CH_3COOCH_3 + excess PhMgBr rightarrow text{p | vedclass

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(A) $CH_3COOCH_3$ (methyl acetate) reacts with two equivalents of $PhMgBr$ (phenylmagnesium bromide). The first equivalent of $PhMgBr$ performs a nucl

Vedclass · Accepted Answer

$1,1$-diphenylethanol

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(A) $CH_3COOCH_3$ (methyl acetate) reacts with two equivalents of $PhMgBr$ (phenylmagnesium bromide). The first equivalent of $PhMgBr$ performs a nucleophilic acyl substitution on the ester to form acetophenone ($CH_3COCH_3$ is incorrect in the prompt,it should be $CH_3COOCH_3$). The second equivalent of $PhMgBr$ attacks the carbonyl group of the formed acetophenone to produce an alkoxide intermediate. Upon acid hydrolysis $(H^+)$,the alkoxide is protonated to yield $1,1$-diphenylethanol $(CH_3C(Ph)_2OH)$. Reaction sequence: $CH_3COOCH_3 + PhMgBr \rightarrow CH_3COPh + CH_3OMgBr$ $CH_3COPh + PhMgBr$ $\rightarrow CH_3C(Ph)_2OMgBr$ $\xrightarrow{H^+} CH_3C(Ph)_2OH$

Column-$1$ (Reaction of carbonyl compound with)	Column-$2$ (Product)
$A$. Hydroxylamine	$I$. Hydrazone
$B$. Alcohol	$II$. Schiff's base (Substituted imine)
$C$. Hydrazine	$III$. Oxime
$D$. Amine	$IV$. Ketal

$CH_3COOCH_3 +$ excess $PhMgBr$ $\rightarrow \text{product}$ $\xrightarrow{H^+} X$. The product $X$ is:

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Match the following:
Column-$1$ (Reaction of carbonyl compound with) Column-$2$ (Product)
$A$. Hydroxylamine $I$. Hydrazone
$B$. Alcohol $II$. Schiff's base (Substituted imine)
$C$. Hydrazine $III$. Oxime
$D$. Amine $IV$. Ketal
The correct match is:

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