
(CH,),CHOCH(CH,),
+ 2AgCI
MIXED ETHERS 1.
Williamson Synthesis
R'OR a1koxide
+ Na' + X-
X = C1, Br, I, OS02R, OS02Ar
1"
Problem 14.3 Specify and account for your choice of an alkoxide and an alkyl halide to prepare the following 4 ethers by the Williamson reaction: (a) C,H,OC(CH,),, (6) (CH,),CHOCH,CH=CH,. (a) C,H,X
+ Na'
-OC(CH,),
(6) (CH,),CHO - N a t
292
+ XCH,CH=CH,
CHAP. 141
293
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
The 2" and 3" alkyl halides readily undergo E2 eliminations with strongly basic alkoxides to form alkenes. Hence, to prepare mixed ethers such as ( a ) and (6) the 1" alkyl groups should come from RX and the 2" and 3" alkyl groups should come from the alkoxide. Problem 14.4 Show how dimethyl sulfate, MeOSO,OMe, is used in place of alkyl halides in the Williamson 4 syntheses of methyl ethers.
Alkyl sulfates are conjugate bases of the very strongly acidic alkyl sulfuric acids and are very good leaving groups. Dimethyl sulfate is less expensive than CH,I, the only liquid methyl halide at room temperature. Liquids are easier to use than gases in laboratory syntheses.
CF
: R Q : T + 'Me:OSO,OMe
ROMe + MeOSO,
Dimethyl sulfate (very toxic)
2.
Alkoxymercuration-demercuration (Section 13.2 for ROH preparation)
Whereas mercuration-demercuration of alkenes in the presence of water gives alcohols, in alcohol solvents (free from H,O) ethers result. These reactions in the presence of nucleophilic solvents such as water and alcohols are examples of solvomercuration. The mercuric salts usually used are the acetate, Hg(OAc), (-OAc is an abbreviation for - O C C H , ) or the trifluoracetate,
-
Hg(OCOCF, 1,. RCH=CH2
+ R'OH + Hg(OC-CF,), II
tl
RCHCH2
I I
NaBH,
RCHCH,
I I
R'O HgOCOCF, R'O H
0
Problem 14.5 Suggest a mechanism consistent with the following observations for the solvomercuration of RCH=CH, with R'OH in the presence of Hg(OAc),, leading to the formation of RCH(OR')CH,Hg(OAc): (i) no rearrangement, ( i i ) Markovnikov addition, (iii) anti addition, and (iv) reaction with nucleophilic solvents. 4
Absence of rearrangement excludes a carbocation intermediate. Stereoselective anti addition is reminiscent of a bromonium-ion-type intermediate, in this case a three-membered ring mercurinium ion. The anti regioselective Markovnikov addition requires an S, 1-type backside attack by the nucleophilic solvent on the more substituted C of the ring-the C bearing more of the partial + charge (6 '). Frontside attack is blocked by the large HgOAc group. OR' OAc RCH=CH,
/
+ Hg-
\
OAC
I
-HOR'
RCH
+
'*
' 1 ,
I
)II HgOAc
-H
'
RCHCH, I HgOAc I
Although the mercuration step is stereospecific, the reductive demercuration step is not, and therefore neither is the overall reaction. Problem 14.6 Give the alkene and alcohol needed to prepare the following ethers by alkoxymercuration( c ) 1-phenyl-1-ethoxypropane,(d) demercuration: ( a ) diisopropyl ether, ( 6 ) 1-methyl-1-methoxycyclopentane, di-t-butyl ether. 4 ( a ) CH,=CHCH,
and CH,CH(OH)CH, give (CH,),CHOCH(CH,),.
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
294
(c)
P h C H d H C H , and CH,CH,OH give PhCH( OCH,CH,)CH,CH,.
[CHAP. 14
Although each double-bonded C is
2", ROH preferentially bonds with the more positively charged benzylic C. (d) Ethers with two 3" alkyl groups cannot be synthesized in decent yields because of severe steric hindrance.
Problem 14.7 Give ( a ) SN2 and (6) (a)
(1)
sN1
mechanisms for formation of ROR from ROH in conc. H,SO,.
-"
ROH + H,SO,base
R:OHi acid I +
(2) R O H + R OH -R
(6) (1)
R+H,O
conjugate acid of an ether
+ HSOh (or ROH)+
Same as (1) of ( a )
(3) ROH+R'+ROR
4
base
x
*a
R'
+ HSO,
+
I
ROR +
!SO,(or RO€
(2) R:OHi -R+
+H20
(4) Same as (3) of
(a)
H Problem 14.8 Compare the mechanisms for the formation of an ether by intermolecular dehydration of ( a ) l", 4 (6) Y,and (c) 2" alcohols. ( a ) For
1" alcohols the mechanism is SN2,with alcohol as the attacking nucleophile and water as the leaving group. There would be no rearrangements. (6) The mechanism for 3" alcohols is SN1. However, a 3" carbocation such as Me,C' cannot react with Me,COH, the parent 3" alcohol, or any other 3" ROH because of severe steric hindrance. It can react with a 1" RCH,OH, if such an alcohol is present. Me,C' (c)
+ RCH,OH-
-H+
Me,COCH,R
(a mixed ether)
The 3" carbocation can also readily eliminate H' to give an alkene, Me,C=CH2. 2" alcohols react either way. Rearrangements may occur when they react by the SN1 mechanism, because the intermediate is a carbocation.
Problem 14.9 Although benzyl alcohol, C,H,CH,OH, is a 1" alcohol, it undergoes intermolecular dehydration by both SN2 and SN1 mechanisms. Explain. 4
The protonated alcohol loses water readily to form the benzyl carbocation, which is stabilized by charge dispersion to the ring.
Problem 14.10 List the ethers formed in the reaction between concentrated H2S0,, and equimolar quantities 4 of ethanol and ( a ) methanol, (6) tert-butanol. ( a ) These
1" alcohols react by sN2 mechanisms to give a mixture of three ethers: C,H,OC,H, 2C,H,OH, CH,OCH, from 2CH,OH, and C,H,OCH, from C,H,OH and CH,OH. (6) This is an SN1 reaction.
Reaction between (CH,),C'
and (CH,),COH is sterically hindered and occurs much less readily.
from
295
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
CHAP. 141
Problem 14.11 Explain why a good yield of a mixed ether can be obtained from a mixture of H,C=CHCH,OH and 2-propanol. 4
The ether is H,C==CHCH,OCH(CH,),. H,C==CHCH,OH forms the stable carbocation H , M H C H l , which then reacts with 2-propanol. Problem 14.12 Use any needed starting material to synthesize the following ethers, selecting from among intermolecular dehydration, Williamson synthesis, and alkoxymercuration-demercuration.Justify your choice of method.
(c) dicyclohexyl ether
(6) CH,CH,CHOCH,CH,CH,
(a ) CH3(CH&,OCH,C H,
I
4
CH3
+ C,H,O-Na'. Since alkoxymercuration is a Markovnikov addition, it cannot be used to prepare an ether in which both R's are 1". Unless one of the R's can form a stable R', intermolecular dehydration cannot be used to synthesize a mixed ether.
( a ) Use Williamson synthesis; CH,(CH,),Cl
4
3
2
1
CH,CH,CH=CH,
+ Hg(OCOCF3), + n-C3H70H
4
1
NaBH,
2
CH,CH,CH-OC,H,-n I
'kH,HgOCOCF,
4
3
2
1
C H 3C H 2C H(CH3)OC447- II
This is better than Williamson synthesis because there is no competing elimination reaction. (c)
Dehydration; Cyclohexyl4H-
H2S04
(Cyclohexyl),O. This is a simple ether.
Problem 14.13 Show steps for the following syntheses: (a) (CH,),CH-O-CH,CH( OH)CH,OH from propylene; (6) CH,CH,-4-CH,C6H,NO,-p from toluene and aliphatic alcohols. 4 ( a ) The glycol is prepared by mild oxidation of the unsaturated ether, as in Problem 14.11.
CHJCHCH3 '/
CH,CH=CH,
/
I
OH
NP
CH3CHCHJ
I
0Isopropoxide anion
+
CICH,CH=CH, Ally1 chloride
c1- +
CH3
I
t
1
KM d"n 0 4
/ \CH-O-CH,CH=CH, CH3
Isopropyl ally1 ether
This method requires fewer steps than does p-NO,C,H,CH,O-
OH OH
CH3\
I
CH--O--CH,CH-CH,
I
/
CH3
+ C,H,Br.
Problem 14.14 (R)-2-Octanol and its ethyl ether are levorotatory. Predict the configuration and sign of rotation of the ethyl ether prepared from this alcohol by: ( a ) reacting with Na and then C,H,Br; (6) reacting in a solvent of low dielectric constant with concentrated HBr and then with C,H,O-Na'. 4 (a) No bond to the chiral C of the alcohol is broken in this reaction; hence the R configuration is unchanged but rotation is indeterminant. (6) These conditions for the reaction of the alcohol with HBr favor an S,2
296
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
[CHAP. 14
mechanism and the chiral C is inverted. Attack by RO- is also SN2 and the net result of two inversions is retention of configuration. Again rotation is unpredictable.
CH, (RI
CH, (3
CH3 (RI
14.3 CHEMICAL PROPERTIES
BRONSTED AND LEWIS BASICITY Ethers are among the most unreactive functional groups toward reagents used in organic chemistry (for an exceptional subclass, see Section 14.6). This property, together with their ability to dissolve nonpolar compounds, makes them good solvents for organic compounds. Because of the unshared pairs of electrons -6is a basic site, and ethers are protonated by strong acids, such as conc. H2S0,, to form oxonium cations, R 2 0 H + .Ethers also react with Lewis acids. Problem 14.15 ( a ) Why do ethers dissolve in cold concentrated H,SO, and separate out when water is added to the solution? (6) Why are ethers used as solvents for BF, and the Grignard reagent? 4
-x
( a ) Water is a stronger base than ether and removes the proton from the protonated ether.
Step 1 (solution) Step 2 (dissolution)
ROR + H,SO, base,
acid,
R,OH'
+ H,O+
acid
,
R R' acid
,
+ HSOa base,
ROR + H 3 0 +
base
base,
acid,
( 6 ) BF, and RMgX are Lewis acids which share a pair of electrons on the U of ethers. X R R + + / \o: MgzfO: R,O:BF, 2R,O: + R'MgX/ I R,O: + BF,R
'
* * \
Notice that two ether molecules coordinate with one Mg atom.
CLEAVAGE
-
Ethers are cleaved by concentrated HI (ROR + HIare formed (ROR + 2HI 2RI).
R'
ROH + RI). With excess HI, 2 mol of RI
Problem 14.16 Identify the ethers that are cleaved with excess HI to yield ( a ) (CH,),CI and CH,CH,CH,l, 4 (6) cyclohexyl and methyl iodides, (c) I(CH2)51.
( 4 (CH,),COCH,CH,CH, (6)
(It-..
(c) The diiodide with 1's on the terminal C's indicates a cyclic ether,
Problem 14.17 ( a ) Show how the cleavage of ethers with HI can proceed by an SN2or an SN1 mechanism. (6) Why is HI a better reagent than HBr for this type of reaction? (c) Why do reactions with excess HI afford two moles of RI? 4
CHAP. 141
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
Step 1
+
+ HI +R-0-R' I
R-0-R'
297
+ I-
H
base,
acid,
acid,
base,
-0
Step 2 for sN2
+ R-0-R' Z R + HOR' I
I-
(R is 10)
H Step 2 for
sNi
Step 3 for sN1
I
RYR' R'
aR'
+ I-
__+
+ ROH
(R is 3")
RI
( 6 ) HI is a stronger acid than HBr and gives a greater concentration of the oxonium ion
H
I
R-Q-R
I - is also a better nucleophile in the SN2reactions than is Br-.
( c ) The first-formed ROH reacts in typical fashion with HI to give RI.
Problem 14.18 Account for the following observations:
4
The high polarity of the solvent (H,O) in reaction (2) favors an SN1 mechanism giving the 3" R'. H
I+
CH,OC(CH,),
-
CH30H + (CH,),C+
2(CH,),CI
The low polarity of solvent (ether) in reaction (1) favors the SN2mechanism and the nucleophile, I-, attacks the 1" C of CH,.
I-
+ CH, OC(CH,),
-
CH31 + HOC(CH,),
Problem 14.19 Why does ArOR cleave to give RI and ArOH rather than ArI and ROH?
4
S,2 attack by I- does not occur readily on a C of a benzene ring nor does C,HJ form by an S,1 reaction. Therefore, ArI cannot be a product.
FREE-RADICALSUBSTITUTION H
Ethers readily undergo free-radical substitution of an H on the
C,
I I
Problem 14.20 Can resonance account for the preferential radical substitution at the a-C of ethers? 4
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
298
[CHAP. 14
The intermediate (I! radical RCHOR is not stabilized by delocalization of electron density by the adjacent 0 through extended T bonding. One resonance structure would have 9 electrons on 0.
++
[RL-O-R
' I
RC=O-R
H
Problem 14.21 ( a ) Give a mechanism for the formation of the explosive solid hydroperoxides, e.g.
OOH
I
RCHOCH,R from ethers and 0,. (6)Why should ethers be purified before distillation? (U)
Initiation Step
RCH,OCH,R
Propagation Step 1
RCHOCH,R
Propagation Step 2
R HOCH,R
c
4
+ *Q-O.+RCHOCH,R
+
+ H ..a ..*
RCHOCH,R
bo
+ RCH,OCH,R+
RCHOCH,R
L
0-0.
+ RCHOCH,R
H
(6) An ether may contain hydroperoxides which concentrate as the ether is distilled and which may then explode. Ethers are often purified by mixing with FeSO, solution, which reduces the hydroperoxides to the nonexplosive alcohols (ROOH-* ROH). Problem 14.22 Does peroxide formation occur more rapidly with (RCH,),O or (R,CH),O?
4
With (R,CH),O because the 2" radical is more stable and forms faster.
ELECTROPHILIC SUBSTITUTION OF ARYL ETHERS
The --OR group is a moderately activating op-directing group (Problem 11.8). Problem 14.23 Give the main products of ( a ) mononitration of p-methylphenetole, (6) monobromination of p-me thoxyphenol. 4
( a ) Since OR is a stronger activating group than R groups, the NO, attacks ortho to - O C , H , ; the product is 2-nitro-4-methylphenetole. (6) OH is a stronger activating group than OR groups; the product is 2-bromo-4methoxyphenol.
14.4 CYCLIC ETHERS
The chemistry of cyclic ethers such as those shown below is similar to that of open chain ethers.
Q Tetrahydro!iran (Oxacyclopentane)
0 Tetrahydropyran(THP) (Oxacyclohexane)
Pyran
1.4-Dioxane (1,4-Dioxacyclohexane)
Problem 14.24 Which alcohol would undergo dehydration to give (a) THP?(6) 1,4-dioxane?
4
( a ) Since the product is a cyclic ether, the starting material must be a diol with OH groups on the terminal C's. The diol must have five CH, groups to match the number in THP.The alcohol used in HO(CH,), OH, 1,Spentanediol. This is an intramolecular dehydration to form an ether.
CHAP. 141
299
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
( 6 ) 1,4-Dioxane has two ether groups, requiring dehydration between two pairs of OH groups. Again the starting alcohol must be a diol, but now the dehydration is intermolecular. The alcohol used is ethylene glycol, HOCH,CH, OH. Problem 14.25 ( a ) With the aid of the mechanism show why DHP,unlike typical alkenes, readily undergoes the following reaction:
8
+ROH-
Hi
O
2.3-Dihydro-4H-pyran
O
a tetrahydropyranyl
(DHP)
R (THP)
ether
(6) Why do THP ethers, unlike ordinary ethers, cleave under mildly aqueous acidic conditions?
4
The H' adds to C=C to generate a carbocation with the positive charge on the C that is a to the U of the ring. This is a fairly stable cation because the positive charge is stabilized by delocalization of electron density from the 0 atom.
The nucleophilic site (-0-)of ROH then bonds to the C' of the carbocation, forming an onium ion of the ether, which loses a proton to the solvent alcohol (ROH) and becomes the ether product. Aqueous acid reverses the reaction of ( a ) , reforming the same intermediate carbocation. This loses a proton to give the C = C , rather than reacting with water to give the very unstable alcohol-analog of the ether. Problem 14.26 Since THP ethers are, like most ethers, stable in base, their formation can be used to protect the OH group from reacting under basic conditions. Using this fact, show how to convert HOCH,CH,Cl to HOCH,CH,D via the Grignard reagent. 4
HOCH,CH,Cl cannot be converted directly to the Grignard because of the presence of the acidic OH group; the product obtained would be HOCH,CH,. The desired reaction is achieved by protecting the OH group as shown schematically: DHP
+ HOCH,CH,Cl-
C~CH,CH,O-THP+
DCH,CH,O-THP-
DCH,CH,OH
+ DHP
To generalize on Problem 14.26, a good protecting group (i) is easily attached, (ii) permits the desired chemistry to occur, and (iii) is easily removed. Other methods for protecting OH groups involve benzyl and silyl ethers: ROH + BrCH,Ph%
-AgBr
R O C H , P h z ROH + CH,Ph a benzyl ether
pyridine
ROH + C l - S i ( C H 3 ) , ~ ROSi(CH,),% Chlorotrimethylsilane
ROH + HOSi(CH,),
a trimethylsilyl ether
Crown ethers are large-ring cyclic ethers with several 0 atoms. A typical example is 18-crown-6 ether, Fig. 14-1(u). The first number in the name is the total number of atoms in the ring, the second number is the number of 0 atoms. Crown ethers are excellent solvaters of cations of salts through €ormation of ion-dipole bonds. 18-Crown-6 ether strongly complexes and traps K', [from, e.g., KF, as shown in Fig. 14-1(b)].
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
300
[CHAP. 14
....
:F:
(b)
Fig. 14-1 Problem 14.27 Suggest two important synthetic uses of crown ethers.
4
(1) They enable inorganic salts to be used in nonpolar solvents, a media with which salts are typically incompatible. (2) The cation of the salt is complexed in the center of the crown ether, leaving the anion “bare” and enhanced in reactivity. These effects of crown ethers are similar to those achieved with phase-transfer agents. The “bare” anion is also present when polar aprotic solvents are used.
14.5 SUMMARY OF ETHER CHEMISTRY
PROPERTIES
PREPARATION
+ H,SO,+
2ROH + H 2 S 0 , 2.
-
1. Basicity
1. Intermolecular Dehydration
+ BF,
S,2 Displacement (Williamson) RO- + RX 1”
H
I
ROR+HSO, R,OBF,
2. Cleavage
\ \3.
+ 2HI+2RI
or RI
Substitution on
+ R’I
C
3. AIkoxymercuration-demercuration RCH=CH,
+ R’OH+
RCH(OR’)CH,
B. EPOXIDES 14.6 INTRODUCTION
Unique among cyclic ethers are those with three-membered rings, the epoxides or oxiranes. Their large ring strains make them highly reactive.
14.7 SYNTHESIS 1.
0
II
From Alkenes and a Peroxyacid, RCOOH (Problem 6.34) (Mainly, rn-Chloroperoxybenzoic Acid)
Problem 14.28 ( a ) Give the structural formula of the epoxide formed when rn-chloroperoxybenzoic acid reacts with (i) cis-2-butene and (ii) trans-2-butene. (The epoxides are stereoisomers.) (6) What can you say about the stereochemistry of the epoxidation? (c) Why is a carbocation not an intermediate? 4
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
CHAP. 141
H H...C&
I
8:+
30 1
,o,
.C'C.
1
.'H
CH.3 ' C H ,
CH,
cis-2,3-Dimethyloxirane
cis-2-Butene
(0
c=o 0
-
trans-2 Bute ne
(ii)
rrans-2.3-Dimethyloxirane
The stereochemistry of the alkene is retained in the epoxide. The reaction is a stereospecific cis addition. The cis- and rrans-alkenes would give the same carbocation, which would go on to give the same product(s). The mechanism probably involves a one-step transfer of the 0 to the double bond, without intermediates.
From Halohydrins by Intramolecular SN2-Type Reaction Halohydrins, formed by electrophilic addition of HO-Cl(Br) to alkenes (Problem 6.33), are treated with base to give epoxides.
H
H
intermediate alkoxide
Problem 14.29 Why does tram-2-chlorocyclohexanol give a very good yield of 1,2-epoxycyclohexane, but the cis isomer gives no epoxide? 4
The nucleophilic 0 - group displaces the C1 atom (as Cl-) by an intramolecular SN2-typeprocess which requires a backside attack. In the trans isomer, the 0 - and C1 are properly positioned for such a displacement, and the epoxide is formed. In the cis isomer, backside attack cannot occur and the epoxide is not formed. This role of the 0 - ,called neighboring-group participation, always leads to an inversion of configuration if the attacked C is a chiral center (stereocenter).
14.8 CHEMISTRY 1.
sN2 Ring-Opening Problem 13.10(d) illustrates such a reaction.
Problem 14.30 Outline the sN2 mechanism for acid- and base-catalyzed addition to ethylene oxide and give the structural formulas of the products of addition of the following: ( a ) H,O, (b) CH,OH, ( c ) CH,NH,, (d) CH,CH,SH. 4
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
302
[CHAP. 14
In acid, 0 is first protonated.
The protonated epoxide can also react with nucleophilic solvents such as CH,OH.
In base, the ring is cleaved by attack of the nucleophile on the less substituted C to form an alkoxide anion, which is then protonated. Reactivity is attributed to the highly strained three-membered ring, which is readily cleaved.
( a ) HOCH, CH,OH,
(6) CH ,OCH,CH,OH , (C) CH ,NHCH,CH, OH, ( d ) CH ,CH ,SCH,CH,OH.
Base-induced ring-openings require a strong base because the strongly basic 0 - is displaced as part of the alkoxide. Acid-induced ring-openings are achieved with weak bases, such as nucleophilic solvents, because now the very weakly basic OH, formed by protonation of the 0 atom, is displaced as part of the alcohol portion of the product. Problem 14.31 ( a ) Give the product of the SN2-typeaddition of C,H,MgBr to ethylene oxide. ( 6 ) What is the 4 synthetic utility of the reaction of Grignard reagents and ethylene oxide?
(4 1-Butanol
(6) It is a good method for extending the R group of the Grignard by --CH,CH,OH in one step. Problem 14.32 Account for the product from the following reaction of the 14C-labeledchloroepoxide:
CH,O- + H2'4C--CHCH,Cl---, \ / 0
CH,0'4CH,CH-CH, \ /
+ C14
0
SN2 attack by CH,O-on the (less s~bstituted)'~C gives an intermediate alkoxide, CH30'4CH,CHCH2Cl, I that then displaces C1- by another SN2reaction, forming the new epoxide. 0-
2.
S,1 Ring-Opening
In acid the protonated epoxide may undergo ring-opening to give an intermediate carbocation. Problem 14.33 Outline mechanisms to account for the different isomers formed from reaction of
(CHJ2C-CHz
'd
with CH,OH in acidic (H') and in basic (CH,O-) media.
4
CHAP. 14)
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
303
CH,O- reacts by an SN2 mechanism attacking the less substituted C.
Isobutylene oxide
In acid the SN1 mechanism produces the more stable 3" R', and the nucleophilic solvent forms a bond with the more substituted C. (CH3)2C-CH2 5 __* (CH,),&H,OH (a 3" R ' )
\0I
(CH,),C-tH,
(CH,),C6H20H
ACH,
AH 1" R '
0 Problem 14.34 Account for the fact that (R)-CH3CH2CdAH, reacts with CH30H in acid to give the product with inversion and very little racemization. 4
When the protonated epoxide undergoes ring-opening, the CH,OH molecule attacks from the backside of the C'. The nearby, newly formed OH group hasn't moved out of the way and blocks approach from the frontside. This leads to inversion at the chiral carbon. Since there was no change in group priorities, the configuration in the product is (S).
14.9 SUMMARY OF EPOXIDE CHEMISTRY
PREPARATION
PROPERTIES
1. Alkene Oxidation by Peroxyacids ( a ) Alkene oxidation by peroxyacids
2. Williamson Reaction (Halohydrin and Base)
1. Acid-Catalyzed Cleavage
\-
I\ \\ 2.
Base-Catalyzed Cleavage
RO-
LI -C-OH 'I
3. Grignard Reagent
+ 1.
RMgX, 2. H20---* R-L-L-OH I t
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
304
[CHAP. 14
C. GLYCOLS 14.10 PREPARATION OF 1,2-GLYCOLS
1.
Oxidation of Alkenes (see Table 6-1 and Problems 6.34 and 6.35)
2.
Hydrolysis of vic-Dihalides and Halohydrins R-CH-CH2
I
CI
H 0 OH-
I
CI
HzO. OH-
R-CH-CH2-R-CH-CH2
I
I
OH
1,2-haloalcohol (haloh ydrin)
Hydrolysis of Epoxides RCH-CH2
RCH-CH2
I
‘ 0 ’
4.
I
OH
C1
oic-dihalide
3.
I
OH
I
OH OH
Reductive Dimerization of Carbonyl Compounds
Symmetrical 1,2-glycols, known as pinacols, are prepared by bimolecular reduction of aldehydes or ketones.
R’
R’ R-C-R’
+ R-C-R‘
0
0
II
II
Mgin
I
Mg2+
I C-R 0I
R‘
H,O __*
R‘
I I
I R-C-C-R I
+ Mg(OH),
OH OH a pinacol
Problem 14.35 What compounds would you use to prepare 2,3-diphenyl-2,3-butanediol,
C,H,-C(OHFC(OH)C,H, AH3
AH3
by ( a ) halide hydrolysis and (6) reductive dimerization of a carbonyl compound? (a) C,H,C(CH,)ClC(CH,)ClC,H,
14.11 1.
or C,H,C(CH,)ClC(CH,)OHC,H,,
(6) C,H,COCH,
UNIQUE REACTIONS OF GLYCOLS
Periodic Acid (HIO,) or Lead Tetracetate (Pb(OAc),), Oxidative Cleavage
,added OH’s,
A 1” OH yields H,C=O; a 2” OH an aldehyde, RCHO; a 3” OH a ketone, R,C=O. In polyols, if two vicinal OH’s are termed an “adjacency,” the number of moles of HIO, consumed is the number of such adjacencies.
CHAP. 141
305
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
Problem 14.36 Give the products and the number of moles of HIO, consumed in the reaction with 4 2,4-dimethyl-2,3,4,5-hexanetetrol. Indicate the adjacencies with zigzag lines.
H
CH, H
CH,
I CH,-L+#+#+#
3 mol of HIO,
I-
Acetaldeh yde
1-
H20
CH3
H
L o
L
OH
Acetic acid
Note that the middle <-OH'S
O
I
AH
Acetone
Hz0
Formic acid
are oxidized to a --COOH because C - 4 bonds on both sides are cleaved.
2. Pinacol Rearrangement Acidification of glycols produces an aldehyde or a ketone by rearrangement. There are four steps: (1) protonation of an OH; (2) loss of H 2 0 to form an R'; (3) 1,2-shift of :H, :R, or :Ar to form a more stable cation; (4) loss of H' to give product.
CHJ CH3
I
CH3-C-C-CH3
I
OH
I
I
CH3 CH3
-$+
I
CH3-C-C-CH3
OH
I
-H20 (2)
I I OH2 OH
CH3-$-
pinacol
y*
I OH
a 3" R'
CH3
CH3
I CHJ-C-C-CH3 I II
I + CHJ-C-C-CH3 I I
CH3 :OH
CH3 :I)H
protonated ketone
- CH,
-CH3
-H+
(3)
CH3
I
CHj-C-C-CH3
I
II
CH3 0 pinacolonc
more stable cation
With unsymmetrical glycols, the product obtained is determined mainly by which OH is lost as H 2 0 to give the more stable R', and thereafter by which group migrates. The order of migratory aptitude is A r > H or R. Problem 14.37 Give the structural formula for the major product from the pinacol rearrangement of 1,1,2-triphenyl-1,2-propanediol.Indicate the protonated OH and the migrating group 4
OH" OHb CH
I
'-1
C-C-CH, C6H5
I
&3HS
Loss of OH" yields the more stable (C,H,),C'-C(oH)(C,H,)CH,.
(C6H5),C-C-CH3
8
C,H, rather than CH, migrates to form
306
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
[CHAP. 14
the major product. Migration of CH, would give
which also arises from the loss of OHb.
3. Reaction with Carbonyl Compounds (see Section 15.3)
Carbonyl compounds react with glycols in anhydrous acid to form 1,3-cyclic ethers. These are called acetals if formed from aldehydes, and ketals if formed from ketones. RC=O
+ HOCH,CH,OHdry”’
h ’ )
+H,O R
X H(R1)
an acetal (a ketal)
14.12 SUMMARY OF GLYCOL CHEMISTRY PREPARA TlON
PROPERTIES
1. Alkene Oxidation
1. Ester Formation
+ R’COX +
OH RCH-CHR
I.
RCOJH
RCH-CHR
RCH-CHR
I
I
R’COO
I
OOCR’
OH RCH-CHR
KMnOI
RCH-CHR
____*
2. Oxidation
OH OH
+ KMnO, + HIO,
2. Hydrolysis Dihalogen:
RC H-C HR
xI
__*
2RCOOH
2 RCH-0
xI
Halohydrin: RCH-CHR
1
_I_*
1
-
OH-
I
1
3. Pinacoi Rearrangement
R O
+ HZSO,
II
I
+ RC-C-R
I
R
3. Reductive Dimerization of Carbonyl Compounds
Aldehyde: RCHO
4.
xt”
Cyclic Acetal or Ketal
H
4. Stereochemistry of Glycol Formation
H
(a) From Cyclic Alkenes Cycloalkenes
RcojfiA trans-
R
Go
cis-
( b ) From Symmetrical Alkenes raccis- A I ke nes
m eso- /
+\,c=o+
t ra n s- A I ke ne s
CHAP. 141
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
D.
307
THIOETHERS
14.13 INTRODUCTION
Thioethers are the sulfur analogs of ethers, with the generic formula RSR. They are also known as sulfides. 14.14 PREPARATION
-
Thioethers are prepared mainly by Williamson-type S,2 displacements of RS- and R X or R’OS0,Ar (aryl sulfonates). RS- is formed from the acidic thiol RSH, with NaOH as the base. /
CH,S- Na’
\ P + CH,CH,-Br-
CH,SCH,CH,
+ NaBr
Ethyl methyl sulfide
Problem 14.38 Explain why the reaction of HS- and RX is little used to prepare RSH.
4
Sulfur atoms in molecules and ions are very good nucleophilic sites. Hence, once formed in base, RSH yields RS-, which reacts with RX to give the thioether. For this reason thiourea is used with RX to give thiols (Problem 13.29). Problem 14.39 Give the expected principal organic products from the following reactions:
+ (CH3)2CHCH2CH2CH2BrICH2CH21+ HSCH2CH2SH-
OH-
( a ) C,H,SH
(6)
OH-
(c) Na2S( 1 mol) + BrCH,CH,CH,CH,Br+
4
14.15 CHEMISTRY
The chemistries of RSR and ROR are quite different, as indicated by the following reactions which are not observed for ROR. 1.
-
Reaction with R’X to Give Stable Sulfonium Salts, R,R’S:+ XMe$:
+ CH,CH,:Br-
Ethers do not undergo this reaction, because 1.
[Me,SCH,CH,)’ BrDimethylethylsulfonium bromide
is a much weaker nucleophilic site than
Reduction (Hydrogenolysis)
The reaction requires the Raney nickel catalyst (H, is adsorbed on Ni). R-SR’
+ H2-Raney Ni RH + R‘H + H,S
-&.
308
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
[CHAP. 14
Problem 14.40 Write the condensed formulas for the two isomeric thioethers giving methane and propane on hydrogenolysis. 4
Methane must come from a methyl group; propane can come from either a --CH,CH,CH, or a -€H(CH,), group. The two isomers are CH,SCH,CH,CH, and CH,SCH(CH,),. Problem 14.41 What structural features must be present in order for an individual thioether to give a single alkane on hydrogenolysis? 4
The thioether must have the same two R groups or it must be cyclic.
3. Oxidation at the S
T h e products are sulfoxides with o n e 0 and sulfones with two 0 atoms.
Problem 14.42 Why are sulfones with an H on an a C (e.g. CH,SO,Ph) soluble in bases?
4
Removal of the a H leaves a carbanion whose unshared pair of electrons induce the negative charge that can be delocalized to each of the sulfone 0 atoms: 0 7 -
Na’ OC2H3 + ;H,SO,Ph-CH,SPh
It It
“I
or H,C+Ph
0
Na’
+ C,H,OH
0
Problem 14.43 Explain why sulfonium salts and sulfoxides having different R or Ar groups are resolvable into enantiomers. 4
The S atom in each of these molecules has three o bonds and an unshared pair of e - ’ s . According to the HON rule (Section 2.3), these S atoms use sp3 HO’s. If all the attached groups are different, the S is a chiral center. Notice in the figure below that chirality prevails even though one of the sp3 HO’s houses an unshared pair of e - ’ s . The fact that these species are resolvable indicates that their molecules do not undergo inversion of configuration, notwithstanding the fact that a lone pair of e - ’ s is present. Such rigidity of configuration is characteristic of third-period elements (S, P), but not of second-period elements (C, N).
chiral sulfoniurn ion
chiral sulfone
Fig. 14-2
CHAP. 141
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
309
Problem 14.44 Since ( C H , ) , w : is an ambident nucleophile, reaction with CH,I could give [(CH,),S'=O]I- or [(CH3),S=0--CH,]I-. ( a ) What type of spectroscopy can be used to distinguish 4 between the two products? (6) Predict the major product. ( a ) Use nmr spectroscopy. In [ ( C H 3 ) & 0 ] I - all H's are equivalent and a single peak is observed. Note that
$3 has eyen numbers of protons and neutrons and so shows no nuclear spin absorption. [(CH;),S=O - C H , ] I - has two different kinds of H's and therefore two peaks would be observed. (6) Since S is a far better nucleophile than 0, [(CH,),S-V]I- is the almost exclusive product. 32
Supplementary Problems Problem 14.45 Give the structural formula and IUPAC name for terr-butyl ether, (c) 12-crown-4 ether. ( a ) CH,CH,CH,-O-CH=CHCH,
CH,
(a)
n-propyl propenyl ether, (6) isobutyl 4
1-(n-Propoxy)-1-propene
CH,
I
I
(6) CH,CHCH,-O-C
I
2-Isobutoxy-2-methylpropane
CH, Problem 14.46 Give the structural formulas for ( a ) ethylene glycol, (6) propylene glycol, and (c) trimethylene glycol. 4 ( a ) HOCH,CH,OH,
(6) CH,CHOHCH,OH,
Problem 14.47 Account for the fact that the C--o--C H U H bond angle in water [112" versus 105"]?
(c) HOCH,CH,CH,OH.
bond angle in dimethyl ether is greater than the
4
The repulsive van der Waals forces between the two CH, groups in dimethyl ether are greater than those between the two H's in water because the methyl groups are larger than the H's and have more electrons. Problem 14.48 Distinguish between an ether and an alcohol by ( a ) chemical tests, (6) spectral methods. 4
1" and 2" alcohols are oxidizable and give positive tests with CrO, in acid (orange color turns green). All alcohols of moderate molecular weights evolve H, on addition of Na. Dry ethers are negative to both tests. (6) The ir spectra of alcohols, but not ethers, show an 0-H stretching band at about 3500 cm-'. Comparing the ir spectra is the best method for distinguishing between these functional groups. (a)
Problem 14.49 What products are formed on HI cleavage of ( a ) 1,4-dioxane? (6) 1,3-dioxane? (a)
1,2-diiodoethane
(6)
excess
-CH,I,
Problem 14.50 Why is di-r-butyl ether very easily cleaved by HI?
4
+ I(CH2),I 4
On treatment with HI, the ether is protonated. This oxonium ion cleaves readily to give t-butyl alcohol and the relatively stable 1-butyl carbocation. Iodide ion adds to the carbocation, and the alcohol reacts with HI; both give r-butyl iodide.
[CHAP. 14
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
310
CH,
CH,
Problem 14.51 Give a chemical test to distinguish C,H,, from (C,H,),O.
4
Unlike C S H I Z(C,H,),O , is basic and dissolves in concentrated H,SO,.
(C,H,),O
+ H,SO,-(C,H,),OH’
+ HSO;
Problem 14.52 Outline the mechanism for acid- and base-catalyzed additions to ethylene oxide and give the structural formulas of the products of addition of the following: (a) H,O, (b) CH,OH, ( c ) CH,NH,, (d) CH ,CH ,SH. 4
In acid, 0 is first protonated.
The protonated epoxide can also react with nucleophilic solvents such as CH,OH.
In base, the ring is cleaved by attack of the nucleophile on the less substituted C to form an alkoxide anion, which is then protonated. Reactivity is attributed to the highly strained three-membered ring, which is readily cleaved. R B
(a) HOCH,CH,OH,
(6) CH,OCH,CH,OH,
(c) CH,NHCH2CH20H, ( d ) CH3CH2SCH2CH,0H*
Problem 14.53 Supply structures for compounds (A) through (F). H,C=CH2
+ (A)-CICH2CH,OH~(B)-(C) H2S04
a l c KOH
HOCl
(CH,),CBr+ alc. K O H + ( D ) - ( E ) z ( F )
4
(A) HOCl
(B) ClCH,CH,-O-CH,CH,CI
(C) H,C=CH-O-CH--LH,
(D) (CH,),C=CH,
(E) (CH,),C--CH,CI AH
(F) (CH3)2c,7H2 0
Formation of (F), isobutylene oxide, is an internal S,2 reaction. (CH,),C-CH,CI
AH
~~~~
’
[
‘I
(CH ) C-CH2-CI
IAl’
(CH3)ZC-CHZ
\/ 0
+ Cl-
CHAP. 141
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
311
Problem 14.54 Are the mle peaks 102, 87, and 59 (base peak) consistent for n-butyl ethyl ether (A) or methyl 4 n-pentyl ether (B)? Give the structure of the fragments which justify your answer. 8'
B
B
~ H , O ~ H , C H , C H , C H , C H , (B)
C H , ~ H , O ~ H , C H , C H , C H ,(A)
The parent P' is mle = 102, the molecular weight of the ether. The other peaks arise as follows:
102 - 15(CH,)
102 - 43(C,H,)
= 87
= 59
ions of ethers occur mainly at the C " - C a bonds. (A) fits these data for
Fragmentations of P' H,&-OCH,CH,CH,CH,
( m l e = 87; CU'-4?'
(mle = 59; C " 4 ' cleavage).
cleavage) and CH,CH,&H, +
Cleavage of the C B - C bonds in (B) would give a cation, CH,O=CH, ( m l e = 4 5 ) , but this peak was not observed. Problem 14.55 Prepare the following ethers starting with benzene, toluene, phenol (C,H,OH), cyclohexanol, any aliphatic compound of three C's or less and any solvent or inorganic reagent: ( a ) dibenzyl ether, ( 6 ) di-n-butyl ether, (c) ethyl isopropyl ether, (d) cyclohexyl methyl ether, ( e )p-nitrophenyl ethyl ether, ( f )divinyl ether (8) diphenyl ether. 4 <'I2
C,H,CH,l,,hI'
0 /\ + CH2CH2
-1-MgBr
CH,CH,Br
OH-
C,H,CH,Cl-
::
H20
"ZS04
C,H,CH,OH-,,,b'
]CH,CHi-CH,CH,OH
+ Na'OCH(CH,),--+
(C,H,CH,),O
: ;1'
* (CH,C H2CH,CH2),0
CH,CH,OCH(CH,),
(1")
Use the 1" RX to minimize the competing E2 elimination reaction or use CH,CH=CH,
+ CH,CH,OH-
Hg( OCOCF,)
NaOH
C,H,O-Na'
-
CH,CH(OCH,CH,)CH,HgOCOCF,
-
C,H,,OH + C H z N , ~ C , H , , O C H + , N, C,H,OH-
+ Na'BrNaBH4
product
CH31
or C , H , , O H ~ C , H , , O - N a ' + C , H , , O C H ,
C H Br
C,H,OC,H,-
HN03 H2s04
p-NO,C,H,OC,H,
Williamson synthesis of an aryl alkyl ether requires the Ar to be part of the nucleophile ArO- and not the halide, since ArX does not readily undergo SN2displacements. Note that since ArOH is much more acidic than ROH, it is converted to ArO- by OH- instead of by Na as required for ROH. See Problem 14.53, compounds (A), (B) and (C). Vinyl alcohol, H,C=CHOH, cannot be used as a starting material because it is not stable and rearranges to CH,CHO. The double bond must be introduced after the ether bond is formed. no solvent
Phenols do not undergo intermolecular dehydration. Although aryl halides cannot be used as substrates in typical Williamson syntheses, they do undergo a modified Williamson-type synthesis at higher temperature in the presence of Cu.
-
Problem 14.56 Prepare ethylene glycol from the following compounds: ( a ) ethylene, (6) ethylene oxide, (c) I ,2-dichloroethane. 4 dil a q K M n 0 4 ( a ) Oxidation: H,C=CH, HOCH24H,OH
( 6 ) Acid hydrolysis
CH,-CH,
\/
H 0 H'
(c) Alkaline hydrolysis: CICH,-CH,CI
HOCH,-CH,OH
-
H 0 OH-
HOCH,-CH,OH
Problem 14.57 Give structures of the major reaction products of l-methyl-l,2-epoxycyclohexanewith ( a ) ZH,O- Na' in CH,OH, ( 6 ) H,SO, in CH,OH. 4
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
312
[CHAP. 14
Problem 14.58 One mole of compound C,H,,O, reacts with 4mol of HIO,. The moles of products are: 1 H,CO, 1 CH,CH=O, and 3 HCOOH. Supply a possible structure for C,H,,O,. 4 There are 4 adjacencies. The aldehydes, H,C==O and CH,CHO, are formed by oxidation of the terminal C - O H groups. The C - O H groups in the middle of the polyol sequences have two adjacencies and cleave to HCOOH. The sequence H H H-'
H
I C-OH I I 0 0
Formaldehyde
gives the structure
H
H
I C-OH It 0
I C-OH I 0
A-CH,
I
3 moles of formic acid
H
H
I
H
I
I
AH
AH
Acetaldehyde
H
H
bH
AH
I
H-C+C+C-+C+C-CH, AH
0
I
I
Problem 14.59 Outline the steps and give the product of pinacol rearrangement of: (a) 3-phenyl-1,2propanediol, (6) 2,3-diphenyl-2,3-butanediol. 4 2-
C6H$H2-CH-CH2
(4
1"
J-B
::::'&
C,H $. H2-q H-C-H
AH AH
AH
I-
C,H,CH2CH2-C-H
C,H,CH,-CH2-6-H
II
AH
The 2" OH is protonated and lost as H,O in preference to the 1" OH. CH, CH3
(b)
I
1
CH, CH,
I
H+\ C,H,-c-C-c,H,
t
CH, CH,
-
H 0
I
OH +OH H
t
w -1-
2C,H,-C-c+-C,H,
CbHs
CH3
I It I
CH,-c-c-c6H5 C6H5
CH, CH,
I I
t 1
+C-C-C,H, OH
C6HS
Since the symmetrical pinacol can give only a single R', the product is determined by the greater migratory aptitude of C,H,. Problem 14.60 Show how ethylene oxide is used to manufacture the following water soluble organic solvents:
CHAP. 141
313
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
(a) Carbitol
(b) Diethylene glycol (HOCH,CH,OCH,CH,OH)
(C,H,OCH,CH,OCH,CH,OH) H
I
( d ) 1,6Dioxane
( c ) Diethanolamine (HOCH,CH,NCH,CH,OH)
( a1
-.
C,H,OH
4
/n H~C-CH~ iy
+ H2C-CH2 -% C2HsOCH2CH20HH+ C2H,0CH2CH,0CH2CH20H
L d‘
..n
H 2 0 + H*C,-CH2 -% HOCH2CH20HH+ HOCH2CH20CH2CH20H
Problem 14.61 Ethers, especially those with more than one ether linkage, are also named by the 0x8 method. The ether 0 ’ s are counted as C’s in determining the longest hydrocarbon chain. The 0 is designated by the prefix 0x8-, and a number indicates its position. Use this method to name: (a)
(CH,),COCH,CH(CH,),
( 6 ) C2HsOCH2CH20CH2CH20H ( c )
CH, I
(a )
2
1
Tetrahydrofuran
H 3
4
5
CJ
1
6
rn (4
O W 0
Dioxane
.
2,2,5-Trimethyl-3-oxahexane
( 6 ) CH,CH, OCH, CH, OCH, CH, OH 3,6-Dioxa-1-octanol. (c)
Oxacyclopentane ,
(d) 1,4-Dioxacyclohexane,
Problem 14.62 Outline mechanisms to account for the different isomers formed from reaction of
withCH30H in acidic (H’) and in basic (CH30-) media. C H 3 0 - reacts by an S,2 mechanism attacking the less substituted C.
Isobutylene oxide
4
314
ETHERS, EPOXIDES, GLYCOLS, AND THIOETHERS
-
[CHAP. 14
In acid, the S,1 mechanism produces the more stable 3" R'.
\0/
(a 3" R ' )
(CH,),&H,OH
(CH3)2C-CH2
(CH3)2C-CH2
JOL
?I
CH&H
(CH,),C--bH, I
I -
H+
(CH,),C&H,OH I
OH
dCH,
1" R '
Problem 14.63 Write structures for products of the reaction of: (a) H3C-CH-CH2
\ 0/
+ CH,NH,
( 6 ) CnH5CH-CH2
\ 0I
+ dry HCI
4
(by SNt) +
Reaction (6) goes via the stable benzyl-type R', C,H,CHCH,OH.
-
Problem 14.64 Prepare mustard gas, (CICH,CH,),S, from ethylene.
H,C=CH,
+ S,Cl,
(ClCH,CH,)S
4
+S
Problem 14.65 A compound, C,H,O,, gives a negative test with HIO,. List all possible structures and show how ir and nmr spectroscopy can distinguish among them. (Note that gem-diols can be disregarded since they are usually not stable.) 4
There are no degrees of unsaturation and hence no rings or multiple bonds. The 0 ' s must be present as C U H and/or C - 4 - C . The compound can be a diol, a hydroxyether or a diether. A negative test with HIO, rules out a vic-diol. Possible structures are: a diol, 'H0CH:CHiCH;OH' (A); two hydroxyethers, 'H0CH:CH:OCH: (B) and 'HOCHiOCHiCH: (C); and a diether (an acetal), CH,OCH,OCH, (D). (D) is pinpointed by ir; it has no OH, there is no 0-H stretch and peaks are not observed at greater than 2950cm-'. (A) can be differentiated from (B) and (C) by nmr. (A) has only three kinds of equivalent H's, as labeled, while (B) and (C) each have four. In dimethyl sulfoxide, the nmr spectrum of (C) shows all H peaks to be split: H3, a quartet, couples H4, a triplet; H2, a doublet, couples H', a triplet. The nmr spectrum of (B) in DMSO shows a sharp singlet for H4integrating for three H's. Other differences may be observed but those described above are sufficient for identification. The DMSO used is deuterated, (CD,),SO, to prevent interference with the spectrum.
Chapter 15 Carbonyl Compounds: Aldehydes and Ketones 15.1 INTRODUCTION AND NOMENCLATURE
Carbonyl compounds have only H, R, or Ar groups attached to the carbonyl group. >&O Aldehydes have at least one H bonded to the carbonyl group; ketones have only R’s or Ar’s.
ALDEHYDES IUPAC names the longest continuous chain including the C of -CH=O and replaces -e of the alkane name by the suffix -al. The C of CHO is number 1. For compounds with two -CHO groups, the suffix -dial is added to the alkane name. When other functional groups have naming priority, - C H O is called formyl. Common names replace the suffix -ic (-oic or axylic) and the word acid of the corresponding carboxylic acids by -aldehyde. Locations of substituents on chains are designated by Greek letters; e.g.
The terminal C of a long chain is designated w (omega). The compound is named as an aldehyde (or carbaldehyde) whenever - C H O is attached to a ring [see Problem 15.4(c)J. KETONES Common names use the names of R or Ar as separate words, along with the word ketone. The IUPAC system replaces the -e of the name of the longest chain by the suffix -one. In molecules with functional groups, such as - C O O H , that have a higher naming priority, the carbonyl group is indicated by the prefix keto-. Thus, C H , - C O - C H , - C H , - C O O H is 4ketopentanoic acid. Groups like
-C-R(Ar)
0 II
are called acyl groups; e.g. -& H 3 is the acetyl group. Phenyl ketones are often named as the acyl group followed by the suffix -phenone [see Problem 15,1(e)]. Problem 15.1 Give the common and IUPAC names for (a) CH,CHO, (6) (CH,),CHCH,CHO, ( c ) 4 CH,CH2CH,CHClCH0, (d) (CH,),CHCOCH,, (e) CH,CH,COC,H,, ( f ) H,C==CHCOCH,. (a)
Acetaldehyde (from acetic acid), ethanal;
315
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
316
[CHAP. 15
P-methylbutyraldehyde, 3-methylbutanal; (c) a-chlorovaleraldehyde, 2-chloropentanal; ( d ) methyl isopropyl ketone, 3-methyl-2-butanone; (e) ethyl phenyl ketone, 1-phenyl-1-propanone(propiophenone); ( f) methyl vinyl ketone, 3-buten-2-one. The C=O group has numbering priority over the C-C group. Problem 15.2 Give structural formulas for ( a ) methyl isobutyl ketone, (6) phenylacetaldehyde, (c) 2-methyl4 3-pentanone, ( d ) 3-hexenal, (e) p -chloropropionaldehyde. (U)
CH,-C-CH
b
CHCH,
( b ) C,H,CH2-C=0
(c) CH3CH2-C-CHCH,
HI
2AH3
H
I
( d ) CH,CH,CH=CHCH,C=O
AH3
H
I
( e ) ClCH,CH,C=O
Problem 15.3 Write IUPAC and common names for ( a ) (CH,),CCHO, (6) C 6 H , C H = C H - C H 0 , (c) 4 HOCH,CH,CH--V, (d) (CH,),C=CHCOCH,, (e) CH,CH,CH(CH,)COCH,CH,.
( a ) 2,2-dimethylpropanal, a,a-dimethylpropionaldehyde or trimethylacetaldehyde; (6) 3-phenylpropenal, cinnamaldehyde; (c) 3-hydroxypropanal, p - hydroxypropionaldehyde; ( d ) 4-methyl-3-penten-2-one, mesityl oxide; ( e ) 4-rnethyl-3-hexanone7ethyl sec-butyl ketone. Problem 15.4 Name the OHCC,H,SO,H, (c) H,C-HO,
following compounds: ( a ) ( d ) o-BrC,H,CHO.
OHCCH,CH,CH,CH(CH,)CHO,
(6)
p-
4
( a ) 2-Methyl-1,6-hexanedial. ( 6 ) -SO,H takes priority over --CHO; thus p-formylbenzenesulfonic acid. (c) The corresponding acid is a cyclopropanecarboxylic acid, and -0xyfic acid is replaced by -aldehyde: 2-methylcyclopropanecarbaldehyde. (6) The -oic acid in benzoic acid is replaced by -aldehyde: 0-6romobenzaldehyde (also called 2-bromobenzenecarbaldehyde). Problem 15.5 ( a ) Draw (i) an atomic orbital representation of the carbonyl group and (ii) resonance 4 structures. (6) What is the major difference between the C=O and C==C groups? ( a ) (i) The C uses sp2 HO's and its three
(T bonds are coplanar, with bond angles near 120". Each pair of unshared electrons is in a nonbonding (n) orbital. The 'TT bond formed by lateral overlap of p AO's of C and 0 is in a plane perpendicular to the plane of the (T bonds.
(T.. (T
(ii) In this case the polar resonance structure makes a considerable contribution to the hybrid and has a profound effect on the chemistry of the C=O group. (6) The C=C group has no significant polar character and its 7r bond acts as a nucleophilic site. The polarity of the 'TT bond in C=O causes the C to be an electrophilic site and the 0 to be a nucleophilic site. Problem 15.6 Account for the following: ( a ) n-butyl alcohol boils at 118°C and n-butyraldehyde boils at 76 "C, yet their molecular weights are close, 74 and 72, respectiyely; (6) the C=O bond (0.122 nm) is shorter than the C - 4 (0.141 nm) bond; (c) the dipole moment of propanal (2.52 D) is greater than that of 1-butene (0.3D); ( d ) carbonyl compounds are more soluble in water than the corresponding alkanes. 4
CHAP. 151
317
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
(a) H-bonding between alcohol molecules is responsible for the higher boiling point. (b) The sharing of two pairs of electrons in c----O causes the double bond to be shorter and stronger. ( c ) The polar contributing structure [Problem 15.5(a)(ii)] induces the large dipole moment of the aldehyde. (d) H-bonding between carbonyl oxygen and water renders carbonyl compounds more water-soluble than hydrocarbons.
Problem 15.7 Compare aldehydes and ketones as to stability and reactivity.
4
As was the case for alkenes, alkyl substituents lower the enthalpy of the unsaturated molecule. Hence, ketones, with two R's, have lower enthalpies than aldehydes, with one R. The electron-releasing R's diminish the electrophilicity of the carbonyl C, lessening the chemical reactivity of ketones. Furthermore, the R's, especially large bulky ones, make approach of reactants to the C more difficult. Problem 15.8 Draw up a table of corresponding sequential oxidation levels of hydrocarbons and organic Cl. 0, and N compounds. 4
See Table 15-1.
Table 15-1 OXIDATION
I
Hydrocarbon Halogen compounds Oxygen compounds Nitrogen compounds
CH,CH,
+
-
H,C=CH, CH,CH,CI
HC'-"CH CH,CHCl ,
CH,CCl,
CCl,
-
CH,CH,OH
CH,CH----O
CH,COOH
CO,
-
CH,CH,NH,
CH,CH=NH
CH,C%N
H , N a N
f
REDUCTION
15.2 PREPARATION As shown in Table 15-1, -0 is at an oxidation level between 4 H O H and - C O O H . Hence, aldehydes, RCHO, and ketones, R,C=O, are made by oxidizing the corresponding 1" RCH,OH and 2" R,CHOH, respectively. RCHO, but not R,C=O, can also be prepared by reducing the corresponding RCOOH or its derivative RCOX. Hydrolysis (overall reaction with water) of the other groups in the same oxidation level as T O . (-C=C--, --CC12-, and -+NH) will also give the T - 0 group.
BY OXIDATION 1.
1" RCH,OH+
RCHO and 2" R,CHOH+
R,CO. See Section 13.3
Alcohols are the most important precursors in the synthesis of carbonyl compounds, being readily available. More complex alcohols are prepared by reaction of Grignard reagents with simpler carbonyl compounds. Ordinarily MnO, and Cr,O;- in acid are used to oxidize 2" R,CHOH to R,CO. However, to oxidize 1" RCH,OH to RCHO without allowing the ready oxidation of RCHO to RCOOH, requires special reagents. These include: ( a ) pyridinium chlorochromate (PCC),
(o>NH' (Cr0,Cl)-
(the best method)
318
CARBONYL COMPOUNDS: ALDEHYDES A N D KETONES
[CHAP. 15
( 6 ) hot Cu (only with easily vaporized ROH); (c) MnO, (mild, only with allylic-type, RCH==CHCH,OH, or benzylic-type, ArCH,OH, alcohols); (d) Na,Cr,O,lH,SO,/acetone (Jones reagent, which may permit RCOOH). 2.
1"RCH,X (X=Cl, Br, 1) or -OSO,R'(Ar)
+ Me,S=O-
RCHO
To prevent overoxidation of aldehydes, the very mild oxidant dimethyl sulfoxide or DMSO, is used to react with 1" halides or sulfonates to give aldehydes. These reactants are in the same oxidation level as alcohols:
0
CH,(CH,),I
I' + CH,SCH,
HCOj,
+ CH,SCH, + HI
CH,(CH,),CHO
Problem 15.9 Suggest a mechanism for the reaction of RCH,Cl with DMSO.
4
A C-0 bond is formed and the CI- is displaced in Step 1 by an S,2 attack. The C=O bond results from an E2 /3-elimination of H' and Me,S, a good leaving group as indicated in Step 2.
Step 1 Me2S'+;\CH&
Me,S'-O-CHR
I
+ C1-
H an alkoxysulfonium salt
Step 2 Me$' Y
H
R
base
7 M H R + Me,S
H 3. AIkyl Arenes:
ArCH,-
ArCHO, ArCH,R-
ArCOR
Benzylic CH, and CH, groups can be oxidized to groups at the same oxidation level as M These groups are then hydrolyzed to the C=O group. ArCH,CH,
ArCH3
""'6"" c*
NBS
ArCBr,CH,
H O
.
ArCOCH,
0
I/
Arc
O-h-CH3
\
H+
ArCHO
+ 2CH3COOH
O T C H 3 a gemdiacetate
4. Alkylboranes
(See Problem 6.24( f ) for hydroboration.)
The vinyl C with more H's is converted into -0. Alkenes can also be transformed into dialkyl carbonyls by a carbonylation-oxidation procedure.
CHAP. 151
319
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
(WO yield)
5. Oxidative Cleavages Ozonolysis of alkenes (end of Section 6.4) and cleavage of glycols (Section 14.11) afford carbonyl compounds. These reactions, once used for structure determinations, have been superseded by spectral methods. BY REDUCTIONS OF ACID DERNATNES, RCOX, OR NITRILES, R C k N Acid chlorides, R(Ar)COCl, are reduced to R(Ar)CHO by H,/Pd(S), a moderate catalyst that does not reduce RCHO to RCH,OH (Rosenmund reduction). Acid chlorides, esters (R(Ar)COOR), and nitriles ( R h N )are reduced with lithium tri-t-butoxyaluminum hydride, LiAlH[OC( CH,),], , at very low temperatures, followed by H,O. The net reaction is a displacement of X - by :H-, RCHO + C1-
RCOCl+ :H-See Section 16.3 for preparation of acid derivatives.
