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CONTENTS
Correlation of Chapters in GOB 6/e and OB 4/e
iv
Suggested Course Outlines
v - xv
Answers to Even-Numbered Problems
1-87
Bettelheim, Brown & March Correlation of Chapters in General, Organic & Biochemistry, 6/e and Organic & Biochemistry, 4/e
Chapter title
GOB 6/e
OB4/e
Matter, Energy, and Measurement Atoms Chemical Bonds Chemical Reactions Gases, Liquids, and Solids Solutions and Colloids Reaction Rates and Equilibrium Acids and Bases Nuclear Chemistry Organic Chemistry Alkanes and Cycloalkanes Alkenes and Alkynes Alcohols, Ethers, and Thiols Benzene and Its Derivatives Chirality Amines Aldehydes and Ketones Carboxylic Acids, Anhydrides, Esters, Amides Carbohydrates Lipids Proteins Enzymes Chemical Communication: Neurotransmitters and Hormones Nucleotides, Nucleic Acids, and Heredity Gene Expression and Protein Synthesis Bioenergetics: How the Body Converts Food to Energy Specific Catabolic Pathways: Carbohydrate, Lipid, and Protein Metabolism Biosynthetic Pathways Nutrition and Digestion Immunochemistry Body Fluids
Chap. 1 Chap. 2 Chap. 3 Chap. 4 Chap. 5 Chap. 6 Chap. 7 Chap. 8 Chap. 9 Chap. 10 Chap. 1 1 Chap. 12 Chap. 13 Chap. 14 Chap. 15 Chap. 16 Chap. 17 Chap. 18 Chap. 19 Chap. 20 Chap. 2 1 Chap. 22
Chap. Chap. Chap. Chap. Chap. Chap. Chap. Chap. Chap. Chap. Chap. Chap. Chap.
Chap. 23 Chap. 24 Chap. 25
Chap. 14 Chap. 15 Chap. 16
Chap. 26
Chap. 17
Chap. Chap. Chap. Chap. Chap.
Chap. 18 Chap. 19 Chap. 20 Chap. 21 Chap. 22
iv
27 28 29 30 31
1 2 3 4 5 6 7 8 9 10 11 12 13
SUGGESTED COURSE OUTLINES Introduction to General, Organic & Biochemistry, 6/e
The first column, 3Q, lists sessions in a three-quarter course (28 sessions per quarter). The second column, 2S, lists sessions in a two-semester course (up to 43 per semester). The third column, 2Q, lists sessions in a two-quarter course (up to 4 1 per quarter).
Session 2S 3Q
2Q
� �
�
TOPICS
1
1
1
Scientific method, measurements
1. 1-1.4
2
2
2
Unit conversions, states of matter
1.5 - 1.6
3
3
3
Density, energy, heat
1.7 - 1-9
4
4
4
Classification of matter, atomic theory, inside the atom
2. 1- 2.4
5
5
5
Periodic table, electronic structure
2.5 - 2.6
6
6
6
Electronic configuration and periodic table, size of atoms
2.7 -2.8
TEXT SECTION
First hour exam
7 8
7
7
Ions, ionic bonding
3. 1- 3.3
9
8
8
Covalent bonding and shapes of molecules
3.4 -3.5
10
9
9
Factors in bonding, polyatomic ions, inorganic nomenclature
3.6-3.8
10
10
First hour exam
11
11
11
Formula weight, moles, equation balancing
4. 1-4.4
12
12
12
Stoichiometry, percent yield, ionic reactions
4.5 -4.7
13
13
13
Oxidation-reduction, heat of reaction
4.8-4.9 v
� �
�
TOPICS
TEXT SECTION
Second hour exam
14 15
14
14
Organization of matter, gases, pressure, gas laws
5. 1- 5.4
16
15
15
Gas laws, intermolecular forces
5.5 - 5.7
17
16
16
Liquids, vapor pressure, solids, phase changes
5.8- 5. 11
18
17
17
Types of solutions, solubility
6.1- 6.4
19
18
18
Concentration units, water as a solvent
6.5 - 6.6
20
19
19
Colloids, colligative properties, osmosis, dialysis
6.7 - 6.8
21
20
20
Second or third hour exam
22
21
21
Kinetics, molecular collisions, activation energy
7.1- 7.3
23
22
22
Factors affecting rates, equilibrium
7.4 - 7.5
24
23
23
Equilibrium constants, Le Chatelier's principle
7.6 - 7.8
25
24
24
Acid and base strength, Brpnsted-Lowry theory, K. 8. 1 -8.4
26
25
25
Properties of acids, Kw, pH
8.5 -8.8
27
26
26
Hydrolysis, buffers, titrations, normality
8.9 - 8. 11
Final exam
28 2nd
Q 1
27
27
Radioactivity, natural transmutation
9.1 - 9.4
2
28
28
Half-life, characteristics of radiation, dosimetry nuclear energy, medical effects
9.5 -9.8
29
29
Third hour exam
3
30
30
Sources and structures of organic compounds
10. 1- 10.3
4
31
31
Functional groups
10.4
vi
� �
�
TOPICS
TEXT SECTION
5
32
32
Alkanes: structure, constitutional isomerism, and nomenclature; cycloalkanes
1 1. 1- 1 1.5
6
33
33
Alkanes and cycloalkanes: IUPAC, shapes, properties, reactions
1 1.6- 1 1.11
First hour exam
7 8
34
34
Alkenes
12.1- 12.3
9
35
35
Terpenes, polymerization of ethylene, and substituted ethylenes
12.4-12.5
10
36
36
Additions to alkenes, polymeriz. of ethylene
12.6- 12.7
11
37
37
Alcohols: structure, nomenclature, properties
13.1-13.4
12
38
38
Reactions of alcohols, ethers, thiols
13.5- 13.7
Second hour exam
13 14
39
39
Benzene: structure and nomenclature
14.1- 14.3
15
40
40
Benzene reactions and derivatives; phenols
14.4-14.5
Review
41
Final exam
42
41
2nd S
�
1
1
Chirality
15.1-15.4
17
2
2
Stereocenters, optical activity
15.4- 15.6
18
3
3
Amines
16.1-16.6
20
4
4
Aldehydes and ketones: structure, bonding, physical properties
17.1-17.4
21
5
5
Keto-enol tautomerism, oxidation, reduction
17.5- 17.8
16
Third hour exam
19
vii
�
�
TOPICS
22
2nd S
6
6
Carboxylic acids and anhydrides
18. 1- 18.3
23
7
7
Esters, amides, interconversion
18.4- 18.6
24
8
8
Phosphoric acid, step-growth polymerization
18.7 - 18.8
TEXT SECTION
Review
9 25
10
9
Fourth hour exam
26
11
10
Monosaccharides
19. 1- 19.5
27
12
11
Disaccharides, Oligosaccharides, Polysaccharides
19.6- 19.8
Final Exam
28
� 13 1
12 14
Fats, complex lipids
20. 1- 20.4
Fats, complex lipids, membranes
20. 1- 20.8
Membranes, glycerophospholipids, sphingolipids, glycolipids
20.5 - 20.8
2
15
13
Steroids
20.9 -20. 12
3
16
14
Amino acids, zwitterions, cysteine
2 1. 1- 2 1.4
4
17
15
Peptides, primary and secondary structures of proteins
2 1.5 - 21.8
5
18
16
Tertiary and quaternary structure of proteins, glycoproteins, denaturation
2 1.9- 2 1. 1 1
6
19
17
First or second hour exam
7
20
18
Enzymes: classification and common terms
22. 1- 22.4
8
21
19
Mechanism, regulation, medical uses
22.5 - 22.7
viii
� � �
TOPICS
9
22
20
Neurotransmitters and hormones
23.1-23.4
10
23
21
Adrenergic, peptidergic, steroid messengers
23.5 -23.7
24
22
Second hour exam
11
25
23
Nucleic acids, structure of DNA and RNA
24.1-24.3
12
26
24
RNA; genes, exons, introns, cloning
24.5 -24.6
13
27
25
Gene expression, transmission, transcription
25.1-25.5
14
28
26
Translation, protein synthesis, recomb. DNA
25.6-25.9
15
29
27
Cells, mitochondria, common catabolic pathway
26.1-26.4
16
30
28
Citric acid cycle, oxidative phosphorylation
26.5 -26.8
17
31
29
Specific catabolic pathways
27.1-27.4
18
32
30
Fatty acids, stearic acid, ketone bodies
27.5 -27.7
19
33
31
Catabolism
27.8-27.10
20
34
32
Third hour exam
21
35
33
Biosynthesis of carbohydrates, fatty acids
28.1-28.3
22
36
34
Biosynthesis of membrane lipids, amino acids
28.4-28.5
23
37
35
Nutrition and digestion
29.1-29.9
24
38
36
Immune system, antigens
30.1-30.3
25
39
37
Immunoglobulins, T -Cells, cytokines
30.4-30.6
26
40
38
Blood functions
31.1-31.4
27
41
39
Kidney functions, blood pressure
31.5 -31.8
Review
42 28
43
TEXT SECTION
40
Final exam
ix
SUGGESTED COURSE OUTLINE Introduction to Organic & Biochemistry, 4/e: 2 quarters
The first column, 2Q, lists sessions in a two-quarter course, up to 28 sessions per quarter.
Session
ID
TOPICS
1
Sources and structures of organic compounds
1 . 1- 1.3
2
Functional groups
1.4
3
Alkanes: structure, constitutional isomerism, and nomenclature; cycloalkanes
2. 1-2.5
4
Alkanes and cycloalkanes: IUPAC, shapes, properties, reactions
2.6-2. 11
5
Alkenes
3. 1-3.3
6
Terpenes, polymerization of ethylene, and substituted ethylenes
3.4-3.5
7
Additions to alkenes, polymerization of ethylene
3.6-3.7
8
First hour exam
TEXT SECTION
9
Alcohols: structure, nomenclature, properties
4. 1-4.4
10
Reactions of alcohols, ethers, thiols
4.5 -4.7
11
Benzene: structure and nomenclature
5.1 - 5.3
12
Benzene reactions and derivatives; phenols
5.4-5.5
13
Chirality
6. 1-6.4
14
Stereocenters, optical activity
6.4 - 6.6
15
Second hour exam
16
Amines
x
7. 1- 7.6
2Q
TOPICS
17
Aldehydes and ketones
8.1-8.4
18
Aldehydes and ketones
8.5 -8.8
19
Carboxylic acids and anhydrides
9.1-9.3
20
Esters, amides, interconversion
9.4-9.6
21
Phosphoric acid, step-growth polymerization
9.7 -9.8
22
Third hour exam
23
Monosaccharides
10.1-10.4
24
Reactions of monosaccharides
10.5
25
Disaccharides, oligosaccharides, polysaccharides
10.6-10.8
26
Review
27
Final exam
TEXT SECTION
� 1
Fats: structure, properties, complex lipids
11.1-11.4
2
Membranes, glycerophospholipids, sphingolipids, glycolipids
11.5-11.8
3
Steroids
11.9-11.12
4
Amino acids, zwitterions, cysteine
12.1-12.7
5
Peptides, structures of proteins, glycoproteins
12.8-12.11
6
First hour exam
7
Enzymes: classification and common terms
13.1-13.4
Mechanism, regulation, medical uses
13.5 -13.7
9
Neurotransmitters and hormones
14.1-14.4
8
xi
SUGGESTED COURSE OUTLINE Introduction to Organic & Biochemistry, 4/e: 1 semester
The first column, IS, lists sessions in a one-semester course, up to 43 sessions. Session
18. 1
2
TOPICS
Organic compounds Alkanes and cycloalkanes
3
Alkanes and cycloalkanes
5
Terpenes, reactions, polymerization
4
Alkenes
6
Alcohols
7
Reactions of alcohols, ethers, thiols
8
First hour exam
9
Benzene
11
Optical activity, significance of chirality
10
Stereocenters and enantiomers
12
Amines
14
Second hour exam
13
Aldehydes and ketones
15
Carboxylic acids
17
Monosaccharides
16 18
Carboxylic esters, amides, functional groups
Disaccharides, oligosaccharides, polysaccharides
TEXT SECTION 1. 1- 1.4
2. 1-2.5
2.6-2. 1 1
3. 1-3.3 3.4-3.7 4.1-4.4 4.5 -4.7
5.1-5.5
6.1-6.3
6.4-6.6
7. 1-7.6 8. 1-8.8 9. 1-9.3 9.4-9.8
10. 1- 10.4 10.5- 10.8
xiii
IS.
TOPICS
19
Lipids
11. 1-11.8
20
Steroids
11.9 - 1 1.12
21
Amino acids, zwitterions, cysteine
12. 1- 12.4
22
Peptides, protein structures
12.5 - 12.8
23
Enzyme classification
13. 1-13.4
24
Enzyme activity
13.5 - 13.7
25
Third hour exam
26
Chemical messengers
14.1-14.4
27
Adrenergic, peptidergic, and steroid messengers
14.5 - 14.7
28
Nucleic acids
15.1 - 15.3
29
DNA replication, RNA, genes, exons, introns, cloning
15.4-15.6
30
Transcription and translation
16.1- 16.5
31
Gene regulation, mutation, recombinant DNA
16.6- 16.8
32
Cells, mitochondria, common catabolic pathway
17.1- 17.4
33
Phosphorylation, energy yield, conversion
17.5 - 17.8
34
Glycosis, energy yield, catabolism
18. 1-18.5
35
Ketone bodies, catabolism
18.6- 18.10
36
Fourth hour exam
37
Biosynthesis of carbohydrates, fatty acids
19. 1- 19.3
38
Biosynthesis of membrane lipids, amino acids
19.4- 19.5
39
Nutrition and digestion
20. 1-20.9
40
Immune systems, antigens
2 1. 1- 21.3
xiv
TEXT SECTION
1S.
TOPICS
41
Immunoglobulins, T Cells, cytokines
43
Final exam
42
Body Fluids
TEXT SECTION 2 1.4-2 1.6
22. 1-22.8
xv
ANSWERS TO EVEN-NUMBERED END-OF-CHAPTER PROBLEMS "Introduction to General, Organic, and Biochemistry", sixth edition
Chapter 1
Matter, Energy, and Measurements
1.12
(a) anything that has mass and takes up space (b) the science that deals with matter
1 . 14
No, some of them are made by serendipity, or chance observations
1.16
2 (a) 3 .75 x 1 0 1 (b) 6.29 x 1 0 4 (c) 9. 1 x 1 0-
1.18
(a) 6.48 x lO \b) 1 .6 x 1 05
(c) 4.69 x1 05 (d) 2.8 x 1 0-15
1 .20
(a) 1 .3 1 x 1 0 5 (b) 9.4 X 1 0
3 (c) 5 . 1 37 X 1 0-
1 .22
4.45x 1 06
1 .24
(a) 4
1 .26
(a) 92 (b) 7.3 (c) 0.68
(d) 0.0032
1 .28
(a) 1 .53
(b) 2.2
(c) 0.00048
1 .30
(a) gram
(b) meter
(c) liter
1 .32
(a) 20 mm
(b) 1 inch
(c) 1 mile
1 .34
See Sec 1 .4
1 .36
(a) 77°F, 298 K
1 .3 8
(a) 0.0964 L (b) 27.5 em (g) 44 mL (f) 3290 cc (k) 83 400 mm(l) 0.361 g
1 .40
50 mileslhr
1 .42
3 . 1 2 g/mL
1 .44
35.9 mL
1 .46
22 1 g
1 .48
(a) oxygen 1 .492 gIL; carbon dioxide 1 .977 giL (b) Yes. The carbon dioxide smothers flames.
(b) 1
(c) 3
4
(d) 5
(e) 5
(f) 1
(c) 45 700 g (h) 0.7 1 1 kg
(d) 2.8 x 1 0-7
(g) 3 (e) 5.9
(d) 4.75 m (i) 63 .7 cc
(e) 2 1 .64 mL G) 73 OOOmg
Chapter 1
Solutions - for 1M
Matter, Energy, and Measurements
1.50
Kinetic energy is the energy of motion. Potential energy is stored energy.
