,Solution manual For Essential Biochemistry
5th Edition Charlotte W. Pratt
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, Solutions
Chapter 1 11. a. 11-Cyclohexylundecanoate is a lipid. b. Selenocysteine is an
amino acid. c. Galactose is a monosaccharide. d. O6-Methylguano-
1. a. carboxylic acid; b. amine; c. ester; d. alcohol. sine monophosphate is a nucleotide.
2. a. ether; b. phosphoric acid ester; c. thiol; d. ketone. 12. The compound contains both a lipid and a monosaccharide.
3. a. aldehyde; b. imine; c. thioester; d. diphosphoric acid. [From Malyarenko, T.V. et al., Molecules 23, 1077 (2018).]
4. a. alcohol; b. aldehyde; c. carboxylic acid. 13. a. C and H plus some O. b. C, H, and O. c. C, H, O, and N plus
5. small amounts of S.
Hydroxyl group
OH 14. It is a lipid (it is actually lecithin). It is mostly C and H, with
Imino group relatively little O and only one N and one P. It has too little O to be
a carbohydrate, too little N to be a protein, and too little P to be a
N nucleic acid.
CH3
O 15. You should measure the nitrogen content, since this would indi-
cate the presence of protein (neither lipids nor carbohydrates contain
Ether linkage appreciable amounts of nitrogen).
[From Li, S.-Y. et al., Eur. J. Med. Chem. 71, 36–45 (2014).] 16. You could add the compound that contains the most nitrogen,
6. Ester linkage compound B, which is melamine. (Melamine is a substance that in
OH Carboxyl
O group the past was added to some pet foods and milk products from China
HO CH O O so that they would appear to contain more protein. Melamine is toxic
C
C CH C OH to pets and children.) Compound C is an amino acid, so it would
H2
C C already be present in protein-containing food.
Hydroxyl N
group HO OH 17. A diet high in protein results in a high urea concentration, since
Nicotinic acid (niacin)
Vitamin C urea is the body’s method of ridding itself of extra nitrogen. Nitrogen
is found in proteins but is not found in significant amounts in lipids
O Carbonyl group
or carbohydrates. A low-protein diet provides the patient with just
O CH3
enough protein for tissue repair and growth. In the absence of excess
Ether linkage H3C CH3
protein consumption, urea production decreases, and this puts less
H3C
O (CH2 CH C CH2)10H strain on the patient’s weakened kidneys.
Carbonyl group O
18. Asn has an amido group and Cys has a sulfhydryl group.
Coenzyme Q 19. Serine has a hydroxyl group and lysine does not.
7. Sulfhydryl 20. a.
HN CH2 CH2 SH O
group
+H
C O 3N CH C O−
CH2 CH2
CH2
Amido group O
−O
NH Amino group P O−
C O NH2 O
Hydroxyl HO C H N b. O
N
group
H3C C CH3 O O +H
3N CH C O−
N N
CH2 O P O P O CH2 O CH2
O− −O
H H CH2
H H
HC OH
Phosphoanhydride Hydroxyl
O OH
linkage group CH2
−O P O−
NH+
3
Phosphoryl
O
group c. O
8. A thioester linkage is formed. +H
3N CH2 C O−
9. Amino acids, monosaccharides, nucleotides, and lipids are the (CH2)4
four types of biological small molecules. Amino acids, monosaccha-
NH
rides, and nucleotides can form polymers of proteins, polysaccha-
rides, and nucleic acids, respectively. C O
10. a. N-Acetylglucosamine is a monosaccharide. b. CMP is a nucleo- CH3 1
tide. c. Homocysteine is an amino acid. d. Cholesteryl ester is a lipid.
,2 SOLU TION S
21. O 34. a. decrease; b. increase.
Carbonyl 35. The polymeric molecule is more ordered and thus has less
group C H
entropy. A mixture of constituent monomers has a large number of
H OH different arrangements (like the balls scattered on a pool table) and
HO H Hydroxyl group thus has greater entropy.
H OH 36. Entropy increases as the reactants (7 molecules) are converted to
products (12 molecules).
H OH
37. The dissolution of ammonium nitrate in water is a highly endo-
CH2 OH thermic process, as indicated by the positive value of ∆H. This means
that when ammonium nitrate dissolves in water, the system absorbs
22. a. Fructose has the same molecular formula, C 6 H 12 O 6 , as
heat from the surroundings and the surroundings become cold. The
glucose. b. Fructose is a ketone, whereas glucose is an aldehyde.
plastic bag containing the ammonium nitrate becomes cold and can
23. Uracil has a carbonyl functional group, whereas cytosine has an be used as a cold pack to treat an injury.
amino functional group.
38. The dissolution of calcium chloride in water is a highly exother-
24. Nucleotides consist of a five-carbon sugar, a nitrogenous ring, mic process, as indicated by the negative value of ∆H. This means
and one or more phosphoryl groups linked covalently together. that when calcium chloride dissolves in water, the system loses heat
25. a. glycosidic; b. peptide; c. phosphoester. to the surroundings and the surroundings become warm. The plastic
26. Aspartame consists of two amino acids linked by a peptide bond. bag holding the calcium chloride solution becomes warm and can be
used as a hot pack by the camper at cold temperatures.
27. As described in the text, palmitate and cholesterol are highly non-
polar and are therefore insoluble in water. Both are highly aliphatic. 39. The dissolution of urea in water is an endothermic process and
Alanine is water soluble because its amino group and c arboxylate has a positive ∆H value. In order for the process to be spontaneous,
groups are ionized, which render the molecule “saltlike.” Glucose the process must also have a positive ∆S value in order for the free
is also water soluble because its carbonyl group and many hydroxyl energy change of the process to be negative. Solutions have a higher
groups are able to form hydrogen bonds with water. degree of entropy than the solvent and solute alone.
28. Glucose has several hydroxyl groups and is a polar molecule. 40. As combustion reactions proceed with a release of free energy,
As such, it will have difficulty crossing the nonpolar membrane. the reaction shown in Problem 36 is expected to be exergonic.
The 2,4-dinitrophenol molecule consists of a substituted ben- 41. Calculate ∆H and ∆S, as described in Sample Calculation 1.1:
zene ring and has a greater nonpolar character. Of the two molecules,
∆H = HB − HA
the 2 ,4-dinitrophenol will traverse the membrane more easily.
∆H = 60 kJ · mol−1 − 54 kJ · mol−1
29. DNA forms a more regular structure because DNA consists of
∆H = 6 kJ · mol−1
only four different nucleotides, whereas proteins are made up of as
many as 20 different amino acids. In addition, the 20 amino acids ∆S = SB − SA
have much more individual variation in their structures than do ∆S = 43 J · K−1 · mol−1 − 22 J · K−1 · mol−1
the four nucleotides. Both of these factors result in a more regular ∆S = 21 J · K−1 · mol−1
structure for DNA. The cellular role of DNA relies on the sequence
42. a. ∆G = ∆H − T∆S
of the nucleotides that make up the DNA, not on the overall shape
∆G = (6000 J · mol−1) − (4 + 273 K) (21 J · K−1 · mol−1)
of the DNA molecule itself. Proteins, on the other hand, fold into
unique shapes, as illustrated by endothelin in Figure 1.4. The ability ∆G = 180 J · mol−1
of proteins to fold into a wide variety of shapes means that proteins The reaction is not favorable at 4°C.
can also serve a wide variety of biochemical roles in the cell. Accord- b. ∆G = (6000 J · mol−1) − (37 + 273 K) (21 J · K−1 · mol−1)
ing to Table 1.2, the major roles of proteins in the cell are to carry out
∆G = −510 J · mol−1
metabolic reactions and to support cellular structures.
The reaction is favorable at 37°C.
30. Polysaccharides serve as fuel-storage molecules and can also
43. ∆G = ∆H − T∆S
serve as structural support for the cell.
0 > 15,000 J · mol−1 − (T) (51 J · K−1 · mol−1)
31. Pancreatic amylase is unable to digest the glycosidic bonds that −15,000 > −(T) (51 K−1)
link the glucose residues in cellulose. Figure 1.6 shows the structural
15,000 < (T) (51 K−1)
differences between starch and cellulose. Pancreatic amylase binds to
294 K < T
starch prior to catalyzing the hydrolysis of the glycosidic bond; thus
The reaction is favorable at temperatures of 21°C and higher.
the enzyme and the starch must have shapes that are complemen-
tary. The enzyme would be unable to bind to the cellulose, whose 44. Process a. is always spontaneous; processes b. and c. are likely
structure is very different from that of starch. to be spontaneous, depending on the temperature, and process
d. is never spontaneous.
