The product A in the following reaction is: | vedclass

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(A) The reaction involves the alkylation of an unsymmetrical ketone,$2$-methylcyclohexanone,using a strong,sterically hindered base,$LDA$ (Lithium dii

Vedclass · Accepted Answer

$2-$ethyl$-6-$methylcyclohexanone

Vedclass · Answer

(A) The reaction involves the alkylation of an unsymmetrical ketone,$2$-methylcyclohexanone,using a strong,sterically hindered base,$LDA$ (Lithium diisopropylamide). $LDA$ is a kinetic base that selectively removes a proton from the less sterically hindered $\alpha$-carbon. In $2$-methylcyclohexanone,there are two types of $\alpha$-carbons: one is the tertiary carbon (substituted with a methyl group) and the other is the secondary carbon ($-CH_2-$ group). $LDA$ removes a proton from the less hindered $-CH_2-$ $\alpha$-carbon to form the kinetic enolate. Subsequent reaction of this kinetic enolate with the alkylating agent,ethyl iodide $(CH_3-CH_2-I)$,leads to the alkylation at the less substituted $\alpha$-carbon. Therefore,the major product $A$ is $2$-ethyl-$6$-methylcyclohexanone.

The product $A$ in the following reaction is:

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