1. 52
3 (a) 2.1 x 10 cal
1.54
(b)
1.56
13 .6 g
1.58
to generate heat by muscle action
1.60
(a) volume (f) heat
1.62
56 kg, assuming that the room is empty
1.64
1.24 x 10 10 erg
1.66
Kinetic: (b) (d) (e) Potential: (a) (c)
1.68
(a) The moon has a much lower gravity than the earth. (b) No
1.70
See Section 1.9
1.72
(a) mercury
1. 74
The quart is the better buy.
1.76
(c)
1.78
No
2
4 (c) 9.9 x10 cal
(b) volume (c) mass (d) heat (h) velocity (g) temperature
(b) 162°C
(c) carbon
3 (d) 2.5 x 10 cal
(e) density
Chapter 2
Atoms
Chapter 2
Atoms
Solutions - for 1M
2.8
Elements: (a) (f) (g) Compounds: (b) (k) Mixtures: (c) (d) (e) (h) (i) G)
2. 1 0
(a) oxygen (f) sulfur
2. 1 2
See Sec 2.3
2. 1 4
No. C O and CO2 are different compounds, and each obeys the law o f constant composition
2. 1 6
(a) 4 8 (b) 1 90 (c) 79 (d) 244
2. 1 8
(a) scandium (b) titanium
2.20
(a) 7
(b) 4 1 (c) 1 1 2 (d) 50
2.22
(a) 22
(b) 1 04
2.26
(a) a particle with an unequal number of protons and electrons (hence it has a charge) (b) atoms with the same number of protons, but different numbers of neutrons.
2.28
(a) 1 7e, 1 7p, 20n (d) 54e, 53p, 74n
2.30
1 .022 amu
2.32
As, P, N; I,
2.34
(a) aluminum (b) arsenic
2.36
(a) 2
2.3 8
2.40
(b) 8
(b) lead (g) iron
F;
(c) calcium (d) sodium (e) carbon (h) hydrogen (i) potassium
(c) silver
(d) thorium
(e) argon
(c) 3 5
(d) 128
(e) 7
(b) 1 8e, 1 9p, 20n (e) 44e, 46p, 60n
(f) 23 8
(c) 80e, 82p, 1 25n (f) 74e, 74p, 1 1 2n
Ne, He; Mg, Ca, Ba; K, Li (c) gallium
(d) aluminum
(c) 1 8
In all three cases the outer-shell configurations are the same, except for the shell number: 2 4 2 4 2 2 2 (a)Na: 3 s 1 Cs: 6s 1 (b) 2S Sp Te: 5s 5p (c) C: 2S 2p2 Ge: 3 s 4p Each has a different number of filled inner shells 2 2 (a) F: l s 2s 2p5 ; 2 3 4 (b) S : I S 1 2s 2p6 3s 3p ; 2 2 6 2 3 (c) P : I s 2s 2p 3 s 3p ;
2.42
7s
2.44
(a) smaller
2 I s 2s22p6 2 2 2 2 S -: l s 2s 2p6 3s 3p6 2 3 p -: Same as S F:
(b) greater 3
Chapter 2 2.46 2.48 2.50
Lowest-to-highest; K, Li, AI, B, C, CI, 0, N, Ar, (b) K
: K :
Some chemicals are; some are not; and for many it depends on the dose. Since everything is made of chemicals, many chemicals must be nonpoisonous.
Calcium is an essential element in human bones and teeth. Since strontium behaves much like calcium, strontium-90 gets into our bones and teeth and gives off radioactivity for many years directly into our bodies.
2.52
by hammering it.
2.54
(a) I s (b) 2s2p
2.56
(a) S2pl (b)
2.58
Solutions - for 1M
Atoms
(Kr)
(c) 3s3p3d
(d) 4s4p4d4f
S2p5
2.60
(a) P
2.64
2 (a) Ne (b) K+ (c) Br- (d) Ba +
2.66
6 1 .3%
2.68
6.0 xl021 protons
(b) K (c) Na (d) N (e) Br (f) Ag (g) Ca (h) C (i) Sn G) Zn
3 (e)H- (f) Sc +
2.70
Any state with an energy higher than the ground state. An atom can be in only one unique excited state at any time.
2.72
54 protons, 77 neutrons
4
Chapter 3
Chemical Bonding
Chapter 3
Chemical Bonding
Solutions - for 1M
3 . 16 (a) Lithium, Li, must lose 1 electron. (b) Chlorine, CI, must gain 1 electron. (c) (d) (e) (f) (g)
Phosphorus, P, which is below N in the Periodic Table, must gain 3 electrons. Aluminum, AI, must lose 3 electrons. Strontium, Sr, which is below Be, Mg, and Ca in the Periodic Table, must lose 2 electrons. Sulfur, S, which is below ° in the Periodic Table, must gain 2 electrons. Silicon, Si, which is below C in the Periodic Table, must either lose or gain 4 electrons.
(h) Oxygen,
°
must gain 2. 1-,
3 .18 Stable ions are (a)
(c) Na+ , (d) S2 - , and (1) Cs+
3.20 To form ionic bonds, carbon would have to become either C4+ or C4 - , and silicon would have to become either Si4+ or Si4 - . All would have charges that are too concentrated for these small ions. 3.22 Each sodium ion, Na+ , is surrounded by six chloride ions, and each chloride ion, Cl -, is surrounded by six sodium ions. 3.24 In completing this table, be certain to balance charges in each ionic compound. 3 F04 BrC104ro/(j2ti)3Li + LiBr 2 Ca + GaBr2 3 Co + Co Br3 KBr K+ 2 eu + CuBr2
Li3P04
LiC I04
Li20
Li N�
Ga(CI04)2
GaO
Ca(N�)2 GaS04
Co(O 04)3
Co2 03 Co(N03h Co2(S04)3 CoP04
KCI 04
K20
KN0:3
Cu(O 04h
QD
Cu(N03)2 CuS04
3 .26 (a) Na+ and Br (d) K+ and H 2P04 -
Li2S04
K2S04
or
LiOH
Ga3(P04)2 Ca(OHh
K3P04
Co(OH) 3 �H
Cu3(P04h Cu(OH)2
(b) Fe2+ and S0 3 2- (c) Mg2+ and P0 4 3 (e) Na + and HC0 3 - (f) Ba2+ and N0 3 -
3 .28 A covalent bond is formed when two atoms share one, two, or three pairs of electrons. 3 .30 (a) In a single bond, two atoms share one pair of electrons. (b) In a double bond, they share two pairs of electrons and (c) in a triple bond, they share three pairs of electrons. 5
Chapter 3
Solutions - for 1M
Chemical Bonding
3.32 (a) Incorrect. The left carbon has five bonds. (b) Incorrect. The carbon in the middle has only three bonds. (c) Incorrect. The second carbon from the right has only three bonds and the oxygen on the right has only one bond. (d) Incorrect. Fluorine has two bonds. (e) Correct. (f) Incorrect. The second carbon from the left has five bonds. 3.34 In these Lewis structures, the symbol of the atom represents the nucleus and all filled valence shells. Only electrons in the valence shell are shown. (a) K-
•
(b ): Se: •
(c) -N• ••
••
(d) : I••
(e) :Ar: --
(f)-Be-
3.36 Each atom in these molecules has a complete valence shell. •• •• -- -HH b ) H-.S-H ( a):Br-Br: N-N(c) ( I I -- - . (e)
:C=N:
(f)
H H
� H-ff-H H
(g) :N=N :
(g)
:0••
•• ••
(d)
'-+-N=N-H
.. .. (h) :0=0:
3 . 3 8 A bromine atom, Br, contains seven electrons in its valence shell. A bromine molecule contains two bromine atoms bonded by a single bond. A bromide ion is a bromine atom that has gained one electron in its valence shell; it has a complete octet and a negative charge. ( a) : Br --
(b) : .Br-Br: . .. ••
••
(c) :.Br: -
3 .40 (a) Linear, with bond angles of 1 80°. (b) Trigonal planar, with bond angles of approximately 1 20°. (c) Tetrahedral, with bond angles of approximately 1 09.5°. 3 .42 Electrons are shifted toward the more electronegative atom, which in each part is (g) 0 (f) S (d) CI (e) C (a) CI (b) 0 (c) 0 3 .44 Parts (a), (b), and (d) are polar molecules
6
Chapter 3
Solutions - for 1M
Chemical Bonding
3 .46 Given are differences in electronegativity between the bonded atoms and the classification of the bond. (a) Br-Br 2.8 - 2.8 0.0; nonpolar covalent (b) Br- CI 3 .0 - 2.8 0.2; nonpolar covalent (c) H - CI 3 .0 2.1 0.9; polar covalent (d) Sr-F 4.0 - 1.0 3 .0; ionic =
=
-
=
=
(e) Si -H 2.1 - 1.8 0.3 nonpolar covalent (f) C-O 3 .0 2.5 0.5; polar covalent (g) N -N 3 .0 - 3.0 0.0; nonpolar covalent (h) Cs- CI 3 . 0 - 0.7 2.3 ; ionic =
-
=
=
=
3 .48 (a) Bicarbonate ion
(b) Nitrite ion (c) Sulfate ion (e) Dihydrogen phosphate ion
(d) Hydrogen sulfate or bisulfate ion 3 .50 (a) KBr (b) CaO (e) Li2 S04 (i) Sr(H2 P04h
(f) Fe2 S 3 (j) Ag2 C0 3
(c) Pb(OHh
(d) CU 3 (P0 4h
(g) NH4 HS0 3 (k) SrCl2
(h) Mg(CH 3 COOh (1) Ba(Mn04) 2
(m) HgO 3 .52 Tin(II) fluoride or stannous fluoride. Its molecular formula is SnF 2 . 3 .54 Sodium iodide, Nal 3 .56 Aluminum hydroxide, AI(OH)), and magnesium hydroxide, Mg(OHh. 3 .58 Given here is a common form of each, but not the only form. (a) P as calcium phosphate (b) Mg as magnesium oxide (c) K as potassium chloride (d) Fe as Ferrous fumarate (e) Ca as calcium carbonate (f) Zn as zinc oxide (g) Mn as manganese sulfate (h) Ti as titanium dioxide (i) Si as sodium silicates, silicon dioxide (k) B as sodium borate, potassium borate (m) Cr as chromium chloride (0) Se as sodium selenate (q) Ni as nickel(II) sulfate
(j) Cu as cupric oxide
(1) Mo as sodium molybdate
(n) I as potassium iodide
(P) V as sodium metavanadate (r) Sn as tin(II) chloride 7
Chapter 4
Chemical Reactions
Chapter 4
Chemical Reactions
Solutions - for 1M
4.16
(a) 70.9 amu (b) 39.9 amu (c) 123.9 amu (d) 28.0 amu (e) 4.0amu
4.18
(a) 2 moles
4.20
(a) 18.1 moles 0 atoms (d) 0.40 moles Hg atoms
4.22
The same; that is, j ust about 2: 1
4.24
10 molecules
4.26.
6.1 x 10 18 molecules of cholesterol
4.28
(b) 11.5 moles (c) 1 mole (b) 1.23 moles Br atoms (e) 3 . 8 x 10-5 mole N atoms
(c) 4.2 x 1 04 moles 0 atoms
(a) h + H2 --7 2HI (b) 2AI + 302 --7 Ah0 3 (c) 2Na + Ch --7 2Na CI (d) CaC03 --7 C02 + CaO
4.30
(a) the study of weight relationships in chemical reactions (b) a solution in which the solvent is water
4.32
9.47 x 10-
4
4.34
138 g C2�0
4.36
7.5 g C02
4.38
0.13 g C
4.40
(a) Benzene is the limiting reagent. (b) 121 g bromobenzene is formed
4.42
The percentage yield was 1 15%. Above 100% yield implies faulty techniques of preparation.
4.44
85 .2%
4.46
Equations (a) and (b) are already balanced. (c) 2Sc3+ (aq) + 3 S04 2- (aq) --7 SC2(S04) 3 (S) 2 2 (d) Sn + (aq) + 2Fe + (aq) --7 2Fe 3+ (aq)+ Sn(s) (e) 2K(s) + 2H20(1) --7 2K+ (aq) + 20H- (aq) + H2 (g)
4.48
Fe2+ (aq) + 20H-(aq) --7 Fe (0H)2(s)
9
Chapter 4
Solut ions - for 1M
Chemical Reactions
4.52
(a) (c) and (d)
4.54
Oxidat ion is (1) loss of electrons (2) ga in of oxygen or loss of hydrogen. Reduction is (l ) gain of electrons (2) loss of oxygen or gain of hydrogen.
4.56
(a) Pb gets oxidized ; Ag + gets reduced (b) Ag + is the oxidizing agent, Pb is the reducing agent.
4.58
Endothermic means it absorbs heat, endergon ic means it absorbs energy in any form not just heat
4.60
1 9 600 cal will be given off
4.62
88.6 g iron
4.64
Calcium, phosphate, and hydroxide
4.66
The CH 4 is oxidized to CO2. The H202 is reduced to H20.
4.68
A mole of lead (207.2 g) has more mass than a mole of silver ( 1 97.0 g)
4.70
2 3 (a) 1 9.2 moles Fe 0
4.72
2 Cd + + 2C r � CdC h(s) ; NH/ and N0 3 ' are spectator ions.
4.74
3 (b) 1.6 1 x 1 0 g Co
893 amu
4.76
H2 is in excess
4.78
(a) An exergonic process is one that gives off energy in any form, not necessarily heat. (b) An endergonic process is the opposite
4.80 4.82
10
An
ion that is present, but does not participate in a react ion.
(a) C 4H S + 602 � 4C02 + 4H20 (b) C 4H s is oxidized and 02 is reduced. (c) O2 is the oxidiz ing agent and C 4 H S the reducing agent.
Chapter 5
Gases, Liquids and Solids
Chapter 5
Gases, Liquids and Solids
Solutions - for 1M
5.12
See Sec 5 .2
5.14
(a) 1 0.3 1 atm (b) 0.033 atm (c) 0.759 atm (d) 1 atm
5. 1 6
73
5. 1 8
342° C
5 .20
6.2 L
5.22
The temperature remains 27° C
5 .24
4.35 atm
5 .26
The gas must be heated to 480 K or 207°C
5.28
1 .97 atm
5.30
(a) 2.33 moles (b) no
5.32
2.8 L
5 .34
-96° C
5.36
(a) 24 moles
5.38
S02 : 2.86 gIL CH4: 0.7 1 4 gI L F2 : 1 .70 gIL
5 .40
2 2.68 x 1 0 4 molecules
5.42
NH3 first, ClOHI 8 02 last
5 .44
MW=36 amu
5.46
(b)
5 .48
(a) H-bond (b) London dispersion forces (c) Dipole-dipole. HF will have the highest surface tension
5.50
hexane
5 .52
(a) 80 mm Hg (b) 1 30 mm Hg (c) 3 80 mm Hg
mm
Hg
(b) 770 g
11
Chapter 5
Gases, Liquids and Solids
Solutions - for 1M
5.54
(a) HC1< HBr < HI Boiling points increase with molecular weight. (b) O2 < HCl < H2 0 2 Here the molecular weights are about the same, but the strength of interaction increases: London < dipole < hydrogen bond.
5.56
See Sec 5 . 1 1
5.58
sublimation
5.60
The ice will sublime and becomes vapor
5 .62
Soot
5.64
The pulsating blood starts to flow in the lower arm; the blood pressure is equal to the external pressure applied. This is the systolic pressure.
5 .66
CO is bound to hemoglobin and does not allow it to carry oxygen. Oxygen under high pressure in the hyperbaric chamber is dissolved in the plasma and is carried to the tissues without the aid of hemoglobin.
5.68
Because the growth of ice crystals killed the affected cells
5.70
Liquid water and water vapor will coexist
5.72
Because it has a diameter in the nanometer range
5.74
(c) water
5.76
There is a lot of empty space between gas molecules, so electromagnetic radiation (including visible light) passes through without much interaction.
5 .78
NH3, because it has a lower molecular weight. 1 .68 times faster.
