Name_______________________________ 215 F07-Final exam Page 2 I. (15 points) Using the pKa information given on page 12, match the following isoelectric point (pI) values (3.08, 6.00, and 10.76) to their corresponding amino acids by writing the correct isoelectric point value in the box beside each amino acid.

Arg pI = Glu pI = Val pI =15

II. (18 points) Draw the full structure (including stereochemistry) for the tripeptide Cys-Pro-Arg (all L amino acids) in the predominant form it would exist at its isoelectric point. Use the Fischer projections.

Structure at the isoelectric point: The predominant form of this tripeptide at pH 1has a net charge of: (circle one)

+2 +1 0 -1 -2

8

5

The predominant form of this tripeptide at pH 11has a net charge of: (circle one)

+2 +1 0 -1 -25

III. (20 points) Draw in the boxes below the full structures (including stereochemistry) of the products from each of the following reactions.

H3N NH

++

+H

O

O

HO

NO2

FO2N

N(CH2CH3)3

1.

2. HCl, H2O, Δ

6 4HCl saltNitro-containing product

HN OO

O H

ONH

CO2dimethyl-

formamide(solvent)

C14H10

(2)

55

(1)

Name______________________________ 215 F07-Final exam Page 3 IV. (20 points) Consider the esterification reaction (A to B) shown below.

+ +Ph OH

OHOCH(CH3)2 Ph

O CH(CH3)2O

H2OHCl

A B (1) (16 points) Which of the following would result in an increase in the amount of ester B formed?

If the following is done:

Yield of B would increase (circle one)

Excess HOCH(CH3)2 is added.

Yes No

NaOCH(CH3)2 is used in place of HOCH(CH3)2

Yes No

HOCH(CH3)2 is removed during the reaction

Yes No

Dilute aqueous HCl is used instead of concentrated HCl

Yes No

If isotopically labeledis used, the 18O label would end up in :

H-18OCH(CH3)2 (circle one)

B H2O4

(2)

V. (32 points) Treatment of L-phenylalanine with excess methanol in the presence of 1.5 mol equiv of gaseous HCl at room temperature produces the HCl salt of L-phenylalanine methyl ester in 95% yield. Provide in the box below a step-by-step mechanism, using the curved arrow convention, for the formation of this methyl ester HCl salt from L-phenylalanine. You may use H-B and B- as a general acid and its conjugate base, respectively.

Ph ONH3

O

Ph OCH3NH3

OHCl (gas, 1.5 mol equiv)

HOCH3room teperature, 5 hr 95%Cl

Ph ONH3

OH B

Ph OCH3NH3

O

B32

Name_______________________________ 215 F07-Final exam Page 4

VI. (18 points) A levulinyl group[CH3C(=O)CH2CH2C(=O)-] is an extremely versatile hydroxyl-protecting group. A team of scientists at The Scripps Research Institute in La Jolla, CA, reported in 2003 that the levulinate group (indicated by a rectangular box) in differentially protected disaccharide C can be deprotected selectively to provide alcohol D in quantitative yield [Proc. Natl. Acad. Sci., USA 2003, 100, 797].

O OO

O

O

O O O

SNHO

OO

OO

OOCCl3

O OO

O

O

O O O

SNHOH

OO

OOCCl3C D

(1) The methods used to achieve this selective deprotection take advantage of the difference in electrophilicity between aldehyde/ketone and ester carbonyl groups. Among these methods known, the use of NaBH4 is most convenient. In the reaction shown below, the levulinate ester of cyclohexanol, E, is treated with NaBH4 to afford cyclohexanol (F). The by-product (G) from this reaction has the molecular formula of C5H8O2. Provide in the box below the structure of this compound.

O

O

O

E

HO

6

+NaBH4CH3OH

0 °C

G (C5H8O2)

F

(2) When a levulinate such as E is treated with hydrazine (NH2-NH2) in acetic acid, the levulinate-protected alcohol undergoes facile deprotection (see below). Draw in the box provided the structure of the by-product H. It might be easier if you solve # (3) before this question.

O

O

O

E

HO

6

+CH3COOH

room temperature 5 min

H (C5H8N2O)

NH2-NH2

F

(3) The reaction of levulinate E with hydrazine first produces its hydrazone derivative, which further undergoes an intramolecular reaction to produce H. Draw in the box below the structure of this hydrazone derivative of E.

6

Name_______________________________ 215 F07-Final exam Page 5 VII. (22 points) Provide in the box below a step-by-step mechanism for the following reaction using the curved arrow convention. Use B- and B-H for the base and the conjugate acid, respectively. Make sure to show the mechanism for the catalytic cycle, i.e., regeneration of B- [or –OC(CH3)3]. You do not need to rationalize the stereochemistry of the new chiral centers created [Tetrahedron Lett. 2007, 48, 4683].

OCH2CH3

OOH

Ph O

HOC(CH3)3K

(catalytic)OCH2CH3

OOO

Ph

tetrahydrofuran (solvent) 69%

22

OCH2CH3

OOH :B

OCH2CH3

OOO

Ph

VIII. (22 points) Provide in the box below a step-by-step mechanism for the following acid-catalyzed reaction using the curved arrow convention. Use H-B and B- for the acid and the conjugate base, respectively.

