Conformational Analysis

Transcription

1 Conformational Analysis C01 3 C C 3 is the most stable by 0.9 kcal/mole C02 K eq = K 1-1 * K 2 = * = 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: (1) - C 3 (2) - C 3 no syn-pentane syn-pentane C06 (a) These two structures both have identical energies, and are the lowest-energy compounds. The structure on the right, though, has the incorrect () stereochemistry. The structure on the left has the proper () stereochemistry. t-u C 3 t-u C07 (a) (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 iology 27 Practice Problems olutions Page 1

2 (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 bond torsion angle C09 ecause = 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: 3 C 3 C 3 C 3 C C11 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 iology 27 Practice Problems olutions Page 2

3 3 C 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. C 3 C C14 (a) A is optically active. (b) t u t u t u t u A (c) A is more likely to form an intramolecular hydrogen bond. Chemistry and Chemical iology 27 Practice Problems olutions Page 3

4 C15 The conformation on the left possesses no A 1,3 strain, whereas the compound on the right does have A 1,3 strain between the starred atoms. Thus, the conformation on the left is more stable. ' ' ' ' * * Chemistry and Chemical iology 27 Practice Problems olutions Page 4

5 C16 C C C one 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. syn-pentane! C C C syn-pentane! A C nly 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. orleucine, 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. C C (/) (/) a syn-pentane at one of these locations no syn-pentane Chemistry and Chemical iology 27 Practice Problems olutions Page 5

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. Val-X: Leu-X: syn-pentane + 2 no 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 syn-pentane; slower + 2 no syn-pentane; faster Chemistry and Chemical iology 27 Practice Problems olutions Page 6

7 C19 (1) (2) 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. ote that there are two viewing modes, which give rise either to the pair on the left or the pair on the right. and or and Chemistry and Chemical iology 27 Practice Problems olutions Page 7

8 C21 χ3 + 3 C - χ1 χ2 C23 C24 2 C- 2 C- amino acid X Chemistry and Chemical iology 27 Practice Problems olutions Page 8

9 2 C- 2 C- amino acid Z C26 D Chemistry and Chemical iology 27 Practice Problems olutions Page 9

10 on-protease Protein Degradation D01 Pro-Val-Ala-Gly D C 6 5 D04 The amino acid whose Edman degradation results in hydantoin X is: - C 2 + The mechanism whereby hydantoin X forms is: 2 2 C hydantoin X D05 Point II tells us that the protein has no free -terminus, so it is cyclic. ow overlay the fragments: Chemistry and Chemical iology 27 Practice Problems olutions Page 10

11 Trp-e-Lys-Gln-t Tyr-Asp-t Gln-e-Ile-Ala-t e-lys-gln-t-tyr Asp-t-Gln-e Ile-Ala-t-Trp o the cyclic peptide is: --Trp-e-Lys-Gln-t-Tyr-Asp-t-Gln-e-Ile-Ala-t D06 (a) A -terminal glutamine can form a lactam. This has the same effect that, for instance, acetylating the -terminus of an amino acid has it dramatically lowers the rate because the mechanism requires that the -terminal nitrogen have two protons which can be removed. (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) (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. product of cis amino acid would-be product of trans amino acid too strained Chemistry and Chemical iology 27 Practice Problems olutions Page 11

12 D10 (a) (b) therwise, they would be unable to distinguish -Lys-Val-Asp- from -Ala-Val-Lys-. D12 (a) 2 (b) 2 PT of lysine PT of glycine D D14 (a) Chemistry and Chemical iology 27 Practice Problems olutions Page 12

13 2 (b) (c) G D I V L G I A G C 2 (d) [M + 2] 2+ (the doubly-charged full-length peptide) D15 (a) The b-ions start from the -terminus. The b 3 ion is GM for this peptide. (b) Chemistry and Chemical iology 27 Practice Problems olutions Page 13

14 (c) D16 The y 4 ion is the 4 C-terminal amino acids VWGK for this peptide. The mass of the y 4 ion includes the residue masses of V, W, G, and K and 3 for the -terminal, the C-terminal, and the additional for the positive charge. In this case, the mass is 489.3kD. For the y 5 ion, the mass is 603.3kD. Chemistry and Chemical iology 27 Practice Problems olutions Page 14

15 Protein ynthesis E01 X X Y X Y Y E02 - (a) Cy C Cy + Cy Cy ' 2 Chemistry and Chemical iology 27 Practice Problems olutions Page 15 ' Cy Cy (b) The oc group prevents the monomer being added from polymerizing with itself. All nucleophilic side chains will need to be protected (i.e. K, Y,, ). (c) + Cy C DCU ' Cy

16 E04 (a) This synthesis was attempted in the C direction, which causes epimerization / racemization. [This mechanism is now worked in lecture.] 3 C (b) 2 DCC - 3 C TFA 3 C 2 DCC - 3 C TFA 3 C 2 E05 (a) II (b) I (c) I E07 er-e-arg- 2 C - - oc C + C - oc + er-e-arg Cyc Cyc + er-e-arg 2 - C + er-e-arg TFA, F er-e-arg Chemistry and Chemical iology 27 Practice Problems olutions Page 16

17 Chemistry and Chemical iology 27 Practice Problems olutions Page 17

18 E08 (a) Peptide Peptide 2 Peptide Peptide 2 Peptide 1 Peptide 2 Peptide 1 + Peptide 2 (b) They are not a problem. The first step of the above mechanism is reversible. nly the -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 C direction, which normally causes racemization / epimerization of the sterocenter.glycine, and only glycine, is achiral, so racemization / epimerization cannot occur. (d) ormally, racemization will occur in this sort of synthesis. When proline is used though, the following intermediate is too strained to exist: 2 (ecall 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) ince it cannot exist, the racemization step cannot happen, which in turn means that there are no problems associated with this synthesis. Chemistry and Chemical iology 27 Practice Problems olutions Page 18

19 E09 (a) 1. Couple - to solid support 2. Deprotect with TFA 3. Couple to -P - with DCC 4. Deprotect with TFA 5. Couple to P - with DCC 6. Deprotect with TFA 7. Couple to -P - with DCC 8. Deprotect with TFA 9. Cleave and deprotect side chains with F (b) elow p=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: 2 (c) Cyclization via attack of the -terminal thiol group on the thioester and rearrangement to the amide (native chemical ligation). At higher concentrations, polymerization will occur. Chemistry and Chemical iology 27 Practice Problems olutions Page 19

20 E11 Peptide ta Ade 2 3 C C 3 C 3 ta 3 C C 3 C 3 ta Ade peptide E12 (a) 2 intermediate esin 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 -terminal amino group to attack the ester functionality. o other amino acid can have such a geometry because they cannot have cis bonds. E13 (a) Chemistry and Chemical iology 27 Practice Problems olutions Page 20

21 3 + C C C - reverse can happen at either face 3 + C - (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 iology 27 Practice Problems olutions Page 21

22 E14 (a) (b) Chemistry and Chemical iology 27 Practice Problems olutions Page 22