32. Cellulose cannot be digested in mammals and therefore the energy
yield is 0 kilocalories per gram. Although both starch and cellulose 45. Since the process is spontaneous, ∆G must be negative. As the
are polymers of glucose, the glucose residues are linked differently in value of ∆H is positive, the value of ∆S must be sufficiently positive
the two molecules, and pancreatic amylase is unable to hydrolyze the in order to result in a negative value of ∆G. This is consistent with
glycosidic bonds in cellulose (see Solution 31). C
ellulose provides no the observation that ions in solution have a greater entropy than
energy to the diet but is an important component of the diet as fiber. ionic solids.
33. A positive entropy change indicates that the system has become ∆G = ∆H − T∆S
more disordered; a negative entropy change indicates that the system 0 > 26.4 kJ · mol−1 − (273 + 25 K) (∆S)
has become more ordered: a. negative; b. positive; c. positive; −26.4 kJ · mol−1 > − (298 K) (∆S)
d. positive; e. negative. 89 J · K−1 · mol−1 > ∆S
, SO LUTI O NS 3
∆S must be greater than 89 J · K−1 · mol−1 to be consistent with the b.
prediction. GAP + Pi + NAD+ → 1,3BPG + NADH ΔG = 6.7 kJ · mol−1
46. Since the process is spontaneous, ∆G must be negative. As the 1,3BPG + ADP → 3PG + ATP ΔG = −18.8 kJ · mol−1
value of ∆H is negative, the value of ∆S could be either positive or _______________________________________________________________
negative as long as the value of ∆G is negative. As ions in solution GAP + Pi + NAD+ + ADP → 3PG + NADH + ATP ΔG = −12.1 kJ · mol−1
have a greater entropy than ionic solids, ∆S is predicted to be positive.
The coupled reaction is spontaneous because the ∆G value is
∆G = ∆H − T∆S negative.
0 > −81 kJ · mol−1 − (273 + 25 K) (∆S)
81 kJ · mol−1 > − (298 K) (∆S) 53. C (most oxidized), A, B (most reduced).
−272 J · K−1 · mol−1 > ∆S 54. a. reduction; b. oxidation.
∆S could be any positive value or it could be a negative number with 55. a. oxidized; b. oxidized; c. oxidized; d. reduced.
a magnitude less than −272 J · K−1 · mol−1. However, given that the
56. a. oxidizing agent; b. oxidizing agent; c. oxidizing agent;
process involves the dissolution of an ionic salt to form ions in solu-
d. reducing agent.
tion, ∆S must be positive.
57. Retinoic acid is most oxidized, followed by retinal and then reti-
47. ∆G = ∆H − T∆S
nol, which is most reduced.
0 > −14.3 kJ · mol−1 − (273 + 25 K) (∆S)
14.3 kJ · mol−1 > − (298 K) (∆S) 58. The coenzyme NAD+ is reduced to NADH, which means that GAP
−48 J · K−1 · mol−1 > ∆S is oxidized to 1,3BPG. Therefore 1,3BPG is more oxidized than GAP.
∆S could be any positive value or it could be a negative number 59. a. Palmitate’s carbon atoms, which have the formula —CH2—,
with a magnitude less than −48 J · K−1 · mol−1. are more reduced than CO2, so their reoxidation to CO2 releases free
48. ∆G = ∆H − T∆S energy. b. Because the —CH2— groups of palmitate are more reduced
−63 kJ · mol−1 = ∆H − (273 + 25 K) (190 J · K−1 · mol−1) than those of glucose (—HCOH—), their conversion to the fully oxidized
∆H = −63 kJ · mol−1 + 56.6 kJ · mol−1 CO2 would be even more thermodynamically favorable (have a larger
∆H = −6.4 kJ · mol−1 negative value of ∆G) than the conversion of glucose carbons to CO2.
The reaction releases heat to the surroundings. Therefore, palmitate carbons provide more energy than glucose carbons.
49. a. Entropy decreases when the antibody–protein complex binds 60. The complete oxidation of stearate to CO2 yields more energy
because the value of ∆S is negative. because 17 of the 18 carbons of stearate are fully reduced. The con-
version of these carbons to CO2 provides more free energy than some
b. ∆G = ∆H − T∆S of the carbons of α-linolenate, which participate in double bonds and
∆G = −87,900 J · mol−1 − (298 K) (−118 J · K−1 · mol−1) are therefore already partially oxidized.
∆G = −52.7 kJ · mol−1
T he negative value of ∆G indicates that the complex forms 61. The folded protein is more ordered than the random coil. How-
spontaneously. ever, the solution in which the protein folding takes place is less
ordered. When the entire system is considered, a loss of entropy
c. The second antibody binds to cytochrome c more readily than the
results, consistent with the laws of thermodynamics.
first because the change in free energy of binding is a more negative
value. [From Raman, C.S. et al., Biochemistry 34, 5831–5838 (1995).] 62. When the substrate binds to the enzyme, entropy decreases.
However, this binding has a negative enthalpy change because favor-
50. a. The reaction releases heat to the surroundings because the
able interactions form between the enzyme and the substrate. When
value of ∆H is negative.
both the enthalpy change and the entropy change are considered,
b. ∆G = ∆H − T∆S the process occurs with an overall decrease in free energy, consistent
−17,200 J · mol−1 = −9500 J · mol−1 − (37 + 273 K) (∆S) with the laws of thermodynamics.
∆S = 25 J · K−1 · mol−1 63. Morphological differences, which are useful for classifying large
The positive value of ∆S indicates that the reaction proceeds with an organisms, are not useful for bacteria, which often look alike. Fur-
increase in entropy. thermore, microscopic organisms do not leave an easily interpreted
c. The ∆H term makes a greater contribution to the ∆G value. This imprint in the fossil record, as vertebrates do. Thus, molecular infor-
indicates that the reaction is spontaneous largely because the reac- mation is often the only means for tracing the evolutionary history
tion is exothermic. of bacteria.
51. a. The conversion of glucose to glucose-6-phosphate is not favor- 64. It is difficult to envision how a single engulfment event could
able because the ∆G value for the reaction is positive, indicating an have given rise to a stable and heritable association of the eukary-
endergonic process. otic host and the bacterial dependent within a single generation. It
b. If the two reactions are coupled, the overall reaction is the sum of is much more likely that natural selection gradually promoted the
the two individual reactions. The ∆G value would be the sum of the interdependence of the cells. Over many generations, genetic infor-
∆G values for the two individual reactions. mation supporting the association would have become widespread.
ATP + glucose → ADP + glucose-6-phosphate 65. A B C
∆G = −16.7 kJ · mol−1
Coupling the conversion of glucose to glucose-6-phosphate with the
hydrolysis of ATP converts an unfavorable reaction to a favorable
reaction. The ∆G value of the coupled reaction is negative, which
indicates that the reaction as written is favorable.
52. a. The reaction is not favorable because the ∆G value for the 66. a. H15 and H7 are closely related, as are H4 and H14. b. H4
reaction is positive, indicating an endergonic process. and H14 are most closely related to H3.
,4 SOLU TION S
67. The antibiotics kill some of the bacteria normally growing in the
intestine, creating an opportunity for the pathogenic species to grow.
68. The patient’s microbiota modifies the drug to convert it to an
active form.
, Solutions
Chapter 2 6. Arrows point toward hydrogen acceptors and away from hydrogen
donors. [From Kubiny, H., in 3D QSAR in Drug Design: Volume 1:
1. The water molecule is not perfectly tetrahedral because the elec- Theory Methods and Application, Springer Science & Business Media
trons in the nonbonding orbitals repel the electrons in the bonding (1993).]
orbitals more than the bonding electrons repel each other. The angle
between the bonding orbitals is therefore slightly less than 109°. H H H3 C R H R
N N O N
2. Because the partial negative charges are arranged symmetrically N H N
N N
(and the shape of the molecule is linear), the molecule as a whole is
not polar. H H
N N+ N N N N
MTX DHF
δ−
δ + −
δ H H H H
O C O
7. Identical hydrogen bonding patterns in the two molecules are
3. Water has the higher boiling point because, although each mol-
shown as open arrows in Solution 6.
ecule has the same geometry and can form hydrogen bonds with its
neighbors, the hydrogen bonds formed between water molecules are 8. [From Taylor, R., Acta Cryst B 73, 474–488 (2017).]
stronger than those formed between H2S molecules. The electronega-
tivity difference between H and O is greater than that between H and F H
N
S and results in greater differences in the partial charges on the atoms
in the water molecule. O N O
4. Water has the highest melting point because each water mole- H
cule forms hydrogen bonds with four neighboring water molecules
and hydrogen bonds are among the strongest intermolecular forces.
