Q58. - Organic Chemistry

Question

Question: Draw a stepwise mechanism for the following reaction.

Step-by-Step Solution

Answer

Answer

 

Stepwise mechanism

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Step 1: Carbonyl groups forming enolate

The carbonyl groups are electron-withdrawing groups and so any hydrogen(s) present at $\alpha$ -position are highly acidic in nature.

 

Those acidic $\alpha$ -H can easily be removed in presence of a base and with the formation of a stable enolate.

 

The CH₂  group containing acidic hydrogen(s) is called the active methylene group.

Step 2: Mechanism

                            The given reaction

  

Step-by-step mechanism:

 

• The compound has an active methylene group which is located in between two strong electron-withdrawing -COOEt groups. The base -OEt easily takes it to generate a stable enolate.

                           Formation of stable enolate

 

• The enolate attacks at the epoxide and cleaves the ring by breaking the C-O bond to generate an alkoxide ion.

                                 Generation of alkoxide ion

 

• The negatively charged oxygen atom attacks at the C=O center with the removal of ethoxide anion to form a five-member cyclic ester, called lactone.

                   Formation of lactone