Ph-C equiv N xrightarrow[{Partial hydrolysis | vedclass

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(C) Step $1$: Partial hydrolysis of benzonitrile $(Ph-CN)$ with $H_3O^\oplus$ yields benzamide $(Ph-CONH_2)$ as product $(A)$.Step $2$: The reaction o

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$Ph-NH_2$

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(C) Step $1$: Partial hydrolysis of benzonitrile $(Ph-CN)$ with $H_3O^\oplus$ yields benzamide $(Ph-CONH_2)$ as product $(A)$.Step $2$: The reaction of benzamide $(Ph-CONH_2)$ with $Br_2 + KOH$ is the Hofmann bromamide degradation reaction.Step $3$: This reaction converts an amide into a primary amine with one carbon atom less than the original amide.Step $4$: Thus,$Ph-CONH_2$ is converted to aniline $(Ph-NH_2)$,which is product $(B)$.

$Ph-C \equiv N$ $\xrightarrow[{Partial \ hydrolysis}]{H_3O^\oplus} (A)$ $\xrightarrow{Br_2 + KOH} (B)$
Product $(B)$ is

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$Ph-C \equiv N$ $\xrightarrow[{Partial \ hydrolysis}]{H_3O^\oplus} (A)$ $\xrightarrow{Br_2 + KOH} (B)$Product $(B)$ is