Ammonia is also capable of forming hydrogen bonds, but they are 9. [From Puschner, P. et al., J. Vet. Diagn. Invest. 19, 616–624 (2007).]
not as strong (due to the smaller electronegativity difference between
hydrogen and nitrogen). Methane cannot form hydrogen bonds; the O
molecules are attracted to their neighbors only via weak London dis-
HN NH
persion forces.
5. The arrows point toward hydrogen acceptors and away from O N O
hydrogen donors. H H H
N N N
H H
N N
O CH2 O N
+H H H
3N CH C N CH C O CH3
Melamine cyanurate
CH2 H
Aspartame
COO− 10. [From Chen, D. et al., Sci. Adv. 2, e1501240 (2016).]
O H
H COOH
N
N DHI
O
O N N
Uric acid
O N O
H H
O
H H H
H2N S NH2
O N O
O CYA
Sulfanilamide HN NH
O
1
,2 SOLU TION S
11. The greater an atom’s electronegativity, the more polar its bond 20. Water is unique in that its liquid form is more dense than its solid
with H and the greater its ability to act as a hydrogen bond acceptor. form. The weight of the skater puts pressure on the thin blade of the
Thus, N, O, and F, which have relatively high electronegativities, can ice skate. The ice melts under the blade because of this increased pres-
act as hydrogen bond acceptors, whereas C and S, whose electronega- sure. A higher pressure favors the liquid form of water over the solid
tivities are only slightly greater than that of hydrogen, cannot. form because the liquid form is more dense and takes up less volume.
12. Compound A does not form hydrogen bonds (the molecule has a 21. The positively charged ammonium ion is surrounded by a shell
hydrogen bond acceptor but no hydrogen bond donor). Compounds B of water molecules that are oriented so that the partially negatively
and C form hydrogen bonds as shown because each molecule contains charged oxygen atoms interact with the positive charge on the ammo-
at least one hydrogen bond donor and a hydrogen bond acceptor. The nium ion. Similarly, the negatively charged sulfate ion is hydrated
molecules in D do not form hydrogen bonds with each other because with water molecules oriented so that the partially positively charged
ethyl chloride lacks both a hydrogen bond donor and a hydrogen bond hydrogen atoms interact with the negative charge on the sulfate
acceptor. The molecules in E do because ammonia has a hydrogen anion. (Not shown in the diagram is the fact that the ammonium ions
bond donor and diethyl ether has a hydrogen bond acceptor: outnumber the sulfate ions by a 2:1 ratio. Also note that the exact
number of water molecules shown is unimportant.)
B N
N H H
H O O
H H
N O H H
H H
N
O
+
NH4 O O 2–
SO4 O
H H H H H
H H
C H3C CH2 O H O H H
H H
O O
H3C CH2 O H
H H
E H N H
H O CH2 CH3 22. Methanol, which has the highest dielectric constant, would be
H2C the best solvent for the cationic NH 4+ . The polarity of the alcohols,
CH3 which all contain a primary —OH group, varies with the size of the
hydrocarbon portion. 1-Butanol, with the largest hydrophobic group,
13. There are many possibilities. One example is is the least polar and therefore has the lowest dielectric constant.
H O 23. Structure A depicts a polar compound, while structure B depicts
R
δ− δ+ δ− δ+ an ionic compound similar to a salt like sodium chloride. This is
R C H O C more consistent with glycine’s physical properties as a white crystal-
R
H line solid with a high melting point. While structure A could be water
soluble because of its ability to form hydrogen bonds, the high solu-
14. CH3 bility of glycine in water is more consistent with an ionic compound
N H
whose positively and negatively charged groups are hydrated in aque-
N N O N ous solution by water molecules.
H N
N N
24. a. Surface tension is defined as the force that must be applied to
O
H surface molecules in a liquid so that they may experience the same
Adenine Thymine forces as the molecules in the interior of the liquid. The surface tension
of water is greater than that of ethanol because the strength and num-
15. a. van der Waals forces (dipole–dipole interactions); b. hydrogen ber of water’s intermolecular forces (hydrogen bonds) are both greater.
bonding; c. van der Waals forces (London dispersion forces); Ethanol’s —OH group also forms hydrogen bonds, but the hydro-
d. ionic interactions. carbon portion of the molecule cannot interact favorably with water
16. a. Hydrogen bonding; b. ionic interactions; c. van der Waals and weaker London dispersion forces form instead. b. The kinetic
forces (London dispersion forces). energy of the water molecules increases when temperature increases.
17. Solubility in water decreases as the number of carbons in the Intermolecular forces are weaker in strength as a consequence of the
alcohol increases. The hydroxyl group of the alcohol is able to form increased molecular motion. Because surface tension increases when
hydrogen bonds with water, but water cannot interact favorably with the strength of intermolecular forces increases as described in part a,
the hydrocarbon chain. Increasing the length of the chain increases surface tension decreases when temperature increases.
the number of potentially unfavorable interactions of the alcohol 25. The waxed car is a hydrophobic surface. To minimize its interac-
with water and solubility decreases as a result. tion with the hydrophobic molecules (wax), each water drop minimizes
18. Solubility in water and dielectric constants are correlated; low its surface area by becoming a sphere (the geometrical shape with the
molecular weight alcohols are miscible with water and have a high lowest possible ratio of surface to volume). Water does not bead on
dielectric constant; high molecular weight alcohols are not very solu- glass, because the glass presents a hydrophilic surface with which the
ble in water and have a low dielectric constant. water molecules can interact. This allows the water to spread out.
19. Aquatic organisms that live in the pond are able to survive the 26. The paper clip, although composed of a metal with a greater den-
winter. Since the water at the bottom of the pond remains in the liq- sity than water, floats due to the strong hydrogen bonding that occurs
uid form instead of freezing, the organisms are able to move around. among water molecules on the surface of the liquid. Soap disrupts
The ice on top of the pond also serves as an insulating layer from the these strong intermolecular forces and as a result the paper clip sinks
cold winter air. to the bottom of the container.
, SO LUTI O NS 3
27. Polar and nonpolar regions of the detergents are indicated. CH2CH3 O
CH3 Nonpolar H3C (CH2)3 CH CH2 O C CH2 O Polar
tails H3C (CH2)3 CH CH2 O C CH S O head
H3C (CH2)15 N+ CH3
CH2CH3 O O
CH3
Nonpolar Nonpolar AOT
Polar Polar
CH3
Hexadecyl- OH b. The protein, which contains numerous polar groups, interacts
CH3 O–
trimethylammonium with the polar AOT groups in the micelle interior.
CH3 O
Isooctane
HO OH Water
Cholate
Protein
28. The polar portion (boxed) of the hexadecyltrimethylammonium 34.
ion (see Solution 27) is able to form favorable dipole–dipole interac-
O
tions with water. The polar portion (circled) of the cholate molecule
(see Solution 27) is able to form both favorable dipole–dipole interac- a. Nonpolar tail H3C (CH2)11 O S O– Na+ Polar head
tions and hydrogen bonds with water, with both oxygens serving as O
hydrogen bond acceptors.
29. Polar and nonpolar regions of the detergents are indicated. b. O– Na+
O S O
Polar Polar +Na –O O O O– Na+
(CH2)11O
CH2OH S S
O
O O O
O
O
(C
H3C (CH2)9 P CH3 (CH2)7 CH3
11
H O
2)
H
H2
H
) 11
(C
CH3 OH H
Nonpolar HO H
Nonpolar
O CH3 CH3 H3C O
+Na –O
Dimethyldecylphosphine H OH S O(H2C)11 CH3 H3C (CH2)11O S O– Na+
oxide O O
CH3 CH H3C
3
n-Octylglucoside
(C
11
2)
(CH2)11O
H2
H
30. The polar portions (circled) of both detergents (see Solution 29)
(C
) 11
O O
O
O
are able to form favorable dipole–dipole interactions with water. In S S
+Na –O
addition, both polar portions can form a hydrogen bond with water, O O O– Na+
with the carbonyl group of dimethyldecylphosphine oxide serving as O S O
a hydrogen bond donor and the hydroxyl groups of n-octylglucoside O– Na+
serving as both hydrogen bond donors and hydrogen bond acceptors.
