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960

CHAPTER 19

Reactions of Benzene and Substituted Benzenes

S UMMARY OF REACTIONS 1. Electrophilic aromatic substitution reactions a. Halogenation (Section 19.4). The mechanisms are shown on pages 927 and 928. + Br2

Br

FeBr3

+ HBr + I2

2 + Cl2

I

H2O2 H2SO4

2

+ 2 H+

Cl

FeCl3

+ HCl

b. Nitration, sulfonation, and desulfonation (Sections 19.5 and 19.6). The mechanisms are shown on pages 929, 930, and 931. H2SO4

+ HNO3

Δ

+ H2SO4

NO2 + H2O

SO3H + H2O

c. Friedel–Crafts acylation and alkylation (Sections 19.7 and 19.8). The mechanisms are shown on pages 932 and 933. O O +

R

C

C

1. AlCl3 2. H2O

Cl

+ RCl

AlCl3

R

+ HCl

R + HCl

excess

d. Formation of benzaldehyde by a Gatterman–Koch reaction (Section 19.8) O CO

+

HCl

high pressure AlCl3/CuCl

+

C

H

2. Reduction of a carbonyl group to a methylene group (Section 19.9) O C

R

H2 Pd/C

R

Zn(Hg), HCl, Δ Clemmensen reduction

CH2R

O C

CH2R

O C

R

H2NNH2, HO−, Δ Wolff–Kishner reduction

CH2R


Summary of Reactions

3. Alkyation via coupling reactions (Section 19.10) a. Alkylation by a Suzuki reaction Br

OR +

R

R

PdL2 HO−

B OR

b. Alkylation with an organocuprate Br

R +

+

R2CuLi

+ LiBr

RCu

4. Reactions of a substituent on a benzene ring (Section 19.12) CH3

CH2Br

Br2

CH2Z

Z−

Z− = a nucleophile

hn

CH3

NO2

COOH

H2CrO4 Δ

NH2

H2 Pd/C

5. Reaction of aniline with nitrous acid (Section 19.23). The mechanism is shown on page 966. NH2

+

N

NaNO2, HCl 0 °C

N Cl−

6. Replacement of a diazonium group (Section 19.21) +

N

+

N

+

N

N Br−

N Cl−

+

+

N

Cl

CuCl

N −

Cl

N

Br

CuBr

CuC

N Cl−

HBF4 Δ

N Cl−

KI

N

C

F

I

+ N2

+ N2

N

+ N2

+ BF3 + N2

+ N2

961


962

CHAPTER 19

Reactions of Benzene and Substituted Benzenes +

N

N Cl

+

N

+

N

OH

H3O+ Δ

+ HCl + N2

N

OH

Cu2O Cu(NO3)2, H2O

Cl− N

H

H3PO2

Cl−

+ N2

+ N2

7. Formation of an azo compound (Section 19.22). The mechanism is shown on page 964. OH

+

N

+

N Cl−

HO

N N

8. Nucleophilic aromatic substitution reactions (Section 19.24). The mechanism is shown on page 969. Br

OCH3 +

Δ

CH3O−

+

Br−

NO2

NO2

P R OBLEMS 46. Draw the structure for each of the following: a. phenol b. benzyl phenyl ether c. benzonitrile

d. benzaldehyde e. anisole f. styrene

g. toluene h. tert-butylbenzene i. benzyl chloride

47. Name the following: COOH CH

d.

a.

CH2

CH3

g.

O2N

Br

Br

OCH3 Br

b. Br

e.

h. CH3 CH2CH3

Br OH

c. H3C

CH2CH3

SO3H CH3

i.

f. Cl

Cl

Cl N N


Problems

963

48. Provide the necessary reagents next to the arrows.

O Br

Cl

NO2

SO3H

NH2

BrCHCH3

CH2CH3

CCH3

COOH

HOCHCH3

CH2CH3

CH2CH2Br N +

CH

CH2

N CH2CH2OH

F

Cl

Br

I

N

OH

C

CH2NH2

49. Draw the structure for each of the following: a. m-ethylphenol d. o-bromoaniline b. p-nitrobenzenesulfonic acid e. 4-bromo-1-chloro-2-methylbenzene c. (E)-2-phenyl-2-pentene f. m-chlorostyrene

g. o-nitroanisole h. 2,4-dichloromethylbenzene i. m-chlorobenzoic acid

50. For each of the statements in Column I, choose a substituent from Column II that fits the description for the compound on the right: Column I

Column II

a. Z donates electrons by hyperconjugation and does not donate or withdraw electrons by resonance.

OH

b. Z withdraws electrons inductively and withdraws electrons by resonance.

Br

c. Z deactivates the ring and directs ortho–para.

+

d. Z withdraws electrons inductively, donates electrons by resonance, and activates the ring.

CH2CH3

e. Z withdraws electrons inductively and does not donate or withdraw electrons by resonance.

NO2

NH3

Z


964

CHAPTER 19

Reactions of Benzene and Substituted Benzenes

51. What product would be obtained from the reaction of excess benzene with each of the following reagents? b. 1-chloro-2,2-dimethylpropane + AlCl3 c. dichloromethane + AlCl3 a. isobutyl chloride + AlCl3 52. Draw the product(s) of each of the following reactions: a. benzoic acid + HNO3/H2SO4 b. isopropylbenzene + Cl2 + FeCl3 c. p-xylene + acetyl chloride + AlCl3 followed by H2O d. o-methylaniline + benzenediazonium chloride

e. f. g. h.

cyclohexyl phenyl ether + Br2 phenol + H2SO4 + Δ ethylbenzene + Br2/FeBr3 m-xylene + Na2Cr2O7 + HCl + Δ

53. Rank the following substituted anilines in order from most basic to least basic: O NH2

CH3

CH3O

CH3C

NH2

Br

NH2

NH2

54. For each horizontal row of substituted benzenes, indicate a. the one that would be the most reactive in an electrophilic aromatic substitution reaction. b. the one that would be the least reactive in an electrophilic aromatic substitution reaction. c. the one that would yield the highest percentage of meta product in an electrophilic aromatic substitution reaction. CH3

CHF2

+

CF3

+

N(CH3)3

+

CH2N(CH3)3

CH2CH2N(CH3)3

O OCH2CH3

CH2OCH3

COCH3

55. The compound with the 1H NMR spectrum shown here is known to be highly reactive toward electrophilic aromatic substitution. Identify the compound.

3

2

10

9

8

7

6

5

d (ppm)

3

4

2

1

frequency

56. Draw the product of each of the following reactions: a.

Cl

AlCl3

Cl

b.

AlCl3

+ Cl

0


Problems

57. Show how the following compounds could be synthesized from benzene: d. 1-phenylpentane a. m-chlorobenzenesulfonic acid e. m-bromobenzoic acid b. m-chloroethylbenzene f. m-hydroxybenzoic acid c. m-bromobenzonitrile

965

g. p-cresol h. benzyl alcohol i. benzylamine

58. How many electrophilic aromatic substitution products are obtained from the chlorination of a. o-xylene? b. p-xylene? c. m-xylene? 59. Arrange the following groups of compounds in order from most reactive to least reactive toward electrophilic aromatic substitution: a. benzene, ethylbenzene, chlorobenzene, nitrobenzene, anisole b. 1-chloro-2,4-dinitrobenzene, 2,4-dinitrophenol, 1-methyl-2,4-dinitrobenzene c. toluene, p-cresol, benzene, p-xylene d. benzene, benzoic acid, phenol, propylbenzene e. p-methylnitrobenzene, 2-chloro-1-methyl-4-nitrobenzene, 1-methyl-2,4-dinitrobenzene, p-chloromethylbenzene f. bromobenzene, chlorobenzene, fluorobenzene, iodobenzene 60. What are the products of the following reactions? O

a.

CH3

C

+ HNO3

O CH3

b.

d.

H2SO4 Δ

1. Mg/Et2O 2. D2O

+ Br2

N

e.

CH3

f.

CF3

1. 2. 3. 4.

Br2, hn Mg/Et2O ethylene oxide HCl

Br O OCH3

c.

1. AlCl3 O 2. H2O

+

FeCl3

+ Cl2

O 61. Describe two ways to prepare anisole from benzene. 62. For each of the following compounds, indicate the ring carbon that would be nitrated if the compound is treated with HNO3/H2SO4: O

OCH3

NO2 a.

C

e.

c.

CH3 CH3

g.

Br

COOH

NO2

Cl O b. CH3

O

C

C O

CH3

Cl

OH

OH

h.

f.

d.

O

CH3 COOH

63. Show two ways that the following compound could be synthesized: O C CH3 64. Why is anisole nitrated more rapidly than thioanisole under the same conditions? OCH3 anisole

SCH3 thioanisole

Cl


966

CHAPTER 19

Reactions of Benzene and Substituted Benzenes

65. If anisole is allowed to sit in D2O that contains a small amount of D2SO4, what products will be formed? 66. Which of the following compounds will react with HBr more rapidly? CH3

CH

CH2

or

CH3O

CH

CH2

67. An aromatic hydrocarbon with a molecular formula of C13H20 has an 1H NMR spectrum with a signal at ~7 ppm that integrates to 5H. It also has two singlets; one of the singlets has 1.5 times the area of the second. What is the structure of the aromatic hydrocarbon? 68. The following tertiary alkyl bromides undergo an SN1 reaction in aqueous acetone to form the corresponding tertiary alcohols. List the alkyl bromides in order from most reactive to least reactive. Br

Br

Br

Br

Br

CH3CCH3

CH3CCH3

CH3CCH3

CH3CCH3

CH3CCH3

CH2CH2CH3

OCH2CH3

SO3H

A

B

C

OCCH3

ClCHCH3

O D

E

69. Show how the following compounds could be synthesized from benzene: a. N,N,N-trimethylanilinium iodide b. 2-methyl-4-nitrophenol c. p-benzylchlorobenzene d. benzyl methyl ether e. p-nitroaniline f. m-bromoiodobenzene g. p-dideuteriobenzene h. p-nitro-N-methylaniline i. 1-bromo-3-nitrobenzene 70. Use the four compounds shown here to answer the following questions: COOH

COOH

COOH

F pKa = 4.2

COOH

Cl

pKa = 3.3

pKa = 2.9

Br pKa = 2.8

a. Why are the ortho-halo-substituted benzoic acids stronger acids than benzoic acid? b. Why is o-fluorobenzoic acid the weakest of the ortho-halo-substituted benzoic acids? c. Why do o-chlorobenzoic acid and o-bromobenzoic acid have similar pKa values? 71. a. List the following esters in order from most reactive to least reactive in the first slow step of a nucleophilic addition–elimination reaction (formation of the tetrahedral intermediate): O C CH3

O O A

C CH3

O O B

C CH3

O C

O

CH3 C

CH3

Cl

O D

b. List the same esters in order from most reactive to least reactive in the second slow step of a nucleophilic addition–elimination reaction (collapse of the tetrahedral intermediate).


Problems

967

72. A mixture of 0.10 mol benzene and 0.10 mol p-xylene was allowed to react with 0.10 mol nitronium ion until all the nitronium ion was gone. Two products were obtained: 0.002 mol of one and 0.098 mol of the other. a. What was the major product? b. Why was more of one product obtained than of the other? 73. What are the products of the following reactions? O a.

Cl

b.

1. AlCl3

Cl

2. H2O

c.

1. AlCl3 2. H2O

O

CH2Cl

+

1. AlCl3 2. H2O

74. Benzene underwent a Friedel–Crafts acylation followed by a Clemmensen reduction. The product gave the following 1H NMR spectrum. What acyl chloride was used in the Friedel–Crafts acylation?

3 2 2 2

5

10

9

8

7

6

5

d (ppm)

4

3

2

1

0

frequency

75. Would m-xylene or p-xylene react more rapidly with Cl2 + FeCl3? Explain your answer. 76. What products would be obtained from the reaction of the following compounds with H2CrO4 + ⌬ ? CH2CH2CH2CH3

CH2CH3 a.

CH3 c.

b. CHCH3

CH3

C(CH3)3

CH3 77. Which set of underlined hydrogens will have its 1H NMR signal at a higher frequency? a.

CH3CH2CH3

or

b.

CH3OCH2CH3

CH3CH

CH2

or

CH3OCH

CH2

78. Friedel–Crafts alkylations can be carried out with carbocations formed from reactions other than the reaction of an alkyl halide with AlCl3. Propose a mechanism for the following reaction:

+

HF


968

CHAPTER 19

Reactions of Benzene and Substituted Benzenes

79. Show how the following compounds could be prepared from the given starting materials. You can use any necessary organic or inorganic reagents. O a.

CH3

C

OH

CH3

b.

NHCH3

O

80. A chemist isolated an aromatic compound with molecular formula C6H4Br2. He treated this compound with nitric acid and sulfuric acid and isolated three different isomers with molecular formula C6H3Br2NO2. What was the structure of the original compound? 81. List the following compounds in order from largest K eq to smallest K eq for hydrate formation: O

O C

C

CH3

O C

CH3

Cl

O C

CH3

CH3

CH3O

O2N

82. a. Describe four ways the following reaction could be carried out. O

b. Describe three ways the following reaction could be carried out. O

83. Propose a mechanism for each of the following reactions:

a.

b.

HCl

CH

CH2

84. How could you prepare the following compounds with benzene as one of the starting materials? O a.

b.

C

85. Describe how naphthalene could be prepared from the given starting material. O C CH2CH2CH2

Cl

?

CH3 HCl


Problems

969

86. Identify A–J: O 1. CH3CCl AlCl3 2. H2O

HNO3 H2SO4

A

B

H2NNH2 HO−, Δ D

H2CrO4

Δ

C

Br2

CH3O−

E

h

HI

F

G

Δ

(CH3)3CO−

H

HBr

I

peroxide

−C

N

J

87. Using resonance contributors for the intermediate carbocation, explain why a phenyl group is an ortho–para director.

+

Cl2

FeCl3

+

biphenyl

Cl

Cl

88. The pKa values of a few ortho-, meta-, and para-substituted benzoic acids are shown here: COOH Cl

COOH

COOH

COOH NO2

COOH

Cl

NO2 Cl

pKa = 2.94

pKa = 3.83

COOH

pKa = 3.99

COOH NH2

pKa = 2.17

COOH

pKa = 3.49

NO2

pKa = 3.44

COOH

NH2 pKa = 4.95

pKa = 4.73

NH2

pKa = 4.89

The relative pKa values depend on the substituent. For chloro-substituted benzoic acids, the ortho isomer is the most acidic and the para isomer is the least acidic; for nitro-substituted benzoic acids, the ortho isomer is the most acidic and the meta isomer is the least acidic; and for amino-substituted benzoic acids, the meta isomer is the most acidic and the ortho isomer is the least acidic. Explain these relative acidities. c. NH2: meta 7 para 7 ortho a. Cl: ortho 7 meta 7 para b. NO2: ortho 7 para 7 meta


970

CHAPTER 19

Reactions of Benzene and Substituted Benzenes

89. When heated with chromic acid, compound A forms benzoic acid. Identify compound A from its 1H NMR spectrum.

6

1

5

3.0

8

7

6

2.9

5

2.8 PPM

4

3

d (ppm)

2

1

0

frequency

90. Describe two synthetic routes for the preparation of p-methoxyaniline from benzene. 91. Which is a more stable intermediate in each pair? OH

CH3

a.

or

+

H

NO2

NO2 b.

+

H

NO2

HO

NO2 −

or

Cl OH

Cl

92. What reagents would be required to carry out the following transformations? O

O

O

O O−

H

O

93. Show how the following compounds could be prepared from benzene: CH3 a.

b. CH3C

CH2

OCH3 Br

O c.

SO3H

C CH3

CH3CH2

CH2CH(CH3)2

NO2

C(CH3)3

d.

Br

94. An unknown compound reacts with ethyl chloride and aluminum trichloride to form a compound that has the following 1H NMR spectrum. What is the structure of the compound?

3

2

2

10

9

8

7

6

5

d (ppm)

4

frequency

3

2

1

0


971

Problems

95. How could you distinguish the following compounds, using a. their infrared spectra? b. their 1H NMR spectra? O CH2OH

CH2OCH3

CH2OH

A

B

C

C

O

O C

OH

D

C

H

E

O C

OCH3

F

CH3

G

96. p-Fluoronitrobenzene is more reactive toward hydroxide ion than is p-chloronitrobenzene. What does this tell you about the ratedetermining step for nucleophilic aromatic substitution? 97. a. Explain why the following reaction leads to the products shown: OH NaNO2 HCl

CH3CHCH2NH2

CH3CCH3 + CH3C

CH3

CH2

CH3

CH3

b. What product would be obtained from the following reaction? OH NH2 CH3

C

C

NaNO2 HCl

CH3

CH3 CH3 98. Describe how mescaline could be synthesized from benzene. The structure of mescaline is given on page 908. 99. Propose a mechanism for the following reaction that explains why the configuration of the asymmetric center in the reactant is retained in the product: COO− O −

O NaNO2 HCl

C

OCCH2CH2

H NH2

O H COO−

100. Explain why hydroxide ion catalyzes the reaction of piperidine with 2,4-dinitroanisole but has no effect on the reaction of piperidine with 1-chloro-2,4-dinitrobenzene.

N H piperidine

101. Propose a reasonable mechanism for each of the following reactions: CH2Br OH

a.

+

Δ

O

O

O

NO2

NO2 O NO2

O b.

H3C

S OH

O− Na+

O CH3O− Na+

NO2

H3C

NO2

S O NO2


972

CHAPTER 19

Reactions of Benzene and Substituted Benzenes

102. What are the products of the following reactions? CH3

O a.

HOOC b.

1. AlCl3 2. H2O

O +

O 1. SOCl2 2. AlCl3 3. H2O

O

103. Tyramine is an alkaloid found in mistletoe and ripe cheese. Dopamine is a neurotransmitter involved in the regulation of the central nervous system. HO HO

CH2CH2NH2

HO

CH2CH2NH2

tyramine

a. b. c. d. e.

dopamine

How can tyramine be prepared from b-phenylethylamine? How can dopamine be prepared from tyramine? Give two ways to prepare b-phenylethylamine from b-phenylethyl chloride. How can b-phenylethylamine be prepared from benzyl chloride? How can b-phenylethylamine be prepared from benzaldehyde?

104. Describe how 3-methyl-1-phenyl-3-pentanol can be prepared from benzene. You can use any inorganic reagents and solvents, and any organic reagents provided they contain no more than two carbons. 105. a. How could aspirin be synthesized from benzene? b. Ibuprofen is the active ingredient in pain relievers such as Advil, Motrin, and Nuprin. How could ibuprofen be synthesized from benzene? c. Acetaminophen is the active ingredient in Tylenol. How could acetominophen be synthesized from benzene? O

OH H N

HO

O

O

O aspirin

O

OH

acetaminophen

ibuprofen

106. a. Ketoprofen, like ibuprofen, is an anti-inflammatory analgesic. How could ketoprofen be synthesized from the given starting material? O CH3

CH

C

OH

CH3 O

O ketoprofen

b. Ketoprofen and ibuprofen both have a propanoic acid substituent (see Problem 105). Explain why the identical subunits are synthesized in different ways. 107. Show how Novocain, a painkiller used frequently by dentists, can be prepared from benzene and compounds containing no more than four carbons. O O H2N

NovocainÂŽ

N


Problems

973

108. Show how lidocaine, one of the most widely used injectable anesthetics, can be prepared from benzene and compounds containing no more than four carbons. H3C NH

N O

H3C

lidocaine

109. Saccharin, an artificial sweetener, is about 300 times sweeter than sucrose. Describe how saccharin could be prepared from benzene. O C NH S O saccharin

O


T U T O R I A L Enhanced by

SYNTHESIS AND RETROSYNTHETIC ANALYSIS Organic synthesis is the preparation of organic compounds from other organic compounds. The word synthesis comes from the Greek word synthesis, which means “a putting together.” You have been introduced to many aspects of organic synthesis, and have had the opportunity to design the synthesis of a lot of organic compounds. In this tutorial, we will examine some of the strategies chemists use when designing a synthesis. The most important factor in designing a synthesis is to have a good command of organic reactions. The more reactions you know, the better your chances of coming up with a useful synthesis. The guiding factor in planning a synthesis is to keep it as simple as possible. The simpler the synthetic plan, the greater the chance it will be successful.

CHANGING THE FUNCTIONAL GROUP The first thing to do when designing a synthesis is to compare both the carbon skeleton and the positions of the functional group in the reactant and the product. If they both are the same, then all you need to determine is how to convert the functional group in the reactant to the functional group in the product. For examples in addition to those below, review Problem 70 in Chapter 6, Problem 32 in Chapter 7, Problem 36 in Chapter 9, Problem 52 in Chapter 11, Problem 56 in Chapter 17, and Problem 61 in Chapter 18. HO−

O

HO−

Br

Br

NaOCl 0 °C CH3COOH

HO−

CH3O−

Br2 hN

DBN

H2 Pd

OH

OCH3

Br2

H2 Lindlar catalyst

Br Br

H2 Pd

−NH 2 excess

Notice that HO− is used as the base to form a double bond between C-2 and C-3, whereas a bulky base (DBN) is needed to put the double bond between C-1 and C-2. Only two SN2 reactions are shown here (using CH3O− and HO−), but many other products could be synthesized just by changing the nucleophile. Notice, too, how single, double, and triple bonds can be interconverted.

FUNCTIONALIZING A CARBON Recall that a carbon can be functionalized by a radical reaction. Br2 hN

974

Br

NBS, ¢ peroxide

Br


CHANGING THE POSITION OF THE FUNCTIONAL GROUP If the carbon skeleton has not changed but the position of the functional group has, then you need to consider reactions that will change the position of the functional group. Recall that in the electrophilic addition reactions shown here, ~Br and R2BH (which is replaced by OH) are electrophiles, so they add to the sp2 carbon bonded to the most hydrogens. 1. BHR2/THF

DBN

2. H2O2, HO-, H2O

Br

OH

HBr peroxide

Br

CHANGING THE CARBON SKELETON If the carbon skeleton has changed but the number of carbons has not, then you need to consider reactions that will form a carbocation intermediate, because you know that a carbocation will rearrange if it can form a more stable carbocation. H2SO4 H2O

PROBLEM 1

H2O

+

OH

What reagents are required to convert the reactant into the product?

a.

e.

b.

f.

OH g.

c. OH

OH d.

Br

h. Br

OH

ADDING ONE CARBON TO THE CARBON SKELETON There are several ways to create a product with one more carbon than the reactant. The method you choose will depend on the functional group that you want to end up with in the product.

975


−C

Br

N

N

H2

NH2

Raney nickel HCl H2O ¢

Mg Et2O

O 1. CO2

MgBr

OH

2. HCl 1. H2C O 2. HCl NaOCl

OH

O

CH3COOH

ADDING MORE THAN ONE CARBON TO THE CARBON SKELETON There are many different ways to increase the carbon skeleton by more than one carbon. Acetylide ions, epoxides, Grignard reagents, aldol additions, Wittig reactions, and coupling reactions are just some of the methods that can be used. Common methods used to form new C—C bonds are summarized in Appendix IV.

PROBLEM 2

Starting with bromocyclohexane, how can each of the following compounds be prepared? OH

a.

COOH

b.

CH2OH

CH2NH2

c.

CH2CH2OH

d.

e.

CHCH3

Describe how the following compounds could be prepared from compounds containing no more than six carbons. (You can also use triphenylphosphine.)

PROBLEM 3

O

O a.

b.

c.

H

d.

USING RETROSYNTHETIC ANALYSIS TO CREATE A FUNCTIONAL GROUP When you know what functional group you want to create, you can try to remember the various ways it can be synthesized. For example, a ketone can be synthesized by the acid-catalyzed addition of water to an alkyne, hydroboration–oxidation of an alkyne, oxidation of a secondary alcohol, and ozonolysis of an alkene. Notice that ozonolysis decreases the number of carbons in a molecule.

976


OH

H2O

H2SO4

NaOCl

O

CH3COOH 0 °C

1. O3, −78 °C 2. (CH3)2S

1. R2BH/THF − 2. H2O2, HO , H2O

In addition, methyl ketones can be synthesized by the acetoacetic ester synthesis, aromatic ketones can be synthesized by a Friedel–Crafts acylation, and a cyclic ketone, when treated with diazomethane, forms the next-size-larger cyclic ketone. Unless you have an exceptional memory, recalling all the methods you have learned to synthesize a particular functional group might be challenging. Therefore, they are listed for you in Appendix III.

a. an ether?

How many ways can you recall to synthesize b. an aldehyde? c. an alkene? d. an amine?

PROBLEM 5

Describe three ways to synthesize the following compound:

PROBLEM 4

O

USING DISCONNECTIONS IN RETROSYNTHETIC ANALYSIS We have seen that a disconnection can be a useful step in a retrosynthetic analysis (Sections 17.17 and 18.20). Recall that a disconnection involves breaking a bond to give two fragments and then adding a positive charge to the end of one fragment and a negative charge to the end of the other. If the following compound is disconnected at bond a, then we can see that the target molecule can be prepared from the reaction of cyclohexanone with a Grignard reagent. HO a CH2

CH

HO

CH2

+

a disconnection

+

CH2

CH

CH2

Mg

O

BrCH2

CH

CH2

On the other hand, if the compound is disconnected at bond b, then we see that an epoxide is the synthetic equivalent for the electrophile and an organocuprate is the synthetic equivalent for the nucleophile.

977


HO

CH2

b

CH

CH2

HO

+

CH2

a disconnection

+

CH CH2

LiCu(CH CH2

CH2)2

O BrCH

CH2

P R O B L E M 6 Do a retrosynthetic analysis on each of the following compounds, ending with the given starting materials:

a. OH O +

b.

+

C H

O

H

O

USING THE RELATIVE POSITIONS OF TWO FUNCTIONAL GROUPS TO DESIGN A SYNTHESIS If a compound has two functional groups, then the relative positions of the two groups can provide a valuable hint as to how to approach the synthesis. For example, the following synthesis forms a 1,2-dioxygenated compound. O

−C

N

HCl

OH C

N

OH O

HCl H2O ¢

OH a 1,2-dioxygenated compound

Retrosynthetic analysis shows that a 1,3-dioxygenated compound can be formed by an aldol addition. OH

O

HO a disconnection

H

+

O

O −

O H

H

H

a 1,3-dioxygenated compound

O

O

O

O

a disconnection

− +

OH

OH

O

a 1,3-dioxygenated compound

Disconnection of a 1,4-dioxgenated compound shows it can be synthesized by nucleophilic attack of a negatively charged a-carbon (an enolate ion) on a compound with a positively charged a-carbon. 978


O A

+

O

OCH3 O

OCH3 O

O

B

a 1,4-dioxygenated compound

+

OCH3 O

An a-brominated carbonyl compound is the synthetic equivalent for the positively charged a-carbon and an enamine is the synthetic equivalent for the negatively charged a-carbon. An enamine avoids the requirement for a strong base that could remove a proton from the brominated carbonyl compound rather than from the nonbrominated carbonyl carbon. Because esters cannot form enamines, path A is the preferred disconnection. Fortunately, the iminium ion hydrolyzes more readily than the ester. O

trace acid

+

N Br

N H

N+

OCH3

HCl

OCH3

O

O OCH3

H2O

O

O

Disconnection of a 1,5-dioxgenated compound shows it can be synthesized by nucleophilic attack of a negatively charged a-carbon on a carbonyl compound with a positively charged b-carbon. O

O

O

+

O

a 1,5-dioxygenated compound

An a, b-unsaturated carbonyl compound is the synthetic equivalent of a compound with a positively charged b-carbon. And we have seen that an enamine is the synthetic equivalent for the negatively charged a-carbon.

N

+

N

O

O

HCl

+

O

O

H2O

P R O B L E M 7 Describe how the following compound could be synthesized from compounds containing no more than six carbons.

O O

Disconnection of a 1,6-dioxgenated compound shows it can be synthesized by nucleophilic attack of a negatively charged a-carbon on a carbonyl compound with a positively charged g-carbon. O

O

O

+

O

a 1,6-dioxygenated compound

979


There is no synthetic equivalent for a carbonyl compound with a positively charged g-carbon, so we have to consider another route. Recognizing that a 1,6-dioxygenated compound can be prepared by oxidative cleavage of a cyclohexene provides an easy route to the target molecule, because a compound with a six-membered ring can be readily prepared by a Diels–Alder reaction. O

O

P R O B L E M 8 Do a retrosynthetic analysis on each of the following compounds, ending with available starting materials.

OH a.

O

O H

e. H

c.

H

O OH

OH

b.

OH

d. OH

Use retrosynthetic analysis to plan a synthesis of the following target molecules. The only carbon-containing compounds available for the syntheses are cyclohexanol, ethanol, and carbon dioxide.

PROBLEM 9

O H

a. H

OH b.

c.

d. OH

O

How could you synthesize the following compounds from starting materials containing no more than four carbons?

PROBLEM 10

OH a.

OH

b.

c.

d.

Br

OH

Show how the following compounds could be synthesized from the given starting materials:

PROBLEM 11

O

OH

Br

a.

O b. OH

980

CH3Br

O OH


EXAMPLES OF MULTISTEP ORGANIC SYNTHESIS To give you an idea of the kind of thinking required for the synthesis of a complicated molecule, we will look at the synthesis of lysergic acid, which was done by R. B. Woodward, and the synthesis of caryophyllene, which was done by E. J. Corey. Woodward (in 1965) and Corey (in 1990) both received the Nobel Prize for their contributions to synthetic organic chemistry. Lysergic acid was first synthesized by Woodward in 1954. The diethylamide of lysergic acid (LSD) is a better known compound because of its hallucinogenic properties. It is a tribute to Woodward’s ability as a synthetic chemist that the next synthesis of lysergic acid was not accomplished until 1969. Lysergic acid has an indole ring. Because indole is unstable under acidic conditions (Section 20.5), Woodward designed a synthesis that did not form the indole portion until the final step.

O

1. Br2, H+, H2O

OH O

O

3. HCl, H2O

C6H5

N

NCH3

HNCH3

3. H2O

C6H5

O

2. O O

1. SOCl2 2. AlCl3

O

C6H5

N O

N O NaOCH3, CH3OH

N O

C NCH3

HN

O NCH3

1. H2O, HO−, Δ 2. Raney Ni

NCH3

1. NaBH4 2. SOCl2 3. NaCN

C6H5

N

C6H5

N

lysergic acid

O

+

H2O

O

• The starting material is a dihydroindole carboxylic acid with its nitrogen protected by being converted into an amide. • An intramolecular Friedel–Crafts reaction forms the third ring. • Acid-catalyzed bromination of the a-carbon is followed by an SN2 substitution of the bromine by an amine and removal of the ketone’s protecting group. • An aldol condensation forms the fourth ring. • The ketone carbonyl group is reduced to an OH group, which is activated by thionyl chloride and replaced by a cyano group. • Both the cyano group and the protecting amide group are hydrolyzed. • The last step is the reverse of hydrogenation of a double bond. It is carried out by using a typical hydrogenation catalyst in the absence of hydrogen. Caryophyllene is an oil found in cloves. The most important strategic element in  its synthesis was the realization that the nine-membered ring could be formed by  fragmenting a six-membered ring fused to a five-membered ring. The problem then  became designing the synthesis of the compound that would undergo fragmentation. 981


H

H hN pentane

+

H 1. NaH 2. CH3l

1. NaH 2. (CH3O)2CO

H

O

H

O

O

COOCH3

CH3 H +− Li C

H

H base

CH3 O H HO

COO− HO−,

Δ

H

O COOCH3

2. H+, H2O 3. CrO3

NaBH4

COOCH3

H CH3 OH

TsCl pyridine

H HO

+ CO2

C OH

CCH(OCH3)2

H

H HO

H

O

H CH3 O

CH3

a lactone

H2O

CCH(OCH3)2

1. H2, Pd/C

CH3 H

COOCH3

O

CH3 OTs H HO base

H

H CH3

H CH2

(C6H5)3P

H CH3

CH2

CH3

epimerization

H O

H O

caryophyllene

• The first step is a photochemical [2+2] cycloaddition reaction (Section 28.4). • One a-hydrogen of the resulting ketone is substituted with a methoxycarbonyl group, and the second a-hydrogen on the same carbon is substituted with a methyl group. (Today we would use LDA, not NaH.) • An acetylide ion is used to add a three-carbon fragment to the carbonyl group of the ketone. • The triple bond is reduced, the acetal is hydrolyzed, and the resulting aldehyde is oxidized to a carboxylic acid that forms a lactone. • A Claisen condensation and hydrolysis of the lactone forms the 3-oxocarboxylic acid that is decarboxylated to give the cyclic ketone. • Reduction of the ketone forms an alcohol. • Because the secondary alcohol is more reactive than the sterically hindered tertiary alcohol, the desired tosylate is formed. • The stage is now set for the desired fragmentation reaction. This reaction involves removing a proton from the OH group, which allows the oxygen to form a carbonyl group and eliminate the tosyl group. • Epimerization of the asymmetric center adjacent to the carbonyl group occurs once the ketone has been formed. • A Wittig reaction forms the exocyclic double bond.

982


Answers to Problems on Synthesis and Retrosynthetic Analysis PROBLEM 1 Na/NH3(liq)

a.

−78 °C H2

b.

1. Br2

e.

2. 2 −NH2

1. NBS, peroxide, ¢

f.

2. tert-BuO−

Lindlar catalyst

OH c.

d.

H2SO4

H2SO4

¢

H2O

Br

− tert-BuO −

HBr

Br CH3O−

HBr

g.

OH

Br

OH

OTs TsCl

h.

tert-BuO−

1. R2BH2 2. H2O2, HO

OH

PROBLEM 2

a.

Br

b.

Br

−C

N N

COOH

¢

Mg

MgBr

Br

−C

1. H2C

O

CH2OH

2. H3O+

Et2O

c.

HCl, H2O

N

H2

N

CH2NH2

Raney Nickel

d.

Br

CuLi

1. Li

O

CH2CH2OH

1. 2. HCl

2. CuI/THF 2

OH e.

Br

Mg Et2O

MgBr

1. CH3CH 2. H3O+

O

CHCH3

983


PROBLEM 3

a. An alkene can be made in one step by a Wittig reaction. O

(C6H5)3P

CH3(CH2)3Li

+

(C6H5)3P

(C6H5)3P

Br

b. The alkyne can prepared from 1-hexyne and ethyl bromide as shown here or it can be prepared from 1-butyne and butyl bromide. 1. −NH2 2. CH3CH2Br

c. The desired aldehyde has eight carbons and a structure that suggests it can be prepared by an aldol condensation using an aldehyde with four carbons. O

OH

HO−

O H3O

H

H

H

¢

O

H2

d.

O

+

MCPBA

Lindlar catalyst the product of the synthesis in part b

PROBLEM 4

See Appendix III on page A-8.

PROBLEM 5

O

O

1.

H 2. H3O+, Δ

O

O 1. AlCl3

+

Cl 2. H O 2

O−

O H

+

MgBr

O PCC CH2Cl2

Mg Et2O

Br

984

O

O LDA


PROBLEM 6

+

a.

OH

OH

MgBr O

Br OH 1. O3

O

2. (CH3)2S

OH

b. O

OH

+

OH

O

CuLi

O

Br

OH

2

Br

MgBr

H2C O

OH

Because the target molecule is a 1,5-dioxygenated compound, it can be synthesized from a negatively charged a-carbon (using an enamine as the synthetic equivalent) and an a,b-unsaturated ketone. PROBLEM 7

O

O

trace acid

+ N H

N

O

O +

N

HCl

O

H2O

985


PROBLEM 8

OH a.

O

+

H

OH

O

O

O

OH

H

H

+

H

O

O

+

b.

O

O

OH

Br O

N

OH

continue as in b

c.

N

Br O

OH OH

OH

O

O

O

d.

+

O −

O N

O H

e. H O

PROBLEM 9

O

OH

H

a. H O

b.

CuLi

Br

Br

OH

2

986


OH

O

c.

2

O

MgBr

O

Br OH +

OH OH

MgBr + CO2

O d.

2

O

OH

MgBr

Br O OH +

OH OH

OH

Br

MgBr +

CO2

PROBLEM 10

O

O a.

H

H2

H

OH

Raney nickel

O

O b.

H2

NH2NH2

HO−, ¢

c.

Pd/C

O ¢

1. O3, −78 °C 2. (CH3)2S

H

H O

1. 2 CH3MgBr 2. H3O+

OH OH

987


Br d.

¢

NBS, ¢ peroxide

PROBLEM 11

O +

a.

OH CH3

CH3MgBr

CH3

H2SO4

CH3

MCPBA

O

¢

Mg Et2O

Br

CH3Br

Mg

MgBr

Et2O

O−

OH H2O

O b.

O

O

HO−

+

H2O ¢

MgBr

Mg Et2O

O−

H2O

PBr3

OH

Br

OH

for Organic Chemistry MasteringChemistry tutorials guide you through the toughest topics in chemistry with self-paced tutorials that provide individualized coaching. These assignable, in-depth tutorials are designed to coach you with hints and feedback specific to your individual misconceptions. For additional practice on Synthesis and Retrosynthetic Analysis, go to MasteringChemistry where the following tutorials are available: • Synthesis and Retrosynthetic Analysis: Functional Groups • Synthesis and Retrosynthetic Analysis: Carbon Chain • Synthesis and Retrosynthetic Analysis: Retrosynthesis of 2-Pentanone Using Reactions of Carbonyl Compounds

988


20

More About Amines • Reactions of Heterocyclic Compounds

Auletta

In 1994, a team from UC Santa Cruz reported the structures of milnamides A and B, compounds the team had obtained from the sponge Auletta. Subsequently, other researchers in various countries isolated these compounds from three other coral reef sponges. Collaborative research between scientists at the University of British Columbia and Wyeth Pharmaceuticals lead to an explanation of why breaking the saturated heterocyclic ring of milnamide A resulted in a compound (milnamide B) with greater potency against solid tumor cancer cells. Using milnamide B as a lead compound, chemists at Eisai Inc. designed and synthesized E7974 (see page 518). This compound shows broad activity against a variety of human tumors and exhibits better efficacy than the classic anticancer drug, 5-fluorouracil (see page 1162). It is now being evaluated as an anticancer drug in Phase I clinical trials (see page 301). Many believe this sponge-inspired compound may someday be a blockbuster drug. CH3

O

N

NCH3

N H

N

CH3

O CO2H

O

N

CH3

CH3

milnamide A

A

NCH3

N H

N O

milnamide B

CH3

O CO2H

N

N H

N

CO2H

O

E7974

mines, compounds in which one or more of the hydrogens of ammonia (NH3) have been replaced by an alkyl group, are among some of the most abundant compounds in the biological world. We will come to appreciate their biological importance as we explore the structures and properties of amino acids and proteins in Chapter  22; as we study how enzymes catalyze reactions in Chapter 23; as we investigate the ways in which coenzymes—compounds derived from vitamins—help enzymes catalyze reactions in Chapter 24; and as we learn about nucleic acids (DNA and RNA) in Chapter 26. Amines are exceedingly important compounds to organic chemists—far too important to leave until the end of a course in organic chemistry. You have, therefore, already studied many aspects of amines and their chemistry. For example, you have seen that the nitrogen 989


990

CHAPTER 20

More About Amines • Reactions of Heterocyclic Compounds

in amines is sp3 hybridized with the lone pair residing in an sp3 orbital (Section 3.8), and you know that amines invert rapidly at room temperature through a transition state in which the sp3 nitrogen becomes an sp2 nitrogen (Section 4.16). We also have examined the physical properties of amines—their hydrogen-bonding properties, boiling points, and solubilities (Section 3.9)—and we learned how amines are named (Section 3.7). Most importantly, we have seen that the lone-pair electrons of the nitrogen atom cause amines to react as bases (that is, to share their lone pair with a proton) and as nucleophiles (that is, to share their lone pair with an atom other than a proton). an amine is a base

R

+

NH2

H

Br

+

R

NH3

R

NH2

+

Br−

an amine is a nucleophile

R

NH2

+

CH3

Br

+

CH3 +

Br−

In this chapter, we will revisit some of these topics and look at other aspects of amines and their chemistry that we have not considered previously. Some amines are heterocyclic compounds (or heterocycles)—cyclic compounds in which one or more of the atoms of the ring are heteroatoms (Section 8.10). A variety of atoms, such as N, O, S, Se, P, Si, B, and As, can be incorporated into ring structures. Heterocycles are an extraordinarily important class of compounds, making up more than half of all known organic compounds. Most drugs, most vitamins (Chapter 24), and many natural products are heterocycles. In this chapter, we will consider the most prevalent heterocyclic compounds—the ones containing the heteroatom N, O, or S.

20.1 MORE ABOUT AMINE NOMENCLATURE In Section 3.7, we saw that amines are classified as primary, secondary, or tertiary, depending on whether one, two, or three hydrogens of ammonia have been replaced by an alkyl group. We also saw that amines have both common and systematic names. Common names are obtained by citing the names of the alkyl substituents attached to the nitrogen (in alphabetical order) followed by “amine.” Systematic names employ “amine” as a functional group suffix. N H

NH2 a primary amine 1-pentanamine pentylamine

systematic name: common name:

N

a secondary amine N-ethyl-1-butanamine butylethylamine

a tertiary amine N-ethyl-N-methyl-1-propanamine ethylmethylpropylamine

A saturated cyclic amine—one without any double bonds—can be named as a cycloalkane, using the prefix aza to denote the nitrogen atom. There are, however, other acceptable names. Some of the more commonly used names are shown here. Notice that heterocyclic rings are numbered to give the heteroatom the lowest possible number.

H N azacyclopropane aziridine

NH azacyclobutane azetidine

N H

N H

azacyclopentane pyrrolidine

azacyclohexane piperidine

N H

CH3

2-methylazacyclohexane 2-methylpiperidine

N CH2CH3 N-ethylazacyclopentane N-ethylpyrrolidine


More About the Acid–Base Properties of Amines

991

Saturated heterocycles with oxygen and sulfur heteroatoms are named similarly. The prefix for oxygen is oxa and the prefix for sulfur is thia. O O

O

S

oxacyclopropane oxirane ethylene oxide

thiacyclopropane thiirane

oxacyclobutane oxetane

O

O

O

oxacyclopentane

oxacyclohexane

1,4-dioxacyclohexane

tetrahydrofuran

tetrahydropyran

1,4-dioxane

PROBLEM 1♦

Name the following compounds: H N CH3 a. CH3

CH3

S c.

e. CH3

CH2CH3

O

CH3

CH3 d.

b.

f.

