conformational analysissites.fas.harvard.edu/~chem27/psets/2006/ps2unassigned.pdf · chemistry and...

Chemistry and Chemical Biology 27 Practice Problems Solutions

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.


(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.


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.


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


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


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


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


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


NH2

COO-H

Me

MeH H

MeH

H

amino acid Z

H2N

HCOO-Me

C26

N

Me

MeMe

Me

H DMe


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:


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


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)


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)


(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.


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


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


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.


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


5. Couple to O-

ONH

O

O with DCC

NN

P


7. Couple to O-

ONH

O

O with DCC

S-P


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.


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)


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.


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

conformational analysissites.fas.harvard.edu/~chem27/psets/2006/ps2unassigned.pdf · chemistry and...

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