5 .80
HCI
5 .84
See Sec 5 .9
12
Chapter 6
Solutions and Colloids
Chapter 6
Solutions and Colloids
Solutions - for 1M
6. 1 4
Glucose is the solute and water the solvent
6. 1 6
(a) wine, vinegar (b) honey, tea (c) oxygen in rivers, household ammonia (d) air
6. 1 8
0.32 g aspartic acid in 1 1 5 mL water may have been close to the solubility limit. On standing, some of the water has evaporated and the solution became saturated and eventually the solute precipitated out of the solution.
6.20
(a) water (b) diethyl ether (c) water
6.22
( 1 ) Heat the mixture. (2) Add more water.
6.24
supersaturated
6.26
The beer is bottled under pressure. The solubility of C02 in water increases with increasing pressure. When the bottle is opened, the pressure decreases to 1 atm, decreasing the solubility of C02. Hence the beer loses C02 (goes flat).
6.28
(a) Dissolve 26.6 g of H2 S in 473 .4 g of water. (b) Dissolve 1 . 5 1 g of benzene in 340.5 g toluene (c) Dissolve 4.3 g dimethyl sulfoxide in 8.2 g of acetone
6.30
In each case, to the given weight of solute add enough water to make the total volume specified. (a) 9.0 g NaCI; 250 mL total volume (b) 3 1 g glycine; 625 mL total volume (c) 5.96 g Na2S04; 43 .5 mL total volume (d) 1 0.9 g acetone; 5 1 8 mL total volume
6.32 6.34
To 1 .0 g NaOH add enough water to make 250 mL of solution. (a) 1 .7 M (b) 0.333
6.36
69J.1L
6.3 8
0.0305 M
6.40
M
0.23% w/v and 0.055
(c) 0.030 M
M
6.42
(a) 72.5 g (b) 42.0 g (c) 0.35 g
6.44
To 530 mL of 95% ethanol add 4.47 L of fruit juice.
6.46
0.875 ppm; 875 ppb
13
Chapter 6
Solutions and Colloids
Solutions - for 1M
6.48
3.66 mg
6.50
Strong: (a)(c) Weak: (d) Nonelectrolyte: (b) (e)
6.52
A water molecule forms a hydrogen bond with the OH group of ethanol.
6.54
between 1 and 1 000 nm
6.56
Colloidal particles are larger than about 1 nm. The largest molecules in any gas are much smaller than that.
6.58
The very large surface area associated with these particles of sizes in nm
6.60
-5.25°C
14
.
Chapter 7
Reaction Rates and Equilibrium
Chapter 7
Reaction Rates and Equilibrium
7. 1 0
Solutions - for 1M
Endothermic. The increase in temperature drives an endothermic reaction toward the products.
The change in concentration of a reactant or product per unit time. Moles per liter per minute. 3 6. 1 4 x 1 0- moleslL per min 7. 1 2
Set up (c); it provides more head on collisions.
7. 1 4
/ transition state
7. 1 6
The higher the concentration, the more molecules there are in the same volume, so there are more collisions per unit time.
7. 1 8
As the reaction goes on, the concentrations of reactants decrease. In most cases the rate is directly proportional to the concentrations of the reactants.
7.20
The nature of the reactants, the concentration, the presence of a catalyst
7.22
Crushing the marble increases the surface are. The rate of reaction . will increase.
. 7.24
No; no. We can start with any proportions of reactants, so at equilibrium they won't have equal concentrations.
7.26
(a) C02+ H2 0 � H2 C0 3 (b) P4 0IO � P4+502
7.28
0. 877 M
7.30
No. Equilibrium would be reached very quickly, but at equilibrium there would be virtually no NH2- or + H 3 0+.
7.32
the set of conditions at which K = 1 .72
7.34
(a) right (b) right (c) left (d) left (e) no shift
(c) 3HF + PF 3 � 3F2+ PH3 (d) 2CaO � 2Ca + 02
15
Chapter 7
Solutions
Reaction Rates and Equilibrium
-
for 1M
7.3 6
(a) The equilibrium will shift to the right. (b) The equilibrium constant stays the same.
7.3 8
yes, a change in temperature
7.40
not directly by heat, but by mobilizing the immune defense mechanism
7.42
See Box 7B.
7.44
The silver chloride is converted to silver and chlorine (the equation is given in Box 7D).
7.46
At -5°C the rate is 0.70 moles of HCI per liter per second: at 45°C the rate is 22.4 moles of HCI per liter per second.
7.48 (a)
K =
2 [Po/-] [H20] 2 [OW] [H2P04-]
(c)
K =
[OH-] [H3U'] 2 [H20]
(b)
K =
[H3U'] [N03-] [H20] [HN03]
7.50
0.026M
7.52
Increase the pressure (or decrease the volume, which is the same thing), thus increasing the concentration.
7. 54
The activation energy must be less.
T
T
less than 1 00 kcal/mole
�
1 00 kcal/mole
�
A
7.56
1 .76Mlmin
7.58
No; see Figure 7.5
16
Chapter 8
Acids and Bases
Chapter 8
Acids and Bases
8. 1 6
Solutions - for 1M
(a) HP + H20 � H3 0+ + P(b) HBr + H20 � H 3 0+ + Br- ( 1 00%) (c) H2 S03 + H20 � H3 0+ + HS03 (d) H2 S04 + H20 � H3 0+ + HS04- ( 1 00%) (e) HC0 3- + H20 � H3 0+ + col(b) H2B0 3 (c) r (d) OH3 (g) P04 (h) HPO/- (i) HS-
8. 1 8
(a) HS04(f) NH2-
8.20
(a) diprotic and amphiprotic (c) monoprotic (e) monoprotic and amphiprotic
8.22
add to the answer: in (a), (c) and (d) the SA and SB are on the left side; in (b), (e) and (f) they are on the right side
8.24
(a) and (b)
8.26
(a)
10-3 M
(b)
10-1 0 M
(c)
(b) monoprotic and amphiprotic (d) monoprotic and amphiprotic (f) diprotic
10-7 M
8.28
(a) 8 (b) 1 0 (c) 2 (d) 0 (e) 7 Acidic: (c)(d) Basic: (a)(b) Neutral: (e)
8.30
(a) 8.5 (b) 1 .2 (c) 1 1 . 1 (d) 6.3 Acidic: (b) (d) Basic: (a)(c)
8.32
(d)
10-15 M
2 2 [OH-] (a) 1 0-1 M (b) 1 0- M (c) 1 0-9 M pOH (a) 1 2 (b) 2 (c) 9 (d) 3 =
=
8.34
Acidic: MgS04, Ca(N0 3)2, N�CI Basic: Na3 P04, K2C0 3 , BaF
8.36
Acidic: (b)(c)(e)(h)
8.38
(a) H 3 0+ + CH3 COO- � CH3 COOH + H20 (a) OH-+ CH3 COOH � CH3 COO-+ H20
8.40
3 .75 (pKa of formic acid)
8.42
(a) 4.85 (b) The 1
Basic: (a)(f)
3
(d) 1 0-
M
Neutral: (d)(g)
M solution has ten times the buffer
capacity of the 0. 1
M solution. 17
Chapter 8
Solutions - for 1M
Acids and Bases
8.44
to determine the acid or base concentration of a solution
8.46
(a) 1 .29
8.48
0.8 N
(b) 0.82
(c) 3 . 80
(d) 0.0942
8.50
(a) To 1 2 g of NaOH (s) add enough water to make a total volume of 400.0 mL. (b) To 6. 1 g of Ba(OH)2 add enough water to make a total volume of 1 .0 L.
8.52
0.57 N
8.54 8.56
0 .900 N
(a) 0.3 M
(a) 1 .42M
(a) 1 .5 7M
8.58
Mg(OH)2
8.60
strong bases
8.62
sodium bicarbonate
8.64
See Box 8B.
8.66
1 42 g
8.68
0.7M
8.70
(a) 0
8.72
(a)
8.74
0.900M
8.76
The acid strength of an acid is inversely proportional to the base strength of its conjugate base. Therefore the Ka value of the conjugate acid of any base tells its base strength, and a separate table of Kb values is not needed.
8.78 8.80
8.82
18
(b) 1 x 1 00
=
1 molelliter
CH3COOH is the stronger acid and H2C03 right.
is the weaker acid. The equilibrium will lie to the
(a) H 3 0+ + F (b) OH-+HF NH3 + HCI � NH4+ + CIAll the NH3 will be converted to NH4+ by the above reaction. None will remain.
Solutions - for 1M
Chapter 9
Nuclear Chemistry
Chapter 9
Nuclear Chemistry
9.8
9. 1 0
9. 1 2
By accident, a photographic plate placed near a radioactive rock became blackened.
2 (a) 4.0 x 1 0-5 cm or 4.0 x 1 0 nm, blue light 7 (b) 3.0 cm or 3 .0 x 1 0 nm, microwaves 2 (c) 2.7 x 1 0-5 cm or 2.7 x 1 0- nm, ultraviolet radiation (d) 2.0 x 1 0-8 cm or 0.20 nm, Longest wavelength: (a)
Highest energy: (c)
9. 1 4
(a) nitrogen-1 4
9. 1 6
europium-lSI
9. 1 8
vanadium-5 1
9.20
carbon- 1 2
9.22
(a)
I�Be
�
I� B
+
_� e
(b)
1��Eu
�
1��Eu
+
(c)
I:�TI
�
1:� Hg
+
y
(d)
2�:PU
(b) phosphorus-33
�
2�i
u
+
(c) lithium 9
(d) calcium-39
beta gamma
+� e
positron
iHe
alpha
9.24
gamma rays
9.26
a diagram that shows how much radioactive material is left after a certain time
9.28
1 6 mg
9.30
There are no ways.
9.32
(a) There will be little change after 2 h. The activity will be 24 mCi and 8.9 x 1 08 counts per minute.
9.34
Gamma rays; they have the most penetrating power.
9.36
30 meters
9.38
0.63 cm
9.40
2500 Bq and 0.068 J!Ci
3
19
Chapter 9
Solutions - for 1M
Nuclear Chemistry
(c )strontium-85 (e) mercury- I 97 , technetium-99m
9.42
(a) cobalt-60 (b) selenium-75 (d) carbon- I I , technetium-99m
9.44
iodine- 1 25 prostate cancer; iodine 1 3 1 thyroid cancer.
9.46
4a; lb; 2c; 3d
9.48
helium
9.50
a neutron (�n)
9.52
lithium-7
9.54
about 1 1 , 500 years ago
9.56
about 5730 years
9.58
They protect either by compounds that absorb the radiation such as oxybenzone or by compounds that reflect the radiation, such as titanium oxide.
9.60
Yes, they can detect small differences in the densities of tissues.
9.62
The reaction between a deuterium and a tritium nucleus produces a helium nucleus and a neutron. The total mass of the helium nucleus and neutron is less than the total mass of the deuterium and tritium nuclei. The difference in mass is converted to energy.
9.64
potassium-40
9.66
28 s
9.68
It is eliminated from the body long before the 602 days have passed.
9.70
A neutron is converted to a proton by emitting an electron.
9.72
2 1 .4 X 1 0- rnrem
9.74
curium-246
9.76
the lowest possible energy state
9.78
zirconium-90
20
Chapter 1 0
Solutions - for 1M
Organic Chemistry
Chapter 10 Organic C hemistry
1 0.6 Under each Lewis structure is its name and the number of its valence electrons. ( a)
•• ••
H-o-o-H •• ••
Hydro gen peroxide (14)
(d)
( e)
HJ HJ H-y-Q-yH H H
(h)
••
••
H H-p-t;-J- H H H
H (f) H-P-�: H
J
Iji Iji
( m)
9
• · ••
H-.Q-o-g-H
Carbonic acid (24)
••
(i)
H-y:y-H H H
Ethylene (12)
(1)
••
:�
H-G-H
Formaldehyde (12) .
Carbon dioxide (16)
(n )
••
Chloromethane (14) .
H-y-y-H H H
(k) •• 0=0=0 ••
·
I
Methylamine (14)
(j) H-a=o-H
•
Methanol (14)
••
Ethane (14)
••
••
(c)
Dimethyl ether (20)
Acetylene (10)
HJ H-y-R-H H
H-tt-rt- H H H
Hydrazine (14)
HJ H-P-.� H H
Methanethiol (14)
(g)
(b)
•• ••
H, :0: IT" H-yC-Q- H H ••
Acetic acid (24)
1 0. 8 In stable organic compounds, carbon must have four covalent bonds to other atoms. In (a), carbon has five bonds to other atoms. In (b), one carbon has four bonds to other atoms, but the second carbon has bonds to other atoms. 1 0. 1 0 Each carbon in benzene is surrounded by three regions of electron density. Therefore, predict all H-C-C and C-C-C bond angles to be 1 20°. A molecule of benzene is planar; all 1 2 atoms lie in the same plane. 1 0. 1 2 Wohler converted two inorganic compounds, an organic compound.
amm
onium chloride and silver cyanate, into urea, 21
Chapter 1 0
Solutions - for 1M
Organic Chemistry
1 0. 1 4 Hydrogens are added to each structural formula. Under each is its molecular formula. (a)
Y H3 CH3-D-i=D-i-CH -CH3 CS H12
(d)
� D-i3 -y H-C-H CH3 C4HaO
(e)
(g)
( h)
(b)
� CH3-D-i2-CH2-C.(}i C4Ha02
y H3 CH3f-C H 2-CH 2-NH2 CH3 Cs H1S N
pi f? CH3-CH-CH2-C.(}i C4Ha03
( c)
(f)
(i)
f( CH3-D-i2-C -CH3 C4HaO
� CH3fH -C D-H N-t 2 C:3H7N02 CH�CH-CH2-D-i C:3HsO
1 0. 1 6 Tertiary (3°) as applied to alcohols means that the carbon bearing the -OH group is bonded to three other carbon atoms. 1 0. 1 8 Tertiary (30) as applied to amines means that the nitrogen atom of the amine is bonded to three carbon atoms. 1 0.20 (a) One hydroxyl group and a carboxyl group. The hydroxyl group is a secondary alcohol. (b) Two hydroxyl groups. Each is a primary alcohol. (c) One amino group and one carboxyl group. The amino group is a primary amine. (d) Two hydroxyl groups and one carbonyl group. The hydroxyl groups belong to one primary alcohol and one secondary alcohol. The carbonyl group belongs to an aldehyde. (e) One carbonyl group and one carboxyl group. The carbonyl group belongs to a ketone (f) Two amino groups. Both are primary amines
22
------- ----
Chapter 1 0
Solutions - for 1M
Organic Chemistry
1 0.22 Here are structural formulas for each part. (a) The four primary alcohols:
yH3
CH 3-CH .cH 2 .Q-f2 -CH Z-Oi pi 3 CH 3 y -Oi2 -Oi 0i3
CH 3.Q-f2 -CH 2.Q-f2-CH 2 .Q-f yH3 CH 3.Q-f2 -C H .Q-f2 -Oi (b) The three secondary alcohols:
p-i
CH 3.Q-f-CH 2 .Q-f2 -CH3 (c) The one tertiary alcohol:
p-i
CH 3 f -Oi2 -CH3 CH3 1 0.24 Here are the eight carboxylic acids of molecular formula C 6H 12 0 2 .
f?
f?
f?
CH3-CH 2 -CH2-CH z-CH 2 -C-a-i C H 3-Oi2 -C H 2 yH -C.(]-I CH 3.Q-12 -yH .(]-I2 -C Q-I CH3 CH3 H3? � f? f? C H 3 fH -CH 2-CH 2 -C Q-I 0i3 -y H -«H -C.(]-I 0i3 -CH 2 y -C-a-i 0i3 CH3 H3C CH3
�
0i3 -CH 2 yH -C Q-I CH 2.Q-f3 1 0.26 Taxol was discovered in a survey of indigenous plants sponsored by the National Cancer Institute with the goal of discovering new chemicals for fighting cancer.
