9-2
n H aJHUa1d ----
NOS \f3d
Lh -E -O N8SI
LUT ONS MANUAL Jan William Simek
California Polytechnic State University
OR
C CHEMISTRY SIXTH EDITION
L. G. Wade, Jr. ".�JL.-':'
Prentice Hall Upper Saddle River, NJ
07458
Assistant Editor: Carole Snyder Project Manager: Kristen Kaiser Executive Editor: Nicole Folchetti Executive Managing Editor: Kathleen Schiaparelli Assistant Managing Editor: Becca Richter Production Editor: Kathryn O'Neill Supplement Cover Manager: Paul Gourhan Supplement Cover Designer: Joanne Alexandris Manufacturing Buyer: Ilene Kahn Cover Image Credit: Joseph Galluccio (2004)
PEARSON
© 2006 Pearson Education, Inc.
Prentice Hall
Pearson Prentice Hall
Pearson Education, Inc. Upper Saddle River, NJ 07458
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Printed in the United States of America 10
9
ISBN
8
7
6
5
0-13-147882-6
Pearson Education Ltd., London Pearson Education Australia Pty. Ltd., Sydney Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada, Inc., Toronto Pearson Educaci6n de Mexico, S.A. de c.y. Pearson Education-Japan, Tokyo Pearson Education Malaysia, Pte. Ltd.
TABLE OF CONTENTS
Preface
........................................................................................................................................
v
Symbols and Abbreviations ...................................................................................................... vii
Chapter 1
Introduction and Review
Chapter 2
Structure and Properties of Organic Molecules
Chapter 3
Structure and Stereochemistry of Alkanes
Chapter 4
The Study of Chemical Reactions
Chapter 5
Stereochemistry
Chapter 6
Alkyl Halides: Nucleophilic Substitution and Elimination
Chapter 7
Structure and Synthesis of Alkenes
Chapter 8
Reactions of Alkenes
Chapter 9
Alkynes
Chapter 10
Structure and Synthesis of Alcohols
Chapter 11
Reactions of Alcohols
Chapter 12
Infrared Spectroscopy and Mass Spectrometry
Chapter 13
Nuclear Magnetic Resonance Spectroscopy
Chapter 14
Ethers, Epoxides, and Sulfides
Chapter 15
Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy
Chapter 16
Aromatic Compounds
Chapter 17
Reactions of Aromatic Compounds
Chapter 18
Ketones and Aldehydes
Chapter 19
Amines
Chapter 20
Carboxylic Acids
Chapter 21
Carboxylic Acid Derivatives
Chapter 22
Condensations and Alpha Substitutions of Carbonyl Compounds
Chapter 23
Carbohydrates and Nucleic Acids
Chapter 24
Amino Acids, Peptides, and Proteins
Chapter 25
Lipids
Chapter 26
Synthetic Polymers
Appendix I:
Summary of IUPAC Nomenclature
Appendix 2:
..... . ...
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.
... . ... . ................ . ....... ................. . ...
.......... ............
.
......................... ........................
....................................
.
.....................
.
. . . .................... . . ............................... .............
.
.... . ... .......... . ... . . . ........................
.
....... . ...
.
...
.
. .. .
. ... ..
..
.....
.
................. . ....... .......
.
.................................. ...........
.
...... . . ........................ . ......................
.................
.............. .................
. .
..... . .......
.
.
..........
......... ..........
..... . .... . ......................
.................. . ................................ ............ ............
........ . ............ ...........
....... . . .............
.. .
....
. .. .. .
.
..
.........
.... . ...... .....
.. . ...
.. .
.
. . . . ....
.
................................. . .......... .............. ..............
...... ........
.
. ..........
... . ..... . ........ . . ..............................................
. . . . ......... . . . ....... . . .. . .
.................. . ...
..
.
........ . ...........
.
............
............... . . . ................................
............................................ ......... . ............
.............. .... . ................. ............................. ..................
. . . ........... .......................................................................
.
....... .....................................................................................
... . .................
. . ..
...................................................... ......................
... . ... . . ...........
................................ . ........ ........ ...................
.
.......
.
...........
............................... ............
.
.
.............................. . .......... ...................................... ....
. ... . ....
.
. . ............... .
.
.......... .......
.
.. . ....... .................................
. . . . ........ .......... . ................. .......... ................
Summary of Acidity and Basicity
... . . . .......
iii
.
......
.
.
.
... ........ .................. . . . . ............
1
25 45 59 79 99
135 159 191 211 229 259 271 299 317 341 365 401 439 469 499 537 585 617 645 659
675 689
PREFACE Hints for Passing Organic Chemistry Do you want to pass your course in organic chemistry? Here is my best advice, based on over thirty years of observing students learning organic chemistry: Hint #1: Do the problems. It seems straightforward, but humans, including students, try to take the easy way out until they discover there is no short-cut. Unless you have a measured IQ above 200 and comfortably cruise in the top 1 % of your class, do the problems. Usually your teacher (professor or teaching assistant) will recommend certain ones; try to do all those recommended. If you do half of them, you will be half-prepared at test time. (Do you w ant your surgeon coming to your appendectomy having practiced only ha l/the procedure?) And w hen you do the problems, keep this Solutions Manual CLOSED. A void looking at my answer before you write y our answer-your trying and struggling with the problem is the most valuable part of the problem. Discovery is a major part of learning. Remember that the primary goal of doing these problems is not just getting the right answer, but understanding the material well enough to get right answers to the questions you haven't seen yet. Hint # 2 : Keep up . Getting behind in your work in a course that moves as quickly as this one is the Kiss of Death. For most students, organic chemistry is the most rigorous intellectual challenge they have faced so far in their studies. Some are taken by surprise at the diligence it requires. Don't think that you can study all of the material the couple of days before the exam-well, you can, but you won't pass. Study organic chemistry like a foreign language: try to do some every day so that the freshly trained neurons stay sharp. Hint #3: Get h elp when y ou need it. Use your teacher's office hours w hen you have difficulty. Many schools have tutoring centers (in which organic chemistry is a popular offering). Here's a secret: absolutely the best way to cement this material in your brain is to get together with a few of your fellow students and make up problems for each other, then correct and discuss them. When you write the problems, you will gain great insight into what this is all about.
Purpose of this Solutions Manual So w hat is the point of this Solutions Manual? First, I can't do your studying for you. Second, since I am not leaning over your shoulder as you write your answers, I can't give you direct feedback on what you write and think-the print medium is limited in its usefulness. What I can do for you is: I) provide correct answers; the publishers, Professor Wade, Professor Kantorowski (my reviewer), and I have gone to great lengths to assure that what I have written is correct, for we all understand how it can shake a student's confidence to discover that the answer book flubbed up; 2) provide a considerable degree of rigor; beyond the fundamental requirement of correctness, I have tried to flesh out these answers, being complete but succinct; 3) provide insight into how to solve a problem and into where the sticky intellectual points are. Insight is the toughest to accomplish, but over the years, I have come to understand where students have trouble, so I have tried to anticipate your questions and to add enough detail so that the concept, as well as the answer, is clear. It is difficult for students to understand or acknowledge that their teachers are human (some are more human than others). Since I am human (despite what my students might report), I can and do make mistakes. If there are mistakes in this book, they are my sole responsibility, and I am sorry. If you find one, PLEASE let me know so that it can be corrected in future printings. Nip it in the bud.
What's New in this edition? Better answers! Part of my goal in this edition has been to add more explanatory material to clarify how to arrive at the answer. The possibility of more than one answer to a problem has been The IUPAC Nomenclature appendix has been expanded to include bicyclics, heteroatom noted. replacements, and the Cahn-Ingold-Prelog system of stereochemical designation. Better graphics! The print medium is very limited in its ability to convey three-dimensional structural i�formation, a problem that has plagued organic chemists for over a century. I have added some graphiCS created .1I1 the software, Chem3D®, to try to show atoms in space where that information is a key part of the solution. In drawing NMR spectra, representational line drawings have replaced rudimentary attempts at drawing peaks from previous editions. Better jokes? Too much to hope for. v
Some Web Stuff Prentice-Hall maintains a w eb site dedicated to the Wade text: try www.prenhall.com/wade. Two essential web sites providing spectra are listed on the bottom of p. 270. Acknowledgments No project of this scope is ever done alone. These are team efforts, and there are several people w ho have assisted and facilitated in one fashion or another w ho deserve my thanks. Professor L. G. Wade, Jr., your textbook author, is a remarkable person. He has gone to extraordinary lengths to make the textbook as clear, organized, informative and insightful as possible. He has solicited and followed my suggestions on his text, and his comments on my solutions have been perceptive and valuable. We agreed early on that our primary goal is to help the students learn a fascinating and challenging subject, and all of our efforts have been directed toward that goal. I have appreciated our collaboration. My new colleague, Dr. Eric Kantorowski, has reviewed the entire manuscript for accuracy and style. His diligence, attention to detail and chemical w isdom have made this a better manual. Eric stands on the shoulders of previous reviewers who scoured earlier editions for errors: Jessica Gilman, Dr. Kristen Meisenheimer, and Dr. Dan Mattern. Mr. Richard King has offered numerous suggestions on how to clarify murky explanations. I am grateful to them all. The people at Prentice-Hall have made this project possible. Good books would not exist without their dedication, professionalism, and experience. Among the many people who contributed are: Lee Englander, w ho connected me with this project; Nicole Fo1chetti, Advanced Chemistry Editor; and Kristen Kaiser and Carole Snyder, Project Managers. The entire manuscript was produced using ChemDraw®, the remarkable software for drawing chemical structures developed by CambridgeSoft Corp., Cambridge, MA . We, the users of sophisticated software like ChemDraw, are the beneficiaries of the intelligence and creativity of the people in the computer industry. We are fortunate that they are so smart. Finally, I appreciate my friends who supported me throughout this project, most notably my good friend of almost forty years, Judy Lang. The students are too numerous to list, but it is for them that all this happens. Jan William Simek Department of Chemistry and Biochemistry Cal Poly State University San Luis Obispo, CA 93407 Email:
[email protected]
DEDICATION To my inspirational chemistry teachers:
Joe Plaskas, w ho made the batter; Kurt Kaufman, w ho baked the cake; Carl Djerassi, w ho put on the icing; and to my parents:
Ervin J. and Imilda
B.
Simek,
who had the original concept.
vi
SYMBOLS AND ABBREVIA TIONS
Below is a list of symbols and abbreviations used in this Solutions Manual, consistent with those used in the textbook by Wade. (Do not expect all of these to make sense to you now. You will learn them throughout your study of organic chemistry.) BONDS a single bond a double bond a triple bond a bond in three dimensions, coming out of the paper toward the reader a bond in three dimensions, going behind the paper away from the reader a stretched bond, in the process of forming or breaking -
111111111111
ARROWS in a reaction, shows direction from reactants to products signifies equilibrium (not to be confused with resonance) signifies resonance (not to be confused with equilibrium) shows direction of electron movement: the arrowhead with one barb shows movement of one electron; the arrowhead with two barbs shows movement of a pair of electrons shows polarity of a bond or molecule, the arrowhead signifying the more negative end of the dipole SUBSTITUENT GROUPS Me a methyl group, CH3 Et an ethyl group, CH2 CH3 Pr a propyl group, a three carbon group (two possible arrangements) Bu a butyl group, a four carbon group (four possible arrangements) the general abbreviation for an alkyl group (or any substituent group not under scrutiny) R Ph a phenyl group, the name of a benzene ring as a substituent, represented: -
..
Ar
..
<»
or
the general name for an aromatic group
0-
continued on next page vii
Symbols and Abbreviations, continued SUBSTITUENT GROUPS, continued Ac
an acetyl group:
Cy
a cyclohexyl group:
Ts
o
II
CH3 - C -
0-
o
tosyl, or p-toluenesulfonyl group: CH3 � "==1 - So
o II Boc a t-butoxycarbonyl group (amino acid and peptide chemistry): (CH3)3C -0 - C II
Z,
or a carbobenzoxy (benzyloxycarbonyl) group (amino acid and peptide chemistry): Cbz
REAGENTS AND SOLVENTS DCC
dicyclohexylcarbodiimide
DMSO dimethylsulfoxide ether
o
o- N= C= N -Q
II
S H3C ....... 'CH3
diethyl ether, CH3CH20CH 2 CH3 Cl
<>
MCPBA meta-chloroperoxyhenzo;c ac;d MVK
NBS
o
methyl vinyl ketone H3C N
-bromosuccinimide O
continued on next page
.......
g�
O � N I
Br
viii
0
g - 0 - OH
0
< )- CH2-0 - g -
Symbols and Abbreviations, continued REAGENTS AND SOLVENTS, continued pyridinium chlorochromate, Cr03· HCI N PCC •
Sia2BH
disiamylborane
THF
tetrahydrofuran
� ')
CH3 H H H CH 3 I I I I I H-C-C-B-C-C-H I I I I CH3 CH 3 CH3 CH3
o
SPECTROSCOPY infrared spectroscopy IR nuclear magnetic resonance spectroscopy NMR mass spectrometry MS ultraviolet spectroscopy UV parts per million, a unit used in NMR ppm hertz, cycles per second, a unit of frequency Hz megahertz, millions of cycles per second MHz tetramethylsilane, (CH3) 4Si, the reference compound in NMR TMS s, d, t, q singlet, doublet, triplet, quartet, referring to the number of peaks an NMR absorption gives nm nanometers, 10-9 meters (usually used as a unit of wavelength) mJz mass-to-charge ratio, in mass spectrometry 8 in NMR, chemical shift value, measured in ppm A wavelength frequency v OTHER a,
ax
e,eq HOMO LUMO NR 0,
m,p
axial (in chair forms of cyclohexane) equatorial (in chair forms of cyclohexane) highest occupied molecular orbital lowest unoccupied molecular orbital no reaction artha, meta, para (positions on an aromatic ring) when written over an arrow: "heat"; when written before a letter: "change in" partial positive charge, partial negative charge energy from electromagnetic radiation (light) specific rotation at the D line of sodium (589 nm) ix
CHAPTER I-INTRODUCTION A ND REVIEW
1-1
1-2
(a)
(e)
P 1 s 2 2s 2 2p6 3s 2 3px 1 3py 1 3pz 1 1 s 2 2s2 2p6 3s 1 S 1 s 2 2s 2 2p6 3s 2 3px 2 3py 13pz 1 Mg 1 s 2 2s2 2p6 3s 2 Is 2 2s2 2p6 3s 2 3px 1 CI 1 s 2 2s2 2p6 3s 2 3px2 3py2 3pz 1 AI Ar 1 s 2 2s 2 2p6 3s 2 3px 2 3 py 2 3 pz 2 Si 1 s 2 2s 2 2p6 3s 2 3px 1 3py I In this book, lines between atom symbols represent covalent bonds between those atoms. Nonbonding electrons are indicated with dots. H H H Na
H - N- H I H
(b)
H H I I H - C - C-N - H I I I H H H
(c)
H-O-H
H H I I (f) H -C - O - C - H I •• I H H
1-3
(a)
(i)
H :0: H I I I H - C - C - C-H I I I H H H
:N
N:
(g)
• •
H I
(h)
+
H - •• O-H I H
(d)
I I I H-C-C-C-H I I I H H H
H H I I H- C - C - F: I I H H (j ) : F- B - F: I -:F:
H -B- H I H
.
.
The compounds in (i) and (j ) are unusual in that boron does not have an octet of electrons-normal for boron because it has only three valence electrons. (b)
H-C
N:
(c)
H - O - N= O
(d)
O == C = O
:0 :
(e) (i) 1-4
II H - C =N - H (h) H - N=N- H (f) H - C - O - H (g) H - C == C - C I : I I I •• H H H H H I I H - C == C - C - H (j ) H - C == C == C - H (k) H - C C - C - H I I I I I I H H H H H H
(a) G}J -NQ) (e)
8
H - C =N - H I H
(b) (f)
H-C
NeD
C)J(D •
·8 8
II H - C-O - H
88(.:)
(c)
H - O - N= O
(g)
H - C == C - C leD I I 8 H H
0)
There are no unshared electron pairs in parts (i), (j ) , and (k). 1
0)
8
(d)
0) 0) o == c=o
(h)
H - N=N- H
0)
0)
88
1-5
The symbols "8+" and "8-" indicate bond polarity by showing partial charge. (In the arrow symbolism, the arrow should point to the partial negative charge.) �
�
(a) C -C 1
�
�
8-
8+
(b)
C- O
�
�
(c)
C-N
(g)
N-O
8+
8-
(h)
N-S
1-6
8-
8+
(i) N - B
Non-zero formal charges are shown beside the atoms. H H H H H 1 1+ 1 1 C1: H N H H C N - C - H :Cl: (b) (a) H - C - O-H : (c) 1 1 1+ 1 1 1 H H H H H H In (b) and (c), the chlorine is present as chloride ion. There is no covalent bond between chlorine and other atoms in the formula. H H I 1(g) Na+ H - B - H (e) H - C - H (d) Na+ : O - C - H (f) H - C-H I 1 1 I H H H H
1+
-
•
. .
. .
+
•
1+
-
H H H, H 1- / (h) Na+ H - B - C N: (i) H -/C - O - C - H U) H - O-N-H I I H H H H :F-B-F: H ,H1/ H :F: As shown in (d), (g), (h), and (k), alkali metals like sodium and C potassium form only ionic bonds, 1 /H never covalent bonds. · O - C - C '- H H - C = O+- H (I) 1 I H H /C1' H H H 1-7 Resonance forms in which all atoms have full octets are the most significant contributors. In resonance forms, ALL ATOMS KEEP THEIR POSITIONS-ONLY ELECTRONS ARE SHOWN IN DIFFERENT POSITIONS. :0: :0: : 0: 1 " 1 :O - C=O O = C - O: (a) : O - C - O: •
•
1 _+ ' 1 •
•
. .
-
•
•
•
• •
. . -
-
..
:0 : " (b) : O - N - O: +
. . -
..
. .
. .
•
. . -
..
..
. .
•
-
..
:0 : 1 : O - N=O
-
-
+
: 0: I O=N - O:
...
2
+
•
•
-
1-7
continued
(c) :O-N=O (d)
..
H-C=C-C-H
I
I
. .
O=N-O:
..
I
+
H H H
..
-
.. H-C-C=C-H
I
I
+
I
H H H
H - C = C - C - H .. .. H - C - C=C - H I I I I I I H H H H H H (f) Sulfur can have up to 12 electrons around it because it has d orbitals accessible . :0 : : 0: : 0: I I II O==S-O: .. :O - S - O: .. .. :O - S=O
(e)
. .
. . -
-
II
II
:0 :
: 0:
: 0:
:0: II :O - S=O I :0:
-
(g)
H I H-C-H H :0I : I I H - C - C+ \ I :0 : H I H - C-H I H
II
..
..
..
..
•
•
H I H-C-H I H + 0: 1/ I .. H- C- C \ I H :0 : I H-C-H I H
:0: I O==S==O I :0 : .
..
..
..
.
-
..
/
:0 : II O=S- O: I :0 : • •
.
-
.
H I H - C -H H :0I : I I H-C-C \\ I H + 0: I H- C- H I H
1-8 Major resonance contributors would have the lowest energy. The most important factors are: maximize full octets; maximize bonds; put negative charge on electronegative atoms; minimize charge separation. + (a) H - C - N=O .. .. H - C - N-O: .. .. H - C=N - O: (negative charge on electronegative I I I II I I H :0 : H :0 : H :0 : atoms) minor mInor major •
•
-
. .-
+
+
3
1-8 continued H H H + + + \ + \ \ C - C = N - O: C = C - N - O: C=C - N=O (b) / / / 1 1 1 11 1 1 H H H H :0 : H :0 : H :0 : minor major major These two forms have equivalent energy and are major because they have full octets, more bonds, and less charge separation than the minor contributor. + (c) H - C- O-H H - C = O - H .. 1 1 H H mmor major (octets, more bonds) •
(e)
(f)
+
+
H - C=N=N: 1 H major (negative charge on electronegative atom) ..
H - C - C N: 1 H minor
..
.-
•
H - C - N N: 1 H minor H - C=C=N: 1 H major (negative charge on electronegative atom) +
+
H - N - C - C=C - N - H " 1 1 1 1 1 H H H H H mmor
t
+
•
H-N=C - C=C - N - H 1 1 1 1 1 H H H H H major these two forms are major contributors because all atoms have full octets +
H - N - C = C - C - N - H .. 1 1 1 1 1 H H H H H mmor • •
•
H- N - C= C - C = N - H 1 1 1 1 1 H H H H H major
:0: :0 : :0 : :0 : :0 : :0 : II II 1 II II 1 H - C - C-- C - H----- H - C = C - C - H ----- H - C - C = C - H 1 1 1 H H H major mITIor major these two have equivalent energy and are major because the negative charge is on the more electronegative oxygen atom . . -
(g)
.
•
+
(d)
..
..
• •
4
1 -8 continued :0 : II (h) H - C - N - H I
major
.. :0 : + I H - C =N H -
.... .. II----I .. �
H
(no charge separation)
-
I
minor
H
1-9 Your Lewis structures may appear different from these. As long as the atoms are connected in the same order and by the same type of bond, they are equivalent structures. For now, the exact placement of the atoms on the page is not significant. A Lewis structure is "complete" with unshared electron pairs shown. H H H H H H H H H I I I I I I I I I (b) H - C - C - C - Cl: (a) H - C - C - C - C - C - C - H I I I I I I I I I H /C H H H H H /C H H H1 ' H H H1 ' H H H :0: H H :0 : H I II I I I II / (d) H - C - C - C-H (c) H-C - C - C - C=C \ I I I I I H H H H H H Always be alert for the implied double or triple bond. Remember that C has to have four bonds, nitrogen has three bonds, oxygen has two bonds, and hydrogen has one bond. The only exceptions to these valence rules are structures with formal charges. H :0: II I (e) H - C - C - C N: I H
(f)
H ,H1 / H H , CI :0: II H -/ C - C - C - O - H I H C / H H1 ' H
H H :0: H H I I II I I (g) H-C - C - C - C - C - H I I I I H H H H 1 - 10 Complete Lewis structures show all atoms, bonds, and unshared electron pairs. (a) H H H (b) (c) H H H ,HI H /H I/H 1/ I I H , /, C C-H H C C H C C ,C / II \\ H - C , C /- H C H H , I I / , N. . /C ' C-H I 1/ H -/ C�' N '�C ,- H H H H -C - C I'H I H H H H H I \ I ..../.. C , ,..... C ,...... H H 0 H H • •
_
• •
5
(e )
1-10 continued
· · H ,, / H · 0I I· H " /C /C, H H - C ,C II I H-C , ...... C, H / / C"- H H H
(f)
H H H " /,, C /.... ,...-0 H - C ' C/·· I I H -/ C, /' /.C, C H H I H •
•
H H :0 : H H H (h) :0 : H I II I I II I I I H ' /C:--.., /,C-C - H H-C - C-C-C-C-H c "c I I I I I I H II H H H H C C, H/ , C 'i H I H 1-1 1 There is often more than one correct way to write condensed structural formulas. You must often make inferences about what a condensed formula means according to valence rules, especially in structures with C=O as shown in parts (a) and (d). (b) (CH3hCHCH2 CH20 H (c) (CH3hCHCH2 CH(OH)CH3 (a) CH3COCH2 CH2 CH3 0 or CH3CH(CH3)CH2 CH(OH)CH3 or CH3CH(CH3)CH2 CH20H
(g)
(the
has a double bond to the
(d) CH3CH2CH(CH3)CH2 CHO
c arbon preceding it)
(the 0 has a doubl e bond to the
c arbon preceding it)
1-1 2 If the percent values do not sum to 100%, the remainder must be oxygen. Assume 100 g of sample; percents then translate directly to grams of each element. There are usually many possible structures for a molecular formula. Yours may be different from the examples shown here. some possible structures: (a) 40.0 gC = 3.33 moles C 3.33 moles = 1 C 1 2.0 g/mole H ° H I I II 6.67 g H HO C-C-C - OH 3.33 moles = 1.98 2 H = 6.60 moles H 1.0 1 g/mole I I H H 53.33 gO = 3.33 moles ° 3.33 moles = 1 ° 16.0 g/mole OH empirical formula c::::::> empirical weight 30 ==
�
molecular weight = 90, three times the empirical weight =
three times the empirical formula = molecular formula 6
=
=
c:::::>
I C3H603 I
A
Ho OH MANY other structures possible.
1 - 1 2 continued (b) 32.0 g C 1 2.0 g/mole 2.67 moles C 1.34 moles 1.99 2 C 6.67 g H 5H 4.93 6.60 moles H -:- 1.34 moles l.01 g/mole IS.7 g N 14.0 g/mole 1.34 moles N -:- 1.34 moles 1 N 42.6 g a . 16.0 g/mole - 2.66 moles a -;- 1.34 moles 1 .99 2 a empirical formula I C2H sN02 I c:::::::> empirical weight 75 =
=
some possible structures:
::=
H
=
molecular weight empirical formula (c)
molecular formula
=
C2HsN02
=
37.2 g C 1 2.0 g/mole 3. 1 0 moles C -:- 1 .55 moles 2 C 7.7 5 g H 1.0 1 g/mole 7.67 moles H -:- 1 .55 moles 4.95 55.0 g CI . 35 .45 g/mole - 1 .55 moles Cl -;- 1.55 moles - 1 CI =
=
=
=
empirical formula molecular weight
=
· 1 .� ��/�:l e
3��4� ��ole
=
=
=
I C2H sCl I
=
molecular formula
3.20 moles C 4.75 moles H 1 .60 moles CI
empirical formula
=
=
-:-
I C2H 3Cl I
1 .60 moles 1 .60 moles c:::::::>
=
2.97
I
H
I
5H
There is only one structure possible with this molecular formula:
c:::::::>
=
64.46
H
I
I
H - C- C==C- C- H
7
I
I
I
H C l Cl H Cl C]
1 Cl
=
I
H
some possible structures: H H I
=
I
H - C - C - Cl I
3H
empirical weight
H
I
H
=
62.45
molecular weight 1 25, twice the empirical weight c:::::::> twice the empirical formula molecular formula ""-C4 -H - -6- C-12--,-o. =
I
MANY other structures possible.
2C =
H
/
I C2HsCI I
1.60 moles -:-
a
II
H
=
c:::::::> empirical weight
64, same as the empirical weight
=
empirical formula 3S.4 g C 12.0 g/mole
::=
\
N- C- O - C- H
_
_
(d)
c:::::::>
75, same as the empirical weight
=
H
::=
=
I
I
H H
=
_
I
H- C- C-N02
=
=
H
I
Cl
U �
MANY other Cl structures possi ble.
1-13 1 mole HEr (a) 5 .00 g HBr x 80.9 g HBr
=
0.06 1 8 moles HEr 0.0618 moles
1 00 mL
pH
(b)
=
-
H30 +
0.06 1 8 moles HBr
0.06 1 8 moles H 30 + (100% dissociated) x
10gIO [H30+]
1000
mL
1L
=
-
0.6 1 8 moles H30 + 1 L solution
=
logl o (0.6 1 8)
=
�
0.0375 moles NaOH 1.50 g NaOH x 1 mole NaOH 40.0 g NaOH 0.0375 moles -OH (100% dissociated) 0.0375 moles NaOH =
1 000 mL 0.75 moles -OH 1 L solution lL 14 1 x 1 0-14 1.33 x 10-14 [H 3 0+] 1 x 1 0- = 0.75 [-OH] pH -loglO [H30+] -logl o (1.33 x 10- 14 )
0.0375 moles -OH 50. mL
x
=
=
=
=
=
0.75 M
(the number of decimal places in a pH value is the number of significant figures)
=
1-14 (a) By definition, an acid is any species that can donate a proton. Ammonia has a proton bonded to nitrogen, so ammonia can be an acid (although a very weak one). A base is a proton acceptor, that is, it must have a pair of electrons to share with a proton; in theory, any atom with an unshared electron pair can be a base. The nitrogen in ammonia has an unshared electron pair so ammonia is basic. In water, ammonia is too weak an acid to give up its proton; instead, it acts as a base and pulls a proton from water to a small extent. (b) water as an acid: H 2 O + NH3 -OH + NH4+ --
H 2 O + HCl water as a base: (c) methanol as an acid: CH30H + NH3 methanol as a base: CH30H
+
H 3 O+
ClCH3O+
NH
4+ CH3OH2 + + HS04-
H2 SO4
8
+
1-15 (a) HCOOH stronger acid pKa 3.76 (b) CH3COOweaker base (c) CH30H stronger acid pKa 1 5.5
(f)
1-16
NaNH2 stronger base
+
..
--
--
HCN FAVORS weaker PRODUCTS acid pK a 9.22
+
CH 3 COOH stronger acid pKa 4 . 74
+
CH30- Na+ weaker base
.-
--
HCN stronger acid pK a 9.22
+
HCOOweaker base
.-
--
CH30H weaker acid pK a 1 5 .5
+
(d) Na+ -OCH3 stronger base (e)
-CN stronger base
+
.-
HOCH 3 weaker acid pK a 15.5
+
+
CH3O- FAVORS stronger REA CTANTS base FAVORS NH3 weaker PRODUCTS acid pK a 33 NaCN weaker base
FAVORS
PRODUCTS
FAVORS HCl + H2O H3O+ + CIPRODUCTS stronger stronger weaker weaker acid base base acid The first reaction in Table 1-5 shows the Keq for this reaction is 160, favoring products. --
.-
H30+ + CH3OH2O + CH30H FAVORS PRODUCTS stronger weaker stronger weaker acid base base acid pK a -1 .7 pK a 1 5.5 The seventh reaction in Table 1-5 shows the Keq for the reverse of this reaction is 3 .2 x 1 0-16. Therefore, Keq for this reaction as written must be the inverse, or 3. 1 x 1 0 15, strongly favoring products. --
.-
:0: II CH 3 - C - O - H
:0: II CH 3 - C - O-H +\ H Protonation of the double-bonded oxygen gives three resonance forms (as shown in Solved Problem 1-5(c)); protonation of the single-bonded oxygen gives only one. In general, the more resonance forms a species has, the more stable it is, so the proton would bond to the oxygen that gives a more stable species, that is, the double-bonded oxygen. • •
9
1-l7 In Solved Problem 1-4, the structures of ethanol and methylamine are shown to be similar to methanol and ammonia, respectively. We must infer that their acid-base properties are also similar. (a) This problem can be viewed in two ways. 1 ) Quantitatively, the pKa values determine the order of acidity. 2) Qualitatively, the stabilities of the conjugate bases determine the order of acidity (see Solved Problem 1-4 for structures): the conjugate base of acetic acid, acetate ion, is resonance-stabilized, so acetic acid is the most acidic; the conjugate base of ethanol has a negative charge on a very electronegative oxygen atom; the conjugate base of methylamine has a negative charge on a mildly electronegative nitrogen atom and is therefore the least stabilized, so methylamine is the least acidic. acetic acid > ethanol > methylamine pK a 4.74 pK a "" 1 5.5 pK a 33 weakest acid strongest acid (b) Ethoxide ion is the conjugate base of ethanol, so it must be a stronger base than ethanol; Solved Problem 1-4 indicates ethoxide is analogous to hydroxide in base strength. Methylamine has pKb 3.36. The basicity of methylamine is between the basicity of ethoxide ion and ethanol. ethoxide ion > methylamine > ethanol weakest base strongest base ""
1-1 8 Curved arrows show electron movement, as described in text section 1-14. stronger acid
__
stronger base
.. CH3CH2-�: conjugate base weaker base
equilibrium favors PRODUCTS
'0' '11' � � (b) CH3CH2 - C - 0 - H + CH3 - N - CH3 I stronger acid H stronger base .
•
•
•
(c)
• •
•
•
equilibrium favors PRODUCTS
i-·· :�: ..
:O - S - O - H II :0 : • •
•
•
. .
CH3 - �+- CH3 H conjugate acid weaker acid 1
--
equilibrium favors PRODUCTS
O' .. � 0. '11' CH3 - O - H + H - O - S - O - H II stronger base :0·: stronger acid
CH 3-� -H H conjugate acid weaker acid
H
•
•
+
--
H 1+ .. CH 3 - O-H conjugate acid weaker acid •
conjugate base weaker base +
•
:0 : II O==S - O-H I : 0: conjugate base, weaker base
..
� --I.. l-. ... ....
. .
10
.
.
•
•
:0:- r ,
..
O==S - O - H II :0: • •
1-1 8 continued (d) - .. � 0· Na+ :O - H + H-S-H .. stronger acid stronger base (e)
---
_ equilibrium favors PRODUCTS __
H � /' I • •CH3 - �� H + CH3-O: stronger base H stronger acid "
+
..
.. H - O - H + Na+ :S-H conjugate acid conjugate base weaker base weaker acid CH3 - � - H H conjugate base weaker base
---
equilibrium favors PRODUCTS
i
+
stronger base
stronger acid
conjugate acid weaker acid
equilibrium favors PRODUCTS
(g)
:0: II . � CH 3 - C - O - H .. weaker acid
:0: - . . II :O-S-CH 3 • • II :0 :
~ +
� ... . ......f---l
equilibrium favors REACTANTS
+
:0: II • CH 3 -C - •.0: .
:0 : II 0==S-CH3 •• I :0 : weaker base
:0: : 0: II I CH 3 -C -�: ...... ... CH 3-C = � conjugate base stronger base
CH 3 - O-H •• conjugate acid weaker acid • :•0: � I •• CH -C =O .... . .. . 3 conjugate base •• weaker base _ -... . ......f--
-
:0:-
•• I 0==S-CH3 • • II :0 :
}
:0 : II H - O-S-CH3 • • II :0 : conjugate acid stronger acid 1-1 9 Solutions for (a) and (b) are presented in the Solved Problem in the text. Here, the newly formed bonds are shown in bold. H I •• /" - CH 3 H-B-H (c) H - B - H + CH 3-O " 1'1+ CH 3-O - CH 3 acid H �base f---l�
_
(d)
oI I : c CH 3-C� -H :O - H +
acid
•• ••
base
••
.. :0: I CH 3-r - H :O - H 11
t
1 - 19 continued (e) Bronsted-Lowry--c-proton transfer
� :�:
:�:
i
_}
H :0:
H - C - C - H +/.. : O - H �==� H --..�C - C - H ....- H - b == b - H ) k � base acid .
(f)
..�
CH3 - � - H H base
+
()
.. .
CH3 - S=!: acid
+
•
.
•
.-
+
..
H - O-H
:Cl :
1-20 Learning organic chemistry is similar to learning a foreign language: new vocabulary, new grammar (the reactions), some new concepts, and even a new alphabet (the symbolism of chemistry). This type of definition question is intended to help you review the vocabulary and concepts in each chapter. All of the definitions and examples are presented in the Glossary and in the chapter, so this Solutions Manual will not repeat them. Use these questions to evaluate your comprehension and to guide your review of the important concepts in the chapter. 1-21 (a) CARBON!
(b) oxygen
1-22 valence e- --- 1 H Li
2 Be
3 B
(c) phosphorus
(d) chlorine
4
5
6
7
C
N
0 S
F Cl Br I
p
8
He (2e-) Ne
1-23 (a) ionic only (b) covalent (H-O-) and ionic (Na+ -OH) (c) covalent (H-C and C-Li), but the C-Li bond is strongly polarized (d) covalent only (e) covalent (H-C and C-O-) and ionic (Na+ -OCH3) (f)
covalent (H-C and C=O and C-O- ) and ionic (HC02 - Na+)
12
(g) covalent only
1-24
: Cl
Cl:
:.. Cl "
:.Cl.
Cl:
:.Cl.
Cl: N (b) P (a) N ...... . . p ........ : Cl I Cl: I I : .c!'" I Cl: . :CI: :Cl: :Cl: :Cl: CANNOT EXIST NCls violates the octet rule; nitrogen can have no more than eight electrons (or four atoms) around it. Phosphorus, a third-row element, can have more than eight electrons because phosphorus can used orbitals in bonding, so PCls is a stable, isolable compound. 1-25 Your Lewis structures may look different from these. As long as the atoms are connected in the same order and by the same type of bond, they are equivalent structures. For now, the exact placement of the atoms on the page is not significant. H ,H1/ H H , C1 / H (a) H - N - N-H (c) H - C - N+- C "- H :CI : (b) H - N=N - H I I H H / CI H H / H H1" H .. ,../ ..
•
/ ..
Cl:
,../ ..
•
•
,
/ ..
•
/'
. •
. . -
(d)
H I H - C - C N: I H
(g)
:0: II H - O-S-O-H II :0 :
(e)
H :0: II I H-C - C - H I H
(h)
H I H - C - N=C = O I H
H H :N H I II I H-C-C-C-H I I H H /
1-26 (a)
(k)
H H :0: I II I (i) H - C-O-S-O-C - H I II I H H :0 :
H H H ,,1/ H " CI H -/ C-C - N=O I H C H /H1" H
H H :0: I I II H - C-C = C-C-C = C-C - O-H I I I I I I H H H H H H .
H :0: H I I II (f) H - C-S-C - H I I H H
H :0: H :0: I II I II (b) :N C-C - C - C - C-H I I H H
•
13
1-26 continued H - O: H :0: .
(c)
.
I
I
I
I
II
H-C = C-C-C-C-O-H
I
I
H
H
••
H H
1-27 In each set below, the second structure is a more correct line formula. Since chemists are human (surprise!), they will take shortcuts where possible; the first structure in each pair uses a common abbreviation, either COOH or CHO. Make sure you understand that COOH does not stand for C-O-O-H. Likewise for CHO. (a)
� /'... � /' � �
.....
OR
� COOH OH � OH OH
1 -28 (a)
OR
o
I
0
o
CHO
OR
H
0
H H H H H H H 1 1 1 1 1 I I H - C - C - C - C - H and H-C - C - C-H 1 1 I I 1 I I H C H H H H H H /H1 ' H H
(b)
� CHO H N - C TIY
N C
o
OH
OR (c)
(b)
COOH
H
I
H
1
H - C - C - N: I I I H H H
and
H
H I H-C-N-C-H I I H H H
I
I
14
these are the only two possibilities, but your structures may appear different-making models will help you visualize these structures
these are the only two possibilities, but your structures may appear different-making models will help you visualize these structures
1 -28 continued (c) There are several other possibilities as well. Your answer may be correct even if it does not appear here. Check with others in your study group. H H H H H H H H H I I I I I I I I I :O - C-C - O - C - H :O - C - C-C - O: :O-C-C - C - H I I I I I I I I I I I I I H H H H H H H H H H H :0: H I H H :0: :0 : These are the only three I II / \ (d) H - C-C - H H-C = C-O - H H - C - C - H structures with this molecular formula. I I I I I H H H H H H H H H H H H H H 1 -29 I I I I I I I I I (a) only three O - C-C-C - H H - C - C - O - C - H H - C-C - C - H possible I I I I I I I I I I structures H H H H H H H H 0 H I HOCH2 CH2 CH 3 CH3CH2 0CH 3 H CH3 CH(OH)CH3 (b) There are several other possibilities as well. H H H 0 H 0 H H I I I I II II I I H - C= C - C - H H -C-C-C- H H - C - C - C-H H - C=C-C - O - H I I I I I I I I I I H H H 0 H H H H H H I CH3 CH2 CHO CH 3 COCH 3 H2 C = CHCH 2 0H H H H 2 C =C(OH)CH3 I H - C = C - O-C - H I I I H H H H2 C = CHOCH 3 1-30 General rule: molecular formulas of stable hydrocarbons must have an even number of hydrogens. The formula CH2 does not have enough atoms to bond with the four orbitals of carbon. H H one carbon: I I H C -H H C H C = C H H C-C -H two carbons: I I I I I H-C-H C2 H 4 H H C2 H6 H H I H CH4 H H H H H I I I I I H - C = C-C - H H-C - C - C - H three carbons: H -C C - C - H I I I I I I I C3H6 H H H C 3 HS H H H C3 H 4 H 15
H H (b) H ..... � _ � _H \ H I H ..../. C ,0 0/ C ".... N H H I H �
H H H :0: I I I " H - N - C-C-C-C-O - H I I I I H H H H
(e)
H H (g) I \ C-C \ :0: H, " II \ H -/C-C C-S==O \ / I H C = C :0: I \ H HI H
H H \ I /C, ,....H C, :N II H H\ C I HI H / H - C , C - H H HH I I,.... I \ C H - C-C /H I 1 " 0O0- C/- C-H H H \ " C_HH I \ H H :0: ·0 0, , ·0 (h) II H H , /C, I /C,oo C C O-H I \\ /C-C-H I H H
(c) C4H9NO (h) C6H60 3
1 -32 (a) CsHsN (b) C4H9N (g) C7Hs0 3 S (f) C9H 1SO
1 -33 (a) 1 00% - 62.0% C - 1 0.4% H 27.6% oxygen 62.0 g C . 1 2.0 g/mole - 5. l7 moles C -;- l.73 moles - 2.99 3 C 1 0.4 g H l.01 g/mole 1 0.3 moles H -;- l.73 moles 5.95 6 H
(c) some possible structures-MANY other structures possible: H H H , "" C, ' H C-O-H H-CI I H-C " 'c" C-O-H , H '" H H H H H H H H ° 1��06g7�ole l.73 moles ° -;- l.73 moles 1 ° H-C-C-C-C-C-C-O-H (b) empirical formula C 3H60 c:::::> empirical weight 58 H H H H H H H H H H 0 molecular weight = 1 17, about double the empirical weight H-O-C-C-C-C-C - C-H q double the empirical formula molecular formula H H H H H C 6H120 2 H H H H 0 H H-C-C-C-C-C-O-C-H H H H H H =
_
_
=
=
=
=
=
I
I
_
==
=
=
I
16
I
I
I
I
I
I
I
I
I
I
I
1
I
I
I
I
I
I
I
1
I
I
I
I
II
I
I
I
I
=
II
II
I I
1-34 Non-zero formal charges are shown by the atoms. + + (a) H - C = N = N : H - C - N - N:
H ,H1 / H (b) H C 1 1 , +1 H H H -/ C - N - O: H H I H H C I 1 + + 1 (c) H - C = C-C - H (d) H - C-N=O (e) H - C-O -C-H H /1' H H 1 I +1 1 1 1 1 1 H C H H H H H :0: H /H1 ' H 1-35 The symbols "8+" and "8-" indicate bond polarity by showing partial charge. Electronegativity differences greater than or equal to 0.5 are considered large. • •
.... ..I----I.. �
-
• •
. . -
8+
8-
8-
8+
8+
8-
8+
8+
(c)
C-Li large
8-
8+
(h)
N-H large
(g) C - Mg large
(f) C-B large
8-
8-
(b) C - H small
(a) C - CI large
8+
8-
8-
8+
(d) C - N small
(i) O - H large
8+
8-
8+
8-
(e) C - O large (j )
C - Br small
1-36 Resonance forms must have atoms in identical positions. If any atom moves position, it is a different structure. (a) different compounds-a hydrogen atom has changed position (b) resonance forms-only the position of electrons is different (c) resonance forms-only the position of electrons is different (d) resonance forms-only the position of electrons is different (e) different compounds-a hydrogen atom has changed position (f) resonance forms-only the position of electrons is different (g) resonance forms-only the position of electrons is different (h) different compounds-a hydrogen atom has changed position (i) resonance forms-only the position of electrons is different (j ) resonance forms-only the position of electrons is different 1-37 (a) H :0: H :0: I
II
H - C-C - C - H 1 I H H
(b) :0: II H - C - C = C-C - H I I I H H H
..
..
i
I
. .
H - C - C=C - H 1 I H H
:0:
:0:
II
� .... .. I-. --J..
H - C-C - C = C - H 1 I I H H H 17
I
.... .. f-. --J � ..
H - C = C-C = C - H 1 1 I H H H
1-37 continued (c)
<>
C H2
0� /;
(d)
4
•
C H2
(f)
(g)
-
+
( )
�:
4
CH,
0/ + �
CH2
.. . 0---0 0+ <> •
�
�
. 0 ..
0.
----
-:
0=
/-
.0 .
< > �: -o� + ( > CN- - CN+-H + 0 .. 0 0 -H
+
4
Q :0 :
.. o ..
+
..
_
H
•
•
0
(h)
.
..
CH,
+
+
(e)
+ O
.�_
�
0+
----
Q Q +
:0 :
:0 : ..
-
(i) CH 3 - C = C - C = C - C - CH3 I I I I I H H H H H
+
+
CH3 - C = C - C - C = C-CH 3 / HI HI HI HI HI
-----
CH3 - C C = C - C = C -CH3 I I I I I H H H H H (j) no resonance fonns-the charge must be on an atom next to a double or triple bond, or next to a non bonded pair of electrons, in order for resonance to delocalize the charge -
18
1-38 .. (a) O==S-O: +
(b) 0=0-0: +
..
•
..
•
.
.
:O-S==O +
..
O==S==O
•
..
:0-0=0 +
(c) The last resonance form of S0 2 has no equivalent form in 03. Sulfur, a third row element, can have more than eight electrons around it because of d orbitals, whereas oxygen, a second row element, must adhere strictly to the octet rule.
1-39 (a)
#3
� ..
#2 NH � II / CH 3 - N - C - NH 2 I H I , H+to#3 �H+to#2 � #1
•
H NH I II CH3 - N - C - NH 2 I H no other resonance forms
•
• •
NH II CH3-N - C - NH3 I H no other resonance forms
+
NH2 II CH 3 - � - C - NH2 H
+
+
t
NH2 NH2 NH2 I I + I CH3 - N = C - NH2 CH 3 - N - C - NH 2 CH3 - N - C = NH2 I I I H H H (b) Protonation at nitrogen#3 gives four resonance forms that delocalize the positive charge over all three nitrogens and a carbon-a very stable condition. Nitrogen#3 will be protonated preferentially, which we interpret as being more basic. +
....... t---'l...
1 -40 (a) CH 3 - C - C N: I H minor :0:
...... t----I ....
I (b) CH3 - C==C - C - CH3 I I H H minor +
....... t---'l....
+
CH3 -C == C = N: I H major (negative charge on electronegative atom) . :0 : I CH 3 -C - C ==C - CH 3 I I H H mmor • •
.
.... .. t----I...
+
19
...... t----I...
:0 : II CH3 - C - C ==C - CH 3 I I H H major-full octets, no charge separation
1 -40 continued (c) :0 : :0: II -. . II CH3 - C - ? - C - CH3
.. : 0: ... .... . 1-----1 .. .
I
II
I
..
....1-----1 . .. .
H
major
II
I
I
CH3 - C - C == C - CH3 H
�-----------
minor
.. :0:
:0:
CH3 - C == C - C - CH3
\
H
:0:
}
------�--_/
major
Y
negative charge on electronegative atoms-equal energy + + • - + CH3 - C - C == C - N == 0 - CH3 - C == C - C• - N = 0 - CH3 - C == C - C == N - 0 :
(d)
I
I
H
I
H
I
••
I
H :0:
I
H
I
H
I
H :0 :
••
I
H
I
H
I
••
:0 :
major-negative charge on electronegative atoms
minor
minor
I
H
NOTE: The two structures below are resonance forms, varying from the first two structures in part (d) by the different positions of the double bonds in the N02. Usually, chemists omit drawing the second form of the N02 group although we all understand that its presence is implied. It is good idea to draw all the resonance forms until they become second nature. The importance o/ understanding resonance forms cannot be overemphasized.
+
••
CH3 - C - C == C - N - 0:
I
H
I
H
I
II
H :0:
II
••
NH2
II
CH3CH2 - C - NH2 .......I---J.�
+
\..major-full octets
y
I +
CH3CH2 - C == NH2
major-full octet:;
equal energy +
CH3 - C - CH3
I
+ I
H
no resonance stabilization +
CH2 == C - C - CH3 ......f---l . ., �
I
H
I
H
+
••
I •• H more stable-resonance stabilized
CH3 - C - 0 - CH3 ......I---l.,� CH3 - C == 0 - CH3
H
�)
I
H :0:
+
CH3CH2 - C - NH2 ......I---J.,..
1 -4 1 (a)
I
H
NH2
I
mmor
I
H
NH2
(e)
- + CH3 - C == C - •C• - N - 0:
..
....f-. -. �
+
H
CH2 - C == C - CH3
I
H
more stable-resonance stabilized
I
I
H
H
I + I
CH2 == C - C - CH2 H
no resonance stabilization
20
1-4 1 continued (c) H - C - CH3 \ H
H - C - C - N: \ H
...
H - C = C = N: \ H
..
more stable-reson ance stabi l i zed
no resonance stabi li zati on
+
CH c './ 2 \ H C ...:/
C
C 'H \ H no resonance stabilization H \ CH3 - C - CH3 I CH3 - C - CH 3
more stable-resonance stabilized (e) CH3 - N - CH 3 1 CH3 - C - CH 3
CH3 - � - CH 3 CH3 - C - CH3 +
.... ...I-. --I.. �
+
+
no resonance stabilization
more stable-resonance stabilized
1-42 These pK a values from the text, Table 1-5, and Appendix 5 provide the answers. The lower the pK a ' the stronger the acid. most acidic least acidic < < NH3 < 33 1-43 Conjugate bases of the weakest acids will be the strongest bases. The pKa values of the conjugate acids are listed here. (The relative order of the first two was determined from the pK a values of sulfuric acid and protonated acedic acid in Appendix 5 of the textbook.) least basic most basic
from �6. 1
from �5
from 1 5.5
from 4.74
from 1 5.7
from 33
1 -44 (a) pK a loglO K a = log lO (5.2 x 10 5 ) 4.3 for phenylacetic acid for propionic acid, pKa 4.87: K a 10 4 .87 1 .35 x 1 0 5 (b) phenylacetic acid is 3.9 times stronger than propionic acid 5 .2 x 1 0 5 3.9 1.35 x l0 - S (c) CH2 COO- + CH3CH 2 COOH ... CH 2 COOH + CH3CH2COO weaker acid stronger acid Equ i l i bri u m favors the weaker acid and base. In this reaction, reactants are favored. =
�
�
=
-
=
-
=
-
-
=
<>
-- < > 21
1-45 The newly formed bond is shown in bold. .. ('. . .. CH 3 - � - CH3 + -CI CH (a) CH 3 - .O: 3 . .: . � electrophile nucleophile Lewis acid Lewis base (b) CH3 - O - CH 3 + H - O - H +1 ) •• H 3 c �ucleoPhile Lewis base e I ectrophOI Ie Lewis acid (c)
c �: H - C-H
nucleophile Lewis base
1
H-C-H 1 H - N+- H 1 H H 1 CH 3 +-N - CH2CH3 1 H
electrophile Lewis acid
� ' ,, ' �
"
+
..:CI :
+
:0:
(e) CH 3-C - CH 3 + H-O-S-OH �. " . . nucleophile :0: Lewis base electrophile Lewis acid
(g )
H +� - H CH3
.. -
�
·0·
+
:0:
:N - H '----/ 1 H electrophile nucleophile Lewis acid Lewis base +
CH 3 - ? - CH3
..:CI :
+
--t..�
-
:O - H :0: " " CH3 -C-CH 3 + :O-S-OH .. " .. :0:
CI 1+ :Cl - Al - Cl 1 nucleophile electrophile Cl Lewis base Lewis acid This may also be written in two steps: association of the CI with AI, and a second step where the C-Cl bond breaks. .. : 0: I .. CH 3 - C - CH2 + : O - H .. CH 3 - C == CH 2 + H - O - H •• � J nucleophile '-- I � H Lewis base electrophile Lewis acid
:R� -
---t�
22
1 -45 continu�
(h)
F-B - F I F electrophile Lewis acid
CH2 = CH2 nucleophile Lewis base
+
�
(i) BF3 - CH2 - CH2 electrophile Lewis acid 1 -46 (a) H2S04
+
+
<)
o
HS0 4-
g- O- H
(d) (CH3hN - H
+
-OH
(e) HO - C - OH
+
2 -0H
oII
(f)
H20
+
(g) HCOOH
NH3 +
CH 30-
CH3COOH
+
CH 3 COO-
(CH3hN:
+
+
.. BF3 - CH2 - CH2 - CH2 - CH2
CH2 = CH2 nucleophile Lewis base
CH3COO-
+
(b) CH3 COOH (e)
•
F + -I F - f - CH2 - CH2 F
+
+
(CH 3 hN
+
< )
�OH
(CH3hN:
oII
+
+
H
o g- o-
+
H20
H20
-O-C - OHO-
-
+
2 H20
+NH4
HCOO-
CH30H
+
1 -47 CH3CH2 - 0- Li+ + CH4 (a) CH3 CH2 - O - H + CH3-Li (b) The conjugate acid of CH3Li is CH4 Table 1 -5 gives the pKa of CH4 as > 40, one of the weakest acids known. The conjugate base of one of the weakest acids known must be one of the strongest bases known. -----l.�
23
1 -48 From the amounts of CO2 and H20 generated, the milligrams of C and H in the original sample can be determined, thus giving by difference the amount of oxygen in the 5.00 mg sample. From these values, the empirical formula and empirical weight can be calculated. (a) how much carbon in 1 4.54 mg CO2 14.54 mg CO2 x
1 mmole C 1 mmole CO2 x 1 mmole CO2 44.0 1 mg CO2
X
12.0 1 mg C 1 mmole C
how much h)::drogen in 3.97 mg H2Q 1 .008 mg H 2 mmoles H 1 mmole H2O X x 3.97 mg H20 x 1 mmole H 1 8.016 mg H2O 1 mmole H2O
=
3.968 mg C
=
0.444 mg H
how much ox):: gen in 5.00 mg estradiol 5.00 mg estradiol 3 .968 mg C 0.444 mg H = 0.59 mg ° calculate empirical formula 3.968 mg C = 0.3304 mmoles C 0.037 mmoles = 8.93 9 C 1 2.01 mg/mole -
-
==
0.444 mg H 1.008 mg/mole
=
0.440 mmoles H
0.037 mmoles
0.59 mg ° 16.00 mg/mole
=
0.037 mmoles °
0.037 mmoles
empirical formula
=
�
=
=
1 1 .9 "" 1 2 H 1°
empirical weight
(b) molecular weight = 272, exactly twice the empirical weight twice the empirical formula = molecular formula =
24
=
1 36
CHAPTER 2-STRUCTURE AND PROPERTIES OF ORGANIC MOLECULES
2- 1 The fundamental principle of organic chemistry is that a molecule's chemical and physical properties depend on the molecule's structure: the structure-function or structure-reactivity correlation. It is essential that you understand the three-dimensional nature of organic molecules, and there is no better device to assist you than a molecular model set. You are strongly encouraged to use models regularly when reading the text and working the problems. (a) requires use of models H H (b) The wedge bonds represent bonds coming out of the plane of the paper ,f H" C """ H toward you. C /" C /" ./ The dashed bonds represent bonds going behind the plane of the paper.
J\ J\./ H H H H (b) The electrostatic potential map for 2-2 (a) The hybridization of oxygen is sp3 since it has water shows that the hydrogens have two sigma bonds and two pairs of nonbonding electrons. low electron potential (blue), and the 3 The reason that the bond angle of 1 04.5° is less than the area of the unshared electron pairs in sp perfect tetrahedral angle of 1 09.5° is that the lone pairs in orbitals has high electron potential (red). the two sp3 orbitals are repelling each other more strongly than the electron pairs in the sigma bonds, high electron thereby compressing the bond angle. potential (red) O repulsion low electron O ,�) potential O (blue) O "" H . H ) compression 2-3 Each double-bonded atom is sp2 hybridized with bond angles about 1220°; geometry around sp2 atoms is trigonal planar. In (a), all four carbons and the two hydrogens on the sp carbons are all in one plane. 3 Each carbon on the end is sp hybridized with tetrahedral geometry and bond angles about 109°. In (b), the two3 carbons, the nitrogen, and the two hydrogens on the sp2 carbon are all in one plane. The CH3 carbon is sp hybridized with tetrahedral geometry and bond angles about 109°.
(0'
(0
(b)
2-4 The hybridization of the nitrogen and the triple-bonded carbon are sp, giving linear geometry (C-C-N are linear) and a bond angle around the triple-bonded carbon of 1 80°. The CH3 carbon is sp 3 hybridized, tetrahedral, with bond angles about 109°.
25
2-5 (a) linear, bond angle 1 80°
• •
• •
o == C == o +2 + Sp Sp Sp+ 2 • •
• •
(b) all atoms are sp3 ; tetrahedral geometry and bond angles of 1 09° around each atom not a bqnd-;-shows not a bond-s how s � � lon� pair gomg '. H H lone Rair coming \ • , behmd paper "'--- / out 01 paper I I H H 0 , H-C-O-C-H ....... , ..... C C" I I A '" l , H H H H H H (c) all atoms are sp3 ; tetrahedral geometry and bond angles of 1 09° around each atom H H H . . H �' ..... H / C H -/" C - N - C "- H not a bond-shows I I ........-- lone H H pair going behind paper C H '/ N .. / 1' '-... H H H C /' , " " J '=:.," C-H H H J \H H (d) trigonal planar around the carbon, bond angles 1 20°; tetrahedral around the single-bonded oxygen, bond angle 109° :0: 3 II ,.....-- ·. 0 ·. sp all atoms in C r I H II one plane / "°/ H sp2 H - C - O - H • •
• •
�
• •
• •
..
• •
(e) carbon and nitrogen both sp, linear, bond angle 1 80° H - C N:
all three atoms in a line
(0
trigonal planar around the sp2 carbons, bond angles 1 20°; around the sp3 carbon, tetrahedral geometry and 1 09° angles sp2 H 1\ H H \C == C/ I H - C - C == C - H
I
H
I
H
I
H
/
H/C " \,' H sp3 H
\
H
(g) trigonal planar, bond angle about 1 20° (the other resonance form
.0. == .0.- .0.-·
• •
+
• •
-
of ozone shows that BOTH end oxygens must be sp 2_ see Solved Problem 2-8) 26
2-6 Carbon-2 is sp hybridized. If the p orbitals making the pi bond between C-l and C-2 are in the plane of the paper (putting the hydrogens in front of and behind the paper), then the other p orbital on C-2 must be perpendicular to the plane of the paper, making the pi bond between C-2 and C-3 perpendicular to the paper. This necessarily places the hydrogens on C-3 in the plane of the paper. (Models will surely help.) model of perpendicular n bonds
H H " '" 1 2 3/ 'C == C == C ' "' H t H sp 2-7 For clarity, electrons in sigma bonds are not shown. (a) carbon and oxygen are both sp2 hybridized
One pair of electrons on oxygen is always in an sp2 orbital. The other pair of electrons is shown in a p orbital in the first resonance form, and in a pi bond in the second resonance form.
empty orbital (b) oxygen and both carbons are sp2 hybridized H 1 200 H \\ C == .O. . 1..
H -· C 1 2�\H
.... ..I---l.. �
\C - O: .. .. II
H-C
\H
27
2-7 continued (c) the nitrogen and the carbon bonded to it are sp hybridized; the left carbon is sp2 1 800
H \ 1 200 c /
H
� .N:.
(
H
II
�/
�
H
c-c
N:
oil
(d) the boron and the oxygens bonded to it are sp2 hybridized H - O: H \ I B-O: I H-O . .: • •
P
..
II
H H - O: \ I B=O: I H - O. : +
Oe Q( (j"" " " '" 0 O�o 06 � O SP'
H-O
'' B
P
H-O
sp2
P
-
P
0
II
�
.
empty p
P H
sp'
-
H B - O· : .. I H-O . .:
oG� " " ",,: :n
(.:) p VH
H_
+
_
sp2
�
•
sP2
{J'Q"" '" " " ,\JD P
..
',
_
�QD O ! Vt
-
• •
28
H - O: H \ I B-O: II H - .O. +
�'
B - O ""'"
sp'
' ' �O "
\\
H-O
" � � �/ O "
+
Oe
/.:\ Sp 2
H-o
I
H-O
H
sp
• •
2-8 Very commonly in organic chemistry, we have to determine whether two structures are the same or different, and if they are different, what structural features are different. In order for two structures to be the same, all bonding connections have to be identical, and in the case of double bonds, the groups must be on the same side of the double bond in both structures. (A good exercise to do with your study group is to draw two structures and ask if they are the same; or draw one structure and ask how to draw a different compound.) (a) different compounds; H and CH3 on one carbon of the double bond, and CH3 and CH2 CH3 on the other carbon-same in both structures; drawing a plane through the p orbitals shows the H and CH3 are on the same side of the double bond in the first structure, and the H and the CH2CH3 are on the same side in the second structure, so they are DIFFERENT compounds
H 3 e' \
" CH3 /
- - - - - - - e ::E-C- - - - - - - - - -
I
H"
.. ... .. ... ..
\
" CH2CH3
these are DIFFERENT
(b) same compound; in the structure on the right, the right carbon has been rotated, but the bonding is identical between the two structures (c) different compounds; H and B r on one carbon, F and Ci on the other carbon in both structures; H and CI on the same side of the plane through the C=C in the first structure, and H and F on the same side of the plane through the C=C in the second structure, so they are DIFFERENT compounds (d) same compound: in the structure on the right, the right carbon has been rotated 1 200 2-9 (b) CH3 - N : ".. CH3 :N H H H " (a) NOT INTER " " , C CONVERTI LE C B H-C-C=N-C-H CH3 ",.. ....... H CH3 ",.. ....... H / , � I , '-. 3 sp two CH3's on opposite two CH3's on the same H \ 2 H sides of the C=N side of the C=N sp compare
compare
(c) the CH3 on the N is on the same side as another CH3 no matter how it is drawn-only one possible structure and
2- 10 (a)
".. CH3 :N " C ",.. " CH3
CH3 H" F / C = C "/ H F
. . groups on one (b) no cis-trans iSOmeriSm . . two IdentIcal (c) no cis.. -tram, Isomensm . . } carb on 0 f th e d ou bl e b on d (d) no cis-trans Isomensm trans
(e)
QC = C CH3 H
... ,
...
CIS
H
and
H QC � c, ... CH3 H "-
trans
29
"cis " and "trans " not defined for this example
2-1 1 Models will be helpful here. (a) cis-trans isomers-the first is trans, the second is cis (b) constitutional isomers-the carbon skeleton is different (c) constitutional isomers-the bromines are on different carbons in the first structure, on the same carbon in the second structure (d) same compound-just flipped over (e) same compound-just rotated (f) same compound-just rotated (g) not isomers-different molecular formulas (h) constitutional isomers-the double bond has changed position (i) same compound-just reversed (j) constitutional isomers-the CH3 groups are in different relative positions (k) constitutional isomers-the double bond is in a different position relative to the CH3 2- 1 2 (a) 2.4 D
4.8 x 8 x 1 .2 1 A 0.4 1 , or 4 1 % of a positive charge on carbon and 4 1 % of a negative charge on oxygen :0: (b) I .. ..... C ....... R + R B Resonance form A must be the major contributor. If B were the major contributor, the value of the charge separation would be between 0.5 and 1.0. Even though B is "minor", it is quite significant, explaining in part the high polarity of the C=O. 8
=
=
..
.....
2- 1 3 B oth NH3 and NF3 have a pair of nonbonding electrons on the nitrogen. In NH3, the direction of polarization of the N-H bonds is toward the nitrogen; thus, all three bond polarities and the lone pair polarity reinforce each other. In NF3 , on the other hand, the direction of polarization of the N-F bonds is away from the nitrogen; the three bond polarities cancel the lone pair polarity, so the net result is a very small molecular dipole moment.
polarities reinforce; large dipole moment
polarities oppose; small dipole moment
1 O·
� N� \\ ' H H 2- 1 4 Some magnitudes of dipole moments are difficult to predict; however, the direction of the dipole should be straightforward, in most cases. Actual values of molecular dipole moments are given in parentheses. (Each halogen atom has three non bonded electron pairs, not shown below.) The C-H is usually considered non-polar. (a) LCi H (b) H �, +- H it" / large dipole ( 1 . 8 1 ) C-F H � C � large dipole ( 1 .54) / XX: CI H .. .. I I net net H
....
30
2-14 continued (c)
F
(d)
�" cL / net dipole 0 �� �F =
each end oxygen has one-half negative charge as it is the composite of two resonance forms; see net solution to 1 -38(b) small dipole (0.52)
Qo 1
(e)
T
0�, . � 0 'x�o " " o�
(g)
(i)
ro or 0; 13 1 � C1 / ' CH3 net
H
1 \)
1 \," ' CH 3
/ N ,�
H3C (k)
,/
CH3
net
1
net (f)
(I)
\\ � Il-F F
-
..
net large dipole (2.95) large dipole
net
/ net
large dipole (l .45)
----+
CI
N CD
I
\
small dipole (0.67)
F
� �
H-C
(h)
large dipole (2.72)
large dipole ( 1 .70)
e C
� B I -
net dipole
=
0
(m)
tt�
� N" "
H "'" \\ " H H
net dipole = 0 net dipole 0 In (k) through (m), the symmetry of the molecule allows the individual bond dipoles to cancel. =
2- 1 5 With chlorines on the same side of the double bond, the bond dipole moments reinforce each other, resulting in a large net dipole. With chlorines on opposite sides of the double bond, the bond dipole moments exactly cancel each other, resulting in a zero net dipole. large net dipole
C� �CI H
/
C == C
'"
H
net dipole
1 31
=
0
(hydrogen bonds shown as wavy bond)
2- 1 7 (a) (CH3hCHCH2CH2CH(CH3h has less branching and boils at a higher temperature than (CH3hCC(CH3h . (b) CH3(CH2)sCH2 0 H can form hydrogen bonds and will boil at a much higher temperature than CH3(CH2) 6CH3 which cannot form hydrogen bonds. (c) HOCH2(CH2)4CH2 0 H can form hydrogen bonds at both ends and has no branching; it will boil at a much higher temperature than (CH3hCCH(OH)CH3 . Cd) (CH3CH2CH2hNH has an N-H bond and can form hydrogen bonds; it will boil at a higher temperature than (CH3CH2hN which cannot form hydrogen bonds. (e) The second compound shown (B) has the higher boiling point for two reasons: B has a higher molecular weight than A ; and B , a primary amine with two N-H bonds, has more opportunity for forming hydrogen bonds than A, a secondary amine with only one N-H bond. 2- 1 8 (a) CH3CH2 0 CH2CH3 can form hydrogen bonds with water and i s more soluble than CH3CH2CH2CH2CH3 which cannot form hydrogen bonds with water.
(b) CH3CH2NHCH3 is more water soluble because it can form hydrogen bonds; CH3CH2CH2CH3 cannot form hydrogen bonds. (c) CH3CH20 H is more soluble in water. The polar O-H group forms hydrogen bonds with water, overcoming the resistance of the non-polar CH3CH2 group toward entering the water. In CH3CH2CH2CH2 0 H, however, the hydrogen bonding from only one OH group cannot carry a four-carbon chain into the water; this substance is only slightly soluble in water. (d) B oth compounds form hydrogen bonds with water at the double-bonded oxygen, but only the smaller molecule (CH3COCH3) dissolves. The cyclic compound has too many non-polar CH2 groups to dissolve.
32
2-19
(a)
H H H H H I I I I I H-C-C-C-C-C-H I I I I I H H H H H
(b)
alkane (Usually, we use the tenn "alkane" only when no other groups are present.)
(d)
H I H-C-C
I
H-C H
(g)
'
"
H =
I
C-C-H /
I
(e)
C-H
H C-C I I 'H H H
H
H H H I I I I H, C C-C-C-H H - C � "C � I I I I I H C H H - C .. . C I H' H C� I H H I H H
(h)
•
(a)
0
H H I I " H-C-C-C-H I
(d)
I
aldehyde H
H H H H I I I I H-C-C-O-C-C-H I I I I H H H H
(b) (e)
ketone
I
I
H
H H " C .. H I , C - C :: C - C - H H-C� I I I H-C C .. H H " C � ... H H ' " H C=C-H I I H H H
,
0
H
H H H I I I I H -C-C-C-C-H I I I I H H H H
alcohol
0
H" H H C .. , " -O-H C-C H-C� I H -, C .. C .- H � H ,c. H H H H
,
I
carboxylic acid
(h)
(g)
H H H I I I H-C-C-C-C-H I I H H H-C-C' I I H H H I
(f)
H
H "
H'
c;
H "C ,
/
H
H H H I I I H - C - C :: C - C - C - C - H I I I I H H H H I
alkyne
H H H I I I C .. H, C=C-H C� " C� I " C C � H , "C � ' H I H
aromatic hydrocarbon and alkene
(i)
alkyne, alkene, cycloalkane
H
ether
H
•
cycloalkene
2-20
alkene
(c)
cyc10alkane
cycloalkyne
H'
H H H I I I H-C-C=C-C-C-H I I I I I H H H H H
I;I
H
aldehyde 33
H
aromatic hydrocarbon and cycloalkene (c) (f)
0
H H H I I I " H-C-C-C-C-H I I I H H H
ketone
H H -' C � I
0
H
" C '- H I
H-C C-H H , ..., c .� . H H H
ether
o " -H C
�.
H .C - C . I H H I H H
'c '
,H ' C o:. " C � " C - H I I II C C C H ' " c � " C �_ , H I I H H C
H
alcohol
2-2 1 (a)
H
H
0
1
H-C-C-C-N-C-H 1
1
1
1
H
H
H
11
1
1
H
H
(b)
H
H 1
1
1
H 1
H-C-C-N-C-C-H 1
1
H
H
1
H
1
H
H
1
�)
H
H
(e)
H
I
H I
I
H-C-C-O-C-C-H
H'
I
I
1
I
H
H
H
(f)
H 1
11
1
1
H
H
H
I
0
1
H-C-C-C-O-C-H
amine
amide
H
1
H
C
H
'
1
H
H
ester
H 1
1
1
H
H - C - C - C - C :: N 1
1
1
H
H
H
H
nitri Ie
ether
I;I H
H
(g)
H 'C ' H I
I
H
I
0
I
U)
(i)
II
,H H C. 1 N-C- H H-C ' I 1 H H -, C .... ... C .- H C H
H-C-C-C-C-C-O-H I
H
H
'
I
C
H
'
I
I
H
carbox ylic acid
(k)
o
H
11
H
" ... C .. N - C '- H H-C I
I
"
H
H
H
H,
H
1
H -, C .... ... C - H H "C H H H
cyclic ester
(I)
H
H
H,
H -' c '
C
ketone and ether ,H
11
1
I
H
0
H 1
.N-C-C-H
H -, C ....
,
H
... C .- H C H H H
1
H
"
cyclic amide
(m)
"
H
amine a
H
H -'C '
C
I
H-C ' H
H
' C '- H
.�'
1
C-H ' H
H-C-H 1
amide
H
ketone and amine
(0 )
(n) I
H"
H
H-C I
1
H -, C .... ... C - H H "C H H H
,
C
.H
H " C - C :: N 1
H -, C ' ' C - H "C H H H H
,
cyclic ester
,
nitrile
ketone
2-22 When the identity of a func ti onal group depends on several atoms, all of those atoms should be circled. For example, an ether is an oxygen between two c arbons, so the oxygen and both c arbons should be circled. A ketone is a carbony l group between two other carbons , so all those atoms should be circled. (a)
€2 §9 alkene
H CH3 2
(b)
€ Y 0
ether
34
3
(c)
CH3
(g Il C-H
aldehyde
2-22
continued
g
(e)
6 c:3)
(f)
H
II H C - NH
(d)
CH
�
amine
q u i te a few ways that carbon and hydrogen ato m s c an go together to form alkyl and a ry l groups . So when you see this s y m bo l , you s h ould know that it represents ONl.. Y some combination of carbon and h ydrogen atoms-except w h en it includes other ato m s .
(g) (i)
(h)
alkene
CH3
Please refer to solution
(
8 ketone
aromatic
(k)
2-23 2-24
carboxylic acid
represent alkyl and ary l groups . As you w i l l s ee in t h e c o urse of your s tudy , there are
(this also looks like an aldeh yde, b u t an amide h as higher "priority " as y o u wi l l s e e l ater)
(j)
~
S uggested by student R i c h ard King: R i s the symbol that organic chemists use to
amide
aromatic
II
R C - O- H
CH3
�
R
1-20, page 12 of this Solutions Manual .
The examples here are representative. Your examples may be di fferent and stil l c orrect. What is important in thi s problem i s to h ave the same functional group. (a) alkane: hydrocarbon wi th all single bonds ; can be acyclic (no ring) or cyclic
(b) alkene : contains a c arbon-carbon double bond
(d) alcohol : contai ns an
(e) ether: contains an ox ygen between two carbons
H H H I I I H - C-C - C - H I I I H H H
OH group on a carbon
H H I I H - C - C - O- H I I H H
H I H - C - C == C - H I I I H H H H H I I H - C - O-C - H I I H H 35
(c) alkyne: contains a c arbon-carbon triple bond
H I H - C-C -C - H I H (f) ketone : conatins a carbonyl group between two carbons
H 0 H I II I H-C-C-C-H I I H H
2-24 continued II
I
I
H
H
I
I
(j ) ester: contains a carbonyl group with an O-C on one side a
I
1/
I
I
H
I
I
I
I
H (I ) amide : contains a carbonyl group with a nitrogen on one side
I
a
I
II
H
I
I
H
H-C-C-N-H
H
H
(m) nitrile: contains the carbon-ni trogen triple bond:
2-25 2-26
0
H-C-C-O-H
H - C - N - H or R group
H
II
H
(k) ami n e : contains a nitrogen bonded to one , two, or three carbons H H or R group
H-C-C-O-C-H H
I
o
H-C-C-C-H
H
(i ) carboxylic acid: contains a c arbonyl group with an OH group on one side
(h) aromatic hydrocarbon : a cyclic hydrocarbon with alternating double and single bonds
(g) aldehyde: contains a carbonyl group with a hydrogen on one side H H a
H3 C - C
N
Models show that the tetrahedral geometry of CH2 CI 2 precl udes stereoisomers . H
(a)
/
"
H
C I H - _\ - H C C /
\
(b) Cyclopropane must have 60° bond angles compared with the usual sp 3 bond angle of 109 . 5° in an acyclic molecule.
H H (c) Li ke a bent spring, bonds that deviate from their normal angles or positions are highly strained. Cycl opropane i s reactive because breaking the ring relieves the strain .
2-27
f)
(a)
(b)
H
0_
(c)
?:}'0
+
./ 0 " " " ..... \ H H
H
H /' H H '::, C
H
""
C
/
J '\.'
H
""
", N
I
--
�/
H
N
H
Sp 3 , no bond angle because oxygen is bonded to only one atom (d)
(J
both sp 3 , all
�
� ==
1 09°
beh i nd the p l ane of the paper
�� ) \� J '\.H
C, - H
H
H
3 a l l sp , al l
==
1 09°
angles around sp 3 atoms angles around sp2 carbon
==
36
==
109° 120°
2-27 continued (h)
beh i nd the p l ane
plane of the
of the pape
p aper
� (j {) H
angles around sp 3 atom angles around sp 2 atoms
�
�
;
i n front o f the
/
a
"'"
I
(i)
...-/
�H H
, ,
C
H
1 09° 1 20°
2-28
For cl ari ty in these picture s , b o n d s between hydrogen and a n sp 3 atom are n o t l abeled; these bonds are s-sp 3 overlap.
(b)
I ;;\'..
A\(A/· ·
H-C-C-O-H
1 09° (e)
(f)
2
( ) g
H
109°
37
7/I-(�Yfp: � C- 1'H
t t
sp 3 _ sp 2
H
sp 2 _ sp 3
2
H
1 09°
sp 3 _sp 2
sp 2 -s
� OJ � �< f"' ; o,, � �
(i)
I
sp 3 _sp 3
H
___
C
1 09°
I
II
120°
2 2 --- sp -sp
1 20 ° \ H / C � C. / C � H / ,,\ � 2 � ' H\ Sp -S H
+
p-p
sp 2 _ sp 3 2-29 The second resonance form of formamide is a minor but significant resonance contri butor. It shows that the nitrogen-carbon bond h as some double bond character, requiring that the nitrogen be sp 2 h ybridized with bond angles approaching 1 20°.
:0: II
°
° °
+
/
sp 2
H-C=N-H I H
H-C-N-H I H
2-30
01
° ° °
(a) The major resonance contributor shows a carbon-carbon double bond, suggesting that both carbons are sp 2 hybridized with trigonal p l an ar geometry . The CH3 carbon i s sp 3 h ybridized with tetrahedral geometry .
� :R:
H-C-C-C-H I I H H
:
�
sp 3 / 1
... ..I-----;. �
?�
Sp 2
H-C-C=C-H I H H m�or
�nor
(b) The major resonance contri butor shows a carbon-nitrogen double bond, suggesting that all three carbons and the nitrogen are sp 2 h ybridized wi th tri gonal pl anar geometry . +
H-N-C=C-C-H I I I I H H H H
..
• •
+
.. H - N - C - C = C - H .. I I I I H H H H minor
minor
+
•
H-N=C-C=C-H I I I I H H H H major
2-3 1 In (c) and (d), the un shadowed p orbitals are vertical and parallel. The s hadowed p orbitals are perpendicula; and hori zontal . (aJ
Q\2
H3C ,,, H3C
M
"-
(bJ
. KN V n h� � (d)
_QOYoQ?o O[Q[J riJJ
C H3
2-32 (b) The cop l an ar atoms i n the structures to the left and below are marked w i th asterisks.
(a) cis
(c)
H *
HJC
"
QD M
(d) ,, ,
�C - C , "
,,
eH,CH] *
trans
H
There are sti l l six copl anar atoms.
2-33
Collinear atoms are marked with asteri sks.
2-34 (a) no cis-trans i someri sm
(b)
(c) no cis-trans i someri sm
0
If/
(d) Theoretical l y , cyclopentene could show c i s-trans i somerism. In reality, the trans form is too unstable to exist because of the necessity of stretched bonds and deformed bond angles. trans-CycJopentene has never been detected · . c ls "trans "--not possible bec ause of ring strain /J
V
these are cis- trans isomers , but the designation of cis and trans to spec i fic structures is not defined because of four different groups on the double bond
(e)
39
2-35 (a) consti tutional i somers-the c arbon skeleton s are different (b) constitutional i somers-the position of the c h l orine atom has changed (c) c i s-trans i somers-the first is cis, the second i s trans (d) constitutional i somers-the c arbon skeleton s are different (e) c i s-trans i somers-the first is trans, the second is cis (f) same compound-rotation of the first structure gi ves the second (g) c i s-trans isomers-the fi rst is cis, the second is trans (h) consti tutional isomers-the position of the double bond relative to the ketone h as changed (while it i s true that the first double bond i s cis and the second is trans, in order to have cis-trans i somers, the rest of the structure must be i dentical)
CO2 i s linear; its bond dipoles c ancel, so i t has no net dipole . S02 i s bent, s o i ts bond dipoles do not
2-36 c ance l .
t: o��"o:
..
O == C == O net di pole moment
=
..
..
0
..
net dipole moment
2-37 Some magni tudes of dipole moments are di fficult to predict; however, the direction of the dipole should be straightforw ard in most cases. Actual values of molecular dipole moments are given in parentheses. (The bond is usually consi dered non-pol ar. )
C-H
(a)
� 'x/ ...... CH3 '--J
N U1 H / 'CH3 \
/
or
net
net
l arge dipolemoment
l arge dipole moment
(b)
� �
CH3 - C I
N CJ �
net l arge dipole moment ( 3 . 96)
/Br ��
B r IllIII / C B
�
(c )
(d)
Br
net dipole moment
=
I
I
net
CH2 CH2 " / 2
CH
moderate dipole moment net dipole moment
=
1
net
l arge dipole moment (2 .89)
(0 4N" �b ", � / 2
R1 C
CH3/ 'CH3
0
e lectron pairs on bromines are not shown
2
0; /0 (g)
CH2
/CI
CH; 'C / I 2 /C1 1 / net CH "" 'CH2 H
moderate dipole moment electron pairs on c h l or i n e are not shown
0 40
2-38 Diethyl ether and I -butanol each have one oxygen , so each c an form hydrogen bonds with w ater (water supplies the H for hydrogen bonding with dieth yl ether) ; their water solubi lities should be similar. The boiling point of I -butanol is much higher because these molecules can hydrogen bond with each other, thus requiring more energy to separate one molecule from another. Diethy l ether molecules cannot hydrogen bond with each other, so it is rel ati vely easy to separate them . CH3CH 2
-
0
-
CH 2CH3
CH3CH2CH2CH2 - O H I -butanol can hydrogen bond with water can hydrogen bond with i tself
diethy l ether can hydrogen bond with water cannot h ydrogen bond with i tself 2-39
C
N - CH]
N-meth y I pyrro Ii di ne b.p. 8 1 °C
tetrahydropyran b.p. 8 8 DC
piperi di ne b . p . 1 06DC
o-
OH
cyc!opentanol b.p. 14 1 °C
(a) Piperi dine has an N-H bond, so it can hydrogen bond with other molecules of i tself. N-Methylpyrrolidine h as no N-H, so it cannot hydrogen bond and will require less energy (lower boi ling point) to separate one molecule from another. (b) Two effects need to be explained: 1 ) Why does cyclopen tanol h ave a higher boi ling point than tetrahydropyran ? and 2) Why do the oxygen compounds have a greater difference i n boi ling points than the analogous nitrogen compounds ? The answer to the first question i s the same as in (a) : cyclopentanol c an h y drogen bond with its neighbors while tetrahydropyran c annot . The answer to the second questi on l i es in the text, Table 2-1, that shows the bond dipole moments for C-O and H-O are much greater than C-N and H-N; bonds to oxygen are more pol arized, with greater charge separati on than bonds to ni trogen . How i s thi s reflected in the data? The boi ling points of tetrahydropyran ( 8 8 DC ) and N-methy l pyrrolidine (8 1 °C) are close ; tetrahydropyran molecules would h ave a slightly stronger dipole dipole attraction , and tetrahydropyran is a little less " branched" than N-methy l pyrrolidine, so it is reasonable that tetrahydropyran boi l s at a s l i ghtly higher temperature. The l arge difference comes when comparing the boi ling points of cyclopentanol (141 DC) and piperidine ( 1 06°C). The greater polari ty of O-H versus N-H is refl ected in a more negati ve oxygen (more electronegati ve than ni trogen) and a more posi ti ve hydrogen , resulti n g i n a much stronger intermolecular attraction . The conclusion i s that hydrogen bonding due to O-H i s much stronger than that due to N-H. 2 40 (a) can hydrogen bond with i tself and with water (b) can hydrogen bond onl y with w ater (c) can hydrogen bond wi th i tself and with water (d) can hydrogen bond only with w ater (e) cannot hydrogen bond (f) cannot hydrogen bond -
2-4 1
(g) (h) (i) (j) (k) (I)
can can can can can can
hydrogen bond only with water hydrogen bond w i th i tself and wi th water hydrogen bond only with water hydrogen bond onl y with water hydrogen bond only with water hydrogen bond with itself and with water
Higher-boi ling compounds are li sted.
( a) CH3CH(OH)CH3 can form h ydrogen bonds with other i dentical molecules
(b) CH3CH 2 CH2 C H2 CH3 has a h i gher molecular weight than CH3CH2C H 2CH3 (c) CH3CH2CH 2 CH2 CH3 has less branching than (CH3hCHCH 2 CH3 (d) CH3CH 2 CH 2 CH 2 CH 2 Cl h as a hi gher molecular weight AND dipole-dipole interaction compared with CH3CH 2 CH 2 C H2CH3 41
2-42 (a)
(b)
ether ether
alkene aldehyde
(c)
(d) ketone
(e)
(f)
ester (cyclic)
aromatic
amide (cyclic)
alkene
(h)
(g)
amme
2-43
.
:0:
:0:
II
CH3
/C'
t
CH3
CH3
" S .......
/ •• /
ester
.
:0 : CH3
2 sp -planar
f----J .. .. .... ..
CH 3
I S .......
/ •• /
+
CH 3
tsp3 -tetrahedral
The key to thi s problem is understanding that sulfur has a lone pair of electrons . The second resonance form shows four pairs of electron s around the sulfur atom, an electroni c configuration requiring sp 3 2 hybri dizati on . S u l fur in DMSO c annot be sp l i ke carbon in acetone, s o we would expect sulfur's geometry to be pyramidal (the four electron pairs around sulfur require tetrahedral geometry , but the three atoms around sulfur define i ts shape as pyrami dal).
42
2-44
(a) penicillin G
CQ) ([) _ II '1
thioether ("thio" means sulfur repl aces oxygen)
amide
\\ \
II
C -N
CH 2
__
aromatic
(b) dopamine aromatic
�
)
©
CH 2
I
H2
(In l ater c h apters , you wi l l learn that the OH group on a benzene ring is a speci al functional group called a "phenol ". For now, it fi ts the broad definition of an alcohol . )
amIne
CH 2
)
(c) thyroxine
aromatic
�
CH 2
@e . amme
c arbox ylic acid
aromatic
(d) testosterone
43
CHAPTER 3-STRUCTURE AND STEREOCHEMISTRY OF ALKANES
3- 1 (a) C nH2n+2 where n
=
25 gives C 2sHs2
Note to the student: The IUP AC system of nomencl ature has a wel l defined set of rules determining how structures are named. You w i l l fi nd a summary of these rules as Appendix 1 i n thi s Solutions Manual . 3-2 Use hyphens to separate letters from numbers. Longest chains may not a l w ays be written left to right. (a) 3 -methylpentane (always fi nd the longest chai n ; it may not be written in a straight line) (b) 2-bromo-3-methylpentane (al w ays find the longest chai n) (c) 5-ethyl-2-methyl-4-propylheptane ("When there are two longest chains of equal length, use the chain w i th the greater number of substituents . " ) (d) 4-isopropyl-2-methyldecane 3-3 Thi s Solutions Manual w i l l present l ine formulas where a question asks for an answer including a structure. If you use condensed structural formulas instead, be sure that you are able to " translate" one structure type into the other. (a)
(b)
(c)
(d)
3 -4 Separate numbers from numbers with commas . (a) 2-methylbutane (b) 2,2-di methylpropane
(c) 3 -ethyl-2-methylhexane (d) 2,4-di methylhexane
(e) 3 -ethyl-2,2,4,5-tetramethylhexane (f) 4-t-butyl-3-methylheptane
3-5 (Hints: systematize your approach to these problems. For the i somers of a six carbon formula, for example, start with the i somer containing all six carbons in a strai ght chain , then the i somers containing a fi ve-carbon chain, then a four-carbon chain, etc. Carefully check your answers to AVOID DUPLICATE STRUCTURES . ) (a) n-hexane
2,2-di methylbutane
2-methylpentane
3-methylpentane
2,3 -di methylbutane 45
3-5 continued (b)
� n-heptane
2-methylhexane
T�
3 , 3 -di methylpentane
2,3-di methylpentane
2 ,2-dimethylpentane
3-methylhexane
��
2,4-di methylpentane
3 -ethylpentane
2,2,3 -trimethylbutane
3-6 For thi s problem, the carbon numbers in the substituents are indicated in italics. (a)
CH3
(b)
CH3 1
-CH2CHCH3
-CHCH3 1
(d)
1
2
I-methylethyl common name
(c)
I
=
2
1
-CHCH2CH3
3
2-methylpropyl common name isobutyl
isopropyl
CH3 2 0/
=
/1C
CH3 30 l,l- di methylethyl common name = r-butyl or rerr-butyl 3-7 (b)
3 -8 Once the number of c arbons is determined, C nH2n+2 gi ves the formula. (a) C IOHn (b) C 1sH32 3-9 (a) (lowest b.p.) hexane
<
(b) (CH3hC-C(CH3h
<
octane
<
decane (highest b.p.) -molecular weight
CH3CH2C(CH3hCH2CH2CH3
(b) octane
<
(lowest m . p . )
octane -branching
(highest b.p.)
(lowest b.p.) 3-10 (a) (lowest m.p . ) hexane
<
<
octane
<
decane (highest m.p.) -molecu l ar weight
CH3CH2C(CH3hC H2CH2CH3
<
(CH3hC-C(CH3h -branching (hi ghest m . p . )
46
2
3
I-methylpropyl common name = sec-butyl
CH3 .-! � CH3
(a)
1
3- 1 1
------------------------., --------------------------1 5 .0 kJ/mole : (3.6 kcal/mole)
%��l
H H
4.2 kJ/mole + 2 H-H
x
x
%��l
H
5 .4 kJ/mole H-CH3
dihedral angle 8
1 20 °
CH3
H staggered
ec lipsed
1
� 60°
0°
eclipsed
1 5 .0 kJ/mole ( 3 . 6 kcaI/mole)
=
3-1 2
A l l energy values are per mole .
,, 1 2.
0°
42H H
e
&: H
5
54 H H3 C
ec lipsed 5
H
�2
H
'fJ H CH3 staggered
H3
H3C
�
H
e
300°
dihedral 360° angle 8
� ,
H3 C 1 2.5 CH3
1 2.5 1
& H
HC .H4 3
,,
240°
1 80°
J!I4. 2
4. 2
ifH.4 3
ecli psed 5
ecli psed
� CH3 H 'fJ H 3.8
CH3
3.8
H3C
H
'fJ H
H staggered
CH3 staggered
(Note that the lowest energy conformers at 60° and 1 80° have at l east one CH3-CH3 interaction = 3.8 kJ (0.9 kcal) higher than ethane.)
Relative energies on the graph above were calculated using these values from the text: 3 .8 kJ/mole (0.9 kcaI/mole) for a CHrCH3 gauche (staggered) interaction ; 4 . 2 kJ/mole (l.0 kcaI/mole) for a H-H ec l ipsed i nteraction ; 5 .4 kJ/mole ( 1 . 3 kcaI/mole) for a H -CH} eclipsed interaction; 1 2.5 kJ/mole (3 . 0 kcaI/mole) for a CHrCH} ecli psed interaction . These values in kJ/mole are noted on each structure and are summed to give the energy value on the graph . S l i gh t variation between thi s graph and t h e one i n the text are due t o rounding. 47
all bonds are staggered - bold bonds are coming toward the reader from the plane of the paper
H
dashed bonds are coming toward the reader from the plane
3- 1 4 (a) 3-sec-butyl-I,I-dimethylcyc l opentane (the 5-membered ring gi ves the base name) (b) 3-cyclopropyl-I,I-dimethylcyclohexane (the 6-membered ring gi ves the base name) (c) 4-cyc l obutylnonane (the chain is l onger than the ring)
3- 1 6
o
(a) no cis-trans isomeri sm possible
(b) and
�H 3 �H CH3 trans
cis
----.Y �
(d) and
CHJ 2
of
the paper
----.Y �
H
and
2
H CH 3 cis trans 3- 1 7 In (a) and (b), numbering of the ring is determined by the first group alphabetically being assigned to ring c arbon 1 . CIS
trans
(a) cis-I-methyl-3-propylcyclobutane ("m" comes before "p"-practice that alph abet!) (b) tral1s-I-t-butyl-3-ethylcyclohexane (the prefi xes t, s, and 11 are ignored in assigning alphabetic al priority) (c) trans- l ,2-dimethylcyclopropane (either carbon with a CH3 could be c arbon-I; the same name results) 3- 1 8 Combustion of the cis i somer gives off more energy, so cis- 1 ,2-di methylcyclopropane must start at a higher energy than the trans i somer. The Newman projecti on of the c i s i somer shows the two methyls are eclipsed with each other; in the trans i somer, the methyls are sti ll eclipsed, but with hydrogens, not each other-a lower energy. "'\ more strain == CHj ) hi gher energy CH3
�
cis
HH
48
�
3-19 trans-1 ,2-Dimethylcyc l obutane i s more stable than cis because the two methyls can be farther apart
when trans , as shown in the New m an projections . ""\ more strain = CH ; ) hi gher energy CH3 cis
HH In the 1 ,3 -dimethylcyc l obutanes, however, the cis allows the methyls to be farther from other atoms and therefore more stable than the trans. H
A
CHl
CH3 � H
trans
3-20
axi al only
equatori al only
showing both axial and equatori a l
3 - 2 1 The abbrevi ation for a methyl group, CH3 , is "Me ". Ethyl i s "Et", propyl is "Pr", and butyl is "B u". �
(a)
Me
�
(b) H
H
H Me
H
Me Me al l methyls axial (all H's equatori al )
H H all methyls equatorial (al l H's axial)
Note that axial groups alternate up and down around the rin g . 3 -22 Carbons 4 a n d 6 are b e h i n d the circ les.
49
3-23 The isopropyl group can rotate so that i ts hydrogen i s near the axial hydrogen s on c arbons 3 and 5,
s i milar to a methyl group's hydrogen, and therefore simi l ar to a methyl group i n energy. The t-butyl group, however, must point a methyl group toward the hydrogens on carbon s 3 and 5 , giving severe diaxial interactions, causing the energy of thi s conformer to j ump dramatic al l y .
}JJ
5=H 3 C Hl
i sopropylcyclohexane
t-buty Icyclohexane
3-24 The most stable conformers have substituents equatori al.
�
CH3
H
(b)
CH]
H
3-25
(a) cis -
equatori al , axial
(b) trans
H
H
EQUAL ENERGY
p:J
H axi al , equatori al
H
�
�
H3C
�
CH 3
H equatorial , equatori al lower energy
axi al , axial h igher energy
(c) The trans i somer is more stable because BOTH substituents can be in the preferred equatori al positions. 3-26 Positions
cis
trans
1,2
(e,a) or (a,e)
(e,e) or (a,a)
1,3
(e,e) or (a,a)
(e,a) or (a,e)
1 ,4
(e,a) or (a,e)
(e,e) or (a,a)
50
3-27 The more stable confonner p l aces the l arger group equatori al . (a)
CH2CH3
c;tH7
4�H]
CH3
�CH2CH]
-----
H
(b)
CH3 CH2CH3 H
(c)
(CH 3hCH
H
more stable
H
� CH3 CH2CH'
-----
more stable H
H]
H 3C , C -CH
...
--
H
H
� CH2CH] H
more stable
H
(d)
more stable H 3-28 The key to detennining c i s and trans around a cyclohexane ring i s to see w hether a substituent group is "up " or "down" relative to the H at the same carbon. Two "up" groups or two "down" groups will be c i s ; one "up " and one "down" w i l l be trans. Thi s works independent o f t h e confonnati on t h e molecule i s in ! (a) cis- l ,3-dimethylcyclohexane (d) cis- 1 ,3-di methylcyc l ohexane (b) cis- l ,4-dimethylcyclohexane (e) cis- l ,3-dimethylcyclohexane (c ) trans-I,2-dimethylcyclohexane (f) trans-I,4-dimethy l c yclohexane
H
(b)
;------f- CCCH3h � CH3
H
( c ) B ulky substi tuents like t-butyl adopt equatorial rather than axial posi tions, even i f that means altering the c onformation of the ring. The tw i st boat conformati on allows both bulky substituents to be "equatorial".
H
H
CH 3 I C -CH 2 CH 3 I CH 3
51
3-30 The nomenclatur e of bicyclic alkanes is summarized in Appendix l . (a) bicycl o[3 . 1 .0]hexane (b) bicyc l o[3 . 3 . 1 ]nonane (c) bicycl o[2 . 2 . 2]octane 3-3 1 Using models is essential for thi s problem. H H
rotate picture
----
(d) bicyclo[3 . 1 . 1 ]heptane
•
H
from text Fi gure 3-29 3-32 Please refer to solution 1 -20, page 1 2 of this Solutions Manual . 3-33 (a) The third structure i s 2-methylpropane (i sobutane). The other four structures are all n-butane. Remember that a compound's i dentity is determined by how the atoms are connected, not by the position of the atoms when a structure is drawn on a page. (b) The two structures at the top-left and bottom-left are both cis-2-butene. The two structures at the top center and bottom-center are both I-butene. The unique structure at the upper right i s trans-2-butene . The unique structure at the lower right i s 2-methylpropene. (c) The first two structures are both cis- l ,2-dimethylcyclopentane. The next two structures are both trans-l,2-di methylcyc lopentane. The l ast structure i s different from all the others, cis- l ,3dimethylcyc lopentane. (d)
Anal ysi s of the structures shows that some double bonds begin at c arbon-2 and some at c arbon-3 of the longest chain . The three structures l abeled A are the same, with the double bond trans; B is a geometric i somer (cis) of A. C and Dare constituti onal i somers of the others.
B
A
D
(e) Naming the structures shows that three of the structures are trans- l ,4dimethylcyclohexane, two are the cis i somer, and one is cis1,3-dimethylcyclohexane. Al though a structure may be shown in two different conformations , it sti l l represents only one compound.
H3C
('(.�
� H
:
��h
3
�CH3H
cis- l A-dimethyl
� H
H3C
trans- l ,4-di methy I
H
M
CH3 cis- I,3-dimethyl
H3C H trans-I,4-di methyl 52
-PH
H
CH3
CH3
cis- l ,4-dimethy I
® CH3 H
trans- l ,4-di methyl
3-34 Line formul as are shown. ( a)
�
(b)
(d)
(g)
(Cl
;or:
H"
(j)
" H
�
(h)
(k)
8
(Cl
I)
(
�
01 0-0
"
(i )
d
6
CH3
or
,
""
"CH2CH3
otxj
-
CI
(I)
B
'H
" H' 3 - 3 5 There are many possible answers to each of these problems. The ones shown here are examples of c orrect answers. Your answers may be different AND correct . Check your answers i n your study group. (a)
or 3-methyl heptane
2-methyl heptane
4-methyl heptane
(b)
4 , 5 -diethyldecane
3 , 5-diethyldecane
(Any combination is correct except using posi tion numbers 1 or 2 or 9 or 1 0 . Why won't these work?)
6
CH3
(c)
Q
CH2 CH3
CH2 CH3
cis- 1 ,2-diethylcycloheptane
cis- 1 ,3 -diethylcycloheptane
(d) only two possible an swers CH3
O
,,,,CH3
trans-l,2-dimethy l c y c l opentane
[)
CH2CH3
cis- 1 ,4-di ethylcycloheptane
"" CH3
trans- 1 ,3 -dimethylcyclopentane
53
NOT 1 ,5 NOT 1 ,6 NOT 1 ,7
3 - 3 5 continued
Other ri n g sizes are possible, although they must have 6 or more c arbons to be longer than the 5 c arbons of the substituent chain .
(e)
(2,3 -dimethylpentyl )cycloheptane
(2,3-di methylpenty I )cyc looctane
(f)
Any combination where the number of c arbons in the bridges sums to 8 will work. (Two carbons are the bridgehead carbons . ) bicyclo[4.4.0]decane bicyclo[3 . 3 .2] decane
3-36
HO - CH 2CH 2CH 3
HO - CH 2CH 2CH 2CH 2CH 3
HO - C H2CH 2CH 2CH 3
HO - CH 2CH 2CH 2CH 2C H 2CH 3
3-37 (a) (b) (c) (d)
3 -ethyl-2,2,6-trimethylheptane 3-ethyl-2,6,7-trimeth y loctane 3 ,7 -diethyl-2,2,8-trimethyldec ane 2-ethyl-l, l -di methylcyclobutane
(e) (f) (g) (h)
bicyclo[4 . 1 .0] h eptane cis- l -ethyl-3-propylcyc lopentane ( 1 , I-di ethylpropyl)cyclohexane cis-l-ethyl-4-isopropylcyclodecane
3 - 3 8 There are ei ghteen i somers of C sH IS. Here are ei ght of them. Yours may be different from the ones shown . An easy w ay to compare is to n ame yours and see if the n ames match.
�
2-meth y Iheptane
Il-octane
�
2 , 2,4-tri methylpentane 3-39 (a)
�
�
~
2 , 3 ,4-tri methy Ipentane
2 , 3 -dimethylhexane
T
T
*
3 -ethylhexane
3-ethyl-3 -methy l pentane 2,2,3,3-tetrameth y I butane (b)
4 4
6 correct name : 3 -methylhexane (longer chain) (c)
correct n ame : 3 -ethyl-2-methy lhexane (more branching with this numbering)
�I
(d)
CI
correct name: 2 ,2-dimethylbutane (include a position number for each substituent, regardless of redundancies)
correct name: 2-ch l oro-3 -methy l hexane (begin numbering at end c l osest to substi tuent) 54
3-39 continued
(f)
(e)
c orrect name: sec-butylcyclohexane or ( I-methylpropyl)cyclohexane (the l onger chain or rin g is the base name)
c orrect n ame: 1 ,2-diethylcyclopentane (position numbers are the l owest possible)
3 -40 (a) n-Octane has a h i gher boi l i n g point than 2,2,3 -trimethylpentane because linear molecules boil h igher than branched molecules of the s ame molecular weight (increased van der Waals interaction). (b) 2-Methylnonane has a h igher boi ling point than n-heptane because it has a signifi c antly higher molecular wei ght than n-heptane. (c) n-Nonane boi l s hi gher than 2,2,5 -tri methylhexane for the s ame reason as in (a). 3-4 1 The point of attach ment i s shown by the bold bond at the left of each structure . -CH 2CHCH 2CH3 I 1° C H 3 2-methylbutyl
-CH2CH2CH 2CH 2CH 3 1° ll-pentyl
CH 3 -CH2CH2CHCH3 I 1° CH3 3-methy lbutyl (i sopenty l )
-CHCH2CH 3 2° I CH 2CH 3 l -ethylpropyl
I
CH3 -CHCHCH 3 I 2° CH3 1 ,2-di meth y l propyl
I
-C -CH 2CH3 3° I CH 3 l , l -dimethylpropyl (t-pentyl )
CH3 I
-CH 2 - C -CH 3 I 1° CH3 2 , 2-dimethylpropyJ (neo-pentyl)
3 -42 In each case, put the l argest groups on adjacent c arbons in anti positions to make the most stable conformations. (a) 3 -methylpentane
2
carbon-3 cannot b e seen; it is behind carbon-2
�
H
C-2 i s the front c arbon with H, H , and CH3 C-3 i s the back c arbon with H , CH 3, and CH2CH 3
55
3-42 continued (b) 3,3-dimethylhexane
carbon-4 cannot be seen; it is behind carbon-3
C-3 is the front carbon w i th CH3, CH3, and CH2CH3 C-4 i s the back c arbon w i th H, H, and CH2CH3 3-43 (a)
M: �
axial H
ax al
equatorial
CH3 equatori al
CH3 axi al
CH3 equatori al (b)
I�
more stable (lower energy)
H equatorial
CH3 axial
less stable (higher energy)
(c) From Section 3- 1 4 of the text, each gauche interaction raises the energy 3.8 kJ/mole (0.9 kcallmole), and each axial methyl has two gauche i nteractions, so the energy is: 2 methyls x 2 interactions per methyl x 3.8 kJ/mole per interaction = 1 5 .2 kJ/mole (3.6 kcallmole) (d) The steric strain from the 1 ,3-di axi al interaction of the methyls must be the difference between the total energy and the energy due to gauche i nteractions: 23 kJ/mole - 1 5 .2 kJ/mole = 7.8 kJ/mole (5.4 kcallmole - 3.6 kcal/mole = 1 .8 kcallmole) 3-44 The more stable conformer p l aces the l arger group equatori a l . (a)
�7C a
CH(CH3 )2 H'
��
a CH2CH3
-�
H
H
(b)
H3 )2
fE:f
a CH CH3 2
H
H
H
CH(CH3h e
more stable
� H
-�
CH e I li CH3 more stable
a CH2CH3
(c)
�
CH(CHJh e
� �
--
CH e
H H more stable 56
CH2CHJ e
3-44 continued (d)
H-- �
H2 CH ]
H
a
H CH 3 more stable
�
a CH3
(e)
H
more stable H
H
�H 2C H 3
3-45 (Usi ng models is essential to this problem.) In both cis- and trans-decalin, the cyclohexane rings can be i n chair conformati ons. The relati ve energies w i l l depend on the number of axial substi tuents.
trans no axial substituents MORE STABLE
cis one axial substituent
3-46 chair form of glucose-al l substituents equatori al
I
OH CH2 HO
HO OH OH H
H (without ri ng H's shown)
57
CHAPTER 4-THE STUDY OF CHEMIC AL REACTIONS
4- 1 (a)
H H H I
I
I
I
I
I
H-C -C-C-H I
C I' CI - C Q Q +
IV
---
:�
CI-C·
I
.. :I .
H
H
I
I
(d)
H H H
H H H
(a)
I
I
H-C-C-C·
4-2
H
H
(c)
(b)
+
C I -C I
HU U u
---
C
CI- · I
+
H-CI
H H I
CI-C-C1 I
+ CI'
H
(b) Free-radical halogenati on substitutes a halogen atom for a hydrogen . Even i f a molecule has only one type of hydrogen, substituti on of the first of these hydrogens forms a new compound. Any remaining hydrogens in this product c an compete with the initi al reactant for the avai l able h a logen . Thus, chlorinati on of methane, CH4, produces a l l possible substitution products: CH3C I , C H2 C I 2 , CHCI3 , and CCI4. If a molecule has different types of hydrogens, the reaction can generate a mixture of the possible substitution products. (c) Production of CCI4 or C H3CI can be controlled by alteri ng the rati o of C H4 to C12. To produce CCI4, use an excess of CI 2 and let the reaction proceed unti l all C-H bonds h ave been repl aced with C-C1 bonds . Producing CH3CI is more chal lenging because the reaction tends to proceed past the first substi tution. By using a very l arge excess of CH4 to C1 2 , perhaps 1 00 to 1 or even more, a chlorine atom i s more likel y t o fi nd a CH4 molecule than it i s t o find a CH3C I , s o o n l y a s m a l l amount of CH4 i s transformed to CH3C1 by the time the C I 2 runs out, with almost no CH 2 Cl 2 bei ng produced. 4-3 (a) Thi s mechani sm requires that one photon of l i ght be added for each CH3 CI generated, a quantum yield of 1. The actual quantum yield is several hundred or thousand. The h i g h q uantum yield suggests a chain reaction, but thi s mechani sm is not a chain; it has no propagation step s . ( b ) Thi s mechan i sm conflicts w i th a t least t w o experimental observations. First, t h e energy o f l i ght requi red to break a H-CH3 bond i s 4 3 5 kJ/mole ( 1 04 kcallmole, from Table 4-2) ; the energy of l i ght determined by experi ment to initiate the reaction i s onl y 251 kJ/mole of photons (60 kcallmole of photons), much less than the energy needed to break thi s H-C bond. Second, as in (a), each CH3Cl produced would require one photon of l i ght, a quantum yield of 1 , i nstead of the actual number of several hundred or thousand. As in (a), there i s no provi sion for a chain process, since al l the radicals generated are also consumed in the mechan i s m . 4-4 (a) The twelve hydrogens of cyclohexane are all on equivalent 2° c arbons. Replacement of any one of the twelve will lead to the same product, c h lorocyclohexane. n-Hexane, h owever, h as h ydrogens in three different posi tions: on c arbon- l (equi valent to carbon-6) , carbon-2 (equi valent to c arbon-5), and carbon-3 (equi valent to carbon-4). Monoch lorination of n-hexane w i l l produce a mi xture of a l l three possible isomers : 1 - , 2-, and 3-chlorohexane.
59
4-4 continued (b) The best conversion of cyclohexane to chlorocyclohexane would require the ratio of cyclohexanel chlorine to be a l arge number. If the ratio were small , as the concentration of c h lorocyclohexane increased during the reaction, chlorine would begin to substitute for a second hydrogen of chlorocyclohexane, generating unwanted products. The goal is to have chlorine attack a molecule of cyclohex ane before it ever encounters a molecule of c h lorocyclohexane, so the concentration of cyclohexane should be kept hi gh. 4-5 (a)
- 2.1 kJ/mole - 2100 Ilmole ='
e -IlGOIRT
=
e
� -(-2100 J/mole)/«8.314 J/K-mole)· (298 K»
e
2100 I 2478
=
e
0.847
Keq
=
2.3
=
degrees kelvin
[ill
=
[CH3S H] [HBr] (b)
K
=
[CH3B r] [H 2 S] [CH3B r]
initi al concentrati ons: final concentrations
Keq
o
=
=
2.3
1 -x
x 0 x
=
1
x =
(l- x) (l- x )
1 .3 x
2
-
-
4.6 x + 2.3
2
1 - 2x + x 2
o
o
x
x
x
c::::> x 2
> x (using quadratic equation)
=
2.3 x
2
0.60 ,
=
-
1
4.6 x + 2.3
-
x
=
0 .40
4-6 2 acetone � di acetone Assume that the initial concentration of acetone is 1 molar, and 5% of the acetone is converted to diacetone. NOTE THE MOLE RATIO. [acetone] [diacetone] 1 M 0 . 95 M
initial concentrations: final concentrations:
6.Go
[di acetone] 2 [acetone] =
=
I
-
=
2 .303 RT 10glO Keq + 8.9 kJ/mole
o
0.025 M
0.025 (0.95 ) =
-
2
2.303 «8 . 3 1 4 J/K-mole) 0 (298K)) o l og(0 .028)
(+ 2. 1 kc al/mole)
I
4-7 6.5° w i l l be negative since two molecules are combined into one, a loss of freedom of motion . Since 6.so is negati ve, - T6.So is positi ve; but 6.Go is a l arge negati ve n umber since the reaction goes to completion . Therefore, /'li{0 must also be a large negative number, necessari ly l arger in absol ute value than 6.Go. We can explain this by formation of two strong C-H bonds (4 1 0 kJ/mole each ) after breaking a strong H-H bond (435 kJI mole) and a WEAKER C=C pi bond. 60
4-8
(a) 6.5° i s positive--one molecule became two smaller molecules w i th greater freedom of motion (b) 6.5° is negative-two smaller molecules combined into one l arger molecule w i th less freedom of motion (c) 6.5° cannot be predic ted since the number of molec ules i n reactants and products i s the same
4-9
(a)
h� Cl - C 1
2
hv
(2) + ii (3(4)) Cl+ + (5) + + (1)
initi ation
-
"�,,
Cl·
propagati on
h
H-CH 2 CH3
� h• CH
Cl -
C l·
-
2 CH3
H-Cl +
-
+
C H2 CH3
•
Cl - CH2 C H3
Cl·
Cl - CI Cl Cl· Cl- C H 2 CH3 • CH2CH3 Cl· • CH2 CH 3 - CH 3CH 2 CH 2 CH3 CH3CH 2• •
termination
(b)
(2) (3)
(6)
step ( 1 )
break Cl-Cl
step
break H-CH 2 CH3 make H-Cl step
!ili0 !ili0 !ili0
break Cl-Cl make Cl-CH2CH3 step (3 )
!ili0 !ili0 /�lr
step
!ili0
(2)
+ 242 + 431 +242339
=
=
-
=
kJ/mole
(+ 58 (+ 98 (+ 5881
kcal/mole)
4 1 0 kJ/mole kcal/mole) kJ/mol e (- 1 03 kcal/mole)
-2 1 kJ/mole (-5 kcaUmole)
=
=
kJ/mole kJ/mole (-
-
=
kcallmole) kcal/mol e)
-97 kJ/mole (-2 3 kcaUmole)
(c) 6.HO for the reaction is the sum of the !ili0 val ues of the individual propagation steps:
(
- 2 1 kJ/mole - 5 kcal/mole
4- 1 0 (a)
(2) i (1)
initiation propagati on
(3)
++ -23
- 97 kJ/mole
=
kc allmole
hv Br - B r ,,�� H-CH 3 B r· h�
h
+
+
�h
B r- Br
•
=
CH3
break B r-Br
!ili0
=
step (2)
break H-CHl make H-B r step (2)
!ili0 !ili0 M-I0
=
(3)
(3)
break B r-Br make B r-CH3 step
(++
!ili0 !ili0 M-I0
+ + -24
2
H-Br + Br-CH3 + + 192 (+ 46 + 435368 (+ 88 + 192293 (+ B r·
• CH3
-
step ( 1 )
step
1 1 8 kJ/mole - 28 kcaUmole)
=
B r·
-
kJ/mole
=
kcallmole)
-
1 04 kcal/mole) kJ/mole kcallmole) kJ/mole ( -
-
46 kcal/mole) kJ/mole kJ/mole ( - 70 kcal/mole)
+67 kJ/mole (+16 kcaUmole)
=
= = =
-1 0 1 kJ/mole (-24 kcaUmole)
(b) 6.HO for the reaction is the sum of the I'IJJO val ues of the individual propagation steps: 67 kJ/mole
1 6 kcallmole
-
101
- 34 kJ/mole kJ/mole kcallmole = - 8 kcaUmole) =
61
4- 1 1 (a) first order: the exponent of [ (CH3hCCI ] in the rate l aw = 1 (b) zeroth order: [CH30H ] does not appear in the rate law ( i ts exponent is zero) (c) first order: the sum of the exponents in the rate law = 1 + 0 = 1 4- 1 2 (a) first order: the exponent of [cyclohexene] i n the rate l aw = 1 (b) second order: the exponent of [ B r2 ] in the rate law = 2 (c) third order: the sum of the exponents in the rate law = 1 + 2 = 3 4- 1 3 (a) the reaction rate depends o n neither [ethylene] nor [hydrogen] , s o i t i s zeroth order in both species. The overall reaction must be zeroth order. (b) rate = kr (c) The rate law does not depend on the concentration of the reactants. It must depend, therefore, on the only other chemical present, the catal yst. Apparently, whatever is happening on the surface of the catalyst determines the rate, regardless of the concentrations of the two gases. Increasing the surface area of the catalyst, or simply adding more c atalyst, would accelerate the reaction . 4- 1 4 (a)
t
+ 1 3 kJ/mole
HCI
•
+
CH 3
CH 4
+
CI ·
....................... �-----
reaction coordinate (b) Ea = + 1 3 kJ/mole ( + 3 kcal/mole) (c) !ili0 = - 4 kJ/mole ( - 1 kcal/mole) 4- 1 5
Cl 2 +
•
CH 3
_____
(a)
t
./......
.......... .
i
....... .,. +
k ca IImole
- 1 09 kJ/mole
reaction coordinate (b) reverse: CH 3C I + CI · (c) reverse: Ea
4
=
-----
C l2
+
----•
CH 3
+ 1 09 kJ/mole + + 4 kJ/mole = + 1 1 3 kJ/mole ( + 26 kcallmole + + 1 kcallmole = + 27 kcaVmole) 62
4- 1 6
numbers are kJ/mol e
(a)
t + 1 92
reaction coordinate -(b) The step leading to the hi ghest energy transition state i s rate-l i mi ti n g . In this mechanism, the first propagation step is rate- l i mi ting:
CH4 [t-------�; 1 (2) [H-�tt'- -H---- -�; r . H 8 + . 8 / [H-�-------- ] B r· +
(c)
HBr +
--
1 ( )
+
(3)
Br - - - -- - B r
•
CH3
"8· " mean s parti al radical character on the atom )
( d ) W O for the reaction i s the sum of the WO values o f the individual propagation steps (refer t o the solution to 4- 1 0 (a) and (b»: + 67 kJ/mole + - 1 0 1 kJ/mole = - 34 kJ/mole ( + 1 6 kcallmole + kcallmole = - 8 kcaUmole)
24
4- 1 7 (a)
initiation
2 (2) I H-CH3 - H CH3 - I-CH3 CH3 1 IH-CH3 -I H-I II--CIH3 (1)
hv hn I- I r\�n •
propagation
(3)
(2)
step ( 1 )
break
step
break make
step (3 )
break make
I·
-I
+
h� h +' I�I
+ I·
•
Mfo
step (2)
step (3 )
WO WO MIo
WO WO MIo
+ .
=
= =
=
=
=
=
+ 1 5 1 kJ/mole ( + 36 kcal/mole)
+ 4 35 kJ/mole (+ 1 04 kcal/mole) - 297 kJ/mole ( - 7 kcallmole) 1 + 138 kJ/mole ( + 3 3 kcaUmole) + 1 5 1 kJ/mole ( + 36 kcallmole) - 234 kJ/mole (- 56 kc al/mole) -8 3 kJ/mole ( -20 kcaUmole) 63
4- 1 7 continued (b) Mia for the reaction is the sum of the Mia values of the i ndividual propagation steps : + 1 3 8 kllmole + - 8 3 kllmole = (+ 33 kcallmole + - 20 kcallmole
55 kJ/rnole + 1 3 kcaVrnole)
+
=
(c) Iodination of methane is unfavorable for both kinetic and thermodynamic reasons . Kinetical l y , the rate of the first propagation step must be very slow because it is very endothermi c ; the acti vation energy must be at least + 1 3 8 kllmole. Thermodynamically, the overall reaction is endothermic, so an equil ibri um would favor reactants, not products ; there i s no energy decrease to dri ve the reaction to products . 4-18 Propane has six primary hydrogens and two secondary hydrogen s , a rati o of 3 : 1 . I f primary and secondary hydrogens were repl aced by chlorine at equal rates , the chloropropane i somers would reflect the same 3: 1 ratio, that is, 7 5 % l -c h loropropane and 25% 2-chloropropane.
