Match the LIST-I with LIST-II.
| LIST-I (Reaction) | LIST-II (Name) |
| A. Benzonitrile \( \xrightarrow{\ce{Na(Hg)/C2H5OH}} \) Phenylmethanamine | I. Reduction |
| B. Aniline + Chloroform + Sodium hydroxide \( \xrightarrow{\Delta} \) Isocyanobenzene | II. Benzoylation reaction |
| C. Aniline + Benzoyl chloride \( \xrightarrow{\ce{NaOH}} \) N-Phenylbenzamide | III. Hinsberg’s test |
| D. Benzenesulphonyl chloride + N,N diethylamine → N,N-Diethylbenzenesulphonamide | IV. Carbylamine reaction |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-I, B-IV, C-II, D-III
The common reactions used to make or identify amines can be recognised from the starting material and the reagent:
Reduction: nitriles, amides and nitro compounds are reduced to amines by \( \ce{LiAlH4} \), \( \ce{H2} \)/Ni or sodium amalgam/ethanol (reduction of nitriles by Na(Hg)/\( \ce{C2H5OH} \) is called the Mendius reduction).
Carbylamine reaction: a primary amine heated with chloroform and alkali gives a foul-smelling isocyanide.
Acylation / benzoylation: amines react with acid chlorides; with benzoyl chloride the product is an N-substituted benzamide. A base removes the HCl formed.
Hinsberg’s test: benzenesulphonyl chloride reacts with 1° and 2° amines to form sulphonamides, and is used to tell 1°, 2° and 3° amines apart.
A. Benzonitrile \( \xrightarrow{\ce{Na(Hg)/C2H5OH}} \) Phenylmethanamine — This matches I. Reduction. Nascent hydrogen from sodium amalgam and ethanol reduces the C≡N group to \( \ce{CH2NH2} \).
\( \displaystyle \ce{C6H5-C#N} \xrightarrow{\ce{Na(Hg)/C2H5OH}} \ce{C6H5-CH2-NH2} \)
B. Aniline + Chloroform + Sodium hydroxide \( \xrightarrow{\Delta} \) Isocyanobenzene — This matches IV. Carbylamine reaction. The primary amine is converted into an isocyanide, detected by its offensive smell.
\( \displaystyle \ce{C6H5NH2 + CHCl3 + 3NaOH} \xrightarrow{\Delta} \ce{C6H5NC + 3NaCl + 3H2O} \)
C. Aniline + Benzoyl chloride \( \xrightarrow{\ce{NaOH}} \) N-Phenylbenzamide — This matches II. Benzoylation reaction. The \( \ce{C6H5CO} \)– group replaces a hydrogen on nitrogen; NaOH neutralises the HCl formed and drives the reaction forward.
\( \displaystyle \ce{C6H5NH2 + C6H5COCl} \xrightarrow{\ce{NaOH}} \ce{C6H5NHCOC6H5 + HCl} \)
D. Benzenesulphonyl chloride + N,N diethylamine → N,N-Diethylbenzenesulphonamide — This matches III. Hinsberg’s test. Benzenesulphonyl chloride is Hinsberg’s reagent; with the secondary amine diethylamine it gives a sulphonamide with no H on nitrogen, which is insoluble in alkali.
\( \displaystyle \ce{C6H5SO2Cl + (C2H5)2NH -> C6H5SO2N(C2H5)2 + HCl} \)
Shortcut: The pairs A–I and B–IV appear in Options (1) and (2), and C–II then leaves only Option (1).