23
Chapter 1 0
Solutions - for IM
Organic Chemistry
�
1 0.28 Arrows point to atoms and show bond angles about each atom . 1Q 50 0 120° ./ 109.5°
�
; "
(a) CH3- CH 2 109.5°
(d)
1J? 0
/"
(b)
0i3-
109.5°
�3-6='c43
rr'/
2 -C-H
1 0.30 Following are structural formulas for each compound.
b'
(a) 0i3 -CH 2 .fri
(b) CH 3.fri2 -C-H
b'
(c) C H 3-C -CH3
1 0.32 Here are structural formulas for the three tertiary amines of molecular formula C S H 1 3 N.
?"i 3
CH3-N-CH 2.fri2 -CH3
?"i 3 CH3-N- YH .Q-f3
pi 3
0i3 -CH 2 -N -CH 2 -CH3
CH3
1 0.34 The O-H bond is the most polar. C-C and C=C bonds are the least polar.
24
Chapter 1 1
Solutions - for 1M
Alkanes and Cycloalkanes
Chapter 1 1 Alkanes and Cycloalkanes
1 1 . 1 0 Statements (a) and (b) are true. Statements (c) and (d) are false.
1 1 . 1 2 Structural formulas (a) & (g) and (d) & (e) represent identical compounds. Structural formulas (b) & (c) are one set of constitutional isomers. Structural formulas (f) & (h) are another set of constitutional isomers.
1 1 . 1 4 (a) The four alcohols of molecular formula C4H I 00 are:
Yl-i
CH3CH20i2CHp-i
D-i3CHCH20i3
yH3 CH30iCH20H
(b) The three tertiary (3°) amines of molecular formula C S H 1 3 N are:
C(H3 CH3NCH2CH2D-i3
C(H3 D-i3�CH3 D-i3
(c) The two aldehydes of molecular formula C 4HgO are:
ft
CH3CH2D-i2 D-i (d) The three ketones of molecular formula C S H 1 0 0 are:
h'
CH3CCH2CH2D-i3
h'
CH3CX(1-D-i3 CH3
(e) The four carboxylic acids of molecular formula C S H 1 0 0 2 are:
Q
CH3CH20i2CH�
h'
D-i 3yHCH2COH CH3
1 1 . 1 6 2-Methylpropane and 2-methylbutane.
25
Chapter 1 1
Solutions - for 1M
Alkanes and Cycloalkanes
1 1 . 1 8 Following are structural fonnulas for each alkane or cycloalkane: (a)
(c)
(e)
yH3 �H3 a-t3 c;x:H2D-CH2CH3 CH3 �H2a-t3 a-t3 ?-Ct-ItJl"i� H2a-t2 CH3 H3 C H3C CH3
(b)
(d)
a-t3 CH2CH2� CHp-i2CH2CH3 CH3a-tCH3
(f)
(h)
1 1 .20 (a) Propanone (b) Pentanal (d) Cyclohexene (e) Cyclohexanone
yH3 a-t3 c;n-t3 CH3 �H3 a-t3 c;n-t2 CH2�1-O-i 2CH2a-t 2 CH3 CH2a-t2CH2CH3 CH3 �H3 a-t3 CH2�H2CH3 CH3 CH2a-t3 r--r � CH2a-t3
(c) Decanoic acid (f) Cyclobutanol
1 1 .22 A confonnation is any three-dimensional arrangement of atoms in a molecule that results from rotation about a single bond. In this chapter we are concerned only with confonnations that result from rotation about carbon-carbon single bonds. 1 1 .24 The cis and trans isomers of 1 ,2-dimethylcyclopropane are drawn here in two ways. In the left drawing for each isomer, the ring is in the plane of the paper and methyl groups are above or below the plane. In the right drawing for each isomer, the ring is perpendicular to the plane of the paper and methyl groups are in the plane of the paper.
cis-l,2-Dimethylcyclopropane
trans-l,2-Dimethylcyclopropane
1 1 .26 When equatorial, a group is as far away as possible from other atoms of the cyclohexane ring. When axial, it comes very close to two hydrogen atoms on the same side of the ring. 26
Chapter 1 1
Solutions - for 1M
Alkanes and Cycloalkanes
1 1 .28 In the following drawing, all hydrogens are shown.
20 H
4
!JA--_ .�6
1 1 .3 0 Heptane, C 7 H 1 6 , has a boiling point of 98°C and a molecular weight of 1 00. Its molecular weight is approximately 5 . 5 times that of water. Although considerably smaller, water molecules are associated in the liquid by relatively strong forces of hydrogen bonding while the much larger heptane molecules are associated only by relatively weak London dispersion forces. 1 1 .32 Alkanes are insoluble in water. 1 1 .34 Boiling points of unbranched alkanes are related to their size; the greater their size the greater their boiling point. The relative increase in size per CH2 group is greatest between CH4 and CH3 CH 3 , and becomes progressively smaller as the molecular weight of the alkane increases. . Therefore, the increase in boiling point per added CH2 group is greatest between CH4 and CH3 CH 3 , and becomes progressively smaller as molecular weight increases. 1 1 .3 6 On a gram-per-gram basis, methane is the better source of heat energy.
Hydrocarbon
CH4 CH� CH 2CH�
Component of
Heat of Heat of combustion combusion (kcal/ g) (kcal/mol )
natural gas
212
13.3
LPG
530
12.0
1 1 . 3 8 Octane rating indicates the relative smoothness with which a gasoline blend burns. The higher the octane rating, the less engine knocking. 1 1 .40 It will produce even more engine knocking than heptane.
27
Chapter 1 1
Alkanes and Cycloalkanes
Solutions - for 1M
1 1 .42 (a) The longest chain is pentane. Its IUPAC name is 2-methylpentane. (b) The pentane chain is numbered incorrectly. Its IUPAC name is 2-methylpentane. (c) The longest chain is pentane. Its IUPAC name is 3-ethyl-3-methylpentane. (d) (e) (t). (g)
The longest chain is hexane. Its IUPAC name is 3 ,4-dimethylhexane. The longest chain is heptane. Its IUPAC name is 4-methylheptane. The longest chain is octane. Its IUPAC name is 3-ethyl-3-methyloctane. The ring is numbered incorrectly. Its IUPAC name is 1 , I -dimethylcyclopropane.
(h) The ring is numbered incorrectly. Its IUPAC name is l -ethyl-3-methylcyclohexane. 1 1 .44 (a) The -OH group on ring A is equatorial. (b) The -CH3 group at the junction of rings AlB is axial to both rings A and B. (c) The -CH 3 group at the junction of rings CID is axial to ring C. 1 1 .46 It is a liquid at room temperature.
28
Chapter 1 2
Solutions - for 1M
Alkenes and Alkynes
Chapter 12 Alkenes and All{)'nes
1 2. 1 2 In ethane, each carbon is surrounded by four regions of electron density; bond angles are 1 0-9. S o. In ethylene, each carbon is surrounded by three regions of electron density; bond angles are 1 20°. 1 2. 1 4 A saturated hydrocarbon contains only carbon-carbon single bonds. contains one or more carbon-carbon double or triple bonds. 1 2. 1 6 (a) I -Heptene
An
unsaturated hydrocarbon
(b) 1 ,4,4-Trimethylcyclopentene
( c) 1 ,3 -Dimethy lcyclohexene (e) 1 -0ctyne
(d) 2,4-Dimethyl-2-pentene (f) 2,2-Dimethyl-3-hexyne
1 2. 1 8 Both (a) fumaric and (b) aconitic acid have a trans configuration of the main carbon chain. 1 2.20 (a) Four alkenes of molecular formula C S H 1 0 do not show cis-trans isomerism.
CH2=D-CH20i2CH3
yH3 CH2=CCH2CH3
I-Pentene
2-Methyl-l-butene
yH3 CH2=D-C1-O-i3 '
3-Methyl-l-butene
yH3 . 0i3C=C1-O-i3
2-Methyl-2-butene .
(b) Only one alkene of molecular formula C S H 1 0 shows cis-traps isomerism.
H3q
H3C(.
,H
,G=Q
H
,(rQ
CH2D-i 3
H
trans-2-Pentene
o
d0i3
Cyclopentane
Methylcyclobutane
,D-i 2 CH3 H
cis-2-Pentene
J;0i3
(c) Four cycloalkanes of molecular formula C SH I O do not show cis-trans isomerism.
H3
X
0i3
l,l-Dimethyl cyclopropane
Ethyl cyclopropane
29
Chapter 1 2
Solutions - for 1M
Alkenes and Alkynes
qCH3
«�3
cis-l,2-Dimethylcyclopropane
trans-l,2-Dimethyl- . cyclopropane
(d) Only one cyc10alkane of molecular formula C SH 1 0 shows cis-trans isomerism.
CH3
D-i 3
1 2.22 The three isoprene units in farnesol are shown enclosed in boxes.
H3
H3
�
�
CH2G-i
1 2.24 Shown are the two different coiling patterns of famesol that might give the carbon skeleton of santonin. Points where the carbon skeleton of famesol are cross-linked are shown by dashed bonds.
H3
H
o
o
1 2.26 In an addition reaction of alkenes, the double bond is broken and two new groups are added,
�r
each forming a single bond to a different carbon of what was the double bond.
CH3CH=CH2
30
+
H3r
.. CH3a-i CH3
Chapter 1 2
12.28 (a)
Alkenes and Alkynes
O�
12.3 0 (a)(
r.H�
(b)
��
Solutions - for IM
O: c�
H3CH 2C; CH 2CH 3
and
).H�
2°
+
CH:3 % CH-� %
�
- CH3
CH3CH2CH-<4HCH3
30 (b)
( c) CH3 ( D-I2 ) 5
and
+
CH3CH2CH2-CHCH3
Both are 2° and equally stable (e)
( e) [
j
cr CH3 and Q Cl-l3
ctRr�r r.H�
12.32 (a)
12.36
(a)
9-1
CH3CH -CH2CH2CH3
(b)
(d )
+
<;l-I
D-i 3CH2 -CI-O-I2CH3
[< !- ] Cl-l 3
and
OOi�
r+ �
'--+-�
CH3 �CH=CH�CH3
D-I3
(b)
dCl-b I
or
----fH -.l 2 0
31
Chapter 1 2
Solutions - for 1M
Alkenes and Alkynes
1 2 .42 Pheromone is a chemical substance used by a member of a species to communicate with another member of the same species. 1 2.44
l .Ox
1 0- 1 2
g
of 1 1 - tetradecenyl acetate is equal to 2.4
x
1 0 9 molecules.
1 2.46 The all-trans isomer. 1 2.48 (a) V
=
poly(vinyl chloride) (b) PP
=
polypropylene(c) PS
=
polystyrene
1 2.50 (a) There are five alkenes with this carbon skeleton.
yH3 D-i 3C=CI-O-f2CH3
yH3 CH2=CXl-t2CH2CH3
2-Methyl-l-pentene
2-Methyl-2-pentene
<;rt3 CH3C�
C(H3 D-i 3C�
ICH3 ,a= q H H
,H ,a= q CH3 H
trans-4-Methyl-2-pentene
cis-4-Methyl-2-pentene
4-Methyl-l-pentene.
(b) There are two alkenes with this carbon skeleton.
yH3 CH2=ql-iCH3 D-i 3
2,3-Dimethyl-l-butene
2,3-Dimethyl-2-butene .
(c) There is only one alkene with this carbon skeleton.
9"i3 CH3YGH=CH2 D-i 3
3,3-Dimethyl-l-butene . 32
Chapter 1 2
Solutions - for 1M
Alkenes and Alkynes
1 2.52 The carbon skeletons are almost identical. The difference is that the 2nd carbon from each end of lycopene is bonded to the 7th carbon from its end to form six-membered rings
12.54 (a)
0
(b)
0- 0;3
(c)
Br O-Br �
1 2.56 Reagents are shown over each arrow.
6
r.t-I�
O o-a;� H2S04 ,
(d)
HI .
()=c�
0- 1
33
--
� -- --�--- ----- ------
Chapter 1 3 Alcohols, Ethers, and Thiols
Solutions - for 1M
Chapter 13 Alcohols, Ethers, and Thiols
1 3 .8 The difference is in the number of carbon atoms bonded to the carbon bearing the -OH group. For a primary alcohol, there is one. For a secondary alcohol there are two and for a tertiary alcohol, there are three. 1 3 . 1 0 (a) I -Pentanol
(b) 1 ,3-Propanediol
(c) 2-Buten- I -ol (e) cis- l ,2-Cyclohexanediol (f) I -Butanethiol
(d) 3-Methyl- l -butanol
1 3 . 1 2 (a) Dicyclopentyl ether
(b) Dipentyl ether
(c) Diisopropyl ether
1 3 . 1 4 In order of increasing boiling point, they are:
Cl-i3 D-i2 D-i 3
D-i 3 D-i2 OH
-42°C
78°C
D-i 3 Cl-i2 D-i2 D-i2 OH 1 1 7°C
1 3 . 1 6 I -Butanol has the higher boiling point. 1 3 . 1 8 Evaporation of a liquid from the surface of the skin cools because heat is absorbed from the skin in converting molecules from the liquid state to the gaseous state. 2-Propanol (isopropyl alcohol), which has a boiling point of 82°C, absorbs heat from the skin, evaporates rapidly, and has a cooling effect. 2-Hexanol, which has a boiling point of 1 40°C, does also absorbs heat from the surface of the skin but, because of its higher boiling point, evaporates much more slowly and does not have the same cooling effect as 2-propanol. 1 3 .20 The more soluble compound is circled. (a) (c)
34
[o-t3 o-tj CH (hI CH3�HJ CH3��3 CH3CH2CH2SH [CH3CH2CH20"1 or
3CXl-i 3 or
or
Solutions - for 1M
Chapter 1 3 Alcohols, Ethers, and Thiols
1 3.22 For three parts, there are two constitutional isomers that will give the desired alcohol. For two parts, only one alkene will give the desired alcohol. (b)
(l( Q-I3
0
or
(y CH2 V
Q-I3CH2CH=C�2CH3
( c) (e)
0
1 3.24 The first reaction is an acid-catalyzed dehydration, the second is an oxidation. (a)
CH3CH2CH2CHp-i
Hl;�i.
Q-I3CH2CH=CH2
+
H20
�
Q-I3CH2CH200H 1 3.26 The first two reactions are oxidations, the third is an acid-catalyzed dehydration, and the fourth is an acid-catalyzed hydration. (a)
h'
CH3(CH 2 ) 6 CDH
(c)
(y
V
Q-l 3
(d)
0
Oi
1 3.28 (a) The three function groups are a thiol group, a primary amino group, and a carboxyl group. (b) Oxidation of the thiol group gives a disulfide (-S-S-) group.
h'
fj>
HX � CH2S-SCH2«HCQ-I N-I 2 NH2 1 3.30 (a) Acid-catalyzed dehydration of I -propanol to propene followed by acid-catalyzed hydration to propene to 2-propanol.
Q-I3CH2CH2� CH3CH=CH2
+
H20
H2004
� Q-I3CH=CH2
+
H20
H2OO4• Q-I3CCH3 �
35
Solutions - for 1M
Chapter 1 3 Alcohols, Ethers, and Thiols
(b) Acid catalyzed hydration of propene to I -propanol followed by oxidation gives propanone.
CH3CH=CH2
+
C?-i
H20
D-i 3CHC H3
<jl-I H2SO4• CH3C t-O-t3
K2Cr207 b' CH3a-i CH3 H2SO;
1 3.32 Each can be prepared from I -propanol (circled) as shown in this flow chart.
1 3 .34 Nitroglycerin was discovered in 1 847, and is a pale yellow, oily liquid. 1 3 .3 6 One of the products the body derives from metabolism of nitroglycerin is nitric oxide, which causes the coronary artery to dilate, thus relieving angina. 1 3 .38 Disulfiram inhibits the body's normal metabolism of acetaldehyde, which accumulates with all sorts of unpleasant side effects. 1 3 .40 Dichromate ion is reddish-orange, chromium(III) ion is green. When breath containing ethanol is passed through a solution containing dichromate ion, ethanol is oxidized and dichromate ion is reduced to green chromium(III) ion. 1 3 .42 Diethyl ether is easy to use and causes excellent muscle relaxation. Blood pressure, pulse rate, and respiration are usually only slightly affected. Its chief drawbacks are its irritating effect on the respiratory passages and its aftereffect of nausea. Further, when mixed with air in the right proportions, it is explosive.
36
Solutions - for 1M
Chapter 1 3 Alcohols, Ethers, and Thiols
1 3 .44 The IUPAC name of halothane is 2-bromo-2-chloro- l , I , I -trifluoroethane. In this instance, the two-carbon chain is numbered from the end that gives the halogen substituents the lower set of numbers. 1 3 .46 The five ethers of molecular fonnula C S H 1 2 0 and their common names are:
0i30(}i2CH2CH20i3
pi3 0i30(}i2CH0i3
9"1 3 �30(}i(}i2CH3
Butyl methyl ether
Isobutyl methyl ether
sec-Butyl methyl ether .
�3 0i30 <;CH3 CH3
(}i 3CH 20 0i2CH20i3
�3 �3CH20(}i(}i3
tert-Butyl methylether
Ethyl propyl ether
Ethyl isopropyl ether
1 3 .48 Arranged in order of increasing boiling point, they are:
CH3CH20i2CH2CH2CH3
(}i3CH2CH2CH�H20H
HCXl-i2CH2CH2CHp-i
Hexane (MW 86 g/mol) bp 69°C
l-Pentanol (MW 88 g /mol) bp 138°C
l,4-Butanediol (MW 90 g/mol) bp 230°C
1 3 .50 The solvent with the greater solubility in water is circled.
( CH3(}i2OH] (b ) (}i3 (}i2OCH2 (}i3 (c) ( CH3CH20CH2CH3] or CH3(CH2b CH3 (a)
CH2C!2
or
or
[CH3CH20H ]
37
Solutions - for 1M
Chapter 1 4 Benzene and its Derivatives Chapter 14 Benzene and its Derivatives
1 4.2 An aromatic compound, as typified by benzene, is cyclic with a closed loop of six electrons arising from three carbon-carbon double or triple bonds. Alternatively, they may arise from two double bonds and an unshared pair of electrons on a nitrogen atom of the ring. 1 4.4 The simplest alkane contains one carbon, the simplest alkene and alkyne each contain only two carbons, and the simplest arene contains six carbons. (a)
Q-i4 Methane
(c)
(b ) Q-i2 =Q-i2
HG=D-i Ethyne (Acetylene )
Ethene ( Ethylene )
(d)
0
Benzene
14.6 A molecule of l ,4-dichlorobenzene is planar; that is, all carbon, hydrogen, and chlorine atoms lie in one plane. Because of its flatness, and because each atom of the ring has only one other group bonded to it, there is no possibility for cis-trans isomerism in a benzene ring
�a YD-i2CH3
1 4.8 Following are structural formulas for each compound. (a)
(b)
� Q-i3 Y Br
(I)
� /a
1 4. 1 0 The three possible products are :
l) 38
Qa
a-{ �OH a
Solutions - for 1M
Chapter 1 4 Benzene and its Derivatives
14. 1 2 (a) Nitration using HN0 3 1H2 S04 followed by sulfonation using H2 S04 . Or these steps may be performed in the reverse order. (b) Chlorination using Cl2IFeCl 3 followed by bromination using Br21FeC1 3 . Or these steps may be performed in reverse order. 1 4. 1 4 The letters DDT are derived from DichloroDiphenylTrichloroethane 1 4. 1 6 Given its lack of polarity, expect it to be insoluble in water. 1 4. 1 8 According to one dictionary definition, the term biodegrade means "to reduce to a lower organic type" . As used by us in this text, to be biodegraded means that organi,? molecules are broken apart by normal reaction pathways present in the environment to smaller molecules that can be assimilated into the environment without harm to the environment. 1 4.20 Iodine, or to be more precise, iodide ion, comes primarily from the sea. There are high concentrations of it in sea water, and in plants grown close to the sea where it has been deposited in the soil. Because of the uneven distribution of this element in the environment, iodide deficiency is more common in inland regions than in coastal regions. 14.22 There is one ether group and one phenol group in capsaicin. an ether
9 ��CH 3 H 0i3
1 4.24 Body odor is due to the formation of certain chemicals by the action of skin bacteria on the natural body secretions that are a part of perspiration. 14.26 Trichlosan is insoluble in water, but readily soluble in aqueous sodium hydroxide as its water soluble sodium salt.
39
Solutions - for 1M
Chapter 1 4 Benzene and its Derivatives
1 4.28 Following are structural fonnulas for each compound.
rn
(a)
�
(d) o,,
(c)
�
�
q
(f)
1 4.30 Following are structural fonnulas for the products. N� Br- � Br � � (c) (a) (b) "" I "" I
6
40
0-1 3
�
Br
�)
Br.
ai
Solutions - for 1M
Chapter 1 5 Chirality
Chapter 15 Chirality
1 5.8 Compounds (a), (c), and (d) contain stereocenters, here marked by asterisks. (a)
? D-i3 Q-\Q-\2D-i2 Q-\3
(c)
*
? D-i2=D-iD-iCH3
1 5. 1 0 Following are mirror images of each compound. ( a)
"
H30\\'" H
�
CH3 H
t-O�
�H (c) H2� �?� H � (?"' N-i2 CH3 D-i 3 (e ) "
;" �
_ _
(dl
�
D-i2D-i CH3 *
g-iO
: �H D-i2a-i
_ _
H Q ��
(� cx: H;o
).
Oi t-O
Qt-O
(b)� : ;..oII D-i CH 2 D-i
" " 1/
CXl) H Hen:;
(d)
*
yl 91
1 5 . 1 2 The groups in each set are arranged in order of increasing priority. (a) -H -CH 3 -CH2 0H -OH (b) -H -CH 3 -COOH -NH2 (c) -CH 3 -COOH -CH2 SH -NH2
,v .J. f �". f"'.J.(
1 �R S�N-C H3
1 5 . 1 4 The atom of highest priority on each stereocenter is shown along with its configuration.
.
0
0 H'''
_
. CH3
� II
41
Solutions - for 1M
Chapter 1 5 Chirality
1 5 . 1 6 Each stereocenter is labeled with an asterisk and the number of possible stereoisomers shown below the formula.
CH3 p-ty H:OOi HO a-t 4 *
(a)
2
*
2
(b)
=
� 2 - croH :fi-cco-t
4 H:rOi ccx:1-i -
2
2
=
(f)
(DOi
( h)
1 5 . 1 8 A racemic mixture is an equal mixture of enantiomers. It will not rotate the plane of polarized light. 1 5 .20 The physiologically form of ibuprofen is the S enantiomer. The R enantiomer has no biological activity, but is converted over time to the active S enantiomer. 1 5 .22 Of the eight carboxylic acids of molecular formula C 6 H 1 2 0 2 , only three are chiral.
t(
CH3CH20i2 P-COi CH3 *
42
f?
o-t 3CH2Y'-o-I 2 CXl-i CH3 *
�
CH3pi�o-t H3C CH3 *
Solutions - for 1M
Chapter 1 5 Chirality
1 5 .24 The eight stereo centers are shown by asterisks.
013
1 5.26 (a) In this chair conformation of glucose, carbons 1 , 2, 3 , 4, and 5 are stereocenters. (b) There are 2 5 32 stereoisomers possible. (c) Because enantiomers always occur in pairs, there are 1 6 pairs of enantiomers possible. =
1 5.28 The only chiral compounds are (e) menthol which has 2 3 8 stereoisomers, (h) halothane which has 2 1 2 stereoisomers, and (k) thyroxine which has 2 1 2 stereoisomers. =
=
=
1 5.30 There are eight stereo centers in triamcinolone acetonide; 2 8
=
256 stereoisomers are possible.
Triamcinolone acetonide
43
Solutions - for 1M
Chapter 1 6 Amines
Chapter 16 Amines
1 6.6 Primary, secondary, and tertiary amines have one, two, and three carbon groups respectively bonded to the amine nitrogen. Primary, secondary, and tertiary alcohols have one, two, and three carbon groups respectively bonded to the carbon bearing the -OH group. 1 6.8 In pyrrole, the unshared pair on nitrogen and the second bond of each double bond provide the aromatic sextet. 1 6. 1 0 Each amine is identified by type. primary aliphatic amine -,
H
(a )
�
I
I
( H2N-Q-� Q-I2a;3 primary aromatic amine
D-i2CH2NH2
(b) heterocyclic aromatic amine
/"" tertiary � Cjl"i 3 , aliphatic amine N HD-i( CH2bN(G.2 H 5h
secondary aro�ati amme (c )
C
heterocyclic aromatic amine
1 6. 1 2 For the structural formulas for these eight amines, review your answer to Problem 1 0.25. There are four primary amines of this molecular formula, three secondary amines, and one tertiary amine.
3 yH3 yH3 9H 1 0 amines: CH3CH2CH2CHlf'i2 CH3CH2D-iNH2 D-i3CHCH2NH2 CH3yN H2 CH3 3 <jl"i 2 0 amines: D-i 3CH2CH2NHCH3 CH3CH�H3 D-i3CH2N I-O-i2CH3
44
Solutions - for 1M
Chapter 1 6 Amines
1 6. 1 4 The intermolecular attractive force due to hydrogen bonding is greater between alcohols than it is between amines because of the greater polarity of an O-H bond compared with an N-H bond. 1 6. 1 6 (a) Ethyl arnmonium chloride (b) Diethylammonium chloride (c) Anilinium hydrogen sulfate 1 6. 1 8 In each part, the stronger base is the aliphatic amine; the aromatic amine is the weaker base. The stronger base is circled.
(C)
N
Y 0-1
NH2 or
3
1 6.20 Following are structural formulas for the product of each acid-base reaction. H 3 � 3 I (a) 0-13 CO(c) PhCH2 a-NI t-O-I 3 (b) PhCH2 o-I N H + C I " 3 + I H H
b' Q
f
H9J4 -
:(J
1 6.22 (a) The primary aliphatic amine is the stronger base and forms the salt with HCl. The salt is named pyridoxamine hydrochloride. 2 NH 3+ CI H .p 0-I 2 Q-1
H3
I
1 6.24 Tamoxifen contains three aromatic (benzene) rings, one carbon-carbon double bond, one ether, and one tertiary amine. Two stereoisomers are possible, a pair of cis-trans isomers. 1 6.26 Possible negative effects are long periods of sleeplessness, loss of weight, and paranoia.
45
Chapter 1 6 Arnines
Solutions - for 1M
1 6.28 Both coniine and nicotine have one stereocenter, and two possible stereoisomers (one pair of enantiomers). 1 6. 3 0 The four stereocenters of cocaine are marked with asterisks. Following is the structural formula of the salt formed by reaction of cocaine with HCl. H 3G.. + H c r
1 6.32 Neither Librium nor Valium is chiral. 1 6.34 Morphine contains one tertiary aliphatic amine, one phenolic hydroxyl group, one secondary alcohol, and one ether. 1 6.36 In addition, each contains a benzene ring bonded to the quaternary carbon. 1 6. 3 8 Their salts are more soluble in water and in body fluids, and are more stable (less reactive) toward oxidation by atmospheric oxygen. 1 6.40 Butane, the least polar molecule has the lowest boiling point and I -propanol, the most polar molecule, has the highest boiling point. CH3CH2CH 2 CH 3 -O.5°C 1 6.42 The alcohol will be unchanged. The amine will react with HCI to form a salt. CH 3CH 2 0i 2 NH3+ CI 1 6.44 (a) Procaine has no stereocenters; it is achiral. (b) The tertiary aliphatic amine is the more basic nitrogen, and the one that reacts with HCI to form a salt. H CI� H2 CX:X:; H 20i 2 0i 2 CH3
�
�
CH20i 3
Procaine hydrochloride (Novocaine)
46
Solutions - for 1M
Chapter 1 7 Aldehydes and Ketones
Chapter 17 Aldehydes and Ketones
1 7. 1 0 Each is prepared by oxidation of the starting alcohol by a mixture of potassium dichromate in sulfuric acid.
1 7. 1 2 Compounds (b), (c), (d), and (f) contain carbonyl groups.
1 7 . 1 6 (a) 4-Heptanone
(b) 2-Methylcyc1opentanone
(c) trans-2-Methyl-2-butenal
(d) (S)-2:-Hydroxypropanal
(e) 1 -Phenyl-2-propanone
(f) Hexanedial
1 7. 1 8 Because it is a hydrogen bond acceptor, the carbonyl group is hydrophilic; it increases solubility in water. The hydrocarbon portions of each molecule are hydrophobic; they decrease solubility in water. For acetone, the hydrophilic character of the carbonyl group outweighs the hydrophobic character of its two methyl group, and acetone is soluble in water. For 4-heptanone, the hydrophobic character of the two propyl groups outweighs the hydrophilic character of its carbonyl group, and 4heptanone is insoluble in water.
47
� �-��-��� � _ �_�__ _ _ _ _ _ _ _ _ _ _ _ ___ -C �
Solutions - for 1M
Chapter 1 7 Aldehydes and Ketones
1 7.20 Acetaldehyde fonns hydrogen bonds with water primarily through its carbonyl oxygen. o- H
H
\
,"
0+0= 05 /
H 3C
,
I
J�+
,.
�
+
o- H
1 7.22 A hemiacetal contains a carbon atom bonded to one -OH group and one -OR group, where R may be an alkyl or aryl group. An acetal contains a carbon atom bonded to two -OR groups, where R may be alkyl or aryl. 1 7.24 (a) A hemiacetal
(b) (f)
(e) An acetal
An
An
acetal (c) Neither
(d) A hemiacetal
acetal
1 7.26 Following are structural formulas for the products of each hydrolysis.
9
( a) CH3CH2�H2 a-t 3
+
2 CH3CH2Q;
(b) CH3�9 Oi � _ (d )
C�
+
H
CH3CHp-i
1 7.28 For (a) and (c), there is only one enol form. For (b) and (d), there are two enol forms.
p-t
9"i
(e)
( }-�
<;J-f
(b) CH 2=CCH2CH 3 and CH3C=OiCH 3
(a ) CH3CH= CH
a-c H 3
(d )
&
a-t 3
and
&
CH 3
1 7.30 Aldehydes are oxidized by these reagent to carboxylic acids. Ketones are not oxidized under these conditions. Secondary alcohols are oxidized to ketones.
�
( a ) CH3CH2a-t 2CXl-i
0- 9
(b) � #
OOH
(c) No reaction
( d)
1 7.32 Treat each with Tollens' reagent. Only pentanal will give a silver mirror. 48
00
Solutions - for 1M
Chapter 1 7 Aldehydes and Ketones
1 7.34 Reduction adds one hydrogen to the carbonyl carbon and another to the carbonyl oxygen. Reduction of - CHO must, therefore, give -CH2 0H, a primary alcohol. Reduction of R2 C=O must give R2 CHOH, which is a secondary alcohol. 1 7.36 (a) Following is its structural formula. (b) Because it has two hydroxyl groups and one carbonyl group, all of which can interact with water molecules by hydrogen bonding, predict that it is soluble in water. Its reduction gives 1 ,2,3-propanetriol, better known as glycerol.
f?
Ha::l-i 2CCH 2a-t
+
H2
1,3-Dihydroxy-2-propanone (Dihydroxyacetone)
metal catalyst
?i
I-CC H2a-iCH20H
•
1,2,3-Propanetriol (Glycerol )
1 7. 3 8 The first reaction is a reduction. There is no reaction in (b) or (c); ketones are not oxidized by these reagents.
( }-� CHa
1 7.40 Compounds (a), (b), (c), and (e) can be prepared by reduction of an appropriate aldehyde or ketone. Compound (d) cannot be prepared by this method.
�Ha
(b)
( )-�
� -H
( e)
�O
1 7.42 Reagents for each conversion are shown along the reaction arrows.
f?