H -- 30
(b)
H (d)
CH3
t
lo
H H
H
H H
CH3- �-CH3
t
H
(c)
I
H
H
--
I
30
CH3- C-CH 2CH3
t CH3 t t lo t 20 1 0
t
10
10
10
H H H
OR H
(e)
H
H
H
all are 20 H all are 20 H
30 al l are 20 H except bri dgehead H's (labeled 30)
4-20 3 a H abstraction
Cl- +
H I
H -Cl
CH3-C-CH 3 I
+
CH3-C-CH3 I
CH3
CH3 break 30 H -C(CH3h make H-Cl overall 3 a H abstraction 10 H abstraction
Mia Mia
Mr
= = =
+ 3 81 kllmole ( + 9 1 kcallmole) - 43 1 kllmole ( - 1 03 kcal/mol e) -50 kJ/rnole ( -1 2 kcaVrnole)
H
H
I
I
CI- + H -CH -C-CH 3 2 I CH3
break 10 H-CH2CH(CH3h make H-CI overall 10 H abstraction
H -CI
+ -CH -C-CH3
2
I
CH3 Mia
!J..Ho
/).H0
= = =
+ 4 1 0 kllmole (+ 98 kcallmole) - 43 1 kllmole ( - 1 03 kcallmole) -2 1 kJ/mole ( -5 kcaVrnole)
(Note: Mia for abstraction of a 1 0 H from both ethane and propane are + 4 1 0 kllmole (+ 98 kcallmole). It i s reasonable to use this s ame value for abstracti on of the 10 H in i sobutan e . ) 64
4-20 continued t l
H
C o +
H-C H2-
1° radi= c-21al; kJ/mole (- 5 kcal/mole)
I
� -CH3
CH3
DJ{0
3° radi= c-50al; kJ/mole (- 12 kcaUmole)
reaction coordinate .. 3° radiclealadiisnmore cal, it ilseadireasonabl totSihenince1f°erradithatcfalothr.efoactrmiivnatgitohneenergy g to thnegat e 3° iradive tchalanis lowerforthfoanrmtheingactthiev1at°ioradin energy ng to e 4-21Methylbutane can produce four mono-chloro isomers. To calculate the relative amount of each in the 2-product xtuofre,hydrogen. multiply tEach he numbers ofamount hydrogensdividwhied cbyh coulthe sumd leadoftalolththate product tiwimesl provi the reactde tihveity forpercentthat tofmiypeeach rel a t i v e amount s in the product mixture. I'1Ho
Mlo
DJ{0
(6 6.1°H0 rel) xat(irveacte amount ivity 1.0) (3 3.0 relxat(irveeactamount ivity 1.0) 23.6.50 x 100= 3° ) xl (reactivity 5.5) (2 2°H) x (reactl ivity 4.5) 23.3.05 x 100 = :5.5 relative amount = 9.0 relative amount � � x 100 � :� x 100 2 2 4-(a)22When n-heptane is burned, only 1° and 2° radicals can be formed (from either or bond clisooct eavageane).(2These areimethihgylhpenergy, unstoawbl)eisradiburned, cals whi3° radich rapicalsdlcany forbemfotormed her product st.herWhen or , 2 , 4 t r ent a ne, bel f r om ei The 3°ionraditranslcalastaresetololweress "knocki in energyng.th"an 1° or 2°, relatively stable, with lowered reactivbond ity. Slcloeweravage.combust isooctane (2,2,4-tnmethylpentane) Any indicated bond cleavage produce a 3° radical. 2 CH3
CH3
I
I
CICH2 - CHCH2CH3
CH3
CH3 I
I
=
CH3-C-CH2CH3
CH3· CH· CHCH3
I
(l
26%
CH3 - CHCH2CH2Cl lOH)
=
I
C
C
=
H
1 3%
=
23%
=
38%
C-H
C-C
C-H
C-C
CH3
I JV\
H
I JV'
C - � -CH - C -CH 3 C H 3 - �? -+ CH 3
will
I
CH3
65
4-22 continued
CH3
CH3 I
(b)
H 3 C - C- O - H
I
I
H 3 C- C- O-
o R -an alkyl radi cal
+
CH3
I
R-H
+
CH3
When the alcohol hydrogen i s abstracted from t-butyl alcohol , a rel ati vely stable t-butoxy radical ( (CH3)3C- O 0 ) is produced. Thi s low energy radical i s slower to react than alkyl radicals , moderating the reaction and producing less " knocking." 4-23 (a) 1 ° H abstraction F0
CH3CH 2 CH3
+
break 1 ° H - CH2 CH 2 CH3 make H-F overall 1 ° H abstraction 2° H abstraction F0
M-[0 M-[0
Mlo
= = =
+ 4 1 0 kJ/mole ( + 98 kcaJ/mole) - 569 kJ/moJe (- 1 36 kcal/mole)
-159 kJ/mole ( -38 kcallmole)
H-F
CH3CH2 CH3
+
+
H-F
break 2° H - CH(CH3)2 make H-F overal l 2° H abstraction
MiO Mio Mlo
= = =
+ 397 kJ/mole ( + 95 kcaJ/mole) 569 kJ/mole (- 1 36 kcal/mole)
-
- 172 kJ/mole ( -41 kcallmole)
(b) Fluorination is extremely exothermic and is likely to be i ndi scrimi nate in which hydrogens are abstracted. (In fact, C- C bonds are also broken during fluorination.) (c) Free radical fl uorination is extremely exothermi c . In exothermi c reacti ons, the transition states resemble the starting material s m ore than the products, so while the 1 ° and 2 ° radicals differ by about 1 3 kJ/mole (3 kcal/mole) , the trans i ti on states w i l l differ by only a tiny amount. For fluorination, then , the rate of abstraction for 1 ° and 2° h ydrogen s wil l be virtuall y identical. Produc t rati os will depend on statistical factors onl y . Fl uorination is difficult to contro l , but if propane were monofluorinated, t h e product mixture would reflect the rati o of the types of h y drogens : six 1 ° H to two 2° H , or 3 : 1 ratio, giving 75% 1 -fluoropropane and 25% 2-fluoropropane .
H
4-24
I
H
H I
I
CH 3- C- C- CH 3 I
in i ti ation
propagation
B r Br
I
CH 3- C-C-CH 3
I
I
CH3CH 3 Mechan i sm
Br I
CH3CH3
nn
Br-Br
hv
--
I
I
2 Br o
���-->�
H
I
CH3CH3
H
CH3- C- C- CH 3 + B r o � HB r + I
I
CH 3- C- C-CH 3
I
CH 3 C H 3
I
.� nn
I
I
CH 3- C- C- CH 3 + Br-Br
H �
I
•
I
I
CH 3- C- C- CH 3
I
CH3CH 3 Br I
CH 3 - C-C- CH 3 I
I
C�C�
C� C� 66
+
Br 0
4-25 (a) Mechanism initiation
nn
H
Br 0
�
IU
H
2
---
a
propagation
i i
hv
B r- B r
..
-
r-'�
�H UH
6=: C(
B,·
+
..
•
\n B,-B,
+
H
H
A U
H
-
+ HBr
H
Br
B,·
+
(b) Energy calcul ation uses the value for the al lylic C-H bond from Table 4-2. . First propagatIOn step
Second propagation step
break allylic H-CH[ri ng] make H-B r overall allylic H abstraction
WO WO M-/0
bre ak B r-Br make 2° C-B r overall C-B r formati on
MfD WO M-/0
[
= =
= = =
+ 364 kJ/mole (+ 8 7 kcallmole) - 368 kJ/mole ( - 88 kc allmole)
-4 kJ/mole ( -1 kcaUmole)
+ 1 92 kJ/mole (+ 46 kcallmole) 28 5 kJ/mole ( - 68 kcallmole) - 93 kJ/mole ( -22 kcaUmole) -
The first propagation step is rate-limiting.
slmclu;c of Ihc transHi on slateo
H
0-
80
8
OJ
- - - H - - - - - - - - - B,
t
(c) The Hammond postul ate tel ls us that, in an exothermic reaction, the transition state is closer to the reactants in energy and in structure . Since the first propagation step is exothermic (although not by much), the transition state is c loser to cyclohexene + bromine radic al . This is indicated in the transi tion state structure by showing the H closer to the C th an to the Br. (d) A bromine radical wi l l abstract the hydrogen with the lowest bond dissoc i ation energy at the fastest rate . The al lylic hydrogen of cyc lohexene is more easi ly abstracted than a hydrogen of cyclohexane because the radical produced is stabilized by resonance. (Energy values below are per mole . )
6 6
aliphatic: 397 kJ (95 kcal ) - H H
SLOWER
4-26
. .
H H
. .
-
FASTER
3 64 kJ (87 kcal) : allylic
. .
. .
¢x ¢x ~ if -----
OCH3 B HA radical
I
•
-----
-----
h-
•
:::--..
OCH3
OCH3
67
I
OCH3
4-27
R- O· J
J¢Ct
() �-0 H3C
I
h
0
R
..
R- O
,H
·0 +
II
H3C
"TT
�
I
l-R
CH3 stabilized by resonance Vitamin E l ess reactive than R O· 4-28 The triphenylmethyl c ati on is so stable because of the delocalization of the charge. The more resonance forms a species h as, the more stable it will be. CH3
� o O +0 ------c c +O O � b Q= b b /' < }-co ---- < }-c0 ---- o-c0 ---- < }-c0 + +U b b b+ + � < }-cQ+ ---- < }-cP ---- < }-+0 c b b b C
II
(Note : these resonance forms do not include the simple benzene resonance forms as shown below; they are signifi cant, but repetiti ve, so for simplicity, they are not drawn here . )
'I �
0-+/ C
_
4-29
most stabl e (c)
>
,
Ph ----
Ph
(b)
+
II
C
,
----
(a) least stable +
(b) CH3-CHCHCH3
I
many more
Ph
+
(c ) CH3 -CCH2CH3
4-30
>
� 0-+/
Ph
(a) CH3 -CHCH2CH2
I
I
CH3
CH3
CH3
3°
2°
1°
most stable (c) .
(c) CH3 -CCH2CH3 I
>
(b)
>
(a) least stable .
.
(b) CH3-CHCHCH3
(a) CH3 -CHCH2CH2
I
CH3
CH3
3°
2°
I
CH3 68
1°
4-3 1
O: H :
II ( I
� II -
H 3 C- C- C- C- C H3
I
..
: O-H
+
•
-- H20 +
•
base
H
II
..
:0:
:0 :
-
: 0:
:0 :
I
II
II
..
:0:
-
:0:
II
I
C /C C .... ____ / C .... ", C , ", C ___ H 3 C/ ' C / ' CH 3 H 3 C ..... ' C ..... ' CH 3 H 3 C ..... ' c/ ' CH 3 -
•
•
I
I
H
H
--
I
H
H-B ase +
4-34 Please refer to sol ution 1 -20, page 1 2 of this Soluti ons Manual . 4-35
transiti on states
/�
(a) and (c)
t
t
-
---------
-------- -
l r
(b) Mio is negati ve (decreases), so the reaction is exothermic . (d) The fi rst transi tion state detennines the rate since it is the hi ghest energy point. The structure of the fi rst transi ti on state resembles the structure of the intennedi ate since the energy of the transition state is c losest to the energy of the intennediate .
E Q
products reacti on c oordi nate 4-36
t
acti vation energy
--
� tranSitIOn .. state
reactants products reaction coordinate -69
4-37
t
�
energy of h i ghest transition state detennines rate
t
--- --------------------
reaction coordinate
t-Jl0 is posi tive
-
4-38 The rate law is first order with respect to the c oncentrations of hydrogen ion and of t-butyl alcohol , zeroth order with respect to the concentration of c h lori de ion, second order overall . rate = kr [(CH3)3COH 1 [ H + 1 1° (c)
H
H H
1°
3°
�
(d)
H
�
CH3
H
1°
�
�--+-- H ......t--'r--
H
(e)
1°
all are 2° H except for the two types l abeled
�
H
H
all are 2° H except as l abeled
4-40 (a) break H-CH2CH3 and I-I, make I-CH2CH3 and H-I kllmole: ( + 4 1 0 + + 1 5 1 ) + ( - 222 + - 297) = + 42 kJ/mole kcallmole: ( + 98 + + 36) + ( - 53 + - 7 1 ) = + 1 0 kcallmole (b) break CH3CH2-CI and H-I, make CH3CHrI and H-CI kllmole: ( + 3 3 9 + + 297) + ( - 222 + - 43 1 ) = -1 7 kJ/mole kcal/mole: ( + 8 1 + + 7 1 ) + ( - 53 + - 1 03) = - 4 kcallmole
70
3°
H
all are 2 ° H except for the two types l abeled
H 3°
---
4-40 continued
(c) break (CH3hC-OH and H-C l , m ake (CH3hC-Cl and H-OH kJ/mol e : ( + 381 + + 431) + ( - 331 + 498) = -1 7 kJ/mole kc al/mole: ( + 9 1 + + 103) + (-79 + -119)= -4kcallmole -
(d) break CH3CH2-CH3 and H-H, make CH3CHrH and H-CH3 kJ/mole: ( + 356 + + 435) + ( - 410 + -435)= -S4kJ/mole kcallmole: ( + 85 + + 104) + ( - 9 8 + -104) = - 1 3 kcallmole (e) break CH3CHrOH and H-B r, make CH3CHrB r and H-OH kllmole: ( + 381 + + 368) + ( -285 + 498) = 34kJ/mole kcal/mole: ( + 91 + + 88) + ( - 68 + - 119)= 8 kcallmole -
-
-
4-41 Numbers are bond dissoci ation energies in kcallmole in the top line and kl/mole in the bottom line.
< >- C
H2
CH2 = CH C H2
>
85 356
(a)
(b)
(CH3hC
87 364
most stable 4-42
>
o ey
_
ey ey
_
>
CH3CH2
91
95
98
381
397
410
>
CH3 104 435
least stable Only o n e product; chlorination would work. B romination o n a 20 carbon would not be predicted to be a h i gh-yielding process .
Cl
C HP
CH3
(CH3)zCH
>
Ch�
+
Hl
Q
+
CH1
Cl Chlorination would produce four constituti onal isomers and would not be a good method to make only one of these. Monobromination at the 30 carbon would gi ve a reasonab l e yield. H H H CI H H (c)
�
CH3 - -� -CH3 I
I
I
I
I
I
CH 3 -C-C-CH 3
-
+
CH3 CH3
CH3 CH3
I
I
I
I
CH 3 -C-C-CH 2 Cl CH3 CH3
Chlori nation would produce two consti tuti onal i somers and would not be a good method to make only one of these . Monobromination would be selective for the 30 c arbon and would give an excellent yield. CH3 CH3 (d)
I
CH 3 -C-C-CH 3 I
CH3 CH3
I
-
I
CH3 C H3 4-43
initiati on
(1) (2)
propagation
I
I
I
I
Only one product; chlorination would gi ve a high yield. Mon obromination would be very difficult si nce all hydrogens are on 1° c arbons .
CH 3-C-C-CH 2 CI CH3 CH3
�l CI
�
� 6)
�
:
H
(3)
Cl 71
O
;C
+ C l· Termination steps are any two radicals combining.
4-44 (a)
CH2=CH-CH2 .
. f----l.� ......
.
:0: II
(c)
CH3 -C-O·
(d)
H
O I
• •
CH3-C=O
.� ......f----l .
• •
H
H H
:0: I
.... ..f----l . .�
4-45 (a) Mechanism
nn
initiation
Br- B r
propagation
(e)
�H UH hv
H
H
H
�H U
f----l.� .......
H
cY
H
2 B r·
---
JH3C yy' CH3 � -1
WId·)
H3C � CH3 � H � '=Jj.J _
The 3 ° allylic H is abstracted selecti vely (faster than any other type In the molecule) , fonrung an intermedi ate represented by two non-equivalent resonance forms. Parti al radic al c h aracter on two different carbons of the allylic radical leads to two different product s .
Br
_
+ HBr
H3C� CH3
� j �
·
W
W
l f
second propagation step
+
Br· (b) There are two reasons why the H shown i s the one that is abstracted by bromine radical: the H is 3° and it is allylic, that i s , neighbori n g a double bond. Both of these factors stabi l i ze the radical that is created by removing the H atom. 4-46 Where mixtures are possible, only the major product is shown . (a)
(b )
(c)
0
�
�
rY B r V
� CH� U CH3H C H3 -C-C-CH3 CH3 CH3
3 ° hydrogen abstracted selecti vely, faster than 2° or 1 °
�
I
I
I
I
only one product possible
-
CH3 Br CH3-C-C-CH3 CH3 CH3 I
I
I
I
72
3 ° hydrogen abstracted selectively, faster than 1 °
4-46 continued (d)
Br
00
-
decalin
(e)
�
(f)�
-
CO �
CHCH3 I I U
3 ° hydrogen abstracted selectively , faster than 2° or 1 °
+
Br
� Br
Q¢-OQ+ cb Br
CH3C\
4-47 (a) As is produced, it c an c ompete with generate CHCI3, etc.
+ +
CH4CH3 CCl2· CCH2C\ C 3 CCl2 CH2Cl2 Cl· Cl2 CHCl3 Cl C 3 Cl2 C 3C
RroRagation steRs +
•
l H
•
l
l
+
•
+
+
•
•
CHCl2 + C l
+
•
..... .. .. ...
both 2°-formed in equal amounts
from resonance-stabi l i zed benzylic radical
Br
h
(h)
3' hydrogen abstracted selectively. faster than 2'
CH4
l +
key to answering this question correctly.
for avai l able Cl ., generatin g
+++ CH3CC ·2C + +CHCl2CCC·· +·
HCl ClCH3 HCl CH2C\2 HCl CHCl3 HC CCl4
H
A l l hydrogens at the starred positions are equi valent and allylic. The from the lower ri ght carbon has been removed to make an intermedi ate w i th two non-equivalent resonance forms, giving the two products shown . Drawing the resonance fonns is the
CH2Cl2.
This can
•
•
l H
l
l
•
+
•
l C l3 Cl
(b) To maxi mize H l and minimize formation of polychloromethanes, the ratio of methane to chlorine must be kept high (see problem 4-2). To guarantee that all hydrogens are replaced with chlorine to produce CCI4, the ratio of chlorine to methane must be kept high.
73
4-48 (a) n-Pentane c an produce three monochloro isomers. To c alculate the relative amount of each in the product mixture, multiply the numbers of hydrogens whic h could lead to that product times the reactivity for that type of hydrogen . Each relative amount divided by the sum of all the amounts will provide the percent of each in the product mixture.
CI
CI CI- CH2CH2CH2CH2CH3 ( 6 10 H) x (reactivity 1 .0) 6.0 relative amount
=
I
CH3-CHCH2CH2CH3 ( 4 20 H) x (reactivity 4 . 5 ) 1 8 .0 relative amount
=
total amount
(b)
6.0 x 1 00
3 3 .0
=
1 8%
i�:�
x 1 00
=
=
I
CH3CH2 - C HCH2CH3 ( 2 20 H) x (reactivity 4 . 5 ) 9.0 relative amount
=
3 3 .0 55%
9 .0 3 3 .0
x 1 00
=
27%
4-49 (a) The second propagation step in the chlorination of methane is highly exothermic (i1Ho = 1 09 kJ/mole ( - 26 kc al/mole». The transition state resembles the reactants, that is, the CI-CI bond will be slightly stretched and the CI-CH3 bond will j ust be starting to form. 8· 8· CI - - - - - - C I - - - - - - - - - - - - - - - CH 3 weaker stronger -
(b) The second propagation step in the bromination of methane is highly exothermic (i1Ho = - 1 0 1 kJ/mo\e ( - 24 kcal/mole». The transition state resembles the reactants, that is, the B r-Br bond will be slightl y stretched and the B r-CH3 bond will j ust be starting to form. Br - - - - - - Br stronger
- - - - - - - - - •..
weaker
- CH3
74
4-50
1 1) 1<
Two mech anisms are possib l e depending on whether HO' reacts with c h l orine or with cyc!opentane.
Mech an ism
initi ation
(2)
HO-OH HO' + Cl -Cl
f)
Cl· +
(4)
propagation
f(2))
Mechanism 2 initi ation
propagation
C l -Cl
:>0 '0
H
C l -Cl +
210 30 (24
Cl
2
'0
+
>O
H ----
H-OH H
---
CI
+
>O
50 2
+
'O
+ C l·
H
H-Cl
----
Cl'
+
HO'
H
:>0
Cl' +
----
---
:>0 '0
Cl'
+
H H-Cl
----
H
+
HO' +
f)
HO'
HO-Cl
----
HO-OH
(4)
2
---
D
2,
(2()2)
1
The energies of the propagation steps determine which mech anism is fol l owed . The bond dissociation energy of HO-Cl is about kJ/mole (about kcallmole), making initiation step in mechanism is exothennic by endothennic by about kJ/mo l e (about 8 kcallmole) . In mechanism i n i ti ation step about kJ/mole kcallmol e ) ; mechanism is preferred.
101
4-51
�� Cl -C-H I
4-52
H
1400
+
:O-H
U·
---
H2O
+
�l_ C Cl -C: I
c=�>
1; 0 1400
fj, G
Assuming fj, H is about the same at
=
I
. .
+ :Cl :
H
a c arbene
K, the equi l i brium constant is
=
'c:
H
This critical equation is the key to this problem: 11 G At
Cl ---
1400
-137 kJ/mole
K
=
therefore:
=
-137,1400000
11 H - T 11 S
>
K as i t is at calorimeter temperature:
=
JIK-mole
-98 J/K-mole (-23 callK-mole)
This is a l arge decrease in entropy, consistent with two molec ules combi ning into one. 75
4-53 Assume that chlorine atoms (radic als) are sti l l generated in the initi ati on reaction . Focus on the propagation steps. B ond dissoci ation energies are gi ven below the bonds, in kJ/mol e (kcal/mole). Cl o
+ H-CH3 435 ( 1 04)
CI - C1 242 ( 5 8 )
+
-
o CH3
H-CI + 0 CH3 43 1 ( 1 03) C I - CH3 + Cl o 3 5 1 (84)
-
+ 4 kJ/mole ( + 1 kcal/mole)
!1H
=
!1H
= -
1 09 kJ/mole ( - 26 kcal/mole)
What happens when the different radical species react with iodi ne? Cl o
+ I-I 1 5 1 (36)
+ I-I 1 5 1 (36)
o CH3
-
I - CI 2 1 1 (50)
01
+
1 - CH3 + I 0 234 (56)
-
!1H
=
- 60 kJ/mole ( - 14 kcal/mole)
!1H
=
- 8 3 kJ/mole ( - 20 kcal/mole)
Compare the second reaction in each pair: methyl radical reacting with c h lorine is more exothermic than methyl radical reacting with i odine ; this does not explain how iodine prevents the chlorination reaction . Compare the first reaction in eac h pair: chlorine atom reacti n g with i odine is very exothermic whereas c h l orine atom reacting with methane is slightly endothermic . Here is the key: chlorine atoms w i l l be scavenged by i odine before they h ave a chance to react with methane. Without chlorine atoms, the reaction comes to a dead stop.
4-54 (a) initiation
fl
nn
hv
B r- B r
f' � "
(2)
Br 0 +
(3)
�
propagation
H
2 Br 0
-
H-SnB u3
� B'
H-Br
-----
O'
o SnBu3 H
+ 1 . snBU j
O· (j-
-
H
(4)
+
+
B r - SnBu3
H
--.. ":":\ +
H-SnB u 3
-
n
H
+
o
SnBu3
(b) A l l energies are in kJ/mole. The abbreviation "cy" stands for the cyclohexane ring . Step 2 : break H-S n , make H-Br: +3 1 0 + - 368
=
- 5 8 kJ/mol e
Step 3 : break cy-Br, make B r-Sn: +28 5 + - 5 5 2
=
- 2 6 7 kJ/mol e WOW !
S tep 4 : break H-S n , make c y-H: + 3 1 0 + - 397
=
- 87 kJ/mole
The sum of the two propagation steps are : - 267 + - 87
76
=
- 354 kJ/mole -a h ugely exothermic reaction .
4-55 Mechanism 1 : (a) CI
°
03 ---
+
(b) 2 CIO
°
---
CIO
+
C I - O - O-Cl
hv --- 02
(c) CI - O - O - Cl
°
2 CI -
+
The biggest problem i n Mechanism 1 lies in step (b). The concentration of CI atoms is very smal l, so at any given time, the c oncentrati on of C I O wi l l be very smal l . The probabi lity o f two C I O radicals finding each other to form CIOOCI i s virtual l y zero. Even though thi s mechani sm shows a catalytic cycle with Clo (starting the mechanism and being regenerated at the end), the middle step makes it highly unl i kely.
Mechanism 2 :
�
(d) 03 (e) CI
03
+
°
(f) CIO
02
°
°
+
°
+
CIO
---
---
02
°
+
+
02 Cl o
S tep (d) i s the " light" reaction that occurs naturally in dayl i ght. At night, the reacti on reverses and regenerates ozone.
Step (f) i s the crucial step . A l ow concentration of CIO will find a rel ati vely high concentration of ° atoms because the " l i ght" reaction i s producing ° atoms in relative abundance. Clo i s regenerated and begins propagation step (e), continuing the c atalytic cycle.
Mechanism 2 i s beli eved to be the dominant mechani sm in ozone depletion . Mechanism 1 can be discounted because of the l ow probability of step (b) occurring, because two species i n very low, catalytic concentrati on are required to fin d each other in order for the step to occ ur. 4-56 (a)
[
H_
�
S.
I_
_ __ - H -
_ _ _ _ _
�:
---
r
In each c ase, the bond from carbon to H (D) i s breaking and the bond from H (D) to CI is forming.
\ C2Hs-C I + DCI ) + \C2H4DCI + H C I ;
�
7 o
1
9 %
D repl acement: 7 % 7 1 D = 7 (reacti vity factor) H repl acement : 9 3 % 7 5 H = 1 8 .6 (re activity factor) relative reacti vity of H : D abstraction = 1 8 .6 7 7 = 2 . 7 Each hydrogen i s abstracted 2 . 7 times faster than deuterium.
(c) In both reactions of chlorine wi th ei ther meth ane or ethane, the first propagation step i s rate-limiting. The reaction of chlorine atom with methane is endothennic by 4 kJ/mole ( 1 kcal/mole), while for ethane thi s step is exothennic by 2 1 kJ/mole (5 kcallmole). By the Hammond Postulate, differences in activation energy are most pronounced in endothennic reactions where the transition states most resemble the products. Therefore, a c hange in the methane molecule causes a greater c h ange in its transition state energy than the same change in the ethane molecule causes in its transition state energy . Deuteri um wi l l be abstracted more slowly in both meth ane and ethane, but the rate effect wi l l be more pronounced in methane than in ethane . 77
C H A PTER 5-STEREOCHEMISTRY
Note to the student: S tereochemi stry is the study of molecular structure and reactions i n three dimension s . Mol ecu l ar models wi l l b e especially hel pful in t h i s chapter. The best test of whether a h o u se h o l d object is chiral is whether it would be used equally wel l by a left or ri ght-handed person . The c h i ral obj ects are the corkscrew, the writing des k , the screw-c ap bottle (only for refi l ling, however; i n use, it would not be chiral ) , the rifle and the knotted rope; the corkscrew, the bottle top, and the rope each have a twi st in one direction, and the ri fle and desk are c learly made for ri ght handed users. A l l the other objects are achiral and would feel equi valent to ri ght- or left-handed users . 5-1
5-2 (a) cis achiralidentical mirror i mages
(b) trans H3
�
H
mirror chiral enantiomers
CH3
mirror (c) cis first, then trans
achiralidentical mirror i mages
mirror
achiralidentical mirror i mages
mIrror CH3
chiralenan ti omers
I
(d)
" ,C H " � '-... CH2CH3 Br mIrror
79
5-2 continued (e) chiral enantiomers
a
a
mirror
(f)
\J(j 0" '"
°
'"
chiral enantiomers
mirror 5 - 3 Asymmetric c arbon atoms are starred. Br
I
(a)
(b)
OH
I
no asymmetric carbons s ame structure
OH
I
no asymmetric c arbons same structure
(c) enanti omers
(d)
(e)
OH
I
CI
OH
I
H
6
no asymmetri c c arbons same structure
no asymmetri c c arbon same structure
80
5-3 continued
COOH
(f)
*
/
*
"' H
~ *
CI
~ *
H
CI
CI
(i )
~ *
same structure *
CI
CI
� *
l *
Q " '"
*
H H
" /
C == C
"
H
CH3
enantiomers
H
CI
CI
jL7i � CI
U)
enantiomers
*
" ,C H " � " CH3 H2 N
NH2
CI
(h)
I
JC ,"
H3C
(g)
COOH
Q
H
' ,..
H3C
/
C == C
no asymmetric c arbonssame structure
CI
/
"
H
enantiomers
H
(k) CH3
enantiomers
H3C
5-4 You may have chosen to in erchange two groups different from the ones shown here . The type of isomer produced w i l l sti l l be the same as listed here . Interchanging an y t w o groups around a c hirality center wi l l create a n enantiomer of t h e first structure. Interc h anging the B r and the H creates an enantiomer of the structure in Fi gure 5 - 5 .
Interchanging the ethyl and the isopropyl creates an en anti orner of the structure i n Figure 5 - 5 .
O n a double bond, interc hanging the two groups o n O NE o f the stereocenters w i l l create the other geometric (cis-trans) i somer. However, interc h anging the two groups on B OTH of the stereocenters w i l l gi v e the origi nal structure. H " ", CH3 C original interchange H and structure II CH3 on top stereocenter is cis to produce trans '" C ....... H CH 3 81
5-5 (a)
(b)
H
""
" ,
H
(f)
CHO 1
CH2 Cl 1
(d)
H'
H" " C, J CH3 Cl
H
Br
Br
chiral-no plane of symmetry
chiral-no plane of symmetry
plane o f symmetry
plane of symmetry
(e)
m Br
(c)
H " "JC ' CH OH 2 HO
COOH 1
(h)
p l ane of symmetry
I H" I C, J
H 2N
chiral-no pl ane of symmetry
CH3
chiral-no pl ane of s ymmetry
thi s view is from the right side of the structure as drawn in the text
plane of symmetry
5-6 AL WAYS place the 4th priority group away from you. Then determine if the sequence 1 �2�3 i s c l ockwise (R) o r counter-clockwi se (S). (There i s a Problem-S olving Hint near the e n d of section 5-3 in the text that descri bes what to do when the 4th priority group is c l osest to you . ) rotate CH3 u p
(a)
(b)
R
only one asymmetric carbon
3 CH 3
1
*1
C
S
2
(c)
Br � :: � CH 2 C
H 4
(d)
S
(g)
SU2
- *:
H
: *
Cl
....-
� 1
3h
1 *C 3 C == C ..... -: � CH 3 / I H H H
H
"
R
4
(f)
S
-
-
Cl
Cl
(i) S
(h)
2
R
3
H 82
4
O� H C"
� �� I
( H3COhHC
3 /- CH2 '/ C -. H CH(CH3h
" '"
4
I,
see next page for an explanation of part (i)
5 -6 continued
� - CH3
Part (i) deserves some explan ation . The di fference between groups 1 and 2 h i n ge on what is on the "extra" oxygen .
CH(OCH3h
I
�
O�
H - C - O - CH3
I
t
C
H ,/
�
higher priority
*
s
OH
OH
1
./ C " , CH3CH 2 CH 2..... " H CH3 (f)
./JC " , CH 3 ./ �" H
NH 2
(g)
(h)
1
R
COOH
1*
'Y C
H" NH2'"
R
" CH3
I;I� � S
� I;I R *� * S
C;: l � * � S
� C;: l R �*
R *
S
*
" ,C " H "� CH2 CH 2 CH3 H3C
COOH
S
Cl
Cl
H
Cl
(j ) II
H
, /
R
0 /(
c == c
"-
H-C
I
_
lower priority
5 -7 There are no asymmetric c arbons in 5-3 (a), (b) , (d), (e) , or (i). (c)
o-c � I 0/
Cl
Cl
Cl
Q� '" "
H CH3
H
CH3
CH3
*
H
S
/
H
C == C " C 3 H
(k)
R 83
R
I m ag in a ry .
5-8 2.0 g 1 1 0 m L
=
0 . 20 g/mL ; 1 00 mm
+ 1 .74° (0.20) ( 1 ) 5-9 0 . 5 g I 10 mL
=
=
=
1 dm
+ 8 . 7 ° for (+ )-gl yceraldehyde
0 .05 g/mL ; 20 cm
- 5 .0° (0.05) (2)
=
-
=
2 dm
50° for (-)-epinephrine
5- 1 0 Measure using a solution of about one-fourth the concentration of the first. The value w i l l be either + 45° or 45°, which gives the s i gn of the rotation. -
5-1 1 Whether a sample i s dextrorotatory (abbrevi ated " ( +)" ) or levorotatory (abbrevi ated " ( -)") i s determined experi mental ly by a pol arimeter. Except for the molecule glyceraldehyde , there is no direct, universal correlati on between direction of optical rotation (( +) and (-)) and designation of confi guration (R and S). In other words, one dextrorotatory compound might have R configuration while a differen t dextrorotatory compound might h ave S configuration. (a) Yes , both of these are determined experimental l y : the (+) or (-) by the polari meter and the smell by the nose. (b) No, R or S cannot be determined by either the pol arimeter or the nose. (c) The drawings show that (+)-carvone from caraway has the S confi g uration and (-)-carvone from spearmint has the R confi guration.
o
o
(For fun , ask your instructor if you can smell the two enantiomers of carvone. S ome people are unable, presumably for genetic reasons , to di stinguish the fragrance of the two enantiomers . )
3
(+)-carvone (caraway seed)
(-)-carvone (spearmint) CH3
CH3
I
C" / " "" H CH3CH2 OH
+
I
./ C " , � " OH CH3CH2 ./ H
(R)-2-butanol (S)- 2-butanol one-third of mi xture two-thi rds of mi xture Chapter 6 will explain how these mixtures come about. For thi s problem, the S enantiomer accounts for 66 . 7 % of the 2-butanol i n the mi xture and the rest, 3 3 . 3 % , is the R en anti orner. Therefore, the excess of one enantiomer over the racemic mixture must be 3 3 . 3 % of the S, the enantiomeric exces s . (All of the R i s "canceled" b y an equal amount of the S, algebraic ally a s wel l a s in optical rotation . ) (R)-2-bromobutane
The optical rotation o f pure (S)-2-butanol i s + 1 3 . 5 ° . The optical rotation o f thi s mi xture i s : 3 3 . 3 % x ( + l 3 . 5 °) = + 4 . 5 ° 84
5- 1 3 The rotation of pure (+)-2-butanol i s + 1 3 . 5 ° observed rotation -----rotation of pure enantiomer
+ 0.45 ° x 1 00% + 1 3 .5 °
=
To calculate percentages of (+) and (-) :
-
(+) + ( ) (+)
(-)
=
=
( -)
3 .3 %
(+ )
(+)
(+) (-) 5-14
(a)
= =
=
= =
3 . 3 % optical purity 3 . 3 % e.e. = excess o f (+) over (-)
(two equations in two unknowns)
1 00%
2
.
=
1 03 . 3 %
- (+) ( 1 00% - (+)) 1 00 %
=
3 .3%
5 1 .6% (rounded) 48 .4%
Drawing Newman projections is the clearest way to determine symmetry of conformations.
,,-h H Br � CI H H
chiral optically acti ve
}
'' Br ',C I ,
(b)
H
Plane includes Br and CI
(c)
I
chiral optically acti ve
I
H
I
I
I
H
I
B r - C - C - Cl
(d) H
� k� C 2
H
B, l(:y �: 20 H H
Br
plane of symmetry-not optically acti ve despite the presence of two asymmetric c �rbons , ,
Br
*
C I CH2 - C - CI
H H no asymmetric carbons (e)
,,-h H B r � CI CI H
plane of symmetry containing B r-C-C-CI ; not optically active H
H
H
�CH2 ,,-h H H �CH20 B r H H H
no plane of symmetry optically active (other chair form is equi valent-no pl ane of symmetry)
Br
(f)
�
Br 1 H
�
- - "'
H 4
Br
plane of symmetry through C - l and C-4-not optically acti ve
no asymmetric c arbons Part (2) Predictions of optic al acti v i ty based on asymmetri c centers give the same answers as predictions based on the most symmetric conformation .
85
5- 1 5 (a)
H
'-
"
CI
/
" H
(b)
\,\
C == C == C ' 'CI
no asymmetric carbons , but the molecule i s chiral (an allene ) ; the drawing below i s a three dimensional picture of the allene in (a) showing there is no plane of symmetry because the substituents of an al lene are in different planes
no asymmetric carbons , but the molecule i s chiral (an allene)
(d) H
"-
CI
(f)
(e)
/
C == C
n o asymmetric carbons ; thi s allene has a plane of symmetry between the two methyls (the plane of the paper), including all the other atoms because the two pi bonds of an allene are perpendicular, the Cl i s i n the plane of the paper and the pl ane of symmetry goes through it; not a chiral molecule
/
H
/
C == C
"
/
H
"-
H H planar molecule-no asymmetric c arbons; not a chiral molecule (g)
o
0
H
" '-j
�JB r
�/
p l ane o f symmetry bi sectin g t h e molecule; no asymmetric c arbons ; not a chiral molecule
Br no asymmetric carbons , but the molecule is chiral due to restricted rotation; the drawing below is a three dimensional picture showing that the rings are perpendicular (hydrogens are not shown)
(h)
two asymmetric c arbon s ; a chiral compound 86
H H no asymmetric carbons , and the groups are not l arge enough to restrict rotation ; not a chiral compound
5- 1 6 (a) H
+
OH
HO
+
H
CH3
+
H
Br
CH2CH3
s ame
(c)
+
HO
+ CH3
CH3
H
CH2CH3
(R)-2-butanol
H
CH3
+
enantiomer
enantiomer
+
+
+
CH3
1I
H
HO
CH2CH3
H
CH2CH3
CH2CH3
same
e n anti omer
Br
CH2CH3
CH3
OH
H
COOH
CH2CH3
CH3
+ same
enantiomer
enantiomer
Br
COOH HO
OH
CH3
CH3
+
Rules for Fischer projections:
CH3
H
COOH
COOH
H
OH
1 . Interchanging any two groups an odd number of times (once, three times, etc . ) makes a n enantiomer. Interchanging any two groups an even number of times (e.g. twice) returns to the original stereoi somer. 2. Rotating the structure by 90 0 makes the enantiomer. Rotating by 1 800 returns to the original stereoisomer. (The second rule is an application of the first. Prove this to yourself.)
CH3
same
5- 1 7 (a) HO
+
CH20H H
CH3
5- 1 8 1 80 0 rotation of the right structure does not give left structure ; n o plane of symmetry: chiral�nantiomers
(a)
CH20H
- - - - -Br
+
H f' on the left ; also has plane of symmetry: same structure
1 80 0 rotation of the ri ght structure gi ves same structure as
CH20H
(c )
Br +�� , Br
plane of symmetcy same structure
(:H 3 ,
87
mirror
5- 1 8 continued
CHO
CHO
H
(d)
H
+ +
OH
HO
OH
HO
H H
Ho H
H
$
CH20H
i: i�
H n 1 80 0 rotation of the right structure gi ves same structure as HO - - - - - - - - - - - - - - - - - - - on the left; also has p l ane of s y mmetry : same structure H HO
OH
CH20H
CH20H
(f)
1 80 0 rotation of the right structure does not give left structure; no plane of symmetry : chiral--enantiomers
H
CH20H
CH20H
(e)
+ +
H HO
OH
+�:
+
1 80 0 rotation of the right structure does not gi ve left structure; no plane of symmetry : chiral--enantiomers
H
CH20H
CH20H
5 - 1 9 If the Fischer projection is drawn correctly, the most oxidi zed c arbon w i l l be at the top; thi s is the carbon with the greatest number of bonds to oxygen . Then the numbering goes from the top down. (d) 2R, 3R (e) 2 S, 3R (numbering down)
(a) R (b) no chiral center (c) no chiral center
(f) 2 R, 3R
(g) R (h) S (i) S
5 -20 (a) enantiomers--confi gurations at both asymmetri c carbons inverted (b) diastereomers--confi guration at only one asymmetric c arbon inverted (c) diastereomers--con fi guration at only one asymmetric c arbon inverted (the left c arbon) (d) constitutional i somers-C=C shifted position (e) enantiomers--c h iral , mirror i mages (f) diastereomers--configuration at only one asymmetri c carbon inverted (the top one) (g) enanti omers--confi guration at all asymmetric carbons inverted (h) same compound-superimposable mirror images (hard question ! use a mode l ) (i) diastereomers--c onfiguration a t o n l y one chirali ty center (the nitrogen) inverted 5 -2 1 (a)
' --
'
:;
u
l �: CH 3 "
CI '
U
plane of s y mmetry
CH3 meso structure not optical l y active
, l -f , H : B *
�
r
, '
,
Cl
*
H3C
CH3
plane of symmetry 88
5-2 1 continued
CH3
Br
(b)
Cl
+ +
Cl
Cl
Br
B,
CH3
+
Br
Cl B
lrCl
C�
-
+:
*
C
�
H3C
CH3
-
f I Br
H
H
B
�
*
*
H3C
CH3
=*
CH3
50: 5 0 racemic mixture-not opti c al l y acti ve, although each enanti omer by itself would be optically acti ve :
H
(c)
H
(
� planes of
=t= tH3
,
H
i HH 1 i f '---LJ/ / : \ H3C
symmetry �
H
CC H3
:
not chiral not opti c al l y acti ve
(d)
OH
(a)
H
+
H
CH3 A
(f)
�/
H
HOH2 C
*
(g)
+
Cl
Cl
COOH
CH3
+ +
B
C
H
H
H
Cl
CH3
enantiomers : A and B ; C and D diastereomers : A and C ; A and D ; B and C; B a nd D
89
OH H
*
CH3 1 CH3
not optically active superimposable on its mirror image technical l y , thi s i s a meso structure (may require models ! )
COOH
HO
CH 2 0H
*
CH3
OH
'\
plane of s ymmetry
*
HO
COOH
H? H
�
H
CH3
optical l y acti ve
5-22
*
B H �
optically active
H0 -T- H
CH 2 0H
Br
*
CH 2 0H
H � OH H � OH HO H � OH
HO
CH3
CHO
(e)
t
CH 2 0 H
H
mesonot optically active
COOH
HO H
+ + CH3 D
H Cl
5-22 continued
(b)
-�
-$�:-- =;��rY
H HO
+ +
HO
OH
H
H
enantiomers: F and G diastereomers : E and F; E and G
+ +
H OH
COOH
COOH
COOH E
G
F
COOH
COOH
COOH
H ---If-- OH
HO -I-- H
HO --ll-- H
OH
HO -I-- H
H ---If-- OH
HO ---lf-- H
H --I- Br
B r --lf-- H
H --II-- Br
B r --ll-- H
COOH H
COOH
COOH
COOH
I
J
K
COOH
COOH
COOH
COOH
COOH (c)
COOH
COOH
COOH
H
HO -f-- H
H --If-- OH
HO
H
H
H --I- Br
OH
H ---lf-- OH
H --I- OH HO --ll-- H
Br --lf-- H
HO r--I-- H H --II-- OH
H --I- Br
Br --lf-- H
COOH COOH COOH M N L enantiomers: H and I ; J and K ; L and M; N and 0 di astereomers : any pair which is not enantiomeric
COOH o
cm
(d)
p
Q
enantiomers : P and Q diastereomers : P and T; P and U; Q and T; Q and U; T and U
rti
H
'
H3
H
plane of symmetry MESO
plane of symmetry MESO
T
U
90
5-23 Any diastereomeric pair could be separated by a physical process l i ke disti l l ation or crystal lization . Di astereomers are found in p arts (a) , (b) , and (d) . The structures in (c) are enantiomers ; they could not be separated by normal physical means . 5 -24
0 H HO (
o
+CH3��
r enantiomers
'
�
H H �' 0 CH3 HO H H OH COOH
+
OH H COOH
(S)-2-butyl (R,R)-tartrate
mirror
(R)-2-butyl (S,S)-tartrate
CH2CH3 H+O CH3 H--If--- O H HO--l�H COOH
di astereomers
\...
�
di astereomers
o
o
HO--lf--- H H 0H COOH ----i...--
(R)-2-butyl (R,R)-tartrate
mirror
(S)-2-butyl (S,S)-tartrate
5 -2 5 Please refer t o solution 1 -20, page 1 2 of t h i s Solutions Manual . 5-26
H CH20H I* H+OH ,C",-" " C / CH3 CH3 HO
(a)
(b)
l
''
......
B j H :,
R chiral
(e)
S
H
- - - ... - - -
R
- - - - - _ .. .
* Br
/C H 2 B r
meso ; achiral
R c h i ral
(f) plane of symmetry
iB:,H H/i OH
sH
(g)
s
/*
B S
CH2
Br
R
chiral
91
B'
*
CH3
chiral
(h) S
CH3
H'" * OH
s�
H
R
*
OH
OH /CH2CH3 *
chiral
5 -26 continued (i) Br
"
(
"
C=C=C' �
(j )
Br
CI C chiral molecule, but no chiral centers
(Y *
I
(k)
Br
"- s
o-
Bf
achiral
chiral
plane of symmetry
(m)
(I)
pl ane of symmetry meso; achiral C n)
chiral
(0)
R
s
R
(NH2 i s group 1 , CH3 i s group 4)
chiral
chiral
c hiral 5 -27 (a)
<;:H3
en anti orner
* =
H - C - CI
c::====�>
CH2CH3 no plane of symmetry no diastereomer chiral structure
<;:H3
en anti orner
* =
CI - C - H
>
CH2CH3
chiral structure no plane of symmetry no diastereomer chiral structure chiral structure
(c)
* -
Br - C - H
*I
Br - C - H CH2CH2CH3
en anti orner
>
<;:H3 = H - C - Br *
*
I
H - C - Br CH2CHzCH3 chiral structure
*
-
H - C - Br *
I
Br - C - H CH2CH2CH3
no plane of symmetry chiral structure chiral structure c::====�> (inverting two groups on the diastereomer bottom asymmetric c arbon i nstead of the top one would also gi ve a diastereomer) 92
continued (d) Br
Br
enanti omer H chistructralure H not chistrrauctl ; ure H di a st e reomer (i n ver t i n g t w o gr o ups on t h e nochiplralanestructof symmet r y r i g ht asymmet r i c car b on i n st ure ofdiatstheereomer left one) would also giveeada (invertitom nasymmet g two groups onbonthe HO-C-H di a st e reomer a e pl of n bot r i c car � one woul) d H-C-OH ialnsstoegiadvofe athdieatstopereomer H .: C OH symmetry CH1CH3 CH1CH3 chiral structure noa mesoenantstioructmerure, not chiml racemic mixture of enantiomers; each is chiml with no plane of symmetry (f) ( H2CH3 � H2CH� HO-C-H H-C-OH H-C-OH HO-C-H �stereomer H2CH3 diastereCH2CH3 omer H-C-OH symmet pane H .:. C OH this mesorystructure is a diastereomer of each of the CH2CH enantiomers; it is not chiml (d� 1I (a) CH70H (b) CHO O +�{ 11 + 011 H + Br H + OH CH3 CH3 CH3
5-27
c::=====�>
>
* -
� ':': � ��
_
>
_ _ _ _
_
_
-------------
*
*
1
------------
�
�
*
* -
*
1
*
+
�
� : �
�
- - - - -
* :: �
t
1
- - - - - - - - -
I
0
_
3
5 -2 8
93
f
m eso
5-29 Your drawings may look different from these and sti l l be correct. Check configuration by assigning R and S to be sure . (a) CH3
COOH
�
CHO
(b)
"H NH2
HOCH 2
�
(d)
" OH H
5-30 (a) s ame (meso)-pl ane of symmetry , superimposable (b) enantiomers--configuration inverted at both asymmetric carbon atoms (c) enanti omers--configuration i nverted at both asymmetric carbon atoms (d) en anti omers-solve this problem by switching two groups at a time to put the groups in the same positions as in the first structure; it takes three switches to make the i dentical compound, so they are enantiomers; an even n umber of switches would prove they are the same structure (e) enanti omers--confi guration inverted at both asymmetric carbon atoms (f) diastereomers--configuration i n verted at onl y one asymmetric carbon (g) enantiomers--configuration inverted at both asymmetric carbon atoms (h) same compound-rotate the right structure 1 80° around a hori zontal axi s and it becomes the left structure 5-3 1 Drawing the enantiomer of a chiral structure i s as easy as drawing i ts mirror i mage. (b) Br
+
CHO (c)
CHO H
CH2 0 H
(e)
HO
H
HO
H
HO
H
(d)
CH3
,H "" C == C == C ' ' / Br "
H
o
(f) plane of symmetryno en anti orner 5-32 (a) 1 .00 g / 20.0 mL
=
0.050 g/mL ;
(0.0500) ( 2.00) (b)
0.050 g / 2.0 mL
=
=
20.0 c m
2 .00 dm
=
- 1 2.5°
0 . 0 2 5 g/mL ;
2.0 cm
=
0 . 20 dm
+ 0.043° (0.025) (0.20) 5 - 3 3 The 32% of the mixture that is (-)-tartaric aci d wi ll cancel the optical rotation of the 32% of the m i x ture that i s (+)-tartaric acid, leavi ng only (68 - 32) = 36% of the mi xture as excess (+)-tartari c aci d to gi ve meas urabl e optical rotation . The specific rotation w i l l therefore be only 36% of the rotation of pure (+)-tartaric aci d : (+ 1 2 . 0°) x 3 6 % = + 4 . 3 ° 94
5-34
(b) Rotation of the enantiomer will be equal in magnitude, opposite in sign:
(c) The rotation
- 7 95° --' - 15.90°
x
-7.95° 100%
is what percent of =
50% e.e.
-
- 15.90°.
15.900?
There is 50% excess of (R)-2-iodobutane over the racemic mixture; that is, another 25% must be R and 25% must be S. The total composition is 75% (R)-(-)-2-iodobutane and 25% (S)-(+)-2-iodobutane. 5-35 All structures in this problem are chiral.
C al H H
+
OH OH CH20H
+ A
CHO HO H H HO CH20H
+ + B
CHO H OH HO H CH20H
+ + C
CHO H HO OH H CH20H
+ + D
enantiomers: A and B; C and D diastereomers: A and C; A and D; Band C; Band
D
(b)
E
F
G
enantiomers: E and F; G and H diastereomers: E and G; E and H; F and G; F and H
95
H
5-35 continued (c) This structure is a challenge to visualize. A model helps. One way to approach this problem is to assign R and 5 configurations. Each arrow shows a change at one asymmetric carbon. CH3 H3C CH3 H CH3 H H \ S H H � L H H
�
H3C H
.0.
H
H
�
.0.
CH3 H
H
H
H CH3 H
H CH3 H3C
H
H3C
H3C
H
Summary R5S5 (K) is the enantiomer of RRR5 (M) 5555 (L) is the enantiomer of RRRR (P) The structures with two R carbons and two 5 carbons (J, N, and 0) have special symmetry. J is a meso structure; it has chirality centers and is superimposable on its mirror image-see Problem 5-20(h). Nand 0 are enantiomers, and are diastereomers o f all of the other structures. Give yourself a gold star if you got this correct!
5-36
s� .Ix. �� S*
A
meso
diastereomers
15,3R equivalent to 1R,35
B
meso
C
R*
chiral 1R,3R
15,3R equivalent to 1R,35
enantiomers
D chiral 15,35
C and Dare enantiomers. All other pairs are diastereomers.
*Structures A and B are both meso structures, but they are clearly different from each other. How can they be distinguished? One of the advanced rules of the Cahn-Ingold-Prelog system says: When two groups attached to an asymmetric carbon differ only in their absolute configuration, then the neighboring (R) stereocenter takes priority. Now the configuration of the central asymmetric carbon can be assigned: 5 for structure A, and R for structure B. This rule also applies to problem 5-37. (Thanks to Dr. Kantorowski for this explanation.)
96
5-37 This problem is similar to 5-36. (a)
*
H
2
H
3
5 *
H
4
R
s CH3
H
Br
Br
Br
H
*
5 *
R
Br
Br
H
H
Br
H
CH3
� '-diastereomers
*
H
H
Br
Br
Br
Br
*
CH3
B
C
meso
meso
chiral
25,4R equivalent to 2R,45
2 5,4R equivalent to 2R,45
2R,4R
A
(b)
Br
CH3
CH3
CH3
[CH3
'--enantiomers�
*
Br H
*
H
CH3 D chiral 2 5,45
The configuration of carbon-3 in A and B can be assigned according the rule described above in the solution to problem 5-36: C-3 in A is 5, and C-3 in B is R. (c) According to the IUPAC designation described in text Section 5-2B, a chirality center is "any atom holding a set of ligands in a spatial arrangement which is not superimposable on its mirror image." An asymmetric carbon must have four different groups on it, but in A and B, C-3 has two groups that are identical (except for their stereochemistry). C-3 holds its groups in a spatial arrangement that is superimposable on its mirror image, so it is not a chirality center. But it is stereogenic: in structure A, interchanging the Hand Br at C-3 gives structure B, a diastereomer of A; therefore, C-3 is stereogenic. (d) In structure C or D, C-3 is not stereogenic. Inverting the H and the Br, then rotating the structure 180°, shows that the same structure is formed. Therefore, interchanging two atoms at C-3 does not give a stereoisomer, so C-3 does not fit the definition of a stereogenic center. 5-38 The Cahn-Ingold-Prelog priorities o f the groups are the circled numbers in (a). (a) H CH3 H CH3 H " H2 .. H C CH CH H C " .......:: ' ' Pt CH 3 C '/" "c/ CH3 "c/ "c/ /" ' H / '' H' CH3 H CH3 H H
10 (2)1
I
'
'
lev 0.1
\
o 0
(R)-3,4-dimethyl-l-pentene
I
\
0 0
(5)-2,3-dimethylpentane
(b) The reaction did not occur at the asymmetric carbon atom, so the configuration has not changed the reaction went with retention o f configuration at the asymmetric carbon. (c) The name changed because the priority o f groups in the Cahn-Ingold-Prelog system of nomenclature changed. When the alkene became an ethyl group, its priority changed from the highest priority group to priority 2. (We will revisit this anomaly in problem 6-21(c).) (d) There is no general correlation between R and 5 designation and the physical property of optical rotation. Professor Wade's poetic couplet makes an important point: do not confuse an object and its properties with the name for that object. (Scholars of Shakespeare have come to believe that this quote from Juliet is a veiled reference to designation of R,5 configuration versus optical rotation of a chiral molecule. Shakespeare was way ahead of his time.) 97
5-39 (a) The product has no asymmetric carbon atoms but it has three stereocenters: the carbon with the OH, plus both carbons o f the double bond. Interchange of two bonds on any o f these makes the enantiomer. (b) The product is an example of a chiral compound with no asymmetric carbons. Like the allenes, it is classified as an "extended tetrahedron"; that is, it has four groups that extend from the rigid molecule in four different directions. ( A model will help.) In this structure, the plane containing the COOH and carbons of the double bond is perpendicular to the plane bisecting the OH and H and carbon that they are on. Since the compound is chiral, it is capable of being optically active. COOH H HO H (c) As shown in text Figure 5-16, Section 5-6, catalytic hydrogenation that creates a new chirality center creates a racemic mixture (both enantiomers in a 1: 1 ratio). A racemic mixture is not optically active. In contrast, by using a chiral enzyme to reduce the ketone to the alcohol (as in part (b)), an excess of one enantiomer was produced, so the product was optically active.
98
In
CHAPTER 6-ALKYL HALIDES: NUCLEOPHILIC SUBSTITUTION AND ELIMINATION
6-1
problems like part (a), draw out the whole structure to detect double bonds. (c) aryl halide (d) alkyl halide
(a) vinyl halide (b) alkyl halide 6-2 (a)
,, H H
(b)
,,
Br
J-C-I
Br -C- Br
�
Br
(f)
(e) vinyl halide (f) aryl halide
� I-C-H ,, ~ FX '-rH2F + (d)
(Cl
(e)
I
Br
Br
(g)
H
(hl
Cl
Br
or (CH3hCCI
6-3 IUPAC name; common name; degree of halogen-bearing carbon l-chloro-2-methylpropane; isobutyl chloride; 1° diiodomethane; methylene iodide; methyl 1,I-dichloroethane; no common name; 1° 2-bromo-l,1,l-trichloroethane; no common name; all 1° trichloromethane; chloroform; methyl (f) 2-bromo-2-methylpropane; t-butyl bromide; 3° (g) 2-bromobutane; sec-butyl bromide; 2° (h) l-chloro-2-methylbutane; no common name; 1° (i) cis-l-bromo-2-chlorocyclobutane; no common name; both 2° U) 3-bromo-4-methylhexane; no common name; 2° (k) 4-fluoro-l,1-dimethylcyclohexane; no common name; 2° (I) trans-l,3-dichlorocyclopentane; no common name; all 2° (a) (b) (c) (d) (e)
6-4
Cl
CI
Cl
CI
CI Cl
Cl
> Cl
0 Kepone®
6-5 From the text, Section 2-9B, the bond dipole moment depends not only on bond length but also on char ge separation, which in tum depends on the difference in electronegativities of the two atoms connected by the bond. Because chlorine's electronegativity (3.2) is si gnificantly higher than iodine's (2.7), the C-Cl bond dipole is greater than that of C-I, despite C-I being a longer bond. 6-6 (a) n-Butyl bromide has a hi gher molecular weight and less branching, and boils at a higher temperature than isopropyl bromide. (b) t-Butyl bromide has a higher molecular weight and a larger halogen, and despite its greater branching, boils at a higher temperature than isopropyl chloride. (c) n-Butyl bromide has a hi gher molecular weight and a larger halogen, and boils at a higher temperature than n-butyl chloride. 6-7 From Table 3-2, the density of hexane is 0.66; it will float on the water layer (d 1.00). From Table 6-2, the density of chloroform is 1.50; water will float on the chloroform. Water is i mmiscible with many organic compounds; whether water is the top layer or bottom layer depends on whether the other material is more dense or less dense than water. (This is an i mportant consideration to remember in lab procedures.) 6-8 (a) Step (1) is initiation; steps (2) and (3) are propagation. (1)
-
nn
Br
(2) H2C (3) H2C
Br
--
2 Br·
hv
= � Cr\ - + Br � HBr + { = � - C n Br - Br � = HC - . � \ = HC - I + CH2
CH2
H2C
•
H2C
+
..
H2 "
CH2
Hi - �
=
CH2
}
Br·
Br
- (+ 368) - 4 kJ/mole kcallmole: + 87 - (+ 88) - 1 kcallmole Step (3): break Br-Br, make allylic C-Br: kJ/mole: + 192 - (+ 280) - 88 kJ/mole kcallmole: + 46 - (+ 67) - 2 1 kcal/mole overall - 4 + - 88 - 92 kJ/mole (- 1 + - 2 1 - 22 kcal/mole ) This is a very exothermic reaction; it is reasonable to expect a s mall activation energy in step ( 1),
(b) Step (2): break allyJic C-H, make H-Br: kJ/mole: + 364
=
=
=
=
M!
=
=
=
so this reaction should be very rapid. 6-9
(a) propagation steps H3C
,
H3C
/
C��)
C=C
/CH2
,
+ Bre
CH3
repeats chain mechanism
100
Br· +
+
6-9
continued
The resonance-stabilized allylic radical intermediate has radical character on both the 1 ° and 3° carbons, so bromine can bond to either o f these carbons producing two isomeric products. (b) Allylic bromination of cyclohexene gives 3-bromocyclohex-l-ene regardless of whether there is an
allylic shift. Either pathway leads to the same product. If one of the ring carbons were somehow
marked or labeled, then the two products can be distinguished. (We will see in following chapters how labeling is done experimentally.)
o 6- 10 (a)
o-
N BS
Br
+
y
the second structure from an allylic shift is identical to the first structure-only one compound is produced
Br CH3
CH3
I
I
CH3-C-CH3 I
CH3
CH3-C-CH2CI I
hv
CH3
This compound has only one type of hydrogen-only one monochlorine isomer can be produced. CH3
CH3
CH3- � -H
CH3- �- Br I
I
(b)
CH2CH3
hv
CH2CH3
Bromination has a strong preference for abstracting hydrogens (like 3°) that give stable radical intermediates.
hv (or NBS) Bromine atom will abstract the hydrogen giving the most stable radical; in this case, the radical intermediate will be stabilized by resonance with the benzene ring.
(d)
CO
Br
hv (or N BS)
OJ
the second structure from an allylic shift is identical to the first structure-only one compound is produced
+
Br
Bromine atom will abstract the hydrogen giving the most stable radical; in this case, the radical intermediate will be stabilized by resonance with the benzene ring.
6- 1 1 (a) substitution-Br is replaced (b) elimination-H and OH are lost (c) elimination-both Br atoms are lost 6- 12 (a) CH3 (CH2)4CH2 - OCH2CH3
(b)
CH3 (CH2)4CH2 - eN 101
6- 13 The rate law is first order in both 1-bromobutane, C4H9Br, and methoxide ion. If the concentration of C4H9Br is lowered to one-fifth the original value, the rate must decrease to one-fifth; if the concentration of methoxide is doubled, the rate must also double. Thus, the rate must decrease to 0.02 molelL per second: rate
=
( 0.05 molelL per second ) x original rate
( 0. 1 M ) ( 0.5 M )
x
( 2.0 M ) ( l.0 M )
=
0.02 molelL per second new rate
change in NaOCH3
change in C4H9Br
A completely different way to answer this problem is to solve for the rate constant k, then put in new values for the concentrations. rate = k [C4H9Br] [NaOCH3] � 0.05 mole L-I sec -I = k (0.5 mol L-I) (l.0 mol L-I) �
rate constant k = 0.1 L mol-I sec-I rate = k [CH3IJ [NaOH] = (0. 1 L mol-I sec-I) (0.1 mol L-I) (2.0 mol L-I)
=
0.02 mole L-I sec -I
6- 14 Organic and inorganic products are shown here for completeness. +
(a) (CH3)3C - 0 - CH2CH3
KEr +
+
Br
+
NaI
(f)
+
-
(c) (CH3hCHCH2 - NH3
(e)
NaCI
�I
�F
+
-
NH4 Br
+ NaCI +
K CI
(lS-crown-6 is the catalyst and does not change; CH 3CN is the solvent)
6- 15 All reactions in this problem follow the same pattern; the only difference is the nucleophile (-:Nuc). Only the nucleophile is listed below. (Cations like Na+ or K+ accompany the nucleophile but are simply spectator ions and do not take part in the reaction; they are not shown here.)
� CI l-chlorobutane (a) HO-
(f) -O CH2CH3
(b)
F-
+
-:Nuc
----
� Nuc
from K F/lS-crown-6
(d)
+
CI-
-CN
(e) H C_C-
(g) excess NH3 (or - NH2)
6- 1 6 (a) ( CH3 CH2)2NH is a better nucleophile-Iess hindered (b) (CH3hS is a better nucleophile- S is larger, more polarizable than 0 (c) PH:, is a better nucleophile-P is larger, more polarizable than N (d) CH3 S- is a better nucleophile-anions are better than neutral atoms of the sa me element (e) ( CH3)3N is a better nucleophile-Iess electronegative than oxygen, better able to donate an electron pair (f) CH3S- is a better nucleophile-anions are generally better than neutral atoms, and S is larger and more polarizable than 0 (g) CH3CH2CH20- is a better nucleophile-Iess branching, less steric hindrance (h) 1- is a better nucleophile-Iarger, more polarizable 102
..
6- 17 A mechanism must show
electron movement.
H
H
3 + CH3-O-CH • •
H+
�
1",
+.. CH3-0-CH3 + :Br: • •
1
-
CH3-�: + CH3-!3[:
---
�
Protonation converts OCH3 to a good leaving group. 6- 18 The type of carbon with the halide, and relative leaving group ability o f the halide, determine the reactivity. methyl iodide> methyl chloride> ethyl chloride> isopropyl bromide» .
neopentyl bromide, t-buty I'IOd'd 1 e
most reactive
}
least. reaCllve
Predicting the relative order of neopentyl bro mide and t-butyl iodide would be difficult because both would be extremely slow. 6- 19 In all cases, the less hindered structure is the better SN2 substrate. (a) 2-methyl- 1-iodopropane ( 1° versus 3°) (b) cyclohexyl bromide (2° versus 3°) (c) isopropyl bromide (no substituent on neighboring carbon) (d) 2-chlorobutane (even though this is a 2° halide, it is easier to attack than the 1° neopentyl type in 2,2dimethyl-1-chlorobutane-see below) CH3 a neopentyl halide1 (e) isopropyl iodide (same reason as in (d)) hindered to backside attack I /c"" - C, � CH 3 by neighboring methyl groups
H , "\. C Lt13 H H "'--- -:Nuc
6-20 All SN2 reactions occur with inversion of configuration at carbon. (a)
8-
Br
--P< :
H
H
H08-
trans
H
CH,
transition state
HO:
(b)
: Br:
inversion
cis
-:CN
�
+
R
103
..
-
:Br:
6-20 continued
H
H .-
1
H
;···
H3C _
(d)
CH3CH2
c� I � , H" � CH3 F :SH Br
CH3 � :
(f)
S
H- �: I
H
�/ �
H3CO
.
�-
-
H
,Br
..
-
: Br:
c::::::>
H
H
CH3
+ +
I
CH}
CH2CH}
.
,
..
H
+
: Br:
• •
(e)
H
�I
� ,�
+
r. .B • •
, � R
c::::::>
H
��
� ID C;CI
C
-
+
:CJ :
---II"�
6-2 1 (a) The best leaving groups are the weakest bases. Bromide ion is so weak it is not considered at all basic; it is an excellent leaving group. Fluoride is moderately basic, by far the most basic of the halides. It is a terrible leaving group. Bromide is many orders of magnitude better than fluoride in leaving group ability. (b)
inverted, but still named S
H"� "'" : CH} . F
H
:
, I CH} �� I OCH3 I
+
..
: Br:
transition state (c) As noted on the structure above, the configuration is inverted even though the designations o f the configuration for both the starting material and the product are S; the oxygen of the product has a lower priority than the bromine it replaces. Refer to the solution to problem 5-38, p. 97, for the caution about con fusing absolute configuration with the designation of configuration. Cd) The result is perfectly consistent with the SN2 mechanis m. Even though both the reactant and the product have the S designation, the con figuration has been inverted: the nomenclature priority of fluorine changes from second (after bromine) in the reactant to first (before oxygen) in the product. While the designation may be misleading, the structure shows with certainty that an inversion has occurred. 104
6-22
.. -
�
:Br:
slow
planar carbocation
+ +
CH3CH20H2 Be (or si mply HBr)
6-23 The structure that can form the more stable carbocation will under go SN1 faster. (a) 2-bromopropane: will form a 2° carbocation (b) 2-bromo-2-methylbutane: will form a 3° carbocation (c) allyl bromide is faster than n-propyl bromide: allyl bromide can for m a resonance-stabilized intermediate. (d) 2-bromopropane: will form a 2° carbocation (e) 2-iodo-2-methylbutane is faster than t-butyl chloride (iodide is a better leaving group than chloride) (f) 2-bromo-2-methylbutane (3°) is faster than ethyl iodide (1°); although iodide is a somewhat better leaving group, the difference between 3° and 1° carbocation stability dominates
6-24 Ionization is the rate-determining step in SN l. Anything that stabilizes the intermediate will speed the reaction. Both of these compounds form resonance-stabilized intermediates.
(x H
I
benzylic +
6-25
CH3
-T
H I
T
H
�H VH
---
( )=
CH2 ---
allylic
O
CHz
+
H
t:..
- HCH3
H
H
•
Br-
+
+ I
CH3 - C - CHCH3
+
C
I
I
CH2CH3
---
H
- R-
�
H:1 H CH3 - � - CHCH3 ......f'CH3 - � - CHCH3 / .. ..---... ('I. :?: CH3 CH3CH2- 9: H-+? CH3
( Br
H
� < }- �
I
CH2CH3
105
CH2CH3
H2
6-26 I t is i mportant to analyze the structure of carbocations to consider if migration of any groups from
adjacent carbons w i l l lead to a more stable carbocation . As a general rule, if rearrangement would lead to a more stable carbocation , a carbocation w i l l rearrange. (Beginning with this problem, only those unshared electons pairs involved in a partic ular step w i l l be shown. )
-
CH3
CH3 I
(a) CH3-C
I
1\
CH3
y y I
1- +
CH-CH3
+
CH3- - -CH3 CH3 H
I�
nucleophilic attack on unrearranged carbocation
c��
CH3 :O-CH3
? ? !
I
CH3- - -CH3
. .
CH3 :O-CH3
HRCH3
� -----l..
I
I
I
I
CH3- C - C-CH3
CH3 H
CH3 H unrearranged product
nucleophilic attack after carbocation rearrangement
CH3 I
+
CH3 -C-C-CH 3
�
I�I
CH3 H Methyl shift to the
2° carbocation forms a more stable 3° carbocation.
cj:
nucleophilic attack on unrearranged carbocation
rl<�",--;: H
H
H
:R - CH,CH
CH2CH3
d=�:
I
;
C H3
CH 3
unrearranged product
106
6-26(b) continued nuc leophilic attack after carbocation rearrangement
H
C)) H
�� H :..b�CH2CH] rf-
hydride shift � ___
30
�
CH3
Hydride shift to the 2° carbocation forms a more stable 3° carbocation .
rearranged product
H
(c ) H
H
H
:(j
H H H H OH �O·=C-CH3 I H ... . 3
o-H
nucleophi lic attack on unrearranged carbocation
H
0I
unrearranged product
I
• •
:0 :
II
O-C-CH
III
• •
?-H
O=C-CH3 +
Note: braces are used to indicate the ONE chemical species represented by multiple resonance forms.
o-. H
.... ..I----J .. �
I
•
•
?
: -H
O-C-CH3 .
+
...
:0:
II
3
HO-C-CH •
•
The most basic species in a mixture i s the most likely to remove a proton . I n this reaction , acetic acid is more basic than iodide ion.
107
H H H H H Cj H �-C-CH3H C D D
H y)
6-2 6(c) continued nucleophilic attack after carbocation rearrangement
H H
H
I
H
+
H
H
hydride shift
..� ---....
H
HO : Cc;?: H -
H H
H
-
II
:0 :
O..
II
-
H+
CH 3
..... f--------
rearranged product
3
..
:I
O-
: 0:
I
allylic-resonance-stabilized
(removes as on p. 107)
H
H
H �==C-CH3
I
�
plus two other resonance forms as shown on p. 107
Comments on 6-26(c) (1) The hydride shift to a 20 carbocation generates an allylic, resonance-stabilized 20 carbocation. (2) The double-bonded oxygen o f acetic acid is more nucleophilic because of the resonance forms it can have after attack. ( See Solved Problem 1-5 and Problem 1- 16 in the text.) (3) Attack on only one carbon of the allylic carbocation is shown. In reality, both positive carbons would be attacked in equal amounts, but they would give the identical product in this case. In other compounds, however, attack on the different carbons might give different products. ALWAYS CONSIDER ALL POS SIBILITIES.
(d)
ct H
0 CH2-I
The 1 0 carbocation initially formed is very unstable; some chemists believe that rearrangement occurs at the same time as the leaving group leaves. At most, the 10 carbocation has a very short lifetime.
---l"�
r+CH2 N CH :�� CH2CH,; hydride shift followed by nucleophilic attack
'------' 3
V
---l"�
V
-
108
6-26 (d) continued alkyl migration (ring expansion) followed by nucleophilic attack H 2°
+ �I
H
CH 2 \
: 0 -CH2CH3
-.-.------�.�
CH 2
/ :..0-CH2CH3 I
H
6-27
CH3
0
(a) CH3 - �-0-C-CH3 I
"
(e) CH
(b) CH3-CHCH20CH3 I
CH 2CH3
CH3
SN2 , 1 °
S N 1 , 3°
o
CH2CHl
SN 1 , weak nucleophile
SN 1 , 3°
6-28
CH 3 CH2
CH3 /
I
C
CH3
"'H Br
(R)-2-bromobutane
I
H 2 0,�
�
SN2-
CH 3 CH 2
inversion
C"" / ,'OH H
(S)-2-butanol
H2 0,�
S N l
CH3
I
racemization
+
\..
S N2, strong nucleophile
CH 3 CH2
(R)-2-butanol
, OH
/C"",
H
�
(S)-2-butano
y
50 : 50 mixture-racemic
If SN 1 , which gives racemization, occurs exactly twice as fast as SN2, which gives inversion, then the racemic mixture (50 : 50 R + S) is 66.7% of the mixture and the rest, 3 3 . 3%, is the S enantiomer from SN2. Therefore, the excess o f one enantiomer over the racemic mixture must be 3 3 . 3%, the enantiomeric excess. (In the racemic mixture, the Rand S "cancel " each other algebraically as well as in optical rotation.) The optical rotation o f pure (S)-2-butanol is + l 3.5° . The optical rotation o f this mixture is: 3 3 . 3% x + l 3.5° = + 4.5° 109
6-29
(a) Methyl shift may occur simultaneously with ionization. The lifetime of 1 ° carbocations is exceedingly short; some chemists believe that they are only a transition state to the rearranged product. CH3 1 ° methyl CH3 30 + I I shift CH3- -CHFHl • CH3-C H2 Ag I +
�
�J H3
I�
H 2 0:
CH3 I
CH3 - C-CH 2CH3 I
...
:O-H
(b) Alkyl shift may occur simultaneously with ionization. CH 2
\I
� Ag+
---I.�
Ag I
CH2
+
alkyl shift ring expansion
..
..
6-30
UBf
(a)
H
:Br: + CH30H2 +
The SN 1 mechanism begins with ionization to form a carbocation, attack of a nucleophile, and in the case of ROH nucleophiles, removal of a proton by a base to form a neutral product. (b) In the E l reaction, the solvent (methanol, in this case) serves two functions: it aids the ionization process by solvating both the leaving group (bromide) and the carbocation; and second, it serves as a base to remove the proton from a carbon adjacent to the carbocation in order to form the carbon-carbon double bond. The SN1 mechanism adds a third function to the solvent: the first step is the same as in E l , ionization to form the carbocation; the second step has the solvent acting as a nucleophile-this step is different from E l ; third, the solvent acts like a base and removes a proton, although from an oxygen (SN1) and not a carbon (El). Solvents are versatile! 110
6-3 1 Thi s solution does not show complete mechanisms. Rather, it shows the general sequence of a l l four pathw ays leading t o fi ve different products.
;on;z.
2-bromo-3-methylbutane
;/
_ H i\
________
- - - - - - - - -...- . - - - - - - - - - �. --
-
-
-
-
-
-
---
----
-
-
-
-
-
Five unique products shown in the boxes. One alkene can be produced from either the r or the 30 carbocation.
-
.......... -_..............-_..............
¥
CH3
hydride shift
... .
.. .. ___ ............................. . ..... __ I
�
H3C
+
rearranged 30 carboc ation
CH3
.. " CH3
H3C
� �"'C H2 T H
H1C
_________
two possible E 1 products
6 32
111
:
----------------______ 1
H
CHl
____________
two possible E1 products
6-33 Substi tution products are shown first, then elimination products .
(a)
(c )
� ) +
+
with rearrangement
without rearrangement
o
(d)
6-34 Et is the abbreviation for ethy l , so EtOH is ethanol .
0\3 Cf � r �Et : Br : ··
EtOH
-----l....
�
•
• •
H
0 ..,J +
•
�? : V Et
Et
+ EtOH2 +
E l -MINOR product more substituted alkene H
H " / � C -H • • J HOEt
cr
+ EtOH2 +
without rearrangement
··
I
HOEt
E I -MAJOR product more substituted alkene
6-35
Cll3
CH
CH 3
• •
S N 1 product:
with rearrangement
112
. .
..
+ EtOH2 +
6-36
yl
NaOCH2CH3
..
+
E2-minor product-less substituted alkene (monosubstituted)
Br
6-37
NaO C H 3
Br
(a)
�
�
..
�
+
NaOMe
Br
�
NaO H
Br
(c )
(d)
~ H"'
0 Br
'
"
...
�
+
�
+
minor al kenemonosubsti tuted
CH]
NaOEt
H
...
H atoms that are
removed are shown
a
E2-major product-morc substituted alkene (trisubstituted)
�
+
�
~
O'
mi nor alkenedisubstituted
CH]
major alkenetrisubstituted
Br
E2-minor
113
OCH2CH3
SN2-probably very small amount because of steric hindrance
OCR,
� No SN2 product wi l l be formed; S N2 c annot occ ur at a 3° carbon.
+
OH
~
S N2 product-very small amount because 2° carbon is hindered
major al kenetrisubstituted
CH]
yl
S N2 product-some w i l l be formed on a 2° c arbon
mi nor alkenedisubstituted
major alkenetrisubsti tuted (cis + trans)
...
+
major alkenedisubstituted (cis + trans)
mi nor alkenemonosubsti tuted
(b)
�
(rCH] OEt -
+
S N2 product showing inversion at the c arbon-some of this product wil l be formed on a 2° carbon
6-38 In systems where free rotation is possible, the to be abstracted by the base and the leaving group (Br here) must be anti-coplanar. The E2 mechanism is a concerted, one-step mechanism, so the arrangement of the other groups around the carbons in the starting material is retained i n the product; there is no intermediate to allow t i me for rotation of groups . (Models w i l l help.)
H
8-
r:RCH3 Ph H H' � r/ C-C, / 3 ""CH � Ph
: 9CH3 • •
:t:
, CH3. - Ph "" "'C- C"'
H�
-
Br
transition state
H and Br anti -coplanar
�
�----------
(
- - - - - - -
=
:0
trans
Ph
only geometric isomer possible from a concerted mechanism
�----------�
stretched bond being broken or forme
orbital picture of this reaction
H
•••
H� : OCH3
,,
H
I
• •
...--
--..�
H
P h --+---=--+--
Ph trans
Br
Br transition state
The other diastereomer has two groups interchanged on the back c arbon of the New man projection , where it must give the cis-alkene.
H
Ph
H
H
Ph •
Ph
CH3
Ph Br
cis
114
CH3
pi bond
6-39 lin this problem, all new internal alkenes fonn cis i somers as wel l as the trans i somers shown. Section 1
Br
� Br
�
C
� � OH
H3
.
NaOCH3 · CH30H
OCH3
� S N2
El
=<
+� E2
+� E2
CH3
NaOCH3 ----j.. �
+� +� El
SN1
H2C
CH30H
E2
CH3
Section 2
+�
� Br
E2
A gN03 � Br
+�
CH30H .. �
El
+� El
Section 3
+ Br CH3
H3C
CH3
+ Br CH3
H3C
C H3
•
=<
CH3
NaOCH3 H2C
CH30H
E2
CH3
+ OCH3 CH3
CH30H •
�
H3C
CH3
+
SN 1
=<
CH3
H2C
CH3
El
Section 4
Br
� Br
�
NaOCH3 " CH30H � CH30H
OCH3
� SN2
+�
.� +
E2
+� E2
OCH3
SN1
115
El
+� E1
6-39 continued Section 5
Br
�
I
�aI -.
�
Br
�
�
+
E2
� E2
6-40 (a) Ethoxide is a strong base/nucleophi le-second-order conditions. The I ° bromide favors substitution over e l i mination, so SN2 will predominate over E2.
�
substitution-major
elimination-minor
(b) Methoxide i s a strong base/nucleophi l e-second-order conditions. The 2° chloride will undergo SN2 by backside attack as well as E2 to make a mixture of alkenes. OCH3
�
e l i mination minor alkene
substi tution
elimination major alkene (cis + trans)
(c) Ethoxide is a strong base/nucleophile-second-order conditions. The 3° bromide is h indered and c annot undergo SN2 by backside attack. E2 is the only route possible. CH3 - C=CH2
I
CH3 (d) Heating in ethanol i s conditions for solvolysis, an SN I reaction.
OCH2 CH3 I CH 3 -C-CH3
I
CH 3 (e) Hydroxide is a strong base/nucieophile-second-order condi tions. The 1° iodide i s more l i kely to undergo SN2 than E2, but both products wi ll be observed. CH 3 - CHCH20H
I
CH3 -C = CH2
I
C�
C�
substitution-major
elimination-minor
(f) S i l ver ni trate i n ethanol/water is ionizing conditions for I ° alkyl halides that w i ll lead to rearrangement
followed by substitution on the 3° carbocation. OCH2CH3
C H3 -
? I
-
CH3
CH3
from ethanol as n ucieophile
OH
? I
CH3- -CH3
116
CH3
from w ater as n ucieophile
6-40 conti nued (g) S i l ver nitrate in ethanoVwater is ionizing conditions for 1 ° alkyl halides that w i l l lead to rearrangement fol lowed by substi tution on the 30 c arbocation. OH
from ethanol as
OCH2CH 3
nucleophile I CH3-C -CH2CH3 I
from w ater as
nucleophile
I
CH3-C-CH2CH3
I
CH3
CH3
CH6OCH3
( h ) Heating a 30 halide in methanol is quintessential first-order conditions, ei ther E l or S N I (sol volysis).
\.
substitution (S N l )
C 6 V
(!,"cc) /
elimination (E l )
(i) Ethoxide in ethanol on a 30 halide will lead to E2 elimination; there w i l l be no s ubstituti on. CH3
A U
major alkene tri substituted
minor alkene disubstituted
6-41 Please refer to solution 1 -20, page 1 2 , of this Sol utions Manual . (a)
6-42
(e)
I""
� CI
0-
CI
(d)
CI-C - CH20H I
CI I
I
(f)
CI
H-C-H I
CI I
6-43
(a) 2-bromo-2-methylpentane (b) l -chloro- l -methylcyc lohexane (c) 1 , I -dichloro-3-f1uorocyc loheptane
(g)
CI
(h)
C1 -C H I
-
CI I
CI
�
o
'
(i)
I-
CH2CH3 I
� -CH3 CH3
(d) 4-(2-bromoethy 1)- 3-(fI uoromethy 1)-2-methy I heptane (e) 4,4-dichloro-5-cyc lopropyl-l- iodoheptane (f) cis- I ,2-dich loro- l -methylcyc lohexane
6-44 Ease of backside attack (less steric hindrance) decides which undergoes S N2 faster i n all these examples except (b). (a)
�CI
faster than
(b)
�I
faster than
� CI
Primary R-X reacts faster than 2° R-X.
�CI 117
Iodide a better leaving group than chloride.
6-44
(c)
continued Cl
faster than
�
Br faster than
(d)
�
(e)
OCH2Ci
faster than
(f)
�Br
faster than
Less branching on a neighboring carbon.
� Cl
Br
00
Same neighboring branching, so 1° faster than 2°. Primary R-X reacts faster than 2° R-X.
OCI XBr
Less branching on a neighboring carbon.
Formation of the more stable carbocation decides which undergoes SN 1 faster in all these examples except (d). Cl Cl (a) faster than
6-45
(b) (c)
30+
�
� 2° Cl
OBr
faster than
2°
� Cl 1°
faster than
OCH2Br
2°
(d) (e)
OJ X
r
1°
faster than faster than
3°
(t)
Q Br
2° allylic
faster than
OCI
113
~ Br
Q 2°
Br
1 18
2°
(leaving group ability)
6-46 For SN2, reactions should be designed such that the nucleophile attacks the least highly substituted alkyl halide. ("X" stands for a halide: Cl, Br, or I.)
(a)
(b) (c)
(d) (e) (f)
(g)
Q-
+
HO-
-
-
CL OH
Q-
X
HO-
--� .. -
CH2X
+
0 O:.A
X
�
NH3 excess
-
-
-
Q-
0
CH20H
SCH2CHl some
0 0
Q-
U OH
also produced
OH
CH2NH2
H2C = CHCH2CN
-
HC:::C
6-47
(a)
CH2X
(1)
CH3
CH - 0 / CH3 \
(2)
CH3
�
-
;."
/ U CH3 2°
+
CH-O-CH2CH3
\
�
__
CH3
t
/
1°
�-
CH-X
\
CH3
cfCH2CH3
this bond formed
CH3
I CH2CH3
CH-O-CH2CH3
\
..
CH3
/
t
this bond formed Synthesis ( 1 ) would give a better yield of the desired ether product. ( 1) uses SN2 attack of a nuc\eophilc on a 1° carbon, while (2) requires attack on a more hindered 2° carbon. Reaction (2) would give a lower yield of substitution, with more elimination. (b) CANNOT DO S N2 ON A 3° CARBON! CH3 � umps into H '\ I CH3-C-Cl + -OCH3 I C� 6-47 (b) continued on next page b
CH2 II CH3-C I C�
before it can find C
..
119
elimination (E2) competes
6-47 (b) continued Better to do SN2 on a methyl carbon: CH 3 CH3 -
? I
-O
� +
CH 3
I
CH3 - C-O - CH3
C H3 - X U
I
CH3
CH3
6-48 (a) SN2-second order: reaction rate doubles (b) SN2-second order: reaction rate increases six times (c) Virtually all reaction rates, including this one, increase with a temperature increase. 6-49 This is an SN I reaction; the rate law depends only on the substrate concentration, not on the nucleophile concentration. (a) no change in rate
(b) the rate triples, dependent only on [ t-butyl bromide ] (c) Virtually all reaction rates, including this one, increase with a temperature increase.
6-50 The key to this problem is that iodide ion is both an excellent nucleophile AND leaving group. Substitution on chlorocyclohexane is faster with iodide than with cyanide (see Table 6-3 for relative nucleophilicities). Once iodocyclohexane is formed, substitution by cyanide is much faster on iodocyclohexane than on chlorocyclohexane because iodide is a better leaving group than chloride. So fast reactions involving iodide replace a slower single reaction, resulting in an overall rate increase.
0-
o-
+
Cl
Cl
+
-CN
-I
moderate rate
fast ----
�.. �
0-
..
0-
1
+
-CN
CN
fast
� � ..
----
0-
CN +
-
1
recycles-only a small amount needed (catalyst)
6-51 (a)
(c)
(d)
rearrangement 6-52 (a)
120
rearrangement
two
(b)
6-52
continued
(i)
·
0: H�(
H
equivalent resonance fonns
H
H
H
.��"
equivalent resonance fonns
}
(iv)
H
H
� H
i
loss of bromide gives unstahle 10 H 2 carhocation Ihat quickly rearranges to a 2° allylic carbocation
1/��· .Jy�,,}
equivalent resonance fonns-same as from part (iii)
(c) Only one substitution product arises from equivalent resonance fonns. (ii)
(i)
(iii)
(iv)
OCH2 CH3
� cis + trans
OCH 2 CH3
� cis + trans
same as part (iii)
+
6-53
most stable
al\ylic (+ on 3° and 2° C)
>0>0> al\ylic (+ on 1 ° and 2° C)
12 1
least stable
6-54
CHX H H -Z 2,0
X
H
0 I t
U I t CH3
CH3
+
I"
hydride shift
hydride shift
H ( C
+ H2
C(
OR
I alkyl shiftt nng expansIOn
H
6 (j
6-55 Reactions would also give some elimination products; only the substitution products are shown here.
(a)
<;:H3 N::C-C-H CH2CH3
SN2 gives inversion: only product:
6-56
(a)
(d) (g)
CH3CH20CH2CH3 C:::CH CH3(CH2)8CH2 I
Q
(b)
<;:H3 H-C-OH H-<;:-CH3 CH2CH3 S2 I
(c)
<;:H2CH3 CH3CH20 <;:-CH3 CH(CH3h -
-
solvolysis ,
N only with inversion
+
SNl,
<;:H2CH3 CH3-<;:-OCH2CH3 CH(CH3h -
racerruzation
(b)
< }-CH2CH2CN
(c)
< }-SCH2CH3
(e)
< �N+- CH3
(f)
(CH3hC -CH2CH2NH2
(h)
HO "'0"" CH3
122
1-
substitution
6-57
H
Br
+ +
CH 3
B, H
B'
H
+
+
H
+
racemic mixture
Br
CH3
Regardless of which bromine is substituted on each molecule, the same mixture of products results.
-----y---� 25,35
"----
HO Br
+ + CH 3
2R,35
elimination
H Br -:.
Br ::: H
'-Y / '"
KOH
H
H
+
H
H
2R,3R
+
Each of the substitution products has one chiral center inverted from the starting material. The mechanism that accounts for inversion is SN2 . If an SN 1 process were occurring, the product mixture would also contain 2R,3R and 25,35 diastereomers. Their absence argues agai nst an SN 1 process occurring here.
OH
+
racemic mixture
rotate..
-)
Br
CH 3
25,3R
B
S=HB,3
�
:M
H H CH 3 H and Br anti -coplanar
HO
'
H'-') �
trans
The other enantiomer gives the same product (you should prove this to yourself).
(a)
The absence of cis product is evidence that only the E2 elimination is occurring in one step through an anti-coplanar transition state with no chance of rotation. If El had been occuring, rotation around the carbocation intermediate would have been possible, leading to both the cis and trans products.
6-58
+ 15.58°
x 100%
=
. . . 98% of ongmal optIcal activity
=
98% e.e.
Thus, 98% of the 5 enantiomer and 2% racemic mixture gives an overall composition of 99% 5 and 1% R. +15.90°
(b) The 1% of radioactive iodide has produced exactly 1 % of the R enantiomer. Each substitution must occur with inversion, a classic S N2 mechanism.
(a) An SN 2 mechanism with inversion will convert R to its enantiomer, 5. An accumulation of excess 5 does not occur because it can also react with bromide, regenerating R. The system approaches a racemic mixture at equilibrium.
6-59
��Be
f\ :: Br -C-H CH2CH3 R
+
S N2
inversion
�H3
.. 123
H-C-Br CH2CH3 5
+ Br-
6-59 continued (b) In order to undergo substitution and therefore inversion, HO- wouid have to be the leaving group, but H O- is never a leaving group in SN2. No reaction can occur. (c) Once the OH is protonated, it can leave as H20. Racemization occurs in the SN1 mechanism because of the planar, achiral carbocation intermediate which "erases" all stereochemistry of the starting material. Racemization occurs in the SN2 mechanism by establishing an equilibrium of Rand S enantiomers, as explained in 6-59(a). HO-C-H .. ::
H+
--
\:::)
HO-C-H
+
H-C-OH
racemic mixture
The carbocation produced in this E l elimination is and will not rearrange. The product ratio follows the Zaitsev rule.
30
minor (b)
(c)
trace-from unrearranged 20 carbocation
trace amount
from either unrearranged 20 carbocation or rearranged carbocation
30
The carbocation produced in this E l elimination is 20 and can either eliminate to give the first two alkenes , or can rearrange by from rearranged a hydride shift to a 30 carbocation which 30 carbocation would produce the last two products. The amounts of the last two products are not predictable as they are both trisubstituted, but the first product will certainly be the least.
¥+�+Jy
trace-from unrearranged 20 carbocation
major from rearranged 30 carbocation
minor from rearranged 30 carbocation
124
The carbocation produced in this E l elimination is 2 0 and can either eliminate to give the first alkene, or can rearrange by a methyl shift to a 30 carbocation which would produce the last two products. The middle product is major as it is tetrasubstituted versus disubstituted for the last structure and monosubstituted for the first structure.
6-61 The allylic carbocation has two resonance forms showing that two carbons share the positive charge. The ethanol nucleophile can attack either of these carbons, giving the S N 1 products; or loss of an adjacent H will give the El product.
O'
n CH2-B,
H
O' +
SNI
..
t a
CH2
H 1 :�-CH2CH3 •
+H
H
---
H 1 :O-CH2CH3 ..
�H2
H
El
- Br-
['t CH2 �H
6-62
H
•
H
"--./
H
U
(a)
SN I
aI
�
H
..
.OH .
...
H
...
+H
H
(1+
H � '---- : 0 -CH2CH3 .. ... _
CH2 '
OcH 2CH3 ('y + � . Z-CH2CH1 H I H •
/ ' ..
C;(
}
O'
CH2OCH2CHl
H
('yCH2 � OCH 2CHl H
�
H2
H
H
H
H
-
CH2
CH2 - OCH2CH3
-H2O: ... OH H
-H2O: ... OH H
··�H+ OSh U OH ..
a
..
··/'"W C);,.+1
H
S N2
I
w /'" r:! C); 9: +1 0' H
(b)
:O-CH2CH3 • •
1 O'
H2
+OH 1 H
..
'�
."!3r.
�
125
H
Q �: 0': (J U H O �
H
H
...
• •
H
:ii,: "--./ ..
B'
+
H
...
H20:
B'
6-63 NBS generates bromine which produces bromine radical. Bromine radical abstracts an allylic hydrogen, resulting in a resonance-stabilized a\lylic radical. The a\lylic radical can bond to bromine at either of the two carbons with radical character.
H H2C=C-C-CH3 H CH3 I
I
I
+ Br-
--
HEr +
continues propagation •
1H2C=C-C-CH3 H CH3 I
I
Br I
H2C=C-C-CH3 H CH3
Br +
I
I
a\lylic
f----I�.... ..
� Br2 +
H2C-T ��H CH3 =
CH,
Br I
H2C-C=C-CH3 H CH3 I
I
6-64 The bromine radical from NBS will abstract whichever hydrogen produces the most stable intermediate; in this structure, that is a benzylic hydrogen, giving the resonance-stabilized benzylic radical.
< }�HCH)
HBr +
�
1< }CHCH).
�
•
.... ..f----I�_
Br
< }CHCH3
�
Br 2 +
Br·
<
OCHCH)
1>= CHCH3
(Even though three carbons of the ring have some radical character, these are minm resonance contributors. The product is most stable when the ring has all three double bonds intact, necessitating that the bromine bond to the benzylic carbon.)
6-65 Two related factors could explain this observation. First, as carbocation stability increases, the leaving group will be less tightly held by the carbocation for stabilization; the more stable carbocations are more "free" in solution, meaning more exposed. Second, more stable carbocations will have longer lifetimes, allowing the leaving group to drift off in the solvent, leading to more possibility for the incoming nucleophile to attack from the side that the leaving group just left. The less stable carbocations hold tightly to their leaving groups, preventing nucleophiles from attacking this side. Backside attack with inversion is the preferred stereochemical route in this case. 6-66
a CH3 ----.. ionization
�
Br
(t� CH3 _--=---H
Br
+
rearrangement
2° carbocation 126
�
3° carbocation
mechanisms continued on next page
}
6-66 continued
substitution on 20 carbocation
6
H �H CH3
:6-CH • •
3
CH3
..
:O-CH 3 • •
-I ... �
----
substitution on rearranged 30 carbocation
a _
CH3
H I
/ �-
:O-CH 3 • •
....
elimination from rearranged 30 carbocation
H 6-67
(a)
one step
E2-
(� r
CH3-CH-CH3
'-= :OH
--
OH I CH3-CH-CH3 + Br-
(b) In the E2 reaction, a C-H bond is broken. When D is substituted for H, a C-D bond is broken, slowing the reaction. In the SN2 reaction, no C-H (C-D) bond is broken, so the rate is unchanged. (c) These are first-order reactions. The slow, rate-determining step is the first step in each mechanism. Br Explanation on next page H l + slow... .. I, fast :Br: H2C-CH-CH3 -- H2C=CH-CH3 E l H2C-CH-CH3
(
H
�
I
SN I
(BI r
H3C-CH-CH3
�
o: + slow ... H3 C-CH-CH3 -fast + Br127
6-67 (c) continued
The only mechanism of these two involving C-H bond c leavage is the E l , but the C-H c leavage does NOT occur in the slow, rate-determining step. Kinetic isotope effects are observed only when C-H (C-- D) bond cleavage occurs in the rate-determining step. Thus, we would expect to observe no change in rate for the deuterium-substituted molecules in the E l or S N I mechanisms. (In fact, this technique of measuring isotope effects is one of the most useful tools chemists have for determining what mechanism a reaction fol lows . )
6-68 Both products are formed through E 2 reactions. The difference is whether a D o r a n H is removed by the base. As explained in Problem 6-67 , C-D cleavage can be up to 7 times slower than C-H cleavage, so the product from C-H c leavage should be formed about 7 times as fast. Thi s rate preference is reflected in the 7 : 1 product mixture . ("Ph" is the abbreviation for a benzene ring.)
D H"
"
.:-!f �H
" ,C - C
- D, Br
H ..
"
"
/
/
C=C
Br Ph Ph requires C-D bond cleavage; slow; minor product H
"
,H .:-� H
" C-C
Ph " 'I
- H, Br
'
'
H H
...
Br D requires C-H bond c leavage; 7 times faster; major product
6-69 The energy, and therefore the structure, of the transition state determines the rate of a reaction. Any
---I"'� [
l
factor which lowers the energy of the transition state will speed the reaction.
R N R" "� : R \
�
n
R' - X
R N��---- R , - -R" �� R
__ _
8x
transition statedeveloping charge
R N+_ ' R" �� R R
x
-
This example of S N2 is unusual in that the nucleophi le is a neutral molecule-it is not negatively charged. The transition state is beginning to show the positive and negative charges of the products (ions), so the transition state is more charged than the reactants. The polar transition state will be stabilized in a more polar solvent through dipole-dipole interactions, so the rate of reaction will be enhanced in a polar solvent.
t
,
reaction coordinate
\,
---
polar sol vent lower energy, faster reactIon
___
128
The problem is how to explain this reaction: : NEt2 Et2 N : I I HO- � H2C - CH - CH2CH3 H2C - CH - CH2CH3
6-70
Solution
I Cl
I
OH
1
+
Cl-
2
facts 1) second order, but several thousand times faster than similar second order reactions without the NEt2 group 2) NEt 2 group migrates
Clearly, the NEt2 group is i nvolved. The nitrogen is a nucleophile and can do an internal nucleophi lic substitution (S N i), a very fast reaction for entropy reasons because two different molecules do not have to come together. very fast
3
The slower step is attack of HO- on intermediate 3; the N is a good leaving group because it has a positive charge. Where will HO- attack 3? On the less substituted carbon, in typical SN2 fashion. :NEt2 I
H 2C - CH - CH2CH3 OH I
2
This overall reaction is fast because of the neighboring group assistance in fonning 3. It is second order because the HO- group and 3 collide in the slow step (not the only step, however). And the NEt2 group "migrates", although in two steps. 6-71 The symmetry of this molecule is crucial.
(a)
Ph H
Ph
I I I CH3 - C - C - C - CH3 I I I H
Br H
Regardless of which adjacent H is removed by t-butoxide, the product will be 2,4-diphenyl-2-pentene.
(b) Here are a Newman projection, a three-dimensional representation and a Fischer projection of the required diastereomer. On both carbons 2 and 4, the H has to be anti-coplanar with the bromine while leaving the other groups to give the same product. Not coincidentally, the correct diastereomer is a meso structure. H Ph Br H CH3 H 2 H3C H Ph Ph
Br H and Br are anti-coplanar
Ph
�
129
H3C
CH3
Ph
Br
Ph
3
CH3
H
H
6-72
"
(a)
Ph "
phenyl
Ph I CH3CHCHCH3 I =
NaOCH3
...
E2
Ph I CH3CHCH = CHz
Ph
CH3-C = CHCH3 + I
major Br __ (b) H and Br must be anti -coplanar in the transition state ,H H
2R,3R
f K
Ph CH3 \ \\
Ph
Ph
H
Br
Br
H
� � H
CH3
CH3
------
Ph I
II C" C CH3""- - . 'CH3 \\ \\
" 1
methyls
CH3
Br
th �
CH3
Na OCH3
\\ \\
H
(c)
CH3
�
CH3 ---{
Ph
V -+ r..
rru nor
Ph
'r
H
H
';(
H
CIS
CH3 CH3
Ph " \\ CH3 II := C C" CH3 CH3....... 'H '1
,\
methyls trans
2S,3 R
(d) The 2S, 3S is the mirror i mage of 2R, 3R; it would give the mirror image of the alkene that 2R, 3R produced (with two methyl groups cis). The alkene product i s planar, not chiral , so i ts mirror image is the same : the 2S, 3S and the 2R, 3 R g i ve the same alkene.
130
6-73 All five products (boxed) come from rearranged carbocations. Rearrangement, which may occur simultaneously with i on ization, can occur by hydride shift to the 3° meth y lcyclopentyl cation, or by ri ng
6
expansion to the cyclohexy l cation.
H
n
C H 2 - Br
-
Br
..
6
+
�H� � 6 (y �
hydride shift
H
�� H 2
H2
�·· H H�CH3 .. H
___
&H
\ 6. b: :\ X' � H
CH3
C
CH3
A ��CH3.. U U +
C H2
-
H CH3
H6
6
CH3
alkyl shift (ri ng expansion)
)q � --
(J'-- U H
H
H
�
O o
H
CH3
H
�
O +
H
�
H C H3 ..
..
II
(t.� H3 � qC HOCH 3 •
• •
-
O
CH3
6-74 Begin with a structure of (S)-2-bromo-2-tluorobutane . Since there i s no H on C-2, the lowest priority group must be the CH3 . The Br has highest priority, then F, then C H2 CH3, and CH3 i s fOUlth . Sodium methoxide i s a strong base and nucleophi le, so the reaction must be second order, E2 or SN2 . (a) H
H3C
� Br
CH3 NaOCH3 ----l.. �
F
H
H3C
Br
ffi H3C '<;Y- F
H3C
( \/ ,
H
H
13 1
H
NaOCH 3
-----
H 3C
H3C
( '> / )
-
6-74 continued In regular structural formulas, the reaction would give three products i nc l uding the stereoisomers shown above .
� Br
+
F
(5)
(b)
3
4
l� Br
3
4
l�
NaOCH3
F
F
OCH3
In these structures, the numbers 1 to 4 indicate the group's priori ty in the Cahn-Ingold-Prelog system.
(5) (5) A cursory analysis of the designation of configuration would suggest to the uncritical mind that this reaction proceeded w ith retention of configurati on-but that would be w rong ! You know by now that a careful analysis i s required. In the Cahn-Ingol d-Prelog system, the F i n the starting material was priority group 2, but i n the product, because Br has left, F is now the fi rst priority group. So even though the designation of configuration suggests retention of configuation, the molecule has actual ly undergone invers i on as would be expected with an S N 2 reaction . (See the solution to problem 6-2 1 for a s i mi l ar example . )
6-75 (a) Only the propagation steps are shown . NB S provides a low concentratin o f Br2 which generates bromine radical in ultraviolet l i ght. In the starting material, all 8 allyJic hydrogens are equ ivalent.
H
H
I
Br · recycles
I
Br - Br '---I
Br +
Br +
H Br ·
H + Br ·
(b) The first step in thi s first-order solvolysis is ionizati on , followed by rearra ngement. H H hydride C 30H H shift + ---....... B r + H II
0?Cr CH2 H
_
mechanism continued on next page
-!CrCH2 132
allylic !
6-75 (b) continued
El
--:
H
H3 H + H �
�
SN I
H
� � ,
- CH2
t
& H
CH2
CH2
+
(j-
CH30H • •
�H2
{y
H3C � � H .O - H � CH3� H CH3� H CH 2 . .
·· � 1 I
Cf
H
..
H
•
(:Y H
H3C
}
�
•
CH2
'3
_ CH 2
U+
CH \�H2
OCH3
\
CH3•OH •
+
X
� H� 1+
CH30H2
•
� C1 H2 U
+
CH1 0H2 . +
(c) This first-order solvolysis generates an allylic carbocation in the ionization step.
EtOH
..
Br
•
•
�
H
! Et�H
Et ,• • / H
H
+
�
��
Et H
The El is shown on the next page.
H
• •
?H)
,..,
CD ! Et�H
Et
OEt
OEt
l 33
+
EtOH2
+
co
These are the SN 1 products.
+
Et OH2
+
6-75 (c ) continued El H
H
+
EtO H
EtOH 2 +
..
Q<> :_ (0)
(d) The first step in this first-order solvolysis is ionization, fol lowed by rearrangement. C 30H _ __ �
H
Br
Br . rearrangement by alkyl shift
-
Q(>�� Q<>
CH3�: �
� � Qj
' H CH3
H
cp � H3C
:O •
•
�
)
CP OCH3
E l on rearranged carbocation
G;1 H
�
�
CH3 H
H�
((\
+
�
H
134
H
CH30H2 +
OCH 1
SN 1 product from unrearranged carbocation
H3C
CH3� H
H�
O- H
j 'Y
SN 1 on rearranged carbocation
� � H � + �.�.
:
+
CH30H 2 +
CHAPTER 7-STRUCTURE AND SYNTHESIS OF ALKENES 7- 1 The number of elements of unsaturation in a hydrocarbon formula is given by: 2(#C) + 2 - (#H) 2 2(6) + 2 - (1 2) 2
=
1 element of un saturation
(b) Many examples are possible. Yours may not match these, but all must have either a double bond or a ring, that is, one element of unsaturation .
o
2(4) + 2 - (6) 2
7-2
=
2 elements of un saturation H
H3 C
/ C = C = CH2
1:
7 -3 Hundreds of examples of C4�NOCI are possible. Yours may not match these, but all must contain two elements of unsaturation .
�
CI � O y NH2
\d
CI
N - C = C - CI
�
OH
CI
�
C
U
N =O
N
OH
7-4 Many examples of these formulas are possible. Yours may not match these, but correct answers must have the same number of elements of unsaturation.
�
(a) C3H4CI2
=>
C3�
=
C4HgO
=>
CI � CI
Cl
Cl (b)
1
C4H g
=
� o
1
o
OH
� 13S
7 -4 continued (c) C4H402 => C4H4 HC
(d) C s H s N0 2
=>
3
=
II
C - C - OCH3
C s sH s
[r
°
4
=
0 , + ,.... 0 '
°
(e) C6H3NClBr HC
=>
C
C - N H,
C6 s H S
=
N
N
°
5
o Br
C - C == C - C I I B r NH
I
C
CH
Cl
)U
U
N == C == CH2
Br
Cl
O
C-N
A
Cl
Note to the student: The IUP AC system of nomenclature is undergoing many changes, most notably i n the placement of position numbers. The new system places the position number close to the functional group designation, which is what this Solutions Manual will attempt to follow ; however, you should be able to use and recognize names in either the old or the new style. Ask your instructor which system to use. 7-5
(a) (b) (c) (d) (e)
4-methylpent- l -ene 2-ethylhex- l -ene penta- l ,4-diene penta- l ,2,4-triene 2,5-di methy\cyclopenta- l ,3 -diene
(f) 4-vinylcyclohex- l -ene (" 1" is optional) (g) 3-phenylprop- l -ene (" 1 " i s optional) (h) trans-3 ,4-dimethy\cyclopent- l -ene (" 1 " is optional)
(i) 7-methy\enecyclohepta- l ,3 , 5-triene
7-6 (b), (e), and (f) do not show cis, trans isomerism
(a)
� (Z)-hex-3-ene (cis-hex -3 -ene)
( c)
\d\
(2Z,4Z)-hexa-2,4-diene cis, cis-hexa-2,4-diene (d)
;=C
(Z)-3-methylpent-2-ene
�
\
(E)-hCX -enc (trans-hex-3-ene)
(2Z,4E)-hexa-2,4-diene cis, trans-hexa-2,4-diene
(2E,4E)-hexa-2,4-diene trans, trans-hexa-2,4-diene " Cis" and "trans" are not clear for this example;
"E' and Z are unambiguous. "
(E)-3-methylpent-2-ene 136
'
7-7 (a)
� IL -
2,3 -dimethylpent-2-ene (neither cis nor trans)
CI
(d)
'0
5 -chlorocyclohexa- l ,3 -diene (positions of double bonds need to be specified)
(c)
(b)
3 -ethylhexa- l ,4-diene (cis or trans not specified; the vinyl group is part of the main chain)
I -methylcyclopentenc
(f)
(e )
cis-3 ,4-dimethylcyclohexene (could also have drawn trans)
(E)-2,5-dibromo-3 -ethylpent-2 -ene (cis does not apply)
7-8 (a)
'x
CI
Br (E)-3-bromo-2-chloropent-2-ene (b)
(2E,4E)-3-ethylhexa-2,4-diene
Br
/=C
(2)-3 -bromo-2-chloropen t-2-ene
(2Z,4E)-
(2E,42)-
(c ) no geometric i somers (d) (2)-penta- l ,3 -diene
(E)-penta- l ,3-diene
(e )
(E)-4-t-butyl-5-methyloct-4-ene
(2)-4-t-butyl-5 -meth yloct -4-ene
1 37
(2Z,42)-
7 -8 continued
(f) CI
CI
CI
CI
(2E,5£)-3 ,7-dichloroocta- 2 , 5 -diene
(2Z, 5 £)-3, 7 -dich loroocta-2 , 5 -diene
CI
CI
(2E,5Z)-3, 7 -dich loroocta-2 , 5 -diene
(2Z,5Z)-3,7 -dichloroocta-2 ,S -diene
(g) no geometric i somers (an E double bond would be too highly strained) (h)
W
W
(£)-cyc lodecene
(Z)-cyc Iodecene
(i)
W
( I E,S£)-cyc Iodeca- l , S-diene
C)
W
( 1 Z,S£)-cycI odeca- l ,5-diene ( I Z,5Z)-cyclodeca- l ,5 -diene
7-9 From Table 7- 1 , approximate heats of h ydrogenation can be determined for similarly substituted alkenes. The energy difference is approxi mately 6 kJ/mole ( 1 .4 kcal/mole), the more highly substituted alkene being more stable.
gem-disubstituted 1 1 7 kJ/mole (28.0 kcal/mole)
tetrasubstituted 1 1 1 kJ/mole (26.6 kcal/mole)
7- 1 0 Use the relati ve values i n Figure 7-7 . (a)
2 x (trans-disubstituted - cis-di substituted)
=
2 x (22 - 1 8) = 8 kJ/mole more stable for trans, trans (2 x (S . 2 - 4.2) = 2 kcal/mole) (b)
gem-di substituted - monosubstituted 2-methylbut- l -ene i s more stable
=
20 1 1 = 9 kJ/mole (4. 8 - 2.7 = 2. 1 kcallmole) -
138
7- lO continued (c)
trisubstituted - gem-disubstituted
=
25 - 20 = 5 kJ/mole (5 .9 - 4.8 = 1 . 1 kcallmole)
2-methylbut-2-ene i s more stable (d)
tetrasubstituted - gem-disubstituted
o(CH3 CH3
=
26 - 20 = 6 kJ/mole (6.2 - 4 . 8 = 1 .4 kcallmole)
2,3-di methylbut-2-ene i s more stable 7- 1 1 (a)
strained but stable
(b) could not exi st-ring s i ze must be 8 atoms or greater to include trans double bond (c )
Despite the ambiguity of this name, we must assume that the trans refers to the two methyls since thi s compound can exist, rather than the trans refeni ng to the alkene, a molecule which could not exist.
(d) stable-trans in l O-membered ring (e) unstable at room temperature�annot have trans alkene i n 7-membered ring (possibly i solable at very low temperature-thi s type of experiment is one of the challenges chemi sts attack with gusto) (f) stable-alkene not at bridgehead (g) unstable-violation of B redt's Rule (alkene at bridgehead in 6-membered ring) (h) stable-alkene at bridgehead i n 8-membered ring (i) unstable-violation of B redt's Rule (alkene at bridgehead i n 7-membered ring) 7- 1 2 (a) The di bromo compound should boi l at a higher temperature because of its much l arger molecular weight. (b) The cis should boil at a h i gher temperature than the trans as the trans has a zero dipole moment and therefore no dipole-dipole interactions . (c) 1 ,2-Dichlorocyclohexene should boi l a t a hi gher temperature because o f i ts much l arger molecular weight and larger dipole moment than cyclohexene. 7- 1 3 (a)
0 O�3
NaOH
•
acetone mmor E2
major E2
139
7 - 1 3 continued (b)
Cl
(c)
( d)
0
E2
E2
major
minor
20
NaOCH3 ..
CH30H
hindered base gives Hofmann product as major isomer
6] oE2 +
major product i s difficult to predict for 2° halides
S N2
NaOC(CH3h ..
(CH3hCOH
0
NaOC(CH3h i s a bulky base and a poor nucleophi le, minimizing SN2
E2
-
7-14
B ase:
=
-, H
� Ph H
"'�
" C-C
/
H3C " Ph
�Br
-
H3C " " " "" H C === C " "'" ' Ph Ph cis
7- 1 5 (a )
substitution product-since the substrate is a neopentyl halide and highly hindered, the SN2 substitution is slow, and elimination is favored
1 40
E2
elimination
':r--f(
7 - 1 5 continued (b) H
jPhH
"" Ph " Br
(CH3 CH2) 3 N :
Br meso
���
(c)
Ph
Ph
C=C " / H Br only alkene isomer E2 Br
/
"
Cb HO
(d) NaOH
..
acetone
+ minor substi tution products
Ph
C=C " / H Ph only alkene i somer E2
B Br Ph one enantiomer of the d, l pair
(c)
/
"
+
minor s ubstitution products
H
SN2 i s the only mechanism possi ble because no H can get coplan ar w i th C l ; an eli mination product would violate B redfs Rule
See Appendix 1 i n thi s manual for numbering and naming bicyc l ic systems.
Cl
H H Models show that the H on C-3 cannot be anti-coplanar with the Ci on C-2. Thus, this E2 elimi nation must occ ur with a syn-coplanar orientation: the D must be removed as the Cl leave s . 7- 1 6 As shown i n Solved Problem 7 - 3 , the H and the Br must have trans-diaxial orientation for the E 2 reaction t o occur. I n part (a), t h e c i s i somer h a s the methyl i n equatorial position, t h e NaOCH3 can remove a hydrogen from either C - 2 or C-6, giving a mixture of alkenes where the most highly substituted i somer is the maj or product (Zaitsev). In part (b), the trans i somer has the methyl in the axial posi tion at C-2 , so no elimination can occur to C-2. The only possible elimination orientation i s toward C-6.
�l
Br
(a)
I
� H
.. eIther H can be removed
( b)
6
NaOCH3 ..
C H3 0H
cj(.
Br
.
H3C
only this H can be removed
6
H
+
major
H
1
c5 c5
NaOCH3 ..
CH3 0H
c5
Zaitsev orientation
minor
only alkene formed; stereochemistry of E2 precludes other i somer from forming 14 1
7- 1 7 E2 elimination requires that the H and the leaving group be anti-coplanar; in a chair cyclohexane, this requires that the two groups be trans diaxial. However, when the bromine atom is in an axial position, there are no hydrogens in axial positions on adj acent c arbons, so no elimination c an occur.
c�
hydrogens lrans diaxial lo lhe Br
CH3
H
7- 1 8
B r -- axial
(X)'" � H
'
(D H
Br
(a)
=> � � T -1 H
Br
C:::::>
H
KOH
Br axial
+
�
= Br � T � atOrial H
H
(b) S howing the chair form of the decalins makes the answer clear. The top i somer locks the H and the O r i nto a trans-diaxial conformation--optimum for E 2 elimination. The bottom i somer h a s B r equatori al where it is exceedingly slow to eli minate .
7-19
c{(;
Mode ls are a b i g help for this problem. Br
(a)
•
H
NaOCH3 ..
CH3
the only H trans diaxial
(b)
the only elimination product H
H
D (from elimination of H and B r)
H
+
both trans diax ial gives two products
H
H (from elimination of D and B r) 142
=
h W : H
H
H
7-20
(
-
Ph H
�-C �/ C,
B
0
�
o
" " " Ph --- Ph " ," " CC H """ - ' H
---
H
Ph
cis
transition state
both Br anti-coplanar
�
(-
- - - - - -
=
:0
only geometric i somer possible from a concerted mechanism
�----------�
�----------
stretched bond being broken or forme Br - - - - - - - r
Br
H
0
:I:
Br
"':'" "
/ /,
0 :.10: o
.....-
Ph
H Ph H -+----;;---1-- Ph
�
--
pi bond
.. �
Ph .
Br
cis
Br transition state
7-2 1 (5, 5)
H
( :.10:
o
o
r
o o
0 •
Br H H3C � " , C ::..::..: C , " ,H 8- , " CH3 Br
H - �/ B r 3C � C-C / X"�" H /, B CH3 0
, II , C-C '\\" , H ___ H """" - 'CH3 H3C cis
transition state
both Br anti -coplanar
------------��---------r. stretched bond being broken or forme:0 - - - - - -
=
H
H3C
Br
:Qx
Br - - - - - - - I
Br
H3C
..
only geometric isomer possible from a concerted mechanism
H3C
H3C
,
,
H
:j: .......
H3C H3C
H cis
Br transition state 143
pi bond
7-22 (a)
�H Br
NaI acetone
Br
o
cyclohcxene
(b )
cis-hept-3 -ene
CH3 r"fyH CH2CH3 Br
(d) NaI acetone
C-l
The two bromi ne atoms must be anti coplanar in order to e l i minate . Onl y the bromines at and C-2 fit that requirement; the bromines at and C-3 cannot eliminate .
C-2
(5)- 3-bromo- l -ethyl-3-meth y Icyc lohexene
(e)
rotation
Br
Br
NaI, acetone
.
9 H �
H
b
In the first conformation shown , the bromine atoms are not coplanar and c annot eli minate. Rotati on in this l arge ring can place the two bromines anti-coplanar, generating a trans al kene in the ring. (Use models ! )
trans-cyclodecene
7 -23 The stereochemical requirement of E2 elimination is anti-coplanar; in cyc lohexanes , thi s translates to trans-diaxial . B oth dibromides are trans, but because the t-butyl group must be in an equatorial position , only the left molecule c an h ave the bromines diaxial. The one on the right h as both bromines loc ked into equatori al positions, from which they cannot undergo E2 elimination.
(CH3hC
# � H Br
---
trans-diaxialcan do E2
(CH3hC 1 44
H
�
Br
r HI
trans-diequatorial--cannot do E2
CH CH 3 ��42U CH_3_ .. � 3 � : o 3 CH 2 CH � : -CH + 2 CH ? . ? 0 cj.. _ H CH:2CH3 (Br
II
O-CH 2 CH 3 H H H 2 CH 2 C " � � r' CH 2 CH : R · H � (j H�:�H -CH2CH3 ] CH .. () C CH 3 H 7 �:�-CH 2 CH 3 + .. rY 3 + 3 . . : Gf C O-CH 2 CH 3 : (t, � � d H H �- 2 H H :O-CH2CH3 CH2CH} ° H CH3 � . ] C () H H 7 �-CH 2 CH 3 � + H H 2CH3 CH] H H �:�-CH a CH3H H ° H 3 ] C :O-CH2CH..3 r:f; / V h< O �
\. •
I
�. I
(b)
...
Br
�
,
(I
�
J �
I
..
c,
�
I
I
..
carbocation rearrangement
°2 _ H
I
same product mi xture as in PaIt (a)
• •
145
� 0H • • Hf -OS03H (J __
(a)
�
__
__ •
(f
H ..0 -H 1+
___
/).
+ G /).
___
cis + trans major isomer
(X •oH• •• _H-OS03H CfR-H
(c)
H 1+
� t_�__
___ A Ll
__
CH3
•
from unrearranged 2° carbocation
-.
HS04 H20 H/-OS03H ...� H H H 1 C � I) H ,\ �H -OS03o/ } OS03H /
7 -25 In these mechanisms, the base removing the final proton is shown as that water removes this proton.
f
It i s equally possible
•
+
!
� mi nor isomer
rearrangeh ydride shift •
CH3
H
H
ct H
from rearranged 3° carbocation
greatest amount 146
least amount
7-26 (a) f..Go
tJ0J{
=
=
=
-
Tf..So
+
116,000 Ilmol - 298 K ( Il7 1IK·mol)
+
81 ,100 Ilmol
=
+
8l. l kl/mol (+19.3 kcaVmole)
f..Go is positive, the reaction is disfavored at 25°C (b)
f..GIOOO
=
=
+ 116,000 Ilmol - 1273 K (117 lIK·mol )
- 3 2,900 Ilmol
=
- 3 2 .9 klima I (- 8.0 kcal/mole)
f..G is negati ve , the reaction is favored at 1000°C 7-27
(a) basic and nucleophilic mechanism: B a(OH}z is a strong base (b) ac idic and electrophilic mechanism: the catalyst is H+ (c) free radical chain reacti on: the catalyst is a peroxide that ini tiates free radical reactions (d) acidic and electrophi l i c mechanism: the catalyst B F3 i s a strong Lewis aci d
7-28 (a)
H � � H - OS03H I .. CH3 - C - C - C-OH I I I H H H
I
H
I
I
I
H
•
•
•
•
I
I
I
I
H
I
H
H
H
H
H
I
H I CH3-C-C - C-OH
I
....
I
I
+
\.::;
I
+
H H I 1+ CH3-C==C - CH3 .. CH 3-C - C-C- H ..I--CH3-C-C-C I...) + I H ) :\. I I-.-A I E+Z '----..- H H H H H This 1 ° c arbocation may or may not exist. It i s shown for clarity. H
(b)
H
H
i
H
H
H
hydride shift -
H
(j-
""
147
OCH3
+
NaBr
7-29
(a)
( b)
() H
9.H H - OP03H2
� ------;..
6 \..
1H rl! vo H H
.. �"
(J
�
+
c5-�� c5
H00S03H .
Two possible rearrange ments
H 1 ( CH2 H-C-H I) H C f7 � H + H H,o:.
0 CJ +
• •
H H20
H
..
H H20: •
•
(X I
o+ c5
2. Alkyl shift-ring expansion
+
CH2
H- �J
H 1 H�"\ C rl + C 'H H,o�
O O ......
-
(c ) carbocation formation
�
HO: � H 0oS03�
H H
• •
H � ,/'-.. �I�'H
T
H
'
contin ued on next page
148
H H
Thi s 10 carbocation mayor may not exist. It is shown for clarity
�20·
�
1. Hydride shift
......
"
+
H
("'1+
CH2-�H
G c
+
H2 0
7-29 (c) continued
without rearrangement
H HI
r:-.
H2O:
��?H : �H� C H H2O: +
H
1
H
H�
H ~ 1
-
H
H
..
H
w ith hydride shift
�
..
E+Z
H
H H
H� +(� .. H20: H HH -- .-.------- H H �:
./"'--r�
H H
El product from unrearranged carbocation
rearrangement by alkyl shift
El on rearranged c arboc ation
H 149
7-29 (d) continued E1 on rearranged carbocation
� �H20 :. <:;LJ
00
+
H,o'
+
�
7-30 Please refer to solution 1 -20, page 1 2 of this Solutions Manual.
