Question:Draw the product formed from a Claisen reaction with the given starting materials using –OEt, EtOH.

The reaction between two ester groups to form a $\mathit{\beta }$-keto ester is known as the Claisen reaction.

The reaction is performed in the presence of an alkoxide base. When the reaction occurs between two different esters, where only one of them has an $\mathit{\alpha }$-hydrogen, it is known as the crossed-Claisen reaction.

Claisen reaction

Crossed-Claisen reaction

The mechanism of the crossed-Claisen reaction is the same as that of the Claisen reaction.

The reaction proceeds via the abstraction of a proton from the $\alpha$-carbon of the carbonyl group, which leads to the formation of the enolate intermediate. In the last step, the -OEt group acts as the leaving group, and hence, the $\beta$-keto ester is formed.

The mechanism of the reaction is given below.

Crossed-Claisen reaction mechanism

a. The proton abstraction takes place in 2-propanone, and the nucleophilic attack takes place at the carbonyl carbon center of the ester. The product is given below.

Product of reaction a

b.

The product is shown below.

Product of reaction b

c.

Here, the ring system undergoes proton abstraction, and the nucleophilic attack takes place at the carbonyl carbon of the methyl formate.

Product of reaction c

d.

Here, chlorine acts as a better leaving group than the -OEt group. Hence, the product is as follows.

Product of reaction d