Match List I with List II List I List II I. Hoffmann Bromamide degrad…

The correct answer is Option (2) → A-II, B-IV, C-I, D-III

Each compound in List I is matched by the reaction it typically undergoes or the preparation it is used in: aromatic primary amines form diazonium salts and azo dyes, primary amines give the carbylamine test, amides undergo Hofmann degradation, and phthalimide is the starting material of the Gabriel synthesis.

• A. Aniline (structure A) — This matches II. Azo dye formation. Aniline is diazotised with \( \ce{NaNO2} \)/HCl at 273–278 K, and the benzenediazonium chloride couples with phenol (or another arylamine) to give an azo dye.

      \( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{\text{273-278 K}} \ce{C6H5N2^+Cl^- + NaCl + 2H2O} \)

      \( \displaystyle \ce{C6H5N2^+Cl^- + C6H5OH} \xrightarrow{\ce{OH^-}} \ce{C6H5-N=N-C6H4OH + HCl} \)  (p-hydroxyazobenzene, orange)

• B. CH₃NH₂ — This matches IV. Carbylamine reaction. Methylamine is a primary amine and gives methyl isocyanide with chloroform and alcoholic KOH. (Aliphatic diazonium salts are unstable, so it cannot give an azo dye.)

      \( \displaystyle \ce{CH3NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{CH3NC + 3KCl + 3H2O} \)

• C. CH₃CH₂CONH₂ — This matches I. Hoffmann Bromamide degradation reaction. Propanamide with \( \ce{Br2} \) and NaOH gives ethanamine, with one carbon atom less.

      \( \ce{CH3CH2CONH2 + Br2 + 4NaOH -> CH3CH2NH2 + Na2CO3 + 2NaBr + 2H2O} \)

• D. Phthalimide (structure D) — This matches III. Gabriel phthalimide synthesis. Phthalimide with ethanolic KOH gives potassium phthalimide, which with an alkyl halide and then alkaline hydrolysis gives a primary amine.

      \( \displaystyle \ce{C6H4(CO)2NH} \xrightarrow{\ce{KOH}\text{ (alc.)}} \ce{C6H4(CO)2N^-K^+} \xrightarrow{\ce{R-X}} \ce{C6H4(CO)2N-R} \xrightarrow{\ce{NaOH}\text{ (aq)}} \ce{R-NH2} \)

Shortcut: C-I (the amide undergoes Hofmann degradation) appears only in Option (2).