�
7 -3 1 (a)
(c)
(d)
Br
(h)
(g)
(f)
Br
� r;
Br
0 1 (")
(e)
o
Z E �
7 -32 (a) 2-ethylpent-l-ene (number the longest chain containing the double bond) (b) 3-ethylpent-2-en e (c ) (3E,6E)-octa-l,3,6-triene (d) (E)-4-ethylhept-3 -ene (e) l-cyc lohexy lcyclohexa-l,3-diene (f) (3Z, 5E)-6-ch loro-3-(chloromethyl)octa-l ,3,5 -triene 7 -33
(a) E
(b) nei ther-two methyl groups on one carbon
�
7-34 (a)
F
(E)-l-fluoroprop-l-ene
(Z)-l-fluoro prop-l-ene
0 F
A
�F
(c) Z
F
3-fluoroprop-l-ene
2-fluoroprop- l -ene
(d) Z
[>-F
fluorocyclopropane
(b) C4H 7 B r has one element of unsaturation, but no rings are permitted in the problem, so all the isomers must have one double bond. Only four i somers are possible with four c arbons and one double bond, so
Y y Br
H
�� H �
Br
Br
. B r - �/'o...'�
Br �" � �
�Br
Br
(b) Cholesterol,
C27H460,
four must be rings"
Br
� Br
Br
� �
Br
has fi ve elements of unsaturation . If only one of those is a pi bond, the other 150
7-3 5
(a) (b)
~ ~ cis,trans
trans, trans
2£,4£
;=\' trans,cis
�
2£,4Z
2Z,4Z
2Z,4£
cis,cis
This problem is i ntended to show the difficulty of using cis-trans nomenclature with any but the simplest alkenes. Cis and trans are ambi guous: the first alkene in part ( a) is cis if the two similar substituents are considered, but trans if the chain is considered. The £-Z nomenc l ature i s unambiguous and i s preferred for a l l four o f these i somers. 7-36 (a) and (d) have no geometric i somers (b)
(c) trans-pent-2-ene (E)-pent-2-ene
cis-pent-2-ene (Z)-pent-2-ene
y
B r cis- 1 ,2-dibromopropene � 1(Z)- 1 ,2-dibromopropene
trans- 1 ,2-dibromopropene (E)- 1 ,2-di bromopropene
Br
Br
(f)
trans,trans-hexa-2 ,4-diene (2£,4E)-hexa-2,4-diene
7-37 F (a)
1
>===<
H
(c)
Cl 1 Cl A
Br
7-3 8 ( a)
H
o
Br
cis,trans-hexa-2,4-diene (2Z,4E)-hexa-2,4-diene
H
F
>===<
cis,cis-hexa-2,4-diene (2Z,4Z)-hexa-2 ,4-diene
F
F H di pole moment
Br
1
Br
Br
H
CH3
>===<
(b) =
cis-hcx-3-ene (Z)-hex-3-ene
trans-hex -3-ene (E)-hex -3-ene
Br
(e)
�
H
0
CH3
>===<
Br
dipole moment
=
0
Cl 1 Cl >===<
H H larger dipole moment (no bromines opposing the dipole of the chlorines) ( b) maj or
+�
minor
15 1
(c)
major
mi nor
7 - 3 8 continued (d)
mmor
major
major H H CH3-C-C-CH3 H OH
7-39 Only (a), (b) and (d). (a)
alkene i somers are shown. Minor alkene i somers would also be produced in parts
I
I
I
I
E+Z
II II CH3 CH3 II II 3 CH3-C-C-CH H H CH3-C-C-CH3
(c)
Br
(d)
H
7 -40
(a)
�
-
hindered base
Br
Br
+ Zn
� ---....,..
E+Z
NaOH,
N aI, acetone Zn,
tl
(d)
7-41
0
�
only product
HOC(CH3h + ZnBr2
tl
..
Br
+ NaBr +
Br
Br
Q'
H20
H H CH3COOH CH3-C=C-CH3 I I
OR CH3COOH Q'0H H2S04,
(c)
en
Br
'"
(b)
+
H
NaOC(CH3h
(b)
(a)
H H I I CH3-C=C-CH3
..
KOC(CH3h
..
2
Br ' hv ..
Q' (b)
Br
0
0 0
KOC(CH3h ..
0
+ � + J \ �---�,,�---� major
1\
152
minor
7-4 1 continued
(c)
�
+
�
(d)
� )l U + U major
(e)
� � U +U maj or
rrunor
rrunor
The E2 mechanis m requires anti-coplanar orientation of R and B r. 7-42 The bromides are shown here . Chlorides or iodides would also work.
�
(a)
Br
7-43
(b)
Br
Y
(c)
�
(d)
Br
cSf Br
( a) There are two reasons w h y alcohols do not dehydrate with strong base. The potential leaving group, hydroxide, i s i tself a strong base and therefore a terrible leaving group . Second, the strong base deprotonates the -OR faster than any other reaction can occur, consuming the base and making the leaving group anionic and therefore even worse.
(b) A halide i s already a decent leaving group. S i nce halides are extremely weak bases , the halogen atom is not easily protonated, and even if it were , the leaving group abi l i ty is not significantly enhanced. The hard step is to remove the adj acent R, something only a strong base can do-and strong bases will not be present under strong acid condi tions. 7-44 (a)
�
(b)
(c)
(d)
without rearrangement
)==I
�H
rearrangement
maJor Zaitsev
153
5 continued 7-4with rearrangement hydride shift
7- 4
(a)
major Zaitsev
61, /' ......
/'
major
(c )
+
"y
� +�
(d)
mInor
maj or
minor
maj or
minor
only product from E2anti -coplanar is possi ble only from carbon wi thout CH3 by rem
+
(b)
(e)
C
3
)
D
minor
major
N y
��
ax
l
(CH3hC -.£:::::i CI eq
The conformation must be considered because the leaving group must be in an axi al position. The t-butyl group is so large that it must be equatori al , locking the conformation into the chair shown. As a result, only the Cl at position is axial, so that is the one that must leave, not the one at position 3. Two isomers are possible but it is difficult to say which would be formed in greater amount.
7- 4 7
H
� .�H� .
V
etcH 0I
4
H
H'
,
'-
H
�
- H20
HSO4- ..
-
.. --�
_
OSO
ct
3H H
154
()+C
H
/
� _ _
+
'\ H
H
... '
•
H 20 . " �
El works well because only one carbocation and only one alkene are possible. Substitution i s not a problem here. The only nuc leophiles are water, which would s i mply form starting material by a reverse of the dehydration , and bisulfate anion. Bisulfate anion is an extremely weak base and poor nucleophile; if it did attack the carbocation , the unstable product would quickly re-ionize, with no net change, bac k to the carbocation.
7-48
I
I
I
I
I
�
CH 3-C-C-CH3 HO
:9 o
I
CH3 CH3
CH3 CH3
I
1+
CH 3-C-C-CH 3
----
H+
HO
--I"�
"'9oH2
I
l
I
+
CH3 -C-C-CH3 HO
C H3
II
I�I
CH3 CH3
-H20
methyl shift
CH3 + I CH 3 -C-C-CH 3
I
CH 3-C-C-CH 3
I
: 0 : CH3
I
: OH CH3
The dri ving force for thi s rearrangement i s the great stabil ity of the resonance-stabi lized, protonated c arbonyl group. 7 -49 NBS generates bromine w hich produces bromine radical. B romine radical abstracts an allylic hydrogen, resulting i n a resonance-stabi l ized aUylic radical. The allylic radical can bond to bromine at either of the two carbons with radical character. See the soluti on to problem 6-63. NBS c::=>
CC \.-I
H
hv
B r2
H2 Br
�� +
0
1a
2 Br
____
0
CH 2 HBr
<;:, H
� a
reCYCI
B ro
Br
7-50
Ph � CHCH3
H
X
, CH {" " Ph H
�
25,3R
\
H
Br
CH 2
+
NaOEt
-----l.�
CH3
+
H
Ph
+ '-r----/ � CH2 CH3
Ph
B
H -CHCH 3 Ph
!
�
JYJ
A = E,Z mixture
�
Br
+
Ph
Ph
X
, CH {" " Ph 25 5 ,3
Ph
Ph A
=
Ph
+
E,Z mi xture
Ph
E2 dehydrohalogenation requires anti-coplanar arrangement of H and Br, so specific ci s-trans I s ome rs (8 or C) are generated depending on the stereochemistry of the starting material . Removing a h ydrogen from C-4 (achiral) will give about the s ame mixture of cis and trans (A) from either diastereomer. 155
�
7 - 51
by 4 kJ/mote
more stable than
~
more stable than
by 16 kJ/mole
Steric crowdi ng by the (-butyl group is responsible for the energy difference. In cis-but-2-ene, the two methyl groups h ave only s l i ght i n teraction. However, in the 4,4-dimethylpent-2-enes , the l arger size of the t-butyl group crowds the methyl group in the cis isomer, increasing its strai n and therefore i ts energy. 7-52
() C1
endocyclic \. tri substituted
V
V ()'
endocyclic \tri substituted
exocyclic disubstituted )
9 kJ/mole
V
exocyclic tri substituted)
5 kJ/mole
A standard pri nciple of science is to compare experiments which differ by only one vari able. Changing more than one vari able c louds the interpretation, possibly to the poi nt of invalidating the experi ment. The first set of structures compares endo and exocyclic double bonds, but the degree of substitution on the alkene i s also di fferent, so this comparison is not valid-we are not isolating si mply the exo or endocyclic effect. The second pai r is a much better measure of endo versus exocyclic stability because both alkenes are tri substituted, so the degree of substitution plays no part in the energy values. Thus , 5 kJ/mole is a better value. 7-53 (a) CH3CH 2CH =CH 2 (b) No reaction-the two bromines are not trans and therefore cannot be trans-diaxial .
(e) (d)
0 (X)
bromines are diequatorial cannot undergo E2 H
H
No reaction-the bromines are trans , but they are diequatorial because of the locked conformation of the trans-decalin s ystem. E2 can occur only when the bromines are trans diaxial.
156
E2, the two groups to be eliminated must be coplanar. In confonnationally mobile systems like 7-54 Inmolecules, acyclic or in cyclohexanes, anti coplanar is the preferred orientation where the H and leaving group are 1 80°apart. In rigid systems like-norbornanes, however, SYN-coplanar (angle 0°) is the only possible orientation and E2 will occur, although at a slower rate than anti-coplanar. The structure having the H and the Cl syn-coplanar is the trans, which undergoes the E2 elimination. (It is possible that the other H and Cl eliminate from the trans isomer; the results from this reaction cannot distinguish between these two possibilities.)
H
Cl
Cl
H Cly'/syn �oplanar
H Cl extremely slow to eliminate H and C\ not coplanar It is interesting to note that even though three-membered rings are more strained than four membered 7-55 three rings, -membered rings are far more common in nature than four-membered rings. Rearrangement from a four-membered ring to something else, especially a larger ring, will happen quickly. .
trans
without rearrangement H� + 2 H20:.. � CH y--
unstable 1 ° carbocation- short lifetime if it exists at all
mmor
with rearrangement-hydride shift H H20: (P�H2 3° carbocation, but still in a strained 4-membered ring
•
157
minor minor
7-55 continued with rearrangement-alkyl shift-ring expansion ��H2 H
-�.
C "......-A H2
("�-H�. O[ H H2C-C-H H20: I H H 2° carbocation on 5-membered ring HOORAY! H ('xCH/'RiI . .
.� �
t
-H 0
2
MAJOR
carbocation-terrible!-will rearrange; can't do hydride shift, must do alkyl shift ring expansion
1°
=
of adjacent abstraction gives-bridgehead Halkene violates Bredfs Rule
though it is 3°; this is an unstable carbocation even bridgehead carbons cannot be sp 2 (planar-try to make a model), so this carbocation does a hydride shift to 2°, more stable carbocation a
t hydride shift
this 2° carbocation loses an adjacent H to form an alkene; can't form at bridgehead (Bredt's Rule) only one other choice
158
8-1 Major products are produced in greatest amount; they are not necessarily the only products produced. Br (c) (b) Cl (d) (a) Br H B' +H �H
CHAPTER 8-REACTIONS OF ALKENES
8- 2 H H
H
H
H H
H+
---
H
H
O H
+
allylic
produced in about equal amounts (mixture of cis and trans) H H
-t� H H-C+ H H .Q H H H H :Br: 1 .. H H H -O- H Br H H H ·
\
+
1-bromobut-2-ene 3-bromobut-l-ene Because the allylic carbocation has partial positive charge at two carbons, the bromide nuc\eophile can bond at either electrophilic carbon, giving two products. 8-3 � II II II L1 2 H3C-C-O· (a) initiation steps H3C-C-0-v' O-C-CH3 •• °
o
+ H3C-C-O· H-Br "�JG °
II
propagation steps
�
~ C� _)
+
• Br
(this radical has another resonance fonn but it is not gennane to this mechanism)
°
-
+
• Br
the 3° radical is more stable than the 1° radical
-
���
II H3C-C-OH •• °
-
159
B' • Br k 1-bromo-2-methylpropane +
8-3 continued (b) initiation steps H2 H 3C-C -0· • • • •
+
+ • Br
H-Br
,,--JG
propagation steps the 3° radical is more stable than the 2° radical
+
•
Br
I-bromo-2-methylcyclopentane (cis + trans)
2
�
(c) initiation steps
-
+ H
propagation steps
I
-
/C Ph' ·
y Br I
� I H
Ph
H
•
Br
the benzylic radical is stabilized by resonance, and more stable than the aliphatic radical
Br
+
• Br
2-bromo-l-phenylpropane (recall that "Ph" is the abbreviation for "phenyl") 8-4 (a)
(c)
�
(y OH
(d)
OH
�
HBr ROOR H2SO4 �
H2SO4 �
•
•
•
(b)
Br�
0
�
HBr
HBr ROOR
•
160
..
� Br
�
C)'
~ Br
HBr
..
Note: A good synthesis uses major products as intermediates, not minor products. Knowing orientation of addition and elimination is critical to using reactions correctly.
CH3 H �
CH3 H H * � H ----"" HH CH3 CH3 �'--O-H CH3 CH3 + "'" + 1 H 2° H HI
8-5
CH3 + I CH3 C CH3 CH3/ H
CH3 tf CH3 CH3
't -H.. CH3 \[)oo CH3
8-6
(a)
:O I H
\f '-..
:O- H H� I H
>=<.H
CH
(a)
CH3
(b)
CH3
CH3 CH3tr H CH2CH3
3°
CH3 jf CH3 CH3
U
0
1
3
,CH3 CH3/ "1' CH3
O
H30+
H30+
+
0d:J
from 3° benzylic carbocation
CH3
(
CH3 Af CH3 gOAC
CH2CH3 :
/ I CH3 1f 0 0
CH3
i
�
H\
ACO :
NaBH4
+
H
(c)
..
..
HI
2, 3-dimethyl-2-butene
+ Hg(OAc)
CH3 I
0
�H H'
( b)
from 3° carbocation 8-7
+
..
4 ....
'6H
CH3 + I CH3 / C � CH3
.. CHJ ° O -H O HO / ')H I / H H k� H 2,3-dimethy-1 2-butanol
CH3 I CH3 ;C CH3
proton removal
..
•
nucleophilic attack by water
CH3
methyl shift
16 1
CH3
gOAC
H CH2CH3 o
I
H
H3C-CII
°
"Ac" = acetyl
II H3C-C'-O o
"OAc" or "AcO" = acetate
8-8
CH 3
(f'::
HO
(a)
(d)
"
(OAC)
(b)
�
&,
(b)
(c)
Hg(OAc}z CH30H
KOH,�
.-
.-
Hg(OAc}z
8-10
�
(a) , (b)
6
(e) , (f) 8-11
(a)
(b)
(c)
..
� �
� Br
..
..
B H3
°
THF
-----;.-�
H2O
N aB H 4
..
OH
C
B H2
H202
.-
Hg(OAc}z
NaBH 4
..
KOH,�
..
.-
Hg(OAc)
o.
H2 02
HO-
,
H O
� OH
.-
HO-
HaB:
H202
I
l
Hg(OAC) " OCH3
Using an acid-catalyzed hydration In part (c) would initially form a 2° carbocation that would quickly rearrange to 3" on carbon-3. The desired product would not be synthesized.
�
�
HO-
H2 O
�
.-
B H2
BH3 0TH F
O
:
OCH 3
Hg(OAc}z
�
B H3 0T HF
OCH3
II
OCH J
NaBH4
6
CI
(c)
H
B H3 0THF
�
(c), (d )
U
O-
CI
8-9
(a)
X
H
H20 2
.-
�
HaO: OH
C
.-
HO-
HO�
� OH
..
�
B H3
°
THF .-
H2 0 2 H O-
162
.-
~ OH
8-12 Instead of borane attacking the bottom face of I-methylcycJopentene, it i s equally likely to attack the top face, leading to the enantiomer.
s=<:.
6
BH3 · T HF ..
H
8-13 (a)
CH3
(b)
,'QH
H
V � +
OH
OH mi nor-steric hindrance to attack of B H 3
major
-
CH3 S
8-14
(a)
Et
Et
>=< H
Et
B H3 · THF
Me
H
Z
S
� � K
'' Me
Et
" OH H )
enantiomers R
( b)
Et
H
>=< Et
Me
S
� � �K�
Et
B H3 • THF
H
Me'
Et
�---------..
E
H
V
enantiomers
The enantiomeric pair produced from the Z-a l kene i s diastereomeric with the other enantiomeric pair produced from the E-al kene. Hydroboration-oxidation is stereospeci fic , that i s, eac h a l kene gives J specific set of stereoi somers, not a random mi xture.
8- 15 (a)
(b)
CO
CO
Hg (O Ach
NaBH 4 ...
...
H2 O
CO OH
B H3 · THF ..
H 202 HO -
..
cp OH
163
continued C (c) o
8 -15
8-16
c5 CH3 CH3
�� empty p orbital in \bottom "'-- :Br: planar carbocation � CH3 Br The planar carbocation is responsible for non-stereoselectivity. The bromide nuc\eophile can attack from the top or bottom, leading to a mixture of stereoisomers. The addition is therefore a mixture of syn and anti addition. 8-17 During bromine addition to either the cis- or trans-alkene, two new chiral centers are being formed. Neither alkene (nor bromine) is optically active, so the product cannot be optically active. The cis-but-2-ene gives two chiral products, a racemic mixture. However, trans-but-2-ene, because of its symmetry, gives only one meso product which can never be chiral. The "optical inactivity" is built into this symmetric molecule. This can be seen by following what happens to the configuration of the chiral centers from the intermediates to products, below. (The key lies in the symmetry of the intermediate and inversion of configuration when bromide attacks.) IDENTICAL-SYMMETRIC .
�Br " H H"'�M H3 GH3 �' a Be nuc\eophile could : r OR :Br : , �: (
��--------� "\
�------�
attack at either carbon CH3 H-
.
+
l
J
+
H",:\
Br+
H CH3 Br Br Br H f + '---<' Br + H '" ", H 'H � / Br Br CH3 CH3 CH3 2S,3S ENANTIOMERS 2R,3R continued on next page
-+
':r-(
H �· · I
-+
H Br H + Br CH3
--------------------------------------------------______________________________________ u __________ u __ •
164
8-17
continued
ENANTIOMERIC � TRANS \ CH � CH3 ( B! H U "" ,U " \B/ bromide attack on this 1V\""11"'CH+ H'/CH3 3 r bsame romonium ion gives the + results a Be nucleophile could :Br: OR :Br: attack at either carbon �------�
,
�( t
�---------
'
••
J
H H CH H Br Br )1-J, Br Br H CH3 � + H + Br Br + H \'\� / �" ' H CH 3 Br Br H CH3 CH3 CH3 2R,3S ID ENTICAL-MESO 2R,3S CONCLUSION: anti addition of a symmetric reagent to a symmetric cis-alkene gives racemic product, while anti addition to a trans-alkene gives meso product. (We will see shortly that syn addition to a cis alkene gives meso product, and syn addition to a trans-alkene gives racemic product. Stay tuned.)
+
+
Enantiomers of chiral products are also produced but not shown. Br .. � + :Br: (a) Br-Br 8-18
(b) (c)
(Hex)
+
+ 0�. Br\ + Br-Br --- , U )
-
___
Br
0 Br "'Br
�e1-I
'>---f/ : :Br '" Br H"'/ . � " It "1 H" Et �' Hex . Et
Bromide will attack the other carbon of the bromonium ion as well.
165
�
continued (d) Three new asymmetric carbons are produced in this reaction. All stereoisomers will be produced with the restriction that the two adjacent chlorines on the ring must be trans. Cl Cl 8-18
-
+
--G--I<
+
Cl-Cl
..
l
:Cl:
j
:Cl:
h r�Cl I Cl� CH3
C CH �
8-19 The trans product results from water attacking the bromonium ion from the face opposite the bromine. Equal amounts of the two enantiomers result from the equal probability that water will attack either C-l or C-2. B, - �r <:v +
Q
H
H
Q�
water will do nucleophilic attack at either carbon .. H20:
a
+
equal amounts of enantiomers enantiomer of product H 8-5 °\O+_ H �\ H in Solved Problem :OH O
from Solved Problem 8-5 H �O O-H the bromonium ion : -H � �� "",CH3 CH ... shown here is the CH3 .. 3 �+ U enantiomer of the one Br \ \ Br > Br> shown in Solved \ Problem 8-5 H H from Solved Problem 8-6: the bromonium ion shown is meso asHit has a plane of symmetry; attack by the nucleophile at the other carbon from what is shown in the text will create the enantiomer H H :Cl: "'Br + .. ,Br 8-20
0
a',
-
//� � show products from HI;I � both nucleophiles attacking at this carbon
oR
0
I
-
H
U
0 0
a
�;
Br ' " a H
OH
'"' H
H
these are the enantiomers of the structures shown in the text 166
8-21 The chiral products shown here will be racemic mixtures. H (b) 1 (C)H3C
-;;61
plus enantiomer (e) CH3 ,OH Rr + 8-22 (a)
(y
�
O
H
HO
CH3 ,Cl
plus enantiomers
CH
(c)
Br
(y Rr
CI2 H2O H2SO4
l
(b)
C
..
~
..
6
C1
H � plus enantiomer
_
C
CI
6 0
..
..
KOH
(y (c) C):)
CI2
(d)
CH3 ,OH
!t,.
,----< '
H3C l-!
Cl
\"H"",'CH-�
HO/
plus enantiomer
. 6 . .. CI
CI2
H2O
0H
Cl
8-23 (a) � (d) -0-< (b) � 8-24 Limonene, CIOH16, has three elements of unsaturation. Upon catalytic hydrogenation, the product, CIOH20, has one element of unsaturation. Two elements of unsaturation have been removed by hydrogenation-these must have been pi bonds, either two double bonds or one triple bond. The one remaining unsaturation must be a ring. Thus, limonene must have one ring and either two double bonds or one triple bond. (The structure of limonene is shown in the text in Problem 8-23(d), and the hydrogenation product is shown above in the solution to 8-23(d).) 8-25 The BINAP ligand is an example of a conformationally hindered biphenyl as described in text section 5 -9A and Figure 5 -1 7 . The groups are too large to permit rotation around the single bond connecting the rings, so the molecules are locked into one chiral twist or its mirror image. "rear" !t,.
H2O
"front" "front" naphthalene naphthalene ring nng These simplified three-dimensional drawings of the enantiomers show that the two naphthalene rings arc twisted almost perpendicular to each other, and the large -P(Phh substituents prevent interconversion of these mirror images. 167
8- 26
(a)
(b) (c)
8- 27
Methylene inserts into the circled bonds.
S
c5
+
� + :CH2 �H
~
d
+
( b)
(a)
BOTTOM VIEW site of origi nal double bond
(c)
__
the original 6-membered ring is shown in bold bonds
SIDE VIEW
8- 28 (a)
( b) (c)
(a)
8- 29
F 0
OH
6 0
CH21 2 Zn(Cu) CH2Br2 50% NaOH (aq) •
R
•
H2SO4 L!
•
(b)
Br
�H
0 0
Br CHCl3 50% NaOH (aq) •
(c)
168
cjCl Cl
CHo
(d)
CQ H
"
,
o
H
(a)
Rc
' =o
8-30
fa . H V ' 0l...''\
H "C ( Ct" H Et" cis 'Et "
'1
•
,
\"
ENANTIOMERS
A"Et" H
H""Et,
r--------A.......
'
J-
Et �"
(b)
:O: �
HO
'
Rc
, =o
EtH
I, II/
+
HEt \"
'o' H ,v ('\t "Et c" " "c Et" trans - 'H ,
OH
,\\
fa l.... H
'I
\
•
:O:�
, " H' A�" t Et
IDENTICAL-MESO
"
Et
�Ofl floyfl
HO�H" Et HEt"'/ �OH stereochemistry shown in Newman projections: +
H'�f\]\ E;H t H-O: OR :O-H H tI H "
H+
Etf!
"
I
Ofl
,---<" H
H O�Et +
2;.
fl0
+
Et •
• •
:O-H
)(
'>-,!-I ---Y
HEt"7 "OH2
+
---
E
r�--------A------�\
� ( ) "' I
-H +
"
"
-
•
H+
+
---
,
----�
:O u-
U (\�" Et H'�1\ t H-O: OR :O-H H + H "
"
.
,
.
I
,
"
.. I
El f! H Et flO Ofl � - H + ,--< + H 0. / "HEt HEt"'/ "O H 2 +
•
2
-
• •
"
Q)H
• •
OH OH Et H rotate Et H .. H Et M H � : £ Et H Et H � Et OH trans MESO Remember the lesson from Problem 8-17: anti addition of a symmetric reagent to a symmetric cis-alkene gives racemic product, while anti addition to a trans-alkene gives meso product. This fits the definition of a stereospecific reaction, where different stereoisomers of the starting material (cis and trans) are converted into different stereoisomers of product (a dl-pair and meso form). v x:
---
---
H�
H-
169
Et
8-31
R
G0, c t )� ----/' H
" " "
8-32
(a)
' C=O
t...H 0'
trans
==
I
C
•
''''CH1
�
8-33
�
H
/
"" ,
� \-
o
H
CH 3 CH2 �
OH
Ho
� '-/ / ��CH2CH3 HO
�
O
HO",
H " 'O H
H+
t
CH2CH 3
H
�
� \. or )
H
�
H
OH
- H+
HO
M
C-O
II
�
(d) � rotate.
cj
\
o
<
R=
C-O
o
� o
,
II
All chiral products are racemic mixtures. (b) o
(c)
R
g
2
+
anti addition trans-alkene gives meso product to
H H
CH2CH 3
t�is is the (e) � trans product � OH OH
H
H
CIS
�
+�
--- 1-Y
enantiomers 8-34 All these reactions begin with achiral reagents; therefore, all the chiral products are racemic. (c) H (b) (a) H
H
---
H
• •
H
--
CH 3 -O:
CH3 0
H
H
I
H
(d)
(): CH 3
HO
OH
H
OH
H
'"
o:�:
H CH2CH3
=f��
HO
H
OH
H
(e)
CH 3
�
OH
OH
H same as (d)!
HO
H HO
(f)
'"
H CH2CH 3
=f�� H
Refer to the observation in the solution to Problem 8-35 on the next page. CH3
HO
H
H
CH3
170
H
CH3
H0
Hl
+
OH
same as (c)!
HO
+� CH3
X �
OCH3
HO
/ . � "'CH2CH 3 ==:> HO
CH1
a
=f�H
� CH2CH l
HO
H CH3
l
H
8-35
�
(a)
HO
HO
(c)
(d)
� �
OH
�
(b)�
CH3C03H .. H2 O
rotate
Os04
rotate
H20 2
..
..
..
meso
OH
� HO
racemic (d,l)
OH
� HO
meso
racemic (d,l)
OH
Have you noticed yet? For symmetric alkenes and symmetric reagents (addition of two identical X groups): cis-alkene + syn addition
cis-alkene + anti addition
meso
----7 ----7
trans-alkene + syn addition
trans-alkene + anti addition
racemic
----7
racemic
----7
meso
Assume that cis/synlmeso are "same", and trans/anti! racemic are "opposite". Then any combination can be predicted, just like math!
+1
+1 -1 -1
+1
x
+1
=
x
-
1
=
-1
x
+1
=
-1
=
+1
x
-1
8-36 Solve these ozonolysis problems by working backwards, that is, by "reattaching" the two carbons of the new carbonyl groups into alkenes. Here's a hint. When you cut a circular piece of string, you still have only one piece. When you cut a linear piece of string, you have two pieces. Same with molecules. If ozonolysis forms only one product with two carbonyls, the alkene had to have been in a ring. If ozonolysis gives two molecules, the alkene had to have been in a chain. (a) two carbonyls from ozonolysis are in a chain, so alkene had to have been in a nng
(b) two carbonyls from ozonolysis are in two different products, so alkene had to have been in a chain, not a ring
(c) two carbonyls from ozonolysis are in two different products, so alkene had to have been in a chain, not a ring
E or Z of alkene cannot be determined from products
8 - 37 (a)
�
03
�
Mc,�
�o
(b)
o� H
+
� OH
O
17 1
8-37 continued
0=0
(c )
U
(d)
(e)
(f)
03
Me2S
Me2S
03
•
•
=0 0 �H 0
•
•
U U
KMn04
t!.
cold
�OH
•
0
0
•
0
0
0
KM n04
+
d
·aH
1I110H
8-38 The representation for a generic acid w i l l be H-B, where B is the conj ugate base.
:x:r+}
+/ J. C
/ B:-
H P Te\ � /, -
8-39 Catalytic B F3 reacts with trace amounts of water to form the probable catalyst: -
F I
+1
dimer
_ _
catalyst
etc. ...
tetramer
trimer 172
H
I
8-40 H-B is used here to symbolize a generic acid.
-
:B
�
alkenes
polymers
�
(colored)
Note : the dashed bond symbol is used here to i ndicate the continuation of a polymer chai n .
8-41
H
I ------
H
I
+
C-C ·
I
H
\
H
/
n/)\...( \H �
H
I
H
I
H
H
Ph
Ph
H
I
------ C-C-C-C ·
C=C
I �
Ph
I
H
H
I
I
I
I
10 radi cal , and not resonance-stabilized this orientation is not observed
Orientation of addition always generates the more stable intermediate; the energy difference between a 10 radical (shown above) and a benzylic radical is huge. The phenyl substituents must necessarily be on alternating carbons because the orientation of attack is always the same-not a random process.
8-42 For clarity, the new bonds formed in this mechanism are shown in bold .
\
H RO ·
+
H
2 RO·
RO - OR
/
C=C
.!)\...( \H ��
---t.�
H
C C �'"o"+-",, , /V \ I ( H H
I
H
H
H
H
H
/
\
..
RO-C-C.
I
I
H
I
H
H
H
H
H
I
I
I
I
I
c / \ l /
I
H
I
H
I
H
I
H
I
H
H
H
H
H
H
H
I
I
I
I
I
I
I
etc. ---- RO-C-C-C-C-C-C·
173
)k)
RO-C-C-C-C ·
H
H
I
H
\H
8-43
II o
\ n/
H
COCH3
C=C
HO :
.. �/ H
---
\
Me
. .-
:0 : II H C-OCH3 I _I 11: HO-C-C : I I H Me
\
:0 : I H C-OCH3 I II HO-C-C I I H Me
.
o
II COCH3
\ n/ C=C
\
/ H
Me . .
· 0· · · COOMe 11 H H C-OCH3 I I _I HO - C-C-C-C : I I I I H Me H Me
I
•
•
!
· · · 0· COOMe I H H C-OCH3 I I II OO - C - C-C - C I I I I H Me H Me
I
ele.
COOMe COOMe COOMe COOMe COOMe Plexiglas
"
Me
Me
Me
Me
Me
D ashed bonds mean that the chain continues.
8 -44
0
0 II
H3C-C-00H ..
H30+
OH
a
" <X ;Y '
Na
-
OH
O
B r� �
o
<X F o
<X � "'o
O
4
MCPBA
OH
174
OH
B r2 -
hv
�
8-45 In the spirit of this problem, all starting materials will have six carbons or fewer and only one carboncarbon double bond. Reagents may have other atoms. (a) Na . ::
O:::J O::�-:y H BC�
major product from resonance stabilized radical on 10 and 30 C
�
anti-Markovnikov syn stereochemistry
� 0: � CH,I, 0: H o
(b)
0
(c)
0
Bhvr ey Br rB BC-p KOH Br 2
(or
•
NBS)
2
C
�
3
Zn,
CuCI
Bhv·r 0- Br
( a)
(d)
KOH
2
•
-
NBS)
MCPS: (or
only substit is possible
,A/ ° � T
H
Ao 1 +
0
175
+ Na
-
0
-
H2 0
�:3 _aCH3 " " , BH OH 2 H intermediate Cl L5F (g) r--peroxides do not affect HCl addition
(c)
(f)
on
0H
./
�
Please refer to solution 1-20, page 12 of this Solutions Manual.
Br (el e/'
:
o-
� \{ 0OH
hi
8-47
CH3
eyC::N
� 8-46
o
q II
,
V
continued CH3 '" O'OH ""OH
8-47 (h )
CH3 (i) ('f. IIOH V ""OH
QH
(I)
(0)
q; HgOAc intennediate
(a) initiation RO-OR 2 RO· :JnBr .. ROH + H RO.0 propagation 8-48
r-,ri.
----
-
Br·
H
+�
r�
H
+
(n)
CO OH
(p)
q;Cl
Br·
Br
HH L" +� C �H--C:S03; Q '--- :9-H
( b)
H
176
-..� •
+
•
Br
8-48 continued
H
H-nBr � � n
(c)
•
I
.I
(d)
H
1(..1- � � :Br: ..
fH
'Y' Br ----... Br HO- H-C-Br --- :C-Br --- Br,
+
......
,.--.10..1
I
Br I
f"
Br I
�
+
(fit:: I :B�: .
.
:CBrz B'
+ � H CI H��CH3 X+C � CH3..O-H HOC;3 � OCH3 H � •
I
•
_
B_r_-B" ; r:J � '--- .Br. •
•
•
:
& . �� :Ci: +
F�'
•
Br
� Br Br
� CI
. � CH30 -H H�CH3 Jy OCH • •
177
�
.
t
Br
'J'
I
OR
(f) � �
•
aA
..
H JY :C1: H--Q� HXH� f/ C H
H
H
hydride shift
Br
� 00 � Brz V
(e)
OR
'fH....,C... +
.
3
continued (g) H � 0 . _H H-B H J... - < T 8-48
H 3C
•
Ph
Recall that "Ph" is the abbreviation for phenyl.
(a)
8-49
(b)
(0)
(d) (c)
(f)
(g) (h)
(i)
(f (f (f (f (f
(f
(f
Hg(OAc)z
..
HEr ROOR
Hg(OAc)z
..
NaBH4
NaBH4
..
.
OS04 H 20 2 or cold, dilute KMn04 ..
Zn(Cu)
..
50%
NaOH (aq)
� OH OH V � CI OH V
� Br V 'Br 178
�H H
Ph
Ph
HQ "
8-50 (a)
CH3
-Q--("
�
OH CH3
(e)
~
HO
H (d)
(g)
CH3
,
OH
OH
B
-0-<
(h)
8
Br OH
(j)
'
CH3
(m)
D--f
(c)
~ H
(f)
+ CO2
0H
OH
",
CH3
-b-\
(i)
Br
-b-f
Ci Cl
"
�
CH,o
(I)
CH3
� H
D-f Br
C I�
(k)
+ CH2=O
HO
H
Cl
Br
HO "
D--f0
(b)
0
Br
OCH3
C 3
8-51 Each monomer h as two c arbons in the backbone, so the substi tuents on the monomer w i l l repeat every two carbons in the polymer. Dashed bonds indicate continuation of the polymer chai n . line formula representation Cl
� F
F
Cl I
N
C
�
I
Cl
Cl
Cl
Cl
Cl
Cl
I
I
--- --- -CH2CHCH2 CHCH2CHCH2CH -- --
pol yvinyl chloride, PVC
0 F -;-III
Cl
F F
F
F F F I I I I I I ---- C - C - C - C - C -C---I I I I I I F F F F polytetrafluoroethylene, PTFE, Teflon
F
F
N
III
N
III
N
III
N
III
C C C C I I I I --- ----CH2CHCH2CHCH2CHCH2CH - - - -polyacry lonitrile , Orion 179
-
-
F
F
F F
N
F
F
F
F
N
F
F
F
F N
III
III
III
C
C
C
F
F
F
F N
III
C
F F
.. ..
8-52 Without divinylbenzene, individual chains of polystyrene are able to sl ide past one another. Addition of di vinylbenzene during polymerization forms bridges, or "crossl i nks", between chains, adding strength and rigidity to the polymer. Divinylbenzene and similar molecu les are called crosslinkjng agents.
two polystyrene chains crossli nked by a divinylbenzene monomer shown in the dashed oval
H CH / 3 \ H CH �\.. ,C/ vC\ I r H CH� RO-CI -C. I HI CH 3
8-5 3 A peroxy radical i s shown as the initi ator. Newly formed bonds are shown in bold.
"-f--'
_-I.�
etc. ---
8-54
'C-OCH2CH3 I H2C=CH 0,
ethyl acrylate
8 - 55 In each case, the compound (boxed) that produces the more stable c arbocation is more reactive. (a)
(b)
0
3"
6
180
(c)
�
allylic
�
8-56 Once the bromonium ion i s formed, i t can be attacked by either nuc\eophile, bromide or chloride, leadi ng to the mi xture of products.
Q
Br
--
~
Br �
H
H
H
Br
��-� � 'i---Y
�..
:Cl :
H
H
Cl
8-57 Two possible orientati ons of attack of bromi ne radical are possible: (A) anti -Markovnikov
>=
+
•
Br
--
(B) Markovnikov
>=
+
•
Br
--
\ .
C,
/ Br 3 ° radical
'l-
r
� H2
HBr
.-
h
+
•
Br
Br
HBr
�
y;
+
•
Br
H
1° radical
The first step in the mechani s m i s endothermic and rate determi ning. The 3° radical produced in anti Markovnikov attac k (A) of bromine radical is several kl/mole more stable than the 1° radical generated by Markovnikov attac k ( B ) . The Hammond Postulate tells us that it is reasonable to assume that the activation energy for anti -Markovnikov addition is lower than for Markovnikov addition. This defines the first half of the energy di agram. The relative stabi l i ties of the final products are somewhat difficult to predict. (Remember that stabi lity of final products does not necessari l y reflect rel ative stabi lities of intermedi ates; this is why a thermodynamic product can be different from a ki netic product.) From bond dissoc i ation energies (kJ/mole) in Table 4-2 : anti -Markovnikov
Markovnikov
H t0 3°C
3 81
H to 1 ° C
410
Br to 1° C
285 666 kJ/mo\e
Br t0 3°C
272 682 kl/mole
If it takes more energy to break bonds in the Markovnikov product, it must be lower in energy, therefore, more stable-OPPOSITE OF STABILITY OF THE INTERMEDIATES! Now we are ready to construct the energy di agram; see the next page .
18 1
8-57 continued
t
Markovnikov
, ,'/
,
,
,
- ..
,
,
anti-Markovnikov
reaction
--
It is the anti-Markovnikov product that is the kinetic product, not the thermodynamic product; the anti Markovnikov product is obtai ned since its rate-determining step has the lower activation energy.
GS
8 - 58 Recall these facts about ozonolysis: each alkene c leaved by ozone produces two carbony l groups; a n alkene i n a chain produces t w o separate products; a n alkene i n a ring produces one product in which the two c arbonyls are connected. (a)
O
H
+
(b)
CH2=O
O
+
0
�
H o CHO
0 H
0
8-59 (a)
(b)
CY
0
¢
(d)
H 0
CHO
CH3C0 3H ..
H2 O
CH3
U
'OH
OH
O
(or cold, dilute KMn04) cis + syn (cis double bond plus syn stereochemistry of addition)
OH
OH
trans + anti (trans double bond plus anti stereochemistry of addition)
182
8-59 continued
B r2
(c )
� Br � Br
..
�
1
This structure shows a trans alkene in a to-membered ring, just the rotated view of the structure to the right.
Br2
(X
..
trans-cyclodecene
o
(d)
C l2
C!
tate around C-2
�'" : H
� 2
Br
"" OH
CO
(e)
anti addition of B r 2 requires t ra ns alkene to gi ve meso product
~
B H3 - THF ..
�
OH
(f)
� V-/
CO
or
8 -60
or
� �
Hg (OA c h� Na BH _ _ _ _ _ -l.. t, CH30H
A ) Unknown X, CSH9Br, h as one element of unsaturation. X reacts with neither bromi ne nor KMn04, so the unsaturation in X cannot be an alkene; it must be a ri ng. B ) Upon treatment with strong base (t-butoxide), X loses H and Br to give Y, CSHg, which does react with bromine and KMn0 4 ; it must have an alkene and a ring. Only one isomer is formed. C) Catalytic hydrogenation of Y gives methylcyc lobutane. This is a B IG c l ue because it gi ves the carbon skeleton of the unknown. Y must have a double bond in the methylcyclobutane skeleton, and X must have a B r on the methylc yc lobutane skeleton . D ) Ozonolysis of Y gi ves a dialdehyde Z, CSHg0 2' which contains all the original carbons, so the alkene c leaved in the ozonolysis had to be in the ring. Let's consider the possible answers for X and see if each fits the information.
.---:f' > U
KO-t-B u
1) I
if this is X
ozonolysis
>
d'
°
+
O=CH2
DOES NOT FIT OZONOLYSIS RESULTS so this c annot be X and Y
this must be Y; only one product
more possi bilities on the next page 183
d KO-t-Bu
8-60 continued 2)
Br
>
1
if thi s i s X
c(
3)
i f this
Br is X
4) P Br
>
O t -
-
�
Q
+
major
d
u
0(
Y would be a mi xture of alkenes , but the e l i mi nation gives only one product. We already saw in ex ample 1 that the exocyclic double bond does not fi t the ozonolysis resul ts so this structure cannot be X.
0(
Y would be a mixture of alkenes, but the e l i mination gi ves only one product, so this structure of X i s not consistent with the i n formation provided.
minor
major
KO-t- B u
K
d
+
minor
+Q
l
ozOnOIYSiS
SAME COMPOUND ; this must be Y ; on ly one product
if thi s i s X
>
r<
o
Z, a dialdehyde
The correct structures for X, Y, and Z are gi ven in the fourth possibi l i ty. The on ly structural feature of X that remains undetermined is whether it is the cis or trans i somer.
trans
'(S)(
8-6 1 The clue to the structure of a-pinene is the ozonolysis. Worki ng backwards shows the alkene position . o
bac kwards
o
>
� became carbony l carbons a-pinene
'fit �� T {iX � �o {L:(
After ozonolysis, the two carbonyls are sti l l connected; the al kene must have been in a ring, reconnect the two carbony l c arbons with a double bond.
~ ~
�
BT
'
B r2 ..
or
, I I CH3 Br
CH3
A
B r2 ..
H2O
A
Br
BT
H2SO4
'OH CH3
�
Zaitsev product
..
B
PhC03H
C
H 30+
..
"
OH
..
"
D
E
184
OH
so
8-62 The two products from pennanganate oxidation must have been connected by a double bond at the carbonyl carbons. Whether the alkene was E or Z cannot be determi ned by this experiment. CH3(CH2) 1 2 CH = CH(CH2hCH3 shown here as Z which is the naturally-occurri ng isomer 8-63 Unknown X
must have three alkenes in this skeleton
Pt
CH2 = O +
Unknown X
� o
H
0
o
+
A/ H o
0
There are several ways to attac k a problem like this. One is the trial-and-error method, that is, put double bonds in all possible positions unti l the ozonolysis products match. There are times when the trial-and-error method is useful (as in s i mple problems where the number of possibilities is few), but thi s is not one of them. Let's try logic. Analyze the ozonolysis products carefully-what do you see ? There are only two meth yl groups , so one of the three termi nal carbons i n the skeleton (C-8, C-9, or C- l O) has to be a =CH2 . Do w e know which terminal c arbon has the double bond? Yes, we can deduce that. If C- l O were double-bonded to C-4, then after ozonolysis, C - 8 and C-9 must stil l be attached to C-7 . However, in the ozonolysis products , there is no branched chain, that i s , no combination of C-8 C-9 C-7 + C- l . What if C-7 had a double bond to C- l ? Then we would have acetone, C 3COC 3, as an ozonolysis product-we don 't. Thus, we can't have a double bond from C-4 to C- l O. One of the other terminal carbons (C- 8 ) must have a double bond to C-7 .
H
8�
9
� 6
H
+
+
10
5
The other two double bonds have to be in the ring, but where? The products do not have branched chains, so double bonds must appear at both C- l and C-4. There are only two possibi l i ties for this requirement. I
H)---
or
}{}-
II
Ozonolysis of I would g i ve fragments contai ning one carbon, two carbons, and seven carbons. Ozonolysis of II would gi ve fragments containing one carbon , four carbons, and five carbons. Aha! Our mystery structure must be II. (Editorial comment: Sc ience i s more than a collection of facts . The application of observation and logic to solve problems by deduction and inference are critical scientific skills, ones that distinguish humans from algae . )
185
8 -64 In this type of problem, begin by detennining which bonds are broken and which are fonned. These w i l l always give c l ues as to w hat is happening.
H+
formed
1
goes to most e lectronegati ve atom
+ HOq� .. H
protonated epoxide opens to gIve the most stable carbocation (3()
\ 3 0 carbocation looks
for electrons, fi nds them at nearby alkene, formi ng a 6-membered ring (yes ! )-leaves a 3 0 carbocation
+
8-65 See the solution to Problem 8 - 3 5 for simpl ified examples of these reactions . (a)
(b)
(c)
(d)
HOOC J HOOe
OH HO � H HooeH1 ""eOOH OH Hooe � " '" IleOOH HO H
eOOH
trans
,,
fCOOH
7--< �
OH '---< IH HO/ �/eOOH
eOOH
eOOH
trans
Hooe�
(COOH (COOH
syn � racemic
OS04
186
anti
cis + anti
OH HO � "1 �/HlleOOH Hooe H
+
cis
+
�
syn
�
meso
racemic
�
meso
8-66
B03 - THF
u
..
HO
� H �nB
-
H 20 2
CH3
a
111 0
..
I I I OH H
Cf.
8 -67 By now , these rearrangements should not be so " unexpected" .
U"
"":
H 20 I �C H + H CH3 H
�
(Y �
�
alkyl mi gration with ring expansion gives 3° carbocation in 6-membered ring--carbocation nirvan a !
H
CH3 •
Br
: !lr:
C H3
0 +
H H
' CH]
You must be asking yourself, "Why didn't the methyl group migrate?" To which you answered by drawmg the carbocation that would have been formed:
I;I
O\}
C+
2°
� H CH3 I "
QC<1''3°
H --
H
C H3 CH3
The new carbocation is i ndeed 3 ° , but it is only in a 5-membered ri ng, not quite as stable as in a 6membered ring. In all probability, some of the product from methyl migration would be formed, but the 6membered ring would be the major product.
8-68 Each alkene w i l l produce two carbonyls upon ozonolysis or permanganate oxidation. Oxidation of the unknown generated four c arbonyls, so the unknown must have had two alkenes . There is only one possibi l i ty for their position s .
CO H
the unknown
a�COOH H
H
KMn04 ..
/j.
:: COOH H
187
+
C OOH I COOH
8-69 (a) Fumarase catalyzes the addition of H and OH, a hydration reaction . (b) Fumaric aci d i s planar and cannot be chiral. Malic acid does have a chiral center and is chiral. The enzyme-catalyzed reaction produces only the S enantiomer, so the product must be optically acti ve. (c) One of the fundamental rules of stereochemistry i s that optical! y inacti ve starting materials produce optically inactive products . Sulfuric-acid-catalyzed hydration would produce a racemic mixture of malic acid, that is, equal amounts of R and S . (d) If the product is opticall y active , then either the starting materi als or the catalyst were chiral . We know that water and fumaric acid are not chiral, so we must infer that fumarase is chiral. (e) The D and the OD are on the " s ame side" of the Fischer proj ection ( someti mes cal led the "erythro " stereoi somer) . These are produced from either: (1) s y n addition t o cis alkenes, o r (2) anti addition to tralls alkenes. We know that fumaric ac id is tralls , so the addition of D and OD must necessari l y be anti. (f) Hydroboration i s a syn addition . H ooe " " H
1 . B D3 • THF
,H
' == e '" ....... e '
eOOH
2. D 2 0 2 , DO-
D ..
H ,
DOOe
" H
�
(note that OH exchanges with D in DO- )
OD
""
H eOOD
+
DOOe
+�
D
eOOD
eO D "
D
eOOD D
,>-1(, R H
�
OD
eOOD
+�
eOOD
As expected, tralls alkene plus syn addition puts the two groups on the "opposite" side of the Fi scher proj ecti on (sometimes cal led " threo" ) .
�. 0
8-70 (a)
Hg(OAc)
H
Hg(OAc )
+ Hg(OAc) -. .
AcO :
mercunntum ion
Br
Br - Br
..
bromonium ion
188
Br
8-7 1 The addition of BH3 to an alkene is reversible. Given heat and time, the borane will eventually " walk" its way to the end of the chain through a series of addition-elimination cycles. The most stable alkylborane has the boron on the end c arbon; eventually, the series of equilibria lead to that product which is oxidized to the primary alcohol.
..
..
...
�
----
most stable
189
H
H2
8-72 First, we explain how the mixture of stereoisomers results, then why. We have seen many times that the bridged halonium ion permits attack of the nucIeophile only from the opposi te side.
Q
expected:
H
Ci-CI
---
Ph
CI�
W H
�:S:1:
�
~ H
Cl trans only
Ph
A mixture of cis and trans could result only if attack of chloride were possible from both top and hottom, somethi ng possible only if a carbocation existed at this carbon. actual:
'"
0� H H
--� +� CI H
Ph
H
trans
Ph
cis
This picture of the p orbitals of benzene show resonance overlap with the p orbital of the carbocation. The chloride nucIeophile can form a bond to the positive carbon from either the top or the bottom.
Why does a carbocation exist here? Not only is it 3°, it is also next to a benzene ring (benzylic) and therefore resonance-stabilized. This resonance stabilization would be forfeited in a halonium ion
intermediate.
0"::::
r'�/
�CI CI
aCI
' Ph
IIi
H
//
:Ci : 'Ci CI
CC"'H Ph
+
. . -
+
�
C 0:CI 190
�
�
CHAPTER 9-ALKYNES
9-1 Other structures are possible i n each c ase. CH3CH2CH2CH2C:::CH
CH3CH2C:::CCH2CH3
0-
HC:::C - CH2CH = CHCH2CH2CH3
C:::CH
9-2 New IUPAC names are gi ven. The asterisk ( * ) denotes acetylenic hydrogens of terminal alkynes. (a)
(b)
*
CH3 - C:::C-CH3 but-2-yne
CH3CH2 -C:::C - H but- l -yne *
CH3CH2 - C::: C - CH3 pent-2-yne
CH3CH2CH2 - C:::C - H pent- l -yne
CH3 I * CH3CH - C:::C - H 3-methylbut-l-yne
9-3 The decomposition reaction below is exothermic (/)J{0 - 234 kJ/mole) as well as having an increase in entropy. Thermodynamical l y, at 1 500°C, an increase in entropy will have a large effect on AG (remember AG /)J{ - TAS). Kinetically, almost any activation energy barrier will be overcome at 1 500°C. Acetylene would likely decompose i nto its elements: =
=
HC::CH
1 500°
�
2 C
+ H2
9-4 Adding sodi um amide to the mixture will produce the sodium salt of hex- l -yne, leaving hex- l -ene untouched. Distillation wi ll remove the hex- l -ene, leaving the non-volatile salt behind.
}
� CHCH2CH2CH2CH3 NaNH2.. J �H� CHCH2CH2CH2CH3 1 Na+ _ C C - CH2CH2CH2CH3 H -C C-CH2CH2CH2CH3 CH
=
•
=
non-volatile salt
9-5 The key to this problem is to understand that a proton donor will react only with the conjugate base of a weaker acid. See Appendix 2 at the end of this Solutions Manual for an in-depth discussion of acidity. (a) H -C:::C-H
+ NaNH2
--
(b) H-C:::C -H
+ CH3Li
--
H -C:::C : Na+ + NH3 H-C:::C : Li + + CH4
(c) no reaction: NaOCH3 i s not a strong enough base (d) no reaction: NaOH is not a strong enough base (e) H-C:::C : Na+ + CH30H -- H-C:::C - H + NaOCH3 (opposite of (c)) (0 H -C:::C : Na+ + H20 -- H-C:::C-H + NaOH (opposite of (d)) (g) no reaction: H - C :::C : Na+ is not a strong enough base (h) no reaction: NaNH2 is not a strong enough base
191
H-C:::C-H NaNH2 H-C:::C : Na+ CH3CH2Br H-C:::C-CH2CH3 � NaNH2 CH3(CH2)S-C:::C-CH2CH3 CH3(CH2)SBr Na+ :C:::C-CH2CH3 (a) H-C:::C-H NaNH2 CH3CH2CH2CH2Br H-C:::C-CH2CH2CH2CH3 NaNH2.. CH3-C:::C-CH2 2CH3 (b) H-C:::C-H NaNH2 H-C::: C -CH CH CH 2 2 3 CH3I CH3CH2CH2Br NaNH2 CH3CH2-C:::C-CH2CH3 (c) H-C:::C-H 2) NaNH2 .. H-C::: C -CH2CH3 CH3CH2Br CH3CH2Br (d) cannot be synthesized by an reacti o n-woul d requi r e at t a ck on a al k yl hali d e r/ 2° �� _ CH3-C::: C : Nale;+ CHCH2CH3 fr CH3-C:::C-CHCH2CH3 stmustrongbenucleophi CH3 second order CH3 low yields; not practical NaNH2 CH3-C::: C -H (e) H-C:::C-H NaNH2 CH3I BrCH2CHCH3 CH3-C::: C-CH2CHCH3 CH3 CH3 H-C:::C-H NaNH2 H-C:::: C-?H2 Br(CH2)8Br Br� � NaNH2 Na+ -,C-=C-CH2 H2C-C:::C-CH2 � Br� Intwilramol e cul a r cycl i z ati o n of large ri n gs must be intramolecular and not intennolecular.be carried out in dilute solution so the last displacement
9-6
-
..
•
..
9-7
1)
..
2)
1)
1)
•
2)
CH
2)
1)
1)
•
2)
S N2
20
+
I
1)
2)
(f)
1)
..
1)
2)
..
I
I
..
2)
..
....!---.
r'.
�.
S N2
9-8
H20 H-C:::C-CH20H (a) H-C:::C-H NaNH2 H2C=O 1)
2)
..
----l.�
192
9-8 contin ued
9H H 0 NaNH 2 2 H3C-C-C::C-H H-C:::C-H Ph yCH3 Ph OH NaNH NaNH 2 2 H-C:::C-H CH3I CH3-C:::C-H 2) HCCH2CH2CH3.- CH3-C:::C-CHCH2CH2CH3 H20 OH NaNH2 NaNH2 H-C:::C-H 2) CH3I CH3-C:::C-H 2) CH3CCH2CH3 CH3-C:::C -CCH2CH3 CH3 H20 H (Br H H H � -oo:�H H-C=C-C-C�C� H -C �C-CH2CH2CH3 :OH C=C I) Br H�oo Br{) \CH2CH2CH3 1 t: H HI ) H HI H-C C-CH2CH3 HO-H H-C=C=C-CH2CH3 H-C=C-C-CH2CH3 U� H 1t :OH H � H-OH H-C=C=C-CH2CH3 H-C-C=- C-CH2CH3 H-C-C=C-CH2CH3 H H H 1)
(b)
..
2)
----l.-�
I
o
1)
(c)
2)
1)
�
I
II o
3)
1)
(d)
1)
�
I
�
II o
3)
.
I
9-9
1
1
1
1
1
..
=(=
---
�
-
0 0
1
_
_
}
-0 0
�
1
-
=
R
o
Ii
8.3 1 4 JIK-mole
( !1G)
[internal] [terminal] 193
Keg
=
75
e
U
�
(
}
/
_
�
0 0
9-1 0 To determine the equilibrium constant in the reaction: !J.G = RT In tenninal alkyne � internal alkyne !J.G - 1 7.0 kJ/mole Keg = (- 4.0 kcallmole) RT e =
1
0 0
0 0
1
_
�
-
0 0
i
,1
\
0 0
I
-
I
-(-17,000) (8.314)(473)
)
98.7% i nternal 1 .3 % tenninal
=
e4.32
=
75
i
9-11 (a) This isomerization is the reverse of the mechanism in the solution to 9-9.
H �- H2 - . H-C-C C-CH2CH3 .. H-C-C C-CH2CH3 H H H H H-C C-C-CH2CH3 H-C==C==C-CH2CH3 H -U2 � H H-C C-C-CH2CH3 H KOH, • •
I)
•
.
N
1
i
.
1
1-
�
• •
l \""'
}
1
_
• •
NH
----
-
• •
I( H
�
.
HJH H-( C==C==C-CH2CH3
:� �H • •
} 2
H-C==C==C-CH2CHJ 1 H
I
1
�
1 I
(b) All steps in part (a) are reversible. Wi th a weaker base like an equilibrium mixture of pent-I yne and pent-2-yne would result. With the strong base NaNH2, however, the final tenninal alkyne is deprotonated to give the acetylide ion:
H-C C -CH2CH2CH3
+
NaNH2
-
Na+
:C C-CH2CH2CH3 NH3 +
==
Because pent-l -yne is about l O pK units more acidic than ammonia, this deprotonation is not reversible. The acetylide ion is produced and can't go back. Le Chatelier's Principle tells us that the reaction will try to replace the pent- l -yne that is being removed from the reaction mixture, so eventually all of the pent-2-yne will be drawn i nto the pent- l -yne anion "sink". (c) Using the weaker base pent-2-yne predominating.
KOH 200°C at
will restore the equili brium between the two alkyne isomers wi th
9- 12 (a) ( 1)
Br 1
Br 1
H3C-T-T -CH3 KOH H3C-C C-CH3 H H CH3CH==CHCH3 CCl4.. CH3CH-CHCH3 ---
L1
Br I
Br2'
(2) (b) (1)
(2)
Br I
Br I
1
1
Br 1
NaNH2
H-C-C-(CH2hCH3 1500 H-C C-(CH2hCH3 H H CH2 CH(CH2)sCH3 Br2' CCl4 H2C-CH(CH2hCH3 ==
•
Br 1
•
194
Br 1
9- 12
(c)
continued KOH
(1)
Br Br Br2' CCl4 CH3CH -CH(CH2)4CH3
(2)
---
"
� -I..
Br
(d)
at insteadcoulofdNaNH have 2 been used
(1)
KOH
H
Br
(2)
H 9- 1 3
Lindlar catalyst Na (c)
Wcis
Br
NaNH2
H Oneo convert of the oneusefulstereoi "tricsks"omerof iorgani c chemiHavi stry nisg tahepaiabilr ofity treacti n t o another. ons opposi like thetetwsteoreochemist reductionsryshown inuseful parts (a), because and (b)a that give i s very intermediateThi(thes princi alkyne)ple canis usedbe transformed eicommon ther stereoisomer. again in partin(d).to 195
200°C
continued
9- 13
(d)
Br
~ Br
�
trans
KOH
-C H2
I ,
C�
Licatalndlaryst
H H I \ C == C
cis / L to add onlcouly oned addequitovthealenttriofplebromine, always avoiwasdiinngexcess. an excessIf theof bromi ne,is because twothe bromi molTheeculne,goaletsheofisfibromine bond i f bromine alkyne added toto I r st drops of al k yne wi l encount e r a l a rge excess of bromine. nst e ad, addi n g bromine the alkyne will always ensure an excess of alkyne and should give a good yield of dibromo product. 9- 14
H
-H � :B�: I
+
CH3CH2CH2 - C == C
2° 1°
betcarbocat ter thanion
2°
carbocation and
�
0 0
resonance-stabilized CI
9- 16
I
H
I
H3C - C - C - (CH2)4CH3
i
I
Cl
I
H
+
H
CI
H
}
I I H3 C - C - C- (CH2)4CH3 I I Cl
(b)of The second addition occurs to make the carbocation intermediate at the carbon with the halogen because :ci: � -- k y+ - - ---- -�-k· l. On no stab·lIlzat resonance
0
stabilization.
:C 1: H I I C==C
0
:C1: H
I
I
I
+
-C-C H
H+
I
H
I
.... .. t-----�� -t
196
: I: H II I C- -
resonance stabilization
9-17
initi ation: propagation: Br.�
hv 2 H-Br j -CH2CH2CH3
RO·
RO - OR RO •
J
+
---
RO-
H Br H-C C -CH2CH2CH3 Br +
•
==
I
•
The radi c al i s more stabl e than The anti Markovni k ov orientati o n occurs because t h e bromi n e radi c al fiatrtstacks(seefirtsthetosolmake utiontheto most stable radical, which is contrary to electrophilic addition where the attacks 2°
1°.
H+
9-15).
9- 18
(b) C C-(CH2hCH3 HBr HBr H-
ROOR
H -C==C-(CH2hCH3 CI Cl -C==C-(CH2hCH3 Br H H C C -(CH2hCH3 Br Br Br H-C-C-(CH2)3CH3 Br Br H
-
I
I
I
I
==
I
H
I
I
I
I
I
Licatnadlarlyst (or Na, NH3) H Br 2HBr H-C-C-(CH2)3CH3 H Br I
I
I
I
197
E
+ Z
E
+ Z
9- 19
r
+
CH3-
=c
)
!
H2 :
1
1+
;)(�(
CH2CII3
O
both 2° vinyl
!
0
H
!
)
!
H
0
O
H2 :
H2 :
)
'f 'f
CH3-
-CH2CH3
�
R
: H
H :
-CH2CH3
H
t
H+
H+
H
A
I)
I)
t
-CH2CH3
CH3-C == C-CH2CH3 1 +1 HO: H
�H
(
=
H2 :
:OH
'f 'f
CH3-
+
carbocations
CH3-C ==C - CH2CH3 H
CH3-
H
+
+ resonance1 CH3 - C - C-CH2CH3 CH3-C-C-CH2CH3 stabilized carbocations , '",\ r' 1
1
H
t
:O-H
H-O : H
..; �
Hi
(resonance forms not shown)
�. 0 t H2
:
:
H
H
I
I
CH3-C-C-CH2 CH 3
CH3-C-C-CH2CH3
I
II 0 3-pentanone
II 1 0 H 2-pentanone
H
The role of the mercury catalyst is not shown in thi s mechanism. As a Lewis acid, it may act like the proton in the first step, helping to form vinyl cations; the mercury is replaced when acid is added. -H + ---I....
CH3-C ==C - CH2CH3
I
+Hg
198
I
OH
9-20 (a) But-2-yne is symmetric. Either orientation produces the same product.
CH3-C
Sia2BH .. C-CH3
CH3
CH3
/
\
C=C
/
\
H
BSia2
H202 .. HO-
CH3 \
CH3
/
C=C / \ H OH
HO-
---I"�
°
CH3CHz
II
-
C
-
(b) Pent-2-yne is not symmetric. Different orientations of attack will lead to different products on any unsymmetrical internal alkyne. CH3-C C-CH2CH3
�H � S
CH3
CH2CH3
\
/
/
\
C=C
H
CH3
�
H202 , HOCH2CH3
CH3
�
CH2CH3
/
H
°
II CH3 -C -CH2CH2CH3
II
°
CH3CCH2CH2CH2CH3
(2)
°
(b)
H202 , HO-
HO-
°
(1)
H
\
HO
II CH3CH2-C -CH2CH3
(a)
\
C=C
/
°
9-2 1
�
\
C=C \ H OH /
C=C
Sia2B
B Siaz
CH2CH3
/
\
/
/
\
CH3
II ( 1 ) CH3CCH2CH2CH2CH3
II HCCH2CH2CH2CH2CH3 °
+
II CH3CH2CCH2CH2CH3
(2) same mixture as in (b) ( 1) °
(c)
II ( 1) CH3CH2CCH2CH2CH3
(d)
( 1)
�O VV
°
(2)
II CH3CH2CCH2CH2CH3
(2)
�O
VV 199
CH3
9-22 CH3 H I I ----1 .. 1 H -C-C-BH)I I CH3 CH3
(a)
\
CH3
H
/
\
+
C==C
/
H3C
CH3
CH3 H H H CH3 I I I I I H-C-C-B-C-C- H I I I I CH3 CH3 CH3 CH3 disiamylborane, Sia2BH (b) There is too much steric hindrance in Sia2BH for the third B-H to add across another alkene. The reagent can add to alkynes because alkynes are linear and attack is not hindered by bulky substituents. 9-23
o II
o
0 II
(a) (1) HO-C-C-(CH2hCH3 oxidation of a tenninal alkyne with neutral KMn04 produces the ketone and carboxylic acid without cleaving the carbon-carbon bond o 0 II II (b) (1) CH3 C-C - CH2CH2CH3 o II
11
(2)
II
CO 2 + HO-C -(CH2hCH3 oxidation of a tenninal alkyne with wann, basic KMn04 cleaves the carbon-carbon bond, producing the carboxylic acid and carbon dioxide o
II
(d) (1) CH3CH-C-C-CH2CH3 1 CH3 (e) ( 1)
II
CH3 - C-O H + HO -C-CH2CH2CH3 o
(2)
0 II
0
II
0
(c) (1) CH3CH2-C-C-CH2CH3 o
(2)
(2 )
(2)
II
CH3CH2-C-O H o II
0 II
CH3CH-C-O H + HO -C -CH2CH3 1 CH3 o
OH
HO o
9-24 (a) CH3 -C:::C- (CH2)4 -C:::C-CH3
200
9-25 When proposing syntheses, begin by analyzing the target molecule, looking for smaller pieces that can be combined to make the desired compound. This i s especially true for targets that have more carbons than the starting materials ; immediately, you will know that a c arbon-carbon bond forming reaction will be necessary. People who succeed at synthesis know the re a ctions-there i s no shortcut. Practice the reactions for each functional group until they become automatic. (a) analysis of target
from acetylene
,,- -, -.
... I\:'��\�i;��>,:': ( ./'""'-.r.f .
3° acetyl enic alcoho ls made from acetylide plus ketones \/ "
forward direction: H-C=:C-H
+ NaNH2
----
�'--' -
,
•
'
OH -
H-C=:C:
,
:
:
...
from alkylation of acetyhde
' •• _---'
8r
put on less reactive group first
�
Na+
......f------
�
H-C=C�
t
..
NaNH2
Na+ -:C=:C�
o
anal ysis of target: cyclopropanes are made by carbene insertion into alkencs; to get cis substitution around cyclopropane, stereochemistry of alkene must be cis; cis alkene comes from catalytic hydrogenation of an alkyne NaNH2 H-C=:C-H
..
------I�
CH31 ----
H3C-C =: C-H
NaNH2 ..
CH3CH2Br .. H3C-C=:C-CH2CH} H2 CH212 ..
(c)
)" 0 /(
H
CH3CH2
CH2CH2CH3
H
Z n(C u)
H
I ,
20 1
H
>=<
H3C
analysis of target: epoxides are made by direct epoxidation of alkenes; to get trans substitution around epoxide, stereochemistry of alkene must be trans; trans alkene comes from sodium/ammonia reduction of an alkyne
9-26 Please refer to solution 1-20, page 12 of this Solutions Manual.
Lindlar catalyst
CH2CH3
9-27 (a)
CH3CH2-C::C -(CH2)4CH3
o-
(d)
C::C-H
�'
(f)
(b)
H3C-C::C-(CH2)4CH3
(e)
CH3CH2-C:: C - CHCH2CH2CH3
( )-
C::C-H
I
CH3
I
CH3CH-C::C - (CH2hCH3
(g)
Br
(h)
H3C
C::C-CHCH2CH3
\
I
I
\
I
C=C
OH
H-C::C-CH=CH2
(k)
CH2CH3
H
H
(j)
H - C:: C - CH2 - C::C - CH2CH3
(i)
(c)
H-C::C
CH3 H "
-(
C=CH2 , H
9-28 (a) ethylmethylacetylene (b) phenyl acetylene
(c) sec-butyl-n-propylacetylene (d) sec-butyl-t-butylacetylene
9-29 (a) 4-phenylpent-2-yne (b) 4,4-dibromopent-2-yne
(c) 2,6,6-trimethylhept-3-yne (d) (E)-3-methylhept-2-en-4-yne
(e) 3-methylhex-4-yn-3-ol (f) cycloheptylprop-I-yne
9-30 terminal alkynes
internal alkynes
(a)
CH3 - C::C-CH2CH2CH3
acetylide ions
H - C::C - (CH2hCH3
C:: C - (CH2hCH3
hex- l -yne
hex-2-yne CH3CH2 - C:: C - CH2CH3
H - C::C - CHCH2CH3
- C:: C-CHCH2CH3.
I
I
CH3
hex-3-yne
CH3
3-methylpent- l -yne
CH3-C::C-CHCH3
NaNH2 C::C-CH2CHCH3
H - C :: C-CH2CHCH3 I
I
CH3
I
CH3
CH3
4-methylpent- l -yne
4-methylpent-2-yne
CH3
CH3
I
I
H -C::C - C - CH3
C::C-C-CH3.
I
I
CH3
CH3
3,3-dimethylbut- l -yn
�
(b) All four terminal alkynes will be deprotonated with sodium amide. 9-3 1
(R�-: �a b .C==CH
CH3CH2-C H
"
CHCl
HC
OH
H20 ---
•
I
NH2 CH
I
CH3CH2 - C-C-CH
202
HC � " CHCI
ethchlorvynol
9-32 H-C:::C-H
NaNH2
CH3(CH2hBr
..
CH3(CH2h
(CH2)12CH3 I
\
C=C
/
\
H2
(a)
"1
Lindlar catalyst
Cl I (b) H3C-C-CH2CH2CH3
Cl I CH2 = CCH2CH2CH3
CH3(CH2)12Br
CH3(CH2h -C::: C - (CH2) 12CH3
.... .. 1-------
H H muscalure 9-33
NaNH2
CHiCH2h-C::: C -H
..
I
(c)
CH3CH2CH2CH2CH3
(f)
H -
Cl H (d)
(g)
G)
o 0 II
II
Na+
I
NaNH2
I
1500
Br Br (b)
I
KOH
I
2000
H3C - C -CH2CH3 Br
(c)
0 (k) H3C - C - CH2CH2CH3
: C ::: C-CH2CH2CH3
H3C - C -CH2CH3
_
CH3CH2-C=C-H
0 CO2 + HO - C - CH2CH2CH3 II
H2O
I
(i) H2C = CHCH2CH2CH1
H-C:::C -CH2CH3
H3C-C::: C-CH3
..
NaNH2
..
..
I
0 II (I) H - C - CH2CH2CH2CH3
II
-
Br
B r Br
I
C=C \ I Br CH2CH2CH3
(h)
HO - C-C-CH2CH2CH3
9-34 (a)
(e)
CH2 = CHCH2CH2CH3
Br
\
_
..
CH3CH2-C=C:
�
Na+
CH3CH2CH2CH2Br
CH3CH2 -C:: C -CH2CH2CH2CH3 H3C (d)
\
B r2
C=C
I
\
\
I
H
H (e)
H
I
CH2CH2CH3
H
C=C
I
H3C
---
\
CH2CH2CH3
CCl4
B r2 ---
CCl4
Br B r I
I
KOH
•
I
2000
CH3C -CCH2CH2CH3 H H
Br Br I
I
•
I
CH3C -CCH2CH2CH3 H H
1 ) NaNH2 1 500 ..
2) H2O 203
..
H3C -C::: C -CH2CH2CH3 this product could contain minor amounts of 3-hexyne from rearrangement
H - C == C - CH2CH2CH2CH3
9-34 continued
Hz
(f)
..
Lindlar catalyst Na
(g)
NH3
..
H z0 (h) HC:: C - CHzCHzCHzCH3 ___---i"� HZ S04 HgS04
[
W
cis
W
trans
��
CH2 = H2CH2cH2cH3 unstable enol .
KOH
I
CH3C-CCH2CH2CH3 •
-
�
H3C - CH2CH2CH2C H]
this product could contain minor amounts of 3-hexyne from rearrangement
Br B r I
1
I
---J"�
200°
H H
H3C - C:: C -CHzCH2CH3 major Hz H3C
Lindlar •I catalyst
CHzCH2CH3
\
I
I
\
C=C
9-35 Br
�
KOH .. 200°
HC
�
\
Br
H
+
j
�
Y
'--------..
only the terminal alkyne reacts with NaNH2 and acetone
OH
+
OH CompoundD b.p. 1 40- 1 50°C under vacuum 204
H
from rearrangementcould contain 3-hexyne
J
/
� o
t
MixtureA
2
�
t
Mhture B
J� � CompoundC b.p. 80-84°C
9-36 elimination on 3° halide (c)
CH3CH2-C:::C - CH20H (after H20 workup)
OH (e)
I
Na+
CH3CH2 -C::: C - CHCH2CH2CH3 (after H20 workup) OH
(g)
-0-0
I
CH3CH2 - C::: C - C -CH2CH3 I
(after H20 workup)
CH3
9-37 NaNH2 (a)
HC::C-H
(b)
HC::C-H
•
HC::C:
NaNH2 •
CH3CH2CH2B r
HC::C:
--=----"'--..::... .. .
�
HC::C -CH2CH2CH3 NaNH2
... ..f-----.
CH3I
(c)
H3C-C:: C - CH2CH2CH3 synthesized in part (b)
(d)
H3C -C:: C - CH2CH2CH3
H3 C
H2
\
C=C
•
Lindlar catalys t Na
H3C-C:: C - CH2CH2CH3 synthesized in part (b)
(f)
HC = C- CH2CH2CH2CH3
H
H H3C
H
\
I
I
\
C=C
H
synthesized in part (b)
(e)
CH2CH2CH3
I
CH2CH2CH3
2 equiv. H2 Pt Br I
2HBr
H3C -C -CH2CH2CH2CH3 I
synthesized in part (a)
Bf
205
continued Sia2BH H-CCH2CH2CH2CH3 (g) H-C::fromC-CH2CH2CH3 (b) H 0 H3C-CCH2CH2CH3 (h) from (b) HgS04 (i) HC::C-H NaNH2 HC:: C: 9-37
o
1)
H-C::C-CH2CH2CH3
I I
2
°
•
II
H2S04
.. � ---t
catLindlal yastr alproduce kene mustthe be ciproduct s to from anti addition NaNH2 HC::C: Na+ HC::C-H
Review the stereochemistry in the solution to Problem 8-35, p. 171 of this Solutions Manual.
(±)
U)
1)
� .. -t
2)
Alternatively,attedrans-but-2-ene d be c acid. anti-hydroxyl with aqueouscoulperaceti Review the stereochemistry in the solution to Problem 8-35, p. 171 of this Solutions Manual.
9-38
NaNH2 H3C-C::C-CH3 CH31 H2 catLindlalyastr 1 H3C OS04 C=C CH3 H H almesokeneproduct must befromcis tosynproduce additiothen
---l"�
--
\
I
I
\
meso dX PtH2 o-CH2CH2CH2CH3 � the fact that five equi v alents of hydrogen are consumed says thatskelemust 2 carbon ton have fi ve pi bonds in the above O O I'� 0 \11O O 'I? 'I? H-C-CH2CH2 -C-C-H H-C-C-H H-C-C-OH H-C-OH from C::C carboxylic acids alkyne carbonyls alkenes ene ithese s cis orresultransts. cannot o-CH=CH-C::CH beWhetdetheerrmithneedalkfrom Compound
T�:� o II
6
5
X
II
I I
+
'-.
+
�
�
� 3
X
2
206
II
I I
+ '"'-__----. ) y�----� 1
9-39 Compound ozonolysis CH3(CH2)4-C-H CH3-C-CH2-C-OH HO-C-H from alkene from alkyne Compound CH3(CH2)4CH C -CH2 -C C-H Whether thedetalkeenerminised cannot be CH3 from this information. 9-40 All four syntheses in this problem begin with the same reaction of benzyl bromide with acetylide ion: HC::CH .. HC::C: � CH2Br .. PhCH2-C::CH .. PhCH2-C::C: use this benzyl I bromide Br � PhCH2-C::C � 6-phenylhex-l-en-4-yne allyl bromide (b) PhCH2-C::C: Br....../ .. PhCH2-C::C� P aS04.. Ph '>={ ethyl bromide H pentH - 2-ene quinoline cis-l-phenyl Lindlar catalyst Na Ph (c) PhCH2-C::C: Br....../ ethyl bromide X phenylpent-2-ene (d) Thestructure, diol witalh tthough he twothisOHonegroupsis notonmeso the same sidethine tthoep Fischer projom egroup ction isarethdifferent e equival. eStntilof, iat meso because and bot t gives aorclueby anas tanto itsi addit syntihonesitso. aThetrans"meso" bond, doublediolbond.can Webe formed saw thebysameeitherthinga synin taddit he solution itono atocis double Ph '>={ OS04 Ph H30+ Hcis H syn addition anti addition product product CH2Ph from part (b) from part (c) H + OH H + OH CH2CH3 racemic Z
�
\
II
I I
y
Z:
N��
=
o
o
o
o
II
II
J
\.'-__......
,-__ -.;)
y
==
I
E or
Z
N��
_
6
helow
?'
�
+
+
\
dlB
+
trans-l-
9-37 (j).
..
trans
207
9-4 1
a2BH (a) CH��-C�C-H SiH202, HO(b) �.. reaction �H-O-Et '" R�C= \{ -==:-l R-C C-H :O-Et H :O Et � O�Et H +O�E' R-�-t react R �-� � 'c=/ ion2 R/ H H H H ! H20 : H H : O�Et H O-Et H :o�H+ R-C-C-H R-C-C-H .. R-C-C-H H :O�H H20: H OH H OH ilJ ! - HOEt H H R -C-C"+-H R-C-C-H R-?-? -H II H :OVH H : O-H H (c) alkyne ROR-C=CH The R-C CH clorbiosertalthiats cleloeserctronsto aarenucltoeusthethannuclaeus,p orbithetalmoreis, asstapble. An s 2 orbitalsoaren iselmore ongatstedablaway from the nucl e us. An sp 2sp carbaniORon carbani 3 an sp carbani o n because t h e 2sp carbanion has es than electron pair alkene ROiwhis clcohseris tonlo thye positsivcharact e nuclcharact eeusr. The tehranandspin2andancarbani spthe3 carbanion R-C-CH2 to form because of its relative stabil i ty. on easier R-C=CH2 H 3sp HcarbaniORon 1)
•
2)
f\
{
+
-
•
1
-..
\ y\ /
1
...
/
\H
�
•
1
1
1
1
1
• •
1
1
1
1
1
•
H
1
-----<
.... ..f----;.�
1
°
•
1
1
I)
1
1
1
+
..
}
1
33%
1
•
25%
1
I
208
is
Diols orareantmadei-dihydroxyl by two react ionsviafrom Chapteusing r anda peroxyacid revisited in and water. (d): either syn-dihydroxyl attioon witusehinorgani Os04' a t i on an epoxide As t h is probl e m says c reagents, the solution shown here wil l use OS04. Recall the stereochemical requirements of syn addition as outlined in this Solutions Manual, p. Problem cis-alkene addition meso citrsans--alkaenelkene additadditionion racemic racemic trans-alkene addition meso Part synthesis on will .have to occur on the cis-alkene. (b) wil(a)l asks requiforre synthe addit ion toofthethetrameso ns-alisomer, kene to sogivsyne thaddit e iproduct (a) NaNH2 HC::C : Na+ HC::C-H "! i OS04 Lindlalyastr -Zc::cJcat H H alproduce kene mustthe meso be cisproduct to from synion addit ion,withso reduct is done theLindlciasralcatkenealyst to produce (b) NaNH2 HC::C-H NaNH2 HC::C : Na+ "! i OS04 Na NH3 -Zc::cJalkene mustthe be tproduct rans to from produce syn addit so reduction is done withion,Na!NH3 9-42
8-35:
8
9-4 1
17 1,
+ syn + anti + syn + anti
�
�
�
�
(±) (±)
Part
(±)
..
..
f
\
..
..
..
(±)
209
Bf
Bf
This synthesis begins the same as the solution to problem HC::CH NaNH2.. HC::C: CH2Br .. PhCH2-C:::CH .. PhCH2-C:::C : benzylde bromi The anion will add across the carbonyl group of the aldehyde: H30+ H --: C PhCH2-C::: lf acid-catalyzed dehydration H2S04, 9-40:
9-43
01
NaNH2
_� +
�
°
+
•
MCPBA
210
1
Ll
CHAPTER 10-STRUCTURE AND SYNTHESIS OF A LCOHOLS
10-1 Please seepropan-2-ol the note on p. 136 of this Solutions Manual regardi nghplylcycl acementohexan-l-ol of positio("n 1"numbers. (d) t ra ns-2-met is optional) (a)(b) 2-phenyl (e) (E) 2-chl o ro-3-met h yl p ent-2-en-l-01 5-bromohept a n-2-ol (c) 4-methylcyclohex-3-en-l-ol (" 1" is optional) (2R,3S)-2-bromohexan-3-ol IUPAC name first, then common name. 10-2 (c)(d) l-cycl oylbutbutylapn-l-01; ropan-2-01; noycommon name (a)(b) cyclopropanol ; cyclopropyl 3-met h i s opent l al c ohol 2-methylpropan-2-01; t-butylalalcohol cohol (also isoamyl alcohol) 10-3 Only constitutional isomers are requested, not stereoiOHsomers, and only structures with an alcohol group. (a) C3H O �OH propan-l-ol A propan-2-01 (b) C4HlOO � � �OH � OH OH OH butan-I-ol butan-2-01 2-methylpropan-I-ol 2-methylpropan-2-01 (c) C4H O has one element of unsaturati on, either a double bond or ring. HO i>OH OH d [>-Jmethanol l-methylcycl �opropanol 2-methylcyclopropanol cyclobutanol cyclopropyl cis or t ra ns OH * HO� � OH � � OH but-3-en-I-ol but-3-en-2-01 but-I-en-2-01 but-I-en-I-ol (E or Z) J:H * �OH � �OH OH but(E-2-en-l-ol but-2-en-2-01 2-methyl2-methylprop-2-en-l-01 or Z) (E or Z) prop-l-en-I-ol (d) C3H40 hasortwaothree-membered elements of unsatrinugratandi on,asodoubleache bond. structureAllmuststructures have eimust ther contai a triplen bond, or (Intwothe doubl e bonds, an OH. name, the "e" is dropped from "yne" because it follow by a vowel in "01".) HO-C::C-CH3 HC::C-CHOH2 H2C=C=C!I y. 0H * prop-l-yn-l-ol prop-2-yn-I-ol OH propa-l,2-dien-I-ol cycloprop-l-en-I-ol cycl'\Zoprop-2-en-l-ol *Thesealcompounds wistthructure the OHwibonded direct ly to the carbon-carbon doublis cale lbond areynolcal. lThese ed "enols" or "vinyl c ohol s . " The t h OH on a carbon-carbon t ri pl e bond e d an are unstable, although the structures are legitimate. (f)
g
g
a
* OH
211
1 0-4 (a) 8, 8-dimethy lnonane-2,7 -diol (b) octane-l,8-diol (c) cis-cyclohex-2-ene-l,4-diol (d) 3-cyclopentylheptane-2,4-diol (e) trans-cyclobutane-l,3 -diol 1 0-5 There are four structural features to consider when determining solubility in water: 1) molecules with fewer carbons will be more soluble in water (assuming other things being equal); 2) branched or otherwise compact structures are more soluble than linear structures; 3 ) more hydrogen-bonding groups will increase solubility; 4) an ionic form of a compound will be more soluble in water than the nonionic form. (a) Cyclohexanol is more soluble because its alkyl group is more compact than in I-hexanol. (b) 4-Methylphenol is more soluble because its hydrocarbon portion is more compact than in I-heptanol , and phenols form particul arly strong hydrogen bonds with water. (c) 3-Ethylhexan-3-ol is more soluble because its alkyl portion is more spherical than in octan-2-01. (d) Cyclooctane-l,4-diol is more soluble because it has two OH groups which can h ydrogen bond with water, whereas hexan-2-ol has only one OH group. (The ratio of carbons to OH is 4 to 1 in the former compound and 6 to 1 in the latter; the smaller thi s ratio, the more soluble. ) (e) These are enantiomers and wil l have identical solubility. 10-6 Dimethylamine molecules can hydrogen bond among themselves so it takes more energy (higher temperature) to separate them from each other. Tri methylamine has no N-H and cannot hydrogen bond, so it takes less energy to separate these molecules from each other, despite its higher molecular weight. 1 0-7 See Appendix 2 at the back of thi s Solutions Manual for a review of acidity and basicity. (a) Methanol is more acidic than t-butyl alcohol. The greater the substituti on, the lower the acidity. (b) 2-Chloropropan-l-ol is more aci dic because the electron-withdrawing chlorine atom is closer to the OH group than in 3-chloropropan-l-ol. (c) 2,2-Dichloroethanol i s more acidic because two electron-withdrawing chlorine atoms increase acidity more than just the one chlorine in 2-chloroethanol. (d) 2,2-Difluoropropan-l-ol is more acidic because fluorine is more electronegati ve than chlorine; the stronger the electron-withdrawing group, the more acidic the alcohol. 10-8 most acidic sulfuric acid
»
2-chloroethanol
>
>
water
ethanol
>
>
t-butyl alcohol
ammonia
>
least acidic hexane
(CH3hCOH Sulfuric acid is one of the strongest acids known. On the other extreme, alkanes like hexane are the least acidic compounds. The N-H bond in ammonia is less acidic than any O-H bond. Among the four compounds with O-H bonds, the tertiary alcohols are the least acidic. Water i s more acidic than most alcohols inc luding ethanol. However, if a strong electron-withdrawing substituent l i ke chlorine is near the alcohol group, the acidity increases enough so that it is more acidic than water. (Determining exactly where water appears in this list i s the most difficult part.) 10-9 Resonance forms of phenoxide anion show the negati ve charge delocalized onto the ring only at carbons 2, 4, and 6:
HQ 66' : 0:
5::::-'" 4
: 0:
: 0:
3
..
..
.
::::-...
..
..
6 CH
2 12
O
..
..
.
h-
6 : 0:
: 0:
H
...
..
10-9 continued Nitro group at position 2 . .
:0: :0:
:0: :0:
Nitro at position 2 delocalizes negative charge.
Nitro group at position 3 :0 :
Nitro at position 3 cannot delocalize negati ve charge at position 2 or 4no resonance stabilization.
:0 :
Nitro group at position 4
:0:
:0 :
..
..
Nitro at position 4 delocalizes negative charge.
-
• •
.."N , +
. •
-
:0'" 0:
Only when the nitro group i s at one of the negative carbons will the nitro have a stabilizing effect (via resonance). Thus, 2-nitrophenol and 4-nitrophenol are substantially more acidic than phenol itself, but 3nitrophenol is only slightl y more acidic than phenol (due to the inducti ve effect). OR
10- 10 A
� �
(a) Structure A is a phenol because the OH i s bonded to a benzene ring. As a phenol, it will be acidic enough to react with sodium hydroxide to generate a phenoxide ion that w ill be fairly soluble in water. Structure B is a 2° benzylic alcohol, not a phenol, not acidic enough to react w ith NaOH. (b) Both of these organic compounds will be soluble in an organic solvent like dichloromethane. Shaking this organic solution with aqueous sodium hydroxide will ionize the phenol A, making it more polar and water soluble; i t w i l l be extracted from the organic layer i nto the water l ayer, while the alcohol will remain in the organic solvent. Separating these i mmiscible solvents w i l l separate the original compounds. The alcohol c an be retrieved by evaporating the organic solvent. The phenol can be i solated by acidifying the basic aqueous solution and filtering if the phenol i s a solid, or separating the layers if the phenol is a liquid. 2 13
1 0- 1 1 The Grignard reaction needs a solvent containing an ether functional group: (b), (t), (g), and (h) are possible solvents. Dimethyl ether, (b), i s a gas at room temperature, however, so it would have to be liquefied at low temperature for it to be a useful solvent.
�Li
10- 1 2 (b)
(a) CH3CH2MgBr
(c) F
+ LiI
-D-
�+ Li
MgBr
(d)
LiCI
1 0- 1 3 Any of three halides-chloride, bromide, iodide, but not fluoride-can be used. Ether is the typical solvent for Grignard reactions. (a)
(b)
(c)
o � o-
+
MgCl
H \ C=O I H
ether H30+
H
ether H30+
+
MgBr
\
C=O
I
..
--
..
--
H
H
MgI
\
+
C=O
I
ether H30+ ...
--
H Note: the alternative arrow symbolism could also be used, where the two steps are numbered around one arrow:
O
� o-
CH2OH
,
'ether " ,
2) H30+
OH
NO! Me BAD! This means that water is present with ether during the Grignard reaction.
1 ) ether ...
CH2OH
OK ,
'ce
'
'
•
3 +' ,}'fo
"
1 0- 1 4 Any of three halides-chloride, bromide, iodide, but not fluoride-can be used. Grignard reactions are always performed in ether solvent; ether is not shown here. (a) two methods
I H In MgCl
0
H
+
y 0
+ CH3MgI
H30+ ...
--
H30+ --
...
�
(b) two methods
o 1.&
MgB,
� 0
+
Ib: Ib: OH
0
H
�
H30+ --
..
OH
H
+ IMg -......./
H30+ --
..
214
~
Where two methods can be used to form the target compound, the newly formed bond is shown in bold.
1 0- 1 4 continued (c)
o
�
OH
o
MgCI
+
H
dD
10- 1 5 Grignard reactions are always performed in ether. Here, the ether i s not shown. (a) Any of the three bonds shown in bold can be formed by adding a Grignard reagent across a ketone. (ii) (i) (iii) 0 (i ) CH3C 2MgBr CH3C 2CH2MgBr
: O�
� + PhMgBr
Ph
� Ph :Ph MgBr + "-- 'c = 0 I Ph
(b)
: �O
,\ ?:;01 '-("') ( Ph 111
Ph Ph
H30+ --
(ii )
I
Ph -C-OH
�
I
Ph
(d) two methods �MgCI \ Cy � + Cy
>= O
Cy
=
H30+ --
�
+ Cy-MgCl
cyclohexyl
:C!0 II CH3 -C� MgBr
10- 16
---
?) CH3-C-CI :
I
(This is just a nucleophilic substitution where CI is the leaving group. The unusual feature is that it occurs at a carbonyl carbon.)
�
- CI-
�
Ph
:0 II
� l
-Ph
Ph-MgBr
OH I CH3- -Ph
TPh
215
�:o:I work-up step
CH -C-Ph 3 I Ph
ketone intermediate
1 0- 17 Acid chlorides or esters will work as starting materials in these reactions. The typical solvent for Grignard reactions i s ether; i t i s not shown here. 0 II
+ 2 PhMgBr
Ph-C-Cl
(a)
Jy
(b)
Ph
H30+
I
Ph-C-OH
•
�
I
Ph H30+
+ 2 CH3CH2MgI
OCH]
�
•
0
OH
o II
+ 2
Ph-C-Cl
(c)
10- 18
:qD II
(a) H-C
o
�
o
./'0...
.
MgCI
. .
:?j
MgB r ---
+ 2 CH3CH2MgBr
0 II (ii) HC-OEt
+ 2
0 II (iii) HC-OEt
+ 2
��
<}
MgBr +
� OH
H30+ •
�
MgB r
� MgBr
H30+ �
•
H30+ �
10- 19 Ether i s the typical solvent in Grignard reactions.
<}
aldehyde intermediate
H-C-OEt
OH I �C� I H
II (b) (i) HC-OEt
(a)
~
o
U
•
� OR
� OR
V
216
10- 19 continued (b)
(c)
�
MgCl
�
gl
10-20 (a)
HC
_
C:
0
U
+
0
U
+
�OH H30+ � OH
H30+
�
----
----
�
�D ?'"O
(b)
10-2 1 There are more than one synthetic route to each structure; the ones shown here are representative. The new bonds formed are shown here in bold. Your answers may be different and still be correct. (a) � Br
Li
CuI (�C
- ----
2
(b)
� Br
Li
( �C
CuI
-----
2
(c)
�
Br
Li
�
UL i
�
I (�C
ULi
(�C
ULi
Cu
2
�
( ()l U
U Li
-----
Br
�
(
�
CJ"
Alternatively, coupli ng lithium dicyclohexy!cuprate with I-bromobutane would also work. As this mechanism i s not a typical S N 2, it i s not as susceptible to steric hindrance like acetylide ion substitution or a similar S N 2 reaction. (d)
�
Br
Li
CuI
-----
2
10-22 These reactions are acid-base reactions in which an acidic proton (or deuteron) is transferred to a basic c arbon in either a Grignard reagent or an alkyllithium. (a)
+
CH3D Mg(OD)I
(b)
+
CH3CH2CH2CH3 LiOCH2CH3 0 CH3C-OLi (d)
+
217
o
II
10-23 Grignard reagents are incompatible with acidic hydrogens and with electrophi lic, polarized multiple bonds like C=O, N02, etc. (a) As Grignard reagent is formed, it would instantaneously be protonated by the N-H present in other molecules of the same substance. (b) As Grignard reagent is formed, it would i mmediately attack the ester functional group present in other molecules of the same substance. (c) Care must be taken in how reagents are written above and below arrows. If reagents are numbered " 1. . .. 2. ... etc.", it means they are added in separate steps, the same as writing reagents over separate arrows. If reagents written around an arrow are not numbered, it means they are added all at once in the same mixture. In thi s problem, the ketone is added in the presence of aqueous acid. The acid will immediately proton ate and destroy the Grignard reagent before reaction with the ketone can occur. (d) The ethy l Grignard reagent will be immediately protonated and consumed by the OH. Thi s reaction could be made to work, however, by adding two equivalents of ethyl Grignard reagent-the first to consume the OH proton, the second to add across the ketone. Aqueous acid will then protonate both oxygens.
6 OH
10-24 Sodium borohydride does not reduce esters. (b) no reaction
(a) CH3(CH2) gCH20H
(c)
o
(e) HO·
Y
�OCH3
Y
(I) c,tcc
OH
no reaction (PhCOO- before acid work-up)
1 lf °
(d)
OH
6 OH
10-25 Lithium aluminum hydride reduces esters as well as other carbonyl groups. (a) CH3(CH2) 8CH20H (e) HO
+ HOCH3
(b) CH3CH2CH20H
� OH y + HOCH3
(c)
PhCH20H
(d)
� OH HO �
HO
(f)
OH 10-26 (a)
�H o
OR
� OH o
OR
(b)
� o
�
OR
NaBH4 CH30H
�OH
•
OR
1) LiAlH4
2) H30+
1) LiAIH4
2) H30+
•
1) LiAIH4
2) H30+ NaBH4
CH30H
•
OH
•
� 218
OR
1) LiAIH4
2) H30+
•
•
1 0-26 continued (c)
~
NaBH4 CH30H
..
OH
0
(d)
W
~
OR
O
NaBH4 CH30H
..
C�X:f
1) LiAIH4 2) H30+
..
LiAlH4 will NOT give the desired product. LiAIH4 will also reduce the ester in addition to the ketone.
OH
0
1 0-27 Approximate pKa values are shown below each compound. Refer to text Tables 1-5, 9-2, and 1 0-3, and Appendix 5 at the back of the text. CH3S03H > CH3COOH > CH3SH > CH30H > CH3C:: CH > CH3NH2 > C H3CH3 < 0 4.74 "" 1 0.5 15.5 25 "" 35 50 least acidic
most acidic 1 0-28 (a) 4-methylpentane-2-thiol (b) (Z)-2,3-dimethylpent-2-ene- l -thiol (" 1" is optional) (c) cyclohex-2-ene- l -thiol (" 1" is optional) 10-29
�
HBr ROOR
�
Br
� Br
�o NBS
� ----l..
OR
�
NaSH
..
NaSH
..
c
�
� SH
(see Problem 8-2)
1 0-30 Please refer to solution 1-20, page 12 of this Sol utions Manual. 1 0-3 1 (a) 5-methyl-4-n-propylheptan-2-01; 2° (b) 4-(I-bromoethyl)heptan-3-01; 2° (c) (E)-4,5-dimethylhex-3 -en-l-01; 1° (d) 3-bromocyclohex-3-en- l -01; 2° (" 1" is optional) (e) cis-4-chlorocyclohex-2-en-l-ol; 2° (" 1" is optional) (f) 6-chloro-3-phenyloctan-3-01; 3° (g) (l-cyclopentenyl)methanol; 1° 1 0-32 (a) 4-chloro- l -phenylhexane- l ,5-diol (b) trans-cyclohexane-l,2-diol (c) 3-nitrophenol (d) 4-bromo-2-chlorophenol 219
SH
3-methylbutane-I-thiol 2-butene-l-thiol (but-2-ene-l-thiol)
OQ -6 &
1 0-33 (a)
I
0
OH
OH
6
(c)
(b)
C-OH
(f)
OH
(g)
¢
�
� 6
OH
OH
(d)
HO (h) H
I
SH
H
U)
(i )
�H OH
(e)
OH
CH3S-SCH3 OH
(k)
~
10-34 (a) Hexan- l -01 will boil at a higher temperature as it is less branched than 3,3-dimethylbutan - l -01. (b) Hexan-2-01 will boil at a higher temperature because its molecules hydrogen bond with each other, whereas molecules of hexan-2-one have no intermolecular hydrogen bonding. (c) Hexane- l ,5-diol will boil at a higher temperature as it has two OH groups for hydrogen bonding. Hexan-2-ol has only one group for hydrogen bonding. (d) Hexan-2-ol will boil at a higher temperature because it has a higher molecular weight than pentan-2-01. All other structural features of the two molecules are the same, so they should have the same intermolecular forces. 1 0-35 (a) 3-Chlorophenol is more acidic than cyclopentanol. In general, phenols are many orders of magnitude more acidic than alcohols. (b) 2-Chlorocyclohexanol is slightly more acidic than cyclohexanol; the proximity of the electronegati ve chlorine to the OH increases its acidity. (c) CyciohexanecarboxyJic acid is more acidic than cyclohexanol. In general , carboxylic acids are many orders of magnitude more acidic than alcohols. (d) 2,2-Dichlorobutan- l -ol is more acidic than butan-I-ol because of the two electron-withdrawing substituents near the acidic functional group. 1 0-36 (a) Propan-2-ol is the most soluble in water as it has the fewest carbons and the most branching. (b) Cyclohexane- l ,2-diol is the most soluble as it has two OH groups for hydrogen bonding. Cyclohexanol has only one OH group; chlorocyc lohexane cannot hydrogen bond and is the least soluble. (c) Cyclohexanol is the most soluble as it can hydrogen bond. Chlorocyclohexane cannot hydrogen bond, and 4-methylcyclohexanol has the added hydrophobic methyl group, decreasing its water solubil ity. 1 0-37 (a)
(b)
�
U
OH
Hg(OAch H2O BH3' THF
•
•
NaBH4
H202 HO-
•
�
l N ({ .-
"
"
OH
220
continued BH3 H202 OH (c)� HO- ~ OH Hg( O Ach NaBH4 (d)� H2O � OH (a) OH (b)� JyPh OH V (OHd) Thigroup,s prothbleefmirsconft thinugsetshata lhappens ot of peoplis eth. atWhenthe GraiGrgnarigdnarredactresagentby reimovis addedng tthoeaH+compound from thet0-.hat has an et h er 0 CH3 M g1 CH4 + o�ly oned requieagentvaleaddednt of! �0- +MgI If a second equivalent (or excess) 00H (Gngnar sAciis, dthhydriens addeditolysisadd Iisf added no morbefe Groriegnaracidd reagent H30+ CH3etherMg1 atbefofwitlGrhoergieigketvhydrnare othne.deorlediyagent o l . mathydreorilaylsiiss,rtehcoveren theed.starting Mg 3 H H H 00H ( ' 0 p'. X 3 0+MgI UOH Ph Ph Ph (e) Q--'OH Ph+OH ( g 6 0H )Ph Ph crtu OH � HO�OCH3 HO�OH (i) V (k)� V OH d:) HO OH (n) HO 1 0-37
- THF
�
�
�
..
10-38
(e)
o
o
1
+
�
will
a
(f)
0
(l)
(h)
U)
�
(m)
(0)
221
ndicating that the Grignard ws are sacihownd isiadded. reactionAlislalGrloiwedgnardtorepractoceed,ions arande runtheninietn haersescondolventst.ep,Twodiluarrte oaqueous H 0 H3 + ? (a) � H BrMg -..../ Mgether CH20.. �Br -OH OH 0 Hp+ (c) )lH BrMg-Q 0 ~ ' B + 0 H3 Mgether U (YbH (d) O -' R (e) O etMgher CH20 H30+ VOH i l CH3 COCH2 O OH (g ) � H ~ 0 (h ) � XMgC C-CH3 agent e r d nar g i Gr a s i XMgC=CCH3 , y l al c Techni of compound c i l a anomet g or an s i t i e becaus H-C=C-CH3 RMgX usual e h t n i made not s i t i , However m. u i magnes d nar g i Gr al n i m r e t e h t g n i nat o t o depr by made s i t i n; o i h s a f alrekagentyl or wheralkenyle R is alkyne as shown. BH3 -THF (a) H CH2 + 0 H3 0 Mg Ph (b ) Ph �OH 'CI ether 10-39
o
+
---
(b)
.. �
�
+
..
..
' h
..
---
..
..
--
o
(f)
o
+
��------� '\ -
�------�
(
any
+
1 0-40
..
_
/'....
........",
--
..
..
222
1 0-40
continued
OH
o
eq. H2 d (c) 6 ) ( 6 Pt .. 6 6 OH o �OEt (e) �OEt o o OH o H30+.. � OH � OEt o OH (a) V MgBr CH,O _e_th_e--;r.. � V OH (b) V HO� � V NaOH.. o � OH � Mgether ( d) (e ) '-- Br NaSH �SH � Br� ( 1 \CULi � Br .� j :"� I the streacingtdhandof tbashe acie, sods torhethsiedebasofetsh. eThe sequattrongerionTheaciwidtposihandthetiosweaker tnroofngertheacibasequideandlwiiblrbasiualmwe canayswil berbeeactdetfatvoreormined gievdeatthequibye weaker librium. See Appendix in this Solutions Manual for a review of acidity. CH20H <weaker'}-o(a) CH3stroCngerH20� <stronger'}- OH CH3weaker base acid acid base °
°
1
( f)
10-4 1
+
�
�
r-==\. ---f
+
�
(f)
'
+
1 0-42
2
+
+
223
products favored
KOH Cl-Q- OH Cl OH CH3 O c6
H20 Cl-Q- 0-K' Cl
1 0-42 continued (b) +
h
a OH
stronger base
KOH
+
p roducts favored
products favored
+
stronger base
(CH3hCOH CH3CH20(CH3hCOH HO-
(CH3hCO- CH3CH2OH reactants favored +
+
stronger acid
stronger base
H20
weaker acid
weaker base
stronger acid
weaker base
weaker acid
(e)
weaker base
+
h
stronger acid
(d)
weaker acid
stronger acid
stronger base
(c)
+
weaker acid
weaker base
products favored +
stronger base
(g)
stronger acid
KOH CH3CH20H +
weaker base
weaker base
weaker acid
products favored
stronger acid
weaker acid
stronger base
�OH OR LiAIH4 (a) � H OR � OH LiAJH4 OR � OR 1 ) LiAJH4 H30+ reactants favored
1 0-43
I)
°
1)
•
°
o
2)
•
224
•
continued0
OH LiH3A0lH+ 4... (b) � CH3NaBH40H OR � 0 NaBH, . OH � CH30H ~ OR LiH3A0lH+ 4 OH 0 NaBH4 Li A l H 4 (d) ... OR CH3 0 H H3 0 + OO 00 OH NaBH4 OEt CH30H � OEt 0 0 OH LiH3A0IH+',. HOCH2CH3 OEt OH Q 0 (biooxed)n" in fworromkisntgarbackwar ting matedrsiaislstwofosisixxcarcarbbonson or fewer. ThefragmentproTheducts whigoalhaschiscoultocarsyntdbbeonshjesio,isonzedetthhienelatoagiGrirgcetalgnarcompound "didscronnect bond, and Brdouble bonds aro e madeMgBr from alcoholsewhiacticohn.arThee thebesprot ductwaystoofmakeGrignarepoxid redactes iiosnsfro. m the double Methger ... OH y O H30+ VD H 4 t MCPBA cro 1 0-43
1)
..
2)
I
(C)
1)
2)
,&
...
1)
...
(e) ('
'\
(f) /
"
2)
r---<
...
1)
+
2)
II
10-44
I
,&
12
I
h
o
+
--
�
H�
225
I
,&
2 S0
v
steps are reversible.
10-45 All
i
� }
H HO H - H20.. ./CHO··I · .. .. H-O H20 :.. :0: A A X
H � HO><�
� I+
• •
The symbol H-B represents a generic acid, where B-is the conjugate bas0 e. � H H� OH H- { H :OH -H } : B v f (a) H H � H -t- C� H H lI T\. H .
1 0-46
n
/ + .......
0:
/
�
---
H
(b ) (C )
OH H H
Q(
1 0-47
I
�
H
\.
..B
r
Q(
6MgBr
H : O -H _ • •
? , OH
�� H
" -
I
0: H
A
B
I)� 1 2) H,o' OH
�
C -......
H-B
H
•
�
• •
°-...:::
6 c
H2S04 .. /).
D
+
>-
. .
-...: :: 2 Mg .. .. ° ° et h er 6'
� o: cr � r crH � + H-B ao : : : H J H�' H a +
• •
�
H
H
E
226
H
+ :O-H .
exH2cesst �P 2 Br2..
isobutylcyclohexane
F
.
10-v4i8ngThiforscemechani srmeactisiosinmiislarreltioefclofeavagering softraitnheinepoxithe 4-demiember n ethyleednecycloxiicdeetbyherGr, whiignarchdisrewhyagentitswi. Thel driunder f o r t h e go a Grignard reaction whereas most other ethers wil not. R "MgX .D . --MgX �U 10-49 When mixtures of isomers can resCH3ult, only the major product is shown. CH3 (y O 1) CH3Mg1 OOH HZS04 o- CH3 HPtz .. 2) H3O+ U t Hz�04 t KM�04 � CH3 (J=O OH t 1 )H3CH3OM+ g1 tOH Hz�04 (J-O Q1 "1 "1" 0 t 2) Mg,cycloetpentheranone t PhC03H 3) H3BrO+ HEr U 0 he deskunkion tnhge smiulxfuturr; ealilsofhydrthesogene fupernctiooxinalde.grThioupsolars are eacioxididci.zed tThe1o0-s5trs0uoctdiTheuurmesmosbihavicart nibmgonatporone,etainttswbasro,eactoricaenough tnthreine toxygens can wash them away. sulfenic acitodisonize these acids, smakiulfinnicg acithemds water soluble whersulfeotnihec acisoapds �O H O H O H H O z z z z z z P SH thiol �OHS � S S-OH 3-� methylbutane-lHzOz �S HzOz � S'OH HzOz �SH S-OH OH (2-butbut-2e-ne-lene-l--thtihoilol) +
+
...
A '"
A
B
"
"
G
D
2)
F
C
I)
..
E
..
...
I
..
I
...
II
II
°
...
227
I I
°
II
°
...
�° II � I I
°
1 0- 5 1
�0 1 ) Mgether �MgBr H30+ H � Br OH t KOH, H20 S 04 , + H2 �OH 21 ) + Na 03 ) � Br 2 � �O- Na+ Br H G� + �O C � Br �O O� KOH, � H H �C::C� plus other alkynes 1 NaNH , 150° C + H20 2 ) O� I)Sia2BH H -C::C ---"" ......., 7- HBrnon-l-+yne H Br Na+-C =C� � Br �
----..
2)
D
..
E
fj,
A
F
�
B
/' Br
+
fj,
2)
/"'..
..
1
/"'..
I
2
J
CH2CH31
L
o
�
HO- CH3C -C=C� I
K
228
.......
/"...
.......,
((ab)) botoxihdatreiacton,iooxinsdarateiooxin, dreatductionsion, oxidation ((cd)) onereductcariobn:on C-O is oxidiiszedreplandacedone carC-Hbon is reduced-no net change (e) oxineitdhateriooxin (adddiatiotnionnorofrXeduct2) ion-the C stil has two bonds to 0 ((hg)) neineitthherer oxioxiddatatiioonn nornor rreeductductiioonn ((aelddiimtiinoatniofon of H20) (i) oxithedfatirsiot n:reactaddiionngisanoxi0dattoioeachn as cara newbonC-Oof thebonddoublise fbondormed to each carbon of the alkene; the shasecondonerbondeactiotno ioxygen s neither oxidation nor reduction, as H20 is added to the epoxide, and each carbon ion: OHH-Bareisadded, addedsiontthhere efirisstnoreactnetiooxin, dandatioBn inors replreaductcediobyn.0(Ninottehe st(hkeatcond) neiin ftruhencteractoxiiioonaln;datovergrioonupsanorl , onlirnevolducty Hand two cartboonsthelfikuenctalikoenesnal groroupalkiynes,s clasbotifiehdcarasboxionsdathaveion toro bereduction.) oxidized or reduced before thevingchange OH 0 0 (a) H2Cr04 6 6H 6 PCC no reaction (b) no reaction H2Cr04 o 0 0 OH (c) cY H2Cr04 PCC cY cY 0 (d) no reaction H2Cr04 6 no reaction (e) no reaction H2Cr04 0 PCC no reaction 0 C r04 H2 no reaction H3C -C-OH PCC no reaction 0 H2Cr04 CH3CH2OH PCC H3C -C-H 0 (g) H3C -C-OH 0 0 C r 0 4 H2 (h) H3C -C-OH H3C -C-H PCC no reaction C H A PTER l l-REA CTIONS OF ALCOHOLS
1 1-1
by
(f)
HX)
(j )
sti l l
net
1 1-2
pcc
..
..
�
..
�
pcc
..
..
(f)
..
II
II
�
�
II
�
..
..
�
�
II
�
229
II
troxiansdfoizred.medDMSO to the allodsehyde, and a fralomcoholthe sluolsfeusr;oneit is clearly (hydra) Aogen alincfoholonnilnogseaskettwoone;hydreachogensalcwhen ohol i s e s an oxygen ous, oxalto CO;yl chlhowever oride under, thegnetoes diefrefduced. sepctroporon tt(hiIoefnatyoueleiomentngotretactshiosnioeoxaln:twoneyo,l youchlCoidirsidoxidethidseiz"prednootblchange". oeCO2m corandrectthlye.)otToherbeC irisgroreduced ction shows that dimethyl sulfide reduces an ozonide. In the process, dimethyl sulfide (oxib)diTextzed toseDMSO. (enera) Dehydr othgenatis reaciontiodoesn, ornotioccur atmodynami C-eictaherl y: unfthaervoreaibls ae,hiwighthki[)'nGet>ic barriTheer (alathiergposh actsiibviatlitiyonis g y f o r t i s t h er ) supporC (tseedebytextthseefctaicotnthat theCre)verandsethrereaceftoioren has(cat[)'alGytic O.hydrThiogenats makesion tofheaquescarbtonylion of) kisnpetonticsaneous at c-a academi prwiolceiedmprmusovet wibethusienlcreseaslyinslgow.temperature for virtually all reactions, so both the (rhydrebact)andoiogenatn(cth)atioKinetcnaandnnoticsdehydr orgenat ion.n rTheeactansionswerwilisgothatfatshteerrm. Iodynamics n this case,wihowever , tthhies quesreacttiioonn asis howthe to fteampervor thaetudehydr o genat i o n e act i o l f a vor es[)'5ti>matesitnhrceeationesMi>raimolsed.esiculThencee tikeyhseconver prisotducthetefduketndament oto tnewo:platuhlserthydrheerfomroeodynami gen, [)'is5lecs equatstAtablieloontwhan[)'teGmperthe sMiataturtri-engTflSalcoholC. )We, Mi. Alcanso. all, so [)'negat G > O.ive.AtFora hitghhe enough overdomiwnathelesmbecaus Mi, ande T[)'iGs swio slmbecome reactionteinmperquesattuioren,, tthheis-T[)'mus5t tbeermthewicasl begie at n to C. (a) AND HOPCC (b) all three reagents give the same ketoneCH3pro(ducCHt2wi)4 Y" th a secondary alcohol (c) - H Na2H2CS04r207 AND PCC OOH 1 1 -3
10
20
a
is
8- 1 5B
1 1 -4
250
1)
O.
2)
250
<
1 0- 1 1
is
=
0
0
-
T
<
O.
(250
3000
1 1 -5
--.....,.�
°
•
•
(d) all three reagent><s givHe no reaction with a tertiary alcohol NR ( n o r e act i o n ) V � -
-.-
230
Note-PCC to the st(puyrdentidin: iuForm chlsimoplroichrcityo,matthise)bookto oxiwidlizeuse talhescohole stasndarto alddlehydes; aboratory methods of oxidation: -H2 C3,r0H2S04, 4 (chromiacetc acioned)(Jtoonesoxidreizagente )altcooholoxidsitzoecarbaloxylcoholicsacitodkets; ones. -cr 0 Under are legi, andtimatCole; floinrsexampl e,orSwemPCC oxiwildatoxiiodnizwore aks about welo a lketasone PCCas welinsltasathndechrprthoeatparmicotahtacieriondchoi.ofIfalcyouesdehydes r e agent al c ohol t consult the table in the text befohavere Proablquesem tion about the appropriateness of a reagent you choose, 0 (a) �OH PCC � H 0 C r 0 4 H2 (b ) � OH cr03 � (c) �OH H2SO4 acetone 0 0 PCC (d ) �H OH 0 C r 0 4 H2 (e) OH � OH (y oH cr03 &0 BH3• H2SO4 H2SO4 oH - H2O 6 H202, HOacetone esent toitihandl etenzyme hanolA, chrmolso roeeniquiculc aleres.csoholmoriec ethashanolinduced"antidmorotee" ADH moleculenzesymeto actto asbe aprcompet ve inhie blairtgoer tamount o "tie up"s thiemextbibreda OH OH CH3-C-CH o0 0 o pyruvaldehyde pyruvicCH3-C-COH pyr u vi c aci d d is a norof mglalucosmeteabol("biloteod sugar") in the bracieakdown 1°
1°
2°
2°
as
1 1 -2.
1 1 -6
•
� OH
•
•
�
•
�
(f)
..
THF
1)
Ll
•
•
2)
•
of
1 1 -7
1 1 -8
[0]
..
I I
II
[0]
231
II
I I
"Ts" �� CH3 Ts � -�� CH3 (a) CH3CH2 - OTs KO - C-CH3 CH3
1 1 -9 From this problem on,
will refer to the "tosyl" or "p-toluenesulfonyl" group:
o
I
+
I
�
CH3 CH3CH20 - CH3 CH3 I
C
(E2 is also possible with this hindered base; the product would be ethylene, �
+
+
(c)
R
(d)
cS
�
(e)
1 1 - 10
NH
OT
s
..
c5
+
�
+
"
S
excess
..
-
Na+ :C=CH
+
---l.�
6
(c)
(d)
inversion--SN2
+ NH
�C=CH NaOTs +
OT NaBr � OH � OH TsCI � OTs NH3 � OH TsCi �OTs NaOCH2CH3 � O -...../ � OH TsCI � OTs KCN � CN TsCI .. pyridine
�
s
•
excess
(b)
s
� I NaOTs �OTS NaI H TsO H NaCN � � NaOTs + H3 -0TS NH2 T, NH3 3 + , OTs
(b)
(a)
OT CH2=CH2)
+ K
-
I
pyridine
pyridine
pyridine
� Br
�
•
•
..
•
•
•
NH2
1 1-1 1
(a)Q- CH20H TsCl - d-
pyn me
"
o
Q-CH20TS OR Q-CH20-�-o-CH1 232
o
1 1 - 1 1 continued
( b)
0-
CH20Ts
( d)
( c)
Pt minor
maj or
� O O 6� B
Br
H
o
o ro 0
+
0
----
o
t
0
: Br :
nBr � o� oo HV
·
1 1-13
CH3 � H3C -
t - OH I
n
H - C�
0 0
CH3
� �H2 +
"-- :B�: o
V
0
�+
CH I
H 3 C - C - OH2 I
� V Sr
--=-
0 0
_
+ H3 C - C
CH3
H20
----'l.�
I
:
:Cl
CH 3 I
+-- H3C - C - Cl
I �
CH3
+ H20
_
I
CH3
CH3
carbocation intermediate 1 1 - 1 4 The two standard qualitati ve tests are:
1 ) chromic acid-distinguishes 3 ° alcohol from either 1 ° or 2° R
R
+ OH R
H2Cr04 (orange)
•
R ( or H)
n o reaction (stays orange)
R
+ OH H
233
H2Cr04 (orange) •
R(or H) I
R-C=O
+
erJ+
blue-green
1 1 - 1 4 continued 2) Lucas test-<listinguishes 1 ° from 2° from 3° alcohol by the rate of reaction
R
R
OH + R
+ HCl
soluble R R
2°
+ OH
+ HCl
+ OH
+ HCl
H soluble H
1°
R
H soluble
ZnCl2
ZnCl2
R
..
R
Cl + R
+ CI
insoluble-" c loudy in
+ H2O
insoluble-"cloudy" in 1 -5 minutes
+ H2O
insoluble-"cloudy" in > 6 mi nutes ( no observable reaction at room temp. )
II
I minute
R
..
R
H
ZnCl2
<
+ H20
H
..
R
+ CI H
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - . - - - - - - - - - - - - - - _ . .
OH
(a) Lucas: H2Cr04:
(b) Lucas: H2Cr04:
A
c loudy i n 1 -5 min . i mmediate blue-green OH
cloudy in < 1 min . n o reaction-stays orange
cloudy in 1 - 5 min . i mmediate blue-green
no reaction no reaction-stays orange
A
�
(c) Lucas: H2Cr04:
Lucas: H2Cr04: (e ) Lucas: H2Cr04:
�
OH
cloudy in 1 -5 min . no reaction D OES NOT DISTINGUISH-i mmediate blue-green for both � OH
(d)
o
� OH
no reaction cloudy in < 1 min. * * D OES NOT D ISTINGUISHi mmediate blue-green for both
�
+OH
( * *Remember that allylic catioIlS are reson �nce stabilized and are about as stable as 1° cations. Thus, they will react as fast as 3 u i n the Lucas test, even though they may be 1 °. Be careful to notice subtle but important structural features !)
no reaction cloudy in < 1 min. D OES NOT D ISTINGUISH-stays orange for both methyl shift .. ---1
3° � jI' ,CH3 : Br : I � C / 1 .... C H2 +
• •
-
Br
�
1° Even though 1 ° , the neopentyl carbon is hindered to backside attack, so SN2 cannot occur easi ly. Instead, an SN 1 mechanism occurs, with rearrangement. 234
a= H
This 3 ° c arbocation is planar at the C+ so that the C l - can approach from the top or bottom gi ving both the cis and trans isomers.