(a) CsHsCCH 2 CH3
(b)
H2 /M etal •
<jl-i
CS HSC t-Oi 2 CH3
H 2S04. heat
CsHsCH=Q-ICH3
()=O
1 7.44 LD 50 is the dose of a chemical required to kill one -half of a test- animal group in 1 4 days. 1 7.46 It will require 1 5 . 1 mglkg x 0.35 kg = 5.3 mg of AS2 0 3 per mouse.
49
Solutions - for 1M
Chapter 1 7 Aldehydes and Ketones
�
1 7.48 The aldehyde and ketone groups in each are circled.
( a)
[J� �
(d)
2cH2
3
� �
�)
(e)
[Rl v ""
H2 Cl-l 3
�
C 3 (f) H
\ #
�
� C
<;l-I
y
(c) CH3CCH2 CH3 CH3
9
( f) 0-I3C«= C H2 0-1 3
1 7.52 The intermolecular attraction due to hydrogen bonding between molecules of I -propanol is stronger than the attraction between molecules of propanal. 1 7.54 Shown in the equation are structural formulas for the equilibrium products.
f? fH
a-rn r- --- CH3
-
.....
An a-hydroxyaldehyde
An enediol
An a-hydroxyketone.
1 7.56 Only pentanal will reduce Tollens' reagent and give a silver mirror.
50
Solutions - for 1M
Chapter 1 8 Carboxylic Acids and Their Derivatives
Chapter 18 Carboxylic Acids and Their Derivatives
1 8. 1 0 (a) 3 ,7-Dimethyloctanoic acid (b) 2-Aminopentanoic acid (c) Hexanoic acid (d) 2-Hydroxybutanedioic acid 1 8. 1 2 Following are structural formulas for each salt. (a)
0- 9CO-
Na+
� #
9 (b) Cl-i 3CO- L i +
9
�a-t
CO- Na+
(e)
1 8. 1 4 One of the carboxyl groups in this salt is present as the anion, the other as -COOH.
9 9
t-a:;-CO- K+ 1 8. 1 6 Hydrogen bonding is between the carbonyl oxygen of one carboxyl group and the hydroxyl group of the other. If you draw this molecule correctly to show this the internal hydrogen bonding, you will see the hydrogen-bonded part of the molecule has the form of a six-membered ring.
9
9 r H� H., O �O\� H2
�
hyd rogen b onding
cS-
1 8. 1 8 In order of increasing boiling point, they are: . (a)
p
CH3(CH 2) 6 Cl-i
CH3 ( Cl-i 2 ) 6 Cl-i 2OH
bp 171 °C (b)
CH3CH2CCH2CH3 bp 35°C
bp 239 °C
P
CH3CH:Pi2CH20H
CH3CH2COi
b p 117 °C
bp 141°C
1 8 .20 Following are structural formulas for the indicated starting material. (a)
CH3 (CH 2) 4 CH20H Cs H 1 40
(b )
fj>
CH3 ( Cl-i 2 ) 4 Cl-i C6 H 1 20
(c)
I-O CH2 (Cl-i 2 )4 Cl-i2a-J Cs H 1 4 02 51
j
Chapter 1 8 Carboxylic Acids and Their Derivatives
Solutions - for 1M
1 8.22 (a) At pH 2.0 or lower (even more acidic) acetic acid is present in solution entirely in its un ionized form, CH 3 COOH. When enough NaOH or other strong base is added to bring the pH of the solution to 4.0 - 5 .0, which is approximately equal to the pKa of acetic acid (PKa 4.76), then there are approximately equal amounts of CH 3 COOH and CH 3 COO - present in solution. When more =
NaOH is added to the solution to bring its pH to 8.0 (basic), acetic acid is present entirely as CH 3 COO - . 1 8.24 The pKa of ascorbic acid is 4. 1 0. In a solution whose pH is 4. 1 0, ascorbic acid is be present 50% as un-ionized ascorbic acid and 50% as ascorbate anion. When the pH of the solution is 7.3 5-7.45, which is slightly on the basic side, ascorbic acid is present primarily as ascorbate anion. 1 8.26 Parts (a), (b), and (c) are acid base reactions. Part (d) is Fischer esterification. (b ) PhCH2Cm- Na+ (a ) PhCH2COO- Na+ + CO2 (c)
PhCH2CDO- NH 4+
(d)
PhCH2CmCH3
1 8.28 Following is the structural formula of methyl salicylate.
CC � �
f?c..
OCH3
a;
1 8.3 0 Following is a structural formula of 5-hydroxypentanoic acid and the five-membered lactone (cyclic ester) it forms.
1 8.32 (a) Benzoic anhydride (b) Methyl decanoate (c) N-Methylhexanamide (d) 4-Aminobenzamide (e) Cyclopentyl acetate (f) Ethyl 3 -oxobutanoate 1 8.34 Acetic acid, which has the possibility for intermolecular association by hydrogen bonding, has the higher boiling point. 1 8.3 6 Only the carboxylic acid (a) will produce bubbles of C0 2 .
52
Solutions - for 1M
Chapter 1 8 Carboxylic Acids and Their Derivatives
0-oqm2 f? )4 C H3
1 8. 3 8 Following are equations for the synthesis of each ester by Fischer esterification. (a)
o-m
+
f?
I-CC (CH2)4 m3
H+
•
+
H20
1 8.42 Following is a balanced equation for this synthesis of phenacetin.
99
CH3CXXX:; H3
+
� OCH2CH3 . H21� ..
•
9
� CCH2CH3 CH3CI� ...
9
CH3cm
Phenacetin
4-Ethoxyaniline
Acetic anhydride
+
1 8.44 Saponification is the hydrolysis of an ester using aqueous NaOH or KOH to give the sodium or potassium salt of the carboxylic acid and an alcohol. Following is a balanced equation for the saponification of methyl acetate.
9
m3�H3
+
Nam
Methyl acetate
Sodium acetate
1 8.46 Each reaction brings about hydrolysis of the amide bond. Each product is shown as it would exist under the specified reaction conditions. (a)
(b )
0-90\H2 � #
0-9I"'t.LJ2 �
\.\
"
II
VI .....
+
+
HCI NaQH
0CO H � # 9 H2O . . 0-9 CO-
H2O
•
� #
+
NH 4+ C I "
Na+
+
NH 3
53
•
Chapter 1 8 Carboxylic Acids and Their Derivatives
Solutions - for 1M
1 8.48 First convert 3-methylbenzoic acid to its ethyl ester by Fischer esterification, and then treat the ester with diethylamine to give Deet.
9 H3� COH
V
+
H0Q-I2CH3
+
HN ( a-i2CH3b�
()
9
H3�(J\J ( a-i2 a-i 3 ) 2
+
Hc:x:H2CH3
Deet
1 8.50 Following are structural formulas for the moni-, di-, and triesters.
9 Ho- r- 0Cl-i2CH3 OH Ethyl phosphate
CH3CH2o- r-c:x:H 2CH3 0Q-I2CH3
Diethyl phosphate
Triethyl phosphate
1 8.52 Two molecules of water are split out.
9 Ho- r- OH OH
+
9 Ho- r- OH OH
+
�
� Ho- r- oa-i2CH3 0Cl-i2CH3
b' Ho- r- OH OH
•
9 9 9 Ho- r- o- r- o- � OH OH OH OH
+
2 H20
1 8 .54 The compound is salicin. Removal of the glucose unit and oxidation of the primary alcohol to a carboxylic acid gives salicylic acid. 1 8.56 Both aspirin and ibuprofen contain a carboxylic acid and a benzene ring. Both aspirin and naproxen also contain a carboxylic acid and a benzene ring; the case of naproxen the benzene ring is a part of a naphthalene ring. 1 8.58 Both contain the carbon-oxygen skeleton of coumarin.
54
1
I
Chapter 1 8 Carboxylic Acids and Their Derivatives
Solutions - for 1M
1 8.60 Hydrogen bonding with water is possible through the oxygen of the carbonyl group and each hydrogen of each amide group. Only one of the possible N-H------0 hydrogen bonds is shown.
,H
+ ,H H-,,\ 0cS-11111111 cSH-O + ,0= �cS-111111cSH"\ 0
�
H
1 8.62 The �-lactam portion of penicillin is indicated by an arrow, and three stereo centers are marked with asterisks.
H
vn� Q<� � -lactam rin g
mOH
1 8.64 The three functional groups in procaine are a primary aromatic amine, an ester, and a tertiary aliphatic amine. 1 8.66 The functional groups in lidocaine are an amide and a tertiary aliphatic amine. The functional groups in etidocaine are also an amide and a tertiary aliphatic amine. 1 8.68 (a) Its IUPAC name is 3,5-tetradecadienoic acid. (b) Each carbon-carbon double bond has the possibility for cis-trans isomerism and, therefore, there are four cis-trans isomers. 1 8.70 There are two carbon-carbon double bonds in the six-carbon chain, each with a trans configuration. The carboxyl group is present as its anion. The structural formula of this salt is drawn in two ways here.
�
�
H 3c' c' I H
�
�
c' I
H
P
c...
0- K+
55
J
Solutions - for 1M
Chapter 1 8 Carboxylic Acids and Their Derivatives
1 8.72 Following are structural formulas for each compound arranged in order of increasing boiling point.
f?
CH 3 ( CH 2) S Q-I
9
CH3 (CH2) S CH20H
CH 3(CH 2) 4 CQ-I
1-Heptanol bp 1 76°C
Hexanoic acid bp 205°C
Heptanal b p 153 °C
1 8.74 Primarily as un-ionized lactic acid. 1 8.76 Add each to an aqueous solution of sodium bicarbonate, NaHC0 3 . Only the octanoic acid will react with this solution. It will dissolve in aqueous NaHC0 3 as a water-soluble salt with the evolution of carbon dioxide. There will be no reaction with ethyl octanoate. 1 8.78 First convert p-aminobenzoic acid to its ethyl ester by Fischer esterification, and then treat the ester with 2-diethylaminoethanol to give procaine. Step 1 : Formation of the ethyl ester
9
� COH H 2 ,,1�
+
CH3CH2Q-1
esterification p-Aminob enzoic acid Step 2: Conversion of the ethyl ester to an amide
H 2 1,,�9 � COCH2Q-13
+
HCXJ-i 2CH2N (CH2Q-13h
Ethyl p-aminob enzoate
Ethyl p-aminob enzoate
�
2-Diethylaminoethanol
��f?
H 2 1 � COCH2CH2 N ( CH2 CH3 )2 Procaine (Novocaine)
1 8 .80 Hydrolysis gives one mole of 2,3-dihydroxypropanoic acid and two moles of phosphoric acid. At pH 7.35 - 7.45, the carboxyl group is present as its anion, and phosphoric acid is present as its dianion
o II
o
CO-
gHOH I
2 CH 0H
56
+
2HO- � -0I
0-
Chapter 1 9 Carbohydrates
Solutions - for 1M
Chapter 19 Carbohydrates
1 9.6 Carbohydrates make up about three fourths of the dry weight of plants. Less than 1 percent of the body weight of humans is made up of carbohydrates. 1 9. 8 The designations D and L refer to the configuration of the stereocenter farthest from the aldehyde or ketone group on a monosaccharide chain. When the monosaccharide chain is drawn as a Fischer projection, a D-monosaccharide has the -OH on this carbon on the right; an L monosaccharide has it on the left.
1 9. 1 0 (a)
An
aldose is a monosaccharide in which its carbonyl group is present as an aldehyde.
(b) A ketose is a monosaccharide in which its carbonyl group is present as a ketone. (c)
a1dopentose is a five-carbon monosaccharide in which the carbonyl group is present as an aldehyde.
An
(d) A ketohexose is a six-carbon monosaccharide in which the carbonyl group is present as a ketone. 1 9. 1 2 Compounds (a) and (c) are D-monosaccharides. Compound (b) is an L-monosaccharide. 1 9. 1 4 An anomeric carbon is the hemiacetal or acetal carbon in the cyclic form of a monosaccharide. Said another way, it was the carbonyl carbon in the open-chain form of the monosaccharide. 1 9. 1 6 The anomeric carbon of a 2-ketohexose is carbon 2. 1 9. 1 8 The designation ocmeans that the -OH on the anomeric carbon of a cyclic hemiacetal is on the same side of the ring as the terminal -CH2 0H group. The designation f3 means that it is on the opposite side of the ring from the terminal -CH2 0H group.
Solutions - for 1M
Chapter 1 9 Carbohydrates
1 9.20 First compare each Haworth projection with that of t3-D-glucose. Compound (a) differs in configuration at carbon 3 . Compound (b) differs in configuration at carbons 2 and 3 . 0 m m
2
2
m
a-i
..
OH Q-i
OH
..
m m m 2Q-i
OH a-i
D-Allose
H 2 a-i m
..
(b )
H
GO
..
m m 2Q-i
OH H
OH H
1 9.22 During mutarotation,
an
D-Altrose
ocor t3 form of a carbohydrate is converted to an equilibrium mixture of
the two. Mutarotation can be detected by observing the change in optical activity over time as the two forms equilibrate 1 9.24 D-Glucose. 1 9.26 In reduction of the ketone group of D-fructose, the new -OH group may be either on the right or the left in the Fischer projection.
yH2a-i
CH2a-i
H20i H H Q-i m CH20i
D-Fructose
D-Mannitol
i� 58
H2 / N i
..
+
H
2m m H m m o-t 2 m
D-Glucitol
Solutions - for 1M
Chapter 1 9 Carbohydrates
1 9.28 Ribitol is the reduction product of D-Ribose. �-D-Ribose- l -phosphate is the phosphoric ester of the OH group on the anomeric carbon of �-D-ribofuranose.
$
CH20i Ribitol
Q-I HO �-D-rib ofuranose-l-phosphate . (�-D-ribose-l-phosphate )
1 9. 3 0 No, glycosides cannot undergo mutarotation because the anomeric carbon is not free to interconvert between oc and � configurations via the open-chain aldehyde or ketone. 1 9.32 (a) Both monosaccharide units are D-glucose. (d) It is a reducing sugar. D- glucose unit D-glucose unit y-carbon forming the hemiacetal
1 9.34 (a) No, it is not a reducing sugar. Both anomeric carbons participate in forming the glycosidic bond between the two monosaccharide units. (b) No, it will not undergo mutarotation for the reason given in (a). (c) Both monosaccharides are units of D- glucose. 1 9.36 Cellulose fibers are insoluble in water because the strength of hydrogen bonding of a cellulose molecule in the fiber with surface water molecules is not sufficient to overcome the intermolecular forces that hold it in the fiber.
59
Solutions - for 1M
Chapter 1 9 Carbohydrates
1 9. 3 8 (a) In these structural formula, the CH 3 CO (the acetyl group) is abbreviated Ac 2 CJ-i
N-i Ac
OH ( (3)
( (3 )
H H
NHAc
Following are chair and Haworth structures for this repeating disaccharide. (3-1,4-glycosidic bond
N-i Ac H
N HAc
1 9.40 Its lubricating power decreases. 1 9.42 With maturation, children develop an enzyme capable of metabolizing galactose. They are, thus, able to tolerate galactose as they mature. 1 9.44 The difference is two hydrogens. L-ascorbic acid is the reduced form of this vitamin; L dehydroascorbic acid is its oxidized form. The L indicates that each has the L configuration at carbon 5 . 1 9.46 L-Ascorbic acid i s biological reducing agent. 1 9.48 The reaction catalyzed by glucose oxidase is oxidation by 0 2 of the anomeric carbon of (3-D glucose to a carboxyl group. 1 9. 5 0 Mixing bloods of these types will result in coagulation. 1 9.52
60
Lubrication of joints and the gel that maintains the retina in its proper position.
Solutions - for 1M
Chapter 1 9 Carbohydrates
1 9.54 Consult Table 1 9. 1 for the structural fonnula of D-altrose and draw it. Then replace the -OH groups on carbons 2 and 6 by hydrogens.
H Cl-i Cl-i Cl-i CH2Cl-i
H
CH CH CH
6(;H3
2,6-Dideoxy-D-altrose (D-Digitose )
D-Altrose
1 9.56 Drawn first is a Haworth projection of oc-D-glucopyranose, and then of the solutions to part (a) and (b).
