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(B) $1$. The reaction of $4$-chloroacetophenone with $NaCN$ and $HOAc$ (cyanohydrin formation) yields $P$,which is $2-(4-chlorophenyl)-2-hydroxypropan

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(B) $1$. The reaction of $4$-chloroacetophenone with $NaCN$ and $HOAc$ (cyanohydrin formation) yields $P$,which is $2-(4-chlorophenyl)-2-hydroxypropanenitrile$.$2$. The reaction of $P$ with $EtOH$ in the presence of $H^+$ (acid-catalyzed esterification/hydrolysis) converts the nitrile group to an ester,yielding $Q$,which is ethyl $2-(4-chlorophenyl)-2-hydroxypropanoate$.$3$. The reaction of $Q$ with $2$ equivalents of $MeMgBr$ followed by $H_3O^+$ (Grignard reaction with an ester) replaces the ester group with a tertiary alcohol group,yielding $R$,which is $1-(4-chlorophenyl)-1,2-dihydroxy-2-methylpropane$ (or $2-(4-chlorophenyl)-3-methylbutane-2,3-diol$).

$R$ formed in the following sequence of reaction is:

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