The product 'D' in the following reaction is:
$$CH_3CN \xrightarrow{H_3O^+} A \xrightarrow{NH_3, \Delta} B \xr

Question

The product 'D' in the following reaction is:
$$CH_3CN \xrightarrow{H_3O^+} A \xrightarrow{NH_3, \Delta} B \xrightarrow{NaOBr} C \xrightarrow{NaNO_2/HCl} D$$

Accepted Answer

The reaction sequence proceeds as follows:
1.  **Hydrolysis (A):** Methyl cyanide ($CH_3CN$) undergoes acid hydrolysis ($H_3O^+$) to form **Ethanoic acid** ($CH_3COOH$) [A].
    $$CH_3CN + 2H_2O + H^+ \rightarrow CH_3COOH + NH_4^+$$
2.  **Amide Formation (B):** Heating ethanoic acid with ammonia ($NH_3, \Delta$) removes water to form **Ethanamide** ($CH_3CONH_2$) [B].
    $$CH_3COOH + NH_3 \rightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O$$
3.  **Hoffmann Bromamide Degradation (C):** Reaction with sodium hypobromite ($NaOBr$ or $Br_2/NaOH$) converts the amide to a primary amine with one less carbon atom. Thus, **Methylamine** ($CH_3NH_2$) [C] is formed.
    $$CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$$
4.  **Reaction with Nitrous Acid (D):** Primary aliphatic amines react with nitrous acid ($NaNO_2 + HCl$) to form unstable diazonium salts which liberate nitrogen gas and form alcohols. Thus, **Methanol** ($CH_3OH$) [D] is the final product.
    $$CH_3NH_2 + HNO_2 \rightarrow [CH_3N_2^+Cl^-] \xrightarrow{H_2O} CH_3OH + N_2 + HCl$$

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