Match List-I with List-II
| Name reaction | Chemical reaction |
| (A) Hinsberg’s reaction | (I) R–NH2 + CHCl3 + 3KOH \( \xrightarrow{\text{Heat}} \) R–NC + 3KCl + 3H2O |
| (B) Gabriel phthalimide | (II) Ph-SO2Cl + C2H5-NH2 → PhSO2NH-C2H5 |
| (C) Hofmann bromamide degradation | (III)  |
| (D) Carbylamine reaction | (IV) R-CONH2 + Br2 + 4NaOH → R-NH2 +Na2CO3 + H2O + 2NaBr |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
These four name reactions belong to two groups:
Preparation of primary amines: the Gabriel phthalimide synthesis (alkyl halide and potassium phthalimide, then hydrolysis) and the Hofmann bromamide degradation (amide with Br2 and NaOH; the amine has one carbon fewer).
Tests for amines: Hinsberg’s test with benzenesulphonyl chloride (the sulphonamide of a 1° amine dissolves in alkali, that of a 2° amine does not, and a 3° amine does not react) and the carbylamine test (only 1° amines give a foul-smelling isocyanide).
(A) Hinsberg’s reaction — This matches (II) Ph-SO2Cl + C2H5-NH2 → PhSO2NH-C2H5. The primary amine gives N-ethylbenzenesulphonamide, whose N–H hydrogen is acidic, so it dissolves in alkali.
\( \displaystyle \ce{C6H5SO2Cl + C2H5NH2 -> C6H5SO2NHC2H5 + HCl} \)
(B) Gabriel phthalimide — This matches (III) the reaction of phthalimide with KOH and R-X. Phthalimide is converted into its potassium salt, which is alkylated by R–X; alkaline hydrolysis then releases a pure primary amine.
\( \displaystyle \ce{C6H4(CO)2NH} \xrightarrow{\text{ethanolic }\ce{KOH}} \ce{C6H4(CO)2N^-K^+} \xrightarrow{\ce{R-X}} \ce{C6H4(CO)2N-R} \xrightarrow{\text{aq. }\ce{NaOH}} \ce{C6H4(COONa)2 + R-NH2} \)
(C) Hofmann bromamide degradation — This matches (IV) R-CONH2 + Br2 + 4NaOH → R-NH2 +Na2CO3 + H2O + 2NaBr. The amide loses its carbonyl carbon as carbonate, so the amine has one carbon atom fewer than the amide.
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
(D) Carbylamine reaction — This matches (I) R–NH2 + CHCl3 + 3KOH → R–NC + 3KCl + 3H2O. Heating a primary amine with chloroform and ethanolic KOH gives a foul-smelling isocyanide.
\( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\text{heat}} \ce{R-NC + 3KCl + 3H2O} \)
Shortcut: The pair (D)–(I) appears only in Option (3).
Note: It seems that there is a mistake in the original NTA paper, which prints H2O in reaction (IV) when it should read 2H2O, and which shows the whole Gabriel synthesis in reaction (III) as one step with KOH giving the disodium salt. The actual steps are:
\( \displaystyle \text{phthalimide} \xrightarrow{\text{ethanolic }\ce{KOH}} \text{potassium phthalimide} \xrightarrow{\ce{R-X}} \text{N-alkylphthalimide} \)
\( \displaystyle \text{N-alkylphthalimide} \xrightarrow{\text{aq. }\ce{NaOH}} \text{sodium phthalate} + \ce{R-NH2} \)