Conformational Analysis
C01
O
H
H
N
is the most stable by 0.9 kcal/mole
H 3C
H
C H3
C02
Keq = K1 -1 * K2 = 0.45-1 * 0.048 = 0.11
C04
The intermediate in the reaction of 2 has an unfavorable syn-pentane interaction, whereas the
intermediate in the reaction of 1 does not:
HO
O-
(1)
HO O-
(2)
OCH3
OCH3
no syn-pentane
C06
syn-pentane
(a) These two structures both have identical energies, and are the lowest-energy compounds. The
structure on the right, though, has the incorrect (R) stereochemistry. The structure on the left has the
proper (S) stereochemistry.
H
H
t-Bu
CH3
H
C07
Me
t-Bu
H
(a)
Ph
H
O
H
N
H3N +
O
H
H
N
H
O
H
N
O
OH
HO
-O
O
(b) That is also the case in this strand — the hydrophobic residues are on one side, and the
hydrophilic residues are on the other.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 1
(c) This allows all the hydrophilic residues to be exposed to solvent by placing them on the same
face which is exposed to solvent, and it also allows all the hydrophobic residues to avoid contact with
the solvent by placing them on the same face which is shielded from the solvent.
C08
relative
potential
energy
0°
60°
120°
180°
240°
300°
360°
bond torsion angle
C09
Because R = H for glycine, in the ß-strand, there are two conformations possible for glycine, whereas
all other amino acids have only one.
C10
All four of the following conformations have 5 gauche interactions:
H
H
H3N
Me
COO
Me H3N
H
H
COO
H 3N
Me
COO
Me H3 N
H
H
Me
Me
H
H
Me
C11
COO
Me
Although the conformation in which threonine is found is by itself higher in energy, it allows for the
hydrogen bonding as depicted below to occcur, which is sufficiently favorable to permit the
conformation.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 2
H
N
H
O
H
H 3C
H
O
threonine
C12
1 and 3 are enantiomers and therefore have identical energies. They each have two additional gauche
interactions compared to 2, so the energy difference is ~1.8 kcal / mole.
H
CH 3
H
H
H
CH 3
H
H
Me
3
2
1
C14
H
(a) A is optically active.
(b)
tBu
HO
HO
tBu
H
H
H
HO
OH
H
tBu
tBu
A
B
(c) A is more likely to form an intramolecular hydrogen bond.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 3
C15
The conformation on the left possesses no A1,3 strain, whereas the compound on the right does
have A1,3 strain between the starred atoms. Thus, the conformation on the left is more stable.
R'
O
R
O
N
O
N
R'
R
O
H
R'
O
R
O
*
N
O
N
R'
R
O
*
H
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 4
C16
H
HN
H
CO
HN
H
CO
HN
CO
Me
Me
Me
None of these conformations will predominate in solution as all have 4 gauche interactions.
To determine the lowest-energy rotamer of the other amino acid, we can graft on a methyl group to
each of the above rotamers and then compare their overall energies.
H
H
H
syn-pentane!
HN
CO
HN
Me
CO
Me
Me
CO
HN
Me
H
Me
Me
syn-pentane!
A
B
C
Only B avoids a syn-pentane interaction, so it is the lowest-energy conformer.
C17
Leucine cannot have a carbon in the forbidden position, because the substitution on the carbon after it
means that a syn-pentane interaction must result. Norleucine, on the other hand, has no branching, so
there is no possibility for syn-pentane interactions. This means that for norleucine, unlike leucine,
there is a minor population with a carbon in the forbidden position. The amount is still minor for
norleucine because there are two gauche interactions, as compared to a single gauche interaction in
the lowest-energy conformation for norleucine.
H
HN
H
(Me/H)
Me
H
CO
H
(H/Me)
a syn-pentane at one
of these locations
HN
CO
H
H
H
Me
H
no syn-pentane
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 5
C18
(a) Formation of the tetrahedral intermediate creates a syn-pentane in the case of Val-X hydrolysis.
This is avoided in the case of Leu-X by placing the α-carbon in the “upper left” position.