I
BY HYDROLYSIS AND HYDRATION OF COMPOUNDS AT --C=O OXIDATION LEVEL 1. -CX2-,
d ( O C O R ) , , and 4 ( 0 R ) , (Acetal or Ketal)
These groupings are hydrolizable to the
&group. (See Section 15.4 for acetal chemistry.)
2. Alkynes See Section 8.2 for direct hydration and for net hydration through formation of vinylboranes by hydroboration. Problem 15.10 Which is the only aldehyde that can be prepared by HgS0,-catalyzed hydration of an alkyne? Since the addition of H,O to c=C is Markovnikov regiospecific, R-H Only H m H is hydrated to give an aldehyde, CH,CHO.
or R-R
4
must give ketones.
BY FRIEDEL-CRAFT ACYLATIONS OR FORMYLATIONS OF ARENES
i l R
Friedel-Crafts acylations of arenes with RCOCl or anhydrides ( R C - M R ) AlCl, give good yields of ketones. Problem 15.11 Suggest a mechanism for acylation of ArH with RCOCl in AlCl,. The mechanism is similar to that of alkylation: (1)
RCOCl
+ AICI,
+
__I*
RC-0:
+ AlCl;
acylonium ion
in the presence of
4
320
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
[CHAP. 15
Problem 15.12 Can formylation of an arene, ArH, with an acid chloride be employed to prepare ArCHO? 4
No. The needed acid chloride is the hypothetical “formyl chloride,” HCOC1. But this compound cannot be realized; attempts to prepare it from formic acid (HCOOH + SOCl,) yield only mixtures of HCl and carbon monoxide, :C=O :.
Arenes can be formylated by generating the active intermediate, :&--H, from reagents other than HCOCI. The Gatterman-Koch reaction uses a high-pressure gaseous mixture of CO and HCI. CO + HCI
AlCl
CuCl
d - H
ArH
ki
~
ArCHO
BY ACYLATION OR HYDROFORMYLATION OF ALKENES 1.
0 x 0 Process
This is an industrial hydroformylation for synthesizing aliphatic aldehydes, RCHO.
+ CO + H,
RCH=CH,
2.
C‘O2(CO)”
CHO
RCH,-CH,CHO
I
+ RCH-CH,
Acylation
This is a Markovnikov addition initiated by RC’=O:,
an acylonium cation.
BY COUPLING REACTIONS 1.
Carboxylic Acids and Their Derivatives, with Organometallics
( a ) R‘--C--CI
II
+ R,CuLi-
R‘
R (cf. Corey-House reaction, Section 4.3)
30
0 an acid chloride
a ketone
a nitrile
(c) R ’ 3 - O H
+ 2RLi-
0 a carboxylic acid
-I?-”
R’
0
cyclohexyl phenyl ketone
+ RH + 2LioH
a ketone
Problem 15.13 ( a ) Why doesn’t reaction of RMgX with R’COCI give a ketone? (6) Account for the different behaviors of RMgX and R,CuLi. (c) What is the relationship between the reactivity of an organometallic and 4 the activity of the metal? (a) The ketone RCOR’ is formed initially, but once formed, since it is more reactive than RCOCI, it reacts further with RMgX to give the 3” alcohol R’R,COH. (6) The C-to-Mg bond has much more ionic character than has the C-to-Cu bond. Therefore, the R group in RMgX is more like R:- and is much more reactive. (c) The more active the metal, the more apt it is to carry a + charge and the more apt is the C to carry a - charge.
CHAP. 151
32 1
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
2. Alkylation of 1,3-Dithianes with 1" RX or ROS0,Ar 1,3-Dithiane, prepared from H,C=O and 1,3-propanedithiol, HSCH,CH,CH,SH [Problem group and then hydrolyzed to give the aldehyde. The acidity of this group results from the delocalization of the negative charge of the carbanion to each S by p-d 7~ bonding (Section 3.11). 15.14(c)], can be alkylated at the acidic -S-CH,-S-
n
n-butyl LI
x H H 1,3-Dithiane
f\lS
RX
S
HgC12, M e O H
H
xH R
carbanion
alkylated product
__*
RCHO
7
H20
Ketones can be prepared by ( a ) dialkylating 1,3-dithiane before hydrolysis, or (b) forming the dithiane of RCHO and then monoalkylating. Problem 15.14 Synthesize: ( a ) p-methoxybenzaldehyde from benzene; (6) cyclohexylethanal by hydroboration and oxidation; ( c ) phenylacetaldehyde, using 1,3-dithiane; ( d ) phenyl n-propyl ketone from a dithiane; (e) 4 cyclohexyl phenyl ketone from PhCOOH and RLi; (f)2-heptanone, using a cuprate
&&-@
C1
0-Na'
C6H6-&-
OCH,
OCH,
CH=O
Cyclohexylacetylene
unstable enol
1,3-Dithiane
C,H,CH=O
/CH,SH
+ CH,
\
CH,SH
(f?) C6HsCOOH-C,H,COOLiOH
-n ,
,q
1 BuLi _____, 2 ~ I - C ~ H ~ X
'xS C6HS H Li'
anion
1 eH30+ x c e s s c y c l o h e x y l lithium
H30+
kS
C,H,
'
c
6
-
S
J
0
0 (f)
CH,(CH,),COOH-
PCl
1. ( C H ~ ) ~ C U L I
II
II
0
n-C,H,
H
C,H,-C-n-C,H,
C H 3 ( C H z ) 4 C O C CH,(CH,),CCH, l ~
322
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
[CHAP. 15
BY PINACOL-PINACOLONE REARRANGEMENT (Section 14.11) Problem 15.15 What products are formed in the following reactions? ( a ) CH,CH,OH, Cr,O:-, H'; (b) CH,CHOHCH,, Cr,O:-, H t (60°C); (c) CH,COCI, LiAl(0-t-C,H,),H; (d) CH,COCI, C,H,, AICI,; (e) CH,COCI, C,H,NO,, AICI,; 4 ( a ) CH,CHO (some oxidation to CH,COOH occurs).
H
0
I
II
(b) CH3-C-CH3
(c) CH3-C=0
(d) C,H,COCH,. (e) No reaction; acylation like alkylation does not occur because NO, deactivates the ring.
Problem 15.16 Show the substances needed to prepare the following compounds by the indicated reactions: ( a)
CH,CH,CCH,CH,C,H,
8
(c) (a)
(Grignard)
( b) C,H,CH,CH=CH-C-CH,C,H,
8
2,4-CI,C,H,COC,HS
(Acylation of an alkene)
(Friedel-Crafts acylation)
4
R' C k N + RMgX. The carbonyl C in RCOR' and one alkyl group (R') come from R'-C%N; the other R from RMgX. The two possible combinations are: CH,CH,C%N
+ CIMgCH,CH,C,H,
or CH,CH,MgBr
+ N=CCH,CH,C,H,
( b ) The R attached to C=C is part of the alkene. d R ' comes from R'COCI.
-
+ CICOCH,C,H,
C,H,CH,CH=CH,
2,4-Cl2C,H,COC1 + C,H,
AIC13
BF3
__*
product
product
C,H,COCl and 1,3-C,H,CI2 cannot react with AICI, because the two aryl Cl's deactivate the ring.
Problem 15.17 Use hydroboration to prepare ( a ) cyclohexanone, (b) dicyclohexyl ketone, (c) pentanal , (d) n-butyl cyclopentyl ketone. 4
(4
2 Cyclohexene + B,H,
(Cyclohexyl),BH (c)
-
+(Cyclohexyl),BH 1 C0.H20
2
HzO?.OH-
Cr03
H
Cyclohexanone
Dicyclohexyl ketone
1-Pentyne + B , H , - - + ( C H 3 C H , C H , C H = C H ) , B ~Pentanal "2%
With mixed alkylboranes, the major ketone formed has the 1"R. Likewise, 2"R's predominate over 3".
Problem 15.18 Prepare the following compounds from benzene, toluene, and alcohols of four or fewer C's: ( a ) 2-methylpropanal (isobutyraldehyde), (6) p-chlorobenzaldehyde, (c) p-nitrobenzophenone ( p-NO,C,H,COC,H,), (d) benzyl methyl ketone, (e) p-methylbenzaldehyde. 4
323
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
CHAP. 151
.:,
(CH,),CHCH,OH
( a1
(CH,),CHCHO
(RCHOis not oxidized further.) 1.
(61
C6H5CH3 HN03
(clC6H5CH3
H*=,
CI
Cr03. acetic anhydride
p-C1C6H4CH3 .,
H30+
b
p-ClC,H,CHO
KMn04
' p-02NC6H4CH3 H+
-
We cannot acylate C,H,NO, with C,H,COCI because NO, deactivates the ring. H,C -CHI
( d ) C6H6
B*2
Fe
C,H,Br 5 C,H,MgBr
C,H,CH,
(e)
' 0 '
KMn04
C6H5CH2CH20H___*
+ CO,HCl
NC13
cucl
b
p-CH,C,H,CHO
15.3 OXIDATION AND REDUCTION
OXIDATION 1. To Carboxylic Acids
Aldehydes undergo the oxidation; R-CH4
K M n 0 4 or K 2 C r 2 0 7 . H +
+ R-COOH
A mild oxidant is Tollens' reagent, Ag(NH,)i (from Ag' and NH,)
+ 2Ag(NH,)f + 30H-
R-C-H
CI
R-COO-
+ 2H20 + 4NH3 + 2Ag
(mirror)
Formation of the shiny Ag mirror is a positive test for aldehydes. The RCHO must be soluble in aqueous alcohol. This mild oxidant permits - C H O to be oxidized in a molecule having groups more difficult to oxidize, such as 1" or 2" OH'S. Ketones resist mild oxidation, but with strong oxidants at high temperatures they undergo cleavage of C - C bonds on either side of the carbonyl group to give a mixture of carboxylic acids. RCH+-C*CH,R'
II
a RCOOH + R'CH,COOH + RCH,COOH from cleavage of bond (a)
+ R'COOH
from cleavage of bond ( b )
2. Via the Haloform Reaction
Methyl ketones,
CH3$R
are readily oxidized by NaOI (NaOH + I,) to iodoform, CHI,, and RCOO-Na'. 13.48.)
(See Problem
324
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
HzSO
Ar4-R
1I
0
[CHAP. 15
(very little Ar-C-0-R)
Ar-0-C-R
1I
1I
0
0 ester
When an aryl alkyl ketone is oxidized, the R remains attached to the carbonyl carbon and Ar is bonded to 0 of the ester group. REDUCTION 1.
To Alcohols by Metal Hydrides or H,/catalyst H:- (from NaBH,)
+
>c=o-
I
H - C - O - H--d--OH ~
I
I
2. To Methylene 'CH,
>*O-
(also see Problem 15.39)
/ 0
II
Zn-Hg + HCI C (1or HlNNHl+KOH (WOW-)*
R-c-R'
RCH2R
ii
The Clemmensen reaction is used mainly with aryl alkyl ketones, Ar R+ArCH,R. 0
Problem 15.19 Give the products of reaction for ( a ) benzaldehyde + Tollens' reagent; (6) cyclohexanone + HNO,, heat; ( c ) acetaldehyde + dilute KMnO,; (d) phenylacetaldehyde + LiAIH,; (e) methyl vinyl ketone + HJNi; ( f ) methyl vinyl ketone + NaBH,; ( g ) cyclohexanone + C,H,MgBr and then H,O+ ; (h) methyl ethyl 4 ketone + strong oxidant; (i) methyl ethyl ketone + Ag(NH,);. ( a ) C,H,COO-NH:, Ago (6) HOOC(CH,),COOH, (c) CH,COOH, ( d ) C,H,CH,CH,OH, (e) CH,-CH(OH)CH,CH, (C==O and C=C are reduced), ( f ) CH,-CH(OH)CH=CH, (only C=O is reduced, not --c----c-).
-
(b)
(a)
hot =id K M d d
CH3+C+CHZCH3
( h)
b
(i) no reaction.
CO,
+ CH,CH,COOH' +- CH3COOH from cleavage at (a 1
from cleavage at ( b1
3. Disproportionation. Cannizzaro Reaction Aldehydes with no H on the a C undergo self-redox (disproportionation) in hot concentrated alkali. 2HCHO2C6H5CHOC,H5CH0 + HCHOalways used
50% N a O H
heat
50% N a O H heat
50% N a O H heat
CH,OH
+ HCOO-Na'
C6H5CH20H4- C6H5COO-Na'
C6H5CH20H+ HCOO-Na' always formed
(crossed-Cannizzaro)
CHAP. 151
325
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
Problem 15.20 Devise a mechanism for the Cannizzaro reaction from the reactions
2ArCDO-
OHH2O
ArCOO-
+ ArCD,OH
2ArCHO-
OD
~
D?O
ArCOO-
+ ArCH,OH
4
The D’s from OD- and D,O (solvent) are not found in the products. The molecule of ArCDO that is oxidized must transfer its D to the molecule that is reduced. A role must also be assigned to OH -.
Ar-C-eOH-
D
-
0-H
,,I
Ar-C-0
@-/
-
D f
I
Ar-C=-O
0Ar-C-0
+ Ar-C-0-
I
-+ Ar-C=O
D
I + Ar-C-OH
Problem 15.21 For the Cannizzaro reaction, indicate (a) why the reaction cannot be used with aldehydes having an a H, - C H C H O ; (6) the role of OH- and OD- (Problem 15.20); (c) the reaction product with ethanedial, O = C H - C H = O ; (d) the reaction products of a crossed-Cannizzaro reaction between (i) formal4 dehyde and benzaldehyde, (ii) benzaldehyde and p-chlorobenzaldehyde.
An a H is acidic and is removed by OH-, leaving a carbanion that undergoes other reactions. (6) They are strong nucleophiles that attack the electrophilic C of C=O to give a tetrahedral intermediate. This intermediate reestablishes the resonance-stabilized C=O group by transferring an :H- to the M of another aldehyde molecule.
(a)
An internal Cannizzaro yields hydroxyacetic acid, HOCH,COOH. (i) H,CO is mainly attacked by OH- because it is more electrophilic than PhCHO, whose Ph group (d) delocalizes the electron deficiency of the C of C k O . (ii) There is little difference in the reactivities of the two aldehydes, and both sets of products are found PhCOOH and PhCH,OH mixed with pClC,H,COOH and p-CIC,H,CH,OH. (c)
15.4 ADDITION REACTIONS OF NUCLEOPHILES TO >C=O
The C of the carbonyl group is electrophilic,
[Problem 2.26(b)]and initially forms a bond with strong nucleophiles.
Electrophile
Nucleophile (strong)
Transition State
Nu
Nu
For example, :Nu- can be :R‘-of R‘MgX or :H- of NaBH,. With :NuH,,the adduct loses water to give -€=Nu.
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
326
&O
I
[CHAP. 15
+ :NuH,+ unstable adduct
:NuH, is most often a 1" amine, RNH,, or one of its derivatives, such as HONH, (hydroxylamine). Acid increases the rate of addition of weak nucleophiles by first protonating the 0 of C=O, thereby enhancing the electrophilicity of the C of C=O.
Transition state
The reactivity of the carbonyl group decreases with increasing size of R's and with electron donation by R. Electron-attracting R's increase the reactivity of C=O. Problem 15.22 ( a ) Explain why HCI and HBr, which add to C=C, fail to add to c----O. (6) In general, do 4 electrophilic reagents such as: HX,Cl,, and Br, add to carbonyl compounds? ( a ) Additions to C=O are reversible reactions. Furthermore, the C=O group has considerable resonance energy. Therefore, in the equilibrium between the carbonyl compound and the adduct, R,C=O is favored if the nucleophile is a good leaving group; C1- and Br- are good leaving groups. (6) No.
Problem 15.23 The order of reactivity in nucleophilic addition is
CH,=O > RCH-0
7 R2C=0
Account for this order in terms of steric and electronic factors.
> RC Y
A-
4
A change from a trigonal sp2 to a tetrahedral sp3 C in the transition state is accompanied by crowding of the four groups on C. Crowding and destabilization of the transition state is in the order
CH,=O
b
lowers the enthalpy of the ground state, raises A H S and decreases the reactivity of C=O toward nucleophilic attack. Hence, acid derivatives RCOY, in which
are less reactive than RCHO or R,CO. P m 15.24 Explain the order of reactivity ArCH2COR > R,C=O > ArCOR > Ar2C0 in nucleophilic addition. 4
CHAP. 151
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
327
When attached to C = O , Ar's, like -Y: (Problem 15.23), are electron-releasing by extended 7r bonding (resonance) and deactivate C 4 . Two Ar's are more deactivating than one Ar. In ArCH,COR only the electron-withdrawing inductive effect of Ar prevails; consequently, ArCH, increases the reactivity of C = O . Problem 15.25 The rate of addition of HCN to R , C 4 to form a cyanohydrin, R,C(OH)CN (a 24 hydroxynitrile), is increased by adding a trace of NaCN. Explain.
-
The rate-controlling step is the addition of CN-. R,C-OT
RC ,$$CN-
I
HCN
R,C-OH I
+ CN-
(regenerated catalyst)
Problem 15.26 Why is cyanohydrin formation useful in synthesis?
4
The cyanohydrin not only adds an additional C at the site of the C = O but also introduces two new functional groups, OH and CN, which can be used to introduce other functional groups. The OH can be used to form an alkene (C=C), an ether (-RO), or a halogen compound (C-X); the C k N can be reduced to an amine (CH,NH,), be hydrolyzed to a carboxyl (COOH) group, or react with Grignard reagents if the O H is protected. Problem 15.27 NaHSO, reacts with RCHO in EtOH to give a solid adduct. (a) Write an equation for the reaction. (b) Explain why only RCHO, methyl ketones (RCOCH,) and cyclic ketones react. (c) If the carbonyl compound can be regenerated on treating the adduct with acid or base, explain how this reaction with NaHSO, 4 can be used to separate RCHO from noncarbonyl compounds such as RCH,OH. (a) HSO, can protonate RCHO.
SOTNa'
k
k
k
Sodium bisulfite adduct (solid)
A C-S bond is formed because S is a more nucleophilic site than 0. SO:- is a large ion and reacts only if C=O is not sterically hindered, as is the case for RCHO, RCOCH,, and cyclic ketones. (c) The solid adduct is filtered from the ethanolic solution of unreacted RCH,OH and then is decomposed by acid or base:
(b)
I;f
I RC-SOqNa'
so2
H+ /
f
+ RCH
(extracted with ether)
so;-
AH
Problem 15.28 Write the formula for the solid derivative formed when an aldehyde or ketone reacts with each of the following ammonia derivatives:
H (U)
I
H-N-OH Hydroxylamine
H
I
(6) H-N-NHC6H5 Phcnylhydrazinc
H
I
(c) H-N-NHCONH, Stmicarbazide
Since these nucleophiles are of the :NuH, type, addition is followed by dehydration.
4
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
328
(a)
G = -OH;
\c=N-oH
(Oxime).
/
( 6 ) G = -NHC,H,; ( c ) G = -NHCONH,;
[CHAP. 15
\C=NNHC,H,
(Phenylhydrazone).
/
'C=NNHCONH,
/
(Semicarbazone).
The melting points of these solid derivatives are used to identify carbonyl compounds. Problem 15.29 Why do carbonyl compounds having an a H react with R,NH (2") to yield eneamines, --C==C-NR,,
I I
but give imines, 'C-C=NR,
'I
I
with RNH, (l")?
4
After protonation of the 0, the nucleophilic RNH, adds to the C and the adduct loses H', to give the
cubinolrmine. Dehydration proceeds by protonation of the 0 of OH, loss of H,O, and then loss of H', to give
the imine.
Carbinolarnine
r
1
L
iminium ion
J
imine
The carbinolamine formed from R2NH lacks an H on N, and its dehydration involves instead loss of the acidic (Y H to give the resonance-stabilized eneamine.
Carbinolamine
enearnine
Problem 15.30 Explain why formation of oximes and other ammonia derivatives requires slightly acidic media (pH = 3.5) for maximum rate'while basic or more highly acid conditions lower the rate. 4 The carbonyl group becomes more electrophilic and reactive when converted by acid to its conjugate acid, )CLOH An acidic medium is needed for protonation of the 0 and to aid in the dehydration. In more strongly acid solutions ( p H < 3 . 5 ) the amine is protonated to give the unreactive RNHf. In basic solutions there is no protonation of C d . Problem 15.31 Reaction of 1 mole of semicarbazide with a mixture of 1 mol each of cyclohexanone and benzaldehyde precipitates cyclohexanone semicarbazone, but after a few hours the precipitate is benzaldehyde semicarbazone. Explain. 4 The C = O of cyclohexanone is not deactivated by the electron-releasing C,H, and does not suffer from steric hinderance. The semicarbazone of cyclohexanone is the kinetically controlled product. Conjugation makes
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
CHAP. 151
329
HzNNHCONHz
+
more stable thermodynamic product
less stable kinetic product
Fig. 15-1
PhCH=NNHCONH, more stable, and its formation is thermodynamically controlled. In such reversible reactions the equilibrium shifts to the more stable product (Fig. 15-1). Problem 15.32 Give structures for the product(s) of the following reactions: (a)
1-Butanal + C,H,NHNH,
(6) 2-Butanone + [H,NOH]CI-, NaC,H,O,
4
NOH ( a ) CH,(CH,),CH=NNHC,H,.
II
(6) CH,CH,C--CH,:
Problem 15.33 Symmetrical ketones, R,C=O, ketones may form two isomeric oximes. Explain.
The
T
C,H,O;
(acetate) frees H,NOH from its salt.
form a single oxime, but aldehydes and unsymmetrical 4
bond in
‘C=N
/
/
prevents free rotation, and therefore geometric isomerism occurs if the groups on the carbonyl C are dissimilar. The old terms syn and anti are also used in place of cis and trans, respectively.
R
H
\C/ ‘OH trans ( a n t i )
/
R
N:
HO
R’ ‘C/
II
It
/
N:
HO
cis ( s y n )
Problem 15.34 Why are oximes more acidic than hydroxylamine? Bf
R
4
Loss of H’ from H,NOH gives the conjugate base, H,NO-, with the charge localized on 0. Delocalization charge by extended T bonding can occur in the conjugate base of the oxime,
is a weaker base, and its conjugate acid, the oxime, is much more acidic.
330
[CHAP. 15
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
15.5 ADDITIONS OF ALCOHOLS: ACETAL AND KETAL FORMATION
H
I
RI-C-0
H
+ 2ROH
dry HCI
W
I
+ H20
RI-C-OR
I
H,O+
OR an acetal (gem-diether)
In H,O', R'CHO is regenerated because acetals undergo acid-catalyzed cleavage much more easily than do ethers. Since acetals are stable in neutral or basic media, they are used to protect the - C H - 0 group. Unhindered ketones form ketals, R,C( OR'),. RSH forms thioacetals, RCH( SR'),, and thioketals, R,C(SR'),. Problem 15.35 Give mechanisms for ( a ) acid-catalyzed acetal formation,
H
H
I
+ R'OH
R-C-0
R'OH
I I
+ HZO
R-C-0-R'
0-R'
0-R'
hemiacetal
acetal
(6) base induced hemiacetal formation with OR- in ROH.
H (U)
4
H
I
R-C=O
+H+
hemiacetal
From the protonated hemiacetal the mechanism is similar to that for the formation of ethers from alcohols [Problem 14.7(6)].
H H
I
R-C-YH
I
H
-H20
AR
H
I R-C+ c----+ R C :OR' I +aR'
H H
-R--(i-OR'
1 n . M - I
I I
OR
+
protonated acetal
protonated hemiacetal
H -H+ +H
I
R-7-OR OR acetal
Problem 15.36 Show how a C=O group can be protected by acetal formation in the conversion of OHCCH,-H to OHCCHIC--=CCH,. 4 The introduction of CH, requires that the terminal alkyne C first become a carbanion and then be methylated. Such a carbanion, acting like the R group of RMgX, would react with the C=O group of another molecule before it could be methylated. To prevent this, C = O is protected by acetal formation before the carbanion is formed. The acetal is stable under the basic conditions of the methylation reactions. The aldehyde is later unmasked by acid-catalyzed hydrolysis. Step 1 Protection of
c=o as acetal:
OCHCH,G=CH+ HOCH
CH O H
L>cHCH*-H
CHAP. 15)
331
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
Step 2 Alkylation of W H :
Step 3 Unmasking of C=O group:
L;
CHCH,mCH,=
OCHCH,-CH,
+ HO(CH,),OH
Problem 15.37 In acid, most aldehydes form nonisolable hydrates (gem-diols). Two exceptions are the stable chloral hydrate, CI,CCH( OH), , and ninhydrin,
( a ) Given the bond energies 749, 464 and 360kJ/mol for
c----O, 0-H and C 4 , respectively, show why the 4 equilibrium typically lies toward the carbonyl compound. (6) Account for the exceptions. Calculating AH for
we obtain (749+ 2(464)]+[2(-360)+ 2(-464)]=AH (C-0) (O-H) cleavages
(C-0)
(O-H)
formations
cndothcnnic
exothcnnic
or AH = +29 kJ/mol. Hydrate formation is endothermic and not favored. The carbonyl side is also favored by entropy because two molecules, 'C=O
/
and H,O
are more random than 1 gem-diol molecule. Strong electron-withdrawinggroups on an a C destabilize an adjacent carbonyl group because of repulsion of adjacent + charges. Hydrate formation overcomes the forces 'of repulsion.
c1
+a+
a+
C1
H
Cl+-y-(f=O
npulsion fmm adjacent + chagcs
a-
Cl
OH
+ H,O ~ Z C L A - A - ( ) H I
(!I H Chloral hydrate less npulsion
Hydration of the middle carbonyl group of ninhydrin removes both pairs of repulsions.
'robkm 15.38 Write structures for the cyclic ketals or acetals prepared from ( a ) butanal 3) cyclohexanone + ethylene glycol.
+ 1,3-propanediol, 4
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
332
I
Problem 15.39 Convert --C=O
[CHAP. 15
to --CH2- via formation of an intermediate thioacetal or thioketal.
4
a thioketal
Problem 15.40 Show steps in the synthesis of cyclooctyne, the smallest ring with a triple bond, from 4 C2H ,OOC( CH ),COOC2H ? .
The IJ-diester is converted to an eight-membered ring acyloin, which is then changed to the alkyne.
0 0 C,H,-O-C-(CH~,-C-OC,H, It II
-%
0;". zz: 0a OH Cyclooctanone
an acyloin
1,l-Dichlorocyclooctane
Cyclooctyne (major)
1,2-CycIooctadiene (minor), an allene
15.6 ATTACK BY YLIDES; WlTTlG REACTION A carbanion C can form a p-d 7r bond (Section 3.11) with an adjacent P or S. The resulting charge delocalization is especially effective if P or S, furnishing the empty d orbital, also has a + charge. Carbanions with these characteristics are called ylides, e.g.
I+ I
-p-c---
..-
I
I
=c--
I
..-
-s-c-..+ I
I
c--*
-s=cI
I
The Wittig reaction uses P ylides to change 0 of the carbonyl group to
The carbanion portion of the ylide replaces the 0.
The ylide is prepared in two steps from RX.
n
Ph3P: + R C H 2 y X -% [Ph3kH,R] Xa phosphine
C,H+i+
Ph,;CHR
+ C,Hlo + LiX-
CHAP. 151
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
333
Sulfur ylides react with aldehydes and ketones to form epoxides (oxiranes):
H
H
The sulfur ylide is formed from the sulfonium salt,
with a strong base, such as sodium dimethyloxosulfonium methylide,
[ 7.1 CH3S-CH2
Na’
Problem 15.41 Which alkenes are formed from the following ylide-carbonyl compound pairs? ( a ) 2-butanone and CH,CH,CH,CH=P(C,H,),, (6) acetophenone and (C,H,),P=CH,, (c) benzaldehyde and C,H,--CH=P(C,H,), , (d) cyclohexanone and (C,H,),P=C(CH,),. (Disregard stereochemistry.) 4
The boxed portions below come from the ylide.
CH3
CH3
I
I
(a) CH,CH2C=ICHCH,CH2CH3]
( b ) C,H,-C=m\ n
Problem 15.42 Give structures of the ylide and carbonyl compound needed to prepare:
H +-
Ph,PCHCH3
I
+ C,H,C=O
H or
I
+ CH3C=0
Ph&C,H,
The cis-tram geometry of the alkene is influenced by the nature of the substituents, solvent and dissolved salts. Polar protic or aprotic solvents favor the cis isomer.
(61 (c1
0
+ Ph3iCH,
+ Ph,iCHC,H,
or
0
or
iPh,
+ 0-CH,
CH,CH,’CH(CH,), +!Ph,
(d)
(CH,),C=O
+
-
+ Ph,S-C(CHJ,
+ C,H,CHO
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
334
15.7 MISCELLANEOUS REACTIONS 1. Conversion to Dihalides
0
0
+
PClJ
[CHAP. 15
- m;
+ POC19
2. Reformatsky Reaction
Ketones or aldehydes can be reacted to form P-hydroxyesters.
R R'
'C-0
+ Zn + Brk-COOEt I
R 1. dryether 2. H1O
\
via
OH a p - hydroxyester
R' and R may also be H or Ar. Problem 15.43 Use the Reformatsky reaction to prepare (a ) (CH,)2C(OH)CH2COOC2H,
(b) PhC(OH)CHCOOC,H,
&H, LH,
( c ) PhC=CCOOH
4
LH, LH,
The structure in the box comes from the carbonyl compound (acceptor); the structure in the oval comes from the a - bromoester (carbanion source).
@ z C a
- EH, CH,-&
+ Zn + BrCH,COOC,H,
( b)
( c1
Product from ( b) .* PhC-C-COOH $
I
I
CH, CH,
3. Reactions of the Aldehydic H The chemistry of the aldehydic H,except for oxidation to OH, is meager. The C-H bond can be homolytically cleaved by participation of a free radical.
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
CHAP. 151
335
Problem 15.44 Propanal reacts with 1-butene in the presence of uv or free-radical initiators (peroxides, sources of RO.) to give CH,CH,COCH,CH,CH,CH,. Give steps for a likely mechanism. 4
7
The net reaction is in addition of H
I
RO. 2 CH,CH,C=O
Step 1
CH,CH,C=O
Step 2
r ? n CH,CH,C- F + H,CII-CHCH,CH,
Step 3
CH,CH,COCH,CHCH,CH,
II
-+
CH,CH,C-CH,CHCH,CH,
r-3+ CH,CH,
8
-0
-
CH,CH,COCH,CH,CH,CH,
+ CH3CH2C=0
Step 1 is the initiation step. Steps 2 and 3 propagate the chain.
Supplementary Problems Problem 15.45 ( a ) What properties identify a carbonyl group of aldehydes and ketones? (b) How can 4 aldehydes and ketones be distinguished? ( a ) A carbonyl group (1) forms derivatives with substituted ammonia compounds such as H,NOH, (2) forms sodium bisulfite adduct with NaHSO,, (3) shows strong ir absorption at 1690-1760cm-' (c=O stretching frequency), (4) shows weak n-v* absorption in uv at 289nm. ( 6 ) The H - C bond in RCHO has a unique ir absorption at 2720 cm-'. In nmr the H of CHO has a very downfield peak at S = 9-10 ppm. RCHO gives a positive Tollens' test.
Problem 15.46 What are the similarities and differences between C=O and C = C bonds?
4
Both undergo addition reactions. They differ in that the C of C=O is more electrophilic than a C of C=C, because 0 is more electronegative than C. Consequently, the C of C=O reacts with nucleophiles. The C = C is nucleophilic and adds mainly electrophiles. Problem 15.47 Give another acceptable name for each of the following: ( a ) dimethyl ketone, (b) l-phenyl-24 butanone, (c) ethyl isopropyl ketone, (d) dibenzyl ketone, (e) vinyl ethyl ketone. ( a ) acetone or propanone, (b) benzyl ethyl ketone, (c) 2-methyl-3-pentanone, (d) 1,3-diphenyl-2-propanone, (e) 1-penten-3-one.
Problem 15.48 Identify the substances (I) through (V).
(I) + H,
Pd (BsSO
CH,
I
(CH,),CH-CHO
+ NaOI
CH,-C-C-CH, &H3
)
! ! I
-
(11) + (111) (continued on pg. 338)
336
[CHAP. 15
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
15.8 SUMMARY OF ALDEHYDE CHEMISTRY
PREPARATION
PROPERTIES
1. Aliphatic Aldehydes
1. Aliphatic Aldehydes
-
(a) Oxidation
( a ) Carbonyl Addition
1" Alcohols:
+ H, Pd CH,CH,CH,OH + HCN+ CH,CH,CH(OH)CN + R M g X - - + CH,CH,CHR(OMgX) + 2ROH (HX)+ CH,CH,CH(OR),
CH,CH,CH,OH 1" Alkyl Halides: DMSO
CH,CH,CH21 Vinyl boranes: CH,CH=CH,
base
3(CH,CH,CH,),B
( b ) Hydrolysis NaOH
CH,CH,CHX,
+ Ar,P=CR-
1
CH,CH,CH=CR
-
( 6 ) Carbonyl Oxygen Replacement
+ PCl,
CH,CH,CHCl,
(c) Reduction CH ,CH,COCI
CH,CH,COOH
Pd I B a S 0 4 / H,
( d ) Free Radical ( R L O ) + Alkene
CH ,CH,COOR LIAIH(~-BuO),
CH,CH,CN
+ Ag
(with H,C=CH,)
2.
2. Aromatic Aldehydes
Aromatic Aldehydes
(a) Reduction
(a) Oxidation ArCH,OH + CrO, (Ac,O) ArCH,X + DMSO (NaOH)
/
\
/
\\ \
( b ) Hydrolysis x2
ArCH, --+ArCHX,
\\
NaOH
+Ar-CH-0
/As
+ H,/Pd-
ArCH,OH
( b ) Carbonyl Addition with aliphatic aldehydes, with HCN, H,NOH, RMgX, etc.
(c) Oxidation ArCOX + H, or ArCOX
( d ) Formylation
RC,H5
'I I
+ LiAIH(t-BuO),
+ CO + HCl (AlC1,)
ArCOOH
\+
( d ) Canniuaro Reaction
\
NaOH-
ArCH,OH
+ ArCOOH
1
337
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
CHAP. 15)
SUMMARY OF KETONE CHEMISTRY PREPARAnoN
PROPERTIES
Aliphatic and Aromatic Ketones
1. Additions to Carbonyl Group
C,H,CHOHCH, ArCHOHCH,
(a) Oxidation
ArCHOHCH,
CN-
C,H,CHOH(CN)CH, ArCHOH(CN)CH, a cyanohydrin
C,H,CR(OH)CH,
0
Ar-H ( c ) Acid-Derivative Reduction (C H , ) ~ L I C ~
ArCOCl H,C=CH ArH
9
C,H,CCH or ArCCH,
, +
( C,H,C(=NOH)CH, ArC(=NOH)CH,
H2NoH
an oxime
EtC( OH)(SO,Na)CH, + Name, + ArC(OH)(SO,Na)CH,
2
{ Arc( OR),CH, EtC(OR),CH,
( d ) From 1,3=Dithianes
Reduction to Methylene
RX + R’X (two steps) 2-Ar-1,3-dithiane + RX
\\ \\
3. Oxidation
Alicyclic Ketones (a) Oxidation
0
-OH?
CH3-
Cr207
CH,-
BaO heat
-
(CH,),C=O
-;F -,so
7’
(c) Acyloin Reaction E t O O C ( C H , ) , , C O O E t ~(CH,),C=O
‘ 4 H o H
+ NaOX-CHX,
+
C,H,COOH ArCOOH
(cyclic) 4.
+ CH,COOH
Replacement Reactions
(a) Formation of gem-Dichlorides
(d ) From Cycyloalkenes
Alkylborane Oxidation:
C, H, COONa
( 6 ) Strong Oxidants
( 6 ) Decarboxylation of Dicarboxylic Acids
HOOC(CH,),COOH-
(a) Haloform for Methyl Ketones
(6) Wittig Reaction +
ArJP=CR2
----*
338
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
(IV) + H 2 0
HaSO H S O ,
CH3CH2-C-CH3
II
CH2CH,
1('
"2.304 -H20
[CHAP. 15
0
' C H 3 C H 2 - 1 IK i - c H 2 c H 3
CH3 (a)
(I)
(CH,),CHC=O
I
( b ) CH3-C-COO-Na'
cI1
(II), CHI, (111)
AH3
( c ) H-C-C-CH,CH,
(IV)
or CH,-C-C-CH,
( d ) (CH,CH2),C-C(CH2CH,), AH
(V)
AH
Problem 15.49 By rapid test-tube reactions distinguish between ( a ) pentanal and diethyl ketone. (b) diethyl ketone and methyl n-propyl ketone, ( c ) pentanal and 2,2-dimethylpropanal, ( d ) 2-pentanol and 2-pentanone. 4
( a ) Pentanal, an aldehyde, gives a positive Tollens' test (Ag mirror). (b) Only the methyl ketone gives CHI, (yellow precipitate) on treatment with NaOI (iodoform test). ( c ) Unlike pentanal, 2,2-dimethylpropanal has no a H and so does not undergo an aldol condensation. Pentanal in base gives a colored solution. ( d ) Only the ketone 2-pentanone gives a solid oxime with H,NOH. Additionally, 2-pentanol is oxidized by CrO, (color change is from orange-red to green). Both give a positive iodoform test.
Problem 15.50 Show steps in the following syntheses: ( a ) acetyl chloride to acetal (CH,CH(OC,H,),), ( b ) ally1 chloride to acrolein (propenal), ( c ) ethanol to 2-butene. 4
(4
(4
H,C=CHCH,Cl CH,CH,OH
CH,CH,Br
CH SOCH
1.
&CH,CHO
Ph3P
H,C=CHCHO CH,CH=PPh,
Problem 15.51 Use benzene and any aliphatic and inorganic compounds to prepare ( a ) 1,l-diphenylethanol, (6) 4,4-diphenyl-3-hexanone. 4 (a) The desired 3" alcohol is made by reaction of a Grignard with a ketone by two possible combinations:
(C,H,),CO
+ CH,MgBr
or C,H,COCH,
+ C,H,MgBr
Since it is easier to make C,H,COCH, than (C,H,),CO from C,H,, the latter pair is used. Benzene is used to prepare both intermediate products.
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
CHAP. 151
339
4”
(6) The 4” C of CH,CH,COCPh,CH,CH, is adjacent to C = O , and this suggests a pinacol rearrangement of CH,CH2CPh-CPhCH2CH3 AH
AH
which is made from CH,CH,COPh as follows: C,H,
+ CH,CH,COCI
Ph
I
aPhCOCH,CH,
CH,CH,-i;i;CH,CH,
Ph
I
product
Problem 15.52 Use butyl alcohols and any inorganic materials to prepare 2-methyl-4-heptanone.
4
The indicated bond
1 CH3CH,CH2C-CH,CH(CH3)2 II
is formed from 2 four-carbon compounds by a Girgnard reaction. CH3
I
CH3CH2CH2CH,0H CH,-CHCH,Br n-Butyl alcohol
I
Cu.heat
CH,CH2CH2CH=0
lm
CH3
I aCH3-CH-CH20H Isobutyl alcohol
*
CH3
I
+ CH,-CHCH,MgBr
2.
H20.H’
CH3CH2CH2
KMd,
product
Problem 15.53 Compound (A), C,H,Cl,, is hydrolyzed to compound (B), C,H,O, which gives an oxime and 4 a negative Tollens’ test. What is the structure of (A)? (B) is a carbonyl compound because it forms an oxime, and therefore (A) is a gem-dihalide.
I
-CC12 Since (B) does not reduce Tollens’ reagent, (B) is not an aldehyde, but must be a ketone. The only ketone with
4 C’s is CH,COCH,CH,, and (A) is CH,CCl,CH,CH,.
Problem 15.54 Compound (A), CSH,OO,forms a phenylhydrazone, gives negative Tollens’ and iodoform tests and is reduced to pentane. What is the compound? 4
Phenylhydrazone formation indicates a carbonyl Compound. Since the negative Tollens’ test rules out an aldehyde, (A) must be a ketone. A negative iodoform test rules out the CH,C----O group, and the reduction product, pentane, establishes the C’s to be in a continuous chain. The compound is CH3CH,COCH,CH,. Problem 15.55 A compound (C,H,O,) is reduced to pentane. With H,NOH it forms a dioxime and also gives positive iodoform and Tollens’ tests. Deduce its structure. 4
Reduction to pentane indicates 5 C’s in a continuous chain. The dioxime shows two carbonyl groups. The positive CHI, test points to
340
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
[CHAP. 15
while the positive Tollens' test establishes a --CH=O. The compound is CH3-C-CHzCHz-CHO
8
Problem 15.56 The Grignard reagent of RBr(1) with CH,CH,CHO gives a 2" alcohol (11), which is converted to R'Br (III), whose Grignard reagent is hydrolyzed to an alkane (IV). (IV) is also produced by coupling (I). What are the compounds (I), (11), (III), and (IV)? 4
Since CH,CH,CHO reacts with the Grignard of (I) to give (11) after hydrolysis, (11) must be an alkyl ethyl carbinol H H RMgBr
-
I
CH~CHZCHO
HOH
bMgBr
I
cH3CHzY-R6H
Grignard of (I)
(11)
The conversion of (11) to (IV) is H
H
H
I
___* HOH
CH,CH,CH,-R
MgBr (IV) must be symmetrical, since it is formed by coupling (I). R is therefore --CH,CH,CH,. (I) is CH,CH,CH,Br. (IV) is n-hexane. (11) is CH,CH,CH(OH)CH,CH,CH,. (111) is CH,CH,CHBrCH,CH,CH,. Problem 15.57 Carry out the following synthesis: 4
Common oxidants cannot be used, because they oxidize the C-C bond. Since the starting compound is a methyl ketone, its reaction with NaOI removes CH, and converts.
0
II
-C-CH,
to --COO-, which is then acidified. Problem 15.58 Translate the following description into a chemical equation: Friedel-Crafts acylation of resorcinol( 1,3-dihydroxybenzene) with CH,( CH,),COCl produces a compound which on Clemmensen reduc4 tion yields the important antiseptic, hexylresorcinol.
Problem 15.59 Treatment of benzaldehyde with HCN produces a mixture of two isomers that cannot be separated by very careful fractional distillation. Explain. 4
Formation of benzaldehyde cyanohydrin creates a chiral C and produces a racemic mixture, which cannot be separated by fractional distillation.
CHAP. 151
34 1
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
H
H
I C,H,-C=O
I*
+ HCN
C,H,-C-OH
AN Problem 15.60 Explain the following reaction: C H C-C=O
7O1 HI
+ NaOH
Phen ylgl yoxal
-
C H CHC-0-Na'
4
II
,I
HO 0
sodium salt of mandelic acid
This is an internal crossed-Cannizzaro reaction: the keto group is reduced and the --CHO is oxidized. Problem 15.61 Prepare 1-phenyl-1-( p-bromopheny1)-1-propanol from benzoic acid, bromobenzene and ethanol. 4
The compound is a 3" alcohol, conveniently made from a ketone and a Grignard reagent as shown.
+
C,H,OH
C,H,Br
=Zr* C,H,MgBr
Problem 15.62 Convert cinnamaldehyde, C,H,-€H=CH--CH=O, propane, C,H,CHBrCHBrCH,CI.
to l-phenyl-l,2-dibromo-3-chloro4
We must add Br, to c=C and convert --CHO to -CH,CI. Since Br, oxidizes --CHO to - C O O H , --CHO must be converted to CH,Cl before adding Br,. 1 NaBH4
C,H,-CH=CH-CH-*~
PCl,
C,H,-CH=CH-CH,OH-----*
C,H,--CH=CH-CH,CI
2C,H,--CH--CH-CH,Cl Ar
$r
Problem 15.63 Compounds "labeled" at various positions by isotopes such as 14C (radioactive), D (deuterium) and " 0 are used in studying reaction mechanisms. Suggest a possible synthesis of each of the labeled compounds below, using I4CH,OH as the source of I4C, D,O as the source of D, and H,"O as the source of " 0 . Once a 14C-labeled compound is made, it can be used in ensuing syntheses. Use any other unlabeled compounds. ( a ) CH,I4CH,OH, (6) I4CH,CH,OH, (c) 14CH,CH,CH0, (d) C,H,14CH0, (e) 14CH,CHDOH, ( f ) CH,CHI80. 4 ( a ) The 1" alcohol with a labeled carbinol C suggests a Grignard reaction with 14
I . CHjMgBr
cu
C H , 0 H = H , ' 4 C - - 0 ~ (6) Now the Grignard reagent is labeled instead of H,CO. I4CH3OH--!%I4CH3Br
zo*
'*CH,MgBr
H,'4C=0.
CH,'~CH,OH
&I4CH3CH,OH I.
CHO
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
342
(c 1
-
14CH,MgBr [see (611 + H2C-CH2 \/
2.
n,o+
''CH,CH2CH20H
[CHAP. 15
314CH,CH2CH0
0
(e)
D on carbinol C is best introduced by reduction of a --CHO group with a D-labeled reductant. I4CH,CH,OH from (6) 5 heat I4CH,CHO then DJPt or LiAID, --+ ''CH,CHDOD---,
4 0
D of OD is easily exchanged with excess H,O. ( f ) Add CH,CHO to excess Hz'*Owith a trace of HC1. H,
H+
+ CH,CHO
CH,CHI80H
l80
OH hydrate
1
CH,CHI80
14CH,CHDOH
+ H20
The unstable half-labeled hydrate can lose H,O to give CH,CH'*O. Problem 15.64 Isopropyl chloride is treated with triphenylphosphine (Ph,P) and then with NaOEt. CH,CHO is added to the reaction product to give a compound, C,H,,. When C,H,, is treated with diborane and then 4 CrO,, a ketone is obtained. Give the structural formula for C,H,, and the name of the ketone. The series of reactions is:
-
Formation of Ylide (CH,),CHCI
+ Ph,P+
+
[(CH,),CH-PPh,]CI-
NaOEt
(CH,),C=PPh,
+ EtOH + Na'CI-
Wittig Reaction H (CH,),C=PPh,
+CH,d=O-
H (CH,),
Anti-Markovnikov Hydroboration-Oxidation I.
BHJ
(CH
3-Methyl-2-butanone
Problem 15.65 Trace the oxidation of 2-propanol to acetone by infrared spectroscopy. Observe the disappearance of the 0-H stretching band at about 1720 cm-I.
4
stretching band at 3600cm-* and the appearance of the C=O
Problem 15.66 Deduce the structure of a compound, C,H,O, with the following spectral data: (a) Electronic absorption at Amax = 213 nm, erns, = 7100 and A,,, = 320 nm, em,, = 27. (6) Infrared bands, among others, at 3000, 2900, 1675 (most intense) and 1602 cm-'. ( c ) Nmr singlet at S = 2.1 ppm (3 H's), three multiplets each integrating for 1 H at S = 5.0-6.0 ppm. 4 The formula C,H,O indicates two degrees of unsaturation and may represent an alkyne or some combination of two rings, C=C and C=O groups. (a)
A,, at 213 nm comes from the T + T * transition. It is more intense than the A,,, at 320 nm from the n-* T * transition. Both peaks are shifted to higher wavelengths than normal (190 and 280 nm, respectively), thus indicating an a,P-unsaturated carbonyl compound. The 2 degrees of unsaturation are a C-C and a C 4 .
CHAP. 151
343
CARBONYL COMPOUNDS: ALDEHYDES AND KETONES
( b ) The given peaks and their bonds are 3000cm-', sp2 C-H; 2900cm-', sp3 C-H; 1675cm-', c----O (probably conjugated to C=C); 1602cm-', C=C. All are stretching vibrations. Absence of a band at 2720cm-' means no aldehyde H. The compound is probably a ketone. (c) The singlet at 6 = 2.1 ppm is from a
0
II
-C-CH, There are also three nonequivalent vinylic H's (6 = 5.0-6.0 ppm) which intercouple. The compound is
H'
'c=c
/
Hd
/ \
Hh C-CH;
I
0
shown with nonequivalent H's. Problem 15.67 A compound, C,Hl,O, has a strong ir band at about 1700cm-I. The nmr shows no peak at 6 = 9-10 ppm. The mass spectrum shows the base peak (most intense) at mle = 57 and nothing at m / e = 43 or m / e = 71. What is the compound? 4
The strong ir band at 1700cm-' indicates a C=O, accounting for the one degree of unsaturation. The absence of a signal at 6 = 9-10 ppm means no 0
1I
-C-H proton. The compound is a ketone, not an aldehyde. Nmr is the best way to differentiate between a ketone and an aldehyde. Carbonyl compounds undergo fragmentation to give stable acylium ions: R-C=O:
I
.+
R' (P')
-
R-C=O; + .R' (or R~-c--o~ + .RI an acyliurn ion
The possible ketones are CH,CH,CCH,CH, 0 II (A)
a
CH,CCH,CH,CH,
CH,CCH(CH,),
b
(B) (C) Compounds (B) and (C) would both give some CH,C=O:' (mle = 43) and C , H , m : ' ( m l e = 71). These peaks were absent; therefore (A), which fragments to CH,CH,(+O:' ( m l e = 57), is the compound.
-
Problem 15.68 Give the steps in the preparation of DDT, (p-ClC,H,),CHCCl,, acetaldehyde) and chlorobenzene in the presence of H,SO,.
Cl3CCHO-
"2'O4
+
Cl,CCH(OH)-
PhCl
p-Cl--C,H,CHOHCCl~
"ZS04
-
from choral (trichloro+
p-CI--C,H,CHCCl,
PhCl
DDT
4
Chapter 16 Carboxylic Acids and Their Derivatives 16.1 INTRODUCTION AND NOMENCLATURE
Carboxylic acids (RCOOH or ArCOOH) have the carboxyl group, ----€--OH, which is an acyl II group, R-C-, bonded to OH. 0
II
Common names, such as formic (ant) and butyric (butter) acids, are based on the natural source of the acid. The positions of substituent groups are shown by Greek letters LY,p, y, S, etc. Some have names derived from acetic acid, e.g., (CH,),CCOOH and C,H,CH,COOH, are trimethylacetic acid and phenylacetic acid, respectively. Occasionally they are named as carboxylic acids, e.g.
(coo. is cyclohexanecarboxylic acid. For IUPAC names replace the -e of the corresponding alkane with -oic acid: thus, CH,CH,COOH is propanoic acid. The C’s are numbered; the C of COOH is numbered 1. C,HSCOOH is benzoic acid. Dicarboxylic acids contain two COOH groups and are named by adding the suffix -dioic and the word acid to the name of the longest chain with the two COOH’s. Problem 16.1 Give a derived and IUPAC name for the following carboxylic acids. Note the common names. (a) CH,( CH,),COOH (caproic acid); (6) (CH ,),CCOOH (pivalic acid); (c) (CH,),CHCH,CH,COOH (y-methylvaleric acid); (d) C,H,CH,CH,COOH (P-phenylpropionic acid); (e) (CH,),C(OH)COOH (a4 hydroxyisobutyric acid); ( f ) HOOC(CH,)2COOH (succinic acid) (no derived name).
To get the IUPAC name, find the longest chain of C’s including the C from COOH, as shown below by a horizontal line. To get the derived name, find and name the groups attached to the a C:
(a 1
m
O
O
H
]
n-Butylacetic acid, Hexanoic acid
(6 C’s in longest chain)
7H3 €Hd€€OOH
I Trimethylacetic acid, 2,2-Dimethylpropanoic acid
Isobutylacetic acid, CMethylpentanoic acid
Benzylacetic acid, 3-Phenylpropanoic acid 344
CHAP. 161
(e 1
(f)
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
345
-3
Dimethylhydroxyacetic acid, 2-Hydroxy-2-methylpropanoic acid O H -
Butanedioic acid
Problem 16.2 Name the following aromatic carboxylic acids.
(a) p-nitrobenzoic acid; (6) 3,5-dibromobenzoic acid; (c) rn-formylbenzoic acid (COOH takes priority over CHO, wherefore this compound is named as an acid, not as an aldehyde); (d) o-methylbenzoic acid, but more commonly called o-toluic acid (from toluene).
Problem 16.3 Account for the following physical properties of carboxylic acids. (a) Only RCOOH’s with five or fewer C’s are soluble in water, but many with six or more C’s dissolve in alcohols. (6) Acetic acid in the vapor state has a molecular weight of 120 not 60. ( c ) Their boiling and melting points are higher than those of corresponding alcohols. 4 (a) RCOOH dissolves because the H of COOH can H-bond with H,O. The R portion is nonpolar and hydrophobic; this effect predominates as R gets large (over five C’s). Alcohols are less polar than water and are less antagonistic toward the less polar carboxylic acids of higher C content. (6) CH,COOH typically undergoes dimeric intermolecular H-bonding.
CH3-C
//O---Ho//\C--CH, \OH---O
(c) The intermolecular forces are greater for carboxylic acids. Problem 16.4 Write resonance structures for the COOH group and show how these and orbital hybridization account for: (a) polarity and dipole moments (1.7-1.9D) of carboxylic acids; (b) their low reactivity toward 4 nucleophilic additions, as compared to carbonyl compounds. (a) The C of COOH uses sp‘ HO‘s to form the three coplanar U bonds. A p A 0 of the 0 of the OH, accommodating a pair of electrons, overlaps with the 72 bond of C=O. In this extended T system there is negative charge on the lone 0 and positive charge on the other 0; the charge separation results in greater polarity and dipole moments.
(6) The electron deficiency of the C of C=O, observed in carbonyl compounds, is greatly diminished in the C of --COOH by the attached OH group. 0
II
Carboxylic acid derivatives (or acyl derivatives), RC-G, have the OH replaced by another electronegative functional group, G, such that it can be hydrolyzed back to the acid: R(Arv-4
II
0
+ H,O+
a carboxylic acid derivative
R(Ar)--C-OH
It
0
a carboxylic acid
+ HG
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
346
General Formula
I
Type
Table 16-1
I
II
0
Ester
II
CH,C-OCH,CH,
II
0
0 R-C--R'
I
Acid anhydride CH,C-OCCH,
I
0
0 1
R-C-N0
A
I
II
0
Name
Example
R-C-CI*
RX-OR'
[CHAP. 16
Change
Acetyl chloride or Ethanoyl chloride
-ic acid to -yl chloride
Ethyl acetate or Ethyl ethanoate
Cite alkyl group attached to 0; then change -ic acid to -ate
Acetic anhydride
acid to anhydride
Acetamide or Et hanamide N-Methylpropanamide
-ic or -oic acid to -amide or -carboxylic acid to carboxamide
0
I
Amide
CH,C-N-H
II I
O H CH,CH ?CNHCH
II
0 ~~~~~
R-4-N-C-R'
Imide
CH,CO-NH-OCCH,CH,
Acetyl propiony 1 imide
-ic or -oic acid to -imide
RBN+
Nitrile
CH,C%N
Acetonitrile or Ethanenitrile
-ic or -oic acid to nitrile or add -nitrile to alkane name
*Some acid bromides are known. Although nitriles have no acyl group, they are grouped with acid derivatives because they are readily hydrolyzed to RCOOH.
CHAP. 161
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
347
The common ones are given in Table 16-1, with conventions of nomenclature that involve changes of the name of the corresponding carboxylic acid; the G group is shown in bold type. Problem 16.5 Name the following acid derivatives:
c
(a) C,H,COCl
(4 CH,CH,CH,C-OCH,CH, It
(6) CH,CH, ,OCCH,CH,
II
0 0
(4
(4 C,HSCH,CNH,
It
It
It
( 3 3 - Z0 H 2
0
0
(g) HCN(CH,),
(f1
C,HsC--OC,Hs
4
( h ) C,H,COCCH,
It It
0
0 0
(a) Benzoyl chloride. (6) Propionic (or propanoic) anhydride. ( c ) Ethyl butyrate (or butanoate). ( d ) Phenylacetamide. (e) Phenyl benzoate. ( f ) Cyclohexanecarboxamide. ( g) N,N-Dimethylformamide. (h) Acetic benzoic anhydride. Problem 16.6 Give structural formulas for the following acid derivatives: ( a ) propionitrile, (6) isopropyl-24 fluorobutanoate, (c) 3-phenylhexanoyl chloride, (d) 3-chloropropyl benzoate. ( a ) CH,CH,C%N
(6) CH,CH,CHFCOOCH(CH,),
(c)
CH,CH,CH,
(d) C,H,COOCH,CH,CH2C1
HCH,COCl 'iC,H,
16.2 PREPARATION OF CARBOXYLIC ACIDS 1. Oxidation of 1" Alcohols, Aldehydes, and Arenes
R--CH,OH-
ArCHR,
2.
-
KMnO,
0-
RCHO-
KMn04 H+
KMn0,
RCOOH
ArCOOH
Oxidative Cleavage of Alkenes and Alkynes "NO,
HOOCCH,CH,CH,CH,COOH Adipic acid
Cyclohexene
a dicarboxylic acid
Problem 16.7 Account for the fact that on oxidative cleavage all substituted alkynes give carboxylic acids, whereas some alkenes give ketones. 4 Imagine that the net effect of the oxidation is replacement of each bond and H on the multiple-bonded C by OH. Intermediates with several OH'S on C are unstable and lose H,O, leaving c=O. [R,C(OH),J~R,C=O
and
[RC(OH),]-
RCOOH
Thus 101
R-R
__*
[RC(OH),] -HZO\ RCOOH
Alkenes with the structural unit R2C=C- would give [R,C(OH),]-
I
-H20
R2C=0
Multiple-bonded C's bonded only to H's would form [C(OH),), which loses two molecules of H 2 0 to give CO,.
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
348
[CHAP. 16
3. Grignard Reagent and CO, I
R-MgX
+ R-C=O
+>C=%
H20
-O( I MgX)'
4.
bH
a carboxylate salt
-
Hydrolysis of Acid Derivatives and Nitriles
R4-G
II
+ H,O+
0
5.