22

HB

p-TsOH (catalytic)O

O

H H HO

O

O

O

H H

HO

O

acetone (solvent)66%

Name_______________________________ 215 F07-Final exam Page 6 IX. (39 points) Complete the following transformations as indicated by the data provided. Clearly indicate the stereochemistry where applicable.

HH3NCH2OH

O OCl O

H2C Ph

O

N(CH2CH3)3

H3O+ to pH ~1

Which amino acid is this? Use its 3-letterabbreviation as your answer.

answer4

1.

2.

,

N H

O O CH2

PhO

O

N H

OHN H

O OH

HH

3 with added DCC 5

4

5

(1)

(2)

(3) [J. Org. Chem. 2004, 69, 1038]

PhCH2 O

LiCu(CH3)2THF (solvent)

1. n-BuLi THF (solvent)

Cl OCH3

O2.

3

1.

2. aq NH4Cl

4C14H18O3 (Z-isomer)(4) [J. Org. Chem. 2007, 72, 9541]

OAcOAcO

OAcOAc

O OCH3HBr (gas)

CH2Cl2(solvent)

Cs

4

SN2

α-anomer C20H22O11

Ac =H3C

O+ CsBr

3

PhO

O

(acetyl) 4

Name _______________________________ 215 F07-Final exam Page 7 X. (34 points) Complete the following transformations as indicated by the data provided. Clearly indicate the stereochemistry where applicable. Sequential experimental steps must be numbered.

+ +

++

(2) [Tetrahedron 2007, 63, 5855]

O

CH2OO

NH

(2 mol equiv)

C12H17NO34

1. KH THF (solvent)2. PhCH2Br

4+ KBr(3) [J. Org. Chem. 2007, 72, 5943]

HOHO

H

H

O OH3CH3C

OCH3

OCH3 H

H

p-TsOHtoluene, Δ

2 HOCH3

5C8H18O4

(4) [Tetrahedron Lett. 2007, 48, 5623]

PhCH2O O

OH O2 LDA

THF(solvent)-30 °C 4

dianion5

diastereomeric mixture at a new chiral center (show that stereochemistry with

CH2CH3

OH

H

1.

(1 mol equiv)

2.aq NH4Cl

4 4

OCH3

OCH3H3CH3C2 HOCH3

diastereomers

p-TsOH

)

C9H18OS24

O H

S

SHLi

(1)

THF (solvent)

1.

2. aq NH4Cl

The Reagent List Shown below is a list of key reagents (not always the whole recipe) that may be useful for solving questions on the final exam. reagent classification or specialized use ____________________________________________________________________________________ From OsO4 oxidation Chem 210 KMnO4 oxidation peroxyacid epoxidation (e.g., 3- or meta-chloroperoxybenzoic acid) i. O3; ii. (CH3)2S or Zn ozonolysis i. O3; ii. H2O2 ozonolysis NaNH2 base, nucleophile NaH base KOC(CH3)3 bulky base H2/Pd hydrogenation H2/Pd, BaSO4, quinoline hydrogenation i. BH3 or 9-BBN; ii. H2O2, NaOH hydroboration PBr3 e.g., R-OH → R-Br SOCl2 e.g., R-OH → R-Cl

p-CH3C6H4SO2Cl (p-TsCl) tosylate formation CH3SO2Cl (MsCl) mesylate formation N-bromosuccinimide (NBS)

N

O

O

Br

source of electrophilic Br =========================================================================== Ch. 13 C5H5NH+•ClCrO3- (PCC) oxidant CrO3/H2SO4/H2O/acetone oxidant i. ClC(=O)-C(=O)Cl, (CH3)2SO; ii. N(CH2CH3)3 oxidant Ch. 14 NaBH4 nucleophilic hydride LiAlH4 nucleophilic hydride diisobutylaluminum hydride (DIBAL) nucleophilic hydride RMgX nucleophilic carbon RLi nucelophilic carbon 4-CH3C6H4SO3H (p-TsOH) (pKa ~ -1) organic-soluble acid Raney Ni desulfurization HSCH2CH2SH; HSCH3 thioacetal/thioketal formation BF3•O(CH2CH3)2 Lewis acid H2NNH2 hydrazone formation H2NOH oxime formation

Ch. 15 SOCl2 R-C(=O)OH → R-C(=O)Cl Ag2O, NaOH, H2O2 oxidant -CH2 +N2 (diazomethane) e.g., R-C(=O)OH → R-C(=O)OCH3 Ch. 17 LiN[CH(CH3)2] (LDA) bulky base K2CO3 weak base Amino acids & proteins C6H11N=C=NC6H11 RC(=O)OH activation (DCC; dicyclohexylcarbodiimide) PhCH2C(=O)Cl (benzyl chloroformate) N-protection ROC(=O)N3 (R = tert-butyl) N-protection [tert-butyl-O-C(=O)-]2O (Boc2O) N-protection

H3CCO O O

C

CH3

H3C

H3CO O CH3

CH3

(9-fluoromethyl)methoxycarbonyl chloride N-protection (Fmoc)

CH2

OCl

O

H2, Pd/C catalytic hydrogenolysis Sanger’s reagent N-terminus AA determination

NO2

FO2N