Chapter 25
Amino Acids, Peptides, and Proteins
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 25. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• The ___________________ of an amino acid is the pH at which the concentration
of the zwitterionic form reaches its maximum value.
• Peptides are comprised of amino acid _________ joined by peptide bonds.
• Peptide bonds experience restricted rotation, giving rise to two possible
conformations, called _______ and _______. The _______ conformation is more
stable.
• Cysteine residues are uniquely capable of being joined to one another via
______________ bridges.
666
CHAPTER 25
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 25. The answers appear in the section entitled
SkillBuilder Review.
25.1 Determining the Predominant Form of an Amino Acid at a Specific pH
CONSIDER THE FOLLOWING AMINO ACID,
AND DRAW THE FORM THAT PREDOMINATES
AT PHYSIOLOGICAL pH.
OH
O
NH
2
H
2
N
25.2 Using the Amidomalonate Synthesis
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
25.3 Drawing a Peptide
DRAW A
BOND-LINE
STRUCTURE FOR
THE TRIPEPTIDE
Phe-Val-Trp.
CHAPTER 25
667
25.5 Planning the Synthesis of a Dipeptide
IDENTIFY ALL REAGENTS NECESSARY TO PREPARE THE DIPEPTIDE Ala-Gly:
25.6 Preparing a Peptide using the Merrifield Synthesis
Ile
Gly
Leu
Ala
Phe
POLYMER
IDENTIFY ALL REAGENTS NECESSARY TO PREPARE THE PENTAPEPTIDE Ile-Gly-Leu-Ala-Phe:
Review of Reactions
Identify the reagents necessary to achieve each of the following transformations. To
verify that your answers are correct, look in your textbook at the end of Chapter 25. The
answers appear in the section entitled Review of Reactions.
Analysis of Amino Acids
R COOH
NH
2
O
O
OH
OH
N
O
O
O
ORCHO
CO
2
H
2
O
+ +
668
CHAPTER 25
Synthesis of Amino Acids
OH
O
ROH
O
R
B
r
OH
O
R
NH
2
Analysis of Amino Acids
N
O
H
NH
2
R
NH
H
PEPTIDE PEPTIDE
+
N
NH
O
S
R
Ph
Synthesis of Peptides
OH
O
H
2
N
N
H
O
+
CHAPTER 25
669
Solutions
25.1. In each case, the chirality center has the R configuration.
25.2.
a)
OH
O
NH
2
b)
OH
O
NH
2
N
H
c)
OH
O
NH
2
S
d)
OH
O
NH
2
25.3.
a) Pro, Phe, Trp, Tyr, and His
25.4.
a)
H
3
C
NHH
O
O
b)
NH
H
O
O
c)
NHH H
O
O
HO
25.5. Arginine has a basic side chain, while asparagine does not. At a pH of 11,
arginine exists predominantly in a form in which the side chain is protonated. Therefore,
it can serve as a proton donor.
25.6. Tyrosine possesses a phenolic proton which is more readily deprotonated because
deprotonation forms a resonance-stabilized phenolate ion. In contrast, deprotonation of
670
CHAPTER 25
25.7.
a) 2.77 b) 5.98 c) 9.74 d) 6.30
25.8.
25.9. Leucine and isoleucine
25.10. The pI of Phe = 5.48, the pI of Trp = 6.11, and the pI of Leu = 6.00.
25.11.