31. Compounds A and D are amphiphilic, compound B is nonpolar, c. The hydrophobic grease can move into the hydrophobic core of
and compounds C and E are polar. the water-soluble soap micelle. The “dissolved” grease can then be
washed away with the micelle.
32. Compound A has a polar head and a nonpolar tail as indicated
and can form a micelle (see Fig. 2.9). Compound D has a polar head 35. a. The nonpolar core of the lipid bilayer helps prevent the pas-
and two nonpolar tails as indicated and can form a bilayer (see Fig. sage of water since the polar water molecules cannot easily penetrate
2.10). Compounds B, C, and E form neither micelles nor bilayers. the hydrophobic core of the bilayer. b. Most human cells are sur-
rounded by a fluid containing about 150 mM Na+ and slightly less
Cl− (see Fig. 2.12). A solution containing 150 mM NaCl mimics the
CH3
extracellular fluid and therefore helps maintain the isolated cells in
A H3C (CH2)11 N+ CH2COO– Polar head near-normal conditions. If the cells were placed in pure water, water
CH3 would tend to enter the cells by osmosis; this might cause the cells to
Nonpolar tail burst.
36. In reverse osmosis, water moves from an area of low concen-
D O tration (high solute concentration) to an area of high concentration
CH2 O C (CH2)11 CH3 Nonpolar (low solute concentration). This movement is opposite that described
HC O C (CH2)11 CH3 tails for osmosis in Problem 35. This is a non-spontaneous process that
requires an input of energy in order to proceed, unlike osmosis,
HO CH2 O Polar head which occurs spontaneously, without input of energy.
37. Compounds a and b are polar and d is ionic; as these substances
33. a. In the nonpolar solvent, AOT’s polar head group faces the are highly hydrated, none of them would be able to cross a lipid
interior of the micelle and its nonpolar tails face the solvent: bilayer. Compound c is nonpolar and would be able to cross a bilayer.
, 4 SOLU TION S
38. Vesicles consist of a lipid bilayer that closes up to enclose an aqueous 48.
compartment. The polar drug readily dissolves in this aqueous compart- Acid, base,
ment. Delivery to the cell is accomplished when the vesicle membrane or neutral? pH [H+] (M) [OH−] (M)
fuses with the cell membrane, releasing the drug into the cytosol.
Pancreatic juice base 7.89 1.3 × 10−8 7.7 × 10−7
39. Substances present at high concentration move to an area of low −8
Blood base 7.42 3.8 × 10 2.6 × 10−7
concentration spontaneously, or “down” a concentration gradient in a
process that increases their entropy. The export of Na+ ions from the Saliva neutral 7.00 1.0 × 10−7 1.0 × 10−7
cell requires that the sodium ions be transported from an area of low Urine acid 6.80 1.6 × 10 −7
6.3 × 10−8
concentration to an area of high concentration. The same is true for
Gastric juice acid 2.10 7.9 × 10−3 1.3 × 10−12
potassium transport. Thus, these processes are not spontaneous and
an input of cellular energy is required to accomplish the transport. 49. The stomach contents have a low pH due to the contribution of
40. The amount of Na+ (atomic weight 23 g · mol−1) lost in 15 min- gastric juice (pH 1.5–3.0). When the partially digested material enters
utes, assuming a fluid loss rate of 2 L per hour and a sweat Na+ con- the small intestine, the addition of pancreatic juice (pH 7.8–8.0) neu-
centration of 50 mM (Box 2.B), is tralizes the acid and increases the pH.
1000 mg Na+ __________
23 g ___________ 50. The carbonate ions accept protons from water and form hydrox-
1 oz chips
0.25 h × ___ 0.05 mol
2 L × ________ × ____
×
× ide ions (as shown in the equation below), resulting in basic urine.
h L mol g Na+ 200 mg Na+
= 2.9 oz chips 3 (aq) + H2 O(l) ⇌
CO 2− HCO − −
3 (aq) + OH (aq)
It would take 2.9 ounces of potato chips (about a handful) to replace 51. a. C 2O2− 2− 2− 2− 3− 3− 2−
4 b. SO 3 c. HPO 4 d. CO 3 e. AsO 4 f. PO 4 g. O 2
the lost sodium ions.
52. a. H 2C2 O4 b. H 2SO 3 c. H 3PO 4 d. H 2CO 3 e. H 2AsO − −
4 f. H 2PO 4
41. a. In a high-solute medium, the cytoplasm loses water and there- g. H 2O2
fore its volume decreases. b. In a low-solute medium, the cytoplasm
gains water and therefore its volume increases. 53. a. The final concentration of HNO3 is (0.020 L)(1.0 M) ÷ 0.520
L = 0.038 M. Since HNO3 is a strong acid and dissociates completely,
42. E. coli accumulates water when grown in a low-salt medium. How-
the added [H+] is equal to [HNO3]. (The existing hydrogen ion con-
ever, regulation of water content only would cause a large increase in
centration in the water itself, 1.0 × 10−7 M, can be ignored because it
cytoplasmic volume. To avoid this large increase in volume, E. coli also
is much smaller than the hydrogen ion concentration contributed by
exports K+ ions. The opposite occurs when E. coli is grown in a high-
the nitric acid.)
salt medium: The cytoplasmic water content is decreased, but cyto-
plasmic osmolarity increases as E. coli imports K+ ions. [From Record, pH = −log [H+]
M.T. et al., Trends Biochem. Sci. 23, 143–148 (1998).] pH = −log (0.038)
43. Since the molecular mass of H2O is 18.0 g · mol−1, a given volume pH = 1.4
(for example, 1 L or 1000 g) has a molar concentration of 1000 g · L−1 ÷
18.0 g · mol−1 = 55.5 M. By definition, a liter of water at pH 7.0 has b. The final concentration of KOH is (0.015 L)(1.0 M) ÷ 0.515 L =
a hydrogen ion concentration of 1.0 × 10−7 M. Therefore, the ratio of 0.029 M. Since KOH dissociates completely, the added [OH−] is equal
[H2O] to [H+] is 55.5 M ÷ (1.0 × 10−7 M) = 5.55 × 108. to the [KOH]. (The existing hydroxide ion concentration in the water
itself, 1.0 × 10−7 M, can be ignored because it is much smaller than
[ H+]= [ OH−]= 1.47 × 1
44. 0−14
[ H+] = √
___________ the hydroxide ion concentration contributed by the KOH.)
1. 47 × 1 0−14
[H+]= 1.21 × 1 0−7M
K w = 1.0 × 10 −14 = [H +][OH −]
−14
pH = − log(1.21 × 1 0−7) = 6.92
1.0 × 1−0
[H +] = _________
[OH ]
45. The HCl is a strong acid and dissociates completely. This means −14
that the concentration of hydrogen ions contributed by the HCl is 1.0 × 10
[H +] = _________
(0.029 M)
1.0 × 10−9 M. However, the concentration of the hydrogen ions con-
tributed by the dissociation of water is 100-fold greater than this: [H +] = 3.4 × 10 −13 M
1.0 × 10−7 M. The concentration of the hydrogen ions contributed by pH = −log [H +]
the HCl is negligible in comparison. Therefore, the pH of the solution
pH = −log (3.4 × 10 −13)
is equal to 7.0.
pH = 12.5
46. The pH of this solution is 7.0 (see Solution 45). The concen-
tration of hydroxide ions contributed by the dissociation of water 54. a. The final concentration of HCl is (0.0015 L)(3.0 M) ÷ 1.0015
is 100-fold greater than that contributed by the dissociation of the L = 0.0045 M. Since HCl is a strong acid and dissociates completely,
dilute NaOH. the added [H+] is equal to [HCl]. (The existing hydrogen ion concen-
47. tration in the water itself, 1.0 × 10−7 M, can be ignored because it
− is much smaller than the hydrogen ion concentration contributed by
the hydrochloric acid.)
pH = −log [H+]
pH = −log (0.0045)
pH = 2.3
5th Edition Charlotte W. Pratt
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, Solutions
Chapter 1 11. a. 11-Cyclohexylundecanoate is a lipid. b. Selenocysteine is an
amino acid. c. Galactose is a monosaccharide. d. O6-Methylguano-
1. a. carboxylic acid; b. amine; c. ester; d. alcohol. sine monophosphate is a nucleotide.
