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Chemistry and Chemical Biology 27 Practice Problems Solutions Page 1
Conformational Analysis
C01
H
NH
CH3H3C
O H
is the most stable by 0.9 kcal/mole
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 theintermediate in the reaction of 1 does not:
OCH3 OCH3(2)(1)
HO O- HO O-
no syn-pentane syn-pentane
C06 (a) These two structures both have identical energies, and are the lowest-energy compounds. Thestructure on the right, though, has the incorrect (R) stereochemistry. The structure on the left has theproper (S) stereochemistry.
H
H
t-BuMe
H
H
CH3t-Bu
C07 (a)
H3N+H
O
HN
H
O
NH
H
O
HN
HO-
OPh
HO
-O O
(b) That is also the case in this strand — the hydrophobic residues are on one side, and thehydrophilic residues are on the other.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 2
(c) This allows all the hydrophilic residues to be exposed to solvent by placing them on the sameface which is exposed to solvent, and it also allows all the hydrophobic residues to avoid contact withthe solvent by placing them on the same face which is shielded from the solvent.
C08
0° 60° 120° 180° 240° 300° 360°
bond torsion angle
relativepotentialenergy
C09 Because R = H for glycine, in the ß-strand, there are two conformations possible for glycine, whereasall other amino acids have only one.
C10 All four of the following conformations have 5 gauche interactions:
H
COOH3N
H
Me
Me
H
COOH3N
H
Me
Me
H
COOH3N
H
Me
H
COOH3N
H
Me
Me Me
C11 Although the conformation in which threonine is found is by itself higher in energy, it allows for thehydrogen bonding as depicted below to occcur, which is sufficiently favorable to permit theconformation.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 3
NH
H3C H
O
threonine
H
O
H
C12 1 and 3 are enantiomers and therefore have identical energies. They each have two additional gaucheinteractions compared to 2, so the energy difference is ~1.8 kcal / mole.
Me
H
H CH3
H
H
H H
H
CH3
1 2 3
C14 (a) A is optically active.
(b)
tBu
OHH
HO
tBu
HtBu
HHO
HO
tBu
H
A B
(c) A is more likely to form an intramolecular hydrogen bond.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 4
C15 The conformation on the left possesses no A1,3 strain, whereas the compound on the right doeshave A1,3 strain between the starred atoms. Thus, the conformation on the left is more stable.
N
O
R'
O R
N
O
R'
O R
*
*
N
H
O
R'
O
R
N
H
O
R'
O
R
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 5
C16
Me
H
HN CO
Me
H
HN CO
H
HN CO
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 toeach of the above rotamers and then compare their overall energies.
Me
HN CO
H
Me
HN CO
H
Me
HN CO
H
A B C
Me
HMe
syn-pentane!
syn-pentane!Me
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 itmeans that a syn-pentane interaction must result. Norleucine, on the other hand, has no branching, sothere 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 fornorleucine because there are two gauche interactions, as compared to a single gauche interaction inthe lowest-energy conformation for norleucine.
H N C O
H
H H
(H/Me)(Me/H)Me
a syn-pentane at oneof these locations
H N C O
H
H H
HHMe
no syn-pentane
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 6
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.
HN
H
Me Me
HO
HN
OH2
H+
HN
H
Me Me
HOH
HN+OH2
syn-pentane
Val-X:
HN
H
H
HO
HN
OH2
H+
HN
H
H
HOH
HN+OH2
Leu-X:
Me
Me
Me
Meno 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 occurswhen the tetrahedral intermediate is formed.
HN
H
H
HO
HN
OH2
H+
HN
H
H
HOH
HN+OH2
Me
Me
Me
Meno syn-pentane;
faster
HN
H
O
HN
OH2
H+H
H
Me
HMe
HN
H
HH
Me
HMe
OH
HN
+OH2
syn-pentane;slower
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 7
C19
Me
SS
Me
Me
(1) (2) SS Me
There is a larger difference in energy between the conformers of (1) as compared to (2). The axialmethyl in (1) suffers from two gauche interactions (~1.8 kcal / mol). In the case of (2), the carbon-sulfur bonds are longer than the analogous carbon-carbon bonds in (1) (1.81 Å compared to 1.54Å). 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 eitherto the pair on the left or the pair on the right.