CH3
C - CH3 +
approach : C l : from above H
0=
: Cl :
approach from below
CH3
;: Cl CH 3
1 1-17 CH 3
0 " ,
1 1-18
3
1 1 - 19
( a)
o�
" H
• •
�
---
•
OH
+
PBrJ
SOCl2
-
CH3
6 -
CH3
·H
+ C 2°
�
-
hydride shift
Br
6
+
(from Hel) CH I .� CH +C ' : Cl : "
O
•
•
•
'
CI
O
P(OHh
retention
OT s
OH
( b)
3
"
3°
Cl
OH
6
-�
H3
G
H
ZnCl2
�
TsC I
pyridine
6
Cl NaCI
SN2-inversion
23 5
o
Another possible answer would be to use PCl 3 .
(a) OH1 : �\S =O0 +0H1 -1S ,- O· D +O-S=O �I I) . . ·· · .. (C . Cl X I .. + 0 O O
1 1 -20
Cl
D
I
I
:o - s = o :
·0
S02
j
+
..
:Cl :
+
{
I
.
-
--
al
- . .
• • -
D
I
• •
H
Cl
• •
A· � tH
allylic! �
�0 }
- � - S = R:
r. D "
+
-
l e b 0 0+
� - HCI
:Cl :
Cl
O I
D
te carbocata "firoene icars alblocatylic,iovern".y Thestablnucle andeophirelaltiicvelchly oloring-de lcanived.attaItckcanany tcar(hber)beTheonforwiekeyesthcposiapeis thttatihveethicharoeninpaitgee,rmedi r andnot jabecome ust the one closest. Since two carbons have partial positive charge, two products result. (a) OH ZnCIHCl 2 no reaction unles heated, then �CI HEr �Br PBr3 �Br P12 SOCI2 1 1 -2 1
�
...
�
�
-
�
�I
�CI
236
IOH ZnCIHCl 2 IClr B
(C) �OH ZnClHCl 2 �C
1 1 -2 1 continued (b)
..
� � �I 12 SOCI2 �C
�
HE r
PBr3
-
1SOCl2 2
P -
-
I II I C I
HBr
Br
PBr3
�
..
�OH ZHCl 2 nCI
1 °, neopentyl
HB r
•
•
l
P
no reaction unless heated, then S N I -rearrangement
�Br
� 12 �I -
Br
..
(poor reaction on 3°)
SN2-minor (hindered)
PBr3
Br
(poor reaction on 3°)
-
(d)
l
Br
P -
SOCIl. �Cl 237
+
CI
(poor reaction on }O)
�
Br
�
l
S N l -rearrangemen t
11-21 continued (e)
carbocat ionfromi ntermedi ate CI HCI can be att a cked ei t h er 'eY ZnClz side by chloride ,Br SN2 with inversion of configuration " HEr � U � ,Br SN2 with inversion of configuration U � " I SN2 with in version of configuration Iz U SOClz �CI retention of confi guration U SN 1 ;
+
..
"
p
"
11-22 OH (a) � (b)
(c)
�
OH
�
+
major major
OH HzS04 ,
major
HZS04 ,
(cis
+
+
minor
trans)
�
Jx OH (d) U
(e)
rrunor +
(cis + trans)
major
minor
major
minor
mmor
�
238
trace
11-2 3
(y O: .. V H I
°
� II�
P, ' Cl'l Cl Cl
•
�N -CI+ II l �- f �-.� V
0
°
H
goo d leavi ng gro up
�
..
lo �
O
C
E2
0
cy clohex ene was fonn ed withouta car boca tion in ter media te
""i
..
Cl
°
• •
I
Cl
• •
• •
II I
N
+
H
!O
: O-P-CI
I+
- -CI
o
H
1+
° +
H
N
+
o
Cl this produc tca n reac t with two morea lcohols ot b eco me leavi ng groups in the elimi na tion E2
11 2- 4 B oth mecha nis ms b egin with protona tion of the oxy gen. H
H
H
� H+ . • H-C-C-O-H • .
1
1
1
..
1
H
1
H
H
1+
1
H-C-C-O-H
1
H
H
1
• •
H
O ne mecha nis m inv olv es a nother molecule of etha nola ctingas a bas e, giv ing elimi na tion. H
H
1
H
1
1+
T� T� i{q
H- .-A
·
H
-H
H H / \ C=C \ / H H
•
CH2CH3
H20
+
The other mecha nis m inv olv es a nother molec ule of etha nola ctingas a nuc leophi le, givi ngs ubs it tution. H
1
H
1
H
1+
H
H
�
+ I)
� U·
H-C-C-O-H
1
H
• •
.. CH3CH2-O-CH2CH3 ·· • • HOCH2CH3 ··
ItqCH2CH3 .. CH3CH2 -
O-CH2 CH3
11-2 5 A n equi mola r mi xtur e of metha nola nd etha nol would produc ea ll ht ree possib le ethers . T he diffi culty ins epa ra ting thes e compounds would pr eclude this method fromb ein ga prac tica l route toa ny one of them. T his method si prac tica lo nly fors ymmetri c ethers , tha tis , wherebo tha kl y l groups ar e identica l. CH3CH20H
+
HOCH3
H+ tl.
..
H20
+
CH3CH20CH3 239
+
CH30CH3
+
CH3CHzOCH2CH3
' � H H CH3 9 H20 CH3-O-CH3 .H CH3-O-CH3 3CH -O-H H+ CH3 -O-H CH-3-O-H � H HC H (a) X�H ' H+ X� H H 0 2 : #"--. . a &H H 2 0 VH H VH H-OH a :CHl H20: OCH3 H H H+ H H H O O : : d � OCH3 (cjOCH3 O H H /·· CH -CH30H HH H+ H : aH aH H l H20: a
11-2 6
• •
,"",1+
�
•
• •
�I+
•
• •
• •
•
• •
• •
11-27
• •
�
�
I+
+
�
1
y---..
• •
•
0.
•
(b)
+
9.CH3
--
+
• •
1+
..
• •
-
�
H
-
er
CH2 '-"h!' H � H+ O O - H2 0 H H �
(C� CH2-�H
�
----;.�
+1
1°
+
..
(1: � L) 240 �
H
H(5
hydride shift_
HC H,I/.H )b
�H � H 20 ) +
H
�+6
(!j (d) a�H,- Ol+-
11 2- 7 conti nued
�
CH
H3C
w
H
il - ''""' Hz O
) " 7
CH3
'--C+
H3c
(a) n ++0 CH,
H3C
w· H3C
n -T-rfHO
HO :"OH "
�
· H3C
CH3
OH + 1 H
M ethy ls hift
H'C
-H20
H
HO
H�:
CH3
CH3
. rmg
H20:
contraction
n -t- 1
-CH3
HO
� � Q-� : QCH
1
;"C
_
___
H2C
11 -2 8
o
CH3
'" H R+ CH3
CH3
-
QCHl 0
"- ""
CH3
H20:
A lky ls hift ri ng contraction
H3C
-W---HO
C-CH3 +
Q
""
H,C,+ C
1
H-O:
4
CH3
H'C ·
� -0
'"
H-O+
241
CH3
H20:
H3C
o
CH3
0H oH (b) oH.f + H H 2 .. CX;; OH cYf G:O-H O<�C � /\ ���; +OH �-H o H2··0 : / a a a 30 and doubly benzylic carbocation
1 1-2 8 co ntinued
H
�
1+
-
°�
• •
Ph
Ph
Ph
Ph
\
+
� Ph
Ph
U
Ph
•
/
f'-.-- /
/
"
+
1
rIng expa ns oi •n
°
H3C",C'H +H� if II
(a )
(c)
11 -3 1 a( ) (c)
(b)
2
°
Ph
°
(d)
O-H [c � CH +
/
• •
-
1
1
S imila r ot the pina co l rea rra ngement, this mecha nis m invo lves a ca rbo ca tio n next ot a na lco ho l, with rea rra ngement ot a p ro ot na ted ca rbo ny l . Reliefof so meri ngs tra in in the cyclo propa ne si a na dded a dva nta geof ht e rea rran gement. 11 -30
Ph
Ph
OH OH0 H+ H C H V, / �C-C-H C==C H H H H
� �
+
.----- -
Ph
�
Ph
11 2- 9
nS oi n
H
�
3
C
[�-CH3
� HS�4-
1
H
c:fCH] 0 (Y"' CH2=0 +
�: °
° II
+O-H ij�
° II
Cl-C-CH2CH2CH3 + HOCH2CHzCH3 (b) CH3(CH2hOH Cl-CCH2CH3 0 H3C-<, }-OH + CI-CCH(CH3h [>-OH Cl-C-0 +
° II
(d)
242
+
II
� !J
11-32
0: : ..
: 0:
II
: O-S- O CH3
.. II
... .. __--i.. �
I
-
0==S- O CH3
•• II
.... ..f-. -l.. �
:0 :
:0 :
· 0· • II . 0==S- O CH3
••
I ..
:0 :
11-33 Proton transfer ( acid- base) reactions are much faster than almost any other reaction. M et ho xide wi ll act as ab ase and remove a proton fromt he oxyg en much fast er than methoxide will act as a nuci eophile and displace water. CH 3CH2 -O H + H+ ---
11-34
( a)
o-
-OH
o-
O CH2CH3
(b) T here are two problems with this att emptedb imolecular dehydr at ion. F irst, all three possible ether combinations of cyci ohexanol and ethanol would be produced. S econd, hot sulfuric acid aret he co nditions for dehydrating secondary al cohols like cyci ohexanol, so elimination would comp ete with substituti on. 11-35
(a) W hat the student did: N a+ 0, H
X
X
CH3 CH2- O ""
H
+ TsO- CH2CH3 H3 C CH2CH3 C H2 CH3 H3 C sodium (S) -2- butoxide (S)-2- ethoxybutane T he product also has the S config uration , not the R. W hy? T he sub stitution is indeed anSN2 rea ct ion, b ut the substitution did not take place at the chiral center, sot he config uration of the starting material si retained, not inverted. � ---� ..
,
( b) T here are two ways to make (R) 2- - ethoxybutane. S tart with (R)-2-b utanol, mak e ht e anion, and substitute on ethyl tosylate simi lar to part ( a), or do anSN2 inversion at the chiral center of (S)-2- butanol . SN2 works better at 1° carb ons s o the for mer method would b e preferred t o the latter. ( c) T his is nott he optimum methodb ecause it req uiresSN2 at a2° carbon, as discussedi n part b( ).
X
HO
H
CH2 CH3 H3 C (S)-2-b utanol
•
TsCI pyri dine
T
S-
O
X
X
H
H
H 3C CH2CH3 (S)-2- butyl tosylate no inversion yet 243
low temperature ( high temp. favors el mi i natio n)
O CH2 CH3
CH2CH3 H3 C (R)-2- ethox yb utan e INVERSI O N!
11-36
OH o-� + CH/�"'0-S03CH3 ---o-� OCH3 :0-S03CH3 CH2 C H3 SOCI 2 CH3 C H2MgBr CH3CH2-COH-CH2CH3 CH3CH2-C-OH • CH3CH2-C-CI H30+ � H2S04 BH3 • CH2CH3 H202, HO- CH2CH3 CH3CH C-CH2CH3 CH3CHCHCH2CH3 OH cr03 OH b( ) CH3CH2CH20H PCC CH3CH2-C-H CH3H30C+ H2MgBr.. � H2S04• � (alo- 0H H+ HC03H OH o H30+ U ' O H OH HCI CI (blo- 0H cr03 O H3 0 + Q--\ Q--\ H2PBr3SO: Mg,CH3CH2MgBr'" CH3CH20H CH3CH2Br O H OH Br MgBr � � 3 Ho+ g U H+ M � ��� T H2S04o- Br Mg OR:o-cr030H � � �! ���A + H , o ) -- .. Q-f; 0H rn 0 ) - �gBr O-(;;C o- 0H � O- Br ��. 0 CH3Br HJ Q-f; CH30H �r3 o� -
1 1-37
�
-
0: N a + .. �
°
o
1) 2
II
II
+
-
(a)
"
I
•
2)
I
I
2) ......I-. ---
I
°
----
TIIP
1)
I
=
°
1)
II
acetone
2)
11-38
--f!.
any peroxy acid canb e usedt o fOlm the epox ide whi ch is c l eaved to the trans dia l in aqueous acid
•
"
_
� U
___
(c)
---
� /' ether
---
ether
--
ether
�
.. ---
from Dart (b ) ----'�
�
-
�
(d)
ether
from (c) see the nex t pag efor an al tern ati ve endi ng 244
_
rom the previous page � H30+ 0--* CH30H SN (d) contcontinOHuedinuedf� I /"'yBr � OH H2S04 �OCH3. ether fmm are severexamplal possie mayblebecomdibf ienratentionsandofsGrtil igbnare cord rreeactct. ions on aldehydes or ketones. This is e) There.e Your (exampl CH3CH2CH20H �C O '> Br etMgher h H)O' �OH cr03 o- 0H PB,) H2 S 04 �o �OH H+ Br2 �Br etMgher �MgBr � CH3SNCIH20H �'; OCH2CH3" H2S04 �OH Please refer to solution page of this Solutions Manual. (a) �OH SOCl2 - �CI (b) Q- 0H SOCl2 o- CI PBr3 -�Br PBr3 - o-Br P12 P12 0-1
1 1- 38
°
/"'y U
U
____
eel
/"'y U
one
/'..
_ ____
___
L\
___
1 1- 39
_
_ ____
�
1-2 0,
hv
____
___
H 3 0+
12
1 1- 40
2°
..
1°
---�I
(c) oH HCl - 01 HBr - 0' -0
..
3°
Cd) (jOH SOCl2 &CI PBr3 &Rr P12 &1 ..
..
III
-
245
" (a) � COOH Q
11-41
(b)
�
(e)
R (f ro min version )
R
(f)
(e)
Cl Q
cJ
(g) �Br
lfH °
(d)
(h)
CH20MgBr bCH3CH3 H
(I) C;-H (i) + OCH] � CH30H 0 (m) �O� � major � mInor EtOH (y 0- Na+ CH)CII,Br (y 0CH2CH3 (y 0H (a) V ether synthesis V V Br OH Mg d NaOH. d� V �H (( d , t t H30+ H2S04 OH MgBr H� � cr03 0 H SO , ; p' O H � OH H � (d) VOH �o 2) CH3H30C+ H2MgBr .. U U)
+
(k)
+
(n)
+
+
11-42
•
( b)
W li li amso n
�
M gB r
+
y
..
(C )
B rM -E.. ether
t:,.
1)
�
ether
°
°
II
•
2)
�
Sr
1)
246
a ceton e
Major product for each reaction is shown. (a)� (c)� (dl O (el O 0 Note that (d), (e), and produce the alkene. (c) V COOCH2CH3 (a) CH3CH2CH2COOCH3 P-OHCH3 (d) CH3CH20-OCH2 HN CI+ (y 0-�-CH3 Cl-�-CH3 pyridine V 0 + a chlmetohrianesul de fonyl (a) HO", �H SOCl2 C� S-retention HO', -.;H pyriTsCdiIne TSO� KEr R-inversion (c) HO", �H pyrTsCidiIne TSO� R-inversion
11-43
cis
+
trans-rearranged
(b)� cis
+
cis
trans
+
(t)
trans
rearranged
( f) ,
same
11-44
I
h-
o II I
o
11-45
II
I
o II
..
o
11-46
/� _
S
(b)
.
..
/"-..../ "-..../ _
S
/"-.� .../ _
S
Alt�rnatlvely, PBr3 c()uld be
.
.
S
..
S
used,
could use NaOH if kept cold to avoid elimination
247
inued ts alcohols to bromides without rearrangement because no carbocation intermediate is cont3 conver PBr produced. OH Br � PBr3 � qU H q U OH �o U PBf3• Bf OH V' V 0 0H 0°- Na CH3.I d) H OH SOCI2 l l ct- • TspyriCdiIn..e OR HCH3 QCH3H �CH3H SN2 CH3H qCH3' H OH HI04 q. oH (e) ct H2S04 CK. 2) Me203 S OH Y CH2 COOH H 0 H2 C r 4 0 a a g -t-o- 0H HCr2S0O43 • +0 0 acetone -t-o- 0H H2SO4 -to �I TsCI OH aI P,I2 .. a CH3 cis OH pyridine � CH3 cis OTs
1 1- 47 (b)
11-48
·
pec.
(a)
(b)
(e)
�
(
•
inversion
retention
�
( f)
(
..
...
..
)
(h)
(i)
1)
I
•
�
..
..
(j)
248
yaH PBr3 )J""Br yaH SOCl2 yCi yaH ZnClHCl2 QCI and yaH HBr QBr and yaH TsCl, pyridine )J""Br O H � (a) �OH cloudy in min. no reaction O H (b) cloudy� � immediiante blumie-gnr.een nocloudyreactiinon-stmin.ays orange (c) Q-OH o climomediudy iante blumie grn.een nono rreeactactiioonn-stays orange (d) Q- OH 0 0 iclmmoudyediiante blumie grn.een nono rreeactactiioon-sn tays orange (e ) clooc udy in min. Lucas: no reaction
(a) (b ) ( c) (d) (e )
1 1 -49
..
inversion
•
retention
SNi
•
cis
trans
SNi
•
cis
trans
inversion, SN2
1)
•
2) NaBr
1 1- 50
1- 5
Lucas:
OH
1 -5
<
1
<
1
1 -5
1-5
249
1 (b){ o:s
1 1-5 1
\
H
H
...
/
p =c\
(a)
H
-
•
: 0:
0:
� ' cO
..
...
: 0:
:::::-....
..
�
cO
•
t
: 0:
•
...
. .
-
:00
..
•
..
� ' c6 �
.
.
. .
: 0:
00
•
: 0:
: 0:
c6
H
: 0:
: 0:
t
H
. .
. .
cO
}
/ \:c-c / \\
H
: 0:
...
cO
•
�
�
�
The las t three res onance for ms ar es imi lar to the firs t three; the change si that the electr ons ar e shown in al tern ate pos itions in the benzeneri ng. To ber igorous ly cor rect, thes e threer es onance for ms sho uld be included, but mos t chemis st would not wr ite thems ince they do notr eveal extr a char ge del ocalization; unders tand that they woulds till bes ignificant, even if not wri tten with the others.
(c)
II
: O-S- CH3
•
II
.
� OH A
.... ... f---l.�
: 0:
1 1-5 2
I
: 0:
: 0:
: 0:
� Br C
I
0 ==S- CH 3
•
.
II
.... .. f---l.�
: 0: Mg • ether
D
� M gBr +
� 0 B
250
II
:0 :
0==S- CH3
•
.
I
---
E
O M gBr
0'
�Br M g __ --i.� U ether
X
w
v 1 1- 54
[��+ f)]J,
00
�
M gBr
HO
era
+
CH 00 _-_H_2-i�� �5
10
alkyl shift-. . n ng expansIO n
recall that c arbocations probably do not exist; this could be considered a transition state
N OTE :
O-D
_----;.�
ri ng exp ansion fr om 1 ° car bocation to 2°, resonance stabilized c ar bocation
T he mi gration directly above does N OT occur as the cationpr oduced is notr esonance- stabilized. An alternative mechanism could be proposed: protonate the ring oxygen, open the ring to a 2° carbocation followed by a hydride shift to a resonance-stabilized cation, ring closure, and dehydration.
r?\
:96
o
�
t OH j ( H: 9 II:� b
hydride sh i ft to resonance stabilized cation
OH
-
2°
car bocation
. .
H+ off o ne 0, H+ on the other ° ..
251
�90. H
H-O+
1 1 -5 5
.
( a) ("f OH
V
( b)
.
'----A Z
•
nCl2
-
� o -z nCI 2 V H r ,.,
-
.
V � OH • •
H i" QS 0 3H
�
'---'tf
Cl- + H20
9+
Cf(+
�
H -Cl
+ H20
H
�H2
C
----;.�
__
.
OH2
,-- .
.. � �
_
• •
H
\ �c� \ . � + H20:
c( ) A ll thr ee of ht e produc ts go thr ough a c omm on carboc ati oni ntermediate. H ... ... H � H�
� •
•
� OH
H dS 03H •
t-
r'0 � � �+ V A.. H V ; C ' H
H
Onc e ht i sc arboc ati on si formed, removal of adjac ent protons produces thec ompounds sh own.
Cl
��
+ HO - Z- n Cl 2
I n ht i s presentati on of the mec hanis m, ht e rearrangementi s sh ownc onc urr ently with c leavage of ht e C- O bond wi th no 1 0 c arbocati oni nter medi ate.
r'0
oc/ :ci: 0 }\ H
'"
-
OH
OH
-
�
�
.. H dS 03H +
�: O ... H
f
�
i OU+r:(ll-_H � CO ) �J ! 00 H20:
ci?
C
Y + 00 � H20 :
+ ' ' 0) H ·· H
T h;s;s the pc oduc t fc om a p; nacol re arr ange ment. 252
ZnCI2 ' .
CH3 (a) 00Hcr03 a CH3MgBr o-CH3 0-°1 H2Saceto0ne4 MgBr t PBr 3 CH30H CH3Br p�HO CH3OH HOI OR: alternati"ve end" ("{1 H2 S 4 � 0 V d Griaftegrnarhydrd prolyosductis of SN1 OH (b) 0oH PBr3 0 Br e� Mg o MgBr .. HO+3 if r 0 3 PCC 2 4 H SO CH3CH20H H � acetone� if MgBr 0- MgBr' o CH3 (c) OH H2CrS003 4 I oE . � . ( b f r o m ) V acetone orshouswne theparS t (meta) hod �OH PBr3.. Br etMgher ! MgBr EtOH, H+ H3 cr03 0 .. . C C .. . .. .. .. .. .. Et O H OH H 4 0 � H2S � twoerRMgX cang a addaltcooholan H2Cr04 t acet o ne OH est , maki n 1t1) H3Br0M+g Mg PBr3 HO grwiothupstw;oseeequisovlualtieontn Rto CH.,OH problem OH (e) MgBr VOH fO rom H30+ 11-56
a
a
..
..
Mg,
ether
+
� I
---
109
...
..
a
..
a
�
°
a
� I
I
•
II
I
�
10 "
(d)
2
2
�
-
2
� +
a II
•
a
3°
-..../
2)
___ �
--..../
10-18
a
l)� ..
(b)
2)
253
..
N
1
a II
1 1 -56 conti nued (f) HO�
V �
OH
(g)
HO"
H
Mg
P B r3 ..
ether
P CC..
B rM g�
�o ..
H
�
K CN
TsCI .. pyri di ne TS O" H
o ..
B r2 -
hv
OS 04
�
H' NC
..
Br
6
PCC
Mg
-
ether
CH 3CH2B r P B r3
t
CH 3CH20H I I -57 F or a compli cated synthesi s il ke this, begi n by work ing backwar ds. Tr y to fi gur e out where the carbon framework came from; in thi s problem we are restricted to alcohols containing five or fewer carbons. The dashed boxes show the fragments that must be assembled. The most practical way of formi ng car bon- carbon bondsi s by G ri gnard reacti ons. The epoxide must be f ormed from an alkene, and the alkene must have come fromd ehydration of an alcohol produced in a Gri gnard reaction.
�o�<>,,l, ,H
maj or i somer
1. 2.
°
�
TsCl, p KO H,1\
BfM g�. y
t
)
H30:
!W
0:
254
\
MgB r
L�
2. Mg
avoid c arbocation conditions to prevent rearrangement
Mg ether
o-
0H
cr 03 , H2S 04, aceto ne
synthesese, thsufe fNer fcondi rom tthioensmiofsconcept iioonntcannot hat incompat ibtlhe rtheagent scondi or conditionstioofns can(sao)diBotco-umehmetxiofsth.thoxiIesnedthe.pseudoe fTheirst exampl i o ni z at exi s t wi e tert,iawoulryincartdhbegiocatsveecondithoen desiistnetp.hreed(fTiprrhestosductitreopnywoulwiisththoutdatnottthheewaifsodiirtstarustomeundpmetbylohitngoxiselenough f o r t h e sodit-butnuymtlhbemetrosemicondhoxide dinreemetactto beihoanoln,added f , t h e s o l v ol y s i s of d e. ) e acidtiiconscondiof tthioensseofcondthesfteirp.st sItfebpasinicwhisodichumthemetalchoholoxidies werproteoadded nated artoethe Ilfuric aciibdleswioluthtiothn,e tbhasie itcnhscondi isnucompat alizat.ion would give methanol, sodium sulfate, and the starting alcohol. No reacttiaontnaonneousthe alacicohold-baswoule neutdroccur SN1 solvolysis conditions war m Several synthetic sequences are possible for the second synthesis. �OH Na �O-Na+ CH31 �O/ OR �OH pyrTsCIidine �OTs NaOCH3 �O/ OR�OH PBr3 �Br NaOCH3 �O/ Compound OH al-muscoholt;becan'a t be° oral2°lylaliccasoholthiswiwoulth andalgikvene;e a posnoirteivacte Lucas ion witthestLucas leads to a Compound possi -musbltebe a cyclic ether, not an alcohol and not an alkene; other isomers of cyclic ethers 6 J5 TS NaOCH3 Ts C I "-J5 pyridinetworhiskresactfinie,on NOcannotREACTIdo anOSN!N2 reaction but wai t . . . . Theare twWiolreiaasmsonsonwhyetherthsyntis tohsylesiasteiscannot an SN2unerdisplgoacement ofreacta leioavin. nFig rgrsto, upbacksiby andealatktaoxickdcannot e ion. Theroccure an S 2 N de of thunder e bridggehead bbriecausdgeheade thecarbackbonsicannot o invercarsionbonbecausis bleockedof thebyconsthetotrahinertsbriofdtthhgehead. eisbricardbgedonSecond, rcannot ing sytsihnteeverm.t � whimechanich issrmequired in the OTS H� atbblackstoackedckidise H -§r OTs \:::J::t' side view side view 1 1 -5 8
S
SN2
1
(b)
11-59
�
X:
..
.-
.-
.-
.-
..
1
10
Y:
___
..
---l'-�
SN2
255
11-61 LetThe'saxibegialnalbycoholconsisidoxieridnigzedthetefnacttism. es as fast the equatorial alcohol. (In the olden days, this observation was used as evidence suggesting the stereochemistry of a ring alcohol.) as
Second, i t i s known t h at t h e oxi d at i o n occur s i n t w o s t e ps : 1 f o r m at i o n of t h e chr o mat e est e r ; and ) 2) loss of H and chromate to form the C=O. Let's look at each mechanism. ' S . TER : FA � H p2 St e � ( ( . :7 ; � 0 ��� 0,\.Cr03H I
A
X
A
OH
H
H
EQU TORI L
� A
HO
A
H
aboutin ththeeseaxiasyl spostemsiti?onWebecausknowe anythatgrsuobsuptiattuentthesaxiarealmorposeitisotanblhase the So whatoriadol posweitknow i o n t h an equat diequataxiaolriinaltechrractoimatonse. estSoewhatr woulifdStfoerpm fwerasteer tthheanratthee-laxiimiatlinchrg sotematp? eWeestewoulr; sindceexpectthis isthcontat thraery to whatis ratethliemdatitinag,shweow,woulStepd expect is not tlhikeelappry toobeachraofte-tlhime basitineg.(pHowrobablabouty watStere)pt2?o the equatthe eloimriainl ation hydrhydrstericoogengenconges((eaquatxitiaolnochrassoci riaolmatchraeotematestd wieert)esthwoulsucher).daMorbelarfgaestovere egrrot,hup.tanhe taxiThiheaapprsl iests consioeachr issmoroftenttheewimotbasthietvhatteoertdehleatotaxiivleeavearlatedues ofto rmechani eaction sfrmomis exper immienttin.g.Thus, it is reasonable to conclude that the second step of the r a t e l i 11-(<0 I C Cl / H Cl Cl H Cl 0 a) 62 oH01 ° �' : I: H C o 1O ,O O -p (y' oo p O O <00-P=CI O QVO-P=O o Cl - 'Cl O C l (Cl Ell H H Cl Cl H (';( �H o-p=o :o-f=o CI CI LiZ \.finH �> continued on next page in
1 ,3-
1
If
1
�
\ /
0
__
00
00
�
I
0
+
0
�
I") 00
1
1
00 00
" 00 00
�
__
+
1
H
256
00
H
�:N
00
I 1-62
(a) co ntinued H
CI
G5H J� N) • •
E2
'0
• •
O( H
•
H
I C O-P=O C
H N
+1
+
0
1
+
1
l
H3� POC�I3 H3CK:;> NOTH3C �H h�h H H OH H
(b)
.....-
•
pyrid ine
Z aitsev
Therem ust be a stereo chemi cal requirem ent in this elimi nation. I f theS ay z t eff alkene is not p roduced because them ethy l gro up is trans ot thel eaving group, then the H and the leaving gro upm ust be trans and the el m i i nationm ust be anti- the characteri stic stereochem istryo f E2 elimi natio n. This evidence differentiates between the two po ssibilities inp ar t (a).
H -CH3 0 .. + uRCH3�H · U�HQCH3 1 Qtc5� } H2� � H O H+ O H2 f) 0 --0 r; : O H 9.) :OH O H H H
11-63
(a)
HO ..
1+
°
°
: 01
+ -H
t
H
H
--
---
• •
+
•
'---
•
� • •
H
• •
• •
.. \
---.-
..
•
• •
'---"" H
H 0 H2 H H..V 0 � ·9t H OCH3 � ('l H .. �-� H H+kO � 0 O
•
°
+
• •
'
H
°
It is equally likely for pro ot nation to occur first on the ringo xy gen, fol ol wed by ri ng opening, ht en replacemento f by water. D r. K anto rowski suggests this altern ative. H e and I will arm wrestle ot determi ne whi ch mec hanism is co rrec _
�
2
��+
'
O
� H
257
H0
°
H
8SZ
H� \ O� I)
H
(Il �qtJ
..
(q)
p;}nUllUO::l £9-11
CHAPTER 12-INFRARED SPECTROSCOPY AND MASS SPECTROMETRY
See p. 270 for some useful web sites with infrared and mass spectra.
12- 1 The tabl e is com pleted by r eco gniz ing that: (V) (A) v
(cm·l)
A Jl ( m)
=
1 0,000
4000
3300
3003
2198
1700
1640
1600
400
2 .50
3.03
3. 3 3
4.55
5.88
6.10
6.25
2 5 .0
12-2 I n general,o nly bo nds with di pol e mom ents will have an I R abso rptio n . H-C
C-H
t t
H-C
t
C-C�
yes
no y es H I
k ytes
H3C-C-H
�C-C
/
\
CH 3 I C=C \ I H H
t y es
t t
no y es
y es
H 3C
C-C�
\
I
C H3
H
t
C=C \ I H H
no y es (weak) 12- 3 (a) al kene: C=C at1 640 cm-I , =C-H at 3080 cm-l (b) al kane: no peaks indicating spo r sp2 car bo ns pr esent ( c) This IR sho wsmo re thano ne gro up. There is atermi nal alky ne sho wn by: C=C at 2 100 cmI - , =C-H at 3300 cm! - . These signals i ndicate an aro matic hy dro carbo n as well: =C- H at 3050 cm! - , C=C at16 00 cm-! . 12-4 (a) 2° ami ne, R- N H-R : o ne peak at 3300 cm-! indicates an N-H bo nd; this spectr um also sho ws a C=C at1 640 cm-! (b) carbo xy lic acid: the extremely bro ad abso rptio n in the 25 00-35 00 cml - range, w ith a "sho ul der" aro und 2 500-2 700 cm-! , and a C=O at1 7 10 cm! - , ar e com pel ling evidence for a carbo xy lic acid (c) al co hol: stro ng, bro ad O- H at 3330 cm-! 12 -5 ( a) co nj ugatedk eto ne: the small p eak at 3030 cm! - suggests =C-H, and the stro ng peak at 1 685 cm-! IS co nsistent with a keto ne co nj ugated wi th the alkene. The C=C is indicated by a very sm all peak aro und 1 600 cm! - . (b) ester: the C=O absor ptio n at 1 73 8 cm! - (higher than the keto ne's 1 7 1 0 cm-! ),i n co nj unctio n with the stro ng C-O at 12 00 cm-! , poi nts ot an ester (c) ami de: the two peaks at 3 160- 3360 cm-! ar el ikely ot be an NH2 gro up; the stro ngp eak at 1 640 cm-! is ot o stro ng fo r an al kene, so itm ust be a differ ent yt peo f C=X, in this case a C=O , so ol w because it is part o f an a mi de 12 -6 (a) The sm al l peak at1 642 cm-! indicates a C=C, co nsistent with the=C- H at 3080 cm-! . This appears ot be a simpl e al kene. (b) The stro ng absorp tio n at 1 69 1 cm-! i s unmi stakably a c=o. The smaller p eak at 1 62 6 cm-! indicates a C=C, pro bably co nj ugated with the C= O . The two peaks at 2 7 12 cm-! and at 2 8 14 cm-! r epresent H-C= O co nfirm ing that this is an al dehy de. 259
1 2- 6 conti nued (c) The str ong peak at 1 650 cm-' is C=C, probably conjugated as it is unusually str ong. The 1 703 em' peak appear s to be a conj ugated C=O, undeni ably a car boxy lic acid because of the str ong , br oad O-H absor pti on from2 400-3 400 cm-' . (d) The C=O absor ption at 1 742 cm' - coupled with C-O at 122 0 cm-' suggest an ester. The smallp eak at 1 604 cm-' , p eaks above3 000 cm-', and peaks in the 600- 800 cm-' regioni ndicate a benz ener ing. 12 -7 (a) TheM an dM +2 p eaks of equal inten sity i dentify the presen ce of bromi ne. The mass ofM ( 1 56 ) mi nus the weight of the lighter isotope of br omi ne (79) gi ves the mass of the rest of the molecule: 1 56 79 = 77. The C6HS (p hen yl) group weighs 77; this compound isb romob enz ene, C6HSBr . (b) The mJz 12 7 peak shows that iodine is presen t. The mol ecular ion mi nus iodine gives ther emainde r of ht e molec ul e: 15 6 -12 7 = 2 9 . The C2HS (ethyl) group weighs 2 9; this compound is iodoethane, C2HSI. c( ) The M an d M +2 peaks have relative intensities of about3 : 1 , a sure sign of chlor ine. T he mass of M mi nus the mass of the lighter i sotope of chl orin e gi ves the mass of ther emainder of the molecul e: 90 35 = 55. A fragmen t of mass 55 is not on e of the com mon al ky l groups (IS, 2 9, 43 , 57, etc.,i ncreasi ng in ni cremen ts of 1 4 mass units (CH2)), so the presence of an atoml ike oxy gen must be consider ed. I n addit oi n to the chlorin e atom, mass 5 5 coul d be C4H7 or C3H30 . P ossible molecular formulas ar e C4H7CI or -
-
C3H3ClO.
(d) The odd-mass molec ul ar ion in dicates the p resenc e of an oddn umber of nitrogen atoms (alway s beg ni by assumi ng onen itrogen). The rest of the mol ecule must be: 1 1 5 - 1 4 = 1 0 1; this is most likely C7H 17 The formula C7H17N is the correct formula of a molecul e withn o elements of un saturation. The seven carbon s probably incl ude alky l gr oups like ethyl or p ropy l or isopr opy .l 12- 8 Recall that radical s are not detected ni mass spectr ometry; only positively - char ged ions ar e detected. CH3 + I CH3-CH-CH2 +
12 -9
[
J. j r � �
CH
CH3- H
57
43
CH3
CH3 I CH3-CH
-
+
85
mJz
mass 2 9 r adicals not detected
mJz 57
CH3
CH2 - H
CH2-CH3
+
CH3 I CH2-CH-CH3
mJz 43
1 00
CH3 I CH3- <;H
-
mass 57 CH3 I CH2-CH-CH3
+
+
mass43
-
mJz
CH3 CH3 I I CH3-CH-CH2-CH mJz
260
85
+
+
57 CH3
mass
15
1 2- 1 0 The molecular weig ht of each isomer is 1 16g/ mole, so the molecular ion app ear s at rnJz 1 1 6. The left h al f of each str ucture is the same; loss of a three car bon radicalg ives a stabiliz ed cation , each with rnJz 73:
iH2C:O� .. H2C";�t iH2C: �'" H2C,,+�} .. .° .
•
..
..
rnlz73
·
°.
rnlz73
.
W here the two stru ctures differ is ni the alpha-cleavage on ht e right side of the oxy gen . A lpha-cleavage on the left structure loses two carbon s, whereas alpha- cleavage on the right structure loses on yl on e carbon.
[ �I o0-/-r CH2CH3 t alphacleavage
loss of
{�..O.. �H2 ... �O�.. CH2} C9H200,
[ �o*- r I alphat cleavag e J �..,�, R + 1
loss of
...
.
CH3 �+��,}_ R
rnJz 87 rnJz 1 0 1 1 2- 1 1 2,6-D imethy lheptan -4-01 , has molecular weight 144. The hig hest mass peak at 1 26 the molecular ion , but rather is the loss of water ( 1 8) from the molecular ion .
[�r
-
rnlz 1 44
H,o+
(CH3)
[�r
is
not
rnlz 1 26
from thef ragmen t of rnJz 1 26 . This is called ally lic Th e peak at rnJz III is loss of an other 1 5 cleavage; i t gen erates a 2°, ally lic, reson an ce-stabiliz ed carbocation.
[ � r-i�+- �}+ CH)
rnJz 1 26 rnJz III The peak at rnlz 87 results fromfrag men tation on on e side of the alcohol :
�:����!C
H}
[�r {JJH )ji + � rnlz 1 44
_
mlz 87 reson an ce-stabiliz ed
.
mass 57
1 2- 1 2 P lease ref er to solution 1 -20,p age 1 2 of thisS olution s M an ual . 1 2- 1 3 D ivide then umbers ni to 1 0,000 to arrive at the an swer. (b) 2959 cm1 -
(a) 1 603 cm-1 1 2- 1 4 (a)
H C==C CH2CH3 H t 'H ,
/
/
1 660 cm-1
or
(c) 1 709 cm1 -
(d) 1 739 cm-1
CH2 C H3 O==C H t '
(e) 22 1 2 cm-1
/
1 7 10 cm1 -
stronger absorption-larger
261
dipole
(f) 3 300 cm-1
1 2- 14 continued H
I
\H t 1660 cm-1
C=C
(b)
H
H
I
\H t 1 660 cm1 -
"N=C
I
\H t 1 640 cm1 -
I
stronger absorption-larger
H
CH2CH3
(c)
OCH2CH3
\
C=C
or
I
H
H
CH2CH3
\
CH2CH3
\
H
I
\H t 1 660 cm-1
C=C
or
/
d ipole
I
stronger absorption-
larg er dipole H
\
\H t 1 660 cm1 -
C=C
(d) H 3C
H
CH3
I
H
I
/
\H t 1 660 cm1 -
C=C
or
I
CH2CH3
\
stronger absorption
(no dipole moment)
-
1 2- 1 5 H3C
\
H3C
I
30003 100 cm1
and
t 1 660 cm1
I
\
CH3
O�
1 620 cm-l
� .
CH3
\
I
1 660 cm-1 moder ate i ntensity
weak or non- existent (b)
<'c=cl H t CH(CH3h H
CH3
C=C
(a)
larg er dipole
and
0 �
1 645 cm�1
conj ug ated
not conj ug ated
(c) both car bony ls show str ong absor ptions ar ound 1 7 10 cm-1 o II
+
o
CH3(CH2)3 - C H
(d)
0\
II
CH3(CH2h - C - CH3
and
2700-2800 c m-1 two small peaks
o y.
H
1200
3300 cm1 br oad, str ong
c m-1
and
a� 262
1 7 1O cm�1 str ong
12 - 1 5c ontinu ed
C 3 300 cm1
CH3(CH2h - C:: C - H
t
(e )
CH3(CH2)6 -C:: N
and
t
22 00-2 300c m1 moder ate tostr ong int ens ti y
2 100-22 00c m-1 weak to moder at e intens ti y
( f) bot hc ar bonyls s hows rt ong abs or pt ions ar ou nd 1 7 1 0c m-1 � 3300c m-1 fH br oad,str ong 0 ' 0 o 1\ 1\ I and CH3CH2CH2 - C - OH CH3 - CH - CH2 - C - H
t
2 � 00 cm1 ver y br oad
2 700-2 800 cmi t wos mall peaks 1 7 1O cm-i � 0
0/ 1 650 cm-1 II
CH3CH2CH2 -C -
(g)
�-� o/ 3300 cm1
II
CH3CH2 - C -CH2CH3
and
two peaks
o�
12 - 1 6 (a)
II
H
\
(b)
C-OH
t
/
.......-
C=C 2 400-3400c \ H CH3 /
o�
1 700 cm1 -
II
1 7 1 5c m-I
CH3-CH-C-CH3
I
m-1
CH3
1 640c m-1
tV 3000- 3 1 00 cm-I
(c )
�
V,
CH2 -C
"-
12 - 1 7 (a)
[
CH3
�71
CH3 � H 43
j
mlz 86
t
(d)
N
1 600 cmI -
CH2CH2CH3
a/" I
22 o cm1
./
H
3000-3 1 00 cmi 3400 cmi H
N ..... V
\ CH2CH3 , ,
"'=::
�,
T 2 900 -3000 cmI
1 600c m-I
1
+
CH3
•
•
I
+ CH
- CH2CH2CH3
mlz
71
CH3 •
I
CH3-CH +
263
mlz 43
[
b�� �
1 2- 1 7c ontinued
(b) CH3 -CH=
CH2CH3
9
(c )
cH
H
CH3
b
CH,- H-CH1
45
rnJz 1 02
[
CH3
�
CH3 -CH==CH - H -CH3
rnJz 84
j
�i
----l..
]
rnJz 98- allylic c leavage
[ Jb��j-
r
CH3_CH=
1
t alpha-
c leavage
..
----I�
+
·· i�
H'
rearr ange ..
F c�
•
H'
rnJz 43
G
•
rnJz 1 29
+ OH CH3 II I ___ CH - CH2 - CH - CH3
O-
rnJz 85
� =c �
H3
264
rnJz 45 H C
o
T he tropylium ion si a ch ar ac teris tic fragment fr om phen ylalkanes.
m1z 9 1
CH
1O-
CH,
CH3- H
�H3
H3
rnJz 69
b }
:OH
.
��
rnJz 1 44 alpha-c leavage
�
CH3_ H_
alpha c leavage
/CH3 HC+
rnJz 77
-
:OH CH3 I I + CH - CH2 - CH -CH3
benzylc ation
rnJz 1 3 4
H3
rnJz 87
(d) [&f-. () �
•
b
H3
}
C
y
r earr an ge ..
H3C, + "CH3 C I CH3
rnJz 57
t
1 2- 1 8 (a )
�CH2+ + H2C' CH3 [�r �CH2+ H2C......./ �CH2+ Hi � H o + CH3 (5 cJ •
-----
mass 29
mJz 85
rnJz114
+
mass 43
mJz 7 1
+
+
(b)
:
mass 57
mJz 5 7
.
-----
mass1 5
rnIz 83
mJz 98
(d)
CH3 CH3-C==CH-CH2+ I
c( )
+ oCH3
--
+j i :OH +OH } OH [ �Hj CH, -CH,CH,CH, �H' gH, +. CH,CH,CH,CH, .
•
31
.
{ CH2=CH- �H2
CH2=CH -CH2-CH2 + CH3 +
0
55
�
----
mass 57
mJz 31
CH,=CH- c::t C� CH, mJz 70
.
-
rnIz 88
-
[
mass 15
.
-
! H20
m1z 69
�
mJz 4 1
H2-CH = CH2 } CH2CH3
-----
mJz 1 2 1 c ontinued o n nex t page
mJz 5 5
+
0
mass 29
Oc+ 1.0
rnJz 77
+
0
NHCH2CH3 mass 44
265
mass 1 5
12- 1 8 (e) c ontinu ed
[a�::rr mlz 121
12- 1 9 (a) Thec har act er si t ci fr equ enc ies of t he OH abs or pt ion andt he C=C abs or ption will indic ate the pr es enc e or abs enc e of t he gr ou ps. A s pectru m with an abs or pt ion ar ou nd 3300c m-i will haves omec yc lohexanpl i n it; i ft hat s ames pec tru m als o has a peak at 1 645c m-i ,t hent hes ample will als oc ont ains omec yc ol he xene. Pur es amples will have peaks r epr es ent at ive of only one of thec ompou nds and not the other . N ote that quantitation of thet woc ompou nds wou ld be ver y diff ci u lt by I R bec aus e thes rt ength of abs orpt ions ar e ver y diff er ent . Us ually, ot her met hods ar eus ed in pr ef er enc et o I Rf or qu antitat ive measur ements .
0\
o y'H
O
3 300c m- i hr oad,str ong
--
1 645c m-1 moder ate
12 00c m-1 (b) M ass s pec trometr yc an be mis leading wit h alc ohols . Usu ally, alc ohols dehydr ate in the inlet s ystem of a mass s pec rt omet er , andt hec har ac ter si t ci peaks obs er ved in the mass s pec tru m ar e thos e of t he alken e, not of t he par ent alc ohol. F or t his par ticu lar analys si , mass s pectr ometr y wou ld beu nr eliable and per haps mis leading. 12-2 0 (a) The "stu dent pr ep"c ompou nd mus t be I- br omobu tane. The mos t obvious featur e of the mass spec rt um si t he pair of peaks atM andM +2 of appr oximat ely equ al heights ,c har act er si tic of a br omine atom. L oss of br omi ne (79)fr om the molecu lar ion at 1 36 gi ves a mass of 57, C4H9, a butyl gr oup. W hic h of t hef our possi ble bu tyl gr ou ps? The peaks at 1 07 (loss of 29, C2Hs) and 93 (loss of 43 , C3H7) ar ec ons si tent w ti h a linear c hai n, not a br anc hedc hain. (b) The bas e peak at 57 si s os rt ong bec aus e thec ar bon-halogen bond si the weakes t in the molecu le. Typic ally, loss of halogen si the dominant fr agmentat ion in alkyl halides .
[4 :r �{�
/' Br
r
rnJz 1 36
+
+ �
rnJz 1 07
mass 29 +
H2C.
Br rnJz 93 +
mass 43
/'-../
+
rnJz 57
H2C-..../
+
Br mass 79 •
266
12 -2 1 (a) D euter ui mh as twic e ht e mas s ofh ydr ogen,bu ts imilar s pr ingc ons tant,k . Compar e th e fr equenc y of C-D vibr ation to C-H vibr ationb ys ettingu p ar atio,ch an ging only ht e mass (su bs titu te 2 m for m). =
=
�lt2
=
0.707
0.7 07 vH = 0.7 07 (3000 em-I) 2100 cm-1 (b) The func tional gr oup mos t lik ely to bec onfus ed with a C- D str etc h si the alk yne c( ar bon- c ar bon tn ple bond), whic h appears in ht es amer egion and si of ten ver y weak. 12-22
(
Vo
=
'"
�
m/z 1 14
[+r r*r
r-
�
+
�
m/z 57
1+
a 1°c ar boc at ion- not favor able
a 2°c ar boc at ion-r eas onable
m/z 57
m/z 1 14
�
�.
a 3°c ar boc ation-t he best
m/z 57
m/z 114 The most likely rf agment ation of 2 ,2, 3,3- tetr amet hylbut ane will give a 3°c ar boc at ion,th e mos tst able of the c ommon alkylc at ions. The molecu l ar ion s hould bes mall or non- exis tent whi le m/z 57 si lik elyt o be the bas e peak , wher eas t he molecu lar ion peaks will be mor e pr omi nent for n -oc tane and for 3 ,4- di methylh exane . 12 -2 3 (a) Th e infor mat iont hat this mys teryc ompound si ah ydr oc ar bon mak es inter pr et ni g the mass s pectru m much eas ier . ( It si r elat ivelys i mple to tel l i f ac ompound has c hlori ne, br omine, or nitr ogen by a mass s pec rt u m, but oxygen si di ffic ult to determ ine by mass s pec rt ometr y alone. ) A h ydr oc ar bon with molec ular ion of 1 1 0c an have only 8c ar bons ( 8 x 12 = 96) and 1 4 hydr ogens . The for mula CgHl4 has t wo elements of uns atur at ion. (b) The IR will be us eful in detenni ning whatt he elements ofu ns atur ation ar e. Cyc loalk an es ar e gener ally not dist inguish able inth e IR. An alk en esh ouldh ave an abs or ption ar ound 1 600- 1 650 em-I; none si pr es ent in ht si I R. A n alkynes houldh ave as mall,shar p peak ar ound 22 00 em-I-PRESENT A T 2 12 0 em-I! A s l o, a sh ar p peak ar ound 3 300c m-1 indic ates ah ydr ogen on an alk yn e,s o ht e alk yne si at on e end oft he molec ul e. B ot h elements of uns at ur at ion ar e acc ounted for byt he alk yne.
267
12-2 3 (c ) T he onl y question is how ar e theo ther carb ons arr anged. T hem ass spec rt um sho ws a pr ogr ess oi n of peak s fr om themo lec ular ion at 1 10 to 95 (lo ss of CH3), to 8 1 l(os so f C2Hs), ot 67 (loss o f C3H7). The mass spec rt um suggestsi ti s a il near chain. T he extr a ev idence that hydro genation of the mystery com po und giv es n-o ctanev eri fies that the chain is linear . T he origi nal c ompoundm ustb eo ct- 1- yne. ( d) T he base peak is so str ongb ecause the ion pr oduc ed iss tabil z i edb yr esonanc e. �
m/z 95
�
m/z 8 1
�_
m/z 67
+
m/z l lO
+
+
+
'-.....
m/z 39
c:::=:::>
12 -2 4 (a) and (b) T hem as s spec is consistent with the for mul a of the alkyne, CSHJ4' mass 1 1 0. T he IR isno t c onsistent with the al kyne, howev er . O ften, symmetric all y substituted alk ynes hav e a mi nisc ule C=C peak, so the fac t that the I R does not sho w this peak does not prov e that the alkyne si ab sent. T he m i pOlt ant I ev idenc e in the I R si the signific ant peak at 1 62 0c m- ; this abs orp tion isc har ac teri stic of a conjugat ed diene. I nstead of the al kyneb eing form ed, the reac tion must hav eb een a doubl e el m i ination to the dic nc .
�*+ Br
12 -2 5 (a)
c( )
b( )
II � Q -
,
\
OT
0 g
oII '/
\\
-
"
�
'\
H
I
k
-CH3
I
12 00 cm-J
1 5 80cm-J
268
11--
t
C-C-H
1 600c mI -
17 60cm-I
O , ...,... 1 72 0 cm-1
27 1 0, 2 82 0c m-1
o
12 2- 6 (a)
� Br
Mg ether
..
/""-.... /""-.... ,,/ ........",
- M gB r
A
H30+.. �OH C7HI60 2m - eth yl hexan-2-01 m ol. wt. 116
b( ) T he mol ec ul ar i oni s not vi sib le in the spec rt um. Alc ohols typic ally dehydr atei n the hot inlet system of the mass spec rt om eter , especi all y true for 30 alc ohol s that ar e the easi est type to dehydr ate. T he two fragmentati ons that produc e a resonanc e-stabiliz edc arb oc ation give the m ajor peak s in the spec rt um at mlz 59 and 1 0 1 .
[ � OH l t { ___
�
mass 1 1 6
;<�-H
_�
.. _
H3C CH3
rnIz 59
..
/
�
rnIz 1 01
O-H c(+CH3 . }
12 -2 7 T he unk nownc om pound has peak s in the mass spec at rnIz 1 98, 1 55 , 127 ,7 1 , and 43. I ti s helpf ul that them asses of two of the fr agm ents, 1 5 5 and 43, sum to 198, as do the other two fragm ent masses, 127 and 7 1 . W ec an say wi thc ertai nty that the mol ec ul ari oni s at rnIz 1 98, and that the unk nowni s a rel ativel y sim pl em ol ec ul e wit h twom ai n fragm entati ons. T hi si s a fairl y hi ghm ass for a sim pl ec ompound; som e heavy group mustb e pr esent. W hat is N O T pr esenti s N b ecause of the even mol ec ul ar ionm ass, nor nor B rb ec ause of thel ack ofi sotope peak s, nor a phenyl groupb ec ause of no peak at77 . T he progression of alk yl gr oupm asses: 1 5 ,2 9,43 , 57 ,71 , 85, 99- incl udes two of the peak s, soi t appear s that the unk nownc ontai ns a pr opyl gr oup and a pentyl group (the pr opyl c oul db e part of the pentyl gr oup). T he 12 7 fragm enti sk ey; the fragm ent has thi s m ass,b ut we would expec tm uc hm or e fr agm entati on fr om a ninec arb on pi ec e. T here mustb e som e o ther expl anati on for thi s 127 pe ak.
CI
C9HI9
A nd ther e is! T her ei s one pi ec e-m ore speci ifc al ly, one atom- that hasm ass 127: i odine! I n all prob abil ity, the iodi ne atom i s attac hed to a fr agment of mass7 1 whic hi s ' a pentyl group. W e c annot tell withc ertai nty whati som er i t is, so unl ess ther ei s som e other evidenc e,l et' s pr opose a str ai ght c hai ni somer, 1 i- odopentane.
CsHl1
127
155
I
T he 155 rf agment probabl y has thisb ir dged struc ture b ec ause iodi nei s sobi g: 269
12-2 8 (a)
CH3
CH3 CH3
I
I
I
I
I
HC-C-C-CH3 3 OH OH
s rt ong peak 17 1 5 cm 1
T
broad,s rt ong OH peak around 3300 cm-I
o
HC-C-C-CH3 3 II I / 0 CH3 around
-
OH
str ong peak ar ound 1 690 cm-I
(b)
br oad, strong 011 pe ak ar ound 3 300 em-'
two peak s at 27 00 and 2 800 cm-I
o
(c)
o
0/
t
H
br oad COOH band mi ss ing
ver y br oad 2 500- 3 500 cm-1 with a "shoulder " ar ound 27 00 cm-I
still has a str ong O H at 3300 cm-I butd iffer ent fr om
COOH
I f you wish to find IR spectr a and mass spectr a of comm on compounds, ther e ar e two w eb sites that ar e ver y helpfuL E nter ing a name or molecular form ula will giv e isom er s fr om which t o choose the desir eds rt uctur e and the IR or MS if av ai lable in their d atabase. http:// webbook .nisLg ov/ "NIS T " is the U.S . N ational I ns titute ofS tandards and T echnology. http://www.aist.go.jpIRIODB/SDBS/menu-e.html
T his si fr om the N ational I nstitute of A dvanced I ndustr ialS cience and T echnology of J apan.
270
CHAPTER 13-NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY
A b enzener ing canb e wri tten with thr ee altern ating doub leb ond s or with a cir cle in ther ing. A l l of the carb ons and hydr ogens in an un sub stitutedb enzener ing ar e equivalent,r egar dless of which sy mb olis m is used.
o
equi val entto
The Japanese web site listed at the bottom of p. 270 also gives proton and carbon NMR spectra. Reminder: The word "spectrum" is singular; the word "spectra" is plural.
1 3- 1 ( a)
650 Hz 300 x 1 06 Hz
=
2. 17 x 1 0-6
(b) D iffer ence in magnetic field
=
=
2. 17 ppm dow nfield rf om TMS
7 0,459 gauss x (2 . 17 x 1 0-6)
0. 1 5 3 gauss
=
( c) T he chemical shift does not change w ith field str ength: «52 . 17 atb oth 60 MHz and 300 MHz. ( d) 2( . 17 ppm) x ( 60 MHz) = 2( . 17 x lO-6 ) x ( 60 x 1 06 Hz) = 130Hz
*
1 3-2 N umb er s are chemi cal shi ft values, in ppm, derived fr om Table 1 3 -3 and the A ppendix in the text.
Your predictions should be in the given range, or within 0.5 ppm o/the given value.
( a)
b
( b)
.... CH3 � b
CH3 ' c CH3 \ / b C= C b \ / CH3 H C.... a CH3 \
......
b
�
( c)
�
c p
a c
H
H
c
a b c
= ""
=
= =
«5 5-6 «5 0.9
CH3
�
I
h-
b
�
a b
OC J
b
a b
CH3
= =
«57 2. «52 .3
b
a
( d)
H
b
CH3
H
b
= =
/)7 2. /) 3 .6
� H3 c
b
CH3-C-C::C-H
c
em
a b c
= = =
«52- 5 /)2 . 5 /) 1-2
a
a c
H
b
CH3
C�
H
� --uH
c
�
* H
( e)
......
a b
a
H
�
( f) a
CH2-C-O-H
b
b
�H3
CH3 -C-CH2 Br I
Br
c /) 10- 12 «5 3 ( between tw o deshielding gr oups) /)7 2 . ( Th eh ydr ogens l ab el ed "c" ar e not equival ent. Th ey appear atr oughly the sam e chemi cal shift b ecause the sub stituent is neither stro ngly electr ond onating nor withdr awing.) 271
a
a b
= =
«5 3- 4 «5 1- 2
1 3 -3
(a)
c
b
a
(b )
CH3CH2CH2Cl
1 3 -4
�
(a)
*:
H
3
h
c
a
(c)
CH3CHCH3 I
�H3 b
a
(d)
c
b
H
CH3-C-CH2CH3
I
two types of H
* H
�
C 3
h
c
a three types of H
a
�
H
CH3
t
H d
fiv e types
Br
of H
H
c
H
a b
Cl
three types of H a
b
(b) T he three types of aromatic hydrogens appear in a relativ elys mall s pace around () 7.2 . The signal si complexb ecause all the peaks from the three types of hydrogens ov erlap.
d four types of H
N ote: N MR spectra drawn in thisS olutions M anual w ill repres ent peaks as single lines. T hese lines may not look like "real" peaks,b ut this av oids the problem ofv ar iation am ong spectrometers and printers. I ndiv idual spectra may look different from the ones presented here,b ut all of the i mpor tant information willb e contained in thes e representational spectra. 1 3 -5
c
CH3
c
I I
a
a
II
a II
�
9H
b
H3C - C - O - C - CH 2 - C - CH3
12 3H
CH3
c
�
�
2H -.J I
I
10
I
I
8
9
7
I
6
I
I
I
4
5
8 (ppm)
/
I
3
2
TMS
I
I I
o
1 3 - 6 The three spectra are identified with their str uctures. D ata are giv en as chem ical shiftv alues, with the integration ratios of each peak giv en in parentheses. S pectrum (a) S pectrum (b ) b a a OH H H a = cS 2. 4 ( 1 ) ( 1 H) C a I CH3 - C - C::C - H b = cS 2. 6 ( 1 ) ( 1 H) a = cS 6.8 (2 ) (4H) I c = I) 1 . 5 (6) (6H) b = cS 3.7 (3 ) (6H) c ,o OC J CH3 c
�
b
CH3
a
CH3-� - CH2Br I
Br
H
a
S pectru m (c) b
*
a = b =
I) 3 . 9 cS 9
H
�
a
( 1 ) (2 H ) (Thisi s the compound in Problem 1 3-2 ( f). You 1. (3) (6H) may wish to check your answer to that questio n agains t the spectrum. ) 272
1 3- 7 Chem ical shift values ar e appro ximate and mayv ar y sli ghtly fro m yo ur s . T he splitti ng and integr atio n values sho uldm atch exactly, ho w ever. (a)
b
b
b
b
Q
H3C , /CH3 a HC - O - CH a / , CH3 H3C
12 H
.1!
(To show the pattern of peaks, this multiplet is larger th an it would appear on a real spectrum . )
I
10
(b)
I
9
I
8
I
I
I
6
7
2H TMS
" III, I
T
3
4
5
(5 (ppm)
I
9
I
8
�
6
5H Note : the three types of aromatic protons above
are accidentally
(5 (ppm)
*� a H
-
� /;
H a
equivalent because the
I
4
5
a H
a H
III
I
I
I
7
.1!
(c)
isopropyl group has little
C
2H
I
3
LI
I
9
I
8
I
7
I
6
I
5 (5 (ppm) 273
I
I
o
�
Q IH
CH3 c
4
I
1
I
6H
"
I
I
c CH]
(To show the pattern o f peaks, this m ultiplet is larger than it would appear on a real spectrum . )
shifts.
TMS
2
/ b
H a
effect on their chemical
10
1
I
T o
3H
.1!
I
I
Q
b a c 0 II ClCH2CH2 - C - O CH3
2H
10
T 2
1
3
,
III, I
2
TMS
I
1 I
o
1 3-7 co ntinued
* aH
(d)
c b l!
CH3CH20
4H
9
I
I
5
6
7
8
H
b
a
0 II
b
�
6H
a
8 ( ppm)
I II I
4
TMS
3
0
b
a
II
1 I
I
I
I
o
2
12 4H
( e) c
e
OCH2CH3
12 4H
I
I
T
I
a
� Ii
aH
All four aromatic protons are equivalent .
10
H
c
�
6H
CH3CH20 - C - CH2CH2 - C - OCH2CH3
l!
N o splitting is ob serv ed when chemically equivalent hydr ogens ar e adjacent to each other , as with hydr ogens "b " ab ov e.
4H TMS
I
I
I
9
10
1 3-8 ( aJ
):
8
I
T
T
7
6
5
0
5 2.4 H2
?
8 1 . 8 H2C \
H3C '( 8 1 .2
H 8 5.8
H 8 6 .7 CH3 J
T he signals at 1 . 8 and 2.4 ar e rt ipletsb ecause each seto f pro tons has two neighbo ring hydro gens. T he N + 1 r ule corr ectly pr edicts each ot b e a triplet.
8 ( ppm)
I II
I
I
3
4
I
2
I
I
T
o
b( ) o 8 1 . 5 H2C I 8 1 .65 H2C ,
C H2
8 2. 1
H 8 6. 2 CH3 8 1 .7 allylic
allylic
T he thr ee CH2 gro ups canb e distinguishedb y chemical shift andb y splitting. T he a l l y l i c CH2 w il l be thefarthest downfield o f the thr ee, at 2. 1 . T he signal at 1 . 5 is a rt iplet, so that mustb e the one witho nly two neighbor ing pr oto ns. T he CH2 sho w ing the multiplet at 1 .65 mustb e theo neb etween theo ther two CH2 gr oups. 274
1 3 -9
aH
(a) i!
5H
T he thr ee typeso f aro matic pro ot ns ar e accidentally
9
1
J ", 15 Hz
equivalent.
10
IH
IH
7
a
1 5 Hz
5
6
8 (ppm)
(b)
9H
C=C / \ C(CH3h H b
,......I-..
,......I-..
8
=
�*� } H
�
12
Q
Ha
4
d
3
3H
c
CH30
CH3 / C=C \ / H Cl
I
2
12
b
TMS
0
�
3H
\
i!
IH
a
TMS
I
10
c( )
I
9
e
I
I
8
7
o
(CH3hC
I II
6
I
5
8 (ppm)
9
4
I
3
I
I
I
2
I
o
�
9H Q
3H
�
b
8
I
d
- OCH2CH 3 \ F C =C \ / H H II
a
10
c
T
2H
IH
IH
7
6
TMS
5
8 (ppm )
275
4
3
2
1
o
(x
1 3-9 continued
c H
I
11
The chemical shift of -COOH is variable. It is usually a broad peak.
I
I
9
10
x>=
Q
d
b H
The electron-withdrawing
�
Q 2H
c arbonyl group deshields the
2H
I
O
protons adjacent to it, moving them downfieid from their usual position at
8 7 .2.
I
I
I
I
7
8
6
.&
�C
c H b H
"' OH a
I
I
3
4
5 8 (ppm)
'-'::
3H
TMS
I
I
2
1 I
o
The formula C3H2NCI has three elements of unsaturation. The IR peak at 1 650 cm- I indicates an alkene, while the absorption at 2200 cm- I must be from a ni tri le (not enough carbons left for an alkyne). These two groups account for the three elements of unsaturation. So far, we have:
1 3- 1 0
\
C I C=C
\
/
N
+
2H
+
Cl
The NMR gi ves the coupling constant for the two protons as 1 4 Hz. This large J value shows the two protons as trans (cis, J = 10 Hz; geminal , J = 2 Hz). The structure must be the one in the box .
H C=N 1 \ C=C \ I H Cl
1 3- 1 1 (a)
C3H7CI-no elements of unsaturation; a b c Cl - CH2CH2CH3
a c
= =
<5 3 . 8 <5 1 .3
3 types of protons in the ratio of 2 : 2 : 3 .
(triplet, 2H) ; b (triplet, 3H)
=
<5 2. 1
(multiplet, 2H);
The other NMR signals are two 3H singlets, (b) C9H 1 002-5 elements of unsaturation; four two CH3 groups . One at <5 3.9 must be a CH30 protons in the aromatic region of the NMR i ndicate a disubstituted benzene ; the pair of c=:====�> group. The other at <5 2.4 is most likely a CH3 group on the benzene ring. doublets with J = 8 Hz indicate the substituents are on opposite sides of the ring + OCH3 + C + 0 + (para) CH3 1 element of unsaturation + 3C + 6H + 20 + � 1 e lement of un saturation
< 1>
-Q-
One way to assemble these pieces consistent with the NMR i s : a b H H 0 a = <5 7.9 (doublet, 2H) b = <5 7.2 (doublet, 2H) C 3 - OC 3 c = <5 3 .9 (singlet, 3H) d = <5 2.4 (singlet, 3H) H H 276 a b
�
*g
�
/
(Another plausible structure is to have the methoxy group directly on the ring and to put the carbonyl between the ring and the methyl. Thi s does not fi t the chemical shi ft " a l ues quite as well as the above structure, a s the methyl would appear around <5 2. 1 or 2 . 2 instead o f 2.4.)
�
. . '
J bc =
l .4 Hz
13-13
(a)
c
f
e
d
c=c
\ /
CH3CH2CH2
.
.
1
... .. ..
I
,
1
, .
. .
I � J bc ,
=
1 .4 Hz
JJ----_ --JU o
H
,
� ,
.. .. ..
5. 1
II
a b c d e f
a
C - OH
/
\
H
b
= 8 1 2. 1 (broad singlet, I H) = 8 5 . 8 (doublet, I H) = 8 7 . 1 (multiplet, I H) = 8 2 .2 (quartet, 2H) = 8 1 . 5 (sextet, 2H) = 8 0.9 (triplet, 3H)
(b) The vinyl proton at 8 7 . 1 is He ; it is coupled with Hb and Hd, with two different coupling constants, and Jed ' respectively. The value of Jbe can be measured most precisely from the signal for Hb a t 8 5 . 8 ; the two peaks are separated by about 1 5 Hz, corresponding to 0.05 ppm in a 300 MHz spectrum. The value of Jed appears to be about the standard value 8 Hz, judging from the signal at 8 7 . 1 . The splitting tree would thus appear:
J be
8 7. 1
1
J be = I 5 Hz " ..
. ... , . .. ... ,
J ed = 8 Hz
J ed = 8 Hz
�
�
II
' ,, '
Jed = 8 Hz
over� n NMR
U
277
,
Jed = 8 Hz
�
,
'I
L
13-14 (a)
* -
o
c
d
H
d
H
H
\
C
�
Ii
H
a
II
a
C- H
b
I
C , H
�
c d
= = = =
8 9.7 (doublet, I H) 8 6.7 (doublet of doublets, I H) 8 7.5 (doublet, I H) 8 7.4 (multiple peaks, 5H)
The doublet for He at 8 7.4 OVERLA PS the 5 H peaks of Hd.
b
H
d d (b) J ab can be determined most accurately from Ha at 8 9.7: Jab "" 8 Hz, about the same as "normal " alkyl coupling. be measured from Hb at 8 6.7, as the distance between either the first and third peaks or the second and fourth peaks (see diagram below): J bc "" 18 Hz, about double the "normal" alkyl coupling.
J be can
0 6.7
... .. .. . . . . . . . . . . 1 . . . . .. . . . . . . . J bc 1 8 Hz 1 1
(c )
=
f"��:� �'� '1
r;.:� �'� '1
1 3- 1 5
J
b
H
a
0
II
d
C - CH3
I \ c=c I ,
(a) a b c d
= ""
= =
0 1 .7 0 6.8 0 5 -6 0 2. 1
�
(b) a b c d
= = = =
H
H3C
(c) using J bc
�
L
doublet mu l t ip l e t (two overlapping quartets-see part (c» doublet singlet
c
=
1 5 Hz and J ab = 7 Hz:
.
6.8
J .1:· ;: · · . · · · . . ; : . 1 r . . . . . . . 'I;� ·;: :::·1- · · · · · · · · · · · '1';:'-;' 1-· · · · · · .
(j:�'= 7 Hz �
o
-;.
. · ·
=
�.
�..
(
• • •
�• • • • • ,
'�� : :
� L J � 278
••••
• •,
L
13-16 replace He
�
cis diastereomer
replace Hd
�
trans diastereomer 13-17 (a)
replace Ha
replace Hb
.. non-superimposable mirror i mages = nantiomers; therefore, Ha and Hb arc enantiotopic and are not distingui shable by NMR
�
(b) replace He on the left
�
enantiomers
replace He on the right ..
(c) The 1 3- 1 8 (a)
Hd protons are also enantiotopic.
Xy
Br a
CH3
b
H "
�
(b)
e
e
H
e
H
CH3
-::.,
H H c
d
H b This compound has six types of protons He and Bct are diastereotopi c , as are � and Hr. a = 8 2-5 ; b = 8 3 . 9 ; c , d = 8 1 .6 ; e,f = 8 1 .3
This compound has fi ve types of protons. � and Hd are diastereotopi c . a = � 1 . 5; b = � 3.6; c,d = � 1 . 7; e 8 1.0 =
279
1 3- 1 8 continued d
(c )
H
b
Br
H
a
H
(d)
\
Cl
/
C=C
/
H
c
\
H
c b This compound has three types of protons. Ha and Hb are diastereotopic. a,b = & 5-6 ; c = & 7-8
H H
b This compound has six types of protons. He and Hf are diastereotopic. a,b,c = & 7.2; d & 5.0; e,f = & 3 .6 ==
13-19
.. �
(a)
+ H-B
CH3CH2 - O-H •
(b)
.
•
+:B
') "
CH 3CH2- O-H .
""
• •
-
H
I') ". -
--- CH3CH2 - �-H + .B +
.. �
""
+ H-B
--- CH3CH2-R:
H 1
--- CH3CH2 - �: H
--- CH3CH2- O: . 1
.
+
H-8
+ :B
1 3 -20 The protons from the OH i n ethanol exchange with the deuteriums in D20. Thus, the OH in ethanol is replaced with OD which does not absorb in the NMR. What happens to the H? It becomes H OD which can usually be seen as a broad singlet around & 5 . 25. (If the sol vent is CDCI3, the i mmiscible H OD will float on top of the solvent, out of the spectrometer beam , and its signal will be missi ng.)
R OH CH3CH2 OH
+
D 20 - R OD
+
+
H OD
D20
3H 2H
H OD Peaks from OH are usually broad because of rapid proton transfer. This is especial l y true in water, or HOD. I
10
I
9
I
J
8
7
I
6
TMS
I , I
5 & (ppm)
280
I
4
II
I
3
I
2
I
I
o I
1 3-2 1 (a) The fonnula C4HlO02 h as no e lements of unsaturation , so the oxygens must be alcohol or ether functional groups. The doublet at 8 1 . 2 represents 3H and must be a CH3 next to a CH. The peaks centered at 8 1 .65 integrating to 2H appear to be an uneven quartet and signify a CH2 between two sets of non-equi valent protons; apparently the coupling constants between the non-equi valent protons are not equal , leading to a complicated p attern of overlapping peaks. The remaining five hydrogens appear in four groups of IH, IH, IH, and 2H, between 8 3.7 and 4.2. The 2H multiplet at 8 3.75 i s a CH2 next to 0, with complex spli tting (doublet of doublets) due to diastereotopic neighbors . The IH multiplet at 84.0 is a CH between many neighbors . The two IH signals at 8 4. 1 -4.2 appear to be OH peaks in different envi ronments, partially split by adj acent CH groups .
( r
H
H3-
I
+
-
-CH2-
Of the two possible structures: H
-OH
+
y
I
�
+
-CH2
�
12 .
I
1.65
375
CH3-C-CH2CH20H I
I
OH-- 4.1-42
CH20H
)
H
H
V :: ::
'/
--r===========;:: o
OR
CH3 -C-CH2 OH
-OH
+
HO
H3C
H
::
. %
��
=
OH
The 3-di en onal view shows clearly that the two Hs on a CH2 are not equi valent
H
CORRECT!
this one has no chiral centers and would not display such complex splitting
If the possibility of intramolecular H-bonding is considered, the diffe rences in the environments for each H become even more obvious.
� H
(b) The fonnula C2H7NO has no elements of unsaturation . The N must be an amine and the ° must be an alcohol or ether. Two triplets, each with 2H, are certain to be -CH2CH2- . Since there are no carbons left, the Nand 0, with enough h ydrogens to fi l l their valences, must go on the ends of thi s chain. The rapidly exchanging OH and NH2 protons appear as a broad, 3H hump at 82. 8 . H
HOCH2CH2NH2
8 3.7
1 3-22
�
'-- 82.9 � 3H
!! 3H 12 2H
I
TMS
I
I
I
9
10
8
° II
CH3 -C - OCH2CH3
8 2.05
8 4. 1
I
7
I
6
5 (ppm) I
8
I
4
3 I
I I
2 I
I
1
I
0
The key is what protons are adjacent to the oxygen . The CH30 in methyl propanoate, above, absorbs at 8 3.9, whereas the CH2 next to the carbonyl absorbs at 8 2.2. Thi s i s in contrast to ethyl acetate, at the left. 281
1 3 -23 Ha
=
Hb
=
He
=
Hd
=
d
[) 2.4 (singlet, IH) [) 3.4 (doublet, 2H) [) 1 .8 (multiplet, 1 H) [) 0.9 (doublet, 6H)
I I
CH3 H2 H-O-C -C-H
b
Q
c
1 3 -24 (a) The formula C4H802 has one e le ment of un saturation . The IH singlet at [) 1 2. 1 indicates carboxylic acid. The IH multiplet and the 6H doublet scream isopropyl group. CH3• CH
i
II °
CH-C-OH
(b) The formula C9HlOO has five e lements of unsaturation . The 5 H pattern between [) 7.2-7 .4 i ndicates monosubstituted benzene. The peak at [) 9.85 is unmi stakably an aldehyde, trying to be a triplet because it i s weakly coupled t o a n adj acent CH2. T h e t w o tri plets a t [) 2 . 7 - 3 . 0 are adj acent CH2 groups .
V I
h-
H2 ° c , c H2
Jl H
(c) The formula CSH802 has two elements of unsaturation. A 3 H singlet at [) 2.3 is probabl y a CH3 next to carbonyl. The 2H quartet and the 3H triplet are certain to be ethyl; with the CH2 at [) 2 . 7 , this also appears to be next to carbonyl. II II ° °
CH3CH2 -C-C-CH3
(d) The formula C4H80 has one element of unsaturation, and the signals from [) 5 .0-6.0 i ndicate a vinyl pattern (CH2=CH-). The complex quartet for IH at [) 4.3 is a CH bonded to an alcohol, next to CH3. The OH appears as a IH singlet at [) 2 . 5 , and the CH3 next to CH is a doublet at [) 1 . 3. Put together: OH I
CH2=CH-CH-CH3
but-3 -en-2-ol
(e) The formula C7H160 i s saturated; the oxygen must be an alcohol or an ether, and the broad IH peak at [) 1 . 2 is probably an OH. Let's analyze the spectrum from left to right. The expansion of the IH multiplet at () 1 .7 shows seven peaks-an i sopropy l group! Six of the nine H in the pattern at.5 0.9 must be the doublet from the two methyls from the i sopropyl. The 2H quartet at.5 1 . 5 must be part of an ethy l pattern, from which the methyl trip let must be the other 3H of the pattern at.5 0.9. This leaves only the 3 H singlet at <5 1.1 which must be a CH3 with no neighboring Hs. -OH
+ -CH2CH3
�-
H
+
I
CH3-
+ -CH3 +
1 C
CH3
These pieces can be assembled in only one way :
CH3 I
HO-C-CH(CH3}z I
CH2CH3
282
2,3-dimethylpentan-3-ol
1 3 -25 Chemical shift values are estimates from Figure 1 3 -4 1 and from Appendix lC , except in (c) and (d), where the values are exac t. (a) estimated
Q 8 70
l!
8 1 30 a a C�C
I
b
C
\
'0/
C
b TMS
I
I
I
1 60
1 80
200
I
1 40
I
I
I
I
100 1 20 80 Carbon - 1 3 8 (ppm)
60
I
40
I I
I
0
20
(b) estimated b a /C ' c C c
8 1 25
I
II
Q 8 30
i!
c, / C a C c
s;:.
8 22
b
I
200
I
1 80
TMS
I
1 60
(c) exact values
I
1 40
I
I
I
I
80 1 00 120 Carbon - 1 3 8 (ppm)
60
I
40
Q 8 1 45
C
c CH2 d b,..-: ;' ' CH 3 a C /"
s;:.
8 1 26
I
I
o
20
g 8 26
0
I
�
<5 1 7
I
CH3 e
l!
8 200 TMS
I
200
I
1 80
I
1 60
I
1 40
I
I
I
1 20 1 00 80 Carbon - 1 3 8 (ppm) 283
I
60
I
40
I
20
l I
o
1 3-25 continued (d) exact values
0
� 8 1 35
12 8 1 36
II
/ a \ C=C I c b\ H H
H
f!
8 192
C-H
TMS
I
I
I
1 60
1 80
200
I
1 40
I
I
I
I
I
80 1 00 1 20 Carbon - 1 3 & (ppm)
40
60
I
I
20
I
o
1 3-26
� t
(a) '68--
12 3H
(b) 30 -0- 2 = 15 37 -0- 2 . 7 = 14 8 -0- 1 = 8
'637 0
'630
--
As a general rule, the 1 5 -20 rule works fairly well.
'6209
� 3H
f!
2H
0
II b I
I
9
10 1 3-27
a
/ b
6
I
5 & (ppm)
I
4
I
II
3
I
2
I
1
I
o
'6 1 30
\
'0/
I
f!
C=C
C
7
8
a
(a)
I
I
c
a
I
TMS
H3C-C-CH2-CH3
C
b
12 '670 TMS
I
200
I
1 80
I
1 60
I
1 40
I
I
I
80 1 00 1 20 Carbon - 1 3 & (ppm) 284
I
60
I
40
I
20
1II1 I
o
1 3-27 continued (b)
b a,..,... C, c C C
�
I
II
8 125
C, ,..,... C a C e b
h 8 30
£
8 22
,II,
TMS
I
I
1 80
200
I
1 60
I
I
1 20
1 40
I
I
I
80
1 00
60
I
40
I
20
o I
Carbon - 1 3 8 (ppm) (c)
0
C
c CH2 d b/. ;' CH3 a 'c /'"
h 8 14 5
I
CH3
Q � 8 26 8 1 7
e
�
8 200
£
8 1 26
JII,
TMS
I
I
200
1 80
I
1 60
(d)
I
I
I
60
I
40
I
20
o I
0
h 8 1 36
�
I
I
80 1 00 1 20 Carbon - 1 3 8 (ppm)
1 40
J !t I I
II
H C- H I a \ C=C Ie b\ H H
8 1 92
� 135 £
I
200
I
1 80
I
1 60
I
1 40
I
TMS
I
I
1 20 1 00 80 Carbon - 1 3 8 (ppm) 285
I
60
I
40
I
20
,11, I
0
1 3-28 The ful l carbon spectrum of phenyl propanoate is presented belo w . f
· ·
CH
200
I
1 80
'6 1 2 2
· ·
CH
· · ·
g
J
1 60
e
_
C
I
O f
'6 1 5 1
C
I
·
.d
(5 1 73
I
�
·
'6 1 29 �
12
g/'6 1 26/CH
f
I
1 40
I
I
�
'628
89
CH2
CH3
0
g
O- -CH2CH1 c b a
e
I
80 100 1 20 Carbon - 1 3 8 (ppm)
I
60
I
40
1
TMS
I
20
I
o
The DEPT-90 wi l l show only the methine carbons, i.e . , CH. All other peaks disappear. f
g/'6 1 26/CH �
'6 1 29 CH
(5 1 22 CH
DEPT-90
The DEPT- 1 3 5 w i l l show the methyl, CH3, and methine, CH, peaks pointed up, and the methylene, CH2• peaks pointed down. f
g/'6 1 26/CH
(5 1 29 CH
�
'6 1 22 CH
DEPT-135
CH,
I
TMS
By analyzing the original spectrum, the DEPT-90, and the DEPT- 1 35, each peak in the c arbon NMR can be assigned as to which type of c arbon it represent. 286
-
13 29 S i nce allyl bromide was the s tarting material, it is reasonable to expect the allyl group to be present in the i mpurity: the triplet at 1 1 5 is a =CH2 ' the doublet at 1 38 is =CH- , and the triplet at 63 is a deshielded al iphatic CH2 ; assembling the pieces forms an allyl group. The formul a has c h anged from C3HS Br to C3H60, so OH has replaced the B r. H2C=CH-CH20H 8 115 tnplet
t
---' 8 I
s
doublet
"-
8 63 triplet
Allyl bromide is easi ly hydrolyzed by water, probably an SN 1 process. H20 H2C=CH-CH20H + HBr H2C=CH-CH2Br °
1 3 -30
II
c
C
/Ca ,
\
d 1 3 -3 1
°
/
C-C
I
C C a ' / c C b 13 -32
Two elements of un saturation in C 4�02' one of which i s a carbonyl, and no evidence of a C=C, prove that a ring must be present.
b
b C a / , c C C
II
a = 8 180, singlet b = 8 70, triplet c = 8 28 , triplet d = 8 22, triplet
a = 8 1 28 , doublet b = 8 25, triplet c = 8 23 , triplet
Two elements of unsaturation in C6HIO must be a C=C and a ring . Only three peaks i ndicates symmetry. Using PB r3 instead of H2S04iNaBr would give a hi gher yield of bromoc yclohexane.
Compound 2
Mass spectrum: the molecular ion at mJz 1 36 shows a peak at 1 3 8 of about equal height, indicating a bromine atom is present: 1 36 79 = 5 7 . The fragment at mJz 57 is the base peak ; this fragment is most likely a butyl group, C4�' so a l i ke l y molecular formula is C4H9B r. Infrared spectrum: Notable for the absence of functional groups : no O-H, no N-H, no =C-H, no C=C, no C=O =} most l ikel y an alkyl bromide. NMR spectrum: The 6H doublet at 8 1.0 suggests two CH3's split by an adj acent H-an isopropyl group. The 2H doub let at 8 3 . 2 i s a CH2 between a CH and the B r.
-
Putting the pieces together gives isobutyl bromide.
1 3 -33 The formula C 9H\\B r i ndicates four elements of unsaturation , just enough for a benzene ring. Here is the most accurate method for determining the number of protons per signal from integration values when the total number a/protons is known. Add the integration heights : 4.4 cm + 1 3 .0 cm + 6.7 cm = 24 . 1 cm. Divide by the total number of hydrogens: 24. 1 cm.;. l lH = 2 . 2 cmIH. Each 2.2 cm of integration height = I H, so the ratio of hydrogens is 2 : 6 : 3. The 2H singlet at I) 7.1 means that only two hydrogens remain on the benzene ring, that is, it has 4 substi tuents. The 6H singlet at 82 . 3 must be two CH3's on the benzene ring i n identical environments. The 3H si nglet at <') 2 . 2 is another CH3 in a slightly different environment from the first two. Substitution of the three CH3's and the Br in the most symmetric way leads to the structures on the next page. 287
13-33 continued a H
�
C 3
a second structure is also possible although it is less likely because the Br would probably deshield the Hs labeled "a" to about 7.3-7.4
b CH3
*
a = 07 . 1 (singlet, 2H) b = 0 2 . 3 (singlet, 6H) c = 0 2 . 2 (singlet, 3H)
B'
CH3 b
H a
*
b H3C
�
C 3
b H3C
H a B'
H a
13-34 The numbers in i talics i ndicate the number of peaks in each signal. (a)
(c)
* H
10 C H3 -CH - CH 3 I 2 CH3
7 CH 3-CH - OH 1 I CH3 2
(b)
CH3 - CH2-CCI2 - CH3 4 3 1
(d) H
H
H
7
(e)
C 3
H 1
(assume OH exchanging rapidly-no splitti ng)
Q-CH1 o
2
(all of these benzene H's are accidentally equivalent and do not split each other) 13-35 Consult Appendi x 1 in the text for chemical shift values. Your predictions should be in the given
range, or within 0.5 ppm oflhe given value. (a)
(c)
<}
(b) H
al l at 07 . 2
03 .8
(f)
CH3 - CH2 - C - CH3 0 2.0
02 . 5 o
� g >=< -
H 08. 0
H 07 . 5
02 . 5
CH\
o 06-7 II CH3-CH=CH -C- H o 1 .7 8 5-6 8 9-10
(h) -H
09-10
H 08 .0
The C=O has its strongest deshielding effect at the adjacent H (ortho) and across the ring (para) . The remaining H (meta) i s less deshielded.
02.2
04 . 3 <5 2 - 5 CH-O-CH2 - CH 2 -OH � 03.8 03 .8 CH3 00. 9
II
(j 1.0
CH3- CH2-C- C-H 8 1 .2
03 . 2
o
(g)
(d)
01.6 CH3 - 0-CH2 - CH2 - CH2Cl 03 .4
(e)
all at 0 1.3
288
1 3 -35 continued
CH3 o / II HOOC- CH2- CH2- C-O - CH - 8 4.0
(i)
8 1 0- 1 2
02. 3
U)
" CH3
02 . 3
i /C,,�-�/F=C\ } H
H
01 .3
H
01 . 1
/
H
,
, /
H
H
H
I
C-C I I'
��
0 1 .7 al lylic
8 1 .3
(k) H
� Y:7(
H
� � H
H
07.2
1 3-36
H
l f
(I)
H
I
OH b
� � \ H
al lylic and benzylic a
=
04.0 (septet, 1 H)
b
=
02 . 5 (broad singlet, 1 H) (rapidly exchanging)
c
=
0 1 . 2 (doublet, 6H)
1 3-37 (a) The chemical shift in
ppm
would not change: 04.00.
(b) Coupling constants do not change with field strength: J = 7 Hz, regardless of field strength. (c) At 60 MHz, 04.00 = 4 .00 ppm = (4.00 x 1 0-6) x (60 x 1 06 Hz) = 240 Hz The signal is 240 Hz downfield from TMS in a 60 MHz spectrum. At 300 MHz, (4.00 x 1 0-6) x (300 x 1 0 6 Hz)
=
1 200 Hz
The signal is 1 200 Hz downfield from TMS in a 300 MHz spectrum. Necessarily , 1 200 Hz i s exactly 5 times 240 Hz because 300 MHz i s exactly 5 times 6 0 MHz. They are directly proportional. 1 3-3 8
* H
H
H
� Ii
H
06 . 0
-H
& l .4
8 2. 5 benzyl i c
c a c CH 3 - CH - CH 3
H
R
a b c d
c b d CH2- CH2-O - C - CH3
H
'-----y--' a
289
= = = =
04.5
07 . 2-7 . 3 (multiplet, 5H) 04. 3 (tri p let, 2H) 02.9 (triplet, 2H) 02.0 (singlet, 3H)
13-39
(a)
Q 3H
c a b CH3-O-CH2CH3
£
3H g
2H
1
TMS
III 10
(b)
9
8
7
6
5
8 (ppm)
4
3
o
1
2
Q 3H
0 b a II c (CH3hCH -C-CH3
£
6H
f!
IH
(To show the pattern o f peaks. this multiplet is larger than it would appear on a real spectrum.)
10
(c)
9
8
7
6
5
8 (ppm)
a a b CICH2CH2CH 2Cl
4
j I,
3
TMS
I o
2
I
a
411
I Q 2H
I
TMS
II 10
9
8
7
6
5
8 (ppm) 290
4
3
2
..--
o
13-39 continued
(d)
12 2H
I
7
8
9
6
I
I
I
I
2H
I
I
9
10
02N
12 2H
I
-
6
I
I
Hb
� ;)
Hb
O-C � IH
II
,I I I
5 8 (ppm)
�/
I
I
I
I
3
4
*
I
TMS
I
I
a H
7
8
NH2
5 8 (ppm)
a H
(e)
f!
H b
g 4H very broad and variable
The NH2 group shields nearby protons on benzene.
IH
10
d NH2
����
� IH
f!
b H
o
2
g 6H
d C H1
C C H3 d
(To show the pattern of peaks, this multiplet is larger than it would appear on a real spectrum.)
I
4
I
I
I
3
TMS
1
2
I
o I
(f) S ignal (a) is split into a quartet because of the adjacent CH3 with J 7 Hz. Each of those peaks is then split into a doublet because of the coupling with the trans H, J 15 Hz. This is called a doublet of quartets, and it is drawn here as two quartets . In a real spectrum, these peaks would overlap and would not be a clean doublet of quartets. See the splitting tree for problem =
=
13-15.
I
V �
C C H3
bH
12
f!
IH .
d H3C
,
3 I
�
g 3H TMS
I III I
10
I
9
I
8
I
7
6 I
I
5 8 (ppm) 291
I
4
I
3
I
2
I
I I
o
1 3 40 -
(a) The NMR of I-bromopropane w ould have three sets of signals, whereas the NMR of 2-bromopropane would have only two sets (a septet and a doublet, the typical isopropyl pattern). (b) Each spectrum would have a methyl singlet at 8 2. The left structure would show an ethyl pattern (a 2H quartet and a 3H triplet) , whereas the ri ght structure would exhibit an isopropyl pattern (a IH septet and a 6H doublet) . (c ) The most obvious difference i s the c hemical shift of the CH3 singlet. In the compound on the left, the CH3 singlet would appear at 8 2. 1 , while the compound on the right would show the CH3 singlet at 8 3.8. Refer to the sol ution of 13-22 for the spectrum of the second compound. (d) The spl itting and integration for the peaks in these two compounds would be identical, so the chemical shift must make the difference. As described in text section 1 3-SB, the alkyne is not nearly as deshielding as a carbon y l , so the protons in pent-2-yne would be farther upfield than the protons in butan-2-one, by about O.S ppm. For example, the methyl on the carbonyl would appear near 8 2. 1 while the methyl on the alkyne would appear about 8 1 .7. 13-4 1 The multiplicity in the off-resonance decoupled spectrum is gi ven below each chemical shift: s = si nglet; d = doublet; t = tri plet; q = q uartet. It is often difficult to predict exact chemical shift values; your predictions should be in the ri ght vicinity. There should be no question about the multiplicity and DEPT spectra, however. (a) estimated
q
t
s
.!!
8 1 70 c
I
200
I
1 80
I
I
1 60
I
1 40
I
I
I
80 1 00 1 20 Carbon - 1 3 8 (ppm)
I
60
I
40
I
20
o
I
The DEPT-90 spectrum for ethyl acetate would have no peaks because there are no C H groups.
DEPT-135
292
1 3-4 1
continued
(b) actual values
c b a H2C = CH - CH2Cl 1 1 8 1 34 d t
12
45
t
�
01 1 8
045
CH2
CH2
(CHJ)4Si
g
CH
I
200
I
1 80
I
I
I
1 40
1 60
DEPT-135
CH
M
Carbons b and c are accidentally equivalent.
CH
C
200
1 80
I
I
1 60
DEPT-90 and DEPT- 1 35
Q
1 27 d
o 1 27
0 1 39
I
d
0 1 29 CH
I
1 40
I
I
80
100
60
Carbon - 13 8 (ppm)
DEPT-90
I
r
1 20
CH
(c) actual v&lues
I
TMS
0 1 34
I
1 20
on next page
< c
b
'\
c-b
1 29 d
1 29 d
20
40
r
o
1 39/s
>/H2 H2
'/
I
I
C -C e f 39 33
t
-
Br
f
033 CH2
t
(CHJ)4Si
TMS
I
1 00 Carbon - 1 3 8 293
I
80
(ppm)
I
60
I
40
I
20
I
T
0
1 3-4 1
(c) continued
CH
DEPT-90
CH
DEPT-135
CH
CH2 13-42 The multiplicity of the peaks in this off-resonance decoupled spectrum show two different CR's and a CH3. There is only one way to assemble these pieces with three chlorines. Cl
CI I
I
020-- CH3- CH - CH q CI I � 07 5 060 d
t
d
13-43 There is no evidence for vinyl hydrogens, so the double bond is gone. Integration gives eight hydrogens, so the formula must be C4HsBr2' and the four carbons must be in a straight chain because the starting material was but-2-ene. From the integration, the four carbons must be present as one CH3, one CH, and two CH2 groups. The methyl is split into a doublet, so it must be adjacent to the CH. The two CH2 groups must follow in succession, with two bromine atoms filling the remaining valences. (The spectrum is complex because the asymmetric carbon atom causes the neighboring protons to be diastereotopic.) H I
Br H I
I
Br I
H- C- C- C- C-H I
H
�
d
I
H
a
I
H
c
I
H
�
b
a b c d
= = = =
8 4.3 (sextet, IH) 8 3.6 (triplet, 2H) 8 2.3 (multiplet, 2H) 0l.7 (doublet, 3H)
1 3 -44 There is no evidence for vinyl hydrogens, so the compound must be a small, saturated, oxygen containing molecule. Starting upfield (toward TMS), the first signal is a 3H triplet; this must be a CH 3 next to a CH2. The CH2 could be the signal at 8 l.5 , but it has six peaks: it must have five neighboring hydrogens, a CH3 on one side and a CH2 on the other side. The third carbon must therefore be a CH2; its signal is a quartet at 0 3 .6 , split by a CH2 and an OH. To be so far downfield, the final CH2 must be bonded to oxygen. The remaining I H signal must be from an OH. The compound must be propan-l-ol.
a b c d
= = = =
03 .6 (quartet, 2H) 0 3.2 (triplet, I H) 0 1 . 5 (6 peaks, 2H) 00.9 (triplet, 3H) 294
13-45
CH3
\
/
/
\
C=C
CH3
CH3
+
H
Isomer A
(a)
/
/
\
C=C
\
a b c d
= = = =
d CH3
b CH3
C=C
/
H d H
CH3
/
H \
\
CH2CH3
Isomer B
\
/
C=C
/
f CH3
\
CH2CH3 H e g d Isomer B d = 84.7 (singlet, 2H) e = 82.0 (quartet, 2H) f = 81.7 (singlet, 3H) g = 8 1.0 (triplet, 3H)
CH3 H a c Isomer A 85.2 (quartet, 1H) 81.7 (singlet, 3H) 81.6 (singlet, 3H) 81.5 (doublet, 3H)
(b) With NaOH as base, the more highly substituted alkene, Isomer A, would be expected to predominate- the Zaitsev Rule. With KO-t-Bu as a hindered, bulky base, the less substituted alkene, Isomer B , would predominate (the Hofmann product). 13-46 "Nuclear waste" is comprised of radioactive products from either nuclear reactions, for example, from electrical generating stations powered by nuclear reactors, or residue from medical or scientific studies using radioactive nuclides as therapeutic agents (like iodine for thyroid treatment) or as molecular tracers (carbon-14, tritium H-3, phosphorus-32, nitrogen-I 5, and many others). The physical technique of nuclear magnetic resonance neither uses nor generates any radioactive elements, and does not generate "nuclear waste". (Some people assume that the medical application of NMR, medical resonance imaging or MRl, purposely dropped the word "nuclear" from the technique to avoid the confusion between "nuclear" and "radioactive" .) 13-47
The molecular ion of m/z 117 suggests the presence of an odd number of nitrogens. 3 2 Infrared spectrum: No NH or OH appears. Hydrogens bonded to both sp and sp carbon are indicated around 3000 em-I. The characteristic C=N peak appears at 2250 em-I and aromatic C=C is suggested by the peak at 1600 cm-I. NMR spectrum: Five aromatic protons are shown in the NMR at 8 7.3. A CH2 singlet appears at 8 3.7. Assemble the pieces: Mass spectrum:
* H
H
H
H mass 77 H
+ C H2 +
mass 14
o-�
C:::N
-
mass 26
H
C -C:::N
H I
mass 117 295
13-48 This is a challenging problem, despite the molecule being relatively small.
Mass spectrum:
or fewer.
The molecular ion at 96 suggests no Cl, Br, or N. The molecule must have seven carbons
The dominant functional group peak is at 1685 cm-I, a carbonyl that is conjugated with C=C (lower wavenumber than normal, very intense peak). The presence of an oxygen and a molecular ion of 96 lead to a formula of C6HsO, with three elements of unsaturation, a c=o and one or two c=c. Carbon NMR spectrum: The six peaks show, by chemical shift, one carbonyl carbon (196), two alkene carbons (129, 151), and three aliphatic carbons (23, 26, 36). By off-resonance decoupling multiplicity, the groups are: three CH2 groups, two alkene CH groups, and carbonyl. Infrared spectrum:
Since the structure has one carbonyl and only two alkene carbons, the third element of unsaturation must be a ring. C=O CH2 + CH2 + CH2 + 1 ring C=CI H
I
H
,
Since the structure has no methyl group, and no H2C=, all of the carbons must be included in the ring. The only way these pieces can fit together is in cyclohex-2-enone. Notice that the proton NMR was unnecessary to determine the structure, fortunately, since the HNMR was not easily interpreted except for the two alkene hydrogens; the hydrogen on carbon-2 appears as the doublet at 6.0. >
The mystery mass spec peak at mlz 68 comes from a fragmentation that will be discussed later; it is called a retro-Diels-Alder fragmentation. 13-49 The key to the carbon NMR lies in the symmetry of these structures. C 3 " / b c a CH3 c �/ 3 a " d J" h c ha CH3 . ,/ c b CH3 b " meta-xylene ortho-xylene cyclohex-2-enone
cr
er:
,
,
,
,
(a) In each molecule, the methyl carbons are equivalent, giving one signal in the CNMR. Considering the ring carbons, the symmetry of the structures shows that ortho-xylene would have 3 carbon signals from the ring (total of 4 peaks), meta-xylene would have 4 carbon signals from the ring (total of 5 peaks), and para-xylene would have only 2 carbon signals from the ring (total of 3 peaks). These compounds would be instantly identifiable simply by the number of peaks in the carbon NMR .
(b) The proton NMR would be a completely different problem. Unless the substituent on the benzene ring is moderately electron-withdrawing or donating, the ring protons absorb at roughly the same position. A methyl group has essentially no electronic effect on the ring hydrogens, so while the para isomer would give a clean singlet because all its ring protons are equivalent, the ortho and meta isomers would have only slightly broadened singlets for their proton signals. (Only a very high field NMR, 500 MHz or higher, would be able to distinguish these isomers in the proton NMR.) 296
13-50 (a), (b) and (c) The six isomers are drawn here. Below each structure is the number of proton signals and the number of carbon signals. (Note that splitting patterns would give even more clues in the H proton NMR.) H3 C 'rI" CH3 l-. /' H CH3 H H CH3 H y� >=< >=< 1 6 H H H3C H H3C CH3 H D 2xH 5xH 3xH* 2xH 2xH IxH 3xC 4xC 2xC 2xC 3xC IxC
;L
_
*Rings always present challenges in stereochemistry. When viewed in three dimensions, it becomes apparent that the two hydrogens on a CHz are not equivalent: on each CHz, one H is cis to the methyl and one H is trans to the methyl. These are diastereotopic protons. A more correct answer to part (b) would be four types of protons; whether all four could be distinguished in the NMR is a harder question to answer. For the purpose of this problem, whether it is 3 or 4 types of H does not matter because either one, in combination with three types of carbon, will distinguish it from the other 5 structures. (d) Two types of H and three types of C can be only one isomer: 2-methylpropene (isobutylene). only isomers that would not be distinguished from each other would be cis- and trans-but-2-ene.)
�NMR' �
(The
13-51 For DEPT to be useful in distinguishing isomers, there should be different numbers of CH3, CHz, CH, and C ,ignal, in 6 C signals
one CH3 peak four CHz peaks one CH peak
~
5 C signals one CH3 peak two CHz peaks two CH peaks
5 C signals one CH3 peak two CHz peaks one CH peak one C peak
A carbon N MR could not distinguish isomers Band C, and if there is any overlap in the signals, it might have a difficult time distinguishing A. Fortunately, the patterns that would appear in the DEPT-90 would distinguish B (two CH peaks up) from the other two (l CH peak up), and the DEPT-135 would instantly distinguish A (4 CHz peaks down) from C (2 CH2 peaks down).
13-52
(a) MS or IR could not easily distinguish these isomers: same molecular weight and same functional group. They would give dramaticaIIy different proton and carbon NMRs however. + OCH]
2 singlets in HNMR 3 singlets in CNMR
�
OCHZCH3
4 signals, all with splitting, in HNMR 4 singlets in CNMR
297
13-52 continued
(b) The only technique that would not readily distinguish these isomers would be MS because they have the same molecular weight and would have similar, though not identical, fragmentation patterns. o
)lo�
IR: C=O HNMR: CNMR:
about 1 730 cm-i methyl singlet and ethyl pattern ester C=O about 8 1 70
IR: C=O about 1 7 1 0 HN MR : 3 singlets CNMR:
cm-i
ketone C=O about 8 200
(c) The big winner here is MS : they have different molecular weights, plus the Cl has the two isotope peaks that make a Cl atom easily distinguished. The other techniques would have minor differences and would require having a detailed table of frequencies or chemical shifts to determine which is which.
F-o-OH M+
=
mlz 1 1 2
C'-o-OH
M+ = mlz 1 2 8 , 1 30
298
CHAPTER 14-ETHERS. EPOXIDES AND SULFIDES
1 4- 1 The four sol vents decrease i n polarity in this order: water, ethanol , ethyl ether, and dichloromethane. The three solutes decrease i n polarity in this order: sodium acetate, 2-naphthol, and naphthalene. The guidi ng principle i n determi ning solubi l i ty is, "Like dissolves like." Compounds of s i milar pol arity w i l l dissolve (in) each other. Thus , sodium acetate w i l l dissolve in water, will dissolve only s lightly in ethanol , and wi ll be vi rtual l y insol uble in ethy l ether and dichloromethane. 2-Naphthol w i l l be i n soluble in water, somewhat soluble i n ethanol , and soluble i n ether and dichloromethane. Naphthalene w i l l be i n soluble i n water, partially soluble i n ethano l , and soluble in ethy l ether and dichloromethane. (Actual solubi lities are difficult to predict, but you should be able to predict trends.) 1 4-2
CI
I I
-
CH2CH3
+
/
Cl-A l-O : CI
\
Oxygen shares one of its electron pai rs with aluminum; oxygen is the Lewis base, and aluminum is the Lewis acid. A n oxygen atom with three bonds and one unshared pair has a positive formal charge . An aluminum atom with four bonds has a negative formal charge .
CH2CH3
14-3 The crown ether has two effects on KMn04: first, it makes KMn04 much more soluble in benzene; second, it holds the potassium ion ti ghtly , making the permanganate more avai l able for reaction. Chemists cal l this a "naked an ion" because it is not complexed with solvent molecules.
o
'� -/
,0
Mn
/' '\:,
0'
0
benzene ..
IS-crown-6 a " crown ether"
Please see the note on p. 1 3 of this Solutions Manual regarding placement of position numbers.
1 4-4 IUPAC name first; then common name (see Appendi x I in this Solutions Manual for a summary of IUPAC nomenclature)
(a) (b) (c) (d) (e)
methoxycyc lopropane ; cyc lopropyl methyl ether 2-ethoxypropane; ethyl i sopropyl ether l-chloro-2-methoxyethane; 2-c h loroethyl methyl ether 2-ethoxy-2,3-di methylpentane; no common name 2-t-butoxybutane; sec-butyl t-butyl ether (f) trans-2-methoxycyc lohexan -l-ol; no common name 1 4-5 (a)
+
() o
+
2 H20
The alcohol is ethane-l,2-di o l ; the common name is ethylene glycol. 299
1 4-5 continued (b)
HO
I
H
:o� I
H2C -- CH2
..
H+
:60 IJ
HO
I
o
HO
I
H2C I
--
:0: =
H2
H2C -- CH2
----l�
H2C
()
HO
I
H2C
\.
-
�
HO
: OH
I
I
H2C
CH2
-
+
H20: CH2 I�
)
.. �
H O:
I
H2C-CH2
:O �H
I
I
0
+
I
�
(I
H2C .
+ H30 + -
�
CH2 I
H O:
0
H2C
CH2
I
CH2
I
H2C-CH2
H OH
\
I
:0:
HO
CH2
CH2
H+
I
The mechanism shows that the acid c atalyst is regenerated at the end of the reaction . 1 4-6 (a) dihydropyran (b) 2-chloro-l,4-dioxane (c) 3-isopropylpyran
(d) trans-2,3-diethyloxirane; trans-3,4-epoxyhexane; trans-3-hexene oxide (e) 3-bromo-2-ethoxyfuran (f) 3-bromo-2,2-dimethyloxetane
1 4-7
CH3 I
+
+CH , CH3
CH3CH2CH2CH2 mJz 57
mJz 43
mJz 73
mJz 1 01 300
1 4-8 SN2 reactions, including the Williamson ether synthesis , work best when the nucleophile attacks a 1 ° or methyl c arbon . I nstead of attempting to form the bond from oxygen to the 2° c arbon on the ring, form the bond from oxygen to the 1 ° carbon of the butyl group. The OH must first be transformed into a good leaving group: either a tosyl ate, or one of the hal ides (not fluoride). TsCI
� OH
pyridine
� OTs
..
O�
0- Na+
OH
Q �Q (f
1 4-9 (a)
0H
Na
OH
Na
A
(b)
� OTs
CH3CH2CH2Br
...
..
OH
¢
(c)
CH3I
Na OH
...
..
N0 2
(d)
¢Hl
CH3CH20H
A
CH3 I CH 3-C- OH I CH3
Na Na
Na
...
�
CH3CH2Br
CH3CH2CH2Br
CH3CH 2CH20CH2CH3
...
CH3 C CH 3 - - OCH2Ph I CH3
PhCH2Br
•
CH3CH2CH20CH2CH3
...
--_..
1 4- 1 0 (a)
(1)
� or
� (2 )
0�
N0 2
CH3CH2CH20H
(e)
(f
..
Q
OCH3
CH3I
...
..
OH �
Hg(OAc)2
CH30H
Na
...
•
CH31
NaBH �
OCH3
�
OCH3
...
� 30 1
3-butoxy-l,ldi methy Icyclohex:me
14-10 continued (b) (1)
(2)
(c)
(1) (2)
(d)
(1)
o
Hg(OAch
..
(yoH 6
Na
--
Hg(OAch
(e)
(1)
CH31
..
NaB H4
..
..
Hg(OAch
NaBH4
CH30 H
(2)
(y
Alkoxymercuration i s not practical here ; the product does not h ave Markovnikov orientation .
------�.. �
y
O
OCH 1
CH 1
..
OH
(2) Wi lliamson ether synthesis would give a poor yield of product as the hal ide i s on a 2° c arbon.
(f)
(1)
O
H Na
--
==< o-
Hg(OAch
K�
..
OH
NaBH4
..
( }-o+
(2) Wi l liamson ether synthesis is n o t feasible here. SN2 does n o t work on either a benzene or a 3 ° halide.
14-11 An i mportant pri nciple of synthesis is to avoid mixtures of i somers wherever possible; minimizing separations increases recovery of products. Bimolecular dehydration is a random process . Heating a mixture of ethanol and methanol w ith acid wi l l produce all possible combinations: dimethyl ether, ethyl methyl ether, and diethy l ether. Thi s mixture would be troublesome to separate. 302
1 4- 1 2 Ether formation ..r"'..
CH3CH2CH2-�.H
H
nl '--
CH3CH2CH2 -..O+- H
H+
4
.. H�-CH2CH2CH3 .. �
H20
H
( 1+
CH3CH2CH2 - � - CH2CH2CH3 • •
Dehydration
H
nl+ CH3CHCH2 - .. ° -H I..)
-
H� H2 �
Remember I:!.G = M/- TM? Thermodynamic s of a reaction depend on the s i gn and magnitude of AG. As temperature i ncreases, the entropy term grows i n i mportance. In ether formation, the I:!.S is smal l because two molecules of alcohol give one molecule of ether plus one molecule of w ater-no net change in the n umber of molecules. In dehydration , however, one molecule of alcohol generates one molecule of alkene plus one molecule of water-a l arge i ncrease in entropy. So TI:!.S is more i mportant for dehydration than for ether formation. As temperature increases, the competition will shift toward more dehydration . 1 4- 1 3 (a) This symmetrical ether a t 1 ° carbons could b e produced in good yield b y bimolecular dehydration. (b) This unsymmetrical ether could not be produced in high yield by bimolec ular dehydration . Wil l iamson synthesis would be preferred.
/'-..../ OH
/"'- OH (c) Even though thi s ether i s symmetrical , both carbons are 2°, so bimolecu l ar dehydration would give low yields. Unimolecular dehydration to give alkenes would be the dominant pathway. Alkoxymercuration demercuration is the preferred route.
or
}
H g ( O Ac ) z
� OH
�
303
14- 1 4
0 •
-
H- Br
•
R� "-.;
(Q
Br
---
• •
'4
I
:Br:
£L oH
H-Br
Br
o
(b)
HBr
� O)
HI
_
< }-
OH
�
� OH� I HBr
(e)
1 4- 1 6
�
o
• •
� �r
/
+
BrCH2CH2 - CH- CH2 - Br I CH3
B
Br' 'Br
---
r-\ + �
Br
1_
:'0- B- Br I
H3C
1
Br
!
HOBBr 2 2
+
H20
B( OHh
-
�. O+-B, Br + H3C
+ 2 HBr
•
o xygen, another H+ goes on the other oxygen; several d ifferent scenarios of how this happens could be proposed
304
.Br.
�
I
J
\
�
H'.. H O� .. � Bu + Bu Bu, "' Br H20: 11_ , ' , , .. .. � : O-B-Br :·O-B-Br .. :0-B . . , I� I two proton transfers happen -Sr Br H Br here: one H+ comes off of H' 0
• •
I BU OHI
JL Br
HI
�
(d)
"-.;
H
• •
+
•
•
• •
Ea CH3Br
nucleophilic attach on CH3 faster than on I 0 carbon
Begin by transforming the alcohols into good leaving groups like halides or tosylates: NaSH NaOH PBr3 /'.... /'.... /'.... /'.... • • .......", 'S- N a+ /' /' .......", 'SH - �Br �OH 1 4-17
OH
A
TsCI pyn·d·me
1 A OTs
I
�S �
14- 1 8 The sulfur at the center of mustard gas is an excellent nucleophile, and chloride is a decent leaving group. Sulfur can do in internal nucleophilic subsitution to make a reactive sulfonium salt and the sulfur equivalent of an epoxide. very reactive alkylating agent (a)
�
HN� �CI S inactivated enzyme
Cl� �Cl C '-- S •
•
. .
(b) NaOCI is a powerful oxidizing agent. It oxidizes sulfur to a sulfoxide or more likely a sulfone, either of which is no longer nucleophilic, preventing formation of the cyclic sulfonium salt. NaOCI
Cl�· �Cl S
o
Cl� �Cl S+
+ CI�I�CI S++
• •
I
I
o
o
sulfoxide
14- 1 9
sulfone
Generally, chemists prefer the peroxyacid method of epoxide formation to the halohydrin method. Reactions (a) and (b) show the peroxyacid method, but the halohydrin method could also be used.
(a)
==<
HO
(b)
(c)
(d) (e)
MCPBA
CI
U
CI
D(
..
H2 SO4
'r-
Ph
o
t1
-
BH3 · THF
U
/
Ph
� OH Cl
H2 02 HO-
..
Hg(OAc}z H2 O
MCPBA • CH2CI2
..
NaBH4 ..
NaOH
..
CI
l)
CI
P
Ph
NaOH ..
U OH
� 305
0
0
NaOH
..
�
1 4-20 (a) 1 ) te rt-B utyl hydroperoxide is the o xidizing agent. The (CH3hCOOH contains the 0-0 bond just l i ke a peroxyacid. 2) Diethyl tartrate has two asymmetric carbons and is the source of asymmetry; i ts function is to create a chi ral transition state that is of lowest energy , leading to only one en anti orner of product. Thi s process is c alled chirality transfer. 3) The function of the titanium (IV) isopropoxide is to act as the glue that holds all of the reagents together. The titanium holds an oxygen from each reactant geraniol , t-B uOOH, and diethyl tartrate-and tethers them so that they react together, rather than j ust havi ng them in solution and hoping that they w i l l eventuall y col lide. (b) Al l three reactants are required to make Sharpless epoxidation work, but the key to enantioselective epoxi dation is the chiral molecule, diethyl tartrate . When it complexes (or chelates) with titanium, it forms a l arge structure that is also chiral . As the t-B uOOH and geraniol approach the complex , the steric requirements of the complex allow the approach in one preferred orientation. When the reaction between the alkene and t-B uOOH occ urs , it occ urs preferentially from one face of the alkene , leading to one major stereosiomer of the epoxide. Without the chiral diethyl tartrate in the complex, the alkene could approach from one side j ust as easily as the other, and a racemic mi xture would be formed. (c) Using the enantiomer of diethyl L-tartrate, cal led diethyl D-tartrate, would give exactly the opposite stereochemical results. t-B uOOH
: +�
Ti(Oi-Pr)4
O
O
H
1 4-2 1
R
H Me
,
" ' 'C
'
'). == '" C
�
/ HO
�
'
H trans-but -2-ene
Me
OH
+
�" Me
� "" " HM e
"" '" H Me
" , Me
H
Ho
0
"7
H Me
y
trans
CH3
H30 +
---.
H
)
H
I
l---<.
Me � -:;
.... ..'-.. _
'"OH
�
.. . �
''
'
e
�
'
Me
.
OR : O - H
I
tI
H
OH
+
H
Ho
l
e
K
H
.: � � Q) ' I,
+
)
stereochemistry shown i n Newman projections:
1 V ): H
+
:O -
H-O:
IDENTICAL-MESO
CH3 H
OO
diethyl D-tartrate , H COOC zCH3 the enantiomer of diethy l L-tartrate
· A
C=O
t
H2CH J
� �O I " �" " " '-... OH
OH
:O :
rO ' �' H V O "
+
•
"'Me
t<::>r H3C y H
--H Ii
H
OH
OH
306
C H3
H
'\
\' '\
+
OH
H
rotate
•
H
H3C
H
H3C
MESO
14-2 1
continued H" H "C==C" " Me' 'Me cis-2-butene I,
>
\'
x �
Me
'"
HO
OH
'" H Me
HO +
�---------Y
�'"
H Me
H"'/ Me
OH
/
----------
ENANTIOMER S
stereochemistry shown in Newman projections:
OH
CH3
H
H--+---;;;.....-+-- CH3
rotate ..
C1h H 20: Cli --
--
3
• •
� H � CH I H 3C H
CH R I AL RACEMIC MIXTURE
CIS
1 4-22
RC0 3 H
-I.. �
---
CH3 CH20CH2CH20H Cellosolve®
H2C-CH2 \/ o
1 4-23
Br: / : .. t
anhydrous HBr--only Bc nucleophiles present H 2C ,,/CH2 .. ·0· H 0r '-----A •
•
� ..
H 2C /C 2 " ·0 � -
•
1+ H
1 : O-H
CH2CH 2Br
:Br:
..
---l"�
H
\
Br
� --...."
CH2CH2Br
�I+ H -O-H
BrCH2CH2Br
+
H20
aqueous HBr-many more H 20 nucleophiles than Bc nucleophiles H 2C
CH 2 , / ,:0 : H 0-r '-----A _
H2C
0:O-H . " Q :?+ k H ---
c
307
..
1+ :? H ,
H2C-CH2 I
HO
H
..
H 29.
�
H2C-CH2 I
HO
1
OH
1 4-24 The cyclization of squalene via the epoxide is an excellent (and extraordinary) example o f how Nature uses organic chemistry to its advantage . In one enzymatic step, Nature forms four rings and eight chira l centers ! Out of 256 possible stereoisomers, only one is formed !
•
HO
\
�
1 4-25
� :\.
-
..
� ,,----" :0 - CH3
H
H
•
1 4-26
(a)
(d )
� O�
(JI � �
1 4-27
(a )
O- Na+
H N� OH
(b)
' '" , , ,A A
" '" Et M
H
e
CH30 Me
" '
...,,,, Et , Me
" Me H
"
"
CH30H
H+ • CH30H
� H
(c)
+
:�-'C H 1
OCH3
� s � O-
o
0 _ Na+
C�
Na+
:N == N == N � _' a O N + +
O- K+
X3oH
'>---f/
Ho H0
•
bonds formed
(f)
N
""- /
.
OCH3
H2N �
(e)
OH HI 80 �
o
(d)
H
•
(b)
o
(c)
H �CH3
•
=
Me
' Et " 1 Me
"'- OCH3
� ; �
Me
CH30
H
308
"
OH
Me
H
�
1 4-28 Newly fonned bonds are shown in bold. (a )
I � OH
(b )
OH OH
(c)
1 4-29 Please refer to solution 1 -20, page 1 2 of this Solutions Manual. 1 4-30
�
(a )
/" O
(d)
� O�
(g)
;A
H
(b) � O �
(c)
(e) � O .......
(
(h)
� ''''
� Br
+
y Br
(c) and (d) no reaction
(g)
� HO'"
"
(i)
�
0 H
0
A
H3C
CH3
H
(e) trans-cyclohexene glycol (f) cyclopentyl methyl ether (g) propylene o xide (h) cyclopentene o xide
14-32 (a ) 2-methoxypropan- l -01 (b) etho xybenzene or phenoxyethane (c) metho xycyclopentane (d) 2,2-dimethoxycyclopentan- l -01 (e ) trans- l -methoxy-2-methylcyclohexane
(a )
f)
, HO H
14-3 1 (a) sec-butyl isopropyl ether (b) t-butyl isobutyl ether (c) ethyl phenyl ether (d) chloromethyl n-propyl ether
1 4-33
0 ""-"""
+
H2 O
(e)
(h)
( trans-3-chloro- l ,2-epoxycycloheptane (g) f) trans- l -metho xy- l ,2-epoxybutane ; or, trans-2-ethy 1- 3 -methoxyoxirane (h) 3-bromooxetane (i) 1 ,3-dio xane
(b)
< )-OH
�N
� Br
OH +
HC H3
H
309
CH3CH21
+
� Br
( f)
(i)
+ H2 O
� OH OCH3
-+o�
1 4-33 continued
HO
�
o
(j )
(k)
(n)
� Br
(I)
OH
ry."OCH3 CH3 �1 I 0H H
1 4-34
(a) On long-term exposure to air , ethers form peroxides. Pero xides are explosive when concentrated or heated. (For exactly this reason, ethers should never be distilled to dryness. ) (b ) Peroxide formation can be prevented by excluding oxygen. Ethers can be checked for the presence of peroxides, and peroxides can be destroyed safely by treatment with reducing a gents . 1 4-35
(a ) Beginning with (R)-butan-2-o l and producing the (R) sulfide requires two inversions of configuration. H Br HO H
�
�
R
5
An a lternative approach would be to make the tosylate, displace with chloride or bromide (SN2 with inversion ) , then do a second inversion with Na SCH3.
(b) Synthesis of the (5) isomer directly requires only one inversion . HO H
� R
TsCI
Ts O H
�
pyridine
R
1 4-36
(a ) ./ .-.
molecular ion mlz
mlz 73
1 02
3 10
))� }
........,.,
�
" C H2
1 4-36 continued (b)
.