2. a. ether; b. phosphoric acid ester; c. thiol; d. ketone. 12. The compound contains both a lipid and a monosaccharide.
3. a. aldehyde; b. imine; c. thioester; d. diphosphoric acid. [From Malyarenko, T.V. et al., Molecules 23, 1077 (2018).]
4. a. alcohol; b. aldehyde; c. carboxylic acid. 13. a. C and H plus some O. b. C, H, and O. c. C, H, O, and N plus
5. small amounts of S.
Hydroxyl group
OH 14. It is a lipid (it is actually lecithin). It is mostly C and H, with
Imino group relatively little O and only one N and one P. It has too little O to be
a carbohydrate, too little N to be a protein, and too little P to be a
N nucleic acid.
CH3
O 15. You should measure the nitrogen content, since this would indi-
cate the presence of protein (neither lipids nor carbohydrates contain
Ether linkage appreciable amounts of nitrogen).
[From Li, S.-Y. et al., Eur. J. Med. Chem. 71, 36–45 (2014).] 16. You could add the compound that contains the most nitrogen,
6. Ester linkage compound B, which is melamine. (Melamine is a substance that in
OH Carboxyl
O group the past was added to some pet foods and milk products from China
HO CH O O so that they would appear to contain more protein. Melamine is toxic
C
C CH C OH to pets and children.) Compound C is an amino acid, so it would
H2
C C already be present in protein-containing food.
Hydroxyl N
group HO OH 17. A diet high in protein results in a high urea concentration, since
Nicotinic acid (niacin)
Vitamin C urea is the body’s method of ridding itself of extra nitrogen. Nitrogen
is found in proteins but is not found in significant amounts in lipids
O Carbonyl group
or carbohydrates. A low-protein diet provides the patient with just
O CH3
enough protein for tissue repair and growth. In the absence of excess
Ether linkage H3C CH3
protein consumption, urea production decreases, and this puts less
H3C
O (CH2 CH C CH2)10H strain on the patient’s weakened kidneys.
Carbonyl group O
18. Asn has an amido group and Cys has a sulfhydryl group.
Coenzyme Q 19. Serine has a hydroxyl group and lysine does not.
7. Sulfhydryl 20. a.
HN CH2 CH2 SH O
group
+H
C O 3N CH C O−
CH2 CH2
CH2
Amido group O
−O
NH Amino group P O−
C O NH2 O
Hydroxyl HO C H N b. O
N
group
H3C C CH3 O O +H
3N CH C O−
N N
CH2 O P O P O CH2 O CH2
O− −O
H H CH2
H H
HC OH
Phosphoanhydride Hydroxyl
O OH
linkage group CH2
−O P O−
NH+
3
Phosphoryl
O
group c. O
8. A thioester linkage is formed. +H
3N CH2 C O−
9. Amino acids, monosaccharides, nucleotides, and lipids are the (CH2)4
four types of biological small molecules. Amino acids, monosaccha-
NH
rides, and nucleotides can form polymers of proteins, polysaccha-
rides, and nucleic acids, respectively. C O
10. a. N-Acetylglucosamine is a monosaccharide. b. CMP is a nucleo- CH3 1
tide. c. Homocysteine is an amino acid. d. Cholesteryl ester is a lipid.
,2 SOLU TION S
21. O 34. a. decrease; b. increase.
Carbonyl 35. The polymeric molecule is more ordered and thus has less
group C H
entropy. A mixture of constituent monomers has a large number of
H OH different arrangements (like the balls scattered on a pool table) and
HO H Hydroxyl group thus has greater entropy.
H OH 36. Entropy increases as the reactants (7 molecules) are converted to
products (12 molecules).
H OH
37. The dissolution of ammonium nitrate in water is a highly endo-
CH2 OH thermic process, as indicated by the positive value of ∆H. This means
that when ammonium nitrate dissolves in water, the system absorbs
22. a. Fructose has the same molecular formula, C 6 H 12 O 6 , as
heat from the surroundings and the surroundings become cold. The
glucose. b. Fructose is a ketone, whereas glucose is an aldehyde.
plastic bag containing the ammonium nitrate becomes cold and can
23. Uracil has a carbonyl functional group, whereas cytosine has an be used as a cold pack to treat an injury.
amino functional group.
38. The dissolution of calcium chloride in water is a highly exother-
24. Nucleotides consist of a five-carbon sugar, a nitrogenous ring, mic process, as indicated by the negative value of ∆H. This means
and one or more phosphoryl groups linked covalently together. that when calcium chloride dissolves in water, the system loses heat
25. a. glycosidic; b. peptide; c. phosphoester. to the surroundings and the surroundings become warm. The plastic
26. Aspartame consists of two amino acids linked by a peptide bond. bag holding the calcium chloride solution becomes warm and can be
used as a hot pack by the camper at cold temperatures.
27. As described in the text, palmitate and cholesterol are highly non-
polar and are therefore insoluble in water. Both are highly aliphatic. 39. The dissolution of urea in water is an endothermic process and
Alanine is water soluble because its amino group and c arboxylate has a positive ∆H value. In order for the process to be spontaneous,
groups are ionized, which render the molecule “saltlike.” Glucose the process must also have a positive ∆S value in order for the free
is also water soluble because its carbonyl group and many hydroxyl energy change of the process to be negative. Solutions have a higher
groups are able to form hydrogen bonds with water. degree of entropy than the solvent and solute alone.
28. Glucose has several hydroxyl groups and is a polar molecule. 40. As combustion reactions proceed with a release of free energy,
As such, it will have difficulty crossing the nonpolar membrane. the reaction shown in Problem 36 is expected to be exergonic.
The 2,4-dinitrophenol molecule consists of a substituted ben- 41. Calculate ∆H and ∆S, as described in Sample Calculation 1.1:
zene ring and has a greater nonpolar character. Of the two molecules,
∆H = HB − HA
the 2 ,4-dinitrophenol will traverse the membrane more easily.
∆H = 60 kJ · mol−1 − 54 kJ · mol−1
29. DNA forms a more regular structure because DNA consists of
∆H = 6 kJ · mol−1
only four different nucleotides, whereas proteins are made up of as
many as 20 different amino acids. In addition, the 20 amino acids ∆S = SB − SA
have much more individual variation in their structures than do ∆S = 43 J · K−1 · mol−1 − 22 J · K−1 · mol−1
the four nucleotides. Both of these factors result in a more regular ∆S = 21 J · K−1 · mol−1
structure for DNA. The cellular role of DNA relies on the sequence
42. a. ∆G = ∆H − T∆S
of the nucleotides that make up the DNA, not on the overall shape
∆G = (6000 J · mol−1) − (4 + 273 K) (21 J · K−1 · mol−1)
of the DNA molecule itself. Proteins, on the other hand, fold into
unique shapes, as illustrated by endothelin in Figure 1.4. The ability ∆G = 180 J · mol−1
of proteins to fold into a wide variety of shapes means that proteins The reaction is not favorable at 4°C.
can also serve a wide variety of biochemical roles in the cell. Accord- b. ∆G = (6000 J · mol−1) − (37 + 273 K) (21 J · K−1 · mol−1)
ing to Table 1.2, the major roles of proteins in the cell are to carry out
∆G = −510 J · mol−1
metabolic reactions and to support cellular structures.
The reaction is favorable at 37°C.
30. Polysaccharides serve as fuel-storage molecules and can also
43. ∆G = ∆H − T∆S
serve as structural support for the cell.
0 > 15,000 J · mol−1 − (T) (51 J · K−1 · mol−1)
31. Pancreatic amylase is unable to digest the glycosidic bonds that −15,000 > −(T) (51 K−1)
link the glucose residues in cellulose. Figure 1.6 shows the structural
15,000 < (T) (51 K−1)
differences between starch and cellulose. Pancreatic amylase binds to
294 K < T
starch prior to catalyzing the hydrolysis of the glycosidic bond; thus
The reaction is favorable at temperatures of 21°C and higher.
the enzyme and the starch must have shapes that are complemen-
tary. The enzyme would be unable to bind to the cellulose, whose 44. Process a. is always spontaneous; processes b. and c. are likely
structure is very different from that of starch. to be spontaneous, depending on the temperature, and process
d. is never spontaneous.