+ Mg2++ 2X-
R-C=O
+ HG
R--C-OH
II
(or H,O+
+ Cl-, for G = Cl)
0
Haloform Reaction of Methyl Ketones
Although this reaction is used mainly for structure elucidation, it has synthetic utility when methyl ketones are readily prepared and halides are not. Problem 16.8 ( U ) Why isn't 2-naphthoic acid made from 2-chloronaphthalene? (6) How is 2-naphthoic acid 4 prepared in a haloform reaction? (a) Since most electrophilic substitutions of naphthalene, including halogenation, occur at the 1-position, 2-chloronaphthalene is not a readily accessible starting material. (6) Acetylation with CH,COCI in the presence of the solvent, nitrobenzene, occurs at the 2-position, which allows for the following synthesis: 0
II
2-Naphthoic acid
',-Acetonaphthalene (Methyl p-naphthyl ketone)
( p-Naphthoic acid)
Problem 16.9 Prepare the following acids from alkyl halides or dihalides of fewer C's. ( a ) C,H,CH,COOH, (6) (CH,) ,CCOOH, (c) HOCH,CH,CH,COOH, ( d ) HOOCCHICH,COOH (succinic acid). 4
Replace COOH by X to find the needed alkyl halide. The two methods for RX-*RCOOH are: I
COz
RX( 1"' 2"' or 3") Me\ R M g X - m l =
RCN ERX( 1")
2 H 3 0*
Either method can be used starting with C,H,CH,Br (a l o R X ) . With the 3"(CH,),C--Br, CN cannut be used because elimination rather than substitution would occur. HOCH2CH,CH,Br has an acidic H (0-H); hence the Grignard reaction can't be used. BrCH,CH,Br undergoes dehalogenation with Mg to form an alkene; therefore use the nitrile method. BrCH,CH,Br".,
N=tCH,CH2C+N---,
H'
product
16.3 REACTIONS OF CARBOXYLIC ACIDS
H OF COOH IS ACIDIC RCOOH acid,
+ H,O f- RCOO- + H,O' base
base,
acid,
pK,
=z
5 (Section 3.10)
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
CHAP. 161
349
RCOOH forms carboxylate salts with bases; when R is large, these salts are called soaps. RCOOH + KOH2RCOOH + Na,CO,+
RCOO-K’ 2RCOO-Na’
+ H,O + H,O + CO,
Problem 16.10 Use the concept of charge delocalization by extended n- bonding (resonance) to explain why
( a ) RCOOH (pK, = 5 ) is more acidic than ROH ( p K , = 15), and ( b ) peroxy acids, RCOOH, are much weaker
than RCOOH.
8
4
( a ) It is usually best to account for relative strengths of acids in terms of relative stabilities of their conjugate
bases. The weaker (more stable) base has the stronger acid. Since the electron density in RCOO- is dispersed to both 0’s.
RCOO- is more stable and a weaker base than RO-, whose charge is localized on only one 0. ( b ) There is no way to delocalize the negative charge of the anion, RCOO-, to the C = O group, as can be done for RCO-.
b
b
Problem 16.11 Use the inductive effect (Section 3.11) to account for the following differences in acidity. ( a ) ClCH,COOH > CH,COOH, (6) FCH,COOH > ClCH,COOH, ( c ) CICH,COOH > ClCH,CH,COOH, (d) Me,CCH,COOH > Me,SiCH,COOH, (e) C1,CHCOOH > ClCH,COOH. 4 The influence of the inductive effect on acidity is best understood in terms of the conjugate base, RCOO-, can be summarized as follows:
0 Electron-withdrawing groups stabilize RCOO and strengthen the acid.
0 Electron-donating groups destabilize RCOO and weaken the acid.
Like all halogens, C1 is electronegative, electron-withdrawing, and acid-strengthening. Since, F is more electronegative than C1, it is a better EWG and a better acid-strengthener. Inductive effects diminish as the number of C’s between C1 and the 0’s increases. ClCH,COO- is a weaker base than ClCH,CH,COO-, and so CICH,COOH is the stronger acid. Si is electropositive and is an acid-weakening EDG. Two Cl’s are more electron-withdrawing than one C1. C1,CHCOO- is the weaker base and C1,CHCOOH is the stronger acid.
Resonance and induction, which affect AH, can be used with certainty to explain differences in acidities only when the K , values are different by a factor of at least 10. Smaller differences may be accounted for by solvation, which affects AS. Problem 16.12 Compare and account for the relative acid strengths of the following benzoic acids: C,H,COOH, p-CH,OC,H,COOH, and p-NO,C,H,COOH. 4 0 +/
is an acid-strengthening EWG by extended T bonding and induction; p-NO,C,H,COOH is the ‘0strongest acid. CH,O- is an EDG by extended 7r bonding, an effect that overshadows its electron-withdrawal by induction; p-CH,OC,H,COOH is the weakest acid.
-N
350
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
[CHAP. 16
Acid-strengthening ring substituents also retard electrophilic, and enhance nucleophilic, aromatic substitution. Conversely, the acid-weakening groups accelerate electrophilic, and retard nucleophilic, aromatic substitution. Pl0b)cHn 16.13 Explain why highly branched carboxylic acids such as
are less acidic than unbranched acids.
4
The --CO, group of the branched acid is shielded from solvent molecules and cannot be stabilized by solvation as effectively as can acetate anion. Problem 16.14 Although p - hydroxybenzoic acid is less acidic than benzoic acid, salicyclic (o-hydroxybenzoic) 4 acid (K,, = 105 x 10-5) is 15 times more acidic than benzoic acid. Explain. The enhanced acidity is partly due to very effective H-bonding in the conjugate base, which decreases its basic strength. 0y O - H
-H+ ___,
Problem 16.15 The K, for fumaric acid (trans-butenedioic acid) is greater than that for maleic acid, the c i ~ isomer. Explain by H-bonding. 4 Both dicarboxylic acids have two ionizable H's. The concern is with the second ionization step.
Maleate monoanion (H-bond)
Fumarate monoanion (no H-bond)
Since the second ionizable H of maleate participates in H-bonding, more energy is needed to remove this H because the H-bond must be broken. The maleate monoanion is therefore the weaker acid. In general, H-bonding involving the acidic H has an acid-weakening effect; H-bonding in the conjugate base has an acid-strengthening effect.
NUCLEOPHIUCITY OF CARBOXYLATES
RCOO- acts as a nucleophile in S,2 reactions with RX to give esters. R C - O q ' X
1I
0 a carboxylate
___*
RC-OR' + X -
1I
0 an ester
35 1
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
CHAP. 161
FORMATION OF ACID DERNATNES (OH-
G)
1 . Acyl Chloride (RCOCl) Formation, OH --+ C1
+ PCI,--+3 R--COCI + H,PO,(t) R - C O O H + PCl, --+ R - C O C I + HCl( g) + POCl,( e ) R - C O O H + SOC1,R-COCl + HCl(g) + SO,(g) 3 R-COOH
Thionyl chloride
Reaction with SOCl, is particularly useful because the two gaseous products SO, and HCI are readily separated from RCOCl. 2.
Ester (RCOOR') Formation, OH --+ OR' R-COOH
+ R'OH-
"2SO4
reflux
R-COOR'
+H20
(In dilute acid the reaction reverses.) Problem 16.16 Give the mechanism for the acid-catalyzed esterification of RCOOH with R'OH.
4
In typical fashion, the 0 of C=O is protonated, which increases the electrophilicity of the C of C=O and renders it more easily attacked in the slow step by the weakly nucleophilic R'OH.
1
+
HO-R'
\ RLOH RC-OH+ R'Oti
c-)
I
rlon
HO
RC-OR'
II
I HO
+ H,O+
0
tetrahedral intermediate
The tetrahedral intermediate undergoes a sequence of fast deprotonations and protonations, the end result being the loss of H' and H,O and the formation of the ester.
3. Amide (RCONH,) Formation, OH --+NH, R-COOH
+ NH,-
acid
ammonium salt
RCOOH + R'NH,-
RCOO- R'NH;
1" amine
RCOOH + R'RNH-
heat
R-COO-NH:
RCOO- R'RNHI-
heat
R-CONH, RCONHR'
heat
2" amine
+ H,O
1" amide
+ H,O
2" amide
RCONHR'R"
+ H,O
3" amide
Problem 16.17 Use ethanol to prepare CH,COOC,H,, an important commercial solvent.
4
CH3COOC2H,, ethyl acetate, is the ester of CH,CH,OH and CH,COOH. CH,CH20H is oxidized to CH,COOH. CH,CH,OH-
MnO-/H+
CH,COOH
Ethanol and acetic acid are then refluxed with concentrated H,S04. C 2 H , 0 H + CH,COOH
HzSOJ
heat
CH,COOC2H, + H,O
With added benzene and a trace of acid, the reversible reaction is driven to completion by distilling off H,O as an azeotrope (Problem 13.56).
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
352
[CHAP. 16
REDUCTION OF C=O OF COOH (RCOOH-, RCH,OH)
Acids are best reduced to alcohols by LiAIH,. RCOOH HALOGENATION OF
LY
2 RCH,OH r I . LIAIHJ(ether)
H's. HELL-VOLHARD-ZELINSKY ( H V Z ) REACTION
One or more a H's are replaced by C1 or Br by treating the acid with C1, or Br,, using phosphorus as catalyst. RCHzCOOH
x IP
RCXzCOOH (X = Cl, Br)
RCHCOOH
I
X a -Halogenated acids react like active alkyl halides and are convenient starting materials for preparing other a -substituted acids by nucleophilic displacement of halide anion.
-
x- + R-CH-COOH I
OH
-H+.X-
R-CH-COO
I
NH,'
an a-hydroxy acid
an a-amino acid
REACTION OF COOH. DECARBOXYLATION 1.
Arylcarboxylic Acids
ArCOOH2.
soda lime
ArH (in poor yields) + CO,
0-Keto Acids and 0-Dicarboxylic Acids
1
I
-CO,
0
0
Decarboxylation proceeds readily and in good yields when the C that is p to COOH is a C=O. RC--CH2COOH+
-cn. -
I1
0
-
I
RC-CH,
II
I
O H
a 6-ketocarboxylic acid
I
a ketone
-CO,
I
H O C - C - C O O H ~H--C--COOH
II
0
I
a 6 -dicarboxylic (rnalonic) acid
I
a carhoxylic acid
Problem 16.18 ( a ) Suggest a mechanism for a ready decarboxylation of malonic acid, HOOCCH,COOH, that proceeds through an activated, intramolecular H-bonded, intermediate complex. (6) Give the decarboxylation products of (i) oxalic acid, HOOC-COOH, and (ii) pyruvic acid, CH,COCOOH. 4
353
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
CHAP. 161
end
complex
Acetic acid
2
0 (6) (i)
II
-CO,
H-HM-H
H0-C
H
(ii) CH,--C
-
Formic acid
0
II
CH,-C-H
Acetaldehydc
Problem 16.19 Prepare n-hexyl chloride from n-butylmalonic ester.
n-Butylmalonic ester is hydrolyzed with base and decarboxylated to hexanoic acid. I
n-C,H,CH(COOC,H,),
CIH
n-C,H,CH,COOH
This acid and n-hexyl chloride have the same number of C's. 1. I A I H 4
n-C,H,CH,COOH--------*
3.
n-C,H, ,CH,OH-
SOCI,
n-C,H,,Cl
Conversion of RCOOH to RBr The Hunsdiecker reaction treats heavy-metal (e.g. , Ag' ) carboxylate salts with Br,.
RCOO- Ag+ + Br2-
RBr + CO,
+ AgBr
Problem 16.20 Suggest a typical free-radical mechanism for the Hunsdiecker reaction which requires the initial formation of an acyl hypobromite, R C - O B r , from the Ag' salt and Br,. 4
RC-OBr+
Propagation
(1) RC--O.---+R*+CO,
8
8
RC-0-
8+
Initiation
8
-0Br-
Br-
R-Br
+ RC-0-
b
ELECTROPHILIC SUBSTITUTIONS During aromatic substitution with ArCOOH or ArCOG, the electron-attracting --COOH or -COG group is rneta-directing and deactivating. Problem 16.21 ( a ) Account for the fact that ArCOOH, with a strongly activating substituent ortho or para to COOH, loses CO, during attempted electrophilic substitution. (6) Write the equation for the reaction of p-aminobenzoic acid and Br,. 4
( a ) When the electrophile attacks the ring C bonded to COOH, the intermediate phenonium ion first loses an H' from COOH and then loses CO,, which is a very good leaving group:
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
354
COOH
Br
p-Aminobenzoic acid
16.4
1.3.5-Tribromoaniline
SUMMARY OF CARBOXYLIC ACID CHEMISTRY PREPARATION
--
PROPERTIES
1. Acidic Hydrogen of the -COOH
1. Oxidation
RCH2CH20H+ Cr20:- or MnO,, H’ RCH&H=O
[CHAP. 16
+ HOH
+ Ag(NH3);
RCH,CH=CH,
+ MnO,,
2.
H30++ RCH2COO-
Hydroxyl Group of the -COOH
+ PCI,, SOCI, RCH,COCl + R’OH -!% RCH,COOR’ + NH3 +RCH2COO-NHi &*
H’
2. Grlgnard
RC H2C ON H2 3.
Group of the -COOH
Hydrolysis
+ 1.
RCH,-CONH, + HOH RCHZCOOR’ + HOH (RCHIC0)20 + HOH RCH,COX + HOH RCHzCCl, + 3HOH
LiAIH,,,
2. HOH
__*
RCH2CH20H
+ heat 7 RCH2-CO-CH2R MnO
+ X2 5 RCHXCOOH 5.
-COOH
RCX2COOH
Group
+ N a O H T RCH, + Ag’, Br, RCH,Br --t
16.5 POLYFUNCTIONAL CARBOXYLIC ACIDS
DICARBOXYLJC ACIDS, HOOC(CH,), COOH; CYCLIC ANHYDRIDES The chemistry of dicarboxylic acids depends on the value of n. See Problem 16.18 for decarboxylations of oxalic acid (n = 0) and malonic acid (n = 1). When n = 2 or 3, the diacid forms cyclic anhydrides when heated. When n exceeds 3, acyclic anhydrides, often polymers, are formed. Problem 16.22 Compare the products formed on heating the following dicarboxylic acids: ( a ) succinic acid, ( b ) glutaric acid ( 1,5-pentanedioic acid), (c) longer-chain HOOC( CH,),COOH. 4 ( a ) Intramolecular dehydration and ring formation:
CH2-COOH CH2-COOH I n=2
hc.l
-H20
CHZ--C=O
I /
’ CH,-C=O‘0
Succinic anhydride
CHAP. 161
355
CARBOXYLIC ACIDS A N D THEIR DERIVATIVES
( b ) Intramolecular dehydration and ring formation:
CH2-COOH / CH2 \CH2-COOH n=3 (c)
heat -H*O’
CH2-C-0 / \ 0 Glutaric anhydride CH2 / ‘CH,--C=O
Longer-chain a,o-dicarboxylic acids usually undergo intermolecular dehydration on heating to form long-chain polymeric anhydrides (see Section 16.8). HOOC-(CH-J,-COOH n>3
-ac;
Problem 16.23 Show steps in the following syntheses, using any needed inorganic reagents:
(a) o-Xylene
C O\
(6)
0
II
__*
Tctrahydrof w a n
0
0
II
CIC(CH2),CCI 4
Adipoyl dichloride
acH3 ri,: acooH Phthalic anhydride
R0
CH3
oxidation
COOH
Phthalic acid
a c > O +H,O C O\
(6) The chain is increased from 4 to 6 C’s by forming a dinitrile.
) (
) ( NC
CN
HOOC
COOH
aCIC(CH2)&CI
II
0
HYDROXYACIDS; LACTONES Reactions of hydroxycarboxylic acids, HO(CH,).COOH, also depend on value of n. In acid solutions, y-hydroxycarboxylic acid (n = 3) and 6-hydroxycarboxylic acid (n = 4) form cyclic esters (lactones) with, respectively, five-membered and six-membered rings. Problem 16.24 Write the structure for the lactone formed on heating in the presence of acid ( a ) y 4 hydroxybutyric acid, ( 6 ) 8-hydroxyvaleric acid.
Since an OH and a COOH are present in each compound, intramolecular dehydration gives lactones with 5and 6-membe red rings, respective I y . ,CHZ-CHz H2c/CH2-cH2 \c=o -&b H’C.\O/ ‘c=o ( a) \OH
I
OH
y- Hydroxybutyric acid
y - Butyrolactone
(4-Butanolide)
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
356
[CHAP. 16
6 -Valerolactone (5-Pentanolide)
6-Hydroxyvaleric acid
Problem 16.25 p-Hydroxyacids ( n = 1) readily undergo dehydration but do not yield a lactone. Give the 4 structure of the product and account for its formation.
RCH=CHy=O
8 x ; O
OH A double bond forms because the product is a stable, conjugated a,p-unsaturated acid [Problem 16.28(b)]. Problem 16.26 ( a ) When heated, 2 mol of an a-hydroxyacid ( n = 0) lose 2 mol of H,O to give a cyclic diester (a lactide). Give the structural formulas for two diastereorners obtained from lactic acid, CH,CHOHCOOH, and select the diastereomer which is not resolvable. (6) Synthesize lactic acid from CH,CHO. 4 0
unresol vable center of symmetry
/
0
CH,CHO
CH,
H3O+
CH,CH(OH)COOH
+ NHf
HALOCARBOXYLIC ACIDS (Amino Acids in Chapter 21 .) With the exception of the a-acid, haloacids behave like hydroxyacids. Heating an a-halogenated acid with alcoholic KOH leads to a,p-unsaturated acids when there is a p H in the molecule.
R-CH,CH-COOH
h
alc. KOH
R-CH-CH-COO-K’
-% R-CH-CH-COOH
a,&unsaturated acid salt
Problem 16.27 Prepare malonic acid (propanedioic acid, HOOC--CH,-€OOH)
from CH,COOH.
4
CH,COOH is first converted to CICH,COOH. The acid is changed to its salt to prevent formation of the very poisonous HCN when replacing the Cl by CN. The C=N group is then carefully hydrolyzed with acid instead of base, to prevent decarboxylation.
CH,COOH c12/p_ C1-CH,COOH Chloroacetic acid
C1-CH2CO06a Sodium chloroacetate
N=C-CH,COONa
+
Sodium cyanoacetate
HO+
3 HOOC-CH,-COOH
+ NHf
CHAP. 161
357
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
Problem 16.28 What is the product when each of the following compounds is reacted with aqueous base (NaOH)? ( a ) 2-bromobutanoic acid, ( b ) 3-bromobutanoic acid, (c) 4-bromobutanoic acid, (d) 5-bromo4 pentanoic acid. (In (a) and (6) the initially formed salt is acidified.) ( a ) a-Hydroxyacid by S,2 substitution.
FOH
FBr
CH3CH2 H-COOH
CH3CH2 H-COOH
2-Bromobutanoic acid
2-Hydroxybutanoic acid
( 6 ) Dehydrohalogenation to an a,P-unsaturated acid. The driving force for this easy reaction is the formation of a conjugated system. OH I.
OH-
I
CH,CH=CH-C=O 2-Butenoic acid
Elimination is a typical reaction of 6-substituted carboxylic acid.
F
R HCH,COOH 5 RCH-CHCOOH
(c)
(Y = C1, Br, I, OH, NHJ
y-Haloacids undergo intramolecular SN2-typedisplacement of X- initiated by the nucleophilic carboxylate anion, to yield y-lactones. H2C-0 BrCH,CH,CH,COOH
Na'
0
/
__*
H,C,
\
,C=O
C H*
+ NaBr
y-Butyrolactone
( d ) Similar to part (c) to give a 6-lactone.
Br(CH,),C-0-
It 0 -
+
6-Valerolactone
Problem 16.29 Explain the fact that (R)-CH,CHBrCOO-Na' CH,CHOHCOO -Na '.
reacts with NaOH to give (R)4
On displacing Br by OH, the order of priorities (Section 5.3) is unchanged. Therefore, since both configurations are R, there was retention of configuration rather than the inversion expected from SN2reactions. It is believed that the --COO- group participates in first displacing the Br- with inversion to give an unstable a-lactone. Then O H - attacks the a-lactone with a second inversion to give the product. Two inversions add up to retention. Intervention by an adjacent group is called neighbring-group participation.
unstable a-lactone (S)
358
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
[CHAP. 16
Intramolecular nucleophilic displacements, such as those in lactone formation, have faster reaction rates than intermolecular S,2 reactions because the latter require two species to collide. The neighboring participant is said to furnish anchimeric assistance. 16.6 TRANSACYLATION; INTERCONVERSION OF ACID DERIVATIVES
Transacylation is the transfer of the acyl group from one G group to another, resulting in the formation of various acid derivatives. Figure 16-1 summarizes the transacylation reactions. Notice that the more reactive derivatives are convertible to the less reactive ones. Because acetic anhydride reacts less violently, it is used instead of the more reactive acetyl chloride to make derivatives of acetic acid. In aqueous acid, the four kinds of carboxylic acid derivatives in the figure are hydrolyzed to RCOOH; in base, to RCOO-. R'C--O--CR'
II
II
0 RCOOH
Socl2
p/
II
II
0
1 t R'COOH (1)
\
R C - O H cRC---cl 0
0
R'OH
1. Transanhydride formation-an exchange of anhydrides. 2. Transesterification-an exchange of esters. 3. Also for H,NR' -P RCONHR' and HNR'R"+ RCONR'R"
Fig. 16-1
Problem 16.30 ( a ) Give the mechanism for the reaction of RCOCl with Nu:-. (6) Compare, and explain the difference in the reactivities of RCl and RCOCl with Nu:-. (c) What is the essential difference between nucleophilic attack on C==O of a ketone or aldehyde and of an acid derivative? 4 (a)
Nucleophilic substitutions of RCOG, such as RCOCl, occur in two steps. The first step (addition) resembles nucleophilic addition to ketones and aldehydes (Section 15.4) and the second step (elimination) is loss of G. in this case C1 as CI-. ADD IT10N
ELIMINATION b
React ant (trigonal sp')
Transition State ( C becoming sp')
Intermediate (C is tetrahedral sp', on electronegative 0)
Transition State ( C becoming s p 2 )
Product (trigonal s p 2 )
CHAP. 161
359
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
Alkyl halides are much less reactive than acyl halides in nucleophilic substitution because nucleophilic attack on the tetrahedral C of RX involves a badly crowded transition state. Also, a U bond must be partly broken to permit the attachment of the nucleophile. Nucleophilic attack on C=O of RCOCI involves a relatively unhindered transition state leading to a tetrahedral intermediate. Most transacylations proceed by this mechanism, which is an example of addition-elimination reminiscent of aromatic nucleophilic substitution (Section 11.3). The addition step leading to the tetrahedral intermediate is the same in each case. However, the intermediates from the carbonyl compounds would have to eliminate an R:- (from a ketone) or an H:(from an aldehyde) to restore the C 4 . These are very strong bases and are not eliminated. Instead the intermediate accepts an H’ to give the adduct. The intermediate from RCOG can eliminate G:-as the leaving group. Problem 16.31 Account for the relative reactivities of RCOG with nucleophiles: RCOX > ( R C 0 , ) , 0 > 4 RCOOR’ > RCONH,. ,
Often this order of reactivity is related to the order of “leavability” of G-, which is the reverse of its order of basicity; CI- is the best leaving group and the weakest base.
increasing o r d e r of hasccirv
Although this rationale gives the correct answer, it overlooks an important feature of the mechanism-the step eliminating G - cannot influence the reaction rate because it is faster than the addition step. The first step, the slow nucleophilic addition, determines the reaction rate, and here resonance stabilization of the 4 - G group II is important: R<-G: :&:
-
0
R<-G’
:A-
The greater the degree of resonance stabilization, the less reactive is RCOG. Thus, NH, has the greatest degree and RCONH, is the least reactive, while X has the smallest degree and RCOX is the most reactive. It so happens that the order of resonance stabilization correlates directly with the order of basicity; e.g. -N: of I amide is the best resonance-stabilizer and -N:- is the strongest base. This relationship permits the use of I relative basicities of the leaving groups to give correct answers, albeit for the wrong reasons.
0
II
Problem 16.32 Place acyl azides, R-C-N,, in the order of reactivities, if for hydrazoic acid, HN,, K,, = 2.6 X 10e5.Recall that for CH,COOH, KO= 1.8 x 10-’. 4
Since HN, is only slightly more acidic than CH,COOH, N 3 is slightly less basic and is a slightly better leaving group than CH,COO-. RCON, is less reactive than RCOCI, but a little more reactive than the anhydride R--C--O--C-R.
b
b
Problem 16.33 Use the Hammond principle (Problem 11.8) to explain why strong bases such as OR- (from esters) and NH 3 (from amides) can be leaving groups in nucleophilic transacylations. 4
The elimination step for breaking the C-G bond is exothermic because it reestablishes the resonancestabilized C=O. Its transition state resembles the reactant, in this case the intermediate. Consequently, there is little breaking of the C - G bond in the transition state of the elimination step and the basicity of the leaving group, G - , has little or no influence.
360
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
[CHAP. 16
Problem 16.34 Outline a mechanism for hydrolysis of acid derivatives with ( a ) H , O ' , ( b ) NaOH. (a)
4
Protonation of carbonyl 0 makes C more electrophilic and hence more reactive toward weakly n u cleophilic H,O. +
OH
I
more electrophilic C
0 -OH+2H,O'
+ G (or H G + H , O )
'OH,
[If G is basic, e.g., OR ,we get HG(HOR).]
( b ) Strongly basic O H - readily attacks the carbonyl C. Unlike acid hydrolysis, this reaction is irreversible, because OH removes H ' from --COOH to form resonance-stabilized RCOO-.
OH Problem 16.35 Does each of the following reactions take place easily? Explain.
+ HCI CH,COOH + NH,-CH,CONH, + H,O (CH,CO),O + NaOH----,CH,COOH + CH,COO-Na' CH,COOC,H, + HBr CH,COBr + C,H,OHCH,CONH, + NaOH-CH,COO-Na' + NH, + -OCH, CH,COOCH, + Br--CH,COBr
( a ) CH,COCI
(6) (c)
(d) (e)
(f)
+ H,O-CH,COOH
4
Nucleophilic substitution of acyl compounds takes place readily if the incoming group (Nu:- or Nu:) is a stronger base than the leaving group ( G : - ) or if the final product is a resonance-stabilized RCOO-. (a) Yes. H,O is a stronger base than C1- and reacts vigorously. ( b ) No. NH, reacts with RCOOH to form RCOO-NH,', which does nor react further. Amides are prepared from RCOOH by strongly heating dry RCOO-NH:, because reaction is aided by acid catalysis by NH;. ( c ) Yes. The leaving group RCOO- is a weaker base than OH-. (d) Yes. Br- is a much weaker base than C,H,OH. (e) Yes. Even though NHS is a stronger base than OH-, in basic solution the resonance-stabilized RCOO- is formed, and this shifts the reaction to completion. ( f ) No. Br- is a weaker base than OCH;.
16.7 MORE CHEMISTRY OF ACID DERIVATIVES
ACYL CHLORIDES (see Section 16.3 for preparation of RCOCI) The use of acyl chlorides in Friedel-Craft acylations of benzene rings, as well as their reactions with organometallics and reductions to aldehydes, has been discussed in Section 15.2. Problem 16.36 Give the structure and name of the principal product formed when propionyl chloride reacts with: (a) H,O, (6) C,H,OH, (c) NH,, (d) C,H,(AICI,), (e) (n-C,H,),CuLi, ( f ) aq. NaOH, (g) LiAl(0-tC,H,),, ( h ) H,NOH, (i) CH,NH,, ( j ) Na,O, (sodium peroxide). 4
(a) CH,CH,COOH, propionic acid; (6) CH,CH2COOC,H,, ethyl propionate; (c) CH,CH,CONH,, propanamide; (d) C,H,COCH2CH,, propiophenone or ethyl phenyl ketone; (e) CH,CH2COC,H,, 3-hexanone; ( f ) CH,CH,COO-Na', sodium propionate; (g) CH,CH,CHO, propanal;
0
II
( h ) CH,CH,C-NHOH Propane h ydroxarnic acid
( i ) CH,CH,CONHCH, N-Methylpropanarnide (a substituted amide)
e
0
II
0') CH3CH2 -0-0-C-CH2CH3 Propionyl peroxide
361
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
CHAP. 161
ACID ANHYDRIDES
All carboxylic acids have anhydrides,
R-C-0-C-R
It
II
0
0
but the one most often used is acetic anhydride, prepared as follows:
0 or
-
/
CH3-C=O
Ketene
CH3-COOH
\o
CH,-C=O
CH3COOH
CHFC=O
Acetic anhydride
Heating dicarboxylic acids, HOOC(CH,),COOH (n = 2 or 3), forms cyclic anhydrides by intramolecular dehydration [Problem 16.22(a), ( b ) ] . Anhydrides resemble acid halides in their reactions. Because acetic anhydride reacts less violently, it is often used in place of acetyl chloride. Acid anhydrides can also be used to acylate aromatic rings in electrophilic substitutions. Problem 16.37 Give the products formed when acetic anhydride reacts with ( a ) H,O, (6) NH,, (c) C,H,OH, (d) C,H, with AICI,. 4 ( a ) 2CH,COOH, (6) CH,CONH, + CH,CO,NH,‘. (c) CH,COOC,H, to form acetates. ( d ) C,H,COCH,. Friedel-Crafts acetylation.
+ CH,COOH. This is a good way
Problem 16.38 Give the structural formula and name for the product formed when 1 mol of succinic anhydride, the cyclic anhydride of succinic acid (Problem 16.22), reacts with (a) 1 mol of CH,OH; (6) 2 mol 4 of NH, and (c) 1 mol of C,H, with AICI,.
Products are formed in which half of the anhydride forms the appropriate derivative and the other half becomes a COOH. CH,COOCH, H2-COOH
0/
Methyl hydrogen succinate
CH2-C7‘
I
/
ZNH3
O\
CH2-C
(a)
\
Succinic anhydride
(6) Ammonium succinamate
\ rr’
6*‘“CIJ
CH,COC,H, I CH,COOH
(c)
/3-Benzoylpropionic acid
The monoamides of dicarboxylic acids, HOOC(CH,),CONH, (n = 2,3), form cyclic imides on heating. CH,-CC=O CH2-C=0 \ NHJ / \ CH, q G + CH2 NH, ‘CH2-C-OH C ‘ H -C =O O\
/
,
p
Glutaric anhydride
Glutaramic acid
CH2-C=0 \ NH CH2 / ‘CH2-C=0
Y&
/
Glutarimide
362
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
Phthalic anhydride
0 //
0
U
0
[CHAP. 16
II
Phthalamic acid
Phthalimide
ESTERS The mechanism for esterification given in Problem 16.16 is reversible, the reverse being the mechanism for acid-catalyzed hydrolysis of esters. As an example of the principle of microscopic reversibility, the forward and reverse mechanisms proceed through the same intermediates and transition states. RCOOH+
1 HOR’ or HOAr
ester
h y d r oI
RCOOR’ or
RCOOAr
For the role of steric hindrance, See Problem 16.39. Esters react with the Grignard reagent:
R-C-0
I
OR’
rnw
Mg(0R’)X + OH a 3’ alcohol with at least 2 like R”’s
not isolated
Reduction of esters gives alcohols: RCOOR’ On pyrolysis, esters give alkenes:
I.
RCH2CH20C-R’
II
LiAIH,
RCH20H
+ R‘OH
2RCH=CH2 + R’COOH
0 Problem 16.39 Use the mechanism of esterification to explain the lower rates of both esterification and 4 hydrolysis of esters when the alcohol, the acid, or both have branched substituent groups.
The carbonyl C of RCOOH and RCOOR’ is trigonal sp*-hybridized, but that of the intermediate is tetrahedral sp”-hybridized. If R’ in R‘OH or R in RCOOH is extensively branched, formation of the unavoidably crowded transition state has to occur with greater difficulty and more slowly. Problem 16.40 Write mechanisms for the reactions of RCOOR’ with (a) aqueous OH- to form RCOO- and R’OH (saponification), ( b ) NH, to form RCONH,, (c) ROH in acid, HA, to form a new ester RCOOR (transesterification; Section 16.6). 4
363
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
CHAP. 161
The last step is irreversible and drives the reaction to completion.
0
0-
I
(6)
( c)
II
+ H' __*
R-C-NH,
+ HOR
-A-
To drive the reaction to completion, a large excess of R O H is used, and when R'OH is lower-boiling than KOH, R'OH is removed by distillation. Problem 16.41 Assign numbers from 1 for LEAST to 4 for MOST to show relative rates of alkaline hydrolysis of compounds I through IV and point out the factors determining the rates.
I ( a ) CH,COOCH(CH,), (6) HCOOCH, (c)
O,N-@COOCH,
I1 CH,COOCH, (CH,),CHCOOCH,
CH,COOC(CH,), CH,COOCH,
CH,COOC,H, (CH,),CCOOCH,
CH,O+COOCH,
(O>-COOCH,
C l e C O O C H ,
111
IV
4
See Table 16-2. Table 16-2
Ranks I
I1
I11
IV
Rate-Determining Factors
2
4
1
3
Steric effects (branching on alcohol portion)
4
2
3
1
Steric factor (branching on acid portion)
4
1
2
3
Electron-attracting groups disperse, developing negative charge in transition state and increasing reactivity
Problem 16.42 Acid-catalyzed hydrolysis with H2l80of an ester of an optically active 3" alcohol, RCOOC*R'R"R, yields the partially racemic alcohol containing '*O, R'RR"'C'*OH. Similar hydrolyses of esters of 2" chiral alcohols, RCOOC*HR'R", produce no change in the optical activity of the alcohol, and "0is found in RC''0,H. Explain these observations. 4 Hydrolyses of esters of most 2" and 1" alcohols occur by cleavage of the 0-acyl
bond,
0 R-!tO-EHR'R" Since no bond to C* is broken, no racemization occurs. However, with 3" alcohols, there is an S,1 0-alkyl cleavage,
0 R-!!-O-/-CR'R"R"
364
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
[CHAP. 16
producing RCOOH and a 3" carbocation, ' C R ' R R , which reacts with the solvent (H,"O) R'R"R"C''0H. This alcohol is partially racemized because ' CR'RR"' is partially racemized.
+
+ RR"R"C+
RCOOH
to form
H2"0
R'R"R"C'80H
+ H+
Problem 16.43 Write structures of the organic products for the following reactions:
(R)-CH,COOCH(CH,)CH,CH, C,H,COOC,H, + NH,C,H,COOC,H,
+ H,O-
NdOti
( 6 ) (R)-CH,COOCH(CH,)CH,CH, ( d ) C,H,COOC,H,
-
+ LiAIH,
0
+ n-C,H,OH-
+ H,O+
__*
H*
0
II II cf) C,H,C-0-0-C-C,H,
+ CH,O-Na'
CH,COO-Na' + (R)-HOCH(CH,)CH,CH,. The alcohol is 2"; we find 0-acyl in configuration of alcohol. RCOO-Na' forms in basic solution. CH,COOH + (R)-HOCH(CH,)CH,CH,. Again 0-acyl cleavage occurs. C,H,CONH, + C,H,OH. C,H, COO-n-C, H, + C, H OH. Acid-catalyzed transesterification. C,H,CH,OH + C 2 H , 0 H .
4
4
cleavage and no change
0
It
C6HsC-o-o-Na'
4-
C6H5COOCH3
Sodium peroxybenzoate
Problem 16.44 Supply a structural formula for the alcohol formed from (a) CH,CH,COOCH,
+ ZC,H,MgBr
( 6 ) C,H,CH,COOCH,
+ 2C,H,MgBr
4
Two R or Ar groups bonded to carbinol C come from the Grignard reagent, while the carbonyl C becomes the carbinol C.
Problem 16.45 Methyl esters can be prepared on a small scale from RCOOH and CH,N,. Suggest a mechanism involving S,2 displacement of N,. 4
Diazomethane is a resonance hybrid,
H
I
H
I
+
HC-N=N: and we have
c----,
H
HC-N-N: +
1 .
or
1
+
HC-N='N:
6-
6-
H
RCOOH
-6
acid,
I + + HC --N=?J: -
base:
6-
+RCOO-
?+ + CH,:N-N:
U
base,
acidz
__*
RCOOCH,
+ :N-N:
CHAP. 161
365
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
Problem 16.46
Prepare CH3CH2CH2COOCH2CH2CH2CH3 from n-C4H90H.
4
The acid portion of the ester has the same number of C’s ( 4 ) as does the starting material, the alcohol. Therefore, oxidize the alcohol. The ester is best formed from the acid chloride: . n-C,H,OH
KMnOJ
CH,(CH,),COOH
S<)<’I?
CH 3 ( CH,),COCI
n-<‘41iuoll
product
Problem 16.47 In living cells, alcohols are converted to acetate esters (acetylated) by the thiol ester CH,COS-(CoA), acetyl coenzyme A. CoA is an abbreviation for a very complex piece. Illustrate this reaction using glycerol-1-phosphate. 4
CH,OH LHOH I kH,OPO,H,
tH,OPO,H,
Glycerol-]-phosphate
The reactivity of thiol esters, RC-SR’,
CI
a thiol ester
a phosphatidic acid
Coenzyme A
lies between that of anhydrides and esters.
FATS AND OILS
Fats and oils are mixtures of esters of glycerol, HOCH,CHOHCH,OH, with acyl groups from carboxylic acids, usually with long carbon chains. These triacylglycerols, also called triglycerides, are types of lipids because they are naturally occurring and soluble only in nonpolar solvents. The acyl groups may be identical, or they may be different. Fats are solid esters of saturated carboxylic acids; oils, with the exception of palm and coconut, are esters of unsaturated acids with cis C=C bonds which prevent close-packing of the molecules. Problem 16.48 ( a ) Write a formula for a fat (found in butter) of butanoic acid. ( b ) Alkaline hydrolysis of a fat of a high-molecular-weight acid gives a carboxylate salt (a soap). Write the equation for the reactions of the fat 4 of palmitic acid, n-C,,H,,COOH, with aqueous NaOH.
n-C3H7-C-O-LHz H0 n-C,J43,COO-CH2 (6 )
A
n-Cl5H3,C0O- H n-C15H,,COO-&H,
+ 3NaOH
4
3 n-C,,H,,COO-Na+
+ HOCH2CHOHCHzOH
a soap
Problem 16.49 Two isomeric triglycerides are hydrolyzed to 1 mol of C,,H,,COOH (A) and 2 mol of C,,H,,COOH (B). Reduction of B yields n-stearyl alcohol, CH,(CH,),,CH,OH. Compound A adds 1 mol of and 9-oxononoic acid, H, to give B and is cleaved with 0, to give nonanal, CH,(CH,),CH=O, 4 O=CH(CH,),COOH. What are the structures of the isomeric glycerides?
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
366
[CHAP. 16
Compound B, which reduces to stearyl alcohol, is stearic acid, CH,(CH,),,COOH. Compound A is a straight-chain unsaturated acid with one double bond, and its cleavage products suggest that the double bond is at Cy, making it oleic acid, CH,(CH,),CH=CH(CH,),COOH. With 1 mol of oleic and 2 mol of stearic acid there are two possible structures: CH,-OOC(CH,)
CH,-OOC(CH,),CH=CH(CH,),CH,
I
I
,,CH,
C H - O O C ( CH,),CH=CH( CH,),CH,
+CH--OOC( CH, ) ,&H,
I
I
CH ,-OOC( CH2) I ,CH
I -01eoyI-2,3-distearoylglyceroI
AMIDES
In addition to transacylation reactions and the heating of ammonium carboxylates (Section 16.3), unsubstituted amides may be prepared by careful partial hydrolysis of nitriles: H2O + RC-N
1 . cddH$O,
R--fi--NH2 0
Amides are slowly hydrolyzed under either acidic or basic conditions. The mechanisms are those shown in Problem 16.34. Unsubstituted amides are converted to RCOOH with HNO,. RCNH2
HONO ( N a N 9 + aq.HCI)
RC-OH
+ N2
and are dehydrated to RCN with P4010, RC-NH2
8
PP,O 3 RCN
a nitrilc
In the Hofmann degradation, RCONH, goes to RNH, (Section 18.2) RCONH,
+ Br, + 4KOH-
RNH,
+ K,CO, + 2KBr + 2H,O
Problem 16.50 Use the concepts of charge delocalization and resonance to account for the acidity of imides. (They dissolve in NaOH.) 4 The H on N of the imides is acidic because the negative charge on N of the conjugate base is delocalized to each 0 of the two C=O groups, thereby stabilizing the anion.
or
CHAP. 161
367
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
16.8 SUMMARY OF CARBOXYLIC ACID DERIVATIVE CHEMISTRY PREPARATION 1.
-
PROPERTIES
1. Hydrolysis
-U
Esters
+ H,O'
(a) Carboxylic Acids
CH,COOH
+ CH,OH
CH COOH + CH,OH
2. Saponification + NaOH+ CH,COONa + CH,OH \CH,COONa + CH,I Reduction 3. (b) Acid Chlorides CH,COOCH, -+ (1) LiAIH,, (2) H,O*CH,CH,OH + CH,OH CH,COCI + CH,OH +' (1) 2RMgX, (2) H,OCH,CR,OH + CH,OH (c) Acid Anhydrides 4. Amide Formation (CH,CO),O + CH,OH heat + NH, CH,CONH, + CH,OH 2. Acid Chlorides CH,COCl-+ HOH-CH,COOH + HCI +' ROH--+CH,COOR + HCl SOCI,--,SO, + HCl + I + NH,+CH,CONH, + NHdCI CH,COOH + PC1, P(OH), + HCl + PCI,-, POCI, + HCI + + (1) LiAIH,, (2) H,O+CH,CH,OH + (1) 2RMgX, (2) H,O+ CH,C(OH)R, + ArH(AICI,)+ CH,COAr + HCl 3. Acid Anhydrides + H Pd/ BaSO,/ S CH,CHO (a) Acyclic AIPO, + HOH+ 2CH,COOH CH,COOHH,C=C=O + HOR- CH,COOR + CH,COOH CH&ooNa CH,CoC1---2-_, (CH,CO),O + NH, -+C H, C O NH, + CH,COO-NH:
i
/ \-
-+
-
-
+
( b ) Cyclic
H2C--COOH
I
H,C--COOH
150 "C
H,C-CH0
I
>o
H,C--C\O
4. Amides
+ H,O---, + HOR+ + NH,+
HOOC-CH,CH2-COOH HOOC-CH,CH,--COOR NH:OOC-CH,CH2--CONH,
- +
-
+, H,O+ C1CH,COOH + NH,CI CH,COOH + NH, --+ CH,COO NH,* - + CH,COCl+ NH, CH,COONa + NH, CH,CONH, -+ NaOH(CH,CO),O + NH,+CH,COOCH,COOCH, + NH,+ CH,OH + CH,CN + (1) cold H2S0,, (2) cold H,O
6.9 ANALYTICAL DETECTION OF ACIDS AND DERIVATIVES
NEUTRALIZATION EQUIVALENT
Carboxylic acids dissolve in Na,CO,, thereby evolving CO,. The neutralization equivalent or equivalent weight of a carboxylic acid is determined by titration with standard base; it is the number of grams of acid neutralized by one equivalent of base. If 40.00 mL of a 0. IOON base is needed to
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
368
[CHAP. 16
neutralize 0.500g of an unknown acid, the number of equivalents of base is eq = 0.00400 eq 40*00mL x 0.100 L 1O00 mL/L The neutralization equivalent is found by dividing the weight of acid by the number of equivalents of base, i.e. 0.500 g = 125g/eq 0.00400 eq Problem 16.51 A carboxylic acid has GMW = 118, and 169.6 mL of 1.0oO N KOH neutralizes 1O.Og of the acid. When heated, 1 mol of this acid loses 1 mol of H,O without loss of CO,. What is the acid? 4 The volume of base (0.1696 L) multiplied by the normality of the base (1.000eq/L) gives 0.1696 as the number of equivalents of acid titrated. Since the weight of acid is 10.0g, one equivalent of acid weighs 10.0 g 0.1696 eq
= 59.0 gleq
Because the molecular weight of the acid ( 1 18) is twice this equivalent weight, there must be two equivalents per mole. The number o f equivalents gives the number of ionizable hydrogens. This carboxylic acid has two COOH groups. The two COOH's weigh 90 g, leaving 118 g - 90 g = 28 g (two C's and four H's) as the weight of the rest of the molecule. Since no CO, is lost on heating, the COOH groups must be on separate C's. The compound is succinic acid, HOOC--CH,-CH,-COOH.
SPECTROSCOPIC METHODS 1.
Infrared
In the H-bonded dimeric state [Problem 16.3(6)] RCOOH has a strong 0-H stretching band at 2500-3000 cm - '. The strong C=O absorptions are at 1700-1725 cm-' for aliphatic, and 16701700cm-' for aromatic acids. See Table 12-1 for key absorptions of acid derivatives (G = X, OH, OR', OCOR, N-).
I
2.
Nmr
The H of COOH is weak1 shielded and absorbs downfield at S = 10.5-12.0ppm. See Tables Y 12-4 and 12-6 for proton and ' - C nmr chemical shifts, respectively. 3.
Mass Spectra Carboxylic acids and their derivatives are cleaved into stable acylium ions and free radicals.
0:
It
.o 1I
+
R--C-G-If,R--CtG+R--C
0'
111
+ -G
I ( G = C I , O H , O R , OCOR, N-)
Like other carbonyl compounds, carboxylic acids undergo p cleavage and y H transfer.
CHAP. 161
369
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
16.10 CARBONIC ACID DERIVATIVES Problem 16.52 The following carboxylic acids are unstable and their decomposition products are shown in parentheses: carbonic acid, (HO),C==O (CO, + H,O); carbamic acid, H,NCOOH (CO, + NH,); and chlorocarbonic acid, CICOOH (CO, + HCI). Indicate how the stable compounds below are derived from one or more of these unstable acids. Name those for which a common name is not given. ( a ) Cl,C=O
( b ) (H,N),C=O
A
Urea
Phosgene
( d ) (CH,O),C=O
(c) CICOCH,
(e) H,NC-OCH,
8
( f ) HN=C=O
4
lsocyanic acid
( a ) The acid chloride of chlorocarbonic acid; ( b ) amide of carbamic acid; (c) ester of chlorocarbonic acid, methyl chlorocarbonate; ( d ) diester of carbonic acid, methyl carbonate; (e) ester of carbamic acid, methyl carbamate (called a urethane); ( f ) dehydrated carbamic acid.
Problem 16.53 Give the name and formula for the organic product formed from 1 mol of COCI, and 2 moles of (a) C,H,NH,, (6) C,H,OH. 4 COCI, behaves like a diacid chloride. 0
II
0
(a ) C,H,NHCNHC,H,
II
N,N‘-Diethylurea
(6) C,H,OCOC,H,
Ethyl carbonate
Problem 16.54 What products are formed when 1 mol of urea reacts with (a) 1 mol, (6) a second mol of 4 methyl acetate (or acetyl chloride)? 0
II H2N--C-NH2 + CH,COOCH,
-
0 - C H OH
II
( b ) CH ,COOCH3
CH,CO--NH<-NH,
-CH>OH
Acet ylurea
(0)
CH,C&NH--C-NH--COCH,
II
0
+ (CH,CO),N--C-NH, II 0
N,N’-Diacetylurea a ureide
N.N-Diacetylurea
Problem 16.55 Barbiturates are sedative-hypnotic varieties of 5,5-dialkyl substituted barbituric acids. Write the reaction for the formation of Verona1 (5,5-diethylbarbituric acid) from the condensation of urea with diethylmalonic ester. [See Problem 17.13(a)]. 4 C2H,\ C2H,’
C
/ COOEt
‘COOEt
+
H-NH H-NH’
\C=O-2EtOH
+
C A \
C,H/
Pen tobarbi tal is 5-ethyl-5-(1-methylbutyl)-barbituric acid.
C
,CO-NH,
‘CO-NH’
c=o
S,S-Diethylbarbituric acid
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
370
[CHAP. 16
16.11 SUMMARY OF CARBONIC-ACID-DERIVATIVE CHEMISTRY
PREPARA TION
PROPERTIES
1. Esters (U)
1. Ammonolysis
Phosgene
+ NH,+2ROH
/
,4
Cl-CO--Cl+ 2HOR (6) Alkyl Halide
2RX + Ag,CO,
+ O=C(NH,),
Saponification
+ 2NaOH-
2ROH + Na,CO,
3. fransacylate + HOH+ ROH
CI-CO--cl+ R O H - R O - - C - d +
/
(b) Carbonyl Chloride CO + Cl,
/ + ROH-
2. Acid Halides
(a) Chlorocarbonic Ester
\+
3. Carbamic Acid Derivatives
HC1+ RO--CO--OR NH,Cl + R-0-NH,
NH,+
+H
CI--CWl-
>
+ ROH-
+ CO, + HCl
O H d 2HC1+ CO, 2HC1+ RO-CO-OR
N H , - - + N H 4 C1+ H,N--CO-NH,
(a) Ester (Urethane)
RM0-OR RO--CO--CI
+ NH,+ NH,
R M O - N H , -+ HOH-
ROH + CO,
+ NH,
Alkyl Carbamate
(6) Chloride
Cl--CO--Cl+ 4.
NH: C 1 - d H,N-CO--Cl-
Carbamide (Urea)
Carbamyl Chloride
4.
NH; NCO-
Miscellaneous
+ H,O'
(6) Calcium
+ H,SO, + H,O
\+
2NH,'
5. Ureides (Acylureas)
+ H,N--CO-NH,
--+
HX
R N H ~ X -+ K+NCO--RNH--CO-NH,
(a) Alkyl Orthocarbonates
C(OR), + NH, Amine and Cyanamide RNH,
+ H,N--CN-
+ Na,CO,
H,N--C0-NH-€0-NH2
+ NaOBr-
+ H,O + HNCO N, + CO, + NaBr
CH,O-
HOCH,NH--CO-NH,
NH
II
H,N--C-NH, NH
I1
goes on to polymer
+ R--CO-NH-CO-NH,
6. Alkylureas
7. Guanidines (Iminoureas)
NH,
+ NH: + X-
N,
(c) Phosgene
R-CO-X
X- -CO,
O + ::H-
0
Cl-CO--cl+
+ HCl
H-N-C-0
A
( a ) Ammonium Cyanate
CaCN,
60 "C
RNH4-NH,
37 1
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
CHAP. 161
16.12 SYNTHETIC CONDENSATION POLYMERS
Condensation polymers are prepared by reactions in which the monomeric units are joined by intermolecular elimination of small molecules such as water and alcohol. Among the most important kinds are polyesters and polyamides. Polyurethanes are addition polymers of acid derivatives. Problem 16.56 Indicate the reactions involved and show the structures of the following condensation polymers obtained from the indicated reactants: (a) Nylon 66 from adipic acid and hexamethylene diamine; ( b ) Nylon 6 from e-caprolactam; ( c ) Dacron from methyl terephthalate and ethylene glycol; (d) Glyptal from glycerol and 4 terephthalic acid; (e) polyurethane from diisocyanates and ethylene glycol. (a) Nylon 66 is a polyamide produced by reaction of both COOH groups of adipic acid with both NH, groups of hexamethylene diamine; -CONH- bonds are formed by H,O elimination. The initial reaction gives a nylon salt, which is then heated.
+ H,N(CH,),NH,mp;-,
HOOC(CH,),COOH
--C--(CH,),--C-NH--(
II
CH2),-NH--C--(CH,),<-NH-(
II
0
- 0 0 C ( C H 2 ) , C 0 0 - 'H,N(CH,),NHf -!!%
II
0
0
0
CH2)6-NH-
II
Poly( hexamethylene adipamide)
( b ) Nylon 6 is also a polyamide, but is made from the monomer e-caprolactam, which is a cyclic amide of
e-aminocaproic acid. Heat opens the lactam ring to give the amino acid salt, which forms amide bonds with other molecules by eliminating water. 0
E -Caprolactam
,
mer
mer
mer 9
-NH--(CH,),--C-NH--(
CH2)s<-NH+CH2)5<-
II
II
0
0
II
0
Poly( 6-aminohexanoic acid amide)
(c)
Dacron is a condensation polyester formed by a transesterification reaction between dimethyl terephthalate and ethylene glycol.
HO--CH,CH,--OH
+ H3COOC-
W
O
C
H
,
+HMH,CH,--OH heat
Dimet hyl terephthalate
-CHJOH U
mer
Poly(ethy1ene terephthalate)
(d) Glyptal is also a polyester condensation product, but glycerol (HOCH,CHOHCH,OH) produces a cross-linked thermosetting resin. In the first stage a linear polymer is formed with the more reactive primary OH groups.
372
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
/O\
C=O
+ HO--CH,CHCH,OH + O=C I
/o\
[CHAP. 16
C=O + HO-CH,CHCH,OH
I
OH
OH
~
H20
O=& Phthalic anhydride mer
C-O--CHz-CH--CH,+
-C
I
OH
OH
The free 2” OH’S are then cross-linked with more molecules of phthalic anhydride. ( e ) Urethanes [Problem 16.52(e)] are made by the rapid exothermic reaction of an isocyanate with an alcohol or phenol.
R-N=C%
[
+ HQ-R’+ U.
’-]
R-N=C~R’
H
I
O
II
-+R-N--C-OR’
Polyurethanes are formed from a diol (e.g., HOCH,CH,OH) and a diisocyanate, a compound with two -N=C=O groups (e.g., toluene diisocyanate).
HOCH,CH,OH
+ O=C=N
N=C=O
+ HOCH,CH,OH + O - C = N
Toluene Diisocyanate
16.13 DERIVATIVES OF SULFONIC ACIDS
Sulfonic acids, R(Ar)SO,H, form derivatives similar to those of carboxylic acids (see Table 16-3). These are sulfonyl chlorides, sulfonates (esters), and sulfonamides. The transsulfonylation reactions are similar to the transacylation reactions, except that the ester and amide cannot be made directly from the acid. See Problem 13.17 for preparation of sulfonyl chlorides and esters and Problem 13.18 for use of sulfonate esters as substrates in S,1 and S,2 reactions. Sulfonate esters can be prepared from optically active alcohols without inversion of configuration of the chiral carbinol C. The reason is that reaction involves cleavage of the H 4 bond of the alcohol. ArSO,-CI
+H-U---&
: \- Arso2-
d +HCI :\
Reduction of sulfonyl chlorides with Zn and acid yields first sulfinic acids and then thiophenols. Zn HCI
Zn HCI
C , H , S O 2 C 1 A C , H , S O , H A C,HSSH Benzenesulfinic acid
CHAP. 161
373
CARBOXYLIC ACIDS AND THEIR DERIVATIVES Table 16-3. Comparison of Sulfonic and Carboxylic Acid Chemistry ~~
~~
Sulfonic
Carboxylic
Acids
Ar(R)S03H
RC0,H
1. Acid strength 2. Formation of derivatives 3. Nucleophilic displacement on anion (occurs only with ArSO,) 4. Solubility in H,O
Strong Indirect (ArS0,Cl)
Weak Direct
By OH-, CN-
None
Soluble
Insoluble, except acids of low MW
Esters 1. Preparation 2. Hydrolysis with H,"O
R'COOR From R'COOH, R'COC1 or R'--COC-R'
3. Reaction with nucleophiles
ArS0,OR From ArS0,Cl ArS0,H + RI'OH Cleavage of alkyl-oxygen bond At alkyl C with inversion (like RX)
Acid Chlorides
ArS0,CI
RCOCI
1. Formation of acids, esters and amides 2. Reduction
Slow; requires base To sulfinic acid, ArS0,H (Zn, HCI), and thiophenols, ArSH
Rapid (ArCOC1 requires base) To RCHO
Amides
ArSO,NH,
RCONH,
1. Hydrolysis
Only by acids; slow
2. Formation from acyl halides 3. Acidity of H on N
Slow
By acids or bases; rapid Rapid
Forms salts with OH-
N o salt formation
R'CO''0H + ROH Cleavage of acyloxygen bond At acyl C with retention or occasionally racemization of R. Intermediate is sp' and has an octet
b8
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
374
Problem 16.57 Name (a) C,H,SO,OCH,,
(6) C,H,SO,NH,,
[CHAP. 16
( c ) p-BrC,H,SO,CI.
4
(a) methyl benzenesulfonate, (6) benzenesulfonamide, (c) p-bromobenzenesulfonyl chloride (brosyl chloride). Problem 16.58 Give the reaction product of PhS0,H and ( a ) aqueous NaOH, (b) fusion with NaOH, ( c ) 4 SOCI,, ( d ) CH,OH, (e) HNO, + H,SO,. ( a ) PhSO, Na +.
(6) PhO- Na *,
(c) PhSO,CI,
(d) no reaction,
(e) rn-nitrobenzenesulfonic acid.
Problem 16.59 Give the product formed when PhS0,CI is treated with ( a ) phenol, (6) aniline, (c) water, ( d ) excess Zn and HCI. 4 (a) phenyl benzenesulfonate, PhS0,OPh; (6) N-phenylbenzenesulfonamide, PhS0,NHPh; (c) benzenesulfonic acid, PhSO,OH, ( d ) thiophenol, C,H,SH. Problem 16.60 Prepare ( a ) tosylamide (Ts = tosyl = p-CH,C,H,SO,-) PhSO,CI, (c) o-methylthiophenol from PhCH,. or
from toluene, (b) PhCOOH from
4
ClSOlH
Problem 16.61 Write structures for compounds (A) through (C) in the synthesis of saccharin.
0
Saccharin (A)
CIS0,H
SOZNHT
4
(C) KMnO,
Problem 16.62 Prepare methyl 2-methylpropanesulfonate from 1-chloro-2-methylpropane.
