Q.75. - Organic Chemistry

Question

Question: Sharpless epoxidation of allylic alcohol X forms compound Y. Treatment of Y with NaOH and ${\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{5}}\mathit{S}\mathit{H}$  in an alcohol–water mixture forms Z. Identify the structure of Y and draw a mechanism for the conversion of Y to Z. Account for the stereochemistry of the stereogenic centers in Z. Z has been used as an intermediate in the synthesis of chiral carbohydrates.

Step-by-Step Solution

Answer

Answer

Step 1: Sharpless epoxidation

The enantiomeric selective method to prepare 2,3-epoxy alcohols from $\mathbf{1}\mathbf{°}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathbf{2}\mathbf{°}$ allylic alcohols is known as Sharpless epoxidation.

Step 2: Attack of oxygen

In the presence of (+) DET complex, oxygen adds from the upper side of the face of allyl alcohol and vice-versa. 

Step 3: Stepwise mechanism

X undergoes Sharpless epoxidation to form Y. The structure of Y is given hereunder.

In the conversion of Y to form Z, the anti-attack of base NaOH takes place and the epoxide ring opens up. 

In this process, the diol is formed. The mechanism for the conversion of X to Y and Y to Z, along with the stereochemistry of Z, is shown hereunder.