32. Cellulose cannot be digested in mammals and therefore the energy
yield is 0 kilocalories per gram. Although both starch and cellulose 45. Since the process is spontaneous, ∆G must be negative. As the
are polymers of glucose, the glucose residues are linked differently in value of ∆H is positive, the value of ∆S must be sufficiently positive
the two molecules, and pancreatic amylase is unable to hydrolyze the in order to result in a negative value of ∆G. This is consistent with
glycosidic bonds in cellulose (see Solution 31). C
ellulose provides no the observation that ions in solution have a greater entropy than
energy to the diet but is an important component of the diet as fiber. ionic solids.
33. A positive entropy change indicates that the system has become ∆G = ∆H − T∆S
more disordered; a negative entropy change indicates that the system 0 > 26.4 kJ · mol−1 − (273 + 25 K) (∆S)
has become more ordered: a. negative; b. positive; c. positive; −26.4 kJ · mol−1 > − (298 K) (∆S)
d. positive; e. negative. 89 J · K−1 · mol−1 > ∆S
, SO LUTI O NS 3
∆S must be greater than 89 J · K−1 · mol−1 to be consistent with the b.
prediction. GAP + Pi + NAD+ → 1,3BPG + NADH ΔG = 6.7 kJ · mol−1
46. Since the process is spontaneous, ∆G must be negative. As the 1,3BPG + ADP → 3PG + ATP ΔG = −18.8 kJ · mol−1
value of ∆H is negative, the value of ∆S could be either positive or _______________________________________________________________
negative as long as the value of ∆G is negative. As ions in solution GAP + Pi + NAD+ + ADP → 3PG + NADH + ATP ΔG = −12.1 kJ · mol−1
have a greater entropy than ionic solids, ∆S is predicted to be positive.
The coupled reaction is spontaneous because the ∆G value is
∆G = ∆H − T∆S negative.
0 > −81 kJ · mol−1 − (273 + 25 K) (∆S)
81 kJ · mol−1 > − (298 K) (∆S) 53. C (most oxidized), A, B (most reduced).
−272 J · K−1 · mol−1 > ∆S 54. a. reduction; b. oxidation.
∆S could be any positive value or it could be a negative number with 55. a. oxidized; b. oxidized; c. oxidized; d. reduced.
a magnitude less than −272 J · K−1 · mol−1. However, given that the
56. a. oxidizing agent; b. oxidizing agent; c. oxidizing agent;
process involves the dissolution of an ionic salt to form ions in solu-
d. reducing agent.
tion, ∆S must be positive.
57. Retinoic acid is most oxidized, followed by retinal and then reti-
47. ∆G = ∆H − T∆S
nol, which is most reduced.
0 > −14.3 kJ · mol−1 − (273 + 25 K) (∆S)
14.3 kJ · mol−1 > − (298 K) (∆S) 58. The coenzyme NAD+ is reduced to NADH, which means that GAP
−48 J · K−1 · mol−1 > ∆S is oxidized to 1,3BPG. Therefore 1,3BPG is more oxidized than GAP.
∆S could be any positive value or it could be a negative number 59. a. Palmitate’s carbon atoms, which have the formula —CH2—,
with a magnitude less than −48 J · K−1 · mol−1. are more reduced than CO2, so their reoxidation to CO2 releases free
48. ∆G = ∆H − T∆S energy. b. Because the —CH2— groups of palmitate are more reduced
−63 kJ · mol−1 = ∆H − (273 + 25 K) (190 J · K−1 · mol−1) than those of glucose (—HCOH—), their conversion to the fully oxidized
∆H = −63 kJ · mol−1 + 56.6 kJ · mol−1 CO2 would be even more thermodynamically favorable (have a larger
∆H = −6.4 kJ · mol−1 negative value of ∆G) than the conversion of glucose carbons to CO2.
The reaction releases heat to the surroundings. Therefore, palmitate carbons provide more energy than glucose carbons.
49. a. Entropy decreases when the antibody–protein complex binds 60. The complete oxidation of stearate to CO2 yields more energy
because the value of ∆S is negative. because 17 of the 18 carbons of stearate are fully reduced. The con-
version of these carbons to CO2 provides more free energy than some
b. ∆G = ∆H − T∆S of the carbons of α-linolenate, which participate in double bonds and
∆G = −87,900 J · mol−1 − (298 K) (−118 J · K−1 · mol−1) are therefore already partially oxidized.
∆G = −52.7 kJ · mol−1
T he negative value of ∆G indicates that the complex forms 61. The folded protein is more ordered than the random coil. How-
spontaneously. ever, the solution in which the protein folding takes place is less
ordered. When the entire system is considered, a loss of entropy
c. The second antibody binds to cytochrome c more readily than the
results, consistent with the laws of thermodynamics.
first because the change in free energy of binding is a more negative
value. [From Raman, C.S. et al., Biochemistry 34, 5831–5838 (1995).] 62. When the substrate binds to the enzyme, entropy decreases.
However, this binding has a negative enthalpy change because favor-
50. a. The reaction releases heat to the surroundings because the
able interactions form between the enzyme and the substrate. When
value of ∆H is negative.
both the enthalpy change and the entropy change are considered,
b. ∆G = ∆H − T∆S the process occurs with an overall decrease in free energy, consistent
−17,200 J · mol−1 = −9500 J · mol−1 − (37 + 273 K) (∆S) with the laws of thermodynamics.
∆S = 25 J · K−1 · mol−1 63. Morphological differences, which are useful for classifying large
The positive value of ∆S indicates that the reaction proceeds with an organisms, are not useful for bacteria, which often look alike. Fur-
increase in entropy. thermore, microscopic organisms do not leave an easily interpreted
c. The ∆H term makes a greater contribution to the ∆G value. This imprint in the fossil record, as vertebrates do. Thus, molecular infor-
indicates that the reaction is spontaneous largely because the reac- mation is often the only means for tracing the evolutionary history
tion is exothermic. of bacteria.
51. a. The conversion of glucose to glucose-6-phosphate is not favor- 64. It is difficult to envision how a single engulfment event could
able because the ∆G value for the reaction is positive, indicating an have given rise to a stable and heritable association of the eukary-
endergonic process. otic host and the bacterial dependent within a single generation. It
b. If the two reactions are coupled, the overall reaction is the sum of is much more likely that natural selection gradually promoted the
the two individual reactions. The ∆G value would be the sum of the interdependence of the cells. Over many generations, genetic infor-
∆G values for the two individual reactions. mation supporting the association would have become widespread.
ATP + glucose → ADP + glucose-6-phosphate 65. A B C
∆G = −16.7 kJ · mol−1
Coupling the conversion of glucose to glucose-6-phosphate with the
hydrolysis of ATP converts an unfavorable reaction to a favorable
reaction. The ∆G value of the coupled reaction is negative, which
indicates that the reaction as written is favorable.
52. a. The reaction is not favorable because the ∆G value for the 66. a. H15 and H7 are closely related, as are H4 and H14. b. H4
reaction is positive, indicating an endergonic process. and H14 are most closely related to H3.
,4 SOLU TION S
67. The antibiotics kill some of the bacteria normally growing in the
intestine, creating an opportunity for the pathogenic species to grow.
68. The patient’s microbiota modifies the drug to convert it to an
active form.
, Solutions
Chapter 2 6. Arrows point toward hydrogen acceptors and away from hydrogen
donors. [From Kubiny, H., in 3D QSAR in Drug Design: Volume 1:
1. The water molecule is not perfectly tetrahedral because the elec- Theory Methods and Application, Springer Science & Business Media
trons in the nonbonding orbitals repel the electrons in the bonding (1993).]
orbitals more than the bonding electrons repel each other. The angle
between the bonding orbitals is therefore slightly less than 109°. H H H3 C R H R
N N O N
2. Because the partial negative charges are arranged symmetrically N H N
N N
(and the shape of the molecule is linear), the molecule as a whole is
not polar. H H
N N+ N N N N
MTX DHF
δ−
δ + −
δ H H H H
O C O
7. Identical hydrogen bonding patterns in the two molecules are
3. Water has the higher boiling point because, although each mol-
shown as open arrows in Solution 6.
ecule has the same geometry and can form hydrogen bonds with its
neighbors, the hydrogen bonds formed between water molecules are 8. [From Taylor, R., Acta Cryst B 73, 474–488 (2017).]
stronger than those formed between H2S molecules. The electronega-
tivity difference between H and O is greater than that between H and F H
N
S and results in greater differences in the partial charges on the atoms
in the water molecule. O N O
4. Water has the highest melting point because each water mole- H
cule forms hydrogen bonds with four neighboring water molecules
and hydrogen bonds are among the strongest intermolecular forces.