H
O
25.12.
b)
OH
O
1) Br
2
, PBr
3
2) H
2
OOH
O
NH
2
(racemic)
3) excess NH
3
CHAPTER 25
671
25.13.
a)
COOH
NH
2
L
e
u
c
i
n
e
b)
COOH
NH
2
V
a
l
i
n
e
c)
COOH
NH
2
Phenylalanine
d)
COOHH
2
N
Glycine
25.14.
b)
EtO OEt
O O
N
O
H
OH
O
NH
2
1) NaOEt
2) CH
3
Br
3) H
3
O
+
, heat
25.15.
a)
OH
O
NH
2
a
l
a
n
i
n
e
b)
OH
O
NH
2
v
a
l
i
n
e
c)
OH
O
NH
2
l
e
u
c
i
n
e
672
CHAPTER 25
25.17.
a)
H
O1) NH
4
Cl, NaCN
2) H
3
O
+
OH
O
NH
2
S
S
c)
H
O1) NH
4
Cl, NaCN
2) H
3
O
+
OH
O
NH
2
25.18.
a)
H
3
C H
O1) NH
4
Cl, NaCN
2) H
3
O
+
OH
O
H
3
C
NH
2
a
l
a
n
i
n
e
c)
H
O1) NH
4
Cl, NaCN
2) H
3
O
+
OH
O
NH
2
v
a
l
i
n
e
25.19.
O
CHAPTER 25
673
25.20. Glycine does not possess a chirality center, so the use of a chiral catalyst is
unnecessary. Also, there is no alkene that would lead to glycine upon hydrogenation.
25.21.
a)
H
2
N
O
H
NN
H
O
OH
O
b)
H
2
N
O
H
NN
H
OH
N
O
HS
OH
O
COOH
25.23. Cys-Tyr-Leu
25.25.
25.26. Steric hindrance results from the phenyl groups:
H
2
N O
NH
OH
O
674
CHAPTER 25
25.27.
H
2
N
O
N
H
OH
O
25.28.
a)
H
2
N
O
H
NO
O
HOOC
25.29.
O
O
ONH
2
D-Glutamic
acid
L-Leucine L-Isoleucine
L-Lysine
Not naturally occurring
CHAPTER 25
675
25.30. An Edman degradation will remove the amino acid residue at the N terminus, and
25.31.
Met-Phe-Val-Ala-Tyr-Lys-Pro-Val-Ile-Leu-Arg-Trp-His-Phe-Met-Cys-Arg-Gly-Pro-
Phe-Ala-Val
25.34.
a)
Trp
(Boc)
2
O
TrpBoc
DCC TrpBoc Met OCH
3
1) CF
3
COOH
2) NaOH, H
2
OTrp Met
676
CHAPTER 25
25.35.
IleBoc
(Boc)
2
O
Ile
CHAPTER 25
677
25.36.
678
CHAPTER 25
25.37.
a)
Cl
O
O
H
N
Boc
Ph
OH
O
H
N
Boc
Ph
POLYMER
1)
2) CF
3
COOH
6) CF
3
COOH
9) HF
7)
8) CF
3
COOH
, DCC
Phe-Leu-Val-Phe
b)
Cl
O
O
H
N
Boc
POLYMER
1)
2) CF3COOH
Ala-Val-Leu-Ile
CHAPTER 25
679
25.38. (N terminus) Val-Ala–Phe (C terminus)
25.39. The regions that contain repeating glycine and/or alanine units are the most likely
25.40.
a)
NHH H
O
O
b)
NHH H
O
O
N
H
c)
NHH H
O
O
H
2
N
O
d)
NH
H
O
O
25.41. When applying the Cahn-Ingold-Prelog convention for assigning the
25.42.
C
HO O
25.43.
a) Isoleucine and threonine
b) Isoleucine = 2S,3S. Threonine = 2S,3R
25.44.
O
O
O
O
25.45. The protonated form below is highly stabilized by resonance, which spreads the
positive charge over all three nitrogen atoms.
H H
680
CHAPTER 25
25.46. The protonated form below is aromatic. In contrast, protonation of the other
25.47.
NHH H
OH
O
HO
O
NHH H
O
O
HO
O
25.48.
NHH
O
O
O
O
N
N
25.49.
a) 6.02 b) 5.41 c) 7.58 d) 3.22
25.50. Lysozyme is likely to be comprised primarily of amino acid residues that contain
25.51.
