Consider the following reaction: C_6H_5NO_2 x | 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 $2$: Aniline $(C_6H_5NH_2)$ reacts with benzoyl chloride $(C_6H_5COCl)$ in the presence of a base like $NaOH$ (Schotten-Baumann reaction) to form 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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