O

HN N H

CH2CH3

20.2 MORE ABOUT THE ACID–BASE PROPERTIES OF AMINES Amines are the most common organic bases. We have seen that ammonium ions have pKa values of about 10 (Section 2.3) and anilinium ions have pKa values of about 5 (Sections 8.15 and 19.16). And we have seen that the greater acidity of anilinium ions compared with ammonium ions is due to the greater stability of the conjugate bases of the anilinium ions as a result of electron delocalization. Amines have very high pKa values. For example, the pKa of methylamine is 40. +

+

CH3CH2CH2NH3 an ammonium ion pKa = 10.8

NH3 an anilinium ion pKa = 4.58

CH3NH2 an amine pKa = 40

Saturated amine heterocycles containing five or more atoms have physical and chemical properties like those of acyclic amines. For example, pyrrolidine, piperidine, and morpholine are typical secondary amines, and N-methylpyrrolidine and quinuclidine are typical tertiary amines. The conjugate acids of these amines have the pKa values expected for ammonium ions. O +

+

H

+

H

H

+

N

N

N

H

H

N H

H

CH3

N

+

H the ammonium ions of:

pyrrolidine pKa = 11.27

piperidine pKa = 11.12

morpholine pKa = 9.28

N-methylpyrrolidine pKa = 10.32

quinuclidine pKa = 11.38


992

CHAPTER 20

More About Amines • Reactions of Heterocyclic Compounds

Atropine Atropine is a naturally occurring heterocyclic compound found in Jimson weed and nightshade (Atropa belladonna). The R isomer is the one nature synthesizes, but it racemizes during isolation. Atropine has a variety of medicinal usages, thus landing it on the list of drugs the World Health Organization deems needed for basic health care. O

OH

O CH3

Atropa belladonna

N

H

Atropine blocks acetylcholine receptor sites, so it lowers the activity of all muscles regulated by the parasympathetic nervous system. It is used to treat low heart rates, spasms from the gastrointestinal tract, and tremors associated with Parkinson’s disease. It also reduces the secretions of many organs, it is used as an antidote to organophosphate poisoning (see page 776), and it dilates pupils. The Romans used atropine in combination with opium as an anesthetic. During the Renaissance, women enhanced their appearance by using the juice from the berries of nightshade to dilate their pupils. Cleopatra was reported to do this as well. Note that belladonna is Italian for “beautiful woman.”

PROBLEM 2♦

Why is the conjugate acid of morpholine more acidic than the conjugate acid of piperidine?

PROBLEM 3♦

a. Draw the structure of 3-quinuclidinone. b. What is the approximate pKa of its conjugate acid? c. Which has a lower pKa value, the conjugate acid of 3-bromoquinuclidine or the conjugate acid of 3-chloroquinuclidine?

P R O B L E M 4 Solved

Explain the difference in the pKa values of the piperidinium and aziridinium ions:

+

+

N H

N H

piperidinium ion pKa = 11.1

H

H

aziridinium ion pKa = 8.0

The piperidinium ion has a pKa value expected for an ammonium ion, whereas the pKa value of the aziridinium ion is considerably lower. The internal bond angle of the three-membered ring is smaller than usual, which causes the external bond angles to be larger than usual. Because of the larger bond angles, the orbital used by nitrogen to overlap the orbital of hydrogen has more s character than that of a typical sp3 nitrogen (Section 1.15). This makes nitrogen in the aziridinium ion more electronegative, which lowers the pKa (Section 2.7).


Amines React as Bases and as Nucleophiles

20.3 AMINES REACT AS BASES AND AS NUCLEOPHILES We have seen that the leaving group of an amine (-NH2) is such a strong base that amines cannot undergo the substitution and elimination reactions that alkyl halides, alcohols, and ethers do (Section 11.9). The relative reactivities of these compounds—each with an electron-withdrawing group bonded to an sp3 carbon—can be appreciated by comparing the pKa values of the conjugate acids of their leaving groups, keeping in mind that the weaker the acid, the stronger its conjugate base. And, when comparing bases of the same type, the stronger the base, the poorer it is as a leaving group (Section 9.2). relative reactivities most reactive

>

RCH2F

strongest acid, has the weakest conjugate base

RCH2OH

HF

H2O

pKa = 3.2

>

RCH2OCH3

RCH2NH2

RCH2OH

pKa = 15.7

NH3

pKa = 15.5

pKa = 36

We have seen that amines react as bases in proton transfer reactions and in elimination reactions (Sections 2.3, 11.2, and 12.4). The lone pair on the nitrogen of an amine makes it nucleophilic as well as basic. We have seen that amines react as nucleophiles in a number of different reactions; for example, in nucleophilic substitution reactions that alkylate the amine (Section 9.2) +

CH3CH2Br

+

CH3NH2

CH3CH2NHCH3 + HBr

CH3CH2NH2CH3 −

methylamine

ethylmethylamine

Br

and in nucleophilic addition–elimination reactions that acylate the amine (Sections 16.8, 16.9, and 16.20). O CH3CH2

C

O +

Cl

2 CH3NH2 methylamine

CH3CH2

C

+

NHCH3

+

CH3NH3Cl−

an amide

We have also seen that aldehydes and ketones react with primary amines to form imines and with secondary amines to form enamines (Section 17.10)

O

+

trace acid

H2NCH2 primary amine benzylamine

O

+

secondary amine pyrrolidine

+

N

H2O

an enamine

and amines are nucleophiles in conjugate addition reactions (Section 17.18). O

O +

C CH3CH

CH

H

H2O

an imine

trace acid

HN

+

NCH2

CH3NH2

C CH3CH

CH2

NHCH3

H

least reactive weakest acid, has the strongest conjugate base

993


994

CHAPTER 20

More About Amines • Reactions of Heterocyclic Compounds

We have also seen that primary arylamines react with nitrous acid to form stable arenediazonium salts (Section 19.23). Arenediazonium salts are useful to synthetic chemists because the diazonium group can be replaced by a variety of nucleophiles.

NH2

N + N Cl−

HCl NaNO2 0 °C

Nu

Nu−

+

N2

+

Cl−

an arenediazonium salt

PROBLEM 5

Draw the product of each of the following reactions: O C a.

trace acid

CH3 + CH3CH2CH2NH2

O C b. CH3

Cl + 2

c.

NH2

N H

1. HCl, NaNO2, 0 °C 2. H2O, Cu2O, Cu(NO3)2

O d.

C

CH3 +

CH3CH2NHCH2CH3

trace acid

20.4 THE SYNTHESIS OF AMINES The many different ways to synthesize amines and the mechanisms of these reactions have been discussed in several different places in this book. These reactions are summarized and the references to where they are discussed can be found on page A-9 of Appendix III.

20.5 AROMATIC FIVE-MEMBERED-RING HETEROCYCLES We will begin the discussion of aromatic heterocyclic compounds by looking at those that have a five-membered ring.

Pyrrole, Furan, and Thiophene Pyrrole, furan, and thiophene are heterocycles with five-membered rings. We have seen that these compounds are aromatic because they are cyclic and planar, every atom in the ring has a p orbital, and the p cloud contains three pairs of p electrons (Sections 8.8 and 8.10).

N H

O

S

pyrrole

furan

thiophene


Aromatic Five-Membered-Ring Heterocycles these electrons are part of the p cloud

these electrons are part of the p cloud

H

N

these electrons are in an sp2 orbital perpendicular to the p orbitals

O

orbital structure of pyrrole

orbital structure of furan

Pyrrole is an extremely weak base because the electrons shown as a lone pair in the structure are part of the p cloud. That is, the nitrogen donates the lone-pair into the fivemembered ring (as shown by the resonance contributors). Therefore, protonating pyrrole destroys its aromaticity. As a result, the conjugate acid of pyrrole is a very strong acid (pKa = −3.8). −

N H

+N

+N

H

+N

H

H

+N

H

resonance contributors of pyrrole d− d−

d−

d+N

d−

H

resonance hybrid

PROBLEM 6

Draw arrows to show the movement of electrons in going from one resonance contributor to the next in pyrrole.

Pyrrole has a dipole moment of 1.80 D (Section 1.16). The saturated amine with a five-membered ring—pyrrolidine—has a slightly smaller dipole moment of 1.57 D, but as we see from the electrostatic potential maps, the two dipole moments are in opposite directions. (The red area is in the ring in pyrrole and on the nitrogen in pyrrolidine.) The dipole moment in pyrrolidine is due to inductive electron withdrawal by the nitrogen. Apparently, the ability of pyrrole’s nitrogen to donate electrons into the ring by resonance more than makes up for its inductive electron withdrawal (Section 19.14).

pyrrole

N H

N H

m = 1.80 D

m = 1.57 D

pyrrolidine

In Section 8.6, we saw a compound’s delocalization energy increases as the resonance contributors become more stable and more nearly equivalent. The delocalization energies of pyrrole, furan, and thiophene are not as great as the delocalization energies of benzene or the cyclopentadienyl anion, each a compound for which the resonance contributors are all equivalent.

995


996

CHAPTER 20

More About Amines • Reactions of Heterocyclic Compounds relative delocalization energies of some aromatic compounds

>

>

>

>

S

N H

O

Thiophene, with the least electronegative heteroatom, has the greatest delocalization energy of the three, and furan, with the most electronegative heteroatom, has the smallest delocalization energy. This is what we would expect, because the resonance contributors with a positive charge on the heteroatom are the most stable for the compound with the least electronegative heteroatom and the least stable for the compound with the most electronegative heteroatom. Because pyrrole, furan, and thiophene are aromatic, they undergo electrophilic aromatic substitution reactions.

O

+ Br2

+ HBr

Br

O

2-bromofuran

N H

CH3

+ Br2

Br

CH3

N H

+ HBr

2-bromo-5-methylpyrrole

Notice that the mechanism of the reaction is the same as the mechanism for electrophilic aromatic substitution of benzene (Section 19.3). MECHANISM FOR ELECTROPHILIC AROMATIC SUBSTITUTION

B

+

N H ■ ■

Pyrrole, furan, and thiophene undergo electrophilic aromatic substitution, preferentially at C-2.

+ Y+

slow

H N H

fast

Y

N H

Y

+ HB+

The electrophile adds to the pyrrole ring. A base (:B) in the reaction mixture removes the proton from the carbon that formed the bond with the electrophile.

Substitution occurs preferentially at C-2 because the intermediate obtained by adding a substituent to this position is more stable than the intermediate obtained by adding a substituent to C-3 (Figure 20.1). Both intermediates have a relatively stable resonance contributor in which all the atoms (except H) have complete octets. The intermediate resulting from C-2 substitution of pyrrole has two additional resonance contributors, whereas the intermediate resulting from C-3 substitution has only one additional resonance contributor.

+

Y

2-position

N H ▶ Figure 20.1 Structures of the intermediates that would be formed from the reaction of an electrophile with pyrrole at C-2 and C-3.

N H

+

Y+

H

Y

+

Y N H

N H

+

H

relatively stable

Y H

3-position

H

Y +N H

H +N

H

relatively stable


Aromatic Five-Membered-Ring Heterocycles

997

If both positions adjacent to the heteroatom are occupied, electrophilic substitution will take place at C-3. Br CH3

CH3 + Br2

S

CH3

CH3

S

+ HBr

3-bromo-2,5-dimethylthiophene

Pyrrole, furan, and thiophene are all more reactive than benzene toward electrophilic aromatic substitution because they are better able to stabilize the positive charge on the carbocation intermediate, since the lone pair on the heteroatom can donate electrons into the ring by resonance (Figure 20.1). Furan is not as reactive as pyrrole in electrophilic aromatic substitution reactions. The oxygen of furan is more electronegative than the nitrogen of pyrrole, so oxygen is not as effective as nitrogen in stabilizing the carbocation. Thiophene is less reactive than furan because the 3p orbital of sulfur overlaps less effectively than the 2p orbital of nitrogen or oxygen with the 2p orbital of carbon. The electrostatic potential maps in the margin illustrate the different electron densities of the three rings. relative reactivity toward electrophilic aromatic substitution

>

>

Pyrrole, furan, and thiophene are more reactive than benzene toward electrophilic aromatic substitution.

>

N H

O

S

pyrrole

furan

thiophene

benzene

The relative reactivities of pyrrole, furan, and thiophene are reflected in the Lewis acid required to catalyze a Friedel–Crafts acylation (Section 19.7). Benzene requires AlCl3, a relatively strong Lewis acid. Thiophene is more reactive than benzene, so it can undergo a Friedel–Crafts reaction using SnCl4, a weaker Lewis acid. An even weaker Lewis acid, BF3, can be used when the reactant is furan. Pyrrole is so reactive that an anhydride is used instead of a more reactive acyl chloride, and no catalyst is necessary. O

O +

CH3

C

C

1. AlCl3 2. H2O

Cl

CH3

+ HCl

phenylethanone pyrrole

O

S

+

CH3

C

1. SnCl4

Cl

2. H2O

CH3

C

S

+ HCl

O 2-acetylthiophene

O

O

+

CH3

C

furan

Cl

1. BF3/THF 2. H2O

O

C

CH3

+ HCl

O 2-acetylfuran

O N H

+

CH3

C

O O

C

O CH3

N H

C

CH3 +

O 2-acetylpyrrole

CH3

C

OH

thiophene


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The resonance hybrid of pyrrole (see page 995) shows that there is a partial positive charge on the nitrogen and a partial negative charge on each of the carbons. As a result, pyrrole is protonated on C-2 rather than on nitrogen. This should be expected because a proton is an electrophile, and we have just seen that electrophiles attach to the C-2 position of pyrrole. + H+

N H

H +N

H

H

pKa = −3.8

Pyrrole is unstable in strongly acidic solutions because, once protonated, it can readily polymerize. H

H N H

+N H

+N H

H

H N H

polymer

H

The sp2 nitrogen in pyrrole is more electronegative than the sp3 nitrogen in a saturated amine (Section 2.6). As a result, pyrrole (pKa ~ 17) is more acidic than the analogous saturated amine (pKa ~ 36). The partial positive charge on the nitrogen atom (apparent in the structure of the resonance hybrid) also contributes significantly to pyrrole’s increased acidity.

N

N H

+ H+

N H

pKa = ~17

N

+ H+

pKa = ~36

The pKa values of several nitrogen-containing heterocycles are listed in Table 20.1.

Table 20.1 The pKa Values of Several Nitrogen-Containing Heterocycles

H H +N

H

HN +

+N

pKa = 1.0

+

N H

pKa = 2.5

N

N

N H

N

H

pKa = −2.4

N

HN +N

H

NH

pKa = 6.8

+

N

H

pKa = −3.8

H

NH

+

H

pKa = 8.0

H

H

pKa = 11.1

pKa = 14.4

N+ H pKa = 4.85

N H pKa ∼17

N+ H pKa = 5.16

N H pKa ∼36

PROBLEM 7

When pyrrole is added to a dilute solution of D2SO4 in D2O, 2-deuteriopyrrole is formed. Propose a mechanism to account for the formation of this compound.


Aromatic Six-Membered-Ring Heterocycles

PROBLEM-SOLVING STRATEGY

Determining Relative Basicity

Rank the following compounds in order from least basic to most basic. NH2 N H

N

N H

First, we need to see whether any of the compounds will lose their aromaticity if they are protonated. Such compounds will be very difficult to protonate, so they will be very weak bases. Pyrrole is one such compound. Next, we need to see whether any of the compounds has a lone pair that can be delocalized. A delocalized lone pair will be harder to protonate than one that is localized. Aniline is such a compound. Finally, we need to look at the hybridization of the nitrogens. The nitrogen of the unsaturated six-membered ring is sp2 hybridized, whereas the nitrogen of the saturated six-membered ring is sp3 hybridized. We know that sp2 nitrogens are more electronegative and therefore harder to protonate than sp3 nitrogens. Thus, the order of basicity is NH2 <

<

<

N H

N

N H

Now use the strategy you have just learned to solve Problem 8. PROBLEM 8â&#x2122;Ś

Explain why cyclopentadiene (pKa = 15) is more acidic than pyrrole (pK a âŹ&#x192;17), even though nitrogen is more electronegative than carbon.

20.6 AROMATIC SIX-MEMBERED-RING HETEROCYCLES Now we will look at heterocycles that have a six-membered aromatic ring.

Pyridine As we have seen, pyridine is an aromatic compound with a nitrogen in place of one of the carbons in a benzene ring (Section 8.10).

N

4 5 6

3

N

these electrons are in an sp2 orbital perpendicular to the p orbitals

2

1

pyridine

orbital structure of pyridine

The pyridinium ion is a stronger acid than a typical ammonium ion because its proton is attached to an sp2 nitrogen, which is more electronegative than an sp3 nitrogen (Section 2.6).

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+ H+ N

N+ H

sp2

pyridine

pyridinium ion pKa = 5.16

+ H+ sp3

H

N +

N H

H

piperidine

piperidinium ion pKa = 11.12

The dipole moment of pyridine is 1.57 D. As the electrostatic potential map indicates, the electron-withdrawing nitrogen is the negative end of the dipole.

N m = 1.57 D benzene

Pyridine undergoes reactions characteristic of tertiary amines. For example, pyridine undergoes an SN2 reaction with an alkyl halide (Section 9.2). + CH3

I I−

+N

N

CH3 N-methylpyridinium iodide pyridine P R O B L E M 9 Solved

Will an amide be the final product obtained from the reaction of an acyl chloride with pyridine in an aqueous solution? Explain your answer. Solution The positively charged nitrogen in the initially formed carbonyl compound causes pyridine to be an excellent leaving group. Therefore, the compound hydrolyzes rapidly. As a result, the final product of the reaction is a carboxylic acid. (If the final pH of the solution is greater than the pKa of the carboxylic acid, the carboxylic acid will be predominantly in its basic form.)

O

O C R

Cl

+

C

+

C

N

R

N

O H2O

R

+

O−

N

Pyridine is aromatic. Like benzene, it has two uncharged resonance contributors. Because of the electron-withdrawing nitrogen, pyridine has three charged resonance contributors that benzene does not have. + +

N

N

N

N

N

+

resonance contributors of pyridine

Because it is aromatic, pyridine (like benzene) undergoes electrophilic aromatic substitution reactions.


Aromatic Six-Membered-Ring Heterocycles MECHANISM FOR ELECTROPHILIC AROMATIC SUBSTITUTION

+

slow

+ Y+

Y

fast

Y

N ■

B

H

N

+ HB+

N

The electrophile adds to the pyridine ring. A base (:B) in the reaction mixture removes the proton from the carbon that formed the bond with the electrophile.

Electrophilic aromatic substitution of pyridine takes place at C-3 because the most stable intermediate is obtained by placing an electrophilic substituent at that position (Figure 20.2). When the substituent is placed at C-2 or C-4, one of the resulting resonance contributors is particularly unstable because its nitrogen atom has an incomplete octet and a positive charge.

+

2-position

+

Y

+

N

N

H

Y

N

H

Y H

least stable

Y

+

+ Y+

3-position

Y H

N

H +

N H

Y

Y

N

N

H

Y

H

Y

+

4-position

H +

+ +

N

N

N

least stable

▲ Figure 20.2 Structures of the intermediates that would be formed from the reaction of an electrophile with pyridine.

The electron-withdrawing nitrogen atom makes the intermediate obtained from electrophilic aromatic substitution of pyridine less stable than the carbocation intermediate obtained from electrophilic aromatic substitution of benzene. Pyridine, therefore, is less reactive than benzene. Indeed, it is even less reactive than nitrobenzene. (Recall from Section 19.14 that an electron-withdrawing nitro group strongly deactivates a benzene ring toward electrophilic aromatic substitution.) relative reactivity toward electrophilic aromatic substitution

NO2 >

NO2 >

> N

NO2

Pyridine, therefore, undergoes electrophilic aromatic substitution reactions only under vigorous conditions, and the yields of these reactions are often quite low. If the

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nitrogen becomes protonated under the reaction conditions, the reactivity decreases further because a positively charged nitrogen would make the carbocation intermediate even less stable.

+ Br2

Br

FeBr3 300 °C

+ HBr

N

N 3-bromopyridine 30%

SO3H

1. H2SO4/230 °C 2. HO–

Pyridine undergoes electrophilic aromatic substitution at C-3.

+ H2O

N

N pyridine-3-sulfonic acid 71%

NO2

1. HNO3, H2SO4/300 °C 2. HO–

+ H2O

N

N 3-nitropyridine 22%

We have seen that highly deactivated benzene rings do not undergo Friedel–Crafts alkylation or acylation reactions (Section 19.18). Therefore, pyridine, whose reactivity is similar to that of a highly deactivated benzene ring, does not undergo these reactions either. PROBLEM 10♦

Draw the product formed when pyridine reacts with ethyl bromide.

Pyridine is less reactive than benzene toward electrophilic aromatic substitution and more reactive than benzene toward nucleophilic aromatic substitution.

Since pyridine is less reactive than benzene in electrophilic aromatic substitution reactions, it should not be surprising that pyridine is more reactive than benzene in nucleophilic aromatic substitution reactions. The electron-withdrawing nitrogen atom that destabilizes the intermediate in electrophilic aromatic substitution stabilizes the intermediate in nucleophilic aromatic substitution. Notice that, in nucleophilic aromatic substitution reactions, the ring has a leaving group that can be replaced by a nucleophile.

MECHANISM FOR NUCLEOPHILIC AROMATIC SUBSTITUTION

+ N ■ ■

Z

Y−

slow

fast

N

Y Z

+ Z− N

Y

The nucleophile adds to the ring carbon attached to the leaving group. The leaving group is eliminated.

Nucleophilic aromatic substitution of pyridine takes place at C-2 or C-4 because addition to these positions leads to the most stable intermediate. Only when addition occurs to these positions is a resonance contributor obtained that has the greatest electron density on nitrogen, the most electronegative of the ring atoms (Figure 20.3).


Aromatic Six-Membered-Ring Heterocycles −

2-position

N

Y

N

Z

N

Z

1003

Y

N

Z

Y Z

most stable

Z

Y

3-position

Y −

N

N Z

Z

Y

Y Z

Z

Z

N

N

Z

Y

4-position

N

N

Z

Y

N

N

most stable

◀ Figure 20.3 Structures of the intermediates that would be formed from the reaction of a nucleophile with a substituted pyridine.

If the leaving groups at C-2 and C-4 are different, then the incoming nucleophile will preferentially substitute for the weaker base (the better leaving group). Br

NH2 +

N

Δ

NH2

OCH3

OCH3

N

Pyridine undergoes nucleophilic aromatic substitution at C-2 or C-4.

CH3

CH3 + CH3O− N

+ Br−

Δ

+ Cl−

Cl

OCH3

N

PROBLEM 11

Compare the mechanisms of the following reactions: Cl

NH2 +

NH2

Δ

+ Cl−

N

N

Cl

NH2 +

+ Cl−

NH2

PROBLEM 12

a. Propose a mechanism for the following reaction:

N

KOH/H2O Δ

b. What other product is formed in this reaction?

N H

O


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Substituted pyridines undergo many of the side-chain reactions that substituted benzenes undergo. For example, alkyl-substituted pyridines can be brominated and oxidized.

N

NBS Δ/peroxide

CH2CH3

N

CHCH3 Br

CH3

COOH

H2CrO4 Δ

N

+N

H

When 2- or 4-aminopyridine is diazotized, a-pyridone or g-pyridone is formed. Apparently, the diazonium salt reacts immediately with water to form a hydroxypyridine (Section 19.21). The product of the reaction is a pyridone because the keto form of a hydroxypyridine is more stable than the enol form. (The mechanism for the conversion of a primary amino group into a diazonium group is shown in Section 19.23.)

N

NH2

NaNO2, HCl 0 °C

H2O

+

N

N

N

2-aminopyridine

N

2-hydroxypyridine enol form

Cl

N H

OH

O

a-pyridone keto form

N N+ Cl−

NH2 NaNO2, HCl 0 °C

N

OH

O

N

N H

H2O

N

4-hydroxypyridine enol form

4-aminopyridine

g-pyridone keto form

The electron-withdrawing nitrogen and the ability to delocalize the negative charge causes the hydrogens on carbons attached to the 2- and 4-positions of the pyridine ring to have about the same acidity as the hydrogens attached to the a-carbons of ketones (Section 18.1). −

CH3

CH2

CH2 −

HO−

N

CH2

N

CH2

CH2

N

N−

N

N

Consequently, the hydrogens can be removed by base, and the resulting carbanions can react as nucleophiles. −

O

CH2

CH3

CH

CH

CH

HO−

+ H2O

an aldol condensation

N CH3 N

CH3

N

N CH3

−NH 2

N

CH2 −

CH3Br an SN2 reaction

CH3 + Br− N

CH2CH3


Some Amine Heterocycles Have Important Roles in Nature PROBLEM 13♦

Rank the following compounds in order from easiest to hardest at removing a proton from a methyl group: CH3

CH3 CH3

N

N

+N

I− CH2CH3

20.7 SOME AMINE HETEROCYCLES HAVE IMPORTANT ROLES IN NATURE Proteins are naturally occurring polymers of a-amino acids (Chapter 22). Three of the 20 most common amino acids in proteins contain heterocyclic rings. Proline has a pyrrolidine ring; typtophan has an indole ring; and histidine has an imidazole ring. CH2CHCOO− +

COO−

N +

H

NH3 N

N H

H

proline

CH2CHCOO− NH

tryptophan

+ NH 3

histidine

Imidazole Imidazole, the heterocyclic ring of histidine, is an aromatic compound because it is cyclic and planar, every atom in the ring has a p orbital, and the p cloud contains three pairs of p electrons (Section 8.8). The electrons drawn as lone-pair electrons on N-1 are part of the p cloud because they are in a p orbital, whereas the lone-pair electrons on N-3 are not part of the p cloud because they are in an sp2 orbital that is perpendicular to the p orbitals. 4 3

5

N

NH

2

1

−N

NH

N

+

NH +

N

NH

N

+

NH +

resonance contributors of imidazole

these electrons are in an sp2 orbital perpendicular to the p orbitals

these electrons are part of the p cloud

N

N

H

orbital structure of imidazole

Because the lone-pair electrons in the sp2 orbital are not part of the p cloud, imidazole is protonated in acidic solutions. The conjugate acid of imidazole has a pKa = 6.8. Thus, imidazole exists in both the protonated and unprotonated forms at physiological pH (7.4).

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This is one of the reasons that histidine, the imidazole-containing amino acid, is a catalytic component of many enzymes (see page 1119). HN

NH

+

NH + H+

N

pKa = 6.8

Notice that both protonated imidazole and the imidazole anion have two equivalent resonance contributors. Thus, the two nitrogens become equivalent when imidazole is either protonated or deprotonated. protonated imidazole

HN

NH

+

HN

d+

HN

imidazole anion

NH

N

+

N

d+

d−

N

NH

resonance hybrid

N

N

d−

N

resonance hybrid

Searching for Drugs: An Antihistamine, a Nonsedating Antihistamine, and a Drug for Ulcers When scientists know something about the molecular basis of drug action—such as how a particular drug interacts with a receptor— they can design and synthesize compounds that might have a desired physiological activity. For example, histidine can be decarboxylated in an enzyme-catalyzed reaction to form histamine (see page 1152). When the body produces excess histamine, it causes the symptoms associated with the common cold and allergies. Knowing that the amine is positively charged at physiological pH gave scientists one clue as to how it might interact with its receptor. − +

+

NH3

histidine

N

N H O

NH3

histamine

N

O−

histamine receptor

NH

After an extensive search, the compounds shown here (as well as several others)—called antihistamines—were found to bind to the histamine receptor but not trigger the same response as histamine. Like histamine, these drugs have a protonated amino group. They also have bulky groups that keep histamine from approaching the receptor. antihistamines

Cl

+

O

N H

+H

N

N diphenhydramine Benadryl®

chlorpheniramine Chlortrimetron®

These compounds, however, are able to cross the blood–brain barrier and bind to receptors in the central nervous system. Binding to these receptors causes drowsiness, the well-known side effect associated with these drugs. So now the search was on for compounds


1007

Some Amine Heterocycles Have Important Roles in Nature

that would bind to the histamine receptor but not cross the blood–brain barrier. Because the blood–brain barrier is nonpolar, this was achieved by putting polar groups on the compounds. Allegra, Claritin, and Zyrtec are nonsedating antihistamines. Zyrtec became available as an over-the-counter drug in 2007. nonsedating antihistamines

Cl O

HO

+

HO

O−

+

O

O

+

NH

NH

HN

O

+

NH

OH

N fexofenadine Allegra®

cetirizine Zyrtec®

loratadine Claritin®

In addition to causing allergic responses, excess histamine production by the body also causes the hypersecretion of HCl by the cells of the stomach lining, which leads to the development of ulcers. The antihistamines that prevent allergic responses have no effect on HCl production. This suggested that a second kind of histamine receptor—called a histamine H2-receptor—causes the secretion of HCl. Because 4-methylhistamine was found to cause weak inhibition of HCl secretion, it was used as a lead compound (Section 11.9). More than 500 molecular modifications were performed over a 10-year period before a compound was found that would bind to the histamine H2-receptor. Tagamet, introduced in 1976, was the first drug for the treatment of peptic ulcers. Previously, the only treatment was extensive bed rest, a bland diet, and antacids. Zantac followed in 1981, and by 1988 it became the world’s best-selling prescription drug. Protonix became available in 1994. +

NH3

N N H

CH3

4-methylhistamine

+

HN N H

S

H N

H N

H H

N

CH3

N +

C

O

S

N cimetidine Tagamet®

N H

C

NO2 OCH3 N H

CH3O

ranitidine Zantac®

PROBLEM 14♦

What is the major product of the following reaction? N

NCH3 + Br2

FeBr3

PROBLEM 15♦

List imidazole, pyrrole, and benzene in order from most reactive to least reactive toward electrophilic aromatic substitution.

+

S N

OCH2F

HN N H

O pantropazole Protonix®


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CHAPTER 20

More About Amines • Reactions of Heterocyclic Compounds PROBLEM 16♦

Imidazole boils at 257 °C, whereas N-methylimidazole boils at 199 °C. Explain this difference in boiling points. PROBLEM 17

Why is imidazole a stronger acid (pKa = 14.4) than pyrrole (pKa ' 17)?

PROBLEM 18

What percent of imidazole will be protonated at physiological pH (7.4)?

Purine and Pyrimidine Nucleic acids (DNA and RNA) contain substituted purines and substituted pyrimidines (Section 26.1); DNA contains adenine, guanine, cytosine, and thymine (abbreviated A, G, C, and T); RNA contains adenine, guanine, cytosine, and uracil (A, G, C, and U). Why DNA contains T instead of U is explained in Section 26.10. Unsubstituted purine and pyrimidine are not found in nature. Guanine (a hydroxypurine) and cytosine, uracil, and thymine (hydroxypyrimidines) are more stable in the keto form than in the enol form, so the keto forms are shown here. We will see that the preference for the keto form is crucial for proper base pairing in DNA (Section 26.3). 6 1N 2

7

4

N

5

3N

8

N

2

N9 H

4

3

N

NH2

O

N H

adenine

N

HN H2N

6

pyrimidine

NH2 N

N 1

purine

N

5

N

N H

guanine

N O

cytosine

O

CH3

HN

HN N H

O

O

N H uracil

O

N H thymine

PROBLEM 19♦

Draw guanine and cytosine in the enol form.

PROBLEM 20♦

Why is protonated pyrimidine (pKa = 1.0) more acidic than protonated pyridine (pKa = 5.2)?

Porphyrin Substituted porphyrins are another group of important naturally occurring heterocyclic compounds. A porphyrin ring system consists of four pyrrole rings joined by onecarbon bridges. Heme, which is found in hemoglobin and myoglobin, contains an iron ion (FeII) ligated by the four nitrogens of a porphyrin ring system. Ligation is the sharing of lone-pair electrons with a metal ion.


Some Amine Heterocycles Have Important Roles in Nature −OOCCH CH 2 2

H3C N

HN

C C A C C N HC

NH

N

a porphyrin ring system

H2C

C H

H C

FeII

CH2CH2COO− C C B C N C

CH3

CH

C N N C C C CH3 C D C C C C C H CH3 HC CH2 heme

Hemoglobin has four polypeptide chains and four heme groups (page 1093); myoglobin has one polypeptide chain and one heme group. Hemoglobin is responsible for transporting oxygen to cells and carbon dioxide away from cells, whereas myoglobin is responsible for storing oxygen in cells. The iron atoms in hemoglobin and myoglobin, in addition to being ligated to the four nitrogens of the porphyrin ring, are also ligated to a histidine of the protein component (globin), and the sixth ligand is oxygen or carbon dioxide. Carbon monoxide is about the same size and shape as O2, but CO binds more tightly than O2 to iron. In addition, once a hemoglobin molecule has bound two CO molecules, it can no longer achieve the configuration necessary to bind O2. Consequently, breathing carbon monoxide can be fatal because it interferes with the transport of oxygen through the bloodstream. The extensive conjugated system of heme gives blood its characteristic red color. Its high molar absorptivity (about 160,000 M−1cm−1) allows concentrations as low as 1 * 10 - 8 M to be detected by UV spectroscopy (Section 14.21). The ring system in chlorophyll a, the substance responsible for the green color of plants (its structure is on page 635), is similar to porphyrin but it contains a cyclopentanone ring and one of its pyrrole rings is partially reduced. The metal ion in chlorophyll a is magnesium (MgII). Vitamin B12 also has a ring system similar to porphyrin, but in this case, the metal ion is cobalt (CoIII). Other aspects of vitamin B12’s structure and chemistry are discussed in Section 24.6.

Porphyrin, Bilirubin, and Jaundice The average human body breaks down about 6 g of hemoglobin each day. The protein portion (globin) and the iron are reutilized, but the porphyrin ring is cleaved between the A and B rings to form a linear tetrapyrrole called biliverdin (a green compound). The bridge between the C and D ring is reduced, forming bilirubin (a red-orange compound). You can witness heme degradation by observing the changing colors of a bruise. Enzymes in the large intestine convert bilirubin to urobilinogen. Some urobilinogen is transported to the kidney, where it is converted to urobilin (a yellow compound), which is excreted. This is the compound that gives urine its characteristic color. If more bilirubin is formed than can be metabolized and excreted by the liver, it accumulates in the blood. When its concentration there reaches a certain level, it diffuses into the tissues, giving them a yellow appearance. This condition is known as jaundice.

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20.8 ORGANIZING WHAT WE KNOW ABOUT THE REACTIONS OF ORGANIC COMPOUNDS We have seen that the families of organic compounds can be put into one of four groups, and that all the members of a group react in similar ways. Now that we have finished studying the families in Group IV, let’s review how these compounds react. I R

CH

II CH

R

R

C

C

R

R

CH

CH

CH

CH

R

Z = an atom more C electronegative Z R than C carboxylic acid and carboxylic acid derivatives

OH OR

O

an ether

C

R

a diene

They undergo electrophilic addition reactions.

R

O

These are nucleophiles.

R

IV

O

an alcohol

an alkyne

R

X = F, Cl, Br, I

X

an alkyl halide

an alkene

R

III

benzene

N Z

Z = C or H

pyridine

aldehydes and ketones

R

These are electrophiles.

an epoxide

They undergo nucleophilic addition–elimination reactions or nucleophilic addition reactions.

These are electrophiles. They undergo nucleophilic substitution and/or elimination reactions.

Removal of a hydrogen from an A-carbon forms a nucleophile that can react with electrophiles.

Z

Z = N, O, or S H

pyrrole, furan, thiophene These are nucleophiles. They undergo electrophilic and/or nucleophilic aromatic substitution reactions.

All the compounds in Group IV are aromatic. In order to preserve the aromaticity of the rings, these compounds undergo electrophilic aromatic substitution reactions and/or nucleophilic aromatic substitution reactions. ■

Benzene and substituted benzenes are nucleophiles, so they react with electrophiles in electrophilic aromatic substitution reactions. If the electron density of the benzene ring is reduced by strongly electron-withdrawing groups, a halo-substituted benzene can undergo a nucleophilic aromatic substitution reaction. Pyrrole, furan, and thiophene are much more nucleophilic than benzene, so they are more reactive in electrophilic aromatic substitution reactions. Pyridine is much less nucleophilic than benzene, so it undergoes electrophilic aromatic substitution reactions only under rigorous conditions. However, halo-substituted benzenes readily undergo nucleophilic aromatic substitution reactions.

S OME IMPORTANT THINGS TO REMEMBER ■

Amines are classified as primary, secondary, or tertiary, depending on whether one, two, or three hydrogens of ammonia have been replaced by alkyl groups. Some amines are heterocyclic compounds—cyclic compounds in which one or more of the atoms of the ring is an atom other than carbon.

Heterocyclic rings are numbered so that the heteroatom has the lowest possible number. The lone pair on the nitrogen causes amines to be both bases and nucleophiles. Amines react as bases in acid–base reactions and in elimination reactions.


Summary of Reactions ■

Amines react as nucleophiles in nucleophilic substitution reactions, in nucleophilic addition–elimination reactions, and in conjugate addition reactions. Saturated heterocycles have physical and chemical properties like those of acyclic compounds that contain the same heteroatom. Pyrrole, furan, and thiophene are heterocyclic aromatic compounds that undergo electrophilic aromatic substitution reactions preferentially at C-2. They are more reactive than benzene toward electrophilic aromatic substitution. Protonating pyrrole destroys its aromaticity. Pyrrole polymerizes in strongly acidic solutions. Replacing one of benzene’s carbons with a nitrogen forms pyridine, a heterocyclic aromatic compound that undergoes electrophilic aromatic substitution reactions at

C-3 and nucleophilic aromatic substitution reactions at C-2 or C-4. Pyridine is less reactive than benzene toward electrophilic aromatic substitution and more reactive than benzene toward nucleophilic aromatic substitution. Three amino acids have heterocyclic rings: histidine has an imidazole ring, proline has a pyrrolidine ring, and tryptophan has an indole ring. Nucleic acids (DNA and RNA) contain substituted purines and substituted pyrimidines. Hydroxypurines and hydroxypyrimidines are more stable in the keto form. A porphyrin ring system consists of four pyrrole rings joined by one-carbon bridges; in hemoglobin and myoglobin, the four nitrogen atoms are ligated to FeII. The metal ion in chlorophyll a is magnesium, and the metal ion in vitamin B12 is cobalt.

SUMMARY OF REACTIONS 1. Reactions of amines as nucleophiles (Section 20.3) a. In alkylation reactions (Section 9.2) b. In reactions with a carbonyl group that has a leaving group (Sections 16.8, 16.9, and 16.20) c. In reactions with an aldehyde or ketone to form an imine or an enamine (Section 17.10) d. In conjugate addition reactions (Sections 17.18 and 17.19) e. In reactions with nitrous acid: primary arylamines form arenediazonium salts (Section 19.23) 2. Electrophilic aromatic substitution reactions a. Pyrrole, furan, and thiophene (Section 20.5): the mechanism is shown on page 996. + Br2 O

Br

O

+ HBr

b. Pyridine (Section 20.6): the mechanism is shown on page 1001. Br

FeBr3 300 °C

+ Br2 N

+ HBr N

3. Nucleophilic aromatic substitution reactions of pyridine (Section 20.6). The mechanism is shown on page 1002. + N

Δ

−NH

2

+ Cl−

Cl

N

NH2

4. Reactions of substituents on pyridine (Section 20.6)

N

CH2CH3

NBS Δ/peroxide

N

CHCH3 Br

CH3 N

COOH

H2CrO4 Δ

1011

+N

H


1012

CHAPTER 20

More About Amines • Reactions of Heterocyclic Compounds

N

NaNO2, HCl 0 °C

NH2

2-aminopyridine

CH3

N

N H

OH

2-hydroxypyridine enol form

O

A-pyridone keto form

CH3

−NH

2

N

N

CH3

CH2 −

P R OBLEMS 21. Name each of the following: a.

b.

NH CH3

N H

Cl

c.

CH3

N H

CH3

CH3

d. N H

CH3CH2

22. What are the products of the following reactions? O O

C

+

a.

d.

Cl

N H

O

b.

+ HO− N

Br

e.

1. C N 2. H2, Raney Ni

1. BF3/THF 2. H2O

C CH3

Cl

+ C6H5N

N

CH3

S

Br

i.

+ CH3CH2CH2Br

N

N H

Cl NH2

f.

+

g.

h.

C6H5Li

Δ

N

c. CH3CH2CH2CH2Br

CH3

+

1. −NH2 2. CH3CH2CH2Br

1. Mg/Et2O 2. CO2 3. HCl

23. List the following compounds in order from strongest acid to weakest acid: H N+ +N

+N

H

H

N H

N H

H

N N H

N H

+N

H

H

24. Which of the following compounds is easier to decarboxylate? or N

COH

N

CH2COH O

O

25. Rank the following compounds in order from most reactive to least reactive in an electrophilic aromatic substitution reaction: OCH3

NHCH3

SCH3


Problems

1013

26. One of the following compounds undergoes electrophilic aromatic substitution predominantly at C-3, and one undergoes electrophilic aromatic substitution predominantly at C-4. Which is which?

