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

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. $CH_3NH_2 + HNO_2 \rightarrow [CH_3N_2^+Cl^-] \xrightarrow{H_2O} CH_3OH + N_2 + HCl$