4
Form the C-S bond by using the hydrogen sulfite anion as the nucleophile in an S,2 reaction. S: is a more nucleophilic site than is 0-, even though 0 has a negative charge. Na’ -0-SOH
8
+ (CH,),CHCH,CI+
(CH,),CH CH,SO; Na’
3
0
(CH,),CHCH, Notice that the strong base CH,O-, rather than the much weaker base CH,OH, must be used to form the ester. This is an important difference between the very reactive acyl chloride (RCOCl) and the much less reactive sulfonyl chloride (RS0,CI).
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
CHAP. 161
375
Supplementary Problems Problem 16.63 Write formulas for each of the following: ( a ) phenylacetic acid, (6) phenylethanoic acid, (c) 2-methylpropenoic acid, (d) (E)-butenedioic acid, (e) ethanedioic acid, ( f ) 3-methylbenzenecarboxylic acid.
4
CH ( a ) C,H,CH,COOH
H, ( d)
HOOC
,c=c,
,COOH H
( b ) C,H,CH,COOH
(e)
I’
CH,=C-COOH
(c)
HOOC-COOH
CH’
Problem 16.64 Name the following compounds:
CH3 (U)
I
CH,CH,-C-COOH
( b ) CH,CH=CH-CH=CH-COOH
1
CH3 CH, H
CH3
I
I
I H
1 CH,
I
( c ) CH3-C-C-CH2-COOH
( d ) CH3CH2CH2CH2CHCH2CH-COOH
4
AH,CH3
( a ) dimethylethylacetic acid, 2,2-dimethylbutanoic acid; (6) 2,4-hexadienoic acid (sorbic acid); (c) 3,4dimethylpentanoic acid, P ,y-dimethylvaleric acid; (d) 2-methyl-4-ethyloctanoic acid.
Problem 16.65 Name the following compounds:
( a ) dimethyl a -methylsuccinate (dimethyl 2-methvlbutanedioate), (6) 3-methylphthalic anhydride, ( c ) Nmethylphthalimide, (d) diethyl oxalate (diethyl ethanedioate), (e) N,N-dimethylpropanamide, ( f ) N ,N’dimethylurea, (g) N,N-dimethylurea. 4
Problem 16.66 Use ethanol as the only organic compound to prepare ( a ) HOCH,COOH, CH’CHOHCOOH. ( a ) The acid and alcohol have the same number of C’s.
CH,CH,OH-
KMnOJ H+
CH,COOH-
CIz/ P
CH,CICOOH-
OH-
1
2 H+
HOCH,COOH
(6) Now the acid has one more C. A one-carbon “step-up” is needed before introducing the OH. PCI,
KCN
C,H,0H--+C2H,CI-C2H,CN-
H10’
CH,CH,COOH-
a s in
part ( a )
CH’CHOHCOOH
(6) 4
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
376
[CHAP. 16
Problem 16.67 Prepare 2-methylbutanoic acid from 2-butanol.
4
The precursor of the acid has a halogen on the site of the COOH. 2-Butanol is converted to this halogenated butane.
-
H
I
CH3CHz-F-CH3
SOcl2
OH
H
H
H
I
CH3CHz-(f-CH3 I
CH3CH2-T-CH3
c1
I. 2.
I
CO2
CH3CH2-T-CH3
HjO'
MgCl
COOH
Do not use the nitrile route, because 2-chlorobutane, a 2" halide, may undergo extensive dehydrohalogenation. Problem 16.68 Prepare a -methylbutyric acid from ethanol.
4
Introduce COOH of CH,CH2CH(CH,)COOH through a CI and build up the needed 4-carbon skeleton
c-c-c-c I
CI
by a Grignard reaction. (1)
C,H,OH
3C,H,Br =Zr*C,H,MgBr
(2) C,H,OH
oxid. A,cu
(3) CH,CH=O
-
(use in Step 3)
CH,CH=O
+ C,H,MgBr
(use in Step 3)
CH,CHCH,CH,
CH,CHCH,CH,
-A(MgBr)+
AH
MgCl Problem 16.69 Convert 2-chlorobutanoic acid into 3-chlorobutanoic acid.
4
2-Chlorobutanoic acid is dehydrohalogenated to 2-butenoic acid and HCI is added. H' adds to give a P-carbocation which bonds to Cl- to form the P-chloroacid.
kil
+ I
IH'I CH,CHCHCOOH
CH,CH=CH-COOH
a B-R'
The a-carbocation is not formed because its
+
Fc1
CH, H-CH,-COOH
charge would be next to the positive C of COOH.
o*-
]HI
I II,+ CH,CHCHC-OH +
Problem 16.70 Use isopropyl alcohol as the only organic compound to synthesize ( a ) a-hydroxyisobutyric acid, (6) P - hydroxybutyric acid. 4
COOH
CN (a)
c
CH, H-CH,= OH
CN-.HCN (HCN add.)
cH3-f-CH3
0
I
' CH3-7-cH3
OH
Cyanohydrin
H30+
I
cH3-7-CH3
OH
377
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
CHAP. 161
H2SO)
CH3-fH-CH3 OH __*
CH3-CH=CH,
CH,-
fOHH-CH,CI 7 CH,
CH3 H-CH2-CN
fOH
‘OH-
NHI
fOHHCH2-COOH
Problem 16.71 Write structures for the compounds (A) through (D). 4
This is a Reformatsky reaction (see Problem 15.43).
CH3 CH2-COOC2H5
I
ZnBr
I I
CH3-C-CH2COOC2H5
(CH 3)2C= C HCOOH
(CH,),CHCH,COOH
(C)
(D)
OH (B)
(A)
Problem 16.72 Use simple, rapid, test tube reactions to distinguish among hexane, hexanol and hexanoic acid. 4
Only hexanoic acid liberates CO, from aqueous Na2C0,. Na reacts with hexanol to liberate H,. Hexane is inert. Problem 16.73 Describe the electronic effect of C,H, on acidity, if the acid strengths of C,H,COOH and HCOOH are 6.3 x l O - ’ and 1.7 x 10-4, respectively. 4
The weaker acidity of C,H,COOH shows that the electron-releasing resonance effect of C,H, outweighs its electron-attracting inductive effect. Problem 16.74 What compounds are formed on heating (a) 1,1,2-cyclohexanetricarboxylicacid, and ( b ) 1,1,2-cyclobutanetricarboxylicacid? 4
cis- and transCyclohexane- 1,2-dicarboxylic acid
COOH
COOH
and transanhydride
cis-
//O
“C O\
trans-l.2Cyclobutanedicarboxylic acid cis-anhydride
The fruns-dicarboxylic acid cannot form the anhydride because a five- and a four-membered ring cannot be fused trans.
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
378
[CHAP. 16
Problem 16.75 Write structural formulas for the products formed from reaction of 6-valerolactone with: ( U ) LiAIH,, then H,O; ( b ) NH,; (c) CH,OH and H,SO, catalyst. 4 (U)
HOCH,CH ,CH ,CH ,CH ,OH,
(6) HOCH ,CH ,CH ,CH ,CONH
,
( c ) HOCH,CH ,CH ,CH ,COOCH,.
Problem 16.76 Prepare the mixed anhydride of acetic and propionic acids.
4
Mixed anhydrides,
are made by reacting the acid chloride of one of the acid portions with the carboxylate salt of the other. Use CH,COCI and CH,CH,COONa, or CH,COONa and CH,CH,COCI. For example:
0
II
-
0
II
+ Na’-0-C-CH,CH,
CH,-C-CI
+ Na’Cl-
Problem 16.77 Use “CH,CH,OH to synthesize “CH,CONH,. 1.4
oxrd
C H , C H , O H y
I,
CH,COOH-
PCI,
1.4
4
CH,COCI-
NH,
“CH,CONH,
Problem 16.78 Distinguish by chemical tests ( a ) CH,COCI from (CH,CO),O, (6) nitrobenzene from 4 benzamide. ( a ) With H,O, CH,COCI liberates HCI, which is detected by giving a white precipitate of AgCl on adding AgNO,. ( b ) Refluxing the amide with aqueous NaOH releases NH,, detected by odor and with moist litmus or pH paper.
Problem 16.79 Name the main organic product@) formed in the following reactions:
+ excess CH,CH,CH,CH,MgBr, then H,O+ (CH,),CHCH,CH,OH + C,H,COCI (c) HCOOCH,(CH,),CH, + NH, CH,CH2CH,COCI + CH,CH,COONa ( e ) C,H,COBr + 2C,H,MgBr, then (a)
(6) (d) (a)
C,H,COOCH,
C6H,C(OH)(CH,CH2CH,cH,),
( b ) C6H,COOCH,CH,CH(CH,),
5-Phenyl-S-nonanol
Isoamyl benzoate
0
II
( c ) HCONH, Formamide
H,O’ 4
+ HO(CHd,CH, 1-Hexanol
0
II
( d ) CH,CH,CH,-C-O-C-CH,CH, Butyric propionic anhydride
(e)
(C6H5)3C0H Triphen ylcarbinol
Problem 16.80 Name and write the structures of the main organic products: ( a ) H,C=CHCH,I heated with NaCN; (6) CH,CH,CONH, heated with P,O,,; (c) p-iodobenzyl bromide reacted with CH,COOAg; ( d ) nitration of benzamide. 4 ( a ) ally1 cyanide or 3-butenenitrile, CH,=CH-CH,-CN; (6) propionitrile or ethyl cyanide, CH,CH,CN (an intramolecular dehydration); (c) p-iodobenzyl acetate p- IC,H,CH,OCOCH, ; ( d ) m-nitrobenzamide, rn-NO,C,H,CONH, (carboxylic acid derivatives orient meta during electrophilic substitution).
Problem 16.81 Give steps for the following preparations: ( a ) 1-phenylpropane from P-phenylpropionic acid, ( b ) p - benzoylpropionic acid from benzene and succinic acid. 4
CHAP. 161
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
( a ) The net change is COOH+CH,,
379
a reduction.
aPhCH2CH,CH20H
PhCH2CH2COOH
H IPl
PhCH2CH=CH2 4 PhCH2CH,CH,
H2m4
-H20
@-Phenylpropionicacid
1-Phenylpropane
CH2COOH
CH2-CO heat
CH2COOH
CH2-CO
Succinic acid
o\ /
Succinic anhydride
c6Hg(4ci3)
QJCOCHfH,CWH
Problem 16.82 Identify the substances (A) through (E) in the sequence
C,H,COOH-
PClS
(B)
p40lO
(A)+C,H,CONH,-(C)~ 4
Problem 16.83 Give the products from reaction of benzamide, PhCONH,, with ( a ) LiAlH,, then H,O'; (6) 4 P4HI0;(c) hot aqueous NaOH; (d) hot aqueous HCl. ( a ) PhCH,NH,; (6) PhCN, benzonitrile (phenyl cyanide); ( c ) PhCOONa + NH,; (d) PhCOOH
Problem 16.84 Suggest a synthesis of (CH,),CCH =&H2 from (CH,),CCHOHCH,.
+ NH4Cl. 4
An attempt to dehydrate the alcohol directly would lead to rearrangement of the R' intermediate. The major product would be (CH,),C=C(CH,),. To avoid this, pyrolyze the acetate ester of this alcohol. CH,COCl
+ (CH,),CCHOHCH,-
(CH,),C H C H , A product "a,OCH,
Problem 16.85 Assign numbers from 1 for LEAST to 3 €or MOST to show the relative ease of acidcatalyzed
esterification of:
-I
I1 -
I11 -
CH3
I
( b ) CH3CH20Hby:
CH3-CH-COOH
CH3CH2COOH
(CH,),CCOOH
CH3CHOHCH2CH3
CH,CH2CH20H
CH3
I
( c ) C,H,COOH by:
CH3-C-CH2CH3 AH
Steric factors are chiefly responsible for relative reactivities. I (a) 3
(6) (c)
2 1
I1 1
I11 2
3
1
2
3
4
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
380
[CHAP. 16
Problem 16.86 The "0 of RIXOH appears in the ester, not in the water. when the alcohol reacts with a carboxylic acid. Offer a mechanism consistent with this finding. 4
The 0 from ROH must bond to the C of COOH and the OH of COOH ends up in H,O. The acid catalyst protonates the 0 of C=O to enhance nucleophilic attack by ROH.
0
II
R-C-OH
H' +
II
R' %H
R-Ce->
OH
OH
R-T+-jlH
+R-Y-OH,
I
I
0
II
+ __*
R-C--"O-R'
+ H30+
I8O
I80H
I R'
I
R'
Problem 16.87 When reactive alcohols such as CH,=CHCH2"OH found. Explain.
are esterified in acid, some H,IXO is 4
Protonated ally1 alcohol undergoes an S, 1-type substitution reaction.
CH,=CHCH, "OH
2CH,=CHCH,-'80H
+
I H
S I
-% H,I80
+ CH,=CH;H,
0 t
CH,t
CH,=CHCvy
OH
CH,C-OCH,CH=CH,
--+
b
Problem 16.88 Show how phosgene,
CI-c-CI
b
is used to prepare ( a ) urea, (b) methyl carbonate, (c) ethyl chlorocarbonate, (d) ethyl N-ethylcarbamate (a urethane), (e) ethyl isocyanate (C2H,-N=C=O). 4
COCI,
NH,CONH,
2CH OH
COCI, A COCI,
-c2H50-8-c' C H O H ( I mole)
H
CXI,
C H NH
C,H5N-l-CI I
C H OH
H
C2Hs&-C-OC2H5
0 H
Problem 16.89 What carboxylic acid (A) has a neutralization equivalent (NE) of 52 and decomposes on heating to yield CO, and a carboxylic acid (B) with an NE of 60? 4
The NE o f a carboxylic acid is its equivalent weight (molecular weight divided by the number of COOH groups). Since CO, is lost on heating, there are at least 2 COOH's in acid (A), and loss of CO, can produce a
CHAP. 161
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
38 1
monocarboxylic acid, (B). Since one mole of COOH weighs 45 g, the rest of (B) weighs 60 g - 45 g = 15 g, which is a CH, group. (B) is CH,COOH, and (A) is malonic acid (HOOC--CH,-COOH), whose NE is 104 g 2 eq
= 52 gleq
Problem 16.90 An acyclic compound, C 6 H I 2 O 2has , strong ir bands at 1740, 1250, and 1060cm-', and no bands at frequencies greater than 2950cm-I. The nmr spectrum has two singlets at 6 = 3.4 ( 1 H ) and 6 = 1.0 4 ( 3 H ) . What is the compound?
The one degree of unsaturation is due to a carbonyl group, indicated by the ir band at 1740cm-I. The lack of bands above 2950cm-I shows the absence of an O H group. Hence the compound is not an alcohol or a carboxylic acid. That the compound is probably an ester is revealed by the ir bands at 1250 and 1060cm-' ( C - 0 stretch). Two singlets in the nmr means two kinds of H's. The integration of 1 : 3 means the 12 H's are in the ratio of 3 : 9. The signal at 6 = 3.4 indicates a C H , 4 group. The nine equivalent protons at S = 1.0 are present in three CH,'s not attached to an electron-withdrawing group. A r-butyl group, (CH,),C--, fits these requirements. The compound is (CH,),CCOOCH,, methyl trimethylacetate (methyl pivalate). Problem 16.91 Predict the base (most prominent) peak in the mass spectrum of the compound in Problem 16.90. 4
Parent ions of esters resemble those of other acid derivatives and carboxylic acids in that they cleave into an acylium ion.
(CH,),CC-OCH,
[
A
I
(CH,),CC=Ot
+ OCH,
or CH,C-
+ &COCH,
The base peak should be m l e = 8 5 or 59. Problem 16.92 Prepare 0-bromotoluene from toluene without using direct bromination (which gives a large amount of the unwanted p-isomer). 4
Use -SO,H
as a blocking group and then remove it by desulfonation.
Toluene
p-Toluenesulfonic acid
Problem 16.93 Identify the compound C,H,,O,SCI(A), and also compounds (B), (C), and ( D ) , in the reactions: AgCl(s)
4
T : -1-
water soluble (B)
Br C H 3 0 C H 3 c-- CHQCH, CH3 (D) 1 -Bromo-2,4,6trimethylbenzene
CH1 (C) Mesitylene
-
H30+
C,H,, (C)
Br2
Fe
%
one monobromo derivative (D) 4
S03H SO*Cl C H , ~ C H t , CH3QCH, CH3
(B) 2.4.6-Trimethylbenzenesulfonic acid
CHi (A) 2,4,6-Trimethyl-
benzenesulfonyl chloride
382
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
[CHAP. 16
Problem 16.94 ( U ) (S)-sec-C,H,OH, which has an optical rotation of +13.8', is reacted with tosyl chloride (see Problem 16.60) and the product is saponified. (b) Another sample of this alcohol is treated with benzoyl chloride and the product is also hydrolyzed with base. What is the rotation of the sec-C,H,OH from each reaction? 4
(b)
A
A
A
(S)(+13.8")
(S)(Tosyl ester)
(R)(- 13.8")
+ 13.8'. Reaction with benzoyl chloride causes no change in configuration about the chiral C of the alcohol. Hydrolysis of PhCOOR occurs by attack at the carbonyl group, with retention of alcohol configuration.
Problem 16.95 Reaction of the (R)-tosyl ester of a-phenylethyl alcohol (C,H,CHMeOTs) with CH,COOH 4 yields a mixture of ( S ) - and ruc-acetate. Explain. Inversion and racemization products show operation of an SN! reaction. The SN1 pathway is possible because the alcohol forms the stable 2" benzylic carbocation, C,H,CHCH,. Problem 16.96 Write structural formulas for the organic compounds (1) through (XI) and show the stereochemistry of (X) and (XI). (Write Ts for the tosyl group.)
ArS0,Cl
+ CH,OH
__+
(1)-
H
I*O+
(11) + (111)
C6H,S02Cl + n-C4H90H +(VII) ______* (VIII) + (IX) PhCH2M.Br
t,,
C6H13
TsCI
+ (S)-HO-
(I) ArS020CH3
-H
+(X)
(11) ArSO,H
CH COO-
3(XI)
4
(111) CH,"OH
Problem 16.97 Why are sulfonic acid derivatives less reactive toward nucleophilic substitutions than are the corresponding acyl derivatives? 4
Attack by Nu: - on the trigonal acyi C leads to a stable, uncrowded, tetrahedral intermediate (and transition state), with an octet of e-'s on C. The sulfonyl S is already tetrahedral, and attack by Nu:- gives a less stable, more crowded, intermediate (and transition state), with a pentavalent S having ten e-'s, as shown:
CHAP. 161
383
CARBOXYLIC ACIDS AND THEIR DERIVATIVES
-[
0 Ar-k-G
A
+ Nu:
N 5 f ]
pentavalent intermediate
Problem 16.98 ( a ) Show the mers of the following condensation polymers, formed from the two indicated (tough and optically monomers: (i) Polycarbonate (Lexan), phosgene, CI2C=O + (HOC,H,),C(CH,), clear); ( i i ) Kodel, dimethyl terephthalate
+ HOCH2--
0
-CH,OH
(strong fibers); (iii) Aramid, (terephthalic acid) + p-H,NC,H,NH, (tire cords). (6) What small molecules are split off during growth of the polymer chain? (c) What type of functional group links the mer units? 4 -CH20-€-
b
( 6 ) HCI
(c) ester
(iii)
Problem 16.99 Which would have the higher A,,, C,H,CH=CHCOOH? Why? Cinnamic acid, because it has a more extended
7r
value in the uv, benzoic acid or cinnamic acid, 4
system.
Problem 16.100 Tell how propanal and propanoic acid can be distinguished by their
spectra.
(a) ir, (6) nmr, (c) cmr
4
Propanal has a medium-sharp stretching band at 2720cm-' due to the aldehydic H. Propanoic acid has a very broad 0-H band at 2500-3300 cm- I. The C=O bands for both are in the same general region, about 1700cm-'. Propanal gives a triplet near 6 = 9.7 ppm (aldehydic H), a triplet near 0.9 (CH,), and a multiplet near 2.5 (CH,). Propanoic acid has a singlet in the range 10-13 (acidic H), a triplet near 0.9 (CH,) and a quartet near 2.5 (CH,). Carbonyl C's are sp'-hybridized and attached to electronegative 0 ' s and hence suffer little shielding. They absorb farther downfield than any other kind of C: 190-220 for carbonyl compounds (propanal) and somewhat less downfield, 150-185, for carboxylic acids (propanoic acid).
Chapter 17 Carbanion-Enolates and Enols 17.1 ACIDITY OF H’s a TO C 4 ; TAUTOMERISM
GENERAL Three base-catalyzed reactions of such a H’s are shown in Fig. 17-1. Since the reactions have the same rate expression, they have the same rate-determining step: the removal of an cy H to form a stabilized carbanion-enolate anion. The name of this anion indicates that the resonance hybrid has negative charge on C (carbanion) and on 0 (enolate).
D
I I
CH 3CH2CCHO
H”
H
*I I
OH-
CH,CH,-CCHO
rncemization
B
I I
CH3CH2CCH0
rate = k[RCHO][Base]
CH3
racemic
optically active
X
I
C H 3CH2CCHO
I
CH3
racemic
CH3
I
CH3CH2C=C+
lA
CH3
-I
or CH3CH2-C==C=0
H acid,
base
’
I H
+ H:B
I
base,
acid,
stable carbanion-enofate ion
Stabilization of the anion by charge delocalization causes the a H of carbonyl compounds to be more acidic than H’s of alkanes. Problem 17.1 17-1.
Show how the stable carbanion-enolate anion reacts to give the three products shown in Fig.
4
384
CARBANION-ENOLATES AND ENOLS
CHAP. 171
385
Table 17-1 Other Stable Carbanions
Reason for Stability
p-p
IT
bond
p-p
IT
bond
p-d
7r
bond
~
c1
cI1
c1 sp hybrid
CH,C=CC
H
H
or
H
)QH
H
H H
w: H
-
H2E-Q-N02
H 2 c = p 4 H * c Q N 0 *
NO,
-
RSXH-SR
Me2G*H,
Ph,P
t*
-
aroma tici t y
p-p
7r
bond
p-p
IT
bond
p-d
IT
bond
N' /\ 0- 0 (RS==C=SR]-
Me,S=CH,
Ph,P=CH,
Electrostatic attraction and p-d IT bond
Electrostatic attraction and p - d 7r bond
386
CARBANION-ENOLATES AND ENOLS
[CHAP. 17
H-D exchange. The anion accepts a D from D,O, regenerating the catalyst O D [anion]-
+ D,O+
(deuterated product) + OD-
Racemization. The negatively charged C of the anion is no longer chiral, as it was in the reactant aldehyde, because the extended 7r bond requires a flattening of the conjugated portion of the anion of which this C is a part. Return of an H (or D ) can occur equally well from the top face, to give one enantiomer, or from the bottom face, to give an equal amount of the other enantiomer; the product is racemic. Halogenation. [anion] -
+ X, ---+ (halogenated
product)
+ X-
All three reactions are fast and, therefore, are not involved in the rate expression.
Groups other than C=O enhance the acidity of an H. Recall the 1,3-dithianes [see text preceding Problem 15.141 and see Table 17-1, which also indicates why the negative charge on the anionic C is stable. TAUTOMERISM
H' may return to C - , to give the more stable carbonyl compound (the keto structure), or to 0-,to give the less stable enol.
H
- -C=C-OH +H+
A
I
H+
keto (weaker acid)
carbanion-enolate
I
enol (stronger acid) less common
more common
With few exceptions (Problem 17.6) the keto structure rather than the enol is isolated from reactions (see Section 8.2, item 4). Structural isomers existing in rapid equilibrium are tautomers and the equilibrium reaction is tautomerism. The above is a keto-enol tautomerism. Problem 17.2 Why is the keto form so much more stable (by about 46-59 kJ/mol) than the enol form? 4
-A++:-)
Th resonance en rgy of the carbonyl group (-&=O: is greater than that of the enol (--d=z-&-H ++-&-&=O'---H). Furthermore, the enol is a much stronger acid than the keto form. When equilibrium is established'between the two forms, the weaker acid, the keto form, predominates. Problem 17.3 Compare the mechanisms for ( a ) base-catalyzed and (6) acid-catalyzed keto-enol tautomerism.
Problem 17.4 Racemization, D-exchange, and bromination of carbonyl compounds are also acid-catalyzed. ( a ) Suggest reasonable mechanisms in which enol is an intermediate. (6) In terms of your mechanisms, are the 4 rate expressions of these reactions the same? (c) Why do enols not add X, as do alkenes?
CHAP. 171
387
CARBANION-ENOLATES AND ENOLS
Again we have:
H
I
R’-C-CR”
A
+ H:X
.=-
Step 1.
X
“-A
oxonium ion
enol
Racemization
enol
(from b )
(from a )
Deuterium Exchange
b
(from b )
(from a )
The D that forms a bond to C more likely comes from the solvent, D 2 0 , than from the O D group.
Bromination
+ Br:Br h
-Br-
Br
R’-C-C-R” I
I
R
{+\ -,t”. BrO
-H+
R
R O + Rtt[-RN
Br
rac-oxonium ion
(6) Since enol formation is rate-determining, these 3 reactions have the same rate expression, rate= k[carbonyl compound][ H ‘1. ( c ) The intermediate in the reaction of C=C with X, is a halogenonium ion in which C has sufficient electron deficiency to permit it to react with Br-. Since the oxonium ion intermediate from the enol has most of its + charge on 0, not on C, a second C-X bond does not form. The oxonium ion is the conjugate acid (a very strong acid) of the carbonyl compound (a very weak base) and loses H’ to give the keto product.
Problem 17.5 ( a ) Show the tautomers of each of the following compounds, which are written as the more stable form: (1) CH,CHO, (2) C,H,COCH,, (3) CH,NO,, (4) Me,C=NOH, and (5) CH,CH=NCH,. (b) Which two enols are in equilibrium with (i) 2-butanone, and (ii) l-phenyl-2-butanone? Which is more stable? 4
( a ) The grouping needed for tautomerism,
encircled in each case.
keto
enol
X-Y-Z-H
(a triple bond could also exist between X and Y),is
keto
enol
388
CARBANION-ENOLATES AND ENOLS
H
I
(3) CH-c-N+)
I
1
H
*
\
H
H 0-
[CHAP. 17
0-
nitro form
aci form
oxime
nitroso
H I
imine
H,C=C
(i)
(6 )
enamine
and CH,--C=CHCH,
I
I
OH
OH
The latter is more stable because it has a more substituted double bond. C,H,CH=CCH2CH,
(ii)
I
and C,H,CH,C--LHCH,
I
OH
OH
The former is more stable because the C=C is conjugated with the benzene ring. Problem 17.6 C,H,OH, an enol, is much more stable than its keto isomer, cyclohexa-2,4-diene-l-one.
626 \
Phenol (enol)
Cyclohexa-2,4diene-l-one(keto)
Explain this exception.
4
Phenol has a stable aromatic ring. Problem 17.7 ( a ) Write the structural formulas for the stable keto and enol tautomers of ethyl acetoacetate. (b) Why is this enol much more stable than that of a simple ketone? (c) How can the enol be chemically 4 detected?
CH3CCHzC-OCzHs
II 0
CH3C=CHC-OC2Hs
I
II 0
II
kefo form
( b ) There is a stable conjugated C=C<=O linkage; moreover, intramolecular H-bonding (chelation) adds stability to the enol. (c) The enol decolorizes a solution of Br, in CCI,. Problem 17.8 Reaction of 1 mol each of Br, and PhCOCH,CH, in basic solution yields 0.5 mol of PhCOCBr,CH, and 0.5 mol of unreacted PhCOCH,CH,. Explain. 4
Substitution by one Br gives PhCOCHBrCH,. The electron-withdrawing Br increases the acidity of the remaining a H, which reacts more rapidly than, and is substituted before, the H’s on the unbrominated ketone.
-
more acidic
less acidic
f
PhCOCH,CH, (0.5 mol)
0 5 mol H r ?
(h
0 5 mol Br2
[ P h C O C H B r C H , ] v PhCOCBr,CH, (0.5mol) isolated
not
(0.5 mol)
CHAP. 171
389
CARBANION-ENOLATES AND ENOLS
Problem 17.9 Describe the formation of CHI, from reaction of C,H,COCH, with NaOH and I, (NaOI).
4
First, in this haloform reaction (Section 16.2, Item 5),
-
0
It
12. NaOH
C6H$-cH3
0
II
C6H5C-c13
Then, OH- adds to the carbonyl group, and 1,C:- is eliminated because this anion is stabilized by electron-withdrawal by the 3 1’s. Finally, H’-exchange occurs.
:o:
0
+ :OH-
R-C-CI,
/
17.2 ALKYLATION OF SIMPLE CARBANION-ENOLATES
GENERAL
Carbanion-enolates are nucleophiles that react with alkyl halides (or sulfonates) by typical S,2 reactions. Carbanion-enolates are best formed using lithium diisopropylamide (LDA) , (i-Pr)*N- Li’, in tetrahydrofuran. This base is very strong and converts all the substrate to the anion. Furthermore, it is too sterically hindered to react with RX. -CH-C=O
I
I
+ (i-Pr),N-
Li’
-
+
x C = = O c . )< -C :+:-
[ I I
I I
carbanion-enolate
1+
(i-Pr),NH
Diisopropylamine
Ketones, esters and nitriles but not aldehydes, which take a different route also can be reacted with LDA. See Table 17-l(b) for stabilization of negative charge on a C cy to a e N group. Since the carbanion-enolates are ambident ions with two different nucleophilic sites, they can be alkylated at C or at 0.
O R
/-“-FRx
II I
an alkyl ketone (C-alkylation)
a vinyl ether (0-alky lation)
0-alkylation reduces the yield of the more usually desired C-alkylation product. Other drawbacks to the synthetic utility of this reaction are: (1) di- and tri-alkylation produces mixtures if more than a single H is present on the a C; (2) ketones with H’s on more than one a C will give a mixture of alkylation products. Problem 17.10 (a) Use carbanion-enolate alkylations to synthesize: (i) 2-ethylbutanenitrile, (ii) 3-phenyl-2pentanone, (iii) 2-benzylcyclopentanone. (6) Why can the unsymmetrical ketone in part (ii) be alkylated in good yield? 4
CARBANION-ENOLATES AND ENOLS
390
CH3CH2CH2-C=N-
[CHAP. 17
(f2H5
I
CH3CH2CH-C%N
LDA
2 C2H71
(ii)
C,H,--CH,--CO--CH,
I
LDA
2 CZHd
C*H, I C,H,CH-CO--CH,
0
0
(iii)
(6) The benzylic C forms the carbanion PheHCOCH, because it is more stabilized through charge delocalization to the benzene ring.
EUAMINE ALKYLA TIONS I This reaction, designed by Gilbert Stork, fosters monoalkylation. Enamines &-N[see Problem 17.5(~)(5)], of ketones are monoalkylated with reactive halides, such as benzyl and allyl, in good yield at the a C. The enamines are made from the ketone and preferably a 2" amine, R,NH.
Cyclohexanone
Pyrrolidine (2" amine)
an enamine
CH2Ph + HZO ___)
+
CHzPh an immonium compound
Enamines also can be acylated on the
cy
C with acid chlorides.
Problem 17.11 Use enamines in the following conversions: ( a ) cyclohexanone to (i) 2-allylcyclohexanone and (ii) 2-acetylcyclohexanone, and (6) 3-pentanone to 2-methyl-1-phenyl-3-pentanone. 4 ( a ) Pyrrolidine is used to convert cyclohexanone into an enamine, which
H,C=CHCH,CI and in (ii) is acylated with CH,COCI.
CH,CH=CH, I
in (i) is alkylated with CH,CH=CH, \
'H+
O----C--CH,
O=C--CH, a B-diketone
CHAP. 171
39 1
CARBANION-ENOLATES AND ENOLS
17.3 ALKYLATION OF STABLE CARBANION-ENOLATES
The acidity of an H is greatly enhanced when the C to which it is bonded is a to two C=O groups: I
I
I
O=C-CH-C=O. The negative charge on the a C of such a P-dicarbonyl compound is now delocalized over two C=O groups: [O=G=C=C==O] -.
I
l
l
Problem 17.12 Compare the relative acid strengths of 2,4-pentanedione, CH,COCH,COCH, (I); ethyl acetoacetate, CH,COCH,COOC,H, (11); and diethyl malonate, C,H,OOCCH,COOC,H, (111). Explain your ranking. 4
I > I1 > 111. All three compounds afford resonance-stabilized carbanions. However, COOEt has an electron-releasing 0 bonded to carbonyl C, which decreases resonance stabilization. There are two COOEt groups in I11 and one in 11, while I has only ketonic carbonyl groups.
The compounds I11 and I1 are useful substrates for the synthesis of carboxylic acids and ketones, respectively. MALONIC ESTER SYNTHESIS OF R-SUBSTITUTED ACETIC ACIDS, RCH,COOH OR R(R') CHCOOH (R' could =R)
-
Step 1 A carbanion is formed with strong base (often NaOEt in EtOH).
+ haOEt
Et00C-CH2-COOEt Malonic ester (Diethyl rnalonate )
+ EtOH
[EtOOC-CH-COOEtINa' or
-
0
0
Step 2 The carbanion is alkylated by S,2 reactions with unhindered RX or ROTS.
-
-
R
Et00C-CH-COOEt I
[EtOOC-CH-COOEt]N>R$
+ Na'X-
For dialkylacetic acids, the second H of the a C is similarly replaced with another R or a different R' group. Et00C-CH-COOEt
I
R
NiOEt
--
[EtOOC-CR-COOEtINa'
RX
R
I I
Et00C-C-COOEt
R
+ :X-
392
CARBANION-ENOLATES AND ENOLS
[CHAP. 17
of the substituted malonic ester gives the malonic acid, which undergoes decarboxylation (loss of CO,) to form a substituted acetic acid. Step 3 Hydrolysis
Et00C-CH-COOEt
k
,. OH-’
hydrolysis
[ n
HOOCCHCOOH
-[ 2. H,O+
R’ Et00C-C-COOEt I
hydrolysis
I
R
R’
1
HOOC-C-COOH
a
RCH2COOH
]a
R‘-CH-COOH
I
R
For a general carboxylic acid, the parts are assembled as follows:
m 1
from alkyl halides
TRX and R’X
from malonic ester
where H* replaced COOEt.
Problem 17.13 Use malonic ester to prepare ( a ) 2-ethylbutanoic acid, ( b ) 3-methylbutanoic acid, (c) 4 2-methylbutanoic acid, ( d ) trimethylacetic acid. The alkyl groups attached to the a C are introduced by the alkyl halides. (a)
In 2-ethylbutanoic acid
CH,CH H C O O ~ C - from malonic ester r ( 2 I
from RX’s-
the circled R’s are both CH,CH,. Therefore each a H is replaced sequentially by an ethyl group (R= R’ = Et), using CH,CH,Br.
CH2(COOC,H,),
+ Na’OC,H;
-
+C2HjBr
C2H,0H + [CH(C0OC2H,),1 y
:cz,:*
~~CH(COOC,H,),
(6) Only (CH,),CH-Br
CH2(C00C2H5)2
aq;z-*
~~IC(COOC,H,),
is needed for a single alkylation (see product below). I. 2.
NaOEt i-PrBr
’
CH,-C H -C H(COOC2H,),
-CH(COO-), (c)
[C,H,le(COOC2HS),Na’
H Oi
CH,-CH
aqiz-’
-CH(COOH),
‘ i CH,-CH
-CH,COOH
To obtain
Alkylate first with CH,CH,Br and then with CH,I. The larger R is introduced first to minimize steric hindrance in the second alkylation step.
CHAP. 171
393
CARBANION-ENOLATES AND ENOLS
CH,(COOEt),
m]CH(COOEt),
(d) Trialkylacetic acids cannot be prepared from malonic ester. The product prepared from malonic ester must have at least one Q H, which replaces the lost COOH. Problem 17.14
Prepare an alkyl succinic acid: @-CH-COOH
4
Alkylsuccinic acids are disubstituted acetic acids as shown above. Alkylate with RX and then with BrCH,COOC,H, to give
R-C(COOC,H,), AH,COOC,H,
-
Hydrolysis and acidification of this ester yields a tricarboxylic acid which is heated and loses CO, from one of the gem-COOH’s to form an alkylsuccinic acid.
R-C(COOC,H,),
R-C(COOH),
AH,COOC2H5 Problem 17.15 Use CH,CH,CH,OH valeramide, CH,CH,CH,CH,CONH, .
heat
+
R-CH-COOH
CH,COOH I
and H,C(COOC,H,),
&H,-COOH as the only organic reagents to synthesize 4
Since valeric acid is n-propylacetic acid,
e C H , C O O H CH,CH,CH,Br is used to alkylate malonic ester.
Problem 17.16 Synthesize cyclopentanecarboxylic acid from malonic ester.
H
>C(COOR), H’
CH,CH, \ I C(COOR),/ CH, I H CH,
I
Br
L
-
NaOEl ---+
4
CH,CH,Br
C( COOR), ______+
H’
NaObt
CH2CH2, 1 CH c ( c 0 0 R ) 2 j I ’/&a+ CH,
$‘
CH2CH2~C(COOR),-I
I
CH,CH,
’
NaOH
2 H30i
CARBANION-ENOLATES AND ENOLS
394
ACETOACETIC ESTER
(ME)
[CHAP. 17
SYNTHESIS
1. Carbanion Formation Acetoacetic ester is acidic (pK, = 10.2) and forms a resonance-stabilized carbanion whose negative charge is delocalized over one C and two 0 ’ s .
O
H
It I I
-
0
II
+
+ NaOC2H5# C2H50H+
CH3-C-C-C-OC2H5 H
2.
Alkylation As with malonic ester (Problem 17.13)’ either one or two R’s can be introduced in acetoacetic
-
ester.
R
I
-
ia(CH3COCHCOOC2H5)5 CH3COCHCOOC2H5
W2HS
R
R
I
I I R’
CH3COCCOOC2H5 3 CH3COCCOOC2H5
3. Hydrolysis and Decarboxylation
Dilute acid or base hydrolyzes the COOC,H, group and forms acetoacetic acids, which decarboxylate to methyl ketones.
O
H
O H
0
II I II CH3-C-C-C-OC2H5 I
3 C2H5OH + H O+
4
R
II I I
CH3-C-C-H
+ CO,
R
This sequence of steps can be used to synthesize methyl ketones. For a general methyl ketone, the parts are assembled as follows:
‘
from the ester
from
alkyl halides
The H* replaced COOEt. Problem 17.17 Prepare 3-methyl-2-pentanone from acetoacetic ester. In the product
H
4
CHAP. 17)
CARBANION-ENOLATES AND ENOLS
395
the - C H , and -CH,CH, attached t o m a r e introduced by alkylation of the carbanion of acetoacetic ester with appropriate alkyl halides; in this case use BrCH,CH, and then CH,I. CH,CH, H
-
n -
+
I
NaOEt
CH3CH2Br
CH,COEHCOOC,H,
CH3 I CH,CO COOC,H,-
1. hydrolysis
'iCH,CH,
I
CH,COCHCOOC,H,
1 N~OEI
2. CH3I
TH3
CH,COCH
I
CH,CH,
Methylethylacetoacetic ester
3-Methyl-2-pentanone
Problem 17.18 Prepare 2,5-hexanedione (a y-diketone) from acetoacetic ester and any alkylating reagent, 4
Monobromoacetone is the alkylating agent. 0
0 CH,-CO-CH,
0
II
CH,-C-CH,-COOEt-
NaOEt
II Na' CH,-C
0
0
L
II I
BrCH2-CO-CH3
-
* CH,-C-CH,
1 . NaOH
COOEt
2. H ~ O +
0
0
I II CH,-C-CH2CH,4-CH,
J
COOH
Problem 17.19 Use AAE (acetoacetic ester) synthesis to prepare the p-diketone 2,4hexanedione, CH,COCH,COCH,CH,. 4
The group attached to the a C of AAE is COCH,CH,. This acyl group is introduced with CH,CH2COCI. Because acyl halides react with ethanol, aprotic solvents are used. The carbanion is prepared with :H- from NaH. n n n -
~
-
' - NaCl
r
CH,-C-CH
L
0
0 \
-
1
CH,~-CH',
L
0
___I+
COOEt
2. H ~ O +
0
II
II
CH,-C-CH2-C4H,CH,
COOH
J
-
Acetoacetic ester is converted to a dianion by 2 moles of a very strong base. CH,COCH,COOC,H,
2 LDA
:CH,COCHCOOC,H, a dianion
When treated with 1 mol of 1" RX, the more basic terminal carbanion is alkylated, not the less basic nterior carbanion. The remaining carbanion-enolate can be protonated. Droblem 17.20 Devise a synthesis of C,H,CH,CH,COCH,COOH
from acetoacetic ester.
4
Since the terminal methyl group of acetoacetic ester is alkylated, its dianion is reacted with C,H,CH2CI.
[CHAP. 17
CARBANION-ENOLATES AND ENOLS
396
0
II
CH,-C--CH,COOEt
2 1.DA
0
It
-
z
:CH,-C--CH-COOEt
-
C,H,CH,CH,-C~HCOOEt
H30
+ C Ci CHZCl
C,H,CH2CH2-CO-CH,-COOH
+
17.4 NUCLEOPHILIC ADDITION TO CONJUGATED CARBONYL COMPOUNDS: MICHAEL 3,4=ADDITION
a,@-Unsaturated carbonyl compounds add nucleophiles at the @ C, leaving a - charge at the cy C. This intermediate is a stable carbanion-enolate. These 3,4-Michael additions compete with addition to the carbonyl group (1,2-addition; Section 15.4).
+ CN-
C H CH-CH,-C-C,H5
II
5~
CN
0
Michael addition product
Ph
C~H~CH=CH-C-C~HS
II
I
maior
4-
0
Ph
trace
Ph PhLi more
Ph
I
I
C ~ H ~ C H C H ~ C C-k ~C~HS-CH=CH-CC~H~ HS
A*
0 II
ionic
minor
major
3,4-product
1.2-product
Stable carbanion-enolates can also serve as the nucleophile. Problem 17.21 ( a ) Show the nucleophilic addition of the malonate carbanion to methyl vinyl ketone and the 4 formation of the final keto product. (b) Show how this product can be converted to a 6-ketoacid.
0
(a)
CII .A *? CH,-C-CH==CH, + :CH(COOR), OH
-
1
CH,-C=CH-CH,-CH(COOR), 0
II
(6) CH,-C--CH,CH,--CH(COOR),
I
CH,-C=CHCH,--CH(COOR),
I
enol
-0 0
Na0H
2
H,O
'
HOE1
II
CH,--C-CH,CH,--CH(COOR),
1
-
keto
0
II
1- CO2
CH,--C-CH,CH,-CH(COOH), 0
II
CH ,*CH~CH
,CH ,COOH
5-Oxohexanoic acid
397
CARBANION-ENOLATES AND ENOLS
CHAP. 171
Problem 17.22 Cyanoethylation is the replacement of an acidic a H of a carbonyl compound by a 4 --CH,CH,CN group, using acrylonitrile (CH,=CHCN) and base. Illustrate with cyclohexanone.
17.5 CONDENSATIONS
A condensation reaction leads to a product with a new C--C bond. Most often the new bond results from a nucleophilic addition of a reasonably stable carbanion-enolate to the C=O group (acceptor) of an aldehyde; less frequently the C=O group belongs to a ketone or acid derivative. Another acceptor is the C==N group of a nitrile. 0-
0
0
"-&1 1
carbanion-enolate
new bond
acceptor
ALDOL CONDENSATION The addition of the nucleophilic carbanion-enolate, usually of an aldehyde, to the C=O group of its parent compound is called an aldol condensation. The product is a P-hydroxycarbonyl compound. In a mixed aldol condensation the carbanion-enolate of an aldehyde or ketone adds to the C=O group of a molecule other than its parent. The more general condensation diagramed above is termed an aldol-type condensation. Since the C, not the 0, is the more reactive site in the hybrid, the enolate Zontributing structure is usually omitted when writing equations for these reactions. This is done x e n though the enolate is the more stable and makes the major contribution. Net Reactions
I
H O Propanal
II
2-Methyl-3-hydroxypentanal
I I
H OH Acetone
Diacetone alcohol (4-Hydroxy4me th yl-2-pentanone)
Aldol condensations are reversible, and with ketones the equilibrium is unfavorable for the :ondensation product. To effect condensations of ketones, the product is continuously removed from he basic catalyst. p -Hydroxycarbonyl compounds are readily dehydrated to give a,/3 -unsaturated marbony1 compounds. With Ar on the /3 carbon, only the dehydrated product is isolated. 'roblem 17.23 Suggest a mechanism for the OH--catalyzed aldol condensation of acetaldehyde.
4
398
(CHAP. 17
CARBANION-ENOLATES AND ENOLS
stop 1
.D
H-0:
H
H
J
+ H:CHI
C="-
I__ H 2 0 + :(f-C=O:
H
H H
-
stop 2 alkoxide ion of the @-hydrox yaldehyde
HH H
I I
Stop 3
I
H H ry
+ OH-
CH3-&-A-C=0
HA 17.24 Aldehydes and ketones also undergo acid-cat alyzed aldol condensations. Devise a mechanism 4 for this reaction in which an enol is an intermediate. OH
protonatcd carbonyl compound (elcctrophlc)
OH
enol (nucleophile)
H-C-
aldol
I I I I
Pl0M.m 17.25 Write structural formulas for the p- hydroxycarbonyl compounds and their dehydration products formed by aldol condensations of: ( a ) butanal, ( 6 ) phenylacetaldehyde, (c) diethyl ketone, ( d ) 4 cyclohexanone , (e) benzaldehyde. Accsptor
Carbanion Source
H
H
H
AH
A2H5
+CH3CH2CH2C-&-(!%0 I
(U)
3 H
A
CH3CH2CH2 =C-CH=O mixture of geometric isomers
Acceptor H
Carbanion Source
H
H
-
H
H
PhC€12&-&-&=0
AH bh
H
3PhCH, 'C=C €/
FHO b h
399
CARBANION-ENOLATES AND ENOLS
CHAP. 171
CH,CH,
-
CH,CH, CH3CH2CI
H
(c) CH3CH2{F>-‘C fi HC , H,
0
I
-H20
C-C-CH,CH,
bH
H3 0
-
H
8
kH3
CH,CH,
I
CH3CH,C=C-CCH,CH3 AH3
I !
0
( e ) No aldol condensation, because C,H,CHO has no a H. Aldol condensations require dilute NaOH at room temperatures. With concentrated NaOH at higher temperatures, C,H,CHO undergoes the Cannizzaro reaction (Problem 15.20). Problem 17.26 Some condensations have little synthetic value because they give mixtures of P-hydroxycarbonyl compounds, Illustrate with ( a ) a mixture of two aldehydes, each with an a H; (b) an unsymmetrical 4 ketone having an H on each a C . ( a ) There are four possible products. Each aldehyde reacts with itself to give two products. If one aldehyde
reacts as carbanion and the other as acceptor, and vice versa, there are two more products. Acceptor Carbanion Source
H
‘ RCH2{Fi-C=0
1
H
R ’ C H , i F ! - - - C R’ =I O HI
I
+ R’CH,C-CbH
H __*
1
I
H
I
A H k
H
Mixed aldol
H
+RCH,C-C-C=O
H
0
Selfaldol
1
H
H
H
H
d’I
H
H
bH
IL
H
H
I I RCH,C-C-C=O
H C=O
A
( 6 ) Two products are possible. There is only one acceptor, but two different carbanions can be formed-one
from each a C.
Acceptor
Carbanion Source
RICH, H R C H , { F ~ - ( ( - IC H , R ~ 0
0
-
R‘CH, H RCH,C-C-C-CH2R’ I I b H k
8
CARBANION-ENOLATES AND ENOLS
400
-
RCH, H R C H 2 - { F C - C - C IH 2 R
It
I
R‘
0
[CHAP. 17
R’CH, H RCH,C-C-C-CH,R I I
I
II
I
OH R
0
0
Problem 17.27 Mixed aldol condensations are useful if ( a ) one of the two aldehydes has no a H, (b) a symmetrical ketone reacts with RCHO. Explain and illustrate. 4 ( a ) The aldehyde with n o a H, e.g., H 2 C 0 and C,H,CHO, is only a carbanion acceptor, so that only two
products are possible.
Acceptor
Carbanion Source
-
H
H
R ~ - - H - ~ ~ , ~ ~ ~ - ~ - - -R’CH,-C-C-C-H - H I
Selfaldol
H
I
(!!)
bH
For reasonably good yields of mixed aldol product, the aldehyde with the a H should be added slowly to a large amount of the one with no a H. ( b ) Ketones are poor carbanion acceptors but are carbanion sources. With symmetrical ketones and an RCHO having an a H, two products can be formed: (1) the self-aldol of RCHO and (2) the mixed aldol. If RCHO has no a H, only the mixed aldol results. As in part ( a ) , the correct sequence of addition can give a good yield of the mixed aldol products. Problem 17.28 Give the product for the reaction of C,HSCH=O in dilute base with ( a ) CH,CHO, (6) CH ,COCH, . 4
H (a)
I
(!)I
C6H5C-CH2C-H bH
H PhCH-CHCHO
(b)
+
PhCH=CHCCH,
II
0 tram
tram
Problem 17.29 Show how the following compounds are made from CH,CH,CHO. Do not repeat the synthesis of any compound needed in ensuing syntheses. (a)
CH ,CH,CH==C( CH,)CHO
(d) CH,CH,CH,CH(CH,)CH,OH
(6) CH ,CH,CH,CH(CH,)CHO (e)
CH,CH,CH,CH(CH,),
(c) CH,CH,CH=C( CH,)CH,OH
(f)
CH,CH,CHCH(CH,)COOH
I
OH
4
Each product has six C’s, which is twice the number of C’s in CH,CH,CHO. This suggests an aldol condensation as the first step.
(4
CH,CH,CHO OH-\ CH,CH,CHCHCHO
: :A
CH,CH,CH=C(CH,)CHO
(A)
HA LH, (6) - C H O can be protected by acetal formation to prevent its reduction when reducing C=C of (A).
CHAP. 171
(A)
C H OH
CARBANION-ENOLATES AND ENOLS
CH,CH,CH=C( CH,)CH( OCH,),
2CH,CH,CH,CH(
401
~,-d HCI
CH,)CH( OCH,),
CH,CH,CH,CH( CH,)CHO
-
(There are specific catalysts that permit reduction only of C=C.) ( c ) CHO is selectively reduced by NaBH,. (A)
(4 (4
NaBH4 H2/ Pt
(A)(A)T
HZNNH2. O H -
CH,CH,CH==C(CH,)CH,OH
CH,CH,CH,CH( CH,)CH,OH
CH,CH,CH=C(CH,),
CH,CH,CH,CH(CH,),
(f) Tollens' reagent, Ag(NH,),', is a specific oxidant for CHO-COOH. 1 Ag(NHj)t
CH,CH,CH( OH)CH( CH,)CHO 7 CH,CH,CH( OH)CH( CH,)COOH r
P
Problem 17.30 Crotonaldehyde (CH,CH
Crotonaldehyde has C==C conjugated with C=O. On removal of the y H by base, the delocalized to 0.
- charge on C is
W
The nucleophilic carbanion adds to the carbonyl group of acetaldehyde.
H
I
H H H H
H H H H H
I I I I
H O
-H 0 A product
CH,-
I
H Problem 17.31 Use aldol condensations to synthesize the following useful compounds from cheap and readily available compounds: ( a ) the food preservative sorbic acid, CH,CH=CH-CH=CH-COOH; (6) 2-ethyl-lhexanol; (c) 2-ethyl-l,3-hexanediol, an insect repellant; (d) the humectant pentaerythritol, C( CH,OH), . 4 (U)
2 CH,CH=O-
OH-H20
CH,CH=CHCH=O
= CH3CH0
mild
CH,CH=CHCH=CHCH=O-
( b ) CH,CH,CH,CH,OH
aCH,CH,CH,CH=O
oxid.
CH,(CH=CH),COOH
C2H5
I
OH-* CH3CH2CH2 H-CHCH-0
*
-H20
EH C2Hs
1
CH,CH,CH,CH=C-CH=O [ c ) As in
C*H5
I
CH,CH,CH,CH,-CH-CH,OH
(6) to
T2Hs I
CH,CH,CH, H-CH-CH-0
6,
CH,CH,CH,
I
402
CARBANION-ENOLATES AND ENOLS
[CHAP. 17
One mole of CH,CHO undergoes aldol condensation with 3 mol of H,CO. A fourth mole of H,CO then reacts with the product by a crossed-Cannizzaro reaction.
CH,OH C.(OH)2
CH,OH ,;2TH20H I
CHyO
HOCH2- b H -CH=O 20H
OH-
+H
r -
HOCH2-
Problem 17.32 Which of the following alkanes can be synthesized from a self-aldol condensation product of an aldehyde [see Problem 17.31(a)] or a symmetrical ketone? (a) CH,CH,CH,CH,CH(CH,)CH,CH,CH,,(6) CH,CH,CH,CH,CH( CH,)CH,CH,, ( c ) (CH,),CHCH,CH,CH,, (4 (CH,),CHCH,C(CH,), (e)
+
4
(CH,),CHCH,CH,CH,CH,CH,,( f ) (CH,CH,),CHCH(CH,)CH2CH2CH,. The general formula for the aldol product from RR’CHCHO is
R
CH-C-C-CHO
The arrow points to the formed bond, and the a and C=O C’s are in the rectangle. The alkane is
There is always a terminal CH, in this four-carbon sequence. From RR‘CHCOCHRR’ the products are OH
11
R’ R’
I I RCHC-CCCHR I \ Ill
__*
R’RCH RO I
I R’
R’
Each half must have the same skeleton of C’s. Note that R and/or R’ can also be Ar or H. The alkane must always have an even number of C’s (twice the number of C’s of the carbonyl compound). ( a ) No. There is an odd number of C’s in the alkane.
Yes. The four-carbon sequence has a terminal CH,, and each half has the same sequence of C’s. Use RR‘CHCHO, where R = H and R’ = CH,CH,.
CH,CH,CH,CHO
CH,CH,CH, H HCHO H I !H,CH,
-
alkane
CH3
1 1
CH,-CH-CH,CH,CH, Yes. Each half has the same skeleton of C’s. A ketone is needed; R = R’ = H.
CARBANION-ENOLATES AND ENOLS
CHAP. 171
1,
(CH3),C=0 OH-_ (CH,), -CH, l-CH3
403
-
alkane
(4 Yes. Use RR’CHCHO, R = R’ = CH,.
#-g
CH, H CH,
-i;FHO-
(CH,),CHCHO OH-_ CH3-
alkane
H H
No. The formed bond is not part of a four-carbon sequence with a terminal CH,, and the two halves do not have the same skeleton of C’s; one half is branched, the other half is not. CH,CH,
(f)
f
!--CH,CH,CH,
&H,CH, bH3 Yes. Each half has the same skeleton of C’s (5C’s in a row). Therefore, use a symmetrical ketone with 5 C’s in a row (R = H, R’ = CH,). OH
0
II
-
H O
I
II
I-C-CH2CH,
CH,CH,CCH,CH, OH;c * $ H j H2CH3
alkane
H,
Problem 17.33 Give the structural formulas for the products of the aldol-type condensations indicated in Table 17-2. See Table 17-1(a)-( g ) for the carbanions of these condensations. 4
Acceptor
PhCHO
PhCHO
Me,CO
Me,CO
Me,CO
OH-
OH-
OH-
NH,
OH-
Base
Ph,CO NHj
NHR,
r
H
(a) *PhCH=CHNO,
(6) *PhCH=CHCN
H
(e)
CMe,
( f ) Ph,C=CPh,
‘The more stable tram product.
I
Ph,CH,
(4 Me C - C C l , AH
(g) *PhCH=CH-
PhCHO
NO* ( d ) Me,C-C=CCH,
bH
CARBANION-ENOLATES AND ENOLS
404
[CHAP. 17
Problem 17.34 Give structures of the products from the following condensations: (U)
(c) (e)
p-CH,C,H,CHO
7
+ (CH,CH,C),O
(6) Cyclohexanone + CH,CH2N0,
CH3CH2COO- Na')
+ C,H,CH,C%Nz OHCH,COCH, + 2C,H,CHOC,H,CHO
(d)
c H 3coo -
OH-
NH P
4
This is a Perkin condensation.
H O
H
H H O
0
H
*
H
I
H
I
-CN
II
C6H5
L I/
C \c,H5
(bulky C,H,'s are rram)
r
L
]-H20
0
II
I
I
.c6H'!L!6H5
(e)
OH-
Benzophenone + Cyclopentadiene
+ N=LCH,COOCH, (U)
-
1
OH
1
1
Diphenylfulvene
Each CH, of (CH,),CO reacts with one PhCHO.
[
H
1
I
Ph CH, CH, -Ph EH
( f ) This is the Cope reaction.
FH
-2H20
PhCH-CHCCH-CHPh
II
0
Problem 17.35 In the Knoevenagel reaction, aldehydes or ketones condense with compounds having a reactive CH, between two C=O groups. The cocatalysts are borh a weak base (RCOO-) and a weak acid (R,NHl). 4 Outline the reaction between C,H,CH=O and H,C(COOEt),.
CHAP. 171
1-
405
CARBANION-ENOLATES AND ENOLS
H
COOC,H,
C6H5-c I + H0 1I
f
-H
H _____* MC~NH;
CHjCOO-
COOC,H,
1 6 H 5 - 1I F 1 - 1H
OOC,H,
OOC,H,
Ethyl malonate
Problem 17.36 Prepare tram-cinnamic acid, C,H,CH--LHCOOH, 17.34(a ) ] ,
PhCHO
+ (CH,CO),O
Benzaldehyde
Acetic anhydride
COOC,H, C,H,-CH=C
I
I
COOC,H,
by a Perkin condensation [Problem 4
Nz* PhCH-CHCOOH
Dehydration of the P-hydroxyester occurs on workup because the resulting C=C is conjugated with Ph. Problem 17.37 The C of the a N group is an electrophilic site capable of being attacked by a carbanion. Show how nitriles like C H , C H , e N undergo an aldol-type condensation (Thorpe reaction) with hindered bases. 4
LiOH
+ CH,CH,
c
CH3
I
-CH-CN
NH
-
CH3
I
H 0'
-CHCN
RT
+ NH,
cH3cH2!
an iminonitrile
CLAISEN CONDENSATION: ACYLATION OF CARBANION-ENOLATES
In a Claisen condensation the carbanion-enolate of an ester adds to the C=O group of its parent ester. The addition is followed by loss of the OR group of the ester to give a P-ketoester. In a mixed Claisen condensation the carbanion-enolate adds to the C=O group of a molecule other than its parent.