H
H
MeMeH
Me
Me
Me
Me
H
MeMe
H
andor
Me
HMe
H
H
MeMeMe
H
Me
Me
H
H
MeMe
Me
and
Me
HMe
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 8
C21
H3N COO-
H
Me
H
Me
H
Me
HMe
+
χ1
χ2
χ3
C23
NH
O H
HN
OMe
H
Me
H
H
Me
Me
C24
NH2
Me
H
COO-H
H
Me
H2N
HCOO-Me
H
amino acid X
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 9
NH2
COO-H
Me
MeH H
MeH
H
amino acid Z
H2N
HCOO-Me
C26
N
Me
MeMe
Me
H DMe
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 10
Non-Protease Protein Degradation
D01 Pro-Val-Ala-Gly
D02
NH2O
RN N-C6H5
S
H O+
D04 The amino acid whose Edman degradation results in hydantoin X is:
H2N
-OOC
+
H
The mechanism whereby hydantoin X forms is:
NH
N C SHPh
H NHRO
N
H NHROS
HNPh
N
SHN
PhH
ONHR
N
SHN
PhH
O H OH2
N
SHN
PhH
OO H
H HH2O H
HOH2
NH
HS
HNPh HO
OH OH2
NH
NS
PhO H
HO
HOH2
H+H2O
H
H2O
+ H2NR
N
OH2
HN
S
OH2
PhH2O
H2O
OOH2
hydantoin X
H+ H
OH2
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 11
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, acetylatingthe N-terminus of an amino acid has — it dramatically lowers the rate because the mechanism requires thatthe 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 firstcyclic intermediate would have negative charges on both oxygens of the former carboxylate group,which is unfavorable.
D09 (a)
HN
NPh
OS
(b) The cis amino acid may react to produce the expected product, as there is no unusual stericstrain. In the trans case, though, the first cyclic intermediate would have to have a trans six-memberedring, which is too sterically strained to occur.
N
S OHNPh
product of cisamino acid
would-be product of transamino acid — too strained
O
N
S
NH
Ph
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 12
D10 (a)
H3N+
O
HN
O
NH O
O-
OHN
(b) Otherwise, they would be unable to distinguish -Lys-Val-Asp- from -Ala-Val-Lys- .
D12
(a)
O NH
O
HN
O
NH
O
HN
O
NH
NH
NO
S
Ph
H2N
(b)
H2N
NH
NO
S
Ph
PTH oflysine
NH
NO
S
Ph
PTH ofglycine
D13
+H3NHN N
HNH
HN O
O
O O
O
D14 (a)
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 13
H2NHN
O
NH2
O
O OH(b)
+H3NHN N
H
HN N
H
HN OH
O OH
OH
O
O
O
O
NH
NHH2N(c)
NH2- N S G D I V N L G S I A G R – CO2H
(d) [M + 2H]2+ (the doubly-charged full-length peptide)
D15 (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 14
(c)
D16
The y4 ion is the 4 C-terminal amino acids – VWGK for this peptide. The mass of the y4 ion includes theresidue masses of V, W, G, and K and H3O for the N-terminal H, the C-terminal –OH, and the additional Hfor 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 15
Protein Synthesis
E01
O
NH
OH B
O
HN
NH O
HN
-O O
O
O
NH2
O
NH
X
O
HN
O
OH B
Y
HN
O
O-O
OH O
NHNH2
X
O
O
O
NH
O
Y
H OHB
H2NO
O-
O
HO
X
ONH
OHO
Y
+
H
HO
H OH
HOH
E02 (a)
O
O-RCN
N
Cy
CyH OH
O
N NHH
Cy Cy
O
NHR'R
DCU
+O
ORN
HN
CyH
Cy
–H+
R' NH2
O
ORN
HN
Cy
CyR'
NH
H
(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)
O
ONH
R
H+
RNH2 CO2++
O
ONH
R
H
H+
H2O
+H OH2
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 16
E04 (a) This synthesis was attempted in the N → C direction, which causes epimerization / racemization.[This mechanism is now worked in lecture.]
