(1) (16 points)

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I. (42 points) (1) (16 points) The titration curve for L-lysine is shown below. Provide (i) the main structures in equilibrium at each of points A and B indicated below ...
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215 F10-Final Exam

Page 2

I. (42 points) (1) (16 points) The titration curve for L-lysine is shown below. Provide (i) the main structures in equilibrium at each of points A and B indicated below and (ii) the structure at the isoelectric point. Draw the structures using the Fischer projections.

(i) Structures in equilibrium at point A: O

O

OH

H3N

H 3N

O

O

H 3N

H

O

O

H3N

H

H3N

(ii) Structure at the isoelectric

Structures in equilibrium at point B:

H

O

O

H2N

H 3N

6

point:

H

H 3N

O

H 2N

H

H3N

6

4

(2) (16 points) Draw the full structure (including stereochemistry) for the tri-peptide Tyr-Phe-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 1 has a net charge of: (circle one)

O O O H2N

H N H

H N H

O

+2

H

+1

0

-1

-2 4

NH2 N H

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

NH2

+2 OH

+1

0

-1

-2

8

4

(3) (10 points) Complete the following reaction sequence by providing in each of the boxes the structure of the product [J. Org. Chem. 2010, 75, 3027]. O Ph

O O N O

O

O

2. N(CH2CH3)3

Ph

O

1. 4M aq HCl

O Ph

O

N

O

HO

O

N

DCC

NH

O

O N

O 5 + CO2 +

5 +

H N

H N O

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215 F10-Final Exam Page 3

II. (42 points) (1) (16 points) Treatment of enone A with NaOH/H2O results in the formation of cyclohexanone (C) and acetone (D) via β-hydroxy ketone intermediate B. Provide in the box below a stepby-step mechanism using the curved arrow convention for the following transformation. You do not need to show Na+ for each step. OH NaOH/H2O

NaOH/H2O O

A

O

B

O

D

Mechanism for B -> C and D:

Na OH

O H

B

O OH

HO

A

O

C

Mechanism for A -> B:

O

+

H

O

OH

Na

O

O 2 pts for the intermediate 2 pts for each set of mech. arrows

B

O

O D

H OH

OH

6

C

O

2 pts for each intermediate 2 pts for each set of mech. arrows

O 10

(2) (26 points) Provide in the box below a step-by-step mechanism using the curved arrow convention for the following transformation. You do not need to show Na+ for each step. O NaOCH2CH3

CH3CH2O O

O

HOCH2CH3

Na

Mechanism: O

O

+ enantiomer

OCH2CH3

O

CH3CH2O

CH3CH2O O

O

O OCH2CH3

OCH2CH3

OCH2CH3

CH3CH2O O

Na

OCH2CH3

O O

CH3CH2O H

CH3CH2O O

H

OCH2CH3

CH3CH2O

O

H OCH2CH3

2 pts for each intermediate 2 pts for each set of mech. arrows

O CH3CH2O

O O

OCH2CH3

O

OCH2CH3 H OCH2CH3

O Na

O

OCH2CH3 26

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215 F10-Final Exam Page 4

III. (56 points) Provide in the boxes below a step-by-step mechanism using the curved arrow convention for each of the following transformations. You may use H-A as your Brønsted acid and A- as its conjugate base. (1) (26 points) O

H N O 2N

OH

N H

HCl/H2O

O

NO2

O

H N O2N

OH

+

H N H H

Cl

NO2

OH O

Mechanism: H A H N O2N

H N

O O2N

O

NO2

O2N

NO2

OH

OH2

O

O2N

NO2

OH

N H

O

A H

O

OH H NO2

H

H

H

H

A H N

OH

NO2

OH

N

H

O

H N O2N

N

O

A

OH H OH

O

OH

N H

H A

OH

H N

2 pts for each intermediate 2 pts for each set of mech. arrows

O2N

NO2

OH

H N

OH

N H

OH

N H

H N

H

O

O2N H

Cl

O

H

OH

NO2

OH

N H

O

O

26

(2) (30 points) [Org. Lett. 2010, 12, 904] OCH3 HO

OCH3

OH OH

HO

O

TsOH (catalytic)

+ 2 CH3OH

O

Mechanism: H

H A

OCH3

OCH3

HO

H

A

OH

OCH3

OCH3

O

CH3O HO

OH

HO

O OH

OCH3

CH3O

OH HO

H A

O OH

CH3 O H HO

O OH HO

O O

2 pts for each intermediate 2 pts for each set of mech. arrows

H A

HO

O O 30

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215 F10-Exam Exam

Page 5

IV. (38 points) Shown below is a trisaccharide recently isolated from Securidaca welwitschii [Helv. Chim. Acta 2010, 93, 2237]. Answer the following questions about this trisaccharide, 1.