-
Step 1 Formation of stabilized a-carbanion.
RCH,C-OR'
II
8:
+
-OR'
-ROH
- 1
Step 2 Nucleophilic attack by a-carbanion on C=O of ester and displacement of -OR'
n + RCH2C-OR';==--
RCHCOOR'
-RO-
RCH2-f-THCOOR O R
CARBANION-ENOLATES AND ENOLS
406
--
[CHAP. 17
This step is reminiscent of a transacylation (Section 16.6). Step 3 The only irreversible step completes the reaction by forming a stable carbanion where negative charge is delocalized to both 0's.
+ -OR'
RCH2f-YH-COR O R
8
-
-
-R'OH
-
a 8-ketoester
Acid is then added to neutralize the carbanion salt.
-
Na+
-CRCOOR']
5 RCH&CHRCOOR' + Na'Cl-
CI
IRCH2fi0 Na' salt if NaOR is used as base
a 8-ketoester
Problem 17.38 Write structural formulas for the products from the reaction of C,H,O-Na' following esters: ( a ) CH,CH,COOC2H,
with the
(6) C,HsCOOC2Hs + CH3COOC2H, ( c ) C6HSCH2COOC2HS+ O=C(OC2H,),
4
In these Claisen condensations -OC2H, is displaced from the COOC,H, group by the a-carbanion formed from another ester molecule. Mixed Claisen condensations are feasible only if one of the esters has no a H.
H O
0
I II CH3CHCIC;-i'Tl;3C-OC.H, II
(a 1
2
O
H
A -C H 0- CH3CH2-C-
-C--OC2Hs
F
CH3
s
0
0
II I II
H
0
(has no a H)
0
1I
H O
I I1
C,HsO-CTTq-C-OC2Hs k2HS
O -C
H
1I I
H 0-
0
II
A C2HsO-C-~-C--OC2Hs I C A
k6HS
(has no a H)
a substituted malonic ester
Problem 17.39 Ethyl pimelate, C,H,OOC( CH,),COOC,H,, reacts with C,H,O- Na' (Dieckmann condensation) to form a cyclic keto ester, C,H,,O,. Supply a mechanism for its formation and compare the yields in 4 ethanol and ether as solvents. We have: 0
H
I
EtOOC-C-H y
I
II
C-OEt \ CH2 +etO-_ EtOH
I
2 ,CH2 CH2
Ethyl pimelate
(7
C-OEt
=2\
+ EtOOC-CH CH2 I I c\H* ,CH2 CH2
407
CARBANION-ENOLATES AND ENOLS
CHAP. 17)
0
II
EtOOC-CH
CH2
I
I
CH CH2 \ 2/ CH2
-
-
2-carbetJloxycyclohexanone
The net reaction is C2H,00C(CH2),COOC2H, product + HOC2H,. Since the reaction is reversible, yields are greater in ether than in alcohol because alcohol is a product (Le Chatelier principle).
Intramolecular Claisen cyclizations occur with ethyl adipate and pimelate because five- and six-membered rings are formed.
Supplementary Problems Problem 17.40 Account for the fact that tricyanomethane, (CN),CH, is a strong acid (KO= 1).
The conjugate base, (CN),C:-, is extremely weak because its negative charge is delocalized, by extended bonding, to the N of each CN group. Hence the acid is strong.
4 7r
Problem 17.41 Acetone reacts with LDA in THF and then with trimethylsilyl chloride, (CH,),SiCl, at -78 “C, to give an enolsilane. (a) Give equations for the reactions. (b) Why does 0- rather than C-silylation occur?
b
4
[
CH,C=CH2 *
( a ) CH,CCH,LDA
(b) The 0-Si bond is much stronger than the C-Si Si(d) atoms.
bond because of p-d
7~
bonding between the 0 ( p ) and
Problem 17.42 Enolsilanes (Problem 17.41) are stable in base but are hydrolyzed in aqueous acid to reform the carbonyl compound. Explain how this chemistry can be used to “protect” a carbonyl group (prevent its reaction) when it is desired to react a functional group elsewhere in the carbonyl compound under basic 4 :ondi tions.
Form the enolsilane, carry out the reaction on the other functional group (e.g., replace a C1- by an OH-) and then hydrolyze the silyl ether in aqueous HCI. Problem 17.43 Show how acetone can be converted into 4-phenyl-2-butanone using enamine alkylation. 4
0
cH3--fi--CH3+ 0
I
H
pc1+ N=c
CH3
408
CARBANION-ENOLATES AND ENOLS
Problem 17.44 Prepare 4-methyl-1-hepten-5-one from (CH,CH,),C=O
[CHAP. 17
and other needed compounds.
4
0
has an ally1 group substituted on the a C of diethyl ketone. This substitution is best achieved through the enamine reaction. CH,CH?-C=O CH3CH1-C=0
I
'ZH5
+
n
3
h
/
HN
U
__*
CH3CLC=N
1I
1I
C2H5
3 U
ClCH2CHwCHz
___I__*
an enamine
Problem 17.45 Prepare 3-phenylpropenoic acid from malonic ester and C,H,CH,Br.
C,H,CH,CH,COOH
Br?/ P
C,H,CH,CH--COOH Jr
alc K O H
4
C,H,CH=CH-COOH Cinnamic acid
Problem 17.46 Can 3,3-dimethyl-2-butanone (pinacolone), CH,COC(CH,),, be synthesized from acetoacetic 4
ester?
No. Pinacolone has a trimethylated a C. Acetoacetic ester is used for preparing mono- or dialkylated methyl ketones. The compound is made by a pinacol-pinacolone rearrangement of (CH,),COHCOH( CH,), . Problem 17.47 Can the following ketones be prepared by the acetoacetic ester synthesis? Explain. (a) (6) CH,COCH,C( CH,),. 4
CH,COCH,C,H,,
(a) No. C,H,Br is aromatic and does not react in an SN2displacement. (6) No. BrC(CH,), is a 3" bromide which undergoes elimination rather than substitution. Problem 17.48 Use acetoacetic ester (AAE) and any needed alkyl halide or dihalide to prepare: (a) CH,COCH,CH,COCH,, (b) cyclobutyl methyl ketone, (c) CH,COCH2CH,CH,COCH,, and (d) 1,3-diacetyIcyclopentane. 4
The portion of all these compounds that comes from
AAE
is
H CH
C-
AI -
Encircle this portion and obtain the rest of the molecule from the alkyl halide(s)
CHAP. 171
409
CARBANION-ENOLATES AND ENOLS
Bond two molecules of
at the acidic CH, group of each with NaOEt and I,.
AAE
0
II
i: tzNaoEt* CH,C
2CH,COCH2COOEt
HCOOEt
:,'t:o".-- CH,CCH,CH,CCH, b oII +
+ CO,
We need three C's in the halogen compound, and the two terminal C's are bonded to the acidic CH, group of AAE. The halide is BrCH,CH,CH,Br.
COOEt
I /H
CHJC-C
0 I'
+
H '
COOEt
BrCH2 \ /CH2 % BrCH2
Join two molecules of BrCH,Br.
AAE
through the acidic CH, of each with two molecules of NaOEt and 1 molecule of
0 COOEt
& A d
CH3
H
H
CH,CO-kH
2NaOE1
CH3C-C
8
COOEt
___*
CH3CO-
kOOEt
CH,CO-kH
k H 2
I.
NaOH
CH3CO-
kOOEt
2 H 2 + 2C02 + 2EtOH
I
H
0
H$c*
Two molecules of AAE are first bonded together with one molecule of BrCH,CH,Br and then the ring is closed with BrCH,Br.
P
0 COOEt
CH,C-C?TOoEt
CH+!-M CH2
I
C HII 3 IC - F d 2 0 COOEt
2NaOEt ___,
I. 2.
CH3C-CH 0 &OOEt
NaOE1 CH~B~~'
e
410
[CHAP. 17
CARBANION-ENOLATES AND ENOLS
Problem 17.49 The Robinson “annelation” reaction for synthesizing fused rings uses Michael addition followed by intramolecular aldol condensation. Illustrate with cyclohexanone and methyl vinyl ketone, . I CH,=CHCOCH,.
Problem 17.50 Prepare
(t
CH,
I
0,
C H=C-N
from simple acyclic compounds.
The ‘C(OCH,),
/
4
is an electron-withdrawing group which activates the dienophile.
Problem 17.51 Give the ester or combination of esters needed to prepare the following by a Claisen condensation.
(a) C6H5CH2CH2C-CHCOOEt
b
(6) EtOC-C-CHCOOEt
88
(!X2C,H,
( c ) H-C-CHCOOEt
LH3
. I
L6H5
In the Claisen condensation the bond formed is between the carbonyl C and the C that is a to COOR. Work backwards by breaking this C--C bond and adding OR to the carbonyl C and adding H to the other C. Mixed Claisens are practical if one ester has no a H.
0
0
(a)
C,H,CH,CH,-C+
II
HCOOEt
C,H,(CHJ,-C
+H
HCOOEt
l1QH2c6H5
t
&1H2c6H5
( b1
W2H5
Ethyl 3-phenylpropanoate
-C+CHCOOEt /&H3
4
i.
EtSC-C
b
OEt
+H
CH-COOEt
bokH3
Ethyl oxalate
Ethyl propanoate
CHAP. 171
411
CARBANION-ENOLATES AND ENOLS
0
It
H-C-kHCOOEt
+
6.
--"" H - C A
: :? N E
Ethyl formate
Ethyl phenylacetate
Problem 17.52 In the biochemical conversion of the sugar glucose to ethanol (alcoholic fermentation) a key step is
H OH OH H203POCH2-~-~-F-~-CH20P0,H, I l l H
OH
I
H
Fructose 1,6diphosphate
Dihydroxyacetone phosphate (11)
(1)
Glyceraldehyde 3-phosphate (111)
Formulate this reaction as a reversal of an aldol condensation (retroaldol condensation).
4
(I) is a p-hydroxyketone. Loss of a proton from the C ' 4 H affords an alkoxide (IV) that undergoes a retroaldol condensation by cleavage of the C - C ' bond. H la
POCH2C-C-
Oj OH
I
-CCH20P03H2
!IAX$B
H
4-+
H203POCH2C-
:
!I b H
(111)
Chapter 18 Amines 18.1 NOMENCLATURE AND PHYSICAL PROPERTIES
Amines are alkyl o r aryl derivatives of NH,. Replacing one, two, or three H's of NH, gives primary (l"),secondary (2") and tertiary (3") amines, respectively.
(
R' RNH2
(e.g. CH3NH2)
R3N
RNH
NH3
R
R
ArNH
ArNR'
Ammonia
ArNH2 (C6HSNH2)
2" amine
1" amine
CH3 )
CH,NCH,
3" amine
Amines are named by adding the suffix-amine to the name of ( a ) the alkyl group attached to N or (b) the longest alkane chain. The terminal e in the name of the parent alkane is dropped when "amine" follows but not when, for example, "diamine" follows [see Problem 18.l(d)]. Thus, CH,CH(NH2)CH2CH, is named sec-butylamine or 2-butanamine. Amines, especially with other functional groups, are named by considering amino, N-alkylamino and N ,N-dialkylamino as substituents on the parent molecule; N indicates substitution on nitrogen. Aromatic and cyclic amines often have common names such as aniline (benzenamine), C,H,NH,; p-toluidine, p-CH,C,H,NH,; and piperidine [Problem 18.1( g)]. Like the oxa method for naming ethers (Problem 14.61), the aza method is used for amines. Di-n-propylamine, CH,CH,CH,NHCH,CH,CH, , is 4-azaheptane and piperidine is azacyclohexane. The four H's of NH: can be replaced to give a quaternary (4") tetraalkyl (tetraaryl) ammonium ion. CH3
I
C,HSCH2N+CH3 OH-
I
(Triton-B)
CH3 is benzyltrimethylammonium hydroxide. Problem 18.1 Name and classify the following amines:
(4
(CH,),CNH,
(')
( d ) H,NCH,CH,CH,NH,
(CH3)2NCH(CH3)2
( e ) CH,NHCH( CH,)CH,CH,
(g)
LJ
(')
C6H5N(CH3)2
( f ) CH3NHCH2CH,NHCH,
( h ) CH,CH,CH,N(CH,)fCl-
4
N
I
CH3 ( a ) t-butylamine or 2-methyl-2-propanamine, 1"; (6) dimethylisopropylamine, 3"; ( c ) N,N-dimethylaniline, 3"; ( d ) 1,3-~ropanediamine (or trimethylenediamine), both 1"; (e) 2-(N-methylamino)butane, 2"; ( f ) 2 3 diazahexane, both 2"; ( g ) 3-amino-N-methylpiperidineor l-methyl-3-amino-l-azacyclohexane, 1" (N of NH,) and 3"; ( h ) n-propyltrimethylammonium chloride, 4".
412
413
AMINES
CHAP. 181
Problem 18.2 Give names for (a) CH,NHCH,, (6) CH,NHCH(CH,),, (c) CH,CH,CH(NH,)COOH,
'N(CH,),Br-
NHCH,
NHCH~ ( a ) dimethylamine, (6) methylisopropylamine or 2-(N-methylamino)propane, (c) 2-aminobutanoic acid, (d) N-methyl-rn-toluidine or 3-(N-methylamino)toluene, (e) trimethylanilinium bromide, ( f ) 3,4'-N ,N'methylaminobiphenyl (note the use of N and N' to designate the different N's on the separate rings).
Problem 18.3 Predict the orders of ( a ) boiling points, and (6) solubilities in water, for 1"' 2", and 3" amines of identical molecular weights. 4
Both physical properties depend on the ability of the amino group to form H-bonds. (a)
Intermolecular H-bonding, as shown in Fig. 18-1 with four molecules of RNH,, influences the boiling point .
I
H I
I I
I
I
H I 1 I R-N-H
H I
/
H
Fig. 18-1
The more H's on N, the greater is the extent of H-bonding, the greater is the intermolecular attraction and the higher is the boiling point. A 1" amine, with two H's, can crosslink as in Fig. 18-1; a single amine molecule can H-bond with three other molecules. A 2" amine molecule can H-bond only with two other molecules. A 3" amine, has no H's on N and cannot form intermolecular H-bonds. The decreasing order of boiling points is RNH, (1") > R,NH (2") > R,N (3"). ( b ) Water solubility depends on H-bonding between the amine and H,O. Either the H of H,O bonds with the N of the amine or the H on N bonds with the 0 of H,O: -N
I - - - H-0-H I
or
-N-HI
- - - OH,
All three kinds of amines exhibit the first type of H-bonding, which is thus a constant factor. The more H's on N, the more extensive is the second kind of H-bond and the more soluble is the amine. Thus the order of water solubility is RNH, (1") > R,NH (2") > R,N (3"). Problem 18.4 Does n-propylamine or 1-propanol have the higher boiling point?
4
Since N is less electronegative than 0 (3.1 < 3 . 3 , amines form weaker H-bonds than do alcohols of similar molecular weights. The less effective intermolecular attraction causes the amine to have a lower boiling point than the alcohol (49 "C < 97 "C).
AMINES
414
[CHAP. I8
18.2 PREPARATION
BY NUCLEOPHIUC DISPLACEMENTS 1. Alkylation of NH,, RNH, and R,NH with RX RX + NH, __* RNH 3 X -
Step 1
(an sN2 reaction)
an ammonium salt
RNHiX-
WP2 Di-, tri-, and tetraalkylation:
+ NH,+RNH,
RX
RX
RNH,-R,NH-R,N-R,N+X-HX -HX 2"
1"
+ NHaXRX
3"
-HX
4"
ammonium salt
When RX = MeI, the sequence is called exhaustive methylation. Problem 18.5 ( a ) What complication arises in the synthesis of a 1" amine from the reaction of RX and NH,? (6) How is this complication avoided? ( c ) Which halides cannot be used in this synthesis? 4 ( a ) Overalkylation: di- and tri-alkylation can occur since both RNH, and R,NH, once formed, can react with more RX. (6) Use a large excess of NH, to increase the probability of RX colliding with NH, to give RNH,, rather than with RNH, to give R,NH. ( c ) 3" RX undergoes elimination rather than SN2displacement. Aryl halides do not undergo S,2 reactions. Some ArX compounds with properly substituted rings undergo nucleophilic aromatic substitution (Section 11.3) with amines.
Problem 18.6 Describe an efficient industrial preparation of allylamine, H,G=CHCH,NH,, from propene, 4 chlorine, and ammonia.
Propene is readily monochlorinated to ally1 chloride, and the reactive C1 undergoes nucleophilic substitution with NH,. H,C=CH-CH,
C'2
H , C = C H - C H , C l s H,C=CH-CH,NH,
2. Alkylation of Imides; Gabriel synthesis of 1" Amines
COO-K'
+ RNHZ COO- K' Phthalimide
Phthalimide anion
Phthalhydrazide
See Problem 16.50 for the acidity of imides.
CHAP. 181
415
AMINES
REDUCTION OF N-CONTAINING COMPOUNDS 1. Nitro Compounds I . Zn, HCI; 2. OHor H2/Pt
C6HSN02
N itrobenzene
NO,
'
C6HSNH2
Aniline
NO,
m-Dinitrobenzene
rn-Nitroaniline
-
LAlH
RCN 4 RCH2NH2
2. Nitriles
RC-NR2
3. Amides
I
0
UAIH,
RCH2NR2 (R
=
H, alkyl, aryl)
NOH
0
NH,
4. Oximes Cyclohex ylamine
Oxime
RX + N;
5. Azides
-X-R
LiAlH o r H,IPI
N
or Na. EIOH
, RNH, ~
alkyl azide
a i d e ion
1" amine
This method is superior to the reaction of NH, and RX for the preparation of RNH, because no polyalkylation occurs. However, alkyl azides are explosive and must be carefully handled. Rather than being isolated, they should be kept in solution and used as soon as made. 6. Reductive Amination of Carbonyl Compounds
CH,CH=O-
NH
H /NI
H20
[CH,CH=NH]-
CH,CH,NH2
(NH,
an imine
-
1" amine)
H /Ni (1" amine 2"amine) + CH,CH,NH, 4 CH,CH,CH,NHCH,CH, + -OH (2"amine- 3" amine) R,C=O + R;NH- H 2 / N i [R,C=NR;]R,CH-NR; RNH, + 2 H,C=O + 2HCOOHRN(CH,), + 2 H 2 0 + 2C0, (Dimethylation of 1" amine)
CH,CH,CHO
-P
HOFMANN DEGRADATION OF AMIDES
RCONH,
+ Br, + 4KOH+
RNH,
Mechanism: Step 1
R-i-NH,
+ K,CO, + 2KBr + 2H,O
OBr-
(Brz. OH3
0
R-f-N-Br
I
O H N-bromoamide
(The amine has one less C than the amide.)
+ OH-
416
AMINES
H Step 2
R-C-N-BrI
II
+ OH-
0
-
[CHAP. 18
-
[R-t-kBr
CO:
+ H20 ..
-
N-bromoamide anion
Step 3 (electrondeficient N
Step 4
20H-
Step 5
-
+ R-N=C=O
“20
R-NH2
Problem 18.7 Synthesize 2-phenylethanamine, PhCH,CH,NH,, reduction method.
PhCH=CH,
tlBr
peroxide
PhCH,CH,Br-
NaN,
+ CO:-
(hydrolysis)
1” amine
from styrene, P h C H d H , , by the azide
PhCH,CH,N,
4
Na
PhCH,CH,NH,
Problem 18.8 Prepare ethylamine by ( a ) Gabriel synthesis, (b) alkyl halide amination, ( c ) nitrile reduction, (d) reductive amination, (e) Hofmann degradation. 4
0
0
II
(a)
+ C,H,NH,
a ) N K’:
0
0
(4)
COOH
II
II
CH,CH,CONH,
CH,CH,NH,
Problem 18.9 Prepare p-toluidine from toluene.
4
CH3 I
This is the best way of substituting an NH, on a phenyl ring.
-
Problem 18.10 Prepare CH,NHCH,CH, from compounds with one or two C’s.
CH,CHO
+ CH,NH,
ti?) N I
CH,CH,NHCH,
CHAP. 18)
417
AMINES
-
Problem 18.11 Prepare C,H,N(CH,),
from C,H,NH,.
2CH O H
C,H,NH,
4
C,H,N(CH,),
(Industrial method)
With CHJ, C,H,N(CH,)fI- is a by-product. H,CO and HCOOH cannot be used because the intermediate electrophile, H,COH, can attack the activated ring to give p-H,NC,H,CH,OH. Problem 18.12 Synthesize the following compounds from n-C ,H,,COOH and inorganic reagents:
(4 (d)
C,,H,,NH, NH2
I
1 2H2S--CH--C
I3 H 2 7
(4 Cl 2H25NH2
(b)
C13H27NH2
(e)
C12H2SNHC13H27
(f)
(C13H27)2NH
Do not repeat preparation of any needed compound. First, note the change, if any, in carbon content.
-
(a) Chain length is increased by one C by reducing RCH,CN (R = n-C,,H,,), prepared from RCH,Br and CN-. n-C,,H,,COOH-
LiAIHJ
PBr
n - C I 2 H , , C H , O H ~n-C,,H,,CH,Br+
KCN
n-C,,C2,CN
LiAIHJ
n-C,,H,,NH,
(6) The chain length is unchanged. SOCI?
~-CI,H2,COOH-
n-C,,H,,COCI+
NH,
LiAIH,
~z-C,,H,,CONH,-----*
n-Cl,H2,NH2
( c ) Chain length is decreased by one C; use the Hofmann degradation. NaOBr
~-C,,H,,CONH, ___* n-C,,H2,NH2
I-'
(d) The C content is doubled. The 1" amine is made from the corresponding ketone.
n-C,,H&OCI
n-C 1 2H,5CH 2 [see part ( a ) ]
I
LI
[see p a t
+
(n-C1 3
27 2
@)I
cuLi
n-C12Ha
"-C12H,Z
'c=o
n-C 13H27
;ZIb
n-C13H27
H ,
A
NH2
-
(e) Secondary amines may be prepared by reductive amination of an aldehyde (from RCOCl reduction), using a 1" amine. H , , Pd / B a S O J n - C 1 H 75N H n-C ,sCOCI n-C H2,CH0 n-C ,H2,NHC ,H,, H2/Ni 7
7
(f) The acyl halide C,,H,,COCI from (e) is reacted with the amine C,,H,,NH, from (b), to form an amide 0
that is then reduced.
C12H25COCl'
II
H2Nc13H27+
LIAIH,
c12H,,--C-NHc,,H,7-----+(C,,H,,),NH
Problem 18.13 There is no change in the configuration of the chiral C in sec-butylamine formed from the Hofmann degradation of (S)-2-methylbutanamide. Explain. 4
:R migrates with its electron pair to the electron-deficient :N, and configuration is retained because C--C is being broken at the same time that C-N is being formed in the transition state.
Intermediate
Transition State
AMINES
418
[CHAP. 18
Problem 18.14 A rearrangement of R, from C to an electron-deficient N, occurs in the following reactions. The substrates and conditions are given. Indicate how the intermediate is formed and give the structure of each product. (a)
Cwtius, RC-N,
II
(an acylazide) with heat
(6) Lessen, RC-NHOH
Schmidt, RCOOH
+ HN,
with H,SO,
(a hydroxamic acid) with base
8
(c)
or
Beckmann, R-C--R'
ll
with strong acid
4
NOH
Intermediate
R-C-N-OH
,,OH-H+ + H,O + 2.
+O=C=N-R
H20 OH-)
RNH,
+ C@-
-0H-
Intermediate (c)
R-f-R' NOH
R-fi-R'
+H,O
:N?H,
+ R-
$-R,O=C-R'
[\:i+ ] -R'
__*
+ H20
R-N:
Intermediate
I
R-N-H
The group tram to the OH (R) migrates as H,O leaves. 18.3 CHEMICAL PROPERTIES
STEREOCHEMISTRY
Amines with three different substituents and an unshared pair of electrons have enantiomers. However, in most cases an :NRR'R"-type amine cannot be resolved. The amine undergoes a very rapid nitrogen inversion similar to that for a C undergoing an S,2 reaction (Fig. 18-2).
transition state
Fig. 18-2
Simple carbanions, - - -):
-, behave similarly. chiral, (ii) resolvable? Give reasons in each case.
C6HSN(CH3)(C2H5) CH3
I
CH3CH2CH-N(C H3)(C2H,)
4
CHAP. 181
419
AMINES
(a) Not chiral and not resolvable, because N has two identical groups (C,H,). (6) Chiral, but the low energy barrier (25 kJ/mol) to inversion of configuration prevents resolution of its enantiomers. ( c ) Chiral and resolvable. N has four different substituents. The absence of an unshared pair of electrons on N prevents inversion as in part (6). (d) Chiral and resolvable. An asymmetric C is present.
CH,CH:bHN(CH,)(C,H,) Problem 18.16 Explain why 1,2,2-trimethylaziridine has isolable enantiomers.
4
This 3" amine is chiral because N has three different ligands and an unshared pair of e-'s. Unlike typical amines this molecule cannot undergo nitrogen inversion. The 3-membered ring requires ring angles of approximately 60" and restrains the N atom from attaining bond angles of 120°, as needed for the inversion transition state.
BASICITY AND SALT FORMATION [see Problem 3.25(b)] The lone pair of electrons on the N atom of amines accounts for their base strength and nucleophilicity. They abstract protons from water, react with Lewis acids, and attack electrophilic sites such as carbonyl carbon. Problem 18.17 ( a ) Why does aqueous CH,NH, turn litmus blue? (6) Why does C,H,NH, dissolve in aqueous HCl? 4 ( a ) Methylamine (pK, = 3.36) is a weak base, but a stronger base than water.
CH,NH, base,
+ H,O acid,
-
CH,NHl
+ OH-
acid,
base
( 6 ) A water-soluble salt forms.
C,H,NH,
+ H,O+ + Cl-+C,H,NHfCl-
+ H,O
Anilinium chloride
Problem 18.18 Account for the following observations. ( a ) In the gas phase the order of increasing basicity is NH, < CH,NH, < (CH,),NH < (CH,),N. (6) In water the order is NH, < CH,NH, = (CH,),N < (CH3)ZNH. 4 ( a ) Me, a typical alkyl group, has an electron-donating inductive effect and is base-strengthening. Basicity
increases with the increase in the number of Me groups. (6) Water draws the acid-base equilibrium to the right by H-bonding more with the ammonium cation than with the free base. The greater the number of H's in the ammonium cation, the greater the shift to the right and the increase in basicity. This effect should make NH, the strongest base and Me,N the weakest base. However, opposing the shift is the enhanced inductive effect of the increasing number of Me groups. The inductive effect predominates for MeNH, and Me,NH. In Me,N, the loss of an H more than cancels out the additional inductive effect of the third Me. Consequently, the 3" amine is less basic in water than the 2" amine.
It is not unusual that relative basicities in gas phase and water differ because of the solvation effects in H,O.
420
[CHAP. 18
AMINES
PrObl8m 18.19 (a) Assign numbers from 1 for LEAST to 4 for MOST to indicate relative base strengths of (i) C,H,NH,, (ii) (C,H,),NH, (iii) (C,H,),N, (iv) NH,. (6) Explain. 4 ( a ) (i) 3, (ii) 2, (iii) 1, (iv) 4, (b) C,H,NH2, an aromatic amine, is much less basic than NH, because the electron density on N is delocalized to the ring, mainly to the ortho and para positions [Fig. 18-3(a)]. With more phenyls bonded to N there is more delocalization and weaker basicity.
/ - -
N+
- \
....
\
Fig. 18-3 Problem 18.20 Compare the reactions of (CH,),N and (C,H,),N with BF,.
We have:
F
(CH,),N: Lewis base
F
I+ 8 : F +(CH,),N:B:F I f: F I
+
Lewis acid
(C,H,),N does not react, because the electron pair on N needed to bond to B is delocalized to 3 benzene rings. Problem 18.21 Assign numbers from 1 for following:
I ( a ) CH,NH-Na'
LEAST
I1
to 4 for MOST to indicate relative base strengths of the
111 (i-C,H,),N
C2H5NH2
IV CH,CONH2 4
I 4
I1 3
I11 2
IV 1
CH,NH, is a stronger base than NH, because the electron-donating effect of the Me group increases the electron density on N , the basic site. The three bulky i-propyl groups on N cause steric strain, but with an unshared electron pair on N , this strain is partially relieved by increasing the normal C-N--C bond angle (109") to about 112". If the unshared electron pair forms a bond to H, as in R,NH+, relief of strain by angle expansion is prevented. With bulky R groups, 3" amines therefore resist forming a fourth bond and suffer a decrease in basicity. Acyl R - C = O groups are strongly electron-withdrawing and base-weakening because electron density from N can be delocalized to 0 of the carbonyl group by extended T bonding [Fig. 18-3(6)]. I 3
I1 1
I11
IV
2
4
The strongly electron-attracting NO, group decreases electron density on N. It thus also decreases base strength by an inductive effect in the meta position, and to a greater extent, by both extended 7r bonding and inductive effects, in ortho and para positions.
CHAP. 181
42 1
AMINES
Since OCH, is electron-donating through extended 7r bonding, it increases electron density on N and the base strength of the amine because the ring accepts less electron density from N. Problem 18.22 Account for the following order of decreasing basicity:
RN H~> R N = C H R ~> R ~ N :
4
The hybrid atomic orbitals used by N to accommodate the lone pair of electrons in the above compounds are RNH,(sp3), RN=CHR'(sp'), RC%N:(sp). The nitrile (RCN)N has the most s character and is the least basic. The 1" amine has the least s character and is the most basic [Problems 8.3 and 8.5(b)].
REACTION WITH NITROUS ACID, HONO 1. Primary Amines (Diazonium Ion Formation) ( a ) Aromatic (ArNH,).
C,H, NH, ( 6 ) Aliphatic (RNH,). HONO
CH3CHZCHzNH2 ~
HONO
HCI, 5 "C
+
Benzenediazonium chloride
[C,H,-N=N:]Cl-
~ [CH3CH2CH2NJ+CIl r (unstable)
CH3CH=CH2
CH3CHCH3 CH3CHCH3
I
I c1
OH 1-Propanol
1-Chloropropane
Propene
2-Propanol
2-Chloropropane
This reaction of RNH, has no synthetic utility, bu11 the appearance of N, gas signals the presence of NH,. Problem 18.23 ( a ) Outline the steps in the formation of a diazonium cation from nitrous acid and a 1" amine. (b) Why are diazonium ions of aromatic amines more stable than those of aliphatic amines? 4 ( a ) Initially a nitrosonium cation is formed from HNO, in acid solution.
H Nitrosonium cation
Nucleophilic attack by the electron pair of the amine on the nitrosonium ion produces an N-nitrosamine. It undergoes a series of proton transfers that produce an OH group on the second N-N bond. H
H
H * *
I
k - *
.. ..
H
I
-H+
k R-N=N-OH-L
+H20
N-Nitrosamine + H Of
a diazoic acid
R-N=N:~+-H-
-H 0
+
R-NEN: Diazonium cation
AMINES
422
[CHAP. 18
(6) Aryldiazonium cations are stabilized by resonance involving the release of electrons from the ortho and puru positions of the ring.
2. Secondary Amines (Nitrosation) Ar(R)NH (or R2NH) + HONO-
Ar(R)N-NO
+ H,O
(or R,N-NO)
an N-nitrosamine (insoluble in acid)
N-Nitrosamines are cancer-causing agents (carcinogens). 3. Tertiary Amines
No useful reaction except for N ,N-dialkyl arylamines.
Problem 18.24 Prepare p-H,NC,H,N( CH,), from C,H,N( CH,), N(CH3)2
HONO
.
4
product
~
-H20
ON
With HN03,polynitration would occur since NMe, is very activating. NO' is less electrophilic than NO;.
REACTIONS WITH CARBOXYUC ACID DERWATWES; TRANSACYLATIONS 2
(R'COhO
R'CONHR
+ HCl
R'CONHR
+ R'COOH
REACTIONS WITH OTHER EUCTROPHILIC REAGENTS R'CH4
C1-
F
-C1
-H 0 + RNH, 2 R'CH-NR
+ RNH,
-
2HC1 + RNH-
e
(Schiff base, imine or azomethine)
-NHR
-
(symmetrical disubstituted urea)
0
II
R'NH-C-NHR
an isocyanate
(unsymmetrical disubstituted urea)
S
R'-N=C--C an isothiocyanate
+ HINR
II
__*
R'NH-C-NHR
(a thiourea)
423
AMINES
CHAP. 181
NUCLEOPHILIC DISPLACEMENTS 1.
Carbylamine Reactions of 1" Amines RNH2 + CHC13 + 3KOH
+ 3KC1+ 3H20
R-%:
__*
an isocyanide (foul smelling)
1: ] [
Nucleophilic RNH, attacks electrophilic intermediate [:CC12] [Problem 7.56(c)].
H
R-N+<:-
2. Hinsberg Reaction
+
+2KOH
-R-N&: -2HCI
O H
II I
Na+OH-
H 2 0 + Na' soluble in water
0
(neutral)
(No noticeable reaction) Problem 18.25 How can the Hinsberg test be used to distinguish among liquid RNH,, R,NH, and R,N?
R,N does not react; RNH, reacts to give a water solution of [C,H,SO,NR]Na'; solid precipitate, C,H,SO,NR,.
4
R,NH reacts to give a
Problem 18.26 Outline two laboratory preparations of sym-diphenylurea.
4
Problem 18.27 Condensation of C,H,NH, with C,H,CH=O yields compound (A), which is hydrogenated to compound (B). What are compounds (A) and (B)? 4
C,H,NH,
+ O=CHC,H,
C,H,N=cHC,H,
C,H,NHCH,C,H,
Benzalaniline (A)
N-Benzylaniline ( B )
OXIDATION
-
Oxidations of 1" and 2" amines give mixtures of product and, therefore, are not synthetically useful. 3" amines are cleanly oxidized to 3" amine oxides. R3N
H202
R3N-Q: +
-
a 3" amine oxide
AMINES
424
[CHAP. 18
18.4 REACTIONS OF QUATERNARY AMMONIUM SALTS
FORMATION OF 4' AMMONIUM HYDROXIDES 2 R 4 N + X - + Ag,O
+ H,O-
2R4NtOH- + 2AgX very strong buses; like NaOH
HOFMANN ELIMINATION OF QUATERNARY HYDROXIDES
[(CH,),NCH(CH,)CH2CH,]+OHA (CH,),N + H,C=CHCH,CH, s-Butyltrimethylarnmonium hydroxide
+ H,O
1-Butene
This E2 elimination (Table 7-3) give the less substituted alkene (Hofmann product) rather than the more substituted alkene (Saytzeff product; Section 6.3). Problem 18.28 Compare and account for the products obtained from thermal decomposition of ( a ) [(CH,),N +(C,H,)loH-, ( 6 ) (CH,),"OH-. 4
---c3-
H,C=CH,, (CH,),N, H,O. Alkenes are formed from C2Hsand larger R groups having an H on the p C.
(a)
H:O:
+ CH,:N(CH,),
HOCH, A Alkene formation is impossible with four CH,'s on N
+ :N(CH,),
(S,2 reaction)
Problem 18.29 Give the alkene formed from heating
(4 [(c2H s )(CH, 1 2 NCH2CH2CH,IOH -
(6) [CH,CH,CH,CH(CH,)N 'Me,JOH-
4
The less substituted alkene is formed: ( a ) H2C=CH2,not H2C=CHCH,; (b) CH,CH,CH,CH=CH,, CH,CH,CH==CHCH,.
not
Problem 18.30 Deduce the structures of the following amines from the products obtained from exhaustive methylation and Hofmann elimination. ( a ) C,H,,N (A) reacts with 1 mol of CH,I and eventually yields propene. (6) C,H,,N (B) reacts with 2mol of CH,I and gives ethene and a 3" amine. The latter reacts with 4 1 mol of CH,I and eventually gives propene. ( a ) (A) is a 3" amine because it reacts with only 1 mol of CH,I. Since propene is eliminated, C,H, can be n-or bo-; hence (A) is (C,H,)N(CH,),.
(6) (B) reacts with 2 mol of CH,I; it is a 2" amine. Separate formation of C,H, and C,H, shows that the alkyl groups are C,H, and C2H,. (B) is C3H,NHC2H,, where C,H, is n-propyl or isopropyl. Problem 18.31 Outline the reactions and reagents used to establish the structure of 4-methylpyridine by exhaustive methylation and Hofmann elimination. 4
.
CH3
A
FH, CH2H,C"
CH,
3-Methyl-] ,4pentadiene
_ OHI *
Ci.1 CH2 CH1 3-Methyl-l,3pentadiene
more stable (conjugated) diene
CHAP. 181
AMINES
425
COPE ELIMINATION OF 3" AMINE OXIDES
CH3CH2CH2NH
]'
[
CH3CH2CH2NCH3 I-
I
CH3
H
0-
I I CH3-C-CH2-N'CH3 I I H
' l HE;I-----3
0
' I ,C H 2 d - C H 3 I
H
CH3
H202
CH3CH2CH2NCH3
CH3
-
I
CH3
HO
CH3-CH=CH2
Elimination is cis-because pyrolysis of [R,N] + O H -.
of the cyclic transition state-and
I I
:N--CH, CH,
Propylene
3" amine oxide
+
N ,N-Dimethylh ydrox ylamine
requires lower temperatures than
PHASE- TRANSFER CATALYSlS (Problem 7.30) 18.5 RING REACTIONS OF AROMATIC AMINES
-NH,, -NHR, substitution. 1.
and -NR,
strongly activate the benzene ring toward ortho-para electrophilic
Halogenation
PhNH, with Br, (no catalyst) gives tribromination; 2,4,6-tribromoaniline is isolated. For monohalogenation, -NH, is first acetylated, because
CH3-C-N-
1I I
O H
is only moderately activating.
6 0 20 NHCOCHq
Ac20,
Br
NHCOCH,
NH2
ZQ
2,4,6-Tribromoaniline
2. Sulfonation
0-0 50 NH2
NHfHSOi
Anilinium hydrogen sulfate
NHSOJH
Sulfamic acid
so,
Sulfanilic acid a dipolar ion
Problem 18.32 How does the dipolar ion structure of sulfanilic acid account for its ( a ) high melting point, (6) insolubility in H,O and organic solvents, ( c ) solubility in aqueous NaOH, (d) insolubility in aqueous HCI? 4
(a) Sulfanilic acid is ionic. (6) Because it is ionic, it is insoluble in organic solvents. Its insolubility in H,O is typical of dipolar salts. Not all salts dissolve in H,O. ( c ) The weakly acidic NHS transfers H' to O H - to form a soluble salt, p-H2NC,H,SOJNa'. (d) -SO, is too weakly basic to accept H' from strong acids.
AMINES
426
[CHAP. 18
+
Problem 18.33 H,NCH,COO- exists as a dipolar ion whereas p-H,NC,H,COOH does not. Explain.
4
--COOH is too weakly acidic to transfer an H' to the weakly basic -NH, attached to the electronwithdrawing benzene ring. When attached to an aliphatic C, the NH, is sufficiently basic to accept H' from COOH.
3. Nitration To prevent oxidation by HNO, and meta substitution of C,H,NHl, amines are first acetylated.
GMC-' 340
0
0
II
II
NH2
H3
HNO,, H2SO4,
heat
NO*
Aniline
Acetanilide
NO?
p-Nitroacetanilide
p-Nitroaniline
18.6 SPECTRAL PROPERTIES
The N-H stretching and NH, bending frequencies occur in the ir spectrum at 3050-3550cm-' and 1600-1640 cm- ', respectively. In the N-H stretching region, 1" amines and unsubstituted amides show a pair of peaks for a symmetric and an antisymmetric vibration. In nmr, N-H proton signals of amines fall in a wide range ( S = 1-5 ppm) and are often very broad. The signals of N-H protons of amides are even broader, appearing at S = 5-8 ppm. Mass spectra of amines show a,P-cleavage, like alcohols.
Problem 18.34 Distinguish among l", 2", and 3" amines by ir spectroscopy. 1" amine, two N-H
band.
stretching bands; 2" amine, one N-H
4
stretching band; 3" amine, no N-H
stretching
18.7 REACTIONS OF ARYL DlAZONlUM SALTS
DISPLACEMENT REACTIONS
+ HPH,O,
+KI-ArI+N,
+ CuCl (CuBr)
-
-
or NaBH,
ArNfX-
ArH + N, (HPH20, is hypophosphorous acid)
Arc1 (ArBr) HF)+
+ N2
-
(Sandmeyer reaction)
ArNlBFa
+ HOC,H, A ArOC,H, + ArH + CH,CHO + N,
+ CUCN ArCN + N2 cu2+ + NaNO, + NaHCO, orCu20+ + NaHAsO, ArAsO,H,
-
&NO2
+ N2
A
ArF + N,
+ BF,
CHAP. 181
AMINES
427
Problem 18.35 Using C6H6, C6H,CH3, via diazonium salts and any other needed reagents, prepare ( a ) o-chlorotoluene, (b) m-chlorotoluene, (c) 1,3,5-tribromobenzene, ( d ) rn-bromochlorobenzene, (e) p iodotoluene, ( f ) p-dinitrobenzene and ( g ) p-cyanobenzoic acid. Do not repeat the synthesis of intermediate 4 products.
The -NO, is used to block the para position and it also directs metu, so that chlorination will occur only ortho to CH,.
The acetylated -NH, is used to direct CI into its ortho position, which is meta to CH,; it is then removed. Aniline is rapidly and directly tribrominated and the NH, removed.
Sr Aniline
br
Tribromoaniline
No, YOOH
1,3,5-Tribromobenzcne
ko,
No,
YOOH
YOOH
Nyc1-
CN
(8
No,
No,
AMINES
428
[CHAP. 18
COUPLING; FORMATION OF DIARYLAZO COMPOUNDS, Ar-N=N-Ar’
The aryl diazonium cation, ArNi, is a weak electrophile that is nevertheless capable of attacking aromatic rings with strongly activating groups. Diarylazo compounds are thereby produced with no loss of N, . A r N l + C,H,G+
p-G-C,H,-N=N-Ar
a weak
a strongly
an azo compound;
electrophile
activated ring
mainly p a m
(G = OH, NR,, NHR, NH,) (electron-releasing group)
Problem 18.36 Explain the following conditions used in coupling reactions: (a) excess of mineral acid during diazotization of arylamines, (b) weakly acidic medium for coupling with ArNH,, ( c ) weakly basic solution for coupling with ArOH. 4
Acid prevents the coupling reaction +
ArN=N:
+ H2NAr’ +ArN-N-NHAr’
by converting Ar’NH, to its salt, Ar’NHJX-. +
In strong base, rather than coupling, ArNEN reacts with O H - to form A r N = N - O H (a diazoic acid), which reacts further to give a diazotate, ArN=N-0-; neither of these couple. Strong acid converts ArNH, to ArNHi, whose ring is deactivated towards coupling. It turns out that amines couple fastest in mildly acidic solutions. High acidity represses the ionization of ArOH and therefore decreases the concentration of the more reactive ArO-. I n weak base, ArO- is formed and ArN=N--OH is not. Azo compounds readily undergo reductive cleavage to form two aromatic 1” amines. Problem 18.37 Deduce the structures of the azo compounds that yield the indicated aromatic amines on reduction with SnCl,; (a) p-toluidine and p-NH,-N,NdiMeaniline, (b) 1 mol of 4,4’diaminobiphenyl and 2 mol of 2-hydroxy-5-aminobenzoic acid. 4
NH2’s originate from N’s of the cleaved azo bond.
Problem 18.38 Which reactants are coupled to give the azo compounds in Problem 18.37?
CH,C,H, is not active enough to couple with p-Ni-C6H4(b)
+ X-Nl(OXO)N;X
+
N(CH3)2X:
ocz -iH
4
CHAP. 18)
429
AMINES
18.8 SUMMARY OF AMlNE CHEMISTRY PREPARATlON
PROPERTIES 1. Basic
1. Aliphatic Amines
(a) Primary
+ HOH- RCH,NHJ OH+ HX-RCH,NH; x+ BF, --+ RCH,NH,BF,
--/
Alkylation by S,2:
+ NH,
ammonia RCH,X
imide o-C,H4 (CO),N- K '
2.
+ RCH,X
+ Ag'-Ag(RCH,NH,)f
+cu2+
Reduction: ------+
3. Acidic
H-
-
Metal Cations CU(RCH,NH,):+
K-+ RCH,NH- K' + i H 2 I R'MgX--+ + RCH,NHMgX + R'H
RCH,NH, -------+
Electrophilic sites
+ R'CH-0-
RCH,CONH, + Br, + KOH RCH2CON3 RCH,CONHOH + H'
-
( b ) Secondary
Alkylation: RNH,
+ R'X-
Reduction: R-NX: RCH-NR'
( c ) Tertiary
-
1. Electrophilic sites
RR'NH
+ H,+ H,
R'CH=NCH,R
+ R'COX or (R'CO),OR'CONHCH,R + CHC1, + KOH- R - N s : + HNO, N, + ROH + HOH '+ ArSOiCl --ArSO,NHCH,R
+ R'COX + HNO,
or ( R ' c o ) , o - + R'coNHRR' RR'N-NO + ArSO,Cl+ ArS0,NRR'
-
'
RNHCH, RCH,NHR'
R N+ -
A1kylation : 3RX + NH, ( d ) Quarternary
R3N + C,H5X-
RXR4N+X ~ + O ~ - - + R , N - O (R3N--C,H5)+ X-+ A ~ , o + (R,NC,H,)+OH- &C,H,
/
+ R,N
2. Aromatic Amines
(a) Primary Reduction: ArNO,
RX+ ArNHR/ +++HNO,----,ArNzXRCOX, (RCO),O+
*ArNH2/
\
HX
\ + RCH=O---+ ArN-NAr ArNHNHAr
PhNH2
-
+ Br2+ H,SO4
RCH-NAr 2,4,6-Br C H,NH, p-H3k&6H4so;
( b ) Secondary ArNH,
+ RX
ArNHR
HNO,
ArNR,ArNRl XRCONHAr
ArNR(N0)
AMINES
430
[CHAP. 18
Supplementary Problems Problem 18.39 Write a structural formula for ( a ) N,N’-di-p-tolylthiourea, (6) 2,4-xylidine, (c) N-methyl-pnitrosoaniline, (d) 4-ethyl-3’-methylazobenzene. 4
( 6 ) &H3
(c)
GH3 NO
CH3
(d) C 2 H 5 e N = N - @ 3
( a ) 2-phenyl-1-aminoethane or P-phenylethylamine, ( 6 ) methylethylamine, ( c ) 1,3-diaminobutane, (d) 2-aminoethanol, (e) di-terr-butylamine, ( f) 1,Zethanediamine or ethylenediamine, ( g ) p-phenylenediamine or 1,4-benzenediamine.
Problem 18.41 Give the structure of ( a ) 3 4 N-methy1amino)-1-propanol,(6) ethyl 34 N-methy1amino)-24 butenoate, (c) 2-N,Ndimethylaminabutane, (d)allylamine. (U)
(6) CH,C=CHCOOC,H,
CH&H2CH,OH
(c)
I
I NHCH,
CH,CHCH,CH,
I
NHCH,
( d ) CHFCHCHZNH~
N(CH,)*
GcH,6 0 CH’
Problem 18.42 Give the structures and names of 5 aromatic amines with the molecular formula C,H,N.
@H3
CHZNHz
0-,
m-
CH,
and p-Toluidine
Benzylamine
4
N-Methylaniline
Problem 18.43 A compound C,H,N is a 2” amine. Deduce its structure.
4
Since the compound is a secondary amine, there must be two R groups and an H attached to an N: R-N-R
B
Removing N-H from C,H,N, we obtain C,H,, which must be divided between two R groups. These can only be a methyl and an ethyl group. The compound is CH,-N--CH,CH
A
,
Methy lethy lamine
Problem 18.44 Give the product of reaction in each case: ( a ) C,H,Br
+ excess NH,
(6) CH,=CHCN
+ H,/Pt
(c)
n-Butyramide + Br,
+ KOH
Acrylonitrile
(d) Dimethylamine + HONO ( a ) CH,CH,NH,,
( 6 ) CH,CH,CH,NH,,
(e) Ethylamine + CHCl,
(c) CH,CH,CH,NH,,
+ KOH
(d) Me,NN=O,
4 (e) CH,CH,&:
CHAP. 181
43 1
AMINES
Problem 18.45 What is the organic product when n-propylamine is treated with ( a ) PhSO,Cl? (6) excess 4 CH,CH,CH,Cl? ( c ) chlorobenzene? (d) excess CH,I, then Ag,O and heat? ( a ) N-(n-propyl)benzenesulfonamide, PhSO,NHCH,CH,CH,; (6) tetra-n-propylammonium chloride; ( c ) no reaction; (d) propene and trimethylamine.
Probem 18.46 Compare reactions of HNO,(NaNO, (c) N,N-dimethylaniline.
+ HCl) with: ( a ) aniline (at O'C),
( 6 ) N-methylaniline, 4
( a ) Soluble diazonium salt, PhN Cl-. ( 6 ) N-Methyl-N-nitrosoaniline. ( c ) p-Nitroso-N ,N-dimethylaniline.
Problem 18.47 What is the product of catalytic hydrogenation of ( a ) acetone oxime, (6) propane-l,3-dinitrile, (c) propanal and methylamine? 4 ( a ) isopropylamine, (6) 1,5-diaminopentane, (c) CH,CH,CH,NHCH,.
-
Problem 18.48 Show the steps in the following syntheses: Ethylamine
(4
Methyle th ylamine
(6)
Ethylamine-+ Dimethylethylamine
(4
n-Propyl chloride +Isopropylamine
(4
Aniline -+ p-Aminobenzenesulfonamide (Sulfanilamide)
4
H CH,CH,NH,
f
CH3CH2NH2&
CH,CH,CH,CI;;;;-,
alc.
-
CH,CH,NC
CH,CH=CH,
HIPI
HBr
I
2CH,CH,NCH,
H C=O
(6)
(4
CHClj KOH p
CH3CH2N(CH3)2 CH3CHBrCH,-
NH3
CH,CH(NH,)CH,
tjHCOCH3
NH,
S02NH2
Problem 18.49 Outline the steps in the syntheses of the following compounds from C6H6,C6H,CH, and any readily available aliphatic compound: (a) p-aminobenzoic acid, (6) m-nitroacetanilide, (c) l-amino-l-phenylpropane, (d) 4-amino-2-chlorotoluene . 4 (
432
AMINES
[CHAP. 18
Side chain is oxidized when a deactivating group (NO,) rather than an activating group (NH,) is attached to ring.
m-Dinitrobenzene
(4
.
C,H, + CH,CH,COCI-
AlC.1,
m-Nitroaniline
C,H,COCH,CH,
Nti,
C,H,CH(NH,)CH,CH,
Problem 18.50 Use the benzidine rearrangement to synthesize 2,2'-dichlorobiphenyI from benzene and inorganic reagents. 4
2,2'-Dichloro-4,4'-diaminobiphenyl
his-Diazonium salt
2,2'-DichlorobiphenyI
Problem 18.51 Use 0- or p-nitroethylbenzene and any inorganic reagents to synthesize the six isomeric dichloroethylbenzenes. Do not repeat preparations of intermediate products. 4
The -NO, is used as a blocking group by the sequence -NO, -+-NH, -+-N 14 - H or as a source of Cl by -Nf -+--Cl or is converted to -NHAc, whose directive effect supersedes that of C,H,. ( i ) 2,3-Dichloroethylbenzene
(ii) 2,4-DichloroethyIbenzene
433
AMINES
CHAP. 181
2,5-Dichloroethylbenzene
see parr ( i )
2,6-Dichloroethyl benzene
CI2 Fe
1.
2. 3.
Sn.HCI:OHHNOpS T HPH&
3,4-Dichloroethylbenzene Ac20 ___*
6
NHAc
NHAc
1. OH2. H N q . 5 T ' 3. Cucl
CI
3,5-Dichloroethyl benzene
Q' NH2
Problem 18.52 Deduce the structural formula of compound (A), C,H,NO,, which is reduced by Sn in OHto product (B). Strong mineral acid rearranges (B) to an aromatic amine (C), which is treated with HNO, and 4 then HPH,O, to form 3,3'-diethylbiphenyl (D).
Since (D) is a diphenyl compound, (C) is a substituted benzidine formed from a hydrazobenzene (B) in which the C,H,'s are ortho to the two NH,'s. Compound (A) is o-nitroethylbenzene.
Problem 18.53 Deduce a possible structure for each of the following. ( a ) Compound (A), C6H4N204,is insoluble in both dilute acid and base and its dipole moment is zero. ( 6 ) Compound (B), C,H,NO, is insoluble in dilute acid and base. (B) is transformed by KMnO, in H,SO, to compound (C), which is free of N, soluble in aqueous NaHCO, and gives only one mononitro substitution product. (c) Compound (D), C,H,NO,, undergoes vigorous oxidation to form compound (E), C,H,NO,, which is soluble in dilute aqueous NaHCO, and forms two isomeric monochloro substitution products. 4
AMINES
434
[CHAP. 18
Problem 18.54 Use any toluidine and necessary aliphatic or inorganic reagents to synthesize ( a ) 3-nitro-4fluorotoluene, ( b ) 2,2'-dimethyl-4-nitro-4'-aminoazobenzene. 4
For m-H,NC,H,CH, see Problem 18.55(a). Problem 18.55 Using any aliphatic and inorganic reagents, outline the syntheses of ( a ) m-HOC,H4CH3 from toluene , (6) 4-bromo-4'-aminoazobenzene from aniline. 4
CH, N.OH A '
6 0 0 NHAc
NHAc
(6)
&lO,
Br2,
NaOH, hut
Br
Problem 18.56 Prepare ( a ) C,H5D, ( b ) optically active sec-butylbenzene. I
(4
C6H5NH2
(b)
C6H6-C6H5-sec-Bu AIC13
N a N 0 2 . HCI, 5 "C
2. DPH202
* C,H5D or C,H,MgBrI HNO3, H2S04
S C E - BuCl
,. Sn, "Cl racemate
4
+
3 OH-
* 2 0
C,H5D
p-H,N-C,H, -sec- Bu racemate
which is resolved with an optically active carboxylic acid, such as tartaric acid (Problem 5.20). Then deaminate via diazonium salt and HPH,O,. Problem 18.57 Synthesize novocaine, p-H,NC,H,COOCH,CH,NEt,, pound of four or fewer C's.
from toluene and any aliphatic com4
435
AMINES
CHAP. 18)
(1)
EbNH
-
\oFH2
+ H2C-
HOCH,CH,NEh
Problem 18.58 An optically active amine is subjected to exhaustive methylation and Hofmann elimination. The alkene obtained is ozonized and hydrolyzed to give an equimolar mixture of formaldehyde and butanal. What is the amine? 4 The alkene is 1-pentene, CH,==CHCH,CH,CH, (H,C=O O=€HCH,CH,CH,). The amine is chiral; CH,CH(NH,)CH2CH,CH,. The other possibility, H,NCH,CH,CH,CH,, is not chiral. Problem 18.59 Draw a flow sheet to show the separation and recovery in almost quantitative yield of a mixture of the water-insoluble compounds benzaldehyde (A), N,N-dimethylaniline (B), chlorobenzene (C), 4 p-cresol (D), benzoic acid (E). See Fig. 18-4. Ether solution of mixture
I extract with'dil. nq. HCI
Upper e t ' r layer
I
extract with aq. NaOH I
i r d d aq. NnOH, ether
,
(in ether)
Upper ether layer p-CH,C,H,OH (D)
I
*add dry ice!Ca;,
ether
Lower aq. NaHCO, layer C H 00-Na'
Lower H,O layer c6Hs-~~--So3Na
sY
dil. HCI (ether)
I
Upper ether layer C6H5C1 (c)
N a 2 C q (ether)
C6HsCOOH (E)
C,H,CH=O
(A)
Fig, 18-4
Problem 18.60 Synthesize the following compounds from alcohols of four or fewer C's, cyclohexanol and any needed solvents and inorganic reagents. (a) n-hexylamine, (6) triethylamine N-oxide, ( c ) 4-( N4 methylamino)heptane, (d) cyclohexyldimethylamine, (e) cyclopentylamine, ( f ) 6-aminohexanoic acid.
'
NaOH + CO, gives NaHCO,, in which carboxylic acids dissolve but phenols do not.
AMJNES
436
SOCl
(2) n-C4H,0H 4 n-C4H9CI
(1)
(c)
n-BuOH (2)
E,+n-PrCHO
n-PrOH
n-C4H&lgC1
CH,OH
[CHAP. 18
I.
Hq;-H2
2,
H~O+
from(1)
-%CH,Br 'i'; CH,NH, H +
n-PrCI 5 n-PrMgC1
CU
1
(1) CH,OH-H2CO-HCOOH heat Na2Cr2O7
C y c l o h e x a n o l ~ 2 ~ 0Cyclohexanone qr
(e) Cyclohexanol-
Cyclohexene
Ag(NH3);
NH3
7
N.2C'24
H2SOI +
H2C0 from ( 1 ) HCOOH from (,)
KMn04
-
HOOC( CH,),COOH-
(f) Cyclohexanol
I
(n-Pr),CO-(MgCI)'
CYclohexYlamine HZSOJ
'
Cyclohexanone
H2NOH
Cyclohexanone oxime
BaO heat
* Cyclohexyldimethylamine
-
Cyclopentanone
-cg H2S04
NH3
H21Pt
Cyclopentylamine
6-Aminohexanoic acid
Caprolactam
[See Problem 18.14(c).]