N

N

CCH2CH3

NHCH2CH3

O 27. Benzene undergoes electrophilic aromatic substitution reactions with aziridines in the presence of a Lewis acid such as AlCl3. a. What are the major and minor products of the following reaction? CH3 AlCl3

N

+

CH3 b. Would you expect epoxides to undergo similar reactions? 28. The dipole moments of furan and tetrahydrofuran are in the same direction. One compound has a dipole moment of 0.70 D, and the other has a dipole moment of 1.73 D. Which is which? 29. A tertiary amine reacts with hydrogen peroxide to form a tertiary amine oxide. R

R R

N

+

R

HO

OH

R

R

+N

R

+

HO

R

N

+

R

+

H2O

O−

OH a tertiary amine

a tertiary amine oxide

Tertiary amine oxides undergo a reaction similar to the Hofmann elimination reaction (Section 11.10), called a Cope elimination. In this reaction, a tertiary amine oxide, rather than a quaternary ammonium ion, undergoes elimination. A strong base is not needed for a Cope elimination because the amine oxide acts as its own base. CH3 CH3 CH3CH2CH2NCH 2CHCH3 +

Δ

CH3 CH3 CH3CH

CH2 + NCH2CHCH3

O

OH

Does the Cope elimination have an alkene-like transition state or a carbanion-like transition state? 30. What products would be obtained by treating the following tertiary amines with hydrogen peroxide followed by heat? CH3

CH3

a. CH3NCH2CH2CH3

b.

CH3NCH2CH2CH3

c.

CH3CH2NCH2CHCH3 CH3

d. N

CH3

CH3 31. The chemical shifts of the C-2 hydrogen in the spectra of pyrrole, pyridine, and pyrrolidine are 2.82 ppm, 6.42 ppm, and 8.50 ppm. Match each heterocycle with its chemical shift. 32. Explain why protonating aniline has a dramatic effect on the compound’s UV spectrum, whereas protonating pyridine has only a small effect on that compound’s UV spectrum. 33. Explain why pyrrole (pKa ' 17) is a much stronger acid than ammonia (pKa = 36). + H+ N H pKa ∼17

N −

NH3 pKa = 36

NH2 + H+


1014

CHAPTER 20

More About Amines • Reactions of Heterocyclic Compounds

34. Propose a mechanism for the following reaction: + H2C

2 N H

HCl

O

CH2

N H

N H

35. Quinolines, heterocyclic compounds that contain a pyridine ring fused to a benzene ring, are commonly synthesized by a method known as the Skraup synthesis, in which aniline reacts with glycerol under acidic conditions. Nitrobenzene is added to the reaction mixture to serve as an oxidizing agent. The first step in the synthesis is the dehydration of glycerol to propenal.

CH2

CH

CH2

OH

OH

OH

NH2

H2SO4 Δ

CH2

CH

CH

O NO2

+ 2 H2O

glycerol

N quinoline

a. What product would be obtained if para-ethylaniline were used instead of aniline? b. What product would be obtained if 3-hexen-2-one were used instead of glycerol? c. What starting materials are needed for the synthesis of 2,7-diethyl-3-methylquinoline? 36. Propose a mechanism for each of the following reactions: O O a. CH3

HCl H2O Δ

CH3

O

C CH3

C CH3

CH2CH2

b.

O

+ Br2

CH3OH

CH3O

OCH3

O

37. What is the major product of each of the following reactions? O

CH3NCH3

C CH3

a. O

+ HNO3

H2SO4

c.

CH3 1. HO− 2. H2C 3. HCl

e.

+ CH3I N

N

O

CH3 b. S

NO2

+ PCl3

d.

+ Br2

N H

f. N H

O

+ CH3CH2MgBr

38. a. Draw resonance contributors to show why pyridine-N-oxide is more reactive than pyridine toward electrophilic aromatic substitution. b. At what position does pyridine-N-oxide undergo electrophilic aromatic substitution? 39. Propose a mechanism for the following reaction:

+ +N

CH3

C

C CH3

O

O

O O

CH3

O

N

C

O−

CH3

+

C CH3

OH

O

40. Pyrrole reacts with excess para-(N,N-dimethylamino)benzaldehyde to form a highly colored compound. Draw the structure of the colored compound. 41. 2-Phenylindole is prepared from the reaction of acetophenone and phenylhydrazine, a method known as the Fischer indole synthesis. Propose a mechanism for this reaction. (Hint: the reactive intermediate is the enamine tautomer of the phenylhydrazone.) O C CH3

+

NHNH2

trace acid Δ

+ NH3 + H2O N H 2-phenylindole


Problems

1015

42. What starting materials are required to synthesize the following compounds, using the Fischer indole synthesis? (Hint: see Problem 41.) CH2CH3 a.

b.

CH2CH3 N H

c. N H

N H 43. Organic chemists work with tetraphenylporphyrins rather than with porphyrins because tetraphenylporphyrins are much more resistant to air oxidation. Tetraphenylporphyrin can be prepared by the reaction of benzaldehyde with pyrrole. Propose a mechanism for the formation of the ring system shown here:

O C H

BH3/THF

+ N H

NH

N H H N

HN

oxidation

tetraphenylporphyrin


PART SEVEN

Bioorganic Compounds Chapters 21 through 26 discuss the chemistry of bioorganic compounds—organic compounds found in living systems. Many of these compounds are larger than the organic compounds you have seen up to this point, and they often have more than one functional group, but the principles that govern their structure and reactivity are essentially the same as those that govern the structure and reactivity of the compounds that you have been studying. These chapters, therefore, will give you the opportunity to review much of the organic chemistry you have learned as you apply this chemistry to compounds found in the biological world.

C H A P T E R 2 1 The Organic Chemistry of Carbohydrates Chapter 21 introduces you to the organic chemistry of carbohydrates, the most abundant class of compounds in the biological world. First you will learn about the structures and reactions of monosaccharides. Then you will see how monosaccharides are linked to form disaccharides and polysaccharides. Many examples of carbohydrates that are found in nature will be discussed. C H A P T E R 2 2 The Organic Chemistry of Amino Acids, Peptides, and Proteins Chapter 22 starts by looking at the physical properties of amino acids. Then you will see how amino acids are linked to form peptides and proteins. You will also see how proteins are made in the laboratory. Later, when you study Chapter 26, you will be able to compare the way proteins are synthesized in the laboratory with the way they are synthesized in nature. What you learn about protein structure in Chapter 22 will prepare you for understanding how enzymes catalyze chemical reactions, which is discussed in Chapter 23. C H A P T E R 2 3 Catalysis in Organic Reactions and in Enzymatic Reactions Chapter 23 first describes the various ways that organic reactions can be catalyzed and then shows how enzymes employ these same methods to catalyze reactions in cells. C H A P T E R 2 4 The Organic Chemistry of the Coenzymes, Compounds Derived from Vitamins

DNA

Chapter 24 describes the chemistry of the coenzymes—organic compounds that some enzymes need in order to catalyze biological reactions. Coenzymes play a variety of chemical roles: some function as oxidizing and reducing agents, some allow electrons to be delocalized, some activate groups for further reaction, and some provide good nucleophiles or strong bases needed for reactions. Because coenzymes are derived from vitamins, you will see why vitamins are necessary for many of the organic reactions that occur in cells. C H A P T E R 2 5 The Organic Chemistry of Metabolic Pathways • Terpene Biosynthesis Chapter 25 looks at the organic reactions that cells carry out to obtain the energy they need and to synthesize the compounds they require. You will see why many reactions that occur in cells could not occur without ATP. The biosynthesis of terpenes will also be discussed. C H A P T E R 2 6 The Chemistry of the Nucleic Acids Chapter 26 covers the structures and chemistry of nucleosides, nucleotides, and nucleic acids (RNA and DNA). You will see how nucleotides are linked to form nucleic acids, why DNA contains thymine instead of uracil, why DNA does not have the 2¿-OH group that RNA has, how the genetic messages encoded in DNA are transcribed into mRNA and then translated into proteins, and how the sequence of bases in DNA is determined.


21

The Organic Chemistry of Carbohydrates

a field of sugar cane

BIOORGANIC COMPOUNDS ARE ORGANIC COMPOUNDS FOUND IN LIVING SYSTEMS.

The first group of bioorganic compounds we will look at are carbohydrates—the most abundant class of compounds in the biological world, making up more than 50% of the dry weight of the Earth’s biomass. Carbohydrates are important constituents of all living organisms and have a variety of different functions. Some are important structural components of cells; others act as recognition sites on cell surfaces. For example, the first event in our lives was a sperm recognizing a carbohydrate on the outer surface of an egg. Other carbohydrates serve as a major source of metabolic energy. The leaves, fruits, seeds, stems, and roots of plants, for instance, contain carbohydrates that plants use for their own metabolic needs and these also serve the metabolic needs of the animals that eat them.

T

D-glucose

D-fructose

he structures of bioorganic compounds can be quite complex, yet their reactivity is governed by the same principles that govern the reactivity of the comparatively simple organic molecules we have discussed so far. In other words, the organic reactions that chemists carry out in the laboratory are in many ways just like those performed by nature inside a cell. Thus, bioorganic reactions can be thought of as organic reactions that take place in tiny flasks called cells. Although most bioorganic compounds have more complicated structures than those of the organic compounds you are now used to seeing, do not let the structures fool you into thinking that their chemistry must be equally complicated. One reason the structures of bioorganic compounds are more complicated is that they must be able to recognize each other. Much of their structure is for that very purpose, a function called molecular recognition. Early chemists noted that carbohydrates have molecular formulas that make them appear to be hydrates of carbon, Cn(H2O)n—hence the name. Structural studies later revealed that these compounds are not hydrates because they do not contain intact water molecules. Nevertheless, the term carbohydrate persisted. Carbohydrates are polyhydroxy aldehydes such as glucose, polyhydroxy ketones such as fructose, and compounds such as sucrose formed by linking polyhydroxy aldehydes or polyhydroxy ketones together (Section 21.15). The chemical structures of carbohydrates are 1017


1018

CHAPTER 21

The Organic Chemistry of Carbohydrates

commonly represented by Fischer projections. Notice that both glucose and fructose have the molecular formula C6H12O6, consistent with the general formula C6(H2O)6 that made early chemists think that these compounds were hydrates of carbon. Notice, too, that the structures of glucose and fructose differ only at the top two carbons. H

O C

H

OH

HO Recall that all the horizontal bonds in Fischer projections point toward the viewer (Section 4.6).

CH2OH C

H

HO

O H

H

OH

H

OH

H

OH

H

OH

CH2OH

CH2OH Fischer projection D-fructose a polyhydroxy ketone

Fischer projection D-glucose a polyhydroxy aldehyde

The most abundant carbohydrate in nature is glucose. Animals obtain glucose from food that contains glucose, such as plants. Plants produce glucose by photosynthesis. During photosynthesis, plants take up water through their roots and use carbon dioxide from the air to synthesize glucose and oxygen. Because photosynthesis is the reverse of the process used by organisms to obtain energy—specifically, the oxidation of glucose to carbon dioxide and water—plants require energy to carry out photosynthesis. They obtain that energy from sunlight, which is captured by chlorophyll molecules in green plants. Photosynthesis uses the CO2 that animals exhale as waste and generates the O2 that animals inhale to sustain life. Nearly all the oxygen in the atmosphere has been released by photosynthetic processes. C6H12O6 + 6 O2 glucose

oxidation photosynthesis

6 CO2 + 6 H2O

21.1 CLASSIFICATION OF CARBOHYDRATES The terms carbohydrate, saccharide, and sugar are used interchangeably. Saccharide comes from the word for sugar in several early languages (sarkara in Sanskrit, sakcharon in Greek, and saccharum in Latin). Simple carbohydrates are monosaccharides (single sugars); complex carbohydrates contain two or more monosaccharides linked together. Disaccharides have two monosaccharides linked together, oligosaccharides have 3 to 10 (oligos is Greek for “few”), and polysaccharides have 10 or more. Disaccharides, oligosaccharides, and polysaccharides can be broken down to monosaccharides by hydrolysis. a monosaccharide subunit

M

M

M

M

M

M

polysaccharide

M

M

M

hydrolysis

xM monosaccharide

A monosaccharide can be a polyhydroxy aldehyde such as glucose or a polyhydroxy ketone such as fructose. Polyhydroxy aldehydes are called aldoses (“ald” is for aldehyde; “ose” is the suffix for a sugar); polyhydroxy ketones are called ketoses. Monosaccharides are also classified according to the number of carbons they contain: those with three carbons are trioses, those with four carbons are tetroses, those with five carbons are pentoses, and those with six and seven carbons are hexoses and heptoses. Therefore, a six-carbon polyhydroxy aldehyde such as glucose is an aldohexose, whereas a six-carbon polyhydroxy ketone such as fructose is a ketohexose.


The D and L Notation PROBLEM 1♦

Classify the following monosaccharides: CH2OH H

O C

H H H

H

C HO H H H

OH OH OH CH2OH

D-ribose

O H OH OH OH CH2OH

O C

HO HO H H

D-sedoheptulose

H H OH OH CH2OH

D-mannose

21.2 THE D AND L NOTATION The smallest aldose, and the only one whose name does not end in “ose,” is glyceraldehyde, an aldotriose. asymmetric center

O C

HOCH2CH

H

A carbon to which four different groups are attached is an asymmetric center.

OH glyceraldehyde

Because glyceraldehyde has an asymmetric center, it can exist as a pair of enantiomers. We know that the isomer on the left has the R configuration because an arrow drawn from the highest priority substituent (OH) to the next highest priority substituent (HC “ O) is clockwise and the lowest priority group is on a hatched wedge (Section 4.7). The R and S enantiomers drawn as Fischer projections are shown on the right. H

clockwise is R

H

O C

HO

H is on a horizontal bond so counterH clockwise is R C

O C

H

C OH H HOCH2

C

H CH2OH

OH CH2OH

perspective formulas

HO

O

H

C H

OH CH2OH

D-glyceraldehyde

the OH group is on the right

O

H

HO

H CH2OH

L-glyceraldehyde

mirror image of D-glyceraldehyde

O

H

C

C H HO HO H

OH H H OH CH2OH

D-galactose

O C

HO H H HO

H OH OH H CH2OH

L-galactose the OH group mirror image of is on the right D-galactose

H CH2OH

(S)-(−)-glyceraldehyde

Fischer projections

The notations d and l are used to describe the configurations of carbohydrates. In a Fischer projection of a monosaccharide, the carbonyl group is always placed on top (in the case of aldoses) or as close to the top as possible (in the case of ketoses). Examine the Fischer projection of galactose shown below and note that the compound has four asymmetric centers (C-2, C-3, C-4, and C-5). If the OH group attached to the bottommost asymmetric center (the carbon second from the bottom) is on the right, then the compound is a D-sugar. If that same OH group is on the left, then the compound is an L-sugar. Almost all sugars found in nature are d-sugars. The mirror image of a d-sugar is an l-sugar. H

O C

(R)-(+)-glyceraldehyde

(S)-(−)-glyceraldehyde

(R)-(+)-glyceraldehyde

H

O

1019


1020

CHAPTER 21

The Organic Chemistry of Carbohydrates

The common name of the monosaccharide, together with the d or l designation, completely defines its structure, because the configurations of all the asymmetric centers are implicit in the common name. Thus, the structure of l-galactose is obtained by changing the configuration of all the asymmetric centers in d-galactose. Emil Fischer and his colleagues studied carbohydrates in the late nineteenth century, when techniques for determining the configurations of compounds were not available. Fischer arbitrarily assigned the R-configuration to the dextrorotatory isomer of glyceraldehyde that we call d-glyceraldehyde. He turned out to be correct: d-glyceraldehyde is (R)-(+)-glyceraldehyde, and l-glyceraldehyde is (S)-(-)-glyceraldehyde. Like R and S, the symbols d and l indicate the configuration of an asymmetric center, but they do not indicate whether the compound rotates the plane of polarization of polarized light to the right (+) or to the left (-) (Section 4.8). For example, d-glyceraldehyde is dextrorotatory, whereas d-lactic acid is levorotatory. In other words, optical rotation, like melting points or boiling points, is a physical property of a compound, whereas “R, S, d, and l” are conventions humans use to indicate the configuration about an asymmetric center. H

O

HO

C H

O C

H

OH CH2OH

D-(+)-glyceraldehyde

OH CH3

D-(−)-lactic

acid

PROBLEM 2

Draw Fischer projections of l-glucose and l-fructose.

PROBLEM 3♦

Indicate whether each of the following structures is d-glyceraldehyde or l-glyceraldehyde, assuming that the horizontal bonds point toward you and the vertical bonds point away from you (Section 4.6): H

O

H

C a. HOCH2

OH

b. HO

H

CH2OH

CH2OH H C H O

c. HO

C H

O

21.3 THE CONFIGURATIONS OF ALDOSES Aldotetroses have two asymmetric centers and therefore four stereoisomers. Two of the stereoisomers are d-sugars and two are l-sugars. The names of the aldotetroses— erythrose and threose—were used to name the erythro and threo pairs of enantiomers described in Section 4.11. H

O

H

C H H

OH OH CH2OH

D-erythrose

O

H

C HO HO

H H CH2OH

L-erythrose

O

H

C HO H

H OH CH2OH

D-threose

O C

H HO

OH H CH2OH

L-threose

Aldopentoses have three asymmetric centers and therefore 8 stereoisomers (four pairs of enantiomers); aldohexoses have four asymmetric centers and 16 stereoisomers (eight pairs of enantiomers). The structures of the four d-aldopentoses and the eight d-aldohexoses are shown in Table 21.1.


The Configurations of Aldoses

1021

Table 21.1 Configurations of the D-Aldoses

H

O C

H

OH CH2OH

D-glyceraldehyde

H

H

O C

H H

OH OH CH2OH

HO H

D-erythrose

H

H

O OH OH OH CH2OH

HO H H

O

H

C

OH OH OH OH CH2OH

HO H H H

D-allose

H OH OH CH2OH

O

H HO H

D-altrose

H

O C

D-glucose

HO HO H H

OH H OH CH2OH

HO HO H

C

D-mannose

O

H

C

H H OH H H OH OH HO H H OH OH CH2OH CH2OH

HO H HO H

D-gulose

H H OH CH2OH

D-lyxose

H

O

O C

D-xylose

H

C

H H OH OH HO H OH H OH H OH OH CH2OH CH2OH

H

O C

D-arabinose

H

O C

H H H H

H

O C

D-ribose

H

H OH CH2OH

D-threose

C H H H

O C

O

O C

H OH H H OH HO HO H H H OH OH CH2OH CH2OH

D-idose

H

C

D-galactose

HO HO HO H

H H H OH CH2OH

D-talose

Diastereomers that differ in configuration at only one asymmetric center are called epimers. For example, d-ribose and d-arabinose are C-2 epimers because they differ in configuration only at C-2; d-idose and d-talose are C-3 epimers. An epimer is a particular kind of diastereomer. (Recall that diastereomers are stereoisomers that are not enantiomers; Section 4.11.) H

O

H

C H H H

O C

OH OH OH CH2OH

D-ribose

HO H H

H OH OH CH2OH

D-arabinose C-2 epimers

H

O

H

O

C HO H HO H

C

H OH H OH CH2OH

HO HO HO H

D-idose

H H H OH CH2OH

D-talose

C-3 epimers

d-Glucose, d-mannose, and d-galactose are the most common aldohexoses in living systems. An easy way to learn their structures is to memorize the structure of d-glucose and then remember that d-mannose is the C-2 epimer of d-glucose and d-galactose is the C-4 epimer of d-glucose. PROBLEM 4â&#x2122;Ś

a. Are d-erythrose and l-erythrose enantiomers or diastereomers? b. Are l-erythrose and l-threose enantiomers or diastereomers?

D-Mannose is the C-2 epimer of D-glucose. D-Galactose is the C-4 epimer of D-glucose.

Diastereomers are stereoisomers that are not enantiomers.


1022

CHAPTER 21

The Organic Chemistry of Carbohydrates

PROBLEM 5♦

a. What sugar is the C-3 epimer of d-xylose? c. What sugar is the C-4 epimer of l-gulose? b. What sugar is the C-5 epimer of d-allose? d. What sugar is the C-4 epimer of d-lyxose? PROBLEM 6♦

What are the systematic names of the following compounds? Indicate the configuration (R or S) of each asymmetric center. c. d-galactose a. d-glucose d. l-glucose b. d-mannose

21.4 THE CONFIGURATIONS OF KETOSES The structures of the naturally occurring ketoses are shown in Table 21.2—they all have a keto group in the 2-position. A ketose has one less asymmetric center than an aldose with the same number of carbons. Therefore, a ketose has only half as many stereoisomers as an aldose with the same number of carbons. Table 21.2

Configurations of the D-Ketoses

CH2OH C

O

CH2OH dihydroxyacetone

CH2OH C

H

O OH CH2OH

D-erythrulose

CH2OH

CH2OH

C

C

H H

O OH OH CH2OH

HO H

D-ribulose

O OH OH OH CH2OH

D-psicose

CH2OH

HO H H

O H OH OH CH2OH

D-fructose

PROBLEM 7♦

What sugar is the C-3 epimer of d-fructose?

H HO H

O OH H OH CH2OH

CH2OH

C

C

C

H H H

D-xylulose

CH2OH

CH2OH

O H OH CH2OH

D-sorbose

C

HO HO H

O H H OH CH2OH

D-tagatose


The Reactions of Monosaccharides in Basic Solutions PROBLEM 8♦

How many stereoisomers are possible for a. a ketoheptose? b. an aldoheptose?

c. a ketotriose?

21.5 THE REACTIONS OF MONOSACCHARIDES IN BASIC SOLUTIONS In a basic solution, a monosaccharide is converted to a mixture of polyhydroxy aldehydes and polyhydroxy ketones. Let’s look at what happens to d-glucose in a basic solution, beginning with its conversion to its C-2 epimer. MECHANISM FOR THE BASE-CATALYZED EPIMERIZATION OF A MONOSACCHARIDE

HO

H

O

H

O

C H HO H H

C HO

OH H OH OH

H

C

HO H H

CH2OH

H

O C

C-2

HO HO H H

OH H OH OH

H H OH OH

CH2OH

D-glucose

+ HO−

CH2OH

an enolate ion

D-mannose

The base removes a proton from an a-carbon, forming an enolate ion (Section 18.3). Notice that C-2 in the enolate ion is no longer an asymmetric center. When C-2 is reprotonated, the proton can come from the top or the bottom of the planar sp2 carbon, forming both d-glucose and d-mannose (C-2 epimers).

Because the reaction forms a pair of epimers, it is called epimerization. Epimerization changes the configuration of a carbon by removing a proton from it and then reprotonating it. In addition to forming its C-2 epimer in a basic solution, d-glucose also undergoes an enediol rearrangement, which forms d-fructose and other ketohexoses. MECHANISM FOR THE BASE-CATALYZED ENEDIOL REARRANGEMENT OF A MONOSACCHARIDE

HO

H

O C

H HO H H

OH H OH OH CH2OH

■ ■

O

H

OH

HO H

C HO H H

OH H OH OH

CH2OH an enolate ion

C HO H H

H H

OH C

C

D-glucose

H

O H H OH OH

CH2OH an enediol

HO

OH C

C HO H H

CH2OH −

O H OH OH

C HO H H

CH2OH an enolate ion

The base removes a proton from an a-carbon, forming an enolate ion. The enolate ion is protonated to form an enediol. The enediol has two OH groups that can form a carbonyl group. Tautomerization of the OH at C-1 (as in base-catalyzed epimerization shown earlier) re-forms d-glucose or forms d-mannose; tautomerization of the OH group at C-2 forms d-fructose.

O H OH OH

CH2OH D-fructose

+ HO−

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The Organic Chemistry of Carbohydrates

In a basic solution an aldose forms a C-2 epimer and one or more ketoses.

Another enediol rearrangement, initiated by a base removing a proton from C-3 of d-fructose, forms an enediol that can tautomerize to give a ketose with the carbonyl group at C-2 or C-3. Thus, the carbonyl group can be moved up and down the chain. PROBLEM 9

Show how an enediol rearrangement can move the carbonyl carbon of fructose from C-2 to C-3.

PROBLEM 10

Write the mechanism for the base-catalyzed conversion of d-fructose into d-glucose and d-mannose.

PROBLEM 11♦

When d-tagatose is added to a basic aqueous solution, an equilibrium mixture of monosaccharides is obtained, two of which are aldohexoses and two of which are ketohexoses. Identify the aldohexoses and ketohexoses.

21.6 THE OXIDATION–REDUCTION REACTIONS OF MONOSACCHARIDES Because they contain alcohol functional groups and aldehyde or ketone functional groups, the reactions of monosaccharides are an extension of what you have already learned about the reactions of alcohols, aldehydes, and ketones. For example, an aldehyde group in a monosaccharide can be oxidized or reduced and can react with nucleophiles to form imines, hemiacetals, and acetals. As you read this section and those that follow, you will find cross-references to earlier discussions of simpler organic compounds that undergo the same reactions. Go back and look at these earlier discussions when you see a crossreference; it will make learning about carbohydrates a lot easier.

Reduction The carbonyl group in aldoses and ketoses can be reduced by NaBH4 (Section 17.7). The product of the reduction is a polyalcohol, known as an alditol. Reduction of an aldose forms one alditol. For example, the reduction of d-mannose forms d-mannitol, the alditol found in mushrooms, olives, and onions. H

O C

HO HO H H

H H OH OH CH2OH

D-mannose

1. NaBH4 2. H3O+

HO HO H H

CH2OH H H OH OH CH2OH

D-mannitol

an alditol

CH2OH 1. NaBH4 2. H3O+

C

O HO H H OH H OH CH2OH D-fructose

1. NaBH4 2. H3O+

H HO H H

CH2OH OH H OH OH CH2OH

D-glucitol an alditol

The reduction of a ketose forms two alditols because the reaction creates a new asymmetric center in the product. For example, the reduction of d-fructose forms d-mannitol and d-glucitol, the C-2 epimer of d-mannitol. d-Glucitol—also called sorbitol—is about 60% as sweet as sucrose. It is found in plums, pears, cherries, and berries. It is also used as a sugar substitute in the manufacture of candy; we will see why on page 1041.


The Oxidation–Reduction Reactions of Monosaccharides PROBLEM 12♦

What products are obtained from the reduction of a. d-idose? b. d-sorbose? PROBLEM 13♦

a. What other monosaccharide is reduced only to the alditol obtained from the reduction of 1. d-talose? 2. d-glucose? 3. d-galactose? b. What monosaccharide is reduced to two alditols, one of which is the alditol obtained from the reduction of 1. d-talose? 2. d-allose?

Oxidation Aldoses can be distinguished from ketoses by observing what happens to the color of an aqueous solution of Br2 when it is added to the sugar. Br2 is a mild oxidizing agent and easily oxidizes the aldehyde group, but it cannot oxidize ketones or alcohols. Consequently, if a small amount of an aqueous solution of Br2 is added to an unknown monosaccharide, the reddish-brown color of Br2 will disappear if the monosaccharide is an aldose because Br2 will be reduced to Br−, which is colorless. If the red color persists, indicating no reaction with Br2, then the monosaccharide is a ketose. The product of the oxidation reaction is an aldonic acid. H

HO

O C

H HO H H

O C

OH H OH + Br2 red OH CH2OH

H HO H H

H2O

D-glucose

OH H OH OH CH2OH

+

2 Br− colorless

D-gluconic acid an aldonic acid

Both aldoses and ketoses are oxidized to aldonic acids by Tollens reagent (Ag+, NH3, HO−), so Tollens reagent cannot be used to distinguish them. Tollens reagent only oxidizes aldehydes, but since the oxidation reaction is carried out in a basic solution, a ketose is converted into an aldose by an enediol rearrangement (Section 21.5), and the aldose is then oxidized by Tollens reagent. H

CH2OH C HO

HO−

O H

O

O C

H HO

OH H

O C

Ag+, NH3 HO−

H HO

OH H

R

R

R

a ketose

an aldose

a carboxylate ion

Dilute nitric acid (HNO3) is a stronger oxidizing agent than those discussed earlier. It oxidizes aldehydes and primary alcohols, but it does not oxidize secondary alcohols. The product obtained when both the aldehyde and the primary alcohol groups of an aldose are oxidized is called an aldaric acid. H H HO H H

O

HO

C OH H OH OH CH2OH

D-glucose

HNO3 Δ

O C

H HO H H O

OH H OH OH C

OH

D-glucaric

acid an aldaric acid

In an aldonic acid, one end is oxidized. In an aldar ic acid, both ends are oxidized.

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PROBLEM 14♦

a. Name an aldohexose other than d-glucose that is oxidized to d-glucaric acid by nitric acid. b. What is another name for d-glucaric acid? c. Name another pair of aldohexoses that are oxidized to identical aldaric acids.

21.7 LENGTHENING THE CHAIN: THE KILIANI–FISCHER SYNTHESIS The carbon chain of an aldose can be increased by one carbon by a modified Kiliani– Fischer synthesis. Thus, tetroses can be converted into pentoses, and pentoses can be converted into hexoses. STEPS IN THE MODIFIED KILIANI–FISCHER SYNTHESIS

N new asymmetric center

H

HCl

O C

H H

OH + OH CH2OH

D-erythrose

C

H

C H H H

N

OH OH OH CH2OH

H2 Pd/BaSO4

HCl

HO H H

H OH OH CH2OH

O

H H H

C

OH OH OH CH2OH

HCl H2O

H H H

OH + OH + NH4 OH products are CH2OH

D-ribose

N C

H

NH C

H H2 Pd/BaSO4

HO H H

H

NH C H OH OH CH2OH

HCl H2O

O C

HO H H

C-2 epimers

H OH OH CH2OH

+

+

NH4

D-arabinose

In the first step, hydrogen cyanide adds to the carbonyl group (Section 17.6). This reaction converts the carbonyl carbon in the starting material to an asymmetric center. Consequently, two products are formed that differ only in their configuration at C-2. The configurations of the other asymmetric centers do not change because no bond to any of the asymmetric centers is broken during the course of the reaction. The C ‚ N bond is reduced to an imine, using a partially deactivated palladium catalyst so that the imine is not further reduced to an amine (Section 7.9). The two imines are hydrolyzed to two aldoses (Section 17.10).

Notice that the modified Kiliani–Fischer synthesis leads to a pair of C-2 epimers. The two epimers are not obtained in equal amounts because the first step of the reaction produces a pair of diastereomers, and diastereomers are generally formed in unequal amounts (Section 6.15). PROBLEM 15♦ The Kiliani–Fischer synthesis leads to a pair of C-2 epimers.

What monosaccharides would be formed in a modified Kiliani–Fischer synthesis starting with each of the following monosaccharides? a. d-xylose b. l-threose

21.8 SHORTENING THE CHAIN: THE WOHL DEGRADATION The Wohl degradation—the opposite of the Kiliani–Fischer synthesis—shortens an aldose chain by one carbon. Thus, hexoses are converted to pentoses, and pentoses are converted to tetroses.


The Stereochemistry of Glucose: The Fischer Proof

1027

a cyano group STEPS IN THE WOHL DEGRADATION

H

H

O C

H HO H H

OH H OH OH CH2OH

NOH C

H HO H H

trace acid NH2OH

D-glucose

OH H OH OH CH2OH

Ac2O 100 ºC

H AcO H H

N

N

C

C

OAc H OAc OAc CH2OAc

HO− H2O

H HO H H

H

OH H OH OH CH2OH

HO− H2O

an oxime

a pentose

Ac2O =

H OH OH CH2OH

O

O

HO H H

D-arabinose

a hexose

O C

C

C O

CH3

CH3

In the first step, the aldehyde reacts with hydroxylamine to form an oxime ( Section 17.10). Heating with acetic anhydride dehydrates the oxime, forming a nitrile; all the OH groups are converted to esters as a result of reacting with acetic anhydride (Section 16.20). In a basic aqueous solution, all the ester groups are hydrolyzed and the cyano group is eliminated (Sections 16.11 and 17.6).

Measuring the Blood Glucose Levels in Diabetes Glucose in the bloodstream reacts with an NH2 group of hemoglobin to form an imine (Section 17.10) that subsequently undergoes an irreversible rearrangement to a more stable a-aminoketone known as hemoglobin-A1c. H

H

O

H HO H H

OH H OH OH CH2OH

N–hemoglobin C

C NH2_hemoglobin trace acid

H HO H H

OH H OH OH CH2OH

CH2NH–hemoglobin C

rearrangement

D-glucose

O HO H H OH H OH CH2OH hemoglobin-A1c

Insulin is the hormone that regulates the level of glucose—and thus the amount of hemoglobin-A1c—in the blood. Diabetes is a condition in which the body does not produce sufficient insulin, or in which the insulin it produces does not function properly. Because people with untreated diabetes have increased blood glucose levels, they also have a higher concentration of hemoglobin-A1c than people without diabetes. Thus, measuring the hemoglobin-A1c level is a way to determine whether the blood glucose level of a diabetic patient is being controlled. Cataracts, a common complication in diabetes, are caused by the reaction of glucose with the NH2 group of proteins in the lens of the eye. Some think the arterial rigidity common in old age may be attributable to a similar reaction of glucose with the NH2 group of proteins.

PROBLEM 16♦

What two monosaccharides can be degraded to a. d-ribose? b. d-arabinose?

c. l-ribose?

21.9 THE STEREOCHEMISTRY OF GLUCOSE: THE FISCHER PROOF Emil Fischer’s determination of the stereochemistry of glucose, done in 1891, is a example of brilliant reasoning. He chose (+)-glucose for his study because it is the most common monosaccharide found in nature. Fischer knew that (+)-glucose is an aldohexose, but 16 different structures can be written for an aldohexose. Which of them represents the structure of (+)-glucose?


1028

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The 16 stereoisomers of an aldohexose are actually eight pairs of enantiomers, so if we know the structures of one set of eight, we automatically know the structures of the other set of eight. Therefore, Fischer needed to consider only one set of eight. He considered the eight stereoisomers that have their C-5 OH group on the right in the Fischer projection (the stereoisomers shown here, which we now call the d-aldoses). One of these is (+)-glucose, and its mirror image is (-)-glucose. H

O

H

C H H H H

O C

OH OH OH OH CH2OH

HO H H H

H

O

H

C

H

O

O

C

H H OH OH HO H OH H OH H OH OH CH2OH CH2OH

HO HO H H

H

H H OH OH CH2OH

H H HO H

H

O

C

C

OH OH H OH CH2OH

HO H HO H

O

H

C

O C

H H OH OH HO H H HO H H OH OH CH2OH CH2OH

HO HO HO H

H H H OH CH2OH

D-allose

D-altrose

D-glucose

D-mannose

D-gulose

D-idose

D-galactose

D-talose

1

2

3

4

5

6

7

8

Fischer used the following information to determine glucose’s stereochemistry—that is, to determine the configuration of each of its asymmetric centers. 1. When the Kiliani–Fischer synthesis is performed on the sugar known as (-)-arabinose,

Emil Fischer (1852–1919) was born in a village near Cologne, Germany. He became a chemist against the wishes of his father, a successful merchant, who wanted him to enter the family business. Fischer was a professor of chemistry at the Universities of Erlangen, Würzburg, and Berlin and in 1902 he received the Nobel  Prize in Chemistry for his work on sugars. Fischer organized German chemical production during World War I. He lost two of his three sons in that war.

the two sugars known as (+)-glucose and (+)-mannose are obtained. This means that (+)-glucose and (+)-mannose are C-2 epimers. Consequently, (+)-glucose and (+)-mannose have to be one of the following pairs: sugars 1 and 2, 3 and 4, 5 and 6, or 7 and 8. 2. (+)-Glucose and (+)-mannose are both oxidized by nitric acid to optically active aldaric acids. The aldaric acids of sugars 1 and 7 would not be optically active because each has a plane of symmetry. (We saw in Section 4.13 that a compound containing a plane of symmetry is achiral.) Excluding sugars 1 and 7 means that (+)-glucose and (+)-mannose must be sugars 3 and 4 or 5 and 6. 3. Because (+)-glucose and (+)-mannose are the products obtained when the Kiliani– Fischer synthesis is carried out on (-)-arabinose, Fischer knew that if (-)-arabinose has the structure shown below on the left, then (+)-glucose and (+)-mannose are sugars 3 and 4. On the other hand, if (-)-arabinose has the structure shown on the right, then (+)-glucose and (+)-mannose are sugars 5 and 6: H

O

H

C HO H H

H OH OH CH2OH

the structure of (−)-arabinose if (+)-glucose and (+)-mannose are sugars 3 and 4

O C

H HO H

OH H OH CH2OH

the structure of (−)-arabinose if (+)-glucose and (+)-mannose are sugars 5 and 6

When (-)-arabinose is oxidized with nitric acid, it forms an optically active aldaric acid. This means that the aldaric acid does not have a plane of symmetry. Therefore, (-)-arabinose must have the structure shown on the left because the aldaric acid of the sugar on the right would have a plane of symmetry. Thus, (+)-glucose and (+)-mannose are represented by sugars 3 and 4. 4. Now the only question remaining is whether (+)-glucose is sugar 3 or sugar 4. To answer this, Fischer had to develop a chemical method for interchanging the aldehyde and primary alcohol groups of an aldohexose. When he chemically interchanged those groups on the sugar known as (+)-glucose, he obtained an aldohexose that was different from (+)-glucose, but when he interchanged those groups on (+)-mannose, he still had (+)-mannose. Therefore, he was able to conclude that (+)-glucose is sugar 3 because interchanging its aldehyde and primary alcohol groups leads to a different sugar (l-gulose).


The Stereochemistry of Glucose: The Fischer Proof

H

H

O C

H HO H H

OH H OH OH CH2OH

reverse the aldehyde and hydroxymethyl groups

H HO H H O

D-glucose

CH2OH OH H OH OH C H

O C

=

HO HO H HO

H H OH H CH2OH

L-gulose

L-gulose

drawn upside down

If (+)-glucose is sugar 3, then (+)-mannose must be sugar 4. As predicted, when the aldehyde and hydroxymethyl groups of sugar 4 are interchanged, the same sugar is obtained. H

H

O C

HO HO H H

H H OH OH CH2OH

reverse the aldehyde and hydroxymethyl groups

D-mannose

HO HO H H O

CH2OH H H OH OH C H

O C

=

HO HO H H

H H OH OH CH2OH

D-mannose

D-mannose

drawn upside down

Using similar reasoning, Fischer went on to determine the stereochemistry of the other aldohexoses. He received the Nobel Prize in Chemistry in 1902 for this achievement. His original guess that (+)-glucose is a d-sugar was shown to be correct in 1951, using X-ray crystallography and a new technique known as anomalous dispersion, so all of his structures are correct. If he had been wrong and (+)-glucose had been an l-sugar, his contribution to the stereochemistry of aldoses would still have had the same importance, but all his stereochemical assignments would have had to be reversed.

Glucose/Dextrose André Dumas first used the term glucose in 1838 to refer to the sweet compound that comes from honey and grapes. Later, Kekulé (Section 8.1) decided that it should be called dextrose because it was dextrorotatory. When Fischer studied the sugar, he called it glucose, and chemists have called it glucose ever since, although dextrose is often found on food labels.

P R O B L E M 1 7 Solved

Aldohexoses A and B are formed from aldopentose C via a Kilani–Fischer synthesis. Nitric acid oxidizes A to an optically active aldaric acid, B to an optically inactive aldaric acid, and C to an optically active aldaric acid. Wohl degradation of C forms D, which is oxidized by nitric acid to an optically active aldaric acid. Wohl degradation of D forms (+)-glyceraldehyde. Identify A, B, C, and D. Solution This is the kind of problem that should be solved by working backward. The

bottommost asymmetric center in D must have the OH group on the right because D is degraded to (+)-glyceraldehyde. Since D is oxidized to an optically active aldaric acid, D must be d-threose. The two bottommost asymmetric centers in C and D have the same configuration because C is degraded to D. Since C is oxidized to an optically active aldaric acid, C must be d-lyxose. Compounds A and B, therefore, must be d-galactose and d-talose. Because A is oxidized to an optically active aldaric acid, it must be d-talose and B must be d-galactose. PROBLEM 18♦

Identify A, B, C, and D in the preceding problem if D is oxidized to an optically inactive aldaric acid; if A, B, and C are oxidized to optically active aldaric acids; and if interchanging the aldehyde and alcohol groups of A leads to a different sugar.

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21.10 MONOSACCHARIDES FORM CYCLIC HEMIACETALS d-Glucose exists in three different forms: the open-chain form of d-glucose that we have been discussing and two cyclic formsâ&#x20AC;&#x201D;a-d-glucose and b-d-glucose. We know that the two cyclic forms are different because they have different melting points and different specific rotations (Section 4.8). How can d-glucose exist in a cyclic form? In Section 17.12, we saw that an aldehyde reacts with an alcohol to form a hemiacetal. The reaction of the alcohol group bonded to C-5 of d-glucose with the aldehyde group forms two cyclic (six-membered ring) hemiacetals. To see that the OH group on C-5 is in the proper position to attack the aldehyde group, we need to convert the Fischer projection of d-glucose to a flat ring structure. To do this, draw the primary alcohol group up from the back left-hand corner. Groups on the right in a Fischer projection are down in the cyclic structure, and groups on the left in a Fischer projection are up in the cyclic structure. 6

HOCH2 5 O 4

H

1

O

Groups on the right in a Fischer projection are down in a Haworth projection.

H HO H H

HO

6

C

3

HOCH2

2

5 OH OH H H C = 4 OH 4 1 OH O 5 HO 3 2 OH OH 6 CH2OH D-glucose

HOCH2 OH O

0.02%

Groups on the left in a Fischer projection are up in a Haworth projection.