H
Cl-i
a-D-Glucopyranose
H a-D-Mannopyranose
OH H a-D-Gulopyranose
1 9.58 The monosaccharide unit in salicin is D-glucose.
61
Solutions - for 1M
Chapter 20
Lipids
Chapter 20
Lipids
20.2
Fats have a higher energy yield per carbon atom than carbohydrates, the other main form of energy storage in the body.
20.4
There are three. One of them is
This one has oleic acid in the middle. The other two have myristic and palmitic acids in the middle. 20.6
(a)
F
A
St St
[
St
& B
(b) there are seven (c) this is F :
62
F
C
01 01
[
01
rn D
F F
E
01
St
St
01
F
[
St
rn G
Chapter 20 20.8
Solutions - for 1M
Lipids
(a) stearic acid, because it has a greater molecular weight (b) arachidic acid, because saturated acids have higher melting points . than unsaturated ones (see text).
20. 1 0 Among similar compounds melting points increase with increasing molecular weight. 20. 1 2 human fat. 20. 1 4 3 moles 20. 1 6 (a) 1 5
(b) 1 0
(c) 5
20. 1 8 The cis double bonds in the unsaturated fatty acids prevent the regular packing of the long aliphatic portions of the lipid bilayers, making them fluid-like. 20.20 Peripheral membrane proteins only touch the surface of the membrane; integral membrane proteins penetrate the lipid double layer - they are partly or fully embedded in it. 20.22
CH2 - O
o II
--
OH
P -- 0
I
O· 20.24 sphingomyelins, glycolipids 20.26 Fluidity. Cholesterol will not fit into the regular zig-zag packing of the fatty acid chains. Therefore, it disturbs the crystalline (high- melting) portions in a manner similar to the cis double bonds of unsaturated fatty acids. 20.28 in gallstones 20.3 0 the D ring 20.3 2 The LDL binds to the LDL- receptor molecules in the coated pits. It is then taken inside the cells where the enzymes liberate cholesterol. 63
Chapter 20
Lipids
Solutions - for 1M
20.34 ( 1 ) The COCH3 side chain is converted to OH. (2) Ring A loses the CH3 group and becomes aromatic; its 0 is converted to -OH. =
20.3 6 The functional group on C- I I does not participate in the binding o f drug to receptor. 20.3 8 Bile salts are synthesized from cholesterol, so they use up cholesterol which means that less is deposited in the form of plaques. 20.40 (a) 3 alcohol groups; amide; carboxylate ion (b) 3 ketone groups; 2 alcohol groups, C C double bond (c) ketone group; 2 alcohol groups; carboxylic acid; 2 C C bonds (d) 2 alcohol groups; 4 C C double bonds; carboxylic acid =
=
=
20.42 Aspirin also prevents the synthesis of thromboxane which helps the platelet aggregation necessary in clot formation. 20.44 Beeswax is a simple ester of fatty acid and a long chain alcohol. Ear wax contains trigycerides, esters of glycerol and three fatty acids, among other lipid compounds. 20.46 Gap junctions are cluster of proteins in a membrane that form a pore. The small size of the pore filters out large molecules such as proteins. 20.48 Degradation of the complex lipid must start with the monosaccharide at the end of the chain. In Fabry' s disease this is a oc-D- galactose. 20.50 Methenolone has an extra CH3 group and the double bond is in a different place. 20.52 It inhibits the enzyme phospholipase A2. 20.54 Cox- l synthesizes prostaglandins needed for normal physiological functions. COX-2 synthesizes prostaglandins that cause inflammation. 20.5 6 Taurine i s a constituent o f bile salts and as such emulsifies lipids and helps enzymes to break them down. 20. 5 8 I t plays a role i n the implantation o f the fertilized ovum i n the uterus. 20.60 stearic acid, sphingosine, phosphate, and choline 20.62
An
active transporter needs energy derived from an ATP molecule to carry ions across the membranes against a concentration gradient. Facilitated transporter does not need the energy of ATP molecule to carry ions, but they are not against a concentration gradient.
20.64 It is the only one with an aldehyde group instead of a methyl connected to C- 1 3 . 20.66 0.75 g
Chapter 2 1
Proteins
Chapter 21
Proteins
Solutions - for 1M
2 1 .4
fibrous proteins
2 1 .6
nonpolar: (b)(f)(g) polar but neutral: (d) (e) acidic: (c) basic : (a)
21 .8
phenylalanine, tyrosine, tryptophan, and histidine.
2 1 . 1 0 cysteine 2 1 . 1 2 isoleucine, threonine 2 1 . 1 4 (a) It is not chiral (b) It is a dimeric molecule (c) It contains a phenolic group (d) There is no NH2 group; instead the N is bonded to the R. 2 1 . 1 6 Ionic compounds are solids at room temperature. Amino acids, being zwitterions, are ionic compounds. 2 1 . 1 8 At low pH there is a positive charge; at high pH a negative charge; at the isoelectric point both charges are present (zwitterion). 2 1 .20 Aspartic acid is at the N-terminal and histidine is at the C-terminal. 2 1 .22 (a) Val-Phe-Trp-Asn
(b) C-terminal end: Asn N-terminal end: Val
2 1 .24 (a) The series of peptide linkages - C - NH - CH - C- NH - CH-
"
O
I
R
"
O
I
R
(b) the amino acid with the free NH2 group at the end of the peptide chain 2 1 .26 (a) six (b) C - C - N - C
� A
2 1 .28 Proteins contain both positive and negative charges; they can react with acids, taking a proton (the COO- becomes COOH), or a base by donating a proton (NH3 + becomes NH2 ), and the pH will not change much.
2 1 .32 1 8 residues 2 1 .34 (b) 65
Chapter 2 1
Proteins
Solutions - for 1M
2 1 . 3 6 The side chain o f proline i s part o f a ring. This restricts its flexibility. An
2 1 .38
intramolecular bond goes between two parts of the same chain. An intermolecular bond goes between two different molecules. Intermolecular bonds are found in the B-pleated sheet; intramolecular bonds in the a-helix and in some antiparallel pleated sheets.
2 1 .40
The B chains would no longer be connected to the A chains The tertiary structure of both chains would be completely different.
2 1 .42 (a) Fetal hemolglobin does not have beta chains instead it has gamma chains. (b) Fetal hemoglobin.
2 1 .44
2 1 .46 cystine is converted to cysteine 2 1 .48 Ag+ denatures the proteins of bacteria and kills them. 2 1 .5 0 The glutamic acid i s linked to the cysteine by its y-carboxyl group rather than the a oc-carboxyl group. 2 1 .52 Hunger, sweating and poor coordination are symptoms of hypoglycemic awareness signaling low level of blood sugar. 2 1 .54 Hydoxyurea therapy promotes the manufacture of fetal hemoglobin, which does not carry beta chains, and therefore the sickle cell mutations. Cell with fetal hemoglobin do not sickle. 2 1 .56 Normal prion has alpha helical conformation, the amyloid prion has beta-sheet conformation. 2 1 . 5 8 The globular G-actin polymerizes and forms a double helix made o f G-actins. 2 1 .60 The heat produced by the energy of the laser beam seals the bleeding blood vessels. 2 1 .62 no
Chapter 2 1
Solutions - for 1M
Proteins
2 1 .64 Yes. Any change that destroys native protein structure without hydrolyzing the chain is a denaturation.
2 1 .68 No; that number is 2060 = 1 078 , even if we limit it to 60 amino acids. This number is possibly greater than the total number of atoms in the universe. 2 1 .70 20 1 0 = 1 .0 X 1 0 1
3 +
+
2 1 .72 NH3 - CH - COOH
I
CH2CH2 - S - CH3
+
NH3 - CH - COOH
I I CH - CH3 I CH2
CH3
+
NH3 - CH - COOH
I
CH2 - COOH
67
Chapter 22
Enzymes
Chapter 22
Enzymes
Solutions - for 1M
22.2
They are made of ribonucleic acids.
22.4
An
enzyme-catalyzed reactions have lower activation energy than the uncatalyzed reaction.
22.6
trypsin
22. 8
(a)
22. 1 0 (a) deaminations
(b) hydrolyses
22. 1 2 (a) aponenzyme
(b) coenzyme
(c) hydrogen removal
(d) isomerizations
22. 1 4 Hg2+ reacts with -SH groups to form -S-Hg-S- cross links. This changes the shape of active site, resulting in a noncompetitive inhibition. 22. 1 6 No.
An
increase of 1 0· C may decrease enzyme activity or even inactivate the enzyme.
22. 1 8 At 8 S · C the enzyme is irreversibly denatured. At - 1 0· C the structure of the enzyme is not changed; reversible inhibition occurs because not enough energy is available for the reaction. 22.20 (c) 22.22 The other amino acids are necessary to hold the enzyme into exactly that 3-dimensional shape into which the substrate can fit. If significant changes were made in the structure, the enzyme activity would decrease or disappear. 22.24 Yes. This is feedback control. 22.26 It would degrade the proteins of our bodies. 22.28 an enzyme that is active in more than one form 22. 3 0 There has been a heart attack, but no hepatitis. 22.32 the liver 22.34 It is denatured by stomach acid and partially digested, thus becoming inactive. 22 .36 Succinylcholine relaxes muscles, making it easier to insert a breathing tube. 2 2 . 3 8 It has a special enzyme, urease, which converts urea to ammonia. The ammonia neutralizes the acid in the vicinity of the bacteria which lives in that basic protective cloud.
Chapter 22
Solutions - for 1M
Enzymes
2 , 22.40 The Mn + is coordinated by the 0 of glutamic and aspartic acid in the active site and by the carbonyl and carboxyl O-s of the pyruvate. 22.42 para-aminobenzoic acid 22.44 Tylenol inhibits the peroxidase activity and aspirin inhibits the cyclooxygenase activity. 22.46 because dead heart muscles spill their enzyme contents into the serum 22.48 They are both basic amino acids. 22.5 0 Since maximum activity is slightly above p H 7 , the enzyme works best in blood. 22.52 hydro lases 22.54 (a) ethanol dehydrogenase
(b) esterase
22.5 6 Isozymes
69
Solutions - for 1M
Chapter 23
Chemical Communication; Neurotransmitters and Hormones
Chapter 23
Chemical Communication; Neurotransmitters and Hormones
23.2
an electrical signal
23.4
(a)(c)(d)(f) See Sec. 23.2 (b)(e) See Sec 23 . 1
23 .6
Hormone. It has to travel a longer distance from gland to neuron or other cells
23 . 8
(a) (b) small molecules derived from amino acids (c) peptides (d) steroids
23 . 1 0 the choline end 23 . 1 2 C-3 and C-5 23 . 1 4 Glutamic acid is removed by a transporter molecule which brings it back through the presynaptic membrane unto the neuron. 23 . 1 6 NMDA is a D-configuration and aspartic acid is of L-configuration. NMDA, furthermore is substituted on the amino nitrogen with a methyl group. 23. 1 8 A phosphate anhydride bond is hydrolyzed and a phosphate ester bond is formed with the C-3 of ribose. 23.20
o
II
o
o
II
rr - p - o - p - o - p - o - c�
I
o·
I
o· OR OR
II
Chapter 23
Chemical Communication; Neurotransmitters and Hormones
Solutions - for 1M
23.22 The cAMP is destroyed by the enzyme phosphodiesterase. 23 .24 the last step, when the ion gates are opened 23.26 It is oxidized by monoamine oxidases. 23.28 No, the two compounds block different histamine receptors. 23 .30 It blocks the pain receptors. 23 .32 Calmodulin is a calcium binding protein, only when calcium ion is bound to it can calcium activate protein kinaseII. 23 .34 It prevents release of acetylcholine from the presynaptic vessels. 23 .36 It is normally water soluble, but becomes insoluble in people with Alzheimer' s disease. 23 . 3 8 Dopamine cannot penetrate th e blood-brain barrier 23 .40 NO helps to dilate blood vessels; dilated blood vessels in the head cause headache. 23 .42 They contribute to the death of neurons. 23.44 Insulin-dependent patients do not manufacture enough insulin. Non-insulin-dependent have enough insulin, but not enough receptors to which insulin can bind. 23 .46 Aldosterone, a steroid, penetrates the plasma membranes of the target cell and the nucleus, affecting mineral metabolism. 23 .48 Yes, decamethonium is a competitive inhibitor of acetylcholinesterase. 23 .50 In alpha-alanine the NH2 group is on the 2 position of propanoic acid; in beta-alanine it is on 3-position. 23 .52 (Box 24E) The relaxed smooth muscles around the arteries allow more blood to flow. This alleviates angina pains of the heart and temporarily remedies impotence. 23 .54 Transferase synthesizes acetylcholine; esterase breaks it down.
71
Chapter 23 23 .56
HN '/
,
o 0-
II
o
II
II
o
...... ,/
N
II
HzN - � ,/ ...... "' N N o
II
\
I
HN '/
,
/ o --. 'O -
P - O - P - O - P - O - CHz I O· o· O·
I
Solutions - for 1M
Chemical Communication; Neurotransmitters and Hormones
II
,
O· OH OH
II
o
" ,/
N
II
H2N - � ,/ ...... "' N N o
II
,
- O - P - O - CHZ
\ / o +
'0 -
I
O·
o· OH OH
23 .58 It increases the concentration of serotonin by preventing its removal from the receptor.
72
II
p - O'
Chapter 24
Nucleic Acids and Heredity
Chapter 24
Nucleic Acids and Heredity
Solutions - for 1M
24.2
chromosomes
24.4
(a) in the chromosomes (b) outside the chromosomes
24.6
o '0
II
-
P
I
-
o 0
II
-
P
o'
I
-
o
o 0
-
'0
CH2
o'
II
-
P
I
-
o'
0
II
-
P
I
-
0
-
CH2
o'
OH OH
OH OH
No, these are not part of nucleic acids. 24.8
guanine, thymine, uracil, cytosine
24. 1 0 adenine, cytosine, guanine 24. 1 2 A nucleoside contains a sugar and a base. A nucleotide also contains a phosphate 24. 1 4 a chain o f nucleotides (deoxyribose and phosphate units) 24. 1 6 C-3 and C-5 24. 1 8 It allowed the builders of the molecular model to pair A with T and C with G. 24.20 hydrogen bonds between the base pairs 24.22 Because of hydrogen bonding, only A fits opposite T, and only G fits opposite C. 24.24 See Fig. 24.5 24.26 thousands 24.28 They are unwinding proteins. They attach themselves to one strand of the DNA and promote the separation of the two strands of DNA.
73
Chapter 24
Nucleic Acids and Heredity
Solutions - for 1M
24.3 0 i n the 5 ' � 3' 24.32 Ribozymes, they act mostly in self-splicing of introns when nascent mRNA is altered to make mature mRNA 24.34 The information in DNA is transferred to RNA, and is then expressed in the structure of proteins. 24.3 6 outside the nucleus 24.3 8
An exon is a portion of a DNA molecule that codes for proteins; an intron is a portion that does not.
24.40 (a) about 3 percent (b) Some of it codes for RNA, and some are regulatory sequences, but the function of most of it is unknown. 24.42 5' GCAGTAGGCCAC-OH 24.44 The anticancer drug fluorouracil interferes with the methylating enzyme that manufactures thymidine, thereby inhibiting new DNA synthesis. 24.46 The lagging strand of the Okazaki fragments cannot be synthesized at the end of the DNA molecule. 24.48 by a technique called gel electrophoresis 24.50 the bacterium Haemophylus injluenzae 24. 52 Caspases attack the companion of endonuclease. Once the companion is degraded, the endonuclease can penetrate the nucleus and cleave the DNA there. 24.54 Where the substrate, mRNA or tRNA, to be cut binds 24.56 because only two of the four strands of the products are newly formed 24. 5 8 See Sec. 24.2
74
Chapter 25
Protein Synthesis
Chapter 25
Protein Synthesis
Solutions - for 1M
25 .2
The initiation signal is located on the DNA.