HO

A O

HO

OH O

α O

HO

O O

HO

C

OH

β

B

HO

1

OH

(1) (4 points) Label each of the glycosidic bonds in trisaccharide 1 using a small arrow ( → ). (2) (4 points) Classify each glycosidic linkage in 1 as α or β right next to each of the arrows you have shown. (3) (3 points) How many D-sugar units does trisaccharide 1 have? (circle one) 0 1 2 3 (4) (3 points) Is the hemi-acetal OH in trisaccharide 1 α or β?

(circle one)

(5) (3 points) Is trisaccharide 1 a reducing sugar? (circle one)

Yes

α

β

No

(6) (4 points) Describe the linkages between the following sugar units [e.g., α(1 -> 6)]: For A and B: α(1 -> 2)

For B and C: β(1 -> 4)

(7) (17 points) Draw Fischer projections for the open-chain forms of the three carbohydrates produced when trisaccharide 1 is hydrolyzed with a warm dilute aqueous acidic solution. Carbohydrate A

O

Carbohydrate B

O

H

H

H

O

OH

H

OH

HO

H

H

OH

HO

H

HO

H

OH

HO

H

H CH3

Carbohydrate C

H HO H

OH H OH CH2OH

CH3 6

H

6

5

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215 F10-Final Exam

Page 6

V. (32 points) Complete the following reactions by providing in each of the boxes the structure of the reagent, intermediate, or product. Indicate stereochemistry for the product/intermediate and if more than one stereoisomer is formed, draw one structure and write “+ enantiomer” or “+ diastereomer.” (1) [Synthesis 2010, 3004] OCH3

O HO

O

HO

HO

TsOH (catalytic)

O + CH3OH

OH

OH 5

(2) O O Ph

α-anomer

Ph

O

O

O

HBr (gas)

O O

O

Ph

O O

Ph

Ph

CH2Cl2 (solvent)

O Ph

O

O O

Ph

NH2

O

(1 equiv)

O O

Ph

K2CO3

O Br Ph O

O

S N2

5

Ph

O

HO HO

4x

+ NH

O

NaOCH3 (catalytic)

O

OH O

OCH3

Ph

Ph

CH3OH (sovent)

OH

O

O

O O

NH

Ph

O O

O

Ph

5

5

(3) [Org. Lett. 2010, 12, 904] O

H

H

K

O

O

H

H

H

+ KCl

OCH3

OCH3 Cl

H

+ diastereomer

4

(4) [J. Am. Chem. Soc. 2010, 132, 14064]

HO

O

LDA (2 equiv) OCH2CH3

THF, -78 °C

Li

O

O

1. CH3I (1 equiv)

Li OCH2CH3

dilithium dianion

2. 4

Si

O

O OCH2CH3

Cl Si

+ diastereomer + LiI + LiCl

4

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215 F10-Final Exam

Page 7

VI. (30 points) Complete the following reactions by providing in each of the boxes the structure of the product or intermediate. Indicate stereochemistry for the product/intermediate and if more than one stereoisomer is formed, draw one structure and write “+ enantiomer” or “+ diastereomer.” (1) [Synlett 2010, 3011] O N O

O N O

K2CO3

OCH2CH3

trans!

O

O

OCH2CH3

+ enantiomer

5

(2) O

N-CH2CH3

CH3CH2NH2 (excess) OCH2CH3

NHCH2CH3

heat

O

+ H2O + CH3CH2OH

O 5

(3) 1. H O

H O

H3C C C Li O

O

2. aq NH4Cl (protonation)

H

H 3C

C

H O

PCC

O

C HO H

H 3C

O

C

C

O

5

5

(4) O CH3O

O CH3

NaOH CH3OH

+ H

CH3O

Br

OCH3

O CH3O

+ H2O Br

CH3O OCH3

5

(5)

1. OLi OLi OCH2CH3

O

(1 equiv)

O

O

Br O 2. aq NH4Cl (protonation)

O + LiBr

OCH2CH3

O

+ LiOH 5