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 $Sn/HCl$ yields aniline $(C_6H_5NH_2)$ as product $X$.Step $2$: Aniline $(C_6H_5NH_2)$ rea

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Benzanilide

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(B) Step $1$: Reduction of nitrobenzene $(C_6H_5NO_2)$ with $Sn/HCl$ yields aniline $(C_6H_5NH_2)$ as product $X$.Step $2$: Aniline $(C_6H_5NH_2)$ reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base to undergo benzoylation (Schotten-Baumann reaction).Step $3$: The reaction is $C_6H_5NH_2 + C_6H_5COCl \rightarrow C_6H_5NHCOC_6H_5 + HCl$.The product $Y$ is $C_6H_5NHCOC_6H_5$,which is known as 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 $Y$?