Q71E - Organic Chemistry (Mcmurry)

Question

The compound known as Hagemann’s ester is prepared by treatment of a mixture of formaldehyde and ethyl acetoacetate with base, followed by acid-catalyzed decarboxylation.

 

(a) The first step is an aldol-like condensation between ethyl acetoacetate and formaldehyde to yield an α,β-unsaturated product. Write the reaction, and show the structure of the product.

 

(b) The second step is a Michael reaction between ethyl acetoacetate and the unsaturated product of the first step. Show the structure of the product.

Step-by-Step Solution

Answer

In the first step, Ethylacetoacetate has two acidic hydrogen atoms in the methylene group. The base abstracts one of the hydrogen atoms and forms an enolate ion. Then the enolate attacks the carbonyl carbon of formaldehyde and forms a carbanion. Lastly, the carbanion removes a hydroxide ion and gives the α,β-unsaturated product. In the second step, This step is a Michael addition. The enolate of ethyl acetoacetate acts as a nucleophile donor and the α,β-unsaturated product acts as an eletrophile acceptor. Lastly by Michael addition, we get our desired product.

Reaction of the first step

The first step for preparation of Hagemann’s ester is an aldol-like condensation between ethyl acetoacetate and formaldehyde which yields an α,β-unsaturated product.

Product of the first step

The α,β-unsaturated product is

Reaction of the second step

Michael reaction between ethyl acetoacetate and the unsaturated product of the first step is

Product of the second step

Protonation of the carbanion will be the product

Explanation of the solution

For the first step:

 

Ethylacetoacetate has two acidic hydrogen atoms in the methylene group. The base abstracts one of the hydrogen atoms and forms an enolate ion. Then the enolate attacks the carbonyl carbon of formaldehyde and forms a carbanion. Lastly, the carbanion removes a hydroxide ion and gives the α,β-unsaturated product.

 

 

For the second step:

 

This step is a Michael addition. The enolate of ethyl acetoacetate acts as a nucleophile donor and the α,β-unsaturated product acts as an eletrophile acceptor. Lastly by Michael addition, we get our desired product.