+
3 1 0CH3
H
\+
C-O:
/
molecular ion mJz
+
/
�? , mJz 7 1
H
14-37
o: �
�
H+ _
i
0·+ (
� C I
(a
�
I
CH2
-:;:::.CH2
MCPBA � CH2Cl2
�
MCPBA • CH2Cl2
(c �
MCPBA .. CH2Cl2
(b
1 02
�
H
+1
CH "
CH3
• •
II
+. . II
OH I
CH2
CH2 _ ��20 : • •
I
CH3
� � �
311
1)
2)
CH3
CH30H
H+ CH30H
mJz 8 7
OH I
�
•
�
CH2 0H
• •
• •
PhMgBr
NaOCH3
}
mJz 59
CH2 _ CH2 H C/ + / H20 : I -0\" CH3 H
O
H2O
(after H migration)
H
OCH3
� CH
•
mJz 3 1
• •
CH "
•
+
\
OCH3
..
O
+
. .
+1
-
H
:OCH3
C=O
/
-----
: OCH3
� CH
O- H
CH 3
1 4-3 8
H
\
H \
J
OH +-
� Ph OH
H3 0+
�OCH3 OH �OH OCH3
CH3
o A
(X "
0H
"'
G
G
(X " " '0 0 Na -
0
Na �
D
+
CH3Br C
(X
0CH]
�
H
'' '"
0
The student turned in the wrong product ! Three pieces of information are consistent with the desired product: molecular formula C4HIOO; O-H stretch in the IR at 3300 cm-I (although it should be strong, not weak); and mass spectrum fragment at rnJz 59 (loss of CH3). The NMR of the product should have a 9H singlet at 8 1 .0 and a IH singlet between 8 2 and 8 5. Instead, the NMR shows CH3CH2 bonded to oxygen. The student isolated diethyl ether, the typical solvent used in Grignard reactions. 1 4-40
Predicted product 59 CH3 CH]
1 4-4 1
(a)
(b)
H
Isolated product
- OH
CH3 C4HIOO O-H at 3 300 cm-I
U OH
/"-.../ OH \ }-
OH
�OH
Na
-
PBr3
�
NaOH�
U
O- Na+
� Br
\ }-
C4HIOO O-H at 3300 cm-I due to water contamination
}
O- Na+
TsCI � �OTs pyridine
312
O _ U �
� \ }- o �
1 4-4 1 continued
Hg(OAc)z
(c)
� --l..
NaBH4
---
t
(d)
1 ) BH3
•
THF
C
HB r, ROOR
..
..
OH
NaO H 3
Br
..
(e)
1 4-42 In the first sequence, no bond i s broken to the chiral center, so the configuration of the product i s the same as the configuration of the starti ng material .
� [a]o
==
-
H
�CH2CH
Na
�
[a] o
8 . 24°
(Assume the enantiomer shown is levorotatory . )
-Q-
==
-
1 5 .6°
In the second reaction sequence , however, bonds to the chiral carbon are broken twice, so the stereochemistry of each process must be considered. � 0 o II • • OH CI - S H � II � !J - -' ,;; . , � � 0 � .. "-/ "-/ "-/ \ . H 0
. . . ,
H,;�\-�-Q-
� ) ----
undOUbtedly, some E2 produ cts will als � . form In thiS reacHon.
1
s
CH 3 C H2 0 ·•• •
Na+
H,
t :t� o-s
�
-Q-
0- _
0
RETENTION OF CONFIGURATION
SN2-INVERSION
The second sequence invol ves retention fol lowed by inversion, thereby producing the enantiomer of the 2ethoxyoctane generated by the first sequence . The optical rotation of the final product w i l l have equal + 1 5 .6° . magni tude but opposite sign, [a]o ==
313
�
14-43
. .'-
HO :
'\
CH3 /""\
CH3
. '. -
I
I
H -- HO - CH2 - CH - �: H2C \F :0:
O
t
O
CH3 I
I
I
.
.- �
HO - CH 2 - CH - O - CH2 - CH - O - CH 2 - CH - O :
+
HO H � u I
I
I
• •
C�
C�
C�
HO - CH2 - CH - 0 - CH2 - CH - 0 - CH2 - CH - OH
1 4 -44 The text uses
CH3
I
I
I
..
H -- HO - CH2 - CH - 0 - CH2 - CH - � : H2C \F :0: CH3
CH3
CH3
�
CH3
CH3
+
I -
H2C - CH
a
HO-
HA to indicate an acid; A- is the conj ugate base. •
•
.
. .O - H + : ANote that all three carbocation intermediates are 3 ° !
· A
•
OH +
..
"-. H " I
'-.... {... C C+
.
tI
.. OH
H- A
This process resembles the cyclization of squalene oxide to lanostero l . (See the solution to problem L 4-24 . ) I n fact, pharmaceutical synthesis o f steroids uses the same type o f reaction called a "biomimetic cyclization" .
314
(b)
(c)
:" , /0
No bond to the chiral center was broken. Configuration is retained; R stays as R.
\CH2
I
" H H3 C
Attack of w ater gave inversion of configuration at the chiral center; R became S.
R
'--HO : • •
-
R
(d) The difference i n these mechan i s ms lies in where the nuc leophi le attacks. Attack at the chiral c arbon gi ves i n version; attack at the ach i ral carbon retains the configuration at the chiral carbon. These products are enantiomers and must necessari l y have optical rotations of opposite sign. methyl cellosolve
1 4-46
CH30CH2CH20H
To begin, what can be said about methy l cellosolve? Its molecular weight is 76; i ts IR would show C - -O i n the 1 000- 1 200 cm- I region a n d a strong O-H around 3300 em- I ; and i t s NMR would show four sets o f signals in the ratio o f 3 : 2 : 2 : 1 . The unknown has molecular weight 1 34 ; this i s double the weight of methy l cellosolve, minus 1 8 (water) . The IR shows no OH, only ether C-O. The NMR shows no OH, only H-C-O i n the ratio of 3 : 2 : 2 . Apparentl y, t w o molecules of methyl ce lloso l ve have combined in a n acid-catalyzed, bimolecular dehydration .
- ° C H2CH20CH3 ________
CH30CH2CH2 "-
-
1-
.. .
(th i s compound i s c a lled "diethylene glycol dimethy l ether" , or a shortened, common n ame is "di g l y me" )
.. H2
0:
3 15
CH30CH2CH2 �O
�H
- CH2CH20CH3
I)
1 4-47 The fonnula CgHgO has five elements of unsaturation (enough (4) for a benzene ring). The IR is usefu l for what is does not show . There i s neither OH nor C=O, so the oxygen must be an ether functional group. The NMR shows a 5H signal at 8 7 .2, a monosubstituted benzene. No peaks in the 8 4. 5 -6.0 range indicate the absence of an alkene, so the remaining element of un saturation must be a ring. The three protons are non-equi valent, with complex splitting.
<>
+ 2C +
0 ether
+ 3 H + ring
These pieces can be assembled in only one manner consistent with the data.
� H
H
H
(Note that the CH2 hydrogen s are not equi valent (one is cis and one is trans to the phenyl) and therefore have distinct chemical shifts . ) 1 4-48 The key concept is that reagents always go to the less hindered side o f a molecule first. I n this case, the " underneath" side is less hindered; the " top" side has a CH3 hovering over the double bond and approach from the top w i l l be much more difficult, and therefore slower, than approach from underneath .
MCPB A
J
less hindered side
B r2
B
MCPBA approaches from the bottom; the epoxide is fanned from MCPB A w i thout the participation of any other reagent. The epoxide fonns on the less hindered side.
I
fonnation of the bromohydrin begins with fonnation of the bromonium ion
on the less hindered side
2,6-lutidine o
water must attack from the side opposite the bromonium ion , from the TOP
c Br
Epoxides B and C are di astereomers ; they will have different chemical and physical properties. Nucle� philes can react with C much faster than with B for precisely the same reason that explai ned thei r fonnatlOn: approach from underneath i s less hindered and is faster than approach from t h e top. 316
CHAPTER I S-CONJUGATED SYSTEMS, ORBITAL SYMMETRY, AND ULTRAVIOLET SPECTROSCOPY
C=C,
1 5 - 1 Look for: 1 ) the number of double bonds to be hydrogenated-the fewer the smaller the MI; 2) conj ugation-the more conjugated, the more stable , the lower the � ; 3) degree of substitution of the alkenes-the more substituted, the more stable, the lower the !1H .
(a) � < � <�< smallest !1H
� < == C � < � biggest !1H
<
< smallest !1H 1 5 -2 Reminder: conj ugate base.
H
H
H H
H-B H
<
<
bi ggest !1H
i s used to symbolize the general fonn for an acid, that i s , a protonated base;
Hf"B
H H
H
H
H
B-
is the
H
•
H
H
H
H
H
allylic
The key step i s hydride shift from a 2° carbocation to a 2° allyiic, resonance-stabilized carbocation, which can subsequently lose a proton to fonn a conjugated diene.
H
H
3 17
! B: H
H
H
H
H
1 5 -3 (You may wish to refer to problem 2-6.)
(a)
2 3/ " 1 C == C == C " "
H"" H
tsp
orbital picture
H
H
The central carbon atom makes two n bonds with two p orbitals. These p orbitals must necessari l y be perpendicul ar to each other, thereby forcing the groups on the ends of the allene system perpendicular.
(b)
/
H" Cl
" " c == C == C '" "
Cl
mirror
Cl
H
H
non-superimposable mirror images
i
'"
" C == C == C ' ' /
=
\ \\
H CI
enantiomers
1 5-4 Carbocation stabi lity depends on conj ugation (benzylic, allylic), then on degree of carbon carry i ng the positive charge.
H H I \ C-C + II \ H2C CH3 2°
� :�� Jl H
H
1°
more significant contributor
H
C I H
less significant contributor
H
H
H
H H equi valent to the first resonance form 3 18
H3C
Y'�
allyJic
...... H
H3C ,
�
():-= / ==: fv ·
+
two carbon electron deficient so nucleophi le can attack ei ther
�
+
C
�e
H
o •. I
C H2 C H 3
The most basic species in the reaction mixture will remove this proton . The oxygen of ethanol i s more basic than bromide Ion .
6� � (i
1 5 -6
I I YI i C
A gC I
+
�
H
:
C H CH'
'----f- OCH2CH3
H
0.
H
••
��
two c arbons are electron defic ient so nucleophi le c an attack either
H - O - CH 2 C 1 3
H3
�
+
H-O: ..
I
CH2CH3
?,H
319
H
• • '---+-+-
H
0. CH2CH3
CH,CH3
�I
HJ
'-l(
H
1 5 -7
'"' � H - Br ..
(
H H3C / \ /C == C
�
H 3C
H
\
("' 1 +
CH 2 - O 0 0- H
-
H20 H H3C / \ + /C - C
\\
electron deficient so nucleophile c an attack either
00 B r° :-0 0
(b)
� '"' 00
/ \ O0 -/C - C o \\ H/ H3C CH2
same carbocation as i n (a)
{
H - Br ...
�
-
H 20
..
H3C
CH2
0° B0 �0 o0
_
320
two c arbon s are electron deficient so nucleophile can attack either
�
1 5 - 7 con / (c)
H
.....
&{ I � n Br - Br
Br
H2C -
? ? +
-
H
H
.......
H
I II I
Br
allylic
H2C
H2C - C - C == CH2
+
-? ?
+
CH2
==
H
�
.. :Br :
Br
H
I )}
Br
H
:B�:
_
IIII
two carbons are e lectron deficient so nucleoph i le can attack either
Br
Br
H2C - C == C - CH2
H
H
H
II - I I } r f ( i � ) II O II ! I I II II
While the bromonium ion mechanism i s typical for isolated alkenes, the greater stabi lity of the resonance stabilized carbocation w i l l make it the lower energy intermediate for conj ugated systems. C�
(d)
H3C - C == C - CH2
I
H
H
Ag +
.. AgCl
allylic
+
H3C - C == C - CH2 H
H
� �
-
H3
. . • H2O .
H20 ..
+ H- �H ••
• •
?
H3C - C == C - CH2 H
+
v?
+
-
9.
H3C - C == C - CH2 H
_
H
H20 . .
+
OH
H3C - C - C == CH2 H
321
H
two carbons are e lectron defic ient so nucleophile can attack either
H3C - C - C - CH2 H
OH
H
H
H
H2
CH2
-
H
1 5-7 continued (e)
+
H3C -
� (? ?
?� A ? ? :; == CH2
-
H
_
H
g
two c arbons are e lectron defic ient so nucleophile can attack either
same carbocation as in (d) allylic
+
H3 C
H2
-
H
R
== CH2
--
H
? ? f
+
H
==
3C -
-
H
I +
H
}R
H2
,
H2
+
OH
OH
I
I
H3C - C - C == CH2
I
I
H
1 5 -8 (a)
Br
I
H2C - C - C == CH2
I
H
A
( b)
I
.....
H
H
Br
I
H2C - C == C - CH2
+
Br
I
H
+
H
20
H
H
I
I
B
H
� n / B r - Br
I
I
H
I
I
H3C - C == C - CH2
H
Br
Br
+
H . .
-
--
allylic • •
:Br : Br
I
Br
I
I
A
• •
Br
H2C - C - C == CH2 H
I
Br
I
I
+
two carbons are electron deficient so nucleophi le can attack either
H2C - C == C - CH2
B
H
-
:Br:
I
H
I
H
(c) The resonance form A + , which e ventuall y leads to product A , has positi ve charge on a 2° carbon and is a more significant resonance contributor than structure B+ . With greater positive charge on the 2° c arbon than on the 1 ° c arbon , we would expect bromide ion attack on the 2° carbon to have lower activation energy. Therefore, A must be the kinetic p ro d uc t . At higher temperature , however, the last step becomes reversible, and the stabi lity of the products becomes the dominant factor in determining product ratios. As B has a disubstituted alkene whereas A i s only monosubstituted, it is reasonable that B is the major, thermodynamic product at 60° C . 322
1 5 - 8 continued
1 0 } )
(d) At 60° C, ionization of A would lead to the same allylic carbocation as shown in (b), which would give the same product ratio as formation of A and B from butadiene. Br (Br I I - B rH2C - C - C == CH2
Br I H2C -
+
I
I
H
A
I
T T ==
allylic
H
_
Br
I
Br
I
I
A
H
10 %
I
H2C - C == C - CH2
+
I
H
Br
I
H2C - C - C == CH2
B
90 %
I
H
I
H
1 5 -9
(a)
nn
(b)
Br - Br
hv
---
�Q)
HB r +
2 Br -
eX 0: H
H
allylic
H
___
{� n Br - Br .
H
H
�---recycles in chain mechan i s m
Br - +
("Pr" i s the abbreviation for n-propyl , used below . )
NB S generates a low concentration of B r2
� �
a
N - Br
+
H - Br
+
CCl4 -
a 323
I
H
H
Br 15-10
CH2
H
H
:B�: Br
+
Br - Br
H
two carbons are electron deficient so nucleophile can attack eIther
1 5 - 1 0 continued
initiation
nn Br - Br
+
hv
propagation
Br ·
2 Br ·
---
pr -
�� -
�ri?
== CH2
HB ,
--.�
I -hexene
�
rec ycles in c hain mechanism
L--__
__________
+{
P, -
�J
��O
Br .
+
H
I
= CH'
� t---l. .... ..
H
I
Pr - C - C == CH2
I
P,
JJ
�� ) H
I
+
- �H
J
H
I
Pr - C == C - CH2 I Br
Br
The HBr generated i n the propagation s tep combines with NBS t o produce more B r2 ' continuing the chain mechanism. 1 5- 1 1
(a)
U
+
�
B,
Br
(b)
<)
(e)
benzylic radicals are even more stable than allylic
These are the major products from abstraction of a 2° al lylic H.
1 5 - 1 2 B oth halides generate the same allylic carbanion. +
Br
I
H3C - C == C - CH2
� �
�
MgBr
•
�
H3C - C = C - CH2
� �
M
1
1 5- 1 3
(a)
Mg ---l��
ether
MgBr I CH3CHCH3
+
CH3CH == CHCH2B r 324
elI,Br
H
+
I
H3C - C - C == CH2
I
H
I
H
1 5 - 1 3 continued dec - 5 -ene
H2C - CH2CH2 CH,
+
Thi s synthesis could also be performed sequentially.
F
o
Br - CH2
add one-half equivalent 1 5- 1 4
C O OCH3
o
o
CN
C O OCH3
!
C O OCH3 15-15
(b) CH) O +
(d )
O
CH30 OMe
(e)
+
O Me
CH ,
:(
o ,(
O +
NC +
NC
X
O Et
(c )
CN
(f)
CN
15-16
( a)
(b)
(C)
0
Co 0
6. . .
+
�
eN
<
0
+
0
0
CH3
I
. C O O CH3
.. ... . CHO
CH3
racemic mixture; wedge and dashed bonds show relative stereochemistry
325
1 5- 1 7 These structures show the alignment of diene and dienophi le i n the Diels-Alder transition state, leading to 1 ,4-orientation in (a) and 1 ,2-orientation in (b). (b) (a) :0:
this left structure is a VERY minor resonance contributor; however, it explains the orientation for the diene as carbon- l is more negative and carbon-2, a 3°C, is s lightly more positi ve because of methyl group stab i lization
1 5 - 1 8 For clarity, the bonds formed in the Diels-Alder reaction are shown in bold.
(b)
COOCH3
1 5 - 1 9 For a photochemically allowed process, one molecule must use an excited state in which an elec tron has been promoted to the first anti bonding orbital . All orbital interactions between the excited molecule's HOMO* and the other molecule's LUMO must be bonding for the interaction to be al lowed; otherwise, it is a forbidden process. HOMO * excited state o f diene HOMO
�J
anti bonding interaction (destructive overlap)
bonding interaction (constructive overlap
LUMO of dienophi le
In the Diels-Alder cycloaddition, the LUMO of the dienophile and the excited state of the HOMO of the diene ( l abeled HOMO * ) produce one bonding interaction and one anti bondin g i nteraction. Thus, this is a photochemically forbidden process.
326
15-20 For a [4 + 4] cyc\oaddition: (a)
Photochemical
Thermal HOMO
HOMO*
)
(
)
( bondi ng interaction
anti bondi ng i n teraction
both interactions are bonding
one interaction is anti bonding
� allowed
� forbidden
(b) A [4 + 4] cycloaddition is not thermal ly al lowed, but a [4 + 2] (Diels-Alder) i s ! HOMO
bonding
bonding '-interaction �
J� interaction 11:3
LUMO 1 5 -2 1
A
=
EC
l
A
E =
E =
6
x
c
1 0--6 moles I O mL
1 mg x l L
X
4 1 0- ) ( 1 )
=
833
I cm
1 000 mg
=
""
=
1 g
1 000 mL
x
0.50 (6
l
l
-
convert mass to moles:
C =
11:3
x
A
1 mole 1 60 g
--
4 6 x l 0-- M
8
327
=
=
0.50
6 x 1 0--6 moles
1 5-22 (a) 353 nm: a conj ugated tetraene-must have highest absorption maximum among these compounds ; (b) 3 1 3 nm: c losest to the bicyclic conj ugated triene in Table 1 5 -2; the diene i s in a more substituted ri ng, so it is not surprising for the maximum to be slightly higher than 304 nm; (c) 232 nm: simi l ar to 3 -methylenecyc lohexene in Table 1 5 -2; (d) 273 nm: 1 ,3 cyclohexadiene (256 nm) + 2 alkyl substituents (2 x 5 nm) = predicted value of 266 nm; (e) 237 nm: like 3 -methylenecyclohexene (232 nm) + 1 alkyl group (5 nm) = predicted value of 237 n m 1 5-23 Please refer to solution 1 -20, page 12 of this Solutions Manual . 1 5-24 (a) isolated
(b) conj ugated
b 0
3
4
5
1 5 -25 (a)
(d)
� � Br
(f)
Br
(g)
(d) conj ugated 1
W (b)
7
o-Cl
0 5
+
3
(e) conj ugated
�
(e)
"
OH
Br
�
Br
+
� Br
+
B r minornot conj ugated
�
Br
(h)
(i)
328
0
(f) cumulated ( 1 ,2) and conj ugated (2,4) H /C � H 2C = C = C " CH2 H (c)
one eq ui valent of HC l
Br
�
(c) cumulated
Br
+
Br
Br �
OH
15-26 Grignard reactions are performed in ether solvent. For c l arity, the bonds formed are shown in bold.
(a)
(b)
� Br + BrMg � I
N'
+
BrMg
r
•
5 4
15-27
(a)
+
�CH2""'''I--·· �?�CH2 ��H2 C H2
CH2
C H2
• • . ..1-....
(b)
(c)
·6
:0:
:0:
H
H
H
-........ C .....
:0
(d)
a:
o
H
O+I
H
H
329
:0 :
:0 :
:0:
:0:
H
'iC,
H
H
_ 2 H � &
15-27 continued
(e)
(f)
{
V
(g)
15-28
(a)
E=
A = Ee l
E
=
A
3.9 x 10-6 moles 100mL
cl 0.00 10 g x
x
0.74 '"
( 3.9 x 10-5 ) ( 1 )
H
__
(C yO: HC �
l = 1 cm
-
convert mass to moles:
c=
(!=.yO: H
1000mL lL
1 mole
=
255 g
= 3.9
A = 0.74
x
3.9 x 10-6 moles
10-5 M
El
(b) Thi s large value of E could only come from a conj ugated system, eliminating the first structure. The absorption maximum at 235 nm is most l i kely a diene rather than a triene. The most reasonable structure is: compare with:
co
"max = 235 nm
Solved Problem 15-3 330
"max
=
232 nm
Table 15-2
15-29
�
(a)
� trans
+
Br
(b)
+
Br
Br
� cis
CPr" is the abbreviation for n-propy l , used below . )
�
NB S generates a low concentration of Br2 o +
N- Br
I1-B r
--
o
�
o
N-H
+
Br-Br
o
initiation
nn
Br-Br
hv
2 B r·
---
propagation first step is abstraction of allylic hydrogen to generate allylic radical
I
I
H H +
Pr -C
",,--,r
Br·
==
CH2
C
-
HBr
---I"�
+
radical w i l l be a mixture of cis and trans
I-hexene
I
I
H H
Pr- C ==C- �H 2
n�)
Br - Br
�--
recycles in chain mechanism
0 �
Br·
+
I
I
H H I
Pr - C -C
==
CH2
I
+
Pr -C
Br
CIS +
15-30
(a)
(c )
(b )
COOH COOCH3
331
I
H H I
C-CH2
==
Br trans
,- k �
COOH
t
15-30 continued
0=
(e)
I
_
"
,
(f)
eN , "
""
"
",CN
",
""'eN
eN
'
(Note: Yield of the product in (f) would be small or zero due to severe steric interaction in the s-cis conformation of the diene. See Figure 1 5- 1 6 . ) 15-3 1 (a)
1f
H2C
�+ /
k
�
..
0(
more significant contributor: 2°
(b)
more significant contributor: 2°
(c)
(d)
i .. I
:0
{ {
+
�C
H
C - CH3
:0: 0(
• •
CH3-C-NH2
..
0
� H
==
:
II
-
:0:
C H3-C=NH2
(e)
H2 C
0(
I
H H
0(
t t
more significant contributor-negati ve charge on more electronegatIve
more significant contributorno charge separ ation 0
..
H
.
H
H
:0: .
�
� H2C H
:0: H
..
..
H
H
H
most sign i ficant contributornegative c h arge on more electronegati ve atom 332
H
atom
(f)
15-31 continued
most significant contributor---all atoms have ful l octets
..
(a) The absorption at 1630 c m migration of the double bond_
15-32
-
i
more significant contributor negative charge c an be stabilized by electronegati ve ch lorines
..
suggests a conj ugated alkene_ The higher temperature al lowed for
UCH,CH,CH] OCH,CH2CH]
(b)
desired
actual
(c ) expected:
+ [OCH,:f-CH2CH]] --{oCH, doubly allylic mlz 122
actual :
.
�
�
+
.6
mlz 122
(
15-33
(aj
(dj
(
+
�H
(b)
0
+
o
o
+
¢ o
(e)
o
+
CH2 C H3 �CH' } �
�
C'lL Cl
333
mass 43
mlz 79
mlz 93
+ •
+
mass 29
CI CI CI +Q-CI ::;:,- CI CI CI CI*
1 5-33 continued (g)
15-34 (a)
+ Co
0
I
NH
0
S�O CI CI CI + d' : CN CI tb ::;:,CI*CI I f NC 0
(i)
(h)
+
0
H
7
90°
� exo
H
(b) The endo isomer is usually preferred because of secondary p orbital overlap of in the transition state.
C, C,
C=O
C
with the diene
(c) The reasoning i n (b) applies to stabi lization of the transition state of the reaction , not the stability of the product. Arguments based on transition state stabi lity apply to the rate of reaction, i nferring that the endo product is the kinetic product. (d) At 25° the reaction c annot easily reverse, or at least not very rapidly. The endo product i s formed faster and is the maj or product because its transition state is lower i n energy-the reaction is under kinetic control. At 90° the reverse reaction is not as slow and equilibrium is achieved. The exo product is less crowded and therefore more stable-equilibrium control gives the exo as the maj or product.
t
,/
,. ,. . ,
�
,,
."
..
" -'., ,.
'.. � ... endo .
exo transition state-no secondary orbital overlap
.
transition statesecondary orbital overlap stabilization
,.
.,
.
...... ...
endo product
.. exo product-more �----reaction -
334
stable, less crowded
(a) 15-35 Nodes are represented by dashed l ines. 1t * 6
1ts*
1t4*
1t3
1t
(b)
-
-
()
1[6* 1[5*
H :: _H '" H 1[
1
()
(C)
()
0
0
'
0
335
2
1t
l
15-35 continued
(d) Whether the triene i s the HOM O and the alkene is the LUMO, or vice versa, the answer wi ll be the same . Thermal
Photochemical
HOM O*
LUM O
LUMO
one anti bonding interaction �
two bonding interactions �
forbidden
(e)
allowed
o +
o
J H2C== ?- ? == ? - CH2 1 H H H
15-36
(a)
?
?
?
H2C== - - ==CH2 .. H H H
.... ..f--. � ..
(b) Five atomic orbitals will generate five molecular orbitals. (c) The lowest energy molecular orbital has no nodes. Each higher molecular orbital will have one more node , so the fifth molecular orbital will have four nodes .
336
15-36 continued
(d) Nodes are represented by dashed l ines.
11:5*
11:4*
11:3
(non bonding)
11:2
11:1
(e)
__
_1
1L
JL
11: * 5 11:4* 11:3 n,
( f) The HOMO,
11:3,
contains an unpaired electron giving thi s species its radical
character. The HOMO is a non-bonding orbital with lobes only on carbons 1,
i�-�-�} 3 , and 5, consistent with the resonance picture.
n,
337
15-36 conti nued
(g)
--
---
H _H (h)
__
1£5*
Agai n , it is e lectron in
1£4*
1£3
1£3
that determines the character of this species. When the single of the neutral radical is removed, positive charge appears only in
the posi tion(s) which that electron occ upied. That i s , the posi ti ve charge depends
on the now empty 1£3' with empty lobes (positive charge) on carbons 1, 3, and 5,
1£3
consistent wi th the resonance description .
J� 1+
1£2
___
+
�
___
l +f
�
1£] 1£5*
Again, it is
1£3
that determines the character of this spec ies. The negative
charge depends on the filled 1£3' with lobes (negative charge) on carbons 1, 3,
i-·�-�-�;t and 5, consistent with the resonance description.
H H _H
a
A
Mg ..
ether
I
MgBr
H3 C-C-OH
CH3 (b) Use Appendix 3 to predict Amax values . Alkyl substituents are circ led .
transoid cyclic diene = 2 17 n m 4 alkyl groups = 20 n m exocyclic alkene = 5 n m TOTAL = 242 n m desired product
cisoid cyclic diene = 253 nm 3 alkyl groups = 15 nm TOTAL = 268 nm 338
c isoid cyclic diene = 253 11m 4 alkyl groups = 20 nm TOTAL = 273 nm--AHA! actual product B
15-37 continued
(c) hydride shift
B is produced in preference to the other ci soid diene above bec ause Bls diene system is more highly substituted and therefore more stable
�
Q
Hi) :
B
H3C-
? -H CH3
15-38 It is stunni ngly c le ver reactions like this that earned E. J. Corey his Nobel Prize.
�o oD Diels-Alder
1
+
COOCH3
CH3
---
endo product
imagine the diene on top
of the dienophile in the transition state, leading to the endo adduct
new
0= = 0
retro-Diels-Alder
C
o
� dienophi le
same as
o
339
o
COOCH3 amazing!
o
CHAPTER 16-AROMATIC COMPOUNDS
Note: The representation of benzene w ith a circle to represent the 1t system is fine for questions of nomenclature, properties, i somers, and reactions. For questions of mechanism or reacti vity, however, the representation with three alternating double bonds (the Kekule picture) is more informative. For c l arity and consi stency , thi s Solutions Manual will use the Kekule form exc lusively.
o
�
equivalent representati ons of henzene
Kekule form used i n the Solutions M anual
H
16- 1
H H
H .
C CH ' ' .. C · H
H '
H
• •
•
' -C .
C·
•
•
C. •
•
•
•
•
•
H
•
C· C
.
•
C,, . H .C. ' 'C . C· •
•
H
H
1 6-2 All values are per mole . benzene
(a)
-208 kJ
-240 kJ
1 ,4-cyclohexadiene MI
(b) -
(c)
-
7.6 kcal
- 49.8 kcal
cyclohexene
- 1 20 kJ
- 28.6 kcal
- 88 kJ
- 2 1 . 2 kcal
f.,}{
=
1 ,3-cyc lohexadiene
-232 kJ
- 55 .4 kcal
cyc lohexene
- 1 20 kJ
- 28.6 kcal
- 1 1 2 kJ
-26 . 8 kcal
f.,}{
.... C,..C.::::-C' H
H
I
C
H 'c-:::' 'c' I
1----
II
=
H
..... C, -:::.C .... H H C I
{
+
-208 kJ
H I
H
32 kJ
- 57.4 kcal
benzene
1 6-3 (a)
+
=
- 49. 8 kcal
II
..... C.::::- ,..C .... H H C I H
H
H "H "H "C-C C=C ---- II II I I C-C C=C , ,
H
"-
H
/
/
H
H
H
}
341
c::::::>
©
1 6-3 continued (b)
16-5 Figure 1 6-8 shows that the fi rst 3 pairs of electrons are in three bonding molecular orbitals of cyclooctatetraene. Electrons 7 and 8, however, are located in two different nonbonding orbitals . As in cyclobutadiene, a planar c yc looctatetraene is predicted to be a diradical , a particular l y unstable electron configuration.
Models show that the angles between p orbitals on adjacent
1l:
bonds approach 90°.
1 6-7 (a) nonaromatic: internal hydrogens prevent planarity (b) nonaromatic: not al l atoms in the ri ng have a p orbital (c) aromatic: [l4]annulene (d) aromatic: also a [ 1 4]annulene in the outer ring: the internal alkene is not part of the aromatic system 16-8 Azulene satisfies all the criteria for aromaticity, and it has a Huckel number of 1l: electrons: 10. Both heptalene ( 1 2 1[ electrons) and pentalene (8 1l: electrons) are anti aromatic . 1 6-9 (a)
1l:6* 1l:4
1[5 1l:2
(b )
1[7*
nonbonding �!1�!g.x
1[3
1l:6* 1l:4 1[2
1
1l:s*
H H H 1l:1
1
1l:7* 1l:s 1l:3
This electronic c onfiguration is antiaromatic.
342
16-9 continued (c )
These are TOP views. Nodes are shown by dashed li nes.
16- 10 ( aj
____
�y.
____
_
______ _
n2 *
�-----iII[
---
(b)
n2 *
n3*
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - non bonding nI
i s bonding;
n2 *
and
n3*
nI
are antibonding.
(c)
n,
��-------- �------�) \�------Y Y
nI
/
�------�
cation-aromatic
ani on-antiaromatic
343
16- 1 1
(a)
---
(b) n2
n5*
antibonding n3
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - _ .
R"
(c) 7l:z
\.
j
H
y
n1
("
R" n3
7l:z
)
\.
cation-antiaromatic
j�
non bonding
bonding
H
y
n1
rj "
anion-aromatic
16-12 (a) antiaromatic: 8 n electrons, not a Huckel number (b) aromatic : 10 n electrons, a Huckel number (c) aromatic: 18 n electrons, a Huckel n umber (d) antiaromatic: 20 n electrons, not a Huckel number (e) nonaromatic: no cyclic n system (f) aromatic : 18 n electrons, a Huckel number
344
n3
)
The reason r the dipolepicture can begives seenonein aring resonance distributing electrons give each ring electrons. This foresonance a negativeformcharge and thetheother ring a topositive charge.
1 6- 1 3 61t
tropylium ion electrons
aromatic 61t
cycIopentadienyl anion electrons
aromatic 61t
�
�
"\ several resonance f o rms several resonance f o rms delocaJizing thearound ___ delocalizing thethenegative ____ charge positive charge around thenngseven-membered membered ring five "'---- ----�
(
�---------
�----------
�--------�
�--------�
composite resonancewithpicture rings are aromatic chargeshows separationboth that
1 6- 1 4 +
AgBF4
AgCl
--
+
aromatic cyc\opropenium ion CI !KCI The crystalline material soluble / incycIopropenium polar organic solvents tetrafluoroborate. Cl-<Jl carbon.Draw0 resonance forms showing the0carbonyl polarization,leaving0 a positive charge on the carbonyl is
1 6- 1 5
& t
:0I : +C B electronsAROMATIC
21t
0
t .. :0:I
0
electronsAROMATIC
.:0:t. +CI
The three-and sevenmembered rings aromatic;the fivemembered and, reactive. 0 antiaromatic surprisingly,very electronsANTI AROMATIC
41t
61t
345
are
ring
is
nut
:(:
l6-l6
'I �oo o 0
7N 9
H I
N
I
N
1
0
)
0 0
N
3
0
The structure of purine shows two types of nitrogens. One type (N-l, N-3, and N-7 ) has an electronic structure like the nitrogen in pyridine; the pair of electrons is in an sp2 orbital planar with the ring. These electrons are available for bonding, and these three nitrogens are basic. The other type of nitrogen at N-9 h as an electronic structure l i ke the nitrogen of pyrrole; its electron pair is in a p orbital , perpendicular to the ring s ystem, and more importantly, an essential part of the aromatic pi system. With this pair of electrons, the pi system is aromatic and has 10 electrons, a Huckel number, so the electron pair is not avai lable for bondi ng and N-9 is not a basic nitrogen .
l6-l7 (a) The proton N MR o f benzene shows a single peak a t 8 7 . 2; alkene hydrogens absorb a t 84.5-6. The chemical shifts of 2-pyridone are more simi l ar to benzene's absorptions than they are to alkenes. It would be correct to infer that 2-pyridone is aromatic . (b)
H H H�H J:� . H --.. � x):� H N H
--..
""
0:
+
The lone pair of electrons on the nitrogen in the first resonance form is part of the cyclic pi system. The second resonance form shows three alternating double bonds with six electrons in the cyclic pi system, consistent with an aromatic electronic system.
H H x�.Hi�. -.... . Hn�. H H 0: 7
H
N
&7 . 26
0: 0 0
� ...
_
N
These resonance forms show that the hydrogens at positions with greater electron density are shielded, decreasing chemical shift.
These resonance forms show that the hydrogens at positions with greater positive charge are deshielded, increasing c hemical shift.
87.3l
:0 :
This resonance form of thymine shows a cyclic pi system w i th 6 pi electrons , consistent with an aromatic system. Four of these electrons c ame from the two lone pairs of electrons on nitrogens in the first resonance form.
346
1 6- 1 8 (a)
/
(..0 ;�:
(b)
(c )
\
:N �..S : aromatic: 4 n e lectrons in double bonds plus one pair from sulfur
aromatic: 4 n electrons in double bonds plus one pair from oxygen
6 6 .
(e)
0.
�
.o.
.
0
+
H 1 H � , B ........ � H �N N' •
1
•
I
� 9
� 10
anthracene
�
I
.
.
N H
� .... ..f. -I.
_
OCO �
I
(�)�6
� h-
0
.o.
.
.
+
aromati c : c ation on carbon4 indicates an empty p orbital ; two n bonds plus a pair of electrons from oxygen makes a 6 n electron system
sp3
H 1 H +, B ...... + H ..... N' N' 1 1_ 1 , B .;:::. , B ...... H H N + 1 H
, B ....... . , B ...... H H N 1 H 1 6-20 (a )
O
not aromatic : no cyclic n system 3 because of sp carbon
aromatic : resonance fonn shows "push-pul l " of electrons from one to the other, making a cyclic n system with 6 electrons
1 6- 19
d0
not aromatic : no cyclic n system 3 because of sp carbon
(f)
.-
H
(d) H
� --h-
347
aromatic: resonance form shows electron pair from N making a cyclic n system with 6 electrons
Borazole is a non-carbon equivalent of benzene. Each boron is hybridized in Its 2 2 nonnal sp . Each nitrogen is also sp with Its pair of electrons in its p orbital. The system has six
n
electrons in 6 p orbitals--aromatic!
� �
---
� �
1 6-20 continued (b)
anthracene H
n B r� B r �0 __ ----,..
©-c00 �cQo� I� H
..
. r. . .B
H
H
H
plus many other resonance forms
Br
�
�
Br H bromide attack at C-lO leaves two aromatic rings
cis and trans H bromide attack at C-9 leaves two aromatic rings
H
(c) A typical addition of bromine occurs with a bromonium ion intermedi ate which can gi ve only anti addition. Addition of bromine to phenanthrene, however, generates a free c arbocation because the carbocation is benzylic, stabi lized by resonance over two rings. In the second step of the mechanism, bromide nucleophi le can attack either side of the carbocation giving a mixture of cis and trans products. (d)
C
L1
-----
E1
1 6-2 1
l
6 C
chlorobenzene
�Cl �Cl Cl
1 ,2,3-trichlorobenzene
+
H
�'" ,
+
Br
H
HB r
·B r. •
•
--../
H
resonance-stabilized
� V Cl
q
Cl
o �
Cl
a-dic h lorobenzene ( 1 ,2-dichlorobenzene)
Cl
Cl p-dichlorobenzenc ( 1 ,4-dichlorobenzene)
Cl
Cl
Cl Cl
o
�
Cl
1 ,3,5-trichlorobenzene 348
¢
m-dichlorobenzene ( 1 ,3-dichlorobenzene)
Cl
Cl 1 ,2 ,4-tri c hlorobenzene
Cl
(y
Y
CI CI
CI 1,2,3 ,4-tetrachloro benzene
CI � CI CI V CI
CI
� CI
16-2 1 continued
CI Y CI
1 ,2,3 ,5-tetrachloro benzene
CI
1 ,2,4,5-tetrachloro benzene
16-22 (a) fluorobenzene (b) 4-phenylbut-I-yne (c) m-methylphenol , or 3-methylphenol (common name: m-cresol)
< }- O-CH3
CI
� CI
�
CI CI CI Y CI CI
¥ CI CI
1 ,2 ,3 ,4,5-pentach loro benzene
(d) o-ni trostyrene (e) p-bromobenzoic acid, or 4-bromobenzoic acid (f) isopropoxybenzene, or isopropyl phenyl ether
CH3 --<
CI
}- S03H
1 ,2 ,3,4,5,6-hexachloro benzene
(g) 3 ,4-dinitrophenol (h) benzyl ethyl ether, or benzoxyethane, or (ethoxymethyl)benzene, or a-ethoxytoluene
< }-
02N -{
}- OH
1 6-23 These examples are representati ve. Your examples may be different from these and sti l l be correct. (a)
(b)
p-toluenesulfonic acid
methyl phenyl ether or methoxybenzene or anisole
(e)
�CH-OH o-
<
(0
I� .0
diphenylmethanol 16-24
H
H
CH2 +
H
H H
(c)
phenyllithium
CH3CH20 (g)
f<> Br
H
+
..
..
H H
H
..
H ..
H
These are phenols. Thi s is 4-nitrophenol .
� >- CH,OH
Br
1 ,3-dibromo-2-phenylbenzene
H ...... C
(d)
Li
ii I � I I
H
3-ethoxybenzyl alcohol or 3-ethoxyphenylmethanol
H
H
H
..
..
H H
H
349
H
0-:O� 1
16-25
Ij_�
CH-CH ==CH2
Amax
H+
<
..
2 20 nm (strong) 258 nm (weak)
CH-CH ==CH2
i O-
CH=CH-
o�
t
1
--
H20 :
OH
250 nm (strong) 290 nm (weak)
<>
�
H20 :
CH ==CH-CH2
Amax
)+¥3� -H20
_
o-�
•
�
�H-CH=CH2
plus resonance fOnTIS with positive charge on the benzene ring
CH==CH-
?�2
H r c
-
electronic systems with extended conj ugation absorb at longer wavelength
N02 O
�
9 9 6
16-26 Please refer to sol ution 1 -20, page 12 of this Solutions Manual . 1 6- 27 (a)
(0
OH
OCH3
1.0
(b)
(g)
HO
1.0
1.0
CH O
CH30
CH2 OCH3
6 1.0
1
(; )
(h
(m)
Ih
¢ I� .0
CH3
350
CH3
NH2
(d)
9 � Br
(I)
C O OH
NH2
HC=CH2
HC=CH2
(k)
(e)
OCH3
HC=CH2
9 �
1.0
OCH3
}
(e)
.0
N02
H
C. Cl-
0
(j)
-
OCH3
CH3
H (n)
.0
O I .0 �
CH3 (0)
H I
c-
0 II
Cl
Na+
1 6-28 (a) 1 ,2-dichlorobenzene (ortho) (b) 4-nitroanisole (para) (c) 2,3-dibromobenzoic aci d (d) 2,7 -dimethoxynaphthalene 16-29
(e) 3-chlorobenzoic aci d (meta) (0 2 ,4,6-trichlorophenol (g) 2-sec-butylbenzaldehyde (ortho) (h ) cycIopropenium tetrafluoroborate
(5
�
q
CH3
llA
¢
o-xylene
toluene
CH l CH3
CHl CH3
CH3 1 ,2,4-tri methylbenzene
1 ,2,3-tri methylbenzene
£ .&
CHJ p-xylene
C H3
1 ,3,5-trimethylbenzene (common name: mesitylene)
1 6-30 Aromaticity is one of the strongest stabilizing forces in organic molecules. The c ycIopentadienyl system is stabilized in the anion form where it has 6 'It electrons, a Huckel number. The question then becomes: which of the four structures can lose a proton to become aromatic? 3 Whi le the first, third, and fourth structures can lose protons from sp carbons to give resonance stabilized anions, onl y the second structure can make a cycIopentadienide anion. It w i l l lose a proton most easily of these four structures which, by definition, means it is the strongest acid.
1 6-31
� H
H
(a)
H yly H H VH H
H
� �
H
AROMATIC
H H H
H
H
H
H
H
H
H H
H
351
H
1 6-3 1 continued (b)
(c)
6 6
u o
Br
Br
Jl U
+
Br
Br
Br
c::::::=>
+
¢ Br
Br
6 +
flZ); a:;; +
Br
Br
Cl �
a L6 Br
Br
Br
+
Br
\.
�
I
B'
+
B'
Y
�
I
J
different positions of double bonds
+
Br
(ignoring enantiomers) (d) The only structure consistent with three isomers of di bromobenzene is the prism structure, called Ladenburg benzene. It also gives no test for alkenes, consi stent with the behavior of benzene. (Kekule defended his structure by claiming that the "two" structures of ortho-dibromobenzene were rapidly interconverted, equilibrating so quickly that they could never be separated. ) (e) W e n o w know that three- and four-membered ri ngs are the least stable, but this fact was unknown to chemists duri ng the mid- 1 800s when the benzene controversy was raging. Ladenburg benzene has two three-membered rings and three four-membered rings (of which only four of the rings are independent), which we would predict to be unstable. (In fact, the structure has been s ynthesized. Called p rismane, it is NOT aromatic, but rather, is very reactive toward addition reactions. )
352
1 6-32
H
H
A
(a)
H H nonaromatic
(b)
0 U
(c)
N H I
aromatic6 11: electrons
aromatic6 11: elec trons
H
nonaromatic
(d)
N
o
o
0
H
anti aromatic4 11: electrons
o
nonaromatic
1+
H
H
o
o
1-
C
aromatic6 11: e lectrons
[) 0
0
aromatic6 11: electrons
(g)
ill W
antiaromatic12 11: electrons
(B is Sp 2 but donates no e lectrons to the 11: system)
o + N I
H nonaromatic
H aromatic6 11: electrons
(00]0 o
(::] o
o
0
-c
+c
I
I
H 2 if oxygen is sp => anti aromatic (8 11: e lectrons ) ; if oxygen is sp3 => nonaromatic =>
(t)
o
. . antJaromatlc4 11: electrons
I
o
o
nonaromatic
I
N
aromatic6 11: electrons
H
H
B
o
° °
nonaromatic
aromatic6 11: electrons
anti aromatic4 11: electrons
N
(e)
+"N'
C
N
aromatic6 11: electrons
o
aromatic2 11: electrons H
H
aromatic6 11: electrons
not aromatic in either case
+\
� oo H -N/' N- H
I
aromatic6 11: electrons
nonaromatic H
this i s a tough call-it has 10 11: electrons s o could be aromatic , but internal H's might force it out of planarity
aromatic10 11: electrons 353
it
1 6-32 continued (h) H H
+
0
H
H
C
0
0
nonaromatic
2
I
B
C-
anti aromatic8 1t electrons
aromatic6 1t electrons
(B is sp , but
CH3
I
I
donates no electrons to the 1t system)
0
aromatic6 1t electrons
1 6-33 The c lue to azulene i s recognition of the five- and seven-membered rings. To attain aromaticity, a seven-membered c arbon ring must have a positive charge; a five-membered c arbon ring must have a negati ve charge . Drawing a resonance form of azulene shows thi s :
OJ ---. C» -..
--..
--�
ri:�
ance s
} <0@
composite picture
The composite picture shows that the negative charge is concentrated in the five-membered ring, giving rise to the dipole. 16-34 Whether a n itrogen i s strongly basic or weakly basic depends on the location of i ts electron pair. If the electron pai r i s needed for an aromatic 1t system, the nitrogen w i l l not be basic (shown here as "weak 2 3 base" ) . If the electron pair i s i n ei ther an sp or sp orbital, i t i s avai lable for bonding, and the nitrogen is a " strong base" . H (a)
I
�
HN:
"
\
weak base
I
N:
I' strong base
ce·�'\ I
strong base
(c)
(f)
() I
(e)
0 strong base
strong base
�
\
O
N:
I strong base
H
weak
�
/N
0 N
strong base
N / base
H
(d)
(b)
/ • •
N
strong base
g1 weak base
1 6-35 Where a resonance form demonstrates aromaticity, the resonance form is shown. (al
1"-': O +R /.
aromatic
(b) : 0 :
6 +
�
aromatic
:6
(c)
..
:0:
aromatic
(d) B AD
+
0 :0 :
:0 :
NOT aromatic-although it can be drawn , the resonance form o n the left i s NOT a significant contributor because the oxygen does not have a ful l octet; the form on the right shows the correct polarization of the carbony l , but it's still not aromatic because of only 4 1t electrons 354
(e)
0 :0:
aromatic
16-35 continued
�+
�+
+ �,..... not basic
Nitrogens whose e lectrons are needed to complete the aromatic
(D
(g)
o
aromatic , not basic
(k)
•
0..
• • •
-
(I)
(::] . o.
�H
o
6
C
NOT aromatic 3 because of sp carbon ; both N are basic , although the top one is conj ugated, lowering its basicity
O
23
..
..
_ .. ---
-"6
nn Br- Br
·
hv
-
6
CH 2
CH 2
..
..
..
-
2 B r-
6·:·
-
C H2
..
propagation steps on the next page 355
-
NH2
-
N H
a, i C
O:
aromatic ; bottom N is not basic because it has donated its electron pair to make the ri ng aromalic
o
..
,..... basic
tf.: • •
+
+
-
initiation
ic
+
..
-.. �-
"
NOT aromatic: if 0 is sp 2 , then the pi system has 8 e-
NOT aromatic : i f both N and 0 are sp 2 , then the pi system has 8 e- ; N will be basic
aromatic; not basic
3 sp
aromatic
(ml
(j)
�,..... basic
system w i l l not be basic .
C� �
(i )
iC
-
H N
B H
(b)
•
c;):; •
aromatic; N i s not basic but the 0 is a weak base
[j
(a)
9 :0 :
+�
16-36
(h)
1t
(5
1 6-36 continued HEr
propagation
6
+
resonance-stabilized
� r-·CH2
n!
B r - Br
+
1.&
----l.. �
B r·
+
(c) Both reactions are S N2 on primary carbons , but the one at the benzylic carbon occurs faster. In the transition state of S N2, as the nuc leophi le is approaching the carbon and the leaving group is departmg, the electron density resembles that of a p orbital . As such, it can be stabi lized through overlap with the 1'[ 8system of the benzene ring. : Br : • •
H
�
stabi li zation through overl ap
1 6-37 (a)
VlCH�
8-
:O-CH3 • •
+
3 isomers
only 1 isomer
(c ) The original compound had to h ave been meta-dibromobenzene as this is the onl y dibromo isomer that gives three mononitrated products . Br Br Br
�
VlBr
�
�N02 o Vl y +
Br
Br
16-38 (a) The fonnula CgH70C I h as five e lements of unsaturation, probably a benzene ring (4) plus either a double bond or a ring. The IR suggests a conj ugated carbony l at 1 690 c m-1 and an aromatic ring at 1602 cm-I. The NMR shows a total of five aromatic protons, indicating a monosubstituted benzene. singlet at 8 4.7 is a deshielded methylene.
<>
o
+
+
I I
C
CI
}
o-� -
A 2H
C - CH2
o
I
CI
(b) The mass spectral evidence of molecular ion peaks of 1 : 1 i ntensity at 1 84 and 1 86 shows the presence of a bromine atom. The rnlz 1 84 minus 79 for bromine gi ves a mass of 1 05 for the rest of the molecule, which i s about a benzene ring plus two carbons and a few hydrogens . The NMR shows four aromatic hydrogens in a typical para pattern (two doublets), indicati ng a para-disubstituted benzene. The 2H quartet and 3H triplet are characteristic of an ethyl group.
-0-
1 6-39 (a)
+
Br
}
+
� � � �
like the ends of a conjugated diene
Br
-o-
CH2 CH3
Diels·Al der product o new sigma bonds shown in bold
16 40 (a) No, biphenyl is not fused. The rings must share two atoms to be labeled "fused". (b) There are 1 2 n electrons i n biphenyl compared with 10 for naphthalene. (c) B i phenyl has 6 "double bonds ". An i solated alkene releases 1 20 kJ/mole upon h ydrogenation . -
predicted: 6
x
1 20 kJ/mole (28 .6 kcaVmole)
""
observed:
720 kJ/mole ( 1 7 2 kcaUmole) 4 1 8 kJ/mole ( 1 00 kcal/mole)
resonance energy:
302 kJ/mole (72 kcaVmole)
(d) On a "per ring" basis, biphenyl is 302 -:- 2 = 1 5 1 kJ/mole, the same as the value for benzene. Naphthalene's resonance energy is 252 kJ/mole (60 kcaVmole); on a "per ring" basis, naphthalene has only 1 26 kJ/mole of stabilization per ring. This i s consistent with the greater reacti v i ty of naphthalene compared with benzene. In fact, the more fused rings, the lower the resonance energy per ring, and the more reactive the compound. (Refer to Problem 1 6-20. ) 3 2 16-4 1 Two protons are removed from sp carbons to make sp c arbons and to generate a n system with to n electrons. H
H 357
1 6-42
(a)
(c)
�H
l:J"
V- Cl ""--
0
H
CH
aromatic
Ag+
� CH
r>-- +
..
(d)
aromatic
H
(e)
O
�C-H
1-
�::: C
aromatic
r:}
NaOH .
-H
aromatic
H
H
O C
..
C4 H9Li
(f)
comatk
0
Y
o
Ag+
Cl
..
+
()-o • •
• •
0
H
aromatic
1 6-43 These four bases can be aromatic, partially aromatic, or aromatic in a tautomeric form. In other words, aromaticity plays an i mportant role in the chemistry of all four structures. (Only electron pairs involved in the important resonance are shown . ) For part (b) , nitrogens that are basic are denoted by B. Those that are not basic are shown as NB. Note that some nitrogens change depending on the tautomeric form. (a) and (c)
cytosine
aromatic to the extent that this resonance form contributes
..
:0:
{ � � �
B H /
N
NB I
r 7�..:_ l_� :0:
O.
•
• •
H
uracil
continued on the next page
•
+
I
N
NB
O·
• • •
H
aromatic to the extent that this resonance form contributes 358
aromatic
OH
r �� N
BN
B
OH
tautomer aromatic
16-43 continued
�)
H '�
·
•
0
..
aroratic
·
•
N
'
H,NB �BN Jl- N NB
...
. _
__
2
I I �N
BN
B
adenine 16-44 (a) Antiaromatic-on ly 4
n
·.
N
\
H
tautomer ful l y aromatic
aromatic to the extent that this resonance form contributes
guamne
:):.
0
N
\ H
B NH
��
HN� . H,NB"/�B Jl- NB ·.
N
B
'\\ .
H
ful l y aromatic
N NB \/
electrons.
(b) This molecule is electronically equi valent to cyc lobutadiene. Cyclobutadiene is unstable and undergoes a Diels-Alder reaction with another molecule of itself. The t-butyl groups pre vent dimerization by blocking approach of any other molecule. (c) Yes , the n itrogen should be basic. The pair of electrons on the nitrogen i s i n an sp 2 orbital and is not part of the n system. a
b'
'
...
I
NMR,
2
Analysis of structure I shows the three t-butyl groups in unique environments i n relation to the nitrogen. We would expect three different s ignals in the as is observed at _1 1 00 C. Why do signals coalesce as the temperature is increased? Two of the t-butyl groups become equivalent-which two? Most likely, they are a and c that become equi valent as they are symmetric around the n itrogen. But they are not equi valent in structure I-what i s happening here? What must happen is an equilibration between struc tures I and 2, very slow at -1100 C, but very fast at room temperature, faster than the NMR can differentiate. So the signal w h ic h h as coalesced is an average of ' a and a and c and c'. (Thi s type of low temperature NMR experiment is also used to differentiate axi al and equatorial hydrogens on a c yclohexane. ) The NMR data prove that I i s not aromatic , and that I and 2 are i somers, not resonance forms. I f I were aromatic , then a and c would have i dentical NMR signals at all temperatures. 359
is loss of methyl. Mass spectrum: Molecular ion at base peak at Infrared spectrum: The broad peak at cm-l is OH; thymol must be an alcohol. The peak at cm-l suggests an aromatic compound. spectrum: The singlet at is OH; it disappears upon shaking with D20. The 6H doublet at and the IH multiplet at are an isopropyl group, apparently on the benzene ring. H singlet at is a methyl group, also on the benzene ring. Analysis of the aromatic protons suggests the substitution pattern. The three aromatic hydrogens confinn that there are three substituents. The singlet at is a proton between two substituents (no neighboring H's). The doublets at and are ortho hydrogens, splitting each other.
1 6-45
ISO;
135, M
-
15,
3 500
NMR
(') 3 .2
�Y Y
(') 1 .2
(') 4.8
(') 2.3
A3
(5 6 . 75
X
1620
(5 6.5
(5 7. 1
CH]
+
Several isomeric combinations are consistent with the spectra (although the single H giving suggests that either or is the OH group-an OH on a benzene ring shields hydrogens ortho to it, moving them upfield). The structure of thymol is: Z
8 6.5
Z
thymol
¢r:;�
CH3 The final question is how the molecule fragments in the mass spectrometer: H H CH3 'CI CH3 I + OH OH +OH .. I hI hCH3 � ¢r CH3 CH3 CH3 mlz resonance stabilization of this benzylic cation includes fonns with positive charge on three ring carbons and on oxygen (shown) 'c
..
..
..
135
360
+
•
Mass spectrum: Molecular ion at two prominent peaks are (as we (loss of methyl) and shall see, most likely the loss of acetyl, CH3CO). Infrared spectrum: The two most significant peaks are at cm-1 (conjugated carbonyl) and cm-I (aromatic C=C). spectrum: A 3H singlet at is methyl next to a carbonyl, shifted slightly downfield by an aromatic ring. The other signals are seven aromatic protons. The IH at d is a deshielded proton next to a carbonyl. Since there is only one, the carbonyl can have only one neighboring hydrogen. Conclusions: rnJz including H mass for carbons C The fragment C1oH7 is almost certainly a naphthalene. The correct isomer (box) is indicated by the This isomer would would II have two deshielded C-CH3 protons in the
1 6 -46
M
1 70 ;
-
15
M
1680
-
43
1600
8 2.7
NMR
8.7
+
1 27
7
=}
1 20
=}
10
NMR.
o
NMR.
Although all carbons in hexahelicene are sp2, the molecule is not flat. Because of the curvature of system, one end of the molecule has to sit on top of the other end-the carbons and the hydrogens would bump into each other if they tried to occupy the same plane. In other words, the molecule is the beginmng of a spiral. An "upward" spiral is the nonsuperimposable mirror image of a "downward" spiral, so the molecule is chiral and therefore optically active. The magnitude of the optical rotation is extraordinary: it is one of the largest rotations ever recorded. In general, alkanes have small rotations and aromatic compounds have large rotations, so it is reasonable to expect that it is the interaction of plane-polarized light (electromagnetic radiation) with the electrons in the twisted pi system (which can also be considered as having wave properties) that causes this enormous rotation.
16-47
the nng
three-dimensional picture oftwist hexahelicene showing the in the system of six rings
361
The key concept in parts (a)-(c) is that an aromatic product is created. (a) +
16-48
5.:\
U
o· ----- 0 I
:O-H
+ H+ �
I
:O-H
..
a c
+ -. · · 0
..
:O-H
..
..
The protonated gives a resonance-stabilized Protonation of the singly bonded oxygen carbonyl does not generate a resonance-stabilizedcation. product. A
+
6� 6·
I
:O-H
I h.
+ H+
----
:O-H
o
0
c
..
..
I
:O-H
c
·
..
..
The last resonance formnonaromatic shows that theion.cation produced is aromatictheandproducts therefore more stable than the corresponding In this second reaction, are more than in the first reaction which is interpreted as the reactant being more basic than favored (b) +0 B
er (r ••
CI ".......-A
CI- +
C
Ih••
CI ".......-A
O o
D
H
O o
CI- +
+
..
..
+
H
..
..
AROMATIC
H
B
O +0
O Ih-
A.
H
The product from ionizationbutofis alsois stabilized by resonance. The stable. ionizationreaction product that of is not only resonance-stabilized aromatic and therefore more produces astate moreleading stable toproduct will usually fasterleading under milder conditions transition that product will behappen stabilized, to a lower activationbecause energy.the C
362
AROMATIC
A
D
(e)
continued
u
16-48
GH
H+
E
H+
d
( )
catalyst catalyst
F
HO
o
Dehydration of that produces anstable aromatic product is more than the producta more fromstable product reaction that produces will usually happenbecause faster under milder conditions the product transition state leading to that will be stabilized, activation energy. leading to a lower
o . .1)- . :0 +
H20
AROMATIC
� base ---
V phenol
HO
umbelliferone
(Umbelliferone is one of about coumarins isolated from plants, primarily the families of
.
·0
:0
Resonance stabilization ofcharge the phenoxide anion shows the negative distributed over the one para and two ortho carbons .
base --1000
Resonance stabilization of the anion ofsame umbelliferone gives not only the three forms as the phenoxide anion, but informaddition, gives an extra resonance with ( -) charge on a carbon, andcontributor, the most significant resonance another form charge on the other with the () carbonyl oxygen. This phenoxide anion is much more stable than which weof the interpret as material, enhanced acidity starting umbelliferone. In fact,while the phenol pKa ofis umbelliferone is about
.
.
:0
Angiosperms: Fabaceae, Asteraceae, Apiaceae, and Rutaceae.)
10.
:0
7.7
363
F
E. A
Humulon (sometimes spelledathumulone), even though highly resonance stabilized, cannot be A aromatic because the carbon shown the bottom of the ring is tetrahedral and must be sp3 hybridized. ring is aromatic only when all of the atoms in the ring have a p orbital which the sp3 carbon does not. Areyouyoulovefamiliar withownthegood? conceptThe"tough love",is that that sometimes is, sometimesto demonstrate you have to one be stem with the someone for their concept emotion, behavior has to atoms appearlike exactly the and opposite. Granted, thisconflicting is a stretcheffects: to applythey it tocanatoms, but theelectron point that sometimes oxygen nitrogen can have withdraw density bydensity their strong electronegativity (an inductive effect) but atwill the same time, they can donate electron through their resonance effect. This phenomenon prove important in the reactivity of substituted benzenes described in Chapter :0 : N '" pyridine N-oxide pyridine I '" I 16-49
1 6-50
is
yr 4
""":::3
17.
98.60
2
yr�+ • •
4
9 7.25
98.19
""":::3
2
9 7.40
Nitrogen is electronegative, so anit exerts a deshielding effect onit is harder in pyridine. Thewhyeffect diminishes with distance as expected with inductive effect, although to explain H-4 is deshielded more than pyridinethatN-oxide,with an even more electronegative oxygen attached to theweN,can it would be reasonable toonIn expect the hydrogens would be deshielded. This is the case with H-3; infer that the effect H-3 is purely an inductive effect. Evenmust morereflect interesting is thesideshielding effects personality, on andthe donation these chemical shiftsdensity are shifted upfield. This the other of oxygen' of electron through a resonance effect. Drawing the resonance forms clearly shows that the electron density at and (and presumably by resonance, perfectinductive agreementeffect. with the results. As we will seeincreases in Chapterthroughin this mostdonation cases, resonance effectintrumps 9 7.32
9 7.64
H-2
H-3.
H-2
H-2
H-6)
17,
:0. . :
'"
:0 :
+ 11
:0 :
C
¢en QtH I
......:::
9
H-4;
--- I
9
.
......:::
9
.
9
+ II
---
(�J(H I
_C
9
9
H-4 NMR
:0 :
HC�
�¢:H
--- . I �
9
9
Note:example, The practice ofthatapplying human emotions to itinanimate objects iselecctrons. called "anthropomorphizing". For we say an atom is "happy" when has a full octet of In casualfor example, conversation, this gets a point across, but it is not appropriate in rigorous scientific terms, exams. conditons, expected to use the specific terms of science because they are well Under definedmore and formal do not permit sloppywe orarefuzzy concepts. As our equipment technician says: "Don't anthropomorphize computers. They hate that."
on
364
The representation of benzene with a circle to represent the system is fine for questions of nomenclature, properties, isomers, and reactions. For questions of mechanism or however, the representation with three alternating double bonds (the Kekule form picture) is more CHAPTER 17-REACTIONS OF AROMATIC COMPOUNDS 1t
clarity and consistency, this Solutions Manual will use the Kekule
reactivity, informative For exclusively.
1 7- 1
Oc:�
� C I O
H
H
� :!=!H2
I
addition product Sigma H NOT AROMATIC complex While theis addition of waterThus,to theit hassigma complex can be shown in a reasonable mechanism, the product not aromatic. lost the kllmol kcal/mole) of resonance stabilization energy. The addition reaction is not favorable energetically, and substitution prevails. H
1 7 -2
: CI-Cl:
CI AI-Cl CI
I
+
H
+
H
H H
HCI
H H +
H
�
.
CI
H
---
+•
OH
(36
.
+•
•
-----..
:CI: AICI3 t CI :CI-AI-C) :U I CI .
/ "
CI : CI -CI - Al -CI CI .
CII :..CI-CI v·- Al -Cl CII .
H
H
H
)
152
�I
H H
/ ,,+
•
1
1-
CII:Cl-AI-CI CII CI
.
..
+
H H
+
H
H ..
t
H
H
..
H
H
1-
.
365
Cl
C +' H
Cl H
H
1 7-3
Like most heavy metals, thallium is highly toxic and should not be used on breakfast cereal. � �====..
1 7-4
+
Tl(OCOCF3h
:0:
CH3
I gD
:
CF3COO-
I
.. --.. . V H C H3
CH3
N+
CH3
+
H
----
--
N02
CH3
Benzene's sigma complex has positive charge on three carbons. The sigma complex above shows positive charge in one resonance form on a 3° carbon, lending greater stabilization to this sigma complex. The morewillstable the intermediate, the lower the activation energy required to reach it, and the faster reaction 2°
3°
2°
the
be.
1 7 -5
delocalization of the positive charge on the ring
delocalization of the negative charge on the sulfonate group :0: :0: :0: O ==S-Ar :O - ISO == S- Ar I :0: :0: : 0:
•• I •
.
Ar
.......I--J�-
• • II .
•• •• II
. .
----
.
366
("Ar" is the general abbreviation for"aryl" an "aromatic" or group, benzene;abbreviation is the for general angroup "ali.)phatic" or "alkyl" in this case, "R"
1 7-6 (a) The key to electrophilic aromatic substitution lies in the stabi l ity of the sigma complex. When the electrophile bonds at ortho or para positions of ethylbenzene, the positive charge is shared by the 3D carbon with the ethyl group. B onding of the electrophile at the meta position lends no particular advantage because the positive charge in the sigma complex is never adj acent to, and therefore never stabilized by, the eth yl group.
�I
ortho
CH 2 CH3
�
;r - Fe B r3
�
B q-
{
-- a:
CH 2 CH3
�
�
CH +
�
CH 2 CH3 I +C
Q
Br
H
3 D-good
(b) Electrophilic attack on p-xylene gives an intermediate in which only one of the three resonance forms I S stabi lized by a substituent (see the solution to Problem 1 7 -4). m-Xylene, however, i s stabil i zed in two of its three resonance forms. A more stable intermedi ate gives a faster reaction .
1 7-7 For ortho and para attack , the positive charge in the sigma complex can be shared by resonance with the vinyl group. Thi s c annot happen with meta attack because the positive charge i s never adj acent to the vinyl group. (Ortho attack is show n ; para attack gi ves an intermediate with positive charge on the same carbon s . )
�
367
&: }
"extra " resonance form
1 7-8 Attack at only ortho and para positions (not meta) places the positive charge on the carbon with the ethoxy group, where the ethoxy group can stabi lize the positive charge by resonance donation of a lone pair of electrons. (Ortho attack is show n ; para attack gives a simi l ar intermediate . )
"extra " resonance form
1 7 -9 O IthO
c 0 NH & 2 : H B,
�
H H
H�
�
�
Br 6 NH2 NH2 Q Q Br
mCla
�
�
�
H
Br
�
�
H
17-10
Br
6
o
+
�
Br6
�
H "extra " resonance form
H
OCH(CH3 h +
NH2 ' B Q H
+
C
�
Br
-H
"extra " resonance form
H H
+ 1
H
V
�
H
H
para
+ C1
B'
?'
' B � ll) +
NH2
8"
Br2 (X' ___
"
Br
+
¢ Br
H
H
HB r (g)
Br
S ubsti tution generates HBr w hereas the addition does not. If the reaction is performed i n an organic solvent, bubbles of HB r can be observed, and HB r gas escaping into moist air will generate a c loud. If the reaction is performed in water, then adding moi st li tmus paper to test for acid w i l l differentiate the results of the two compounds. 368
1 7- 1 1 (a) Nitration i s performed with n itric aci d and a sulfuric acid catalyst. In strong acid, amines in general, including ani line, are protonated. +
o
(b) The NH 2 group i s a strongly activating ortho,para-director. In acid, however, i t exists as the protonated ammonium ion-a strongly deactivating meta-director. The strongly acidic nitrating mixture itself forces the reaction to be slower. (c) The acetyl group removes some of the electron density from the nitrogen, making it much less basic; the nitrogen of this amide is not protonated under the reaction conditions . The N retains enough electron density to share with the benzene ring, so the NHCOCH3 group is sti l l an activating ortho,para-director, though weaker than NH2 . :0: :0: + I II .. � Ph Ph C - CH3 - N = C - CH3 • •
I
H
H
1 7- 1 2 Ni tronium ion attack at the ortho and para positions places positive charge on the carbon adjacent to the bromine, allowing resonance stabilization by an unshared electron pai r from the bromine. Meta attack does not gi ve a stabilized intermedi ate. ortho + Br: Br :Br : Br I N02 N02 NO l 02 +C
a ::::-...
H CH +
..
..
a
HC +
h
H ..
t) .. .. ::::-... ::::-... G .. Q .. Q .. ..
meta
Br
Br
H
N 02
N 02
H
para
:Br : I +C
Br
H
N0 2
..
H
..
H
•
.. (J
H
N0 2
..
+ Br :
Q
H .
e Br
( C I
h
H
"extra " resonance form
N0 2
H H
.. ..
H N 02 "extra " resonance form
369
•
..
QH Br
H
N0 2
Cl H (b) O; H Ocd a �:- a� : } : �)� d !r H H 1 H
(a) O= !r 17- 13
atom can stabilize positive charge by sharing a pair of electrons. Bromine can this (c ) bromine cationic species, whether from electrophilic addition or electrophilic aromatic substi tution. resonance-stabilized
do
A
(a)
1 7- 1 4
(b)
(c) (e)
N02
+
Br
(x N02 �
OH
~CH3 �
CH3 CH3 (r 'Q :::::-.. Cl trace Cl COOHBr
02 N
+
O2N
+
02 N
N 02
(a)
1 7- 1 5
h ��"g VCH3
+
I
:::::-.. O
OH '(\ :::::-.. CH3 I
+
N 02
�
OH (x trace
�
o,p-d'''''"'
weak o,p-director
370
CH3
N 02
(d)
COOH
~OCH3 �
N02
in any
1 7- 1 5 continued (b)
� Y
02N + N0 2
N02
(c)
OH
n � I
+
N 02
V N02 trace
strong m-director
NO'
Jy
(d)
OCH3
q'I �
I I
N0 2
trace
C - CH3
0 NH -
strong o,p-director
CH3
o
strong o,p-dlrector
V N 02
N0 2
-
(f)
N02
strong o,p-director
C l weak o,p-director
(e)
Jy
Cl weak o,p-director
Cl
Cl
+
weak o,p-director
h
° 2N •
strong m-director
•
CH3 - C - NH
strong o,p-director
'-----y---' acti vating
1/ I
\\ \
-
"
0 C - NH 2
�
strong m-director
deacti vating
1 7- 16 (a) Sigma complex of ortho attack-the phenyl substituent stabilizes positive charge by resonance : H
C + I
I E
E
H
____
H
____
Para attack gives similar stabilization. Meta attack does not permit del ocalization of the positive charge on the phenyl substituent. 371
H
1 7 - 1 6 continued (b)
< ) < }- N02
(i)
(ii)
02 N
\>
trace
OH
�)
+
+
\
(v)
f
N02
\ }- N02
� N02 \ > °2N) > 02 Q-o-
+
N02
OH
OH
+
N
o
(i v)
�>
\> \
02N -{ > < }- C -
(iii )
\>
°2 N
CH3
+
� II
� II
°I I
C - CH3
°2 N
\
f0 � > 2
N
minor
maj or (mi nor amounts of ni tration on the outer rings)
�H
17-17 (a)
f �
��):
+
Al C I4-
AlCl3
-:?
:¢�
..
+
Cc�-n •
372
..
}
1 7- 1 7 continued
+
•
•
CH3 - Cl - AlCI3
v·
B
C1I3
l: a:] ��
..;...-. .., � . . .... H
para I somer also formed by simi l ar mechanism
CH3 :ci : � AICl3 I CH 3 - C - CHCH 3 • I CH3
I
(c )
OCH3
5�H
3
I� CH 3 - C - CHCH 3 I + 2° CH
.. : OCH3 I +c CH3
A lCI 4-
CH3
. ·� .... ..----l
O
1
..
H
5 ..
•
I
CH]
�
methyl 3° CH3 + I shift ---l.� CH3 - C - CHCH3 I CH 3
CH3 + I CH3 - C - CHCH3 I CH3
+
:CI :
\ �� �I
CH(CH3h _
/
CH3
O n l y a small amount of the ortho isomer might be produced as steric interactions will discourage thi s approach path of the electrophi le.
CH3
1 7- 1 8 (a J
(b)
ex : C H3
CH3 - � - OH I
CH3
Ct o o H
+
HF
+
H
--
C. + H
BF3 �
373
-c
u
CH
CH 3
I
CH3
If
�
-
H
1 7 - 1 8 continued (c)
(d) HO - CHCH3 I
+
+
B F3
CHCH3
--
I
CH3
CH3
1 7 - 1 9 In (a), (b), and (d), the electrophi le has rearranged. (a)
-c
U
CH
CH 3
I
CH3
(b) (c) (d) (e)
r; � _
~
0
0
(excess)
6
C(CH3h
(CH3hCCI •
AICI3
(excess) (b)
plus over-alkylation products
gives desired product gi ves desired product, plus ortho i somer; use excess bromobenzene to avoid overalkylation gives desired product, plus ortho i somer gi ves desired product
1 7-2 1 (a)
(c) No reaction : nitrobenzene is too deactivated for the Fri edel Crafts reaction to succeed.
(b)
CH3I •
AICI3
6
C(CH3 h
HN03 •
H 2SO 4
S03 •
H 2SO 4
¢
S03H
374
¢
(separate from ortho, although steric hindrance would prevent much ortho substitution)
N0 2
(separate from ortho)
1 7-2 1 continued (c )
0
•
AlCl3
(excess)
°
1 7-22 (a)
(b) (e )
(d)
0
+
� I
0 0
Q
o ?'
� I
+
(f)
0 ?'
� I
+
II
Cl 2
°
I I
I I
Cl - C
...
II
Cl - C - C(CH3h
•
I I
1 ) AlCl3 •
2) H2 0
H3C0
1 ) AlCl3
-o-
•
2) H20
II
C l - C - C H 2CH2 CH3
C - C(CH3h
2) H2 0
� II
AlCliCuCl
°
� I
..
2) H2O
1 ) AlCl3
-0
CO, HCl
°
1 ) AlCl3
C - CH 2CH(CH)),
II
I I
+
a ° � o ° a c 'Q Cl
C l - C - C(CH3h
0
¢°
(separate from artha)
•
AlCl3
Cl - C - CH 2 CH(CH3h
+
OCH3
(e)
2)
CH3I
CHO
° a ?'
II
C - C(CH 3 ) 3
� I
°c a
Zn(Hg) ' HCl
•
II
..
I ) AICl)
2) H 2 0
�
I
375
- CH,Cll,CH) •
Zn(Hg), HCl
a � I
CH2c(C H)))
a � I
CIl 2CH 2CH2CH3
-:� N: N02
1 7 - 23 Another way of asking this question i s thi s: Why is fluoride ion a good leaving group from A but not from B (either by S N I or S N2) ?
�� :NU A
C
B
Formation of the anionic sigma complex A is the rate-determining (slow) step in nucleophilic aromatic substitution . The loss of fluoride ion occurs in a subsequent fast step where the nature of the leaving group does not affect the overall reaction rate . In the SN 1 or SN2 mechanisms, however, the carbon-fluorine bond is breaking in the rate-determining step, so the poor leaving group ability of fluoride does indeed affect the rate.
t
t
;>,
�
nucleophilic aromatic substitution
SN I
SN2
1 7 -24
Cl
H
376
. . ' ' 0 CH : Cl OCH3 : ] X; I � ;R N N ( .0.... � : �' � y y N02 0.... N+ O0.... N+ 0 O C H R U 02N -o:N � 1 �C. y N02 0...... ..N+ 0 CH ] � : � �2N
17-25
•
..
_ •
�
..
'¢
I
�I
•
.....
....
....
3
{:
-Cl-
.....
(b)
Cl
O
//C;,-
..
Cl
CH3
...H
H H
_
CH3
I
.
I
C_ I
.
....
+
• • -
.....
oell,
I
...
_
�
Cl
...
�
- . •
.... + .....
-
A4�:9.n�,,)l� c�o CH3
CH3
..
...
OCH3 H
C
4
.....
0N Cl OCH3 -0 .... DN :0.... 0: R t -o:N 1� y 0: N 0..: +/1
I
�
CH3
-o o
�
CH3
'L \CH{JLf� coj �
377
H
o H
17-25 continued (c) Br I
Br NH2CH3 HC.
�
.....
CH3NH2
.
..
N02
°2N
I
..
- ........ N ....... :0 + 0 :
-
(H-�HCH3 CH3NH2 -Br• • •
_
-
N02
..
-.. ; N :0 ..
+
NH2NH2 .. N02
-Cl-
0
+11 N - 0"
----..-
_
R
..
"N,
_
0 "+ 0
378
I
•
o II +N
-0"
Cl
+
C I
..
I N, .. � .0 + 0: ..
N, 0" ..... + 0
_
15NH2NH2
...... N ....... 0"""+ 0
+ NH 2NH2
:0:
N, "+ 0 0"
N HNH2
N02
...... N ....... · 0""" + 0 :
+ NH2NH2
N02
..
+
NH2CH3
C
"'+
..
02 N
..
..
NHCH3
CI
O
Br
+
...... N ....... 0 /+ 0
N02
(d)
Br NH2CH3 I .. I .. ..
+
+N -0"
..
Cl
l
-..", N :0 ..
-
+ NH2NH2
_
....... .. + 0: ..
17-26 Assume an acidicClworkup to each of theseCl reactions to produceOHthe phenol, not the phenoxidc ion. (a) Clz from additionHN03 NaO It, 1 elimina�i o n mechani sim;somer 0 AICl3 6 HZS04 NOz onl y t h i s NOz ort h o (b) Cl OH OH Bf 3 Brz B' Clz NaOH 0 AlCl3 6 3500 C 6 (c) CI Cl Cl Br Clz NaOH 0 AlCl3 Cl Q OH vi a benzyne mechani s m OH ortho (d) Clz NaOH AlCI3 OH vi a benzyne mechani s m Cl OH + ort h o CH2CH2CH' (e) o 0= CCHzCHzCH3 CH3CHzCHz-C-C� 61 Zn(Hgl HCl O AICl3 I Clz t AI Cl3 CHzCHzCHzCH3 CHzCHzCHzCH3 CHzCHzCHzCH3 Jr NaOH + ortho V OH via benzyne mechanism OH Cl '7
..
..
¢� ¢ 1'1.
+
..
..
2
¢
..
6 ¢ I�
1'1.
..
+
..
1'1.
II
'7
+
�
¢I �
..
..
q I�
� ¢ & ¢ 6 +
+
�
..
379
1'1.
¢
17-27
from chlorobenzene 17-28 (a) First, theocarboxylic acid proton is neutralized. C"'OH + NH3 + NH/ 0r 1 � V +
NaOH
+
heat
"
--
+ Na+
Q:HOC IH Li/. the negatavoiiveds charge
(b)
�c r?'
U
+
..
H ·
HH
HpluI s + Li+ resonance forms
H plus resonance forms
G" I�
380
II
h
H
o
o
n � y OCH3 H-O� r(_I(� t-B� � OCH3 ------.. lL�) Li7 tL.) / C
�
the carbon beari eldonatectrnoninggthe methoxy group
-NH3ry0C"' O .... C plus other resonance forms ( piresonance us Na. � forms -HCI
H
H
I .. I
resonance forms plus resonance forms
17-29
(b) t i o n of chl o ri n e morepressure C12, CI Xtl addi .. V' ttmiohetxhriteunreg,systofgivsteienmreoig ofa somers first product benzyl ic substformed itutionfrom CI (f) (c ) �,
1t
CI
CI
COOH 17-30 COOH cr COOH (b) ) c ( (a) V I I COOH COOH 17-31 CI-Cl hv 2 CI. initiation CH3 · C-C1 cis + trans
h-
r-,,,
..
¢ I
CI·
CH3 CI-C-C1 I
+
6
II
h-
h-
6 propagat i o n step 1
C. OH o
CI-CI stpropagat ep i o n r\r
..
2
381
CH3 ·C-C1
I
6HC t
I
6
CH3 C-CI
..-..-
NH2
17-32
Br-Br hv 2 Br- initiation CH3 CH3 · C-H �-�-H r'\ r\
Br-(A Vpropagat step 1 ion +
CH3 C-H
�
�
�
I
I
6
CH3 Br-C-H
6
6HC t
CH3 ·C-H
I
I
Br-Br BrI propagation step 17-33 As must statisbeticalcorrect mixtuered forwoulthed numbers give 2 : 3oforeach40%ty: pe60%of hydrogen. to . To calculate the relative reactivities, the percent 44% 56% 28 rel a t i v e react i v i t y 2H 3H 14.7 relative reactivity The reactivity of to is 14.287 1.9 to 1 17-34tioReplns, butacement ofonalatiparomat hatic hydrogens wiis tunfavorabl h bromine ecanbecause be doneof under freehigradih energy cal substofittuhteioaryln condi react i i c carbons t h e very radical. Benzyl�H3ic substitution is usual�Br3l y the only product observed. Br Br cc) B r � (a) (1) Br6 - C-CBr3 (b)(l) -C-CH3 (2) Br6 � I Br ic :--.."'.. I ....:--".. I alhydrogens l fourBrbenzylreplaced I excess Br2 • hv, time +
a:
=
a
�
'/'"
�
nr
..
2
6 �
a
�:
"'-':
=
(2)
h
382
��, � Br Br Br Br
17-35 n CH2-Br
, t � 5 6 6 � 6 6�o HC H +
+
-
-
CH3CH20-CH2
� H ! CH3CH2-A: CH3CH - -CH2
6 CH3CH2-0: I
�
I
..
2 R) +1
6 �
I
17-36 ic cations are stabipropane. lized by resonance and are much more stable than regular alkyl cations. The (a) Benzylis l-bromo-I-phenyl product H H (b) HC [Hl HC-CHCH3 CH3 ,., H-Br moee stable than I benzylic:B. r: Br H HC-CHCH3 I-bromo-I-phenylpropane
(]
..
V
• •
1
6
6 +
�
1
.
-
..
1
1
+
6CH
2°
2°
resonancestabilized
383
a) The combi nateintonatofion.HEr(Recal withl athfreeradiecveral inspeciitiatoesr generat es tbromi nekeneradidetcalesrmiandnesleadsorietnto antatioin.) The (Markovni k ovori at what adds o an al product wi l be 2b romo-l-phenyl p ropane. (b) Assume the free-radical initiator is a peroxide. R�rR 'l RO· RO· H r'Br ROH Br Br HCV HCH3 C: Br more stable than I
17-37
first
6
+
�
2
---
.
+
---
-
6C
·
I
..
2°
2-bromo-l (Br nrecycl phenylpropane chai mechanies isnm) Br· Br gBr M CH2CH20H (a) HC=CH2 3 CHCH3 H3 Mg H+ . L\ --6� ether 6 Br OCH3 C�2CH3 CH3CH2Br CH30H hv 9OCH3 AICl3 �OCH3 ortho Br2NBSor OCH3 OCH1 B' CN HN03 NaCN H2S04 Br2NBhvSor N02 N02 ortN02ho •
17-38
I
(b)
C( )
•
6
6ClI ¢
+
•
¢
6C 3 3 ¢1 I H ¢]C I
,
--
I
•
+
•
¢
+
384
I
¢ 1'1
•
•
0 enzyme H-OH HO-H HO-OH catalysts 0 OH 17-40 0 OH CH] (c) OH Br '(\ (a) OCH2CH3 (b) �I � � I CH3 � I CH3 Q CH3 CH3 Br (e) 0 OH (d)Br OH Br (CHJhC � I CBr3 � CH3 CH 3 ¢t Br C(CH3h 17-41 (a ) o (b) (c) o
17-39 OH
¢
+
..
& II
* �
¢
cQ
17-42 OH
o
I
I
� * +
(f)
�
,
+
+
°
o
OH
'(y
o
Br Br2 Br h Br Br2 Br 6 yBr Br Br C6H30Br3 C6H20Br4 17-43 Please refer to solution 1-20, page 12 of this Solutions Manual. 17-44 (a) C(CH3h (b) CH H2CHJ (c) C(CH3h (d) Br (e) C(CH3h
6
3
-
3C
o
---
6 rearrangement rearrangement 385
6 6
(08H
rearrangement -45 Product i17sopropyl group.s from substitution at the ortho position wil be minor because of the steric bulk of the (a)CH Br CHl ( CH3h ( CH3)2 (d) COOH ( CH3h
Q;
Q O Br 17-46 I
(a)
71 �
0
Cl � AlCl3 °
6
S03H
Q0,.C" CH3
..
o
t:.
...
I
...
°0
...
...
CH(CH3h Br
NBS � Zn(H"g) � -�1 ~ HCl � 1 CH30H °
....
Br from (a) ~ OH (c) CI 1) BH3• THF CH2=CH-CH2 �I 0 AICI3 U 2) H202, HO- � Br OH � Br2' hv I � Mg U H,o+ I or NBS � OH (d) CI Cl2 NaOH NaOH 0 AlCI3 6 350°C 6 CH3CH2Br Cl Cl (e) Cl Cl HN03 Cl2 Cl2 �I 0 AlCl3 6 SO N0 AICI3 N0 2 2 ----
('QCHlh
..
°
...
----
�
...
H ¢ 2
62CH3 ¢r ..
.. 4
..
386
17 -4t6hreecontmetinuedhods
NaBH, � Hg(OAch .. .. Ih Ih OR CI H2O OH � V Mg I � ---;;° CH2MgBr .. + 3 H 0 h et h er V � CH3-CH I : from (c) from (d) O COOH COOH (g) Fe or Sn or Zn n04 RN03 HCl(aq) H2SO4 N02 N02 NH2 OH (h) two possible methods B r M g :r new bond Mg " Br2 hv H O 2 i n bol d .. � or NBS I ether ° OH OR Br MgBr Mg B r 2 _ -l _ 7,:, I AlCl3 6 ether 6+� 0
(0
from (c)
..
2) 89 ¢ 8 2) KM
..
f1
..
..
'
..
--
�
¢
� ..
major
9Br
major COOH COOH RN03 H2SO4 N0 2 ~ Br Br
¢
..
387
�
17 -46 continued
Zn( H g ) KOH �CI (I) 0 FeCi; CI2 -HCI 6 6 AICI3 17-47 (a J) 0 N 1 ;;; (d) noacylreaacttioniodoesn: Frnotiedeloccur-Crafonts ri n gs wi t h st r o ng deact i v at i n g US03 H yN02 gr o ups l i k e N0 2 C(CH3h OCH3 Br (gl CI (h)¢CH3 ( l Br * (e) CH30q0C � 'CHl or YN02 NH2 CH Br CH Br 2 CH3 2 S03H (koCH] (I) COOH (i) Ph-C-� -o-� oCCH2CH3 U) � aCOOH H� N02 0 3 (m) HN-C-CH CH2CH3 � stronger o,p-director than CH3 � .. . CH3 CH3 Co 17-48 Major products are shown. Other isomers are possible. �
I
o
OH
CI
¢ ¢H
.
--
( bl
IJ,.
�I
I
--
(el
�
_
OH
OH
I
� I
basic workup to Isolate free amine
�I
�I
II
II
as (same
H 388
E)
17-50
product mollosseofcular weilgohtss of132H20 stmolarteinculg matar weierigahtl 150 IR spectrum: The dominant peak is the carbonyl at 1710 cm-I. No COOH stretch. NMR pofuadjacent l. In themethyl regioneofnes, 2.CH6-3.2CH2, 2th.ere ospectsignalricum:regis eachTheon wifromsplthitinint7.eggrat3is-7.compl i8onhasvaliciunateteeofgratd but2H;ionththofeesein4H,temustgratso itobehne tirihsnehelgtwmust areThetwaromat be di s ubst i t u t e d. our ofngthae diaromat itcusitegdnalbenzene. s talAl, insodiicnatdiincatg eC-H; two are ishort , with noandattatched NMR:alsoFshowi Carbon hydrogens, s ubst i d are carbons n a carbonyl wo meth� ylenes. I�S\ o (F-� CH2CH2 - C -8 or cO . , , � molecular weight 132 lose 1 H retain The product must be the cyclized ketone, formed in an intramolecular Friedel-Crafts acylation. 17-51 (a) < } CH2CH2COOH
18
=
=
8
8
___
are
c:=:::>
"w.�� H A
H
+
(jl .H � Br H
H
389
..
17(b)-51Assume continuedthe free-radical initiator is a peroxide. R�rdR r'J ROinitiation RO H - Br --- ROH Br- } � � lC. BrH .. -.- (9, � �C BrH � I H )-Br· ro H H H propagat i o n step 1 -
r
C'
---
+
2
+
0
---
I
..
Brpropagat i o n step 17-5� 2 H CI HO:.. m �, � a subst i t u t e d benzyne Attack A H HOm C�n Ho m ..HO:..- 2CO H 0 I l! �, � �I product A Attack B OH 2
+
+
� +
____
/'"
'7
H� product B
lJlJ
----
390
17-53 The electron-w ithdrawing carbonyl group stabilizes the adjacent negative charge. resonance plus forms :0: H-OEt dJ ��::o: � - ... C ��\ OO \ � Na+ plus other resonance forms Na' 0 •
•
+
U
H
'
HcO . H
(e)
(b) �
17-55
(f)
�
h-
H
:� �A)
U H .
(d)
plresonance us forms +
Na·f.
�
H3h
�
�
�
r
H .
°
4 (a)17-5�
°
H
�
< }-Q C + HS04- H30+ H2S04 0 H 2 6 conjugat colorless yellow ed ratnatedessulthfeuriwatc acier dto"dehydrat ees"reverse the alcohol , iproduci ng addi a hignhlg ymoreconjugat er,d,however, colored carbocati otn,oo andConcent prot o prevent t h react o n. Upon wat e t h ere are many water molecules for the acid to protonate, and triphenylmethanol is regenerated.
() -y o� 0H
6
+
2
+
---
391
2
+
17-56
o
17-(a) 57 bromination at C-2
Cl is this (why tishomer?) e major
Br
Cl is this (why tishomer?) e major
( � H -i �'b -zfH - HH� i� ""'C� �
o
three resonance forms :Br: + ,- 0 y Br �H •
bromination at C-3
H C:O � l; YCl
H / ��C Br H
• _
_
�:�!: j
(,.0yH � Br
O(H
Honly two resonance forms
Br opposedy. to only two at stgiabiveslizaninginattetarmedi bilizednatbyiotnhreeat C-2resonance (b) Attackforms resonance ck ataC-3.te staBromi wil occurforms,moreas readil C-2
392
17-58
abbreviate h phenylalanine as ·0 · F F NH2
I
P CH2 - CHCOOH
NH2
I
R
I
II
+ - ../N :0
- ..;N,.. C :0
02 N
...
+N -0/
.
O-/
�
NH2R
� 1. . C
R �
•
...
N, _ 0/ /+ a
N02
+ - / 0
-/H ..
..
� 'O-
� +
I
o-:/�'o-
H2 R
I
..
..
- ../N,.. ; :0 + O.
I
HN-CHCOOH N H2 R
N02
° N 2 ...
N02
393
1(j
- "'0
CH2Ph
° N 2
:0:. .
•
•
I
1
.
. C_
I
+
H2 R
I
/ N, :0 /+ O· . • •
-
)l
17-59
H3C
·. 0·. �
CH3
�
H-C�
i
+ H
H3C
CH3
OH
:Cl :
�
H
>b�: } � 1 H
OH
OH
+OH
CH3 Ho i CH3 Ho i CH3 CH3 CH3 CH3 OH
OH I
HO XCH, CH3
)l
/ ·. 0
l
I
H
OH
OH
I
I
��[J ["1 CH, C-CH3 C-CH3 I CH3 CH3 L (�H3 �
'!?� -�
H 't
r
Ho i CH3 CH3
..
..
I
:CH
"
..
plhuers otresonance forms
OH
OH
Ho
CH3
-O-FQOH CH3 bisphenol A
� C-CH3 H20: � �:� CI H3
.. .. HO
394
plus other resonance forms
17-a) 6Thi0 s is an example of kinetic versus thermodynamic control of a reaction. At low temperature, the (kinet ic product predomimatneatlyes:equalin trathisecase, alC.mostAta10001 : 1 C,mihowever, xture of ortenough ho andenergy para. These twdoedisfomers must be formed at approxi s at 00 i s provi o r t h e to occur rapidly; the large excess of the para isomer indicates the para is more stable, even products from e 00iConreact rat(b)ioTheof product as thethreact whiciohnwaswil runequiinliitbiratal ye asat i1000t is warmed, C. and at 1000 C wil produce the same 17-6 1 0 CH3CCH2CH3 H+ Br2 Mg ether FeBr3 Br MgBr CH3 -C -CH2CH3 (possible alternative syntheses include acylation followed by Grignard) OH desulfonation
though it is formed initially at the same rate as the ortho.
6 Q Q
II
•
--
---
17-62
--
I
from chi oro benzene reactive sites NaOH at 350°C As wetiosawn, leaviin Chapt eris16,olattheed carbons ofrinthgse oncenttheer ends. ring ofBenzyne anthraceneis suchare suscept iivbeleditoenophi electrlophie thatlicthe addi n g t w o benzene a react reluctant anthracene is forced into a Diels-Alder reaction. 17-(a 63 Cl 0- Na+ J CI NaOH CI CICH2COO- Na+ ;Y��H Cl � CI � CI � CI Cl 2,4Cl,5-T cl a, 0 :CC Cl � 2 Cl- CI � 0 � CI TCDD Two c substd folitutloiownsthform a newaddisix-membered (Thoughsnotm.) shownnuclhere,eophithilsicreactaromationiwoul e standard tion-eliminriatniog.n mechani +
+
-
11
•
-
•
+
395
w.
(This is the
compound used to pois on Ukrainian political leader, Boris Yushchenko.)
17(c)-63Tocontmininimuedize fOnTIation of TCDD during synthesis: 1) keep the solutions dilute; 2) avoid high taddemperat ure; 3)ofrepltheahalce ochlacetoroacet an excess ate. ate with a more reactive molecule like bromoacetate or iodoacetate; 4) ToT. The separat2,4e,5TCDD from 2,v4e,5in-Tanataqueous the end ofsoltuhteiosyntn ofhaesiweak s, takebaseadvant aNaHC03. ge of the aciThedicTCDD propertwiiels ofremai2,4,n5T wi l di s sol l i k e e andpitcanatedbefromfilteaqueous red or extsolractuteiodninbytoaddian organi caninsolbeublpreci ng acicdsol. vent like ether or dichloromethane. The 2,4,S-T 17-64 a( ) �:O:�H+ �:OH �O �O 0 plfOnTISus 3 wiresonance ththposie tive charge on benzene ring OH OH OH plfOnTISus resonance wie charge th posi t i v onon tthhee oxygen ring and : OH OH ('i� �O: resonance fOnTIS onplthuebotsoxygen h benzene riphenol ngs, , OH of t h e HO OH and the oxygen in the ring +
.. =-'\ ----.
-
�
o
¢
o
---.J...-J
o
o
th positive charge onplutshresonance e ring and fOnTIS on thewioxygen o
o 396
..
red dianion
(c) 17-65
°
q
CH CH3 3 Br2 Br I� Br H2O Br I� Br AICl3 O S03H para posiS03Htion blocked 17-66 A benzyne must have been generated from the Grignard reagent. ((I Br � �S MgBr 0 + 0 1 1 F �
6
Hfumi2SO4ng
..
¢
..
A
..
+
�
.
\--�----- �----�) V °
397
t
17-67 The intermediate anio� n forcesHtheOHloss of hydroxide. ax H OH CH3 For simplicity, abbreviate I NHCH3 >1 R OHLi + H H H H OH /OH s:, �c c2R c . R ffR .. ..-; a ___ I I .o L/ H 'C:Y HO H"S: H � plus plresonance us other � resonance forms forms tI Li· I
.0
.0
\
•
r
:·
+
I
..
---.-
H H � ,C CH3 y � NHCH3 H .
I
�R
.0
J'-R D u� H
'"cH3CH,o - H-
I
plus resonance forms 17-68 OH OH equiv. OH lcqui v . NaOH CH3I_ � Cl-C(CH3h � C(CH3h BHA, butyl.ted hydroxy.niso\c I AlCl3 YOCH3 OH OCH3 hydroquinone plus other isomer OH OH 2equiv. OH CH3I A Cl- C(CH3: (H3C)3C : c(CH3h BIrr, butyl.ted hydroxytoluenc I 6 AICI; Y AlCl3
¢
¢1
•
.0
CH3
•
V CH3
398
ve the probl i17-69 denticalSolmechani sm.) em by writing the mechanism. (See the solution to problem 17-2S(a) for an Br � l.5(OH Br � I I I attack C- ; HCHydroxi d e at t a ck on C-l put s t h e negat ive charge ongroups, carbonssoththatesedo notaniohave t h e ns are not stabilized. 02 N
J):I �
�B
r
2
"
Br
•
•
-
HO: •
•
02N
N02
..
N02
N02
Br
OH
a only product formed
02 N
N02
..
Hydroxi dcharge e attackononcarbons C-2 putwis tthhe negat i ve niatrbiolgroups, stnegat iizvate charge. ion bythereby delThiocalsiniincreasi ztienrmedi g tnhge athtee is formed preferential y.
399
also stabilized by resonance onto the nitro group
J3t:Cr -�H Br
�
...... N02 02N also stabilized by resonance onto the nitro group
18-1(a) 5-hydroxyhexan-3-one; ethylj3-hydroxypropyl ketone ((bc)) t3-rapns-2-met henyl buthaoxycycl nal;j3-pohenyl butyraldehyde hexanecarbal (or nameif you named this enantiomer); no common name (d) 6,6-dimethylcyclohexa-2,4-dienone;dehyde no common 18-2 O 5 el e ment s of unsat u rat i o n (a) C9Hl O IH5H doubl e t (very smal l coupl i n g const a nt ) at 9. 7 al d ehyde hydrogen, next t o CH mUl t i p l e peaks at 7. 2 -7. 4 monosubst i t u t e d benzene IH multiplet at 3.6 and 3H doublet at 1.4 CHCH3 -CH � -CH<� CH3 Thelookssplliikteinagquart of theet duehydrogen onsplcarbon-2, nextthe tadjacent o the aldCH3. ehyde,Aisclwortoserhexami examininatniog.n ofIn tihtse overal lsho\NS that t o t h e i t i n g from peaks each of the quartandet tishesplmetit hinyltohydrogens two peaks:arethnotis isequiduevtaloetnt,he splso iitt iins gtofrombe expect the aleddehyde hydrogen. The alconstdehydepeak hydrogen t h at t h e coupl i n g ainntgsconst wil notantsbe, thequal . Ifbea hydrogen is separat coupledeltyo, justdifferent neiwoul ghboring hydrogens byitdiinfferent coupl ey must consi d ered as you d by drawi n g a spl g tree for each type of adjacent hydrogen. 5 el e ment s of un sat u rat i o n (b) clCgHgO upeaksteatr of 1974 peaks atcarbonyl 128-145carbon (tmono-or paras ubst i t u t e d benzene ring h e smal l peak hei g ht suggests a ket o ne rat h er t h an an al d ehyde) peak at 26 methyl next to carbonyl or benzene < }- C -CH3 or CH3 -o- C- H more likely also possible 3 A compound has to haveto occur. a hydrogen or other atom) in order for the y rearrangement Butan-2-onoaney carbon hasy posino (y-hydrogen. M18-cLaffert tion-no carbon CH3-C -CH2 -CH3 t CHAPTERl�KETONESANDALDEHYDES
(R,R)
=:}
0
0
+
0
=:}
0
=:}
o II
+
I
=:}
sh:lpe,
=:}
0 0
=:}
=:}
0
=:}
o
o
o II
a
a
J3
401
it
18-4
]
; Lr 1j :+R:C-CH3 :I�C-CH3 �H2CH2CH2CH2CH2CH3 CH3 It 85 rnIz 43 rnIz �+CH2CH2CH2CH2CH2CH3 rnIz 85
[
43
t
---
---
128
McLafferty rearrangement
The first value is the to the second value is the n to The values are approximate. (a) 200 nm; 280 nm; this simple ketone should have values similar to acetone d systnmembase(210)valpluue)s 2plalusk30yl groups valueeofbond310 nm310 nm si(b)mi230lar tonm;Fig310ure nm;18-7:conjugat ketonee(280 nm for (20)the conj230;ugattehdedoubl 360 nm;ngconjugat e bondbase(30)valpluuse 4ofal290kyl groups (40) (280;c) 280simnm;ilar reasoni for the otedhersysttreansim (210) tion, stplaurtsin1gextwirtah doubl an average (d) 270 nm; 350 nm; same as in (c) except only 3 alkyl groups instead of 4 REMINDERS ABOUT PROBLEMS: SYNTHESIS There may be more t h an one l e gi t i m to a synthesis, especial y as the list of reactions getsBegilonger.n your analysis by comparing thateetaapproach rgetoneto tofhe tstheartsmaling lmatnumber erial. ofIf react the product hasformmorecarbon carbons t h an t h e react a nt , you wi l l need t o use i o ns t h at carbonWherebonds.possible, work backwards from the target back to the starting material. 4. KNOW THE REACTIONS. problems. There is no better test of whether you know the reactions than 18-5
n
n*.
n*;
<
=
=
is
=
1.
2.
3.
attempting synthesis
402
18-6 Alclarbon-bondthree target fmolormieculng ereact s in ithions.s problSo far,em thave more thantysipesx carbons, soonsaltlhanswers wil include carbonh ere are t h ree of react i at form carbon-carbon SN2 substitution by an acetylide ion, and the Friedel-Crafts reactions ion) ionon,benzene. (albonds:kylatitohne andGrigacylnardatreact 0 OH 0 f 0 MgB l) H � 0 Bf � ether H2C�; � 2) H30+ � 0 (b) Bf identical sequence as in part (a) I � U ~ 0 OR s metheffiodcusiientngasFriit eidels onl-Craft s acylstepation Cl �... � ithis more y one o AlCl3 U (c) a synthesi s as in part (a) could also be used here OR 0 � Br Mg � MgBr � � L-.l � L-.l 2) H30+ L-.l 6H �C �4 OR (:j' Br Na+ -C=CH (:j' C=CH Hg2+, H20 H2SO4 og 18-7 (a) (J BuLi (J BrCH2CH2Ph (J H30+... PhCH2CH2 -CH0 SXS S C'" S S X S HgCI2 H H PhCH2CH2 H H (b) (J CH3I (J BuLi (J Ph H30+ 0 Ph S ::-C" S SXS � HgCI2 � Ph CH3 H H Br � from
(a)
..
..
o
1)
..
..
...
• •
..
I
..
::- I
II
..
• •
...
...
(a)
403
�
18-7 continued H30+ Ph � Ph (c) rl PhCH2Br SrlS BuLi PhCH2Br SrlS HgCl 2 S' " S X X CH PhCH2 CH2Ph PhCH2 H rl Q- CH2B' rl BuLi PhCH2CH2Br ('j() S:-C"S o-:x S S � Ph H H (a) ('j � � H30+ , HgCl2 � Ph 18-8 0 0 (c) CH3(CH2h -C0 -CH2CH3 ( a) (b) � Ih if 18-9 0 (a) CH3(CH2h -C-CH2CH3 (b) (siPhCH2CN mple SN2) (c) PhCH2C\l-(J MgB' CH3CH2C= � H30+ CCH2CH3 (a) I � Br � O ether O O (b) CH3CH2C::N BrMgCH2CH2CH2CH3 --- H30+ � H30+ (c) CH3(CH2hCOOH 2 CH3CH2Li ----� CH2C -o CH2Br CN MgBr 6 NBS. 6 NaCN. I� o
•
•
•
•
.
•
I
(d)
from (a)
�
-
•
I
from
II
II
o
18-10
II
--
h
+
--
+
( d
6
o
o
' H, o 6 6
::::'--1�---
404
o
h
IS-II (a)
V
CH, H (b) o
V
CH
o
OH (d) � � o
(C)
II
IS-12dirRevi ndersstonepsp.i s402.usualThere more ect routew tehwie remithPhBrfewer ly betareter. often more than one correct way to do syntheses, but ( a ) � . Mg 0 OH � Cr03,H20 � + H30 PhMgBr -..;: :: H • Ih V V H2S04 V V PhBrMg . H30+ � PhMgBr o
•
VV
•
•
(c) 0
•
H,o'.
OH
Li A I H 4 NaBH4 OR � � CH30H VV VV ( s ol v ent ) HC::CCH2CH3 OH t NaNH2 (d) 0 + C::CCH2CH3 H30+ � C�CCH2CHJ � Na H Ih V cetoxyborohydride replIS-13acedThethreetriahydri des. ion is similar to borohydride, BH4-, where three0acetoxy groups have (b) .... C... (a ) C .... C...0 :0: " '- 0 0 ' 0 0 ' 0 ) 0 �O-B-=-H � O - B + R--1' H Na+ H- 0� -O -1/\ \. - Na-+ +. )!.:: � R H / / O O H y y y � 0 H3CCOO-H 1O-H R+ H ..
_
•
•
o II
o II
II
-I
o
•
o
I
• .
405
--
1
• •
I
H
a
ylides. TrimCH3ethylphosphine has a-hydrogensCH3that could be removedCH3by butyl ithium, generating undesired 1+ 1+ BuLi CH3-P-CHR CH2-P-CH 2R R CH3-P-CH 2 1 1 1 CH3 CH3 CH3 wrong ylide desired ylide �+ �h PPh3 : rotation '" I" I I HMe MeH :O---! PPh) / H I I I C - C """ H Ph3PO 1.:1:1 I I I """" ' H', r Me cis-2-butene Me Me �Me/ H Thefashistoen,reochemi is inpverthosphied. nThee oxinuclde emustophielleimitripnathenyle wipthosphi ne mustry.attack the epoxide anti yet the tsrtirpyhenyl h syn geomet (b) :0: rotate and 'H el i m i n at e .. PPh3 7 1 fl o O . trans + PhCHO .. CH2=CHCH-PPh3 PhCH=CH-CH=CH2 (a) CH2=CHCH2Br 1)2) Ph3P BuLi 1) Ph3P... PhCH-PPh3 O=CH-CH=CH2.. PhCH=CH-CH CH2 OR 2) BuLi PhCH=CH -CH2Br 1)2) Ph3P BuLi... PhCH=CH-CH-PPh3 CH2-PPh3 PhCH=CH-CH=O PhCH CH-CH CH2 18-14
1+
�
-
+
• •
..
+
_
.... .. f. -
\ \\
.
" ,
in
. .
"
o
CIS
�
:
,
..
18-16
�
-
(b)
+
=
+
+
�
406
=
=
an
enes form can bethesyntphosphoni hesized ubym tsalwot difromfferentthe Wilesst ihignreact The18-17onesManyshownalkhere deredionsalk(asyl halin itdhe.e previous problem). PhCH=C(CH3)z (a) PhCHzBr 1) Ph3P BuLi.. PhCH-PPh3 )l (b) BuLi CHz-PPh3 (c) PhCHzBr Ph3P.. PhCH-PPh3 PhCH CH-CH 0.. PhCH CH-CH CHPh 2) BuLi O �CHz CHz-PPh3 � U U +
o
+
2)
H3C
CH3
-
+
2)
1)
+
+
18-18 (a)
=
=
=
=
_
+
OH Cl3C-C-H OH " :O: � H-OH OH CH3 -C-CH3 CH3 -C-CH3 HOOH OH I I
I
. .
I
18-19
I
lofeasthydratamounte
I
407
great est amount of hydrat e
+
:0 : H-CN 3 OH CH CH2 -C-H CH3CH2-C-H
18-20
. .
I
I
�
CN
(c) t-Bu-C-t err -Bu
'"""
I
CN
..
I
OH H-CN t B u-C-t B u t B u-C-t B u � :C N CN CN 0 OH (a)� HCN vi' V OHI pcc 0 OH (b) . � � · . H H HCN �H H CN OH (c) CHCOOH u smsivofe charnuclgees.ophiThese lic attspeci ack atescararebonylverycashorrbont-lifvredequentas eachly incharcludegespeciis quiecsklwiythneutbotrhalized Nposioteti:veMechani and negat on tr ansfwierl; ishow n factt,htesehese stepsstearpseasthoccur e fasterstingofatththeewholsamee mechani sm.though In mostyoucahases,ve tbbyheenisaSolrapiadmoni udtioprnsoshedtManual t i m e, even sm separstoodatetlhy.atThethesepraracteice offashowi transfers in one stetop show is legitimstateepsasofloangmechani as it is under st steps.ng these proton ::::..;::: :: � ::=
:o:� '""" I
I
I
I
18-21
II
I
all
two
two
408
(a)
18-22
(J<-:NH; 0:
H
H
H ' ) JW : O h (J :0:\ :o-� d,+NHPh HO+3 . C} NHPh HO+ �20:
(
HI
H20:
two fast prot'---./ on transfers
Ph-C-H
•
•
:o:� 1 ��
(twro fast� proton transfers 1
---•
•
•
Ph-C-H 1 NHCH,
H
� 3
•
:O� 1
��
_
O -H H- Ph-C-H ( 1+
NHCH3 1
+ - H20
Ph-C-H .. H20: i Ph-C-H Ph-�-H } Hplposiusttihvreee charge resonance on theforms benzenewith II
: NCH 3
409
�U�CH3 II
�
NCH3 I
ring
isomers.a double bond is formed, stereochemistry must be considered. The two compounds are t18-he 23andWhenever Z
E
Z
18-25 This mechanism is the reverse of the one shown in 18-22(c) on the previous page. H-O-H 1+ H20: Ph-C-H .. Ph - C - H H30+ Ph-C-H Ph-C-H � + I II I NHCH3 H -NCH3 H NCH3 .NCH3 1 plus th+ree resonance forms with ') t ( Positive charge on the benzene ring H-O: -CH3NH2 H-O: H-O: : II0 : 1 1 I + .. .. Ph-C-H Ph-C-H Ph-C-H Ph-C-H .. I CH3 (I H 0+ 3 H-NHCH3 � NH t H-O+ VII Ph-C-H 18-26 02N NH2-NH -O- N02 A abbreviY. ate "Z" :o � + H-O-H + H30+ Q: H 1 I I 30 ( CH3-C-CH3 CH3-C-CH3 --CH -C-CH 3 CH -C-CH 3 3 '-- :NH2Z I I � H2 0 : + .�
.1 1
j �
•
�
• •
..
..
. .
....f----
+
o
..
\
)
:o:�
-
(k�
two fast proton transfers
NHZ
NHZ
•
•
3
CH3 -�-CH t H-NZ I
..
3
NH N(b) oJ 2
o
I� .b
(a) PhCHO + H2NNH -C -NH2 (c) 0 N-NHPh H 2 oJ NH2 CH3 (e) C(f
18-28
(b)
II
(d)
+
I
N ",NHPh Ph )l Ph
(d)
ho + H2NOH 0 0
02N + NH2-NH � }-N02
0
o
+ ·0· + � :O-H Ph-CH H Ph-CH
:O-H O-H +1 � . Ph-CH•• }-H�CH3 .-----l Ph- 1 + . "---.:. : OCH 3 ) �k H.qCH3 � H-O-H } ,----- :O-H : +1) � � H : 0 � 2 � ili �� Ph-CH ---H+ Ph-CH j Ph CH ,...-W ---j OCH3 OCH3al 1 + �CH3 (�H3 hemiacet 18-29
. . � II
---
II
plus resonance forms with (+) on benzene ring
----
'"
J
plus resonance forms (+ ne ring
_
I
... ....
H..2:0CH3 Ph-CH OCH3 VI I
OCH3 Ph-CH acetal OCH3 I I
41 1
I
CH
I
I
18-30CH3 CH3 1 1
CH31
:000: ""\H 0+ • •
• .
:0,
t
�
18-31
6
(c) 0 0
. .
1
• .
3
..
: O�H +
•
1
6
1
H20:
.
+ 1"
. .
.
.
..f--....
H2O:
:0:
•
6
(b) 0 CH3 -CH + 2 (CH3hCHOH (d) HO + O O HO) II
+ 2 HO OH (e)0 HOD H+ �
+
.
t -CH30H ..
:�
.�H20: ! V CH3 H CH3 H 0 0 o O-H :O : : H� 0+ 0 f
oo:0 / H
CH31
r\
(f)
HO )(l OH II
o
412
�';led lX�U uo p�nu!luo:> WSlUnlj:>�W S!SAI0.1pAlj
t
:9 zH
H
o'0:
o :0 '0 . .....
J�
H-O
J,
'----1
�� HO
H
-
+
H-O: • •
J,
+0 -Q�Q 0 0 0:
•
'----1 +
•
H-O� + '---1
�:o
H
-1 '---(q)
0- Q 0 '0: H 0: 0: J,
'----1
+
• •
-
J,
• •
'----1 +
FHQ Q-Q � :0:
(n)
H
+6 :0·o/�
:0
H
-------
+
contiynsiueds mechanism continued (b)18-33hydrol OH
OH
r + r H-O� .... OH :0, .... OH C C, � rOH ____ � H30+ HO
I\
�6 6-6 (c) The mechanisms of fonnation of an acetal and hydrolysis of an acetal are identical, just the reverse ·0·
((·O �
.. H
+ H
+
order. Thipaths,hasthentothbee reverse true because thave his process iws antheequiidentlibiricalum:minifimthume forward stpatepsh. folThiloswis tmihenimum energy st e ps t o fol l o energy Principle Microscopic Reversibility, text section 8-4A. (d) of
;;o:CO CO-; 9 o 0 � + H
RJ
CJ:::'t /h·· OH
+ O�
18-34
a ( )
;-0-
0
NaBH4 A CH30H V CHO 1
equivalent 1\
HO
00-00 o � ··� � CCl �CCl � '--• •
•
+0
O
• •
famous
:0 I H
H20:
�+ OH '--- R
W ..
1 equivalent
0
a
RH
�
�� OH
OH
\ �
CO . .:
+O 1 H
OH
OH
arecolreduced fastcreaseer thselaneketctivonesity). (keeping tAlAlhidtseehydes react i o n d wil l i n be usedrnatiforvely,thissodireductum itorin;acetseeoxyborohydri problem 18-13.de could
..
OH
414
18-34 continued equivalent 0 0 o 0 o 0 Mg (c) CH2B' HO OH CH R' CH MgB' 2 2 6 H+ a ether a + H� 1
(\
(\
(\
..
..
Theoxygen,last stdehydrat ep protoesnattheesalthceohol, and hydrolyzes the acetal.
18-35 (a) oCOO O HOafter adding H+ : Ag (d) HO COOH HO V
0
\..
t
..
)
V
[MgBr]+
t
HC::C -CH2 -CH2 C (b) DCOOH (c) COO: y o 0 after adding H+
H30
+
- - CH3
415
a II
A gO
j 18-36 :
hydrazone formation
cJ'-
-000 0, + H0 .. •
",-- :NH2NH2
�H V
�
"H"OH
..
�
:OH
" 2 " NHNH2_ (I :OH
two ofastn transfers prot
reduction of the hydrazone H
• • •
O
H--./ ..
..
C ....... N
H-N II
O
C� H --
• •
-·..N ('i2:�IIN V
416
H
NJfNH 2
18-37 (a)
H
CO
(b)
H
�
H
H
(c)
(d)
1\
� V H ' 'H
18-38 Please refer to solution 1-20, page 12 of this Solutions Manual. 18-39 nIgUPplacement namesoffiposirst; ttihoenn numbers. common names. Please see the note on p. 136 of this Solutions Manual regardi (a)(b) hept aan-n-24--oone;ne; metdi-nh-propyl yl n-pentketyloneketone bromo-p2rop--met2h-eylnalhexanal ; no common name (gh)) 4-3-phenyl ( ; ci n namal d ehyde hept (i) hexa-2, 4-dainalenal; ;nonocommon commonname name anal; no simdiplpehenyl commonketoname (j) 3o xopent (d)((ec)) buthept benzophenone; ne ( k ) 3o xocycl o pent a necarbal d ehyde; no common name a nal ; but y ral d ehyde ) ci s 2 , 4 -di m et h yl c ycl o pent a none; no common name (I (18-f) 4propanone; acet o ne accept s "acet o ne") 0 I0n order of increasing equil0 ibrium constant for hydrat0ion: l 2 lofeasthydratamountion est amount ofgreathydrat ion 18-4 1 AC
(lUPAC
II
CH3-C-CH3
C CH
<
o
b2
o
-
II
C-H
}c
I I
H-C-H
<
�
9H
CH3 II I H-C-CH -C-CH3 I f! CH 3
18-42
o
comparison with similarly substituted molecules shown in the text: A By1t--t1t * value (210) plus 3 alkyl groups (30) 240 nm U CH3 n--t1t* base 300-320 nm o
=
417
18-43 C6HJ002 indicates two elements of unsaturation. 1 Thetwo elIRement absorpts ofiounn atsat1708 em· suggests asiketngloene,ts inorthpossi biolyoftw2o: ket3 inodinescatsienacehitghhlerey symmet are two rioxygens ande. u rat i o n. The e rat c mol e cul to carbonyl liThekelysintoglbeet atCH22.15on thise probabl other siydemetof hthyle next carbonyl . , and the singlet at 2.67 integrating to two H3C -C -CH2 -CH2 -C -CH3 Since the molecular formula is double this fragment, the molecule must be twice the fragment. Two questofionshydrogens. arise. WhyWhyis tdon' he inttethgrate twioonmet: hylandenesnotshow4 : 6?splIint tiengratg? iAdjacent on provi,des a hydrogens, not absoluwite th numbers, identical chemical shifts, do not split each other; the signals for ethane or cyclohexane appear as singlets. 4dehyde The formul ndicatievse 5Tolelelement suggests ant18-heal4unknown ormusta ketabeCone,JOa HIketbut2Oone.ainegat ns tesstofpreclunsatudesurattihoen.possiA solbiliidty2,of4-DanNPaldderiehyde;vativtheerefore, Themonosubst NMR ishows the typicatal et7.hyl3 pat5H,temulrn attip8le1.t)0. The tsirinpglleet)t andat 3.7.5i s a CHquartet ), andte fara downfield, t u t e d benzene , but qui 2 apparently deshielded by two groups. Assemble the pieces: 0- CH2 C NMR
(5
(5
>
o
is
o
II
II
2 3
8
(3H,
(
[
o
+
+
18-45 . H ,CH3 " 0 ) ( CH (a) H /C ,�CH CH2 2 rn/z 86 )H>-. �CH3 . (b) � CH CH3/C,�,CH2 CH rn/z 114CH3 +
II
I
+
o
II
I
I
ratio, identical
II
..
..
(5
(5 2
(2H,
H CHCH3 H J:CH2 CH2 rn/z44 mass 42 0....... H . CH/CH3 CH/3 C',CH CH2 mass 42 CH3 z72
r
+
II
+
I
rn/
418
I
+
II
18-45 continued (c) O} ( H
H3C""""
II
C
18-46
CH
. CH3
I
mJz 114
.Jr-.... CH 'c"""" 'CH3
.
+
..