OH OH

2

anomeric carbon (a new asymmetric center)

A-D-glucose 36% a Haworth projection

3

D-glucose

1

OH

OH HO OH

anomeric carbon (a new asymmetric center)

B-D-glucose 64% a Haworth projection

The cyclic hemiacetals shown here are drawn as Haworth projections. In a Haworth projection, the six-membered ring is represented as flat and is viewed edge-on. The ring oxygen is always placed in the back right-hand corner of the ring, with C-1 on the right-hand side, and the primary alcohol group attached to C-5 is drawn up from the back left-hand corner. There are two different cyclic hemiacetals because the carbonyl carbon of the openchain aldehyde becomes a new asymmetric center in the cyclic hemiacetal. If the OH group bonded to the new asymmetric center is down (trans to the primary alcohol group at C-5), then the hemiacetal is a-d-glucose; if the OH group is up (cis to the primary alcohol group at C-5), then the hemiacetal is b-d-glucose. The mechanism for cyclic hemiacetal formation is the same as the mechanism for hemiacetal formation between individual aldehyde and alcohol molecules (Section 17.12). a-d-Glucose and b-d-glucose are anomers. Anomers are two sugars that differ in configuration only at the carbon that was the carbonyl carbon in the open-chain form. This carbon is called the anomeric carbon. The prefixes a- and b- denote the configuration about the anomeric carbon. Because anomers, like epimers, differ in configuration at only one carbon, they too are a particular kind of diastereomer. Notice that the anomeric carbon is the only carbon in the molecule that is bonded to two oxygens. HOCH2 O OH HO

OH OH

A-D-glucose

H

O C

H HO H H

OH H OH OH CH2OH

D-glucose

HOCH2 OH O OH HO OH B-D-glucose


Monosaccharides Form Cyclic Hemiacetals

In an aqueous solution, the open-chain form of d-glucose is in equilibrium with the two cyclic hemiacetals. Because formation of the cyclic hemiacetals proceeds nearly to completion (unlike formation of acyclic hemiacetals), very little glucose is in the openchain form (about 0.02%). Even so, the sugar still undergoes the reactions discussed in previous sections (oxidation, reduction, imine formation, etc.) because the reagents react with the small amount of open-chain aldehyde that is present. As the open-chain compound reacts, the equilibrium shifts to produce more open-chain aldehyde, which can then undergo reaction. Eventually, all the glucose molecules react by way of the openchain form. When crystals of pure a-d-glucose are dissolved in water, the specific rotation gradually changes from +112.2 to +52.7. When crystals of pure b-d-glucose are dissolved in water, the specific rotation gradually changes from +18.7 to +52.7. This change in rotation occurs because in water, the hemiacetal opens to form the aldehyde, and both a-d-glucose and b-d-glucose are formed when the aldehyde recyclizes. Eventually, the three forms of glucose reach equilibrium concentrations. At equilibrium, there is almost twice as much b-d-glucose (64%) as a-d-glucose (36%). The specific rotation of the equilibrium mixture is +52.7. This is why the same specific rotation results whether the crystals originally dissolved in water are a-d-glucose or b-d-glucose or any mixture of the two. A slow change in optical rotation to an equilibrium value is called mutarotation. If an aldose can form a five- or a six-membered-ring, it will exist predominantly as a cyclic hemiacetal in solution. Whether a five- or a six-membered-ring is formed depends on their relative stabilities. d-Ribose is an example of an aldose that forms fivemembered-ring hemiacetals: a-d-ribose and b-d-ribose. The Haworth projection of a five-membered-ring sugar is viewed edge-on, with the ring oxygen pointing away from the viewer. The anomeric carbon is again on the right-hand side of the molecule, and the primary alcohol group is drawn up from the back left-hand corner. Again, notice that the anomeric carbon is the only carbon in the molecule that is bonded to two oxygens. H

1

O

C H H H

2

OH = OH 4 OH 5 CH2OH 3

D-ribose

5

HOCH2 OH 4

1C

1

4

O 3

HOCH2 O

HOCH2 O

H

5

3

2

OH OH

OH

OH anomeric carbon

+

2

OH OH

OH OH

A-D-ribose B-D-ribose Haworth projections

D-ribose

Six-membered-ring sugars are called pyranoses, and five-membered-ring sugars are called furanoses. These names come from pyran and furan, the names of the  cyclic ethers shown in the margin. Consequently, a -d-glucose is also called a -d-glucopyranose, and a -d-ribose is also called a -d-ribofuranose. The prefix “ a ” indicates the configuration about the anomeric carbon, and pyranose or furanose indicates the size of the ring. CH2OH O OH HO

OH OH

A-D-glucose A-D-glucopyranose

HOCH2 O OH OH OH A-D-ribose A-D-ribofuranose

Ketoses also exist in solution predominantly in cyclic forms. For example, d-fructose forms a five-membered-ring hemiketal because its C-5 OH group reacts with its ketone carbonyl group. If the OH group bonded to the new asymmetric center is trans to the primary alcohol group, then the compound is a-d-fructofuranose; if it is cis to the

O pyran

O furan

1031


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primary  alcohol group, the compound is b-d-fructofuranose. Notice that the anomeric carbon is C-2 in ketoses, not C-1 as in aldoses. 6

HOCH2 O 5

HOCH2 O

1

CH2OH

OH anomeric carbon HO CH2OH

HO 2 OH

4

3

HO

HO

A-D-fructofuranose

1

O CH2OH HO 2

6 5

HO

4

3

O OH HO

OH

HO HO

HO

B-D-fructofuranose

A-D-fructopyranose

CH2OH

B-D-fructopyranose

d-Fructose can also form a six-membered ring by using its C-6 OH group. The pyranose form predominates in the monosaccharide, whereas the furanose form predominates when the sugar is part of a disaccharide. (See the structure of sucrose on page 1039.) Haworth projections are useful because they show clearly whether the OH groups on the ring are cis or trans to each other. Five-membered rings are nearly planar, so furanoses are represented fairly accurately by Haworth projections. Haworth projections, however, are structurally misleading for pyranoses because a six-membered ring is not flat—it exists preferentially in a chair conformation (Section 3.12). P R O B L E M 1 9 Solved

4-Hydroxy- and 5-hydroxyaldehydes exist primarily as cyclic hemiacetals. Draw the structure of the cyclic hemiacetal formed by each of the following: a. 4-hydroxybutanal b. 4-hydroxypentanal c. 5-hydroxypentanal d. 4-hydroxyheptanal Solution to 19a Draw the reactant with its alcohol and carbonyl groups on the same side of the molecule, then look to see what size ring will form. Two cyclic products are obtained because the carbonyl carbon of the reactant has been converted into a new asymmetric center in the product.

O

H O

C HOCH2CH2CH2

H

= CH2 CH2

H C

O

CH2

O

H + OH

O

OH new asymmetric center

H

PROBLEM 20

Draw the following sugars using Haworth projections: b. a-d-tagatopyranose a. b-d-galactopyranose

c.

a-l-glucopyranose

PROBLEM 21

d-Glucose most often exists as a pyranose, but it can also exist as a furanose. Draw the Haworth projection of a-d-glucofuranose. PROBLEM 22♦

Draw the anomers of d-erythrofuranose.

21.11 GLUCOSE IS THE MOST STABLE ALDOHEXOSE Drawing d-glucose in its chair conformation shows why it is the most common aldohexose in nature. To convert the Haworth projection of d-glucose into a chair conformer, start by drawing the chair so that the backrest is on the left and the footrest is on the right. Then place the ring oxygen at the back right-hand corner and the primary alcohol group in the equatorial position. The primary alcohol group is the largest of all the substituents, and we know that large substituents are more stable in the equatorial position because there is less steric strain in that position (Section 3.13).


Glucose Is the Most Stable Aldohexose

1033

6

4

CH2OH 5 O 1 OH

HO 3

6

6

CH2OH

= HO

5

4

HO

2 OH

3

OH

equatorial

O 4

2

HO

1

axial

CH2OH OH 5 O 1 = HO OH

HO 3

OH

5

4

HO 2

O

2

3

OH

A-D-glucose chair conformer

equatorial

6

CH2OH

HO

equatorial

OH 1

B-D-glucose chair conformer

Because the OH group bonded to C-4 is trans to the primary alcohol group (this is easily seen in the Haworth projection), the C-4 OH group is also in the equatorial position. (Recall from Section 3.14 that 1,2-diequatorial substituents are trans to one another.) The C-3 OH group is trans to the C-4 OH group, so the C-3 OH group is also in the equatorial position. As you move around the ring, you will find that all the OH substituents in b-d-glucose are in equatorial positions. The axial positions are all occupied by hydrogens, which require little space and therefore experience little steric strain. No other aldohexose exists in such a strain-free conformation. This means that b-d-glucose is the most stable of all the aldohexoses, so we should not be surprised that it is the most prevalent aldohexose in nature. The OH group bonded to the anomeric carbon is in the equatorial position in b-d-glucose, whereas it is in the axial position in a-d-glucose. Therefore, b-d-glucose is more stable than a-d-glucose, so b-d-glucose predominates at equilibrium in an aqueous solution. CH2OH HO

CH2OH

O

HO

HO HO

HO

HO CH

HO

OH

A-D-glucose 36%

CH2OH

OH O

equatorial

O OH

HO HO B-D-glucose 64%

axial

If you remember that all the OH groups in b-d-glucose are in equatorial positions, you will find it easy to draw the chair conformer of any other pyranose. For example, if you want to draw a-d-galactose, you would put all the OH groups in equatorial positions except the OH group at C-4 (because galactose is a C-4 epimer of glucose) and the OH group at C-1 (because you want the a-anomer), which both go in axial positions.

The b-position is up in a Haworth projection and equatorial in a chair conformation.

the OH at C-4 is axial

HO

CH2OH

O

HO OH

OH

the OH at C-1 is axial (a)

A-D-galactose

To draw an l-pyranose, draw the d-pyranose first, and then draw its mirror image. For example, to draw b-l-gulose, first draw b-d-gulose, then draw its mirror image. (Gulose differs from glucose at C-3 and C-4, so the OH groups at these positions go in axial positions.) the OH at C-4 is axial

HO

CH2OH

the OH at C-1 is equatorial (b)

HO the OH at C-3 is axial

O

O OH

OH

B-D-gulose

The a-position is down in a Haworth projection and axial in a chair conformation.

OH HOCH2

HO HO

OH

B-L-gulose


1034

CHAPTER 21

The Organic Chemistry of Carbohydrates

Olestra: Nonfat with Flavor

O

glucose

O

O

Chemists have been searching for ways to reduce the caloric content of foods without decreasing their flavor. Many people who believe that “no fat” is synonymous with “no flavor” think this is a worthy endeavor. Procter and Gamble spent 30 years and more than $2 billion to develop a fat substitute they named Olestra (also called Olean). After reviewing the results of more than 150 studies, in 1996 the U. S. Food and Drug Administration approved the limited use of Olestra in snack foods.

n

n

O

O

O O n

n

O

O O

O

O O O

ester groups are too hindered to be hydrolyzed

O

O

O n

n

O n

O fructose

n

Olestra

Olestra

a fat

courtesy of Procter & Gamble Company

Olestra is a semisynthetic compound because it does not exist in nature, but its components do. Developing a compound that can be made from units that are a normal part of our diet decreases the likelihood that the new compound will be toxic. Olestra is made by esterifying all the OH groups of sucrose (a disaccharide composed of d-glucose and d-fructose; see page 1039) with fatty acids obtained from  cottonseed oil and soybean oil. Therefore, its component parts are table sugar and vegetable oil. Because its ester linkages are too sterically hindered to be hydrolyzed by digestive enzymes, Olestra tastes like fat but it cannot be digested and therefore has no caloric value.

P R O B L E M 2 3 ♦ Solved

Which OH groups are in the axial position in b. b-d-idopyranose? a. b-d-mannopyranose? Solution

to

c.

a-d-allopyranose?

23a All the OH groups in b-d-glucose are in equatorial positions.

Because b-d-mannose is a C-2 epimer of b-d-glucose, only the C-2 OH group of b-d-mannose is in the axial position.

21.12 FORMATION OF GLYCOSIDES In the same way that a hemiacetal reacts with an alcohol to form an acetal (Section 17.12), the cyclic hemiacetal formed by a monosaccharide can react with an alcohol to form two acetals. CH2OH HO

CH2OH

O OH

HO OH

H

B-D-glucose B-D-glucopyranose

CH3CH2OH HCl

HO

a glycosidic bond

O OCH2CH3

HO OH

H

CH2OH +

HO

O H

HO OH

OCH2CH3

ethyl B-D-glucoside ethyl B-D-glucopyranoside

ethyl A-D-glucoside ethyl A-D-glucopyranoside

acetal

acetal

The acetal of a sugar is called a glycoside, and the bond between the anomeric carbon and the alkoxy oxygen is called a glycosidic bond. Glycosides are named by replacing the “e” ending of the sugar’s name with “ide.” Thus, a glycoside of glucose is a


Formation of Glycosides

glucoside, a glycoside of galactose is a galactoside, and so on. If the pyranose or furanose name is used, the acetal is called a pyranoside or a furanoside. Notice that the reaction of a single anomer with an alcohol leads to the formation of both the a- and b-glycosides. The mechanism of the reaction shows why both glycosides are formed. MECHANISM FOR GLYCOSIDE FORMATION

CH2OH HO

O

B+

H

CH2OH HO

OH

HO HO

HO

H OR

HO HO

alcohol approaches from the bottom

+

HO

O H

HO

+

HO

H B

HB+ + HO

CH2OH

CH2OH OR

HO

H

a B-glycoside

■ ■

+

OR H

B

O

HO

H

H + ROH + H2O

CH2OH

O

+ B

alcohol approaches from the top

+

O

an oxocarbenium ion

CH2OH

OH

HO

H

HO

HO

+H

HO

CH2OH HO

O

+

HO

O H

HO HO

+

HB+

OR

an A-glycoside major product

The acid protonates the OH group bonded to the anomeric carbon. A lone pair on the ring oxygen helps eliminate a molecule of water. The anomeric carbon in the resulting oxocarbenium ion is sp2 hybridized, so that part of the molecule is planar. (An oxocarbenium ion has a positive charge that is shared by a carbon and an oxygen.) When the alcohol approaches from the top of the plane, the b-glycoside is formed; when it approaches from the bottom of the plane, the a-glycoside is formed.

Notice that the mechanism is the same as that shown for acetal formation in Section  17.12. Surprisingly, d-glucose forms more a-glycoside than b-glycoside. The reason for this is explained in the next section. PROBLEM 24♦

Draw the products formed when b-d-galactose reacts with ethanol and HCl.

The reaction of a monosaccharide with an amine in the presence of a trace amount of acid is similar to the reaction of a monosaccharide with an alcohol. The product of the reaction is an N-glycoside. An N-glycoside has a nitrogen in place of the

1035


1036

CHAPTER 21

The Organic Chemistry of Carbohydrates

oxygen at the glycosidic linkage. The subunits of DNA and RNA are b-N-glycosides (Section 26.1).

HOCH2 O NH2

OH

trace acid

HO HO

HOCH2 O NH

HOCH2 O +

NH HO HO

HO HO N-phenyl B-D-ribosylamine a B-N-glycoside

N-phenyl A-D-ribosylamine an A-N-glycoside

PROBLEM 25♦

Why is only a trace amount of acid used in the formation of an N-glycoside?

21.13 THE ANOMERIC EFFECT We have seen that b-d-glucose is more stable than a-d-glucose because there is more room for a substituent in the equatorial position. However, the preference of the OH group for the equatorial position is not as large as you might expect. For example, the relative amounts of b-d-glucose and a-d-glucose are 2 : 1 (Section 21.10), but the relative amounts of the OH group of cyclohexanol in the equatorial and axial positions is 5.4 : 1 (Table 3.9 on page 131). When glucose reacts with an alcohol to form a glucoside, the major product is the a-glucoside. Because acetal formation is reversible, the a-glucoside must be more stable than the b-glucoside. The preference of certain substituents bonded to the anomeric carbon for the axial position is called the anomeric effect. What is responsible for the anomeric effect? If the substituent is axial, one of the ring oxygen’s lone pairs is in an orbital that is parallel to the s* antibonding orbital of the C–Z bond. The molecule then can be stabilized by electron delocalization—some of the electron density moves from the sp3 orbital of oxygen into the s* antibonding orbital. If the substituent is equatorial, neither of the orbitals that contain a lone pair is aligned correctly for overlap. axial lone pair

axial lone pair overlapping orbitals

O

equatorial lone pair

O Z Z

21.14 REDUCING AND NONREDUCING SUGARS Because glycosides are acetals, they are not in equilibrium with the open-chain aldehyde (or ketone) in aqueous solutions. Without being in equilibrium with a compound that has a carbonyl group, they cannot be oxidized by Ag+. Glycosides, therefore, are nonreducing sugars—they cannot reduce Ag+.


Disaccharides

Hemiacetals, on the other hand, are in equilibrium with the open-chain sugars in aqueous solution, so they can reduce Ag+. In summary, as long as a sugar has an aldehyde, a ketone, or a hemiacetal group, it can reduce Ag+ and therefore is classified as a reducing sugar. An acetal is a nonreducing sugar. P R O B L E M 2 6 ♦ Solved

1037

A sugar with an aldehyde, a ketone, or a hemiacetal group is a reducing sugar. An acetal is a nonreducing sugar.

Name the following compounds and indicate whether each is a reducing sugar or a nonreducing sugar: HO CH2OH CH2OH O O HO c. HO a. OCH2CH2CH3 OH OH OCH3 HO HO b. HO

HOCH2 O OCH2CH3

CH2OH O OH

d. CH2OH OH OH

OH

Solution to 26a The only OH group in an axial position in part a is the one at C-3. Therefore, this sugar is the C-3 epimer of d-glucose, which is d-allose. The substituent at the anomeric carbon is in the b-position. Thus, the sugar’s name is propyl b-d-alloside or propyl b-d-allopyranoside. Because the sugar is an acetal, it is a nonreducing sugar.

21.15 DISACCHARIDES If the hemiacetal group of a monosaccharide forms an acetal by reacting with an alcohol group of another monosaccharide, the glycoside that is formed is a disaccharide. Disaccharides are compounds that consist of two monosaccharide subunits hooked together by a glycosidic linkage. For example, maltose, a disaccharide obtained from the hydrolysis of starch, contains two d-glucose subunits connected by a glycosidic linkage. This particular linkage is called an  A@1,4⬘-glycosidic linkage because the linkage is between C-1 of one sugar subunit and C-4 of the other, and the oxygen bonded to the anomeric carbon in the glycosidic linkage is in the a-position. The prime superscript indicates that C-4 is not in the same ring as C-1. CH2OH O

HO HO

an acetal 1

HO

4'

O an a-1,4′-glycosidic linkage

CH2OH O can be a or b

HO

HO

OH

the configuration of this carbon is not specified maltose

Notice that the structure of maltose does not specify the configuration of the anomeric carbon that is not an acetal (the anomeric carbon of the subunit on the right marked with a wavy line) because maltose can exist in both the a and b forms. In a-maltose, the OH group bonded to this anomeric carbon is in the axial position. In b-maltose, the OH group is in the equatorial position. Because maltose can exist in both a and b forms, mutarotation occurs when crystals of one form are dissolved in a solvent. Maltose is a reducing sugar because the right-hand subunit is a hemiacetal and therefore is in equilibrium with the open-chain aldehyde that is easily oxidized. Cellobiose, a disaccharide obtained from the hydrolysis of cellulose, also contains two d-glucose subunits. Cellobiose is different from maltose, however, because the two glucose subunits are hooked together by a B@1,4⬘-glycosidic linkage. Thus, the only

Remember that the a-position is axial and the b-position is equatorial when a sugar is shown in a chair conformation.


1038

CHAPTER 21

The Organic Chemistry of Carbohydrates

difference in the structures of maltose and cellobiose is the configuration of  the glycosidic linkage. Like maltose, cellobiose exists in both a and b forms because the OH group bonded to the anomeric carbon not involved in acetal formation can be in either the axial position (in a-cellobiose) or the equatorial position (in b-cellobiose). Cellobiose is a reducing sugar because the subunit on the right is a hemiacetal. CH2OH O

HO HO

a b-1,4′-glycosidic linkage

CH2OH O

O

OH

HO

OH

OH

cellobiose

Lactose is a disaccharide found in milk. It constitutes 4.5% of cow’s milk and 6.5% of human milk by weight. The subunits of lactose are d-galactose and d-glucose. The d-galactose subunit is an acetal, and the d-glucose subunit is a hemiacetal. The subunits are joined by a b@1,4⬘-glycosidic linkage. Because one of the subunits is a hemiacetal, lactose is a reducing sugar and undergoes mutarotation. D-galactose is a C-4 epimer of D-glucose

HO CH2OH O HO

OH

a b-1,4′-glycosidic linkage

CH2OH O

O

D-galactose

HO

OH

OH

D-glucose

lactose

A simple experiment can prove that the hemiacetal group in lactose belongs to the glucose residue and not to the galactose residue. The disaccharide is treated with excess methyl iodide in the presence of Ag2O, reagents that methylate all the OH groups via SN2 reactions. Because the OH group is a relatively poor nucleophile, silver oxide is used to increase the leaving propensity of the iodide ion. The product is then hydrolyzed under acidic conditions. acetal is hydrolyzed

OH CH2OH

O

CH2OH O

HO

CH3I excess

O

HO

OH

OCH3 CH2OCH3 O CH3O

Ag2O

OH

CH3O

OH

acetal

acetal is hydrolyzed

CH2OCH3 O O CH3O CH3O

OCH3 acetal

hemiacetal HCl H2O

did not react with CH3I reacted with CH3I

OCH3 CH2OCH3 O

CH2OCH3 O

HO CH3O

CH3O CH3O

OH

2,3,4,6-tetra-O-methylgalactose

CH3O

OH

2,3,6-tri-O-methylglucose

This treatment hydrolyzes the two acetal groups, but the ethers, formed by methylating the OH groups, are untouched. Identification of the products shows that the glucose


Disaccharides

residue contained the hemiacetal group in the disaccharide because its C-4 OH group was not able to react with methyl iodide (since it was used to form the acetal with galactose). The C-4 OH of galactose, on the other hand, was able to react with methyl iodide.

Lactose Intolerance Lactase is an enzyme that specifically breaks the b@1,4⬘-glycosidic linkage of lactose. Cats and dogs lose their intestinal lactase when they become adults; they are then no longer able to digest lactose. Consequently, when they are fed milk or milk products, the undegraded lactose causes digestive problems such as bloating, abdominal pain, and diarrhea. These problems occur because only monosaccharides can pass into the bloodstream, so lactose, a disaccharide, has to pass undigested into the large intestine. When humans have stomach flu or other intestinal disturbances, they can temporarily lose their lactase, thereby becoming lactose intolerant. About 75% of adults lose their lactase permanently as they mature—explaining why “lactose-free” products are so common. Those intolerant to lactose can take lactase in pill form before eating products that contain lactose. Lactose intolerance is most common in people whose ancestors came from nondairy-producing countries. For example, only 3% of Danes are lactose intolerant, compared with 90% of all Chinese and Japanese and  97% of Thais. This is why you are not likely to find dairy items on Chinese menus.

Galactosemia After lactose is degraded into glucose and galactose, the galactose must be converted into glucose before it can be used by cells. Individuals who do not have the enzyme that converts galactose into glucose have the genetic disease known as galactosemia. Without this enzyme, galactose accumulates in the bloodstream, a condition that can cause mental retardation and even death in infants. Galactosemia is treated by excluding galactose from the diet.

The most common disaccharide, sucrose, is the substance we know as table sugar. Obtained from sugar beets and sugar cane, sucrose consists of a d-glucose subunit and a d-fructose subunit linked by a glycosidic bond between C-1 of glucose (in the a-position) and C-2 of fructose (in the b-position). About 90 million tons of sucrose are produced commercially throughout the world each year. Unlike the other disaccharides that we have looked at, sucrose is not a reducing sugar and does not exhibit mutarotation because its glycosidic bond is between the anomeric carbon of glucose and the anomeric carbon of fructose. Sucrose, therefore, does not have a hemiacetal group, so it is not in equilibrium with the readily oxidized open-chain aldehyde or ketone form in aqueous solution. CH2OH O

HO HO

HO HOCH2

a linkage at glucose

O

O b linkage at fructose

HO CH2OH HO sucrose

a mixture of sucrose, glucose, and fructose

1039


1040

CHAPTER 21

The Organic Chemistry of Carbohydrates

Sucrose has a specific rotation of +66.5. When it is hydrolyzed, the resulting 1 : 1 mixture of glucose and fructose has a specific rotation of −22.0. Because the sign of the rotation changes when sucrose is hydrolyzed, a 1 : 1 mixture of glucose and fructose is called invert sugar. The enzyme that catalyzes the hydrolysis of sucrose is called invertase. Honeybees have invertase, so the honey they produce is a mixture of sucrose, glucose, and fructose. Because fructose is sweeter than sucrose, invert sugar is also sweeter than sucrose. Some “lite” foods contain fructose instead of sucrose, which means that they achieve the same level of sweetness with a lower sugar (lower calorie) content. PROBLEM 27♦

What is the specific rotation of an equilibrium mixture of fructose? (Hint: The specific rotation of an equilibrium mixture of glucose is +52.7.)

21.16 POLYSACCHARIDES Polysaccharides contain as few as 10 or as many as several thousand monosaccharide units joined together by glycosidic linkages. The most common polysaccharides are starch and cellulose. Starch is the major component of flour, potatoes, rice, beans, corn, and peas. It is a mixture of two different polysaccharides: amylose (~20%) and amylopectin (~80%). Amylose is composed of unbranched chains of d-glucose units joined by a@1,4⬘-glycosidic linkages. CH2OH O

O

HO HO CH2OH O

CH2OH O

O

an a-1,6′-glycosidic linkage

HO OH

O

CH2OH

HO

CH2OH

HO

O

an a-1,4′-glycosidic linkage

1

O HO

O

HO HO

CH2OH O HO

three subunits of amylose

HO

O

O

O

6′ CH2

O

O

HO HO

HO

O

five subunits of amylopectin

CH2OH O

O

HO HO

O

Amylopectin is a branched polysaccharide. Like amylose, it is composed of chains of d-glucose units joined by a@1,4⬘-glycosidic linkages. Unlike amylose, however, amylopectin also contains A@1,6⬘-glycosidic linkages. These linkages create branches in the polysaccharide (Figure 21.1). Amylopectin can contain up to 106 glucose units, making it one of the largest molecules found in nature. an a-1,4′-glycosidic bond

an a-1,6′-glycosidic bond

▶ Figure 21.1 Branching in amylopectin. The hexagons represent glucose units. They are joined by a@1,4⬘- and a@1,6⬘-glyclosidic bonds.


Polysaccharides

1041

Cells oxidize d-glucose in the first of a series of processes that provide them with energy (Section 25.7). When animals have more d-glucose than they need for energy, they convert the excess d-glucose into a polymer called glycogen. Glycogen has a structure similar to that of amylopectin, but glycogen has more branches (Figure 21.2). The branch points in glycogen occur about every 10 residues, whereas those in amylopectin occur about every 25 residues. The high degree of branching in glycogen has important physiological consequences. When an animal needs energy, many individual glucose units can be simultaneously removed from the ends of many branches. Plants convert excess d-glucose into starch.

◀ Figure 21.2 amylopectin

A comparison of the branching in amylopectin and glycogen.

glycogen

Why the Dentist Is Right Bacteria found in the mouth have an enzyme that converts sucrose into a polysaccharide called dextran. Dextran is made up of glucose units joined mainly through a@1,3⬘- and a@1,6⬘glycosidic linkages. About 10% of dental plaque is composed of dextran, and bacteria hidden in  the plaque attack tooth enamel. This is the chemical basis for your dentist’s warning not to eat candy. This is also why sorbitol and mannitol are the saccharides added to “sugarless” gum—they cannot be converted to dextran.

Cellulose is the major structural component of plants. Cotton, for example, is composed of about 90% cellulose, and wood is about 50% cellulose. Like amylose, cellulose is composed of unbranched chains of d-glucose units. Unlike amylose, however, the glucose units in cellulose are joined by b-1,4⬘-glycosidic linkages rather than by a@1,4⬘-glycosidic linkages (see page 1038). CH2OH O HO

O

CH2OH O

OH

HO

a b-1,4′-glycosidic linkage

O

CH2OH O

OH

HO

O O

OH

three subunits of cellulose

The different glycosidic linkages in starch and cellulose give these compounds very different physical properties. The a-linkages in starch cause amylose to form a helix that promotes hydrogen bonding of its OH groups to water molecules (Figure 21.3). As a result, starch is soluble in water. On the other hand, the b-linkages in cellulose promote the formation of intramolecular hydrogen bonds. Consequently, these molecules form linear arrays (Figure 21.4), held together by hydrogen bonds between adjacent chains. These large aggregates cause cellulose to be insoluble in water. The strength of these bundles of polymer chains makes cellulose an effective structural material. Processed cellulose is also used for the production of paper and cellophane.

▲ Figure 21.3 The a@1,4⬘-glycosidic linkages in amylose cause it to form a lefthanded helix. Many of its OH groups form hydrogen bonds with water molecules.


1042

CHAPTER 21

The Organic Chemistry of Carbohydrates intramolecular hydrogen bond

note that this ring is inverted compared to that in the structure of cellulose on page 1041

strands of cellulose in a plant cell wall

▲ Figure 21.4 The b@1,4⬘-glycosidic linkages in cellulose form intramolecular hydrogen bonds, which cause the molecules to assemble in linear arrays. (Notice that hydrogens are not shown in the figure.)

All mammals have the enzyme (a-glucosidase) that hydrolyzes the a@1,4⬘-glycosidic linkages that join glucose units in amylose, amylopectin, and glycogen, but they do not have the enzyme (b-glucosidase) that hydrolyzes b@1,4⬘-glycosidic linkages. As a result, mammals cannot obtain the glucose they need by eating cellulose. However, bacteria that possess b-glucosidase inhabit the digestive tracts of grazing animals, so cows can eat grass and horses can eat hay to meet their nutritional requirements for glucose. Termites also harbor bacteria that break down the cellulose in the wood they eat. Chitin (KY-tin) is a polysaccharide that is structurally similar to cellulose. It is the major structural component of the shells of crustaceans (such as lobsters, crabs, and shrimp) and the exoskeletons of insects and other arthropods, and it is also the structural material of fungi. Like cellulose, chitin has b@1,4⬘-glycosidic linkages. Unlike cellulose, chitin has an N-acetylamino group instead of an OH group at the C-2 position. The b@1,4⬘-glycosidic linkages give chitin its structural rigidity. CH2OH O

O

CH2OH O

HO

NH

O

C

CH3

The shell of this bright orange crab from Australia is composed largely of chitin.

Cl

F

O

O

O

C

C

F

F

N H

N H

O

three subunits of chitin

Several different drugs have been developed to help pet owners control fleas. One of these drugs is lufenuron, the active ingredient in Program. Lufenuron interferes with the flea’s production of chitin. The consequences are fatal for the flea because its exoskeleton is composed primarily of chitin.

Cl

lufenuron

F

CF3

O O

HO

CH3

Controlling Fleas

F

CH2OH O

HO

NH C

O

NH C CH3

O

an N-acetylamino group


Some Naturally Occurring Compounds Derived from Carbohydrates

1043

PROBLEM 28♦

What is the main structural difference between a. amylose and cellulose? b. amylose and amylopectin?

c. amylopectin and glycogen? d. cellulose and chitin?

21.17 SOME NATURALLY OCCURRING COMPOUNDS DERIVED FROM CARBOHYDRATES Deoxy sugars are sugars in which one of the OH groups is replaced by a hydrogen (deoxy means “without oxygen”). 2-Deoxyribose is a deoxy sugar that is missing the oxygen at the C-2 position. d-Ribose is the sugar component of ribonucleic acid (RNA), whereas 2-deoxyribose is the sugar component of deoxyribonucleic acid (DNA) (see Section 26.1). 5

HOCH2 O OH

HOCH2 O OH 1

4 3

OH OH

2

OH

B-D-ribose

B-D-2-deoxyribose

In amino sugars, one of the OH groups is replaced by an amino group. N-Acetylglucosamine—the subunit of chitin and one of the subunits of bacterial cell walls (Section 23.12)—is an example of an amino sugar. Some important antibiotics contain amino sugars. For example, two of the three subunits of the antibiotic gentamicin are deoxyamino sugars. Notice that the middle subunit is missing the ring oxygen, so it is not a sugar. CH3CHNHCH3

O

NH2

NH2

O HO

NH2

O

gentamicin an antibiotic

O HO

NH CH3

CH3 OH

Gentamicin is one of several aminoglycoside antibiotics; streptomycin and neomycin are others. The antibiotics work by binding to a bacterial ribosome on which protein synthesis takes place (Section 26.8). As a result, the bacteria are not able to synthesize proteins.

Bacterial Resistance A bacterial strain typically takes 15 to 20 years to become resistant to an antibiotic. For example, penicillin became widely available in 1944 and by 1952, 60% of all Staphylococcus aureus infections were penicillin resistant (see page 759). As a result of the widespread use of the aminoglycoside antibiotics, some bacteria developed enzymes that can acetylate or phosphorylate the OH and NH2 groups of the antibiotic. When this happens, the antibiotic can no longer bind to the bacterial ribosome, so it has no effect on the bacteria.


1044

CHAPTER 21

The Organic Chemistry of Carbohydrates

Until relatively recently, the last discovery of a new class of antibiotics was in the 1970s, so drug resistance became an increasingly important problem in medicinal chemistry. Vancomycin had been the antibiotic of last resort because there were no reported cases of vancomycin-resistant bacteria until 1989, when more and more bacteria became resistant. The approval of Zyvox by the FDA in April 2000 was met with great relief by the medical community because it was the first in a new family of antibiotics—namely, the oxazolidinones. In clinical trials, Zyvox was found to cure 75% of the patients infected with bacteria that had become resistant to all other antibiotics. Another new class of antibiotic became available in 2005 when the FDA approved Cubicin, the first of the cyclic lipopeptide antibiotics. O

O O

C

CH3

N H

N

N

H

O

F

linezolid Zyvox®

Heparin—A Natural Anticoagulant Heparin is a polysaccharide found principally in cells that line arterial walls. Some of its alcohol and amino groups are sulfonated, some of its alcohol groups are oxidized, and some of its amino groups are acetylated. Heparin is released to prevent excessive blood clot formation when an injury occurs. Heparin is widely used clinically—particularly after surgery—to prevent blood from clotting.

O

CH2OSO3− O HO

CO2−

O3SNH

CH2OSO3− O

O

O HO

OSO3−

O

CO2−

HO CH3CNH

heparin

O

O

O

HO

OSO3−

O

l-Ascorbic acid (vitamin C) is synthesized from d-glucose in plants and in the livers of most vertebrates. Primates and guinea pigs do not have the enzymes necessary for the biosynthesis of vitamin C, so they must obtain the vitamin from their diets. STEPS IN THE SYNTHESIS OF L-ASCORBIC ACID

H H HO H H

H

O C OH H OH OH CH2OH

oxidizing enzyme

HO

O C

H HO H H

OH H OH OH

reducing enzyme

H HO H H

C OH

O

D-glucose

O

CH2OH OH H OH OH C OH

HO HO H HO

rotate 180°

CH2OH OH O

H oxidation

O

L-dehydroascorbic

oxidizing enzyme

O

O

HO acid

L-ascorbic

pKa = 4.17

acid

lactonase

CH2OH OH O

O

H H OH H CH2OH

L-gulonic

L-configuration

H

O C

OH

acid vitamin C

H

CH2OH OH O OHHO

H

H

a G-lactone

O


Carbohydrates on Cell Surfaces

■ ■

The biosynthesis of vitamin C involves the conversion of d-glucose to l-gulonic acid, reminiscent of the last step in the Fischer proof. l-Gulonic acid is converted into a g-lactone by the enzyme lactonase. The lactone is oxidized to l-ascorbic acid. The l-designation of ascorbic acid refers to the configuration at C-5, which was C-2 in d-glucose and C-5 in l-gulonic acid.

Although l-ascorbic acid does not have a carboxylic acid group, it is an acidic compound because the pKa of the C-3 OH group is 4.17. l-Ascorbic acid is readily oxidized to l-dehydroascorbic acid, which is also physiologically active. If the lactone ring is opened by hydrolysis, all vitamin C activity is lost. Therefore, not much intact vitamin C survives in food that has been thoroughly cooked. And if food is cooked in water and then drained, the water-soluble vitamin is thrown out with the water!

Vitamin C Vitamin C is an antioxidant because it traps radicals formed in aqueous environments, preventing harmful oxidation reactions the radicals would cause (Section  13.11). Not all the physiological functions of vitamin C are known. However, we do know it is required for collagen fibers to form properly. Collagen is the structural protein of skin, tendons, connective tissue, and bone. Vitamin C is abundant in citrus fruits and tomatoes. When the vitamin is not present in the diet, lesions appear on the skin, severe bleeding occurs about the gums, in the joints, and under the skin, and any wound heals slowly. The condition, known as scurvy, was the first disease to be an English sailor circa 1829 treated by adjusting the diet. British sailors who shipped out to sea after the late 1700s were required to eat limes to prevent scurvy (which is how they came to be called “limeys”). Not until 200 years later did it become known that the substance preventing scurvy was vitamin C. Scorbutus is Latin for “scurvy”; ascorbic, therefore, means “no scurvy.” PROBLEM 29♦

Explain why the C-3 OH group of vitamin C is more acidic than the C-2 OH group.

21.18 CARBOHYDRATES ON CELL SURFACES Many cells have short oligosaccharide chains on their surface that enable the cells to recognize and interact with other cells and with invading viruses and bacteria. These oligosaccharides are linked to the surface of the cell by the reaction of an OH or an NH2 group of a cell-membrane protein with the anomeric carbon of a cyclic sugar. Proteins attached to oligosaccharides are called glycoproteins. The percentage of carbohydrate in glycoproteins is variable; some glycoproteins contain as little as 1% carbohydrate by weight, whereas others contain as much as 80%. glycoproteins

O

CH2OH

CH2OH

O

O O-protein

HO NH C

O NH-protein

HO NH

O

CH3

C

O

CH3

1045


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CHAPTER 21

The Organic Chemistry of Carbohydrates

oligosaccharide cell surface

Type A

Carbohydrates on the surfaces of cells provide a way for cells to recognize one another, serving as points of attachment for other cells, viruses, and toxins. Therefore, surface carbohydrates have been found to play a role in activities as diverse as infection, prevention of infection, fertilization, inflammatory diseases such as rheumatoid arthritis and septic shock, and blood clotting. The fact that several known antibiotics contain amino sugars (Section 21.17) suggests that the antibiotics function by recognizing target cells. Carbohydrate interactions also are involved in the regulation of cell growth, so changes in membrane glycoproteins are thought to be correlated with malignant transformations. Differences in blood type (A, B, or O) are actually differences in the sugars bound to the surfaces of red blood cells. Each type of blood is associated with a different carbohydrate structure (Figure 21.5). Type AB blood is a mixture of type A blood and type B blood. N-acetyl-D-galactosamine

D-galactose

N-acetyl-D-glucosamine

PROTEIN

N-acetyl-D-glucosamine

PROTEIN

N-acetyl-D-glucosamine

PROTEIN

L-fucose

Type B

D-galactose

D-galactose

L-fucose

Type O

D-galactose

L-fucose

▲ Figure 21.5 Blood type is determined by the sugars on the surfaces of red blood cells. Fucose is 6-deoxygalactose.

Antibodies are proteins that are synthesized by the body in response to foreign substances called antigens. Interaction with the antibody causes the antigen to either precipitate or be flagged for destruction by immune system cells. This is why blood cannot be transferred from one person to another unless the blood types of the donor and acceptor are compatible. Otherwise the donated blood will be considered a foreign substance and will provoke an immune response. Looking at Figure 21.5, we can see why the immune system of people with type A blood recognizes type B blood as foreign and vice versa. The immune system of people with type A, B, or AB blood does not recognize type O blood as foreign because the carbohydrate in type O blood is also a component of types A, B, and AB. Thus, anyone can  accept type O blood, so people with that blood type are called universal donors. People with type AB blood can accept types AB, A, B, and O blood, so they are referred to as universal acceptors. PROBLEM 30

Refer to Figure 21.5 to answer the following questions: a. People with type O blood can donate blood to anyone, but they cannot receive blood from everyone. From whom can they not receive blood? b. People with type AB blood can receive blood from anyone, but they cannot give blood to everyone. To whom can they not give blood?


Artificial Sweeteners

21.19 ARTIFICIAL SWEETENERS For a molecule to taste sweet, it must bind to a receptor on a taste bud cell on the tongue. The binding of this molecule causes a nerve impulse to pass from the taste bud to the brain, where the molecule is interpreted as being sweet. Sugars differ in their degree of “sweetness.” Compared with the sweetness of glucose, which is assigned a relative value of 1.00, the sweetness of sucrose is 1.45, and that of fructose, the sweetest of all sugars, is 1.65. Developers of artificial sweeteners must evaluate potential products in terms of several factors—such as toxicity, stability, and cost—in addition to taste. Saccharin (Sweet’N Low), the first synthetic sweetener, was discovered accidentally by Ira Remsen in 1879. One evening he noticed that the dinner rolls initially tasted sweet and then bitter. Because his wife did not notice that the rolls had an unusual taste, Remsen tasted his fingers and found they had the same odd taste. The next day he tasted the chemicals he had been working with the day before and found one that had an extremely sweet taste. (As strange as it may seem today, at one time it was common for chemists to taste compounds in order to characterize them.) He called this compound saccharin; it was eventually found to be about 300 times sweeter than glucose. Notice that, in spite of its name, saccharin is not a saccharide. O

O C O

NH S

CH3CH2O

O

O saccharin

NHCNH2

O

O

Cl

OCCH2CHCNHCHCOCH3 +

NH3

aspartame

NHSO3− Na+ sodium cyclamate

CH2

O

O

acesulfame potassium

dulcin

O

CH3

N−K+ S O

CH2OH

O

HO OH ClCH2 O O HO OH

CH2Cl

sucralose

Because it has little caloric value, saccharin became an important substitute for sucrose when it became commercially available in 1885. The chief nutritional problem in the West was—and still is—the overconsumption of sugar and its consequences: obesity, heart disease, and dental decay. Saccharin is also a boon to people with diabetes, who must limit their consumption of sucrose and glucose. Although the toxicity of saccharin had not been studied carefully when it was first marketed (our current concern with toxicity is a fairly recent development), extensive studies since then have shown saccharin to be harmless. In 1912, saccharin was temporarily banned in the United States, not because of any concern about its toxicity, but because of a concern that people would miss out on the nutritional benefits of sugar. Dulcin was the second synthetic sweetener to be discovered (in 1884). Even though it did not have the bitter, metallic aftertaste associated with saccharin, it never achieved much popularity. Dulcin was taken off the market in 1951 in response to concerns about its toxicity. Sodium cyclamate became a widely used nonnutritive sweetener in the 1950s, but was banned in the United States some 20 years later in response to two studies that appeared to show that large amounts of sodium cyclamate cause liver cancer in mice. Aspartame (NutraSweet, Equal), about 200 times sweeter than sucrose, was approved by the U.S. Food and Drug Administration (FDA) in 1981. Because aspartame contains a

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CHAPTER 21

The Organic Chemistry of Carbohydrates

phenylalanine subunit, it should not be used by people with the genetic disease known as phenylketonuria (PKU) (see page 1186). Acesulfame potassium (Sweet and Safe, Sunette, Sweet One) was approved in 1988. Also called acesulfame-K, it too is about 200 times sweeter than glucose. It has less aftertaste than saccharine and is more stable than aspartame at high temperatures. Sucralose (Splenda), 600 times sweeter than glucose, is the most recently approved (1991) synthetic sweetener. It maintains its sweetness in foods stored for long periods and at temperatures used in baking. Sucralose is made from sucrose by selectively replacing three of sucrose’s OH groups with chlorines. During chlorination, the 4-position of the glucose ring becomes inverted, so sucralose is a galactopyranoside, not a glucopyranoside. Sucralose is the only artificial sweetener that has a carbohydrate-like structure. However, because of the chlorine atoms, the body does not recognize it as a carbohydrate, so it is eliminated from the body instead of being metabolized. The fact that these synthetic sweeteners have such different structures shows that the sensation of sweetness is not induced by a single molecular shape.