Val-X:
Leu-X:
OH2
H
H
N
HN
Me
Me
H
H+
H
H
N
+ OH
2
Me
Me
Me
H
H
N
+ OH
2
HN
H
OH
O
H
H+
HN
H
N
HN
H
Me
H
Me
Me
O
OH2
H
OH
H
no syn-pentane
syn-pentane
(b) Any β-branched amino acid will necessarily face this syn-pentane problem. In the case of leucine
(not β-branched), if the a carbon is placed on the carboxyl side, a syn-pentane interactions occurs
when the tetrahedral intermediate is formed.
OH2
H
HN
H
N
Me
H
Me
Me
H+
H
N
HN
H
O
H
OH2
H
H
O
H+
H
Me
H
HN
H
H
Me
H
N
H
+ OH
Me
2
Me
OH
H
syn-pentane;
slower
HN
H
Me
H
H
N
+ OH
2
OH
no syn-pentane;
faster
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 6
C19
Me
Me
(1)
(2)
S
S
Me
S
Me
S
There is a larger difference in energy between the conformers of (1) as compared to (2). The axial
methyl in (1) suffers from two gauche interactions (~1.8 kcal / mol). In the case of (2), the carbonsulfur bonds are longer than the analogous carbon-carbon bonds in (1) (1.81 Å compared to 1.54
Å). Thus, the gauche interactions are less severe— note the analogy to methionine.
C20
The two conformers are enantiomeric. Note that there are two viewing modes, which give rise either
to the pair on the left or the pair on the right.
H
H
Me
Me
Me
Me
Me
H
H
Me
Me
H
Me
Me
Me
H
or
and
and
H
H
Me
Me
Me
Me
Me
Me
H
H
Me
Me
H
H
Me
Me
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 7
C21
χ3
H
Me
+
COO-
H3 N
Me
Me
H
H
H
Me
χ1
χ2
C23
O
H
H
N
N
H
Me
O
Me
H
H
H
Me
Me
C24
Me
H2 H
N
H
COO-
H
Me
H
N
H2
COOH
Me
amino acid X
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 8
Me
H2 H
N
H
COOMe
Me
COOH
N
H2
H
H
Me
H
amino acid Z
C26
Me
H
D
Me
Me
N
Me
Me
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 9
Non-Protease Protein Degradation
D01
Pro-Val-Ala-Gly
D02
S
O
NH2
N
+
R
H
N-C6H5
O
D04 The amino acid whose Edman degradation results in hydantoin X is:
-
OOC H
H2 N+
The mechanism whereby hydantoin X forms is:
OH2
+
H
O
H
H 2O
H
S
NHR
HN
Ph
N
Ph
N C S
H
O
O
H
NHR
N
Ph
H
OH2
S
NHR +
H
H
N
OH2
H2O
HO
Ph H
N
Ph
OH2
O
H
N
S
OH2
H2 O
O H
H
S
H
N
N
OH2
H
H
H2O
Ph O
N
O H
H S
H
N
H
H2O H
H OH2
O
H
HS
H
N
H
Ph
N
H
N
Ph
S
O
H
N
H OH2
+
H2NR
O
H
N
hydantoin X
D05
Point II tells us that the protein has no free N-terminus, so it is cyclic. Now overlay the fragments:
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 10
Trp-Phe-Lys-Gln-Met Tyr-Asp-Met Gln-Phe-Ile-Ala-Met
Phe-Lys-Gln-Met-Tyr Asp-Met-Gln-Phe Ile-Ala-Met-Trp
So the cyclic peptide is:
--Trp-Phe-Lys-Gln-Met-Tyr-Asp-Met-Gln-Phe-Ile-Ala-Met-|
|
------------------------------------------------------D06 (a) A N-terminal glutamine can form a lactam. This has the same effect that, for instance, acetylating
the N-terminus of an amino acid has — it dramatically lowers the rate because the mechanism requires that
the N-terminal nitrogen have two protons which can be removed.