Ammonia is also capable of forming hydrogen bonds, but they are 9. [From Puschner, P. et al., J. Vet. Diagn. Invest. 19, 616–624 (2007).]
not as strong (due to the smaller electronegativity difference between
hydrogen and nitrogen). Methane cannot form hydrogen bonds; the O
molecules are attracted to their neighbors only via weak London dis-
HN NH
persion forces.
5. The arrows point toward hydrogen acceptors and away from O N O
hydrogen donors. H H H
N N N
H H
N N
O CH2 O N
+H H H
3N CH C N CH C O CH3
Melamine cyanurate
CH2 H
Aspartame
COO− 10. [From Chen, D. et al., Sci. Adv. 2, e1501240 (2016).]
O H
H COOH
N
N DHI
O
O N N
Uric acid
O N O
H H
O
H H H
H2N S NH2
O N O
O CYA
Sulfanilamide HN NH
O
1
,2 SOLU TION S
11. The greater an atom’s electronegativity, the more polar its bond 20. Water is unique in that its liquid form is more dense than its solid
with H and the greater its ability to act as a hydrogen bond acceptor. form. The weight of the skater puts pressure on the thin blade of the
Thus, N, O, and F, which have relatively high electronegativities, can ice skate. The ice melts under the blade because of this increased pres-
act as hydrogen bond acceptors, whereas C and S, whose electronega- sure. A higher pressure favors the liquid form of water over the solid
tivities are only slightly greater than that of hydrogen, cannot. form because the liquid form is more dense and takes up less volume.
12. Compound A does not form hydrogen bonds (the molecule has a 21. The positively charged ammonium ion is surrounded by a shell
hydrogen bond acceptor but no hydrogen bond donor). Compounds B of water molecules that are oriented so that the partially negatively
and C form hydrogen bonds as shown because each molecule contains charged oxygen atoms interact with the positive charge on the ammo-
at least one hydrogen bond donor and a hydrogen bond acceptor. The nium ion. Similarly, the negatively charged sulfate ion is hydrated
molecules in D do not form hydrogen bonds with each other because with water molecules oriented so that the partially positively charged
ethyl chloride lacks both a hydrogen bond donor and a hydrogen bond hydrogen atoms interact with the negative charge on the sulfate
acceptor. The molecules in E do because ammonia has a hydrogen anion. (Not shown in the diagram is the fact that the ammonium ions
bond donor and diethyl ether has a hydrogen bond acceptor: outnumber the sulfate ions by a 2:1 ratio. Also note that the exact
number of water molecules shown is unimportant.)
B N
N H H
H O O
H H
N O H H
H H
N
O
+
NH4 O O 2–
SO4 O
H H H H H
H H
C H3C CH2 O H O H H
H H
O O
H3C CH2 O H
H H
E H N H
H O CH2 CH3 22. Methanol, which has the highest dielectric constant, would be
H2C the best solvent for the cationic NH 4+ . The polarity of the alcohols,
CH3 which all contain a primary —OH group, varies with the size of the
hydrocarbon portion. 1-Butanol, with the largest hydrophobic group,
13. There are many possibilities. One example is is the least polar and therefore has the lowest dielectric constant.
H O 23. Structure A depicts a polar compound, while structure B depicts
R
δ− δ+ δ− δ+ an ionic compound similar to a salt like sodium chloride. This is
R C H O C more consistent with glycine’s physical properties as a white crystal-
R
H line solid with a high melting point. While structure A could be water
soluble because of its ability to form hydrogen bonds, the high solu-
14. CH3 bility of glycine in water is more consistent with an ionic compound
N H
whose positively and negatively charged groups are hydrated in aque-
N N O N ous solution by water molecules.
H N
N N
24. a. Surface tension is defined as the force that must be applied to
O
H surface molecules in a liquid so that they may experience the same
Adenine Thymine forces as the molecules in the interior of the liquid. The surface tension
of water is greater than that of ethanol because the strength and num-
15. a. van der Waals forces (dipole–dipole interactions); b. hydrogen ber of water’s intermolecular forces (hydrogen bonds) are both greater.
bonding; c. van der Waals forces (London dispersion forces); Ethanol’s —OH group also forms hydrogen bonds, but the hydro-
d. ionic interactions. carbon portion of the molecule cannot interact favorably with water
16. a. Hydrogen bonding; b. ionic interactions; c. van der Waals and weaker London dispersion forces form instead. b. The kinetic
forces (London dispersion forces). energy of the water molecules increases when temperature increases.
17. Solubility in water decreases as the number of carbons in the Intermolecular forces are weaker in strength as a consequence of the
alcohol increases. The hydroxyl group of the alcohol is able to form increased molecular motion. Because surface tension increases when
hydrogen bonds with water, but water cannot interact favorably with the strength of intermolecular forces increases as described in part a,
the hydrocarbon chain. Increasing the length of the chain increases surface tension decreases when temperature increases.
the number of potentially unfavorable interactions of the alcohol 25. The waxed car is a hydrophobic surface. To minimize its interac-
with water and solubility decreases as a result. tion with the hydrophobic molecules (wax), each water drop minimizes
18. Solubility in water and dielectric constants are correlated; low its surface area by becoming a sphere (the geometrical shape with the
molecular weight alcohols are miscible with water and have a high lowest possible ratio of surface to volume). Water does not bead on
dielectric constant; high molecular weight alcohols are not very solu- glass, because the glass presents a hydrophilic surface with which the
ble in water and have a low dielectric constant. water molecules can interact. This allows the water to spread out.
19. Aquatic organisms that live in the pond are able to survive the 26. The paper clip, although composed of a metal with a greater den-
winter. Since the water at the bottom of the pond remains in the liq- sity than water, floats due to the strong hydrogen bonding that occurs
uid form instead of freezing, the organisms are able to move around. among water molecules on the surface of the liquid. Soap disrupts
The ice on top of the pond also serves as an insulating layer from the these strong intermolecular forces and as a result the paper clip sinks
cold winter air. to the bottom of the container.
, SO LUTI O NS 3
27. Polar and nonpolar regions of the detergents are indicated. CH2CH3 O
CH3 Nonpolar H3C (CH2)3 CH CH2 O C CH2 O Polar
tails H3C (CH2)3 CH CH2 O C CH S O head
H3C (CH2)15 N+ CH3
CH2CH3 O O
CH3
Nonpolar Nonpolar AOT
Polar Polar
CH3
Hexadecyl- OH b. The protein, which contains numerous polar groups, interacts
CH3 O–
trimethylammonium with the polar AOT groups in the micelle interior.
CH3 O
Isooctane
HO OH Water
Cholate
Protein
28. The polar portion (boxed) of the hexadecyltrimethylammonium 34.
ion (see Solution 27) is able to form favorable dipole–dipole interac-
O
tions with water. The polar portion (circled) of the cholate molecule
(see Solution 27) is able to form both favorable dipole–dipole interac- a. Nonpolar tail H3C (CH2)11 O S O– Na+ Polar head
tions and hydrogen bonds with water, with both oxygens serving as O
hydrogen bond acceptors.
29. Polar and nonpolar regions of the detergents are indicated. b. O– Na+
O S O
Polar Polar +Na –O O O O– Na+
(CH2)11O
CH2OH S S
O
O O O
O
O
(C
H3C (CH2)9 P CH3 (CH2)7 CH3
11
H O
2)
H
H2
H
) 11
(C
CH3 OH H
Nonpolar HO H
Nonpolar
O CH3 CH3 H3C O
+Na –O
Dimethyldecylphosphine H OH S O(H2C)11 CH3 H3C (CH2)11O S O– Na+
oxide O O
CH3 CH H3C
3
n-Octylglucoside
(C
11
2)
(CH2)11O
H2
H
30. The polar portions (circled) of both detergents (see Solution 29)
(C
) 11
O O
O
O
are able to form favorable dipole–dipole interactions with water. In S S
+Na –O
addition, both polar portions can form a hydrogen bond with water, O O O– Na+
with the carbonyl group of dimethyldecylphosphine oxide serving as O S O
a hydrogen bond donor and the hydroxyl groups of n-octylglucoside O– Na+
serving as both hydrogen bond donors and hydrogen bond acceptors.
