Consider the following reaction:C_6H_5NO_2 xr | vedclass

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(B) Step $1$: Reduction of nitrobenzene $(C_6H_5NO_2)$ with $Sn/HCl$ gives aniline $(C_6H_5NH_2)$ as $X$.$C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_

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Benzanilide

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(B) Step $1$: Reduction of nitrobenzene $(C_6H_5NO_2)$ with $Sn/HCl$ gives aniline $(C_6H_5NH_2)$ as $X$.$C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$Step $2$: Reaction of aniline $(X)$ with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base (like $NaOH$) is known as benzoylation,which yields benzanilide $(C_6H_5NHCOC_6H_5)$ as $Y$.$C_6H_5NH_2 + C_6H_5COCl \xrightarrow{NaOH} C_6H_5NHCOC_6H_5 + HCl$Thus,$Y$ is benzanilide.

Consider the following reaction:
$C_6H_5NO_2$ $\xrightarrow{Sn/HCl} X$ $\xrightarrow{C_6H_5COCl} Y + HCl$
What is $Y$?

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Consider the following reaction:$C_6H_5NO_2$ $\xrightarrow{Sn/HCl} X$ $\xrightarrow{C_6H_5COCl} Y + HCl$What is $Y$?