25.4
in the nucleus
25.6
the recognition site (anticodon) and the 3' tenninal to which the specific amino acid is attached
25. 8
327- 2 (for initiation and tennination) = 325
25. 1 0 The amino acid is attached to the 3 ' end of tRNA, the energy comes from the hydrolysis of ATP molecule. 25. 1 2 (a) It supports the mRNA. (b) It provides the A and P sites for proteins synthesis. 25. l 4 DNA and mRNA 25. 1 6 a functional unit containing DNA sequences that are structural genes, regulatory genes and control sites 25. 1 8 electrostatic interactions and hydrogen bonding 25 .20 By cutting out damaged areas and resynthesizing them. 25.22 No; but most carcinogens are also mutagens. 25.24 the ends of a strand of DNA that has been cut by endonucleases to leave several free bases that can pair up with a complementary section. 25 .26 In principal yes, but it has not been achieved as yet. 25.28 It is the protein envelope of the virus. The ingredients come from the host cell. 25.30 (a) reverse transcriptase (b) HIV-l protease 25 .32 It is a site in the primary structure of a protein which has the same amino acid at that position in all species investigated. 25.34 p53 is a tumor suppressor protein. When it undergoes mutation it cannot prevent cancer. The mutated fonns show up in 40% of all cancers. 25 .36 See Sec. 25.6
75
Chapter 25
Protein Synthesis
Solutions - for 1M
25 . 3 8 only by cautioning carriers o f defective genes to select noncarrier partners or not to have children 25.40 (a) a certain circular double-stranded DNA molecule (b) A gene is a double-stranded non-circular DNA molecule, or a portion of such a molecule, that codes for one protein molecule. 25.42 All have a G in the first position of the codon; they have different bases in the second position of the codon and the third position is variable. 25 .44 Ala-Glu-Val-Glu-Val-Trp
76
Solutions - for 1M
Chapter 26
Bioenergetics; How the Body Converts Food to Energy
Chapter 26
Bioenergetics; How the Body Converts Food to Energy
26.2
(a) two (b) the citric acid cycle and oxidative phosphorylation
26.4
See Sec. 26.2
26.6
(a) in the matrix (b) They are imbedded in the inner membranes of the mitochondria
26.8
adenosine and inorganic phosphate
26. 1 0 No, the storage is only for about a minute. 26. 1 2 The two electrons are in the bond connecting the newly arrived hydrogen to the 4 position of the pyridine ring. 26. 1 4 (a) flavin (b) ADP 26. 1 6 In NAD+ it is ribose; in FAD it is ribitol, a sugar alcohol (the reduced form of ribose). 26. 1 8 (a) nicotinamide (b) riboflavin (c) pantothenic acid 26.20 a thiol ester 26.22 (a) the numbers tell how many carbon atoms are in each compound (b) acetate, oxaloacetate, and citrate, in the order given 26.24 steps
77
���� ---- ---- -----
Chapter 26
Bioenergetics; How the Body Converts Food to Energy
Solutions - for 1M
26.40 at the favoprotein, Q enzyme and cytochrome c, stages 26.42 It is lower . 26.44 because energy built into ATP is generated by the osmotic pressure of the accumulated protons across the semipermeable inner membrane of the mitochondrion. 26.46 (a) 36 (b) 6 26.48 nerve conduction 26.50 A compound which allows the H+ ions to diffuse through a membrane without the expenditure of energy. 26.52 In a stroke the brain is temporarily deprived of oxygen. When oxygen is administered (ischemial-reperfusion), dangerous superoxide anions are formed. Superoxide dismutase decomposes the superoxide anions to peroxide and catalase decomposes the peroxide to harmless water and oxygen. 26.54 (a) GTP has guanine instead of adenine. (b) about the same 26.5 6 I n citrate the alcohol group i s a tertiary alcohol, i n isocitrate i t i s a secondary alcohol. 26.5 8 ( 1 ) oc-Ketoglutaric acid i s made by oxidative decarboxylation o f isocitric acid. (2) Oxaloacetic acid is made by oxidation of malic acid. (All these are in the form of their salts.) 26.60 (a) 3 (b) 2 26.62 No. The C02 comes from the alcohol end of the isocitrate, which was originally the oxaloacetate that combined with acetyl CoA.
78
Chapter 27
Specific Catabolic Pathways: Carbohydrate, Lipid, and Protein Metabolism
Chapter 27
Specific Catabolic Pathways: Carbohydrate, Lipid, and P rotein Metabolism
Solutions - for 1M
27.2
glycerol and fatty acids or monoglycerides
27.4
The glucose must be chemically activated before any subsequent step can take place.
27.6
Three; hexokinase, phosphofructokinase, pyruvate kinase
27.8
glyceraldehyde 3-phosphate and dihydroxyacetone phosphate
27. 1 0 oxidative decarboxylation 27. 1 2 (a) energy consuming (b) energy yielding 27. 1 4 (a) 2 (b) O 27. 1 6 (a) 6 (b) 6 (c) 3 6 27. 1 8 Glycolysis is the specific pathway by which the body gets energy from monosaccharides. Glycogenolysis is the pathway in which glycogen breaks down to glucose. 27.20 ATP, because hydrolysis of an anhydride (P-O-P) linkage yields more energy than hydrolysis of an ester (C- O-P) linkage 27.22 (a) the mitochondria (b) they are transported by camitine acyltransferase 27.24 (a) 5 turns (b) 7 turns 27.26 stearic acid. For the oleic acid step 2 is not needed on one turn of the B-oxydation spiral. Therefore, oleic acid yield one less FADH2 molecule and consequently 2 less ATP molecules than stearic acid. 27.28 Fats, because their catabolism produces more ATP molecules than carbohydrates, per carbon atom. 27.3 0 Yes, they c an be metabolized via the common catabolic pathway. 27.3 2 transaminases 27.34 They are toxic. 27.3 6
amm
onia and citrulline
79
Chapter 27
Specific Catabolic Pathways: Carbohydrate, Lipid, and Protein Metabolism
Solutions - for 1M
27. 3 8 (a) NH/ (b) by converting it to urea in the urea cycle 27.40 The breakdown product of heme, bilirubin, is normally removed from the blood stream by the liver, later to be excreted by the gallbladder and intestines. High bilirubin levels in the blood indicate that the liver is malfunctioning because it is not sufficiently removing the bilirubin. 27.42 the globin 27.44 When epinephrine occupies its receptor it activates the ligand binding/ G-protein! adenylate cyclase cascade. This in tum phosphorylates the muscle enzyme, phosphorylase. This form of the enzyme is the active form which cleaves glucose 1 phosphate unit from glycogen, eventually depleting it. 27.46 Yes; she could smell acetone in the patient' s breath indicating ketoacidosis, a sign of diabetes not under control. 27.48 The primary structure is essentially the same in yeast and in humans. 27.50 by ubiquitin targeting 27.52 bilirubin 27.54 (a) Lysine has one more CH2 group. (b) no 27.56 2 moles of ATP 27. 5 8 Glucose and sequentially, glucose 6-phosphate i s shunted to the pentose phosphate pathway which produces ribose5-phosphate. 27.60 PEP is able to transfer phosphate to ADP and form one anhydride linkage, producing ATP in step
80
Chapter 2 8
B iosynthetic Pathways
Chapter 28
B iosynthetic Pathways
28.2
28.4 28.6 28.8
To have flexibility. pathway .
Solutions - for 1M
If a pathway is blocked, a key component can be synthesized through another
Ko, most catabolic reactions take p lac e in the mitochondria; anabolic reactions in the cytosol. (a) carbon dioxide (b) water (c) sunlight by
forming UDP-glucose
28. 1 0
the
28. 1 4
acetyl
same
28. 1 2 (a) the number o f glucose units in the chain (b) a large number; more than a thousand CoA
28. 1 6 No. It is a carrier protein. 28. 1 8 (a) fatty acid synthase (b) It picks up a C2 fragment from ACP, combines it with a C3 fragment attached to another ACP, and forms a C4 compound, releasing C02. 28.20 a thiol group 28.22 (c)(e) 28.24 when excess food is available 28.26 glyceroI 3-phosphate, palmitoyl CoA, lauryl CoA, and serine 28.28 geranyl and farnesyl pyrophosphates. 28.30 oxidative deamination 28.32 NH2 - C - CH2 - C - COO-
II
o
II
0 28.34 The chlorophylls have magnesium, while heme has iron. 28.36 (a) indigo blue (b) oxidation of tryptophan 28.3 8 a thiol, -SH, group 28.40 phenylalanine 28.42 four; one in the formation of glycerol 3-phosphate, one each for the two acyl CoA molecules and the fourth for the activation of choline. 81
Chapter 29
Nutrition and Digestion
Chapter 29
Nutrition and Digestion
29.2
(a) no (b) It is a carboxylic ester.
29.4
No, they are equally nutritious
29.6
bread, cereal, rice, and pasta (starches)
29. 8
the average daily requirement o f calories for a resting body
Solutions - for 1M
29. 1 0 hypertension, cardiovascular disease, and diabetes 29. 1 2 fats 29. 1 4 (a) 2 (b) 1 0 29. 1 6 Add animal proteins to the diet. 29. 1 8 (a) 1 200 to 1 500 mL (b) none 29.20 night blindness 29.22 (a) vitamin B I 2 (b) part of methyl-removing enzyme in folate metabolism 29.24 meat and dairy products 29.26 degradation of the nervous system 29.28 maltose 29.30 No; amylose does not contain branches. 29.32 Trypsin hydrolyzes proteins only at an arginine or lysine residue; HCI does it randomly. 29.34 No. High dosages of niacin have been shown to cause health problems 29.36 (a) Sucralose and acetsulfame-K are not metabolized. (b) Phenylalanine, aspartic acid and methanol. 29. 3 8 Eat some candy o r other sugar-containing food 29.40 pantothenic acid
82
Chapter 29
Nutrition and Digestion
Solutions - for 1M
29.42 They contain the essential amino acids we cannot synthesize ourselves. 29.44 yes; fats, oils, and sweets 29.46 Nuts contain minerals such as Mn, Zn, Mg, and K, vitamins including niacin, biotin, and vitamin E, and are also a rich source of essential fatty acids, carbohydrates, and proteins. 29.48 Pepsin. Removal of part of the stomach reduces natural pepsin production. 29.50 It is a cofactor necessary for the operation of certain enzymes.
83
Chapter 30
Immunochemistry
Chapter 30
Imm unochemistry
Solutions - for 1M
30.2
The adaptive immunity we call the immune system
30.4
Redness, selling, heat and pain
30.6
Both of them originate in the bone marrow.
30.8
Memory cells are specialized T cells, which remain in the blood stream to fight infection the second time around. By remembering the original antigen they speed up and strengthen the immune response.
30. 1 0 No; it is too small of a molecule. 30. 1 2 epitope 30. 1 4 none whatsoever. ABO antigens are polysaccharide antigens. HMC binds only peptide antigens. 30. 1 6 one to two weeks 30. 1 8 (a) IgE (b) It plays a part in allergic reactions. 30.20 the Fab fragments; they carry the variable regions that bind the antigen. 30.22 Many communication molecules, such as cell adhesion molecules, have structures similar to that of immunoglobulins. They form a superfamily of compounds. 30.24 intermolecular attractions: London dispersion forces, dipole-dipole interactions, and hydrogen bonds 30.26 They both belong to the immunoglobulin superfamily; they both bind antigens with the variable (V) region. 30.28 Two polypeptide chains are cross linked by intermolecular disulfide bridges. 30.30 The antigen must be broken down to small peptides which contain the epitope. The peptides must be presented to the TcR complex by major histocompatibility complex molecule, MHC. 30.32 1. TcR 2. MHC and 3. CD. TcR is the receptor on the surface of leukocytes; MHC brings the antigen from the cytoplasm to the surface and presents it to the TcR complex; CD-s act to help to bind MHC to the receptor and also signal the presence of the antigen delivered by MHC. 30.34 Cytokines have their own receptors on the surface of target cells. They do not interact with antigens.
84
Chapter 30
Immunochemistry
Solutions - for 1M
30.36 Chemokines are a subgroup of cytokines, the chemotactic cytokines. As the name implies they attract leukocytes to the site of injury. They do so by binding to special receptors on the leukocytes. These receptors have associated G protein which amplifies the message. 30.38 They are classified by the nature of the secondary structure of the protein; (a) contain only oc-helical structure (b) containing only B-sheet and (c) containing both oc and B-structures 30.40 (Box 30A) The patient's immune system mistook the acetylcholine receptors at the junction of nerve and muscle cells as a foreign body. Antibodies against these receptors are manufactured and these block the signal transmission between neurons and muscle cells. 30.42 (Box 30B) Non-Hodgkin lymphoma cells have a unique marker antigen, CD20, on their surface. Rituxan is an antibody which developed precisely against this marker antigen. Thus when the drugs gets in the lymph and circulates there, it binds to the surface of the lymphoma. 30.44 (Box 30C) He performed a daring experiment, immunizing a child with cow pox and injecting him with smallpox. The child did not get the disease. More likely than not, today he could not do such an experiment legally. Probably animal studies must come first, before one can try out the experiment on humans. 30.46 (Box 30D) Special adhesion molecules, selectins, on the surface of both leukocytes and endothelial cells provide adhesion and subsequent release. This keeps the leukocyte rolling along the walls of the blood vessels. 30.48 the IgA immunoglobulins 30.50 (Box 30D) A monoclonal antibody hones in to only one type of targeted cell, for example, one breast cancer carrying a unique antigen. Polyclonal antibodies have many type of target cells not necessarily only cancer cells. 30.52 No, the light chains contain both C as well as V regions. 30.54 four
85
Chapter 3 1
Body Fluids
Chapter 31
Body Fluids
Solutions - for IM
3 1 .2
about 5%
3 1 .4
erythrocytes, leukocytes, platelets or thrombocytes
3 1 .6
leukocytes
3 1 .8
(a)(b) in the bone marrow (c) in the lymph nodes and spleen (d) in the bone marrow and spleen
3 1 . 1 0 (a) to provide proper osmotic pressure (b) to provide immune reaction (c) blood clotting 3 1 . 1 2 When blood plasma is allowed to stand it will clot. When we squeeze the clot the clear liquid obtained is the serum. 3 1 . 1 4 75 % saturation, which means that, on the average, three 02 molecules are bound. 31.16 4
=
1 00 mm Hg; 3
=
36 mm Hg; 2
=
27 mm Hg; 1
=
1 8 mm Hg; 0
=
0 mm Hg
3 1 . 1 8 (a) It decreases (b) Bohr effect 3 1 .20 four 3 1 .22 Glomureli are the fine blood vessels entering and leaving the Bowman's capsule that is part of the kidneys filtration unit, called nephron. 3 1 .24 urea, creatine, creatinine, hippuric acid, and ammonia 3 1 .26 They are neutralized by HC0 3 - ions in the blood. 3 1 .28 Aldosterone enhances the reabsorption of sodium ions in the kidneys, thus compensating for their loss in the sweat. 3 1 .3 0 Angiotensin, the active form of angiotensinogen, constricts blood vessels, increasing blood pressure, Captopril inhibits formation of angiotensin from angiotensinogen. 3 1 .32 (a) Thrombin cuts fibrinogen and allows the fibrin to form clot (b) Vitamin K is needed for the production of prothrombin (c) Thromboplastic substances activate prothrombin 3 1 .34 (a) blood plasma, interstitial fluid (b) intracellular fluid
86
Chapter 3 1
Body Fluids
Solutions - for 1M
3 1 .3 6 a lower concentration of estrogens, which help absorb calcium from the diet 3 1 .3 8 Sodium ions are surrounded by water molecules (hydration). The lower the [Na+ ] in the body, the less water is needed, decreasing the amount of water and the blood pressure. 3 l .40 creatine 3 l .42 no 3 1 .44 When kidneys fail, no filtration of blood occurs, and urine excretion diminishes. Water accumulates in the interstitial fluid, hence the swelling. 3 1 .46 It converts angiotensinogen to angiotensin, which increases blood pressure.
87
