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Chapter 23: Q58. (page 924)

Question: Draw a stepwise mechanism for the following reaction.

Short Answer

Expert verified

Answer

Stepwise mechanism




Step by step solution

01

 Step 1: Carbonyl groups forming enolate

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

Those acidic α-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.

02

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

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Most popular questions from this chapter

Question: Rank the protons in the labeled CH2groups in order of increasing acidity, and explain why you chose this order.

Question: What alkyl halides are needed to prepare each carboxylic acid using the malonic ester synthesis?

Question: Rank the labeled protons in each compound in order of increasing acidity.

Question: Identify the product M of the following two-step reaction sequence. M was converted to the hallucinogen LSD (Figure 18.4) in several steps.

Question: Which carbonyl compound in each pair exhibits the higher percentage of the enol tautomer?
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