Acceptable Daily Intake The FDA has established an acceptable daily intake (ADI) value for many of the food ingredients it clears for use. The ADI is the amount of the substance a person can consume safely, each day of his or her life. For example, the ADI for acesulfame-K is 15 mg/kg/day. This means that each day a 132-lb person can consume the amount of acesulfame-K that would be found in two gallons of an artificially sweetened beverage. The ADI for sucralose is also 15 mg/kg/day.

S OME IMPORTANT THINGS TO REMEMBER ■

Bioorganic compounds—organic compounds found in living systems—obey the same chemical principles that smaller organic molecules do. Much of the structure of bioorganic compounds exists for the purpose of molecular recognition. Carbohydrates are polyhydroxy aldehydes (aldoses) and polyhydroxy ketones (ketoses), or compounds formed by linking up aldoses and ketoses. The notations d and l describe the configuration of the bottommost asymmetric center of a monosaccharide in a Fischer projection. Most naturally occurring sugars are d-sugars. Naturally occurring ketoses have the ketone group in the 2-position. Epimers differ in configuration at only one asymmetric center: d-mannose is the C-2 epimer of d-glucose and d-galactose is the C-4 epimer of d-glucose. In a basic solution, a monosaccharide is converted to a mixture of polyhydroxy aldehydes and polyhydroxy ketones. Reduction of an aldose forms one alditol; reduction of a ketose forms two alditols.

■ ■

Br2 oxidizes aldoses but not ketoses; Tollens reagent oxidizes both. Aldoses are oxidized to aldonic acids or to aldaric acids. The Kiliani–Fischer synthesis increases the carbon chain of an aldose by one carbon; it forms C-2 epimers. The Wohl degradation decreases the carbon chain by one carbon. The OH groups of monosaccharides react with methyl iodide/silver oxide to form ethers. The aldehyde or keto group of a monosaccharide reacts with one of its OH groups to form cyclic hemiacetals: glucose forms a-d-glucose and b-d-glucose. More b-d-glucose is present than a-d-glucose in an aqueous solution at equilibrium. a-d-Glucose and b-d-glucose are anomers—they differ in configuration only at the anomeric carbon, which is the carbon that was the carbonyl carbon in the openchain form. A slow change in optical rotation to an equilibrium value is called mutarotation. The a-position is axial when a sugar is shown in a chair conformation and down when the sugar is shown in a


Summary of Reactions

Haworth projection; the b-position is equatorial when a sugar is shown in a chair conformation and up when the sugar is shown in a Haworth projection. Six-membered-ring sugars are pyranoses; fivemembered-ring sugars are furanoses. The most abundant monosaccharide in nature is d-glucose. All the OH groups in b-d-glucose are in equatorial positions. A cyclic hemiacetal can react with an alcohol to form an acetal, called a glycoside. If the name “pyranose” or “furanose” is used, the acetal is called a pyranoside or a furanoside, respectively. The bond between the anomeric carbon and the alkoxy oxygen is called a glycosidic bond. The preference for the axial position by certain substituents bonded to the anomeric carbon is called the anomeric effect. If a sugar has an aldehyde, ketone, hemiacetal, or hemiketal group, it is a reducing sugar. Acetals are not reducing sugars. Disaccharides consist of two monosaccharides hooked together by a glycosidic linkage. Maltose has an A@1,4⬘-glycosidic linkage between two glucose subunits; cellobiose has a B@1,4⬘-glycosidic linkage between two glucose subunits.

The most common disaccharide is sucrose; it has a d-glucose subunit and a d-fructose subunit linked by their anomeric carbons. Polysaccharides contain as few as 10 or as many as several thousand monosaccharides joined together by glycosidic linkages. Starch is composed of amylose and amylopectin. Amylose has unbranched chains of d-glucose units joined by a@1,4⬘-glycosidic linkages. Amylopectin also has chains of d-glucose units joined by a@1,4⬘-glycosidic linkages, but it also has a@1,6⬘-glycosidic linkages that create branches. Glycogen is similar to amylopectin but has more branches. Cellulose has unbranched chains of d-glucose units joined by b@1,4⬘-glycosidic linkages. The a-linkages cause amylose to form a helix and be water soluble; the b-linkages allow the molecules of cellulose to form linear arrays and be water insoluble. The surfaces of many cells contain short oligosaccharides that allow the cells to interact with each other. The oligosaccharides are attached to the cell surface by protein groups. Proteins bonded to oligosaccharides are called glycoproteins.

SUMMARY OF REACTIONS 1. Epimerization (Section 21.5). The mechanism of the reaction is shown on page 1023. H

O

H

O

C H

C HO−

OH (CHOH)n

HO

H2O

H (CHOH)n

CH2OH

CH2OH

2. Enediol rearrangement (Section 21.5). The mechanism of the reaction is shown on page 1023. H

O

H

C CHOH (CHOH)n

OH C

HO− H2O

CH2OH

C

CH2OH OH

H2O −

(CHOH)n

HO

CH2OH

C

O

(CHOH)n CH2OH

3. Reduction (Section 21.6) H

CH2OH

O C (CHOH)n CH2OH

CH2OH 1. NaBH4 2. H3O+

1049

C

CH2OH

O

(CHOH)n

(CHOH)n

CH2OH

CH2OH

CHOH 1. NaBH4 2. H3O+

(CHOH)n CH2OH


1050

CHAPTER 21

The Organic Chemistry of Carbohydrates

4. Oxidation (Section 21.6) H

O

C

C

CHOH

Ag+, NH3 HO−

O C

(CHOH)n

Br2 H2O

(CHOH)n + 2 Br−

CH2OH

H

CH2OH

O

HO

C

(CHOH)n + Ag

CH2OH

HO

O

O

CH2OH b. (CHOH)n

c.

CH2OH −

O C

(CHOH)n + Ag

CH2OH

O

H

C

Ag+, NH3 HO−

(CHOH)n

a.

O

O

C

(CHOH)n

d.

HNO3 Δ

(CHOH)n

CH2OH

CH2OH

O C

C O

OH

5. Chain elongation: the Kiliani–Fischer synthesis (Section 21.7) H

O

H

C (CHOH)n

O C

1. NaC N/HCl 2. H2, Pd/BaSO4 3. HCl/H2O

(CHOH)n + 1

CH2OH

CH2OH

6. Chain shortening: the Wohl degradation (Section 21.8) H

O

H

C

O C

1. NH2OH/trace acid 2. Ac2O, 100 ºC 3. HO−, H2O

(CHOH)n CH2OH

(CHOH)n − 1 CH2OH

7. Hemiacetal formation (Section 21.11) CH2OH HO

CH2OH

O

HO

HO HO

HO HO

OH

CH2OH

OH

HO CH

O

O OH

HO HO

8. Glycoside formation (Section 21.12). The mechanism of the reaction is shown on page 1034. CH2OH HO

CH2OH

O

HCl ROH

HO OH

OH

HO

CH2OH

O OR

HO

+ HO

O

HO

OH

P R OBLEMS 31. What product or products are obtained when d-galactose reacts with each of the following substances? a. nitric acid + ⌬ g. 1. hydroxylamine/trace acid d. excess CH3I + Ag2O b. Ag+, NH3, HO2. acetic anhydride/ ⌬ e. Br2 in water c. NaBH4, followed by H3O+ 3. HO-/H2O f. ethanol + HCl

OH

OR


Problems

1051

32. Name the epimers of d-glucose. 33. Identify the sugar in each description. a. An aldopentose that is not d-arabinose forms d-arabinitol when it is reduced with NaBH4. b. A sugar that is not d-altrose forms d-altraric acid when it is oxidized with nitric acid. c. A ketose that, when reduced with NaBH4, forms d-altritol and d-allitol. 34. d-Xylose and d-lyxose are formed when d-threose undergoes a Kilianiâ&#x20AC;&#x201C;Fischer synthesis. d-Xylose is oxidized to an optically inactive aldaric acid, whereas d-lyxose forms an optically active aldaric acid. What are the structures of d-xylose and d-lyxose? 35. Answer the following questions about the eight aldopentoses. a. Which are enantiomers? b. Which are C-2 epimers? c. Which form an optically active compound when oxidized with nitric acid? 36. What is the configuration of each of the asymmetric centers in the Fischer projection of a. d-glucose?

b. d-galactose?

c. d-ribose?

d. d-xylose?

e. d-sorbose?

37. The reaction of d-ribose with one equivalent of methanol plus HCl forms four products. Draw the products. 38. Name the following compounds: HOCH2 O OCH3 a. OH CH2OH OH

HO

CH2OH

b.

HO O

c.

OCH2CH3 HO

OH

CH2OH O OH HO

OCH3

39. A student isolated a monosaccharide and determined that it had a molecular weight of 150. Much to his surprise, he found that it was not optically active. What is the structure of the monosaccharide? 40. Propose a mechanism for the formation of d-allose from d-glucose in a basic solution. 41. Treatment with sodium borohydride converts aldose A into an optically inactive alditol. Wohl degradation of A forms B, whose alditol is optically inactive. Wohl degradation of B forms d-glyceraldehyde. Identify A and B. 42. A hexose was obtained after (+)-glyceraldehyde underwent three successive Kilianiâ&#x20AC;&#x201C;Fischer syntheses. Identify the hexose from the following experimental information: oxidation with nitric acid forms an optically active aldaric acid; a Wohl degradation followed by oxidation with nitric acid forms an optically inactive aldaric acid; and a second Wohl degradation forms erythrose. 43. Draw the mechanism for the interconversion of a-d-glucose and b-d-glucose in dilute HCl. 44. An unknown b-d-aldohexose has only one axial substituent. A Wohl degradation forms a compound which, when treated with sodium borohydride, forms an optically active alditol. This information allows you to arrive at two possible structures for the b-d-aldohexose. What experiment can you carry out to distinguish between the two possibilities? 45. The 1H NMR spectrum of d-glucose in D2O exhibits two high-frequency doublets. What is responsible for these doublets? 46. d-Glucuronic acid is found widely in plants and animals. One of its functions is to detoxify poisonous HO-containing compounds by reacting with them in the liver to form glucuronides. Glucuronides are water soluble and therefore readily excreted. After ingestion of a poison such as turpentine or phenol, the glucuronides of these compounds are found in the urine. Draw the structure of the glucuronide formed by the reaction of b-d-glucuronic acid and phenol.

H HO

COOH O H OH H H

OH H

OH

B-D-glucuronic acid

47. Determine the structure of d-galactose, using arguments similar to those used by Fischer to prove the structure of d-glucose. 48. A d-aldopentose is oxidized by nitric acid to an optically active aldaric acid. A Wohl degradation of the aldopentose leads to a monosaccharide that is oxidized by nitric acid to an optically inactive aldaric acid. Identify the d-aldopentose. 49. Draw the mechanism for the formation of b-maltose from a-d-galactose and b-d-glucose in dilute HCl.


1052

CHAPTER 21

The Organic Chemistry of Carbohydrates

50. Hyaluronic acid, a component of connective tissue, is the fluid that lubricates the joints. It is a polymer of alternating N-acetyld-glucosamine and d-glucuronic acid subunits joined by b@1,3⬘-glycosidic linkages. Draw a short segment of hyaluronic acid. 51. In order to synthesize d-galactose, a student went to the stockroom to get some d-lyxose to use as a starting material. She found that the labels had fallen off the bottles containing d-lyxose and d-xylose. How could she determine which bottle contains d-lyxose? 52. The aldonic acid of d-glucose forms a five-membered ring lactone. Draw its structure. 53. The aldaric acid of d-glucose forms two five-membered ring lactones. Draw their structures. 54. Draw the mechanism for the acid-catalyzed hydrolysis of b-maltose. 55. How many aldaric acids are obtained from the 16 aldohexoses? 56. A hexose is obtained when the residue of a shrub Sterculia setigeria undergoes acid-catalyzed hydrolysis. Identify the hexose from the following experimental information: it undergoes mutarotation; it does not react with Br2; and d-galactonic acid and d-talonic acid are formed when it reacts with Tollens reagent. 57. When d-fructose is dissolved in D2O and the solution is made basic, the d-fructose recovered from the solution has an average of 1.7  deuterium atoms attached to the C-1 carbon per molecule. Show the mechanism that accounts for the incorporation of these deuterium atoms into d-fructose. 58. Draw each of the following compounds: c. a-d-tagatopyranose a. b-d-talopyranose d. b-d-psicofuranose b. a-d-idopyranose

e. f.

b-l-talopyranose a-l-tagatopyranose

59. Calculate the percentages of a-d-glucose and b-d-glucose present at equilibrium from the specific rotations of a-d-glucose, b-d-glucose, and the equilibrium mixture. Compare your values with those given in Section 21.10. (Hint: The specific rotation of the mixture equals the specific rotation of a-d-glucose times the fraction of glucose present in the a-form plus the specific rotation of b-d-glucose times the fraction of glucose present in the b-form.) 60. The specific rotation of a-d-galactose is 150.7° and that of b-d-galactose is 52.8°. When an aqueous mixture that was initially 70% a-d-galactose and 30% b-d-galactose reaches equilibrium, the specific rotation is 80.2°. What is the concentration of the anomers at equilibrium? 61. An unknown disaccharide gives a positive Tollens test. A glycosidase hydrolyzes it to d-galactose and d-mannose. When the disaccharide is treated with methyl iodide and Ag2O and then hydrolyzed with dilute HCl, the products are 2,3,4,6-tetra-O-methylgalactose and 2,3,4-tri-O-methylmannose. Propose a structure for the disaccharide. 62. Predict whether d-altrose exists preferentially as a pyranose or a furanose. (Hint: In the most stable arrangement for a five-membered ring, all the adjacent substituents are trans.) 63. Trehalose, C12H22O11, is a nonreducing sugar that is only 45% as sweet as sugar. When hydrolyzed by aqueous acid or the enzyme maltase, it forms only d-glucose. When it is treated with excess methyl iodide in the presence of Ag2O and then hydrolyzed with water under acidic conditions, only 2,3,4,6-tetra-O-methyl-d-glucose is formed. Draw the structure of trehalose. 64. Propose a mechanism for the rearrangement that converts an a-hydroxyimine into an a-aminoketone in the presence of a trace amount of acid (page 1027). H

N

hemoglobin

CH2NH

C H HO H H

OH H OH OH CH2OH

rearrangement

hemoglobin

C

O HO H H OH H OH CH2OH

65. All the glucose units in dextran have six-membered rings. When a sample of dextran is treated with methyl iodide and Ag2O and the product is hydrolyzed under acidic conditions, the final products are 2,3,4,6-tetra-O-methyl-d-glucose, 2,4,6-tri-O-methyl-d-glucose, 2,3,4-tri-O-methyl-d-glucose, and 2,4-di-O-methyl-d-glucose. Draw a short segment of dextran. 66. When a pyranose is in the chair conformation in which the CH2OH group and the C-1 OH group are both in axial positions, the two groups can react to form an acetal. This is called the anhydro form of the sugar (it has “lost water”). The anhydro form of d-idose is shown here. Explain why about 80% of d-idose exists in the anhydro form in an aqueous solution at 100 ⬚C, but only about 0.1% of d-glucose exists in the anhydro form under the same conditions. O

CH2 O

OH

HO OH

anhydro form of D-idose


22

The Organic Chemistry of Amino Acids, Peptides, and Proteins

Cobwebs, silk, muscles, and wool are all proteins. In this chapter you will find out why muscles and wool can be stretched but cobwebs and silk cannot. You also will see how a reduction reaction followed by an oxidation reaction can make hair (another protein) either straight or curly.

T

he three kinds of polymers prevalent in nature are polysaccharides, proteins, and nucleic acids. We have just looked at polysaccharides (Section 21.16), so now we will turn our attention to proteins and the structurally similar, but shorter, peptides. (Nucleic acids are discussed in Chapter 26.) Peptides and proteins are polymers of amino acids. The amino acids are linked together by amide bonds. An amino acid is a carboxylic acid with a protonated amino group on the a-carbon. amide bonds

O

O

C

C R

CH

O

OH

+

NH3

a protonated A-aminocarboxylic acid an amino acid

H2NCH R

O

C NHCH R′

C NHCH

O−

R′′

amino acids are linked together by amide bonds a tripeptide

Amino acid polymers can be composed of any number of amino acids. A dipeptide contains two amino acids linked together, a tripeptide contains three, an oligopeptide contains 4 to 10, and a polypeptide contains many. Proteins are naturally occurring polypeptides made up of 40 to 4000 amino acids. Proteins serve many functions in biological systems (Table 22.1). 1053


1054

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The Organic Chemistry of Amino Acids, Peptides, and Proteins

Table 22.1

Examples of the Diverse Functions of Proteins in Biological Systems

Structural proteins

These proteins impart strength to biological structures or protect organisms from their environment. For example, collagen is the major component of bones, muscles, and tendons; keratin is the major component of hair, hooves, feathers, fur, and the outer layer of skin.

Protective proteins

Snake venoms and plant toxins are proteins that protect their owners from predators. Blood-clotting proteins protect the vascular system when it is injured. Antibodies and peptide antibiotics protect us from disease.

Enzymes

Enzymes are proteins that catalyze the reactions that occur in cells.

Hormones

Some hormones, compounds that regulate the reactions that occur in living systems, are proteins.

Proteins with physiological functions

These proteins include those that transport and store oxygen in the body, store oxygen in the muscles, and contract muscles.

Proteins can be classified as either fibrous or globular. Fibrous proteins contain long chains of polypeptides arranged in threadlike bundles; these proteins are insoluble in water. Globular proteins tend to have roughly spherical shapes and most are soluble in water. All structural proteins are fibrous proteins; most enzymes are globular proteins.

22.1 THE NOMENCLATURE OF AMINO ACIDS The structures of the 20 most common naturally occurring amino acids and the frequency with which each occurs in proteins are shown in Table 22.2. Other amino acids occur in nature but only infrequently. Notice that the amino acids differ only in the substituent (R) that is attached to the a-carbon. The wide variation in these substituents (called side chains) is what gives proteins their great structural diversity and, as a consequence, their great functional diversity. Notice too that all amino acids except proline contain a primary amino group. Proline contains a secondary amino group incorporated into a five-membered ring.

Table 22.2 The Most Common Naturally Occurring Amino Acids Shown in the Form That Predominates as Physiological pH (7.4)

Formula

Aliphatic side-chain amino acids

O C H

Name

Abbreviations

Average relative abundance in proteins

Glycine

Gly

G

7.5%

Alanine

Ala

A

9.0%

Valine*

Val

V

6.9%

O−

CH +

NH3 O C

CH3

CH +

O−

NH3 O C

CH3CH

CH

CH3 +NH3

O−


The Nomenclature of Amino Acids

Formula

Name

O C CH3CHCH2

Abbreviations

1055

Average relative abundance in proteins

Leucine*

Leu

L

7.5%

Isoleucine*

Ile

I

4.6%

Serine

Ser

S

7.1%

Threonine*

Thr

T

6.0%

Cysteine

Cys

C

2.8%

Methionine*

Met

M

1.7%

Aspartate (aspartic acid)

Asp

D

5.5%

Glutamate (glutamic acid)

Glu

E

6.2%

Asparagine

Asn

N

4.4%

O−

CH +

CH3

NH3 O C

CH3CH2CH

O−

CH

CH3 +NH3 Hydroxy-containing amino acids

O C HOCH2

O−

CH +

NH3 O C

CH3CH OH

CH +

O−

NH3

Sulfur-containing amino acids

O C HSCH2

CH

O−

+

NH3 O C

CH3SCH2CH2

O−

CH +

NH3

Acidic amino acids

O

O

C

C

O

CH2

O−

CH +

NH3

O

O C

C H2N

CH2CH2

CH +

Amides of acidic amino acids

NH3

O

O C

C H2N

CH

CH2

O−

O−

+

NH3 (Continued)


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Formula

O

O C

C H2N

CH2CH2

O C

+

H3NCH2CH2CH2CH2

Glutamine

Gln

Q

3.9%

Lysine*

Lys

K

7.0%

Arginine*

Arg

R

4.7%

Phenylalanine*

Phe

F

3.5%

Tyrosine

Tyr

Y

3.5%

Proline

Pro

P

4.6%

Histidine*

His

H

2.1%

Tryptophan*

Trp

W

1.1%

O−

CH +

NH3

NH2

O

C

C

H2N

Average relative abundance in proteins

NH3

Basic amino acids

+

Abbreviations

O−

CH +

Name

NHCH2CH2CH2

CH

O−

+

NH3

Benzene-containing amino acids

O C CH2

O−

CH +

NH3 O C

HO

CH2

CH

O−

+

NH3

Heterocyclic amino acids

O− N +

H

C HO O C CH2

N

NH

CH +

O−

NH3 O C

CH2

CH

O−

+

NH3

N H *

Essential amino acid

The amino acids are always called by their common names. Often, the name tells you something about the amino acid. For example, glycine got its name from its sweet taste (glykos is Greek for “sweet”), and valine, like valeric acid, has five carbons. Asparagine was first found in asparagus, and tyrosine was isolated from cheese (tyros is Greek for “cheese”).


The Nomenclature of Amino Acids

Dividing the amino acids into classes, as in Table 22.2, makes them easier to learn. The aliphatic side-chain amino acids include glycine, the amino acid in which R = H, and four amino acids with alkyl side chains. Alanine is the amino acid with a methyl side chain, and valine has an isopropyl side chain. Notice that, in spite of its name, isoleucine has a sec-butyl substituent, not an isobutyl substituent. Leucine is the amino acid that has an isobutyl substituent. Each of the amino acids has both a three-letter abbreviation (in most cases, the first three letters of the name) and a single-letter abbreviation. Two amino acid side chains—serine and threonine—contain alcohol groups. Serine is an HO-substituted alanine, and threonine has a branched ethanol substituent. There are also two sulfur-containing amino acids: cysteine is an HS-substituted alanine, and methionine has a 2-(methylthio)ethyl substituent. There are two acidic amino acids (amino acids with two carboxylic acid groups): aspartate and glutamate. Aspartate is a carboxy-substituted alanine, and glutamate has one more methylene group than aspartate. (If their carboxyl groups are protonated, they are called aspartic acid and glutamic acid) Two amino acids—asparagine and glutamine— are amides of the acidic amino acids; asparagine is the amide of aspartate, and glutamine is the amide of glutamate. Notice that the obvious one-letter abbreviations cannot be used for these four amino acids because A and G are used for alanine and glycine. Instead, aspartate and glutamate are abbreviated D and E, and asparagine and glutamine are abbreviated N and Q. There are two basic amino acids (amino acids with two basic nitrogen-containing groups): lysine and arginine. Lysine has an e-amino group, and arginine has a d-guanidino group. At physiological pH, these groups are protonated. Use the e and d to remind you how many methylene groups each amino acid has. +

O +

H3N

e

d

g

b

C

a

CH2CH2CH2CH2CH

NH2 d

g

+

a protonated e-amino group

b

a

CH2CH2CH2CH

NH

H2N

+

NH3

leucine

O

C

O−

glycine

a protonated d-guanidino group

C

O−

NH3

arginine

lysine

Two amino acids—phenylalanine and tyrosine—contain benzene rings. As its name indicates, phenylalanine is phenyl-substituted alanine. Tyrosine is phenylalanine with a para-hydroxy substituent. Proline, histidine, and tryptophan are heterocyclic amino acids. We have noted that proline, with its nitrogen incorporated into a five-membered ring, is the only amino acid that contains a secondary amino group. Histidine is an imidazole-substituted alanine. Imidazole is an aromatic compound because it is cyclic and planar, each of its ring atoms has a p orbital, and it has three pairs of delocalized p electrons (Section 20.7). The pKa of a protonated imidazole ring is 6.0, so the ring can exist in both the acidic form and the basic form at physiological pH (7.4). +

HN

NH

N

protonated imidazole

aspartate

NH + H+

imidazole

Tryptophan is an indole-substituted alanine. Like imidazole, indole is an aromatic compound. Because the lone pair on the nitrogen of indole is needed for the compound’s aromaticity, indole is a very weak base. (The pKa of protonated indole is - 2.4.) Therefore, the ring nitrogen in tryptophan is never protonated under physiological conditions.

N H indole

lysine

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The 10 amino acids denoted in Table 22.2 by asterisks (*) are essential amino acids. Humans must obtain them from their diet because they either cannot synthesize them at all or they cannot synthesize them in adequate amounts. For example, humans must have a dietary source of phenylalanine because they cannot synthesize benzene rings. However, they do not need tyrosine in their diet because they can synthesize it from phenylalanine (Section 25.9). Although humans can synthesize arginine, it is needed for growth in greater amounts than can be synthesized. So arginine is considered an essential amino acid for children but not for adults.

Proteins and Nutrition Proteins are an important component of our diets. Dietary protein is hydrolyzed in the body to individual amino acids. Some of these amino acids are used to synthesize proteins needed by the body, some are broken down (metabolized) to supply energy to the body, and some are used as starting materials for the synthesis of nonprotein compounds that the body needs, such as thyroxine (Section 19.4), adrenaline, and melanin (Section 25.9). Complete proteins (meat, fish, eggs, and milk) contain all 10 essential amino acids. Incomplete proteins contain too little of one or more essential amino acids to support human growth. For example, beans and peas are deficient in methionine, corn is deficient in lysine and tryptophan, and rice is deficient in lysine and threonine. Vegetarians, therefore, must have a diet that includes proteins from different sources.

PROBLEM 1

a. Explain why, when the imidazole ring of histidine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. O C

NH

C

O− + 2 H+

CH2CH N

O CH2CH +

NH2

HN

O−

+

NH3

NH

b. Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton.

C

C H2N

+

O

NH NHCH2CH2CH2CH

O− + 2 H+

H2N

NH2

O

C

C NHCH2CH2CH2CH

NH2

+

O−

NH3

22.2 THE CONFIGURATION OF AMINO ACIDS The a-carbon of all the naturally occurring amino acids (except glycine) is an asymmetric center. Therefore, 19 of the 20 amino acids listed in Table 22.2 can exist as enantiomers. The d and l notation used for monosaccharides (Section 21.2) is also used for amino acids.


The Configuration of Amino Acids

An amino acid drawn in a Fischer projection with the carboxyl group at the top and the R group at the bottom of the vertical axis is a d-amino acid if the amino group is on the right and an l-amino acid if the amino group is on the left. Unlike monosaccharides, where the d isomer is the one found in nature, most amino acids found in nature have the l configuration. To date, d-amino acids have been found only in a few peptide antibiotics and in some small peptides attached to the cell walls of bacteria. (You will see how an l-amino acid can be converted to a d-amino acid in Section 25.4.) H

H

O

C HO

OH CH2OH

D-glyceraldehyde

O H

H CH2OH

L-glyceraldehyde

O C

+

O +

NH3

O C

H3N

H R

R D-amino

acid

L-amino

acid

Why d-sugars and l-amino acids? Although it made no difference which isomer nature “selected” to be synthesized, it was important that only one was selected. For example, proteins that contain both d- and l-amino acids do not fold properly, and without proper folding there can be no catalysis (Section 22.15). It was also important that the same isomer be synthesized by all organisms. For example, since mammals ended up having l-amino acids, l-amino acids must be the isomers synthesized by the organisms that mammals depend on for food.

Amino Acids and Disease The Chamorro people of Guam have a high incidence of a syndrome that resembles amyotrophic lateral sclerosis (ALS, or Lou Gehrig’s disease) with elements of Parkinson’s disease and dementia. This syndrome developed during World War II when, as a result of food shortages, the tribe ate large quantities of Cycas circinalis seeds. These seeds contain b-methylaminol-alanine, an amino acid that binds to cell receptors that bind l-glutamate. When monkeys are given b-methylamino-l-alanine, they develop some of the features of this syndrome. There is hope that, by studying the mechanism of action of b-methylamino-l-alanine, we may gain an understanding of how ALS and Parkinson’s disease arise. COO− C CH3

H + NH3

L-alanine

Naturally occurring amino acids have the L configuration.

O

C H

Naturally occurring monosaccharides have the D configuration.

COO− +

CH3NH2CH2

C

H + NH3

b-methylamino-L-alanine

PROBLEM 2♦

a. Which isomer—(R)-alanine or (S)-alanine—is d-alanine? b. Which isomer—(R)-aspartate or (S)-aspartate—is d-aspartate? c. Can a general statement be made relating R and S to d and l?

COO− O −

C

OCCH2CH2

H NH3

+

L-glutamate

L-alanine an amino acid

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A Peptide Antibiotic Gramicidin S, an antibiotic produced by a strain of bacteria, is a cyclic decapeptide. Notice that one of its 10 amino acids is ornithine, an amino acid not listed in Table 22.2 because it occurs rarely in nature. Ornithine resembles lysine but has one less methylene group in its side chain. Notice that gramicidin S contains two d-amino acids. L-Val L-Pro

L-ornithine

L-Orn

L-Phe

L-Leu

L-Leu

D-Phe

O D-ornithine

D-Orn

D-phenylalanine

C

+

O−

H3NCH2CH2CH2CH

L-Pro

+

L-Val

gramicidin S

NH3

ornithine

P R O B L E M 3 Solved

Threonine has two asymmetric centers and therefore has four stereoisomers. Naturally occurring l-threonine is (2S,3R)-threonine. Which of the following stereoisomers is l-threonine? COO− +

+

COO−

COO−

H

NH3

H3N

H

H

H

OH

HO

H

HO

+

NH3 H

+

COO−

H3N

H

H

OH

CH3

CH3

CH3

CH3

A

B

C

D

Solution Stereoisomer A has the R configuration at both C-2 and C-3 because in both cases

the arrow drawn from the highest to the next-highest-priority substituent is counterclockwise. In both cases, counterclockwise signifies R because the lowest priority substituent (H) is on a horizontal bond (Section 4.7). Therefore, the configuration of (2S,3R)-threonine is the opposite of that in stereoisomer A at C-2 and the same as that in stereoisomer A at C-3. Thus, l-threonine is stereoisomer D. Notice that the +NH3 group is on the left, just as we would expect for the Fischer projection of a naturally occurring l-amino acid.

PROBLEM 4♦

Do any other amino acids in Table 22.2 have more than one asymmetric center?

22.3 THE ACID–BASE PROPERTIES OF AMINO ACIDS Every amino acid has a carboxyl group and an amino group, and each group can exist in an acidic form or a basic form, depending on the pH of the solution in which the amino acid is dissolved. We have seen that compounds exist primarily in their acidic forms (that is, with their protons attached) in solutions that are more acidic than their pKa values, and primarily in their basic forms (that is, without their protons) in solutions that are more basic than their pKa values (Section 2.10).


The Acid–Base Properties of Amino Acids

O

O

C R

CH

C R

OH

CH

O C

O−

R

NH3 + H+

+

pH = 0

O−

NH2 + H+

+

NH3

CH pH = 12

a zwitterion pH = 7

The carboxyl groups of the amino acids have pKa values of approximately 2, and the protonated amino groups have pKa values near 9 (Table 22.3). Both groups, therefore, will be in their acidic forms in a very acidic solution (pH ~ 0). At pH = 7, the pH of the solution is greater than the pKa of the carboxyl group, but less than the pKa of the protonated amino group, so the carboxyl group will be in its basic form and the amino group will be in its acidic form. In a strongly basic solution (pH ~ 12), both groups will be in their basic forms.

Table 22.3 The pKa Values of Amino Acids

pKa A-COOH

pKa A-NH3

Alanine

2.34

9.69

Arginine

2.17

9.04

12.48

Asparagine

2.02

8.84

Amino acid

pKa Side chain

Aspartic acid

2.09

9.82

3.86

Cysteine

1.92

10.46

8.35

Glutamic acid

2.19

9.67

4.25

Glutamine

2.17

9.13

Glycine

2.34

9.60

Histidine

1.82

9.17

6.04

Isoleucine

2.36

9.68

Leucine

2.36

9.60

Lysine

2.18

8.95

10.79

Methionine

2.28

9.21

Phenylalanine

2.16

9.18

Proline

1.99

10.60

Serine

2.21

9.15

Threonine

2.63

9.10

Tryptophan

2.38

9.39

Tyrosine

2.20

9.11

10.07

Valine

2.32

9.62

Notice that an amino acid can never exist as an uncharged compound, regardless of the pH of the solution. To be uncharged, an amino acid would have to lose a proton from an +NH3 group with a pKa of about 9 before it would lose a proton from a COOH group with a pKa of about 2. This is impossible because a weak acid (pKa = 9) cannot lose a proton easier than a strong acid (pKa = 2) can. Therefore, at physiological pH (7.4), an amino acid such as alanine exists as a dipolar ion, called a zwitterion. A zwitterion is a compound that has a negative charge on one atom and a positive charge on a nonadjacent atom. (The name comes from zwitter, German for “hermaphrodite” or “hybrid.”)

1061

The acidic form (with the proton) predominates if the pH of the solution is less than the pKa of the ionizable group, and the basic form (without the proton) predominates if the pH of the solution is greater than the pKa of the ionizable group.

Recall from the Henderson– Hasselbalch equation (Section 2.10) that half the group is in its acidic form and half is in its basic form at pH = pKa.


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CHAPTER 22

The Organic Chemistry of Amino Acids, Peptides, and Proteins PROBLEM 5♦

Alanine has pKa values of 2.34 and 9.69. Therefore, alanine will exist predominately as a zwitterion in an aqueous solution with pH > ____ and pH < ____.

A few amino acids have side chains with ionizable hydrogens (Table 22.3). The protonated imidazole side chain of histidine, for example, has a pKa of 6.04. Histidine, therefore, can exist in four different forms, and the form that predominates depends on the pH of the solution. O

O

C CH2CH +

HN +

NH

NH3

C OH

CH2CH

O C

O−

CH2CH

+

HN +

pH = 0

NH

O C

O−

CH2CH

+

NH3

N

NH

pH = 4

NH3

N

pH = 8

NH

O−

NH2

pH = 12

histidine

PROBLEM 6♦

Why are the carboxylic acid groups of the amino acids so much more acidic (pKa ~ 2) than a carboxylic acid such as acetic acid (pKa = 4.76)? P R O B L E M 7 Solved

Draw the predominant form for each of the following amino acids at physiological pH (7.4): a. aspartate c. glutamine e. arginine b. histidine d. lysine f. tyrosine Solution to 7a Both carboxyl groups are in their basic forms because the pH of the solution is greater than their pKa values. The protonated amino group is in its acidic form because the pH of the solution is less than its pKa value.

O −

O

C O

C CH2CH

O−

+

NH3

PROBLEM 8♦

Draw the predominant form for glutamate in a solution with the following pH: a. 0 b. 3 c. 6 d. 11

PROBLEM 9

a. Why is the pKa of the glutamate side chain greater than the pKa of the aspartate side chain? b. Why is the pKa of the arginine side chain greater than the pKa of the lysine side chain?

22.4 THE ISOELECTRIC POINT The isoelectric point (pI) of an amino acid is the pH at which it has no net charge. In other words, it is the pH at which the amount of positive charge on an amino acid exactly balances the amount of negative charge: pI = pH at which there is no net charge


The Isoelectric Point

The pI of an amino acid that does not have an ionizable side chain—such as alanine—is midway between its two pKa values. This is because half of the molecules have a negatively charged carboxyl group and half have an uncharged carboxyl group at pH = 2.34, whereas half of the molecules have a positively charged amino group and half have an uncharged amino group at pH = 9.69. As the pH increases from 2.34, the carboxyl group of more molecules becomes negatively charged; as the pH decreases from 9.69, the amino group of more molecules becomes positively charged. Therefore, the number of negatively charged groups equals the number of positively charged groups at the intersection (average) of the two pKa values. O pKa = 2.34

C CH3CH

OH

+

NH3 pKa = 9.69

alanine

2.34 + 9.69 12.03 = = 6.02 2 2

pI =

The pI of most amino acids (see Problem 13) that have an ionizable side chain is the average of the pKa values of the similarly ionizing groups (either positively charged groups ionizing to uncharged groups or uncharged groups ionizing to negatively charged groups). For example, the pI of lysine is the average of the pKa values of the two groups that are positively charged in their acidic form and uncharged in their basic form. The pI of glutamic acid, on the other hand, is the average of the pKa values of the two groups that are uncharged in their acidic form and negatively charged in their basic form. O pKa = 2.18

C

+

H3NCH2CH2CH2CH2CH

OH

NH3

lysine

pI =

O

C

C

HO

+

pKa = 10.79

O CH2CH2CH NH3

glutamic acid

pI =

Explain why the pI of lysine is the average of the pKa values of its two protonated amino groups.

PROBLEM 11♦

c. serine

d. aspartate

PROBLEM 12♦

a. b. c. d.

pKa = 9.67

2.19 + 4.25 6.44 = = 3.22 2 2

PROBLEM 10

Calculate the pI of each of the following amino acids: a. asparagine b. arginine

OH

+

pKa = 4.25 pKa = 8.95

8.95 + 10.79 19.74 = = 9.87 2 2

pKa = 2.19

Which amino acid has the lowest pI value? Which amino acid has the highest pI value? Which amino acid has the greatest amount of negative charge at pH = 6.20? Which amino acid has a greater negative charge at pH = 6.20, glycine or methionine?

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PROBLEM 13

Explain why the pI values of tyrosine and cysteine cannot be determined by the method just described.

22.5 SEPARATING AMINO ACIDS A mixture of amino acids can be separated by several different techniques, such as electrophoresis, paper/thin-layer chromatography, and ion-exchange chromatography.

Electrophoresis An amino acid will be positively charged if the pH of the solution is less than the pI of the amino acid and it will be negatively charged if the pH of the solution is greater than the pI of the amino acid.

Electrophoresis can be used to determine the number of amino acids in a mixture. It separates the amino acids on the basis of their pI values. A few drops of a solution of an amino acid mixture are applied to the middle of a piece of filter paper or to a gel. When the paper or the gel is placed in a buffered solution between two electrodes and an electric field is applied (Figure 22.1), an amino acid with a pI greater than the pH of the solution will have an overall positive charge and will migrate toward the cathode (the negative electrode).

cathode

anode

+

+

+

NH2

O C

C H2N

O

NHCH2CH2CH2CH +

NH3

arginine pl = 10.76

O O

C O−

CH3CH +

NH3

alanine pl = 6.02

O−

C

−OCCH

2CH

+

O−

NH3

aspartate pl = 2.98

▲ Figure 22.1 Arginine, alanine, and aspartate are separated by electrophoresis at pH = 5.

The farther the amino acid’s pI is from the pH of the buffer, the more positive the amino acid will be and the farther it will migrate toward the cathode in a given amount of time. An amino acid with a pI less than the pH of the buffer will have an overall negative charge and will migrate toward the anode (the positive electrode). If two molecules have the same charge, the larger one will move more slowly during electrophoresis because the same charge has to move a greater mass. Considering that amino acids are colorless, how can we detect them after they have been separated? After the amino acids have been separated by electrophoresis, the filter paper is painted with a solution of ninhydrin and dried in a warm oven. Most amino acids form a purple product when heated with ninhydrin. The number of amino acids in the mixture is determined by the number of colored spots on the filter paper. The individual amino acids can be identified by their location on the paper compared with a standard. The mechanism for formation of the colored product is shown here, omitting the mechanisms for dehydration, imine formation, and imine hydrolysis. (These mechanisms are shown in Sections 17.10 and 17.11.)


Separating Amino Acids

1065

STEPS IN THE REACTION OF AN AMINO ACID WITH NINHYDRIN TO FORM A COLORED PRODUCT

O

O

O

C

OH

O

OH O

+ H2NCH

ninhydrin

C

O

N

R

+ H2O O

O

O

O

CH

R + H2O

O

an amino acid

O

H +

O

C

H

H

H2O

NH2 O

O−

O N O

C

R − + HO

O

N

C

H

O

R

H + CO2

R O O O −

O HO O H N O

O + H2O

O

O N

O

O

+ H2O

purple-colored product

Loss of water from the hydrate forms a ketone that reacts with the amino acid to form an imine. Decarboxylation occurs because the electrons left behind can be delocalized onto a carbonyl oxygen. Tautomerization followed by hydrolysis of an imine forms the deaminated amino acid and a ninhydrin-amine. Reaction of this amine with another molecule of ninhydrin forms an imine. Loss of a proton forms a highly conjugated (colored) product (Section 14.21). PROBLEM 14♦

What aldehyde is formed when valine is treated with ninhydrin?