HN
O
(b) Acid treatment will open the lactam, regenerating the free carboxylate and the free amino
groups.
(c) Glutamic acid should not have this problem, because it lacks a good leaving group. Also, the first
cyclic intermediate would have negative charges on both oxygens of the former carboxylate group,
which is unfavorable.
D09
(a)
S
Ph
N
O
HN
(b) The cis amino acid may react to produce the expected product, as there is no unusual steric
strain. In the trans case, though, the first cyclic intermediate would have to have a trans six-membered
ring, which is too sterically strained to occur.
Ph
HN
O
S
O
N
product of cis
amino acid
S
Ph
N
H
N
would-be product of trans
amino acid — too strained
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 11
D10
(a)
O
H
N
H
N
H3 N+
O
N
H
O
OO
(b) Otherwise, they would be unable to distinguish -Lys-Val-Asp- from -Ala-Val-Lys- .
D12
(a)
Ph
O
N
S
N
H
H2N
N
H
O
O
H
N
N
H
O
H
N
O
N
H
O
(b)
Ph
O
N
Ph
S
O
N
H
N
S
N
H
PTH of
lysine
H 2N
PTH of
glycine
D13
H
N
+H 3N
O
D14
O
O
N
H
N
H
H
N
O
O
(a)
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 12
NH2
O
H
N
H2N
O
O
OH
(b)
H
+H
H
N
3N
O
O
N
H
OH
H
N
O
O
N
H
H
N
O
OH
O
NH
H 2N
(c)
(d)
D15
NH
NH2 - N S G D I V N L G S I A G R – CO2 H
[M + 2H]2+ (the doubly-charged full-length peptide)
(a) The b-ions start from the N-terminus. The b3 ion is RGM for this peptide.
(b)
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 13
(c)
D16
The y4 ion is the 4 C-terminal amino acids – VWGK for this peptide. The mass of the y4 ion includes the
residue masses of V, W, G, and K and H3 O for the N-terminal H, the C-terminal –OH, and the additional H
for the positive charge. In this case, the mass is 489.3kD. For the y5 ion, the mass is 603.3kD.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 14
Protein Synthesis
E01
B
H
O
-
O O
O
N
H
H
N
N
H
O
H
N
O
O
H
O
H
N
O
NH 2
HO
X
H OH
O
O
N
H
O
H
N
-
O O
O
H
B
B
H OH
Y
HO
OH
O-
H2N
O
O
+
X
HO
O
OH
O
H
O
NH2
O
N
H
O
X
O
N
H
Y
O
N
H
Y
E02
(a)
O
Cy
O
R
O
-
H OH
N
C
N
N
R
–H+
H
H
O
O
Cy
O
R
HN
Cy
N
+
O
NH
Cy
R
Cy
HN
N
H
R'
O
Cy
Cy
R'
R' NH 2
N
H
N
H
Cy
DCU
(b) The Boc group prevents the monomer being added from polymerizing with itself.
All nucleophilic side chains will need to be protected (i.e. K, Y, R, H).
(c)
H+
R
N
H
H2O
O
H
H
O
R
O
N
H
O
+
OH2
RNH 2
+
+
H
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 15
+
CO2
E04 (a) This synthesis was attempted in the N → C direction, which causes epimerization / racemization.
[This mechanism is now worked in lecture.]
