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(B) $1$. The starting material is $4$-chloroacetophenone. Reaction with $NaCN$ followed by $EtOH, H_3O^+$ converts the ketone into an ethyl ester of a

Vedclass · Accepted Answer

$A$ = ethyl $2-$($4$-chlorophenyl)$-2-$hydroxypropanoate,$B$ = $2-$($4$-chlorophenyl)$-3-$methylbutane$-2,3-$diol

Vedclass · Answer

(B) $1$. The starting material is $4$-chloroacetophenone. Reaction with $NaCN$ followed by $EtOH, H_3O^+$ converts the ketone into an ethyl ester of an $\alpha$-hydroxy acid. Specifically,the cyanohydrin intermediate is hydrolyzed and esterified to form ethyl $2-(4-chlorophenyl)-2-hydroxypropanoate$ $(A)$.$2$. Reaction of the ester $(A)$ with excess $MeMgBr$ followed by acidic workup $(H_3O^+)$ involves two successive nucleophilic additions to the ester carbonyl group. The first addition forms a ketone intermediate,and the second addition forms a tertiary alcohol. The final product $B$ is $2-(4-chlorophenyl)-3-methylbutane-2,3-diol$.

Find out the major products $A$ and $B$ from the following reaction sequence.

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