Paper/Thin-Layer Chromatography Paper chromatography once played an important role in biochemical analysis because it provided a method for separating amino acids using very simple equipment. Although more modern techniques are generally employed today, we will describe the principles behind paper chromatography because many of the same principles are employed in modern separation techniques. Paper chromatography separates amino acids on the basis of polarity. A few drops of a solution of an amino acid mixture are applied to the bottom of a strip of filter paper. The edge of the paper is then placed in a solvent. The solvent moves up the paper by capillary

Painting a paper with a solution of ninhydrin allows latent fingerprints (as a consequence of amino acids left behind by the fingers) to be developed.


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Less polar amino acids travel more rapidly if the solvent is less polar than the paper.

action, carrying the amino acids with it. Depending on their polarities, the amino acids have different affinities for the mobile (solvent) and stationary (paper) phases and therefore some travel up the paper farther than others. When a solvent less polar than the paper is employed, the more polar the amino acid, the more strongly it is adsorbed onto the relatively polar paper. The less polar amino acids travel farther up the paper, since they have a greater affinity for the less polar mobile phase. Therefore, when the paper is developed with ninhydrin, the colored spot closest to the origin is the most polar amino acid and the spot farthest away from the origin is the least polar amino acid (Figure 22.2).

Chromatography

least polar amino acid

▶ Figure 22.2 Separation of glutamate, alanine, and leucine by paper chromatography.

Leu

Ala

Glu

origin

most polar amino acid

The most polar amino acids are those with charged side chains, the next most polar are those with side chains that can form hydrogen bonds, and the least polar are those with hydrocarbon side chains. For amino acids with hydrocarbon side chains, the polarity of the amino acid decreases as the size of the alkyl group increases. In other words, leucine [R = –CH2CH(CH3)2] is less polar than valine [R = –CH(CH3)2]. Paper chromatography has largely been replaced by thin-layer chromatography (TLC), which differs from paper chromatography in that TLC uses a plate with a coating of solid material instead of filter paper. How the amino acids separate depend on the solid material and the solvent chosen for the mobile phase. Chromatography separates amino acids based on their polarity and electrophoresis separates them based on their charge. The two techniques can be applied on the same piece of filter paper—a technique called fingerprinting—to give a two-dimensional separation (that is, the amino acids are separated according to both their polarity and their charge. (See Problems 54 and 67.) PROBLEM 15♦

A mixture of seven amino acids (glycine, glutamate, leucine, lysine, alanine, isoleucine, and aspartate) is separated by chromatography. Explain why only six spots show up when the chromatographic plate is coated with ninhydrin and heated.

Ion-Exchange Chromatography A technique called ion-exchange chromatography is able to both separate and identify amino acids and to determine the relative amount of each amino acid in a mixture. This technique employs a column packed with an insoluble resin. A solution of a mixture of amino acids is loaded onto the top of the column, and a series of buffer solutions of increasing pH are poured through the column. The amino acids separate because they flow through the column at different rates. The resin is a chemically inert material with charged groups. The structure of a commonly used resin is shown in Figure 22.3. If a mixture of lysine and glutamate in a solution of pH = 6 were to be loaded onto the column, glutamate would travel down the column rapidly because its negatively charged side chain would be repelled by the negatively charged sulfonic acid groups of the resin. The positively charged side chain of lysine, on the other hand, would cause that amino acid to be retained on the column. This kind of resin is called a cation-exchange resin because it exchanges the Na+ counterions of


Separating Amino Acids

1067

the SO3− groups for the positively charged species that is traveling through the column. In addition, the relatively nonpolar nature of the column causes it to retain nonpolar amino acids longer than polar amino acids. SO3−Na+

SO3−Na+

SO3−Na+

◀ Figure 22.3 SO3−Na+

SO3−Na+

Resins with positively charged groups are called anion-exchange resins because they impede the flow of anions by exchanging their negatively charged counterions for negatively charged species traveling through the column. A common anion-exchange resin, Dowex 1, has i CH2N+(CH3)3Cl− groups in place of the i SO3−Na+ groups in Figure 22.3. An amino acid analyzer is an instrument that automates ion-exchange chromatography. When a solution of amino acids passes through the column of an amino acid analyzer containing a cation-exchange resin, the amino acids move through the column at different rates, depending on their overall charge. The solution that flows out of the column (the effluent) is collected in fractions. These are collected often enough that a different amino acid ends up in each one (Figure 22.4).

A section of a cation-exchange resin. This particular resin is called Dowex 50. Cations bind most strongly to cation-exchange resins. Anions bind most strongly to anion-exchange resins.

mixture of amino acids resin

◀ Figure 22.4 fractions are sequentially collected

If ninhydrin is added to each of the fractions, the concentration of amino acid in each fraction can be determined by the amount of absorbance at 570 nm because the colored compound formed by the reaction of an amino acid with ninhydrin has a lmax of 570 nm (Section 14.19). This information combined with each fraction’s rate of passage through

Separation of amino acids ion-exchange chromatography.

by


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the column allows the identity and relative amount of each amino acid in the mixture to be determined (Figure 22.5).

A typical chromatogram obtained from the separation of a mixture of amino acids using an automated amino acid analyzer.

Absorbance

▶ Figure 22.5

Asp

Thr

pH 5.3 buffer

pH 4.3 buffer

pH 3.3 buffer Ser

Glu

Ala

Met Ile

Gly Val

Leu

His

Tyr

NH3

Phe Lys

Arg

Pro 40

80

120

160

200 240

280 320 330 370 Effluent (mL)

410

450

490 550

590 630

Water Softeners: Examples of Cation-Exchange Chromatography Water-softening systems contain a column packed with a cation-exchange resin that has been flushed with concentrated sodium chloride. When “hard water” (water with high levels of Ca2+ and Mg2+; Section 16.13) passes through the column, the resin binds Ca2+ and Mg2+ more tightly than it binds Na+. Thus, the water-softening system removes Ca2+ and Mg2+ from the water and replaces them with Na+. The resin must be recharged from time to time by being flushed with concentrated sodium chloride, thereby replacing the bound Ca2+ and Mg2+ with Na+.

PROBLEM 16

Why are buffer solutions of increasingly higher pH used to elute the column that generates the chromatogram shown in Figure 22.5? (Elute means wash out with a solvent.) PROBLEM 17

Explain the order of elution (with a buffer of pH 4) of the following pairs of amino acids through a column packed with Dowex 50 (Figure 22.3): a. aspartate before serine c. valine before leucine b. serine before alanine d. tyrosine before phenylalanine PROBLEM 18♦

In what order would histidine, serine, aspartate, and valine be eluted with a buffer of pH 4 from a column containing an anion-exchange resin (Dowex 1)?

22.6 THE SYNTHESIS OF AMINO ACIDS Chemists do not have to rely on nature to produce amino acids; they can synthesize them in the laboratory, using a variety of methods. Some of these methods are described here.

HVZ Reaction Followed by Reaction with Ammonia One of the oldest methods used to synthesize an amino acid is to employ an HVZ reaction to replace an a-hydrogen of a carboxylic acid with a bromine (Section 18.5). The resulting a-bromocarboxylic acid can then undergo an SN2 reaction with ammonia to form the amino acid (Section 9.2). O C RCH2

O

O OH

a carboxylic acid

1. Br2, PBr3 2. H2O

C RCH Br

OH

excess NH3

C RCH +

+

O− + NH4Br−

NH3

an amino acid


The Synthesis of Amino Acids

PROBLEM 19♦

Why is excess ammonia used in the preceding reaction?

Reductive Amination Amino acids can also be synthesized by reductive amination of an a-keto acid (Section 17.10). O

O 1. excess NH3, trace acid 2. H2, Pd/C

C RC

OH

C RCH

O−

+

NH3

O

PROBLEM 20♦

Cells can also convert a-keto acids into amino acids, but because the reagents organic chemists use for this reaction are not available in cells, they carry out this reaction by a different mechanism (Section 24.5). a. What amino acid is obtained from the reductive amination of each of the following metabolic intermediates in a cell? O

O OH

HO O

O

HO

OH

O pyruvic acid

O

O

OH O

A-ketoglutaric acid

oxaloacetic acid

b. What amino acids are obtained from the same metabolic intermediates when the amino acids are synthesized in the laboratory?

N-Phthalimidomalonic Ester Synthesis Amino acids can be synthesized with much better yields than those obtained by the previous two methods via an N-phthalimidomalonic ester synthesis, a method that combines the malonic ester synthesis (Section 18.18) and the Gabriel synthesis (Section 16.18). THE STEPS IN THE N-PHTHALIMIDOMALONIC ESTER SYNTHESIS

O

O

O

RO

OR Br

A-bromomalonic ester

+

N

O O −

+

N

K

O

O

potassium phthalimide

OR OR C

O O RO

H O

N

OR OR C

O ROH

O

N-phthalimidomalonic ester R′

COOH + CO2 + COOH phthalic acid

■ ■

O O

O

+

H3N

OH

HCl, H2O Δ

R′ an amino acid

a-Bromomalonic ester and potassium phthalimide undergo an SN2 reaction. A proton is easily removed from the a-carbon of N-phthalimidomalonic ester because it is flanked by two carbonyl groups. (Section 18.1)

N O

Br

OR OR C R′ O

Br−

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The resulting carbanion undergoes an SN2 reaction with an alkyl halide. Heating in an acidic aqueous solution hydrolyzes the two ester groups and the two amide bonds and decarboxylates the 3-oxocarboxylic acid.

■ ■

A variation of the N-phthalimidomalonic ester synthesis uses acetamidomalonic ester in place of N-phthalimidomalonic ester. O

O

O

RO

OR

RO−

O −

RO

O R′ Br

OR

RO

N

N H

H O

O

O

O

R′ N H

OR O

ROH

HCl, H2O Δ

OH

+ CO2 +

OH R′

acetic acid

Br−

O

+

H3N

an amino acid

acetamidomalonic ester

Strecker Synthesis In the Strecker synthesis, an aldehyde reacts with ammonia to form an imine. An addition reaction with cyanide ion forms an intermediate, which, when hydrolyzed, forms the amino acid (Section 16.19). Compare this reaction with the Kiliani–Fischer synthesis of aldoses in Section 21.7. O R C

O

trace acid NH3

H

R

C

NH

H

an aldehyde

an imine

C N HCl

R

CH +

NH3

C

N

HCl, H2O Δ

C R

CH

OH

+

NH3

an amino acid

PROBLEM 21♦

What amino acid would be formed using the N-phthalimidomalonic ester synthesis when the following alkyl halides are used in the third step? a. CH3CHCH2Br b. CH3SCH2CH2Br CH3 PROBLEM 22♦

What alkyl halide would you use in the acetamidomalonic ester synthesis to prepare a. lysine? b. phenylalanine? PROBLEM 23♦

What amino acid would be formed when the aldehyde used in the Strecker synthesis is a. acetaldehyde? b. 2-methylbutanal? c. 3-methylbutanal?

22.7 THE RESOLUTION OF RACEMIC MIXTURES OF AMINO ACIDS When amino acids are synthesized in nature, only the l-enantiomer is formed (Section 6.17). When amino acids are synthesized in the laboratory, however, the product is a racemic mixture of d and l amino acids. If only one isomer is desired, the enantiomers must be separated, which can be accomplished by means of an enzyme-catalyzed reaction. Because an enzyme is chiral, it will react at a different rate with each of the enantiomers (Section 6.17). For example, pig kidney aminoacylase is an enzyme that catalyzes the hydrolysis of N-acetyl-l-amino acids, but not N-acetyl-d-amino acids.


Peptide Bonds and Disulfide Bonds

Therefore, if the racemic mixture of amino acids is converted to a pair of N-acetylamino acids and the N-acetylated mixture is hydrolyzed with pig kidney aminoacylase, then the products will be the l-amino acid and N-acetyl-d-amino acid, which are easily separated. O C H2NCH

O

C

C

CH3COCCH3

CH3

pig kidney aminoacylase H2O

NHCH

R D-amino

O

O

O

O O

O

C CH3

O

R acid

L-amino

N-acetyl-D-amino acid + N-acetyl-L-amino acid

+ L-amino acid

COO− H + H2N R

H

C

+

CH3

NH

COO−

R N-acetyl-D-amino acid

Because the resolution (separation) of the enantiomers depends on the difference in the rates of reaction of the enzyme with the two N-acetylated compounds, this technique is known as a kinetic resolution. In Section 6.17, we saw that a racemic mixture of amino acids can also be separated by the enzyme d-amino acid oxidase. PROBLEM 24

Pig liver esterase is an enzyme that catalyzes the hydrolysis of esters. It hydrolyzes esters of l-amino acids more rapidly than esters of d-amino acids. How can this enzyme be used to separate a racemic mixture of amino acids?

22.8 PEPTIDE BONDS AND DISULFIDE BONDS Peptide bonds and disulfide bonds are the only covalent bonds that join amino acids together in a peptide or a protein.

Peptide Bonds The amide bonds that link amino acids are called peptide bonds. By convention, peptides and proteins are written with the free amino group (of the N-terminal amino acid) on the left and the free carboxyl group (of the C-terminal amino acid) on the right. O

O

C

+

+

O− + H3NCH

H3NCH R

R

C

+

O− + H3NCH

R′

O−

O C

C NHCH

C

R′′

O

C

+

H3NCH

O

R′

O

NHCH

O− + 2 H2O

R′′

peptide bonds the C-terminal amino acid

the N-terminal amino acid

acid

O

a tripeptide

When the identities of the amino acids in a peptide are known but their sequence is not, the amino acids are written separated by commas. When their sequence is known, the amino acids are written connected by hyphens. For example, in the pentapeptide

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represented on the right, valine is the N-terminal amino acid and histidine  is the C-terminal amino acid. The amino acids are numbered starting with the N-terminal end. Alanine is therefore referred to as Ala 3 because it is the third amino acid from the N-terminal end. Glu, Cys, His, Val, Ala

Val-Cys-Ala-Glu-His

the pentapeptide contains the indicated amino acids, but their sequence is not known

the amino acids in the pentapeptide have the indicated sequence

In naming a peptide, adjective names (ending in “yl”) are used for all the amino acids except the C-terminal amino acid. Thus, the preceding pentapeptide is valylcysteinylalanylglutamylhistidine. Each amino acid has the l configuration unless otherwise specified. A peptide bond has about 40% double-bond character because of electron delocalization (Section 16.2). Steric strain in the cis configuration causes the trans configuration about the amide linkage to be more stable, so the a-carbons of adjacent amino acids are trans to each other.

CH

C CH

O−

R a-carbon

O

C

N

a-carbon R

H

R +

CH

N

R

H

CH

resonance contributors

The partial double-bond character prevents free rotation about the peptide bond, so the carbon and nitrogen atoms of the peptide bond and the two atoms to which each is attached are held rigidly in a plane (Figure 22.6). This regional planarity affects the way a chain of amino acids can fold; this has important implications for the three-dimensional shapes of peptides and proteins (Section 22.14). ▶ Figure 22.6 A segment of a polypeptide chain. Colored squares indicate the plane defined by each peptide bond. Notice that the R groups bonded to the a-carbons are on alternate sides of the peptide backbone.

O

H

R

C

N

O

H

CH

R

C

N

O

H

CH

R

C

N

O

H

CH

C

N

CH

N

C

CH

N

C

CH

N

C

CH

R

H

O

R

H

O

R

H

O

R

PROBLEM 25

Draw the tetrapeptide Ala-Thr-Asp-Asn and indicate the peptide bonds.

PROBLEM 26♦

Using the three-letter abbreviations, write the six tripeptides consisting of Ala, Gly, and Met.

PROBLEM 27

Draw a peptide bond in the cis configuration.

PROBLEM 28♦

Which bonds in the backbone of a peptide can rotate freely?


Peptide Bonds and Disulfide Bonds

1073

Disulfide Bonds When thiols are oxidized under mild conditions, they form a disulfide—a compound with an S—S bond. (Like C—H bonds, the number of S—H bonds decreases in an oxidation reaction and increases in a reduction reaction.) 2R

mild oxidation

SH

RS

SR

a disulfide

a thiol

An oxidizing agent commonly used for this reaction is Br2 (or I2) in a basic solution. MECHANISM FOR THE OXIDATION OF A THIOL TO A DISULFIDE

R

■ ■

SH

HO−

R

H2O

S

Br

Br

R

S

R

Br

S

R

S

R + Br−

S

+ Br−

A thiolate ion attacks the electrophilic bromine of Br2. A second thiolate ion attacks the sulfur and eliminates Br−.

Because thiols can be oxidized to disulfides, disulfides can be reduced to thiols. RS

SR

reduction

2R

a disulfide

SH

a thiol

Disulfides are reduced to thiols.

The amino acid cysteine contains a thiol group, so two cysteine molecules can be oxidized to a disulfide. The disulfide is called cystine. O

O C 2 HSCH2CH

O−

mild oxidation

O

C O

C CHCH2S

SCH2CH

+

+

NH3

+

NH3

cysteine

O−

NH3

cystine

Thiols are oxidized to disulfides.

Two cysteines in a protein can be oxidized to a disulfide, creating a bond known as a disulfide bridge. Disulfide bridges are the only covalent bonds that are found between nonadjacent amino acids in peptides and proteins. They contribute to the overall shape of a protein by linking cysteines found in different parts of the peptide backbone, as shown in Figure 22.7.

SH oxidation

HS

SH

SH

SH

reduction

SS

SS S S

SH polypeptide

disulfide bridges cross-linking portions of a polypeptide

The hormone insulin, synthesized in the pancreas by cells known as the islets of Langerhans, maintains the proper level of glucose in the blood. Insulin is a polypeptide

◀ Figure 22.7 Disulfide bridges cross-linking portions of a polypeptide.


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with two peptide chains; one contains 21 amino acids and the other 30 amino acids. The two chains are connected to each other by two interchain disulfide bridges (between two chains). Insulin also has an intrachain disulfide bridge (within a chain). an intrachain disulfide bridge S

S

A-chain Gly Ile Val Glu Gln Cys Cys Thr Ser Ile Cys Ser Leu Tyr Gln Leu Glu Asn Tyr Cys Asn S S

interchain disulfide bridges

S S

Ala in pigs, Thr in humans

B-chain Phe Val Asn Gln His Leu Cys Gly Ser His Leu Val Glu Ala Leu Tyr Leu Val Cys Gly Glu Arg Gly Phe Phe Tyr Thr Pro Lys Ala insulin

Diabetes Diabetes is the third leading cause of death (heart disease and cancer are the first and second) in the United States. It is caused either by insufficient secretion of insulin (type 1 diabetes) or its inability to stimulate its target cells (type 2 diabetes). Injections of insulin can control some of the symptoms associated with diabetes. Until genetic engineering techniques became available (Section 22.11), pigs were the primary source of insulin for people with diabetes. The insulin was effective, but there were concerns about whether enough could be obtained over the long term for the growing population of diabetics. In addition, the C-terminal amino acid of the B-chain is alanine in pig insulin and threonine in human insulin, which caused some people to have allergic reactions. Now, however, mass quantities of synthetic insulin, chemically identical to human insulin, are produced from genetically modified host cells (Section 26.13).

insulin receptor

Insulin binds to the insulin receptor on the surface of cells, telling the cell to transport glucose from the blood into the cell.

Hair: Straight or Curly? Hair is made up of a protein called keratin that contains an unusually large number of cysteines (about 8% of the amino acids compared to an average of 2.8% for other proteins). These cysteines furnish keratin with many disulfide bridges that preserve its three-dimensional structure. People can alter the structure of their hair (if they think it is either too straight or too curly) by changing the location of these disulfide bridges. This can be accomplished by first applying a reducing agent to the hair to reduce all the disulfide bridges in the protein strands. Then, after rearranging the hair into the desired shape (using curlers to curl it or combing it straight to uncurl it), an oxidizing agent is applied to form new disulfide bridges. The new disulfide bridges hold the hair in its new shape. When this treatment is applied to straight hair, it is called a “permanent.” When it is applied to curly hair, it is called “hair straightening.”

S S

S S S S

SS S S

curly hair

S

S

S

S

S

S

S

S

S

S

S

S

straight hair


Some Interesting Peptides

1075

22.9 SOME INTERESTING PEPTIDES Several peptide hormones, including b-endorphin, leucine enkephalin, and methionine enkephalin are synthesized by the body to control pain. b-Endorphin has a chain of 31 amino acids, whereas the two enkephalins are pentapeptides. The five amino acids at the N-terminal end of b-endorphin are the same as those in methionine enkephalin. These peptides control the body’s sensitivity to pain by binding to receptors in certain brain cells. Part of the their three-dimensional structures must be similar to that of morphine because they bind to the same receptors. The phenomenon known as “runner’s high” that kicks in after vigorous exercise and the relief of pain through acupuncture are thought to be due to the release of these peptides.

Tyr-Gly-Gly-Phe-Leu

Tyr-Gly-Gly-Phe-Met

leucine enkephalin

methionine enkephalin

The nonapeptides bradykinin, vasopressin, and oxytocin are also peptide hormones. Bradykinin inhibits the inflammation of tissues. Vasopressin controls blood pressure by regulating the contraction of smooth muscle; it is also an antidiuretic. Oxytocin induces labor in pregnant women by stimulating the uterine muscle to contract, and it also stimulates milk production in nursing mothers. Vasopressin and oxytocin, like b-endorphin and the enkephalins, also act on the brain. Vasopressin is a “fight-or-flight” hormone, whereas oxytocin has the opposite effect: it calms the body and promotes social bonding. In spite of their very different physiological effects, vasopressin and oxytocin differ only by two amino acids.

bradykinin

Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg

vasopressin

Cys-Tyr-Phe-Gln-Asn-Cys-Pro-Arg-Gly-NH2 S

oxytocin

S

Cys-Tyr-Ile-Gln-Asn-Cys-Pro-Leu-Gly-NH2 S

S

Vasopressin and oxytocin both have an intrachain disulfide bridge, and their C-terminal amino acids contain amide rather than carboxyl groups. Notice that the C-terminal amide group is indicated by writing “NH2” after the name of the C-terminal amino acid. The synthetic sweetener aspartame (or Equal or NutraSweet; see Section 21.19) is about 200 times sweeter than sucrose. Aspartame is the methyl ester of a dipeptide of l-aspartate and l-phenylalanine. The ethyl ester of the same dipeptide is not sweet. If a d-amino acid is substituted for either of the l-amino acids of aspartame, the resulting dipeptide is bitter rather than sweet.

PROBLEM 29♦

What is the configuration about each of the asymmetric centers in aspartame?

O O −

O

+NH

N H 3

OCH3 O

aspartame NutraSweet®


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PROBLEM 30

Glutathione is a tripeptide whose function is to destroy harmful oxidizing agents in the body. Oxidizing agents are thought to be responsible for some of the effects of aging and to play a causative role in cancer. Glutathione removes oxidizing agents by reducing them. In the process, glutathione is oxidized, resulting in the formation of a disulfide bond between two glutathione molecules. An enzyme subsequently reduces the disulfide bond, returning glutathione to its original condition so it can react with another oxidizing agent. O−

O O

O

2 H 3N

H3N

O

H N

+

O−

N H

O HS

N H

O

oxidizing agent

O− O

S

reducing agent

O

O

H N

+

S O

+

H3N

glutathione

H N

N H

oxidized glutathione

O−

O

O−

O

O

oxidized glutathione

a. What amino acids make up glutathione? b. What is unusual about glutathione’s structure? (If you cannot answer this question, draw the  structure you would expect for the tripeptide, and compare your structure with the actual structure.)

22.10 THE STRATEGY OF PEPTIDE BOND SYNTHESIS: N-PROTECTION AND C-ACTIVATION One difficulty in synthesizing a polypeptide is that the amino acids have two functional groups, enabling them to combine in different ways. Suppose, for example, that you wanted to make the dipeptide Gly-Ala. That dipeptide is only one of four possible dipeptides that could be formed by heating a mixture of alanine and glycine. H N

+

H3N O

Gly-Ala

O

CH3 O−

CH3

+

H3N

H N

O

O

H N

+

O−

H3N

O

CH3 O−

O

CH3

Ala-Ala

+

H3N

H N

O O−

O

Gly-Gly

Ala-Gly

If the amino group of the amino acid that is to be on the N-terminal end (in this case, Gly) is protected (Section 17.13), then it will not be available to form a peptide bond. If its carboxyl group is activated before the second amino acid is added, then the amino group of the added amino acid (in this case, Ala) will react with the activated carboxyl group of glycine in preference to reacting with a nonactivated carboxyl group of another alanine. protect

glycine

O

H2N

O O−

activate The N-terminal amino acid must have its amino group protected and its carboxyl group activated.

alanine

H2N

O− CH3

peptide bond is formed between these groups


1077

The Strategy of Peptide Bond Synthesis: N-Protection and C-Activation

The reagent most often used to protect the amino group of an amino acid is di-tert-butyl dicarbonate. Notice that the amino group rather than the carboxylate group of the amino acid reacts with di-tert-butyl dicarbonate because the amino group is a better nucleophile. When glycine reacts with di-tert-butyl dicarbonate in an addition-elimination reaction, the anhydride bond breaks, forming CO2 and tert-butyl alcohol. protecting group

O

O O

(CH3)3CO

O

O OC(CH3)3

O−

+ H2NCH2

di-tert-butyl dicarbonate

O NHCH2

(CH3)3CO

O−

+ CO2 + HOC(CH3)3

N-protected glycine

glycine

The reagent most often used to activate the carboxyl group is dicyclohexylcarbodiimide (DCC). DCC activates a carboxyl group by putting a good leaving group on the carbonyl carbon. activating group

O

O O

NHCH2

(CH3)3CO

N H +

N-protected Gly

C

O

proton transfer

(CH3)3CO

O NHCH2

O

N O

+

C

O NHCH2

(CH3)3CO

N O

C

+NH

N

NH

protected activated dicyclohexylcarbodiimide DCC

After the amino acid’s N-terminal group is protected and its C-terminal group is activated, the second amino acid is added. The unprotected amino group of the second amino acid adds to the activated carboxyl group, forming a tetrahedral intermediate. The C—O bond of the tetrahedral intermediate is easily broken because the bonding electrons are delocalized. Breaking this bond forms dicyclohexylurea, a stable diamide. H B+ O (CH3)3CO

O NHCH2

O

C NH

O H2NCH

O

N

O−

(CH3)3CO

NHCH2

O

H

N

C

O

C

NH

O

B (CH3)3CO

+ NH + H

O NHCH2

O NHCH

new peptide bond

tetrahedral CH3CH intermediate −

+

CO2

CH3 Ala

NH O

C NH

dicyclohexylurea a diamide

CH3

O−


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Amino acids can be added to the growing C-terminal end by repeating the same two steps: activating the carboxyl group of the C-terminal amino acid of the peptide by treating it with DCC and then adding a new amino acid. O (CH3)3CO

O

O NHCH2

NHCH

O O−

CH3

1. DCC

O NHCH

NHCH2

(CH3)3CO

O

O−

NHCH

CH3

2. H2NCHCO−

N-protected Gly–Ala

O

O

CH(CH3)2

N-protected Gly–Ala–Val

CH(CH3)2

When the desired number of amino acids has been added to the chain, the protecting group, known by the acronym t-Boc (tert-butyloxycarbonyl; pronounced “tee-bok”), on the N-terminal amino acid is removed with trifluoroacetic acid  in dichloromethane, a reagent that will not break any other covalent bonds. The protecting group is removed by an elimination reaction, forming isobutylene and carbon  dioxide. (The red arrows represent formation of the tetrahedral intermediate; the blue arrows represent collapse of the tetrahedral intermediate.) t-BOC is an ideal protecting group because it can be removed easily and, since the products formed are gases, they escape, driving the reaction to completion. H OOCCF3 O O

CH3 CH3

C

O

CH2

NHCH

O NHCH

CH3

O

CH3 OH CH Cl CH3 2 2

CH(CH3)2

O

+ HO

C

NHCH2

CH2

O NHCH

H OOCCF3

N-protected Gly–Ala–Val

H CF3COO

NHCH2

O

O NHCH

OH

CH(CH3)2

CH3

O +

CO2 + H3NCH2

O

O NHCH CH3

OOCCF3

O−

NHCH

CH(CH3)2

Gly–Ala–Val

Theoretically, you should be able to make as long a peptide as desired with this technique. Reactions never produce 100% yields, however, and the yields are further decreased during the purification process. (The peptide must be purified after each step of the synthesis to prevent subsequent unwanted reactions with leftover reagents.) Assuming that each amino acid can be added to the growing end of the peptide chain with an 80% yield (a relatively high yield, as you can probably appreciate from your own experience in the laboratory), the overall yield of a nonapeptide such as bradykinin would be only 17%. It is clear that large polypeptides could never be synthesized this way. Number of amino acids

Overall yield

2

3

4

5

6

7

8

9

80%

64%

51%

41%

33%

26%

21%

17%

PROBLEM 31♦

What dipeptides would be formed by heating a mixture of valine and N-protected leucine?


Automated Peptide Synthesis

PROBLEM 32

Suppose you are trying to synthesize the dipeptide Val-Ser. Compare the product that would be obtained if thionyl chloride were used to activate the carboxyl group of N-protected valine with the product that would be obtained if it were activated with DCC. PROBLEM 33

Show the steps in the synthesis of the tetrapeptide Leu-Phe-Ala-Val.

PROBLEM 34♦

a. Calculate the overall yield of bradykinin when the yield for the addition of each amino acid to the chain is 70%. b. What would be the overall yield of a peptide containing 15 amino acids if the yield for the incorporation of each is 80%?

22.11 AUTOMATED PEPTIDE SYNTHESIS In addition to producing low overall yields, the method of peptide synthesis described in Section 22.10 is extremely time consuming because the product must be purified at each step of the synthesis. In 1969, Bruce Merrifield described a method that revolutionized the synthesis of peptides because it provided a much faster way to produce peptides in much higher yields. Furthermore, because it is automated, the synthesis requires fewer hours of direct attention. With this technique, bradykinin was synthesized in 27 hours with an overall yield of 85%. Subsequent refinements in the technique now allow a reasonable yield of a peptide containing 100 amino acids to be synthesized in four days. In the Merrifield method, the synthesis is done on a solid support in a column. The solid support is similar to the one used in ion-exchange chromatography (Section 22.5), except that the benzene rings have chloromethyl substituents instead of sulfonic acid substituents. Before the C-terminal amino acid is added to the solid support, its amino group is protected with t-BOC to prevent the amino group from reacting with the solid support. The C-terminal amino acid is attached to the solid support by means of an SN2 reaction—its carboxyl group attacks a benzyl carbon of the resin, displacing a chloride ion (Section 9.2). After the C-terminal amino acid is attached to the resin, the t-BOC protecting group is removed (Section 22.10). The next amino acid, with its amino group protected with t-BOC and its carboxyl group activated with DCC, is added to the column, and then its protecting group is removed. Then the next N-terminal-protected and C-activated amino acid is added to the column. Thus, the protein is synthesized from the C-terminal end to the N-terminal end. (Proteins are synthesized in nature from the N-terminal end to the C-terminal end; Section 26.9.) Because the process uses a solid support and is automated, Merrifield’s method of protein synthesis is called automated solid-phase peptide synthesis. THE STEPS IN THE MERRIFIELD AUTOMATED SOLID-PHASE SYNTHESIS OF A TRIPEPTIDE

O (CH3)3CO

an SN2 reaction

O NHCH R

N-protected amino acid

O

+ Cl

CH2

solid support

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The Organic Chemistry of Amino Acids, Peptides, and Proteins

O

O CH2

O

NHCH

(CH3)3CO

R O

CH3 CH3

+ CO2 +

C

R

O NHCH

(CH3)3CO

O DCC

OH

O

R

R N-protected and C-activated amino acid

O

O

O

NHCH

(CH3)3CO

NHCH

R

CF3COOH CH2Cl2

O

O

+ CO2 + H2NCH

CH2

NHCH

R

O

CH2

O

R

O

O DCC

OH

NHCH

(CH3)3CO

CH2

O

R

CH3 C

DCC

NHCH

(CH3)3CO

N-protected amino acid

CH3

CH2

O

H2NCH

CH2 O

CF3COOH CH2Cl2

O NHCH

(CH3)3CO

O

DCC

R

R N-protected amino acid

N-protected and C-activated amino acid

O

O NHCH

NHCH

(CH3)3CO

O

O NHCH

R

R

O

CH2

R CF3COOH CH2Cl2

CH3 CH3C

O

O

+ CO2 + H2NCHC

NHCHC

NHCHCO

R

R

CH2

R

O CH2

HF

O

O

H3NCHC

NHCHC

+

R

R

O NHCHCOH + HOCH2 R


An Introduction to Protein Structure

A huge advantage of the Merrifield method of peptide synthesis is that the growing peptide can be purified by washing the column with an appropriate solvent after each step of the procedure. The impurities are washed out of the column because they are not attached to the solid support. Because the peptide is covalently attached to the resin, none of it is lost in the purification step, leading to high yields of purified product. After the required amino acids have been added one by one, the peptide can be removed from the resin by treatment with HF under mild conditions that do not break any peptide bonds. Over time, Merrifield’s technique has been improved so that peptides can be made more rapidly and more efficiently. However, it still cannot begin to compare with nature. A bacterial cell is able to synthesize a protein containing thousands of amino acids in seconds and can simultaneously synthesize thousands of different proteins with no mistakes. Since the early 1980s, it has been possible to synthesize proteins by genetic engineering techniques (Section 26.14). Strands of DNA, introduced into host cells, cause cells to produce large amounts of a desired protein. Genetic engineering techniques have also been useful in synthesizing proteins that differ in one or a few amino acids from a natural protein. Such synthetic proteins have been used, for example, to learn how a change in a single amino acid affects the properties of a protein (Section 23.9). PROBLEM 35

Show the steps in the synthesis of the tetrapeptide in Problem 33, using Merrifield’s method.

22.12 AN INTRODUCTION TO PROTEIN STRUCTURE Proteins are described by four levels of structure, called primary, secondary, tertiary, and quaternary. ■

■ ■

The primary structure of a protein is the sequence of the amino acids in the chain and the location of all the disulfide bridges. Secondary structures are regular conformations assumed by segments of the protein’s backbone when it folds. The tertiary structure is the three-dimensional structure of the entire protein. If a protein has more than one polypeptide chain, it also has a quaternary structure. The quaternary structure is the way the individual polypeptide chains are arranged with respect to one another.

Primary Structure and Taxonomic Relationship When scientists examine the primary structures of proteins that carry out the same function in different organisms, they can correlate the number of amino acid differences in the proteins to the closeness of the taxonomic relationship between the species. For example, cytochrome c, a protein that transfers electrons in biological oxidations, has about 100 amino acids. Yeast cytochrome c differs by 48 amino acids from horse cytochrome c, whereas duck cytochrome c differs by only two amino acids from chicken cytochrome c. Ducks and chickens, therefore, have a much closer taxonomic relationship than horses and yeast. Likewise, the cytochrome c in chickens and turkeys have identical primary structures. Human cytochrome c and chimpanzee cytochrome c are also identical and differ by one amino acid from the cytochrome c of the rhesus monkey.

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22.13 HOW TO DETERMINE THE PRIMARY STRUCTURE OF A POLYPEPTIDE OR A PROTEIN The first step in determining the sequence of amino acids in a polypeptide (or a protein) is to reduce the disulfide bridges in order to obtain a single polypeptide chain. A commonly used reducing agent is 2-mercaptoethanol. Notice that when it reduces a disulfide bridge, 2-mercaptoethanol is oxidized to a disulfide (Section 22.8). Reaction of the protein thiol groups with iodoacetic acid prevents the disulfide bridges from re-forming as a result of oxidation. reducing disulfide bridges

O NHCH

C

NHCH

C I

CH2

CH2 S

+

S

2 HS

SH

OH

2-mercaptoethanol

CH2 HNCH

O

O

+

SH

HO

S

S

OH

OH

O iodoacetic acid

O

NHCH

CH2

C O

S

OH + 2 HI

CH2 C

NHCH

OH

S C

CH2

O

NHCH

O C O

PROBLEM 36

Write the mechanism for the reaction of a cysteine side chain with iodoacetic acid.

Now we need to determine the number and kinds of amino acids in the polypeptide chain. To do this, a sample of the polypeptide is dissolved in 6 M HCl and heated at 100 °C for 24 hours. This treatment hydrolyzes all the amide bonds in the polypeptide, including the amide bonds in the side chains of asparagine and glutamine. protein

6 M HCI 100 °C 24 h

amino acids

The mixture of amino acids is then passed through an amino acid analyzer to identify the amino acids and determine how many of each kind are in the polypeptide (Section 22.5). Because all the asparagines and glutamines have been hydrolyzed to aspartates and glutamates, the number of aspartates or glutamates in the amino acid mixture tells us the number of aspartates plus asparagines or glutamates plus glutamines in the original protein. A separate test must be conducted to distinguish between aspartate and asparagine or between glutamate and glutamine in the original polypeptide. The strongly acidic conditions used for hydrolysis destroy all the tryptophans because the indole ring is unstable in acid (Section 20.5). However, the tryptophan content can be determined by hydroxide-ion-promoted hydrolysis of the peptide. This is not a general method for peptide bond hydrolysis because the strongly basic conditions destroy several amino acids.

Determining the N-Terminal Amino Acid One of the most widely used methods to identify the N-terminal amino acid of a polypeptide is to treat the polypeptide with phenyl isothiocyanate (PITC), more commonly known as Edman’s reagent. Edman’s reagent reacts with the N-terminal amino group, and the resulting thiazolinone derivative is cleaved from the polypeptide under mildly acidic conditions, leaving behind a polypeptide with one less amino acid.


How to Determine the Primary Structure of a Polypeptide or a Protein

B R N +

B

C

H

R NH

COO−

N COO−

NH

S H2N O

O

S

N H

H

phenyl isothiocyanate PITC Edman's reagent

R N NH2

COO−

+

O

N

O S

N H

peptide without the original N-terminal amino acid

R

N H

S

COO−

NH H

B+

a thiazolinone

The thiazolinone rearranges in dilute acid to a more stable phenylthiohydantoin (PTH) (see Problem 72). R H N

R N N H

O

S

HB+

O N

S

a thiazolinone

a PTH-amino acid

Because each amino acid has a different side chain (R), each amino acid forms a  different PTH–amino acid. The particular PTH–amino acid can be identified by chromatography using known standards. An automated instrument known as a sequencer allows about 50 successive Edman degradations of a polypeptide to be performed (100 with more advanced instruments). The entire primary structure cannot be determined in this way, however, because side products accumulate that interfere with the results. PROBLEM 37♦

In determining the primary structure of insulin, what would lead you to conclude that insulin had more than one polypeptide chain?

Determining the C-Terminal Amino Acid The C-terminal amino acid of a polypeptide can be identified using a carboxypeptidase, an enzyme that catalyzes the hydrolysis of the C-terminal peptide bond, thereby cleaving off the C-terminal amino acid. Carboxypeptidase A cleaves off the C-terminal amino acid, as long as it is not arginine or lysine. On the other hand, carboxypeptidase B cleaves off the C-terminal amino acid, only if it is arginine or lysine. Carboxypeptidases are exopeptidases, enzymes that catalyze the hydrolysis of a peptide bond at the end of a peptide chain. site where carboxypeptidase cleaves

O

O

C NHCH R

O

C NHCH R

C NHCH R

O−

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Carboxypeptidases cannot be used to determine the amino acids at the C-terminal end of a peptide by cleaving off the C-terminal amino acids sequentially, because peptide bonds hydrolyze at different rates. For example, if the C-terminal amino acid hydrolyzed slowly and the next one hydrolyzed rapidly, then it would appear that they were being cleaved off at about the same rate.

Partial Hydrolysis Once the N-terminal and C-terminal amino acids have been identified, a sample of the polypeptide can be hydrolyzed under conditions that hydrolyze only some of the peptide bonds—a procedure known as partial hydrolysis. The resulting fragments are separated, and the amino acid composition of each can be determined using electrophoresis or an amino acid analyzer. The process is repeated and the sequence of the original protein can then be deduced by lining up the peptides and looking for regions of overlap. (The Nterminal and C-terminal amino acids of each fragment can also be identified, if needed.)

PROBLEM-SOLVING STRATEGY

Sequencing an Oligopeptide

When a nonapeptide undergoes partial hydrolysis, it forms dipeptides, a tripeptide and two tetrapeptides whose amino acid compositions are shown. Reaction of the intact nonapeptide with Edman’s reagent releases PTH-Leu. What is the sequence of the nonapeptide? 1. Pro, Ser 3. Met, Ala, Leu 5. Glu, Ser, Val, Pro 7. Met, Leu 2. Gly, Glu 4. Gly, Ala 6. Glu, Pro, Gly, Pro 8. His, Val ■

■ ■

Because we know that the N-terminal amino acid is Leu, we need to look for a fragment that contains Leu. Fragment 7 tells us that Met is next to Leu, and fragment 3 tells us that Ala is next to Met. Now we look for another fragment that contains Ala. Fragment 4 contains Ala and tells us that Gly is next to Ala. From fragment 2, we know that Glu comes next; Glu is in both fragments 5 and 6. Fragment 5 has three amino acids we have yet to place in the growing peptide (Ser, Val, Pro), but fragment 6 has only one, so we know from fragment 6 that Pro is the next amino acid. Fragment 1 indicates that the next amino acid is Ser, so we can now use fragment 5. Fragment 5 indicates that the next amino acid is Val, and fragment 8 tells us that His is the last (C-terminal) amino acid. Thus, the amino acid sequence of the nonapeptide is Leu-Met-Ala-Gly-Glu-Pro-Ser-Val-His

Now use the strategy you have just learned to solve Problem 38.

