Aniline on treatment with NaNO_2 / HCl at abo | vedclass

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Question

(A) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2 / HCl$ at $0^\circ C$ to form Benzenediazonium chloride $(A = C_6H_5N_2^+Cl^-)$.Step $2$: $A$

Vedclass · Accepted Answer

$C_6H_5CH_2OH$

Vedclass · Answer

(A) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2 / HCl$ at $0^\circ C$ to form Benzenediazonium chloride $(A = C_6H_5N_2^+Cl^-)$.Step $2$: $A$ reacts with $CuCN / KCN$ to form Benzonitrile $(B = C_6H_5CN)$.Step $3$: $B$ undergoes catalytic hydrogenation with $H_2 / Ni$ to form Benzylamine $(C = C_6H_5CH_2NH_2)$.Step $4$: Benzylamine $(C)$ reacts with nitrous acid $(HNO_2)$ to form Benzyl alcohol $(D = C_6H_5CH_2OH)$ along with $N_2$ gas and $H_2O$.

Aniline on treatment with $NaNO_2 / HCl$ at about $0^\circ C$ gives a compound $A$, which on further treatment with $CuCN / KCN$ gives a compound $B$, which when reacts with $H_2 / Ni$ gives a compound $C$. The compound $C$ on reaction with $HNO_2$ gives a compound $D$. The structure of $D$ is:

Similar Questions

Write $IUPAC$ names of the following compounds and classify them into primary,secondary,and tertiary amines.
$(i)$ $(CH_3)_2CHNH_2$
$(ii)$ $CH_3(CH_2)_2NH_2$
$(iii)$ $CH_3NHCH(CH_3)_2$
$(iv)$ $(CH_3)_3CNH_2$
$(v)$ $C_6H_5NHCH_3$
$(vi)$ $(CH_3CH_2)_2NCH_3$
$(vii)$ $m-BrC_6H_4NH_2$

In which of the following reactions is a tertiary amine formed?

Suppose the following reaction:
The product $D$ and $F$ are related as:

The product $(A)$ of the reaction is:

What are $X$ and $Y$ respectively in the following set of reactions?

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