Problem 18.61 Use simple, rapid, test-tube reactions to distinguish between (a) CH,CONHC,H, and C,H,CONH2, (6) C,H,NH;Cl- and p-ClC,H,NH2, (c) (CH,),N'OHand (CH,),NCH,OH, ( d ) p CH,COC,H,NH, and CH,CONHC,H,. 4 (a) With hot aqueous NaOH, only C,H,CONH, liberates NH,. (6) Aqueous AgNO, precipitates AgCl from C,H,NH;Cl-. (c) CrO, is reduced to green Cr3' by (CH,),NCH,OH. (CH,),N'OH- is strongly basic to litmus. ( d ) Cold dilute HCl dissolves p-CH,COC,H,NH,, which also gives a positive iodoform test with NaOI.
Problem 18.62 Synthesize from benzene, toluene, naphthalene (NpH), and any aliphatic or inorganic compounds (a) a-(p-nitrophenyl)ethylamine, ( 6 ) p-( p-bromophenyl)ethylamine, ( c ) 1-(a-aminomethyl) 4 naphthalene, (6)2-naphthylamine, (e) /3-NpCH2NH2.
c1 I
CHAP. 181
(c)
HNO3
NpH-
437
AMINES
H2S04
1-NpNO,
1
Sn HCI
2 OH-
1-NpNH,
NaNOZ
CH COCl
pP h N 0H 2 , A l CA l i 2-NpCOCH3
1
NaOH,12
LiAlH4
2 H3O
( d ) Naphthalene cannot be nitrated directly at the p-position.
N
I
CuCN
1-NpN C1- ___* 1-NpCN+71-NpCH2NH,
I
SOCI,
-
2 - N p C O O H c 2-NpCONH2
K O H , Br2
2-NpNH2
Problem 18.63 Synthesize from naphthalene and any other reagents: (a) naphthionic acid (4-amino-lnaphthalenesulfonic acid), (6) 4-amino-1-naphthol, (c) 1,3-dinitronaphthalene, (d) 1,4-diaminonaphthaIene, (e) 4 1,2-dinitronaphthalene. Do not repeat the synthesis of any compound.
NH~HSO;
NH,
so;
-SO;
blocks C' position
Problem 18.64 In the presence of C,H,N02, 2 mol of C,H,NH, reacts with 1 mol of p-toluidine to give a triarylmethane that is converted to the dye pararosaniline (Basic Red 9) by reaction with PbO, followed by acid. Show the steps, indicating the function of ( a ) nitrobenzene, (6) PbO,, (c) HCl. 4
0 is eliminated with the paru H's from 2 molecules of C,H,NH, to form p-triaminotriphenylmethane (a leuco base).
( a ) Nitrobenzene oxidizes the CH, of p-toluidine to CHO, whose
( b ) PbO, oxidizes the triphenylmethane to a triphenylmethanol.
438
AMINES
[CHAP. 18
(c) HCI protonates the OH, thus making possible the loss of H,O to form Ar,C', whose + charge is delocalized to the three N's. Delocalization of electrons is responsible for absorption of light in the visible spectrum, thereby producing color.
-
-To@-H,N-C,H,),C'Cl-
(p-H,N-C,H,),C-OH
+
dye
-
(P -H2N-C6HJ2C=( one of several contributing structures
)=iiH2
+ C1-
Problem 18.65 Suggest a structural formula for a compound, C,H,,N (A), that is optically active, dissolves in 4 dilute aqueous HCI, and releases N, with HONO.
The compound has four degrees of unsaturation (if it were a substituted alkane, its formula would be C,H,,N), which indicates the presence of a phenyl ring. (A) dissolves in HCl and therefore is an amine. Since (A) releases N, with HONO, it is a 1" amine. Since (A) is optically active, it has a chiral C. The -NH, cannot be on the ring since the two remaining C's cannot be positioned to give a chiral C. The -NH, must be on the side chain. The compound is C,H,CH( NH,)CH,. Problem 18.66 How can N-methylaniline and o-toluidine be distinguished by ir spectroscopy?
4
o-Toluidine is a 1" amine and has a pair of peaks (symmetric and antisymmetric stretches) in the N-H stretch region. N-Methylaniline is a 2" amine and has only one peak. Problem 18.67 Amines A, B, D, and E each have their parent-ion peaks at mle = 59. The most prominent peaks for each are at m / e values of 44 for A and B, 30 for D, and 58 for E. Give the structure for each amine and for the ion giving rise to the most prominent peak for each. 4
Since the parent peak is m l e = 59, the formula is C,H,N. The major fragmentation of amines is a bond to the a carbons,
-UI!+
A C-€ bond is weaker and breaks more easily and more often than a C-H CH, (m = 1 5 ) ; 59 - 15 = 44. The two isomers are: CH,NHCH,CH, (A)
H-CH, ""-ZH,
--P
I
[CH,NHCH,,'CH,]+
-[ --c
(B)
I
I
H2N H,t'CH, ZH,
Amine D loses CH,CH, (59 - 29 = 30). H2NCH2CH2CH, Amine E loses H (59 - 1 = 58).
1
+
I
[H,NCH,,'CH,CH,]' I
bond. Amines A and B both lose a H
I+
* CH,N=CH2 + C H , +
__*
H2N=CHCH3
-
H2A=CH2
+ .CH3
+ CH,CH,
Problem 18.68 What compound, C,H,NO, has the following nmr spectrum: 8 = 6.5, broad singlet (two H's); 6 = 2.2, quartet (two H's); and 8 = 1.2, triplet (three H's)? 4
CHAP. 181
AMINES
439
The integration ratio 2 : 2 : 3 accounts for the seven H’s. The peaks at S = 2.2 and S = 1.2 are from a as indicated by the splitting, and the group is attached to a C=O group, as shown by the S = 2.2 value for the H’s on the CH,. The broad singlet at S =6.5 are H’s of an amide. The compound is CH,CH,CONH,, propanamide.
--CH,CH,,
Problem 18.69 Determine the structure of a compound (C,H,,NO) which is soluble in dilute HC1 and gives a positive Tollens’ test. Its ir spectrum shows a strong band at 1695cm-’ but no bands in the 3300-3500cm-’ region. The proton-decoupled I3C spectrum shows six signals which would display the following splitting pattern in the coupled ‘,C spectrum: one quartet, two singlets, and three doublets, one doublet being very downfield. 4
The positive Tollens’ test, the band at 1695cm-’, and the downfield doublet are consistent with a CHO group. Basicity and absence of an N-H stretching band indicate a 3” amino group, -N(CH,), (the CH,’s give the quartet). The 5 degrees of unsaturation indicate a benzene ring and a C = O group, and the two singlets show that the ring is disubstituted. Since the other two doublets must arise from the other four ring carbons, there are two pairs of equivalent ring C’s, and the two substituents must be puru. The compound is p-dimethylaminobenzaldehyde,(CH,),NC,H,CHO. Problem 18.70 What compound results from treating the diazonium salt of p-toluidine with copper bronze 4 powder?
4’4‘-dimethylbiphenyl (Gatterman reaction).
Chapter 19 Phenolic Compounds 19.1 INTRODUCTION
NOMENCLATURE Phenols (ArOH) and alcohols (ROH) are similar in properties but they differ sufficiently so that phenols may be considered as a separate homologous series. Problem 19.1 Name the following phenols by the IUPAC system:
OH
Catechol
OH
OH
H ydroquinone
Salicylic acid
4
( a ) hydroxybenzene, (b) m-hydroxytoluene, (c) 1,3-dihydroxybenzene, ( d ) 1,2-dihydroxybenzene7 (e) 1,4-dihydroxybenzene, ( f ) o-hydroxybenzoic acid.
Problem 19.2 Name the following compounds:
GCH3 OCH,
COO-Na'
G OOH C H 3
cYocH2c
( a ) p-methoxyethylbenzene, (b) p-hydroxyacetanilide, (c) y-allylphenol, ( d ) sodium acetylsalicylate (sodium salt of aspirin), (e) ethoxybenzene or phenetole.
Problem 19.3 Write structural formulas for ( a ) p-cresol, ( b ) 2-nitro-4-bromopheno1, ( c ) 4-n-hexylresorcinol, 4 (d) ethyl salicylate, (e) P-naphthol.
PHYSICAL PROPERTIES Problem 19.4 Compared to toluene, phenol ( a ) has a higher boiling point and (b) is more soluble in H 2 0 . 4 Explain. H H I / Ph-O\ ,O Ph-0 ,0-Ph \H,/ H" H ' (a) Intermolecular H-bonding
440
( b ) H-bonding with H,O
CHAP. 191
441
PHENOLIC COMPOUNDS
Problem 19.5 Account for the lower boiling point and decreased H,O solubility of o-nitrophenol and 4 o-hydroxybenzaldehyde as compared with their m and p isomers. In some ortho-substituted phenols, intramolecular H-bonding (chelation) forms a six-membered ring. This inhibits H-bonding with water and reduces solubility in H,O. Since chelation diminishes the intermolecular H-bonding attraction present in the para and meta isomers, the boiling point is decreased.
0-Nitrophenol
o-Hydroxybenzaldeh yde chelation
The greater H,O solubility of the meta and para isomers is due to coassociation with water molecules through H-bonding. Problem 19.6 (a) Why is the C - 0 bond in phenol shorter than in alcohol? (6) Why are the dipole moments of phenol (1.7 D) and methanol (1.6 D) in opposite directions? 4 (a) All C-X bond lengths, including C - 0 , decrease as the s character of the hybrid orbital used by C increases. Hence, the CSPz--O bond (more s character) of phenols is shorter than the C , 3 4 bond (less s character) of alcohols. Additionally, the delocalization of electron density from 0 to the ring by extended 7r bonding engenders some double-bond character in the phenolic C - 0 bond and makes this bond shorter. (6) The delocalization of electron density from 0 to the ring causes the 0 of phenol to be the positive end of the molecular dipole. In alcohols the electronegative 0 is the negative end of the dipole.
19.2 PREPARATION
INDUSTRIAL METHODS 1. Dow Process [by Benzyne Mechanism (Section 11.3)J Sodium phenoxide
Chlorobenzene
Phenol
2. From Cumene Hydroperoxide
C,H,
+ CH,=CHCH,Propene
CH3
H2S04
I 2C,H~C-O-O-H
Cumene
Cumene hydroperoxide
CH3
CH3
CH3
I C,H,-cH I
I
CH3
I I
3 C6H50H+ C-0 H O+
Phenol
CH3
Acetone
Problem 19.7 Give a mechanism for the acid-catalyzed rearrangement of cumene hydroperoxide involving an 4 intermediate with an electron-deficient 0 (like R+). Ph H3C\
H3C'
I
C-0-0-H
H+ __*
H H 3C Ph \ I,C-0-0-H I + H,C' an oxonium ion
-Ph: ___*
electrondeficient intermediate
442
PHENOLIC COMPOUNDS
-
C-0-Ph
H3 C
(CHAP. 19
/
carbocation
H,C \C=O / Hemiacetal
Ace tone
+ PhOH Phenol
The rearrangement of Ph may be synchronous with loss of H,O.
3. Alkali Fusion of Arylsulfonate Salts [Problem 16.58(6)
LABORATORY METHODS 1.
Hydrolysis of Diazonium Salts (Section 18.7)
2.
Aromatic Nucleophilic Substitution of Nitro Aryl Halides
See Section 11.3. 3.
fi
Ring Oxidation (with trifluoroperoxyacetic acid, F,C -0OH)
0 OH
(electrophilic substitution by OH') CH,
CH,
Mesitylene (an activated ring)
Mesitol
Problem 19.8 Outline reactions and reagents for industrial syntheses of the following from benzene, and naphthalene (NpH) and inorganic reagents: ( a ) catechol, ( h ) resorcinol, (c) picric acid (2,4,6-trinitrophenoI), ( d ) P-naphthol ( P - N p O H ) . 4
(c)
0
*:;
6
H2S04 HNO3)
fj NO*
2.4-Dinitrochlorobenzene
OH 2I
N H,O+ eOH)
QN02
N0
H2m4 HN%)
2
2.4-Dinitrophenol
0 2 N 6 N 0 2
NO2
Picric acid
CHAP. 191
443
PHENOLIC COMPOUNDS
Direct nitration of phenol leads to excessive oxidation and destruction of material because HNO, is a strong oxidizing agent and the OH activates the ring.
(4
NPH
OH-
1 . H2S04(140"C) 2 OH-
Problem 19.9 Devise practical laboratory syntheses of the following phenols from benzene or toluene and any inorganic or aliphatic compounds: ( a ) rn-iodophenol, (6) 3-chloro-4-methylphenol, ( c ) 2-bromo-4methylphenol. 4
rn-lodophenol
3-Chloro-dmeth ylphenol
2-Bromo4methylphenol
19.3 CHEMICAL PROPERTIES REACTIONS OF H OF THE OH GROUP 1.
Acidity
Phenols are weak acids (pK, = 10): ArOH
+ H,O
ArO-
+ H,O+
They form salts with aqueous NaOH but not with aqueous NaHCO,. Problem 19.10 Why does aqueous NaHCO, solution dissolve RCOOH but not PhOH?
4
In both cases the product would be carbonic acid (pK, = 6 ) , which is a stronger acid than phenols (pK, = 10) but weaker than carboxylic acids (pK, =4.5). Acid-base equilibria lie toward the weaker acid and weaker base. RCOOH + HCO; stronger acid,
stronger base,
ArOH + HCO;% weaker acid,
weaker base
RCOO-
+ H2C03
weaker base,
weaker acid,
ArO-
+ H2C0,
stronger base,
stronger acid,
PHENOLIC COMPOUNDS
444
[CHAP. 19
Problem 19.11 Account for the considerably greater acid strength of PhOH (pK, = 10) as compared to ROH (pK, = 18). 4
*oo*
The negative charge on the alkoxide anion, RO-, cannot be delocalized, but on PhO- the - charge is delocalized to the ortho and para ring positions as indicated by the starred sites in the resonance hybrid, '.. ..
*
PhO- is therefore a weaker base than RO-, and PhOH is a stronger acid. Problem 19.12 What are the effects of ( a ) electron-attracting and ( 6 ) electron-releasing substituents on the acid strength of phenols? 4 ( a ) Electron-attracting substituents disperse negative charges and therefore stabilize ArO- and increase acidity of ArOH. ( 6 ) Electron-releasing substituents concentrate the negative charge on 0, destabilize ArOand decrease acidity of ArOH. Electron-Attracting
(-NO,,
Electron-relea ring
(-CH,,
-CN, -CHO, -COOH, -NR;,--X)
-CZHJ, -OR, -NRI)
0-
0
A Charge dis rsed; ion stabiLd
A More acidic
R
R
Charge concentrated;
Less
ion destabilized
acidic
Problem 19.13 In terms of resonance and inductive effects, account for the following relative acidities. (U)
p- 02N C , H 40H> m-O,NC,H,OH > C,H,OH
(b)
m-CIC6H40H> p-CIC,H40H > C,H,OH
4
( a ) The -NO,
is electron-withdrawing and acid-strengthening. Its resonance effect, which occurs only from para and ortho positions, predominates over its inductive effect, which occurs also from the rneta position. Other substituents in this cateogry are >C=O
-CN
X 0 0 R
-SO,R
( b ) C1 is electron-withdrawing by induction. This effect diminishes with increasing distance between C1 and OH. The meta is closer than the para po,sition and rn-C1 is more acid-strengthening than p-C1. Other substituents in this category are F, Br, I, NR,. Problem 19.14 Assign numbers from 1 for LEAST to 4 for MOST to indicate the relative acid strengths in the following groups: ( a ) phenol, rn-chlorophenol, rn-nitrophenol, rn-cresol; ( 6 ) phenol, benzoic acid, p-nitrophenol, carbonic acid; (c) phenol, p-chlorophenol, p-nitrophenol, p-cresol; (d) phenol, o-nitrophenol, rnnitrophenol, p-nitrophenol; (e) phenol, p-chlorophenol, 2,4,6-trichlorophenol, 2,4-dichlorophenol; ( f ) phenol, 4 benzyl alcohol, benzenesulfonic acid, benzoic acid. (a) 2, 3, 4, 1. Because
has
+ on N, it has a greater electron-withdrawing
inductive effect than has C1.
PHENOLIC COMPOUNDS
CHAP. 191
445
( b ) 1, 4, 2, 3. ( c ) 2, 3, 4, 1. The resonance effect of p-NO, exceeds the inductive effect of p-CI. p-CH, is electron-releasing. (d) 1, 3, 2, 4. Intramolecular H-bonding makes the o isomer weaker than the p-isomer. (e) 1, 2, 4, 3. Increasing the number of electron-attracting groups increases the acidity. ( f ) 2, 1, 4, 3.
FORMATION OF ETHERS 1.
Williamson Synthesis
+ X-
Ar-0-R'
A r O H Z ArOAr-MH,
+ CH,OSOJ
2. Aromatic Nucleophilic Substitution
2,CDinitrophenetole (2,CDinitrophenyl ethyl ether)
Problem 19.15 Why does cleavage of aryl ethers, ArOR, with HI yield only ArOH and RI?
4
An S,2 displacement on the onium ion of the ether occurs only on the C of the R group, not on the C of the Ar group. ArOR
+ HI-
ArORT-
A
+ArOH
+ RI
FORMATION OF ESTERS
Phenyl esters (RCOOAr) are not formed directly from RCOOH. Instead, acid chlorides or, anhydrides are reacted with ArOH in the presence of strong base. (CH,CO),O
+ C,H,OH + NaOH+
+ CH,COO-
CH,COOC,H,
Na'
+ H,O
Phenyl acetate
C,H,COCl
+ C,H,OH + NaOH-
C,H,COOC,H,
+ Na'
C1-
+ H,O
Phenyl benzoate
OH- converts ArOH to the more nucleophilic ArO- and also neutralizes the acids formed. Problem 19.16 Phenyl acetate undergoes the Fries rearrangement with AICI, to form ortho- and puruhydroxyacetophenone. The ortho isomer is separated from the mixture by its volatility with steam.
0
C-CH3
II
0 o-Hydroxyacetophenone
p-Hydroxyacetophenone
446
PHENOLIC COMPOUNDS
[CHAP. 19
( a ) Account for the volatility in steam of the ortho but not the puru isomer. ( 6 ) Why does the para isomer predominate at low, and the ortho at higher, temperatures? ( c ) Apply this reaction to the synthesis of the antiseptic 4-n-hexylresorcinol, using resorcinol, aliphatic compounds and any needed inorganic reagents. 4
The ortho isomer has a higher vapor pressure because of chelation, O-H---O----C (see Problem 19.5). In the para isomer there is intermolecular H-bonding with H,O. The para isomer (rate-controlled product) is the exclusive product at 25 "C because it has a lower AHSand is formed more rapidly. Its formation is reversible, unlike that of the ortho isomer which is stabilized by chelation. Although it has a higher A H S , the ortho isomer (equilibrium-controlled product) is the chief product at 165°C because it is more stable. Two activating OH groups in meta positions reinforce each other in electrophilic substitution and permit Friedel-Crafts reactions of resorcinol directly with RCOOH and ZnC!, .
Problem 19.17 Alkylation of PhO- with an active alky! halide such as CH,=CHCH,CI gives, in addition to phenyl ally1 ether, some o-allylphenol. Explain. 4 PhO- is an ambident anion with negative charge on both 0 and the ortho C atoms of the ring. Attack by 0 gives the ether; attack by the ortho carbanion gives o-allylphenol.
DISPLACEMENT OF OH GROUP Phenols resemble aryl halides in that the functional group resists displacement. Unlike ROH, phenols do not react with HX,SOCl,, or phosphorus halides. Phenols are reduced to hydrocarbons, but the reaction is used for structure proof and not for synthesis. ArOH
+ Zn A ArH + ZnO
(poor yields)
BUCHERER REACTION FOR INTERCONVERTING P-NH, AND @-OHNAPHTHALENES
&Naphthol
B-Naphthylamine
The ready availability of P-naphthol [see Problem 19.8(d)] makes this a good method for synthesizing P-naphthylamine. REACTIONS OF THE BENZENE RING 1. Hydrogenation
Cyclohexanol
447
PHENOLIC COMPOUNDS
CHAP. 191
2. Oxidation to Quinones
0
0 (KS03)2N0
,
(Fremy’s salt)
0 Phenol
Quinone
3. Electrophilic Substitution The - O H and even more so the 4-(phenoxide) are strongly activating and op-directing.
Phenol
Phenoxide anion
oxonium cations
fairly stable unsaturated ketones
Special mild conditions are needed to achieve electrophilic monosubstitution in phenols because their high reactivity favors both polysubstitution and oxidation. ( a ) Halogenation
0 OH
BrZ.H20+
OH BQB~
(goes via A ~ ono -; monobromophenols)
Br Monobromination is achieved with nonpolar solvents such as CS, to decrease the electrophilicity of Br, and also to minimize phenol ionization.
o-Bromophenol
(6) Nitrosation.
p-Bromophcnol (major)
6 6 -0 0
tautornerire
HONO+ 73T
N-0
p-Nitroso-
phenol
NOH Quinone monoxime
448
PHENOLIC COMPOUNDS
[CHAP. 19
Nitration. Low yields of p-nitrophenol are obtained from direct nitration of PhOH because of ring oxidation. Sulfonation.
o-Phenolsulfonic acid (rate-controlled)
p-Phenolsulfonic acid (equilibrium-controlled)
Diazonium salt coupling to form azophenols (Section 18.7). Ring alkylation.
RX and AICI, give poor yields because AICI, coordinates with 0. Ring acylation. Phenolic ketones are best prepared by the Fries rearrangement (Problem 19.16). Kolbe synthesis of phenolic carboxylic acids. C,H,O
Na'
+ O=C=O-
12s "C h atm
o-HOC,H,COO- Na'-
H 0 '
o-HOC,H,COOH Salicylic acid
Sodium salicylate
Reimer-Tiemann synthesis of phenolic aldehydes.
Condensations with carbonyl compounds; phenol-formaldehyde resins. Acid or base catalyzes electrophilic substitution of carbonyl compounds in ortho and puru positions of phenols to form phenol alcohols (Lederer-Manasse reaction).
RCH-0 1.
OH-(OnPhOH)
2.
H+
I
R-C-H
I
OH base catalysis
acid catalysis
Rearrangements from 0 to ring. (1) Fries rearrangement of phenolic esters to phenolic ketones (Problem 19.16). (2) Claisen rearrangement. The reaction is intrumofecufurand has a cyclic mechanism.
CHAP. 191
PHENOLIC COMPOUNDS
449
(3) Alkyl phenyl ethers.
0-H
Phenetole
o-Ethylphenol
0-H
p-Ethylphenol
Problem 19.18 Outline a mechanism for the ( a ) Kolbe reaction, (b) Reimer-Tiemann reaction. (a)
4
Phenoxide carbanion adds at the electrophilic carbon of CO,.
The conjugated ketonic diene tautomerizes to reform the more stable benzenoid ring.
(b) The electrophile is the carbene :CCl,
.n cl :OH-+ H:C:Cl__* H:OH +
c1
Cl
a b e n d chloride
Problem 19.19 Use phenol and any inorganic or aliphatic reagents to synthesize ( a ) aspirin (acetylsalicyclic acid), (b) oil of wintergreen (methyl salicylate). Do not repeat the synthesis of any compound. 4 OH
OH
OCOCHa
PHENOLIC COMPOUNDS
450
[CHAP. 19
Problem 19.20 Predict the product of the Claisen rearrangement of (a) a l l ~ l - 3 - ' ~ phenyl C ether, (6) 2,6-dimethylphenyl allyl-3-I4Cether. 4 (a) In this concerted intramolecular rearrangement, the ends of the allyl system interchange so that the y C is bonded to the ortho C.
'
0
1
OH
( 6 ) When the ortho position is blocked, the allyl group migrates to the para position by two consecutive rearrangements and the '"C is in the y position of the product.
P M ~ 19.21 I Use phenol and any aliphatic and inorganic compounds to prepare: ( a ) o-chlorornethylphenol, 4 (6) o-methoxybenzyl alcohol, ( c ) p-hydroxybenzaldehyde.
( 6 ) From (a),
o-HOC,H,CH,OH
NaOH
( C H J)2SO4
o-CH,OC,H,CH,OH
Only the phenolic OH is converted to its conjugate base with NaOH.
(4
C,H,OH
+ HCN-
ZnClZ HCI
p-HOC,H,CHO
19.4 ANALYTICAL DETECTION OF PHENOLS
Phenols are soluble in NaOH but not in NaHCO,. With Fe3' they produce complexes whose characteristic colors are green, red, blue, and purple. Infrared stretching bands of phenols are 3200-3600cm-' for the 0-H (like alcohols), but 1230 cm-' for the C-0 (alcohols: 1050-1150 cm-I). Nmr absorption of OH depends on H-bonding and the range is S = 4-12 ppm.
CHAP. 191
45 1
PHENOLIC COMPOUNDS
19.5 SUMMARY OF PHENOLIC CHEMISTRY PREPARATION
PROPERTIES 1. OH Group Reactions
1. Industrial
(a) H of OH
(b) Raschig
Ar-H
+ Fe3'
(c) NaOH Fusion ArS0,Na
-
+ NaOH+
+ HCl + 02+ NaOH+
ArO-Na'
ArO-Fe2'
+H20
(acidity)
(colored)
(d) Cumene Peroxide
+ RCOX-
base
ArO-COR
2. Benzene Ring Reactions (a) Reduction
+ 3H2/Ni-C,H,,0H
-
M 6 H 4 4 p-Quinone
]
(c) Electrophilic Substitution
+ Br2(H,0 sol.)-
2,4,6-Br,C6H20H p-BrC6H40H + H2S04+p - H M 6 H 4 - S o 3 H + dil. HN0,- 0-and p-0,NC,H40H
+ Br2(CS,)
2. Laboratory (a) Diazonium
5 "C
( b ) Sulfonic Acid Salt
+ H,C=O, + RCHO-
+ NaOH
ArSOiNa'
,-N=O,
p-O=C,H4=NOH
+ Ar'NlX-- p-HO-Ar-N-N-Ar' + ROH/H2S04+ p - H W b H 4 - R + CH,=O0- and p -Ho -c6 H,CH2 0H
HCl0- and p-HOC6H,CH2Cl RCH(C6H40H-p)2
--
19.6 SUMMARY OF PHENOLIC ETHERS AND ESTERS
ArO-Na'
+ X-CH,CH=CH,
ArO-Na'
+ X-CH2CH-CH3 0
ArO-Na'
II + X-C-CH3
heat
+ArO-CH,CH=CH, __*
-
ArO-CH2CH(CH3)2
ArO-
P
-R
-
+ AlC13
o-HOC,H4CH2CH=CH2 (A-Ph)
AlClj
p-HO-c6H4C(CH3)3 (Ar=Ph)
p- or o-HO-C6H,
P
-R
(Ar=Ph)
PHENOLIC COMPOUNDS
452
[CHAP. 19
Supplementary Problems Problem 19.22 Name the following compounds:
OH
OH
( a ) rn-ethoxytoluene, (6) methylhydroquinone, (c) p-allylphenol.
Problem 19.23 Write the structure for ( a ) phenoxyacetic acid, (6) phenyl acetate, (c) 2-hydroxy-3-phenylben4 zoic acid, (d) p-phenoxyanisole.
(a) G O C H , C O O H
(6) O O - I - C H 3
0
HOOC,
I
OH
Ether layer (upper)
Aq.
NaOH layer (lower)
I
PhO-Na'. PhCOO-Na'
PhCHIOH
Solution
r PhCOO-Na'
Precipitate
I
PhOH
(add H,O+to regenerate acid) Fig. 19-1
Problem 19.25 What product is formed when p-cresol is reacted with ( a ) (CH,CO),O, ( b ) PhCH,Br and 4 base, ( c ) aqueous NaOH, (d) aqueous NaHCO,, (e) bromine water? ( a ) p-CH,C,H,OCOCH,, p-cresyl acetate; (6) p-CH,C,H,OCH,Ph, p-tolyl benzyl ether; (c) pCH,C,H,O- Na +, sodium p-cresoxide; (d) no reaction; (e) 2,6-dibromo-4-methylphenol.
Problem 19.26 Use simple test tube reactions to distinguish ( a ) p-cresol from p-xylene, (6) salicylic acid from 4 aspirin (acetylsalicylic acid). ( a ) Aqueous NaOH dissolves the cresol. (6) Salicyclic acid is a phenol which gives a color (purple in this case) with FeCI,.
CHAP. 191
453
PHENOLIC COMPOUNDS
Problem 19.27 Identify compounds (A) through (D).
(CPSF4 OH-
' (A)
Zn*HCI
--
(B)
N a N 4 . HCl
c 6 H F
( 0
(D)
4
OH (A)
(B)
p-C,H,0C,H,N02
p-C,H,OC,H,NH,
( p-phenetidine)
(C)
p-C,H,OC,H,N
C1-
Problem 19.28 Prepare ( a ) 2-bromo-4-hydroxytoluene from toluene, (6) 2-hydroxy-5-methylbenzaldehyde 4 from p-toluidine, (c) rn-methoxyaniline from benzenesulfonic acid.
(a)
6 6 6 6 6 6 0 z* 2 +6
z i ,
B
NO2
NO2 % ,:N
(6
B
r
1.
N.OH
hue with NaOH
,
NO2
r
~
~
G0 B,
1.
r
OH
NiCICHCI3,NmOH
G C H O
2. H,O+
OH
0
S03H
2. 3.
B
OH
N;CI-
O N O 2
G
* " N ;
NH2
"0.
NH2
S03H
G
r
":%?,
H+
0 OCH3
OCH3
Zn.HCI
+
NO2
NH2
Problem 19.29 ' Use the Bucherer reaction to prepare 2-(N-methyl)- and 2-(N-pheny1)naphthylamines.
4
CH,NH, and C,H5NH, replace NH,.
Problem 19.30
(U)
Give products, where formed, for reaction of PhNlCl- with (i) a-naphthol, (ii)
p -naphthol, (iii) 4-methyl-1-naphthol, (iv) 1-methyl-2-naphthol. (b) How can these products be used to make
m~ 00 00
the corresponding aminonaphthols?
4
p ) to OH; (iii) ortho ( p ) to OH, since para ( a )position is blocked. (iv) No reaction. An activating @-substituentcannot activate other @ positions.
( a ) Structural formulas are given below. Coupling occurs (i) paru ( a )to OH; (ii) ortho (a, not
@
00
N-NPh
N-NPh (i) 4-PhenylazoI-naphthol
@N=NPh
H &
00
CH3 (ii)
1-Phenylazo2-naphthol
(iii) 4-Methyl-2-phenylazo1-naphthol
(iv)
no reaction
PHENOLIC COMPOUNDS
454
[CHAP. 19
(6) Reduction of the azo compounds with LiAlH, or Na,S,O, or Sn, HCl yields the amines.
NH2 4-Amho- 1-naphthol from (i)
CH, 1-Amino-2-naphthol from ( i i )
2-Amino4mcthyl- 1-naphthol from (iii)
Problem 19.31 From readily available phenolic compounds, synthesize the antioxidant food preservatives ( a ) BHA (tert-butylatedhydroxyanisole, a mixture of 2- and 3-tert-butyl-4-methoxyphenol; (6) BHT (tert-butylated hydroxytoluene). 4 ( a ) p-CH,OC,H,OH + (CH,),C==CH, gives the mixture. This Friedel-Craft monoalkylation occurs both orfho to the OCH, and to the OH group. (6) p-cresol, p-HOC,H,CH,, is tert-butylated with (CH,),CH=CH,.
Problem 19.32 The acid-base indicator phenolphthalein is made by using anhydrous ZnCl ,to condense 2 mol of phenol and one mol of phthalic anhydride by eliminating 1 mol of H,O. What is its foiSmula? 4
OH-
H+
0
0 Phenolphthalein (colorless)
(red)
(colorless)
Problem 19.33 From phenol prepare ( a ) p-benzoquinone, (b) p-benzoquinone dioxime, (c) quinhydrone (a 4 1 : 1 complex of p-benzoquinone and hydroquinone). OH
0
NO
tautomers
NOH
a charge-transfer complex
CHAP. 191
PHENOLIC COMPOUNDS
455
Problem 19.34 HCI adds to p-benzoquinone by a 1,4-addition. Show the steps and predict the structure of the 4 phenolic product.
6ijCl4 0
tautornerire
OH 2-Chloro- 1 ,Mihydroxybenzene
Problem 19.35 Write a structural formula for the product of the Diels-Alder reaction of p-benzoquinone with: ( a ) Butadiene
(J
(b)
(c)
r
4
1.1 '-Bicyclohexadienyl
1,3-CycIohexadiene
OH
- - 0J
OH
OH
V
taulomers
Problem 19.36 The ir OH stretching bands for the three isomeric nitrophenols in KBr pellets and in dilute 4 CCI, solution are identical for the ortho but different for mefa and para isomers. Explain.
In KBr (solid state) the OH for all three isomers is H-bonded. In CCI,, the H-bonds of mefa and para isomers, which are intermolecular, are broken. Their ir OH absorption bands shift to higher frequencies (3325-3520 cm-'). There is no change in the absorption of the ortho isomer (3200cm-'), since intramolecular H-bonds are not broken upon dilution by solvent.
-
Problem 19.37 Show all major products for the following reactions: (U)
C H 3 e O H + CHCI,
+ NaOH
(c) Br O O C C H ,
II
+
70
-c
(6) C,H,CH,OOH
+ acid
4
CzHS 0
0
(a) Reimer-Tiemann reaction;
CH,OCE:' ( b ) This is a 1.2-shift of phenyl to an electron-deficient 0.
C,H,CH,OOH
+ HX
n
c6H5cH20-p?+H
H
-HO
-t
2CH,OC,H,
3
H, W,H,
[6,
1
__*
H&=O
+ C,H,OH
a hemiacetal
( c ) Intramolecular rearrangement gives two products, (A) and (B), and intermolecular rearrangement gives two more products, (C) and (D). These are Fries rearrangements.
PHENOLIC COMPOUNDS
456
(*) G C O C H ,
Br
(c) @OC*H5
[CHAP. 19
(D) G C O C H ,
&0C2H5
C2H5
C2H5
Br
Problem 19.38 ( a ) Compound (A), C,H,O, is insoluble in aqueous NaHCO, but dissolves in NaOH. When treated with bromine water, (A) rapidly forms compound (B), C,H,OBr,. Give structures for (A) and (B). (6) What would (A) have to be if it did not dissolve in NaOH? 4 ( a ) With four degrees of unsaturation, (A) has a benzene ring. From its solubility, (A) must be a phenol with a methyl substituent to account for the seventh C. Since a tribromo compound is formed, (A) must be (6) (A) could not be a phenol. It would have to be an ether, rn-cresol and (B) is 2,4,6-tribromo-3-methylphenol. C,H,OCH, (anisole).
Problem 19.39 A compound (C,,H,,O) dissolves in NaOH but not in NaHCO, solution. It reacts with Br, in water to give C,,,H,,Br,O. The ir spectrum shows a broad band at 3250cm-I and a strong peak at 830cm-’. The proton nmr shows: a singlet at S = 1.3 (9 H), a singlet at S = 4.9 (1 H), a multiplet at 6 = 7.0 (4 H). Give the structure of the compound. 4
The level of acidity, the facile reaction with Br,, the broad band at 3250 cm-’, and the singlet at S = 4.9 suggest a phenolic compound. The multiplet at S = 7.0, the dibromination (not tribromination), and the strong band at 830cm-’ indicate a p-substituted phenol. The singlet at S = 1.3 is typical of a t-butyl group. The compound is p-HOC,H,C(CH,),. Problem 19.40 Dioxin (2,3,7,8-tetrachlordibenzo-p-dioxin),
a very toxic compound, is a by-product of the manufacture of 2,4,5-trichlorophenoI by treating 1,2,4,54 tetrachlorobenzene with NaOH. Suggest a mechanism for the formation of dioxin. NaOH converts 2,4,5-trichlorophenol to its phenoxide anion, which undergoes aromatic nucleophilic displacement with unreacted 1,2,4,5-tetrachIorobenzene.
This product undergoes another aromatic nucleophilic displacement with OH -, to give a phenoxide thaf reacts by an intramolecular displacement to give dioxin.
Chapter 20 Aromatic Heterocyclic Compounds The chemistry of saturated heterocyclic compounds is characteristic of their functional group. For example, nitrogen compounds are amines, oxygen compounds are ethers, sulfur compounds are sulfides. Differences in chemical reactivity are observed for three-membered rings, e.g., epoxides, whose enhanced reactivity is driven by the relief of their severe ring strain. This chapter discusses heterocycles that are aromatic and have unique chemical properties. 20.1 FIVE-MEMIBERED AROMATIC HETEROCYCLES WITH ONE HETEROATOM
NOMENCLATURE; AROMA TICITY The ring index system combines ( 1 ) the prefix oxa- for 0, aza- for N or thia- for S; and (2) a stem for ring size and saturation or unsaturation. These are summarized in Table 20-1. Table 20-1. Ring Index Heterocyclic Nomenclature No N
Ring Size
I
Stem Saturated
Unsaturated
irane etane olane
irene
epane ocane
epine ocin
ete ole
ane
ine
The three most common five-membered aromatic rings are furan, with an 0 atom; pyrrole, with an N atom; and thiophene, with an S atom. Problem 20.1 Name the following compounds, using (i) numbers and (ii) Greek letters. ( a1
HC -CH HC,II ,C-CH3 I1
HC-CH (6 ) H,C-C, II ,C-CH, 11
0
S (c)
HC --C--CHJ HIC-C,
I1
II
(d)
,CH 0
HC-CH Br-C,
I1
11 ,C-COOH
4
N
( a ) 2-methylthiophene (2-methylthiole) or a-methylthiophene, ( b ) 2,5-dimethylfuran (2,s-dimethyloxole) or a , a '-dimethylfuran, (c) 2,4-dimethylfuran or a , /3 '-dimethylfuran (2,4-dimethyloxole), (d) 1-ethyl-5-bromoacid). 2-pyrrolecarboxylic acid or N-ethyl-a - bromo-a '-pyrrolecarboxylic acid (N-ethyl-5-bromazole-2-carboxylic
457
AROMATIC HETEROCYCLIC COMPOUNDS
458
[CHAP. 20
Problem 20.2 Write structures for ( a ) 2-benzoylthiophene, (6) 3-furansulfonic acid, ( c ) a,p ’-dichloropyrrole. 4 SO,H
i
CI
\
H Problem 20.3 Account for the aromaticity of furan, pyrrole and thiophene which are planar molecules with 4 bond angles of 120”.
See Fig. 20-1. The four C’s and the heteroatom Z use sp2-hybridized atomic orbitals to form the U bonds. When Z is 0 or S, one of the unshared pairs of e-’s is in an sp2 HO. Each C has a p orbital with one electron and the heteroatom 2 has a p orbital with two electrons. These five p orbitals are parallel to each other and overlap side-by-side to give a cyclic 7~ system with six p electrons. These compounds are aromatic because six electrons fit Huckel’s 4n + 2 rule, which is extended to include heteroatoms. It is noteworthy that typically these heteroatoms would use s p 3 HO’s for bonding. The exceptional sp2 HO’s lead to a p A 0 for the cyclic aromatic w system. Problem 20.4 (toward 0).
Account for the following dipole moments: furan, 0.7 D (away from 0);tetrahydrofuran, 1.7 D 4
In tetrahydrofuran the greater electronegativity of 0 directs the moment of the C - 0 bond toward 0. In furan, delocalization of an electron pair from 0 makes the ring C’s negative and 0 positive; the moment is away from 0. See Fig. 20-1.
Tetrahydrofuran
Furan
s p 2-s
U
bond
Fig. 20-1
PREPARATION Problem 20.5 Pyrroles, furans, and thiophenes are made by heating 1,4-dicarbonyl compounds with (NH,),CO,, P40,0,and P2S,, respectively. Which dicarbonyl compound is used to prepare ( a ) 3,4-dimethylfuran. (6) 2,5-dimethylthiophene. ( c ) 2,3-dimethylpyrrole? 4
The carbonyl C’s become the a C’s in the heterocyclic compound.
459
AROMATIC HETEROCYCLIC COMPOUNDS
H H
I I
H3c-7-7-cH3 H-C C-H
-
~
H3C-E-y-CH3 H- ,..,C-H
___* p4°10
-"l0
II II
8
0 0 2,3-Dimethylbutanedial
a dienediol
Ace tonylacetone
H2C-cHCH 3 H-C I C-CH3 1
3,CDimethylf uran
2,5-Dimethylthiophene
(NH&CO3
d NL
____*
H
3
I
H 3-Methyl4oxopentanal
2,3-Dimeth ylpyrrole
Problem 20.6 Prepare pyrrole from succinic anhydride.
4
Zn
N
N
H
H
Succinimide
23-Dihydrox ypyrrole
I
Succinic anhydride
I
Problem 20.7 Identify compounds (A) through (D).
CH,COOC,H,
Pyrrole
+ (A) 5C,H,COCH,COOC,H,
N + ' a
(B)
dil. NmOH
(C)
(D)
4
(A) PhCOOEt
Problem 20.8 Use any necessary inorganic reagents to prepare (a) a,a'-dimethylthiophene from 4 CH,COOC,H,, ( 6 ) phenyl 2-thienyl ketone from PhCOOH and OHCCH,CH,CHO.
a ) 2CH3COOC,H,
CH,COCH,COOC,H,
NaOE1 ,2 b
CH3CO-CH-CHCOCH3
I
C,H,OOC
t
COOC,H,
AROMATIC HETEROCYCLIC COMPOUNDS
460
PhCOOH 5 PhCOCI
[CHAP. 20
1
CHEMICAL PROPERTIES Problem 20.9 (a) In terms of relative stability of the intermediate, explain why an electrophile (E') attacks the (Y rather than the p position of pyrrole, furan, and thiophene. (6) Why are these heterocyclics more reactive 4 than C,H, to E'-attack?
( a ) The transition state and the intermediate R' formed by a-attack is a hybrid of three resonance structures which possess less energy; the intermediate from p-attack is less stable and has more energy because it is a hybrid of only two resonance structures. I and I1 are also more stable allylic carbocations; V is not allylic.
major product
c
111
I
I1
V
4
IV I
less stable
more stable
very minor product
( b ) This is ascribed to resonance structure 111, in which Z has + charge and in which all ring atoms have an octet of electrons. These heterocyclics are as reactive as PhOH and PhNH,. Problem 20.10 Explain why pyrrole is not basic.
4
The unshared pair of electrons on N is delocalized in an "aromatic sextet.,' Adding an acid to N could prevent delocalization and destroy the aromaticity. Problem 20.11 Give the type of reaction and the structures and names of the products obtained from: (a) furfural,
O 0C H O a -Furancarboxaldehyde
and concentrated aq. KOH; (6) furan with (i) CH,CO-ONO, (acetyl nitrate), (ii) (CH,CO),O and BF, and then H,O; (c) pyrrole with (i) SO, and pyridine, (ii) CHCI, and KOH, (iii) PhN;Cl-, (iv) Br, and C,H,OH; ( d ) thiophene and ( i ) H,SO,, ( i i ) (CH,CO),O and CH,COONO,, (iii) Br, in benzene. 4 (a)
Cannizzaro reaction;
Potassium furoate
(6)
(i) Nitration; 2-nitrofuran,
Furfuryl alcohol
CHAP. 201
AROMATIC HETEROCYCLIC COMPOUNDS
46 1
(ii) Acetylation; 2-acetylfuran,
(c)
(i) Sulfonation; 2-pyrrolesulfonic acid, O S O , H
N
(H,SO, alone destroys ring).
H
(ii) Reimer-Tiemann formylation ; 2-pyrrolecarboxaldeh yde (2-formyl pyrrole) ,
(iii) Coupling; 2-phenylazopyrrole,
O N = N P h
N H (iv) Bromination; 2,3,4,5tetrabromopyrrole. (d) (i) Sulfonation; thiophene-2-sulfonic acid,
0 S 0 3 H (ii ) Nitration ; 2-nit rot hiop hene ,
(iii) Bromination; 2,5-dibromothiophene. (Thiophene is less reactive than pyrrole and furan.) Problem 20.12 Write structures for the mononitration products of the following compounds and explain their formation: ( a ) 3-nitropyrrole, (6) 3-methoxythiophene, (c) 2-acetylthiophene, (d) 5-methyl-2-methoxy4 thiophene, (e) 5-methylfuran-2-carboxylic acid. ( a ) Nitration at C" to form 2,4-dinitropyrrole. After nitration C5(i) becomes
at C' (ii) would form an intermediate with a
C2,and C" becomes C'. Nitration
+ on C3, which has the electron-attracting -NO,
1
I1
IdocH3
( d ) H,C
2-Ni tro-3-methox ythiophene (Q and ortho to OCH,)
2-Acet yl-5-nitro-
thiophene
(a-attack)
2-Methoxy-3-nitro-5-methylthiophene (ortho to OCH,,a stronger activating group than CH,)
group.
AROMATIC HETEROCYCLIC COMPOUNDS
462
[CHAP. 20
Attack of NO; at C2, followed by elimination of CO, and H'. Problem 20.13 Name the products formed when ( a ) furan, (6) pyrrole are catalytically hydrogenated. (a)
CO)
( tetrahydrofuran, oxacyclopentane)
(6)
(, )
(pyrrolidine, azacyclopentane)
4
N
H Problem 20.14 Give the Diels-Alder product for the reaction of furan and maleic anhydride.
4
Furan is the least aromatic of the five-membered ring heterocyclics and acts as a diene toward strong dienophiles.
Furan
p-0
Maleic anhydride
0
Problem 20.15 Give the products of reaction of pyrrole with (a) I, in aqueous KI; (6) CH,CN followed by hydrolysis; ( c ) CH,MgI. (a)
2,3,4,5-Tetraiodopyrrole
A, (stronger
acid)
+ HCI,
4
(6) a-Acetylpyrrole [see Problem 19.21(c))
B, (stronger base)
B, (weaker base)
A, (weaker acid)
Problem 20.16 Give the sequence of steps for the formation of the open-chain products from reaction of ( a ) 2,5-diethylfuran with acid (H,O' ); ( b ) pyrrole with hydroxylamine, NH,OH+ Cl-. 4 ( a ) This is the reverse of the formation of furans from 1, 4-dicarbonyl compounds [Problem 20.5(a)]. Here the
product is the diketone formed after tautomerization of the intermediate dienediol.
463
AROMATIC HETEROCYCLIC COMPOUNDS
CHAP. 201
00
H,C - CH,
H,C--CH,
(6)
I
I
HC CH
\ /
OH OH
N
H
I
t
HC CH
It
HON
II
NOH
Butanedial dioxime
20.2 SIX-MEMBERED HETEROCYCLES WITH ONE HETEROATOM
The most important example of this category is pyridine (azabenzene), C,H,N: . Problem 20.17 Write the structural formulas and give the names of the isomeric methylpyridines.
4
There are three isomers. CH3
2- or a-Methylpyridine (a-Picoline)
3- or 8-Methylpyridine (B-Picoline)
4- or y-Methylpyridine (y-Picoline)
Problem 20.18 (a) Account for the aromaticity of pyridine, a planar structure with 120" bond angles. (6) IS pyridine basic? Explain. (c) Explain why piperidine (azacyclohexane) is more basic than pyridine. ( d ) Write the 4 equation for the reaction of pyridine with HCI.
( a ) Pyridine (azabenzene) is the nitrogen analog of benzene, and both have the same orbird picture (Figs. 10-1 and 10-2). The three double bonds furnish six p electrons for the delocalized T system, in accordance with Huckel's rule. (6) Yes. Unlike pyrrole, pyridine does not need the unshared pair of electrons on N for its aromatic sextet. The pair of electrons is available for bonding to acids.
y Piper idine
s p hybridized (less s character)
& %s p
hybridized (more s character)
P yridine
The less s character in the orbital holding the unshared pair of electrons, the more basic the site.
Pyridinum chloride a salt
Problem 20.19 Explain why pyridine (a) undergoes electrophilic substitution at the /3 position, and (6) is less 4 reactive than benzene. (a) The R"s formed by attack of E' at the a or y positions of pyridine have resonance structures (I, IV) with
a positive charge on N having a sextet of electrons. These are high-energy structures.
AROMATIC HETEROCYCLIC COMPOUNDS
464
I
a-Attack
I
111
I1
I
[CHAP. 20
y- Attack
IV
V
VI
With p-attack, the + charge in the intermediate is distributed only to C's. A + on C with six electrons is not as unstable as a + on N with six electrons, since N is more electronegative than C. P-Electrophilic substitution gives the more stable intermediate. 0-Attack
(6) N withdraws electrons by induction and destabilizes the R' intermediates formed from pyridine. Also, the N atom reacts with electrophiles to form a pyridinium cation, whose + charge decreases reactivity. Problem 20.20 How do the ' H nmr spectra of pyridine and benzene differ?
4
These are both aromatic compounds, and their ring-H signals are decidedly downfield. As all the H's of benzene are alike, one signal is observed. Pyridine gives three signals (not counting spin-spin coupling): 6-8.5 (two C2 H's), S = 7.06 (two C-?H's) and S = 7.46 (lone C' H). Notice that the C' H-signal is most downfield because the N is electron-withdrawing and less shielding. Problem 20.21 Give structural formulas and the names of the products formed when pyridine reacts with ( a ) 4 Br, at 300°C; ( 6 ) KNO,, H,SO, at 300°C and then KOH; (c) H,SO, at 350°C; (d) CH,COCI/AICI,.
(a)
aBr BrmBr aNo2 (6)
N
N
N
3-Bromo- and 33-Dibromopyridine
N+
I
(KOH neutralizes the formed pyridinum salt)
3-Nitropyridine
( d ) N o reaction. Pyridine is too unreactive to undergo FriedelCrafts acylations.
H 3-P yridinesulfonic acid (a dipolar ion)
Problem 20.22 Compare, and explain the difference between, pyridine and pyrrole with respect to reactivity 4 toward electrophilic substitution.
Pyrrole is more reactive than pyridine because its intermediate is more stable. For both compounds the intermediate has a + on N . However, the pyrrole intermediate is relatively stable because every atom has a complete octet, while the pyridine intermediate is very unstable because N has only six electrons. Problem 20.23 If oxidation of an aromatic ring is an electrophilic attack, predict the product formed when a-phenylpyridine is oxidized. 4
The pyridine ring is less reactive, and the benzene ring is oxidized to a-picolinic acid (a-NC,H,COOH).
CHAP. 201
AROMATIC HETEROCYCLIC COMPOUNDS
465
Problem 20.24 Predict and account for the product obtained and conditions used in nitration of 2aminopyridine. 4
The product is 2-amino-5-nitropyridine because substitution occurs preferentially at the sterically less hindered p position para to NH,. The conditions are milder than those for nitration of pyridine, because NH, is activating. Problem 20.25 Explain why (a) pyridine and NaNH, give a -aminopyridine, (6) 4-chloropyridine and NaOMe 4 give 4-methoxypyridine, ( c ) 3-chloropyridine and NaOMe give no reaction.
Electron-attracting N facilitates attack by strong nucleophiles in a and y positions. The intermediate is a carbanion stabilized by delocalization of - to the electronegative N. The intermediate carbanion readily reverts to a stable aromatic ring by ejecting an H:- in (a) or a :Cl:- in (6).
(c)
P-Nucleophilic attack does not give an intermediate with - on N.
Problem 20.26 Give the product(s) formed when pyridine reacts with (a) BMe,, (6) H,SO,, (c) t-BuBr,
Peroxybenzoic acid
4
+
Pyridine is a typical 3" amine. (a) C,H,N-BMe,; (6) (C,H,NH):SO:-, pyridinium sulfate; ( c ) [C,H,N'-Et]I-, N-ethylpyridinium iodide; (d) C,H,NH'Br- + Me,C----CHMe (the 3" halide undergoes elimination rather than S,2 displacement); (e)
0
Pyridine-N-oxide
N
I
0Problem 20.27 Account for the following orders of reactivity:
(a) Toward H,O+ :
2,6-dimethylpyridine (2,6-lutidine) > pyridine
(b) Toward the Lewis acid BMe,:
pyridine > 2,6-lutidine
4
( a ) Alkyl groups are electron-donating by induction and are base-strengthening. (6) BMe, is bulkier than an H,O+. The Me's at C 2 and C6 flanking the N sterically inhibit the approach of BMe,, causing 2,6-lutidine to be less reactive than pyridine. This is an example of F-strain (Front strain).
AROMATIC HETEROCYCLIC COMPOUNDS
466
[CHAP. 20
Problem 20.28 Compare, and account for the products formed from nitration of pyridine and of pyridine N-oxide. 4 Pyridine is nitrated in the p position only under vigorous conditions [see Problem 20.21(6)]. However, its N-oxide is nitrated readily in the y position; the intermediate formed by such attack is very stable because all atoms have an octet of electrons. The N-oxide behaves like phenoxide ion, C,H,O- [Problem l9.l8(a)].
Problem 20.29 Pyridine N-oxide is converted to pyridine by PCI, or by zinc and acid. Use this reaction for the 4 synthesis of 4-bromopyridine from pyridine.
0Pyridine
Pyridine N-oxide
04-Bromopyridine N-oxide
4-Bromopyridine
Problem 20.30 Account for the fact that the CH,’s of a - and y-picolines (methylpyridines) are more acidic than the CH of toluene. 4 They react with strong bases to form resonance-stabilized anions with - on N.
anion (base I )
u-Picoline (acid, )
(base:)
(acid:)
anion (base )
Problem 20.31 Give the products formed when y-picoline reacts with C,H,Li and then with (a) CO, then 4 H,O’ ; ( h ) C,H,CHO then H,O‘.
AROMATIC HETEROCYCLIC COMPOUNDS
CHAP. 201
467
CH,COOH I
6
/
4-Pyridylacetic acid
N
4-Stilbazole
Problem 20.32 From picolines prepare ( a ) the vitamin niacin (3-pyridinecarboxylic acid), ( b ) the anti4 tuberculosis drug isoniazide (4-pyridinecarboxylic acid hydrazide) .
6 N
3-Picoline
Niacin
0
COOH
KMnO4,
2. I.
H soclz ~ N N H I GHNH2 ~
N
QPicoline
N Isoniazide
4-P yridine-
carboxylic acid
20.3 COMPOUNDS WITH TWO HETEROATOMS
We use the oxa-aza-thiu system to name these compounds; the suffixes -ole and mine indicate five- and six-membered rings, respectively. When there is more than one kind of ring heteroatom, the atom of higher atomic number receives the lower number. Problem 20.33 Name the following compounds:
(4
r y S
(4
0
(4
l 0g
I
H ( a ) 1,3-diazine (pyrimidine); ( b ) 1,3-thiazole; ( c ) 1,4-diazine (pyrazine); ( d ) 1,2-0xazole; (e) imidazole.
H3cc10
Three pyrimidines are among the constituents of nucleic acids:
60, NH,
Cytosine
OH
CLOH Uracil
OH
Thymine
468
AROMATIC HETEROCYCLIC COMPOUNDS
[CHAP. 20
Problem 20.34 Write the tautomeric structures of these pyrimidines.
Cytosine
Uracil
4
Thy mine
Problem 20.35 ( a ) What makes imidazole (Prob. 2 0 . 3 7 ~ )aromatic? (b) Explain why midazole, unlike pyrrole, 4 is basic. Which N is the basic site? ( a ) Imidazole is aromatic because it has a sextet of electrons from the two doubl : bonds and the electron pair on the N bonded to H. (b) The proton acceptor is the N not bonded to H, because its lone pair is not part of the aromatic sextet.
20.4
CONDENSED RING SYSTEMS
Many biologically important heterocyclic compounds have fused (condensed) ring systems. In particular, the purines adenine and guanine are found in DNA (with cytosine, 5-methylcytosine, and thymine) and also in RNA (with cytosine and uracil).
Adenine
Guanine
Quinoline
Isoquinoline
Indole
QUINOLINE ( I-AZANAPHTHALE") Problem 20.36 Which dicarboxylic acid is formed on oxidation of quinoline? The pyridine ring is more stable than the benzene ring [Problem 20.19(6)]. COOH
N Quinoline
COOH
Quinolinic acid
Problem 20.37 Quinoline is prepared by the Skraup reaction of aniline, glycerol and nitrobenzene. Suggest a mechanism involving Michael addition of aniline to an a,P -unsaturated aldehyde, ring closure, and then 4 dehydration and oxidation. The steps in the reaction are: ( 1) Dehydration of glycerol to acrolein (propenal).
H,COHCHOHCH,OHGlycerol
H2S04
H,C=CHCHO Acrolein
+ 2H,O
CHAP. 20)
469
AROMATIC HETEROCYCLIC COMPOUNDS
(2) Michael-type addition (Section 17.4) c
L
J
( 3 ) Ring closure by attack of the electrophilic carbonyl C on the aromatic ring ortho to the electronreleasing -NH. The 2" alcohol formed is dehydrated to 1.2-dihydroquinoline by the strong acid.
1,2-Dihydroquinoline
Quinoline
(4) PhNO, oxidizes the dihydroquinoline to the aromatic compound quinoline. PhNOz is reduced to PhNH,, which then reacts with more acrolein. This often violent reaction is moderated by added FeSO,. Problem 20.38 Give the expected products of (a) monobromination of quinoline, (6) catalytic reduction of 4 quinoline with 2 mol of H,. (a) The more reactive phenyl ring undergoes E +-attack in the
(Y
positions, giving
Br A
Br
I
I
8-Bromoquinoline
5-Bromoquinoline
(b) The more electron-deficient pyridine ring is more easily reduced.
