Aniline in a set of reactions yielded a produ | vedclass

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(D) The reaction sequence is as follows: $1$. Aniline reacts with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form benzene diazonium chloride $(A)$: $C_6H_5NH_2

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$C_6H_5CH_2OH$

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(D) The reaction sequence is as follows: $1$. Aniline reacts with $NaNO_2/HCl$ at $0-5^{\circ}C$ to form benzene diazonium chloride $(A)$: $C_6H_5NH_2 \xrightarrow{NaNO_2/HCl} C_6H_5N_2^+Cl^-$. $2$. Benzene diazonium chloride reacts with $CuCN$ to form benzonitrile $(B)$: $C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN$. $3$. Benzonitrile is reduced by $H_2/Ni$ to form benzylamine $(C)$: $C_6H_5CN \xrightarrow{H_2/Ni} C_6H_5CH_2NH_2$. $4$. Benzylamine reacts with nitrous acid $(HNO_2)$ to form benzyl alcohol $(D)$: $C_6H_5CH_2NH_2 \xrightarrow{HNO_2} C_6H_5CH_2OH$. Thus,the final product $D$ is benzyl alcohol $(C_6H_5CH_2OH)$.

Aniline in a set of reactions yielded a product $D$. The structure of product $D$ would be:
Aniline $\xrightarrow{NaNO_2/HCl} A$ $\xrightarrow{CuCN} B$ $\xrightarrow{H_2/Ni} C$ $\xrightarrow{HNO_2} D$

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Aniline in a set of reactions yielded a product $D$. The structure of product $D$ would be: Aniline $\xrightarrow{NaNO_2/HCl} A$ $\xrightarrow{CuCN} B$ $\xrightarrow{H_2/Ni} C$ $\xrightarrow{HNO_2} D$