31. Compounds A and D are amphiphilic, compound B is nonpolar, c. The hydrophobic grease can move into the hydrophobic core of
and compounds C and E are polar. the water-soluble soap micelle. The “dissolved” grease can then be
washed away with the micelle.
32. Compound A has a polar head and a nonpolar tail as indicated
and can form a micelle (see Fig. 2.9). Compound D has a polar head 35. a. The nonpolar core of the lipid bilayer helps prevent the pas-
and two nonpolar tails as indicated and can form a bilayer (see Fig. sage of water since the polar water molecules cannot easily penetrate
2.10). Compounds B, C, and E form neither micelles nor bilayers. the hydrophobic core of the bilayer. b. Most human cells are sur-
rounded by a fluid containing about 150 mM Na+ and slightly less
Cl− (see Fig. 2.12). A solution containing 150 mM NaCl mimics the
CH3
extracellular fluid and therefore helps maintain the isolated cells in
A H3C (CH2)11 N+ CH2COO– Polar head near-normal conditions. If the cells were placed in pure water, water
CH3 would tend to enter the cells by osmosis; this might cause the cells to
Nonpolar tail burst.
36. In reverse osmosis, water moves from an area of low concen-
D O tration (high solute concentration) to an area of high concentration
CH2 O C (CH2)11 CH3 Nonpolar (low solute concentration). This movement is opposite that described
HC O C (CH2)11 CH3 tails for osmosis in Problem 35. This is a non-spontaneous process that
requires an input of energy in order to proceed, unlike osmosis,
HO CH2 O Polar head which occurs spontaneously, without input of energy.
37. Compounds a and b are polar and d is ionic; as these substances
33. a. In the nonpolar solvent, AOT’s polar head group faces the are highly hydrated, none of them would be able to cross a lipid
interior of the micelle and its nonpolar tails face the solvent: bilayer. Compound c is nonpolar and would be able to cross a bilayer.
, 4 SOLU TION S
38. Vesicles consist of a lipid bilayer that closes up to enclose an aqueous 48.
compartment. The polar drug readily dissolves in this aqueous compart- Acid, base,
ment. Delivery to the cell is accomplished when the vesicle membrane or neutral? pH [H+] (M) [OH−] (M)
fuses with the cell membrane, releasing the drug into the cytosol.
Pancreatic juice base 7.89 1.3 × 10−8 7.7 × 10−7
39. Substances present at high concentration move to an area of low −8
Blood base 7.42 3.8 × 10 2.6 × 10−7
concentration spontaneously, or “down” a concentration gradient in a
process that increases their entropy. The export of Na+ ions from the Saliva neutral 7.00 1.0 × 10−7 1.0 × 10−7
cell requires that the sodium ions be transported from an area of low Urine acid 6.80 1.6 × 10 −7
6.3 × 10−8
concentration to an area of high concentration. The same is true for
Gastric juice acid 2.10 7.9 × 10−3 1.3 × 10−12
potassium transport. Thus, these processes are not spontaneous and
an input of cellular energy is required to accomplish the transport. 49. The stomach contents have a low pH due to the contribution of
40. The amount of Na+ (atomic weight 23 g · mol−1) lost in 15 min- gastric juice (pH 1.5–3.0). When the partially digested material enters
utes, assuming a fluid loss rate of 2 L per hour and a sweat Na+ con- the small intestine, the addition of pancreatic juice (pH 7.8–8.0) neu-
centration of 50 mM (Box 2.B), is tralizes the acid and increases the pH.
1000 mg Na+ __________
23 g ___________ 50. The carbonate ions accept protons from water and form hydrox-
1 oz chips
0.25 h × ___ 0.05 mol
2 L × ________ × ____
×
× ide ions (as shown in the equation below), resulting in basic urine.
h L mol g Na+ 200 mg Na+
= 2.9 oz chips 3 (aq) + H2 O(l) ⇌
CO 2− HCO − −
3 (aq) + OH (aq)
It would take 2.9 ounces of potato chips (about a handful) to replace 51. a. C 2O2− 2− 2− 2− 3− 3− 2−
4 b. SO 3 c. HPO 4 d. CO 3 e. AsO 4 f. PO 4 g. O 2
the lost sodium ions.
52. a. H 2C2 O4 b. H 2SO 3 c. H 3PO 4 d. H 2CO 3 e. H 2AsO − −
4 f. H 2PO 4
41. a. In a high-solute medium, the cytoplasm loses water and there- g. H 2O2
fore its volume decreases. b. In a low-solute medium, the cytoplasm
gains water and therefore its volume increases. 53. a. The final concentration of HNO3 is (0.020 L)(1.0 M) ÷ 0.520
L = 0.038 M. Since HNO3 is a strong acid and dissociates completely,
42. E. coli accumulates water when grown in a low-salt medium. How-
the added [H+] is equal to [HNO3]. (The existing hydrogen ion con-
ever, regulation of water content only would cause a large increase in
centration in the water itself, 1.0 × 10−7 M, can be ignored because it
cytoplasmic volume. To avoid this large increase in volume, E. coli also
is much smaller than the hydrogen ion concentration contributed by
exports K+ ions. The opposite occurs when E. coli is grown in a high-
the nitric acid.)
salt medium: The cytoplasmic water content is decreased, but cyto-
plasmic osmolarity increases as E. coli imports K+ ions. [From Record, pH = −log [H+]
M.T. et al., Trends Biochem. Sci. 23, 143–148 (1998).] pH = −log (0.038)
43. Since the molecular mass of H2O is 18.0 g · mol−1, a given volume pH = 1.4
(for example, 1 L or 1000 g) has a molar concentration of 1000 g · L−1 ÷
18.0 g · mol−1 = 55.5 M. By definition, a liter of water at pH 7.0 has b. The final concentration of KOH is (0.015 L)(1.0 M) ÷ 0.515 L =
a hydrogen ion concentration of 1.0 × 10−7 M. Therefore, the ratio of 0.029 M. Since KOH dissociates completely, the added [OH−] is equal
[H2O] to [H+] is 55.5 M ÷ (1.0 × 10−7 M) = 5.55 × 108. to the [KOH]. (The existing hydroxide ion concentration in the water
itself, 1.0 × 10−7 M, can be ignored because it is much smaller than
[ H+]= [ OH−]= 1.47 × 1
44. 0−14
[ H+] = √
___________ the hydroxide ion concentration contributed by the KOH.)
1. 47 × 1 0−14
[H+]= 1.21 × 1 0−7M
K w = 1.0 × 10 −14 = [H +][OH −]
−14
pH = − log(1.21 × 1 0−7) = 6.92
1.0 × 1−0
[H +] = _________
[OH ]
45. The HCl is a strong acid and dissociates completely. This means −14
that the concentration of hydrogen ions contributed by the HCl is 1.0 × 10
[H +] = _________
(0.029 M)
1.0 × 10−9 M. However, the concentration of the hydrogen ions con-
tributed by the dissociation of water is 100-fold greater than this: [H +] = 3.4 × 10 −13 M
1.0 × 10−7 M. The concentration of the hydrogen ions contributed by pH = −log [H +]
the HCl is negligible in comparison. Therefore, the pH of the solution
pH = −log (3.4 × 10 −13)
is equal to 7.0.
pH = 12.5
46. The pH of this solution is 7.0 (see Solution 45). The concen-
tration of hydroxide ions contributed by the dissociation of water 54. a. The final concentration of HCl is (0.0015 L)(3.0 M) ÷ 1.0015
is 100-fold greater than that contributed by the dissociation of the L = 0.0045 M. Since HCl is a strong acid and dissociates completely,
dilute NaOH. the added [H+] is equal to [HCl]. (The existing hydrogen ion concen-
47. tration in the water itself, 1.0 × 10−7 M, can be ignored because it
− is much smaller than the hydrogen ion concentration contributed by
the hydrochloric acid.)
pH = −log [H+]
pH = −log (0.0045)
pH = 2.3
