Match List I with List II
| List I | List II |
| (A) Carbylamine reaction | (I) forms amine with lesser number of carbon atoms |
| (B) Gabriel Phthalimide synthesis | (II) formation of foul smelling substances |
| (C) Hofmann Bromamide reaction | (III) formation of only primary amine |
| (D) Ammonolysis | (IV) mixture of primary, secondary and tertiary amines |
Choose the correct answers from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-II, B-III, C-I, D-IV
These are the main named reactions of amines: two methods of preparation that give only primary amines (Gabriel, Hofmann), one that gives a mixture (ammonolysis), and one test for primary amines (carbylamine).
(A) Carbylamine reaction — This matches (II) formation of foul smelling substances. A primary amine with chloroform and ethanolic KOH gives a foul-smelling isocyanide:
\( \ce{RNH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{RNC + 3KCl + 3H2O} \)
(B) Gabriel Phthalimide synthesis — This matches (III) formation of only primary amine. Phthalimide carries one N–H, so only one alkyl group can be attached:
phthalimide \( \xrightarrow{\text{KOH (alc)}} \) potassium phthalimide \( \xrightarrow{\ce{R-X}} \) N-alkylphthalimide \( \xrightarrow{\text{NaOH (aq)}} \ce{RNH2} \)
(C) Hofmann Bromamide reaction — This matches (I) forms amine with lesser number of carbon atoms. The carbonyl carbon is lost:
\( \ce{RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O} \)
(D) Ammonolysis — This matches (IV) mixture of primary, secondary and tertiary amines. The amine formed reacts further with the alkyl halide:
\( \ce{RX + NH3 -> RNH2} \xrightarrow{\ce{RX}} \ce{R2NH} \xrightarrow{\ce{RX}} \ce{R3N} \xrightarrow{\ce{RX}} \ce{R4N+X-} \)