H2
[
0
H3C
....... H
mJz 58 I
....... C"
-""':CH2
r •
HO XO X
�O +
mass 56
CHCH3
+
0
II
CHCH3
OH
ATol18-mol4le7nsecultestariniodincofatemJa zket70one.meansThe°a carbonyl derivativmass e andfora negat ive fairly smal(CO)l molhaseculmasse. A28,solsoid semi c28arbazone42, enough 70 onl y with t.wo elements of unsaturation, we The molofecula doubl ar formul canmoreincarbons. fer the presence e bonda isorprobabl a ring yin addition(mtoassthe70);carbonyl I, indicative of a ketone in a small ring. No peak in the 1600-1650 Thecm-IIRregishows a st r ong peak at 1790 cmo n shows t h e absence of an al k ene. The onl y possi b i l i t i e s for a smal l ri n g ket o ne cont a i n i n g f o ur carbons are these: 0
0
�
4<)' 0
A
3
=
C4H60
0
B
The HNdoes show can dias4Htinguitripslhetthatese.3.1;Nothmetis sihylgnaldoublcomeset appears ienttwheo NMR spect r(C-um,2rulandingC-out4) adjaThecent to from t h met h yl e nes the carbonyl by the twoof splhydrogens roughl y a qui, nspltetitbecause it ing by onfourC-3.neigThehborisinggnalprotforons.the methylene at C-3 appears at 2.0, unknownion ofis cycl obutanone,at 1790Thecm-symmet ry inediricsatticedofbysmalthelcarbon NMR rulrinegsstoutraistn rstuctrengturehensThe TheIR absorpt I t h e carbonyl i s charact ri n g ket o nes; carbon-oxygen double bond, increasing its frequency of vibration. (See Section 12-9 in the text.) 18-48The conjugated diene has a maximum at 235 nm (see ved Problem 15-3) and the ketone has a (a) mum at about 237 nm, so the to transition cannotSolbe used to differentiate the compounds. maxi (b) The ketone has an n to transition around 315 nm that the diene cannot have. NMR
MR
3
CH3
&
A.
n*
n
B. 8
B.
n*
41 9
the
18-49 (a) (�l���etal) cetalketal) old: ahemi (hemi
o
6
II
CH3CHCH2 -C -CH3 2
(b)
(c) acetal
(d) (olacetd:alketal) 18-50 (a) G.o: II
,--
3 HC-C-H
+
o
+
CHC 3 Ho , H
..
0H
1\
:0: � I
H2
H
O.
H3C-C-H ..
II
:o� I
H30+
.. H3C-?-H
H
II
HC-C-H 3
l l "----/' U
CH3-CH
I
I
acetal
OCH3
420
0"
0
........CH I
",CH
HN 2
+
..
1+
-Q
H-O-H
(
..
I
HC-C-H 3
NHNHPh
---
H3C-
H-
} NNHPh � -H I
• •
O-H
CH3-CH
=(
+
1+ · :kV I
OCH3
H
H
NH2
+NHNHPh 2 • NHNHPh (� J two fast proton transfers i 0: 2 : NNHPh � NNHPh + I
H 3C-C-H
H30+
O
OH
( imine(s) (h) imine 0=0
OH
+
(f) diether, inert to hydrolysis (g)
HO
: NH2NHPh
2 CH0 3 H
(X (e) acetal
O
H CH3
I
:O-H
•
hemiacetal I
CH3-CH
OCH3
18-(c) 50 continued. C0
3 h P O P C h ) H ,P 2 cr � Ph' ; ' � : U"---- CH2 tPh, �V C�2 V rCH2 /\ /\ /\ 0 �-H ' O -H : O � O-H o � ° ° o (d) :�H30+ 0 o ..
:
+
�
�
-
+
----
.
�
: 9
+
�
..
Hi) � : OH OH <---JOHH H-+/0 �O r'0"� "c/ OH . 0 H 2 + H 30 o ,. 0 : l 0' H H OH : HO� o H1 + (e) � ' O -H : o m H H,O+ � H 02 c N � +H �� H ' O H ' H 0 :�.;-H r:J2-�';� + + O .. C,H 4) <;:,H H H ..H3. EJD2 H H H H . 0. 0 ' ; � O· C, C ,H � 'H H30+ 0 N H 3 H 'H
H
.
...
•
•
-6 r�:. 6
..
HH H /N,
�b
1
N/ 1
•
+
/ .'
--
..
/N,
'/
•
�/+ o
+
�
I
+
H ' · O /
•
/N
�
N
+
/N � ,
•
•
421
..
� .'
I
� H
H
18-51 o OH OH H30+ (a) CH3-CH KCN .. .. HCN CH3-CH-C::N 0=CH3-CH-COOH ,Ph HC o PhCH2Br Ph3P BuLi .. PhCH -PPh3 .. 6 0 0 0 1 equivalent 0 (c) NaBH4 Al t e rnat i v el y , usi n g sodi u m t ri a cet o xyborohydri d e .. sel e ct i v el y reduces t h e al d ehyde; see Probl e m CH30H CHO CH20H o 1I10equi/\val0ent1! �o NaBH4 H OH 30+. 11 (d) 0 � ,H CHPH" Q C C 0'" '0 0'"C'0 Q�1I0 CH3CH2CH2Br PPh3 t tBuLi CHCH2CH3 5112:113 1 equi v al e nt o o (e) 0 HO/\OH.. o CII3CH2CH -PPh� 1130+. H o C C Co 0'" '0 o,..C,0 o 1 equiH2valent.. roo ro Pt OH o ro RaneyH2 .. co- H OH (h) roo NaBH4 CII,oIl· roll I
I
II
+
(b)
..
18-13.
'
LJ
'--1
+
+
,II
LJ
LJ
(
(g) f)
Ni
422
II
18-52 All of these reactions would be acid-catalyzed. (a) q + H2NOH (b ) (c)
o
�
� NH
0
+
H2N 'O
�CHO (d) ro + 0 HO OH 6 + CH30H
O � 2+ LJ
I
1\
�
, C H3 ONH2 O=CCH3 18-53 °2N 0 a( ) (b) C NH2 , O N N } N � H r N/ cf V (c) NOH (d) (e) CH3-CHOCH3 � OCH} L CH2 C H3 OH h ) (g) ( rl H -OCH3 CH < }- CCH2CH3 CH3CH2 (f)
+
\
(c )
o
2
II
1\
I
I
( f)
I I
N
...
SX S
H
423
18-54 (a) (] BuLi SX S H H
..
CH31 (] S, .. ..... S _C H I
S(]S CH3XH C '/ HeH, C:··S ' H ..
• •
..
+
HS �S-H
..
..
/' CH, H2O: :··S ' HH
t
"c
.. H ..... C ..'CH3 2+, assists the hydrolysis in two ways. First, mercuric ion is a Lewis acid of moderate (stcr)engtMercuri c i o n, Hg h, performing the same function as a proton from a protic acid. Theits charge: effectiveven enesscompl of Hge2xed+ aswia Lewi s acifur,d tihs epartmercury ly dueatom t h a sul stil has a positive charge, attracting the sulfur's electrons. S ...... Hg2+ Athisecond expl a nat i o n for mercuri c i o n' s effect i v eness i n hydrol y si s of lms.s liesThiin sthestacompl ex formed betivweleeny removes the ion andHSCHthe2CHtwo2SH sulfromofuracetthateaoequi bl e compl e x effect t l i b ri u m, shi f t i n g t h e equi l i b ri u m t o product . ( A n exampl e S of whose principle? His initials are "Le Chatelier".) Thiols are often referred to as "mercaptans" because of their ability to CAPTure MERcury. 0: II
to
424
18-55 icTheallykeyaffectto thitheis rproblchemistem irsy.understanding that the relative proximity of the two oxygens can dramat 1 , 2 d i oxane Thedescribed: "second"twoisomer oxygens connect e d by a si gma bondbondarei s aeasiperoxi d e. The 0-0 l y cl e aved to gi v e radi c al s . In the presenceradiofcalorgani cions compounds, react can be explosive. Mechanism of acetal hydrolysis
1, 3 d i o xane Thetwo oxygens "third" isbonded omer descrito thbeed: 3 carbon constitute an same sp acetal whiacichd.isSeehydroltheyzed in aqueous mechanism below.
�O - H
C:0
:0 / H
H/C, H I
•
+
o
CH
+
" ' H
---
•
.. .... .. f-. -i �
0 2
l, 4 -di o xane Thean excel"firstle"nti ssolomerventdescri bed:gh (al t hou toxic), thapart ese oxygens are far enough t o act i n depen dently. It is a simple ether.
. .
..
HO
0 ()
} 425
..
..-
"C ' H H2O:
(:'0 •
/
H
" ' H
(a)18-56
(b)
6 (e) N-N-Ph ]
(t)
6H (i)
0)
0
6 PoH
CH ]
(c) NOH
c5
18-57 Thecomesnewfrombondthteoprotcarbonic solcomes oxygen vent. from the or (a) NaBD4 OD (b) NaBD4 OD (c) NaB 0
NaBH4
H2O
II
CH3 - C - CH2CH3
..
..
0 "
CH3 - C - CH2CH3
..
0
H4
"
D 20
..
D 20
(d)
6 (g) no reaction
CH]
(h)
(k) Na+ -(j ON (I)
NaBD4,
OH I
1\
0 er 0
H
HO
.;; C
/
H
COOH
shown in bold below. The new bond
to
CH3 - C - CH2CH3 I
D I
CH3 - C - CH2CH3 I
D I
58 Whito lapproach e hydridefromis a smal legroup, the actsiduealofchemi cal especi e, sthsuppl ythinegsiitd,eAlopposi H4-, itsefaitherlymetlarge,hyl . so iThi18-t prefers t h e l ss hi n dered t h e mol cul e at i s , s forces the oxygen to go to the same side as the methyl, producing the isomer as the major product. O � �fd Q lessndered AI· hiface majproductor H CH3 - C - CH2CH3
..
..
CH3 - C - CH2CH3 I
H
cis
H
H
" /
/
C
H
H
_-,
� H "- I H· H
�
..
H3C
0
H
H+ ---.
H
H
CIS
426
CH3
18-(a) 59
e bond i sic BuLi CHz -PPh3O•O O CH2 lAndoubl essexocycl steablbond.e tihcandoublan endocycl methylenecyclohexane onhexane from cycl ohexanone withoutoutsusiidentghethrienWig t ig (b)reactTheion direstffisciulnttyheinstsyntabilhitesiy ofzinthgemetdoublhyleenecycl bond i si d e t h e ri n g ( e ndocycl i c ) versus (exocyclic). HOoCH] : +. H20. + CH] MgBr mi n or maj o r Amechani dehydratsmiopassi n folnlgowithrough ng the a t cold carbocat i o n i n t e rmedi a t e will gi v e t h e more assubsttheitmajuted,orendocycl icThedoublonley Br CH3 bond product . chance he exocyclis toicdodoublan e o K+ -O-I-Bu bondeliasmofitnhatmaking eimajon usior ntproduct give Hofmann orientgaatibulon k(Cyhaptbaseerto7). minor major 18-60 CI -CCHzCH2CH3 zn(Hg� � (a) • � HCI V 0 Br H3O+ 1 .0 C-CH2CHl (b) V C'N CH3CHzMg ether V Cl OH OCH3 CI2 • 6 NaOH Na0 ,!;l 6 HCl • I) 1 0 AICI3 3500 C 6 CH31 .0 AlCuCICI3 CHO 0::5 0 (d) H Cr 0 2 7 I: Z conc. H2SO: I � c6 0:> AlwittehrnatSOCIivelz,yt,htenhe cyclacidizchledobyridFericouledeld-Cberaftmades acylation with AICI3. +
-
6
�
+
El
HB r
E2
E2
°
° II
•
c5 6 +
° II
•
(e)
�
co
2)
.0
¢l
°
�
•
427
18-61 (a)
(b)
OHCP , (y h H V OCH2CH3 CI, H ()OCH2CH3
(c)
H 0 2 ()N-N-C-NH C-H I
II
0C .0O () Ag + (e) ('I ()CH1 c: � d + c CH3Mg1 H30 r03, H20 � H ether OH H2S04 � (b)�C=CH HgS040 H2H2S04 � + H 0 3 ('I ('I ( 0 SXHS CH31 Sx S CH3 CH2hBr S� HgCI..2 � H CH3 CH3 Hc r03, H20 � H2S04 � � OH excess +CH�Li H30 � + M CH3 g1 H30 C=N ether � (g) 03 ,CH3 O=C Me2S � + CH3 (d)
I
II
(f)
I
18-62 (a)
II
..
o
�
� I
..
o
..
(c)
1) BuLi
..
2)
..
1) BuLi
2)
--��
o
/"'.-.. I I
..
(d)
o
(e)
1)
2)
o
(f) �
o
..
_
�
o
1)
2)
..
o
428
\
18-63
o
�H
PCC
(a) � H
---.-
O
o
�H
(b) �CH2 (c) �C=CH 1)
BuLi
..
�H
HgCl2 o
Cl
(e)
o
� Cl
�H
KOH H20
---.-
(This reaction needs a solvent like THF to keep all reactants in solution.) o
II
(f) � OR
2)
OH H30+ ft OH
� o
(b)
..
�
SOCl,
�
18-64
(a)
LiAIH4
1)
--
o
II �H
OH � ( Ot-BuhH ft
" � Cl
� o
PCC
+
(e)
� 0 +
18-65
(a) ketone: no reaction (b) aldehyde: positive
(c) enol of an aldehyde-tautomerizes to aldehyde in base: pOSitive (d) hemiacetal of an aldehyde in equilibrium with the aldehyde in base: positive (e) acetal-stable in base: no reaction (f) hemiacetal of an aldehyde in equilibrium with the aldehyde in base: positive 429
(c)
� o
A
18-66
J
K.
be deduced from itsioreact ion wibethGrigandnard reactWhationsiswicommon tn-2-one, o both product s stirouctns uisrethofe heptcanan-2-01 ofmustthesebeThereact part ; t h e react ns must t h hept a so heptan-2-one. S(l S ~ X H H3 BuLi ' B 4 ) H2S04 �H� HgS04 H 20 S S �CH3 2 + Hg + MgBr H30 � CH3CHO + H30 OH N 0 � a2Cr207 � H2S04 • MgBr N� PhMgBr+ � 6 0� H30 H3:' ! OH �OOH �Ph �C=CH
C
1
A
1)
G
D
F
E
B
1)
2)
c
A
1)
1)
2)
J
K
I
o
430
bsorption at nm in tbhe spectrum suggest I The very st r ong t o a s a conj u gat e d ket o ne or onyl at emand smal l al k ene at u gat e d car alThedehyde. ba senceTheof peaksconfiatrms this: strong,m I conj shows that the unknown is not an aldehyde. The molecular ion at leads to the molecular formula: b C=C-C-C mass units add car ons and hydrogens mass molecular formula C6H80 elements of unsaturation Two elements of unsaturation are accounted for ibn the enone. The other one is likely a ring. b ) hydrogens. The dou l e t at neiTheghboringshows carbonstwo(twovinyldoupeaks one nei g h ori n g H blet: neighboring H . says that the two hydrogens are on H H 0 C-C=C-C-C C H ring NComo metbinhiylngstarehe piapparent eces: in the so the group of peaks at is most likely CH2 groups. 18-67
11:
IR
UV
225
11:*
2700-2800
e
16 10 em-I.
1690
-
96
o
,
'(
II
96 -64
I
=>
32
64
=
=
NMR
2
8
3
86.0
=
1
..-I--.
I
I
II
NMR,
+
+
1
+
6
1
82.0-2.4
6H
3
b b The mass spectral fragmentation can e explained y a "retro" or reverse Diels-Alder fragmentation: 0 0 H xC H TH 2 CHz �CHz Hz C H CHz H CHz loss of j3-unsaturated carbonyl I3 Ibnecause the of the resonance one of the form vinylthydrogens appears at Thi s i s t y pi c al of an a, hat shows deshielding of the -hydrogen. H H 0 H H :0: C-C=C-C-C C-C-C=C-C +
�
+
.
.
II
+
--
II
28
rnIz 96
rnIz 68
HNMR ,
87.0.
I
13
I
a
II
87.0 --..... +1
.... .. f-. -t .. �
431
13
I
a
I
(a) Building a model wil help visualize this problem. H H " ··H H H :O+-"":CH O CH20H :O +CCH:O CH20H :O-"":CH O CH20H HO OH open-chaiOHn fonn � HO y .y CH20H OH HO OH same as HO � OH OH cyclic fonn-hemiacetal (b) Yes, the cyclic fonn of glucose wil give a positive Tollens test. In the basic solution of the Tollens test, tconcent he hemiratacetion.al iHowever, s in equilibitriiusmthewiopen-chai th the open-chai n aldtehyde wisthwiththe cycl icifonn inevenmuchthough largerthere is n al d ehyde h at react si l v er o n, so open-chai nc fonn fonn present at tanyo replgiavceentthime consumed e, more open-chai of the open-chai n fonn isually all oxionlof tydhieazedcyclsmalbyiclsiamount lfonnver iwion,lofmore cycl i wi l open n fonn. Event carboxylate. Chatelbeier'dragged s Principkilecstkirinkgesandagaiscreami n! ng through the+open-chain fonn to be oxidized to oxidized attoe open-chain fonn NOTAg carboxyl cyclic fonn """"" latargerequiconcent at equilerliconcent brium ration reversible libriumration smal 18-68
• •
y I
w --
I
y I
I
• •
-
-
y I
as
the
Le
•
432
18-69
Any with singroups gle bondsan belongscarbonto thwie acetth twalofamioxygens ly. Ifbonded one of tthoeitoxygen groupketatal as iits came a hemifromacetaal.ket(Tohene.ol) d name for this g;rollp hemi (b) Models wil help. Ignore stereochemistry for the mechanism.
is OR, then the functional group is a hemiacetal. Thus, the functional C-2 is
H
OH
OH
1
HO
oo f' H+ o
*'
\
H
OH
00 o
,...
H
HO
HO
1
I
\.
(00 H 00,... +
HO
OH
OH
HOH2C
OH
OH
OH
OH
---" C-O
HO
OH
H2�
H
OH
HO
OH
HO
0
O� o
HO
same as
OH
433
H
18-70
Recall that "dilute acid" means an aqueous solution, and aqueous acid wil remove acetals. XMgJ 1__)) _ _ _+� � _ -=�. 2 H30 V U + Ht OH 0 OH H C H3C()' 3 OH �aBH4 � H )) CH3MgI + (2removes H30) acetal equi v . OH 0 0 + � cY OH excess Ag no reaction PCC H HO + OH H is identical to j H2Cr04 H20 0 PhNHN NNHPh OH �H
&H I
II
G
H
�
1
(\
--
..
B
(P� A
o
cr j HCIZn
E
c
(Hg)
(i'oH o
F
434
--
D
A
18-7 1
C, H HSfI fI fi H3 .. ' BuLi SH /S S /S S /" ' , , .. -----. C I fu Ph/"C, CH2 Ph HgCI2 Ph/"C CH2Ph PhCH 2 Ph H () H H H )(a H w H � ' C+ OH � d l...... _� H 0"""" � H �H /.. RH 0
II
W
�
__
II
2)
18-72
o
0
0+
1)
--- (1-:
�
Cl:
•
R
0=:OR
(b) This is not an ether, but rather an acetal, stable to base but reactive with aqueous acid.
(C)Q- .. H O:R
W
U
9-: - ROH. { QtC:o H}
H2� ! O O-O-HH -Ct.. H H2.. Q + ('O H O '--· -H · / .. H / . '-... t w
/ " �� +-H .. . 0 H �- � \L) l...... _ : H H2 H Oo
I
I
•
•
o
IIY"\
I :0-
H
+
J
O
• •
435
o
• •
..
:
-
U -H II I :0:
o
H
\
0
H
18-73
(a) First, deduce what functional groups are present in A and B. The IR of A shows no alkene and no carbonyl: the strongest peak is at 1060 cm-1, possible a C-O bond. After acid hydrolysis of A, the IR of B shows a carbonyl at 17 15 cm-1: a ketone. (If it were an aldehyde, it would have aldehyde C-H around 2700-2800 cm-1, absent in the spectrum of B.) What functional group has C-O bonds and is hydrolyzed to a ketone? An acetal (ketal)! ° R'O OR ' '\..
R
C
/ ,
A
II
/
R-C-R
B
R
mol. wt. 1 16
=>
C6H 1 ZOZ There is only one ketone of formula
C4HgO:
butan-2-one.
°
II
H3C - C - CHzCH3
t--... ..
B
A must have the same alkyl groups as B. A has one element of unsaturation and is missing only CZH4 from the partial structure above. The most likely structure is the ethylene ketal. Is this consistent with the NMR?
4H)
03.9 (singlet,
� HzC-CH2 I
o 1.3 (singlet, 3 H)
i
0
\ O
X�
H3C
A
-
CH3
0 1.6
}
00.9 (triplet, 3H)
(quartet, 2H) What about the peaks in the ion at 1 16.
+
'W H3C
MS at rnJz 87 and
WI? The 87 peak is the loss of 29 from the molecular
.
plus two resonance forms with positive charge on the oxygen atoms
CHz-CH3
rnJz116
..... _---I..�
The peak at rnJz 10 1 comes from loss of CH3 from the molecular ion at
1 16.
rnJz 101 436
plus two resonance forms with positive charge on the oxygen atoms
18-73
continued .
"c O r:"
o
:
H
/
"-
Me
1
:
1
C+ / "-
Me
:
-H .... ..1--__ • _
II
C / "-
Me
Et
p
3VH
:0:
.
H20
• •
Et
}
II
�
C / "-
Et
Me
B
Et
18-74 The strong UV absorption at 220 nm indicates a conjugated aldehyde or ketone. The IR shows a strong carbonyl at 1690 em-I, alkene at 1625 em-I, and two peaks at 2720 em-I and 28 10 em-I -aldehyde! o
II
C==C- C-H proton at 89.5 split into
The NMR shows the aldehyde a doublet, so it has one neighboring H. There are only two vinyl protons, so there must be an alkyl group coming off the B carbon: ,..-----,. H H 0 H H 0 I I II The only other NMR signal is a 3H doublet: 18-75
(a)
0
II H P,-
EtO""'" I
......
�
I) R
C OEt 1 R
. -
·B
--
0 II
R
must be methyl.
......
D
� :O. R + " II +
• �
R
II
CH3-C==C-C- H
•
P,EtO""'" I C OEt I
I
I
R-C==C-C- H
.....
R './
C
"' R '
"crotonaldehyde" ·0· .r-" EtO· I ... EtO l.O " �: ; " EtO-P 0 EtO -P � O. •
.
I� I .
--
.
R- C-C - R '
I
R
I
1'\'-- 1 -R-C ..!.. C -R'
R' R
437
(EtOhP02-
+
I
R' / , C==C ' R'
�
I
R
I
R'
18-75
nued OEt� (b) conti(X : -OEt OEt (c) R
+
�
/1
�
+
(i)(EtOhP
Br
�
....
COOMe
.. (Et0}z0P, � .... COOMe � -EtBr �
� COOMe (CH3}zCHCHO MeO _
�
�
I t
(------... ..
(ii)
°
(EtOhP BrCH2COOMe .. (EtOhOP COOMe -EtBr "- � -' -- �) � y COO M eO+
+
------
-----
18-76
(a)
� NM H+ U e2 .. 0
H ('::�( )�.- H \!!.)C+ 1
�
(b) Aminoacetallinkage�� in the dashed bOXCS' 2 /.: : : -olD .----1 N�lo N HO HO : N : W W OH deoxyadenosine deoxycytidinOHe '0 I
:
I
-'
�-----
:
'0
.�----
)
I
1
--
438
((c) The first step in the mechanism in ectron painucla) ri.seosiprTheodtes,oninatthowever, roiogensn of ofthearethamie partDnNe'Asofelaromat ifcor rithnegs,aromat and tihcietyelofecttrhone ripaing.rs (areSeerequi r ed the ion solof tuhteioarn otomatproblicityemof these fornuclaedescript osidwie l not bases. ) Prot o nat i o n of t h e ni t r o gen hlel notacidprotis extonatremele they Nstrandong; dioccur ltherefore ute unlacidethsssewitnucl eoside wil be stable. part
16-42
CHAPTER 19-AMlNES
19-1 These compounds satisfy the criteria for aromaticity (planar, cyclic 1t system, and the Huckel number of 4n + 2 1t electrons): pyrrole, imidazole, indole, pyridine, 2-methylpyridine, pyrimidine, and purine. The systems with 6 1t electrons are: pyrrole, imidazole, pyridine, 2-methylpyridine, and pyrimidine. Th� systems with 1 0 1t electrons are: indole and purine. The other nitrogen heterocycles shown are not aromatic because they do not have cyclic 1t systems. 19-2 (a)
CH3 H3C- �-NH2 CH3 I
(d)
19-3 (a) pentan-2-amine (c) 3-aminophenol (or meta-) (e) trans-cyclopentane-l,2-diamine
(b)
NH2 CH3-CHCHO I
(c)
Q
N H3C/ ...... CH3 (e)
(f)
(b) N-methylbutan-2-amine (d) 3-methylpyrrole (f) cis-3-aminocyclohexanecarbaldehyde
19-4 (a) resolvable: there are two asymmetric carbons; carbon does not invert (b) not resolvable: the nitrogen is free to invert (c) not resolvable: it is symmetric (d) not resolvable: even though the nitrogen is quaternary, one of the groups is a proton which can exchange rapidly, allowing for inversion (e) resolvable: the nitrogen is quaternary and cannot invert when bonded to carbons 19-5 In order of increasing boiling point (increasing intermolecular hydrogen bonding): (a) triethylamine and n-propyl ether have the same b.p. < di-n-propylamine (b) dimethyl ether < dimethylamine < ethanol (c) trimethylamine < diethylamine < diisopropylamine 19-6 Listed in order of increasing basicity. (See Appendix 2 for a discussion of acidity and basicity.) (a) PhNH2 < NH3 < CH3NH2 < NaOH (b) p-nitroaniline < aniline < p-methylaniline (p-toluidine) (c) pyrrole < anil ine < pyridine (d) 3-nitropyrro\e < pyrrole < imidazole 19-7
(a) secondary amine: one peak in the 3200-3400 cm-i region, indicating NH (b) primary amine: two peaks in the 3200-3400 cm-i region, indicating NH2 (c) alcohol: strong, broad peak around 3400 cm-i 439
19-8
A compound with formula
lypes of
H,
with the
NH2
C4H"N
NMR 8l.15,
NMR
has no elements of unsaturation. The proton
appearing as a broad peak at
of hydrogens on the four carbons. The carbon
shows five
meaning that there are four different groups
also shows four carbons, so there is no symmetry in
this structure; that is, it does not contain a t-butyl group or an isopropyl group.
8l.05, 3H H. 8l.35, CHNH2 81.15, 3HCH CH2 2H NH2 -C HC-C3 H H CH, 80.90 H, / 8 . H3C-C� -CH-CH 8 l.05 8l.35 / NH� deshielded by the nitrogen; integration shows it to be one
The multiplet farthest downfield is a There is
a
a
peak at
the broad
multiplet at
80.90. The latter two signals must represent methyl groups next to a
and a
doublet at
triplet at
respectively. So far:
and a
I
The pieces shown above have one carbon too many, so there must be one carbon that is duplicated: the only possible one is the
and the structure reveals itself. 2
8 (multiplet)
(triplet)
3
I
(doublet)
8l.15
"'---r""'"
(multiplet)
(singlet)
(b)
44.7 25.8 4l.0 CH3 0 CH3CH2-N -CH2CH3 CH3C+ H2-CH + H2CH2OH C CH3 �12.51.41/ 7.t9 20l.t 9 10.t 0 63.t 6 --
(c)
I
(d)
II
(An older printing of the text used values of 13.8,47.5, and 58.2. The values shown here are taken from a spectrum.)
19-10 (a)
[H)C+;�2-r��1-CH2CHf iCH-) CH2-t-�H2 - CH)-CH2-�=CH2} j CH3CH229 58 CH3 �
mlz
87
.
+
mlz
mass
+
•
mass
15
(c) The fragmentation in (a) occurs more often than the one in (b) because of stability of the radicals produced along with the iminium ions. Ethyl radical is much more stable than methyl radical, so pathway . (a) tS preferred.
440
19- 1 1 Nitration at the 4-position of pyridine is not observed for the same reason that nitration at the 2position is not observed: the intermediate puts some positive character on an electron-deficient nitrogen, and electronegative nitrogen hates that. (It is important to distinguish this type of positive nitrogen without a complete octet of electrons, from the quaternary nitrogen, also positively charged but with a full octet. It is the number of electrons around atoms that is most important; the charge itself is less important.)
GOOD:
1+ 1
mechanism
�II f
� _J N
+-
D
-N
• •
I
+
�
a�� 0 :�l)
H
o
0
VERYBAD:
- N-
H
NO
--'. --
·
N
H
NO
+N
VERYBAD N does not have an octet of electrons
N
N02
�
6 N
not produced because of unfavorable in termediate
19- 12 Any e lectrophilic attack, including sulfonation, is preferred at the 3-position of pyridine because the intermediate is more stable than the intermediate from attack at either the 2-position or the 4-position. (Resonance forms of the sulfonate group are not shown, but remember that they are important!) The N of pyridine is basic, and in the strong acid mixture, it will be protonated as shown here. That is part of the reason that pyridine is so sluggish to react: the ring already has a positive charge, so attack of an electrophile is slowed.
441
�-o:OoCH3 j;�- o�H �5Hl c �' GOOD
19- 13
Cl
�
I
-Cl
0 0 _ _
__ ..
_N
N
N
�
N
N
19- 14 (a)
� : NH� � _ jlC /
N
-0 0
GOOD
Br
stabilized by induction from nitrogen
� NH2
(jib:, N
NH - �
:
This is a benzyne-type mechanism. (For simplicity above, two steps of benzyne generation are shown as one step: first, a proton is abstracted by amide anion, fol lowed by loss of bromide.) Amide ion is a strong enough base to remove a proton from 3-bromopyridine as it does from a halobenzene. Once a benzyne is generated (two possibilities), the amide ion reacts quickly, forming a mixture of products. Why does the 3-bromo follow this extreme mechanism while the 2-bromo reacts smoothly by the addition e limination mechanism? Stability of the intermediate! Negative charge on the electronegative nitrogen makes for a more stable intermediate in the 2-bromo substitution. No such stabilization is possible in the 3bromo case. 442
19-15
H
..--....
Pr- N-H I
H� n n ..--.... CH3 - I + HC03CH3 - I .. .. Pr- N-H � Pr- N -
I),
H
19- 16
(b)
excess NH3
+
(c)
excess NH3
+
Br
�
---
H2N
�
PhCH2Br
19- 17 ° II
(a) CH3C -NHCH2CH3
19- 18
If the amino group were not protected, it would do a nucleophilic substitution on chlorosulfonic acid. Later in the sequence, this group could not be removed without cleaving the other sulfonamide group.
6�
NH-� ====::> A I Y � J
NH-S03H /
19-19
/
c.
I
�
both sulfonamide,
2NH2
sulfathiazole continued on next page 443
¢
continued OCHl
19-19
+
NH2
6
S02CI
19-20
(b)
(a)�
(d)
H3C,
9'-'::
NH2
NHCOCH3
(e)
,CH3
CXJ
�
HCI H20
h-
!1
S02 NH I
lll
6
t?
9
•
S02 NH I
6
sulfapyridine
+ ,CH3
V \H1
CH3 N
o a CH3 H1C-N�
H3C,
,..CH3 N CH3
I
(c)
,
(I)
+
H2C=CH2
�N-CH]
19-2 1 Orientation of the Cope elimination is similar to Hofmann elimination: the less substituted alkene is the major product. HO' � N (b) H2C=CH2 (a)� � minor major (CH3hNOH + (CH3CH2)2NOH
+ + �
+
(c)
0
+
(CH,hNOH
(d)
H2C=CH2
+0 N OH
+ r0 N�
OH nunor
19-22 The key to this problem is to understand that Hofmann elimination occurs via an E2 mechanism requiring anti coplanar stereochemistry, whereas Cope elimination requires syn coplanar stereochemistry. (a) H(CH3h H CH3 , .:- ...; Hofmann orientation loses a hydrogen from the CH3 and the N(CH3h C � / group to make the less substituted double bond � \ H 2C H
F
444
continued (b) Hofmann elimination
19-22
(CH3) 3N+ \� ,) "C H H3C
"I
5=H(CH3h f�CH3 C
"--\
---
HO: H�
H
",CH(CH3h 'I, IIC::::: C" � "' H3C CH3
..
"
,,\
E
(Saytzeff product-more highly substituted)
Cope elimination
..
,CH3 " " IIC-C � - "' H3C CH(CH3h H"
'1
,\\
/
Z
(Saytzeff product-more highly substituted) 19-23
o
+
Aliphatic diazonium ions are very unstable, rapidly decomposing to carbocations. (b) /' " NO N
(c)
�
19-24
i
(d)
NO I
o
+
N2
6
ClAryl diazomum IOns are relatively stable If kept cold.
The diazonium ion can do aromatic substitution like any other electrophile.
�
� N=NV� N=N � --- -03S -o-oN(CH3h �(CH3h � HO . . .. plus two other resonance forms most slgmflcant resonance contnbutor .. with positive charge on the ring .. :Cl:
03S
_
-
03S
j
H I+
_
_
(or some other base)
-< >- N N -< >- N(CH3h =
methyl orange
445
}
19-25
(a)
(b)
8
HBF4 NaN02 HCI a CI N2+ CICuCI •
a
-
t::..
6
..
from (a) (c)
N2+ CI-
F
6
°
°
II
�
II
HN-CCH3 2 * CH3C-;I : I CH31- CH3'¢rI CH3 H30+ CH3 � I CH3 NaN% 2 HCI CH". CH3 H CI 2 N q CH3 CH3 H3P02 CH3 CH3 �I �I '¢r CH3 CH3 N2 CIB N2 CICuBr
8
(SCCHl
3
NH
�
t::..
+
9'
•
+
(d)
a
..
(h)
+
N2 CI-
a +
N2 CI-
+
r
from (a)
(f)
9'
..
CuCN
..
6 CN
(g)
6
a HO� }-OH +
(e)
a
I
KI
..
+
N2 CI-
H2O a H2SO4 ..
..
t::..
< }-N'Ni }-OH HO 446
6 OH
6
delinNa(CH3COOhBH es for choice of reagent forutioreduct ivethamie imniatnieoorn: iuseminLiiumAliHon4 when tihseoliamteid.ne or oxiAltmernate iGeneral siviselolya, tcateguid.alUse i n sol n when i s not ytic hydrogenation works in most cases. 0 (a) (i' H NOH 0 NH2 H2NOH Li A l H 4 PhCH2 -C -CH3 PhCH2 - C -CH3 H2O PhCH2 -CH -CH3 (c) H CH2Ph 0 Na(CH3COOhBH 0 Ph)l H 0 N,Ph (d) 0 NHPh LiAlH4 H+ H2O 6 6 6 NOH (e) 0 H2NOH LiAlH4 OR H2O 6 6 H 19-26
(b)
II
if
1)
II
..
2)
I
I
..
+
W
2)
I
0 +
+
1)
-.
..
2)
0
Cl �
CI 'l)
---
a
..
2)
1)
..
19-27
(a) H
6
I
I
+
(b)
..
c5
0-r
c�H
LiAlH4 r" 00 H20 0 HN� HNJ) 6 I)UAIH20H� 6 :::::... 1)
..
2)
A
---
A
2)
:::::...
447
le alkylations of each NHnitr2ogen. Use a large excessBrof ammonia to avoid multip� NH3 � excess
19-28
�N:
19-29 (a)
+
____
C
o
o
BrCH2Ph
..
o
(b)�
o
v--t
o
-
:::-...
..
!l
o
I
!l
•
(c)� Br(CH2hCOO- � N: must use anion :::-... I N -(CH2hCOO- NH2NH2 H+ � tthoeavoiphthdalprotimideonataniingon 0 Assume that LiAIH4 or Hz/catalyst can be usedH interchangeably. PhCH2Br NaN3 PhCH2N3 Pt2 PhCH2NH2 � Br � � C = N LiAlH4 � NH2 H20 • V V V (c) 0 LiAIH4 TsCl Ts H20 pyridine � � + NaN3 Li A I H 4 �NH2 H20 �N3 0 C2 NH3 � � � Cl NH2 LiAIH4 + H20 �NH2 OTs��C=N LiH20AIH4 �NH2 � from (c) o
o
o
o
' !l ...
1)
-4_
----
..
2)
o
19-30
(a)
+
..
--
(b)
1) 2)
OH
1) 2)
•
�
OH
O
..
1)
•
2)
OR
OH
�
SO l
..
o
o
-
1)
-
1)
2)
448
..
2)
H2N(CHlhCOOH
continued (e) Br\�H NaN3 H N3 LiAIH4 � � � � inversion Br\�H NaCN �H2NH2 Li A I H 4 � inversion (g) 0 HO� CN H � Li A I H 4 KCN � HCN Tortualreduce nirchangea troaromatblyic. s,Assume the reducia workup ng reagentin base s (H2toplgiusvea tmethe freeal catamialynste, fiornala metproduct al pl.us HCI)can be used vi l y i n t e (a) nHCl o Br Br Br (b) FeHCI 6 o N0orth2o nHCI )) f; U Br U Br from (a) COOH COOH (d) FeHCI Jr 6 U NO CH30 y-:�H :0: �' 0 R-C- � : R-C=N,: R-C N) Br hCH T-C � H H H CH3 H H ( , - H 19-30
SN2-
R
(D
SN2-
R
..
1)
..
1)
NH2
•
S
..
1)
..
S
..
19-3 1
S
•
¢ ¢ •
+
S
2)
19-32 P
2-
�
,
II
,
-
. �.. ,"-,
2
II
.. -
449
----
..
:0: } ,
B,
U
II
.. '
�
mechan;,m cont;nued on next page : �
!
19-32 continued
:OH R-N=C=O .. � --- BrII ..R-N=C-OH R-C=N-Br --R-C-N-Br i :0:- -.. � U· i �U (:0:- "" .. t t t �H OH II � -.. II ?� :OH2 R-N:I - CO2 ----l.. R-NI.R-N-C-OH H""OH.. R-..7� CH HJ H � ..
I
:0:
�
• •
:0:
'-,1
:0:
..
..
+
-
�
.
H
H
'" at a chi r al carbon i s l o st i f t h e carbon goes t h rough a pl a nar Stinteereochemi s t r y � NH2 rmedi a t e , ei t h er a carbocat i o n or a free radi c al . Confi g urat i o n at a chi r al carbontcane thebeleiavinvertngegroup. d durinHowever, g substitutwhen ion bythaenuclcarboneophiretleaifrom thfoure sidpaie rs of opposi n s al l H CH3 electrons, as in this Hofmann rearrangement, it retains its configuration. 1(a)9-34The acyl azide of the Curtius rearrangement is similar to the N-bromoamide of the Hofmann rearrangement hat bothehave an amiydsiesnithtrrough ogen withethcarbami a good cleaciavidngtogroup mimechani gratiosnms.to theinistocyanat and hydrol the amiatntaeched. are idSubsequent entical in botalhkyl (manb) Theor beast leavi"n, gasgroup in thtoe say.Curtius rearrangement is N2 gas, one of the best leaving groups known "to we used (c) ,----------------,, e.g.,
R
.-----------------,
abbreviate as "R"
19-35 Please refer to solution 1-20, page 12 of this Solutions Manual. 6 mary amine; 2,2-dimethylpropan-l-amine, or neopentylamine ((b)19-a)3prisecondary amierocyclne; iN-met heylandpropan-2agroup; mine, or3-niisopropyl minetehylamine t e rt i a ry het c ami n a ni t r o t r opyri d iec oxiammoni umhylio-n;N-N,mNet-dihylmaetnihliylnepioxiperiddeinium iodide (d)(e) tquatertiaeryrnaryaromatheteicrocycl ami n d e; N-et ic amiic nammoni e; N-ethuylm-N-ion;metpyrihyldaininiluimnechloride (f)(g)tteertrtiiaaryryaromat het e rocycl (h) secondary amine; N,4-diethylhexan-3-amine 19-37 ShownH in order of increasing basicity. In sets a-c, the aliphatic amine is the strongest base. (a) Ph-N-Ph < PhNH2 < o NH2 aliphatic amine is the strongest base H H (b) N < N < aliphatiicciamity woul ne isdtbehe lstorstongest base; pyrrol e's aromat i f prot o nat e d 0 pyrrole 0 (c) CNH < HN "N < aliphatiicciamity woul ne isdtbehe lstorstongest base; pyrrol e's \ I aromat i f prot o nat e d pyrrole Basi c i t y i s a measure of t h e abi l i t y to NH2 < N 2 H O (d) O donat e t h e pai r of el e ct r ons on t h e < I I amine. lElikeectN0ron-wi thdrawibasingcity, H3C 02N groups decrease 2 whileinelcrease ectron-donat CH3 basicityin. g groups NH2 cr CH2NH2 aliphatic amine is the strongest base (e) � O NH 2 < I I < I (amides are not basic) (c)
I
I
�
�
h
h
°
h
h
like
h
Congrat ritgyhton! CH3 is onlThey second slightlystelructectruon-donat ingbasibycin(byductabout ion so1 tpKhe first stbecause ructureuNputlaHti2soinstsheleifleeyouctastron-donat elgotectthronisinonedensi t h e re i s more g ofbyresonance. resonance. (See The ltahste stlarstucttoupireciisnthAppendi e strongestx Twobaseofforthisa Manual complet.e)lyThe diortfferent reason: st e ri c i n hi b i t i o n substconjituuentgatsedonwiNthandtheiaromat t forcesicthpie NRsyst2em.out ofEssentplanariial tyy, twihitsh the ribecomes nhg,osometthanhatylalthiinepthateelrferes ecticramine. onwipaithrthonThee itshopropyl epKbN iofs not2-met hyly -alN,ipNhat-diiectamihylannielbecause ine is 6.9th, eabout 2.5icpKrinunig istssomewhat more basic telheanctranionlwiine.thdrawi It is notng asbystinrongductiaobase as a t e rt i ar aromat make the N significantly more basin,cyet. the ortho methyl reduces the pi overlap of the nitrogen's electrons to unit)
N.
N
451
19-38 (a) not resolvable: (c) not resolvable: (e) not (g) not resolvable in conditions proton on N can exchange
planar ric symmet resolvable: symmetric where the pKa he weaker of the aciditywiandpKbth tbasi city. )
19-39 The values of side reaction discussion of (a )
vvablablee:: asymmet rnicversicarbonon is very slow (b) resol resol ni t r ogen i (h)resol resolvvablablee:: asymmet asymmetrriiccninittrrogen, ogen, unabl unablee to iinnvertvert be favored at equilibrium. (See Appendix forThea � �JII + CH3COOpKa I[J + CH3COO pKa (d)
(f)
to
of amines or of the conjugate acids can be obtained from Table 19-3. acid and base will 2
products are favored
N+ I
H
(b)
I[)
..
--
N
H
(c )
d)
H
.N,
reactants are favored
H
"'- 1
o H
(
8.75
I O pKa
+ NH3
�
+
4.60
19-40
(a) PhCH2CH2NH2
Q H
o N
products are favored
+
NH2
pKa
n
OI pKaH N+,H �
+
....
1 1.27
( ) e
19-4 1
(a) PhCH2CH2CH2NH2 (b) �NH2
(e) (D CH3
1 1. 12
�N I
452
e) (
products are favored
cr(\�
d��-
(h)
NH,
retentgiuraton ofion confi CHJ eN, (d
)
o
H2after workup N OI with base �
OH
og
19-4 1 (i)
continued
(I)
U)
NHCH1
(m) /"'N�
CH2NH2 I
�
LiAIH4
..
H20
HO CN
CH2CH2NHCHl
- � HO CH
19-42
(k)
(b) CH1
Q
2NH2
N02
'O
from (a)
(c) CH1
'O from (a)
(d) CH1
'O
0
CN
+
N2 Cl-
Kl
+
N2 Cl-
H2S04 .. H20
�
(f)
CH3
I
'0I �
CH1
•
0
+
°
�
CN
I
o
'0I �
CH3
NHCOCH3 HN03 H2S04
CH3 after workup with base °2N
NH2
I
/
CH3 � OH
° CH3 NH2 CH3C-Cl II
0
o
---
from (a)
(e) CH3
CH3 � CH2NH2
LiAIH4 .. 2) H20
1)
qH
OCH2CH3
+
0
I
(0)
CH3 (a) CH3 � NH2 NaN0 CH3 � N2 Cl2 Cuctt HCl
o
NHCH3
CH3(CH2)3CHCH2CH3 (n)
�
PhCH2CHCH3 (p)
0
=<J Na(CH3COOhBH453
0 ?'
�
I
NH2
°2N
0I �
+ JJH3O
CH1 �
o
!1
� --o
NHCOCH3 (+ isomer)
19-43 This fragmentation is favorable because the iminium ion produced is stabilized by resonance. Also, there are three possible cleavages that give the same ion. Both factors combine to make the cleavage facile, at the expense of the molecular ion.
t?
+ .
CH3
CH3
I
-NH2
CH3 ---
CH3
oCH3 + mass 15
.... ..f-. -l.. �
mlz 58
mlz 73 19-44
(b)
0 O �I
�I
�-o
Cl-C
..
NH2
+
1)
(e)
(I)
I
CH3
58
(a)
I
CN-CH3 C
N-CH3
H202
H202
..
..
/N�o C CH3 /o N� C CH3
..
LiAlH4
t:.
----
(g) CH3 Y'1( COOH SOCl2
o
..
454
(
oH N. "'CH3
+
C = NH2
19-45 The problem restricts the starting materials to six carbons or fewer. Always choose starting materials with as many of the necessary functional groups as possible.
HO-Q-NH2 (a)
phenacetin
Mg
-< )
HO
NBS
O
..
HO HO
19-46
O
HO
HO
(c)
CH2CH2NH2 ..
H2, Pt
dopamine
455
HO
methamphetamine CH2B'
� continued
19-46
(b)
H H 2 2 N� 00=7 Pt
�n C
--
o
H
r'Y:l -
��: !)
I
H
H :.
l bas :
co HH
19-47
(a)a0C 'OH SOCl2
� H+
H NH2
TI
1
- H20
..
1
• •
..... N. co +
I
H
t H2 Pt (X)� � � N) ..N base: �
---
1
R:
NH2
Wco H+O: H HO:..� 1 } H 1
H2 CO ) NH,
�
+
o
o
1
H)
, C ...Cl NH3 o C CH2NH2 Li A I H 4 NH 2 r?'Y r?'Y 2)H20 0 aI (b) c' a CH2NH2 Na H3COOhBH (C H I aI NH3 (c) C LiAIH4 N - CH2CH3 NH CI-C-CH3 CN-C-CH3 2)H20 " C � "
•
I I
:::::-...
I I
1)
..
o I I
:::::-...
+
�
o II
•
:::: :::: : :-...
1)
456
..
continued r-, Na(CH3COOhBH 0- r-, .. N0 0 o HN0 NH H2N � NH2 (e) HO� OH SOCI2... CI �CI � o 0 o 0 / 0 0 LiAl y 2)H20
19-47
(d)
+
_
1)
0
19-48
(a) �Br CH2CH3 (b)
6 �
(c)
+
-N
HNO] .. H2SO4
I
�C
0
0
--
NH2NH2 �NH2 CH] ..
L1
Q QNH2 I
�
� OH Ph
H3
H2N�NH2
N02
SOCI2
...
Sn '" HCI
NH3 � NH2 0 0 Nao� � H2 N � NH2 NH2NH2 ____
� CI Ph
Ph
Ph
(d) �Br
+
Ph
-� o
--
L1
...
Ph
(e)�Br KCN � CN 2 iAOH4"' � NH2 )H2 Ph
...
I
Ph
457
1) L
l
Ph
19-49 (a) When guanidine is protonated, the cation is greatly stabilized by resonance, distributing the positive charge over all atoms (except H):
..
� w ..
NH
II
H, N-C- NH,
1
�
..
NH2
II
..
..
NH2
� I
..
+ H,N-C- NH, �H,N- - NH, �H'N=
�
NH2 I
NH,
• •
• •
NH, I
+
}
H2N-C=NH2 . (b) The unprotonated molecule has a resonance form shown below that the protonated molecule cannot have. Therefore, the unprotonated form is stabilized relative to the protonated form. This greater stabilization of the unprotonated fonn is reflected in weaker basicity.
..-o�
H2N
o· . + // N \
-
..
:0:
(c) Anilines are weaker bases than aliphatic amines because the electron pair on the nitrogen is shared with the ring, stabilizing the system. There is a steric requirement, however: the p orbital on the N must be parallel with the p orbitals on the benzene ring in order for the electrons on N to be distributed into the TC system of the ring.
If the orbital on the nitrogen is forced out of this orientation (by substitution on C-2 and C-6, for example), 3 the electrons are no longer shared with the ring. The nitrogen is hybridized sp (no longer any reason to be 2 sp ), and the electron pair is readily available for bonding � increased basicity. H
\/
H
C ___
As surprising as it sounds, this aniline is about as basic as a tertiary a liphatic amine, except that the aromatic ring substituent is electron-withdrawing by induction, decreasing the basicity slightly. This phenomenon is called "steric inhibition of resonance". We will see more examples in future chapters. A lso, it is the last topic in Appendix 2. (See another example in the solution to 19-37(f).)
H
458
19-50 In this problem, sodium triacetoxyborohydride
(a)
/'.... /'.... ........, -O
/
( b)
TsCI
---
H
pyridine
will be represented as Na(AcOhBH.
KCN . /'... /'....... . /'... /'....... ...., -. OTs --...., -. CN
..
1) LiAIH4
/
/
2)
H20
/'.... /'....
/
........,
........,
_
H C H2 3 --- � .. 3 �NH2 cr03 H2 �H 2 0.. H20 � 0 CH3-CH tpcc CH3CH20H " -H � �CH20H (r � a Na0't V PctV aCH2NHCH2CH2CH] �NH2 (rCOCI ac-" NH2 (rCH3 � I I I PCC
� OH
(c) �OH
TsCI
O
�OTs
HN
Na(AcOhBH
NH
�
NHCH3
•
pyridine
H S 4
O
./"--./ II o
Na(AcOhBH /"'--.../
+ �N
I HN�
•
(ACOhBH
/"'--.../ I
I I
�N�
(d)
CH3
o
C
CH2Br
•
Na(AcOhBH
from (c)
(e)
o
COOH
•
KMn04
�
o
·
---?"
�
NH3
j
�
Br2 NaOH
N02 � NH2 Na .. 0 0
�HNO] � HCI
(I)
0 � I .&
0
�
�OH
AICI3
CI
•
H2Cr04
I
.&
SOCI2
.. ---
0
•
�
.&
H2SO4
°
N02
CI
459
J
zna[g
HCI Clemmensen reduces both
y .&
NH2
NH2
o� y� � � IuY U ,? 5 Y Y �oI °
19-50 continued (g)
'
I
AICl3
/
H2C
OH
19-5 1
'
Zn(Hg
°
Cl
II
CI- CCH2CH2CH3
Zn(Hg)
�
b( ) O Ph
-Q�
19-52
h
HCI
HN03 Fe, HCI • � H 2SO4 h H2N after workup with base
°
(a)
(c)
f)I I
l
S
� '''''
•
'--C
N-CH1
Hofmann elimination
�
N02
HCI
t)
�
HCI
H
Ph
-Q-
NH2
H
I
coniine
460
19-53 unknown X (a) -fishy odor
�
amine
-molecular weight
10 1
�
�
odd number of nitrogens
if one nitrogen and no oxygen, the remainder is C6H)S
Mass spectrum: -fragment at
86
=
M
-
15
=
loss of methyl
�
have this structural piece: CH3 IR spectrum:
-no OH, no NH
�
� a-cleavage N
the compound is likely to -
must be a 3° amine
-no C=O or C=C or C=N or N02 NMR spectrum:
--only a triplet and quartet, integration about
3
:
2
�
ethyl group(s) only
assemble the evidence: C6H)SN, 3° amine, only ethyl in the NMR:
(b) React the triethylamine with HCI. The pure salt is solid and odorless. (CH3CH2hN
+
HCI
+
-.
(CH3CH2hNH salt
CI -
(c) Washing her clothing in dilute acid like vinegar (dilute acetic acid) or dilute HCI would form a water soluble salt as shown in (b). Normal washing will remove the water-soluble salt.
---
yQ o
H
H
(b) Both answers can be found in the resonance forms of the intermediate, in particular, the resonance form that shows the positive charge on the N. This is the major resonance contributor; what is special about it is that every atom has a full octet, the best of all possible conditions. That does not arise in the benzene intermediate, so it must be easier to form the intermediate from pyrrole than from benzene. Also, acylation at the 3-position puts positive charge at the 2-position and on the N, but never on the other side of the ring, so this substitution has only two resonance forms. The intermediate from acylation at the 3-position is therefore not as stable as the intermediate from acylation at the 2-position. 461
19-55 (a) OZN
'Q
000
F
�
�
I
oo N H zR
0 0
:0'" ..
..
+ NH2R
o o F li
_
Y
'c.
N:"
I
0
�
+ ... N....
N02
0'"
y �
0 0
-
0 : 0'" 0
�..
� 1j ... .. ...
0o-; 1
N+
�
_
0 '"
0
+ N....
(b) Why is fluoride ion a good leaving group from A but not from F _ HC
Nu
I
N02
NOz
A
�� A
B
0II
+ N HzR
F
I
-0'"
�
_
N+
+ NH 2R
F
I
00
:0'"
0
C1 N+
I
0 0-
0:
B (either by SNI or SN2)?
:NU
C
Formation of the anionic sigma complex A is the rate-determining (slow) step in nucleophilic aromatic substitution. The loss of fluoride ion occurs in a subsequent fast step where the nature of the Jeaving group does not affect the overall reaction rate. In the SN 1 or SN2 mechanisms, however, the carbon-fluorine hond is breaking in the rate-determining step, so the poor leaving group ability of fluoride does indeed affect the rate.
t
t
>-.
lrl�
________________
nucleophilic aromatic substitution
(c) Amines can act as nucleophiles as long as the electron pair on the N is available for bonding. The initial reactant, methylamine, CH3NHZ' is a very reactive nucleophile. However, once the N is bonded to the benzene ring, the electron pair is delocalized onto the ring, especially with such strong electron withdrawing groups like N02 in the ortho and para positions. The electrons on N are no longer available for bonding so there is no danger of it acting as a nucleophile in another reaction. 462
delocalized ,
N02
19-56
Compound
A
Mass spectrum: -molecular ion at 73
=
=
odd mass
odd number of nitrogens;
if one nitrogen and no oxygen present
- -, t
-base peak at 44 is M - 29 N I
44
�
�
molecular formula C4H11N
this fragment must be present: CH3-
CH2CH3
�
EITHER
OR
�
H-C-CH2CH3
I
H
a-cleavage
2
IR spectrum: -two peaks around 3300 cm-I indicate a 1 ° amine; no indication of oxygen NMR spectrum: -two exchangeable protons suggest NH2 present -I H
A
multiplet at 8 2.8 means a CH-NH2
The structure of A must be the same as 1 above:
Compound B an isomer of A, so its molecular formula must also be C4H11N
IR spectrum:
-only one peak at 3300 cm-I NMR spectrum: -one exchangeable proton -two ethyls present The structure of B must be:
�
�
2° amine
NH
B +
-
CH2 = NCH2CH3
I
H rnJz 58
resonance-stabilized
19-57 (a) The acid-catalyzed condensation of P2P (a controlled substance) with methylamine hydrochloride gives an imine which can be reduced to methamphetamine. The suspect was probably planning to use zinc in muriatic acid (dilute HCl) for the reduction.
0Jl
+
-
Cl
methamphetamine
phenyl-2-propanone, P2P (phenylacetone)
(b) The jury acquitted the defendant on the charge of attempted manufacture of methamphetamine. There were legal problems with possible entrapment, plus the fact that he had never opened the bottle of the starting material. The defendant was convicted on several possession charges, however, and was awarded four years of institutional time to study organic chemistry. 463
1 9- 58 Mass spectrum: -molecular ion at 87
=
odd mass
-if one nitrogen and no oxygens -base peak at mJz 30
�
=
�
odd number of nitrogens present molecular formula CS H13N
R
structure must include this fragment
IR spectrum:
1 -two peaks in the 3300-3400 cm- region
�
t
H H I \ -.. C = N+ \ / H H
CH2NH2 30
mJz 3 0
1 ° amine
NMR spectrum: -singlet at 80.9 for 9H must be a t-butyl group -2H signal at 8 1 .0 exchanges with D20
�
80.9
must be protons on N or
{
°
8 1.0
CIl3_
�
H3 CH2-
� t H3
l
,
82.4 Note that the base peak in the MS arises from cleavage to give these two, relatively stable fragments:
CH3
I
CH3 -C·
1
CH3
+
H H / \ C=N+
/
H
\
H
mJz 30
19-59 (a tough problem)
molecular formula CII Hl6N2 has 5 elements of unsaturation, enough for a benzene ring; no oxygt:ns precludes N02 and amide; if C:::N :: is present, there are not enough elements of unsaturation left for a benzene ring, so benzene and C=N are mutually exclusive
IR spectrum: --one spike around 3300 cm-1 suggests a
2° amine
-no C=N
-CH and C=C regions suggest an aromatic ring Proton NMR spectrum: - 5H multiplet at 8 7.3 indicates a monosubstituted benzene ring (the fact that all the peaks are huddled around 7.3 precludes N being bonded to the ring)
-1 H singlet at 8 2.0 is exchangeable
�
NH of secondary amine
-2H singlet at 83.5 is CH2; the fact that it is so strongly deshielded and unsplit suggests that it is between a nitrogen and the benzene ring
< >-
fragments so far:
CH, -N
+
NH
continued on next page
+
4C
+
8H 464
+
I clement of unsaturation
19-59 continued Carbon NMR spectrum: -four signals around 5 125-138 are the aromatic carbons -the signal at 5 65 is the CH2 bonded to the benzene -the other 4 carbons come as two signals at () 46 and () 55; each is a triplet, so there are two sets of two equivalent CH2 groups, each bonded to N to shift it downfield fragments so far: CH2 CH2-N + NH + CH2 CH2 + CH2 + 1 element of unsaturation
<}
There is no evidence for an alkene in any of the spectra, so the remaining element of unsaturation must be a ring. The simplicity of the spectra indicates a fairly symmetric compound. Assemble the pieces:
19-60 (a) Ph
H
NMR
� N�
Jl
(b)
�H CH3 �a(AcO h BH 0
11
�
V
(
N�
(
HN� 465
1 9-6 1 Not only is substitution at C-2 and C-4 the major products, but substitution occurs under surprisingly mild conditions. Begin by drawing the resonance forms of pyridine N-oxide: ..:0: :0: :0:
:0: 1+
.. ..
0
11+
11+
. .. .. .. .. OR (�) N �
11+
RO - N
CH Resonance forms show that the electron density from the oxygen is distributed at C-2 and C-4; these positions would be the likely places for an electrophile to attack.
-
Resonance forms from electrophilic attack at C-2 .. :0:
a 1++
N
E
.0
H
a .. (:f :0: II +
"
N does not have an octet; not a significant resonance contributor
1+
E
N
_
·. ·0··.
0
N
H "
HC +
GOOD! All atoms have octets.
Resonance forms from electrophilic attack at C-3
( \ .. ·. ·0··. 1+
N
H
+
C H
-
E
1+
Q
H�
,,=
�
1+
H "
-
OR H
E
..
continued on next page
�
N
H "
CH
-
�
E
+CH
H
These two Eresonance forms place twoE positive charges on adjacent atoms-not good.
-
Resonance forms from electrophilic attack at C-4 .... :0: 1+
N'
c� 1+
E
a ..:0:
·. ·0··.
N
0
.
·. ·0·....
:0:
1++
-
0
.. ..
H E N does not have an octet; not a significant resonance contributor
466
:0: 11+
..
.-
. :0: 1+
0 -Q
H E GOOD! All atoms have octets.
HC +
H
E
H
19-61 continued The resonance forms from electrophilic attack at C-3 are bad; only one of the three is a significant contributor, which means that there is not much resonance stabilization. When the electrophile attacks at C-2 or C-4, however, there are two forms that are good plus one great one that has all atoms with full octets. Clearly, attack at C-2 and C-4 give the most stable intermediates and will be the preferred sites of
(f0: �C-!l
attack.
19-62 (aJ
O
�
<>?: 0/9.: H
H
I
I
.... ..f--l . . �
H� NC I!3
CH3 H-N-H - Q.H + I�
O 1�
B- is the conjugate base B: of the acid HB CH3 CH3 I I H-N: H - N: C-OH H C '-""1+ H
,...... d 9.. � d·· � -
�
called an aminal
continued on next page
467
19-62
(b)
cr
.�
continued O:'� H -B
+C
..
-
q
B- is the conjugate base of the acid HB
0
C-OH • •
11
B:
. -H20 . / �H�_ O�:·· + o�:. aD -' �� � 0::0 � (X/O C
o C
/'
I
-
H
OH H�B -
0 C
-
OH
enamme
To this point, everything is the same in the two mechanisms. But now, there is no H on the N to remove to form the imine. The only H that can be removed to form a neutral intermediate is the H onC next to the carbocation.
(c) A secondary amine has only one H to give which it loses in the first half of the mechanism to form the neutral intermediate called an aminal, equivalent to a hemiacetal. In the second half of the mechanism, the H on an adjacent carbon is removed to form the neutral product, the enamine. The type of product depends entirely on whether the amine begins with one or two hydrogen atoms.
468
CHAPTER 20-CARBOXYLIC ACIDS
20-1
CH3 I (a) CH3CH2 CHCOOH (d)
(g)
(
)
COOH
Cl �
'>
H
tH
(c)
OH
cf:�:�
(f)
H
(h)
CI
COOH
N � COOH
.
(1)
CH3
(j) HOOC � COOH Cl
(k)
o
� COOH
(I)
N H2 I COOH HOOC � -
0(y 0 H
V a
20-2 IUPAC name first; then common name. (a) 2-iodo-3-methylpentanoic acid; a-iodo-p-methylvaleric acid (b) (Z)-3,4-dimethylhex-3-enoic acid (c) 2,3-dinitrobenzoic acid; no common name (d) trans-cyclohexane-1,2-dicarboxylic acid; (trans-hexahydrophthalic acid) (e) 2-chlorobenzene-\ ,4-dicarbox ylic acid; 2-chloroterephthalic acid (f) 3-mcthylhexanedioic acid; p-methyladipic acid 20-3 Listed in order of increasing acid strength (weakest acid first). (See Appendix 2 for a review of acidity.) Br CH3 - CCOOH (a) CH3CH2COOH < CH3 - CHCOOH < Br Br The greater the number of electron-withdrawing substituents, the greater the stabilization of the carboxylate ion. (b) CH3CHCH2CH 2COOH < CH3CH2CCH2 COOH < CH3CH2 CH2 CHCOOH I I I Br Br Br The closer the electron-withdrawing group, the greater the stabilization of the carboxylate ion. (c) CH3CH2 COOH < CH3-CHCOOH < CH3 - CHCOOH < CH3-CHCOOH N02 o C�N The stronger the electron-withdrawing effect of the substituent, the greater the stabilization of the carboxylate ion. I
I
I
I
I
469
I
�COOH �CHO � CH20H H2S04 "Y shake with ether and water ether water �COOH isoln etublhere �CH20H �CHO shake with NaOH (aq) ether NaOH (aq) ionized form � CH2OH ca� b oxyl i c t a COON NaOH � byunchanged sol u bl ........",- -CHO acidify with HCl (aq) and ether in e HCl (aq) ether NaCl �COOH evaporate ether t �COOH
20-4
{
/"'-..
./
/"'-..
........",-
of aCId is water
/"'-..
470
20-5 The principle used to separate a carboxylic acid (a stronger acid) from a phenol (a weaker acid) is to neutralize with a weak base (NaHC03), a base strong enough to ionize the stronger acid but not strong enough to ionize the weaker acid.
�-o- OH
< }- co,
0° y
shake with ether and NaHC03 (aq) NaHC03 (aq)
ether
< }- cooNa
A
,-------- -------�\ � OH + CH3
\..
-Q-
y
!
00
)
1. add HCl 2. filter or extract
� COOH � - pure
shake with NaOH (aq)
NaOH (aq)
ether
-Q- oNa ! 2.1. filter add HCl or extract CH3 -Q- 0H
00
!
CH3
evaporate
Q�
pure
471
20-6 The reaction mixture includes the initial reactant, reagent, desired product, and the overoxidation product-not unusual for an organic reaction mixture.
< :N
(a)
eHO COOH er03 H e 2 � H _0- -----------) �
�
�
,,
____________
ether
�eOOH
"N C
y
(
)
shake with ether and water
some compoundS have appreciable solubility in both ether and water
water
o
�
b.p. 137°
eHe 20H
�eHO b.p. 102° e
(b) Pentan-l-ol cannot be removed from pentanal by acid-base extraction. These two remaining products can be separated by distillation, the alcohol having the higher boiling point because of hydrogen bonding.
eOOH
20-7 The has a characteristic IR absorption: a broad peak from 3400-2400 cm-I, with "shoulder" around 2700 cm-I. The carbonyl stretch at 1695 cm-I is a little lower than the standard 1710 cm-I, suggesting conjugation. The strong alkene absorption at 1650 cm-I also suggests it is conjugated. a
472
20-8 (a) The ethyl pattern is obvious: a 3H triplet at b 1.15 and a 2H quartet at b 2.4. The only other peak is the COOH at b 11. 9 (a 2.1 b uni t offset added t o 9. 8). o
II
CH3CH2-C-OH
(b)
o
Q
II
£
H-C-CH2CH3 .i!
£
Q
2H
.i!
IH
1 :1
3H
II II
TMS I
8 7 6 b (p5pm 4 3 2 1 0 ) Theand tmulhe CH2tipletgroup. betweenThese2 andcoupli3ngis const drawnantass area pentprobabl et as ythough it wer, in ewhisplciht equal lhyebyactthuealalspldehyde protteornn unequal case t i t i n g pat wi(c)l Thebe achemi complcealx shimulfttiofpletht.e aldehyde proton is between b 9-10, not as far downfield as the carboxylic acid proton. Also, the aldehyde proton is split into a triplet by the CH2, unlike the COOH proton which always 10 I
9
I
I
<5
I
I
I
I
I
I
I
I
<5
e iins spltheitacibydan. extra proton, so it wil give a multiplet with complex splappearit insg,asinastsieadnglofet.thFie nquartally,etthshown CH2
20-9
:OH + C1 ........ / OH H2C==C
. :OH
• •
\
t
H
.
I
..
..
.. +OH
H2C==C
.. :OH
II
C . / ....... OH
\
H
C ........ h + :I" OH H2C-C\ H I
..
..
+ /C� OH ""' H2C==C \
H
473
[ CH3CH2tCH' H0-OH ]
20-10
ig
87 CH3 m/z1l6
McLafferty rearrangement H3C I1�H� : �ft'C 'OH CH3 m/z1l6
t
:OH CH3masCs H,· H2C==C/C,9,..H CH3 plasushown s resonance im/z n8720-9forms H,0..: H3C,CH CH2 HC/',/.C,OH mass 42 CH3 29
-
+
+
I
\
t
t --
II
I
+
I
mlz74
(a) �C C� orconc.1 ) 03,KMn04 2) H20 �COOH C-:::) conc. KMnO� C:COOH COOH H 0+ CH2CH20HH2Cr04 6CH2COOH (c ) < }- Br etMgher < }- MgBr D H30+ 6 (d) � PB r3 � Br Mg CO2 H30+ COOH � ether COOH (e) conc. KMn04 t"H30+ CH3 COOH Mgether CO2 H30+ �COOH OR KCN �CN H30+ �COOH 20-11
•
(h)
t, ,
3
o
�
-
�
(f)
--
Q
�I
�I
•
--
---
•
¢
---
---
�
•
t,
•
474
�
I
•
�
I
+ .. .. } :O-H :O-H :O-H first intennediate RC-�H RC+- OH RC==�H i :O-H :O-H iY-H second intennediat RC( RC RC :OR.. +OR :OR.. (b) Theomechani smthofataciared-catalreaadylyzedvernucly famieophiliarlitco acylyou.substitution may seem daunting, but it is simpl)' successi n of st e ps mechanileavismsng,havewithsiaxlistt eleps:resfouronanceprotstoanbitlriansfzatioersn t(htrwoownon,intwthoatoff)makes, a nuclthe eophilic atwork.tack,TypiTheandcaalsixlleyavi,sttehpsneseg argroup e labeled in the mechanism below: nuclprotprotooennophioffonl(ereatsonance tacks stabilization) lprotpreaviotoonnngoffongroup leaves (resonance stabilization) i :O-H + :O-H+ :O-H 1 + CH)C-OH.. CH)C-OH. CH3C=OH.. fL 1 / H ........CH2CH3 O-H CH3C-OH .... H2CH3 H '" ..... CH2CH H /R� 3 :O-H O-H + - H20 CH3C:O-H . + CH3C CH3C :RCH2CH3 +RCH2CH3 :�CH2CH3
20-12 (a)
·
II
•
i
•
�
1
�
/
\\
�
1
• •
+
\
�
}
a
whole thing
StepA StepB Step C Step D Step E Step F
• •
II
•
•
_
•
.
�
• •
.. �
StepA
_
•
1
Step B
o
1
StepC
+ AI
StepD
---I��
Step E
1
l
/
\
�
475
/
\\
""0 .....
�
//'0 \
}
of MicroscopiOEtc Rever Applyin(cg) thAle lstsetpsepsasaroute reversi lined onble,thwhie previch isousthepage:reason(atbbrhe ePrviinacitinpgleOCH2CH3 ) sibility applies. -B B is the acid catalyst nucIprotprotooenophin offonl(eresonance attacks stabilization) althoughly removes in protleavionngongroup leaves (resonance stabilization) :hydr- oilsysithserconjeactiuogatns,ewatbase,er usual proton off :O-H 1+ 'OEt
20-12
as
H
StepA Step B StepC StepD StepE Step F
O \ :O-H :O-H c1� H C C+ C 'OEt a a �O:t (f
H+
/
: O-H +1C
cr �
:O- H
�
I
+
CH
HC
+
I
�
:O-H C-O: aA)Et \ � H-Ok �11 :OHQO� EtOH :O-H C-O: same seri s e cr � +· OEt 1). \H offorresonance � ms as above-- - � I ( H Oil H1 II :OH, OH (JC=O 1 + I)' O· ·H2 C-O: aAEt H :O - H H �
I
• .
Step B
• •
St.pC
Step D
-
..
___ _-
StepE
StepF
I
476
+�
I
• •
20-13 For the sake of space in this problem, resonance forms will not be drawn, but remember that they are critical! '0' '0' Ph
XJ �· + ....H+H
Ph
5
Ph � OH H� CH 3 H O ....
+
�
� +
Ph1 OH ....0+ ;» H ·'CH 3 H� CH 3 + ·0 .... H
Ph10 H + ;» H .... "CH 3 H� CH 3 H
•
•
0h-t-OH
t c: + •
•
•
H 'CH 3
h
H+ � H ... 0. ..... H
+ ,-
+H
Ph
o
'CH 3 + -H20 +
Ph / c '" H plus resonance forms O'CH 3 , HERE'S THE DIFFEREN CE! CANNOT LOSE H+ TO HR CH 3+ MAKE CARBON YL H "'" CH 3 �O Ph H O'CH 3 H� CH 3 ..... CH 3 o
p
(+
4
• •
+
Ph
�H o
•
'CH 3
H+ H ...0. ..... H
Ph
+ H� ·0 .... H ).? plus 6 resonance forms +
0 plus resonance forms ).? Ph �H H�CH 3 H O ....
+
X=' '!J
'CH 3 477
Ph
+OH +
,-
o
'CH 3 t -H20
yO� plus 5 resonance forms .\ CAN LOSE H+ 'C � MAKECARBONYL +H� CH 3 o
o
II CH � Ph O"'" 3
TO
20-14 (a)
°
R-C-OH H+ '-
{R-C-OH } :0 : � R-C-OH II
II
�
• •
:0:-
++I
I
..
+I
H
.
H BAD-two adjacent •
positive charges (b) Protonation on the gives only two resonance forms, one of which is bad because of adjacent positive charges. Protonation on the is good because of three resonance forms distributing the positive charge over three atoms, with no additional charge separation.
OH C=O
: 0: � H+ RC-OH
i+
O-H+ : O-H : O-H : + � RC=9 H RC-RH �RC-RH . II
II
�
I
I
• •
..
•
• •
•
t
(c) The carbonyl oxygen is more "basic" because, by definition, it reacts with a proton more readily. It does so because the intermediate it produces is more stable than the intermediate from protonation of the 20-15
(a)
(b)
(c)
OH
Jv COOH � °
CH30H use CH30H as solvent
+
HC -OH useCH30H CH30H as solvent II
+
OH
Jv COOCH3 H20 � remove water with �
+
molecular sieves
°
..
HC-OCH3 remove by II
distillation b.p.32°C ..
�
� �
478
COOCH2CH3
+ H20
remove water with molecular sieves or by distillation
OH.
20-16 The asterisk (" * ") denotes the 18 0 isotope. (a) and (b)
i
+ :O-H
"
.. •
•
CH3C-OII
,.:.---- O-H
• •
•
: O-H �
1+
..
j
CH3C-OH �
.. �
/ 0* H ' " CH3
•
•
�
•
1 .. f
: O-H I + CH3C=OH
O-H
1
..
CH3C-OH 1
0 * CH3
H
1
�
/0*' " CH
CH3C-OH 1+ � O* / ' H " CH3
(c) The 18 0 has two more neutrons, and therefore two more mass units, than 160. The instrument ideally suited to analyze compounds of different mass is the mass spectrometer. .. 20-17 + O-Et : O-Et : o -Et (a) II I I H-C H-C first intermediate: H -C+ \\ + \ \ : O-Et O-E t : O-Et
..
I
----
H
: O-
second intermediate: H-C+
\ : O-Et
----
..
I
+ O-H
----
H-C \\ + O-Et
II •
: O-H ----
•
H-C \ : O-Et
The more resonance forms that can be drawn to represent an intermediate, the more stable the intermediate. The more stable the intermediate, the more easily it can be formed, that is, under milder conditions. These intermediates are highly stabilized due to delocalization of the positive charge over the carbon and both oxygens. A trace of acid is all that is required to initiate this process.
479
20-17
continued (b) :O :� II HC-OEt H+
i
..
� IF
+ :O - H
II HC-OEt
r
:O - Et
I HC-OH
i
. .
:O - H I + HC=OEt
• •
H/R'H t
. .
I
�
. .
:O-H 1+ -O t
OH
H-)5-Et I) HC-OH I
20-18
EtOH
-
..
OH
(a) H 3C
'O �
I
0 C ....
20-19
----
/
HC \ \ +OH
P-
r:-' : O-H +O-H o H //� H-C HC \ OH :OH
----
. .
"
\
OH o
o
(a)
(b)
I I
:O-H . .
:O- H / HC+ \ : RH
}
ar ?'
I �
2
o II
CH -C - Cl LiAl(O-t-BuhH ..
ar ?'
�I
480
2
o II
CH -C - H B214 selectively reduces a carboxylic acid in the presence of a ketone. Alternatively, protecting the ketone as an acetal, reducing the COOH, and removing the protecting group would also be possible but longer.
� :0 : Li+ '\
20-20
HO \ R - C - R' \
.
\
+ H+
R -C- R' :6: .
/
Li+
(b)
O
o: Y Ph
1-
Y
+ O-H I IV R - C - R'
_
H20
: O-H l R- � - R' f +
..
�
RC - OH
20-22 HO
..
·· H20:
-{ � _ H_2-10..
Li
2
H+
a hydrate
\I R - C - R'
+
+1) H H-�-
HR:---./
o
20-21 (a) 0 CH3CH2 - C - OH
HO I R - C - R'
.. Jt? �Cl
:0 �+ Cl If 0
+ � �. � H (-O � rCl ...." '0' •
• •
-
•
Cl
•
1
:o�oj( I'"
'0'
Ph 0 plus resonance forms
�Cl I(
.. P�
-:O '
-{
+ CPh;'0iYCI Cl 0 tt :0.: :0 :
HO, • •
• •
I
.......f--. ..-' H+Oy O � CI Cl .
plus three other resonance forms with positive charge on the benzene ring
� R �� �Cl '0'
IrJ Cl',/\0 ph ° 481
�
o
.. Ph )l Cl +
• •
(
Ph
O=C=O +
Cl
_
0
+
20-23 (a) Co
: II
:0 :
Cl + H-O-CH2 Ph -C� .. CH3
CIIO:
3
(b)
H
�
CH -C-C1
d'
20-24 (a)
(b)
0
+
I
H-N-CH3
II
C
",1+ -
H-OCH2CH3
.. ?j CH3 -C-C1 :
�� R t
:0 :
+1 U
H-N-CH3
I
• •
H
oH
� OH 0
Ph
-
o
� Cl
20-25 Please refer to solution 1-20, page 12 of this Solution Manual. 20-26 (a) 3-phenylpropanoic acid (c) 2-bromo-3-methylbutanoic acid (e) sodium 2-methylbutanoate (g) trans-2-methylcyclopentanecarboxylic acid (i) 7,7-dimethyl-4-oxooctanoic acid
(b) (d) (f) (h)
20-27 (a) f3-phenylpropionic acid (c) a-bromo-f3-methylbutyric acid, or a-bromoisovaleric acid (e) sodium f3-methylbutyrate (g) o-bromobenzoic acid (i) 4-methoxyphthalic acid
(b) a-methylbutyric acid (d) a-methylsuccinic acid (f) /3-aminobutyric acid (h) magnesium oxalate 482
2-methylbutanoic acid 2-methylbutanedioic acid 3-methylbut-2-enoic acid 2,4,6-trinitrobenzoic acid
Ph
-
C-�CH2CH3
-8
20 2 0 (a) CH3 - C - OH
(b)
II
(X �
(e)
o
COOH
(
(c)
COOH 0 ClCH2 -C - OH II
0-
)2
(h)
20-29 Weaker base listed first. (Weaker bases come from
H-t O-
< }- C - O- Na+ 0
stronger
(f)
)
2
Mg2+
0 CH3-C - Ci II
0 (i) FCH2-C-0- Na+ II
conjugate acids.)
(a) ClCH2COO- < CH3COO- < PhO (c) PhCOO- Na+ 20-30 (a) (b) (c)
(X
COOH
7"1 � COOH
CH3
+
2 NaOH
-Q- COOH
(X �
I
COO- Na+ COO- Na+ no reaction
(d) CH3 - CHCOOH + CH3CH2COO- Na+ CH3- CHCOO- Na+ Br Br (e) Na+ -0 COOH + COO- Na+ + HO o 031 0 OH CH3C-OCH2CH3 < < CH3CH2CH2 -C - OH � O� lowest b.p. (b.p. highest b.p. The ester cannot form hydrogen bonds and will be the lowest boiling. The alcohol can form hydrogen bonds. The carboxylic acid forms two hydrogen bonds and boils as the dimer, the highest boiling among these three compounds Listed in order of increasing acidity (weakest acid first): (a) ethanol < phenol < acetic acid EWG electron-withdrawing group (b) acetic acid < chloroacetic acid < p-toluenesulfonic acid Acidity increa�es. with: (c) benzoic acid < m-nitrobenzoic acid < o-nitrobenzoic acid 1. closer proxImIty of EWG (d) butyric acid < (3-bromobutyric acid < a-bromobutyric acid great number of EWG Cl Br increasing strength COOH (electronegativity) of EWG COOH < COOH < (e) 483 .-
I
2-
< }II
-1 �
n-
< }-
-1 �
II
143°C)
(n°C)
.
20-32
..
I
0=
(l62°C)
[yF
} 2.
3.
=
20-33 d derivatives areareoftcommerci en used aasl ya avai test lofablelee),ctroni c effects ofarea series ofmeasured substituentbys:titthey are fairSubsti lAceti y easitcuents lyacisynthesi z ed (or and pK val u es easil y riats ion. a on carbon-2 of aceti c aci d can express onl y an inductive effect ; no resonance effect 3 hybridized and no pi overlap is possible. possibTwo le because is sp the CH2 drawntutedfromacitheds aregivenstronger pKa valthuanes.acetiFirst,c acialldfour substituents are teludeectron wieletctron-wi hdrawintconclusi ghdrawi becausenognsaleflcanfectfourbeinsubsti . Second, the magni of thoen creases in t h e order : OH Cl CN N02 . (It i s al w ays a safe assumpti that nitro is the strongest electron-withdrawing group of all the common substituents.) 20-34 H� OH 0 0 c aci d i s not a carboxyl i c aci d . It i s an exampl e (a)of a Ascorbi " structurel y acidi callecdbecause an ene-diofothel where one carbonyl of the OHgroup. groups HOCH2 iSees unusual adjacent -ene part (c). HO OH (b)of acetiAscorbic aci d , pKa 4.71, is alm ost identical to the aci d i ty acid, acipKadi4.74. '(ol (c)givesThethcemore c H wi l l be t h e one t h at, when removed, di more stable conjugate base. R -<;l(0 _R -}:(O �� -}:(O� R -}:(O_R 1�O :0 OH :0: OH HO OH HO :0: HO 0: t more THREE onl y resonance fOnTIS t d c i aci esonance r fOnTIS e 0 »( R 0 !. : (d)present In the assligthhtle conjugate y basic pHbase, of physi ologiec,alwhose fluid,structure ascorbic canacid is ascorbat :R OH '\..J. beof part represented (c). by any of the three resonance fOnTIS on the left 20-35 (b) 0(a) (C) � (d) 2�COOH � CH2COOH r; � - CH20H � (e) o- COOH Ph (h) COOH (g) H � CH3CH2 -CHCH20H � V-- COOH (i) COOH (k) COOH CHO <
<
<
_
{
. .
_
•
•
. .
}
start here
�
(f)
.
_
. .
two
��
o
�o C
I
484
}
I
.
Q) \
{
,
20-36
(a)�
Mg .. ether
Br
(b)�
0
�H
(c)
conc. KMn04
Ag+ .. NH3 (aq)
� OH
SOCI2
11) LiAlH4 .. 2) H30+
� COOH
KCN
OR
...
,./"-COOH
2
..
�, H30+
0
(d)
H30+ ..
CO2
---
OR H2Cr04 ..
0
� OH 0
�C I
.. � H 0
LiAI(Ot-BuhH
OH
�
J
PCC
CH2N2 CH30H OR ... ... �COOCH3 H+ OR SOCI2 C H3 0 H �COOH .. �COCI .. �COOCH3
(e) �
COOH
O'COOH
(f)
(g)
U 1
h
H30+
...
O' CH2OH
( COOH
+
( (X �
--I
Diels-Alder
OR B2H6 ...
QgHN'
CH2COOH SOCI CH3 NH2 � ... I
(h) CI I
LiAlH4 ..
CI COOH
h
H2
--
PI
0
(X
485
CH3
CI COOH
Hp+
.�
20-37
� }-
COOH
< }- OH
o- CH,oH
----V _
0-,
shake with ether and HCl (aq)
('hCOOH, PhOH, PhCH20H)
+
PhNH3 Clshake with NaOH (aq) and ether ether NaOH (aq) NaCI PhNH2
V
shake with NaHC03 (aq)
!
PhCOONa shake with NaOH (aq)
NaCI ether
! < }-pureCH20H PhCH20H evaporate
evaporate
shake with ether and HCl (aq)
< }-pureNH2
ether evaporate � COOH PhCOOH � _ pure •
NaOH (aq) PhONa
NaCI
PhOH
486
evaporate..
pure
20-38
H "OH OH 1;(t N
(a)
"",
CH3
S
OH
+
racemic (R + S)
0
(b) Isomers which are R,S and S,S are diastereomers.
20-39
TsCI KCN PhCH2CH2CN H30+.. PhCH2CH2COOH (a) PhCH2CH2OH pyridine � P B r3 Mg CO2 H30+ .. PhCH2CH2COOH PhCH2CH2Br ether (b) CH3 Mg CH2 HB ('-r( CH3 r COOH Br ether.. V
!
cr
(f
..
..
---t .. �
c;cr� o
(d)
h
()
..
..
(e)
COO
Br
,
W�
:
2
O�
1\
HO
6 Li
H_30_ +...
_ _
o
dD 487
� COOH
yV o (l
HO OH W, �
..
20-40 (a) Mass spectrum: -mlz 152 ==> molecular ion ==> molecular weight 15 2 -m1z 107 => M 45 => loss of COOH -m1z77 => monosubstituted benzene ring, H
-
-Q K
IR
+
H H spectrum: -3400-2400 cm-i , broad ==> O-H stretch of COOH -1700 cm-i ==> C=O -1240 cm-i ==> C - O -1600 cm-i ==> aromatic C=C NMR spectrum: � 6.8-7.3, two signals in the ratio of 2H to 3H ==> monosubstituted benzene ring � 4.6, 2H singlet ==> CH2 , deshielded
Carbon NMR spectrum: � 170, small peak ==> carbonyl � 115-157 , four peaks ==> monosubstituted benzene ring; deshielded peak indicates oxygen substitution on the ring (b) Fragments indicated in the spectra: COOH mlz 14 mlz 45 and from IR mlz77 This appears deceptively simple. The problem is that the mass of these fragments adds to 136, not 152-we are missing 16 mass units oxygen! Where can the oxygen be? There are only two possibilities: O - CH2COOH CH2-O - COOH
<>
<}
�
<}
How can we differentiate? Mass spectrometry!
( } CHj O - COOH
(
8157
Yat
CH2COOH This structure is consistent with the peak at 157 in the 91 93 J carbonNMR. phenoxyacetic acid The mlz 93 peak in the MS confirms the structure is phenoxyacetic acid. The CH2 is so far downfield in the NMR because it is between two electron-withdrawing groups, the 0 and the COOH. (c) The COOH proton is missing from the proton NMR. Either it is beyond 10 and the NMR was not scanned (unlikely), or the peak was broadened beyond detection because of hydrogen bonding with DMSO. 488
8
20-41 (a)
6 0
a-valemlactone
(b) : O�\
� o r �OH V
t+ .... O
.
. .
0:
20-42 CoII:� Ph
/' C .......
Cl
---
o
6
tt='�:
/
\
O-H
o
Ph-C-O - C-CH3 II :0:
0+
---
..
:0 : I + Ph-C - O=C-CH3 I I Cl :O - H • •
�
:0: .. Ph - CII - O - C - CH3 II :cl · · :� H � -?: plus two other resonance forms • •
� --q CH3
H30+
CH2CO 'l carboxylic Compound 1 acid
---
.
:0: + I Ph - C - O - C - CH3 I I Cl :O - H .
• •
.
(s:
489
CH3 OH alcohol Compound
2
.
Ph-C-O. - C - CH3 I II Ccl :O+ -H •
ester (an ester in a ring is called a lactone)
CH3
acetal
0:
AD:
O=C
II
20-(a) 43
O· o·
CH3
• .
.
H-O :
H-O :
20-(b) 4Compound 3 continued has 8 carbons, and Compound 2 has 6 carbons. Two carbons have been lost: the two 1
carbons of the acetal have been cleaved. (This is the best way to figure out reactions and mechanisms: find out which atoms of the reactant have become which atoms of the product, then determine what bonds have been broken and formed.) (c) Acetals are stable to base, so the acetal must have been cleaved when acid was added. CH3 H30+ oAo 5 � CH3 4 CH2COOH 1
2
(d) The carbons have been numbered above to help you visualize which atoms in the reactant become which atoms in the product. The overall process requires cleavage of the acetal to expose two alcohols. The alcohol at carbon-3 can be found in the product, so it is the primary alcohol at carbon-5 that reacts with the carboxylic acid to form the lactone.
3°
+ 0-H resonance stabilized;
H3C --'( H
leaves the reaction
+
..
: OH
:·0 0 0 .
•
.
OH
� :.O
� '----{
•.
-
. O� +- H
c50:
_
+0
:O - H
�
H - O: H CH3 OH ,
.... .. 11-. -
..
o",-,,-H
CH3 OR 490
..
Q
O-H CH3 two fast OR H+ transfers
•
•
..
.Q
+0
:O-H
CR3" OH
C\
H,+ H-.a �- H :0
.Q·O •
!
CH3
OR H20 -
:O-H C' +
...
CH 1
OR
20-words44 about (A more complete discussion of acidity and electronic effects can be found is Appendix 2.) A few the two types of electronic effects: induction and resonance. Inductive effects are a result of
polarized bonds, usually because of electronegative atom substituents. Resonance effects work through systems, requiring overlap of p orbitals to delocalize electrons. All substituents have an inductive effect compared to hydrogen (the reference). Many groups also have a resonance effect; all that is required to have a resonance effect is that the atom or group have at least one p orbital for overlap. The most interesting groups have both inductive and resonance effects. In such groups, how can we tell the direction of electron movement, that is, whether a group is electron-donating or electron-withdrawing? And do the resonance and inductive effects reinforce or conflict with each other? We can never "tum off' an inductive effect from a resonance effect; that is, any time a substituent is expressing its resonance effect, it is also expressing its inductive effect. We can minimize a group's inductive effect by moving it farther away; inductive effects decrease with distance. The other side of the coin is more accessible to the experimenter: we can "tum off' a resonance effect in order to isolate an inductive effect. We can do this by interrupting a conjugated system by inserting an sp3 -hybridized atom, or by making resonance overlap impossible for steric reasons (steric inhibition of resonance). These three problems are examples of separating inductive effects from resonance effects. (J
11:
11:
(a) and (b) In electrophilic aromatic substitution, the phenyl substituent is an ortho,para-director because it can stabilize the intermediate from electrophilic attack at the ortho and para positions. The phenyl substituent is electron-donating by resonance. plus other resonance forms
+ ..
..
H H H < }-CH2-C-O-H is a stronger acid than H -CH2 -C -0 +
BUT:
E
o
E
E
o II
-
H
The greater acidity of phenylacetic acid shows that the phenyl substituent is electron-withdrawing, thereby stabilizing the product carboxylate's negative charge. Does this contradict what was said above? Yes and no. What is different is that, since there is no p-orbital overlap between the phenyl group and the carboxyl group because of the group in between, the increased acidity must be from a pure inductive effect. This structure isolates the inductive effect (which can't be "turned off') from the resonance effect of the phenyl group. We can conclude three things: (1) phenyl is electron-withdrawing by induction; phenyl is (in this case) electron-donating by resonance; (3) for phenyl, the resonance effect is stronger than the inductive effect (since it is an ortho,para-director).
CH2
(2)
491
o
o
20-44 continued (c) The simpler case first-induction only:
II
II
is a stronger acid than H-CH2 -C-O-H There is no resonance overlap between the methoxy group and the carboxyl group, so this is a pure inductive effect. The methoxy substituent increases the acidity, so methoxy must be electron-withdrawing by induction. This should come as no surprise as oxygen is the second most electronegative element. The anomaly comes in the decreased acidity of 4-methoxybenzoic acid: CH30-CH2 -C-O-H
CH30
o -< >-
C -0 - H
-< >- C
0
is a weaker acid than
H
-
0-H
Through resonance, a pair of electrons from the methoxy oxygen can be donated through the benzene ring to the carboxyl group-a stabilizing effect. However, this electron donation destabilizes the carboxylate anion as there is already a negative charge on the carboxyl group; the resonance donation intensifies the negative charge. Since the product of the equilibrium would be destabilized relative to the starting material, the proton donation would be less favorable, which we define as a weaker acid.
VC O . .
CH3O +
o
:0:
- -H •
-
•
o
Methoxy is another example of a group which is electron-withdrawing by induction but electron-donating by resonance. (d) This problem gives three pieces of data to interpret: II
II
is a weaker acid than H-CH2-C-O-H Interpretation: the methyl group is electron-donating by induction.
(1)
(2)
CH3-CH2-C-O-H
<
�
CH3 0
-
H is a weaker aeid than
CH3
< >-
0 -H
Interpretation: the methyl group is electron-donating by induction. This interpretation is consistent with (1), as expected, since methyl cannot have any resonance effect.
(3)
CH3
� ct is a \4- O-H CH3
stronger
acid than
o-� it -
O-H
(1 )
Interpretation: this is the anomaly. Contradictory to the data in and (2), by putting on two methyl groups, the substituent seems to have become electron-withdrawing instead of electron-donating. How? Quick! Turn the page! 492
20-44 continued Steric inhibition of resonance! In benzoic acid, the phenyl ring and the carboxyl group are all in the same plane, and benzene is able to donate electrons by resonance overlap through parallel p orbitals. This stabilizes the starting acid (and destabilizes the carboxylate anion) and makes the acid weaker than it would be without resonance.
< � C - � -H :0:
�
+
.. .. C-O-H .. 0:9:
plus other resonance forms
Putting substituents at the 2- and 6-positions prevents the carboxyl or carboxylate from coplanarity with the ring. Resonance is interrupted, and now the carboxyl group sees a phenyl substituent which cannot stahdizc the acid through resonance; the stabilization of the acid is lost. At the same time, the electron-withdrawing inductive effect of the benzene ring stabilizes the carboxylate anion. These two effects work together to make this acid unusually strong. (Apparently, the slight electron-donating inductive effect of the methyls is overpowered by the stronger electron-withdrawing inductive effect of the benzene ring.) • " CH3 " H" •
�---�!� o-H H ""
\ \\
\\
CH3 COOH group is perpendicular to the plane of the benzene ring no resonance interaction. 20-45 (a)
•
0
\'
o
�H
•
- t··o
.
���.-----..---
"
this three-dimensional view down the C-C bond between the COOH and the benzene ring shows that COOH is twisted out of the benzene plane
o
� OH
stock bottle students' samples (b) The spectrum of the students' samples shows the carboxylic acid present. Contact with oxygen from the air oxidized the sensitive aldehyde group to the acid. (c) Storing the aldehyde in an inert atmosphere like nitrogen or argon prevents oxidation. Freshly prepared unknowns will avoid the problem.
CH2I2 A
20-46 H h �COOH Ph' I' ' Zn, cuci H"" Ph H (Simmons-Smith reaction, Sec. 8-9A)
IIIIICOOH H
�AH
H ,,,,,
SOCl2
..
(Hofmann rearrangement) Br 2 ' HO-
IIIIINH2" 493
H20
A
H" Ph
",COCI H
�A
H"','
II'''CONH2
H
20-47 Products
are boxed. (a) All steps are reversible in an acid-catalyzed ester hydrolysis. (abbreviating O CH2CH3 as OEt) Step A p oton on (resonance stabilization) H B is the acid catalyst Step B nucleophile attacks :B- is the conjugate base, although in Step C proton off hydrolysis reactions, water usually removes H+ Step 0 proton on Step E leaving group leaves (resonance stabilization) Step proton off F
-
r
.. :O-H 1+ vC'OEt
!
O/��Et :O-H 1
..-----....
:O-H :O-H 1 1 C ac�a �O�t � CH HC+ 0 +
\
(f HC +
:O-H C1 'OEt
:O-HH� :O-H 1 1 + :OH2 .OH2 C-O:� C-O· I H � I " " �:REt V:REt /
\
Step B
same series of resonance ..... forms as above
..E---.� -;
cr� ( I
"/ ,
....""-...
Step
F
494
\
•
•
StepC
\
I
:OA,
V +k� � A-a) �11
StepD
:OH 1- EtOH 1 t;) + H Step E
•
A
..
20-47 continued (b)
-
:0: (c)
.·0//H.�.H
·�..0 / ~ •
\\ OH
�
HI'B.
OH
I
:..IO-H
O�:
�
• •
\ OH
s.�
---
: O-H I"
\
..
HI C)h]
:0��/,...-:H.OH
• •
: OH
OC
c
:0/H..
+
+
•
+
: OH
:0:
+O-H 00: II .....,
·
c
: B-
•
495
0: II
00: c
A cyclic ester is called lactone. Lactones form when the nucleophile is just a few carbons away from the carbonyl electrophile.
OH
a
20-47 continued � (d)
HO
___
/ �,
HO � C /'
C �H � :OH II ..
o
..O"·-
II
o
Esters can be fonned only in acid, not in base. Cyanide substitution is an reaction and requ�res a o or ��rbon with � leaving �roup., The Grignard reaction is less particular about the type of hahde, but ,IS sensitive to, and IncompatIble with, acidic functional groups and other reactive groups. (a) Both methods will work. Mg CO2 H30+ NaCN.. OR CH2COOH CH2Br ether
20-48
SN2
< }0_ ',
1 2°
< }-
---- ---- ---
0" _
SN2
(b) Only Grignard will work. The reaction does not work on unactivated benzene rings. Mg CO2 H30+ 1 1 COOH \ \ Br ether (c) Grignard will fail because of the OH group. The cyanide reaction will work, although an excess of cyanide will need to be added because the first equivalent will deprotonate the phenol.* HO 1 \ CH2Br NaCN.. H3 0+.. HO CH2COOH ____ ____ ____
-0-" -
-o-
tJ.
(d) Grignard will fail because of the OH group. The cyanide reaction will fail because SN2 does not work on unactivated sp2 carbons. In this case, NEITHER method will work. Ho
-o-
0-
Br
(e) Both methods will work, although cyanide substitution on 2°C will be accompanied by elimination. Br Mg CO2 H30+ ether (f)
---- ---- ---
Grignard will fail because of the OH group. The cyanide reaction will work. Since alcohols are much less acidic than phenols, there is no problem with cyanide deprotonating the alcohol. H0 CH2Br NaCN H�O+ H0 CH2COOH .. ..
-o-
-o-
9.1, 16-18,
*
10.0.
The pKa of HCN is and the pKa of phenol is Thus cyanide is strong enough to pull off some of the H from the phenol, although the equilibrium would favor cyanide ion and phenol. The pK;l of secondary alcohols is so there is no chance that cyanide would deprotonate an alcohol. *
CN
+
The side with the weaker acid is favored.
-{ ) pKa 10.0
HO
� H - CN pKa
496
9.1
+
-
o
-{ )
° ' pKa11.6 H H .. H+ H ...... ' H ... 4 ° .... H pKa Hydrogen peroxide isizedfourrelati pK unitve tso(10hydroxi times)de. stronger acid than water, so the hydrof tohperoxide anion,ve HOO, must be stabil Thi s i s from the e el e ct r onegati oxygen bonded to the by i n ducti o n, the negati v e charge is distri b ut e d over both oxygens. The oxygen in hydroxide has to support the full negati ve charge with no delocalization. (b) ° .. H) C Jl-° ... .. H) C �O pKa4.74 H) C )lO .-- H
20-49 (a) H , °
...... 0 ,
+
-
°
-
H+
-0
+
"""""
15.7
_
inductive effect
0-
;
-
°
1 .. 1 H3 C Jl /o-. w
}
+
O H) C .--l� / } pKa 8.2
° H) C )l O ...... O ....... H reason that carboxyl icoacin ofds theare carboxyl so acidicate(o veranio10n wipKthunitwots equival more acientdicresonance than alcoholformss) isinbecause ofThe the resonance stabil i zati whiall ch alpossil atobmsle resonance have octetsworlandds.theThenegatiperoxyacetate ve charge isanionon,thehowever, more electcannot ronegatidelvoecalatoizm-the best of e thea wetnegatnosei veoncharge ontfrigoidthmorni e carbonyl oxygen; that negati v e charge i s stuck out on t h e end oxygen l i ke al the n g. There i s some delocal i zati o n of the el e ct r on densi t y onto the carbonyl , but wi t h al charge separat iperoxyaceti on, this second formis more is a minor resonance contributor .de.ThiIts resonance doesclexplai n, however, why c acid aci d i c than hydrogen peroxi does not come o se t o acetic acid, though. (c) � ...... O yCH3 -
o
W
+
. .
. .
•
°
H) C )l O J Carboxyl the diymbyer,hydrogen that is, tbonding. wo moleculicTheaciesddiaresmboilerhelisdasantightl 8-membered ring widashed th twolinhydrogen bonds as shown wi t h escarbonyl in the diaoxygen gram. Thihasssiworks because the gnifiis cant negative charge, and the H-O bond weak it is a relatively strong acid. The b.because p. is 118°C.
Dodoesperoxyaci ds butboilthere as tharee dithree mer? reasons The autthoor not know, thinatsteadtheyofdohigher not. Fisuggesti rst, thenb.gpthat. is ltohwer (notsuspect l05°C) ey do boil as a team but rather i n divi d uall y . Second, dimberleshown is laikellO-membered rimembered. ng-stillthepossi but l e ss y thanant, 8-the Third and most import elimplectireonid inc nature of the carbonyl group, as places the resonance forms in part (b), lessHnegative on the carbonyl oxygen, and thebondi insg lisessmuch acicharge dilce,sssuggesti n g that the hydrogren strong. 497
20-50 1002 5 elements of unsaturation Spect8ru, mIHA: C9HCOOH 811. 87. 3 , 5H monosubsti t ut e d benzene ri n g (' )- � H-COOH 83. 8 , IH quart e t CHCH3 CH3 81.6, 3H doublet CHCH3 H602 2 elements of unsaturation Spect1r,umIHB: C4COOH 812. H COOH 86. 2 , IHsi n gl e t H-C==C C==C 85. 7 , IH si n gl e t H -C==C 81.9, 3H singlet vinyl CH3 with no H neighbors CH 3 -C=C H CH3 C: Spect ru m C6H 1 0 02 2 el e ments of unsaturation 812. 0 , IH COOH 87.0, IHmultiplet H-C=C-COOH H 85.7, IHdoublet C=C-COOH 82.2-0.8 CH2CH2CH3 must be in doubldueettoatla8rge5.7coupling constant �
�
�
�
�
�
�
\
�
/
�
�
�
�
�
�
I
�
trans
498
/
\
CHAPTER 21-CARBOXYLIC ACID DERIVATIVES
IUPAC name first; then common name (a) isobutyl benzoate (both IUPAC and common) (b) phenyl methanoate; phenyl fonnate (c) methyl 2-phenylpropanoate; methyl a-phenyl propionate (d) N-phenyl-3-methylbutanamide; l3-methylbutyranilide (e) N-benzylethanamide; N-benzylacetamide (f) 3-hydroxybutanenitrile; j3-hydroxybutyronitrile (g) 3-methylbutanoyl bromide; isovaleryl bromide (h) dichloroethanoyl chloride; dichloroacetyl chloride (i) 2-methylpropanoic methanoic anhydride; isobutyric formic anhydride (j) cycIopentyl cycIobutanecarboxylate (both IUP AC and common) (k) 5-hydroxyhexanoic acid lactone; 8-caprolactone (I) N-cycIopentylbenzamide (both IUPAC and common) (m) propanedioic anhydride; malonic anhydride (n) I-hydroxycyclopentanecarbonitrile; cycIopentanone cyanohydrin (0) cis-4-cyanocyclohexanecarboxylic acid; no common name (p) 3-bromobenzoyl chloride; m-bromobenzoyl chloride N-methyl-5-aminoheptanoic acid lactam; no common name (r) N-ethanoylpiperidine; N-acetylpiperidine 21-2 An aldehyde has a C-H absorption (usually 2 peaks) at 2700-2800 cm-I. A carboxylic acid has a strong, broad absorption between 2400-3400 cm-I. The spectrum of methyl benzoate has no peaks in this region. 21-3 The C-O single bond stretch in ethyl octanoate appears at 1170 cm-I, while methyl benzoate shows this absorption at 1120 and 1280 cm-I. 21-4 (a) acid chloride: single C=O peak at 1800 cm-I; no other carbonyl comes so high (b) primary amide: C=O at 1650 cm-I and two N-H peaks between 3200-3400 cm-I (c) anhydride: two C=O absorptions at 1750 and 1820 cm-I
21-1
(q )
21-5
(a) The formula C3HSNO has two elements of unsaturation. The IR spectrum shows two peaks between 3200-3400 cm-I, an NH2 group. The strong peak at 1670 cm-I is a C=O, and the peak at 1 610 cm-I is C=C. This accounts for all of the atoms. The HNMR corroborates the assignment. The multiplet at 0 5.8 is the vinyl next to the carbonyl. The 2H multiplet at 0 6.3 is the vinyl hydrogen pair on carbon-3. The 2H singlet at 0 4.8 is the amide hydrogens. The CNMR confinns the structure: two vinyl carbons and a carbonyl. (b) The formula CSHS02 has two elements of unsaturation. The IR spectrum shows no significant OH, so this compound is neither an alcohol nor a carboxylic acid. The strong peak at 1 730 cm-I is likely an ester carbonyl. The C-O appears between 1 050-1250 cm-I. The IR shows no C=C absorption, so the other element of unsaturation is likely a ring. The carbon NMR spectrum shows the carbonyl carbon at 0 1 71, the C-O carbon at 0 69, and three more carbons in the aliphatic region, but no carbons in the vinyl region between 0 100-150, so there can be no C=c. The proton NMR shows multiplets of 2H at 0 4.3 and 2.5, most likely CH2 groups next to oxygen and carbonyl respectively. The only structure with an ester, four CH2 groups, and a ring, is o-valerolactone:
H
a
IH
499
:0:II -- PhCH2 -O-C-CH3 H� ClA : 0 : II 3 PhCH2 -O-C-CH . (b) ·0· :0:.I . CH3 :0:..I CH3 · CH3 1� C C C ...... ...... . .. . .. .. .. :0'"I eCI :0'" 'Cl :0II/ :0:II �:0:II) Ph-C=O:I Ph-C-O:I -- Ph-C-O:I Ph-C-O.I Cl .......C,CH3 H H H H plus thrdelocal ee re sonance wit h posirintigve charge ized onforms the benzene +I }
• •
.
+
.......
'0
---
+1
.......
+
---
the carbonyl oxygen i s more nucleophilic than the single-bonded oxygen because the product is resonance stabilized
:0I' ....: ... CH3 :0'"....... C .. Ph-C=O: cl-�(A plus alasl theabove resonance forms .· (c) :0: 0 ·0· :0: :0: I I .. I� II �I I ) .. II-CH3 II Ph-C-O:I Cl ...... C, CH3 ---Ph-C+0-C-CH3 --Ph -C-0 -C I}H H HI Cl� �CC\ � :0:II II Ph-C-O-C-CH3 I
.
+
o
1"\
nucleophilic attack by this oxygen does not generate a resonancestabilized intermediate
+
o
• •
500
0 ued n conti 21-6 0 : :0 (d) Ph_ �J o A H
:0 : II
/+ �
�N
---
H
....
0
Ph
:0 :
� HN I
....
Ph
II
:O �
/ H� _
0
0
0 0
p I us
resonance
form
+
leaving group � � N � Ph �--... :0 :
---
---
The l e aving group i s ethoxi d e i o n, CH3CH20-, a very strong never bea base. a leaving group in an SN2base.reactiEthoxi on asdeitwould is too strong
+
H
/
'"
:0
H�
....
:0 :
-
Ph
�
o .
o .
� + HO � 21-7 onal group ("downhill reactions") Reactionswiwhill occur ch goreadi fromlya. more reactive functional group to a less reacti ve functi oride will NOT occur- i t is an "uphil l" transformati o n aciamided chlttoooaciester rided tochlwillamide wi l l occur rapidly (b)(c)(a ) amide NOT occur-another "uphill" transformation acid chldoerideto amide to anhydri e willrapioccurdlyra pidly (d)(e) anhydri will doccur o 21-8 o HOCH2CH3 ---- CH3CH2 - C - OCH2CH3 HCl o HO -{ > �O � + HCl � o 0 (c) < � C Cl HOCH2 --< > ---- < � C - OCH2 � > HCl -<J (d) 0 HCl o- C C l HO ---N I H
Figure 2 1 ·9 is critical!
II
+
+
-
....
+
----
+
+
-
+
+
501
21-8 conti n ued CH3 This on would haved toelimbeinreact keptatioin.cold to aVOi CH3 (e) 0 O =< �H3 rs of 3 HCl butyl alcoholEstareehard CH3-C-Cl HO-C-CH3 --- O-C-CH make. CH3 CH3 o o Cl � Cl � OH --- � 0 � 0� HCl o o 21-9 o o a) H3C-C-Cl HN(CH3h H3C -C -N(CH3h HCl o (b) 0 H3C -C-Cl H2N -{ .> --- H3C -C -NH -{ .> HCl o �C-NH2 HCl O C-CI NH3 --- � � o 0 d) < }- C-Cl HNJ --- < }- C-NJ HCl 21-10 (a) (i) 0 0 0 -Q o -0� H3C -C -0 -C-CH3 HOCH2 Ij_ � H3C -C -OCH2 � II HO-C-CH3 (ii) 0 0 o r-� H3C-C -0-C-CH3 HN'-- � H3C -C-N,-- HO-C-CH3 .. 0 (b) (i) 0.0: 0 :0: II II I'.;) �-CCH3 II � H3C-C-0-CCH3 H3C-C HOCH 2Ph �. . I( H-O-CH2Ph +
II
I
(f)
(
t
+
I
I
to
+ 2
II
+
+
II
---
II
+
+
+
(e)
+
+
(
+
+
II
II
+
II
II
+
II
+
r--
+
II
+
• •
• •
"""
• •
+
+
• •
:0:I I :0:I I 0I I 0II . . H3C-C-0-CH2Ph + H -0-CCH3 " H3C-CI :0-CCH3 .. .. . · H.0 o�CH,Ph / +
502
�
II
2
21-10 (b) continued (ii) :0:I� . . 0I I c�: R H3 HNEt2 3 H3C-C-0-CC � . . -- H3C-C-O-CCH l,.. H t2 t :0: 0 :0: :0-CCH3 H3C-C-NEt2 H-0-CCH3 .. H3C-CI .. · ..
+I
+
-
o
\I
21-11
� · :0: O + CH3C-0-CH2Ph H-,NCH3 �. II
..
:0: CH3C-NHCH3 \I
T.S .+ �
+
+
\I
II
+
.
H - R -CH2Ph .
+
..
�
--
...... f---
• •
+
503
.
\I
:OJ CH37 l,.."q -CH2Ph H-NHCH3 � T.S.+ :0: C� H37 + : R -t CH2Ph H-NHCH3 I leavi n g group I
•
•
+
:0:8- 8i CH3IC----O-CH 2Ph H-NHCH3 I
NE
�.
\I +
• •
o C
:f) )t ��
21-12
. .� �' H
OCH2CH3
o
HOCH2CH3
+
� •
.....--....
:0 : � II H+ Ph - C - OEt
i
-
OH H I "" I + Ph - C - O - Et I O - C�
�
-
•
H' CH3
�
;--... _ \
.. : P.CH2CH3
+ :O - H II .. Ph - C - .OEt .
n
+
+
II
HO - CCH3
� ·ti .. .
O
:O - H I .. - .OEt Ph - C+ ) . CH3P.H
�
.. :O - H I Ph - C = �Et • •
___
OH I H+ Ph - C - OEt .. Ph - C - OEt I "" I + CH30H · '-.. H � - CH3 O - C�
---
I
nCH,CH3
-
• •
---
o
�-
•
.
CH3C - 0 - H
: R - CCH3
21-13
o
� O:
•
II
+
II
• •
0 -· I I O . - CCH3
}
• .
EtOH
. :O - H + O:O - H V H� CH30H g .. I I _ --- Ph Ph - C + --- Ph - C II I I + R - CH3 : R - CH3 :0. . - CH3 .
•
504
o II
Ph - C - OCH3
21-14
�
H 1 �.,p H
;9.� o \
OCH3
;
-Ii'
rR� 0 •
•
OH
OCH3
505
.0. ,HC6s03H 0 • • ·.0e+ .. H /' 0 -""i /'H'
21-15
• •
o
II
II
�
II
'-
-<->-
0ll • • ·.0"•c+• H
0 + .,·. •0"c• H }
/'H' �H' 'II
0-• • H CC�OOH OH OH �H OH I •• •• + � t c CC � O• • - CI - O• •,+... CH3 CC � O-C I - ·O·' ''' CH3 · · HSO; I - o CH3 cc · · � � ... CH3 CH3 CH3 I I I H SO 4 2 COOH H COOH H COOH ·•0 •· � /
I
�
:R,
�
t
�
:<],
�
�
- 0: =C O CC + CH3 I COOH t H � O· :0 0C : HS04 • • II •• + C I • •- , CH3 R aspirin CC I CH3 I CC COOH COOH �
o
:R,
II
two rapid proton transfers
•• H "' OH 0: •• CC -CH3O• •"+� '" CH3 ----. C + CC •O• •• - C + � ·O· - C1 \...0( I C H3 I H3C" .. .. OH COOH H COOH I
'C ..
II
I
I
�
�
�
�
506
I
�
�
-
....
H •• •• I
II
�
21-16 The asterisk will denote (a) 0 :0.) \ 80.
(*)
:0 : 11
�- . .
H3 C - C - O* R
: OH
�
I
..
H3 C-C V - .O*R . I
:R - H
• •
The alcoholcarbon productwicontai s the label tetrahedral th (R)nconfigurati on di, dwinotth none break,i nsothethecarboxylate. configuratioThen is bond retainbetween ed. (b) The products are identical regardless of mechanism. :0: �"'
II
H3 C - C - O*R
,,1 +
CH3 C - O*R I
O- H
�I -
:O - H
H+
---
H+
--
..
HO*R
I
H3 C - C
I
+
: O- H
I
f----I .... .... . I �
: O- H
.
I
..
• •
: O-H
I
• •
+
CH3 C -O*R � CH3 C - O*R � CH3 C = O*R
I
(+
•
H20 :
..
CH3 C - R *R
• •
:O- H
II
� f 9H ,
OH H
I
i +
\ 80
1
....
O- H
..
H3 C -C
II
+ O- H
OH
I
•
the
t
CH3 C - O*R ,, 1 + H20 : '-..."... H - � - H
H20 : .I I.V �
+ O- H
:O-H
I
�..
•
• •
\ 80 and
�
H3 C - C
I
: O-H
---
:0 :
II
H3 C-C - OH
acetic acid
its nucleus thanof mlz 60,Mass spectratheofmoltheseecularproducts would show the hasof2acetimorec acidneutratonsits instandard (molc) The e cular i o n val u e whereas i o n of 2-butanol appear at mlz that 76 instead of between mlz 74, proxygen oving thatand thethe carbonyl heavy isotope of oxygen wentnotwithethbond the albetween cohol demonstrates the bond carbon is br o ken, oxygenTo show and theif thealkylalccarbon. oholueswaswoulchiral orve racemic, measuri ng(Theits optiheavycal actioxygen vity inisotope a polahasrimaeternegliandgible comparing t o known val d pr o i t s confi g urati o n. effect on optical rotation. ) \ 80
\ 60.
507
would Th i S the
lyst ioslydefsisi,nacied dasisausedchemiincalthespecifirsteands thatfourspeeds apsreofactthioenmechani but is notsmconsumed in the ereactd in itohn.e In t(a)thheirdaciAandcatdicalhydr t h st e but i s regenerat aiostn.steInps.theAcibasid cishydrnot consumed; the finthale concent rdateiothnatofinaciitiadl iys atthteackssametheascarbonyl the initiails never concent r at o l y si s , however, hydr o xi regenerat , but it quithceklbasey neutralizes tthheecarboxyl moleculetehofd.e estreAnacter,alioonekn.oximolde leeculavese offromhydrthoexicarbonyl de is consumed; reactionibutc acidoesd. Fornotevery (b) Basian estc hydrer. oAcilysidscatis anotlysireversi ble. Onceis an anequiestliebrrimolum:eculthee miis xhydrtureowilyzedl aliwnaysbase,contthaeicarboxyl ateer,cannot form s , however, n some est ander tunthe iyil tehleyd wiarle never be as hi g h as i n basi c hydrol y si s . Second, l o ng chai n fat t y aci d s ar e not sol u bl e wat y as th.eir sodium salts (soap). Basic hydrolysis is preferred higher yield andionigreatzed;etrheysoluarebilsolity uofblethonle product 21-18 :0: :O :OH · O-H O O °G �
21-17
promotes
catalyze
�
21-19
�
. .
..
. .
:O-H II
i + :O-H + -C-RH R i 21-20 (a) :�� H3C- C NMe2 _:OH
:ttr . .
. .
R -C-NH2
• •
C· � C· · C
:O-H I
�:
+I :O-H R-C-OH
. .
0:
----t��
�' . .
H
. .
. .
• •
�
---l��
..
�
for
:0 :
:0 :
R -C-NH2
• •
II
•
in
. .
+
�
0J 1 H3C- b ==0
. .
508
+
----
0:
�' H . .
21-20 continued (b) :0:I I � H+ H3C - C - NMez
.
.
: OH H I '" I MeC - NMez + I H O-
!
i
--
H+
I
-
..
I O-H
:O - H I H3 C - C + I :O - H
----
..
:O - H I MeC - NMez + ( H zO :
-----
• •
MeC - NMez
� � MeC - NMez
- IJN (CH , h
..
:O+ - H II
�
: OH I
..
MeC -
..
Meb �ez r :O - H
-----
NMez
=
'" I + HzO : H � .O. - H
..
:O - H I H3C - C II +O-H
-----
l ( )
+ OV -H :0 : II I I �HN C H3 Z .. H3C - C - OH H3C - C I :O - H • •
+
21-deprot2 1 onatIn tihoen basiof thcehydrcarbooxyllysisic(21-20( a»th,ethamie stdeep anithaton.driIvnesthteheacireactdic ihydrol on to compl et1-io2n0(isb»th,eprotfinalonatsteiop,ntofhe aci d by y si s ( 2 aminesobythaciat tdheisamiexotneherismnoic landongerit prnucleventeophis thliecreverse nithetrogen . reaction by tying up the pair of electrons on the 21-22 :- H - OH � "
n
Ph - C N : - Ph - C = N · I \�: O . .H :O. . - H
.. i h .. -
:0 : II HO':l H � P -C-N-H Ph - C - N - H II :
)
.?l :
O. .H
(
h
.
Ph - C - NHz IU :O - H
:0:
---l..�
-
:0: II
�
• •
-
..
+ Ph - C :NHz J I Y'\ :O - H �
509
..
+
HzN(C H3 h
Ph - C = N - H I Y'\ :O - Hj
!
-
:0:T
: OH
- Ph - = N - H ..
i.. h
r
: 0 : r - --- h
:0 : II P -C O:
I
P -C O
• •
• •
+
NH3
• •
=
• •
21-23
Hi) :� H � H Ph-C-N-H Ph- - H2 Ph -C- N-H . . H : O -H : : -H . Note: species with positive charge on U -H Ph-C=N H ��+- H
I
+
I
� ..E---l.. ....
I IJr\
<./
0
I
2
II C
----
0
N
carbon adjacent to(notbenzene alswio thave resonance shown) forms positive charge distributed over theh thering. 21-24 er reductism owhere n, a newthe C-H fi(a)rstRsteducti ep andoninoccurs the thiwhen rd step.a newThisC-Hcan bond also beis formed. seen in thiIns estmechani stepsbond are simformed ilarly l ainbeltehd.e (b) Step 1-on reducti
: o C
IS
alStkeoxip 4-workup; de protonation � 2 1-25
(a) � NH
«2 (5CH2CH'
C(
OH HH H I
O
H
Streducti ep 3-on
O� Q.N , H
(d) (' N · H (e) 510
3C
(t) ex>
CH2CH2CH3
.
C H2 NH2
21-26
+ MgI Ph "MgI MgI C=N. H+ C=N+ .. H3C H """----""' H Ph H 0 I H2 : : O1 "\� C-N + I :H 1 I � H CH3 PhI Ph PhI (':H1 - NH3 PhI H + +O =C O-C-N-H O-C+ : : : O-C-N : (I I H CH3 CH'1 HI CH3 H\--... H+ H CH3 H H) � Note: speciadjacent es with tposi tive Charge ! onhavecarbon o benzene al s o forms (notdisshown) :0 : wioverth-thetheresonance posi t i v e charge tributed + Ph '- C-I I - CH3 ring, NH4 + :?j MgBr CH3CH2Q: -C - Cl Ph -MgBr .. CH3CH2 -C-Cl � PhI � MgBrCl + OH H+ - :0 : MgBr :0 : G II CH3CH2 -CI -Ph ..workup CH3CH2-C-Ph .. CH3CH2-C-Ph MgBr -./ Ph I -:::. Ph Ph � Ph
\
/ H3C
/
•
"--."..