PROBLEM 38♦

A decapeptide undergoes partial hydrolysis to give peptides whose amino acid compositions are shown. Reaction of the intact decapeptide with Edman’s reagent releases PTH-Gly. What is the sequence of the decapeptide? 1. Ala, Trp 3. Pro, Val 5. Trp, Ala, Arg 7. Glu, Ala, Leu 2. Val, Pro, Asp 4. Ala, Glu 6. Arg, Gly 8. Met, Pro, Leu, Glu

Hydrolysis Using Endopeptidases A polypeptide can also be partially hydrolyzed using endopeptidases—enzymes that catalyze the hydrolysis of a peptide bond that is not at the end of a peptide chain. Trypsin, chymotrypsin, and elastase are endopeptidases that catalyze the hydrolysis of only the specific peptide bonds listed in Table 22.4. Trypsin, for example, catalyzes the hydrolysis of the peptide bond on the C-side of (meaning, on the right of) positively charged


How to Determine the Primary Structure of a Polypeptide or a Protein

1085

side chains (arginine or lysine). These enzymes belong to the group of enzymes known as digestive enzymes. C-side of lysine

O

O

C NHCH

O

C NHCH

R

CH2

C-side of arginine

O

C NHCH R′

O

C NHCH

C NHCH

R′′

CH2 lysine

O C NHCH

CH2

R′′′

CH2 arginine

CH2

CH2

CH2

NH

NH3

C

+

+

NH2

NH2

Thus, trypsin will catalyze the hydrolysis of three peptide bonds in the following polypeptide, creating a hexapeptide, a dipeptide, and two tripeptides. Ala-Lys-Phe-Gly-Asp-Trp-Ser-Arg-Glu-Val-Arg-Tyr-Leu-His cleavage by trypsin

Table 22.4

Specificity of Peptide or Protein Cleavage

Reagent Chemical reagents Edman’s reagent Cyanogen bromide Exopeptidases* Carboxypeptidase A Carboxypeptidase B Endopeptidases* Trypsin Chymotrypsin Elastase Thermolysin

Specificity removes the N-terminal amino acid hydrolyzes on the C-side of Met removes the C-terminal amino acid (not if it is Arg or Lys) removes the C-terminal amino acid (only if it is Arg or Lys) hydrolyzes on the C-side of Arg and Lys hydrolyzes on the C-side of amino acids that contain aromatic six-membered rings (Phe, Tyr, Trp) hydrolyzes on the C-side of small amino acids (Gly, Ala, Ser, and Val) hydrolyzes on the C-side of Ile, Met, Phe, Trp, Tyr, and Val

*

Cleavage will not occur if Pro is at the hydrolysis site.

Chymotrypsin catalyzes the hydrolysis of the peptide bond on the C-side of amino acids that contain aromatic six-membered rings (Phe, Tyr, Trp). Ala-Lys-Phe-Gly-Asp-Trp-Ser-Arg-Glu-Val-Arg-Tyr-Leu-His cleavage by chymotrypsin

trypsin (see the legend to Figure 22.10)


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Elastase catalyzes the hydrolysis of peptide bonds on the C-side of the two smallest amino acids (Gly, Ala, Ser, and Val). Chymotrypsin and elastase are much less specific than trypsin. (An explanation for the specificity of these enzymes is given in Section 23.9.) Ala-Lys-Phe-Gly-Asp-Trp-Ser-Arg-Glu-Val-Arg-Tyr-Leu-His cleavage by elastase

None of the exopeptidases or endopeptidases that we have discussed will catalyze the hydrolysis of a peptide bond if proline is at the hydrolysis site. These enzymes recognize the appropriate hydrolysis site by its shape and charge, and the cyclic structure of proline causes the hydrolysis site to have an unrecognizable three-dimensional shape. Ala-Lys-Pro

Leu-Phe-Pro

Pro-Phe-Val

trypsin will not cleave

chymotrypsin will not cleave

chymotrypsin will cleave

Cyanogen bromide (BrC ‚ N) hydrolyzes the peptide bond on the C-side of methionine. Cyanogen bromide is more specific than the endopeptidases about which peptide bonds it cleaves, so it provides more reliable information about the primary structure. Because cyanogen bromide is not a protein, it does not recognize the substrate by its shape. As a result, it will still cleave the peptide bond if proline is at the hydrolysis site. Ala-Lys-Phe-Gly-Met-Pro-Ser-Arg-Met-Val-Arg-Tyr-Leu-His cleavage by cyanogen bromide

MECHANISM FOR THE CLEAVAGE OF A PEPTIDE BOND BY CYANOGEN BROMIDE

CH3

Br

C

N CH3 +

S CH2 CH2

N H

NH O

CH2

R

O

C

S

O

NH

N H

+N

O

O H2O, HCl

R +

O + NH3 NH

NH

N

H2O, HCl

O

+ CH3SC

OH

O

O

O

OH

H

NH

O

C-side of methionine

R

O

R

O

CH2

+ Br

N

O

O

The nucleophilic sulfur of methionine attacks the carbon of cyanogen bromide and displaces a bromide ion. Nucleophilic attack by oxygen on the methylene group, resulting in departure of the weakly basic leaving group, forms a five-membered ring (Section 9.2). Acid-catalyzed hydrolysis of the imine cleaves the protein (Section 17.10). Further hydrolysis causes the lactone (a cyclic ester) to open to a carboxyl group and an alcohol group (Section 16.10).


How to Determine the Primary Structure of a Polypeptide or a Protein

The last step in determining the primary structure of a protein is to figure out the location of any disulfide bonds. This is done by hydrolyzing a sample of the protein that has intact disulfide bonds. From a determination of the amino acids in the cysteine-containing fragments, the locations of the disulfide bonds in the protein can be established (Problem 61). PROBLEM 39

Why doesn’t cyanogen bromide cleave on the C-side of cysteine?

PROBLEM 40♦

Indicate the peptides that would result from cleavage by the indicated reagent: a. His-Lys-Leu-Val-Glu-Pro-Arg-Ala-Gly-Ala by trypsin b. Leu-Gly-Ser-Met-Phe-Pro-Tyr-Gly-Val by chymotrypsin

P R O B L E M 4 1 Solved

Determine the amino acid sequence of a polypeptide from the following data: Acid-catalyzed hydrolysis gives Ala, Arg, His, 2 Lys, Leu, 2 Met, Pro, 2 Ser, Thr, and Val. Carboxypeptidase A releases Val. Edman’s reagent releases PTH-Leu. Treatment with cyanogen bromide gives three peptides with the following amino acid compositions: 1. His, Lys, Met, Pro, Ser 2. Thr, Val 3. Ala, Arg, Leu, Lys, Met, Ser Trypsin-catalyzed hydrolysis gives three peptides and a single amino acid: 1. Arg, Leu, Ser 3. Lys 2. Met, Pro, Ser, Thr, Val 4. Ala, His, Lys, Met Solution Acid-catalyzed hydrolysis shows that the polypeptide has 13 amino acids. The N-terminal amino acid is Leu (revealed by Edman’s reagent), and the C-terminal amino acid is Val (revealed by carboxypeptidase A).

Leu ■

Val

Because cyanogen bromide cleaves on the C-side of Met, any peptide containing Met must have Met as its C-terminal amino acid. Thus, the peptide that does not contain Met must be the C-terminal peptide, so we know that the twelfth amino acid is Thr. We know that peptide 3 is the N-terminal peptide because it contains Leu. Because peptide 3 is a hexapeptide, we know that the sixth amino acid in the polypeptide is Met. We also know that the eleventh amino acid is Met because cyanogen bromide cleavage gave the dipeptide Thr, Val. Ala, Arg, Lys, Ser Leu

His, Lys, Pro, Ser Met

Met

Thr

Val

Because trypsin cleaves on the C-side of Arg and Lys, any peptide containing Arg or Lys must have that amino acid as its C-terminal amino acid. Therefore, Arg is the C-terminal amino acid of peptide 1, so we now know that the first three amino acids are Leu-Ser-Arg. We also know that the next two are Lys-Ala because if they were Ala-Lys, then trypsin cleavage would give an Ala, Lys dipeptide. The trypsin data also identify the positions of His and Lys. Pro, Ser Leu

Ser

Arg

Lys

Ala

Met

His

Lys

Met

Thr

Val

Finally, because trypsin successfully cleaves on the C-side of Lys, Pro cannot be adjacent to Lys. Thus, the amino acid sequence of the polypeptide is Leu

Ser

Arg

Lys

Ala

Met

His

Lys

Ser

Pro

Met

Thr

Val

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PROBLEM 42♦

Determine the primary structure of an octapeptide from the following data: Acid-catalyzed hydrolysis gives 2 Arg, Leu, Lys, Met, Phe, Ser, and Tyr. Carboxypeptidase A releases Ser. Edman’s reagent releases Leu. Treatment with cyanogen bromide forms two peptides with the following amino acid compositions: 1. Arg, Phe, Ser 2. Arg, Leu, Lys, Met, Tyr Trypsin-catalyzed hydrolysis forms the following two amino acids and two peptides: 1. Arg 2. Ser 3. Arg, Met, Phe 4. Leu, Lys, Tyr PROBLEM 43

Three peptides were obtained from a trypsin digestion of two different polypeptides. In each case, indicate the possible sequences from the given data and tell what further experiment should be carried out in order to determine the primary structure of the polypeptide. a. 1. Val-Gly-Asp-Lys 2. Leu-Glu-Pro-Ala-Arg 3. Ala-Leu-Gly-Asp b. 1. Val-Leu-Gly-Glu 2. Ala-Glu-Pro-Arg 3. Ala-Met-Gly-Lys

22.14 SECONDARY STRUCTURE Secondary structure describes the repetitive conformations assumed by segments of the backbone chain of a peptide or protein. In other words, the secondary structure describes how segments of the backbone fold. Three factors determine the secondary structure of a segment of protein: ■

The regional planarity about each peptide bond (as a result of the partial double-bond character of the amide bond), which limits the possible conformations of the peptide chain (Section 22.8) Minimizing energy by maximizing the number of peptide groups that engage in hydrogen bonding (that is, that form a hydrogen bond between the carbonyl oxygen of one amino acid and the amide hydrogen of another). R

O

HN C R NH

O

H

R N

R

O

hydrogen bonding between peptide groups

The need for adequate separation between neighboring R groups to avoid steric strain and repulsion of like charges

A-Helix One type of secondary structure is an A-helix. In an a-helix, the backbone of the polypeptide coils around the long axis of the protein molecule. The substituents on the a-carbons of the amino acids protrude outward from the helix, thereby minimizing steric strain (Figure 22.8a). The helix is stabilized by hydrogen bonds—each hydrogen attached to an amide nitrogen is hydrogen bonded to a carbonyl oxygen of an amino acid four amino acids away (Figure 22.8b). Because the amino acids have the l-configuration, the a -helix is a right-handed helix—that is, it rotates in a clockwise direction as it spirals down (Figure 22.8c).


Secondary Structure

Each turn of the helix contains 3.6 amino acids, and the repeat distance of the helix is 5.4 Å. a.

side chain

b.

c.

hydrogen bond

5.4 Å

▲ Figure 22.8 (a) A segment of a protein in an a-helix. (b) The helix is stabilized by hydrogen bonding between peptide groups. (c) Looking at the longitudinal axis of an a-helix.

Not all amino acids are able to fit into an a-helix. Proline, for example, causes a distortion in a helix because the bond between the proline nitrogen and the a-carbon cannot rotate to let proline fit into a helix properly. Similarly, two adjacent amino acids that have more than one substituent on a b-carbon (valine, isoleucine, or threonine) cannot fit into a helix because of steric crowding between the R groups. Finally, two adjacent amino acids with like-charged substituents cannot fit into a helix because of electrostatic repulsion between the R groups. The percentage of amino acids coiled into an a-helix varies from protein to protein, but on average about 25% of the amino acids in a globular protein are in a-helices.

Right-Handed and Left-Handed Helices The a-helix, composed of a chain of l-amino acids, is a right-handed helix. When scientists synthesized a chain of d-amino acids, they found that it folded into a left-handed helix that was the mirror image of the right-handed a-helix. When they synthesized a peptidase that contained only d-amino acids, they found that the enzyme was just as catalytically active as a naturally occurring peptidase with l-amino acids. However, the peptidase that contained d-amino acids would cleave peptide bonds only in polypeptide chains composed of d-amino acids.

B-Pleated Sheet The second type of secondary structure is a B-pleated sheet. In a b-pleated sheet, the polypeptide backbone is extended in a zigzag structure resembling a series of pleats. The hydrogen bonding in a b-pleated sheet occurs between neighboring peptide chains, and these chains can run in the same direction or in opposite directions—called a parallel B -pleated sheet and an antiparallel B-pleated sheet (Figure 22.9). A b-pleated sheet is almost fully extended; the average two amino acid repeat distance is 7.0 Å.

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R

CH

H

N HC

C

O R H

N-terminal

R

CH

H

N HC

O

C

R O

C N

H

CH C

O

R H

N HC

R

N-terminal

R

CH

H

N HC

O

C

R

C

O N

H

CH C

O

R

N

H

HC

R

C-terminal C-terminal Parallel

C-terminal

HC

R

O H

N O

R

C CH N

H

R

C

7.0 Å

N

H O

C

CH C

HC N

R

O H

C

O

N HC

R

R

CH

C-terminal N-terminal Antiparallel

▲ Figure 22.9 Segments of a parallel b-pleated sheet and an antiparallel b-pleated sheet drawn to illustrate their pleated character.

Because the substituents (R) on the a-carbons of the amino acids on adjacent chains are close to each other, the substituents must be small if the chains are to nestle closely enough together to maximize hydrogen-bonding interactions. Silk, for example, contains a large proportion of relatively small amino acids (glycine and alanine) and therefore has large segments of b-pleated sheet. The number of side-by-side strands in a b-pleated sheet ranges from 2 to 15 in a globular protein. The average strand in a b-pleated sheet section of a globular protein contains six amino acids. Wool and the fibrous protein of muscle have secondary structures that are almost all a-helices. Consequently, these proteins can be stretched. In contrast, proteins with secondary structures that are predominantly b-pleated sheets, such as silk and spider webs, cannot be stretched because a b-pleated sheet is almost fully extended already.

Coil Conformation Generally, less than half of the protein’s backbone is arranged in a defined secondary structure—an a-helix or a b-pleated sheet (Figure 22.10). Most of the rest of the protein, though highly ordered, is nonrepetitive and therefore difficult to describe. Many of these ordered polypeptide fragments are said to be in coil or loop conformations.

a-helix

coil conformation b-pleated sheet

▲ Figure 22.10 The backbone structure of an enzyme called ligase (Section 26.5): a b-pleated sheet is indicated by a flat arrow pointing in the N S C direction, an a-helix by a flat helical ribbon, and a coil or loop conformation by a thin tube.


Tertiary Structure

1091

PROBLEM 44♦

How long is an a -helix that contains 74 amino acids? How long is a fully extended peptide chain that contains the same number of amino acids? (The distance between consecutive amino acids in a fully extended chain is 3.5 Å; the repeat distance of an a -helix is 5.4 Å.)

b-Peptides: An Attempt to Improve on Nature b -Peptides are polymers of b -amino acids, so they have backbones that are one carbon longer than the peptides nature synthesizes using a-amino acids. Therefore, each b-amino acid has two carbons to which side chains can be attached. Like a-polypeptides, b-polypeptides fold into relatively stable helical and pleated sheet conformations, so scientists are trying to find out whether biological activity might be possible with such peptides. Recently, a b-peptide with biological activity has been synthesized that mimics the activity of the hormone somatostatin. There is hope that b-polypeptides will provide a source of new drugs and catalysts. Surprisingly, the peptide bonds in b-polypeptides are resistant to the enzymes that catalyze the hydrolysis of peptide bonds in a-polypeptides. This resistance to hydrolysis suggests that a b-polypeptide drug would have a longer duration of action in the bloodstream. a-carbon

O C

+

H3NCH

b-carbon

O−

R

O C

+

H3NCHCH

O−

R R

22.15 TERTIARY STRUCTURE The tertiary structure of a protein is the three-dimensional arrangement of all the atoms in the protein (Figure 22.11). Proteins fold spontaneously in solution to maximize their stability. Every time there is a stabilizing interaction between two atoms, free energy is released. The more free energy released (the more negative the ΔG°), the more stable the protein. Consequently, a protein tends to fold in a way that maximizes the number of stabilizing interactions.

◀ Figure 22.11 The three-dimensional structure of thermolysin (an endopeptidase).


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The stabilizing interactions in a protein include disulfide bonds, hydrogen bonds, electrostatic attractions (attractions between opposite charges), and hydrophobic (van der Waals) interactions. Stabilizing interactions can occur between peptide groups (atoms in the backbone of the protein), between side-chain groups (a-substituents), and between peptide and side-chain groups (Figure 22.12). Because the side-chain groups help determine how a protein folds, the tertiary structure of a protein is determined by its primary structure.

CH2

OH

O

C O

H CH2C NH

hydrogen bond between a sidechain group and a peptide group

H3N+

hydrogen bond between 2 sidechain groups

H

OCH2

Helical Structure

C

hydrogen bond between 2 peptide groups

CH3

hydrophobic interactions

HN

CH3

CH3

CHCH2CH3 CH

O

+

Pleated Sheet Structure

electrostatic attraction

O

(CH2)4 NH3 −OCCH2

CH S S CH2

disulfide bridge

hydrogen bond

hydrogen bond

COO−

▲ Figure 22.12 Stabilizing interactions responsible for the tertiary structure of a protein.

Disulfide bonds are the only covalent bonds that can form when a protein folds. The other bonding interactions that occur in folding are much weaker, but because there are so many of them, they are important in determining how a protein folds. Most proteins exist in aqueous environments, so they tend to fold in a way that exposes the maximum number of polar groups to the surrounding water and buries the nonpolar groups in the protein’s interior, away from water. The hydrophobic interactions between nonpolar groups in the protein increase its stability by increasing the entropy of water molecules. Water molecules that surround nonpolar groups are highly structured. When two nonpolar groups come together, the surface area in contact with water diminishes, decreasing the amount of structured water. Decreasing structure increases entropy, which in turn decreases the free energy, thereby increasing the stability of the protein. (Recall that ΔG° = ΔH° − TΔS°.) The precise mechanism by which proteins fold is still an unanswered question. Protein misfolding has been linked to numerous diseases such as Alzheimer’s disease and Huntington’s disease.


Quaternary Structure

1093

PROBLEM 45

How would a protein that resides in the nonpolar interior of a membrane fold compared with the water-soluble protein just discussed?

Diseases Caused by a Misfolded Protein Bovine spongiform encephalopathy (BSE), commonly known as mad cow disease, is unlike most other diseases because it is not caused by a microorganism. Instead, it caused by a misfolded protein in the brain called a prion. It is not yet known what causes the prion to become misfolded. The misfolded prion causes cells to deteriorate until the brain has a sponge-like appearance. The deterioration causes the loss of mental function, which makes cows with the disease act strangely (hence the name mad cow disease). It is not curable and it is fatal, but it is not contagious. It takes several years from the time of first exposure until the first signs of the disease appear, but then it progresses quickly. There are other diseases caused by misfolded prions that have similar symptoms. Kuru— transmitted through cannibalism—has been found to occur in the Fore people of Papua New Guinea (kuru means “trembling”). Scrapie affects sheep and goats. This disease got its name from the tendency of sheep to scrape off their wool on fences as they lean on them in an attempt to stay upright. It is thought that mad cow disease, first reported in the U.K. in 1985, was caused by cows eating bone meal made from scrapie-infected sheep. The human form of the disease is called Creutzfeldt–Jakob disease (CJD), which is rare and apparently arises spontaneously. The average age of onset of CJD is 64. In 1994, however, several cases of the disease appeared in young adults in the U.K. To date, 200 cases have been reported. This new variant Creutzfeldt–Jakob disease (vCJD) is caused by ingesting meat products of an animal infected with the disease.

22.16 QUATERNARY STRUCTURE Some proteins have more than one polypeptide chain. The individual chains are called subunits. A protein with a single subunit is called a monomer; one with two subunits is called a dimer; one with three subunits is called a trimer; and one with four subunits is called a tetramer. The quaternary structure of a protein describes the way the subunits are arranged with respect to each other. The subunits can be the same or different. Hemoglobin, for example, is a tetramer; it has two different kinds of subunits and each hemoglobin molecule has two of each kind (Figure 22.13). Turn to page 649 to see a protein with seven subunits.

◀ Figure 22.13 The quaternary structure of hemoglobin. The two a-subunits are green and the two b-subunits are purple. The porphyrin rings are blue.


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The subunits of a protein are held together by the same kinds of interactions that hold the individual protein chains in a particular three-dimensional conformation—namely, hydrophobic interactions, hydrogen bonding, and electrostatic attractions. PROBLEM 46♦

a. Which would have the greatest percentage of polar amino acids, a spherical protein, a cigarshaped protein, or a subunit of a hexamer? b. Which would have the smallest percentage of polar amino acids?

22.17 PROTEIN DENATURATION Destroying the highly organized tertiary structure of a protein is called denaturation. Anything that breaks the bonds maintaining the three-dimensional shape of the protein will cause the protein to denature (unfold). Because these bonds are weak, proteins are easily denatured. The following are some of the ways that proteins can be denatured: ■

Changing the pH denatures proteins because it changes the charges on many of the side chains. This disrupts electrostatic attractions and hydrogen bonds. Certain reagents such as urea and guanidine hydrochloride denature proteins by forming hydrogen bonds to protein groups that are stronger than the hydrogen bonds formed between the groups. Organic solvents or detergents such as sodium dodecyl sulfate denature proteins by associating with the nonpolar groups of the protein, thereby disrupting the normal hydrophobic interactions. Proteins can also be denatured by heat or by agitation. Both increase molecular motion, which can disrupt the attractive forces. A well-known example is the change that occurs to the white of an egg when it is heated or whipped.

S OME IMPORTANT THINGS TO REMEMBER ■

Peptides and proteins are polymers of amino acids linked together by peptide (amide) bonds. The amino acids differ only in the substituent attached to the a-carbon. Almost all amino acids found in nature have the l-configuration. The carboxyl groups of the amino acids have pKa values of ~2, and the protonated amino groups have pKa values of ~9. At physiological pH (7.4), an amino acid exists as a zwitterion. The isoelectric point (pI) of an amino acid is the pH at which the amino acid has no net charge. A mixture of amino acids can be separated based on their pI values by electrophoresis or based on their polarities by paper chromatography or thin-layer chromatography.

Separation can also be done by ion-exchange chromatography employing an anion- or a cationexchange resin. An amino acid analyzer is an instrument that automates ion-exchange chromatography. Amino acids can be synthesized by a Hell–Volhard– Zelinski reaction (followed by reaction with excess NH3), a Strecker synthesis, reductive amination, a N-phthalimidomalonic ester synthesis, or an acetamidomalonic ester synthesis. A racemic mixture of amino acids can be separated by a kinetic resolution using an enzyme-catalyzed reaction. Rotation about a peptide bond is restricted because of its partial double-bond character. Two cysteine side chains can be oxidized to a disulfide bridge, the only kind of covalent bond that is found between nonadjacent amino acids.


Problems ■

By convention, peptides and proteins are written with the free amino group (of the N-terminal amino acid) on the left and the free carboxyl group (of the C-terminal amino acid) on the right. To synthesize a peptide bond, the amino group of the N-terminal amino acid must be protected (by t-BOC) and its carboxyl group activated (with DCC). A second amino acid is added to form a dipeptide. Amino acids can be added to the growing C-terminal end by repeating the same two steps: activating the carboxyl group of the C-terminal amino acid with DCC and adding a new amino acid. Automated solid-phase peptide synthesis allows peptides to be made more rapidly and in higher yields. The primary structure of a protein is the sequence of its amino acids and the location of all its disulfide bridges. The N-terminal amino acid can be determined with Edman’s reagent. The C-terminal amino acid can be identified with a carboxypeptidase.

1095

An exopeptidase catalyzes the hydrolysis of a peptide bond at the end of a peptide chain. An endopeptidase catalyzes the hydrolysis of a peptide bond that is not at the end of a peptide chain. The secondary structure of a protein describes how local segments of the protein’s backbone fold. An A-helix and a B-pleated sheet are two types of secondary structure. A protein folds so as to maximize the number of stabilizing interactions—namely, disulfide bonds, hydrogen bonds, electrostatic attractions, and hydrophobic interactions. The tertiary structure of a protein is the threedimensional arrangement of all the atoms in the protein. The quaternary structure of a protein describes the way the peptide chains (subunits) of a protein with more than one peptide chain are arranged with respect to each other.

PROBLEMS 47. Explain why amino acids, unlike most amines and carboxylic acids, are insoluble in diethyl ether. 48. Alanine has pKa values of 2.34 and 9.69. At what pH will alanine exist in the indicated form? O a. H2N

+

O 50%

O

H3N

+

O

O

b. H3N

50%

+

O

100%

O

c. H3N

+

O 50%

O

H3N

OH 50%

49. Show the peptides that would result from cleavage by the indicated reagent: a. Val-Arg-Gly-Met-Arg-Ala-Ser by carboxypeptidase A b. Ser-Phe-Lys-Met-Pro-Ser-Ala-Asp by cyanogen bromide c. Arg-Ser-Pro-Lys-Lys-Ser-Glu-Gly by trypsin 50. Which would have a higher percentage of negative charge at physiological pH (7.4), leucine with pI = 5.98 or asparagine with pI = 5.43? 51. Aspartame has a pI of 5.9. Draw its prevailing form at physiological pH (7.4). 52. Draw the form of aspartate that predominates at the following pH values: a. pH = 1.0 b. pH = 2.6 c. pH = 6.0

d. pH = 11.0

53. A professor was preparing a manuscript for publication in which she reported that the pI of the tripeptide Lys-Lys-Lys was 10.6. One of her students pointed out that there must be an error in her calculations because the pKa of the e-amino group of lysine is 10.8 and the pI of the tripeptide has to be greater than any of its individual pKa values. Was the student correct? 54. A mixture of amino acids that do not separate sufficiently when a single technique is used can often be separated by two-dimensional chromatography. In this technique, the mixture of amino acids is applied to a piece of filter paper and separated by chromatographic techniques. The paper is then rotated 90°, and the amino acids are further separated by electrophoresis, producing a type of chromatogram called a fingerprint. Identify the spots in the fingerprint obtained from a mixture of Ser, Glu, Leu, His, Met, and Thr.

Electrophoresis at pH = 5

+

Chromatography


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55. Determine the amino acid sequence of a polypeptide from the following data: Complete hydrolysis of the peptide yields Arg, 2 Gly, Ile, 3 Leu, 2 Lys, 2 Met, 2 Phe, Pro, Ser, 2 Tyr, and Val. Treatment with Edmanâ&#x20AC;&#x2122;s reagent releases PTH-Gly. Carboxypeptidase A releases Phe. Treatment with cyanogen bromide yields the following three peptides: 1. Gly-Leu-Tyr-Phe-Lys-Ser-Met 2. Gly-Leu-Tyr-Lys-Val-Ile-Arg-Met

3. Leu-Pro-Phe

Treatment with trypsin yields the following four peptides: 1. Gly-Leu-Tyr-Phe-Lys 3. Val-Ile-Arg 2. Ser-Met-Gly-Leu-Tyr-Lys 4. Met-Leu-Pro-Phe 56. Explain the difference in the pKa values of the carboxyl groups of alanine, serine, and cysteine. 57. Which would be a more effective buffer at physiological pH, a solution of 0.1 M glycylglycylglycylglycine or a solution of 0.2 M glycine? 58. Identify the location and type of charge on the hexapeptide Lys-Ser-Asp-Cys-His-Tyr at each of the following pH values: a. pH = 1 b. pH = 5 c. pH = 7 d. pH = 12 59. Draw the product obtained when a lysine side chain in a polypeptide reacts with maleic anhydride. O NHCH

NH

+

O

(CH2)4

O

O

NH2 lysine residue

maleic anhydride

60. After the polypeptide shown here was treated with maleic anhydride, it was hydrolyzed by trypsin. (After a polypeptide is treated with maleic anhydride, trypsin will cleave the polypeptide only on the C-side of arginine.) Gly-Ala-Asp-Ala-Leu-Pro-Gly-Ile-Leu-Val-Arg-Asp-Val-Gly-Lys-Val-Glu-Val-Phe-Glu-Ala-GlyArg-Ala-Glu-Phe-Lys-Glu-Pro-Arg-Leu-Val-Met-Lys-Val-Glu-Gly-Arg-Pro-Val-Gly-Ala-Gly-Leu-Trp a. After a polypeptide is treated with maleic anhydride, why does trypsin no longer cleave it on the C-side of lysine? b. How many fragments are obtained from the polypeptide? c. In what order would the fragments be eluted from an anion-exchange column using a buffer of pH = 5? 61. Treatment of a polypeptide with 2-mercaptoethanol yields two polypeptides with the following primary structures: Val-Met-Tyr-Ala-Cys-Ser-Phe-Ala-Glu-Ser Ser-Cys-Phe-Lys-Cys-Trp-Lys-Tyr-Cys-Phe-Arg-Cys-Ser Treatment of the original intact polypeptide with chymotrypsin yields the following peptides: 1. Ala, Glu, Ser 3. Tyr, Val, Met 5. Ser, Phe, 2 Cys, Lys, Ala, Trp 2. 2 Phe, 2 Cys, Ser 4. Arg, Ser, Cys 6. Tyr, Lys Determine the positions of the disulfide bridges in the original polypeptide. 62. Show how aspartame can be synthesized using DCC. 63. a-Amino acids can be prepared by treating an aldehyde with ammonia/trace acid, followed by hydrogen cyanide, followed by acidcatalyzed hydrolysis. a. Draw the structures of the two intermediates formed in this reaction. b. What amino acid is formed when the aldehyde that is used is 3-methylbutanal? c. What aldehyde would be needed to prepare isoleucine?


Problems

1097

64. Reaction of a polypeptide with carboxypeptidase A releases Met. The polypeptide undergoes partial hydrolysis to give the following peptides. What is the sequence of the polypeptide? 1. Ser, Lys, Trp 4. Leu, Glu, Ser 7. Glu, His 10. Glu, His, Val 2. Gly, His, Ala 5. Met, Ala, Gly 8. Leu, Lys, Trp 11. Trp, Leu, Glu 3. Glu, Val, Ser 6. Ser, Lys, Val 9. Lys, Ser 12. Ala, Met 65. a. How many different octapeptides can be made from the 20 naturally occurring amino acids? b. How many different proteins containing 100 amino acids can be made from the 20 naturally occurring amino acids? 66. Glycine has pKa values of 2.3 and 9.6. Would you expect the pKa values of glycylglycine to be higher or lower than these values? 67. A mixture of 15 amino acids gave the fingerprint shown here (see Problem 54). Identify the spots. (Hint 1: Pro reacts with ninhydrin to produce a yellow color; Phe and Tyr produce a green color. Hint 2: Count the number of spots before you start.) −

Starting mixture:

Electrophoresis at pH = 5

Ala Arg Asp Glu Gly

Ile Leu Met Phe Pro

Ser Thr Trp Tyr Val

Origin +

Chromatography

68. Write the mechanism for the reaction of an amino acid with di-tert-butyl dicarbonate. 69. Dithiothreitol reacts with disulfide bridges in the same way that 2-mercaptoethanol does. With dithiothreitol, however, the equilibrium lies much more to the right. Explain. HO HO

SH + RSSR SH

HO HO

S + 2 RSH S

dithiothreitol

70. Show how valine can be prepared by a. a Hell–Volhard–Zelinski reaction. b. a Strecker synthesis. c. a reductive amination.

d. a N-phthalimidomalonic ester synthesis. e. an acetamidomalonic ester synthesis.

71. A chemist wanted to test his hypothesis that the disulfide bridges that form in many proteins do so after the minimum energy conformation of the protein has been achieved. He treated a sample of an enzyme that contained four disulfide bridges, with 2-mercaptoethanol and then added urea to denature the enzyme. He slowly removed these reagents so that the enzyme could re-fold and re-form the disulfide bridges. The enzyme he recovered had 80% of its original activity. What would be the percent activity in the recovered enzyme if disulfide bridge formation were entirely random rather than determined by the tertiary structure? Does this experiment support his hypothesis? 72. Propose a mechanism for the rearrangement of the thiazoline obtained from the reaction of Edman’s reagent with a peptide to a PTH-amino acid (page 1083). (Hint: Thioesters are susceptible to hydrolysis.)


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73. A normal polypeptide and a mutant of the polypeptide were hydrolyzed by an endopeptidase under the same conditions. The normal and mutant polypeptide differ by one amino acid. The fingerprints of the peptides obtained from the two polypeptides are shown here. What kind of amino acid substitution occurred as a result of the mutation? (That is, is the substituted amino acid more or less polar than the original amino acid? Is its pI lower or higher?) (Hint: Photocopy the fingerprints, cut them out, and overlay them.)

Electrophoresis at pH = 6.5

Electrophoresis at pH = 6.5

Normal +

Paper chromatography

Mutant +

Paper chromatography

74. Determine the amino acid sequence of a polypeptide from the following data: Complete hydrolysis of the peptide yields Ala, Arg, Gly, 2 Lys, Met, Phe, Pro, 2 Ser, Tyr, and Val. Treatment with Edman’s reagent releases PTH-Val. Carboxypeptidase A releases Ala. Treatment with cyanogen bromide yields the following two peptides: 1. Ala, 2 Lys, Phe, Pro, Ser, Tyr 2. Arg, Gly, Met, Ser, Val Treatment with trypsin yields the following three peptides: 1. Gly, Lys, Met, Tyr 2. Ala, Lys, Phe, Pro, Ser

3. Arg, Ser, Val

Treatment with chymotrypsin yields the following three peptides: 1. 2 Lys, Phe, Pro 2. Arg, Gly, Met, Ser, Tyr, Val

3. Ala, Ser


23

Catalysis in Organic Reactions and in Enzymatic Reactions N

lysozyme

F

E C

Asp 52

Glu 35 D

C

B

A

Lysozyme is an enzyme that catalyzes the hydrolysis of bacterial cell walls. C indicates its C-terminal end, N indicates its N-terminal end, and the cell wall is indicated by A–F. Lysozyme’s catalytic groups (Glu 35 and Asp 52) are also shown. (See Section 23.10.)

A

catalyst is a substance that increases the rate of a chemical reaction without itself being consumed or changed in the reaction. In this chapter we will look at the types of catalysts used in organic reactions and the ways in which they provide an energetically more favorable pathway for a reaction. We will then see how the same catalysts are used in reactions that take place in living systems—that is, in enzyme-catalyzed reactions. We will also see why enzymes are extraordinarily good catalysts—they can increase the rate of an intermolecular reaction by as much as 1016. In contrast, rate enhancements achieved by nonbiological catalysts in intermolecular reactions are seldom greater than 10,000-fold. We have seen that the rate of a chemical reaction depends on the energy barrier of the rate-determining step that must be overcome in the process of converting reactants into products. The height of the “energy hill” is indicated by the free energy of activation ( ⌬G [). A catalyst increases the rate of a chemical reaction by providing a pathway with a lower ⌬G [ (Section 5.11). A catalyst can decrease ⌬G [ in one of three ways: 1. The catalyzed and uncatalyzed reactions can have different, but similar, mechanisms, with the catalyst providing a way to make the reactant more reactive (less stable) (Figure 23.1a). 2. The catalyzed and uncatalyzed reactions can have different, but similar, mechanisms, with the catalyst providing a way to make the transition state more stable (Figure 23.1b). 3. The catalyst can completely change the mechanism of the reaction, providing an alternative pathway with a smaller ⌬G [ than that for the uncatalyzed reaction (Figure 23.2).

1099


1100

CHAPTER 23

Catalysis in Organic Reactions and in Enzymatic Reactions

a more stable transition state

b.

a more reactive reactant

Free energy

Free energy

a.

‡ ΔGuncatalyzed

‡ ΔGcatalyzed

‡ ΔGcatalyzed

Progress of the reaction

‡ ΔGuncatalyzed

Progress of the reaction

▲ Figure 23.1 Reaction coordinate diagrams for an uncatalyzed reaction (black) and for a catalyzed reaction (green). (a) The catalyst converts the reactant to a more reactive species. (b) The catalyst stabilizes the transition state.

When we say that a catalyst is neither consumed nor changed by a reaction, we do not mean that it does not participate in the reaction. A catalyst must participate in the reaction if it is going to make it go faster. What we mean is that a catalyst has the same form after the reaction as it had before the reaction. Because the catalyst is not used up during the reaction, only a small amount of the catalyst is needed. (If a catalyst is used up in one step of the reaction, it must be regenerated in a subsequent step.) Therefore, a catalyst is added to a reaction mixture in small amounts, much smaller than the number of moles of reactant (typically .01 to 10% of the number of moles of reactant). We call this a catalytic amount. Notice in Figures 23.1 and 23.2 that the stability of the original reactants and final products is the same in both the catalyzed and corresponding uncatalyzed reactions. In other words, the catalyst does not change the equilibrium constant of the reaction. Because the catalyst does not change the equilibrium constant, it does not change the amount of product formed when the reaction has reached equilibrium. It changes only the rate at which the product is formed. ‡ ΔGcatalyzed

‡ ΔGuncatalyzed

Free energy

a one-step mechanism a two-step mechanism

▶ Figure 23.2 Reaction coordinate diagrams for an uncatalyzed reaction (black) and for a catalyzed reaction (green). The catalyzed reaction takes place by an alternative and energetically more favorable pathway.

Progress of the reaction

PROBLEM 1♦

Which of the following parameters would be different for a reaction carried out in the presence of a catalyst, compared with the same reaction carried out in the absence of a catalyst? (Hint: See Section 5.11.) ⌬G⬚, ⌬H [, Ea, ⌬S[ , ⌬H⬚, K eq, ⌬G [, ⌬S⬚, k rate


Acid Catalysis

1101

23.1 CATALYSIS IN ORGANIC REACTIONS There are several ways a catalyst can provide a more favorable pathway for an organic reaction: ■

■ ■ ■

It can increase the reactivity of an electrophile so that it is more susceptible to reaction with a nucleophile. It can increase the reactivity of a nucleophile. It can increase the leaving propensity of a group by converting it into a weaker base. It can increase the stability of a transition state.

Now we will look at some of the most common catalysts—namely, acid catalysts, base catalysts, nucleophilic catalysts, and metal-ion catalysts—and the ways in which they catalyze the reactions of organic compounds.

23.2 ACID CATALYSIS An acid catalyst increases the rate of a reaction by donating a proton to a reactant. In  the preceding chapters we have seen many examples of acid catalysis. For example, we saw that an acid provides the electrophile needed for the addition of water or an alcohol to an alkene (Section 6.6). We also saw that an alcohol cannot undergo substitution and elimination reactions unless an acid is present to protonate the OH  group, which increases its leaving propensity by making it a weaker base (Section 11.1). To review some of the important ways an acid can catalyze a reaction, let’s look again at the mechanism for the acid-catalyzed hydrolysis of an ester that we first saw in Section  16.10. The reaction has two slow steps: formation of the tetrahedral intermediate and collapse of the tetrahedral intermediate. Donation of a proton to, and removal of a proton from, an electronegative atom such as oxygen are always fast steps. MECHANISM FOR ACID-CATALYZED ESTER HYDROLYSIS +

O R

C

OH

OH H

B+

R

OCH3

C

OCH3

+ H2O

slow

R

C +

OCH3

OH

B

H In an acid-catalyzed reaction, a proton is donated to the reactant.

OH R

C

OCH3 H

OH

O

HB+

+

C R

H

O C

OH

R

OH

B + CH3OH

OH slow

R

C OH

+

OCH3 H

B+


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Catalysis in Organic Reactions and in Enzymatic Reactions

A catalyst must increase the rate of a slow step because increasing the rate of a fast step will not increase the rate of the overall reaction. The acid increases the rates of both slow steps of the hydrolysis reaction. It increases the rate of formation of the tetrahedral intermediate by protonating the carbonyl oxygen, thereby making it more susceptible to nucleophilic addition than an unprotonated carbonyl group would be. Increasing the reactivity of the carbonyl group by protonating it is an example of providing a way to convert the reactant into a more reactive species (Figure 23.1a). A catalyst must increase the rate of a slow step. Increasing the rate of a fast step will not increase the rate of the overall reaction.

acid-catalyzed first slow step

uncatalyzed first slow step

+

OH R

C

O OCH3

+ H2O

R

C

OCH3

+ H2O

more susceptible to nucleophilic addition

The acid increases the rate of the second slow step by decreasing the basicity—and thereby increasing the leaving propensity—of the group that is eliminated when the tetrahedral intermediate collapses. In the presence of an acid, methanol is eliminated; in the absence of an acid, methoxide ion is eliminated. Methanol is a much weaker base than methoxide ion, so methanol is much more easily eliminated. acid-catalyzed second slow step

uncatalyzed second slow step

OH

OH

R

C

+

OCH3 H

OH CH3OH is the leaving group

R

C

OCH3

OH CH3O− is the leaving group

The mechanism for the acid-catalyzed hydrolysis of an ester shows that the reaction can be divided into two distinct parts: formation of a tetrahedral intermediate and collapse of a tetrahedral intermediate. There are three steps in each part. Notice that in each part, the first step is a fast protonation step, the second step is a slow acid-catalyzed step that involves either breaking a p bond or forming a p bond, and the last step is a fast deprotonation step (to regenerate the catalyst).

PROBLEM 2

Compare each of the mechanisms listed here with the mechanism for each part of the acidcatalyzed hydrolysis of an ester, indicating a. similarities. b. differences. 1. acid-catalyzed formation of a hydrate (Section 17.11) 2. acid-catalyzed conversion of an aldehyde into a hemiacetal (Section 17.12) 3. acid-catalyzed conversion of a hemiacetal into an acetal (Section 17.12) 4. acid-catalyzed hydrolysis of an amide (Section 16.16)

There are two types of acid catalysis: specific-acid catalysis and general-acid catalysis. In specific-acid catalysis, the proton is fully transferred to the reactant before the slow step of the reaction begins (Figure 23.3a). In general-acid catalysis, the proton is transferred to the reactant during the slow step of the reaction (Figure 23.3b). Specific-acid and general-acid catalysis increase the rate of a reaction in the same way—by donating a proton in order to make either bond making or bond breaking easier. The two types of


Acid Catalysis

1103

acid catalysis differ only in the extent to which the proton is transferred in the transition state of the slow step of the reaction. proton is transferred to the b. general-acid catalysis reactant during slow step

a. specific-acid catalysis

H

A

Free energy

Free energy

R

RH+

P

P

proton is transferred completely to the reactant before slow step

R + H+

R + HA

Progress of the reaction

Progress of the reaction

▲ Figure 23.3 (a) Reaction coordinate diagram for a specific-acid-catalyzed reaction. The proton is transferred completely to the reactant before the slow step of the reaction begins (R = reactant; P = product). (b) Reaction coordinate diagram for a general-acid-catalyzed reaction. The proton is transferred to the reactant during the slow step of the reaction.

