Looking for classes? Ksquare Career Institute, Bengaluru →
Aniline in a set of reactions yielded a product D
$C_6H_5NH_2 \xrightarrow{NaNO_2,\ HCl} A \xrightarrow{CuCN} B \xrightarrow{H_2/Ni} C \xrightarrow{HNO_2} D$
The structure of the product D would be -
$C_6H_5NH_2 \xrightarrow{NaNO_2,\ HCl} A \xrightarrow{CuCN} B \xrightarrow{H_2/Ni} C \xrightarrow{HNO_2} D$
The structure of the product D would be -
A
$C_6H_5CH_2OH$
B
$C_6H_5CH_2NH_2$
C
$C_6H_5NHOH$
D
$C_6H_5NHCH_2CH_3$
Detailed Solution
Step 1 (diazotisation): $C_6H_5NH_2 \xrightarrow{NaNO_2,\ HCl,\ 273-278\ K} C_6H_5N_2^+Cl^-$ (A, benzenediazonium chloride)
Step 2 (Sandmeyer reaction): $C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN$ (B, benzonitrile) $+ N_2$
Step 3 (reduction): $C_6H_5CN \xrightarrow{H_2/Ni} C_6H_5CH_2NH_2$ (C, benzylamine)
Step 4: benzylamine is a primary aliphatic amine (the $NH_2$ is on a $CH_2$ group, not on the ring). With nitrous acid it gives an unstable aliphatic diazonium salt, which loses nitrogen and gives the alcohol.
$C_6H_5CH_2NH_2 + HNO_2 \rightarrow C_6H_5CH_2OH + N_2 + H_2O$
So D is benzyl alcohol, $C_6H_5CH_2OH$.
Step 2 (Sandmeyer reaction): $C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN$ (B, benzonitrile) $+ N_2$
Step 3 (reduction): $C_6H_5CN \xrightarrow{H_2/Ni} C_6H_5CH_2NH_2$ (C, benzylamine)
Step 4: benzylamine is a primary aliphatic amine (the $NH_2$ is on a $CH_2$ group, not on the ring). With nitrous acid it gives an unstable aliphatic diazonium salt, which loses nitrogen and gives the alcohol.
$C_6H_5CH_2NH_2 + HNO_2 \rightarrow C_6H_5CH_2OH + N_2 + H_2O$
So D is benzyl alcohol, $C_6H_5CH_2OH$.