•
....
\
I
/
\
-
I
\
/
• •
\
I
..
I
I
•
I
{
I
I
I
..
•
H
+
21-27
• •
+
I
V
----I�
� JI'
• •
-
I
511
21-28 0 OH (a) H30+ MgBr -.. � Ph )l OR Ph these alcohols can also be synthesized from ketones: OR OH H30+ .. � � Ph � MgBr Ph OR 0 OH H30+ .. � � PhMgBr Ph OH (b) 0 H30+ HAOR � MgBr __ .. � H 0 (c) CH3C N � MgBr __ H30+.. � ORN ::-.... 0 H30+ �C CH3MgBr -- .. � 21-29 ·0+ L «f0CHJ : 0I I : : 0I I : Et-C �AlCI3 --- AlCI4- i Et-C' .. .. Et � � r + +
2 �
0
+
--
+
+
---
2
+
�
+
l
+
AICI4-(0
o
o
II
II
C-Et
H C-Et ------
------
:OCH3 ..
� CCH2CH3 _ CH30 -o-
+OCH3 ..
o
II
+
HCl AlCl3 +
512
o
II
H
I
6-� C :OCH3 ..
:OCH3 ..
21-30 Ij-� 0
(a )
'
+
21-31
(a ) (i)
o
0\\ C - Cl _
\
o
II
II
� Cl
0 0 HC - 0 - CCH3
+
H2N
AlCl � -
AlCl3
•
-\ .J
< '}-c0l l -1 �
o
r �
0 HC - NH
Zn(Hg) HCl
-\ .J
+
o II
HO - CCH3
HO - CCH3
+
(b) C�: ft H2.N. Ph - 0 - CCH 3 H - C� (i)
+
�
.. 0 :0: I::,) . . I I H-CO. - CCH3 I l,� + H - NHPh
�
:0: II H-C I H
0
. 0 · 01 1 · II H - C - 0 - CCH3
·· 0·. 0 II 1::,) H - C (",,9· · - CCH3 1 H -+O - CH2Ph .
+
HOCH2Ph
�..
• •
�
• •
513
0 .. II :O. . - CCH3
�
:0: II H - C - NHPh (ii)
o
II
�
+
+
0 .. II H - R- CCH�
:0: 0 II II H-C + :0CCH3 ·· ,, 1 + l{0Q� CH?Ph ) - • •
�
� �
O",C 'H
�OH + HC-O-CCH3 HC Cl does not exist, so acetic fonnic anhydride is the most practical way to fonnylate the alcohol. (b) CH3C -O-CCH3 O",C 'CH3 OH + Acetic anhydride is more convenient and less expensive than acetyl chloride. (c) 0=$ (Vl NH2 + NH 3 Vy OH 21- 32
(a)
o
0
o
�
I I
I I
o II
-
�
o I I
I I
o
0
�
II
II
o
(d)
c$
o
The chloride would tend to react at both carbonyls instead of just one; only the anhydride wi l l give acithids product. o
o
�OCH 3 H �O o
o
The acid chloride would tend to react at both carbonyls instead of just one; only the anhydride will give this product.
514
SIS
o
...... J� . H
I
:0
+/
: 0.
.
.
HO
,
I
:0 • •
------
+0
..
+
II��� .� • •
H
HO .
HO
+H
..
HO
HO
• •
�
r
Q:oH
21-34
(a)
I
0 CH3C-O-CCH3 a
II
+
II
COOH Generally, acetic anhydride is the optimum reagent for the preparation of acetate esters. Acetyl chloride would also react with the carboxylic acid to form a mixed anhydride. (b) 0H 0H CH30H I I COOH COOCH3 Fischer esterification works well to prepare simple carboxylic esters. The diazomethane method would also react with the phenol, making the phenyl ether. (c) rh CH2N2 .. rh �COOH �COOCH3 Fischer esterification would make the ester, but in the process, the acidic conditions would risk migrating the double bond into conjugation with the carbonyl group. The diazomethane reaction is run under neutral conditions where double bond migration will not occur. 21-35 Syntheses may have more than one correct approach. (a) 0 OH H30+ ether Ph-C-OCH3 2 PhMgBr Ph -C-Ph Ph (b) 0 OH H30+ ether H-C-OCH2CH3 2 PhCH2MgBr .. H-C-CH2Ph CH2Ph (c) o o Ph -C-OCH3 H2NCH2CH3 Ph -C- NHCH2CH3 h
Q: h
II
+
(e) (f)
..
II
0 Ph -C-OCH3 II
+
+
I
I
II
�
ether.. H30+ 2 PhMgBr
---
OH H-C- Ph Ph
H30+ LiAIH4 ether.. PhCH20H
Ph-C-OCH3 H30+.. a II
I
---
+
0 H -C-OCH2CH3
I
---
+
II
+
h
+
II
(d)
Q:
---
Ph -C-OH a II
+
CH30H 516
I
I
continued pyridine PhCHzC:: N PhCH20H 2)1) TsCl, KCN from (e)
21-35
(g)
•
(h) a PhCHz-C-OCH(CH3h 2 CH3CHzMgBr ether from (g) a (i) How to make an 8-carbon diol 0 from an ester that has no more than 8 carbons? Make the ester a 1 lactone! II
OH PhCH2-C-CHzCH3 CHzCH3 I
•
+
3
0
4
8
I
+
7
H O+ LiAIH4 ether. �
There may be additional correct approaches to these problems.
21-36
(b)
H I
6
a
a
HC -0-CCH3 II
II
•
H C=O I
I
6
5
6
LiAIH4
•
3
vlOH 8
4
5
6
7
OH
CH3 I
�6
Prepare the amide, then dehydrate. (a) a SOClz � NH3 � POCI3 � OH �CI �NHz .�C::N 21-37
•
•
(b)
a
PhC-OH II
a
SOCl2
----l.�
NH3
o
PhC-CI II
o
•
a
PhC-NHz II
21-38
(a)
517
POCI3..
a
PhCH2 -C::N
continued
(b)
2l-38
Fe HCI a)
�
(c)
NaN02 CuCN HCl
..
-�.. �
•
2l-39 (
OH
Cl
�
OH
C ... CH3 II
..
o
•
Ar=
• •
----l.�
¢ Cl
�Brg M ether
!
0 .. CH3-N=C-O. : CH3-N=C=O Ar-OH i � H-O.. Ar �
21-40
NaCN PhCH2CH 2CN
C pyridine CN Ts I
.. . 1+
-
..- .. }
---- CH3-N- C=O I Af ((\J,'two rapid OH proton Ar H Ar +9 I) transfers .
• •
• •
0
518
1
• •
Sevin (carbaryl)
• •
21-41 (a)
(X0>= 0
(b) (c)
("y 0H
o
� OH
carbonate ester (iii) not aromatic (i)
ao
(i) thiolactone (iii) not aromatic
(l
SyS
o
(i) thiocarbonate ester (iii) not aromatic
+
CO2
(lSH SH
imine enediamine (i) a substituted urea (iii) AROMATIC-more easily seen in the resonance form shown (The enediamine product would not be stable in aqueous acid. It would probably tautomerize to an imine, hydrolyze to ammonia and 2-aminoethanal, then polymerize.) (e) At first glance, this AROMATIC compound does not appear to be an acid derivative. Like any enol, however, its tautomer must be considered. H 3 O +.. � H30+ H 0 COOH C' COOH [[' COOH U 0N 0 N NH N OH enamine H H H NH3 imine lactam ..
...
I
I
o
---
2
I
HO NH2 0)( NH \=..! \=..! (i) a carbamate Dr urethane (iii) AROMATIC-more easily seen in the resonance form � ... f----'l .. -
519
+
polymer see part (d)
---
21-42 between carbamic acid and phosphoric acid. It would react (a) Carbamoyl phosphate is a mixed anhydride a urea), with phosphate as the leaving group. (technically, bond amide an form to amine an with easily two nitrogens on either side; this group is a urea derivative. (b) N-Carbamoylaspartate has a carbonylthewith (c) The NH2 group on one end replaces OH of a carboxylic acid on the other end; this reaction is nucleophilic acyl substitution. (d) Orotate is aromatic as can be seen readily in the tautomer. It is called a "pyrimidine base" hecause of its structural similarity to the pyrimidine ring. OH o H oN N��� pyrimidine N� N N COOO� N COO-��h HO �� HJl H 21-43 Please refer to solution 1-20, page 12 of this Solutions Manual. 21-44 (g) benzonitrile (a) 3-methylpentanoyl chloride (b) benzoic formic anhydride (h) 4-phenylbutane nitrile; y-phenylbutyronitrile (c) acetanilide; N-phenylethanamide (i) dimethyl isophthalate, or dimethyl benzene-l,3-dicarboxylale (d) N-methylbenzamide (e) phenyl acetate; phenyl ethanoate (j)(k) N,N-diethyl-3-methylbenzamide 4-hydroxypentanoic acid lactone; y-valerolactone (f) methyl benzoate (I) 3-aminopentanoic acid lactam; �-valerolactam 21-45 (c) a -./\ (b) 0 0 (a) 0 PhC-� � PhC-O-CCH3 PhC-OCH2CH3 H (e) OH (f) a Ph -C- Ph PhC-H Ph 21-46 When a carboxylic acid is treated with a basic reagent, the base removes the acidic proton rather than attacking at the carbonyl (proton transfers are much faster than formation or cleavage of other types of bonds). Once the carboxylate anion is formed, the carbonyl is no longer susceptible to nucleophilic attack: nucleophiles do not attack sites of negative charge. By contrast, in acidic conditions, the protonated carbonyl has a positive charge and is activated to nucleophilic attack. basic conditions a
\
_ _
I
II
II
I
II
II
II
I
o
R-C-OH II
acidic conditions R -C- OH o II
+
+
-OR'
H+
a
R-C-OHOR' anion-not susceptible to nucleophilic attack
---
II
+
OH R -C-OH + rapidly attacked by R'OH nucleophile I
520
_
c(
o 21-47 (h) ( (a)o- o )- O- oC- H (C) I NH -{ r; � O-C-CH3 > OH h _ anhydrides react o only once OH o CH3 OH CH3 .. . c .. . .. . (e) (d) C .... C=O O I (f) o 0 h H3CO amines are more nucleophilic than alcohols 21-48 o 0 o (a) oo r; � NH2 C "'CH NH- C - H ... C ... O .... C ... CH 3 )H HO .... ( 3 _ _ (b) 0 o o 0 C C"'OH SOCl2 C ... O .... C ... CH3 'CI Na + -OOCCH3 II
� if'
II
(} (c) (tH oH II
I
�
h
H 0 �
OH
(}
CI
0
h
(e) 0
0
C
(f) �
(g)
OH OH
-H2O /:!,.
Ag> OH NH 3 (aq)
0
�
A HolloH.
h
0
C
•
0 0
H+
/:!,. OH -H20
X 1\
�
6 \...
II
o
� OCH(CH3h
HOCH(CH3h o
II
h
H:: o Ao
•
� w.d;o
(d)
(}"-':: a�°Y'°
II
II
+
---�
Cl xO
+
II
II
II
+
)C H � r � I
�OH
H+ /:!,. H20 -
...
0
o OH
0
NaBH4 �H ) � COOH�H30H �COOH Y (f)
0
)
LiAlH, A Ac,o A X 2) H30+ Y H+,-H20 Y MeOOC COOMe MeOOC COOMe HOH2C CH20H AcOH2CYCH20Ac I)
•
any ester where ethylene glycol displaces methanol will be reduced with LiAIH4
521
aqueous acid workup removes ketal protecting group
•
21-48 continued (h) MCPBA
0
..
0
OR
Bf,
H20
..
d OH 6 ... • •
...
:0:-
H+, - H2O
:0.) I .. Ph-C-OEt I :O-H
+ :O :� :O-H II II .. H+ PhCPh-C-OEt REt
<
--.. �
------
1
.
+
�
--
I
:O-H
1
II
+ O-H
Note: species with positive charge on carbon adjacent to benzene also have resonance forms (not shown) with the positive charge distributed over the ring.
2. CO2
..
:O-H 1 PhC-OEt (+ ..
•
. :O-H :O-H I Ph-C+ ------ Ph - C
...
f\.
\=�
H20 : ! OH H OH I I . 1 H+ PhC-O-Et PhC-OEt .. PhC-OEt 1 1 1+ H20: H-O-H O-H O-H .. '-.......".. � - EtOH ""
0
HCl 0 � -0 " 1 0 :ci: I "-':: h
-
··
00]
:0: Ph -CII
Ph -: fI�J H-O..
v
i
f\.
...
..
(c)
...
0
CtD] H2SO4
-
OH
6 ,,"COOH Q- COOH 6·,,·CN KOH H2SO4 H2O,/). . o , H Q- Bf HO OH 0- Br d 1. Mg
KCN H2O
Bf
21-49 (a) :0 I Ph-CI�R � \
OH
""
+ O-H� H20 : Ph-CII� ... 1
:O-H
522
.
� +
0
(g)
II
:0:
II
X; '. H
0
� H3 C-C-O -CCH '--' '--' H3
/ � '-'
+ T
R
(a)
'
I
0
'
" " ,
t
H
--
configuration
No bond to the chiral center is broken, so the configuration is retained.
21-50 0 11
Ph-C-O
-0
• •
(b)
I::> II H3 C-C-O-CCH3 I l.,,�' +
. .Xv ... ...
H-°
0
(g)
C-NHCH3
/j
C
'
OCH3
OH
:0:
0
• •
il
O
-
:0:
523
(c)
(h)
'H
I"'J
Ph-C- N0 0 II
H
>:::
� OH
(d)
H-o
C-N
0 II
C (i) 0-
I
COOH 0 11
C-CH3
continued
+ C
21-50
(j)
o II
(k)
a
' O- N NHz
(!�
(m)
CH3 (I) I PhCHzCHCHzNHz
OH
OH
Products after adding dilute acid in the workup: 0 (b) (a) 0 21-51
II
HC-OH
+
�
(c)
HO-Ph
(
(d)
COOH
� OH
(a)
�
H
HO
X
0
0 II
CH2 - 0-C-(CH2)12CH3
CH-OH
CH-O-C-(CHz)12CH3
glycerol
CH2-0 -C-(CHz)12CH3
I I
I I
o
(b) CHz-O-C-(CHz),zCH3
I I
I) LiAIH4
..
2) H20
�
6
CO,HCI
..
AICl3/CuCI
Gatterman-Koch �
OH
"-':
HN03
h
HZS04
...
I I
0
&" I
"-':
C,
H
H2Cr04
...
h
+
para
OH
¢ ¢ Fe
..
HCl
+
tetradecan-1-ol
+
CH-OH
3 CH3(CHzhzCHzOH
CHz-OH
OH
OH
OH
trimyristin
glycerol
CHz-0 -C-(CHZ)I2CH3
6
�
CHz-OH
�
21-53
�
II
CH - 0 -C - (CHZ)I2CH3
,
+
O
CHz-OH
CHz-OH
(b)
HOCHzCH3
HO
OH
21-52
(a)
+
OH
ortho
NOz
NHz
:
II
OH
0
&" I
524
CH3-C
C'
1 equivalent AczO
&
0
..
3
OH
AczO
OH
¢
CH -C-NH II
0
..
I
"-':
h
more
NHz is nucleophilic
\i
C,
0 H
than OH
(a)
21-54
I � OH
o 0 HC- O-CCH3 II
+
II
H30+ .. 2 CH3CH2MgBr -o
OH
�
(e) See the solution to 21.36(b) for one method. Here is another: +
H Jl H
(f) 0- NH2 (g) 0- Mg (h)
CH3O
Br
O
CH30X0
o- MgBr
...
ether
+
two equivalents NH2
C NH2
+
..
CH30H � + ... large excess
+
0 II
H -C-OEt
H
O
I
reductive alkylation.
9H -o 0+ H,o -- -CH
N X0
(
(y CONH2
I
H
o C � "'OCH3
t POCI3
II
'--1
525
+
Ho D
(y C::N
21-55 Diethyl carbonate has two leaving groups on the carbonyl. It can undergo two nucleophilic acyl substitutions, followed by one nucleophilic addition. o PhMgBr H30+ OH (a) 0 PhMgBr
----�.�
� ---'-I..
----�.�
II
+
G:
1
[
II
I
I I
.
+
---
+ C�: CH3C� -NEt3 HO-CH2CH3 .. +
:0:
---
An alternate mechanism explains the same products, and is more likely with hindered bases: oI I � like an E2 � .. H�-CH2CH3 1\ -I H2C)"" C-CI .. Et3NH CIH2C=C=O a ketene H... .---. NEt3 +
I
I
..
•
+
•
526
�
21-57 H (a) H2C � OH
o
(b) Br2 CH30-o� _
---
03 2) Me2S
1)
6
•
..
w !'i
H
O� OH ! Ag+ , NH3 (aq) 0O� OH
Mg CH30-O-- Br ether
---I"�
CO2 H30+
___
�
CHP -O-- CONH2
•
NHJ
CH30-O-- COOH
*1
*
*1
CH20H
COOH
COOH
1) NaOH 1) LiAIH! 7 1 excess .. CH3I CH 30 ::::-"" OCH3 2) H20 CH30 ::::-"" OCH3 HO ::::-"" OH 2)excess OH OCH3 OCH3 TsCl pyridine 7
*1
CH2CH2NH2
7
*1
CH2CN
!
*1
CH20Ts
LiAIH4 KCN .., ::::-" " CH30 ::::-.... OCH3 2) H20 CH30 OCH3 CH30 ::::-.... OCH3 OCH3 OCH3 OCH3 ..,1)
527
+
CH30 -O-- COCI
(c) � CH2Br KCN � CH2CN 1) LiAIH4 � CH2CH2NH2 .. 2) H20 " 0 0 o (d)
SOCI2
21-58 o (a) CH3CH2 0-C-OCH2CH3 (d)
(c)
(b)
II
(e)
0
0'" C ..... 0 II
'---1
CH3 CH3-� -OH I
CH3
0
II
Cl-C-Cl o II
+
� � _
CH3 0-C- �
H �H3
CH3 -�- 0 CH3
�
- C -Cl
21-59 (a) The rate of these nucleophilic acyl substitution reactions is controlled by two factors: stability of the starting material as detennined by the amount of resonance donation from the leaving group into the carbonyl, and the leaving group ability which is determined by the basicity of the leaving group, the least basic being the best leaving group. LEAST STABLE: no
significant sharing of electrons from chlorine
electrons from oxygen are also distributed through the ring and nitro; very little resonance stabilization
o ..)l ..
:0
leaving group ability and basicity: Cl-
weakest base; best leaving group
>
02N
-o-
MOST STABLE: most significant resonance donation of electrons from oxygen
electrons from oxygen are also distributed through the ring; small resonance stabilization
0:
do 0-
>
� }-
O-
>
CH3 O-
negative charge negative charge strongest base; delocalized through delocalized through worst leaving group ring and nitro ring The least stable starting material with the best leaving group will be fastest to react. The most stable starting material with the poorest leaving group will be slowest to react. (b) ( 0: :0: Nuc ··0·· ·
..� II � NUC
:O � O: /
Cl-C
\
C-Cl
/ I 'Cl Cl I
Cl
Cl
C :o
..
Cl-C ' I
/
Cl
X-o:
C-Cl Cl Cl / I \ �\
"
..)l
:0
Cl-C' I Cl Cl /
Nuc
+
Cl-
triphosgene The first step is the standard attack of nucleophile like CH3 0H at the carbonyl carbon to make the tetrahedral intennediate. The second step is key: the collapse of the tetrahedral intennediate produces one equivalent of phosgene. Attack of a second nucleophile of the other side of triphosgene would release a latent (hidden or trapped) equivalent of phosgene from that side too. Thus, the equivalent of three molecules of phosgene are locked into the triphasgene molecule. Eventually, all six positions would be substituted with methanol producing three molecules of dimethyl carbonate for each molecule of triphosgene. Cl;'
a
528
(a)
II o
21-60
H H3C-N-C-OH
methyl isocyanate a carbamic acid-unstable of gaseous products causes a large pressure increase, risking an explosion. CH3-N=C=O
Both
H20
+
�
CH3NH2(g)
�
+
CO2 (g)
these reactions are exothermic. In a closed vessel like an industrial reactor, the production of
.
(b)
0 �
CH3- N=C=O
�
H-OH
•
•
•
1·· ··-
CH3-N=C-O:
1+ "
---
H-O-H ..
} ( \! 3
CH _�r=-c=o
+
I)
H-RH
•
•
!
I
• •
'-H
H - OH •
•
two rapid proton transfers
.. �
II
..
3 .,.../ decompositionV could be proposed either acid or base catalyzed
H-OH H .. .. H C-N-C- O-H
(c)
o I I
OH
o II
phosgene
CI-C-C1
+
as
CI-C-O
�-� � VJ
(a) (i) The repeating functional group is an ester, so the polymer is a polyester (named Kodel®). (ii) hydrolysis products: 21-61
o
HO-
C
-o- C
0
-OH
+
HOCH2
-o-
CH20H
(iii) The monomers could be the same as the hydrolysis products, or else some reactive derivative of the dicarboxylic acid, like an acid chloride or an ester derivative. 6). (b)(i) The repeating functional group is an amide, so the polymer is a polyamide (named 0 (ii) hydrolysis product: Nylon
H2N
II
(iii) The monomer could be the same as the hydrolysis product, but in the polymer industry, the actual monomer used is the lactam shown at the right. � OH
529
continued (c) (i) The repeating functional group is a carbonate, so the polymer is a polycarbonate (named Lexan®). (ii) hydrolysis products: -o- CH3� HO r;_� � � OH CH3 (iii) The phenol monomer would be the same as the hydrolysis product; phosgene or a carbonate ester would be the other monomer. (d) (i) The repeating functional group is an amide, so the polymer is a polyamide. (ii) hydrolysis product: H2N -o- COOH p-aminobenzoic acid, PABA, used in sunscreens 21-61
(iii) The monomer could be the same as the hydrolysis product; a reactive derivative of the acid like an ester could also be used. (e) (i) The repeating functional group is a urethane, so the polymer is a polyurethane. (ii) hydrolysis products: NH2 H2N HOCH2CH20H CO2 I h CH3 (iii) monomers: � N=C=O HOCH2CH20H O=C=N � I � CH3 (f) (i) The repeating functional group is a urea, so the polymer is a polyurea. (ii) hydrolysis products: CO2 NH2(CH2)9NH2 (iii) monomers: ° )l or or an equivalent carbonate ester CI CI +
+
'Ct
+
+
(a) Both structures are 13-lactam antibiotics, a penicillin and a cephalosporin. (b) "Cephalosporin N" has a 5-membered, sulfur-containing ring. This belongs in the penicillin class of antibiotics. 21-62
530
21-63 The rate of a reaction depends on its activation energy, that is, the difference in energy between starting material and the transition state. The transition state in saponification is similar in structure, and therefore in energy, to the tetrahedral intermediate: ·0· � � � OCH3 tetrahedral intermediate
� _ OH I
The tetrahedral carbon has no resonance overlap with the benzene ring, so any resonance effect of a substituent on the ring will have very little influence on the energy of the transition state. What will have a big influence on the activation energy is whether a substituent stabilizes or destabilizes the starting material. Anything that stabilizes the starting material will therefore increase the activation energy, slowing the reaction; anything that destabilizes the starting material will decrease the activation energy, speeding the reaction. energy
__::---
smaller Ea because of destabilized starting material
similar transition state energy
larger Ea because of stabilized starting material
(a) One of the resonance forms of methyl p-nitrobenzoate has a positive charge on the benzene carbon adjacent to the positive carbonyl carbon. This resonance form destabilizes the starting material, lowering the activation energy, speeding the reaction. :0: /8+ poor resonance contributor, -'0· 11/ . ,. + destabilizing the starting + C-C-OCH3 material; no effect in the r -transition state :0: _. . (b) One of the resonance forms of methyl p-methoxybenzoate has all atoms with full octets, and negative charge on the most electronegative atom. This resonance form stabilizes the starting material, increasing the activation energy, slowing the reaction. good resonance contributor, stabilizing the starting material; no effect in the transition state '---- reaction
=C
21-64
�COOH
+
NH3
___
�COO-
A 531
21-65 A singlet at 2.15 is on carbon next to carbonyl, the only type of proton in the compound. The IR spectrum shows no and shows two carbonyl absorptions at high frequency, characteristic of an anhydride. Mass of the molecular ion at 102 proves that the anhydride must be acetic anhydride, a reagent commonly used in aspirin synthesis. ° ° cS
H
OH,
II
""'"
C
II
' / 0
C
'
Acetic anhydride can be disposed of by hydrolyzing (carefully! exothermic!) and neutralizing in aqueous base. 21-66 NMR spectrum: IR spectrum: -sharp spike at 2250 cm-I -triplet and quartet -1750 cm-I -this quartet at 8 4.3 ° } maybe an ester -2 singlet at 8 3.5 isolated -1200 cm-i ° H 3C
===> C
===> C
===> C
CH3
===> CH3CH2
N
===> CH3CH20
=
===>
H
-
+
°
II
+
+
CH2
sum of the masses is 113, consistent with the MS The fragments can be combined in only two possible ways: CH3CH20
°
II
CH3CH20 - C - CH2 C
C
CH2
C
N
°
II
N
CH3CH20CH2 - C - C
N
The proves the structure to be A. If the structure were the between oxygen and the carbonyl would come farther downfield than the of the ethyl (deshielded by oxygen and carbonyl instead of by oxygen alone). As this is not the case, the structure cannot be B. The peak in the mass spectrum at mlz 68 is due to a-cleavage of the ester: B
A
B,
NMR
CH2
CH2
[
CH3CH20
tR
- CH2 C
mlz 113 68
N
]t
..
1J :R:
mlz 68
+ CCH2CN "
532
..
:
?t
I
CCH2CN
Lr
The formula C5�NO has 2 elements of unsaturation. IR spectrum: The strongest peak at 1670 cm-1 comes low in the carbonyl region; in the absence of conjugation (no alkene peak observed), a carbonyl this low is almost certainly an amide. There is one hroad NHlOH s on m d . o H II -C - NI HNMR spectrum: The broad peak at 8 7.55 is exchangeable with D20; this is an amide proton. A broad, 2H peak at 8 3.3 is a CH2 next to nitrogen. A broad, 2H peak at 8 2.4 is a CH2 next to carbonyl. The 4H peak at 8 1.8 is probably two more CH2 groups. There appears to be coupling among these protons bUl it is not resolved enough to be useful for interpretation. This is often the case when the compound is cyclic, with restricted rotation around carbon-carbon bonds, giving non-equivalent (axial and equatorial) hydrogens the same carbon. o H I II CH2 CH2 CH2 - C - N-CH2 The peak at 8 175 is the C=O of the amide. All of the peaks between 8 25 and 8 50 are CNMR spectrum: aliphatic sp3 carbons, no sp2 carbons, so the remaining element of unsaturation cannot be a C=C; it must be a ring. The carbon peak farthest downfield is the carbon adjacent to N. The most consistent structure: o II C, H N/ 21-67
peak in the
region, hinting at the likelihood of a ec
da ry a
i e
on
+
+
O
8-valerolactam
533
21-68 IR spectrum: A strong carbonyl peak at 1720 cm-I, in conjunction withI the C-O peak at 1200 cm-I, suggests the presence of an ester. An alkene peak appears at 1660 cm- . o
C - O-C C=C HNMR spectrum: The typical ethyl pattern stands out: 3H triplet at � 1.25 and 2H quartet at � 4.2. The chemical shift of the CH2 suggests it is bonded to an oxygen. The other groups are: a 3H doublet at () 1.8, likely to be a CH3 next to one a IH doublet at 05.8, a vinyl hydrogen with one neighboring H; and IH multiplet at 06.9, another vinyl H with many neighbors, the far downfield chemical shift suggesting that is beta to the carbonyl. The large coupling constant in the doublet at 05.8 shows that the two vinyl hydlTlgens are trans. 0 /H II C=C CH3-C OCH2CH3 C-O-C 1 If H There is only one possible way to assemble these pieces: o 1.8, d H3C H 05.8, d \ C=C\ / 06.9, m H C-OCH2CH3 01.25, t II + o 04.2, q CNMR spectrum: The six unique carbons are unmistakeable: the c=o of the ester at 0 166; the two vinyl carbons at 0144 (beta to C=O) and at 0123 (alpha to C=O); the CH2-O of the ester at 060; and the two methyls at 018 and at 0 14. Mass spectrum: This structure has mass 114, consistent with the molecular ion. Major fragmentations: H3C H . \ CH3CH20. C=C / \ mass 45 C+ H II :0: plus two other resonance forms + H rnIz 69 H3C \ C=C\ / + C-O=CH2 H plus one other resonance form rnIz 99 g .. II
a
H;
/
+
+
+
/
+
+
---J"�
/
534
/
It
If you solved this problem, put a gold star on your forehead. The formula C6Hg03 indicates 3 elements of unsaturation. IR spectrum: The absence of strong OH peaks shows that the compound is neither an alcohol nor a carboxylic acid. There are two carbonyl absorptions: the one about 1770 cm I is likely a strained cyclic 1150 em-I), 1720 e m (An anhydride also has two peaks, but they are of higher frequency than the ones in this spectrum.) o o II II C-O-C C-C - C Proton NMR spctrum: The NMR shows four types of protons. The 2H triplet at 84.3 is a CH2 group next to an oxygen on one side, with a CH2 on the other. The IH multiplet at 83.7 is also strongly deshielded (probably by two carbonyls), a CH next to a CH2. The 3H singlet at 82.45 is a CH3 on one of the carbonyls. The remaining two hydrogens are highly coupled, a CH2 where the two hydrogens equivalent. There are no vinyl hydrogens (and no alkene carbon in the carbon NMR), so the remaining element of unsaturation must be a ring. Assemble the pieces: o o II II \: 0-CH2CH2CH C - C - CH3 1 rinV 21-69
-
ester (reinforced with the C--O peak around
while the one at
-
I is probably
a
ketone.
arc not
-
+
y
o
+
0
C...... CH3 o A�
K
Carbon NMR:
On each carbon, the "up" hydrogen is cis to the acetyl group, H H H H 1r while the "down" hydrogen is trans. Thus, the two hydrogens on each of these carbons are not equivalent, leading to complex splitting. o 853 0 8200 8173 --g � g / 0/ 'C/ 'C 830 / \ 868 ---C - C ---
•
824
535
CHAPTER 22CONDENSATIONS AND ALPHA SUBSTITUTIONS OF CARBONYL COMPOUNDS
22-1
(a)
O� -
OH CH2 -�==CH2 2
(b) Enol 1 will predominate at equilibrium as its double bond is conjugated with the benzene ring, making it more stable than 2. (c) basic conditions fonning enol 1 H :0: H 0 _ I II I II Ph - C - C - CH3 ...;:::=:=:::: Ph - �-C-CH3 (I -:··OH H ::::
�
1
acidic conditions forming enol 1 H :O :� n H : �- H I II I II H-B.. Ph-C-C-CH3 Ph-C-C-CH3 I I H H
1
basic conditions forming enol 2 o H :0: . II \I (I � :�I-!. PhCH2 - C-CH2 ...;::: =: ::: ::: PhCH2-C-�.H2 ::::
1
----
H :O-H I I Ph-C-C-CH 3 IJ H +
�
}
ri'
H :O-H I I Ph-C==C-CH 3 1
..t----l.� ....
: O-H I PhCH2 - C CH2 ==
2
537
(c) continued acidic conditions forming enol 2 22-1
-H II
�
+ :O
� H B.. PhCH2-C-CH2 PhCH2-C-?H2 I H :O:
II
:O - H I PhCH2-CCH2 + .."-I H . .
---
H
�
B: , I PhCH2-C CH2 =
�
2
0"
22-2
:O:
. .
:O-H
'CH3 H.... I')..
B:
planar enol This planar enol intermediate has lost all chirality. Protonation can occur with equal probability at either face of the pi bond leading to racemic product. H :0/ o R
�" "CH3
• •
(plus one other resonance form for each)
22-3 o
...
racemicU mixture
:0: . .
..
unaffected " by base H"
_ ,
cis
mixture of diastereomers
CIS
0
CH3 CH3
11
H20 +
538
0
H CH3
trans
H
(a)
. :0: :0: I ..- II .. .. H2C=C- CH3 H2C-C-CH3 .. :0: :0: H tH
22-4
.
(b) (c)
H3C
C
:0:
...
:0:
(y
..
)l C. )l CH3 - H3C I .
..
H
22-5
&'''�H
-
..
22-6
H� - .. :OH.. Ph-C�C. - H I H II
H
rl
•
•
.
i
:0: .
.. :0:
V 0
H
CH3 o
.:0.:- t
:0: II I Ph-C - -S: H2" - Ph-C = CH2 .
�
C
0
+
CI-
CI I Ph-C - C-H I CI 01 H) r -:RH ----
o
II
..:0:- !CI }
CI :R: j:l I j e l ... .. Ph-� = �H Ph - Ci CI ... � Ph-C-CH .. l C: 1 CI 1 !l .. :OH � .. o
_
1
CH" 3 -H 3 C
:0:
H & Cf_
"'"'--:OH •
:0:
V �
:0:
o
o
.. :0:
....----. �
_____
'"--" I
-
:0: CI :..0: CI II _ I I I Ph-C-C-Cl .. - Ph-C==C-Cl • •
---
CI I Ph-C-C-Cl I Cl \...CI .( CI
}
-
539
o
II
22-7
o H �-, :OH II rl Cy-C�C-H I H
i
. •
i
Q4 = Cy } o :0:
For this problem, the cyC\ohexyl group is abbreviated "Cy". :0: "
-
•
•
---
I
Cy-C-CH2 ..
1
·.0· -Br o Br " Cy-C" -�- Br .. - B Cy-C-�I H I) b � ..
•
}
(b)
•
HCBr3 bromoform
. .
(a) � COO- Na+ CHCl3 U
.
II
I
I)
Cy-C-C-H
(H -.. :OH lI
:0: Br
I
I
Cy-C=CH
na· Bf �II· I
+
----
rH
(;:1 r
:0: ,, � Cy-C-O -H
+
-
'--../
o Br " I (c) Ph-C-C -CH3 I Br
:CBr3
Coo- Na + CHI3 (precipitate) 22-9 Methyl ketones, and alcohols which are oxidized to methyl ketones, will give a positive iodoform test. All of the compounds in this problem except pentan-3-one (part (d)) will give a positive iodoform test. +
}
·0· CY-C� - C-Br_--::- Cy - C - CBr3 ---A Br r I :RH 0
:0: Br I I Cy - C=C-Br o " Cy - CR:
22-8
Br0r •
.. Cy-C=CH2
• •
�J
:0: Br " -I Cy - C - C - Br'"
Br
..
� U
540
+
H :0""'"
22-10
:0
+
.......
II �cy"
H
o
22-11
(a)
� COOH
22-13
(a) 0
0
Cl2 CH3COOH from Solved Problem 22-2
6
Br
22-12
0
0ly Br
�
•
6
•
�H
�c y Br II
:Br:
0 0-
0
H
�j ci I
.......
BGBr
..
0
Cl KOH
•
L1
6
E2 elimination
(b) no reaction: no a-hydrogen (b) 0
H
I · �?V \ I AcO : H� o 0
03
(c)
0
HO 'Jl'I I( OH Br 0
3 +
541
.......
.
(d) no reaction: no a-hydrogen
..O V �
,
..
J..
..... VI
� II
""0
A
::s-
rQ zj �
n (�+
\l
.
.....
...
.
::c
\
0. . o
I
N
::c
t :s:
(
\
Q={J
+'iJ
.�+. n o =� t ..� G
,
J..
\
Q. t t O Z Z + Q QO t n+-. : 0 � : � � r Q� q<J � 5 ln < ��� C C Q � � Z 0 0 G �;Y ::c
�
Ul � N
n
N N
::c
+n-
::c
�
o
\
I
\.
+
t
�
y
::c
::c
IJ
\ o ::c ::c
::c
::c
J
"
�:q:
�
n \ 0 ::c ::c
::c
)
y
N
\'" ·o-::c
n
::c
::c
o
::c
I
::c
z
t��
O.
.::c
C� 0 z
•
::c
:
N ... CH3 Ph-C-CH3
22-16
(a)
II
22-17
(a) (b)
(b)
I
Any 2° aliphatic amines can be used for this problem.
0
6
+
0
H
(c) 0" PhC-CH 3
H + .. - H2O
I
6
�
22-18
Cd) o-iJ
H3C' N ... CH3 Ph-C=CH2
H I
+
0
n N
6
H+ - H2O
•
1
)
2)
� Br
H30+
..
0
~
� Ph o
0 PhC=CH2 I
0" 1) PhC-CI 2) H30+
•
0" 0" PhC- CH2 -CPh
HO-H ... Y\
In general, the equilibrium in aldol condensations of ketones favors reactants rather than products. There is significant steric hindrance at both carbons with new bonds, so it is reasonable to conclude that this reaction of cyclohexanone would also favor reactants at equilibrium. 543
OH
a
22-19 ( )
o
b
( )
H
Ph 22-20 All the steps in the aldol condensation are reversible. Adding base to diacetone alcohol promoted the reverse aldol reaction. The equilibrium greatly favors acetone. o o CH3 H � 2 CH3 () I ••II I ("I CO 3 ..II CH3 -C-CH2 -C-O· : CH3 -C-CH2-C-O•• : I IU CH3 CH3 +
H II I CH3 -C-CH2 o
i
22-21
+ :O :� :O-H II H - B CH3-g-CH2 CH3 -C-CH2 B HI HI
CH3 H B:II I I CH3-C-CH2-?-O CH3 a
---
� .... ..t---i . ..
}
:O-H •·O-H I CH3 -�+ -CH2 CH3-C==CH2 "-I H�B: �
. .
__
O-H :I
H H3 :R� ? CH3 -C-CH2- C -O� I CH3
CH3-C-CH3 +
H3 -H : ? ? �: CH3 -C-CH2C-O + I·· CH3 carbons of the eiectrophi Ie are shown in bold just to keep track of which carbons
••
+
.... If-.. . -il .. �
come from which molecule
544
(a) acidic conditions 22-22
CH3 H I I H C-C-CH -- C-O : �W I I ·. H CH3 3
II
o
(b) basic conditions o CH3 H II I I H3 C - C - CH-C - O (I I H�- :..OH H O I II H3C-C-C-H .. - .. (I H�:OH
•
{
i
CH3 H :0: CH3 H :0: I I I II - � 1,,1 H3C-C==CH-C-0 ......'---; . .�H3C - C - CH-C-O: I I CH3 CH3 o CH] / II H3C-C-CH=C \ CH3 :RH . .
----l--<
22-23
•
• •
• .
H :0: _I II H3 C-C-C-H ...
� \ CH]CH2 � fQ
•
• •
j
�
• •
..- }
H :0: I I H3C - C == C - H
+
HO CH3 0 n �:O: CH3 0 II I I ... - H CH3CH2 - TI - TI - CI I - H CH3CH2 - C - C- C - H HO I I) H H
t R0 -:
H-O) CH3 :0: HO CH3 I I I I II . .�CH3CH2 -C - C CH3CH2 -C-C - C-H ......'---; IU I H · H
.:0: I C-H
==
545
.
}-
CH3 0 I II CH3CH2 - C == C - C -- H I
H
22-24 (a)
(b) Ph
o
H
22-25 (a) Step 1: carbon skeletons
�
H3C
�
,0
-&-11 C
C-H
�_I •
•
--H
H
H
H
I
·
I
:0:
i(2,2-Dimethylpropanal has no a-hydrogen.) t H-C�C-H ·
•
•
:0: 1 ++1 H
H:0:
.
II
•
H
H -C-C-H .
H-O
H :0:
I I ! 33C-C-C�C-H... (CH) I
a
\"" H-OH
C-H
(CH3hC
Step 4: conversion to final product
i
+
H
_I II
Step 3: nucleophilic attack H HC�� 3 � CH 3 q .
:0:
H-�-C-H
-
-:OH H�··
CH3
H3 C
H
0
H
H
CH3
Step 2: nucleophile generation H 0 H
�0
>
HH H
H3
M Ph CH3
H
(c)
H
H-O: H
� II
•
H
0
C-H (CH3hC � H H\ I-:.OH , H : 0: O H-:
.
}
(I I II (CH) 33C-C-C-C-H I U· H
j
H
I
0 "
(CH3hC-C�C-C-H
Step 5: combine Steps 2, 3, and 4 to complete the mechanism 546
�
22-25
continued
b Step 1: carbon skeletons
( )
Ph
� H
O
H3C H Step 2: nucleophile generation
>
Ph -{
H +
o
Step 3: nucleophilic attack
Step 4: conversion to final product
CH3 0 I II Ph-C==C - C-H I H Step 5: combine Steps 2, 3, and 4 to complete the mechanism 547
This solution presents the sequence of reactions leading to the product, following the fonnat of the Problem-Solving feature. This is not a complete mechanism. Step 2: generation of the nucleophile H :0: H :0: H 0 _I II I II H-b = b-c� H-C-C-CH3 - H-C-C-CH3 .. (I �:OH ·· H Step 3: nucleophilic attack H- O : H 0 �: -� :�: 1 I II __ H-OH Ph-C-C-C-CH3 3 Ph-C-H H-C-C-CH � .. 1 I H H Step 4: dehydration H - O) H 0 H 0 1 II I 1 II Ph-C=C-C-CH3 Ph-C-C-C-CH3 1 I '--- I :OH H H H� ··
22-26
i
•
• •
.. �
•
+
•
The same sequence of steps occurs on the other side. - .. H 0 H 0 H o :RH I II II I " I Ph-C=C-C-CH3 Ph-C-H .. Ph-C=C-C-C=C-Ph I I I H H H final product +
548
:0:
:0: . .
:��
A
l'�
HC-Ph EtO-�
CH3
..
There are three problems with the reaction as shown: 1. Hydrogen on a 3° carbon (structure A) is less acidic than hydrogen on a 2° carbon. The 3° hydrogen will be removed at a slower rate than the 2° hydrogen. 2. Nucleophilic attack by the 3° carbon will be more hindered, and therefore slower, than attack by the 2° carbon. Structure B is quite hindered. 3. Once a normal aldol product is formed, dehydration gives a conjugated system which has great stability. The aldol product C cannot dehydrate because no a-hydrogen remains. Some C will form, but eventually the reverse-aldol process will return C to starting materials which, in tum, will react at the other a-carbon to produce the conjugated system. (This reason is the Kiss of Death for C.) B
(a)
22-28
H
22-29
Ph
yY H 0
22-30
0
H
==�>
I C
Ph
H
(b)
i1n H 0
>Vy
H
==�>
I C
o
Ph
o
Ph
Y 0
H o
- --
The formation of a seven-membered ring is unfavorable for entropy reasons: the farther apart the nucleophile and the electrophile, the harder time they will have finding each other. If the molecule has possibility of forming a 5- or a 7-membered ring, it will almost always prefer to form the 5-membered ring. a
549
:0:
: 0:
22-32
(a)
�)
(c)
II �
a
PhC - CH2CH3 a
H
+
II
a
not feasible: requires condensation of two different aldehydes, each with a-hydrogens
a
CH3CH2C Ph feasible: self-condensation -
feasible: only one reactant has a-hydrogen; cannot use excess benzaldehyde as acetone has two reactive sites
feasible; however, the cyclization from the carbon a to the aldehyde to the ketone carbonyl is also possible
550
a
CH3
feasible: symmetric reagent will give the same product in either direction of cyclization
22-33 (a) and (b)
- .. 0 :RH
starting ( CkJ diketone o
0
o o
22-34 (a) The side reaction with sodium methoxide is transesterification. The starting material, and therefore the product, would be a mixture of methyl and ethyl esters. 0 o II II H3C -C-OCH2CH3 NaOCH3 H3C-C-OCH 3 NaOCH2CH3 (b) Sodium hydroxide would irreversibly saponify the ester, completely stopping the Claisen condensation as the carbonyl no longer has a leaving group attached to it. o II H3C -C-OCH2CH3 + NaOH 22-35 CH3 0 +
:::...;:::=="�
I
II H3 C-C-C-OEt - .. :OEt "( I H '----.-/ • •
CH 3 0 3 2CH - C - C - C - OEt (CH) 1 CH3 II I II
o
explanation on next page
551
+
continued There are two reasons why this reaction gives a poor yield. The nucleophilic carbon in the enolate is 3° and attack is hindered. More important, the final product has no hydrogen on the a-carbon, so the deprotonation by base which is the driving force in other Claisen condensations cannot occur here. What is produced is an equilibrium mixture of product and starting materials; the conversion to product is low.
22-35
22-36
(a)
� OCHJ
(b) Ph Jl Jl ......." '( 'OCHzCH3 Ph o
0
o
0
_
22-37
i
H O H :0: PhCHz �-g-OMe 00 PhCHz 00�-=-g-oMe :OMe I) H �
,-----/ ..
H 0 I PhCH2CHz-C" -C-C" -OMe 1 CHzPh Ph
Ph
- MeO......1-
--
III
�o:OMe o0
o
o0
:0:
OMe o
III
!
OMe
1
�
PhCHz � == b-OMe
o0
(?: f R
:0:
Ph
�
W
(workup)
====
:0:
}
PhCHzCHz-C-C-C-OMe I) 1 MeO CH2Ph
OMe
:0:
Ph
�:D
PhCHzCHz-C - OMe
o
o
III
H
o0
:0:
:0:
:0:
OMe
H 0 " 1 " PhCHzCHz-C-C-C-OMe 1 CH2Ph
552
o
o II
o II
(c) CH3CHCH2 -C -OCH2CH 3 I CH3 This one would be difficult because the alphiJ-carnon hindered.
(b) PhCH2C - OCH3
IS
H� II CH - C -OEt _:9_.E_t ._ � C-OEt
((I
22-39
. .
g
I I C .
This is the final product, after removal of the hydrogen by ethoxide, followed by reprotonation during the workup a-
I) I C
0
:g�
- EtO-
0(
0
-
•
.
II o
a
G{C-OEt
CH - C -OEt C "":-...
H� I : RMe CH - C-OMe ---I I.� C-OMe
This is the final product, after removal of the hydrogen by ethoxide, followed by reprotonation during the workup.
II -�.H - C-O � :0:
C
0-
C
0
Cg o II
CH - C - OEt I r'4 C-OEt
II
.
:0:
�
:0
yH - C-OMe C ""::--...
CH=C-OEt C-OEt
:�)
..CH - C-OMe { C-OMe
o II
j
I
. .
:0:
- MeO-
..
I
CH=C - OMe C -OMe
Cg
j
I
o
CH - C - OMe I" C-OMe
C6) :
553
.
:0:
(c) C(COOCH3 COOCH3
H 0 I " H-C-C-OEt -.. I) H'---.-/ :OEt
22-41
i
Ph
)
H
o
#
�:OEt.
� o
0
H :0: _I II_OEt H-C-C
H
r : 0 �g-OEt
II I II Ph-C-C-C-OEt I o
OEt
Pho/OEt H
�
0
H
:0:
3
..
�
H :0: I I H-C=C-OEt
\
t
(:0: H 0 �I I II -EtO- Ph-C-C-C-OEt I) I EtO H : 0:
:0:
:0:
. .
:0:
:0:
Ph- �'OEt Ph-� Jl'OEt � Jl OEt�Jl T T H H H ---
-
wj H
II I II Ph-C-C-C-OEt I o
+
+
NaOCH3
+
�COOCH3 �COOCH3 NaOCH (mixture of products results) COOCH2CH3 O� NaOCH2CH3 (d) C COOCH2CH3 o The protecting group is necessary to prevent aldol condensation. Aqueous acid workup removes the protecting group.
(b)
(a) not possible by Dieckmann-not a !3-keto ester 22-40
0
H
554
(a)
22-42
� 0
Ph
(c)
Ph
OCH3
� 0
EtO
(b)
0
(d)
0
OEt
� 0
V 0
EtO
II
o
'---.--/
Ph -C-OEt
(c)
+
EtO-C-CH2Ph
22-44
(a)
0
(f
(a) two ways: 22-45
(b)
II
0
H
--------
0
if
CH2CH3
II
o
0
II
CH3CH2CH2CH2"C-OMe
(c)
(b) �
0
V'
0
� VyPh o
OCHl
OR
o
II + CH3CH20-C-C-OCH2CH3 (d) two ways: 0
0� 0
�
II
o 0
�
PhCH
H
0 /'��II
0
- C - OMe + MeO-C-C-OMe II
o
HO
�CII - Ph
CH3CH2-C-CH2CH3
(c)
0
actually present in the enol form:
6
OCH3
�
EtO-C-OEt
�
o
�
OEt
o (d) II (CH3hC - C-OMe +
II
o
o II
Ph
0
(b)
+ CH3CH2-C-OEt
II
OCH3
CH3
22-43 0
+
0
o
Ph plus 2 self-condensation products-a poor choice because both esters have a-hydrogens
CH3
o
(a)
0
II
0
C-OCH2CHl
555
� o
OR
0
OCH2CHl
OCH2CH3
:0:
22-46
(a)
:0:
:0:
:0:
¥ ¥ · .
:0:
:0:
y y · .
H3C OEt H3C OEt H3C Jl C)l OEt I H H H :0: :0: :0 : :0: (b) :0: :0: H3C CHl H3C CH3 H3C Jl C Jl CH3 I H H H :0: :0: :0: (c) .. N :N.� N� �C h� "c,. ;"C )l OEt OEt OEt C I H H H :0: :0: :0 : :0: (d) :0: :0: lI - N" 'c-)l hN N :0/ :0 / CH :0/ CH3 CH3 I H H H (other resonance forms of the nitro group are not shown) 22-47 In the products, the wavy lines cross the bonds that must be made by alkylation, before hydrolysis and decarboxylation produce the substituted acetic acid. 0 0 (a) 0 0 H30+.. .Jl.Jl 1 ) NaOEt EtO � OEt 2) PhCH-,B; EtO' I' �OEt �OH CO2 2 EtOH CH2Ph CH2Ph 0 0 (b) � 0 1) NaOEt.. I) NaOEt 0 0 H30+ .. OEt 2) CH) 2) CH3I .. EtO EtO OEt OH CO2 2 EtOH -----
-----
· .
· .
-----
.
-----
-
+1
-----
· .
0
-----
· .
· .
+
-----
•
•
-
• •
• •
yl yl
0
+
-----
-
• •
• •
3
o
+
II
y
II
�
+
+
+
0 0 (c) � 0 0 0 H30+ I) NaOEt Ph � OH CO2 EtO OEt 2) Ph(CH2);Br EtO V OEt CH2CH2Ph 0 0 (d) � 0 0 H30 + I) 2 NaOEt COOH " O2 EtO OEt 2) Br(CH2)4Br EtO OEt ..
�
556
II
..
II
6
+
+ C
+
''' 2
2
EtOH
EtOH
22-4 8
(a) Only two substituent groups plus a hydrogen atom can appear on the alpha carbon after decarboxylation at the end of the malonic ester synthesis. The product shown has three alkyl groups, so it cannot be made hy malonic ester synthesis. desired product 0 0 0 0 0 H 0 EtO�OEI ==> EIOYOEI ==> H 00H � � C �OH R\ R2 R \ R2 H3C CH3 () Li+ :0: =,i+ :0: '>-- '>-bH + : ' C AOCH3 --- CAOCH Jl' OCH3 + :N\-HN\-: x Li+ / / H3C CH3 H3C CH3 H3C CH3
�
•
�
pKa 24
pKa 40
1
'"
1
'"
(to
}
With such a large difference in pKa values, products are favored 99%. :0: (c) 0 Be r r� H Jl OCH3 LDA -' C A OCH3 H30+� ' X OH VOCH3 /). H3C CH3 H3C CH3 SN2 H3C CH3 H3C CH3 plus resonance form as shown in part (b) 22-49 0 (b) (c) (a) PhCH2CH2 - C - CH3 co, EtOH CO2 EtOH CO2 EtOH 22-50 In the products, the wavy lines indicate the bonds that must be made by alkylation, before hydrolysis and decarboxylation produce the substituted acetone. 0 0 0 0 1) NaOEt (a) + 0 CO2 EtOH EtO� 2) PhCH2Br EtO� H�O . � CH2Ph CH2Ph 0 0 0 (b) 0 0 1) 2 NaOEt H30+� CO2 EtOH EtO� 2) Br(CH2)4Br EtO ()' /). LDA
»
•
o
'"
"
+
cr
+
..
+
o6
+
+
•
•
o
0 (c) 0 0 1) NaOEt EtO� 2) PhCH2Br EtO� CH2Ph ..
\3!'
+
+
+
o 0 NaOEt 2) Br� EIO �Ph ) A �:'�+ 0
1)
+
..
CO2 EtOH +
557
+
+
�
(a) There are two problems with attempting to make this compound by acetoacetic ester synthesis. The acetone "core" of the product is shown in the box. This product would require alkylation at BOTH carbons of the acetone "core" of acetoacetic ester; in reality, only one carbon undergoes alkylation the acetoacetic ester synthesis. Second, it is not possible to do SN2 type reaction on an unsubstituted benzene ring, so neither benzene could be attached by acetoacetic ester synthesis. (b) o o � Br 0 LDA II � . Ph Ph Ph �C Ph � Ph ' � H
22-5 1
in
an
..
plus resonance fonns showing
delocalization of e- into ring and C=O
(c)
n N
0 Ph � Ph
� Br H30+ Ph Ph � Ph ..
� -l..
---
0" Ph - C1H l- �H2 - C - Ph came from Ph - CH > II 0 Ph -C-C - CH3 Ph - CH - C - CH3 Michael acceptor forward direction o �CH O Ph - C - CH2 NaOEt Ph - C� - CH II) � ( COOEt) (COOEt) Ph -C - C - CH3 resonance temporary stabilized ester group
22-52
�
y
"
"
a
"
+
CH2 - C - Ph Michael donor
- . ..
o II
( OOEt� I Ph - CH - CH - C - Ph
Ph - CH t- CH2 - C - Ph
I
�
II
o
CO2 EtOH
+
o
I
Ph - CH - C -CH3 Ph - CH - C - CH3 22-53 First, you might wonder why this sequence does not make the desired product: O 0 0 0 H .� � ft CP H�H 0 .. - +� .. poor yield MVK resonance stabilized The poor yield in this conjugate addition is due primarily to the numerous competing reactions the enolate can self-condense (aldol), can condense with the ketone of MVK (aldol), or can deprotonate the methyl of MVK to generate a new nucleophile. The complex mixture of products makes this route practically useless. (continued on next page) +
+
�
�
---
ketone
558
continued What pennits enamines (or other stabilized enolates) to work are: a) the certainty of which atom is the nucleophile, and b) the lack of self-condensation. Enamines can also do conjugate addition: 22-53
H
MVK.
22-54
o
o
I
o
high yield
The enolate of acetoacetic ester can be used in a Michael addition to an a,l3-unsaturated ketone l i k� 0 0 0 0 � � NaOEt. EtO EtO �� •
/:: � o
CO2 EtOH +
O
o
+
8
y
�
a
: N -' C - CH = CH2 ----- : N = C - CH= CH2 ----- : N C CH - +CH2 lr acrylonitrile Nuc :
22-55 J
+
1
-
- ..
•
•
==
! ( ==
.
n � Nuc N C - CH2 - CH2 .. - H : N - C-CH-?H2 ----- : N C CH ?H1 I Nuc Nuc Nuc
i : o:
}
-.
==
==
-
:0: :0: I II 2 : O - N - CH = CH2 ----- : O - N = CH - CH : O - N+ = CH - CH � I Nuc : Nuc + nitroethylene + (some resonance fOnTIS of the nitro group are not :0: hown) II :0 - N - CH-CH2 :0 - N CH2 CH2 .. I I Nuc Nuc
i
-
-
• •
• •
+
• •
• •
+
+
- ..
•
s
•
•
+
-
-
•
559
•
+
-
2
}
.
22-56 o (a) II PhCH CH - C - OEt
� ==
EtO
o
(c) two ways
H2 = CH - C N : OR 0t
�
(d)
(f)
0
OEt
followed by hydrolysis and decarboxylation
(b) CH2 = CH - C - �
�
EtO
followed by hydrolysis and decarboxylation
0 (1\
�
II
OEt
o
0 followed by hydrolysis and decarboxylation o
CH2 = CH - C N
� �aOEt
PhCH CH - C - OEt
�
0
followed by hydrolysis
o
EtO
o
o o
H3C.... /CH3 0 � II � CH3 CH2 == CH - C - Ph U followed by hydrolysis ==
OEt
OEt + �
o
(could also be synthesized by the Stork enamine reactions)
560
22-57
Step 1: carbon skeleton CH3
comes from
>
o Step 2: nucleophile generation
. .
0:
o These hydrogen atoms more acidic than the other alpha hydrogens because the anion can be stabilized by the benzene ring.
H plus resonance forms with negative charge on the benzene ring
Step 3: nucleophilic attack (Michael addition)
H
continued on next page
. . -
0
561
22-57 continued Step 4: conversion to final product (nucleophile formation)
plus one other (enolate) resonance form
•
(nucleophilic attack)
o
I t
: 0:
o
(base-catalyzed dehydration)
o·
0
..
plus one other (enolate) resonance form Step 5: The complete mechanism is the combination of Steps 2, 3, and 4. Notice that this mechanism is simply described by: 1) Enolate formation, followed by Michael addition; 2) Aldol condensation, followed by dehydration. 22-58
o'" U
Step 1: carbon skeleton
"
o II CH=CH-C-OH
�
Step 2: n leophile generation
II CH2 - C-OCOCH3
D
H
o 0 II II CH3-C-O-C-CH3
comes from =====� >
•
CH3COO-
i
-
: 0:
II
: CH2 - C-OCOCH3 562
-.. }
·
.... ..I----!.�
CH2
•
0
·
� � OCOCH3
OH
22-58 continued
0
� g : O: �H -OAC '-.." I
Step 3: nucleophilic attack II
·0:
II
�
Ph-C -H
: CH2-C-OCOCH3 ---- Ph-CH-CH2-C-OCOCH3
+
Step 4: conversion to final product 0 OH
I
II
Ph-CH-CH-C-OCOCH3
I)
Ph- H -CH2 - -OCOCH3
-
0 _
0
CH3COO-
(OH I
: 0: II Ph -CH - CH-C-OCOCH3 U. plus one other (enola!e) resonance fonn 0
..
_
�
H
II
Ph-CH = CH - C-OCOCH3
�
hydrolysis
: O: II II H+ .. PhCH==CH-C-0-C-CH3
�H+ OH :0: I II PhCH==CH-C-0-C-CH3 .. •• I H20: '"" + / H -0 -H
..(
H20:
I two fast t proton transfers ••
Q
+ OH .O-H I II PhCH==CH-C-0-C-CH3 IV OH plus two other resonance fonns
i
:
O -H II
t
0
II
PhCH==CH - C-0-C-CH3
�
t
• •
:O -H 0 I II PhCH==CH-C-0-C-CH3 + . plus one other resonance fOlm with positive on the other oxygen
I
O -H ... O==C
I
+
C�
o II PhCH==CH - C-OH
i
PhCH==CH -
..
+
t
: OH
� 0� -H -O.
•• /
H20:
?
·· H -O:
PhCH==CH -
II
?
:OH ..
}
plus one other resonance fonn with positive on the other oxygen Step 5: The complete mechanism is the combination of Steps 2, 3, and 4. 563
22-59 The Robinson annulation consists of a Michael addition followed by aldol cyclization with dehydration. In the retrosynthetic direction, disconnect the alkene formed in the aldoVdehydration, then disconnect the Michael addition to discover the reactants. (a) aldol and dehydration forms the a,� double bond:
o
>
o
Michael addition forms a bond to the �' carbon:
;> o
(b) aldol and dehydration forms the a,� double bond:
o
o
o
;>
o
Michael addition forms a bond to the W carbon:
o
o
22-60 Please refer to solution 1-20, page 12 of this Solutions Manual.
S64
22-61 The most acidic hydrogens are shown in boldface. (See Appendix 2 for a review of acidity.) (Braces around resonance forms omitted here.) : 0: :0: o (a)
o
/H
--
·9 .
. .
R:
.
H
H
I
y f::y o: :0 �V--
H
o
o
I
H
:O
. .
--
. .
:0:
:0:
: 0:
H
0:
0:
--
(d)
. .
. .
. .
:0 . .
H
H
. . . .
. .
. .
0:
0:
. .
same enolate as in (b)
�" U
(e) 0
0
:0:
�
COCH
(f)
: 0:
: 0:
Because of the stereochemistry of the double bond, the H atoms on the left side of the ring are not equivalent to those on the right side. However, the two enolates will be equivalent except for the double bond geometry. 565
:0:
22-61 continued
a H " I CH H ' ...... C� .... .. " c c I \ I H H H
H
(g)
'iC ...... 1 ...... CH H2C C I
I H
" I - ........ C� ...... CH H 2C "C
....--
I H
a
H
(h)
---
H
: 0:
H
"
---
" I - ........ C� ...... CH H2C "C
I H
: 0:
" 'iC ....... ;-...... CH .. H2C C
H I
....--
I H same enolate as in (g)
H
" I 'iC ..... .. ;-...... CH .. H2C C
: 0:
I
H
..
H
: 0:
..
:0: I
I
'iC ........... CH H 2C C/'
I
H
H
..
:0: I
I
'iC ........... CH .. H 2C C /'
H
I
11011 0°11
22-62 In order of increasing acidity. The most acidic protons are shown in boldface. (The approximate pKa values are shown for comparison.) See Appendix 2 for a review of acidity. a
H
uCOOCH3
<
a
.6COOCH3
<
(a) pKa 13 .......... .... .. ....
<
ex:
<
(c) pKa 11
_
<
(f) pKa 17-18
a
,.----
,
H
(b) pKa20
(g) pKa25 least acidic
1_
H
0COOII
(d) pKa5
........ ............ ....
fully deprotonated by ethoxide ion
566
a < CN
H
coaCH3
(e) pKa2 most acidic
22-63 o
0
H.
O� '0 �
The enolgationfonnandisbecause stable because of the conj u of i n tramolecular hydrogen-bonding in a six-membered nng. CH .... CH;�Jl M T H enolH keto 0 Inthedielectron-donati carbonyl compounds in general , the weaker ngtheabienollity fonn: of the group G, the more i t wi l l exist in aldehydes G N (ketG ones, areestealmost completely enolized, t h en rs, andcontent. finally amides which have virtually no enol keto
cH
cH3
---
3
H
o
� ) =
H)
22-64
H
enol
TheO wavy0 line lies across the bond fonned in0the aldol condensation. (b
..
H
OH
0 (c) OH 0 O H20 Ph�Ph .. Ph�Ph H
(e) 22-65
H
H
o
_
2
o
The wavy 0line lies across the bond formed in the Claisen condensati0on. 0 O (c) A� (b � ) OCH3 � D o
.....
,.,
o
Cxj o
567
o
c6
89S
H
t
.. ..
H
D - H
1 .. ) H O:
H
o
:.Q:
..
H -OH
�
u
H
H
(e)t9-ll w�lqOJd U! uO!1esu�puo::lloPle)o ws!ue4::l�w (e) 99-ll
69S
OZH-
-----
: 0:
) t�:O-H
0
\Jy"z:
:a
0
O-H
----
: 0:
H
-----...
H
t
:O-H +
H-O
H-O
w� H-O· H-OI:H o
+
•
a
a
a-H V� H :O:
(q)W-ll W�IqOld U! u0!lusu�puo::> 10PIU)0 ws!ump�w (q) p�nu!luo::> 99-ll
22-66 continued (c) mechanism of Claisen condensation in problem 22-65(a)
�o H
..
..
OMe -:OMe
H�
OMe
I I A\ _
•
OMc
4
.
�Me
1 J: OMe /T"
-MeO-
product wiluponbe deprotonated but(thisregenerated acidic workupby) methoxide, (d) mechanism of Ciaisen condensation in problem 22-65(b)
o
o OMe H
'---..-- ..
OMe
:0: ..
..
'" -: OMe
:0:
o
•
~
-MeO-
product willuponbe deprotonated but(thisregenerated acidic workupby) methoxide, 570
:.. 0:
o
All products shown are after acidic workup. C (a) aldol self-condensati o n \YT 22-67
CHO
Ha
H
a
Claisen self-condensation Oll (c) aldol cyclization
(b)
aEt+
a
a
�
Ha- ..
a
(d) mixed Claisen
lf OR lfCH3
+
�
caaEt
t
Ea
-
a
OC
H3C �
Eta-..-
+
E1O
Eta-..
�
a
a
a
a
�
a
a
(e) mixed aldol
COaEt
Ha-
-
a
a
OEl
�
+
�
� Ph lfCH3 H Jl ph (f) enamine acylation-attempt at aldol would give self-condensation Q Ph CI .Jl ph a
a
6
Ha-
+
..
� a
a
+
a
_--1"-_
HP·.
571
a
o
II
(d) 0
o
(c) CH3CH2CH2 hCH3 (b)A CO2 + � + CH3CH20H Ph
0 (a) CH3CH2 -C-CH2CH3 + CO2 + CH30H 22-68
0
.....
(e) Ph� U
o
(f)
V
....
CH3
0 H2CH2CH2CH3 CrY OCH3
o
(g) 6CH2CH2CH2CH) 22-69
(c) reagents: excess (or Br2 or Cl;», (b) reagents: (a) reagents: Br2' H+ NaOH followed byBr2'H20PBr3, (d) 0 PhCH=CHCH3 Ph3P=O Ph-C-H Ph3P-CHCH3 Ph O -Ph NaOH Oyo + < CH3 H H In the products, the wavythe lsubstituted ines indicatacet e thiecbonds and decarboxyl ation produce acid. that must be made by alkylation, before hydrolysis 0 0 (a) � NaOEt .. V0 H30+ �0 EtO OEt EtO OEt OH CO2 EtOH CH2Ph CH2Ph (b) 0 0 0 NaOEt .. Y0 NaOEt EtO�OEt EIO OEI � OEt \.-J Br � O u( CO2 + EtOH + 0H 12
+
II
-
+
..
(e)
..
o
22-70
1)
II
II
1)
o W
+
o
!1
o
"
+
1)
•
2
2
572
o
+ 2
continued NaOEt COOH + cO2 + EtOH (c) a a Br(CH2 )sBr EtO�OEt In the products, the wavy lines indicateacetone. the bonds that must be made by alkylation, before hydrolysis and decarboxyl ation produce the substituted (a) a a NaOEt a NaOEt EtO � EtB: EtO () 2Br Eta + CO2 + EtOH Ih 22-70
II
1) 2
II
•
2)
22-71
� 0
1)
2)
(b) a a EtO �
1) 2
NaOEt
0
EtO
1)
2)
•
CH
�
()' a a
2
6 o
o
~ + cO2 + EtOH
(c) The acetoacetic ester synthesis makes substituted acetone, so where is the acetone in this product? substituted acetone
The substituted acetone canon!be made by the acetoacetic ester synthesis. How can we makesingle the bonddoublto ethisbond? Aldol condensati a a,13
make makeizatibyoalnlddolehydration additionby conjugate : cycl a a a a a NaOH a NaOEt .. EtO � + � Eta A ---
o
.�
573
22-72 These compounds are made by aldol condensations followed by other reactions. The key is to find the skeleton make by the aldol.
(a) Where is the possible a,\3-unsaturated carbonyl in this skeleton? reverse 0 0 OH I II L aldol II PhCH2CH2-CHPh ==> PhCHjCH-C-Ph ===> Ph-C-H
o
forward synthesis II
Ph-C-H
II
+
o
..
NaOH
CH3-C-Ph
II
o
+
0 II CH3-C-Ph
PhCH = CH - C - Ph
(b) The aldol skeleton is not immediately apparent in this formidable product. What can we see from it? Most obvious is the \3-dicarbonyl (I3-ketoester) which we know to be a good nucleophile, capable of substitution or Michael addition. In this case, Michael addition is most likely as the site of attack is \3 to anotho. carbonyl. 0 O Ph Ph '. +
!�"""'\.
O � >Q I
j , \ �-ketoester / OCH3
\
forward synthesis 0
¢o
..
'
Q
� + 0
0
�
OCH3 0 AHA! The aldol product reveals itself. (See the solution to 22-71 (c).)
......... ",/
0
NaOH
\.
0
Ph OCH3
..
Ph
1 ) NaOCH3 2) H+
0
OCH3
��� R +
(c) The key in this product is the a-nitroketone, the equivalent of a \3-dicarbonyl system, capable of doing Michael ,-. addition to the \3-carbon of the other carbonyl.
(¥�> Jly-AJl o AJl +A \'"
".
NO ; '.,
forward synthesis
�o
N 02
aldol product
NaO �
574
reverse aldol
==�>
0
SLS
+ H
..
H
y o o
4d-:)
II
:.. 0:
0R:;
4d -
??[(
HO: ..
H�H
0
t
d_Hi? :R:
4d
: 0:
WJ
H (I HO :
-
(q)
:)
/ ..
·. 0·.
0
-1(--
:.. 0:
: 0:
H£;J
O:� H::J H..
I(
o
(n)
(L-ZZ
22-73 continued (c) Robinson annulations are explained most easily by remembering that the first step is a Michael addition, followed by aldol cyclization with dehydration. OMe
OMe
j;� .c-
:0 ..
o
plus other resonance forms
·1
H
MiChael addition)
two rapid
�
OMe
�
o
.l'mton transrc,
O J
r- H �� C.I· HJ
- .. : RH
n
0
n
H -OH
�
H plus one other resonance form
OMe
H -OH
•
OMe
OMe
..
(dehydration)
o
576
plus one other resonance form
cr-�HP'
22-73 continued
the negative charge in this structure is stabilized by resonance with the carbonyl H H
(d)
�
N �O +
H
H
H
N�\ �;�")
+ OH
H20:
�
H 30+
H
°
H I
H
•
�.
two rapid proton transfers
H20:
H30+ two rapid proton transfers plus one other resonance fonn
R) R:� +
H plus one other resonance fonn -
/'
�o: 2
H
o
+ -H20
H30+
•
o
+
H20:
..
two rapid proton transfers
� c;r.
�H
H20:
577
'OH
+
o
•
l
N-O � �I
o H
°
H H
I
o
�
��
22-74 (a)
H 30+
..
-
o
o
o
H 0+ 3
aldol, then dehydration •
•
22-75
o NaOCH3
---
Br ..
Dieckmann
�
�
alkylation
o
o
578
o
•
hydrolysis, . decarboxylatIOn
H
22-75 continuued
(b) 0
NaO
o
+
Robinson
reduction HO
o
Claisen doing this reaction first blocks this side of the ketone
Ph
Ph
22-76
(a)
:o:� rlJ � '" H-B � OH
anion formed in Claisen used without isolation in the next step
:O- H
I
C
..
---I....
H-B"
..
o
o
9
+ .....
C ..... ..
· plus one oth r resonance form
plus one other resonance form
o
O-H H �H
plus one other resonance form 579
B:! -
.
•
(O-H ..
o
22-76 continued
(;) (\
... ·
H B --.R
OH
I
.. :O -H
H
rb� )
•
01
-
! c: �<>
B:-
..
c
-
:OH
plus other resonance fonns
: 0:
0
•
H
1+
: OH
"'\
1
: O -H
H
1+
OH
i
plus one other resonance fonn
:O -H
..
'1
: OH
�
. .
: O -H
+
1+
H
c
OH
plus one other resonance fonn
t
� -
I
--
" H-B
:��
!
H G Q. H20:) o
0
580
22-76 continued
o
(c) 0
n
,-
o
.. H C_ I
•
C_N. •
____
! HO -.: H H ""
l
+
lr
-C 1
_
+
Ph
(e
)
0
Uo
+
� Ao
(b)
o
581
& .. :0:
____
� CN
22-77 All of these Robinson annulations are catalyzed by NaOH.
(a) CH'
H
(l �o
+
� �o
CH 3
H
22-78
a
a
E'O � OE' -:1iE'· E'O H H�
:0: :0:
OE Y H
:0:
;---
.
.
:0:
:0: t :0: C_� OEt H Ph, C /H EtOOC./ C 'COOEt
+
• •
I
II
plus two other resonance forms
Ph, C /H CO2 H./ C 'COOH +
II
---
[
Ph, C /H HOOC./ C, COOH II
22-79
1
CH2OPO32CH20P032CH20P032(plus one other C==O C== resonance C==O form) .. HO{CH HO-CH HO-CH HC-O�-H .OH HCi 0: HC==O HC-OH HC-OH HC-OH CH20P032CH2OPO32- CH OPO32- CHglycer 0P0322 To identify the retro 2 aldehyde aldol, we must first C==O 3-phosphate locate the HO that is HO-CH2 dihydroxyacetone I
I
I
I
I
_I
I
..
I
• •
I
• • •
�
I
-
• •-
I "\......I e.
I
I
I
I
beta to the C=O.
a
I
phosphate 582
22-80
..
This is an aldol condensation. P stands for a protein chain in this problem. H :O-H 1 1 P-CH2CH2 C J CH
H :O:� 1 II H+ P-CH2CH2 C - CH --1
..
H
!
H :O-H 1 1 + � P-CH2CH2 C-CH 1 + H
this is another molecule of protonated aldehyde
..;).
1 ..7
H
+
H2 0.
plus one other resonance form
..
H :0-H 1 1.....1
P-CH2CH2 C=CH
H H I +1 :OH C==O I) 1 P-CH2CH2CH2-CH· CHCH2CH2-P
H + I two fast :OH C==O-H proton transfers .. 1 1 P-CH2CH2CH2-CH· CHCH2CH2-P plus one other resonance form • •
CHO 1 P-CH2CH2CH2 . CH = CCH2CH2-P
enol serving as a nucleophile
!-
CHO 1 + P-CH2CH2CH2 - CH � CCH7.CHr-P .
..
�-----H
...... 1
22-81
(a) aldol followed by Michael o EtOOC CH� 2 NaOEt � + CH2(COOEth -----l... + CH2(COOEth )l Y aldol H EtOOC NaOEI M;chael
�
EtOOC
yy COOEt
EtOOC
iS
..
COOEt
!
(b) Michael followed by Claisen condensation; hydrolysis and decarboxylation o o 0 EtOOC OEt OEt NaOEt Etooc NaOEt EtOOC ... Michae i Claisen I I o �O
J �
CO2
+
E tO H +
H2 0
p
A � � o
583
+
0
0
22-81 continued
(c) aldol, Michael, aldol cyclization, decarboxylation o EtOOe EtOOe ....... A NaOEt NaOEt • • d Michael � �
(0
�
IT �
\.�
H 0l. H
eOOEt
h o
Etooe
+
EtOH
+
eOOH
584
� !
�
EtOOe
NaOEt aldol EtOOC
e 02
r.
a
eOOEt
CHAPTER 23-CARBOHYDRA TES AND NUCLEIC ACIDS
Reminder about Fischer projections, first introduced in Chapter 5, section 5- 10: vertical bonds are equivalent to dashed bonds, going behind the plane of the paper, and horizontal bonds are equivalent to wedge bonds, coming toward the viewer. o� H CHO "C,.. HO
Fischer projection:
HO
+ +
H
HO-C-H
H
HO-C-H
I
CH2 0H
X �'.'T
A'
�
B_
"4
view from left side
X
X
Y
view from right side
y
X
Y B
y
CHO
H-t--OH
>
'A
"'r.'�
�
.B
CH2 0H
H
HO H
H
HO
A
X
CHO
23-1
CH2 0H
:
°
OH
HO
H
H
OH
HO
H
H
OH
H0
H
OH
HO
H
H
OH
HO
CH2 0H glucose
CH2 0H mirror image
CH2 0H
i =t=
OH H H
CH2 0H mirror image
CH2 0H fructose
All four of these compounds are chiral and optically active. 23-2
(a)
+ +
H H
OH
HO
OH
HO
CH2 0H two asymmetric carbons
+ +
CHO
CHO
=>
+ +
H
H
H
HO
OH H
+ +
CHO
CHO HO H
H OH
CH2 0H CH2 0H CH2 0H four stereoisomers (two pairs of enantiomers) if none are meso 585
23-2 continued
(b)
one chiral center
�
two stereoisomers (enantiomers)
(c) An aldohexose has four chiral carbons and sixteen stereoisomers. A ketohexose has three chiral carbons and eight stereoisomers.
��H
23-3
(a)
23-4
HO H HO HO
-
CH20H CHO H OH H CH20H H
--
L-( )-glucose
23-5
2
(b)
CHO H OH HO H HO H CH20H
CHO 4H-C-OH CH20H :CH1
i H
L-( +)-arabinose
1
1
3
D =R
23-6
HO 23-7
(a)
H H
--
t
CHO Ho H HO H CH20H
Ph
CH3
L-( -)-gJyceraldehyde
2
CHO HO-C-H 4 CH20H L =S �H2CHJ
i OH
3
CH20H Ph
2
CHO HO+H CH20H
L-(+)-erythrose
+ OHNHCH3 H3CHNHo t HH
H CH20H
1
Ho H
CHO HO+H CH20H
CHO H+OH CH20H
CH3
H H3CHN 3
586
+ HOH Ph
CH3
Ph
HO+H H+NHCH3 CH3 4
23-7 (a) continued H
H
Ph
+ =
HO
OH
H3CHN
NHCHl
2
1
Ph
+ =
H
H
H
H3CHN 3
Ph
+ =
OH
H
HO H
4
Ph
+ =
H
NHCH)
(b) Ephedrine is the erythro diastereomer, represented by structures 1 and 2. Structures 3 and 4 represent pseudoephedrine, the threo diastereomer. (c) The Fischer-Rosanoff convention assigns D and L to the configuration of the asymmetric carbon at the bottom of the Fischer projection. Structure 1 is D-ephedrine; structure 2 is L-ephedrine; structure 3 is L· pseudoephedrine; and structure 4 is D-pseudoephedrine. (d) It is not possible to determine which is the (+) or (-) isomer of any compound just by looking at the structure. The only compound that has a direct correlation between the direction of optical rotation and its D or L designation is glyceraldehyde, about which the D and L system was designed. For all other compounds, optical rotation can only be determined by measurement in a polarimeter. 23-8
(a)
H
H
H
H
CHO
(b)
OH
HO
OH
3
HO
H
OH
CH20H
H
H
OH
H 4
H
OH
HO
H
OH
H
CH20H
23-9 D-mannose
HO HO
H
3
OH
H
OH
CH20H
H�t--OH
r:=::�> HO
H
� inversion at bottom chiral center
HO --t=;;... H
CH20H
CHO 1
5
L-series sugar
L-arabinose
---,3,+-_: \ 2
rotare
H ----'i4t-- 0H H
CHO
D-idose
CHO
CH20H
D-xylose
HO
H
D-talose
CHO
HO
(c)
H
HO
OH
D-allose (d)
CHO
2
OH
6CH20H
C:===:::::>
587
�
OH
23-1 0 D-allose HO
1
OH OH H
OH
OH at C-3 is axial 23-11 D-talopyranose
OH groups at C-2 and C-4 are axial H
23-1 2
5
(a) HOH2C
(b)
OH
4
5
HOH2C
H H
OH
OH
4
H
OH
OH
D-ribofuranose
D-arabinofuranose 6
23-1 3 HOH2C
OH
5
CH20H 1
H
H
23-1 4 (a) ex-D-mannopyranose OH
(b) �-D-galactopyranose OH
(c) �-D-allopyranose OH
equatorial =�
HO H H
OH
OH
H
axial =ex
(d) ex-D-arabinofuranose
(e) �-D-ribofuranose H
HOH2C
equatorial =�
HO
OH cis to CH20H =�
HOH2C
trans to
OH
OH
H
OH
CHzOH =ex 588
OH
a
23-15
=
-
fraction of galactose as the a anomer; b
a (+ 150.7°) + b (+ 52.8°) a + b
b
1;
=
=
1
=
=
fraction of galactose as the � anomer
+ 80.2°
a
a (+ 150.7°) + (1 - a) (+ 52.8°)
=
solve for "a" c::====�> a
+ 80.2°
0.28; b
=
..
=
o.n
The equilibrium mixture contains 28% of the a anomer and 72% of the � anomer 23-16
H,
H�'c ....... .. 0:
0 C/ .......
. . �+
� 1 H-C-OH
• • -
HO:
H
•
II C-OH
•
H
OH
erythrose H
.......
, C/
0
1 H-C-OH H
+
H,
H
: + :� +
H
OH
CHzOH
I
0
H
OH
+
OH
CHzOH threose
CHzOH 1 C=O
CHzOH 1 C=O I Y"\
.......
C/
HO-C-H
+
CHzOH erythrose
23-17
OH
CHzOH
CHzOH
The planar enolate can reprotonate from either side, producing a mixture of erythrose and threose.
+
:OH
_I
+ +
HO-C: H H
OH OH
CHzOH
fructose CH,oH I C=O 1 HO-C-H H H
+ +
OH OH
CHzOH
589
�
H r'OH CHzOH 1 C=O
+
:+�: I
H-C-OH
CHzOH
23-18
H
1
CH20H 1 2C==O 31Y\
: + :� +
H
:OH
--
CH20H CH20H 1 I C==O C-O: • • II -1 ---- HO-C HO-C: H H
OH
+ +
OH
H
OH
H
CH20H
6CH20H
fructose
+ +
�
•
.
-
OH OH
CH20H
H
CH20H 1 C-OH
CH20H 1-
:C-OH
� + OH 1
O==C H
+
----
OH
H"OH 1
CH20H 21 H-C-OH 31 O==C H H
+ +
+
� + OH CH20H
CH20H
II
:O-C
!
+
OH
CH20H 1 HO-C-H 1 O==C
OH
H
OH
H
6CH20H
I
�H C-OH HO: I) II ... :O-C
H
CH20H
CH20H
+ +
OH OH
CH20H
590
� + OH
H
+
OH
CHO
23-1 9
X
H HO
CH20H H
OH H
----_ ......
..
H
OH
i
CH20H
CHO
HO H H
1=
CH20H
OH H
H
H
H NaBH4 HO
..
H
OH
H
OH
OH
H
OH
H
OH
COOH
(b)
H
HO
H
HO
H
H
OH
HO
H
H
OH
H
23-22 (a)
COOH
OH CH20H
D-galactonic acid
(b)
COOH
HO
H
H
HO
H
HO
H
OH
HO
H
H H
i OH
COOH mannaric acid
H
H HO
(c) no reactionD-fructose is a ketohexose; only aldoses react
H -+- OH
HO
D-mannonic acid
HO -
CHO L-gulose
COOH
CH20H
HO
H
..--
CH20H D-glucitol
CHO
OH
0H
CH20H D-glucose 23-21 (a)
CH20H
OH
NaBH4 HO
plane of symmetry
The reduction product of D-galactose still has four chiral centers but also has a plane of symmetry; it is a meso compound and is therefore optically inactive.
H
CH20H D-galactose
H
........ _------ .. -
HO
OH
23-20
H
HO
NaBH4
HO-+---H H
OH
OH
OH
COOH galactaric acid (meso) 591
X 1=
H H 0H H
CH20H L-gulose
COOH
CHO
23-23
H
OH
H
H
HO
HN03
CHO
OH
HO
H
H
HO
H
HO H
H
OH
OH
HO
..
H
A
H
OH
H
HN03 HO ..
H
H
OH
H
OH
H
OH
H
COOH mesooptically inactive
CH20H galactose
COOH
CH20H glucose
OH OH
COOH optically active
B
23-24 (a) (b) (c) (d) (e) (f)
not reducing: an acetal ending in "oside" reducing: a hemiacetal ending in "ose" reducing: a hemiacetal ending in "ose" not reducing: an acetal ending in "oside" reducing: one of the rings has a hemiacetal not reducing: all anomeric carbons are in acetal form
23-25 (c)
OCH2CH3 cis to CH20H= B
(d) HOH2C
HO
�
H
axial
23-26
=
�
a
OH
H
�W C�OH
\ ..
It
HO
H I
�\ H
••
�O
HO HO
I
H
H HO
I
H
9• •
�
H
\ � I+
It
HO
C_•O• -H I
�\ 5A �O
HO
H I
H
HO
592
..
- H20
H
�
CH3 H . HO
_OCH
OH
�O
HO HO
H
OH
H
..
OH
OH
H
�O
HO HO
OH
axial = a
I
H
�H3
H
9• •
CH3 H . ·
HO H
H
23-27
OH
OH
HO
2
H30+ ..
HO OH H
CN
CN I ./ HO-CHPh / a cyanohydrin
I
O-CHPh °
H
H
+8
II
Ph-CH
+
HCN is released from amygdalin. HCN is a potent cytotoxic (cell-killing) agent, particularly toxic to nerve cells. 23-28 HOH2C
OH
CH3CH20H
CH20H OH
_
W H20
HOH2C
..
H
H
OH
ethyl B-D-fructofuranoside
Ct and B-D-fructofuranose
+
CH20H
HOH2C
The aglycone in each product is circled.
H OH ethyl Ct-D-fructofuranoside 23-29
OH
HO ..
H
�
.0 : I
•
H
OH
OH
�.
H
HO ..
CH'-0 V SO,eH,
H
H +
593
-OS03CH3
23-30
(a) CH)OH2C
CH20CH)
OCH)
(b)
OCH) O CH) H
H
23-31
(a)
OAc
(b)
OAc
AcOH2C
H AcO
OAc
H 23-32
(a)
CHO
r
H
-J.- OH
--------
l
-----
CHO HO
- -- - '
HO --l-- H
i oH H -t- OH
r--- .. ..
,
:
--- -
HO
,
CH20H
... H.. . --
----
r--------
: HO
H
,
,
°
.. -_ ..
-----
H
,
H
H
OH
H
OH
CH20H
,
,
CH20H
D-g\ucose PhNHNH2 W
~
H
OH
H
OH CH20H
l
D-mannose
H I
C=NNHPh
I
C=NNHPh
iH H -t- OH
r--------·---------
HO
H--i--OH CH20H
594
/
D-fructose
OAc
23-32 continued
(b) CHO H---+-O H
r--------
H
..
I
W
I
I
W
C=NNHPh
HO
HO
HO H
CHO
--
r--------
HO---+-H
iH H + OH
......1--
C=NNHPh PhNHNH2
PhNHNH2
HO
H
HO
H
HO
OH
H
CH20H
CH20H
I
H
r--------
H H OH
CH20H
D-talose
D-galactose D-Talose must be the C-2 epimer of D-galactose.
23-33 Reagents for the Ruff degradation are: 1 . Br2, H20; 2. H202, Fe2(S04h .
CHO
CHO
r-
.. H---+.......... OH --
=t=
HO---+-H H H ..
-
CHO
=t=
HO
r--------
Ruff
HO---+-H
•
0H
H
OH
H
-
•
Ruff
HO
0H
H
OH
H
- -
CH20H
CH20H
_ ... - .. - -_ ... --- -_ ...._ .
---------
----
...
H H 0H OH
CH20H
--
--------- _
D-arabinose
D-glucose
=t=
r--------
------_ .
D-mannose
t
23-34 Reagents for the Ruff degradation are: 1 . Br2' H20; 2. H202, Fe2(S04h .
CHO H HO
I
OH H
HO�H
t
CHO
Ruff
HO
�
HO
H
•
H
H�OH
H�OH
CH20H
CH20H
D-galactose
CHO
Ruff
Ho H
H OH
CH20H
D-Iyxose
D-threose
595
23-35 Reagents for the Ruff degradation are: 1 . Br2' H20; 2. H202, FeiS04h .
CHO
CHO H
OH
H
OH
H
OH
H
OH
..
Ruff
H
OH
H
OH
H
OH
D-allose
HO
CHO H
H
OH
H
OH
...
Ruff
H
OH
H
OH
H
OH CH20H
CH20H
CH20H
23-36
H
HO
CHO
Br2 H2O
..
..
H202
FeiS04h
t
D-altrose
CHO
H H
CRO
CHO
OH
l) HCN 2) H3O+
H
HO
.. 3) Na(Hg)
OH
H
OH
H
OH
+ H
OH
H
OH
H
OH
CH20H
CH20H
CH20H
CHzOH
D-arabinose
D-erythrose
D-arabinose
D--ribose
23-37
H'
CHO HO
H
H
OH
H
OH CH20H
D-arabinose
..
H2NOH· HCI
C=NOH
HO
C
H
H
OH
H
OH
..
AC20
CH20H
HO
N H
H
OH
H
OH CH20H
oxime
cyanohydrin
596
t
CN- + ..
HOH2O
CHO
H H
OH OH
CH2 0H
D-erythrose
*:
23-38 Solve this problem by working backward from (+)-glyceraldehyde.
+
�
H
CHO
H
OH
<�:
==::1
CH20H
H ---If-- OH
H--if--OH ..
HO
H
H ---tl-- ° H
H---If-O - H
OR
CH20H cannot be C
COOH not optically active
(
HO ---If-- H
HN03 ..
COOH
CHO HO ---tl-- H H
HO
HO---tl-- H HN03 �
H�f--OH
CH20H must be C D-Iyxose
COOH optically active
HO -I-- H AND HO ---II-- H
H�I--OH
H---tl-O - H
COOH not optically active
CH20H A D-galactose
HO�I--H H
COOH
/
597
HO_f--H
HN03 •
HO---tl-- H H---If--OH
CH20H
COOH optically active
D-talose
r-----",..
HO---tl--H
OH B
'"
HO---lf--H
H ---tl-- ° H
HO ---tt- H
HO�f-- H
OH
optically active tartaric acid (not meso)
/
H ---tl-- OH
H---tO t- H
HO---lH l--
CHO
H
COOH
��--------� CHO \
�------�
COOH
H
CH20H
CHO
COOH HN03
HO
H
u
+ +
COOH
D-threose D
D-(+)-glyceraldehyde
HO--if--H
CHO
:=t=::
23-39 Solve this problem by working backward from (+)-glyceraldehyde. CHO
+
CHO
H
OH
<¢:: ===::J
: r:
CH20H
CH20H
CH20H
F
D-erythrose
D-(+)-glyceraldehyde
optically inactive erythritol
CHO
CH20H
H-t-OH
H-t--OH
H
OH
H
OH
H
OH
H
OH
CH20H D-ribose
CH20H optically inactive ribitol
E
23-40 (a) HOH2C
(b)
OH
H OH D-fructose
iCH2OCH3 2
CH30H
+
CH30 H
3 4
H
°
H OCH3 OH
6
CH2OCH3
(c) Determining that the open chain form of fructose is a ketone at C-2 with a free OR at C-5 show", that fructose exists as a furanose hemiacetal.
598
23-41
(a)
HC=O I C=O
CH20H
I
CH20H +
H OH methyll3-D-fructofuranoside
(b) HO
� H
D-glyceraldehyde 2H5IO�
OH O
H
H
OH
H30+ ___
OCH3
CH20H
HC=O I C=O I CH20H +
methyl I3-D-fructopyranoside
Production of one equivalent of formic acid, and two fragments containing two and three carbons respectively, proves that the glycoside was in a six-membered ring. (c) In periodic acid oxidation of an aldohexose glycoside, glyceraldehyde is generated from carbons 4,5, and 6. If configuration at the middle carbon is D, that means that carbon-5 of the aldohexose must have had the D configuration. On the other hand, if the isolated glyceraldehyde had the L configuration, then the original aldohexose must have been an L sugar. 23-42
OH
OH
HO
HO
a anomer of maltose
H
13 anomer of maltose
H
599
H
H
OH H
�O
OH H
0
H
OH
B anomer of maltose
H
OH
o
�i n
_ open chain fonn of maltose
H
-
HO
H
OH
oH H
�
OH H
0
o
+
H
AgO (mirror)
23-44 Lactose is a hemiacetal. Therefore, it can mutarotate and is a reducing sugar.
OH
OH
OH o
o
H
H H a anomer of lactose
H
OH
H
OH
H B anomer of lactose
OH H
H
23-45 Gentiobiose is a hemiacetal; in water, the hemiacetal is in equilibrium with the open-chain fonn and can react as an aldehyde. Gentiobiose can mutarotate and is a reducing sugar. 23-46 Trehalose must be two glucose molecules connected by an
OH
a-I, I '-glycoside.
a-D-glucopyranosyl-a D-glucopyranoside -
H H
H
600
0
melibiose = 6 -0-(o.-galactopyranosyl)OH D-glucopyranose
raffinose OH
23-47
galactos
H
H
0.-1,6'
invertase
..
glucose
HO H
H +
HOH2C
OH
H
The lower glycoside linkage is a-I ,2' from the glucose point of view, but f}2,1' from the fructose point of view.
�
O� /0 "C
NH2
I
I
0
H
CA I
H
N,
N I
"
0 II 0-C-CH3
0 0 H I H CCH3 .y 0
eta 2::� I CH3
23-49 cytosine:
H OH fructose (D-fructofuranose)
0 II 0-C-CH3
23-48 cellulose acetate
uracil:
OH
HOH2C
fructose
N
o
0 I H C .y , CH3
0 II 0-C-CH3
o�
0
I I H H H H �C ' C CH3 // " 0/ CH3 //C" 0 CH3 0
OH
..
O
H
OH
guanine:
N� �� JL N I NH2 _ _
601
H
23-50 Aminoglycosides, including nucleosides, are similar to acetals: stable to base, cleaved by acid.
(a)
�RA) HHH I
R
H R,I,R
OH
OH
I
0
•
H 1 R N
OH OH
3° aliphatic amine strong base
CH2
HO-CH2
+
-
OH OH
OH OH
H2 :
I
hen:iacetal form of nbose
CH2
0
••
H
H
•• H :0' H
/ iU
H2 :
..
H�
CH2 H
H
H
O
OH OH
NH2
NH2
H
0
0
N
.
CH2
0
N+
� yl :0'
H
H
� -t )
C1 ? Cl.. � � I
tH2
OH
+
0:::-...
��\) �� 0 ! � R � 0 �
OH
(b)
R2NH2
•
+ • • '---"H "
----I.�
•
+
•
�
hH2 ..�
OH
R:
H
H
.
N
N
H
N
N
H
site of OH OH OH OH OH OH protonation; weak cytidine base adenosine NucJeosides are less rapidly hydrolyzed in aqueous acid because the site of protonation (the N in adenosme, and in cytidine, the oxygen shown with the negative charge in the second resonance form) is much less basic than the aliphatic amine in an aminoglycoside. Nucleosides require stronger acid, or longer time and higher temperature, to be hydrolyzed. This is important in living systems as it would cause genetic damage or even death of an organism if its DNA or RNA were too easily decomposed. Organisms go to great length and expend considerable energy to maintain the structural integrity of their DNA. 23-5\
-- - -:N� t�N-H + + :O: --- -H-N
N
I ribos�1
H . •• N-H-- -:O
H
={
)= N-H ----: · O: � � guanine Ii N
_
.
. .
N
.
.
�� t
N
nbose
f
'\
y\
CH3
I .... . N ..-: N N: ----H I -r--,
N =.!
.
.
+
�
.• 0.• .
.
�
adenine thymine cytosine The polar resonance forms show how the hydrogen bonds are particularly strong. Each oxygen has significant negative charge, and in each pair, one H -N is polarized more strongly because the N has positive charge.
602
23-52 Please refer to solution 1-20, page 12 of this Solutions Manual. 23-53 (a)
CHO H
HO
OH
(c)
(b)
CH20H
HO
H
H
OH
H
OH
OH H
H
CH20H 23-54
OH
OH (b)
(a)
HO
HO
H
OH H
H (c)
(d)
OH
HO OH
OH H
H
23-55 (a) (b) (c) (d) (e) (f) (g)
O-aldohexose (D configuration, aldehyde, 6 carbons) O-aldopentose (0 configuration, aldehyde, 5 carbons) L-ketohexose (L configuration, ketone, 6 carbons) L-aldohexose (L configuration, aldehyde, 6 carbons) O-ketopentose (D configuration, ketone, 5 carbons) L-aldotetrose (L configuration, aldehyde, 4 carbons) 2-acetarnido O-aldohexose (0 configuration, aldehyde, 6 carbons in chain, with acetarnido group at C-2)
603
23-56
H
(a)
0 , C'l I
tr\
0.0HR :
HO-C - H HO
H
H
,.-::0: ,c/ I-
HO-C:
--
II
HO-C
..
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH CH20H
CH20H
CH20H D-mannose
CH20H
I I
o==c HO H H
t1
HO-H
H
HO
H
HO
H
..
"D
0 ,c/o•'o
H
i)
- OH ,oC /
OH ,C/
.
•. I o==C
HO-H ..
H0 -j H
H
H
OH
CH20H
• HO'.. .. ..
II
---
: o-c HO
H ----+- OH
OH
Cl
H
OH
H
H
'C/
�'o0.. C H
. HO
OH
H
H
OH
H
CH20H
I II
H
CH20H
OH
H OH OH
CH20H
D-fructose
enediol
(b) The a isomer has the anomeric OH trans to the CH20H off of C-5. The �-anomer has these groups cis. c
HOH2C
CH20H
�
HOH2C
A
OH
OH
H
�-0- fructofuranose
a-D-fructofuranose
604
23-57
(a) HO
+
CHO
H
+ CN
OH
HCN �
�
H
OH
H
H
CN
+
�
OH
B
CH20H
H
H
Ba(OHh H2O
+
..
H
H
H
HO
OH
+
H OR
COOH (-)-tartaric acid COOH
OH
H
HN03 ---
OH
D
+
H
OH Oll
COOH meso-tartaric acid
CH20H
COOH
COOH (S,S)-(-)-tartaric acid
HO
COOH
OH
+
COOH
OH
CH20H
+
H
+
H
C
COOH
HO
H2O
HO
..
CH20H
H
(b)
+
Ba(OHh
H
A
CH20H
+
COOH
COOH
OH
H
H
H
COOH (R,R)-(+)-tartaric acid
OH OH
COOH (R,S)-meso-tartaric acid
23-58
(a) D-(-)-ribose
(b) D-(+ )-altrose
(c) L-( +)-erythrose
(d) L-(-)-galactose
(e) L-(+)-idose
23-59
(a)
(b) OCH3
OH
H
H
(c) CH30H2C
OCH3
(d)
OH CH20CH3 OH
H
OCH3
H
H
605
23-60 (a)
(b)
OH
HO
OH
H H
CH20H
OH OH
OH
HOH2C
OH
H
HO
°
H
H
H
H °
I
H
HO HO
�
OH H
H
23-61 (a) (b) (c) (d)
methyll3-D-fructofuranoside 3,6-di-O-methyl-I3-D-mannopyranose 4-0-( CI.- D-fructofuranosy I)-I3-D-galactopyranose /3-D-N-acetylgalactopyranosamine, or 2-acetamido-2-deoxy-/3-D-galactopyranose
23-62 These are reducing sugars and would undergo mutarotation: -in problem 23-59: (b) and (c); -in problem 23-60: (a) and (c); -in problem 23-61: (b), (c), and (d) 23-63 (a)
H
HO
HO H
COOH
OH H
H OH
CH20H
(b)
CH20H
CHO
HO HO
HO H
+H H H
OH
CH20H
+
(c)
6: 0
HO
H
HO
H
H
OH
CH20H
606
H +
others
-\4-i \ . -OH
OB
-
H
,.OCH3 H
23-63 continued COO(d)
(e)
CHzOH
H-t-0H
H
OH
HO--f--H
HO
HO--f--H
HO
H
i oH
H
+ (i)
CHzOH
HO H
I
H AcO
H H
�
H
H
OH
�
CH
,0
H CHOIO
OAC
OCH ", 3
H
(j)
CHO
H
H
H
OH
H
HO
H
H
OH
H
H
OH
OH
HO
H
CHzOH
HO
H
HO H
+
HO
H OH H
H
HO
OH
H
OH
H
CHzOH
±
CHO
CHO
HO
H--+-OH HO
�
19}
OAc
OAc
CHzOH
CHzOH
(h)
(f)
CHzOH
CHzOH CHO
(k) H
� OH
HO--f--H
excess HI04 ..
5
HO----t=-H H
II
°
� OH
II
°
H-C-OH from C-l through C-5
+
1
H-C-H from C-6
CHzOH
23-64 Use the milder reagent, CH3VAgzO, when the sugar is in the hemiacetal form; the mild conditions prevent isomerization. When the carbohydrate is present as an acetal (a glycoside), use the more basic reagent, NaOHl(CH3)zS04; an acetal is stable to basic conditions. (a)
CH20CH3
CH3° H H
I
(b)
(c)
CHO
°
H
H
CH30
OCH3
CHO H
H
CH30
OCH3 H
CHzOCH3
bo
+
CH3O-+--H
OCH3
H
OCH3
H
OCH3
H
OH
H
OH
H
OH
H
CH20CH3 using CH3I, AgzO
CHzOCH3 using (CH3)2S04' NaOH 607
I
oCH3 OH
CHzOCH3 CH2OCH3 using (CH3hS04' NaOH
23-64 continued
H CH30 CH30 H
� OCH3
�
(e)
CHO
CHO
(d)
+
H
OCH3
H
H
CH30
H
CH30
H
H
OH
H
H
OH
H
i oH
CH2OCH3 CH2OCH3 using CH3I, Ag20
(f)
=C
CHO
CHO
r
+
OCH3 H
CH30
OCH]
H
OH
H
CH2OCH3
NHCOCH3
CH30
using (CH3hS04' NaOH
H
H
OH
H
OH CH2OCH3
23-65
(a) These D-aldopentoses will give optically active aldaric acids. CHO D-arabinose
CHO
H
HO
H
H
OH
HO
H
H
OH
H
HO
CH20H
+ +
D-Iyxose
OH CH20H
(b) Only D-threose (of the aldotetroses) will give an optically active aldaric acid. CHO
D-threose
HO H
H OH
CH20H
608
X
+
OCH3 H OCH] OH
CH20H using CH31, Ag20
CHO H
H
23-65 continued
(c) X is D-galactose. COOH
CHO
H X:
HO HO H
I !
H
OH H
HN03
HO
H
HO
H
..
H
H
OH
COOH
HO
H H
H
OH
Ruff
CHO
HO
optically inactive
COOH
CHzOH
HO
OH
--!- H
HO -f-- H
HN03 •
H
OH
i OH
optically active
COOH
CHzOH
The other aldohexose that gives an optically inactive aldaric acid is D-allose, with all OH groups on the right side of the Fischer projection. Ruff degradation followed by nitric acid gives an optically inactive aldaric acid, however, so X cannot be D-allose.
t
t
(d) The optically active, five-carbon aldaric acid comes from the optically active pentose, not from the optically inactive, six-carbon aldaric acid. The principle is not violated. CHO
(e) HO
--!- H
HO-f--H
H-f--OH CHzOH
CHO
Ruff
---
Ho H
H
COOH
HN03 •
Ho H
OH
CHzOH D-threose
H OH
COOH (S,S)-tartaric acid optically active
609
23-66 (a)
+
Ho
CHO
H
t
:=+=:: CN
CN
HCN
OH
..
H
H
+
OH
CH20H CH20H CH20H (b) The products are diastereomers with different physical properties. They could be separated by crystallization, distillation, or chromatography. (c) Both products are optically active. Each has two chiral centers and no plane of symmetry. 23-67
�
fc]\
(a) The Tollens reaction is run in aqueous base which isomerizes carbohydrates. H
HO
'¢f=9 \ u H
H
-
(II
C-OH
.. HO:
H--II--OH
HO-I--H
H--II--OH CH20H
H-t-OH
H-It-OH
H-t-OH
H--I--
CH20H plus one other resonance fonn H
HO-f--H H--II--OH
-H ....... /·· C
+
0: ....... / C/ 1:C OH
r
HO'l H
HO-t--H
HO
..
-
H
H---1r--OH
H--ll--OH
H--II--OH
H---1r--OH
H
CH20H D-glucose
CH20H D-mannose
OH CH20H
plus one other resonance fonn
(b) Bromine water is acidic, not basic like the Tollens reagent. Carbohydrates isomerize quickly in base, but only very slowly in acid, so bromine water can oxidize without isomerization.
610
23-68 Tagatose is a monosaccharide, a ketohexose, that is found in the pyranose form.
61 1
Z19
o
HO
-o
--- -'d 'i°
\
H
0
H
Z H:J
/
H
o
I
Z H:J-O-d-OII o
(q) HO 0-
I
0
I
0-
I
H :J-O-d- O-d-O-d-O
Z
\I
0
II
0
\1 o
23-72 Bonds from C to
H
are omitted for simplicity.
NH2
V\
o
A � � N
5'
?
N
o
\
0
\
P -0/
�o
c
O
NH2
,) t ? �
N
I
" N
A
N
{
o
\
�O
\
:.--' - /P 0
o
O
H C J I
?
t
H T
o
0
\
...,0 p"-0""'"
\
o
? �
N
:c O
I
N
,H
A NH2 N
G
°
3'
23-73
�
(a) No, there is no relation between the amount of G and A. (b) Yes, this must be true mathematically. (c) Chargaffs rule must apply only to double-stranded DNA. For each G in one strand, there is a complementary C in the opposing strand, but there is no correlation between G and C in the same strand.
61 3
23-74 Bonds from C to H are omitted for simplicity.
�;:
H
�;:
H
a
CH3
HO
5'
a
OH
l N�
0
2'-deoxythymidine (abbreviated dT)
a
CH3
HO
5'
a
N N+
l N�
0
3'-azido-2',3'-dideoxythymidine (AZT)
II
�'"
No phosphate can attach to the azide group, so synthesis of the DNA chain is terminated.
�;: a
CH3 a a a I II I O -P -O-P-O-p -O I I I a a a -
-
5'
a
-
AZT 5'-triphosphate
N+ N N II
II
61 4
l N�
H 0
H-O-N=O
H+
H0
N=O
23-75 Recall from section 19-17 that nitrous acid is unstable, generating nitrosonium ion. (a)
+
+
2
-
+
AI:�" A
l
+
0
H N
N=R
cytosine (C)
+
two rapid proton transfers
N
�' +
: N=N
tA tA I
-""'::::N
H
N
H 0: 2
�
o
I
H
N
H
H2O
..
o
H� \ ':tO� •
•
I
-""':N :::
H
H+
I N�O H
..
I
-""'::::N
O
H 0: 2 �
OH
NH
...
0
===::
enol tautomer of uracil
(b) In base pairing, cytosine pairs with guanine. If cytosine is converted to uracil, however, each replication will not carry the complement of cytosine (guanine) but instead will carry the complement n1' uracil (adenine). This is the definition of a mutation, where the wrong base is inserted in a nucleic acid chain. (c) In RNA, the transformation of cytosine (C) to uracil (U) is not detected as a problem because U is a base normally found in RNA so it goes unrepaired. In DNA, however, thymine (with an extra methyl group) is used instead of uraci l. If cytosine is diazotized to uracil, the DNA repair enzymes detect it as a mutation and correct it.
615
H
I N�O H
f� r�� NH
if this species exists, its lifetime is very short as water will attack quickly
-
!
Q 6q
23-76
I
(a)
< }-
h
C-O
se
(b) Trityl groups are specific for 1 ° alcohols for steric reasons: the trityl group is so big that even a 2° alcohol is too crowded to react at the central carbon. It is possible for a trityl to go on a 2° alcohol, but the reaction is exceedingly slow and in the presence of a 1 ° alcohol, the reaction is done at the 1 ° alcohol long before the 2° alcohol gets started.
OH
OH
(c) Reactions happen faster when the product or intermediate is stabilized. Each OCH3 group stabilizes the carbocation by resonance as shown here; two OCH3 groups stabilize more than just one, increasing the rate of removal. The color comes from the extended conjugation through all three rings and out onto the OCH3 groups. Compare the DMT structure with that of phenolphthalein, the most common acid base indicator, that turns pink in its ring open form shown here. Phenolphthalein is simply another trityl group with different substituents. + H + CH3 :o� :0'" one of the major resonance contributors showing the delocalization of the positive charge on the oxygen
¢
I I 0-0� Ij_
C
�-!J
<
OCH3
¢} �c� o
23-77
o
�
(a)
acetal
-\H
Ph
phenolphthalein
H
6
(b)
O
o
3
OH
OH
OH
>
Ph
�
O
benzaldehyde OH
OH
�-D-glucose
(c)
The 4- and 6-0H groups of glucose reacted with benzaldehyde to form the acetal. (c) and (d) The chiral center is marked with (*). This stereoisomer is a diastereomer since only one chiral center is inverted. This OH diastereomer puts the phenyl group in an axial position definitely less stable than the structure shown in part (a). Only the product shown in (a) will isolated from this reaction. ,
OH
Only the 2'- and 3'-OH groups are close enough to form this cyclic acetal (ketal) from acetone, called an acetonide. 616
CHAPTER 24-AMINO ACIDS, PEPTIDES, AND PROTEINS
24-1 (a)
COOH
H2N
I "'''CH2Ph � H
���
(d)
H2N
COOH
(c) H2N
COOH
(b)
-
NH
NH2
COOH
H2N
,oH
I "CH2CH2CH2NH- C � H II
I '''''CH2 � H
24-2 (a) The configurations around the asymmetric carbons of (R)-cysteine and (5)-alanine are the same. The designation of configuration changes because sulfur changes the priorities of the side chain and the COOH. 2
(5)-alanine with group priorities shown 1
H2N
COOH
�
� 3
3 ' CH 3 H 4
" "
1
H2N
COOH 2
'"''CH2SH 4H
(R)-cysteine with group priorities shown; note that the COOH and the CH2SH priorities are reversed compared with (5) alanine
(b) Fischer projections show that both (5)-alanine and (R)-cysteine are L-amino acids.
+
H2N
H
CH3
(5)-alanine L-alanine
+
COOH
COOH H2N
H
CH2SH
( R)-cysteine L-cysteine
24-3 In their evolution, plants have needed to be more resourceful than animals in developing biochemical mechanisms for survival. Thus, plants make more of their own required compounds than animals do. The amino acid phenylalanine is produced by plants but required in the diet of mammals. To interfere with a plant's production of phenylalanine is fatal to the plant, but since humans do not produce phenylalanine, glyphosate is virtually non-toxic to us. 24-4 Here is a simple way of determining if a group will be protonated: at solution pH below the group's pKa value, the group will be protonated; at pH higher than the group's pKa value, it will not be protonated. H (a)
r
H
? 1
(b)
H2N - -COOCH(CH3h
(d)
(c)
617
H-
1+
\
C -COOH
�
24-4 continued (e)
+
lysine H
alanine H I
H3N-C-COO-
(i) pH 6
I
CH3
+
I
?
H2N- -COO-
H2N-C-COO-
(CH2)4NH2
CH2COO-
H I
H3N-C-COOH
(iii) pH 2
I
? I
H3N - -COOH
I
I
H
I
. .
H -N -H
I
I
.... . .. t--I.�
H
H -N -C -N -H
I
H
t
1.... . . --I"� ..
I
+
I
CH2COOH
H -N -H
I
H
H3N-C-COOH
+
H -N-H +
+
(CH2)4NH3
. .
H -N==C-N -H
I
H
+
CH3 24-5
I
I
CH3 +
H
H
H2N-C-COO-
I
CH2COO-
+
I
?
H3N- -COO-
H3N -C-COOI (CH2)4NH3
H (ii) pH 11
+
I
aspartic acid H
H
+
H -N-C==N--H I I H H
+
H -N -H
II
H-N -C-N -H
I
I
H H Protonation of the guanidino group gives a resonance-stabilized cation with all octets filled and the positive charge delocalized over three nitrogen atoms. Arginine's strongly basic isoelectric point reflects the unusual basicity of the guanidino group due to this resonance stabilization in the protonated form. (See Problems 139 and 19-49(a).) 24-6
tryptophan
histidine
H
H
H2N-
? I
-COOH
? I
H2N- -COOH
CH2
ro
not bask (like pyrrole)
I
H
CH2
G�
basIc --.. N (like pyridine) • •
not basic (like pyrrole)
The basicity of any nitrogen depends on its electron pair's availability for bonding with a proton. In tryptophan, the nitrogen's electron pair is part of the aromatic 1t system; without this electron pair in the 1t system, the molecule would not be aromatic. Using this electron pair for bonding to a proton would therefore destroy the aromaticity-not a favorable process. In the imidazole ring of histidine, the electron pair of one nitrogen is also part of the aromatic 1t system and is unavailable for bonding; this nitrogen is not basic. The electron pair on the other nitrogen, however, 2 is in an sp orbital available for bonding, and is about as basic as pyridine. 618
24-7 At pH 9.7, alanine (isoelectric point (IEP) 6.0) has a charge of -1 and will migrate to the anode. Lysine (IEP 9.7) is at its isoelectric point and will not move. Aspartic acid (IEP 2.8) has a charge of -2 and will also migrate to the anode, faster than alanine. caili
�
� � ��
nOde
t
t
lysine o
t
alanine -1
aspartic acid -2
24-8 At pH 6.0, tryptophan (IEP 5.9) has a charge of zero and will not migrate. Cysteine (lEP S.O) has a partial negative charge and will move toward the anode. Histidine (IEP 7.6) has a partial positive charge and will move toward the cathode.
�
catho
�
� t
nooc
t
histidine H
� � t
tryptophan +
cysteine
?
H
+
I
H3N - -COO-
\N � I
H
tryptophan
I
CH2Scysteine (partially deprotonated sulfur) (The SH is more acidic than the NH3+ group.)
24-9 (a)
O=C -COOH
I
CH3 (b)
O=C-COOH
H2N -CH-COOH I CH2CH(CH3h
I
CH2CH(CH3h (c)
O=C-COOH
H2N -CH-COOH
I
I
CH20H (d)
O=C-COOH
CH20H NH3
H2, Pd CH2CH2CONH2
I
..
I
H3N -C-COO-
�
histidine (partially protonated imidazole ring)
H
H2N -CH-COOH
I
CH2CH2CONH2 619
24-10 All of these reactions use: first arrow: (1 ) Br2IPBr3' followed by H20 workup; second arrow: (2) excess NH3, followed by neutralizing workup. (a) H2C -COOH
(1)
I
..
Br - CH-COOH
I
.. H2N - CH-COOH I H
(2)
H
H (b) H2C -COOH
..
(1 )
I
CH2CH(CH3h
(c) H2C-COOH
..
(1 )
I
Br -CH-COOH I CH2CH(CH3h
(2)
..
I
H2N-CH-COOH CH2CH(CH3h
I
Br - CH-COOH
I
.. H2N - CH-COOH
(2)
CH(CH3h
CH(CH3h
CH(CH3h
(d) H2C -COOH
(1)
I
..
Br -CH-COOH
I
(2)
..
H2N-CH-COOH
I
CH2CH2COOH
CH2CH2COOH
CH2CH2COOH
In part (d), care must be taken to avoid reaction a to the other COOH. In practice, this would be accomplished by using less than one-half mole of bromine per mole of the diacid. o
I
1 ) NaOEt
I
2) BrCH(CH3h
COOEt
N - CH
I
I I
o
abbreviate COOEt
G�
? (}- H C ?
COOEt 1 ) NaOEt
2) BrCH2Ph-
COOEt
(c)
OOEt
G
COOEt I N - C-CH2Ph
COOEt
N-
H
1 ) NaOEt
COOEt �
!
H30+
H2N - CH-COOH I CH(CH3h
N-CH
(b)
COOEt
N - C-CH(CH3h
..
COOEt
�
o
o
..
2) BrCH2CH2COO\ COOEt .,. salt, not acid-why?
I
COOEt
H30+ �
-
H30+ �
- H2N -CH-COOH I CH2Ph
H2N - CH-COOH
I
CH2CH2COOH
620
24-1 1 continued
0�
(d)
COOEt
- H
1) NaOEt
H30+
H2N - CH-COOH
tJ.
----,t.�
COOEt
I
CH2CH(CH3h
24-12
COOEt
I
COOEt 1) NaOEt
AcNH -CH
I
COOEt
2) BrCH2Ph
I
H30+
I
AcNH - C - CH2Ph
•
tJ.
.. ----,t �
COOEt
H2N -CH - COOH I CH2Ph
acetamidomalonic ester o
24-1 3 (a)
II
PhCH2-CH
NH3, HCN
NH2
I
..
H2N -CH-COOH
PhCH2-CH
I
C
I
CH2Ph
N
(b) While the solvent for the Strecker synthesis is water, the proton acceptor is ammonia and the proton donor is ammonium ion.
�
�
:0:
�+ H - NH3
+NH2
I)
H
+
I
PhCH2-CH
t
:NH2
-H20 .... ..f---
'--../
0NH3
°
H-O - H +Jr\ + 1) H3N -H PhCH2-CH ..
I
NH2
II
I
this protonated imine is rapidly attacked by cyanide nucIeophile
N
PhCH2-CH
I
+ NH2
NH2 (abbreviate as R -C
� :O - H I
I
PhCH2 - CH NH2
C
PhCH2-CH
l
two fast pMon tmnSf
PhCH ,- H
N
on the next page) 621
24-1 3 (b) continued mechanism of acid hydrolysis of the nitrile
+ J .�H+ .. l R-C==N -H .. R-C==N. _
H 1+ ==N - H
1 .�. R-
T
+
JI'
H
I
..
I
20: two rapid OH H proton transfers I �I + .. R - C -N-H
� H+
R - C -N -H
I
HO
•
•
+O. -H
� I I
I
.
I
HO
I
-
i
NH3 ..
H
•
R - C==N-H
H -O \. .. -H
-
H
II
H20:
H -OH H
•
H20:
• •
R -C-N - H
:O - H
+..
l} H
�
R - C==N -H
R -C -N - H '"
:O - H
VI
+
..
_
..
}
�I+
H20:
+
.
H+"'--R - C==N - H
�
I
:O . -H
:O -H
I
• •
• •
R-C - OH + • •
---
:O-H I + R - C==OH
+
H2N - CH - COOH
I
H20: ..1- .... .
CH2Ph
24-1 4 (a)
R I
C-H
�
NH3, HC H20
CH2CH(CH3h (b) 0
II
C-H
I
II
C -H
I
CH(CH3h
I
CH2CH(CH3h NH3, HCN .. H2O
H (c) 0
H2N - CH-CN
H2N -CH-CN
I
H NH3, HCN .. H2O
T
H2N - H-CN CH(CH3h
622
H30+ .. !l
H2N-CH -COOH
H30+ .. !l
H2N-CH-COOH
H30+ !l
..
I
CH2CH(CH3h
I
H
H2N - CH - COOH
I
CH(CH3h
:O�
II
• •
R - C -OH
24-1 5 In acid solution, the free amino acid will be protonated, with a positive charge, and probably soluble in water as are other organic ions. The acylated amino acid, however, is not basic since the nitrogen is present as an amide. In acid solution, the acylated amino acid is neutral and not soluble in water. Water extraction or ion-exchange chromatography (Figure 24-1 1 ) would be practical techniques to separate these compounds. 24-1 6
o
0
+
II
abbreviate H3N -C H - COEt
I
:O :
II
CH2Ph
�
R-C -OEt
II
R-COEt
as
' ' OH � I I' . . H20 : • RC-OEt
H+
-
+
•
?
o
+ I '"
II
•
?
