Match the LIST-I with LIST-II.
| LIST-I Reaction | LIST-II Name of the Reaction/Test |
A.  | I. Gattermann reaction |
| B. RNH2 + CHCl3 + KOH | II. Hinsberg’s test |
C.  | III. Gabriel phthalimide synthesis |
D.  | IV. Isocyanide test |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-IV, C-II, D-I
The named reactions and tests of amines can be grouped by what they are used for:
Preparing primary amines: Gabriel phthalimide synthesis — phthalimide + ethanolic KOH, then an alkyl halide, then alkaline hydrolysis gives a pure 1° amine; Hofmann bromamide degradation — an amide + \( \ce{Br2} \)/NaOH gives a 1° amine with one carbon fewer.
Tests that distinguish amines: carbylamine (isocyanide) test — only 1° amines with CHCl3 and KOH give a foul-smelling isocyanide; Hinsberg’s test — benzenesulphonyl chloride (\( \ce{C6H5SO2Cl} \)) tells 1°, 2° and 3° amines apart.
Replacing the diazo group by halogen: Sandmeyer reaction — diazonium salt + Cu2Cl2/HCl or Cu2Br2/HBr; Gattermann reaction — diazonium salt + copper powder and HCl or HBr. Both give an aryl halide and \( \ce{N2} \).
A. Phthalimide + KOH and R-X — This matches III. Gabriel phthalimide synthesis. KOH forms potassium phthalimide, whose nitrogen anion is alkylated by R–X; hydrolysis then releases the primary amine. Aryl amines cannot be made this way, because aryl halides do not undergo this substitution.
\( \displaystyle \text{phthalimide} \xrightarrow{\text{KOH (alc.)}} \text{K-phthalimide} \xrightarrow{\ce{R-X}} \text{N-alkylphthalimide} \xrightarrow{\ce{NaOH}\text{ (aq)}} \ce{R-NH2} \)
B. \( \ce{RNH2} \) + \( \ce{CHCl3} \) + KOH — This matches IV. Isocyanide test. A primary amine heated with chloroform and ethanolic KOH gives an isocyanide (carbylamine) with a very unpleasant smell; 2° and 3° amines do not respond.
\( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\text{heat}} \ce{R-NC + 3KCl + 3H2O} \)
C. Benzenesulphonyl chloride + \( \ce{C2H5NH2} \) — This matches II. Hinsberg’s test. A primary amine gives an N-alkylbenzenesulphonamide whose N–H is acidic, so it dissolves in alkali.
\( \displaystyle \ce{C6H5SO2Cl + C2H5NH2 -> C6H5SO2NHC2H5 + HCl} \)
D. Benzenediazonium salt + Cu/HCl — This matches I. Gattermann reaction. Copper powder with HCl replaces the –\( \ce{N2^{+}} \) group by chlorine.
\( \displaystyle \ce{C6H5N2^+X^-} \xrightarrow{\ce{Cu}/\ce{HX}} \ce{C6H5X + N2} \)
Shortcut: The pair A–III appears in Options (3) and (4), and B–IV then leaves only Option (3).