PROBLEM 12.27 - Organic Chemistry

Question

Question: Draw the products of each Sharpless epoxidation.

a.

b.

Step-by-Step Solution

Answer

Answer

a.

b.

Sharpless epoxidation

The epoxide is a three-membered ring having an ether linkage between them. The enantioselective conversion of allylic alcohols to epoxides. The general reaction scheme is given below:

Representation of Sharpless epoxidation

Sharpless reagents

The following are the compounds that constitute the Sharpless reagent: tert-butyl hydroperoxide,  (CH₃)₃COOH a titanium catalyst—usually titanium(IV) isopropoxide, $\mathrm{Ti}{\left[\mathrm{OCH}{\left({\mathrm{CH}}_3\right)}_2\right]}_4$ and diethyl tartrate(DET). 

Diethyl tartrate can have two forms, such as (+)-DET and (-)-DET. When the allylic alcohol is placed on the plane in such a way that the hydroxyl group is in the upper right corner, the two isomers have different functions, such that

 (-)-DET: attacks from above the plane;

(+)-DET: attacks from below the plane.

Determination of the products

a.Once the allylic alcohol is placed such that the hydroxyl group is on the right-hand corner, then the (+)-DET attacks from below the plane. The reaction is shown below:

Sharpless epoxidation with (+)-DET

b.When the compound is redrawn, keeping the double bond horizontally and the hydroxyl group is kept at the top right corner, (-)-DET can be seen to attack from above the plane.

Sharpless epoxidation with (-)-DET