(b)
O
O
O
NH2
H3CO
O-
N
H
Ph
O
H3CO
O
O
H
N
N
H
O
Ph
DCC
O
TFA
O
O
Ph
O
H3 CO
NH2
N
H
O
E05
H3CO
DCC
H
N
N
H
O
Ph
O
O
O
H
N
N
H
O
Ph
O
H3 CO
O-
N
H
TFA
N
H
O
O
NH2
O
(a) II
(b) I
(c) I
E07
Ser-Phe-Arg-NH2
COO-
COON
C
N
O-
BocHN
O
O
Ser-Phe-Arg
O
NH
N
H+
O
H
N H
N
H
O
BocHN
H+
B
Ser-Phe-Arg
NH2
N
H
NHCyc
O
O
NCyc
O
N
C
N
O-
H+
O
Ser-Phe-Arg
N
H
H
N
O
O
TFA, HF
Ser-Phe-Arg
N
H
H
N
O
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 16
H+
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 17
E08
(a)
O
H3
Peptide 1
B
N+
Peptide 2
S
H
SR
H
H2 N +
Peptide 1
B
O
Peptide 2
S
O
O
B
H
N
Peptide 1
O
H
HN
O
O
Peptide 2
Peptide 2
Peptide 1
S
SH
H+
O
(b) They are not a problem. The first step of the above mechanism is reversible. Only the N-terminal
amine can “trap” the final product. Thus, if an internal cysteine were to react first, it would
equilibrate and interchange with other cysteines until, by LeChatelier’s principle, only the above
product is formed
(c) This synthesis is being carried out in the N → C direction, which normally causes racemization /
epimerization of the sterocenter.Glycine, and only glycine, is achiral, so racemization / epimerization
cannot occur.
(d) Normally, racemization will occur in this sort of synthesis. When proline is used though, the
following intermediate is too strained to exist:
R2
N
O
O
(Recall that the carbonyl, in the absence of strain, would enolize and then de-enolize back to the
carbonyl with racemization of stereochemistry at the α-carbon) Since it cannot exist, the racemization
step cannot happen, which in turn means that there are no problems associated with this synthesis.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 18
E09
(a)
O
1. Couple
O
S-
N
H
to solid support
O
2. Deprotect with TFA
O-P
O
3. Couple to
O
O-
N
H
4. Deprotect with TFA
O
P
N
N
O
5. Couple to
O
with DCC
O-
N
H
with DCC
O
6. Deprotect with TFA
S-P
O
7. Couple to
O
O-
N
H
with DCC
O
8. Deprotect with TFA
9. Cleave and deprotect side chains with HF
(b) Below pH=6, both the amine group and the thiol group will be protonated. The amine group is
not nucleophilic at all when protonated, and the thiol group is only weakly nucleophilic when
protonated.
The product formed when the product in (a) reacts with an equivalent of benzyl bromide is:
HN
O
H2 N
HS
N
H
N
H
N
O
O
N
H
OH
S
O
(c) Cyclization via attack of the N-terminal thiol group on the thioester and rearrangement to the
amide (native chemical ligation). At higher concentrations, polymerization will occur.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 19
E11
H3 C
tRNA
O
Ade
O
N
HO
O
N
H
N
CH3
N
N
H3 C
N
N
tRNA
HO
tRNA
O
Ade
O
N
N
O
CH3
N
R H
H
R H
Peptide
N
NH OH
O OH
O
OCH3
O
NH OH
O
H
H2 N
OCH3
N
H
O OH
O
H
HN
O
H
R
NH
peptide
E12
(a)
O
N
O
NH2
O
Resin
O
N
NH
O
R
R
intermediate
reaction product
(b) Proline is the only amino acid that can form cis and trans bonds in the intermediate. The cis
form possess a geometry that allows the N-terminal amino group to attack the ester functionality. No
other amino acid can have such a geometry because they cannot have cis bonds.
E13
(a)
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 20
H
N
N
H
+
H3N
B
COO
-
H
N
N
N
NH
reverse can
happen at
either face
H
H3 N+
COO-
H 3N+
COO-
H
N
N
H
+
H3N
COO-
(b) The protecting group cannot be abstracted by base, so the mechanism shown above cannot occur.
Also, because the protecting group will decrease electron density around the π nitrogen (both by
being an electron-withdrawing group and by allowing for delocalization over its aromatic system), the
π nitrogen is unlikely to abstract the proton even without the effect of the now-protected τ nitrogen.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 21
E14
(a)
B
H
B
B
O
H
O
H2S
H
O
SH
S
B
S
SH
H
NH 2
HN
H
S
B
H
B
H2N
B
H
SH
O
H
O
N
H
B
S
HN
(b)
H
N
O
HS
O
NH3+
+H3 N
O
SH
O
N
H
NH3+
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 22