I ,2,3,4Tetrahydroquinoline
Problem 20.39 Give structures for the products of reaction of quinoline with ( a ) HNO,, H,SO,; (b) NaNH,; (c) PhLi. 4 (a) 5-nitro- and 8-nitroquinoline; (b) 2-amino- and 4-aminoquinoline (like pyridine, quinoline undergoes nucleophilic substitution in the 2 and 4 positions); (c) 2-phenylquinoline. Problem 20.40 Outline a mechanism for the Bischler-Napieralski synthesis of 1-methylisoquinoline from N-acetylphenylethylamine by reaction with strong acid and P,O,, and then oxidation of the dihydroisoquinoline intermediate. 4
470
AROMATIC HETEROCYCLIC COMPOUNDS
[CHAP. 20
The mechanism is similar to that of the Skraup synthesis (Problem 20.37) in that carbonyl 0 is protonated and the electrophilic C attacks the benzene ring in cyclization, to be followed by dehydration and oxidation.
Supplementary Problems Problem 20.41 Supply systematic names for:
( a ) 4-phenyl-l,2-oxazole, ( 6 ) 3-methyl-5-bromo-l,2,4-triazine-6-carboxylic acid, (c) 2,4-dimethyl-I ,3thiazole, (d) 1,2,3,4-thiatriazoIe, (e) 2,3-benzazole (indole).
Problem 20.42 Name the following compounds systematically: (a)
11) N
H
(6)
c- 0 0 (d)
(c)
S
0
(e)
(sB S
0
U)
CJ 4
N
( a ) azole (pyrrole), (b) 1,3-thiazole, (c) 2H-oxirine, ( d ) 4H-oxirine (pyran), (e) 1,4dithiazine, ( f )
I ,3-diazine (pyrimidine). 2H- and 4H- are used in (c) and (d) to differentiate the position of the saturated sp3 atom, Common names are given in parentheses.
Problem 20.43 Write structures for ( a ) oxirane, (6) 1,2-oxazole, (c) 1,4-diazine (pyrazine), (d) 1-thia-4-oxa4 6-azocine, (e) 3H-1,2,4-triazole, ( f ) azepane.
N ( a ) H2C-CH2
0 ''
N
H
Problem 20.44 How many thiophenyl-thiophenes (bithienyls) are possible?
2,2'-Bithienyl
2,3'-Bithienyl
3,3'-Bithienyl
4
CHAP. 201
47 1
AROMATIC HETEROCYCLIC COMPOUNDS
Problem 20.45 Which methylquinoline is vigorously oxidized to a tricarboxylic acid that is then dehydrated to 4 a mixture of two carboxylic acid anhydrides?
If the CH, is on the benzene ring, oxidation gives only 2,3-pyridinedicarboxylic acid, regardless of the position of the CH,. If the CH, is on the pyridine ring, there are three possible tricarboxylic acids:
W
C
I
H
WCH3
3
HoococooH I oxid.
oxid.
oxid.
N
N
HOOC
H HOOC o o C ONC O O H
HOOC
- H20
0
0
0
II
II
F D C O O H
4‘
O\
‘
II
N
H
II
0
0
HOOC
N
N
0
Two anhydrides are obtained only from 4-methylquinoline.
Problem 20.46 Identify the compounds represented by Roman numerals.
Quinoline
P -CsHs OrH
Pyrrole (PyH) Furan (FuH)
HNOj
I
+ p-HO3SC6H4N2X-
+I-
(CH,CO),O
(CzHs)zO. BF3
I
PCI
11 A 111
NaOl
Sn HCI
I1 + I11
f u m . H SO
I1A I11
I
I1
Quinoline N-oxide
4-Nitroquinoline N-oxide
4-Nitroquinoline
2-PyN=NC,H4SO3 H-p
2-PyNH2
p-NH,C,H,SO,
2-FuCOCH3
2-FuCOO-Na’
5-H03S-2-FuCOOH
4
I11
Problem 20.47 Prepare ( a ) 3-aminopyridine from P-picoline, (b) 4-aminopyridine from pyridine, ( c ) 8hydroxyquinoline from quinoline, (d) 5-nitro-2-furoic acid from furfural, (e) 2-pyridylacetic acid from pyridine. 4
AROMATIC HETEROCYCLIC COMPOUNDS
472
0Ouinoline
_*,
H2S04
[CHAP. 20
0-
Ouinoline-8-sulfonic acid
OH-
I.
-
.. _ _ _
U
-
(COOH stabilizes the ring towards acid cleavage of the ether bond.)
(4
Pyridine
NHj
2-Aminopyridine
-
2-Bromopyridine + [CH(COOC,H,),l
Na'
__*
nucleophilic nucleophilic displacement
N
I
N a N 0 2 . H ' . 5 "C Cuer
2-Bromopyridine
:::
CH(COOC,H,),
'.
0 CHZCOOH
A. -cq
N N LV
Problem 20.48 ( a ) Explain why pyran [Problem 20.42(d)] is not aromatic. (6) What structural change would theoretically make it aromatic? 4 ( a ) There are six electrons available: four from the two bonds and two from the 0 atom. However, C4 is sp'-hybridized and has no p orbital available for cyclic p orbital overlap. (6) Convert C4 to a carbocation. C4 would now be sp2-hybridized and would have an empty p orbital for cyclic overlap.
Problem 20.49 How can pyridine and piperidine be distinguished by infrared spectroscopy?
4
Piperidine has an N-H bond absorbing at 3500 cm-' and H - C ( s p 3 ) stretch below 3000 cm-'. Pyridine has no N-H; has H--C(sp2) stretch above 3000cm-'; C = C and C=N stretches near 1600 and lSOOcrn-', respectively; aromatic ring vibrations near 1200 and 1050 cm - ; and C-H deformations at 750 cm - The peak at 750 varies with substitution in the pyridine ring. Problem 20.50 How can nmr spectroscopy distinguish among aniline, pyridine and piperidine?
4
The NH, of aniline is electron-donating and shields the aromatic H's; their chemical shift is 8 = 6.5-7.0 (for benzene the chemical shift is S = 7.1). The N of pyridine is electron-withdrawing and weakly shields the aromatic H's (6 = 7.5-8.0). Piperidine is not aromatic and has no signals in these regions. Problem 20.51 From pyridine (PyH), 2-picoline (2-PyMe), and any reagent without the pyridine ring prepare
( a ) 2 -acetylpyridine, ( 6 ) 2-vinylpyridine (c) 2 -cyclopropylpyridine (d) 2 -PyCH,CH,CH,COOH ' (e) 2-
PyC( Me)==CHCH,, ( f ) 2-pyridinecarboxaldebyde. Any synthesized compound can be used in ensuing steps.
4
-
( a ) By a crossed-Claisen condensation
2-Py Me
KMnOI
-
EtOH
2-PyCOOH 7 2-PyCOOEt
2-PyCOCH2COOEt
(6)
(4
OH -
heat
2-PyCOCH3-
C H COOEt
NaBHI
-
-
2[2-PyCOCH2COOH]
2-PyCOCH2COO-
-
2-PyCHOHCH3
2-PyCH=CH2+ CH,N,
uv
PP,"
2-PyCH=CH2
product
- CO2
2-PyCOCH3
CHAP. 201
AROMATIC HETEROCYCLIC COMPOUNDS
( d ) By a Michael addition:
-
2-PyCH=CH2 + (EtOOC),CH Na'
473
--+ 2-PyCHCH2CH(C0OEt),Na'
This a-pyridylcarbanion is stabilized by charge delocalization to the ring N (Problem 20.30). Refluxing the salt in HCl causes decarboxylation and gives the pyridinium salt of the product, which is then neutralized with OH-. ( e ) Use the Wittig synthesis: 2-PyCOMe + Ph,P=CHCH, +products (?runs and cis).
(f)
1. 0
2-PyCH=CH2 3 2-PyCHO 2. Zn, HOAc
Problem 20.52 Suggest a mechanism for the formation of 2-pyridone from the reaction of pyridine with KOH.
4
As with the strongly basic :NHS [Problem 20.25(u)], nucleophilic attack by :OH- at Cz, followed by :H--elimination, yields an enolic intermediate that tautomerizes to 2-pyridone.
Problem 20.53 ( a ) Account for the aromaticity of a- and y-pyridones. ( b ) Explain why a-pyridone (Problem 20.52) predominates over a -pyridinol, especially in the solid state. (The same is true for the y-tautomers.) (c) How can ir spectroscopy show which tautomer -predominates? 4 (a)
See Fig. 20-2 for the cyclic aromatic extended .rr-bonding with six electrons. The carbonyl carbon furnishes an empty p AO, N furnishes a p A 0 with a pair of electrons, and each doubly bonded C furnishes a p A 0 with one electron.
Fig. 20-2
( b ) The N-H forms a strong intermolecular H-bond with 0 of C=O. This bond, repeated throughout the crystalline solid, links molecules in endless helices. ( c ) The ir spectra of the solid (and solution) show a strong C=O stretzhing band.
Chapter 21 Amino Acids, Peptides, Proteins 21.1
INTRODUCTION
This chapter deals with the very important a-amino acids, the building blocks of the proteins that are necessary for the function and structure of living cells. Enzymes, the highly specific biochemical catalysts are proteins. a-Amino acids are dipolar ions (zwitterions), RCH(N+H,)COO-, as is indicated by their crystallinity, high melting point, and solubility in water rather than in nonpolar solvents. The standard (naturally occurring) amino acids are listed in Table 21-1; those marked with an asterisk are essential amino acids that cannot be synthesized in the body and so must be in the diet. They have 1' NH,'s except for proline and hydroxyproline (2'). They have different R groups. Table 21-1. Natural a-Amino Acids
Name
Formula
Symbol
MonoaminomonocarboxyIic Glycine Alanine Valine* Leucine* Isoleucine* Serine Threonine*
I
+
H,NCH,COOH,hCH(CH,)COOH,N+ C H( i-Pr)COOH,N+ CH(i-Bu)COOH,NC H( s-Bu)C 00+ H,NCH(CH,OH)COO+ H,NCH(CHOHCH,)COO-
GlY Ala Val LeU
Ileu Ser Thr
Monoaminodicarboxylic and Amide Derivatives Aspartic acid Asparagine Glutamic acid Glutamine
ASP ASP(NH3 Glu Glu(N H,)
+
HOOCCH,CH( N+H,)COOH,NCOCH,CH(NH,)COO+ HOOC( C H2),C H(N H ,)COOH,NCOCH,CH,CH(kH,)COO-
Diaminomonocarbox y lic ~
+
~~
H ,N( C H,),C H( N H,)COO-+ H,NCH, HCH,CH,CH(NH,)COO-
Lysine* H ydrox ylysine
'iOH
H2N\
+..yC-=N
Arginine*
H (CH 2)3CH(N HJCOO-
H2N
C y steine Cystine Met hionhe*
CySH cysscy Met
+
H,NCH(CH,SH)COO-
-OOCCH(~H,)CH,S-SCH,CH(hH,)COOC H ,SC H ,C H,C H
474
(6H ,)COO-
475
AMINO ACIDS, PEPTIDES, PROTEINS
CHAP. 211
I
Table 21-1. Natural a-Amino Acids (Continued) Name
Phenylalanine* Tyrosine
Formula
Symbol
Phe TYr
+
PhCH,CH(NH,)CO?p-HOC,H,CH,CH( NH3)COO-
Histidine* ~
,
Proline
Hydrox yproline
H o br3 C O O -
H'
'H
&
C H2C HCOO-
Tryptophane*
+
L
3
H
Except for glycine, they are chiral molecules with the configuration at the a C.
L
or S (or R, for cysteine and cystine)
Problem 21.1 (a) Find the absolute configuration of the following L-amino acids:
coo-
coo-
coo-
CH,SH
CH,OH
CH,CH,SCH,
(i) L-Cysteine
(ii) L-Serine
(iii) L-Methionine
( 6 ) Why is L-cysteine (along with L-cystine) unique among the twenty a-amino acids in having the R 4 configuration? (a) Redrawing the Fischer formulas with the lowest-priority ligand (H) away from the viewer and assigning priorities, we obtain:
AMINO ACIDS, PEPTIDES, PROTEINS
476
/
\H
[CHAP. 21
(iii) S
(ii) S
(i) R
( b ) The CH,SH group of cysteine is one of two R's (the other is part of cystine) that outrank COO- because S has a higher atomic weight than 0.
21.2
PREPARATION OF a-AMlNO ACIDS
LABORATORY AND INDUSTRIAL METHODS Problem 21.2 Synthesize leucine, (CH,),CHCH,CH(NH,)COO , by ( a ) Hell-Volhard-Zelinsky reaction (Section 16.3) followed by ammonolysis, (6) Gabriel synthesis, (c) phthalimidomalonic ester synthesis, ( d ) reductive amination of a keto acid, (e) Strecker synthesis (addition of NH,, HCN to RCH=O), ( f ) 4 acet yla m i nomalonic ester . Hr2 P
(CH,),CHCH,CH,COOH-
excess
( C H , ) 2 C H - - € H 2 C H B r C O O H r Leu
QQt;; Q&pT 0
Br
C
cII
COOEt
COOEt HjO+
CH2 I
II
11
CH(C H A
-L&u+Q
COOH COOH
I
CH(CHh Use ester rather than acid to prevent conversion of anion to phthalimide. 0
C H , ( C O O E t ) , z BrCH(COOEt),
-
0
II
1. 2,
PhthN--CH(COOEt)2
COOEt -CH,CHMe,
0 (PhthN -)
(use below)
NaOH HjOf
1. 2.
Ns+OEtBcH2CHMe2
Leu
3. hept
'
+ Phthalicacid + CO2 + 2EtOH
AOOEt
R
NH
,
ME~CHCH~CCOOHT
H
I
Me,CHCH,C=O
LEU
F
H
+ NHf
0'
Me,CHCH, HCN A Leu
NH2 In the presence of NH, an aminonitrile is formed instead of a cyanohydrin.
( f ) H,C(COOEt),
HON=C(COOEt), an oximino
ester
1. 2,
H2IPI
Ac.O
+
AcNHCH(COOEt),
NaOEt BcH2CHM9+
Acet y laminomalonic
ester
COOEt
1.
2. 3,
NaOH HJO"
b
Leu
+ CO, + 2EtOH
AMINO ACIDS, PEPTIDES, PROTEINS
CHAP. 211
477
Problem 21.3 Write structural formulas for compounds (A) through (C) and classify the reactions. 4 2
'CH2COOC2H, (A) 'L=O I
(crossed-Claisen condensation)
*COOC2H, 1
3
4
(B) CH,-C-COOH
I
0 (C) CH,-CH-COO-
I
(ester hydrolysis and decarboxylation of a P-keto acid; see Section 17.3) (reductive amination)
+NH3 Problem 21.4 Outline the preparation of ( a ) methionine from acrolein by the Strecker synthesis [Problem 4 21.2(e)], (6) glutamic acid by the phthalimidomalonic ester synthesis [Problem 21.2(c)].
H NHJ Methionine
Phthalimidomalonic ester
oddition
COOEt
I
CH,CHCOOEt % H3A-
phthN-lOOEt
Glutamic acid
BIOSYNTHESIS: TRANSAMINATION In the transamination reaction, in the presence of transaminase enzymes, (S)-glutamic acid reacts with the a-keto acid analog of the desired a-amino acid to give the desired (S)-amino acid and the keto analog of glutamic acid. HOOCCH,CH,CHCOO-
I NH ;
+ HOOCCH,CCOO-
Glutamic acid
b
transaminase
Oxaloacetic acid
HOOCCH,CH,CCOO-
b
+ HOOCCH,CHCOOI
a-Ketoglutaric acid
NH_:
Aspartic acid
Problem 21.5 (a) What is the relationship between reactants and products in the transamination reaction? (6) Which ketoacid is needed to give (i) alanine? (ii) leucine? (iii) serine? (iv) glutamine? (c) Which amino acids cannot be made by transamination? 4 (0)
Glutamic acid + Oxalacetic acid+ oxidant
reductant
a-Ketoglutaric acid + Aspartic acid reductant
( b ) (i) CH,COCOOH; (ii) (CH,),CHCH,COCOOH; (iii) HOCH,COCOOH;
(iv) H,NCOCH,CH,COCOOH.
oxidantz
AMINO ACIDS, PEPTIDES, PROTEINS
478
(c)
[CHAP. 21
Proline and hydroxyproline; they are 2' amines, and only 1' amines can be made this way.
21.3 CHEMICAL PROPERTIES OF AMINO ACIDS
ACID-BASE (AMPHOTERIC1 PROPERTIES Problem 21.6 Write equilibria showing the amphoteric behavior of an H 2 0 solution of glycine. All monoaminomonocarboxylic acids are slightly acidic. 4 H,O+
+ H,NCH,COO-
-
K,] = 1 0 -2
KOZ=16 X 1 0 - I o
Anion (acts as RNH2)
Ampholyte
H,kCH,COOH
+ OH-
Cation (acts as RCOOH)
Since the solution is acidic, the reaction to the left is favored and the anion predominates. Problem 21.7 ( a ) How does one repress the ionization of an a-amino acid? (6) The pH at which [anion] = [cation] is called the isoelectric point. In an electrolysis experiment, do a-amino acids migrate at the isoelectric point? 4 ( a ) Since the a-amino acid solution is slightly acidic, the anion predominates, and acid is added to repress the ionization. (6) No.
Problem 21.8 Why is H,NCH,COOH (pK,, = 2.4) more acidic than RCH,COOH (pK, t
a -H,N
= 4-5)?
4
increases acidity because of its electron-withdrawing inductive effect.
Problem 21.9 ( a ) Why does sulfanilic (p-aminobenzenesulfonic) acid exist as a dipolar ion, while p 4 aminobenzoic acid does not? (6) Why is sulfanilic acid soluble in bases but not in acids? ( a ) -SO,H is strongly acidic and donates H' to the weakly basic arylamino grpup. ArCOOHjs not acidic enough to transfer H' to the arylamino group. (6) In dipolar sulfanilic acid, p-H,NC,H,SOJ, H,N- is acidic enough to transfer H' to bases to give the soluble anion, p-H,NC,H,SO;, -SO; is too feebly basic and cannot accept H' from acids.
Problem 21.10 Predict whether the isoelectric points for the following a-amino acids are considerably acidic, slightly acidic, or basic; ( a ) alanine, (6) lysine, (c) aspartic acid, (d) cystine, (e) tyrosine. (See Table 21-1 and Problem 21.6.) 4 (a) Alanine is a monoaminomonocarboxylic acid. Its isoelectric point is slightly acidic (pH = 6.0). (6) Lysine is a diaminomonocarboxylic acid.
H,N(CH,),CH(NH2)COO-
+ H 2 0=H,N(CH,),CH(NH,)COO-
+ OH-
Additional base is needed to repress this ionization. The isoelectric point of lysine is basic (pH = 9.6). ( c ) Aspartic acid is a monoaminodicarboxylic acid. HOOCCH,CH(~H,)COO- + H,O-
-OOCCH,CH(hH,)COO-
+ H,O+
A more strongly acidic solution is needed to repress this ionization. Aspartic acid's isoelectric point is strongly acidic (pH = 2.7). ( d ) Cystine is a diaminodicarboxylic acid and behaves like a monoaminomonocarboxylic acid. The isoelectric point is slightly acid (pH = 4.6). (e) Tyrosine is a monoaminomonocarboxylic acid containing a phenolic OH, which however is too weakly acidic to ionize to any significant extent. The isoelectric point is slightly acidic (pH = 5.6).
CHAP. 211
479
AMINO ACIDS, PEPTIDES, PROTEINS
Problem 21.11 Why is the isoelectric point of arginine (pH = 10.7) greater than that of lysine (pH = 9.6)? 4
The second basic site in arginine is an R-substituted guanidino group. NH
II
H,N-C-NHR This group is more basic than the charge-delocalized cation:
E
NH, of lysine because, on accepting H', the guanidino group gives a
Problem 21.12 How can lysine be separated from glycine?
4
The isoelectric points are different: pH = 9.6 for lysine and pH = 5.97 for glycine. An aqueous solution of the mixture is placed between two electrodes, the pH adjusted to either 5.97 or 9.6, and an electric current applied. At pH 5.97, glycine doesn't migrate, while cationic lysine migrates to the cathode, where it is collected. At pH 9.6, lysine doesn't migrate, while anionic glycine migrates to the anode. Problem 21.13 Give the products of the reaction of glycine with ( a ) aqueous KOH, (b) aqueous HCI. ( a ) H,NCH,COO
- K ',
4
(6) CI- H,NCH,COOH.
REACTIONS OF FUNCTIONAL GROUPS Problem 21.14 Show that amino acids have properties typical both of - C O O H and -NH,, by giving the products formed when glycine reacts with ( a ) CH,COCI; (6) NaNO, + HCI(HON0); (c) EtOH + HCI; (d) PhCOCl+ NaOH; (e) Ba(OH),, heat. 4 ( a ) CH,CONHCH,COOH (6) HOCH,COOH + N, (c) C1- H, NCH,COOE t
(d) PhCONHCH,COO - Na +
(e) CH,NH,
+ BaCO, + H,O
(decarboxylation)
Problem 21.15 On being heated, amino acids behave like hydroxyacids (Problems 16.24 an$ 16.25). giving N-analog products. Write the structural formulas for the products formed on heating ( a ) o-RCHNH,COO-, (6)
p-RCHhH,CH,COO-, (c) yH,N(CH,),COO-, (d) 6-H3N(CH,),C00-. (e) E-H~N(CH,),COO-. R O
O R
an a-amino acid
a diketopiperazine
(b) RCH=CHCOO-NH: (c)
d ; H
.heat*
6: N H
+ H,O
H a y-amino acid
a y-lactam (a cyclic amide)
4
[CHAP. 21
AMINO ACIDS, PEPTIDES, PROTEINS
480
mer of Nylon 6
Intramolecular cyclization would give a 7-membered ring, which is formed with difficulty. Hence the more facile intermolecular reaction occurs. Since the substrate is bifunctional, polymerization occurs (see Section 16.12). Problem 21.16 When the methyl ester of rac-alanine is heated, two diastereomeric dimethyldiketopiperazines [Problem 21.15(a)j are obtained. One of them cannot be resolved. Give structures for the two products, and account for their stereochemistry. 4
The cis is resolvable.
CH3
CH3
cis ;racemic
trans ;meso
Problem 21.17 Threonine has two chiral C's. Write Fischer projections for its stereoisomers.
cooH3k&-I H OH \
coo-
cooH$H3 HO
CH3
CH3
H3i$ HO
coo-
OH
CH3
,
,3H3
4
CH3
,
racemate 2 (erythro)
racemate 1 (threo)
Problem 21.18 Write structural formulas for the products of the reactions of: (a) alanine + carbobenzoxy 4 chloride (C,H,CH,OCOCI); (6) tyrosine + aq. Br,, and the product reacted with (CH,),SO, + NaOH.
O H H
I I
I
C,HsCH,OC-N--C-COOH
'
CH30bCH2-CH-C Br
NH2
No \O-Na+
Problem 21.19 Can the acid chloride of an a-amino acid be made by adding SOCI, to the acid?
4
48 1
AMINO ACIDS, PEPTIDES, PROTEINS
CHAP. 211
No. Two or more molecules of amino acid chloride react to give peptides. H,NCHRCOCI
+ H,NCHRCOCI+
H,NCHRCONHCHRCOCI
ANALYTICAL DETECTION Amino acids are separated by chromatography and detected with ninhydrin; they give an extensively charge-delocalized, purple anion. 0
03=N-b 0-
0
+ NH;
CHRCOO----,
II
II
0
II
0
Ninhydrin hydrate
+ R c H o + CO, + H +
0 purple
Problem 21.20 Many a-amino acids react with HONO to give a quantitative volume of N2 (Van Slyke method of analysis). Which amino acids in Table 21-1 cannot be analyzed by this method? 4
The amino acid must have an -NH,. HONO.
Proline and hydroxyproline are 2" amines and do not evolve Nzwith
Problem 21.21 The ir spectra of amino acids have bands near 1400 and 1600cm-' that correspond to the COO- group. Why does a strong peak appear at 1720cm-' in highly acidic solution? 4
The 1400- and 1600-cm- I peaks are caused by the symmetrical and antisymmetrical C - 0 stretching in COO-. Acid converts COO- to COOH, whose C=O produces the 1720-cm-' peak.
21.4
PEPTIDES
INTRODUCTION Problem 21.22 Peptides are polyamides, like the product in Problem 21.15(e). Draw the structural formula for (a) a dipeptide, (h) a polypeptide unit of alanine (indicate the mer). 4 N-terminal acid
C-terminal acid
+
1
,
H3N-(fH-E-Y-TH-COOMe H Me H
I
H
H
I
I
H
I
a polypeptide
Problem 21.23
Name the tripeptide, Ala.Gly.Va1.
4
The full name of the C-terminal amino acid is used. For all other amino acids, the suffix -yl replaces - h e of the name. The name is alanylglycylvaline. Problem 21.24 Draw the structural formula for the pentapeptide glycylglycylalanylphenylalanylleucine (Gly .Gly .Ala. Phe. Leu). 4
482
AMINO ACIDS, PEPTIDES, PROTEINS
[CHAP. 21
+
The sequence starts at the left end with the amino acid having the free a NH, and ends on the right with the amino acid having the free COO-. GlY /
+
L
' -
GlY
Ala
Phe
A
A
f
H3NCH2~--~-CH2~-~-
I
LeU
II
IH3
A
I I LH2
H CH,CH(CH,),
hh Problem 21.25 ( a ) The artificial sweetener aspartame is a synthetic dipeptide, Asp.Phe. How many stereo4 men can aspartame have? (b) Which isomer would result if naturally occurring amino acids were used?
( a ) Since each amino acid component has one stereocenter. aspartame has two different stereocenters. There are four stereomers: SS and RR to give one racemic form, and SR and RS to give another. (6) The naturally occurring amino acids are both S, and their combination would give only the SS stereomer.
STRUCTURE DETERMINATION 1.
Total Hydrolysis to Amino Acids
Peptides (and proteins) are hydrolyzed with acid in amino acid analyzers to give the kind and number of each amino acid constituent. 2. Partial Hydrolysis to Smaller Polypeptide Chains Large peptides are partially hydrolyzed with enzymes (trypsin and chymotropsin) or acid to mixtures of di- and tripeptides. From the composition of these small units we establish the sequence of amino acids in the large polypeptide. Problem 21.26 What products are expected from ( a ) complete and (6) partial hydrolysis of the tetrapeptide 4 Ala .Glu .G1y .Leu? ( a ) The amino acids Ala, Glu, Gly, and Leu. (6) The dipeptides Ala.Glu, Glu.Gly, Gly.Leu and the tripeptides Ala. Glu.Gly and Glu .Gly .Leu.
Problem 21.27 What is the sequence of amino acids in the following peptides? ( a ) a tripeptide Gly,Leu,Asp (commas indicate an unknown sequence) that is partially hydrolyzed to the dipeptides Gly.Leu and Asp.Gly; (6) a heptapeptide Gly,Ser,His, ,Ala, ,Asp that is hydrolyzed to the tripeptides Gly.Ser.Asp, His. Ala.Gly, and 4 Asp. His. Ala.
S i y e Gly is bonded to Leu in one dipeptide and to Asp in the other, Gly must be in the middle. The free -NH, is in aspartic acid in Asp.Gly and the free --COO- is in leucine in Gly.Leu. The sequence is Asp.Gly. Leu. The N-terminal acid in the heptapeptide must be the first amino acid in one of the three tripeptides. The possibilities are Gly,His, or Asp. However, Gly and Asp, which are one of a kind in the heptapeptide, are also found in the third position of different tripeptides. Hence, Gly and Asp cannot be first in the original compound. In the heptapeptide, there are two His units, one of which must be the first amino acid. His is first in His.Ala.Gly and this unit comes from the first three amino acids of the heptapeptide. The C-terminal amino acid must be the last amino acid in one of the three tripeptides. The possibilities are Asp, Gly, or Ala. Since the one-of-a-kind Gly and Asp are multi-positioned in the tripeptides, they cannot be terminal in the heptapeptide. Ala must be the end amino acid in the heptapeptide and the tripeptide Asp.His.Ala comes from the last 3 amino acids. The seventh and middle amino acid is Ser. The tripeptides are arranged in order, and cross lines remove duplicating amino acids. His.Ala.Gly
+ @$.Ser.Mp + Asp.His.Ala = His.Ala.Gly.Ser.Asp.His.Ala
CHAP. 211
483
AMINO ACIDS, PEPTIDES, PROTEINS
SEQUENCING
By lopping off one terminal amino acid at a time from either the free amino end (N-terminal) or the free COOH end (C-terminal), sequencing techniques give the linkage order of the amino acids in the peptide. 1. Sanger method
The N-terminal acid is detected with 2,4-dinitrofluorobenzene
H
H
NO*
peptide
DNFB
(DNFB).
N-(2,4-dinitrophenyl) peptide (N-DNPpeptide)
R
O
H I 11 w-CH-C-OH
(This step is an aromatic nucleophilic displacement.)
+
+
R'
I
H3N-C-COO-
I
H N-( 2,4-dinitrophenyl)amino acid (identified)
unlabeled amino acid
2. Edman Method
The initial reagent for detecting the N-terminal group is an aryl isothiocyanate. R O H R ' O H R " Ar-N-C-S
-!.
I
+ H-N-
11 I
I
I1 I
-C-N-~-C-N-~-COOH H
aryl isothiocyanate
I
*
H
peptide
H
S H
Ar-N-C-yI I1
I
R 0 H R ' O H R"
I -CtN-C-C-N-C-COOH II I I II I I I I H
dry HCI
H
Thiourea-labcled peptide
S
II
R' 0 H
C
Ar-N' 'h-H 0-C-C-R I I
H
Phenylthiohydantoinlabeled amino acid (identified)
+ H3A-(!f
R"
-c-N-rcO " I
H
I
H
peptide less H,N terminal amino acid (can repeat degradation)
AMINO ACIDS, PEPTIDES, PROTEINS
484
[CHAP. 21
3. Carboxypeptidase enzyme is used to detect the C-terminal amino acid.
R 0 H R 0 H R' +
I
I I
I
I I
H3N-C-C-N-C-C-/-N-C-COO-
I
I
H
H
I I
carboxy-
peptidase b
R O H R
H
H
H
R"
+ I + H3N-C-COOI H
+ I I I I I H,N-C-C-N-C-COOI I
Problem 21.28 Determine the amino acid sequence of a peptide which: (i) on complete hydrolysis, separation of the amino acids, and determination of the amounts, shows the composition Le u z, A rg ,Cy SH ,G lu ,I le u ,Val ,Tyr ,Phe (ii) in sequencing studies reacts with DNBF to produce N-DNP tyrosine on hydrolysis, and yields valine with carboxypeptidase; (iii) undergoes partial hydrolysis to the tripeptides Leu.Val.Va1 + Leu.Arg.CySH
+ Tyr.Ileu.Phe + Phe.Glu.Leu + Arg.CySH.Leu
4
(i) There are 10 amino acids in the chain, but since there are 2 mol each of Leu and Val, only 8 are distinct. (ii) Sequencing establishes tyrosine as the first ( l ) , and one of the valines as the last ( l O ) , amino acid in the chain. (iii) From partial hydrolysis, the tripeptides establish the first three amino acids as Tyr(l), Ileu(2), Phe(3); and the last three as Leu(8), Val(9), and Val( 10). The remaining amino acids (4) through (7) are placed from the other tripeptides. There is only one Phe(3) and thus the tripeptide Phe(3).Glu(4).Leu(S) gives the fourth and fifth acids. There are two Leu molecules and one has been identified as Leu(8), so that the second Leu, which is (5) in the chain, is in the tripeptide Leu(5).Arg(6).CySH(7). Thus amino acids (6) and (7) are established. Placing the tripeptides in correct position and eliminating common acids, we have:
The position of CySH(7) is confirmed from the tripeptide Arg(6).CySH(7).Leu(8). The sequence in the peptide is Tyr.Ileu.Phe.Glu.Leu.Arg.CySH.Leu.Va1 .Val.
SYNTHESIS The problem in combining different amino acids in a specific sequence to synthesize polypeptides is to prevent reaction between the COOH (or COX) and NH2 of the same amino acid (Problem 21.19). Blocking the N-Terminal Amino Group
1.
This reaction is blocked by attaching a group (B) to N of NH,. After the desired peptide is constructed, the blocking group is removed without destroying the peptide linkages. The general scheme is: +
H,NCHRCOO-
block NH,
(BPNHCHRCOOH
activate COOH
(Bb-NHCHRCOX
H~~HR'COO-
0 NHCHR'COOH
II
H3NCHRCNHCHR'COO+
485
AMINO ACIDS, PEPTIDES, PROTEINS
CHAP. 21)
Table 21-2
Blocking
(B) Group PhCH,OC=O
I
Unblocking
Reagent
Name
Reagent
PhCH20COCl
Carbobenzoxy chloride
HJPt or cold HBr, HOAc
di-t-Butyl di-carbonate
CF,COOH
(2)
t-BuOC=O
,
(f-Buo% )
I
(Boc)
2°
c
Products PhCH, + CO, + peptide PhCHzBr+ CO, + peptide (CH3),C=CHZ + peptide
+ CO,
2. Activating the COOH The --COOH is activated to form a peptide bond with an N-terminal NH2 by reacting the N-blocked amino acid with ethyl chloroformate, CIC-OEt
8
A mixed anhydride of an alkyl carbonic acid is formed. This technique minimizes racemization of the chiral (Y C. Unactivated COOH groups form peptide bonds directly with N-terminal amino groups in the presence of
(N ,N-dicyclohexylcarbodimideor
cyclohexyl-N=C=N-cyclohexyl
DCC)
a powerful dehydrating agent that forms the urea derivative H--c y cloh e x y I
cy clohex y l-N H-C-N
tl
Problem 21.29 Prepare Gly.Val by the ( a ) carbobenzoxy and (6) Boc methods. Use symbols Z and Boc as in 4 Table 21-2, and X= - 0 C O O E t . 0 0 (U)
Z-CI
+ H,N+ CH,COO-
-
II
ClCOEt
Z-NHCH~COO- -Z-NHCH~C--%
II
H,I Pt
- PhCH3. -CO?
H,N'CH,CONHCH( i-Pr)COO-
(6) Boc-0-Boc+ H,N' CH,COO--Boc-NHCH,COO-. fluoroacetic acid, CF,COOH, is used instead of HJPt.
3.
k i 3 N < . t i ( i .Pr ) C ' O O (va,,ne)
*
N-protected, COOH activated
N-protected
0
2-NHCH,CNHCH(i-Pr)COOH-
II
(product)
Repeat as in ( a ) , changing Z to Boc. Tri-
Merrifield Solid-Phase Synthesis
To expedite peptide syntheses of up to 124 amino acid units, beads of solid polystyrene-with chloromethyl groups, - C H , C I , para to about every hundredth phenyl group-are used in an automated process. The sequence of steps is: (1) attach C-terminal Boc-protected amino acid by
AMINO ACIDS, PEPTIDES, PROTEINS
486
[CHAP. 21
replacing C1 of the CH,C1 group, and wash; (2) remove Boc and wash; (3) add Boc-protected amino acid with the aid of DCC, and wash. Steps (2) and (3) are repeated as many times as is necessary to make the desired peptide. In the final step the completed peptide is detached by hydrolysis of the ester bond that holds the peptide to the solid surface. The average yield for each step exceeds 99%.
Problem 21.30 Describe the Merrifield solid-phase synthesis when carried out in the reverse order of the classical Sanger and Edman methods. 4
A Boc-protected amino acid is first attached to the resin and then the Boc is removed to liberate a free amino group. This is reacted with another Boc-protected amino acid. To synthesize, e.g., Ala.Gly, the Gly is first attached to the resin.
Problem 21.31 Which a-amino acid can act to cross-link peptide chains?
4
Cysteine. Cysteine linkages also occur between distant parts of the same chain.
-
Cross-linking
21.5
s
PROTEINS
INTRODUCTION
,
Proteins with molecular weights in the millions are the major constituents of all living cells. Simple proteins are hydrolyzed only to amino acids. Coqjugated proteins are hydrolyzed to amino acids and nonpeptide substances known as prosthetic groups. These prosthetic groups include nucleic acids of nucleoproteins, carbohydrates of glycoproteins, pigments (such as hemin and chlorophyll) of chromoproteins, and fats or lipids of lipoproteins. PROPERTIES 1.
Amphoteric Properties. Isoelectric Points and Electrophoresis
Proteins have different isoelectric points, and in an electrochemical cell they migrate to one of the electrodes (depending on their charge, size, and shape) at different speeds. This difference in behavior is used in electrophoresis for the separation and analysis of protein mixtures.
CHAP. 211
AMINO ACIDS, PEPTIDES, PROTEINS
487
2. Hydrolysis Proteins are hydrolyzed to a-amino acids by heating with aqueous strong acids or, at room temperature, by digestive enzymes such as trypsin, chymotrypsin, or pepsin. 3.
Denaturation
Heat, strong acids or bases, ethanol, or heavy-metal ions irreversibly alter the secondary structure of proteins (see below). This process, known as denaturation, is exemplified by the heat-induced coagulation and hardening of egg white (albumin). Denaturation destroys the physiological activity of proteins.
STRUCTURE 1. The primary structure consists of the number and sequence of the constituent amino acids. 2. The secondary structure arises from different conformations of the protein chains; these conformations are best determined by X-ray analysis. There are three types. ( a ) The a-helix is a mainly right-handed coiled arrangement maintained by H-bonds between an N-H and O=C that are four peptide bonds apart. (b) The pleated sheet has chains lying side by side and linked through N - H - - - - O K H-bonds. The a C’s rotate slightly out of the plane of the merger (to minimize repulsions between their bulky R groups), which gives rise to the “pleats.” (c) Random structures have no repeating geometric pattern. However, there are structural constraints on the randomness leading to an ordered random orientation. 3. Tertiary structure is determined by any folding of the chains. There are two types. ( a ) Fibrous proteins are water-insoluble, elongated, threadlike helixes (occasionally pleated sheets) made up of chains which are bundled together intermolecularly through N-H - - - O=C H-bonding. They include fibroin (found in silk), keratin (in hair, skin, feathers, etc.), and myosin (in muscle tissue). (b) Globular proteins, or globulins, are folded into compact spheroid shapes such that hydrophilic R groups point outward toward the water solvent and the hydrophobic (lipophilic) R groups turn inward. As a result, globulins can dissolve in or easily emulsify with water. The shape is maintained by intramolecular H-bonding. The secondary structure is a combination of random (always present), helical, and pleated structures. Globulins include all enzymes, antibodies, albumin of eggs, hemoglobin, and many hormones such an insulin. 4. Quaternary structure exists when two or more polypeptide chains are linked only by weak forces of attraction among R groups at the surface of the chains. Such proteins are called oligomers (dimers, trimers, and so on).
Problem 21.32 The position of an amino acid along the chain and the nature of its R group critically influence protein function. This is revealed by a study of sickle-cell anemia, a hereditary disease affecting red blood cells (erythrocytes). Normal hemoglobin, part of disk-shaped erythrocytes, has a glutamic acid residue at position 6, whereas abnormal hemoglobin, part of crescent-shaped (“sickle”) cells, has a valine residue. Explain how this single error, (there are 146 amino acid units) causes sickling. 4 Glutamic acid has an R group with a polar ionic end (--COO-) that extends outward toward the water solvent, giving the normal disklike shape. Valine has a lipophilic R group, i-propyl (isopropyl), which pulls the portion of protein where it occurs inward away from the water, causing sickling.
488
[CHAP. 21
AMINO ACIDS, PEPTIDES, PROTEINS
21.6 SUMMARY OF AMINO ACID CHEMISTRY
PREPARATION
PROPERTIES
1. a-Amino Acids
1. a-Amino Acids
(a) Ammonolysis
(a) Amino Group
+ NaOH -+ H , N - C H R - C O O -
X
I
OH
+ CHR-COOH
Halo Acids: NH,
+ HNO,+
I
CHR-COO-
Gabriel: Qf>N--CHR--COOH
+ R'-C(t-X+ + CH,--V+ Ba(OH),-
R
Malonic Acid: NH,
Na'
I + X-C-COOH
HI
R'--CONCHR--COOH,C=N--CHR--COOH,N--CH,R + BaCO,
AOOH COOH
I
HN--CHR--C----O
+ NH, + H,
Keto Acids: R&O
I
W-CHR-NH
\
( b ) Alkylation
( b ) Carboxyl Group
\+
/
Acetylaminomalonic Ester:
R\/C \ CH3%-N COOR'
I
H +--+ H,&--CHR--COOH + R'OH+ H,~--CHR--COOR' + HOH 0
+'
SOCl,+
II
H,k-CHR--C--CI
Azlactone Synthesis H 0
I
II
R (c) Aminonitrile RCH-0 2.
+ NH,CN
-
@-AminoAcids
NH,
I
RCH-CN
2.
/
(a) Halo Acids, Gabriel, etc. X-CH,CH,COOH + NH, ( b ) Unsaturated Acids H , C= C-C O O H 3.
k
I :N-"\
N=C;
+
H,N-CH,CH,COO-
/
p-Amino Acids
+ HNO,
T H , C - - C - - IC O O H
k
+ NH,'
y- and &Amino Acids
(a) Halo Acids X--CH,CH,CH,--COOH
H,C=C--COOH
-
+ NH3
3.
k
+ NH,
7- and &Amino Acids
H3N--CH2CH2CH2-COO--
heat
NH
+ N,
CHAP. 21)
AMINO ACIDS, PEPTIDES, PROTEINS
489
Supplementary Problems Problem 21.33 (a) How many dipeptides can be synthesized from the amino acids in Table 21-I? (6) How many dipeptides in (a)have the empirical formula (i)X2,(ii)X,Y? ( c ) How many dipeptides are there with (i) two (ii) one essential amino(s)? (6)Give the number of possible proteins with 100 amino acids. (a) There are 23 choices for X, followed by 23 choices for Y. Thus there are 23 x 23 = 529 possible dipeptides. (6) (i) There are 23X,'s. (ii) For X, Y, there are 23 x 22 = 506. (c) There are 10 essential amino acids X* and 13 non-essential ones, (i) lOX: (ii) 10 x 13 = 130 when the H,N terminal amino acid is essential (X?Y) and 13 x 10 = 130 when the COOH terminal amino acid is essential (Y.X*). The total =260. (6)23100. Problem 21.34 Name the tetrapeptide
0 0 H ,N-CH-C-N+H--C-N+H-C-NH-CH~-COO 11 H F H
0
II
-
4
Val.Phe. Ala.Gly, or valylphenyalanylglycine. Problem 21.35 Count the possible amino acid sequences for a tetrapeptide having the empirical formula 4 Gly, ,Ala,. (The number of sequences equals the number of ways of reserving two out of four positions €or glycine, which is 4! G.G.AA, G.A.G.A., G.A.A.G, A.A.G.G, A.G.A.G, A.G.G.A Problem 21.36 Write the structural formula for tyrosylglycylalanine.
4
+
-
Problem 21.37 Identify the compounds (I), (11), (111). (I)
+ H,NCHMeCOOCH,CONHCHMeCOO(11) + HONO+ N, + HOCH,CH(n-Pr)COOH (111) + NH, + H2-MeCH(NH3)COONaOH
Na' 4
(I) ( C H , C 0 ) 2 0 ; (11) H,NCH,CH(n-Pr)COO- ; (111) MeCOCOOH. Problem 21.38 Identify the compounds ( I ) through (IV).
aP
\ N - K +5 (11) /
C
II
0
hydrolysis
+
4
PhCOOH
(111)
AMINO ACIDS, PEPTIDES, PROTEINS
490
[CHAP. 21
(I) ethyl bromoacetate (or any haloacetate); (11) PhthN--CH,-COOC2H, [Problem 21.2(c)]; (111) PhCOCl; (IV) hippuric acid, PhCONHCH,COOH. Problem 21.39
synthesis.
Synthesize valine ( a ) from isobutyl alcohol by a Strecker reaction, (6) by malonic ester
4
I
(a 1
CH, NH2
CH3
CH,
I
NHlCl NaCN
* CH$HCH=O
CHJCHCHZOH
+
I
CH, NH,
I
I
H O+
I
* CH,CH-CHCN 3 CH,CH-CHCOOValine
CH, CH,CHBr
(b)
CH, H
CH3
I
I
&a:CH(COOR)2 +
CH3CHCH(C00R)2
I 2.
3.
NaOH.hea1 HCI Br2/p
4
NH3(xs)
I
t
+ CH3cH--F-c00-+NH, Valine
Problem 21.40 Starting with acetamidomalonic acid, prepare phenylalanine.
COOEt
I
HCNHCOCH,
Loon
::
4
-
COOEt
I
* PhCH2-C-NHCOCH,
'NH,
I
PhCH,-CH-COO-
AOOEt
Problem 21.41 Use p-methoxytoluene and any other needed reagents to prepare tyrosine.
0'' :: '
CH30
C H 3 0 ~ c H 2 B r If,
CH(COOEI)~,
CH,O
CH,-CH(COOEt),
4
i, heal
3
'
$JH3 0cH2cH2c00H CH3O
CH2dHc00-
I.
Br2/P
2.
N H J (excess)*
CH3O NH3BrCH2CHCOOH I
HO
conc. -CHjBr HBr. A
~
-0
NH3
I
CH2CHCOO-
H Tyrosine
p-CH,C,H,OH itself cannot be used because the ring would be brominated and the acidic H of OH would convert basic malonate carbanion to malonic ester. t
Problem 21.42 Calculate the isoelectric point of alanine, given for H,NCHMeCOOH pK,, = 2.3 and pK,, = 9.7. 4
Isoelectric point = 4 (2.3 + 9.7) = 6.0. Problem 21.43 What occurs when an electric current is passed through an aqueous solution, buffered at pH = 6.0, of alanine (6.0), glutamic acid (3.2), and arginine (10.7)? The isoelectric points are shown in parentheses. 4
Glutamic acid is present as an anion and migrates to the anode. Arginine is a cation and migrates to the cathode. Alanine is a dipolar ion and remains uniformly distributed in solution.
CHAP. 211
491
AMINO ACIDS, PEPTIDES, PROTEINS
Problem 21.44 Supply structural formulas for principal organic compounds (I) through (XIV).
02N0N:2
Ph-N=C=S
-
loo-
+ H3N+
H-CH,--CH,-S-CH, +
+ (CH3),CHCH-NH3
(I)
(111)
1(1;)
Loo-
F
2 Ph- COCl+ -0OC H ( C H , ) , N H , ’ Z (IV) NH2
+
NH3
I
C&CH2-cH-cOOHooCOO-
+ (CH3CO)20OH-\ (V)
+ C,H,OH
(VI)
H/”\H
[
=(XI) -
I -
-0OC-CHCHzCOO-
H+
H+
OH-
OH-
(:;I)] (XIII)
02N@H-kH-CH2-CH2SCH3
loo-
PhCO-NH-
H(CHJ4-NH-COPh
HO ON C O O C 2 H , (VI) H’
‘H
H+
OH-
(I)
(IV)
(XIV)
AMINO ACIDS, PEPTIDES, PROTEINS
492
1
(VII) -OOC--SH(CH,),NH,
+kH3
U)
-
+
+
-OOC-CH(CH2),NH3
H0OC-CH(CH2),NH3
I
L
‘NH,
I
NH,
[CHAP. 21
+
(1x1
+
11 (XI)
Problem 21.45
Explain the use of the enzyme deacylase in the resolution of racemic amino acids.
4
The amino group of the racemic amino acid is acetylated with acetic anhydride to give racemic N-acetylamino acid. The deacylase hydrolyzes only the L enantiomer, which is thereby separated from the D enantiomer. The latter is then chemically hydrolyzed to give the free D amino acid. Problem 21.46 Use DNFB
(see Section 21.4, “Sequencing”) to distinguish between Lys.Gly and Gly.Lys.
DNFB
+ H,F;(cH?),CH(NH,)CONHCH,COOLys.Gly
--+
HN(CH,),CHCONHCH,COOH=
I
DNP
I
DNPNH
HN(CH3,CHCOOH
I
The a and
E
NH2’s are bonded to 0 H
DNFB
1I I
COO-
I
DNP
H
DNP
.
I
DNPNH
+ H,NCH,COOGlycine
COOH
I
+
+ H,NCH2C-N-~(CH2),NH3
4
__*
H O+
NH --L
)
DNP
DNP
Gly.Lys
HNCH,COOH
I DNP
+ HOOC-CH(CH,),NH +kH,
I
DNP
After hydrolysis of the DNP-derivative of Gly.Lys, both amino acids have a D N P attached. However, only the amino acid with DNP attached to the a NH, (in this case, glycine) is the NH, terminal amino acid. When a diaminomonocarboxylic acid is part of a peptide chain, but not in an NH, terminal position, it will be derivatized by D N P but not at the a NH,. Problem 21.47 Among the hydrolysis products of the pentapeptide Gly,,Ala,Phe are Ala.Gly and Gly.Ala. 4 Give two possible structures if this substance gives no N, with HONO (Van Slyke method).
Since N, is not obtained, there is no free NH,. The compound must be a cyclic peptide. It has the partial sequence Gly.Ala.Gly. The remaining Gly and Phe units can be placed in two sequences, giving
Gly .Ala.GIy Gly.Ala.Gly . or . Gly.Phe Phe.Gly Problem 21.48 Indicate the products obtained after a second Edman degradation of Ala.Phe.Gly.Val.
4
CHAP. 211
493
AMINO ACIDS, PEPTIDES, PROTEINS
S
II
C
Gly.Val
and
Ar-N
,./ \ NH
I
I
O=C--CHPhe
-
-
Problem 21.49
Prepare glycylalanyltyrosine from the free amino acids by the carbobenzoxy method.
C,H,CH,OCOCI
+ H,NCH,COO-
+
4
alanine
C,H,CH,OCONHCH,COOH~
-GlY
GiY
C,H,CH~OCONHCH,CONHCH(CH,)COOHGlY
Ala
C,H,CH,OCONHCH,CONHCH(CH,)CONH . . . COOH[Dcc-see
page 485-forms
H,i
I
\
GlY
Ala
a peptide from the acid itself.]
Pd
C,H,CH,
+ CO, + Gly.Ala.Tyr
TYr
Use Boc (Table 21-2) as a protecting group to synthesize the tripeptide Ala.Gly.Va1. (Let 4 H CH H CH CH,
Problem 21.50
X- - 0 C O O E t .)
+
I
H,N--C--COO-
A
H
CH,
H3N+--CH2--C00-
Eoc-0-Eoc ______*
HI
I k H O
+
(1-Pr)
I I
H,N-C--COO-
H
CH
H
k Problem 21.51 Account for the stereochemical specificity of enzymes with chiral substrates.
4
Since enzymes are proteins made up of optically active amino acids, enzymes are themselves optically active and therefore react with only one enantiomer of a chiral substrate. Problem 21.52 Explain how the following denature proteins; ( a ) Pb2+and Ag', (6) EtOH, (c) urea, and (d) heat, 4 ( a ) These heavy-metal cations form insoluble salts with the free COO- groups. (6) EtOH interferes with H-bonding by offering its own competing H. (c) Urea is a good H-bond acceptor and interferes with H-bonding in the protein. (d) Heat causes more random conformations in the protein and decreases the conformational population needed for helix formation.
Chapter 22 Carbohydrates and Nucleic Acids 22.1
INTRODUCTION
Carbohydrates (saccharides) are aliphatic polyhydroxyaldehydes (aldoses), polyhydroxyketones (ketoses), or compounds that can be hydrolyzed to them. The suffix -ose denotes this class of compounds. The monosaccharide D-( +)-glucose, an aldohexose, is formed by plants in photosynthesis and is converted to the plysaccharides cellulose and starch. Simple saccharides are called sugars. Polysaccharides are hydrolyzable to monosaccharides; e.g., a mol of trisaccharide gives 3 mol of monosaccharides. Problem 22.1 Classify the following monosaccharides: (a)
( 6 ) HO-CH2-
g-l-CH2-OH
( c ) HO-CH2-
H6H
-
6
H-
6,
H-
H-
6H 6H
H-CH2-OH
6,
H-CH-0 4
6H
The number of C's in a chain is indicated by the prefix -di-, -tri-, etc. (a) ketotriose, (6) aldohexose, (c) ketopen tose . Problem 22.2 ( a ) Write a structural formula for (i) an aldotriose, (ii) a ketohexose, (iii) a deoxypentose. ( b ) Write molecular formulas for (i) a hexose trisaccharide, (ii) a pentose polysaccharide. (c) What is the general formula for most carbohydrates? 4 ( a ) (i)
HO-CH,
(ii)
H--CH=O
(iii)* 0 OH OH OH
OH
TH=O CH2 I HCOH
I
* "Deoxy" means H in place of OH.
HCOH
I
CH20H
(6) Because n moles of C,H,,O, (monosaccharide) = 1 rnol of polysaccharide + (n - I ) rnol of H 2 0
we have: (9 n (c)
= 3,
C,,H,,O,,;
(ii) (C,H,O,),
C,(H,O),. This formula accounts for the name carbohydrate.
Problem 22.3 Deduce the gross structure of glucose from the following data. ( a ) The elementary composition is C=40.0%, H = 6 . 7 % , 0=53.3%. (6) A solution of 18g in 100g of H,O freezes at -1.86"C. (c) Mild oxidation with Br2/H,0, Ag(NH,)i (Tollens' reagent) or Cu" (tartrate) complex (Fehling's test) gives a carboxylic acid with the same number of C's. (4 This acid, an aldonic acid, reacts with HI and P to give n-hexanoic acid. (e) One mole of glucose reacts with 5 rnol of Ac,O to form an ester. (#) Vigorous oxidation of glucose with HNO, gives a dicarboxylic acid (an aldaric acid) with the same number of C's. (g) Reduction of glucose with H,/ Ni or Na/ Hg gives a product that reacts with 6 mol of Ac,O. 4
494
CARBOHYDRATES AND NUCLEIC ACIDS
CHAP. 221
495
(a) The molar amounts of C, H, and 0 in 1OOg of glucose would be
53.3
6.7 - = 6.7 1
4012.0= 3 * 3 3
-= 3.33 16
Thus, C : H : 0 = 1 : 2 : 1, and the empirical formula is CH,O (empirical mol. wt. = 30). (6) 18 g in 100 g of H,O is equivalent to 180 g in 1 kg of H 2 0 . Since Afp(freezing point) = 1.86, this solution is one-molal (Afp = K,,rn), and 180g is the gram molecular weight. The molecular formula is (CH,O),, where 180(molecularweight) =6 30(empirical weight)
n=
Thus glucose is C,H,,O,. (c) Glucose has a HC=O group, which undergoes mild oxidation to COOH. ( d ) The six C’s of glucose are in a chain sequence (there are no branching C’s). (e) Glucose has five OH’s that are acetylated to a pentaacetate. ( f )One of the OH’s is l”, since it was oxidized to a COOH along with CHO. (g) CHO is reduced to CH,OH to give a hexahydroxy compound (an alditol). The gross structure of glucose is HOCH,CHOHCHOHCHOHCHOHCHO.
---
Problem 22.4 ( a ) The monosaccharide
Fructose
HCN
D-(
-)-fructose undergoes the reaction sequence
H30’
HI/P
CH,CH,CH,CH,CHCH,COOH
Classify fructose. (6) Which carboxylic acid is formed when glucose is submitted to the same sequence?
4
( a ) Fructose is a hexose since the final product has seven C’s, one of which came from HCN addition to a
M.The precursor of COOH is CN, which added to C=O. COOH is attached to the C that was originally the C of -0. Fructose is a 2-ketohexose. The sequence is:
-
=O
HO-
HCN
H20H 2-ketohexose
cyanoh ydrin
hydroxy acid
2-Methylhexanoic acid
(6) By this sequence, glucose gives heptanoic acid. Problem 22.5 ( a ) How many chiral C’s are there in a typical (i) aldohexose (C,H,,O,), (ii) 2-ketohexose? (6) How many stereomers should an aldohexose have? 4 ( a ) (i) Four:
6
5
4
3
2
6
I
5
4
(6) There are four different chiral C’s and 2, = 16 stereomers.
22.2 CHEMICAL PROPERTIES OF MONOSACCHARIDES 1.
Oxidation [see Problem 22.3(c) and ( f ) ]
Problem 22.6
I
6
What are the products of reaction of HOCH,(CHOH),CHO (I) and
0 2
II
HO~H,(CHOH),-C-&H,OH with ( a ) Br,
+ H,O?
(6) HNO,? (c) HIO,?
3
2
1
(ii) 3: HOCH,$HOH$HOH$HOHCOCH,OH
HoCH,$HOH$HOH$HOH$HOHCHO
(11)
496
[CHAP. 22
CARBOHYDRATES AND NUCLEIC ACIDS
(a)
( b) (c)
I
I
I
I
HOCH,(CHOH),COOH
N o reaction (ketones are not oxidized)
0 II HOOC(CHOH),-C-COOH
HOOC(CHOH),COOH tH,=O
I1
l6
+ 4HCOOH + HdOOH
(1" OH'S
react)
+ 3HCOOH + 60, + dHz=O
CH,=O
2. Reaction in Base Problem 22.7 ( a ) Explain how in basic solution an equilibrium is established between an aldose, its Cz epimer (a diastereomer with a different configuration at one chiral C) and a 2-ketose. (b) Will fructose give a positive Fehling's test which is done in a basic solution? 4 (a)
In basic solutions, aldoses and ketoses tautomerize to a common intermediate enediol, and the following equilibria are established: achiral, sp'
H-C-0 H I~
chiral, s p 3
U
1 .
C
~
O
I
H
H-C-OH
I C-o
~
I
I
H-C-OH I
H-~--oH
I
H-&OH
I
enediol
aldose
ketose
When the aldose is reformed from the enediol, H' can attack the now achiral C2 from either side of the double bond to give C' epimers. chiral, sp'
H-A-OH
I
I
enediol
I C 2 epimeric aldoses
(6) Since fructose isomerizes to an aldohexose (glucose) in base, it gives a positive test: Cupric tartrate (blue solution)
C ~ , O(red solid)
3. Reduction to Alditols [Problem 22.3(g)J Problem 22.8 Glucose is reduced to a single alditol; fructose is reduced to two epimers, one of which is identical to the alditol from glucose. Explain in terms of configurations. 4
In glucose no new chiral C is formed on going from CHO to CH,OH. In fructose the carbonyl C (C') becomes chiral; there are two configurations and we get epimers.