CH2CH2CONH2
RC -OEt
•
t:J
H
II
OH H
EtOH
�.�---
RC
I
I r\1+
RC-O -Et
l
:O -H
(protonated form in acid solution)
+
•
�
CH2Ph
H3N- H -COO-
I
.. H20 :
RC-OEt
. H20 :
H3N- H - COH
24-1 7
OH
I
H - � - H�
•
plus two other resonance forms
+
OH
O-H
plus two other resonance forms
PhCH20H
+ .. H3N -CH - COOCH2Ph
--i�
---
H+
I
H2, Pd
--
+ H3N -CH-COO-
I
CH2CH2CONH2
CH2CH2CONH2 +
°
24-1 8
II
+
H3N-CH-COO-
I
PhCH20 - C -Cl ..
o
II
H
I
PhCH20C -N-CH-COOH
! I
CH2CH2SCH3
CH2CH2SCH3
PhCH3
+
CO2
+
+
?
H2, Pd
H3N- H-COOCH2CH2SCH3
623
CH3Ph
24-1 9
..
N :0 :
:0 :
1 N
:0:
i
N :0:
:0:
�
:0 :
:0 :
-
-
:0: ..
:0 :
:0 :
:0 :
:0 :
:0 : ..
1
"
:0:
N�
:0 :
:0:
These are the most significant resonance contributors in which the electronegative oxygens carry the negative charge. There are also two other forms in which the negative charge is on the carbons bonded to the nitrogen, plus the usual resonance forms involving the alternate Kekule structures of the benzene rings. 24-20 (a)
(b)
o:
II :
0: 0 II : iI H3N - CH·C�NH·CH·C�NH·CH·C-O: I : I I CH2Ph : HO-CHCH3: CH2CH2SCH3 Phe ' Met Thr +
,
,
o: 0: 0: 0 I I: II : II I I :. H3N -CH . C � NH . CH . C -+- NH . CH . C -7- NH . CH . C - 0+
I
:
CH20H: seryl
:
I
:
I
:
(CH2h :
I
H
:
:
HN - C - NH2 glycyl
II
:
I
CH2Ph phenylalanine
arginyl NH
(c)
0: 0 I I: II H3N -CH . C --f NH . CH . C -:- NH . CH . C -7" NH . CH . C -:- NH . CH . C-0: I : I I : I : I : (CH2h : (CH2h : CH2 (CH2)4 CH3CH2CHCH3 : . I : I : I : I I (isoleucine) SCH3 CONH2 COOH NH2 M (methionine) Q (glutamine) D (aspartic acid) K (lysine) +
0:
0:
0:
I
I
I
I
0:
0:
I
I
II:
0: (d)
+
II:
II :
0:
II :
II :
II :
II :
-
0
II
H3N - CH . C --f NH . CH . C -:- NH . CH . C -7" NH . CH . C � NH CH - C - 0I : I : I : I : I (CH2h : : CH2 :, CHMe 2' CHCH3: CH2 . ' . I Try spelling your '' I '' ' I .' I COOH name in peptides! CHMe2 CH2CH3 OH L (leucine) E (glutamic acid) V (valine) I (isoleucine) S (serine) 624
'
I
•
(a)
II
II
(b)
/C,
HN NPh \ I HC -C
�0
I
CH3
S
S
S
S
24-21
II
/ C,
(c)
HN NPh \ I HC-C=O
I
I
II
NPh I \ C -C
I
H
(CH2)4NH2
24-22
/C,
C
HN NPh \ I HC-C=O
CH(CH3h
II
(d)
/C,
�0
S 1 ) PhNCS
H2N-Ile--7Gln--7peptide
..
/C,
+
HN NPh \ I HC-C=O I CH3-CHCH2CH3
H2N-Gln--7peptide
II
S Step 4
/C,
HN NPh + H2N-peptide \ I HC-C=O I CH2CH2CONH 2
1) PhNCS
H2N-Gln--7peptide
Fy
. I�
24-23 Abbreviate the N-terrninus of the peptide chain as NH2R .
F
(a) This is a nucleophilic aromatic substitution by the addition-elimination mechanism. The presence of two nitro groups makes this reaction feasible under mild conditions.
'Q F
02N
·
;-...... . . NH2R
�
I
.
" :0
-
�
•
•
..
0
·
".
�-C . .
�
+
NH2R
I
· •
N : 0"
-
• •
�..
F
+
NH 2R
I
� ,. N+
+ ,. N _ " '0 0
N02
0·
0"
0
- "N 0 �
+
NH2R
D
11+
I
.. C-
I
1+
'0
_
,.
:0"
N'
• •-
0:
\ I
�
N02
�
I
�
+:r::
--
N02
(b) The main drawback of the Sanger method is that only one amino acid is analyzed per sample of protein. The Edman degradation can usually analyze more than 20 amino acids per sample of protein. 625
I
24-24
I
trypsin
� �
N-tenninus
l
1
r
Tyr-Ile----Gln-Arg-Leu----Gly-Phe-Lys-Asn-Trp--Phe----Gly-Ala-Lys�ly-GIn-GlnoNHz
I
t
C tcnrunu, (amide fonn)
chymotrypsin
Phe-Gln-Asn
24-25
Pro-Arg-Gly·NH2 Cys-Tyr-Phe
Asn-Cys-Pro-Arg
\ Tyr-Phe-Gln-Asn �----------
_/
)
Y
---------
Cys- Tyr-Phe-Gln-Asn-Cys-Pro-Arg-Gly·NH2
24-26 abbreviations used in this problem: o
I
H2N f-CH - COOH Rl
II
°
II
PhCH20C - NH - CH COH
I
CH3 R2
CH3
mechanism of fonnation of Z-Ala
: c�� +
PhCH2O-C-CI
°
II
2 R - COH
o.
H N-R i
�
°
¢:::=l
II
PhCH20 - C-NH-R i
---
626
o
:Cl : • •
II
H
1+
PhCH2O - C-N-R i
( �k +
24-26 continued two possible mechanisms of ethyl chlorofonnate activation mechanism 1
R
00,,--/:��
2 R -C -OH
CI -C -OEt
+
�:?j
II 2 R -C -0 -C -OEt +1) H"",\
CI-C -OEt 1 + :O -H
-
00
o
II
2 I R -C=0
0
� :9!:
o
II 2 R -C-0 -C-OEt II
�:?j I+
C)-C-OEt
Cl-
___ -
2 II R -C -OH :0
J 1
II 2 II R - C -0-C -OEt
G0
0
+ II 2 R - C=0 -C-OEt
I
0
H-O :
00
I
H-O :
00-� i
�
oo il
II 2 R -C -0 -C-OEt
:Cl : . . ..
o
0
0
�200 II
1
0
II
+
3 H N -R
0
----
II
0
PhCH20C - NH - CH - C-NH - CH - C-OH
I
CH3
I
I+ 3 2 " R -C -N -R :0 : H
---
CH(CH3h
(I
+
CO2
��
H
627
0
H-O : 0 + 11 II . . 2 R - C-0 -C -OEt
mechanism of the coupling with valine
o
t
II 2 R - C -0 -C -OEt +
plus two other resonance fonns
o
0
-
: Et
1 \���
24-27
o
°
0
0
II
II
II
II
PhCH20C -NH -CH - C -NH -CH - C-NH-CH-C-OH
o
CH(CH3h
0
II
II
I
I
I
CH3
!
CH2Ph
o
II
Cl-C-OEt 0
0
II
II
Z-N H -CH -C -N H -C H-C-N H -CH-C -0 -C-OEt
I
I
CH(CH3h
CH3
o
I
CH2Ph
t
°
0
0
II
II
glycine
H2NCH2COOH
II
II
Z-NH-CH - C -NH-CH -C-NH -CH -C-NH-CH -C-OH
I
I
CH(CH3h
CH3
°
0
II
I
I
CH2Ph
H
j
1
II
H2N-CH-COOH
0
b
leucine
H2CH(CH3h °
II
II
Z -NH -CH-C-NH -CH - C-NH -CH -C-NH CH-C-NH -CH - COOH
I
I
CH3
CH(CH3h
o
0
/I
-
I
CH2Ph
.
II
I
I
H
CH2CH(CH3h
H2, Pd °
0
II
II
H N -CH - C-NH-CH -C-NH -CH-C -NH -CH -C-NH -CH - COOH 2
I
CH3
I
CH(CH3h
I
CH2Ph
I
H
628
I
CH2CH(CH3h
24-28
II
II
°
°
PhCH20C - Cl
+
I
H2N -CH -COOH
-
I
Z - NH-CH -C-OH CH3 - CHCH2CH3
CH3 -CHCH2CH3 isoleucine
II
o
I
I
II
°
Cl-C-OEt
II
°
Z -NH-CH - C-0-C -OEt CH3 - CHCH2CH3
II
o
I
t
glycine
H2NCH2COOH
II
°
I
Z -NH-CH-C - NH-CH -C-OH CH3 - CHCH2CH3
II
0
I
1
H
II
° CI-C-OEt
II 0
I
II
0
Z -NH CH - C-NH-CH - C-0 -C -OEt -
1
CH3 - CHCH2CH3
II
o
I
H
?
H2N - H -COOH
asparagine
CH2CONH2
II
°
I
I
Z-NH-CH - C-NH-CH - C-NH-CH -COOH CH3 - CHCH2CH3
II
o
I
t
H
CH2CONH2
H2' Pd
II
°
I
I
H2N-CH -C-NH-CH -C-NH-CH -COOH CH3 - CHCH2CH3
629
H
CH2CONH2
24-29 In this problem, "Cy" stands for "cyc1ohexyl".
o-
N=C =N
mechanism
-Q
c=:::::::>
Cy-N =C=N -Cy DCC
a
\
II
CH3C-� :
a
II
1
CH3C-O-C
N --Cy
"
Cy-N=C
a
I
II
:N --Cy .... ..f--. .. � -I
+
H3N -Ph
J-Cy
II
N-Cy
CH3C-O-C
,
N-Cy • • _
plus other resonance forms
a
II
a
+
(I
CH3C-NHPh H
� � +
+
II
Cy-N-C - NH-Cy .. •
•
resonance stabilized
II
a
Cy - NH-C - NH - Cy DCU
630
����
24-30
0
0
II
0
II
0
" -0
II
P
Me3COC - NH - CH - C-NH-CH - C-NH-CH - C -0
I
I
CH3
+
CH(CH3h
I
CH2Ph
CF3COOH 0
0
II
+
0
" -0
II
p
H3N - CH - C-NH-CH - C-NH - CH - C -O
I
I
CH3 DCC o
0
II
•I
CH(CH3h
I
CH2Ph
�
Boc-glycine
Me3COC - NHCH2COOH 0
II
0
0
II
II
"
Me3COC-NH - CH - C-NH-CH-C - NH - CH - C - NH - CH - C -0
I
I
I
H
CH3
+
0
0
" -0
0
0
II
II
H3N -CH - C-NH-CH - C-NH . CH - C - NH - CH - C -0
I
I
H
1
I
CH3 DCC
o
0
II
II
Mc,c
ci1
CH(CH3h
I
er
Boc-leuc;nc
CH2CHMe2
" -0
0
0
II
II
Boc NH-CH-C- NH - CH - C - NH-CH - C-NH-CH - C-NH - CH - C -0
I
CH2CHMe2
I
H
o
I
CH3 0
II
+
I
II
tHF
CH(CH3h
0
I
CH2Ph
0
II
0
II
II
H N - CH - C - NH - CH -C - NH - CH - C - NH-CH-C - NH-CH - C -OH 3
I
CH2CHMe2
I
H
I
CH3
I
CH(CH3h
631
P
CH2Ph
-NH . H . COOH
0
P
CH2Ph
CF3COOH
II
+
CH(CH3h
I
-0
I
CH2Ph
P
24-31
a
CH2Cl
-
II
Me3COC - NH-CH - COO
I
+
�
CH2CONH2 Boc-asparagine
a
a
11
II
--- Me3COC-NH-CH-C-O
I
1
CF3CO H
11
+
H3N-CH -C-O a
II
Me3COC - NHCH2COOH a
a
II
II
I
I
I
l
H
a
11
I
CF3COOH a
11
H3N - CH -C - NH-CH - C-O
I
H a
?
II
Boc-isoleucine Me3COC - NH - H - COOH CH3 - CHCH2CH3 a
a
II
II
II
1
I
CH3 - CHCH2CH3
I
H
a
--0 P
DCC
a
11
I
--0
CH2CONH2
.HF
a
II
+
P
CH2CONH2
Me3COC -NH-CH - C-NH CH -C - NH-CH-C-a -
--0
CH2CONH2
II
I
P
DCC
a +
--0
CH2CONH2
Me3COC-NH-CH - C-NH-CH -C-O
a
P
CH2CONH2
o Boc-glycine
:
--0
a
II
II
H3N-CH -C-NH-CH -C - NH-CH-C-OH
I
CH3 - CHCH2CH3
I
H
632
I
CH2CONH2
P
24-32 Please refer to solution 1-20, page 12 of this Solutions Manual.
24-34 :0 :
:0 :
(a)
:0 :
:0 :
II
0
(c)
CO2
+
N
(d)
?
CH3C -NH- H - COOH
I
� CHO
+
C'x, '
I
N
(CH2)4NHCOCH3
+
H
H
the enzyme does not recognize D amino acids
I
(f) H2N-CH - COOH
H2N-CH - CN I H3C-CHCH2CH3
,
N
I
I
O==C
H L-proline
(e)
y:)
HOOC
COOH
N-acetylD-proline
o
II
Br-CH -C-Br
(g)
H3C-CHCH2CH3
CH3
I
CH2CH(CH3h with a water workup, the acid bromide would become COOH
isoleucine o
I
H2N-CH - COOH
(h)
OR
I
II
H2N-CH - C -NH2
CH2CH(CH3h
CH2CH(CH3h
another possible answer, depending on whether the acid bromide is hydrolyzed before adding ammonia
leucine o
24-35
II
(a)
C-COOH
NH3
I
..
I
H2N-CH- COOH
CH(CH3h (b)
H2C-COOH I
H3C-CHCH2CH3
valine
CH(CH3h H20 ---i ... _
PBr3
---
I
Br-CH-COOH H3C-CHCH2CH3 633
excess NH3 ---
I
H2N-CH - COOH H3C-CHCH2CH3 isoleucine
24-35 continued (c)
o
II
NH3, HCN .. H20
C-H
I
H2N -CH -COOH
I
CH2CH(CH3h leucine
CH2CH(CH3h o
o
COOEt
I
.. 2) B�H2Ph
N-CH
I
o
COOEt
COOEt
H H2N - -COOH
?I
1 ) NaOEt
T
N- -CH2Ph
CH2Ph
COOEt
o
phenylalanine
24-36 (a) H2N-CH -COOH
HOCH(CH3h
+
I
CH3 o
II
(b) H2N-CH -COOH
PhC-CI
+
I
CH3
(pyridine)
o
II
.. PhC- NH - CH -COOH
----1�
I
o
(c) H2N-CH -COOH
I
o
II
PhCH20C-CI
+
CH3
II
--
PhCH20C - NH - CH -COOH
I
CH3
CH3 o
(d) H2N-CH -COOH
I
o
0
II
II
--
Me3COC - 0 - COCMe3
+
II
Me3COC - NH - CH-COOH
I
CH3
24-37
CH3
COOH -
HO-C-H -
CH3
COOH TsCI
-
...
pyridine
TsO-C-H
excess NH3 ...
-
COOH -
H-<;:-NH2 -
CH3
CH3 D-alanine
634
o
24-38
COOEt
I
N-CH
o
1) NaOEt
N-
I
COOEt
o
o
?
COOEt -CH,
COOEt
L\!
,-1H N
H30+
l
H2N -CH - COOH
I
racemic
histidine
'
N
24-39
0
II
NH3, HCN
C-H
H2O
I
H2C
to 0
N H
0
II
0
II
I
I
CH2
CHCH3
CH2SCH3
OH
II
H2N-CH - C-NH-CH - C-OH CH2
OH
CH2SCH3 0
racemic
0
tryptophan
0
II
II
II
H2N-CH - C-NH-CH . C-NH - CH - C-OH
II
N H
I
0
(CH2h I HN-CNH2
to
neutral
I
CHCH3
I
I
H2C
0
II
I
H H2N -C-COOH
I
o
I
..
neutral
H2N-CH - C-NH-CH -C-OH
I
(c)
to N H
o
24-40
(b)
L\
I
H2C
(a)
H30+
H H2N -C -CN
..
==.i
NH
I
CH2
I
I
(CH2)4NH2
COOH
NH 635
basic: two basic side chains and one acidic side chain
24-40 continued acidic: carboxylic acid side chain, and the SH is weakly acidic
(d)
24-41 (a), (b), (c)
(
isoleucine �A CH3
�
glutamine
��
\ 0 II II * CH3CH2-CH - C H - NH - C-CH - CH2CH2- C-NH2 I I * NH - CO - CH2NH2 -- N-terminus C-terminus --- CONH2 \(0
____
I
________�
\
'( glycine
Peptide bonds are denoted with asterisks ( * ) . (d) glycylglutamylisoleucinamide; Gly-GIn-He
NH2
0
o
24-42
•
II
II
H2N -CH . C - NH - CH . C - OCH3
Aspartame:
I
CH2COOH
I
CH2Ph
Ic,-_� ,--___) �'-_--.. ,--_�I
Y
aspartic acid (from Edman degradation)
J
'(
phenylalanine
}
methyl ester no free COOH => no reaction with carboxypeptidase
Aspartame is aspartylphenylalanine methyl ester.
24-43
0 0 0 0 o II II II II II H2N -CH . C - N H - CH . C - N H - CH . C - NH - CH . C - NH - CH . C-OH I I I I I CH2Ph CH3 H CH2 CH3
I
CH2SCH3
'-----r--' '-----r--''-----r--' '----r-" '-----r--' phenylalanine
�
alanine
y
glycine
)
methionine
from Edman degradation
alanine
� from carboxypeptidase
636
24-44 (a)
°
protect the N-tenninus of the first amino acid 0
II
PhCH20C - Cl
+
I
H2N -CH -COOH
�
II
I
Z-NH - CH -C-OH
CH2CHMe2
CH2CHMe2
leucine
o
I
II
1
°
II
activate the C-terrninus
CI-C -OEt °
II
Z -NH - CH -C-0-C-OEt
1
CH2CHMe2 add the next amino acid o
Z
-
I
II
?
H2N - H -COOH
alanine
CH3
°
I
II
NH-CH - C-NH - CH -C -OH
!
CH2CHMe2 activate the C-terrninus
II
0
I
CH3
�
Cl-C-OEt
II
0
I
II
0
Z -NH-CH-C-NH - CH -C-0 -C -OEt
j
CH2CHMe2 add the next amino acid o
I
II
CH3
?
H2N - H - COOH
phenylalanine
CH2Ph
°
I
II
I
Z -NH-CH -C-NH-CH - C-NH - CH -COOH CH2CHMe2 deprotect the N-tenninus o
I
II
t
CH3
CH2Ph
H2, Pd °
I
II
I
H2N-CH -C-NH CH - C - NH -CH - COOH CH2CHMe2 637
-
CH3
CH2Ph
r:
24-44 continued (b) °
II
Me3COC - NH - CH - COOI CH2Ph
� U
II
---
V
+
°
0
CH2Cl
1
Boc-phenylalanine attach C-terminus of N-protected amino acid to polymer support
CH2Ph
-0
deprotect CF3COOH N-terminus
1\ -0
o
+
P
H3N -CH - C-O
o
Boc-alanine
T
1\
Me3COC-NH- H - C-0
I
II
Me3COC - NH -CH - COOH
I
CH3 0
0
II
II
I
CH2Ph
. next ammo DCC a dd . aCId and couple 0
" -0 P
Me 3COC - NH- CH-C -NH-CH-C-O
I
CH3
I
CH2Ph
I
o
t
CF3 COOH 0
II
+
deprotect . N-termmus
H3N-CH-C -NH-CH - C-O
I
CH3
o
Boc-leucine
?
II
Me3COC-NH- H - COOH CH 2CHM e2
0
0
\I
0
II
II
1
I
"
CH2Ph
CH2CHMe2
I
°
1\
0
0
II
+
I
CH3
+
P
add next amino DCC acid and couple
Me3COC - NH-CH - C-NH· CH . C-NH . CH . C - 0
I
-0
CH2Ph HF
-0
deprotect and remove from polymer 0
II
II
H3N -CH . C-NH· CH . C-NH . CH . C-OH
I
CH2CHMe2
I
CH3
638
I
P
CH2Ph
24-45
o
o
protect N-terminus
II
PhCH20C - Cl
H2N -CH-COOH
+
I
II
---
Z - NH-CH-C-OH
I
CH3
CH3 alanine o
II
activate the C-tenninus
0
1
II
Z -NH - CH - C-0-C-OEt
I
CH3
o
II
T
H2N - H - COOH
valine add the next amino acid
CH(CH3h
0
II
Z-NH - CH - C -NH-CH-C-OH I I CH3 CH(CH3h
o
II
!
H2• Pd
dcprolect thc N-terminus 0
II
react the N-terminus of the dipeptide at the left with the N-protected, C-activated tripeptide below
H2N -CH - C-NH-CH - C -OH
I
I
CH3
CH(CH3h
o
II
1
o
0
II
0
II
0
II
II
Z -NH-CH - C-NH - CH - C -NH-CH - C -0 -COEt
I
H3C-CHCH2CH3 0
I
CH2CHMe2
0
II
I
CH2Ph
0
II
II
Z - NH -CH-C-NH-CH-C - NH CH-C -NH-CH-C -NH . CH-COOH
I
H3C-CHCH2CH3
o
�
-
I
CH2CHMe2 H2, Pd
I
CH2Ph
I
CH(CH3h
deprotect the N-terminus 0
0
II
II
I
CH3
0
II
II
H2N -CH - C -NH . CH . C -NH . CH - C -NH . CH - C-NH . CH-COOH
I
H3C -CHCH2CH3 lie
I
CH2CHMe2 Leu
I
I
I
CH2Ph
CH3
CH(CH3h
Phe
Ala
Val
639
24-46 (a) There are two possible sources of ammonia in the hydrolysate. The C-terminus could have been present as the amide instead of the carboxyl, or the glutamic acid could have been present as its amide, glutamine. (b) The C-terminus is present as the amide. The N-terminus is present as the lactam (cyclic amide) combining the amino group with the carboxyl group of the glutamic acid side chain. (c) The fact that hydrolysis does not release ammonia implies that the C-terminus is not an amide. Yet, carboxypeptidase treatment gives no reaction, showing that the C-terminus is not a free carboxyl group. Also, treatment with phenyl isothiocyanate gives no reaction, suggesting no free amine at the N-terminus. The most plausible explanation is that the N-terminus has reacted with the C-terminus to produce a cyclic amide, a lactam. (These large rings, called macrocycIes, are often found in nature as hormones or antibiotics.) 24-47 (a) Lipoic acid is a mild oxidizing agent. In the process of oxidizing another reactant, lipoic acid is reduced. COOH S-S
SH
oxidized form
�
(b) O==C
I
I
I
�
H20
0
II
+
�
�R
RC-H
+
� (CH2)4COOH \ /
HC-CH2S
I
�
HN
H2N -CH-COOH
I
CH2
f--.J
SH
SCH2-CH
I
�R
NH
\not basic 640
RC-OH
+
�
NH
SH
II
---l.�
24-48
basic --- N
I
--
o
S-S
(a) histidine:
�
C==O
I
---l.�
NH
S-S
reduced form
O==C
I
HN
SH
�
C==O
HC -CH2SH HSCH2-CH
(c)
COOH
N
SH
SH
(CHZ)4COOH
24-48 continued (b)
t
N -H
�
..
..
I
N -H
�
..
�
l
� -H
Jl
+N :N :N H H H I I I H H H In the protonated imidazole, the two N's are similar in structure, and both NH groups are acidic. (c)
tj
-H
H+
---
tj
-H
-H+
----
(J
+N I I H H resonancestabilized We usually think of protonation-deprotonation reactions occurring in solution where protons can move with solvent molecules. In an enzyme active-site, there is no "solvent", so there must be another mechanism for movement of protons. Often, conformational changes in the protein will move atoms closer or farther. Histidine serves the function of moving a proton toward or away from a particular site by using its different nitrogens in concert as a proton acceptor and a proton donor. N
24-49 The high isoelectric point suggests a strongly basic side chain as in lysine. The N-CH2 bond in the side chain of arginine is likely to have remained intact during the metabolism. (Can you propose a likely mechanism for this reaction?) H2N-CH - COOH I (CH2h I HN C-NH2 I NH arginine
�
H2N -CH - COOH I (CH2h I + NH2 ornithine
641
o
II
H2N--C-NH2 urea
0
24-50
0
0
II
II
II
H2N - CH -C -NH -CH -C -NH -CH - C-OH
(a) glutathione:
,.-__-,,1
gluta c acid (from Edman degradation) (b) reaction: 2 glutathione
+
'-----y---" CH2SH
�
-.. ""....__ .
H
---
J
. ,--_-,,' . \....._----.
gl cine (from carboxypeptidase)
cysteine
H202
glutathione disulfide
II
structure of glutathione disulfide:
I
I
I
HOOCCH2CH2
0
+
2 H20
0
0
II
II
H2N-CH - C-NH -CH -C - NH -CH - C-OH
I
I
HOOCCH2CH2 HOOCCH2CH2
I
I
I<
CH2S
H I
CH2S
II
I
new S-S bond
H
I
II
II
H2N -CH - C -NH -CH -C -NH -CH -C-OH 0
0
0
24-51 end groups: N-tenninus
Ala ------- lie
chymotrypsin fragments B
A
C-tenninus
Glu-Gly-Tyr (middle)
Ala-Lys-Phe c
N-terminus
�-----y
Arg-(Ser? Leu?)-lIe
y
c-tenninu
Ala-Lys-Phe-Glu-Gly-Tyr-Arg-(Ser? Leu?)-Ile
trypsin fragments D
�
Ala-Lys E
Phe-Glu-Gly-Tyr-Arg F
y
l
}
' - - - - -' se r -Leu lIe
Ala-Lys-Phe-Glu-Gly-Tyr-Arg-Ser-Leu-Ile
642
)
I
(a) One important factor in ester reactivity is the ability of the alkoxy group to leave , and the main factor in determining leaving group ability is the stabilization of the anion . An NHS ester is more reactive than an alkyl ester because the anion R 2NO- has an electron-withdrawing group on the 0-, t hereby distributing the negative charge over two atoms instead of just one. A simple alkyl ester has the full negative charg� on the 24-52
oxygen with nowhere to go, RO-.
o (b» )l R OH
+
- F3C
\
o abbre\iate O-Succ
)
� :0
O=C""'" I R tetrahedral intermediate
J
:0:
-SUCC
----
R
reactive anhydride 0 o
t.�
)l CP3
:O-Succ . .
o R
�
)
:0
�
0
)l o CF]
o I NHS ester Succ o tetrahedral intermediate NHS trifluoroacetate is an amazing reagent. Not only does is activate the carboxylic acid through a mixed an hydride to form an ester under mild conditions , but the NHS leaving group is also the nucleophile that forms an ester that is both stable enough to work with, yet easily reactive w hen it needs to be. Perfect! +
(c)
:� Succ H2NR R )l � ./
+
: 0:
R
----
+
)l N "R H
/
+
� HJ
:0:
R 643
-:O - Succ
)l N "R H
+
HO-Succ
24-53 (a)
COOH
H 2N
� L
' R H
,
NaN0 2 + HC I
•
Y
HONO
)
COO H
COOH
..
nitrous acid makes diazonium ions, Sec. 19-17
N
+� N
"
H
R
intermediate 1 L
NaN ) azide as nucleophile, Sec. 19-21B
•
"
R" N) H intermediate 2 D
COO H
H2 -
Pd
R"
"
H
D
(b) The product has the opposite configuration from the starting because of the stereochemistry of the reactions. Diazotization does not break a bond to the chiral center so the L configuration is retained in Reaction 1. It is well documented that azide substitution is an SN2 process that proceeds with inversion of configuration; this is why this process works. The third reaction does not break or form a bond to the chiral center, so the D configuration is retained.
644
NH 2
CHAPTER 25-LIPIDS
0
25-1
II
CH2 - 0 - C - (CH2 )12CH3 trimyristin
25-2
I � � I CH2 - 0 - C - (CH2 h2CH3 CH - 0 - C - (CH2 )12CH3
0
o
all c i s
II
II
CH2 - 0 -C -(CH2hCH == CH(CH2hCH3
I CH I CH2
-
-
� 0- C -(CH2hCH � 0 -C -(CH2hCH
==
CH(CH2hCH3
==
CH(CH2hCH3
triolein, m.p. -40 C (liquid at room temperature) 25-3
(a)
0
II
2+ 2 CH3(CH2 )16 -C -0- Na+ + C a --
0
(b)
II
2 CH3(CH2 )16 -C
(c)
-
0- Na+ + Mg 2+--
0
II
3+ 3 CH3(CH2h6-C-0 - Na+ + Fe
--
excess
H2
Ni
[ [ [
CH2 - 0 - C - (CH2 )16CHl
� I � I CH2 - 0 - C - (CH2 )16CH 3 CH - 0 - C - (CH2 )16CH3
tristearin, m.p. 72° C (solid at room temperature)
a
II
CH3(CH2 )16-C - 0 a
II
CH3 (CH2 )16 - C - 0 a
II
CH3(CH2 )16-C-0
1 1 1
2
Ca
+ 2 Na+
2
Mg
+ 2 Na+
3
Fe
+
3 Na+
25-4
(a) Both sodium carbonate (its old name is "washing soda") and sodium phosphate will increase lhe pH above 6, so that the carboxyl group of the soap molecule will remain ionized, thus preventing precipitation.
(b) In the presence of calcium, magnesium, and ferric ions, the carboxylate group of soap will form precipitates called " h ard-water scum", or as scientists l abel it, "bathtub ring". B oth c arbonate and phosphate ions will form complexes or precipitates with these cations, thereby preventing precipitation of the soap from solution.
645
25-5
S03
abbreviate
-
o- S03-
as
.lVV\l\l\l\lVVV
-03S
S03-
S03-
25-6 In each structure, the hydrophilic portion is circled. The uncircled part is hydrophobic.
N ' 0(' � (�?_���O�O-../'O�O-../'O�O�?-�(��I I
,
� ',
,
h-
,'+
,
"
'
I
-',
I
,
-- ','
,
'
benzalkonium chloride
,
l-
C
.....
..
.........
_
_
---
_
...
__
...
__
...
__
...
__
-
...
_---------- --
....
...
----
- --- --- ------- ---_
....
_--
-
-
--
-
-
---
---_ ....
_ _ ----
...
---
... -
-
............
---�----
Nonoxynol-9
-O�N
,
"
"
' :" Na+ ,
II
'."
I
0
"
...
..
_
.....
.....
, ,
0"
,
, , , ,
Gardol®
_ ...
646
--
electrophile
pentamer
..
�
NaOH
25-8
o
(a)
II
CH 2-0-C-(CH 2)nCH 3
------
I
CH
I
-
�
0 -C - (CH 2)nCH 3
:?:
CH 2 -O - P=O
I
OH (b) CH 2-0
I
-
o "
C - (CH2)nCH 3
�
CH -O-C-(CH 2)"CH]
I
••
...
.
:�:
CH 2 -O - P - O:
I
: .0: .
• •
: .0: .
-
:0 :
25-9 Estradiol i s a phenol and can be ionized with aqueous N aOH. Testosterone does not have any hydrogens acidic enough to react with NaOH. Treatment of a solution of estradiol and testosterone in organic solvent with aqueous base will extract the phenoxide form of estradiol into the aqueous layer, leaving testosterone in the organic layer. Acidification of the aqueous base will precipitate estradiol which can be filtered. Evaporation of the organic solvent will leave testosterone.
647
25- 1 0 Models may help. Abbreviations: "ax" = axial; "eq" ring junctures are axial to one ring and equatorial to another.
=
equatorial . Note that substituents at cis-fused ax to B eq to A C H3
ax
CH3
(a)
(b) ax to A eq to B
eq
HO
C H3
H
ax
H
ax
CH3
(c)
ax OH
o
H H
OH ax
ax
ax
ax HO
25- 1 1
-
---"" - --
H
0
....... . .... - ..... ...
" -'
"
"
, ,
,
,
,
,
.
.
.
,
,
.
.
.
.
, ,
.
,
H:
, ,
OR':
.
,
,
,
:
,
.
geranial
,
'
.
.
OR
menthol
, . . ,
,
. .
,
,
.
,
.
,
.
, ,
.
, , ,
,
.
,
,
.
,
.........
,
,
.
,
.
.. ...
,
.' .
,
..... - ...........
, . .
o
COOR : .
.
, . .
>-':-ii-+--
,
,
.
.
... ,
-
camphor
-- ........ -
,
.
,
. .....
-
,
,
.
,
,
...
,
,
.
,
. . .
.
. , , ,
.
,
. ,
OR .. ,
, ... ......... ",
.
camphor shown in top view
,
648
"
,
, .
, , .
,
, .
o
.
,
, ,
. ,
,
.
- ..... - ..........
,
o
.
.
abietic acid
,
25- 1 2 B -carotene
.. - .. _- ... _ -..... ,
,�
-
,
,
,
,
.
I
•
�
,
,
--- .. _ ..
_
.. --
�--"--"-- ..
.
,
,
..............
25-13 "'''''''' ..... ,
,
,
.
.
.
. . ,
I
.
. ,
,
,
.
..... ... -
- .....
.. � .. ...... , .
'--
-
.
.
,
.
.
, , ,
.. .............
.
. ,
,
.
. -
"- ..
...
........ ...... ..
_ ---'
..... '
limonene monoterpene
a-famesene sesqui terpene
a-pinene monoterpene
(Limonene can also be circled in the other direction around the ring-see menthol in 25- 1 1.)
..
,
"
\..-----Y
,
, '
.
. .
. .
OR
. .
V
'' --.:. - -...- ' --,:...::...:; -
.
, . .
�
-.... - .. -'
,
.
.
--
zingiberene sesquiterpene
25- 1 4 Please refer to problem 1 -20, page 1 2 of this Solutions Manual. 25- 1 5 (a) triglyceride
(b) synthetic detergent
(c) wax
649
(d) sesquiterpene
(e) steroId
CH3(CH2hCH =CH(CH2hCOO-
25- 1 6 3
(a)
soap
(b) CH2
I I
o
-0
-g
0
-
)U(- - -
Br
(c)
H, H
CH3(CH2h
'
0
Br
--
tristeann .
CH3(CH2hCHO
3
o
)l--1<
II
I I CH2 -0
+
0 II C (CH2h
Bf
_
(e)
R I R I CH2-0 -C - (CH2hCHO
CH3(CH2h
(mixture of diastereomers)
(g)
3
'"
"
H H _
�
"
_
3
Bf
(CH2)7CH ]
_
Br
(CH2hCH3
CH3(CH2hCOOH o
+
I R R I CH2 -0-C - (CH2hCOOH
CH - 0-C - (CH2hCOOH
0 II CH2 -0-C - (CH2h'
I H "'" R I CH2 -0-C - (CH2h
(CH2h C -0 - CH II
�
II
CH - 0 -C - (CH2hCHO
o
H
CH2 -0 - C - (CH2hCOOH
CH2 -0-C - (CH2hCHO
H
II
)l--1< "
II
(f)
'"
0 " II CH2-0-C - (CH2h Bf
(CH2);C-O - CH
(mixture 0f diastereomers)
(d)
glycerol
(CH2)16CH3
CH - O-C-(CH2116 CH3 0 II CH2-0-C-(CH2h6CH3 II
HO CH(CH20Hh
Na+ +
CH3(CH2hCH =CH(CH2hCH20H
+
H
,,
"�'
HOCH (CH20Hh glycerol
650
C
H /
(CH2hCH3 (CH2hCH3
H2 (a) CH3(CH2hCH =CH(CH2)7COO H ----:-Nl
LiAIH4 2) H30+
25-17
H2 CH3(CH2hCH = CH(CH2hCOOH � CH3(CH2)16COOH
(b)
(c) CH3(CH2)16COOH from (b)
+
2)
o
Me2S
excess Br2 PBr3
•
II
o
II
R 6-
I H2C-OH
+
o
H2C-0 -C-(CH2)14CH3
I
II
NaOH
+
0I
t1
..
H2C -OH
I
HC -OH
I
H2C -OH glycerol
651
+
k
CH3(CH2h
Na+
(CH2)6
b:COOH
o
0-C - (CH2)16CH3 II
Na+ oII 0 -C - (CH2hCH = CH (CH2hCH3 Na+
glycerol
R HC-0 -C - (CH2)14CH3 I R H2C -0- -OCH2CH2N(CH3h P
•
HC-OH
H2C-0- P- OCH2CH2NH3
(b)
H20
0 =CH(CH2hCOOH
+
HOOC(CH2hCOOH nonanedioic acid B H H Br
I
t1
HC -0-C - (CH2hCH = CH(CH2hCH3
I
CH3(CH2hCH = 0 nonanal
NaOH H2C-OH ..
H2C-0 -C - (CH2)16CH3
I
t1
KMn04 CH3(CH2)7COOH H20, t1-
CH3(CH2hCH =CH(CH2hCOOH
25-18
CH3(CH2)16COOCH2(CH2)16CH3
----
1 ) 03
(e) CH3(CH2hCH = CH(CH2hCOOH
(f)
H+
HOCH2(CH2)16CH3 from (a)
(d) CH3(CH2hCH =CH(CH2hCOOH
(a)
CH3(CH2)16CH20H
1)
2
Na+
o
0- P-OCH2CH2NH2 I Na+ 0 II
o
Na+
0 -C - (CH2)14CH3 II
o
0 - P- OCH2CH2N(CH3h . I Na+ - 0 II
+
3
25-20 Reagents in parts (a), (b), and (d) would react with alkenes. If both samples contained alkenes, these reagents could not distinguish the samples. Saponification (part (c)), however, is a reaction of an ester, so only the vegetable oil would react, not the hydrocarbon oil mi xture. 25-2 1 (a) Add an aqueous solution of calcium ion or magnesium ion. Sodium stearate will produce a precipitate, while the sulfonate w i l l not precipitate. (b) Beeswax , an ester, can be saponified with NaOH. Paraffin wax is a solid mixture of alkanes and will not react. (c) Myristic acid will dissolve (or be emulsified) in di l ute aqueous base . Trimyristin will remain unaffected. (d) Triolein (an unsaturated oil) will decolorize bromine in CCI4, but tri myristi n (a saturated fat) will not.
o
25-22
II
(a)
CH2 - 0 - C - (CH2 )12CH3 asymmetric carbon
*
I I CH2 -
� � - C - (CH2hCH == CH(CH2hCH3
CH - 0 - C - (CH2hCH - CH(CH2hCH3 _
0
(b)
optically acti ve
o II
CH2 - 0 - C - (CH2hCH == CH(CH2 )7CH3
I � I � CH2 - 0 - C - (CH2hCH == CH(CH2hCH3 CH - 0 - C - (CH2 )12CH3
not optically active ; symmetric
0
25-23
II
CH2 - 0 - C - (CH2 )16CH3 asymmetric carbon
*
I I CH2 -
� � - C - (CH2hCH == CH(CH2hCH3
CH - 0 - C - (C H 2 h C H - C H (C H 2 h CH 3 _
0
652
optically active
25-23 continued
o
CH2-0-g - (CH2)16CH3
(a)
I CH-O _ g (CH2)16CH3 0 ICH2-O-g-(CH2)16CH3 0
not optical ly active
_
(b)
'--1"",.","., "'
�
Br
/
H
Br
CH,(CH2)7
0 --' II (CH2);C-O CH
, l l y acti ve optlca (mixture 0 f d'la stereomers
(c) products are not optically acti ve
HOCH(CH20Hh
+
Na+
glycerol
o
(d)
0 II C1 li2 - O-C(CH2)16CH3
.L.J/ I / CH2),CH, ) CH2 0 C _ (CH2),
�
_
-OOC(CH2)16CH3
JCH - -� - (CH2hCHO I CH2- -� - (CH2)7CHO
+
+
C
0
C
0
• • ••
•• •
H
_
II
CH2-0-C-(CH2)16CH3
Br. ...
2
2
Na+
�
Bf
'\
-OOC(CH2hCH CH(CH2hCH3 ==
0 == CH(CH2hCH3
not optically active
(b) , (c) and (f) are mixtures of stereoisomers,
optically active
25-24 The products in
(e)
(c)
(b)
(a)
Br
Br
H
Br Br
Br
Br
(f)
HO 653
+
CH20 0
+
CO2
25-25 Is it any wonder that Olestra cannot be digested !
oleic
0
0
�
L
o
0
#" #
654
l inolenic
(a) COOH
H OH eq NHCH 2COOH
I
(b)
HO "
Like a soap or detergent, this cholic aci d-glycine combination has a very polar " head" and a slightly polar "tail ". The "tai l " can dissolve non-polar molecules and the polar " head" can carry the complex into polar medi a.
",
25-27
........... ..
- ....
... .,. ...
... ... ---- ..
...
(a) sesquiterpene
,
OR \ .
,
.. - .. - ......
,
, , ,
\ .
, , ,
,
"
,
,
,
.
.
.
.......... _
.
---_ ........ -
, , , , ,
\
I
,
\, --
---
:'
-
-
o
(b) monoterpene
-
. .
..... -- ..
, .
"
, \
\
\
\
\ .
,
.
,
.
,
-
, ,
OR
.
, ,
,
.............
. , ,
. . \
.
\
.
\
,
,
,
, ,
--
(c) sesquiterpene
,
,
.. ... __ ...
............
... - ..... - .....
\ , . .
,
.
,
,
,
,
,
OR
, , .
, , , ,
""
--H -
.
--
. , , , ,
,
,
H
. \ , , ,
, , --.
..........
655
25-27 (d) sesquiterpene
_ ....... ,
,
_
·
, .. -
·
.... - ...........
· ·
,
·
,
,
CH3
,
J OR
,
·
,
,
,
.
:
"
.... --- .............
CH3
· ·
,
· ·
,
,
· ,
CH3
--'
.
.
,
: ' CH3 : ... :
.
,
\
,
.
"
,
,
·
CH3
"
........................... '
I
.............. -
25-28 The fonnula ClsH340 2 has two elements of unsaturation; one is the carbony l , so the other must be an alkene or a ring. Catalytic hydrogenation gi ves stearic acid, so the c arbon cannot include a ring; it must contain an alkene. The products from KMn04 oxidation detennine the location of the alkene: HOOC(CH 2)4COOH
+
HOOC(CH 2)IOCH3 :::::> HOOC(CH2)4CH
==
CH(CH 2)IOCH3
If the alkene were trans, the coupling constant for the vinyl protons would be about 1 5 Hz; a 10 Hz coupling constant indicates a cis alkene. The 7 Hz coupling is from the vinyl H's to the neighbori ng CH2·s.
25-29 COOH
linolenic acid
COOH
linoleic acid-known
COOH NEW COOH NEW
COOH NEW COOH NEW COOH
oleic acid-known
In addition to the new isomers with different positions of the double bonds , there has been growing concern over the trans fatty acids created by isomerizing the naturally occuring cis double bonds. More manufacturers are now listing the percent of "trans fat" on their food labels. 25-30 The cetyl glycoside would be a good emulsifying agent. It has a polar end (glucose) and a non polar end (the hydrocarbon chain), so it can dissolve non-polar molecules, then carry them through pol ar medi a in micelles. This is found in Nature where non-polar molecules such as steroid honnones, antibiotics, and other physiologically-active compounds are c arried through the bloodstream (aqueous) by attaching saccharides (usually mono, di , or tri), making the non-polar group water soluble. 656
25-3 1 (a) Four of these components in Vicks Vapo-Rub® are terpenes . The only one that is not is decane; it is not comprised of i soprene units despite having the correct number of carbons for a monterpene. Three of the four terpenes have had their isoprene units indicated in previous problems; in each of those three, there are two possible w ay s to assign the isoprene units , so those pictures will not be duplicated here. a-pinene;
cineole; eucal yptol
see 25- 1 3
. .
.
.
(
.
.
.
menthol; see 25- 1 1 .
camphor; see 25- 1 1
. .
o
OH
. .....
..
.. __ ........
camphor
decane (b) Vicks Vapo-Rub® must be optically active as it contains four optically active terpenes. ........
25-32
"
, .............. _-
, ·
(a) Of the two, only nepetalactone is a terpene. The other has only 8 c arbons and terpenes must be in multiples of 5 c arbons.
· · ·
"
.
CI:?0 . .
(b) Of the two, only the second is aromatic as can be readily seen in one of the resonance forms.
.
,
o
......... --_ ....
"
::::-...
o
'1 .. ,
,
-
.
,,
.
0
_ _. ... __ __
:-.
o+�o °
::::-...
I
00/ //
o
(c) Each compound i s c leaved with NaOH (aq) to give an enolate that tautomerizes to the more stable keto form.
NaOH -
NaOH
o
a(cooo
657
0: .0
25-33 (a) High temperature and the diradical O 2 molecule strongly suggest a radical mechanism.
,---H
(b) Radical stability follows the order: benzylic> allylic doubly allylic making it a prime site for radical reaction. (c)
R R
•• ·
••
-
·
�
•
initiation
>
3°> 2° > 1°. A radical at C-ll would be
}
COOH
Ll
j'\ .. . . H
�C; propagation step 1
..J
•
R-R
COOH
.. .... . f--l��
plus two allylic resonance forms
.
�� } H
•• • .
0
I
�
•• -
0
••
·
T
H
.. ..
_
_
!
_
COOH
COOH
propagation step 2
:O-OH
��
�
t
I
COOH
••
T-""'=�
COOH
H
recycles to begin propagation step J
foul-smelling aldehydes, ketones, and c arboxylic acids = " rancidity" (d) Antioxidants are molecules that stop the free radical chain mechanism. In each of the two c ases here, abstraction of the phenolic H makes an oxygen radical that is highly stabi lized by resonance, so stable that it does not continue the free radical chain process. It only takes a small amount of antioxidant to prevent the chain mechanism. Interestingly, in breakfast cereals, the antioxidant is usually put in the pl astic bag that the cereal is packaged in , rather than in the food itself. BRA and BHT can also be used directly in food as there is no evidence that they are h armful to humans. :0 : :0 : •
(H3C h C
�
Y
OCH 3
658
B RA radical
•
(CH3hC
�c(CH3h
Y CH 3
BHT radical
CHAPTER 26-SYNTHETIC POLYMERS
Note: In this chapter, the " wavy bond" symbol means the continuation of a polymer chain . .JVVVVVVV"I..
H
I
H
I
H
I
H
I
IVVVV'C -C - C - C·
I
H
I
Ph
I
Ph
I
H
10 radical, and not resonance-stabilized
this orientation is not observed
Orientation of addition always generates the more stable intermediate; the energy difference between a 10 radical (shown above) and a benzylic radical is huge. Moreover, thi s energy difference accumulates with each repetition of the propagation step. The phenyl substituents must necessarily be on alternating carbons because the orientation of attack is always the same-not a random process. 26-2 PhCOO - OOCPh
Ph·
+
2 Ph·
----I .. �
2 CO 2
,� T!)� \H +
CH3 H / \ C=C
H
I
CH3H
I
I
I
I
CH3H
I
I
I
I
CH3
I
Ph-C-C - C-C - C - C.
I
H
H
H
H
H
659
I
H
etc.
26-3 The benzylic hydrogen will be abstracted in preference to a 2° hydrogen because the benzylic radical is both 3° and resonance-stabilized, and the 2° radical is neither. benzylic H I'V'VVV'
\)H1
H
1
H
1 (I 1
1
1
Ph H
1
Ph H
.JVVVV
(k (
C-C-C-C ,JVV\J\A
1
Ph
I
H
new benzylic radical
H I'V'VVV'
I
1
I
I
1
H
+
H
I
I
H
I
I
H-C-C-C-CJVVVV\
1
Ph
1
1
H
1
H
Ph
tenninated chain
Ph
Ph
Ph H
1
1
I
I
C-C-C- C ,JVV\J\A
I
Ph H
1
H H \ rU c==c / \ Ph H
1
1
H
H
Ph H
H
1
H
1
growing polystyrene chain
middle of a polystyrene chain
H
H
1
oC-C-C-CJVVVV\ I I 1 1 Ph H Ph H
+
c-C-C-C ,JVV\J\A
H
1
H2C,
>
H
?
Ph
Ph
/H
Ph
Ph
o
Ph
26-4 Addition occurs with the orientation giving the more stable intennediate. In the case of isobutylene, the growing chain will bond at the less substituted carbon to generate the more highl y substitited carbocation. H I'V'VVV'
I
Me
1+
+
C-C
I
H
1
Me
H Me / \ C==C � .. ----I \ / Me H
H .JVVVV
1
Me H
I
1
I
1
1+
C-C-C-C
1
H
Me H
1
Me
3° carbocation-favored 26-5
H1
H
1
1
(a) chlorine can stabilize a c arbocation intennediate by resonance
.JVVVV
1+
C- C
H :Cl:
..
..
H
1
H
I
IVVV\I'C-c
1
II+
H :Cl
660
H
Me OR
I
Me Me H
1
I
1+
1
1
1
IVVV\I'C-C-C-C
1
H
Me Me H
1 ° c arbocation---disfavored (also more steric hindrance)
26-5 continued (b) CH3 c an stabilize the c arbocation intermediate by induction H
H
1
1+
2° (not the best case imaginable, but still possible)
C -C
IVVVV"
1
1
H
CH3
(c) terrible for cationic polymerization : both substituents are electron-withdrawing and would destabilize the carbocation intermediate H CaaCH3
1
1+
1
1
H
26-6
C
N
benzylic H
IVVVV"
destabilized carbocation
C-C
.JVVVV
1
\.
H
H
1
1
H
1
C -C - C - C.JVVVV\ 1 1 1 1 Ph H Ph H
+
+
H
1
H
1
hydride transfer ---
H
1
1
C-C-C-C.JVVVV\
1
1
Ph
1
H
1
Ph
1
H
H
1
1
H
1
H-C - C-C-C.JVVVV\
1
Ph
H
growing polystyrene chain
middle of a polystyrene chain
H
1
1
H
Ph
1
H
terminated chain +
H IVVVV"
1
Ph
Ph
1
1
1
I
1
H
1
1
..
.JVVVV
H H I \ C==C I \ Ph H
H
H 2C, / H C
/
Ph H I 1 C - C - C -C.JVVVV\
H
H
C -C-C -C.JVVVV\ Ph
Ph
Ph
H
1+
1
1
Ph
H
1
1 +
H
1
H
new henzylic cation
Ph
Ph
Ph
Polystyrene is particularly susceptible to branching because the 3° benzylic c ation produced by a hydride transfer is so stable. In poly(isobutylene), there is no hydrogen on the carbon with the stabi lizing substituents; any hydride transfer would generate a 2° c arbocation at the expense of a 3° c arbocation at the end of a growing chain-this is an increase in energy and therefore unfavorable. 2° H JVVVV
1
"
Me H
1
1
1
C - C - C -C.JVVVV\
1
H
1
1
Me H
Me H
Me
1
+
Me
middle of a poly(isobuty lene) chain
+1
30
1
Me H
1
1
C -C -C -C.JVVVV\
1
1
Me H
1
1
no hydride transfer
Me H
growing poly(isobutylene) chain
661
26-7
. .
:0:
:0 :
II
H I\IVVV'
C-OCH3
I
_I
C -C : I I H H
� .. If-.... -l.. .
JVVVV
C-OCH3
I
II
C-C
I
I
H
0
26-8
I
H
II
H
H COCH3 \ nl ---C=C HO: '---I-" \ H C N
I
I
H
:0 :
:0 : II C-OCH3
_I
HO-C-C:
• •
H
I
C
I
H
�---l"'�
I
HO-C -C
I
N:
H
C-OCH3
II
H
� .... . -l... ..!--
I
I
:0 : II C-OCH,
I
HO-C-C
I
C -N:
H
o
II
C=N: . .
II
COCH3 \ nl C=C I \ H C_N . .-
:0:
:0 :
H
I
COOMe I I H C-OCH3
I
I
_I
HO-C-C-C-C: I I I I H CN H C
_
•
H ...
I
COOMe I H C-OCH3
I
I
"
HO-C-C-C-C
!
N:
I
H
I
I
CN H
I
C
H
I
: 0: COOMe " H C-OCH3
I
I
I
I
II
�HO-C-C-C-C
.... .!-. -l...
N:
I
H
I
CN H
C = N:
etc.
COOMe COOMe COOMe COOMe COOMe
CN
CN
CN
CN
CN
This polymerization goes so quickly because the anionic intermediate is highl y resonance stabilized by the c arbonyl and the cyano groups . A stable intermediate suggests a low activation energy which translates to a fast reaction.
662
26-9 H
(a)
I
rvvVV'
H
H
I (I
H
H
I
- I
C - C -C - C .JVVVV'\
I
I
CN H
I
..JVVVV
I
H
I
I
I
CN H
I
C - C - C - C .JVVVV'\
I
I
I
H
I
I
I
I
CN H
CN H
H
H
+
I
H
I
I
H
I
H-C-C-C-C .JVVVV'\
I
H
CN H
I
growing poly(acrylonitri le) chain
!
CN H
_I
H
I
:C - C -C-C .JVVVV'\
middle of a poly(acrylonitrile) chain
H
H
I
I
CN H
I
CN H
terminated chain H H \ nl C=C I \ H CN H
I
rvvVV'
H
I
CN
CN
CN H
I
I
>
C - C - C - C .JVVVV'\ I I I H CN H
I
H2C
CN
CN
CN
CN
"" H , / C
1-
CN (b) The chain-branching hydride transfer (from a cationic mechanism) or proton transfer (from an aniontc mechanism) ends a less-highly-substituted end of a chain and generates an intermediate on l more-highly s ubstituted middle of a chain (a 3° carbon in these mechanisms). This stabilizes a carbocation, but greater substitution destabilizes a carbanion. B ranching can and does happen in anionic mechanisms, but it is less like ly than in cationic mechanisms.
26- 1 0
isotactic poly(acrylonitrile)
NC
H
NC
H NC
syndiotactic polystyrene
663
H NC
H
26- 1 1
(a)
all trans
(b) The trans double bonds in gutta-percha allow for more ordered packing of the chains, that is, a higher degree of crystallinity. (Recall how cis double bonds in fats and oils lower the melting points because the cis orientation disrupts the ordering of the packing of the chains.) The more crystalline a polymer is, the less elastic it is. 26-12 Whether the alkene i s cis or trans is not specified.
I
Me
I
H
H
Me
H
H
I
I
I
I
I
Me
I
I
H
H
I
� C -- C--C -- C==C -- C �
\
Y isobutylene
I
H
\.
Y
)
isoprene
26-13 The repeating unit in each polymer is boxed.
(a) Nomex®
o
IVV'g
0
0
0
--o- --o- --o- --o- �.NV' � �
g- �
H
� �
�H -g
�
�
g - �H
26- 1 4 Kodel® polyester (only one repeating unit shown)
.NV' 0 - CH 2
-0
� !J
�-o-�
CH 2 - 0 - C
664
C.NV'
� �
H
26-15
o II O-C I
�
\\ -0'/
CHOJVV\ I CH20 I
'\
_
o 11 H2C -O -C
0 II I c-o -rn 0 I 11 H 2C-O -C
-0-0\\ \.
!J
� h
CH,o-"'" I 0II CHO I """" C-O -CH2
0 C H,oII C-O -CH I CH20 �
,
Glycerol is a trifunctional molecule, so not only does it grow in two directions to make a chain, it grows in three directions. All of i ts chains are cross-linked, forming a three-di mensional l attice with very little motion possible. The more c ross-linked the polymer is, the more rigid it is. 26-16 For simplicity in this problem, bisphenol A will be abbreviated as a substituted phenol.
HO
-o-�
b� I � OH
--
CH3
represented as
__
bisphenol A
aC :
mechanism
..
C I > 'C I R .. �OH � � ·
• •
--
:?) 0C
Cl 1'CI o+, H
..
R
-oI� :R: \.OH
-C
l--
R
HCI(g)
�D
H /C, ·CI· /C.'CI CI O �9+ ,
•
•
•
:
•
R
R
H
:0: 1
R
10
o II C O/ 'O
I
O U �
R
-HCI
-
.---CI-
(?¢CI C o/ ' o·
R 665
+1 u......; H
as above
+
I
�
R
R
26-17 B isphenol A i s made by condensing two molecules of phenol with one molecule of acetone, with loss of a molecule of w ater. This is an e\ectrophi lic aromatic subsitution (more specifically, a Friedel-Crafts alkylation), and would require an acid catalyst to generate the carbocation. While a Lewis acid could be used, the mechanism below shows a protic acid.
from acetone
OH
HO
from phenol
from phenol
...
...
B 0O
/
_
---0-
H CH3 � // C� I
HO
'/
-
C-OH I
CH3
plus three resonance forms
plus four resonance forms
666
Q R �
26-18
Ph-N = C
i -+k cf. :0 : .. I Ph-N = C-O-Et �
�
-Et
-
:0 :
II
Ph -N - C - O -Et .. .. f H -O - Et
• •
• •
+ 1)
;i?
H-O-Et
H
I
• •
}
• •
:0 :
II Ph-N-C-O-Et
two rapid proton transfers
H 26-19 Glycerol is a trifunctional alcohol . It uses two of its OH groups in a growing chain. The third group cross-links with another chain. The more cross-linked a polymer, the more rigid it is. 26-20
urethan linkage
I
-O-I� � /; � I r
0
Me
.1\1\1'0
OH
I
T
I
O-C-N
Me
\...
,
H
H
�
) \.
V
0
� N-CI I I
Me
)
y
bisphenol A
.N\r
toluene diisocyanate
26-21 Please refer to solution 1-20, page 12 of this Solutions Manual. 26-22
H
(a) .rvvvv
I
Me
I
I
H
Me
I
H I
Me
I
C-C-C-C-C-C .I\I\I' I
H
I
Me
I
I
H
Me
I
H
I
Me
(b) Polyisobutylene is an addi tion polymer. No small molecule is lost, so this c annot be a condensation polymer. (c) Either cationic poly merization or free-radical polymerization would be appropriate. The carhocation or free-radical intermediate would be 3° and therefore rel atively stable. Anionic polymerization would be inappropriate as there is no electron-withdrawing group to stabilize the anion . 26-23
(a) It is
a polyurethane.
(b) As with all polyurethanes, it is a condensation polymer.
o
(c)
II
.N\r CH2CH2CH2 - N-C- 0 .1\1\1'
I
HOCH2CH2CH2NH2
H 667
+
CO2
26-24
(a) It is a polyester. (b) As with all polyesters, it is a condensation polymer. (c)
Using the dicarboxylic acid instead of the ester would produce water as the small neutral molecule lost in this condensation. 26-25
(a) Ury lon® is a polyurea. (b) A pol yurea is a condensation polymer. (c) .JVV'-
(CH2 )
9
-N I
° -
II C
H
-
N I
JVV'
H20
�
H
26-26
(a) Pol yethylene glycol, abbreviated PEG, is a polyether. (b) PEG is usually made from ethylene oxide (first reaction shown). In theory, PEG could also be made by intermolecular dehydration of ethylene glycol (second reaction shown), but the yields are low and the chains are short. °
n
U
+
HO-
ethylene oxide HO
",OH � "'" �
ethylene glycol
(c) B asic c atalysts are most l ikely as they open the epoxide to generate a new nuc\eophile. Acid catalysts are possible but they risk dehydration and ether cleavage.
668
26-26 continued (d) Mechanism of ethylene oxide polymerization (showing hydroxide as the base):
-0 0
/"""-.... .., 0 .... /"""-.... ' � � 0
HO'
.., 0 .... ... /"""-.... /"""-.... 0 0 /""'0.... .., 0 .... ' '/"""-.� '0 : � � 'v/ 0
f))"
etc.
HO'
:
26-27 (a) Polychloroprene (Neoprene®) is an addition polymer. (b) Polychloroprene comes from the diene, chloroprene, j ust as natural rubber comes from isoprene: H
\
Cl
I
C -C
II
H2C
chloroprene
\\
CH2
26-28 (a)
H
H
1
1
H
H
1
1
oJV\r C- 0 -C - 0 -C - O-C - OoJV\r
1
1
H
H
1
Delrin® (polyformaldehyde)
1
H
H
(b) All of these intermediates are resonance-stabilized.
'---/
H+
H
/
O=C---
\
H
H
•• I ,
H
1+
H-O-C",---, H
1
00
H
H
1
/
H
O=C
\ H
�
1+
H-O- C-O-C
•
1
H
'\0
H
/
O=C 0 0
etc. 4
H
I
\ H
H
I
H 1+
I H
I H
H-O-C-O-C-O-C
I H
trimer
(c) Delrin® is an addition polymer; i nstead of adding across the double bond of an alkene, addition occurs across the double bond of a c arbonyl group. 669
(a)
26-29 cis
trans
(b) Each structure has a f ully conjugated chain. It is reasonable to expect electrons to be able to be transferred through the system, just as resonance effects can work over long distances through conjugated systems. (c) It is not surprising that the conductivity is directional. Electrons must flow along the 'It system of the chain, so if the chains were aligned, conductivity would be greater in the direction parallel to the polymer chains. (It is possible, though less likely, that electrons could pass from the 'It system of one chain to the 'It system of another, that is, perpendicular to the direction of the chain; we would expect reduced conductivity in that direction.) 1t
(a) A Nylon is a polyamide. Amides can be hydrolyzed in aqueous acid, cleaving the polymer chain in the process. 0 0 0 0 H30+ II II II II + .JVV' N -C C - N .JVV' .JVV' NH3 + HO -C .JVV'C - OH + H3N .JVV' I I H H (b) A polyester can be saponified in aqueous base, cleaving the polymer chain in the process . o 0 0 0 II II II II NaOH .JVV'O -C'VVV'C-O.JVV' .. .JVV' OH + 0-C.JVV' C - 0 + HO.JV\r 26-31 0 0 0 II II II (a) CCH3 CCH3 CCH3 I I I H 0 H 0 H 0 H OH H OH H OH H2O I I I I I I I I I I I I .JV\rC -C -C-C -C-C.JV\r .JVV'C -C-C -C-C-C.JV\r H+ or I I I I I I I I I I I I H H H H H H H H H H H H HOpoly(vinyl acetate) poly(vinyl alcohol) (b) A polyester is a condensation polymer in which monomer units are linked through ester groups as part of the polymer chain . Poly(vinyl acetate) is rea lly a substituted polyethylene, an addition polymer, with only carbons in the chain ; the ester groups are in the side chains, not in the polymer backbone . 26-30
+
JV'V\.
---I'-�
_
.-
(c) Hydrolysis of the esters in poly(vinyJ acetate) does not affect the chain because the ester groups do not occur in the chain as they do in Dacron® . (d) Vinyl a lcohol cannot be polymerized because it is unstable, tautomerizing to acetaldehyde. o OH I II H 2C==C H CH3-CH --
670
(a)
OCOC H3
26-32
�
cellulose acetate
o I
COC H3
(b) Cellulose has three O H groups per gl ucose monomer, which form hydrogen bonds with other polar groups. Transforming these O H groups into acetates makes the polymer much less polar and therefore more soluble in organic solvents . (c) The acetone dissolved t he cellulose acetate in the fibers . As the acetone evaporated, t he cellulose acetate remained but no longer had the fibrous, woven structure of cloth. It recrystallized as white fluff . (d) Any article of clothing made from synthetic fibers is susceptible to the ravages of organic solvents. Solvent splashes leave dimples or blotches on Corfam shoes . (Yet, Corfam shoes co uld still provide protection for the toenail polish!) 26-33 OH OH OH OH Bakelite is highly cross-linked through the ortho and para positions of phenol; each phenol can form a chain at two ring positions, then form a branch at the third position.
mechanism
..
H°
l-w
H
plus four resonance forms
-o- i -oH
°H
further coupling at ortho positions leads to cross-linked Bakelite 671
· .0.
26-34 ·
H2N
II
....... C, . . .... � \..
N
H
I
H.J
..NH3..
H2N
7,)
....... c, .. -
N
..
..
H2N
6
..
N:...J I
-HO..
I
H-C-H
H2N
....... C,
eN:
this leads to cross-linking
/""" w
....... C,.........
HN
H2N""'"
I
)
N
A
H2N
...... � 0
II
....C ... ,
C
g
+
N I
H2N
II
.......C,
N
••
H
NH2
C:--,
"0
672
�
.... H
-
N
••
�
a
N I
H
II
....... C,
NH2
plus another resonance form
�
H""'" 'H
II
�
:0:
N
II
....C ... , .. ....
....... C ,
H2N
+
gA-"
:0:
....... C,..
H
:0:
H""'" 'H
II
l
....
from above
II
I
N
+
H2N
:0 :
'N:
N
H-?-H U" � :o: ..
H-C-H
OH
H2N
� cg
H3N-H
OH
....... C,
II
....... C�
:0:
I
C H
o
H2N
..
'-__--_
:0 :
H2N
H
H' /H c
H-C-H
I
....... C,.:....
� H3N: ....... C,...... H
:0:
II
i
H2N
• •
II
;}
..O .
· · .0 .
63 2
� � � � � � -y O
O
O
glycolic acid
lactic acid
o
O
O
glycolic acid
lactic acid
0
O
�
0
glycolic acid
lactic acid
'-----T---" '-----T---" '-----T---" '----r-' '-----T---" '-----T---"
26-36
�
OH
OH OH
OH OH
OH
cellulose cotton
O�
=
polypropylene As we have seen repeatedly through this presentation of organic chemistry, physical and chemical behavior depend on structure. The structure of cotton, i.e. cellulose, has multiple oxygen atoms that form hydrogen bonds with water. When cotton gets wet, it holds onto the water tightly, as you have seen if you have put cotton clothes in a clothes dryer-it takes a long time to dry. Polypropylene is a hydrocarbon with no hydrogen bonding groups; the fiber feels dry because it cannot hold the water the way cotton can. Athletic garments are increasingly using polypropylene because they allow evaporation and cooling during periods of exertion ; cotton is just the opposite .
26-37 (a) This addition polymer is called polyvinylidene chloride, trade name Saran®. It could be made by any of the three mechanism types : radical, cationic, or anionic .
(b) When the substituent is on every fourth carbon, and one double bond in the chain in every 4-carbon unit, the polymer must come from addition across a diene, probably under cationic conditions.
(c) This polyester is a condensation polymer of two monomers a diol and a derivative of phthalic acid, either the anhydride, an O ester, the acid chloride, or the acid itself. Heating the monomers will make the polymer; no catalyst is required if done at high temperature. HO
---<>-monomer
OH
673
Cl
monomer
>=== Cl
H3CO
h 0
one of these is the other monomer
monomer 0
X
X
X
= OH orCl or OR
0
26-3 7 continued (d) This polyamide (Nylon) is a condensation polymer made from two monomers, a diamine and a derivative of succinic acid, either the anhydride, an ester, the acid chloride, or the acid itself. Heating the monomers will make the polymer; no catalyst is required if done at high temperature. H2N� NH2 monomer
(a)
26-3 8
COOH
===
o
�
o
one of these is the other monomer
X=OH or Cl or OR
o
HO .... /"".... ........, ' OH
}- OCH2CH20H hydroxyethyl methacrylate ====CH3 =\
j
polymerize using radical or anionic conditions
polymer
(b) This polymer has a few properties that make it useful as the material in soft, extended-wear contact lenses. First, carboxylic acids usually are crystalline solids with high melting points, but esters and alcohols are low melting, often liquids, so the polymer with this ester is softer than the carboxylic acid or even the methyl ester. (The methyl ester, polymethyl methacrylate or Plexiglas, was the first material used in the original hard contact lenses.) Second, the ability of the free OH to form hydrogen bonds with water makes the contact lens more fluid and less irritating to the cornea. Third, a hidden advantage but very important for ocular health: the fluidity of the contact lens also permits oxygen to go through the lens. Because the cornea does not have a large blood flow, it needs to absorb oxygen from the air to maintain its health, and this enhanced gas permeability permits the contact lens to be worn for days at a time without compromising the health of the cornea. Thanks, polymers!
Note to the student: BO N VOYAGE! I hope you have enjoyed your travels through organic chemistry. Jan Simek
674
Appendix I-Summary of IUPAC Nomenclature ofOq:anic Compounds
The purpose of the IUPAC system of nomenclature is to establish an international standard of naming compounds to facilitate communication. The goal of the system is to give each structure a unique and unambiguous name, and to correlate each name with a unique and unambiguous structure. Introduction
IUPAC nomenclature is based on naming a molecule's longest chain of carbons connected by single bonds, whether in a continuous chain or in a ring. All deviations, either multiple bonds or atoms other than carbon and hydrogen, are indicated by prefixes or suffixes according to a specific set of priorities. I. Fundamental Principle
II. Alkanes and Cycloalkanes (also called "aliphatic" compounds)
Alkanes are the family of saturated hydrocarbons, that is, molecules containing carbon and hydrogen connected by single bonds only. These molecules can be in continuous chains (called linear or acyclic), or in rings (called cyclic or alicyclic). The names of alkanes and cycloalkanes are the root names of organic compounds. Beginning with the five-carbon alkane, the number of carbons in the chain is indicated by the Greek or Latin prefix. Rings are designated by the prefix "cyclo". (In the geometrical symbols for rings, each apex represents a carbon with the number of hydrogens required to fill its valence.) CI C2 C3 C4 Cs C6 C7 Cs C9 C IO Cll
CH4 CH3CH 3 CH3CH2CH3 CH3[CH2hCH3 CH3(CH2hCH3 CH3(CH2]4CH3 CH3(CH2ls CH3 CH3[CH2]6CH3 CH3[CH2hCH3 CH3[CH2lsCH3 CH3[CH2]9CH 3
methane ethane propane butane pentane hexane heptane octane nonane decane undecane
C12 C13 C I4 C2 0 C21 C22 C23 C3 0 C3 1 C40 C so
H
D
cyclopropane
0
cyclohexane
>
H
,
C
,
/ \
H
'C-C'" I
H
CH3(CH2]IOCH3 CH3(CH2]IICH3 CH3(CH2]12CH3 CH3(CH2]ISCH3 CH3[CH2]19CH 3 CH3(CH2hoCH3 CH3[CH2hICH 3 CH3[CH2hsCH3 CH3(CH2h9CH3 CH3(CH2hsCH3 CH3[CH2]4SCH3
\
H
H
D
dodecane tridecane tetradecane icosane henicosane docosane tricosane triacontane hentriacontane tetracontane pentacontane
0
cyclobutane
cyclopentane
0
0
cycloheptane
cyclooctane
The IUPAC system of nomenclature is undergoing many changes, most notably in the placement of position numbers. The new system places the position number close to the functional group designation; however, you should be able to use and recognize names in either the old or the new style. Ask your instructor which system to use. 675
Appendix 1, Summary of IUPAC Nomenclature, continued III. Nomenclature of Molecules Containina: Substituents and Functional Groups
A. Priorities of Substituents and Functional Groups LISTED HERE FROM HIGHEST TO LOWEST PRIORITY, except that the substituents within Group C have equivalent priority . Prefix
Structure 0 II R - C-OH 0 \I R - C-H 0 \I R - C-R
Functional Group
Suffix
Group A-Functional Groups Named By Prefix Or Suffix
Carboxylic Acid
Aldehyde
Ketone
carboxyoxo(formyl) oxo-
hydroxy-
R-O - H
Alcohol
Amine
R-N
Functional Group
Structure
amino-
/
"
-oic acid (-carboxylic acid) -al (carbaldehyde) -one -01
-amine
Group B-Functional Groups Named By Suffix Only
-ene
/ \ C=C \ / -C:::C-
Alkene
Alkyne
Substituent Alkyl (see next page)
-yne
Structure
Group C--Substituent Groups Named By Prefix Only
alkylR -0 Alkoxy alkoxy(alkoxy groups take the name of the alkyl group, like methyl or ethyl, drop the "yl", and add "oxy"; CH 30 is methoxy; CH3CH 20 is ethoxy) Halogen F fluoro CI chloro Br bromo 1iodoR-
Miscellaneous substituents and their prefixes -N0 2 nitro
-
CH=CH2 vinyl
-CH 2CH=CH2 allyl
676
<> phenyl
Appendix 1, Summary of IUPAC Nomenclature, continued Common alkyl groups-replace "ane" ending of alkane name with "yl". A lternate names for complex substituents are given in brackets.
-CH3 methyl
CH3 -CH CH3 I
\
isopropyl [ 1 -methylethyl]
-CH2CH3 ethyl -CH2CH2CH3 propy l (n-propyl)
\
sec-butyl [ I-methylpropyl]
CH3 -C-CH CH3 I
isobutyl [2-methylpropyl]
-CH2CH2CH2CH3 butyl ( -butyl) n
B.
CH3 -CH CH2CH3 I
I
3
t-butyl or tert-butyl [ 1,I-dimethylethyl]
Naming Substit uted Alkanes and Cycloalkanes-Group C Substituents Only
Organic compounds containing substituents from Group are named following this sequence of steps, as indicated on the examples below :
C
·Step l. Find the longest continuous carbon chain . Determine the root name for this parent chain. In cyclic compounds, the ring is usually considered the parent chain, unless it is attached to a longer chain of carbons ; indicate a ring with the prefix "cyclo" before the root name. (When there are two longest chains of equal length, use the chain with the greater number of substituents .) ·Step 2. Number the chain in the direction such that the position number of the first substituent is the smaller number. If the first substituents have the same num ber, then number so that the second substituent has the smaller number, etc. ·Step 3. Determine the name and position number of each substituent. (A substituent on a nitrogen is designated with an "N" instead of a number ; see Section 111.0. 1 . below .) ·Step 4. Indicate the n umber of identical groups by the prefixes di, tri, tetra, etc. ·Step 5. Place the position numbers and names of the substituent groups, in alphabetical order, before the root name. In alphabetizing, ignore prefixes like sec -, tert-, di, tri, etc., b ut include iso and cyclo. Always include a position number for each substit uent, regardless of redundancies.
CH3 CH2CH2CH3 CH2CH2CH3 1 1 1 1 I CH3 -CH -CH 4C -sCH CH2CH3 CH H H T T CH3 -TH H TF 2 3 � _ 1 _ _ _ _3-, 1 1 CI CH3 CH3 CH3 3-bromo-2-chloro-5-ethyl-4,4-dimethyloctane 3-fluor o-4-isopropyl-2-methylheptane H3C - CHCH2CH3 l 2 I-sec-butyl-3-nitrocyclohexane (numbering determined by t he alphabetical order of s ubstituents, 3 N02 "b" comes before "n") I
.l-__
2
3
__
__
-I
-"
__
Br
6 5
a
F
4
677
Appendix 1 , Summary of IUPAC N omenclature, continued C. Naming Molecules Containing Functional Groups from Group B-Su ffix Only
1 . Alkenes-Follow the same steps as for alkanes , except : a. Number the chain of carbons that includes the C= C so that the C=C has the lower position number, since it has a higher priority than any substituents; b. Change "ane" to "ene" and assign a position number to the first carbon of the C=C ; place the position number just before the name of functional group(s); c . Designate geometrical isomers with a cis,trans or E, Z prefix.
F
2 I 4 C H 3C - C = C H2 H / 1 F CH3
\
F
H
I
C ==C - C = CH2 2
\
I
-
4,4-difluoro-3-methylb ut - l -ene
0,
F
I
3
4
H
1 , I-di fluoro-2-methyl buta- l ,3 -diene CH3
5-methylcyclopenta1 ,3-diene 2
3
Special case : When the chain cannot include an alkene , a substituent name is used. See Section V.A.2.a. 2 6
5
4
C = CH2 3 H
3 -vmy Icyc I 0hex- 1-ene .
Numbering must be on EITHER a ring OR a chain, but not both.
2. Alkynes-Follow the same steps as for alkanes , except : a . Number the chain of carbons that includes the C-C so that the alkyne has the lower position number ; b. Change "ane" to "yne" and assign a position number to the first carbon of the C=C ; place the position number just before the name of functional group(s). Note : The Group B f unctional groups (alkene and alkyne) are considered to have equal priority: in a molecule with both an ene and an yne , whichever is closer to the end of the chain determines the direction of numbering. In the case where each would have the same position n umber, the alkene takes the lower number. In the name , "ene" comes before "yne" because of alphabetization . F 2 I H2 \ H -C =CH2 CH - C - C == CH HC = C - C = C - CH 3 HC C -C 3/ 4; H H H F C H3 pent-3-en- l -yne pent- l-en-4-yne 4,4-difluoro-3-methylbut- l-yne ( "yne" closer to end of ("ene " and "yne" have equal priority un less chain) they have the same position n umber, when "ene" takes the lower number) (Notes : 1 . An "e" is dropped if the letter following it is a vowel: "pent-3-en- l-yne" , not "pent-3ene- l-yne". 2 . An "a" is added if inclusion of di , tri, etc. , would put two consonants together: "buta- l ,3 -diene" , not "but- l ,3 -diene".) D. Naming Molecules Containing Functional Groups from Group A-Prefix or Suffix
In naming molecules containing one or more of the functional groups in Group A, the group o f highest priority is indicated by suffix; the others are indicated by prefix, with priority equivalent to any other substituents . The table in Section lIlA. defines the priorities ; they a re discussed on the following pages in order of increasing priority. 678
Appendix 1 , Summary of IUPAC Nomenclature, continued Now that the functional groups and substituents from Groups A , B , and C have been described, a modified set of steps for naming organic compounds can be appli ed to all simple structures :
-Step 1 . Find the highest priority functional group. Detennine and name the longest continuous carbon chain that includes this group. -Step 2 . Number the c hain so that the highest priority functional grou p is assigned the lower number . (The number " I " is often omitted when there is no confusion about where the group must be. Aldehydes and carboxylic acids must be at the first carbon of a chain, so a " I " is rarely used with those functional groups.) -Step 3 . If the carbon chain includes multiple bonds (Group B ) , rep lace "ane" with "ene" for an alkene or "yne" for an alkyne. Designate the position of the multiple bond with the number of the first carbon of the multiple bond . -Step 4. If the molecule includes Group A functional groups, replace the last "e" with the suffix of the highest priority functional group , and include its position number just before the name of the highest priority functional group . -Step 5 . Indicate all Group C substituents, and Group A functional groups of lower priority, with a prefix . Place the prefixes , with appropriate position numbers, in alphabetical order before the root name. 1. Amines : prefix : a mino-; suffix : -amine-substituents on nitrogen denoted by "N' NH2 CH P CH3CH, , /CH 2CH J propan- l -amine
yY V
�
3-methoxycyclohexan- l -amine (" 1 " is optional in this case) 2. Alcohols : prefix : hydroxy- ; suffix : -01 OH I H3C-C -C =CH 2 H H but-3-en-2-01
ethanol
3. Ketones: prefix : oxo- ; suffix : -one (pronounced "own ") o
CH3 -CH -C -CH3 I OH 3-hydroxybutan-2 -one II
o
6
H2C =C -CHCH3 H N,N-diethylbut-3-en-2-amine 0H
c(
NH 2 2-aminocyclobutan- l-01 ( " 1 " is optional in this case)
H3C , /CH3 N o II I CH3 -C - CH2 - C = CH 2 4-(N,N-di methylamino)pent A-co-2-one
cyclohex-3-en- l-one (" 1" is optional in this case) 4. Aldehydes : prefix : oxo-, or formyl- (O=CH ); suffix : -al (abbreviation : --CHO) An aldehyde can only be on carbon I , so the " 1"- is generally omitted from the name. o OH 0 o 0 o " I " " II II C H3 -CH H 2C- � = �-CH CH3CCH2CH2CH HCH ethanal; 4-oxopentanal methanal; 4-hydroxybut -2-enal acetaldehyde formaldehyde 679
Appendix 1, Summary of IUPA C Nomenclature , continued Special case : When the chain cannot include the carbon of the aldehyde, the suffix 2 3 "carbaldehyde " is used: a 4
0-1 1 5
I
6
cyc lohexanecarba ldehyde
CH
5 . Carbox ylic Acids : prefix : carboxy-; suffix : -oic acid (abbreviation : -COOH ) A carboxy lic acid can only be on carbon 1 , so the "1" is genera lly omitted from the name. (Note : Chemists traditionally use , and IUPAC accepts , the names "formic acid" and "acetic acid" in p lace of "methanoic acid" and "ethanoic acid".) a
a
II
methanoic acid; formic acid HC - OH
O �
II
ethanoic acid; acetic acid CH3C - OH
_
a
a
g
II
a
CH
II
I
3
HC - C - C - COOH I CH3
CH 2 - CH - OH I NH2
2-amino-3-phenylpropanoic acid
2 ,2-dimethyl-3 ,4dioxobutanoic acid
Specia l case : When the chain numbering cannot inclu de the carbon of the carboxylic acid, the suffix "carboxy lic acid" is used: CHO 2-formyl-4-oxocyclohexanecarboxylic acid ("formyl" is used to indicate an aldehyde as O COOH a substituent when its carbon cannot h e in the chain numbering)
K, � 5
6
E . Naming Carboxylic Acid Derivatives The six common groups derived from carboxylic acids are , in decreasing priority after carbox ylic acids : s alts , anhydrides, esters , acyl ha lides , amides , and nitriles . 1. Sa lts of Carbox ylic Acids Sa lts are named with cation first , followed by the anion name of the carboxylic acid, where "ic acid" is rep laced by "ate" : acet ic acid becomes acetate butanoic acid becomes butanoate cyc lohexanecarboxylic acid becomes cyc lohexanecarboxy late
NH2
Li + lithium 2-aminopropanoate
Na+ sodium chloroacetate
I
CH3 - CHCOO-
CICH2COO-
2. Anhydrides : "oic acid" is replaced b y "oic anhydride " a
II
alkanoic acid
R - C - OH
a
>
a
II
II
a lkanoic anhydride
R-C-O-C-R
680
CH30
U
COO
ammonium 2-methox y cyclobutanecarboxylate
�� a
a
O
benzoic anhydride
Appendix 1 , Summary of IUPAC Nomenclature, continued 3 . Esters Esters are named as "organic salts" that is, the alkyl name comes first, followed by the name of the carboxylate anion . (common abbreviation : -COOR) CH3 a carbJ,xylate ( � I II a \
a
II
R-C-O�R "alkanoate" . "alky l" "alkyl alkanoate"
II
a
II
H3C - C - O + CH2CH3
ethyl acetate
HO
.
a II
V
a
II
alkanoic acid
R - C - OH
5.
>
Amides : "oic acid" is replaced by "a mide" a
II
alkanoic acid
R - C - OH
< }a
� butanamide
II
a lkanamide
R - C - NH2
Nitriles : "oic acid" is replaced by "enitrile" a
II
alkanoic acid
R - C - OH
I
NH2
�C = N
a
benzoyl chloride
�
(i.: I
CI
0
benzamide
�
NH 2
V:� :
butanenitrile
alkanenitrile
+-0
�
l
C
Amides are notable for their role in biochemistry, i .e . , the special amide bond between two amino acids is called a peptide bond.
6.
CH2COO
cyclohexyl 2-pheny lacetatc
a
a
>
CH3
butanoyl chloride
alkanoyl chloride
:
I
�
II
R - C - Cl
I
isopropyl 2,2-dimethylpropanoa te
C - �H 3
a
CH3
H3 C - C - C - O � CHCH 3
methyl 3-hydroxycyclo pentanecarbox ylate 4. Acyl Halides : "oic acid" is replaced by "oyl halide" vinyl prop-2-enoate
H2C = C - C - 0 7" C = CH2 . H H
:
z nil'iIC
(common spelling di ffers from IUPAC)
"Aromatic" compounds are those derived from benzene and similar ring systems . As with aliphatic nomenclature described above, the process is : determining the root name of the parent ring ; determining priority, name, and position number of substituents ; a nd assembling the name in alphabetical order . Functional group priorities are the same in aliphatic and aromatic nomenclature. See p . 676 for the list of priorities . A . Common Parent Ring Systems (a) IV. Nomenclature of A romatic Compounds
7
benzene
6
0) 8
I
'-'::
�
1
'-'::
�
naphthalene 5
4
2 (t})
3
681
� 8
9
1
5
10
4
6� 3
7
a nthracene
2
Appendix 1 , Summary of IUP AC Nomenclature , continued B. Monosubstituted Benzenes 1 . Most substituents keep their designation , followed by the word "benzene ":
CH3 8
6 6 H CHO OH 3 6 6 6 6 6 6 6 C. �Y � ¢ COOH H0'Q { }OCH3 Ql 6 NH 2 CHO CH] CH3 �OCH2CH3 r(YCI 02 A 02 H O Y Y CH3 N0 2 NH2
nitrobenzene ethylbenzene chlorobenzene 2 . Some common substituents change the root name of the ring . IUPAC accepts these as root names , listed here in decreasing p riority (same as Group A, p . 676): COOH benzene- benzaldehyde phenol benzoic sulfonic acid acid Disubstituted Benzenes 1. Designation of substitution--only three possibilities : X
U ortho1 ,2-
common: IUPAC:
aniline
X
meta1,3 -
toluene
X
�y
(0-)
anisole
(m-)
para- (p-) 1 ,42 . Naming disubstituted benzenes-Priorities from Group A, p . 676 , determine root name and substituents Br
Br
p-dibromobenzene 1 ,4-dibromobenzene
y
I �
�
�
o-methoxybenzaldehyde 2-methoxybenzaldehyde
m-aminobenzoic acid 3 -aminobenzoic acid
D. Polysubstituted Benzenes-must use numbers to indicate substituent position
HN I
VCl
3,4-dichloro-N-methylaniline
N
m-methylphenol 3 -methylphenol
N
2 ,4,6 -trinitrotoluene (TNT) 682
ethyI 4-amino-3-hydroxybenzoate
Appendix 1, Summary of IUPAC Nomenclature, continued E. Aromatic Ketones A special group of aromatic compounds are ketones where the carbonyl is attached to at least one benzene ring. Such compounds are named as "phenones ", the prefix depending on the size and nature of the group on the other side of the carbonyl. These are the common examples :
o
o C
< }-C-CH3
< }- CH2CH3
acetophenone
propiophenone -
benzophenone V. Nomenclature of Bicyclic Compounds "Bicyclic" compounds are those that contain two rings. There are four possible arrangements of t wo rings that depend on how many atoms are shared by the two rings . The first arrangement in which the rings do not share any atoms does not use any special nomenclature, but the other t ypes require method to designate how the rings are put together. Once the ring system is named, then functional groups and substituents follow the standard rules described above. Type 1 . Two rings with no common atoms These follow the standard r ules of choosing one parent ring system and describing the other ring as a substituent . 5 4 ketone is the highest priority functional group, phenyl is substituent 6 c::==�> 3-pheny Icyclohexan- l-one (" 1" could be omitted here) o benzene is the parent ring system as it is larger than cyclopentane and it has three substituents c::=�> l-cyclopentyl-2,3-dinitrobenzene a
y,D
Type 2. T wo rings with one common atom-spiro ring system The ring s ystem in s piro compounds is indicated by the word "spiro" (instead of "cyelo"), followed by brackets indicating how many atoms are contained in each path around the rings , ending with the alkane name describing how many carbons are in the ring systems including the spiro carbon. (If any atoms are not carbons, see section VI.) Numbering follows the s maller path first, passmg through the sprio carbon a nd around the second ring. S
1
5
3
:00,
4
spiro[3.4 ]octane
spiro[4.5]decane 683
6
7
Appendix 1, Summary of IUPAC Nomenclature, continued Substituents and functional groups are indicated in the usual ways . Spiro ring systems are a lways numbered smaller before larger, and numbered in such a way as to give the highest priority functional group the lower position number. I 10 9 o
8
7 r\A 2
6
5 3
l0V
'---+- CH3 CH 3 7 ,7 -dimethylspiro [4.5]decan-2-one
spiro [3.4 ]oct -5-ene
Type 3 . Two rings with t wo common atom-fused ring system Two rings that share two common atoms are called fused rings. This ring system and the next type called bridged rings share the same designation of ring system. Each of the two common atoms i s called a bridgehead atom, and there are three paths between the two bridgehead atoms . In contrast with naming the spiro rings, the longer path is counted first, then the shorter, then the shortest. in fused rings, the shortest path is al ways a zero, meaning zero atoms between the two bridgehead atoms . Numbering starts at a bridgehead, continues around the largest ring, through the other bridgehead and around the shorter ring . (In these structures, bridgeheads are marked with dark circle for clarity .) 2 2 LO
4
bicyclo [2 . i .O]pentane (path of 2 atoms and a path of I atom)
CO 1 3 8 6 4
9
4
bicyclo [4.4.0]decane 7
(path of
5
atoms in
a
9
5
4
6
bicyclo [5.3.0]decane (path of
atoms and
Substituents and functional groups are indicated in the usual ways . Fused rings systems are always numbered larger before smaller, and numbered in such a way as to give the highest priority functional group the lower position number. Type 4. Two rings with more than two common atom-bridged ring system Two rings that share more than t wo common atoms are called bridged rings . Bridged rings "hare the same designation of ring system as Type 3 in which there are three paths between the t w'.) bridgehead atoms. The longer path is counted first, then the medium, then the shortest . Numbering starts at a bridgehead, continues around the largest ring, through the other bridgehead and around the medium path, ending with the shortest path numbered from the original bridgehead atom . (In these structures, bridgeheads are marked with a dark circle for clarity .)
2
64 �
31 5
bicyclo [2. 1 . 1 ]hexane (paths of 2 atoms, I atom, and 1 ato m )
each d irection)
6
�84 51 3
bicyc 10[2 . 2. 2]octane 2
(three paths of 2 atoms)
684
path of 3 atoms)
8
7
3
3
bicyclo [3.2. 1 ]octane (paths of
atoms, 2 atoms,
and 1 atom)
Appendix 1 , Summary of IUPAC Nomenclature, continued 6
� B, �Br 3
2
5
5,5-dibromo bicyclo [2 . 1 . 1 ]hexane
3
3
6
bicyclo [2.2.2]oct-5-en -2-one o
8,8-dimethyl bicyclo [3 .2. 1 ]octan - l-ol
The teon "heteroatom" a pplies to any atom other than carbon or hydrogen . It is common for heteroatoms to appear in location s that are inconvenient to name following ba sic rules, so a simple system called "replacement nomenclature" has been devised . T he fundamental principle is to na me a compound as if it contained only carbons in the skeleton , plus any functional groups or substituents, and then indicate which carbon s are "replaced" by heteroatoms. The prefixes u sed to indicate these substitions are li sted here in decreasing priority and listed in this order in the name: OH Prefix Example Element o oxa nonan- l -01 thia S aza P phospha H2 H H sila Si B '-.,./ Si � S /" B bora '-.,./OH 2-thia-8-aza-4-sila-6 -boranonan- l -ol VI. Replacement Nomenclature of Heteroatoms
N
N ......."..
In the above example , note that the (imaginary) compound no longer has nine carbons, even though the name still includes "nonan" . The heteroatom s have replaced carbon s, but the compound is named as if it still had those carbon s. Where the replacement system i s particularly useful i s in polycyclic compounds. Shown below are three examples of commercially available and synthetically useful reagents that use this system . parent hydrocarbon reagent abbreviation
6
8 7�4
� S
3
6
bicyclo [2.2.2 ]octane
4
� 8 6
I
1 ,4-DiAzaBiCyclo [2 .2.2]Octane
2
��
i(5 3
DABCO
2
( upper case added to explain abbrev iation)
O
b icyclo[5.4.0]undec-7 -ene
4
:)
r� � �N8 6
O
1 ,8-DiazaBicyclo [5.4.0]Undec-7-ene
( upper case added to explain abbreviation)
685
DBU
Appendix
I,
Summary of IUPAC Nomenclature , continued
3�5 9
9 3
B ....
H
9-BBN
6
2
I
8
7
9-BoraBicyclo [3 . 1. 1]Nonane
bicyclo [3 . I. l ]nonane
7
( upper case added to explain abbreviation)
How can we distinguish this structure from its mirror image ?
Is this alkene cis or trans ? I F
VII. Designation of Stereochemistry; Cahn-Ingold-Prelog system
F
,I "1"
I \\' CI Br Br Cl Compounds that exhibit stereoisomerism , whether geometric isomers around double bonds , substituent groups on rings , or molecules with asymmetric tetrahedral atoms (which are almost always carbons) , require a system to designate re lative and absolute orientation of the groups. The terms cis/trans , DIL in carbohydrates and amino acids , and dlffor optically active compounds, are li mited and cannot be used generally , although each still is used in appropriate situations . For example , cis/trans st ill is used to indicate relative positions of substituents around a ring . A system developed by chemists Cahn , Ingold , and Prelog , uses a series of steps to determine group priorities , and a definition of position based on the relative arrangement of the groups. In alkenes , the system is relatively simple : high high low high
>=<
>===<
>===<
low low low high this arrangement is defined this arrangement is defined as as Z =zusammen , together , E = entgegen , opposite , from from both high priority groups the two high priority groups on the same side o f the C=C on opposite sides of the C=C As with alkenes , the orientation around an asymmetic carbon can be only one of two choices . In three dimensions , clockwise and counterclockwise are the only two directions that are definite , and even that description requires a f ixed reference point. To designate configuration , the lowest (fourth) priority group is always placed farthest away from the v iewer ( indicated by a dashed line) , and the group priorities will follow 1 to 2 to 3 in either a clockwise or a counterclockwise direction. to 2 to 3 is as the configuration
1 clockwise, defi ned R = rectus
1
3
)<
1
2
2 686
)<
1 to 2 to 3 is countercIock defined as the S sinister configuration WIse,
3
=
Appendix 1 , Summar y of IUPAC Nomenc lature, continued The only step remaining is t o determine the priority of groups , for which t here is a carefully defined set of rules . Rule 1. Consider the first atom of the group , the point of attachment. Atoms with higher atomic number receive higher priority. Heavier isotopes have higher priority t han lighter isotopes. I
Br
>
>
CI
>
F
Rule 2. If the first atoms of two or more groups are the same , go out to the next atoms to break the tie . One high priority atom takes priority over any number of lower-priority atoms . H CH 3 H H I I I I -C - OH > -C -H > -C -H > -C - H I I I I H H CH 3 CH 3
Rule 3. Treat multiple bonds as if they were all single bonds ; one wil l be to the real atom, the others will be to imaginary atoms. This is the hardest rule to put into practice . This example of an alkyne shows stepwise how to accomplish t his. In the pictures , imaginary atoms (ones that did not start out in the structure, had to invent them) are ind icated by italics.
(
I
C
-- C C - H all atoms are real
>
- C -C -H
I
halfway there-had to add two fake carbons
O-C r:::= ==�> C .... . C ..... R ....... H H R ....... added a fake 0 to the real C , and a fake C to the real 0 II
R -C :: N
0, /
Examples applying the Cahn -Ingold-Prelog sytem
i
compari n g F and C l
low
F
high
CI
'-
I
J t==\ __
E
I
Br
high
low
}
comparing I and B r
687
I
I
C
C
-- C - C - H
>
I
I
final-had to add four fake carbons
C
becomes
o
started with 3 bonds to carbon, so have to rep l ace them
C
C
becomes
we
>
N
C
R -C - N I
I
I
I
added two fake Ns to the C , and two Cs to the N N
C
Appendix 1 , Summary of IUPAC Nomenclature , continued 4 F
,I "1"
' I" Br
2
Often, the hard part of applying the RlS system to asymmetric carbons is orienting the molecule so that the fourth priority group is farthest away. Sometimes it is easier to put the fourth priority group toward you, then take the opposite of what the direction of 1 to 2 to 3 says. CI 1 I 3
� '\) ,
flip Br up so F goes bac k
I"F �CI I 2 Br
4
1
R
3
\1"
R
=
,\\\
Br"
�',/1
"
CI
3 up � Putting the F (4th priority) pointing up toward you, the groups 1 to 2 to 3 appear counterclockwise, but we have to take the opposite, so instead uf S, it is R.
2
688
F
�
Appendix 2 : Summary of Acidity and Basicity
Imagine that you are at a family reunion w here you can observe t he competition for ice cream cones a mong your nieces and nephews . Pretty soon , you formulate a generalization : t he older kids can hold onto their ice cream cones more strongly t han the younger ones. Another way of saying it is t hat t he older ones are less
equilibria between the child with ice cream , and the free ice cream plus the hungry child . The differences in strength could also be quantitated : the larger the hunger factor , pKH ' the less likely the child will give up the ice cream. likely to give up their cones. You could even represent this information i n a table showing a series of
p KH
Approximate pKH Values of Children ---
12
1 2 -year-old with ice cream
---
10
1 O-year-old w ith ice cream
---
8
8 -year-old w ith ice cream
---
6
6-year-old w ith ice cream
---
4
4-year-old w ith ice cream
---
---
---
---
---
Ice cream
+
hungry 1 2-year-old
Ice cream
+
hungry l O- year-old
ice cream
+
hungry 8 -year-old
Ice cream
+
hungry 6-year-old
Ice cream
+
hungry 4-year-old
What good is this table ? It allows anyone to make predictions about what will happen when kids with ice cream are mixed with hungry kids . hungry 1 O-year-old +
4-year-old with ice cream
}?1 ---
---
1 O-year-Old with ice cream +
hungry 4-year-old
If the hungry lO-year -old was left unattended in a room with the 4-year-old with ice cream, which side of this equilibrium would be favored w hen you came back in a few minutes: "reactants" or "products "? More likely than not , there would be an ice cream transfer in your absence ; "products" would be fa vored . You could ha ve predicted this from t he table, and you could generalize : the hungry lO- year-old will be strong enough to rip the ice cream awa y from an y kid lower on the table than the lO-year-old hi mlherself . Let's predict the results of another equilibrium : hungry 6-year-old +
}?i
l 2-year-old with ice cream
---
---
6-year-old with ice cream +
hungry 1 2-year-old
Is the hungry 6-year-old stron g enough to pull the ice cream away from the 1 2 -year-old ? Not in most families . The table shows that the only chance for the hungry 6-year-old is to find a 4-year-old with ice cream . The 12-year -old with ice cream is pretty safe as long as the hunger table doesn't go to hi gher ages.
If you understand this analogy and can make predictions of ice cream transfer using the table, then you can understand how to predict the direction of equilibrium in acid-base reactions. Turn the page . 689
Appendix 2 continued, Summary of Acidity and B asicity
A few generalizations: To set the stage ....
A) This Appendix deals with pro tic acids and bases, called Bronsted-Lowry acids. Similar statements can be made about Lewis aci ds but they are not the focus of this discussion. B ) Values of pKa are measures of equilibrium constants, described further below. Values between 0 and 15 .7 are measured by titration in water solution and are known accuratel y , to within 0. 1 pK unit and sometimes better. Values outside of thi s range cannot be measured in water because of the leveling effect of water, and there is no universal ly accepted method for measuring these pKa values . This lack of a single standard of measurement means that the values below 0 and above 15 .7 should be considered relative, not absolute . If your instructor says the pKa of methane is 46 and this book says it is 50, those should be considered the same value within experimental variation . C) Acidi ty is a thermodynamic property, and the acid equi libri um constant, K a, is a measure of the relative concentrations of species in the protonated and unprotonated form. As most organic acids are weak acids, meaning they are present mostly in the protonated form at equi librium, the Ka < l. Since the pKa = -log Ka, the pKa values will be greater than 0 , with the larger pKa representi ng a weaker acid. If this is not clear, review text section 1 - 13 . D) Acids do not spontaneously spit out a proton ! Despite our way of writing ionization equilibri a as shown on the next page, aci ds do not give up a proton unless a base comes by to take the proton away . The reactions as drawn in the table should be consi dered half-reactions, just as the reactions in the electromotive series were half-reactions for balancing oxidation-reduction reactions in general chemistry. Look at the table on p. 6 9 1 and notice how it looks j ust like the "chi ldren-with-ice-cream" table. We c an use this table to make predictions about equi libri um position in acid-base reactions j ust as we did for the children with ice cream. 1) A base will deprotonate any acid stronger than its conjugate acid. This is the most important principle of predicting acid-base reactions. On the table, this means that any base, hydroxide for example, can react with any acid more acidic than the conj ugate acid of itself, water in our example. So hydroxide is a strong enough base to pull the proton from any of these : bicarbonate ion, a phenol, c arbonic acid, a carboxylic acid, or a sulfonic acid. We can also predict that hydroxide is NOT a strong enough base to react with any acid above water on the table; for example , a mi xture of hydroxide with an alkyne will favor the reactants at equilibrium, with onl y a small amount of products. I. Predicting equilibrium position
reactants favored
HO
-
+
RC _ C - H pKa 25
weaker base
.. ___
weaker acid
H20
+
RC - C :
pKa 15.7
stronger acid
stronger base
2) Another way of predicting the position of an equilibrium is to assign "stronger" and " weaker" to the acid and base on each side of the equation, using the table to determine which is stronger and which i s weaker. Equilibrium will always favor the weaker acid and base. This method will always give the same answer as the principle in # 1 above. To lead into the next section, look again at the table on p. 691 and notice two things: a) with only a couple of exceptions, all the aci dic protons are on either oxygen or c arbon; and b) generalizations can be made about the acidity of functional groups. Learning to correlate acidity with functional group is i mportant in predicting reactivity of the functional group. 690
Appendix 2 continued, Summary of Acidity and Basicity A��roximate �Ka Values of Oq�anic Com�o un ds p Ka I I "" 50 alkane - H H+ + weaker acid
RC I
a lkene
"" 45
a mine
3 5 -40
alkyne ketone and ester
"" 3 5 "" 25
==C
-
/
H
N -H
.. I
R R -C -O - H I
l
t
alcohol canno! be meas ured in water solution
I
1 5 .7
1 0.3
HC03-
I
"" 1 7
I
I
"" 1 6
measured in water solution
+
-N :
H+
R
H R -C -O -H R H R -C - O - H H H2O
H+
== � .
H+
I
"" 1 8
+
H+
RC ::C - H 0 I -C -C - H
..
..
I
I
H+
H+
H -H
"
20-25
R-C:
-
+
H:
stronge r base
/
I
+
RC ::C : 0 I -C - C :
+
R -C - O :
+
H+
+
H+
+
H+
+
H+
+
"
R
I
I
R I
H R -C - O : R H R -C - O : H HOI
I
I I
- - - � - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
phenol
",, 1 0 6.4
carboxylic acid sulfonic acid
stronger acid
Ar - O -H H2C03
0 R -C - O -H I I
4-5 <0
RS02 - O - H 691
..
..
H+
H+ H+
H+
+
+ + +
col-
Ar - O :
HC030 R -C - O : I I
RSO 2 - 0 : • •
weaker base
Appendix
2 continued, Summary of Acidity and Basicity
II. Correlation of Acidity with Functional Group
A. Ox�gen Acids
0 R
"
-
S
I I
0
-
O-
H
pKa < O sulfonic acid
>
0 I I
R
-
C
-
O-
pKa 4- 5
H
>
( }O-H pKa
c arboxylic acid
10.0
>
HO-H pKa
phenol
1 5 .7
water
>
R I
R
-
C I
R
-
O-
pKa
H
16 - 1 9
alcohol
Sulfonic acids are the strongest of the oxygen acids but are not common in organic chemistry. Carboxylic acids, however, are everywhere and are considered the strongest of the common organic oxygen acids. (Note that "strong" and " weak" are relative terms: acetic acid, pKa 4.7 4, was a " weak" acid in general c hemistry in comparison to sulfuric and hydrochloric acids , but acetic acid is a "strong" acid in organic chemistry relati ve to the other oxygen acids. ) Phenols having OR groups on benzene or other aromatic rings are sti l l stronger acids than water. Why are phenols, carboxylic acids, and sulfonic acids stronger acids than water? B ecause their anions are stabi lized by resonance. (Refer to text sections 1 - 13 , 10-6 , and 20-4, especially Figure 20- 1 .) Let's look at that statement in more detai l . Remember that acidity i s a thermodyn amic property ; that i s , acidity equi librium depends o n the difference in energy between the reactants and products. The more the anion is stabilized by resonance, the lower in energy it is, and the less positi ve the 1\G, as shown on the reaction energy diagram: ...
,
,
,
,
- alcohol water
Energy
phenol , , ,
-
- - - . carboxylic acid
oxygen anions stabilized by resonance
sulfonic acid reaction -. Why are alcohols weaker acids than water? There are two effects that contribute, both of which are consi stent with the trend that 1 ° alcohols (pKa 16) are slightly stronger than 2° alcohols (pKa 1 7 ) which are slightly stronger than 3° alcohols (pKa 1 8). Alkyl groups are mildly electron-donating in their inductive effect (more about thi s later) and destabilize the anion , as shown in the energy diagram above. S econd, the more crowded the anion is, the less it can be stabilized by hydrogen bondin g with the solvent.
692
Appendix 2 continued, Summary of Acidity and Basicity B. Carbon Acids When we think of "acids", we do not usually think of protons on carbon, yet carbon acids and the carbanions that come from them are of tremendous importance in organic chemistry. "Unstabilized" carbon acids are those that do not have any substituent to stabilize the anion. Alkanes, with only sp 3 carbons are the weakest acids with pKa around 50. The vinyl carbon in a carbon-carbon
2
double bond is sp hybridized with the electrons of the anion slightly closer to the positive nucleus, leading to some s tabilization of the anion. This type of stabilization is particularly important in alkynes with sp hybridized carbons.
I
I
R -C:
R -C-H
=C
I
I
3 s p hybridized
2
pKa 50
/ ,
=
H
c:
/
RC::C-H
.
sp hybridized
sp hybridized
pKa 45
pKa 25
RC::C:
..
C. Carbon Acids Alpha to Carbonyl (This topic is described in detail in text section 22-2B.) Look at these huge differences in pKa when the acidic group is next to a carbonyl.
o
O- H
H2 O
e=:>
RC
II
-
pKa 16- 18
H
R -C-O-H
0 N-H
H H
e=:>
RC 2 N
-
H
H
II
pKa 35-40
R-C- N-H
0 C- H
H2
RCH2 C
e=:>
H
pKa 50
pKa 16
H2
II
-
pKa 4-5
R-C-C -H
pKa 20
Hydrogens alpha to carbonyl are unusually acidic because of resonance stabilization of the anionic conjugate base.
D. Acidities of Acyl Functional Groups In addition to the significant variation in the acidity of alpha hydrogens depending on which atom the H is bonded to, what is on the other side of the carbonyl also has a dramatic influence. In this case,the stabilization is more important on the starting material,not on the conjugate bas e. See the energy diagram on p. 694.
H I
O II
H2C -C-H pKa
17
no stabilization of starting material
H I
amide
es ter
ketone
aldehyde
O
H
II
I
H2C-C-CH3
H
O
II
H2C- C OCH3 -
pKa 20
••
pKa 25
mild stabilization of starting material by weak electron donation from
CH3
H I
H2C
0 I
+
H2C-C=�CH3
••
II
-
C- N(CH3}z
t
pKa 30
significant resonance stabilization of starting material
693
O
I
H 0 I
+
I
H2C-C N(CH3}z =
s trongest resonance stabilization of starting material
Appendix 2 continued, Summary of Acidity and Basicity
,, , , .-
(
least tabilization
Energy
aldehyde
I
, ,
,,
,
I ,
'
...
,
, " , ...... \ . " , " '" \ ' '�
,
,I'
' ,,/ ,
"/
,
ketone ester amide
greatest stabilization of starting material, largest LlG
__
reaction
--
E. Carbon Acids Between Two Carbonyls (This topic is described in detail in text section 22-15.) While a hydrogen alpha to one carbonyl moves into the pKa 20-25 range for ketones and esters respectively, a hydrogen between two carbonyls (cyano and nitro are similar to carbonyl electronically) is more acidic than water. The increased resonance stabilization of the conjugate base is largely responsible, but there are subtle variations depending on the type of functional group as noted at the bottom of p.
693. 0
0
EtO
Look at the enormous influence of the nitro group.
¥
OEt
N:--... � C
13.5 0 II
H2C' I
+
N
pKa
pKa
10.2
EtO
-
H3CO
o
CH3
H
pKa
+ �" N
10.2
0
-
'0
H3C
0
V
CH3
H
pKa 9,0
+" , "+
0
0
-O,N
H
pKa
0
¥
1 1.2
0
'0
H
C
H
H
pKa
y
0
/N �
N, O
H
5.8
pKa 3.6
III. Correlation of Basicity with Functional Group The bulk of this Appendix is on acidity because many more functional groups are acidic than are basic. Basically (oooh, sorry), only one functional group is basic: amines. There is variation among aliphatic, aromatic, and heteroaromatic amines; these are covered thoroughly in text sections 19-5 and 19-6. One point in the text, just before Table 19-3, deserves emphasis: for any conjugate acid-base pair: pKa + pKb
=
694
14
Appendix 2 continued, Summary of Acidity and Basicity This simple algebraic relationship is very useful: Sample problem. Is triethylamine (pKb 3.24) a strong enough base to deprotonate phenol (pKa
1O.0)?
We need to calculate either the pKa of the conjugate acid of triethylamine or the pKb of the conjugate base of phenol to see which is stronger and weaker. Then we can say with certainty which side of the equilibrium will be favored.
(} pKa
10.0
OH
+
( }o-
Et-N-Et 1
Et pKb 3. 24
pKb
=
4.0
(14 - 10.0) stronger acid
stronger base
weaker base
H +1
+
Et - N -Et 1
Et pKa = 10.76
(14 - 3. 24) weaker acid
Aha! Products are favored at equilibrium, so the correct answer to the question is "Yes,triethylamine is a strong enough base to deprotonate phenol." Try this for fun: How weak must a base be before it does NOT deprotonate phenol? What algebraic rule can you formulate to predict whether any combination of acid and base will favor products or reactants?
IV. Substituent Effects on Acidity So far, we have focused on acidities of different functional groups. Let's tum to more minor,more subtle, structural changes to see what effect substituents will have on the acidity of a group. Primarily, we imply electronic effects as opposed to steric effects, but this Appendix will conclude with a discussion of how steric and electronic effects can work together. A. Classification of Substituents-Induction and Resonance Substituent groups can exert an electronic effect on an acidic functional group in two different ways: through sigma bonds, where this is called an inductive effect, or through p orbitals and pi bonds which is called a resonance effect. Groups can also be electron-donating or electron-withdrawing by either of the mechanisms, so there are four possible categories for groups. Note that a group can appear in more than one category, even in conflicting groups!
a) Electron-donating by induction: only alkyl groups (abbreviated R) have electrons to share by induction; b) Electron-withdrawing by induction: every group that has a more electronegative atom than carbon is in this category; some examples: F, Cl,Br,I, OH,OR,NH2,NHR, NR2, N02, C=O,CN,S03H, CX3 where
X is halogen;
c) Electron-donating by resonance: groups that have electron pairs to share: F, Cl,Br,I, OH,OR,NH2, NHR,NR2; d) Electron-withdrawing by resonance: N02,C=O,CN,S03H.
695
Appendix 2 continued, Summary of Acidity and Basicity B. Generalizations on Electronic Effects on Acidity (refer to text section 20-4B) Electric charge is the key to understanding substituent effects. An acid is always more positive than its conjugate base; in other words, the conjugate base is always more negative than the acid. Electron donating and electron-withdrawing groups will have opposite effects on the acid-base conjugate pair. Electron-donating groups stabilize the more positive acid form and destabilize the more negative conjugate base. From the diagram, it is apparent that electron-donating groups widen the energy gap between reactants and products, making L\G more positive, favoring reactants more than products. In essence, this weakens the acid strength. conjugate base
- "", � ...
- electron-donating substituent no substituent effect
Energy
acid electron-withdrawing substituent
reaction
-
Electron-withdrawing groups destabilize the more positive acid form and stabilize the more negative conjugate base, narrowing the energy gap between reactants and products, making L\G less positive. Products are increased in concentration at equilibrium which we define as a stronger acid. Electron-withdrawing groups increase acid strength; electron-donating groups decrease acid strength.
C. Inductive Effects on Acidity Text section 20-4B gives a thorough explanation of the inductive effect of electron-withdrawing groups on simple carboxylic acids, from which three generalizations arise: A) Acidity increases with stronger electron-withdrawing groups. (See problem 20-33.) B) Acidity increases with greater number of electron-withdrawing groups. C) Acidity increases with closer proximity of the electron-withdrawing group to the acidic group.
We don't usually look to aromatic systems for examples of inductive effects, because the pi system of electrons is ripe for resonance effects. However, in analyzing the resonance forms of phenoxide on the next page, it becomes apparent that the negative charge is never distributed on the meta carbons. Meta substituents cannot exert any resonance stabilization or destabilization; at the meta position, substituents can exert only an inductive effect. The series of phenols demonstrates this phenomenon, consistent with aliphatic carboxylic acids.
696
Appendix 2 continued, Summary of Acidity and Basicity
: 0:
:0:
:0:
6 �-6 '
OH
O
CH3
----
OH
0 �
H
pKa 10.00
pKa 10.09
donating
(] C
meta
OH
0
----
:0:
the delocalized negative charge does not increase electron density at meta position
6-
OH
OH
OCH3
pKa 9.65 withdrawing
0
0
CI
N02
pKa 9.02
pKa 8.39
withdrawing
withdrawing
D. Resonance Effects on Acidity-benzoic acids and phenols (review the solution to problem 20-45) Resonance effects can be expressed with placement of substituents at ortho or para positions, but ortho has the complication of steric effects, so just para substitution is shown here.
�
Electron-withdrawing substituents increase the acidity of benzoic acids and phenols:
I
/
-...;::: �
H
OH
OH
hO
I
COOH
COOH
COOH
-...;::: �
' para
pKa 10.00
OH
¢ ¢ ¢ ¢ ¢ N02
CN
pKa 8.05
pKa 7.95
pKa 7.15
H
pKa 4.20
CN
pKa 3.55
N02
pKa 3.42
Electron-donating by resonance but electron-withdrawing by induction:
There is a group of substituents that donate by resonance but withdraw by induction: alkoxy groups and halogens are the most notable examples, and the acidity data provide an insight into which effect is stronger. COOH
¢
COOH
0 �
OCH3
H
pKa4.20
COOH
¢
OCH3
pKa4. 09
pKa4. 47
meta-Methoxybenzoic acid is stronger than benzoic acid, consistent with electron-withdrawing by induction which is expressed at the meta position. But the para isomer is weaker than benzoic acid; electron donation by resonance has not only compensated for the inductive effect (which is still operative at the para position) but has decreased the acidity even further. Thus, the donating effect by resonance must be stronger than the withdrawing effect by induction for the methoxy group.
See another example on the next page.
697
Appendix 2 continued, Summary of Acidity and Basicity OH
I
"'"
h
61
¢
"'"
I
¢
OH
OH
h
"'"
1
h
F
F
H
pKa 10.00
pKa 9.81
pKa 9.28
meta-Fluorophenol is stronger than phenol, consistent with electron-withdrawing by induction which is expressed at the meta position. The para isomer is still stronger than phenol; electron donation by resonance has not compensated for the inductive effect (which is still operative at the para position). Thus, the donating effect by resonance must be weaker than the withdrawing effect by induction for the fluoro group.
Studying substituent effects on acidity is the standard method of determining whether a group is donating or withdrawing by induction and resonance. E. Proximity Effects of Substituents pKa 10.0
pKa 8.05
pKa 9.19
pKa 9. 90
OH
OH
OH
OH
6
9r
cr
o
B
A
0
D
Three effects influence the pKa values of these substituted phenols. In C, the acetyl group at the meta
position is electron-withdrawing by induction only. In B, the acetyl group at the para position exerts both resonance and inductive effects, both of which are electron-withdrawing, making the acid stronger. In
theory, substituents at the ortho position should be like para, exerting both resonance and inductive effects; in fact, the inductive effect should be stronger because of closer proximity to the acidic group. So
we would predict D to be a stronger acid than B, yet it is not. What other effect is operating?
Structure E shows that because of the proximity of the acetyl group to the OH, intramolecular hydrogen bonding is possible. Hydrogen bonding stabilizes the starting material, lowering the energy of the
starting material and making �G more positive. Intuitively, it should be apparent that a hydrogen held between two oxygens will be more difficult to remove by a base. Also, after the proton has left as shown in structure F, the negative charge on the phenolic oxygen is close to the partial negative charge on the
oxygen of the carbonyl, destabilizing product F, raising its energy, also making AG more positive. The proximity of the acetyl group influences both sides of the equation to make the acid weaker.
intramolecular hydrogen bond
cr
� I
E
"'"
h
0 8-
o
..
F
cr
Here are two more examples where intramolecular hydrogen-bonding influences acidity.
)-
OH
Ho
-o-
eOOH
pK\ 4.6; pK2 9.3
(
Hooe
�
COOH
eOOH
pK\ 2.75; pK2 13.4
pK\ 3.02; pK2 4.38
698
Hooe
eOOH
\::=./
Appendix 2 continued, Summary of Acidity and Basicity F. Steric Inhibition of Resonance Another type of proximity effect arises when the placement of a substituent interferes with the orbital overlap required for resonance stabilization. This can be seen clearly in the acidity of substituted benzoic acids and in the basicity of substituted anilines. Let's analyze this series of carboxylic acids.
COOH COOH COOH OH COOH COOH H3 C CHJ H 6 � � 0 CH3 ¢CH3 I
pKa 3.75
,
pKa 4.27
pKa 4.20
"
�
pKa 3. 46
,
"
h
pKa 3. 21
pKa 4. 34
Formic acid, pKa 3. 75, serves as the reference carboxylic acid. Benzoic acid is weaker because the phenyl group is electron-donating by resonance, stabilizing the protonated form. Methyl substituents are known to be electron-donating by induction, strengthening the electron-donating effect, making the meta- and para substituted acids weaker than benzoic acid. Then come the anomalies. Alkyl groups are electron-donating by induction and should weaken the acids, but the ortho-t-butyl and the 2,6-dimethylbenzoic acids are not only stronger than benzoic acid, they are stronger than formic acid! Something has happened to tum the phenyl group into an electron-withdrawing group. Phenyl is electron-donating by resonance but electron-withdrawing by induction, so what has happened is that the ortho substituents have forced the
out of the plane of the benzene ring so that there is no
COOH COOH
resonance overlap between the benzene ring and the
orbitals. The
COOH
"feels" the benzene ring as
simply an inductive substituent. Resonance has been "inhibited" because of the steric effect of the substituent. •
"CH3 <----.l�CH3
•
•
0
•
group is not parallel with
this three-dimensional view down the bond between the and the benzene ring shows that is twisted out of the benzene plane
the plane of the benzene ring no resonance interaction.
The same phenomenon is observed in substituted anilines. Anilines are usually much weaker bases than aliphatic amines because of resonance overlap of the nitrogen's lone pair of electrons with the pi sy stem of benzene. When that resonance is disrupted, the aniline becomes closer in basicity to an aliphatic amine. (See problem 19-49(c). )
CC COOH COOH CH3 0-'\ N.�H3 � N:�H3 - CH3 CH3 CH3 pKb 8.94
�-
pKb
More examples on the next page.
=::
6-7
(estimated)
699
Appendix 2 continued, Summary of Acidity and Basicity Examples of steric inhibition of resonance:
o
o
strong base similar to aliphatic amine; geometry of bridged ring prevents overlap of N lone pair with carbonyl
amide-not basic because of resonance sharing of N lone pair with carbonyl
�ND
pKb 3.4
pKb 8. 9
3° aromatic amine
� pKb 6.2
3° aliphatic amine
pKb
-
2.3
3° amine, and the N
is bonded to a benzene, but the bridged ring system prevents overlap of N lone pair with benzene
Not only is steric inhibition of resonance important in this example, but so is intramolecular hydrogen bonding in the protonated form. Draw a picture. (yes, negative!)
700