(b)
O
H3CONH2
Ph
O-
ONH
O
DCC
O
H3COHN
PhO
NH
O
TFA
H3CO
ONH
OPh
NH2
O-
ONH
O
DCCH3CO
ONH
OPhHN
ONH
OTFA
H3CO
ONH
OPhHN
ONH2
O
O
O
O
E05 (a) II (b) I (c) I
E07
BocHN
COO-
O
O-
NCN
H+ BocHN
COO-
O
O
N
NH
Ser-Phe-Arg-NH2
H+
Ser-Phe-Arg NH
ONH2
O O-
NCN
H+
Ser-Phe-Arg NH
O HN
O O NCyc
NHCyc
H B
H+
Ser-Phe-Arg NH
O HN
OTFA, HF Ser-Phe-Arg N
H
O HN
O
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 17
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 18
E08 (a)
H3N+
S
HN
SH
SR
O
O
Peptide 1
Peptide 2
Peptide 1 Peptide 2
O
O
HB
S
O
Peptide 1
H2NO
Peptide 2
H+
B
S
O
Peptide 1
HNO
Peptide 2
HB
H+
(b) They are not a problem. The first step of the above mechanism is reversible. Only the N-terminalamine can “trap” the final product. Thus, if an internal cysteine were to react first, it wouldequilibrate and interchange with other cysteines until, by LeChatelier’s principle, only the aboveproduct 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 / epimerizationcannot occur.
(d) Normally, racemization will occur in this sort of synthesis. When proline is used though, thefollowing intermediate is too strained to exist:
O
N
O
R2
(Recall that the carbonyl, in the absence of strain, would enolize and then de-enolize back to thecarbonyl with racemization of stereochemistry at the α-carbon) Since it cannot exist, the racemizationstep cannot happen, which in turn means that there are no problems associated with this synthesis.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 19
E09 (a)
1. Couple
2. Deprotect with TFA
S-
ONH
O
O to solid support
3. Couple to O-
ONH
O
O with DCC
O-P
4. Deprotect with TFA
5. Couple to O-
ONH
O
O with DCC
NN
P
6. Deprotect with TFA
7. Couple to O-
ONH
O
O with DCC
S-P
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 isnot nucleophilic at all when protonated, and the thiol group is only weakly nucleophilic whenprotonated.
The product formed when the product in (a) reacts with an equivalent of benzyl bromide is:
H2N NH
HN N
HS
O
O
O
OOH
NHN
HS
(c) Cyclization via attack of the N-terminal thiol group on the thioester and rearrangement to theamide (native chemical ligation). At higher concentrations, polymerization will occur.
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 20
E11
N
N
NH3C CH3
N
NOHO
OHNHO
H2N
HOCH3
OOtRNA
Ade
OHO
O
R HNPeptide
H
tRNAN
N
NH3C CH3
N
NOHO
OHNHO
HN
HOCH3
NH
ORH
peptide
OOtRNA
Ade
OHO
O
R HNH
H
E12 (a)
NO Resin
O
O NH2
Rintermediate
NO
O NH
R
reaction product
(b) Proline is the only amino acid that can form cis and trans bonds in the intermediate. The cisform possess a geometry that allows the N-terminal amino group to attack the ester functionality. Noother amino acid can have such a geometry because they cannot have cis bonds.
E13 (a)
Chemistry and Chemical Biology 27 Practice Problems Solutions Page 21
H3N+ COO-
H
NN
HB
H3N+ COO-
NHN
H3N+ COO-
H
NN
H
H3N+ COO-
NN
H
H
reverse canhappen ateither face
(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 bybeing an electron-withdrawing group and by allowing for delocalization over its aromatic system), theπ 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 22
E14 (a)
O
SH
H2N
S
O
NH
SH
H
BB H
O
SNH2SH
H
B H
H2SO
S
HN
H
B
B H
HN
SO
B
H
B
BH
(b)
+H3N
O O
HS
NH
NH3+
O
NH3+
HN
O
SH
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