In the examples that follow, notice the difference in the extent to which the proton has been transferred when the nucleophile adds to the reactant. ■

In the specific-acid-catalyzed addition of water to a carbonyl group, the nucleophile adds to a fully protonated carbonyl group. In the general-acid-catalyzed addition of water to a carbonyl group, the carbonyl group becomes protonated as the nucleophile adds to it.

specific-acid-catalyzed addition of water proton has been transferred to the reactant +

O R

C

OH

+

OCH3

+ H

R

C

OH OCH3

+ H2O

slow

R

C

OCH3

+ OH

H

general-acid-catalyzed addition of water proton is being transferred to the reactant

H

B+

B

O R

C

OH OCH3

+ H2O

slow

R

C

OCH3

+ OH

H

In the specific-acid-catalyzed collapse of a tetrahedral intermediate, a fully protonated leaving group is eliminated, whereas in the general-acid-catalyzed collapse

The proton is donated to the reactant before the slow step in a specific-acid-catalyzed reaction, and during the slow step in a generalacid-catalyzed reaction.


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Catalysis in Organic Reactions and in Enzymatic Reactions

of a tetrahedral intermediate, the leaving group picks up a proton as the group is eliminated. specific-acid-catalyzed elimination of the leaving group

OH R

C

OH OCH3 + H+

+

C

R

OH

OCH3 H

+OH

slow

R

C

OH

+ CH3OH

OH proton has been transferred to the reactant

general-acid-catalyzed elimination of the leaving group +OH

OH R

C

slow

OCH3

OH

H

R

B+

C

OH

+ CH3OH B

proton is being transferred to the reactant

A specific-acid catalyst must be a strong enough acid (such as HCl, H3O+) to protonate the reactant fully before the slow step begins. A general-acid catalyst can be a weaker acid because it has only a partially transferred proton in the transition state of the slow step. In the mechanisms shown in previous chapters, we have used strong acids as catalysts, so all the mechanisms are written as specific-acid-catalyzed reactions. Weaker acids could have been used to catalyze many of the reactions, in which case the protonation step and the subsequent slow step would have been shown as a single step. A list of acids and their pKa values is given in Appendix I. PROBLEM 3♦

Are the slow steps for the acid-catalyzed hydrolysis of an ester on page 742 general-acid catalyzed or specific-acid catalyzed?

PROBLEM 4

a. Draw the mechanism for the following reaction if it involves specific-acid catalysis. b. Draw the mechanism if it involves general-acid catalysis. CH3

CH2 HO

O HB+

P R O B L E M 5 Solved

An alcohol will not react with aziridine unless an acid is present. Why is the acid necessary? H N aziridine

+ CH3OH

HCl

+

H3N

OCH3 −

Cl

Solution Although relief of ring strain is sufficient by itself to cause an epoxide to undergo a ring-opening reaction (Section 11.7), it is not sufficient to cause an aziridine to undergo a ring-opening reaction. A negatively charged nitrogen is a stronger base, and therefore a poorer leaving group, than a negatively charged oxygen. An acid, therefore, is needed to protonate the ring nitrogen to make it a better leaving group.


Base Catalysis

1105

23.3 BASE CATALYSIS We have already encountered several base-catalyzed reactions, such as the interconversion of keto and enol tautomers (Section 18.3), the Claisen condensation (Section 18.13), and the enediol rearrangement (Section 21.5). A base catalyst increases the rate of a reaction by removing a proton from the reactant. For example, the dehydration of a hydrate in the presence of hydroxide ion is a base-catalyzed reaction. Hydroxide ion (the base) increases the rate of the reaction by removing a proton from the neutral hydrate. In a base-catalyzed reaction, a proton is removed from the reactant.

specific-base-catalyzed dehydration

O

H

O

HO

ClCH2CCH2Cl

O

ClCH2CCH2Cl

slow

ClCH2

OH + H2O

OH

C

CH2Cl

+

HO−

a hydrate

Removing a proton from the hydrate increases the rate of dehydration by providing a pathway with a more stable transition state. The transition state for the elimination of HO - from a negatively charged tetrahedral intermediate is more stable because a positive charge does not develop on the electronegative oxygen atom, as it does in the transition state for the elimination of HO - from a neutral tetrahedral intermediate. d−

d+

O

OH

C

C

d− OH

d− OH

transition state for elimination of HO− from a negatively charged tetrahedral intermediate

transition state for elimination of HO− from a neutral tetrahedral intermediate

The foregoing base-catalyzed dehydration of a hydrate is an example of specific-base catalysis. In specific-base catalysis, the proton is completely removed from the reactant before the slow step of the reaction begins. In general-base catalysis, on the other hand, the proton is removed from the reactant during the slow step of the reaction. Compare the extent of proton transfer in the slow step of the preceding specific-base-catalyzed dehydration with the extent of proton transfer in the slow step of the following generalbase-catalyzed dehydration: The proton is removed from the reactant before the slow step in a specific-base-catalyzed reaction and during the slow step in a general-base-catalyzed reaction.

general-base-catalyzed dehydration −

O

B O

H

ClCH2CCH2Cl OH

slow

ClCH2

C

CH2Cl

+ HO− + HB

a hydrate

In specific-base catalysis, the base has to be strong enough to remove a proton from the reactant completely before the slow step begins. In general-base catalysis, the base can be weaker because the proton is only partially transferred to the base in the transition state of the slow step. We will see that enzymes catalyze reactions using general-acid and general-base catalysis because at physiological pH (7.4), the concentration of H + (苲1 * 10 - 7 M) is too small for specific-acid catalysis and the concentration of HO - is too small for specific-base catalysis.


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PROBLEM 6

a. Draw the mechanism for the following reaction if it involves specific-base catalysis. b. Draw the mechanism if it involves general-base catalysis. O

O O

B

O + CH3OH

OH

23.4 NUCLEOPHILIC CATALYSIS A nucleophilic catalyst increases the rate of a reaction by reacting as a nucleophile to  form a new covalent bond with the reactant. Nucleophilic catalysis, therefore, is also  called covalent catalysis. A nucleophilic catalyst increases the reaction rate by completely changing the mechanism of the reaction. In the following reaction, iodide ion increases the rate of conversion of ethyl chloride to ethyl alcohol by acting as a nucleophilic catalyst: a nucleophilic catalyst

A nucleophilic catalyst forms a covalent bond with the reactant.

CH3CH2Cl + HO−

I− H2O

CH3CH2OH + Cl−

To understand how iodide ion catalyzes this reaction, we need to compare the mechanisms for the uncatalyzed and catalyzed reactions. In the absence of iodide ion, ethyl chloride is converted to ethyl alcohol in a one-step SN2 reaction: the nucleophile is HOand the leaving group is Cl- .

MECHANISM FOR THE UNCATALYZED REACTION

HO

+ CH3CH2

CH3CH2OH + Cl−

Cl

If iodide ion is present in the reaction mixture, then the reaction takes place by two successive SN2 reactions.

MECHANISM FOR THE IODIDE-ION-CATALYZED REACTION

I− is a better nucleophile than HO− I− is a better leaving group than Cl−

I

HO

+ CH3CH2 −

+ CH3CH2

Cl

CH3CH2I + Cl−

I

CH3CH2OH + I−

The first SN2 reaction in the catalyzed reaction is faster than the uncatalyzed SN2 reaction because in a protic solvent, iodide ion is a better nucleophile than hydroxide ion, the nucleophile in the uncatalyzed reaction (Section 9.2). The second SN2 reaction in the catalyzed reaction is also faster than the uncatalyzed SN2 reaction because iodide ion is a weaker base and therefore a better leaving group than chloride ion, the leaving group in the uncatalyzed reaction.


Metal-Ion Catalysis

Thus, iodide ion increases the rate of formation of ethanol by changing a relatively slow SN2 reaction to one that involves two relatively fast SN2 reactions (Figure 23.2). Iodide ion is a nucleophilic catalyst because it reacts as a nucleophile, forming a covalent bond with the reactant. The iodide ion consumed in the first reaction is regenerated in the second, so it comes out of the reaction unchanged. Another reaction in which a nucleophilic catalyst provides a more favorable pathway by changing the mechanism of the reaction is the imidazole-catalyzed hydrolysis of an ester. a nucleophilic catalyst

N

O + H2O

CH3CO

NH

O

imidazole

CH3COH + HO

phenyl acetate

acetic acid

phenol

Imidazole is a better nucleophile than water, so imidazole reacts faster with the ester than water would. The acyl imidazole that is formed is particularly reactive because the positively charged nitrogen makes imidazole a very good leaving group. Therefore, it is hydrolyzed much more rapidly than the ester would have been. Because formation of the acyl imidazole and its subsequent hydrolysis are both faster than ester hydrolysis, imidazole increases the rate of hydrolysis of the ester. O

O

CH3C

+

O

N

NH

CH3C

O O

CH3C

+N

an ester

NH +

N +

−O

an acyl imidazole

N H

H2O

O +

CH3CO− + HN

NH + HO

23.5 METAL-ION CATALYSIS Metal ions exert their catalytic effect by complexing with atoms that have lone-pair electrons. Metal ions, therefore, are Lewis acids (Section 2.12). A metal ion can increase the rate of a reaction in the following ways: ■

■ ■

It can make a reaction center more susceptible to receiving electrons (that is, more electrophilic) as in A. It can make a leaving group a weaker base, and therefore a better leaving group, as in B. It can increase the rate of a hydrolysis reaction by increasing the nucleophilicity of water, as in C. A

d+

B

Metal O

d+

C

O OH C

C

d+

OCH3 d+

Nu

Metal ions are Lewis acids.

C

Metal

d+

d+

Metal

OH2

metal-bound water

d+

Metal

d−

OH + H+

metal-bound hydroxide ion a better nucleophile

In A and B, the metal ion exerts the same kind of catalytic effect as a proton does. However, a metal ion can be a much more effective catalyst than a proton, because metal

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ions can have a charge greater than +1, and a high concentration of a metal ion can be made available at neutral pH. In C, metal-ion complexation increases water’s nucleophilicity by converting it to metal-bound hydroxide ion. That is, the metal-ion increases water’s tendency to lose a proton as shown by the pKa values listed in Table 23.1. (The pKa of water is 15.7.) Metal-bound hydroxide ion, while not as good a nucleophile as hydroxide ion, is a better nucleophile than water. Metal ions are important catalysts in living systems because hydroxide ion itself would not be available at physiological pH (7.4). Table 23.1 The pKa of Metal-Bound Water

M2ⴙ

pKa

M2ⴙ

pKa

Ca2+

12.7

Co2+

8.9

Cd

11.6

Zn

2+

8.7

Mg2+

11.4

Fe2+

7.2

10.6

2+

6.8

2+

5.7

2+

2+

Mn Ni

2+

Cu

9.4

Be

Now we will look at some examples of metal-ion-catalyzed organic reactions. The decarboxylation of dimethyloxaloacetate can be catalyzed by either Cu2 + or Al3 + . O

O

O−

O

O

Cu2+ or Al3+

+ CO2

O

O

O

dimethyloxaloacetate

In this reaction, the metal ion complexes with two oxygen atoms of the reactant. Complexation increases the rate of decarboxylation by making the carbonyl oxygen more susceptible to receiving the electrons left behind when CO2 is eliminated. Cu2+ stabilizes the negative charge developing on the oxygen +

Cu Od+ d− O

+

Cu O d− d− O

O O

O

+ CO2

O

The hydrolysis of the ester shown next has two slow steps. Zn2 + increases the rate of the first slow step by providing metal-bound hydroxide ion, a better nucleophile than water. Zn2 + increases the rate of the second slow step by decreasing the basicity of the group that is eliminated from the tetrahedral intermediate. O−

O CF3 +

Zn

slow

C

OCH3

Zn

C

OCH3

CF3

C

d+

OH

d−

OH + H+ metal-bound hydroxide ion

+

CF3

O−

+

Zn

O OCH3

CF3

OH +

Zn

− d+

C

OCH3

OH d+

OH2

H2O

+

Zn

decreases the basicity of the leaving group

slow

O

d+

OH2 CF3

C

OH

+

+

Zn

d−

OCH3

H2O

O CF3

C

+

O

+ CH3OH + Zn

d+

OH2


Intramolecular Reactions PROBLEM 7♦

Although metal ions increase the rate of decarboxylation of dimethyloxaloacetate, they have no effect on the rate of decarboxylation of either the monoethyl ester of dimethyloxaloacetate or acetoacetate. Explain why this is so. O

O

O

O

O

O

O− O

O

O

O−

O−

O

dimethyloxaloacetate

monoethyl ester of dimethyloxaloacetate

acetoacetate

PROBLEM 8

Propose a mechanism for the Co2 +-catalyzed hydrolysis of glycinamide. O H2N

NH2

+ H2O

Co2+

O H2N

O−

+

+

NH4

23.6 INTRAMOLECULAR REACTIONS The rate of a chemical reaction is determined by the number of molecular collisions with sufficient energy and with the proper orientation in a given period of time (Section 5.8): rate of reaction =

number of collisions fraction with fraction with * * sufficient energy proper orientation unit of time

Because a catalyst decreases the energy barrier of a reaction, it increases the fraction of collisions that occur with sufficient energy to overcome the barrier. The rate of a reaction can also be increased by increasing the frequency of the collisions, which can be achieved by increasing the concentration of the reactants. In addition, we have seen that an intramolecular reaction that forms a five- or a six-membered ring occurs more readily than the analogous intermolecular reaction. This is because an intramolecular reaction has the advantage of the reacting groups being tied together in the same molecule, which gives them a better chance of finding each other than if they were in two different molecules in a solution of the same concentration (Section 9.8). As a result, the frequency of the collisions increases. If, in addition to being in the same molecule, the reacting groups are arranged in a way that increases the probability that they will collide with each other in the proper orientation, then the rate of the reaction will be further increased. The relative rates shown in Table 23.2 demonstrate the enormous increase in the rate of a reaction when the reacting groups are properly oriented. Rate constants for a series of reactions are generally compared in terms of relative rates because relative rates allow us to see immediately how much faster one reaction is than another. Relative rates are obtained by dividing the rate constant for each of the reactions by the rate constant for the slowest reaction in the series. The slowest reaction in Table 23.2 is an intermolecular reaction; all the others are intramolecular reactions. Because an intramolecular reaction is a first-order reaction (with units of time−1) and an intermolecular reaction is a second-order reaction (with units of M−1 time−1), the relative rates in Table 23.2 have units of molarity (M) (Section 5.9). relative rate =

time-1 first@order rate constant * = M second@order rate constant time-1M-1

Relative rates are also called effective molarities. Effective molarity is the concentration of the reactant that would be required in an intermolecular reaction for it to have the same rate as the intramolecular reaction. In other words, the effective

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Catalysis in Organic Reactions and in Enzymatic Reactions

Table 23.2 Relative Rates of an Intermolecular Reaction and Five Intramolecular Reactions

Reaction

Relative rate

O A CH C 3

O Br + CH3C

O

O O−

O

CH3C

O

C

O−

Br

B

O

−O

+

O R

C

O

Br

1.0

1 × 103 M

Br

O

R

Br

O

C

O−

C

R

R

O

+

−O

2.3 × 104 M R = CH3 1.3 × 106 M R = (CH3)2CH

Br

O

O O

C

O

C

O−

Br +

O

−O

2.2 × 105 M

Br

O

O O C

O O

Br O

E H

−O

O

O

H

CCH3 +

O

O

C

D

O

C

−O

+

Br

1 × 107 M

O

O−

O O O F

C

O

C

O−

O

Br

O

O O

+

−O

Br

5 × 107 M

O

molarity is the advantage given to a reaction by having the reacting groups in the same molecule. In some cases, juxtaposing the reacting groups provides such an enormous increase in rate that the effective molarity is greater than the concentration of the reactant in its solid state! Reaction A, the first reaction in Table 23.2, is an intermolecular reaction between an ester and a carboxylate ion. The second reaction, B, has the same two reacting groups in a single molecule. The rate of the intramolecular reaction is 1000 times faster than the rate of the intermolecular reaction. The reactant in B has four C i C bonds that are free to rotate, whereas the reactant in D has only three such bonds. Conformers in which the large groups are rotated away  from each other are more stable. However, when these groups are pointed away from each other, they are in an unfavorable conformation for reaction. Because


Intramolecular Catalysis

the reactant in D has fewer bonds that are free to rotate, the groups are more apt to be  in a conformation that is favorable for a reaction. Therefore, reaction D is faster than reaction B. the reacting groups are in a favorable conformation for a reaction

O

O C

O

C

O−

Br

C

O

C

O−

Br

O

O four carbon–carbon bonds can rotate

three carbon–carbon bonds can rotate

Reaction C is faster than reaction B because the alkyl substituents of the reactant in C  decrease the available space for the reactive groups to rotate away from each other. Thus, there is a greater probability that the molecule will be in a conformation that has the reacting groups positioned for ring closure. This is called the gem-dialkyl effect because the two alkyl substituents are bonded to the same (geminal) carbon. Comparing the rate when the substituents are methyl groups with the rate when the substituents are isopropyl groups, we see that the rate is further increased when the size of the alkyl groups is increased. The increased rate of reaction of E is due to the double bond that prevents the reacting groups from rotating away from each other. The bicyclic compound in F reacts even faster because the reacting groups are locked in the proper orientation for reaction. PROBLEM 9♦

The relative rate of reaction of the cis alkene (E) is given in Table 23.2. What would you expect the relative rate of reaction of the trans isomer to be?

23.7 INTRAMOLECULAR CATALYSIS Just as having two reacting groups in the same molecule increases the rate of a reaction compared with having the groups in separate molecules, having a reacting group and a catalyst in the same molecule increases the rate of a reaction compared with having them in separate molecules. When a catalyst is part of the reacting molecule, the catalysis is called intramolecular catalysis. Intramolecular general-acid or general-base catalysis, intramolecular nucleophilic catalysis, and intramolecular metal-ion catalysis are all possible. When chlorocyclohexane reacts with an aqueous solution of ethanol, an alcohol and an ether are formed. Two products are formed because there are two nucleophiles (H2O and CH3CH2OH) in the solution. Cl

OH CH3CH2OH H2O

OCH2CH3 +

+ HCl

A 2-thio-substituted chlorocyclohexane undergoes the same reaction. However, the rate of the reaction depends on whether the thio substituent is cis or trans to the chloro substituent. If it is trans, then the 2-thio-substituted compound reacts about 70,000 times

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Catalysis in Organic Reactions and in Enzymatic Reactions

faster than the unsubstituted compound. But if it is cis, the 2-thio-substituted compound reacts at about the same rate as the unsubstituted compound. this is a nucleophilic catalyst but only if it is trans to the Cl

Cl

OH SC6H5

OCH2CH3 SC6H5

SC6H5

CH3CH2OH H2O

+

+ HCl

What accounts for the much faster reaction of the trans-substituted compound? When the thio substituent is trans to the chlorine, the substituent can be an intramolecular nucleophilic catalyst—it displaces the chloro substituent by attacking the back side of the carbon to which the chloro substituent is attached (an SN2 reaction). Back-side attack requires both substituents to be in axial positions, and only the trans isomer can have both of its substituents in axial positions (Section 3.14). Subsequent attack by water or ethanol on the sulfonium ion is rapid because breaking the three-membered ring releases strain and the positively charged sulfur is an excellent leaving group. C6H5

C6H5

C6H5

S

+

C6H5

S

S

S +

Cl

+ OH

HO H + Cl−

H+

OH

H

PROBLEM 10♦

Show all the products, including their configurations, that would be obtained from the reaction illustrated in the preceding diagram.

The rate of hydrolysis of phenyl acetate is increased about 150-fold at neutral pH by putting a carboxylate ion in the ortho position. The ortho-carboxyl-substituted ester is commonly known as aspirin (Section 16.11). In the following reactions, each reactant and product is shown in the form that predominates at physiological pH (7.4). O CH3C

O O

relative rate = 1

+ H2O

O CH3C a catalyst

CH3CO− + HO O

O

relative rate ∼ 150

+ H2O

CH3CO− + HO −

OOC

OOC

The ortho-carboxylate group is an intramolecular general-base catalyst that increases the nucleophilicity of water, thereby increasing the rate of formation of the tetrahedral intermediate. O CH3C

O H

O

H

C O

O

If there are nitro groups on the benzene ring, the ortho -carboxyl substituent acts as an intramolecular nucleophilic catalyst instead of an intramolecular


Catalysis in Biological Reactions

1113

general-base catalyst . In this case, the carboxyl group increases the rate of hydrolysis by converting the ester to an anhydride, which is more rapidly hydrolyzed than an ester ( Section 16.20 ) . O2N

O CH3C

O2N

O

O−

NO2

CH3

C

O

O2N

O C O C

O

NO2 CH3C O

O

NO2

O

H2O

O2N

O CH3CO−

+ HO

C

O

NO2

OOC

O

an ester

an anhydride

P R O B L E M 1 1 Solved

Why is the ortho-carboxyl substituent a general-base catalyst in one reaction and a nucleophilic catalyst in another? Solution Because of its location, the ortho-carboxyl substituent will form a tetrahedral intermediate. If the tetrahedral intermediate’s carboxyl group is a better leaving group than its phenoxy group, then the carboxyl group will be eliminated preferentially (blue arrow). This will re-form the starting material (path A), which then will be hydrolyzed by a general-base-catalyzed mechanism. However, if the phenoxy group is a better leaving group than the carboxyl group, the phenoxy group will be eliminated (green arrow), thereby forming an anhydride (path B), and the reaction will have occurred via nucleophilic catalysis.

CH3C

O2N

O2N

O

O

NO2

A

CH3

C

O

O

O2N

O C O C

B

NO2

O

CH3C O

O

tetrahedral intermediate

O

O

O2 N NO2

H2O

C O

PROBLEM 12♦

Why do the nitro groups change the relative leaving tendencies of the carboxyl and phenoxy groups in the tetrahedral intermediate in Problem 11? PROBLEM 13

Whether the ortho-carboxyl substituent acts as an intramolecular general-base catalyst or as an intramolecular nucleophilic catalyst can be determined by carrying out the hydrolysis of aspirin with 18O-labeled water and determining whether 18O is incorporated into orthocarboxyl-substituted phenol. Explain the results that would be obtained with the two types of catalysis.

23.8 CATALYSIS IN BIOLOGICAL REACTIONS Essentially all organic reactions that occur in cells require a catalyst. Most biological catalysts are enzymes, which are globular proteins (Section 22.0). Each biological reaction is catalyzed by a different enzyme.

Binding the Substrate The reactant of an enzyme-catalyzed reaction is called a substrate. substrate

enzyme

product

CH3CO− + HO O

O

NO2 C O


1114

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Catalysis in Organic Reactions and in Enzymatic Reactions

The enzyme binds its substrate in its active site, a pocket in the cleft of the enzyme. All the bond-making and bond-breaking steps that convert the substrate to the product occur while the substrate is bound to the active site.

enzyme the substrate is bound at the active site

Enzymes differ from nonbiological catalysts in that they are specific for the substrate whose reaction they catalyze (Section 6.17). Enzymes, however, have different degrees of specificity. Some enzymes are specific for a single compound. For example, glucose6-phosphate isomerase catalyzes the isomerization of glucose-6-phosphate only. On the other hand, some enzymes catalyze the reactions of several compounds with similar structures. For instance, hexokinase catalyzes the phosphorylation of any d-hexose. The specificity of an enzyme for its substrate is another example of molecular recognition— the ability of one molecule to recognize another molecule (Section 21.0). The particular amino acid side chains (a-substituents) at the active site are responsible for the enzyme’s specificity. For example, an amino acid with a negatively charged side chain can associate with a positively charged group on the substrate, an amino acid side chain with a hydrogen-bond donor can associate with a hydrogen-bond acceptor on the substrate, and a hydrophobic amino acid side chain can associate with hydrophobic groups on the substrate. In 1894, Emil Fischer proposed the lock-and-key model to account for the specificity of an enzyme for its substrate. This model related the specificity of an enzyme for its substrate to the specificity of a lock for a correctly shaped key.

lock-and-key model

induced-fit model

In 1958, Daniel Koshland proposed the induced-fit model of substrate binding. In this model, the shape of the active site does not become completely complementary to the shape of the substrate until the enzyme has bound the substrate. The energy released as a result of binding the substrate can be used to induce a change in the conformation of the enzyme, leading to more precise binding between the substrate and the active site. An example of induced fit is shown in Figure 23.4.

Catalyzing the Reaction There is no single explanation for the remarkable catalytic ability of enzymes. Each enzyme is unique in the factors it employs to catalyze a reaction. Some of the factors most enzymes have in common are: ■

Reacting groups are brought together at the active site in the proper orientation for reaction. This is analogous to the way proper positioning of reacting groups increases the rate of an intramolecular reaction (Section 23.6). Some of the amino acid side chains of the enzyme serve as acid, base, and nucleophilic catalysts, and many enzymes also have metal ions at their active site that act


The Mechanisms for Two Enzyme-Catalyzed Reactions That Are Reminiscent of Acid-Catalyzed Amide Hydrolysis

1115

◀ Figure 23.4 The structure of hexokinase before binding its substrate is shown in red. The structure of hexokinase after binding its substrate is shown in green.

as catalysts. These species are positioned relative to the substrate precisely where they are needed for catalysis. This factor is analogous to the way intramolecular catalysis by acids, bases, nucleophiles, and metal ions enhances reaction rates (Section 23.7). Amino acid side chains can stabilize transition states and intermediates—by van der Waals interactions, electrostatic interactions, and hydrogen bonding—which makes them easier to form (Figure 23.1b).

Now we will look at the mechanisms of five enzyme-catalyzed reactions. Notice that the modes of catalysis used by enzymes are the same as the modes of catalysis used in organic reactions. Thus, if you refer back to sections referenced throughout this chapter, you will be able to see that much of the organic chemistry you have learned also applies to the reactions of compounds found in the biological world. The remarkable catalytic ability of enzymes stems in part from their ability to use several modes of catalysis in the same reaction.

23.9 THE MECHANISMS FOR TWO ENZYME-CATALYZED REACTIONS THAT ARE REMINISCENT OF ACID-CATALYZED AMIDE HYDROLYSIS The names of most enzymes end in “ase,” and the enzyme’s name tells you something about the reaction it catalyzes. For example, carboxypeptidase A catalyzes the hydrolysis of the C-terminal (carboxy-terminal) peptide bond in polypeptides, releasing the terminal amino acid (Section 22.13). O

O

C NHCH

C NHCH

R

O C NHCH

R′

O− + H2O

R′′ carboxypeptidase A

O C NHCH R

O

O C NHCH R′

O−

C

+

+ H3NCH

O−

R′′

Carboxypeptidase A is a metalloenzyme, which is an enzyme that contains a tightly bound metal ion. The metal ion in carboxypeptidase A is Zn2 + . About one-third of all


1116

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Catalysis in Organic Reactions and in Enzymatic Reactions

enzymes require a metal ion for catalysis; carboxypeptidase A is one of several hundred enzymes known to contain zinc. In bovine pancreatic carboxypeptidase A, Zn2 + is bound to the enzyme at its active site by forming a complex with His 69, Glu 72, and His 196, as well as with a water molecule. (Glu 72 means that, starting from the N-terminal end of the enzyme, Glu is the seventy-second amino acid.) It is common to specify the source of the enzyme because, although carboxypeptidase As from different sources employ the same mechanism, they have slightly different primary structures. PROPOSED MECHANISM FOR THE REACTION CATALYZED BY CARBOXYPEPTIDASE A

Arg 145

C

d+ H2N

Tyr 248

OH

O

Arg 145

d+

NH2 d−

d−

Tyr 248

O

OH

O

C base catalyst

Glu 270

C

O

H2N

O d+

N

O C

d+

this bond is hydrolyzed

HN

NH

C

d+

O

CO− +

+

H3NCHCO−

d+

Tyr 248

O

O

peptide product

OH

O

d+

H2N d+ H2N

O C

C Arg 127

Glu 270

C

NH2

d+

H2N

O

d+

HO C

C Arg 127

H2N

O d+

Zn

Zn d−

Glu 72

d+

d+

N

O C

His 69

O

CHCH2

O

N

d−

CHCH2

d+

d+

d−

O

C

− −

d+

NH2

C +NH3

O

C

d+

H2N

d−

O

the amino acid has been cleaved off the polypeptide

Arg 145

NH2

d−

OH

HN

His 196

CH2

H2N

C

NH

O

Glu 72

Arg 145

Glu 270

N

O

His 69

His 196

CH2

Tyr 248

C Arg 127

d+

d−

N

C

carboxypeptidase A

H2O

d+ H2N

d+

O

NHCHCO− +

d+

H2N

O H C O− HO Zn

Overall Reaction O O C

C

NH

O

Glu 72

His 69

Glu 270

d+

d−

N

O

metal ion

Zn

d+

O

C Arg 127

d+ H2N

H

HN

this represents the rest of the polypeptide substrate

d+

H

active site of the enzyme

d−

d−

acid catalyst CHCH2

NH C

O−

d+

NH2

C

CHCH2

O

C

d+ H2N

hydrophobic pocket

NH

HN

O

N

C His 196

His 69

d+

O− Glu 72

N O

His 196

NH


The Mechanisms for Two Enzyme-Catalyzed Reactions That Are Reminiscent of Acid-Catalyzed Amide Hydrolysis

Several amino acid side chains at the active site of carboxypeptidase A participate in binding the substrate in the optimal position for reaction. Arg 145 forms two hydrogen bonds and Tyr 248 forms one hydrogen bond with the C-terminal carboxyl group of the substrate. (In this example, the C-terminal amino acid is phenylalanine.) The side chain of the C-terminal amino acid is positioned in a hydrophobic pocket, which is why carboxypeptidase A is not active if the C-terminal amino acid is arginine or lysine (Section 22.13). Apparently, the long, positively charged side chains of these amino acids (Table 22.2) cannot fit into the nonpolar pocket. ■

When the substrate binds to the active site, Zn2 + partially complexes with the oxygen of the carbonyl group of the amide that will be hydrolyzed. Thus, Zn2 + polarizes the carbon–oxygen double bond, making the carbonyl carbon more susceptible to nucleophilic addition and stabilizing the negative charge that develops on the oxygen atom in the transition state that leads to the tetrahedral intermediate. Arg 127 also increases the carbonyl group’s electrophilicity and stabilizes the developing negative charge on the oxygen atom. In addition, Zn2 + complexes with water, making it a better nucleophile. Glu 270 is a general base-catalyst, further increasing water’s nucleophilicity. In the next step, Glu 270 is a general-acid catalyst, increasing the leaving tendency of the amino group. When the reaction is over, the amino acid (phenylalanine in this example) and the peptide with one less amino acid dissociate from the enzyme, and another molecule of substrate binds to the active site. The unfavorable electrostatic interaction between the negatively charged carboxyl group of the peptide product and the negatively charged Glu 270 side chain may facilitate the release of the product from the enzyme.

Notice that the protons are being donated and removed during (rather than before) the other bond-making and bond-breaking processes in these enzyme-catalyzed reactions, so the catalysis that occurs is general-acid and general-base catalysis (Sections 23.2 and 23.3). At physiological pH (7.4), the concentration of H + or HO - (⬃ 1 * 10 - 7 M) is too small for specific-acid or specific-base catalysis.

P R O B L E M 1 4 Solved

Which of the following amino acid side chains can aid the departure of a leaving group? O C

+

CH2CH2SCH3

CH(CH3)2

1

2

NH

CH2 N H

3

CH2

OH

4

Solution Side chains 1 and 2 do not have an acidic proton, so they cannot aid the departure of a leaving group by protonating it. Side chains 3 and 4 each have an acidic proton, so they can aid the departure of a leaving group.

PROBLEM 15♦

Which of the following amino acid side chains can help remove a proton from the a-carbon of an aldehyde? O

O

C CH2

O−

NH2 1

2

N

CH2 3

N H

C CH2 4

O−

1117


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CHAPTER 23

Catalysis in Organic Reactions and in Enzymatic Reactions PROBLEM 16♦

Which of the following C-terminal peptide bonds would be more readily cleaved by carboxypeptidase A? Explain your choice. Ser-Ala-Leu or Ser-Ala-Asp

PROBLEM 17

Carboxypeptidase A has esterase activity as well as peptidase activity, so it can hydrolyze ester bonds as well as peptide bonds. When carboxypeptidase A hydrolyzes ester bonds, Glu 270 acts as a nucleophilic catalyst instead of a general-base catalyst. Propose a mechanism for the carboxypeptidase A–catalyzed hydrolysis of an ester bond.

Trypsin, chymotrypsin, and elastase are members of a large group of endopeptidases known collectively as serine proteases. (Recall that an endopeptidase cleaves a peptide bond that is not at the end of a peptide chain; see Section 22.13). They are called proteases because they catalyze the hydrolysis of protein peptide bonds. They are called serine proteases because each one has a serine side chain at the active site that participates in the catalysis. The various serine proteases have similar primary structures, suggesting that they are evolutionarily related. Although they all have the same three catalytic side chains at the active site (that is, Asp, His, and Ser), they have one important difference—namely, the composition of the pocket at the active site that binds the side chain of the amino acid in the peptide bond that undergoes hydrolysis (Figure 23.5). This pocket is what gives the serine proteases their different specificities (Section 22.13). CHCH3 H Gly 216

H −

COO HOCH2

Asp 189

H Gly 226

H

Gly 216

Gly 226

Val 216

CH3 CH3CH

Thr 226

OH

Ser 190 trypsin

chymotrypsin

elastase

▲ Figure 23.5 The binding pockets in trypsin, chymotrypsin, and elastase. The negatively charged aspartate is shown in red, and the relatively nonpolar amino acids are shown in green. The structures of the binding pockets explain why trypsin binds long, positively charged amino acids; chymotrypsin binds flat, nonpolar amino acids; and elastase binds only small amino acids.

The pocket in trypsin is narrow and has a serine and a negatively charged aspartate carboxyl group at its bottom. The shape and charge of the binding pocket cause it to bind long, positively charged amino acid side chains (Lys and Arg). This is why trypsin hydrolyzes only peptide bonds on the C-side of arginine and lysine. The pocket in chymotrypsin is narrow and is lined with nonpolar amino acids, so chymotrypsin cleaves on the C-side of amino acids with flat, nonpolar side chains (Phe, Tyr, and Trp).


The Mechanisms for Two Enzyme-Catalyzed Reactions That Are Reminiscent of Acid-Catalyzed Amide Hydrolysis

1119

In elastase, two glycines on the sides of the pocket in trypsin and in chymotrypsin are replaced by relatively bulky valine and threonine. Consequently, only small amino acids can fit into the pocket. Elastase, therefore, hydrolyzes peptide bonds on the C-side of small amino acids (Gly, Ala, Ser, and Val). The proposed mechanism for bovine chymotrypsin-catalyzed hydrolysis of a peptide bond is shown here. The other serine proteases follow the same mechanism.

PROPOSED MECHANISM FOR THE REACTION CATALYZED BY SERINE PROTEASES Ser 195

active site of the enzyme

His 57

Asp 102

O

base HC catalyst

CH2

O−

HN

stabilizes the developing positive charge on the imidazole ring

N H

CH2 N

nucleophilic catalyst

C H NH CH CH2

His 57

Asp 102

O H N

O

C

CH2

Ser 195

CH2

Gly 193

O

HC

O− H N

O

C O− H +N

H

CH2

CH2 N

oxyanion hole

N

C NH CH CH2

H

acid catalyst

Ser 195 His 57

Asp 102

CH2

O

CH2 −

HN

O chymotrypsin

+ H2O

NH

CHC

CH2

CH

O NH2

O

O H N C

H

base catalyst

Overall Reaction

H

O N

acyl-enzyme intermediate

N

CH2

C O

CHC

HC

Gly 193

Gly 193

CH2

H

+

O− + H3N

CH2 this bond is hydrolyzed

Ser 195 His 57

Asp 102

O

CH2

+

NH3

HC CH2

CH2

O

C O− −

O

HN O

Ser 195

N

H

N H

His 57

Asp 102

H N

CH2 Gly 193

O

HC

N NH2

CH2 note: the last proton transfer is not explicitly shown

O− H N

O

C

CCH

H

CH2

CH2

O− H +N

N

H

C HO

CH CH2

Gly 193


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CHAPTER 23

Catalysis in Organic Reactions and in Enzymatic Reactions ■

As a consequence of binding the flat, nonpolar side chain in the hydrophobic pocket, the amide linkage to be hydrolyzed is positioned very close to Ser 195. His 57 is a general-base catalyst, increasing the nucleophilicity of serine, which adds to the carbonyl group. This step is helped by Asp 102, which does not remove a proton from imidazole but remains as a carboxylate ion, using its negative charge to stabilize the developing positive charge on His 57 and to position the five-membered ring so that its basic N atom is close to the OH group of serine. The stabilization of a charge by an opposite charge is called electrostatic catalysis. Formation of the tetrahedral intermediate causes a slight change in the conformation of the protein. This allows the negatively charged oxygen to slip into a previously unoccupied area of the active site known as the oxyanion hole. Once in the oxyanion hole, the negatively charged oxygen can hydrogen bond with two peptide groups (Gly 193 and Ser 195), which stabilizes the tetrahedral intermediate. In the next step, the tetrahedral intermediate collapses, eliminating the amino group. This is a strongly basic group that cannot be eliminated without the participation of His 57, which acts as a general-acid catalyst. The product of the second step is an acyl-enzyme intermediate because the serine group of the enzyme has been acylated—that is, an acyl group has been put on it. The third step is just like the first step, except that water instead of serine is the nucleophile. Water adds to the acyl group of the acyl-enzyme intermediate, with His 57 acting as a general-base catalyst to increase water’s nucleophilicity, and Asp 102 again stabilizing the positively charged histidine side chain. In the final step, the tetrahedral intermediate collapses, eliminating serine. His 57 is a general-acid catalyst in this step, increasing serine’s leaving propensity. (Because the carboxylic acid is a stronger acid than the amine, the carboxylic acid loses a proton and the amine gains a proton.)

The mechanism shows the importance of histidine as a catalytic group. Because the p K a of the imidazole ring of histidine is close to neutrality ( pK a = 6.0 ), histidine can act as both an acid catalyst and a base catalyst at physiological pH (7.4). Information about the mechanism of an enzyme-catalyzed reaction has been obtained from site-specific mutagenesis, a technique that replaces one amino acid of a protein with another. For example, when Asp 102 of chymotrypsin is replaced with Asn 102, the enzyme’s ability to bind the substrate is unchanged, but its ability to catalyze the reaction decreases to less than 0.05% of the value for the native enzyme. Clearly, Asp 102 must be involved in the catalytic process. We just saw that its role is to position histidine and use its negative charge to stabilize histidine’s positive charge. CH

CH

CH2

CH2

C O

C O−

side chain of an aspartate (Asp) residue

O

NH2

side chain of an asparagine (Asn) residue

PROBLEM 18♦

Arginine and lysine side chains fit into trypsin’s binding pocket (Figure 23.5). One of these side chains forms a direct hydrogen bond with serine and an indirect hydrogen bond (mediated through a water molecule) with aspartate. The other side chain forms direct hydrogen bonds with both serine and aspartate. Which is which?


The Mechanism for an Enzyme-Catalyzed Reaction That Involves Two Sequential SN2 Reactions PROBLEM 19

Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a d-amino acid. Trypsin, for example, cleaves on the C-side of l-Arg and l-Lys, but not on the C-side of d-Arg and d-Lys.

23.10 THE MECHANISM FOR AN ENZYME-CATALYZED REACTION THAT INVOLVES TWO SEQUENTIAL SN2 REACTIONS Lysozyme is an enzyme that destroys bacterial cell walls. These cell walls are composed of alternating N-acetylmuramic acid (NAM) and N-acetylglucosamine (NAG) units linked by b-1,4⬘-glycosidic linkages (Section 21.15). Lysozyme destroys the cell wall by catalyzing the hydrolysis of the NAM–NAG bond. lysozyme-catalyzed hydrolysis occurs here

CH2OH O

O

CH2OH O

HO NAG

CH2OH O

RO

NH C

O

NAM

O

O

O

CH3

CH2OH O

O

HO NAG

CH3

CH2OH HO OH + HO

RO

NH C

O

H2O

COO− CH2OH

NAG

NH C

CH3

R = CH3CH

O

O

HO

NH C

CH3

O

O

NAM

NH C CH3

O NAG

O

O

NH C

O

CH3

The active site of hen egg-white lysozyme binds six sugar residues of the substrate. They are labeled A, B, C, D, E, and F on page 1099 and in Figure 23.6. The many amino acid side chains involved in binding the substrate in the correct position in the active site are also shown in Figure 23.6. The carboxylic acid substituent of the RO group of NAM cannot fit into the binding site for C or E. This means that NAM units must bind at the sites for B, D, and F. Hydrolysis occurs between D and E. Lysozyme has two catalytic groups at the active site: Glu 35 and Asp 52. The discovery that the enzyme-catalyzed reaction takes place with retention of configuration at the anomeric carbon indicates that it cannot be a one-step SN2 reaction. (Recall that an SN2 reaction takes place with inversion of configuration; Section 9.1.) Therefore, the reaction must involve either two sequential SN2 reactions or an SN1 reaction with the enzyme blocking one face of an oxocarbenium ion intermediate to nucleophilic

1121


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Catalysis in Organic Reactions and in Enzymatic Reactions

attack. Although lysozyme was the first enzyme to have its mechanism studied—and it has been studied extensively for over 40 years—only recently have data been obtained that support the mechanism involving two sequential SN2 reactions.

OH HO

O

CH2OH

A

C

O

N

CH3

O

NAG

C −

O

H O

H

O CH2

RO

B NAM

H

O

N C

CH3

O O H

Trp 62

CH2 N

H

O

O

H