ch402 2011 paper - warwick

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Page 1: ch402 2011 paper - Warwick
Page 2: ch402 2011 paper - Warwick

CH402X

(c) An example of the Sharpless asymmetric epoxidation of allylic alcohols is illustrated below

X)H

HO-s>C°2Et 100% conversion>90% ee

HO 'C02Et

(i) Explain the roles of the titanium complex, the peroxide and the diester in thereaction.

[10%]

(ii) Using a Sharpless asymmetric epoxidation as the key step, design a synthesis of themolecule 9 from reagents 10 and 11.

[20%]

Ph

^ ^U PhHN'

HO10 11

(d) In the following reaction, the oc,p-unsaturated aldehyde 12 is reduced enantioselectively to13, using the enantiomerically pure amine 14 as catalyst.

n 10mol%: Ph""^MX^Ph OH 14

H

1.1 equivalents:12 -""N 13

Me

(i) Amine 14 catalyses the reaction by reacting with 12 to form a reactive intermediate.Draw the structure of this intermediate.

[10%]

(ii) Illustrate the mechanism of hydride transfer to the reactive intermediate formedbetween 12 and 14, and explain how the asymmetric reduction is controlled.

[10%]

Section A continues on the next page

Page 3: ch402 2011 paper - Warwick

CH402X

3 . Answer ALL parts.

(a) Using appropriate illustrations of the likely intermediate complex formed, explain why boththe E- and Z-isomers of 15 are hydrogenated to the same major enantiomer of product 16 inthe reaction shown below.

Catalyst:

oA.

Q or

OMe

h

-crO

A

Or OMe H2Z-15 E-15

NHH o

R-16

[25%]

(b) Enantiomerically pure 17 can be prepared using a three step sequence illustrated below:

Catalyst:1©

C02Me

100%eeAsymmetrichydrogenation

V1 4'

OA /C02Me

O ^NMe

(i) Give a mechanism for the conversion of 18 into 19 and explain how asymmetricinduction is achieved in the process.

[25%]

(ii) Provide a mechanism for the conversion of 20 into 17.

(iii) Explain how the process could be modified to generate an enantiomerically purediastereoisomer of 17.

(Question 3 continues on next page)

Section A continues on the next page

Page 4: ch402 2011 paper - Warwick

CH402X 1(c) Using a mechanism, illustrate how the oxazaborolidine 21 catalyses the formation of the

diol product 22 in the reaction shown below.

10mol% Ph Phcatalyst:

PhPh

OB—Me21

HO H

BH3 (2.0 molar equivalents)

Ph

HO H

Ph22

[20%]

(d) Using catalyst 21, propose a synthesis of amine 23 from the starting material 24.

MeN-

MeO MeO

[20%]

End of Section A.Section B continues on the next page