CH,OH
>c=o R'
---
CH,OH
~--foH
+H
CH,OH o k H
The chiral C-', C', and C' [see Problem 22.5(a)] of glucose and fructose have the same configuration. The identical alditols also have the same configuration at C'.
CHAP. 221
CARBOHYDRATES AND NUCLEIC ACIDS
497
4. Kiliani-Fischer Step-up Method (Addition of One Carbon)
CN
H-
COOH
&-OH),
-!%(H-t-oH),
-OH
H-
H-C-0
H-
-OH (!H20H
I
LH20H
LH20H
HCN. NaCN
COOH
CN
I
HO-
-H
I (H-C-OH)3
CH2OH
I
H O*
A
H-A-OH
I
I H-
I
CH2OH aldohexose
H-C=O
CH20H
cyanohydrins
glyconic acids
I
CHzOH I
CH20H
glyconolactones
aldoheptoses
diastereomeric and epimeric
Problem 22.9 Why does the Kiliani-Fischer method give unequal amounts of diastereomeric products?
4
The presence of stereocenters in sugars causes their C = O groups to have diastereotopic faces (Section 5.4) that react at different rates, resulting in unequal amounts of diastereomers.
5. Degradation (Loss of One Carbon) (a)
Ruff degradation:
glyconic acid
aldohexose
aldopentose
( b ) Wohl degradation: Problem 22.10 Outline the steps in the Wohl degradation, which employs a dehydration of an aldose oxime and is thus a reversal of the Kiliani-Fischer step-up method. 4 HC=O
I
HC=NOH
I
HCN is removed from the cyanohydrin by Ag'.
C=N
I
498
6.
[CHAP. 22
CARBOHYDRATES AND NUCLEIC ACIDS
Reaction with Phenylhydrazine. Osazone Formation
CH-0
'
I
I
CHoH
H-C-OH
PhNHNH
I
A
(CHOH)3 AH20H
H-C=N
NHPh7
I
C-NNHPh
PhNHNH
PhN H N H2
d
-PhNH2. - N H J *
(H- A-OH), I 6H20H
H-C-NNHPh
(H-k-OH),
J
LH20H
phen ylh ydrazone
aldohexose
osazone
Problem 22.11 The C'-epimeric aldohexoses glucose and mannose give the same osazone as fructose. Write 4 equations and explain the configurational significance.
Since mannose and glucose are C' epimers, they are identical at the C3,C', C-', and C" portions of the molecule, which are unaltered during osazone formation. The chiral C 2 , which differs in each hexose, loses chirality in the osazone and becomes identical for both mannose and glucose. 2-Ketohexoses give osazones in which the C ' CH,OH is oxidized. Since C ', CJ, C.', and C" of fructose and glucose are identical (Problem 22.8), fructose also gives the same osazone. Identical portions are in the boxes below.
k, CHO
H-
H-C-NNHPh
CHO
-OH
Or
3PhNHNH2
or
H o - l ,
Glucose
=NNHPh
same osazone
Mannose C' epimers
: :1
Problem 22.12 Osazones are converted by PhCHO to 1,2-dicarbonyl compounds called osones. Use this 4 reaction to change glucose to fructose.
H-C=O
H-C-NNHPh
I
3PhNHNH2*
(H-
k="Hph -OH),
H-C=O 1-2 PhCH-NNHPh) phCHO
H20H Glucose
-CHO
22.3
1.
osazone
*
Zn.HAc,
(H-
-OH),
(H-
-OH),
H20H osone
Fructose
rather than C=O of the osone is reduced.
EVIDENCE FOR HEMIACETAL FORMATION AS EXEMPLIFIED WITH GLUCOSE
Some Negative Aldehyde Tests
Glucose does not form a bisulfite adduct or give a magenta color with the Schiff reagent. 2.
Mutarotation
Naturally occurring (+)-glucose is obtained in two forms: mp = 146 "C, [a],= +112" and mp = 150 O C , [a],= + 19". The specific rotation of each changes (mutarotates) in water, and both reach a constant value of +52.7".
499
CARBOHYDRATES AND NUCLEIC ACIDS
CHAP. 221
3. Acetal (Glycoside) Formation Unlike typical aldehydes, 1 mol of glucose reacts with only 1 mol of ROH in dry HCI to give two isomeric acetals (alkyl glycosides). See Problem 15.35. To explain the above results, glucose is assumed to exist as a cyclic hemiacetal in equilibrium with a small amount of the open-chain aldehyde. In the hemiacetal, C' is chiral and two diastereomers (anomers) are possible; they are called a- and p-glucosides. The C of C=O is always the anomeric position. Cyclic hemiacetals form when an OH and a CHO group in the same molecule can form a five- or six-membered ring that is more stable than the open chain. 1
-I
- H-I"-oHb c-
I
aGlucose: [aID=
CHJOH/dVHCI
+112",
C' OH on right side
I CHzOH
Hk
- 0 ~H
I
0
Methyl a-glucoside
8
CHjOHldry HCI
-CH)O-
F
Glucose (open-chain)
-C5
I
@-Glucose:[ a ] , = + 1 9 , C' OH on left side
i i I
Methyl 6-glucoside
Hemiacetal Anomers
Problem 22.13 Why do aldoses give positive Fehling and osazone reactions but negative Schiff and bisulfite tests? 4 These reactions are typical of the -CHO group. The Schiff and bisulfite reactions are reversible, and the equilibrium favors unreacted hemiacetal. Since osazone and Fehling reactions are irreversible, equilibrium is shifted to restore the low concentration (0.02%) of open-chain aldehyde as soon as some reaction occurs, and eventually all the aldose reacts. Problem 22.14 Glycosides do not react with either Fehling's or Tollens' reagents and do not mutarotate. Explain. 4 Glycosides are acetals. They are stable in the basic Fehling's and Tollens' solutions and in aqueous solutions used in mutarotation. Glycosides are nonreducing. Problem 22.15 Calculate the percentages of a- and /3-glucose present in the equilibrium mixture whose rotation is +52.7". 4
Let a and b be the mole fractions of the a- and p-anomers. Solving the simultaneous equations a+b=l
112"a + 19'6 = 52.7" gives
ax
100% = 36.2% and b x 100% = 63.8%.
500
CARBOHYDRATES AND NUCLEIC ACIDS
[CHAP. 22
Problem 22.16 Acetylation of glucose produces two isomeric pentaacetates that do not react with either phenylhydrazine or Tollens' solution. Explain. 4
The C' OH of the hemiacetal is acetylated instead of the OH that forms the ring. Since equilibrium with the open-chain aldehyde is prevented, these reactions are negative.
22.4
STEREOCHEMISTRY AND STRUCTURE PROOF
Problem 22.17 ( +)-Glucose, (+)-mannose and (+)-galactose are among the hexoses isolated after three Kiliani-Fischer step-ups, beginning with D-(+)-glyceraldehyde. Do these hexoses belong in the D or L families? 4
During the step-ups the D configuration of the chiral C' of glyceraldehyde (glycerose) does not change. This C becomes C' of (+)-glucose, (+)-mannose and (+)-galactose. Configuration of C s determines the family, and these monosaccharides are D-sugars. See Fig. 22-1. Notation A = CH-0 (aldehyde)
0
=
CH,OH
=
C=O (ketone)
- = OH
<
D-(+)-Glyceraldehyde a triose
-
OH of uncertain configuration -- - COOH
a D-aldohexose
=
Fig. 22-1
Problem 22.18 Write shorthand formulas for the tetroses prepared from ( a ) L-glycerose, ( 6 ) D-glycerose, by the Kiliani-Fischer synthesis. 4
See Fig. 22-2.
L-Glycerose
L-Erythrose
L-Threose
D-Glycerose
D-Erythrose
D-Threose
Fig. 22-2 Problem 22.19
How can oxidation with HNO, be used to distinguish between D-erythrose and D-threose?
See Fig. 22-3.
4
501
CARBOHYDRATES AND NUCLEIC ACIDS
CHAP. 221
E
r
plane of symmetry
HNO)
COOH meso-Tartaric acid (a glycaric acid; optically inactive)
D-Etythrose
T
t
I
____, HNO,
D-Threose
L
D-Tartaric acid (optically active)
Fig. 22-3
Problem 22.20 Two Ruff degradations of an aldohexose give an aldotetrose that is oxidized by HNO, to meso-tartaric acid. What can be the family configuration of the aldohexose? 4
See Fig. 22-4. The C4and C’ OH’Smust be on the same side. The aldohexose can belong to either the
L family.
D-aldohexose
D- aldotetrose
rneso-Tartaric acid
L- aldotetrose
D
or
Laldohexose
Fig. 22-4
Problem 22.21 Which D-aldohexoses give meso-aldaric acids on oxidation with HNO,?
4
There is a plane of symmetry ( - - - -) in the dicarboxylic acids formed by HNO, oxidation of D-allose and D-galactose. See Fig. 22-5.
i
D-Allose
--
acid
1 1
__-------
-
meso-Galactaric acid Fig. 22-5
HNOj
D-Galactose
CARBOHYDRATES AND NUCLEIC ACIDS
502
-Ruff
Problem 22.22 The reaction sequence on D-allose gives a meso-but optically active-pentaaldaric acid. Assign the structures. HN03
[CHAP. 22
on D-galactose gives an
4
See Fig. 22-6.
f---
D-Allose
D-Ribose
meso
optically active
D-Lyxose
D-Galactose
Fig. 22-6
--
Problem 22.23 D-Allose, D-glucose 5nccp-talose (the C2 epimer of D-galactose) each give a meso-heptaglycaric Kiliani 4 acid after the sequence . Assign all structures. There are three meso-heptaglycaric acids, as shown in Fig. 22-7.
---El T T D-Allose
T
D-Glucose
D-Talose Fig. 22-7
Problem 22.24 What monosaccharide is obtained from a Ruff degradation of D-mannose (Problem 22.1 l ) , the
C2epimer of D-glucose (Problem 22.23)?
4
See Fig. 22-8.
I-[ Ruff
D-Mannose
D-Arabinose
Fig. 22-8
Problem 22.25 What is the structure of the aldopentose D-ribose, a constituent of ribonucleic acids (RNA), if it gives the same osazone as D-arabinose (Problem 22.24) and is reduced to an optically inactive alditol?
4
CARBOHYDRATES AND NUCLEIC ACIDS
CHAP. 221
503
Ribose and arabinose are C 2epimers, since they give the same osazone (Fig. 22-9). HC=NNHPh
f D-Arabinose
I
-
C-NNHPh PhNHNH2,
PhNHNH2
i-
osazone
f
D-Ri bitol
optically inactive
Fig. 22-9 Problem 22.26 The pentoses D-xylose and D-lyxose give the same osazone. Xylose and lyxose are oxidized to an optically inactive and optically active dicarboxylic acid, respectively. Give their shorthand structural formulas. 4
They are D-sugars; the C4 OH is on the right side. They differ from D-arabinose and D-ribose (Problem 22.25), both of which have C3 OH on the right side; therefore D-xylose and D-lyxose have the C3 OH on the left side. Since D-xylose gives the inactive (rneso) dicarboxylic acid, its Cz OH must be on the right side, the same side as C4 OH. C2 OH is on the left side in D-lyxose. See Fig. 22-10. HC-NNHPh
-T
C-NNHPh
-iosazone
Xylaric acid
D-Lyxaric acid
optically inactive
optically active
Fig. 22-10 Problem 22.27 Fischer prepared L-gulose by oxidizing D-glucose to two separable lactones of glucaric acid. These were reduced to lactones of gluconic acid, which were further reduced. Give all the structures in these reactions. 4
i7 1 1-i 5 $ P I - +$-+ { fi $ See Fig. 22-11.
L___, -H20
reduce,
lactone
D-Glucose
0"
L-Gulonic acid
___, duce
D-Glucaric acid
Na-HI,
L-Gulonolactone
-H2",
L-GuIose
N*W,
0" lactone
D-Gluconic acid Fig. 22-11
SGluconolactone
D-Glucose
504
CARBOHYDRATES AND NUCLEIC ACIDS
[CHAP. 22
Problem 22.28 An aldohexose ( I ) is oxidized by HNO, to a meso-aldaric acid (11). Ruff degradation of (I) yields ( I I I ) , which is oxidized to an optically active dicarboxylic acid (IV). Ruff degradation of (111) yields (V), which is oxidized to L-( +)-tartaric acid (VI). Represent compounds ( I ) through (VI). 4
See Fig. 22-12. Since
L-(
+)-tartaric acid (VI) is isolated, ( I ) is an L-sugar.
I
HNO,
Problem 22.29 Deduce whether the hemiacetal ring of glucose is five- or six-membered from the following data and write equations for all steps. Glucose is first monomethylated at C' with MeOH and HCI, then tetramethylated with Me2S0, and NaOH. The pentamethyl derivative is treated with aq. HCI, then vigorously oxidized with HNO, to give 2,3-dimethoxysuccinic acid and 2,3,4-trimethoxyglutaric acid. 4
We have: Glucose
McOft
ltC.1
Methyl glucoside (acetal)
Me,S04
Methyl tetra-0-methyl glucoside (other 4 OH's
HCI
H20
(A)
form ethers)
(Only acetal C' 4 M e is hydrolyzed; the other ether linkages are stable.) The unmethylated OH'S of (A) are those involved in ring formation. They are C ' and CJ if the ring is five-membered (furanose) or C' and C" for a six-membered ring (pyranose). See Fig. 22-13. (Wavy lines denote cleavages.) Vigorous oxidation in the last step causes cleavage on both sides of the 2" H L H
I
the routes being (a) and (6) for the pyranose, and (c) and (d) for the furanose. Only the six-membered ring gives the observed degradation products. Problem 22.30 (a) Draw the chair conformations for a - and p-D-(+)-glucopyranose. The bulky CH,OH is usually assigned an equatorial position. In flat Fischer formulas, OH's on the right are tram to CH,OH, and 4 those on the left are cis. (6) Why are most naturally occurring glycopyranosides p-anomers?
B-W+)-Glucopyranose
a-D(+)-Glucopyranose
(b) Only in P-glucopyranosides are all substituents equatorial. In the a-anomer, the C' --OR is axial.
--H104
Br?. t f 2 0
Problem 22.31 How can the sequence pyranoside? (See Problem 14.36 for HIO, oxidation.)
See Fig. 22-14 for different product distribution.
H,O'
show if a methyl glycoside is a furanoside or a
The acetal formed by HIO, is hydrolyzed by H,O'.
4
505
CARBOHYDRATES AND NUCLEIC ACIDS
CHAP. 221
to,, 'COOH
Me0
H
+ EO, + MeOH
't0Me
TOOH
Me0
2.3.4-Trimetho x ygl U taric acid
'COOH
2t--oMe (A) if 6-membered
aldehyde form
(pyranose)
+ MeOEH,tOOH
~e0-13 *COOH 2,3-Dimethoxysuccinic acid
'COOH
HO-C
H
+OM,
6
5
MeOCH,COOH
MeO{T) 'COOH CH,OMe
.t
OMe 3 CO,
(A) if 5-membered
6CH20Me
(f uranose)
+ MeOCH,COOH
'COOH
Dimethoxy- Methoxymalonic gly ceric acid acid
Fig. 22-13
HLC-OCH,
ZH104
-t-
H~COOH
HO--'C-H
st;-
H-C-OH
I
2
H30t
I- -l-lFk H-
dH2OH
&H,OH
-3-
H0--CHA
LI
H
T
I
H-5C-OH -$%,OH Methyl a-D-glucofuranoside
'4H20H
~ G l y c e r i cacid
H'C-OCH,
HJC-OCH, L
COOH
'
Methyl a-~-glucopyranoside
L
4
H-C-0
HLC
H
+
BrZ.H 2 0
_____3
2H104
~-3c-o
I.
Br2. H 2 0
2.
HjO'
HA-0 'AOOH
[
+
3~~~~
H-*C-OH
+6 HLA=O H,C=O
-
'LOO] Hydroxymalonic acid
Fig. 22-14
-CO*
COOH
I
H-C-OH
k
Glycolic acid
506
CARBOHYDRATES AND NUCLEIC ACIDS
[CHAP. 22
Problem 22.32 Draw the more and less stable conformations of (a) P-D-mannopyranose (Problem 22-11), ( b ) a - D-idopyranose (idose is the C’ epimer of gulose; see Problem 22.27), (c) P-L-glucopyranose (p-L-and p-Dare enantiomers)1 . Explain each choice. 4
eH20H
OH
HO
H
H
OH
more stable
8
H less stable
In the more stable conformation, CH,OH and three OH’s are (e).
n
M
Ho
H
1
O
H
H
OH
HO
W H H OH
OH
H
more stable
less stable
The less stable conformation has CH,OH (e) but four axial OH’s, while the more stable conformation has CH,OH (a) but four equatorial OH’s. &OH HO
H
*
3
H
H OH
1
c H 2 0MH
H 3
OH
4
OH
more stable
H
OH
less stable
The more stable conformation has all groups (e); the less stable has all groups (a). Problem 22.33 Write a structural formula for ( a ) p - and a -D-( -)-fructofuranose, (b) B-D-(-)-fructopyranose. 4
In this ketose the anomeric position is C’, which is part of the C=O group.
k 1 t
OH H a-D-(-)-Fructofuranose
Keto form of D(-)-fruc tose
A
fl-D-(-)-Fructofuranose
CARBOHYDRATES AND NUCLEIC ACIDS
CHAP. 221
507
t
Problem 22.34 ( a ) Write shorthand structures for the three D-2-ketohexosesother than D-fructose. (6) Which one does not give a meso alditol on reduction? ( c ) Which one gives the same osazone as does D-galactose (Problem 22.22)? 4 ( a ) See Fig. 22-15(a). The D family has the C5 OH on the right side. ( 6 ) See Fig. 22-15(6). (c) Compound I11 is D-tagatose, since its C3, C4, and C5 OH’s have the same configuration as the corresponding OH’s in D-galactose.
I I1
I
i
I11
H2/Ni
____*
D-Sorbose (11)
22.5
:j
both optically active
( b)
Fig. 22-15
DISACCHARIDES
Disaccharides have an OH of one rnonosaccharide (the aglycone, denoted A) bonded to the anomeric C of a second rnonosaccharide, B . The disaccharide is a glycoside of B . Problem 22.35 (a) Maltose (I), C,,H,O,,, gives D-glucose with maltase, a specific enzyme for a-glycosides. (b) (I) reduces Fehling’s solution,-undergoes mutarotation, and is oxidized by Br, in H 2 0 to D-maltobionic add, C,,H,,O,,COOH (11,). ( c ) (11) reacts with Me,SO, and NaOH to form an octamethyl derivative (111) that is then hydrolyzed to 2,3,4,6-tetra-O-methyl-~-glucose (IV) and 2,3,5,6-tetra-O-methyl-~-gluconic acid (V). Deduce the structure of maltose. 4 ( a ) Maltose is composed of two glucose molecules with an a-glycosidic linkage. (6) Maltose is a reducing sugar and therefore has one free hemiacetal unit A. The anomeric C’,
-?{=I
0
I
CARBOHYDRATES AND NUCLEIC ACIDS
508
[CHAP. 22
of A is oxidized to COOH to give (11). (c) ( V ) comes from unit A (the aglycone). Unit A is linked through its C' - O H , since this group in (V) is not methylated. (IV) comes from B . The C5- O H of (IV) is not methylated and must be involved in forming a pyranose ring. Unit A is also a pyranose; a furanose is ruled out because the CJ - O H needed for this ring size is part of the glycosidic linkage. The only uncertainty in the structure is the configuration of the anomeric C of A , which is assumed to have the more stable P-configuration. See Fig. 22-16. A
B H
6
HO
-H
H
1'H
OH
(I) p- Anomer of (+)-maltose, 4-0-(a-~-glucopyranosyl)-~-glucopyranose
I
Br2. H 2 0
B
A
A
B
COOH
HO
cH30m H
(11)
D-Maltobionic acid
(CH3)2S04.NaOH
OMe MeO-
r
-
OMe OMe
-OMe
(111)
H
h ydrol y zed bond
H CH20CH3
CH30
CH30
H
CH30
Octa-0-methyl-D-maltobionicacid
H
COOH OCH3
-OMe Me0
3
OMe (IV) (a-anomer)
6
-OH -OMe -OMe
H
(W 2,3,4,6-Tetra-O-methyl-~glucopyranose
Fig. 22-16
H
(V) 2,3,5 ,dTetra-O-methyl-D-
gluconic acid
CHAP. 22)
509
CARBOHYDRATES AND NUCLEIC ACIDS
Problem 22.36 Cellobiose, isolated from the polysaccharide cellulose, has the same chemistry as maltose 4 except that it is hydrolyzed by emulsin. Give the structure of this disaccharide. Unlike maltose, cellobiose is a P-glycoside. (The aglycone unit is turned 180" to permit a reasonable bond angle for the glycosidic linkage.)
Problem 22.37 Deduce the structure of the disaccharide lactose. ( a ) It is hydrolyzed by emulsin to D-glucose and D-galactose. (6) It is a reducing sugar and undergoes mutarotation. (c) It forms an osazone that is hydrolyzed to D-galactose and D-glucosazone. (d) Mild oxidation, methylation and then hydrolysis gives results 4 analogous to those observed with D-maltose. ( a ) Lactose is a p-glycoside made up of D-glucose and D-galactose. (6) It has a free hemiacetal group. (c) Since the glucose unit forms the osazone, it is A that must have the hemiacetal linkage. Lactose is a P-galactoside. (d) Both units are pyranoses joined through the C4 -OH of the glucose unit.
B
A
D-Galactose
D-Glucose
Problem 22.38 Use shorthand formulas to show how osazone formation establishes the glucose unit of lactose to have the hemiacetal linkage (Problem 22.37). 4 Lactosazone is hydrolyzed to D-(+)-galactose and D-glucosazone.
NNHPh NNHPh
F4 Lactosazone
p-anomer of lactose
NNHPh NNHPh H30t ___,
B-D-(+ )-Galactose
D-Glucosazone
Problem 22.39 ( a ) Give the structure of sucrose (cane and beet sugar) from the following information: (1) It is hydrolyzed by maltase or emulsin to a mixture of D-(+)-glucose and D-(-)-fructose. (2) It does not reduce Fehling's solution and does not mutarotate. (3) Methylation and hydrolysis gives 2,3,4,6-tetra-o-methyl-~glucose and a tetramethyb-fructose. (b) What structural features are uncertain? 4
CARBOHYDRATES AND NUCLEIC ACIDS
510
[CHAP. 22
(1) Sucrose has a glucose linked to a fructose unit. The linkage is a to one monosaccharide and /3 to the other. (2) It has no hemiacetal group. The anomeric OH'S on C' of glucose and C2 of fructose are joined to form two acetals. (3) The glucose unit is a pyranose; the C5 - O H is unmethylated. The glycosidic linkages: the linkage to glucose is a and the one to fructose is p. The fructose ring size: it is a furanose. ( +)-Sucrose is a-D-g~ucopyranosyl-~-~-fructofuranoside or P-D-fructofuranosyl-a-Dglucopyranoside.
- Ho*&H c*o
OHao
H Mlucose
=Fructose
H
HO
D-Glucose
DFructose
Problem 22.40 Hydrolysis of (+)-sucrose gives a mixture of D-(+)-glucose ( [ a ) D = 52.7") and D-(-)-fructose 4 ([a],= -92.4') called invert sugar. Calculate the specific rotation of invert sugar. One mole of sucrose produces 1 mol each of glucose and fructose, and the specific rotation is one-half the sum of those of the two monosaccharides; i.e. 4[+52.7" + (-92.4")]
Pt-
=
-19.9" 4
22.41 Write the reactions for methylation and hydrolysis of sucrose.
-OMe 0 MeOOMe
(+)-Sucrose
-OMe
Octa-0-methylsucrose
Me0
[r"r OMe
OMe
2.3.4.6-Tetra0-methyl-D-glucose
+ M
e
o OMe r ] OMe
1,3,4,6-Tetra-Omethyl-D-fructose
22.6 POLYSACCHARIDES Plobkm 22.42 The polysaccharide amylose, the water-soluble component of starch, is hydrolyzed to ( +)-maltose and D-( +)-glucose. Methylated and hydrolyzed, amylose gives mainly 2,3,6-tri-O-methyl-~4 glucopyranose. Deduce the structure of amylose. It has D-glucose units joined by a-glycosidic linkages [Problem 22.35(a)] to the C4of the next unit. This is revealed by isolating maltose, an a-glycoside, and the 2,3,6-tri-O-methyl derivative. The C5 --OH forms the pyranoside ring.
CARBOHYDRATES AND NUCLEIC ACIDS
CHAP. 221
511
Problem 22.43 Methylation and hydrolysis of amylose (Problem 22.42) also gives 0.2-0.4% 2.3.4.6-tetra-04 methyl-D-glucose. Explain the origin of this compound. The end glycosidic glucose unit has a free C' -OH
and gives this tetra-0-methyl derivative.
Problem 22.44 Amylopectin, the water insoluble fraction of starch, behaves like amylose, except that more 2,3,4,6-tetra-O-methyl-~-glucose ( 5 % ) , and an equal amount of 2,3-di-O-methyl-~-glucose,is formed. Deduce 4 the structure of amylopectin. Like amylose, amylopectin has chains of D-glucose units formed by an a-glycosidic linkage to CJ of the next glucose unit. Occasionally along the chain, the C 6 - O H is not free but is used to branch two chains at the aglycone end of one chain. Since more 2,3,4,6-tetra-O-methyI glucose is formed, the amylopectin chains are shorter than those of amylose. 4
1
I
4
-CO-Glucose-a-OC-OC-Glucose-a-OC
-CO-Glucose-a-OC 4
22.7
-db 4
I
1
I
OC-Glucose-a-OC-OC-
4
1
NUCLEIC ACIDS
The nucleic acids, R N A (ribonucleic acid) and DNA (deoxyribonucleic acid), are carbohydrate biopolymers. The repeating sugar in R N A is ribose, and in DNA it is 2-deoxyribose. Problem 22.45 Give structures for ( a ) D-ribose, (b) P-D-ribofuranose, and (c) P-~-2-deoxyribofuranose,and show the numbering of the C's. 4
OH
OH
OH
H
Problem 22.46 Draw structures for the nucleosides of ribose with ( a ) cytosine (Problem 20.34), (6) uridine,
( c ) adenosine, (d) guanine.
4
Nucleosides are glycosides of ribofuranose or deoxyribofuranose whose aglycones are pyrimidine or purine bases. The bases are bonded to C' of the sugar. NH, I
NH,
0
H
II
/
\
NH,
* H O
C OH
d
H OH
CARBOHYDRATES AND NUCLEIC ACIDS
512
[CHAP. 22
Problem 22.47 Write the structural formulas for adenine monophosphate (AMP) and triphosphate (ATP) nucleotides. (ATP is the most important source of energy for biochemical processes in living tissues; the energy comes from the breaking of the high-energy polyphosphate bonds.) 4
Nucleotides are phosphate esters of nucleosides, formed at the CH,OH group of the sugar
y-4
NH,
OH
OH
OH
5’-Adenylic acid (AMP)
Adenosine triphosphate
OH
(ATP)
Nucleic acids are pol meric nucleotides in which phosphate esters link ribose or deoxyribose Y molecules through the C- OH of one and the C-’ OH of the other. In RNA, the aglycone nitrogen bases are cytosine, adenine, guanine, and uracil. In DNA, thymine replaces uracil. The R N A polymer is like that of DNA, except that a CH, replaces the OH group on C 2 of the ribose ring. Problem 22.48 Use the hydrolysis of nucleosides. base
DNA
to show the relationship between base
base
I I I ... phosphate-sugar-phosphate-sugar-phosphat~ugar-phosphate or
DNA
base suiar-phosphate Nucleottde
t
H?O
RNA
base
phosphate
I
+ sugar
DNA,
nucleotides and 4
Ha0
+ H O
sugar + base
Nucleoside
DNA, a constituent of the cell nucleus, consists of two strands of polynucleotides that are coiled to form a double helix. The strands are held together by H-bonding between the nitrogen bases. The pyrimidines always form H-bonds with a specific purine; i.e., cytosine with guanine and thymine with adenine. However, in RNA the pairing is between uracil and adenine. Problem 22.49 Use a schematic diagram like that in Problem 22.48 to show H-bonding (by a dotted line) of two complementary, oppositely directed strands of nucleotides in which all base pairs are H-bonded. Denote the 4 bases A for adenine, T for thymine, C for cytosine, and G for guanine.
0-
B
HO-P-0-ri
A-
T
0
I II bose-0-P-0-ri
A-
0-
G
I
0-
0
I1
bose-O-P-O-ri
A-
A-
CHAP. 221
CARBOHYDRATES AND NUCLEIC ACIDS
513
Through its sequence of nitrogen bases, D N A stores the genetic information for cell function and division, and for biosynthesis of enzymes and other essential proteins. In protein synthesis, the information in the DNA is transcribed onto messenger R N A (mRNA), which moves from the nucleus to the ribosomes in the cytoplasm of the cell. Here the information is transferred to ribosomal RNA (rRNA). Transfer R N A (tRNA) carries amino acids to the surface of the m u , where the protein is “grown.” A specific three-term sequence of bases in the mmA, called a codon, calls up a tRNA carrying the specific amino acid that is to be the next unit in the growing protein chain. For example, the codon cytosine.uracil.guanine is translated as “leucine.” Problem 22.50 Verify that three is the smallest length of sequence possible for codons.
4
There are only four different bases in mmA. Thus there could exist 4 ’ = 4 codons of length 1; 4’ = 16 codons of length 2; 4.’ = 64 codons of length 3; etc. As there are 20 amino acids, a minimum of 20 codons are required; thus the codon length must be at least 3. The existence of 64 -20 =44 excess codons allows a valuable redundancy in the genetic code. It also permits the signaling for the start and end of the protein chain.
Supplementary Problems Problem 22.51
How do epimers and anomers differ?
4
Epimers differ in configuration about a single chiral center in molecules with more than one chiral center. Anomers are epimers in which the chiral site was formerly a carbonyl C. Problem 22.52 What enediol is common to D-fructose and its C.’ epimer, D-allulose?
4
See Fig. 22-17.
i-
D-Allulose
CH,OH
I
koH C-OH
1
D-Fructose
Fig. 22-17
enediol
Problem 22.53 A monosaccharide is treated with HCN, the product hydrolyzed and reduced to a carboxylic acid by heating with HI and P. What carboxylic acid is formed if the monosaccharide is (a) D-glucose, (6) D-mannose, (c) D-fructose, (d) D-arabinose, (e) D-erythrose? 4 ( a ) and (6) heptanoic acid, (c) 2-methylhexanoic acid, (d) hexanoic acid, (e) pentanoic acid.
Problem 22.54 Prepare D-talose from its C 2 epimer, D-galactose (Problem 22.22).
See Fig. 22-18.
4
[CHAP. 22
CARBOHYDRATES AND NUCLEIC ACIDS
514
i i i i pyridin
B'2
I
H P
D-Galactose
\
D-Galactonic acid
BTalonic acid
v
D-Talose
J
separate the epimers
Fig. 22-18 ptobkm a . 5 5
Convert D-glucose to hexa-0-acetyl-D-sorbitol (an alditol polyester).
4
See Fig. 22-19. CH,OAc OAc I.
NIBHI
2.
H,O'
&
-
OAc
-
OAc
&
CH,OAc
Fig. 22-19
Problem 22.56 Give names and structural formulas for the products of the reaction of a-o-mannose (Problem 22.11) with: (a) Ac,O, pyridine; (6) HIO,; ( c ) CH,OH, HCI, then HIO,; (d) H,NOH; (e) CH,OH, HCI, then (CH,),SO,, NaOH, and then aq. HCI; ( f ) HCNIKCN, then H,Ot, and then Na-Hg, CO,; (g) PhNHNH,, then PhCH=O. 4
T
CHO
(6) SHCOOH, CH,O
(a) AcO
cP
( c ) HCOOH
AcoQ
CHzOH
1,2$,4,dPenta-0acet yl-mmannose
T
NOH
DNannose oxime
+ 0
Me0 (e)
Me0
OMe OMe a -2,3,4,6Tetra-0-
methyl-D-mannose
epimeric heptoses
f
lpMannosone (D-Glucosone)
CHAP. 221
CARBOHYDRATES AND NUCLEIC ACIDS
515
Problem 22.57 ( a ) Write shorthand open structures for a ketohexose (I) and an aldohexose (11) which form the same osazone as D-(-)-altrose (111), the C3 epimer of D-mannose, and for the “aric” acid (IV) formed by HNO, oxidation of the D-aldohexoses (111) and D-talose (V).(b) Write shorthand and conformational structures 4 for p-D-altropyranose (VI). See Fig 22-20.
C’, C‘ and C’ have the same configuration in (I), (11) and (111). ( a1
4ti
5
Fig. 22-20 Problem 22.58 Chemically distinguish 2deoxyglucose from 3deoxyglucose.
4
3-Deoxyglucose forms an osazone with 3 mol of PhNHNH,, but 2-deoxyglucose has no C2 OH and forms only a phenylhydrazone. Problem 22.59 What hexose (I) forms a cyanohydrin (11) that is hydrolyzed and then reduced with HI and P to a carboxylic acid (111), where (111) is synthesized from n-PrI and Na’[CC,H,(COOEt),]-? 4 We have: 6
S
4
6
CH3CH2CH21+ N i i + [ p 0 0 E t ) 2 r
5
4
CH3CH,CH2 HCOOH !*26H3
H27H3
2-Ethylpentanoicacid (111)
For COOH to be placed on C3 by the cyanohydrin reaction, C’ must be a carbonyl C in (I). (I) is therefore a 3-ketohexose,
and (11) is
&
i
i
L
OHOH OHOH (A wavy line indicates unknown configuration.)
h
516
CARBOHYDRATES AND NUCLEIC ACIDS
[CHAP. 22
Problem 22.60 Which D-pentose is oxidized to an optically inactive dibasic acid and undergoes a Ruff 4 degradation to a tetrose whose glycaric acid is meso-tartaric acid?
Ribose; see Fig. 22-21.
F1
HNO,
c---
'r - i Y I Run
4
dcu
5
D-Ribose
inactive
D-Erythrose
meso-Tartaric acid
Fig. 22-21
Problem 22.61 Deduce possible structures for the following disaccharides. ( a ) A compound, C,,H,,09, undergoes oxidation with aq. Br,, followed by methylation and hydrolysis by maltase, to yield 2,3,4-tri-Oacid. (6) (+)-Gentiobiose, C,,H,,O I , reduces Fehling's methyl-D-xylose and 2,3-di-O-methyl-~-arabonic solution, undergoes mutarotation and is hydrolyzed by emulsin to D-glucose. Methylation followed by hydrolysis yields 2,3,4,6-tetra-O-methyI-~-glucose and 2,3,4-tri-0-methyl-~-ghcose.( c ) Trehalose, C,,H,,O I I , is nonreducing and is hydrolyzed only to D-glucose by either maltase or aqueous acid. Methylation followed by hydrolysis yields only 2,3,4,6-tetra-O-methyI-~-glucose. (d) Isotrehalose is identical with trehalose except that it 4 is hydrolyzed by either emulsin or maltase.
H) ( * * "
H
H
H
H D-X ylose portion
h
"*EH2 HO
OH
3
H
2 H
H
Ho
HO H
HO
3S
H
HO
H
OH
Z
'H
@
OH
L-Arabinose portion
( a ) 4-0-(a-~-Xylopyranosyl)-~-arabinose
HO
~ H ~ O H
( 6 ) 6-0-(/3 -D-Glucopyranosyl)-D-glucopyranose
H CH,OH
.OH (c )
a-~-Glucopyranos yI-a-D-glucopyranoside
( d ) a-D-Glucopyranosyl-&D-glucopyranoside
Problem 22.62 What is the enantiomer of a - ~ -+)-glucose? (
a-L-(-)-GIucose. In the D family the more dextrorotatory anomer is named a-D-.In the levorotatory anomer is named a - ~ - .
4 L
family the more
Index Ammonium hydrosulfide, 415 Amphoteric substances, 40,47 Amylose, 510 Amylopectin, 511 Anhydrides, 361 cyclic, 354 Aniline, 412 Annelation, 410 Annulenes, 204 Anomers, 499 Anthracene, 207, 209 Anti-addition, 101 Antiaromat icit y , 203 Antibonding orbitals, 13 Anti-elimination, 1298 Arabinose, 503 Arenes, electrophilic substitution, 2 15ff nitrosation, 216 Arenonium ion, 215 Aromatic character, 201 Aromaticity, 202 non-benzenoid, 214 Aromatic reactions, 209 Aromatic substitutions, nucleophilic, 225ff Aromatization, 212 Aryl halides, summary of chemistry, 234 Aspartame, 482 Aspirin, 440, 449 Atomic orbitals, 12 hybridization, 16 Aufbau, 12 Axial bond, 174 Axial interaction, 175 Azabenzene, 463 Aza method for amines, 412 Azanaphthalene, 468 Azepane, 470 Aziridine, 419 Azlactone synthesis, 488 Azo compounds, 428 Azole, 470 Azulene, aromaticity of, 214
AAE, 394 Absolute configuration, 75 Acetals, 306, 330 Acetanilide, 425 Acetoacetic ester, 394 Acidity and structure, 41 Acids, reduction of, 274 Acids and bases, 40 Acrolein, 338, 468 Activation energy, 37 Activation of ring, 218ff Acyl chlorides, 360 Acylation of alkenes, 320 Acylium ion, 343 Acyloin, 332 Acylonium ion, 319 Addition, carbene, 104 free radical, 104 radical-catalyzed, 155 Addition-elimination reaction, 225ff Adenine, 468 mono and triphosphate, 512 Adipic acid, 347 ADP, AMP, and ATP, 512 Alcohols, 279, 317 Aldaric acid, 494 Alditol, 496 Aldonic acid, 494 Aldehydes, oxidation of, 323 reduction of, 324 summary of chemistry, 336 Aldol condensation, 397 Alkadienes, 142ff summary of chemistry, 157 Alkanes, summary of chemistry, 61 Alkenes, addition reactions, 98ff ozonolysis, 106 polymerization, 103ff, 117 summary of chemistry, 108 Alkoxymercuration-demercuration, 292 Alkylation, Friedel-Crafts, 216 Alkylboranes, 270, 318 Alkyl groups, 54 Alkyl halides, dehydrohalogenation, 91 summary of chemistry, 132 Alkynes, partial reduction, 92 summary of chemistry, 157 Allenes, chirality, 163 Allylic substitution, 107 Ambident ions, 127 Amide formation, 35 1 Amines, summary of chemistry, 429 Amino acids, summary of chemistry, 488
Baeger-Villiger reaction, 324 Barbiturates, 369 Base peak in ms, 260 Bases, soft and hard, 122 Basicity and structure, 41 Basic Red 9, 437 Beckmann rearrangement, 418 Benzene, resonance structure, 201 structure, 198
517
518
Benzenonium ion, 215 Benzhydrol, 282 Benzidine rearrangement, 432 Benzyne, 227 BHA and BHT,454 Bicyclic compounds, 166 and nucleophilic substitution, 139 Birch reduction, 209 Bischler-Napieralski reaction, 469 Bithienyl, 470 Boat and chair forms, 175 Boiling point, influences on, 21 Bond dissociation, 35 Bond distances, 149 Bond order, 16 Bond stretching, 245 Bonding orbitals, 13 Boron hydride, reaction with alkynes, 146 Bredt’s Rule, 173 Bromonium ion, 101 N-Bromosuccinimide, 107 Bronsted, 40 Bucherer reaction, 446, 453 Butyl groups, 54 t-Butyl hypochlorite in monochlorination, C-13 nmr, 257ff Cahn-Ingold-Prelog rules, 71 Cane sugar, 509 Cannizzaro reaction, 324 Carbaldehyde, 315 Carbenes, 30 Carbene addition, 104 Carbinolamines, 327 Carbitol, 313 Carbocation rearrangements, 94 Carbocation reactions, 44, 114 Carbonic acid derivatives, 369 Carbonyl group, reduction of, 274 resonance structures, 345 Carboxylic acids, summary of chemistry, 354 Catalyst, phase transfer, 128 Catalytic re-forming 212 Catechol, 440 Cellobiose, 509 Center of symmetry, 68 Chair, and boat form, 175 Charge, formal, 6 Chemical shift, 249ff Chloral, 343 Chloral hydrate, 331 Chlorination of [R]-2-chlorobutane, 79 Chlorination using t- butyl hypochlorite, 65 Chiral center, 69 Chiral centers, more than one, 76ff Chiral stereomer, 68
INDEX
Cholesterol, chirality in, 81 Chromatography, 48 1 Cinnamaldehyde, 341 Cis-trans interconversion, 112 Cis-trans isomerism, 88ff in cyclic compounds, 167 Claisen condensation, 405 rearrangement, 448 Cleavage, oxidative, 118 Clemmensen reduction, 229, 324 Codon, 513 Coenzyme A, 365 Collins reagent, 277 Collision frequency, 37 Color base, 439 Configuration, 71 relative, 75 Conformation, 50 Conformational diastereomers, 79 enantiomers, 79 stereomers, 79 Conformation, cycloalkanes, 170ff energy, 50, 53 Conjugated bonds, 148 Conjugated dienes, stability of, 149 Conjugates, 40, 47 Cope elimination, 404, 425 Copolymers, 104 Corey-House synthesis, 56 Coupling constant, 256 Coupling reaction, 56, 428 Coupling, spin-spin in nmr, 253 Covalent bonds, 5 13 Cracking of alkanes, 57 Cresol, 440 Crown ethers, 299 Cumene hydroperoxide, 441 Cumulated bonds, 148 Curtius rearrangement, 418 Cyanide ion, structure, 26 Cyanoethylation, 397 Cyclic ethers, 298 Cyclization, 179, 183ff, 212 Cycloalkanes, 87 cis-trans isomerism, 167 Cyclooctatetraene, structure, 203 Cyclooctyne, 332 p-Cymene, 228 Cytosine, 467 Dacron, 371 DDT, 343 Decarboxylation, 352 Delocalization, 41 energy, 23 Denaturation, 487
INDEX
2-Deoxyribose, 511 Deuteration of arenes, 216 Deuterium, role in nmr, 255 Dext rorotatory , 69 Diastereomers, 68 conformational, 79 Diastereoselective reactions, 92 Diazine, 467,470 Diazomethane, 66, 364 Diazonium ions, 421 Diazonium salts, 426 Dicarboxylic acids, 354 Diels-Alder reaction, 180, 185, 193, 212, 214, 462 Dienes, polymerization, l5Sff summary of chemistry, 157 Dienophile, 156 Dihalides, dehalogenation , 92 Dimethyl sulfoxide, 318 Dioxane, 298 Dioxin, 456 Dipole-dipole interaction, 20 Dipole moment, 20, 27, 67 Dipoles, 91 induced, 20 Disaccharides, 507ff Displacement, 32 Disproportionation, 324 Dithiane, 321 Dithiazine, 470 DMSO, 318 DNA, 511 Dow Process, 441 El and E2 mechanisms, 129ff Edman method, 483 Electron-dot structures, 2 Electronegativity, 19 Electrophiles, 33, 43, 45 Electrophilic substitution of arenes, 215ff in naphthalene, 223 Elimination, 32 1,2-Elimination, 91 Enantiomers, 68 conditions for, 80 conformational, 77 Eneamines, 328, 390 Enolsilanes, 407 Enthalpy, 34 diagrams, 38, 48 Entropy, 34 Enzymes, 474 Epimer, 496 Epoxides, summary of chemistry, 303 Equatorial bond, 174 Equilibrium and free energy, 36 Equivalent H’s, 49
Erythro form, 82, 85 Erythrose, 501 Esters, formation, 351 inorganic, 275, 285, 289 reduction , 274 Ethers, cleavage, 296 crown, 299 cyclic, 298 silyl, 299 summary of chemistry, 300 tetrahydropyranyl, 299 Excited state, 242 Exclusion Principle, Pauli. 12 Exhaustive methylation, 414, 435 E, Z notation, 88
Fats, 356 Fischer projection, 69 Fluorine, role in nmr, 255 Formal charge, 6 Formic acid, 344 Formulas, condensed, 2 Lewis, 2 Formylation, 319 Formyl group, 315 Free energy, 35 and equilibrium, 36 Free radical, 30 additions, 104 Friedel-Crafts, acylation, 3 19 alkylation, 221 synthesis, 216, 229 Fries rearrangement, 445 D-Fructose, 5 13 F-strain, 465 Fumaric acid, 350 Functional groups, 6, 8, 9 Furan, 457ff Furanose, 504 Furfural, 460
Gabriel synthesis, 414, 476 Galactose, 509, 514 Gatterman reaction, 439 Gatterman-Koch reaction, 320 Gauche conformation, 52, 175. 289 Gentiobiose, 516 Geometric isomerism, 88ff Glucose, 494, 499 Glucosides, 499 Glutamic acid, 477 Glutaric acid, 355 Glyceraldehyde , 75 , 500 Glyceric acid, 75
519
520
Glycerol, 365 Glycerose , 500 Glycols, summary of chemistry, 306 Glycosides, 499 Glyptal, 371 Grignard reaction, 271, 282ff, 348 with C=O groups, 324, 348, 364 limitations , 272 with water and D,O, 56 Grignard reagent, 55 Ground state, 242 Groups, electron-donating and withdrawing, 219ff Guanine, 468 Haloform reaction, 287, 323, 348 Halogen exchange, 120 Halogenation of alkanes, 57, 59 of arenes, 21 1 Halogenonium ion, 387 Halohydrin, 102 Hammond Principle, 2 18 Hard bases, 122 H-bonding, 269, 345, 413 H-counts in nmr, 252 Heat of combustion and stability, 91 Hell-Volhard-Zelinsky reaction, 352, 476 Hemiacetal, 330 Heptane isomers, 65 2-Hexene isomers, 10 Hertz, 242 Hexylresorcinol, 340, 440, 446 Hinsberg reaction, 423 Hofmann, 129 Hofmann degradation, 415 elimination, 424 Huckel's Rule, 202, 205 Hund's Rule, 12 Hunsdiecker reaction, 353 Hybridization, 16 Hybrid orbital number, 18, 19, 31 Hybrid, resonance, 23 Hydration of cyclohexane derivatives, 199 Hydrazine, 4 Hydride shift, 94 Hydroboration, 96 Hydroboration-oxidation, 270, 283, 322 Hydrocarbons, cyclic, 166ff unsaturated, 87 Hydrogenation of alkenes, 55 Hydrogen bond, 20, 27 Hydrogen iodide as cleaving agent, 296 Hydrolysis of alkyl halides, 134 Hydroperoxides in ethers, 298 Hydroquinone , 440 Hydroxy acids, 355 Hydroxylamine, 326 8-Hydroxyquinoline , 472
INDEX
Imidazole, 467 Imines, 328 Iminium ion, 328 Indole, 468 Inductive effect, 127 Infrared spectroscopy , 244ff Infrared absorption peaks [Table], 246 Inhibitors, 37 Initiation, 32 Intermediates, 30, 40, 45 Inorganic esters, 275, 285, 289 Invcrsion, 124, 140 Invert sugar, 510 3-Iodo-2-butano1, configuration, 77 [ R]-3-Iodo-2-chloropropanol, configuration of, 76 Ion-dipole attraction, 22 Ionic bonds, 6 Isobutyl group, 148 Isoelectric point, 478 Isolated bonds, 148 Isomerism, 3 alkyl halides, 119 cis-trans, 88ff geometric, 88ff optical, 69 Isomerization , 212 Isomers of butane, 49 of chloronaphthalene , 209 of heptane, 65 of 2-hexene, 10 of pentane, 49 Isoniazide , 467 Isoprene rule, 188 Isopropyl group, 54 Isoquinoline, 468 Isotope effect, 129 IUPAC, 54 Jones, reagent, 278, 318 Joule, 242 Kekuld, 198 Ketals, 306, 330 Ketones, oxidation of, 323 summary of chemistry, 337 Kiliani-Fischer method, 497 Knoevenagel reaction, 404 Kodel, 383 Kolbe synthesis, 448 Lactic acid, 356 esterification, 71 Lactones, 355 Lactose, 509 LCAO, 13
INDEX
LDA, 389 Lederer-Manasse reaction, 448 Leuco base, 439 Levorotation, 69 Lewis acids and bases, 43 Lewis structural formulas, 2 Lexan, 383 Ligands, 69 Lindlar’s Catalyst , 147 Lithium aluminum hydride, 273 Lithium dialkyl cuprates, 56 Lithium diisopropylamide, 389 London forces, 20 Lossen rearrangement, 418 Lucas test, 278, 288 Maleic acid, 350 Malonic ester, 391 Maltose, 507 Mannose, 498 Markovnikoffs Rule, 98 Mass spectroscopy, 259ff Mechanism, alcohol dehydration, 93 alkane halogenation, 57, 59 benzyne, 227 E l and E2, 129ff S,1 and S,2, 123ff Mercaptans, 280 Merrifield synthesis, 485 Mesitylene, 228, 381, 442 Meso forms, 77 Messenger RNA, 513 Methane, bromination mechanism, 59 Methide shift, 94 Methionine, 477 Methylene in synthesis, 66 Methyl salicylate, 449 Michael addition, 396, 469, 473 Microscopic reversibility, 99 Migratory aptitude, 305 Molecular orbital, 13 Molecularity, 38 Molecules, geometry of, 17 polar, 26 Monochloropentane enantiomers, 74 MO theory and ally1 compounds, 151 and 193-butadiene, 150 and ethene, 150 Mustard gas, 314, Nanometer, 242 Naphthalene, 286 electrophilic substitution in 223 oxidation, 210 NBS, 107 Neighboring-group participation, 301
Neutralization equivalent, 367 Newman projection, 3, 50, 70 Niacin, 467 Ninhydrin, 331, 481 Nitrosation, 422 of arenes, 216 Nitrous acid, 421 Nonbonding orbital, 14 Novocaine , 434 Nuclear magnetic resonance, 248ff Nucleic acids, 511ff Nucleophiles, 33, 43, 45 Nucleophilic displacement, 121 Nucleophilicity of bases, 121 Nucleophilic substitutions, aromatic, 225ff and bicyclic compounds, 139 Nucleosides, 511 Nucleotides, 512 Nylon 6 and, 66, 371
Oil of wintergreen, 449 Oils, 365 Olefins, 87 Optical activity and synthesis, 78 Optical isomerism, 69 Optical purity, 126 Orbitals, 44 atomic, 12 hybrid, 16 Organometallics, 55 Orientation rules, 222 Osazone, 498 Osones, 498 Oxalic acid, 352 Oxa method for ethers, 412 Oxazole, 467, 470 Oxidation, anthracene, 21 1 benzene, 210 naphthalene, 210 phenanthrene, 21 1 Oxidation number, 20, 28 Oxidation-reduction, 32 Oxidative cleavage, 118 Oxirane, 470 Oxirine, 42 0x0 Process, 320 Oxymercuration-demercuration, 271 Ozonolysis of alkenes, 106, 319
Palmitic acid, 365 Paramagnetism, 16 Pararosaniline, 437 Particle size, 37 Pauli Exclusion Principle , 12 PCC, 317
521
522
Peak areas in nmr, 252 Peak base in nmr, 260 Peak-splitting in nmr, 253ff Pentene isomers, structures, 80, 90 Peptides, 481ff Periodic acid, 312 Perkin condensation, 404 Peroxides in ethers, 298 Phase transfer catalyst, 128 Phenanthrene, 207 Phenetidine, 453 Phenetole, 292, 440 Phenolphthalein, 454 Phenols, acidity of, 443 summary of chemistry, 453 3-Phenyl-3-chloro-l-propene, enantiomers of, 75 Phenylhydrazine, 327, 498 Phenylpyridine, 464 Phosgene, 4, 369, 380 Phthalic acid, 355 Phthalimide, 414 Pi bond, 13 Picoline, 463, 466 Picric acid, 442 Pinacolone, 408 Pinacol rearrangement, 305, 312, 322, 339 Piperidine, 463 Planck’s constant, 242 Plane of symmetry, 68 Polar bonds, 26 Polarity, 19 Polarizability, 122 Polar molecules, 26 Polycyclic compounds, 166 Polygon Rule, 205 Polymerization, alkenes, 103ff, 177 dienes, l55ff Polymers, 371 Polysaccharides, 5 lOff Polyurethane, 371 Potassium t-butoxide as a base, 135, 137 Potassium iodide as hydrolysis catalyst, 134 Primary carbon, 54 Primary groups, 54 Priorities of ligands, 71ff Probability factor, 37, 58 Propagation, 32 Proteins, 474ff Pyran, 298 Pyranose, 504 Pyrazine, 467 Pyridine, 463 Pyridinium chlorochromate, 317 Pyrimidine, 467 Pyrolysis of alkanes, 57 Pyrrole, 457fl Pyruvic acid, 352
INDEX
Quaternary ammonium salts, 412 Quaternary carbon, 54 Quinhydrone, 454 Quinoline, 468 Quinones, 447 R and S configurations, 71ff Racemization, 68, 81, 124 Radicals, 30, 60,232 Raney nickel, 307 Rate-controlled reaction, 152 Rate of reaction, 47 Reaction mechanism, 30, 39 Reaction rate, 37, 47 Reaction, spontaneity of, 35 Reactions, addition-elimination, 225ff Reactions, aromatic, 209ff Reactivity-Selectivity Principle, 58, 221 Redox, 32 Redox equations, balancing, 278 Reduction of aromatic compounds, 209 Reduction, Rosenmund, 319 Reductive amination, 415 Reductive hydroboration, 96 Reformatsky reaction, 334 Regioselective reactions, 98 Reimer-Tiemann reaction, 448 Resolution of enantiomers, 68,78 Resonance, 22 in benzene, 201 energy, 23 structures, 28 Resorcinol, 340, 440 Retroaldol condensation, 411 Ribose, 502, 511 Ribosomal RNA, 513 Ring activation, 218fl RNA, 511, 513 Robinson reaction, 410 Rosenmund reduction, 319 Ruff degradation, 497
S and R configuration, 71ff Saccharin, 374 Salicylic acid, 440 Sandmeyer reaction, 426 Sanger method, 483 Saturated compounds, 49 Saytzeff, 92, 129 Schiff base, 422 Schmidt rearrangement, 418 Secondary carbon, 54 Semicarbazide, 327 Sickle-cell anemia, 487
INDEX
Sigma bond, 13 Sigma complex, 215 Silyl ethers, 299 Singlet carbene , 30 Skraup reaction, 468 SN1 and SN2reactions, influences on rates, 136 mechanisms, 123ff Soaps, 349, 365 Sodium bisulfite, 327 Sodium borohydride , 273 Soft bases, 122 Solvation, 22 Solvents, classification, 21 Solvolysis, 123,127 Sorbose, 507 Specific rotation, 69 Spectroscopy, infrared, 244ff mass, 259ff nuclear magnetic resonance , 248ff ultraviolet, 243ff visible, 243 Spin-spin coupling in nmr, 253 Spontaneous reaction, 35 Stability, heat of combustion, 91 of radicals, 60 resonance, 23 Staggered conformation , 50 Stereoisomers, 68 Stereomers, conformational, 79 Stereoselective reactions, 92 Steric hindrance, 220 Steric strain, 52,127 CStilbazole, 467 Stilbene, 228 Strain, torsional, 52 Strecker synthesis, 476 Structure of benzene, 198 cyclooctatetraene, 203 Substitution , allylic, 107 in arenes, 215ff in naphthalene, 223 nucleophilic aromatic, 225ff Succinic acid, 354 Succinic anhydride, 459 Sucrose, 509 Sulfamic acid, 425 Sulfanilamide, 431 Sulfanilic .acid, 425 Sulfonates, 276 Sulfonation of arenes, 217 Sulfhydryl group, 280 Sulfides, 307 Sulfones, 308 Sulfonic acid derivatives, 372 Sulfonium salts, 307 Sulfonyl chlorides, 276 Sulfoxides, 308
Symmetry, center of, 68 plane of, 68 Syn addition, 99 Synthesis of deuterated compounds, 64 Synthesis, use of blocking groups in, 224
Tartaric acid, 50 1, 5 16 Tautomerism, 384ff Terpenes, 188 Tetrabutyl ammonium chlorides, 128 Tetraethyl lead, 65 Tetrahydrofuran [THF], 97,298,458,462 Tetralin, 209 Thermodynamics of reactions, 34,46 Thermodynamic-controlled reactions, 152 Thiazole, 467,470 Thioethers, 307ff Thiols, summary of chemistry, 281 Thiophene, 457ff Thiourea, 422 Threo form, 82,85 Threonine, 480 Threose, 501 Thymine, 467 Tollens’ reagent, 323 Toluidine, 412 Torsional strain, 52 Tosyl group, 291 Transacylation, 358,422 Transfer RNA , 5 13 Transamination, 477 Transition state, 38,124 Triazole, 470 Trifluoracetic acid, 485 Trifluoroperoxyacetic acid, 442 Triglycerides, 365 Triplet carbene, 30 Triton-B, 412 Tschugaev reaction, 286
U1traviole t spect roscopy , 243 Unsaturated hydrocarbons, 87 Uracil, 467 Urea, 369,422 Urethane, 369
van der Waals forces, 20 Veronal, 369 Visible spectroscopy, 243
523
524
Wave number, 242 Wedge projection, 3, 50, 70 Williamson synthesis, 292, 307, 445 Wintergreen, oil of, 449 Wittig reaction, 332 Wohl degradation, 497 Wolff-Kishner reduction, 324 Woodward-Hoffmann Rules, 184
INDEX
Xanthate, 286 ~-Xylose,516 Ylides, 332fi Ziegler method, 183 Zwitterions, 474
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