Match List-I with List-II
| List-I | List-II |
| Chemical reactions | Products |
| (A) Acylation of N-ethylethanamine | (I) Quaternary ammonium salt |
| (B) Benzoylation of methanamine with benzoyl chloride | (II) N,N-Diethylethanamide |
| (C) Alkylation of tertiary amines | (III) N-Ethylbenzenesulphonamide |
| (D) Hinsberg’s reagent with ethylamine | (IV) N-Methylbenzamide |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
Amines react through the lone pair on nitrogen. An acid chloride (acylation, benzoylation) or benzenesulphonyl chloride (Hinsberg’s reagent) replaces an N–H hydrogen, so only primary and secondary amines react; an alkyl halide adds one more alkyl group, so a tertiary amine still reacts and gives a quaternary ammonium salt.
(A) Acylation of N-ethylethanamine — This matches (II) N,N-Diethylethanamide. The secondary amine \( \ce{(C2H5)2NH} \) loses its one N–H hydrogen to the acetyl group.
\( \displaystyle \ce{(C2H5)2NH + CH3COCl} \xrightarrow{\text{pyridine}} \ce{CH3CON(C2H5)2 + HCl} \)
(B) Benzoylation of methanamine with benzoyl chloride — This matches (IV) N-Methylbenzamide. One N–H hydrogen of methanamine is replaced by the benzoyl group.
\( \displaystyle \ce{CH3NH2 + C6H5COCl -> C6H5CONHCH3 + HCl} \)
(C) Alkylation of tertiary amines — This matches (I) Quaternary ammonium salt. The tertiary amine takes a fourth alkyl group on nitrogen.
\( \displaystyle \ce{R3N + R-X -> R4N^+X^-} \)
(D) Hinsberg’s reagent with ethylamine — This matches (III) N-Ethylbenzenesulphonamide. The product still has one acidic N–H hydrogen, so it dissolves in alkali; this is how Hinsberg’s test picks out a primary amine.
\( \displaystyle \ce{C6H5SO2Cl + C2H5NH2 -> C6H5SO2NHC2H5 + HCl} \)
Shortcut: The pair (A)–(II) appears in Options (1) and (4), and (B)–(IV) then leaves only Option (4).