In the given reaction,A can be .... . | vedclass

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(B) The reaction involves the formation of a Grignard reagent from $A$ followed by its reaction with methyl benzoate to form triphenylmethanol. $1$. $

Vedclass · Accepted Answer

bromobenzene

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(B) The reaction involves the formation of a Grignard reagent from $A$ followed by its reaction with methyl benzoate to form triphenylmethanol. $1$. $A + Mg \xrightarrow{THF} A-MgBr$ $2$. The Grignard reagent $(PhMgBr)$ attacks the carbonyl carbon of methyl benzoate $(PhCOOCH_3)$. $3$. This leads to the formation of a ketone intermediate,benzophenone $(Ph_2CO)$,which further reacts with another equivalent of $PhMgBr$ to form the final product,triphenylmethanol $(Ph_3COH)$. Since the product contains three phenyl groups attached to the central carbon,and one comes from the methyl benzoate,the other two must come from the Grignard reagent. Thus,$A$ must be bromobenzene $(C_6H_5Br)$.

List-$I$ Conversion	List-$II$ Reagents,Conditions used
$A$. Chlorobenzene $\rightarrow$ Phenol	$I$. Warm,$H_2O$
$B$. $1$-Chloro-$4$-nitrobenzene $\rightarrow$ $4$-Nitrophenol	$II$. $(a)$ $NaOH, 368 \ K$; $(b)$ $H_3O^+$
$C$. $1$-Chloro-$2,4$-dinitrobenzene $\rightarrow$ $2,4$-Dinitrophenol	$III$. $(a)$ $NaOH, 443 \ K$; $(b)$ $H_3O^+$
$D$. $1$-Chloro-$2,4,6$-trinitrobenzene $\rightarrow$ $2,4,6$-Trinitrophenol	$IV$. $(a)$ $NaOH, 623 \ K, 300 \ atm$; $(b)$ $H_3O^+$

In the given reaction,$A$ can be .... .

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