O
O
H
2
N
O
O
O
O
O
O
OOH
CHAPTER 25
681
25.52.
R
a
c
e
m
i
c
m
i
x
t
u
r
e
25.53.
The pI of Gly = 5.97, the pI of Gln = 5.65, and the pI of Asn = 5.41.
25.54.
a)
H
O
H
O
b)
H
O
c)
H
O
d) no reaction
25.55.
a) Methionine, valine, and glycine.
25.56.
H
O1) NH
4
Cl, NaCN
2) H
3
O
+
OH
O
NH
2
25.57. Alanine can be prepared via the amidomalonate synthesis with higher yields than
25.58. The side chain (R) of glycine is a hydrogen atom (H). Therefore, no alkyl group
needs to be installed at the α position.
682
CHAPTER 25
25.59.
a)
OH
O
1) Br
2
, PBr
3
2) H
2
O
3) excess NH
3
OH
O
NH
2
c)
EtO OEt
O O
N
O
H
1) NaOEt
2) CH
3
I
OH
O
NH
2
3) H
3
O
+
, heat
25.60.
a)
b)
EtO OEt
O O
N
O
H
OH
O
NH
2
1) NaOEt
2)
Br
3) H
3
O
+
, heat
25.61. 20
5
= 3,200,000
25.62.
CHAPTER 25
683
25.63.
1) Leu-Met-Val, 2) Leu-Val-Met, 3) Met-Val-Leu,
4) Met-Leu-Val, 5) Val-Met-Leu, 6) Val-Leu-Met
25.65.
H
2
N
O
H
NN
H
OH
N
O
N
H
OH
N
O
OH
O
HN OH
S COOH
N Terminus C Terminus
25.66.
25.67.
H
3
N
O
H
NO
O
O
O
25.68.
684
CHAPTER 25
25.69.
HO
HO
tyrosine
25.70. It does not react with phenyl isothiocyanate so it must not have a free N terminus.
It must be a cyclic tripeptide:
25.71.
a) Arg + Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg
b) Arg-Pro-Pro-Gly-Phe + Ser-Pro-Phe + Arg
25.72. Phenylalanine
HN
N
O
S
R =
CHAPTER 25
685
25.74. There cannot be any disulfide bridges in this peptide, because it has no cysteine
25.75. Prior to acylation, the nitrogen atom of the amino group is sufficiently
25.76.
a)
H
3
N
O
O
b)
N
HO
OHBoc
c)
H
2
N
O
O
d)
H
3
N
O
OH
25.78.
H
2
N
O
H
NOH
O
CH
3
H
2
N
O
H
NOH
OCH
3
686
CHAPTER 25
25.79.
(Boc)
2
O
Ile
Ile PheBoc
Phe
25.80.
Cl
O
O
H
N
Boc
POLYMER
1)
2) CF
3
COOH
Leu-Val-Ala
25.81.
OH
O
H
N
Boc
25.82. A proline residue cannot be part of an α helix, because it lacks an N-H proton and
CHAPTER 25
687
25.83.
OH
O
H
NO
O
OO
O
OH
O
NO
O
H H
+
25.85. The stabilized enolate ion (formed in the first step) can function as a base, rather
than a nucleophile, giving an E2 reaction:
Br
H
25.86. The lone pair on that nitrogen atom is highly delocalized via resonance and is
participating in aromaticity. Accordingly, the lone pair is not free to function as a
base.
688
CHAPTER 25
25.87.
OH
O
25.88. At low temperature, the barrier to rotation keeps the two methyl groups in
different electronic environments (one is cis to the C=O bond and the other is
25.89.
a) The COOH group does not readily undergo nucleophilic acyl substitution
because the OH group is not a good leaving group. By converting the COOH
25.90.
O
OH
2
N
HCl, H
2
O
heat
OH
O
NH
2
OH
t
h
r
e
o
n
i
n
e
OH
OH
O
H
2
N