In the following sequence of reactions:C_6H_5 | vedclass

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(D) $1$. $C_6H_5-NH_2$ reacts with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ to form benzene diazonium chloride $(A)$,$C_6H_5N_2Cl$.$2$. $(A)$ reacts with $CuC

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$C_6H_5-CH_2-OH$

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(D) $1$. $C_6H_5-NH_2$ reacts with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ to form benzene diazonium chloride $(A)$,$C_6H_5N_2Cl$.$2$. $(A)$ reacts with $CuCN$ (Sandmeyer reaction) to form benzonitrile $(B)$,$C_6H_5-CN$.$3$. $(B)$ undergoes catalytic hydrogenation with $H_2/Ni$ to form benzylamine $(C)$,$C_6H_5-CH_2-NH_2$.$4$. $(C)$ is a primary aliphatic amine,which reacts with $HNO_2$ $(NaNO_2/HCl)$ to form an unstable diazonium salt that decomposes to give benzyl alcohol $(D)$,$C_6H_5-CH_2-OH$.

In the following sequence of reactions:
$C_6H_5-NH_2$ $\xrightarrow[HCl]{NaNO_2} A$ $\xrightarrow{CuCN} B$ $\xrightarrow{H_2/Ni} C$ $\xrightarrow{HNO_2} D$
Product $(D)$ would be:

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In the following sequence of reactions:$C_6H_5-NH_2$ $\xrightarrow[HCl]{NaNO_2} A$ $\xrightarrow{CuCN} B$ $\xrightarrow{H_2/Ni} C$ $\xrightarrow{HNO_2} D$Product $(D)$ would be: