See image — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

Concept: We need a test that gives different results for aniline (aromatic primary amine) versus cyclohexylamine (aliphatic primary amine). Step 1 - Evaluate option (a) Hinsberg test: Both aniline and cyclohexylamine are primary amines. Both react with benzenesulfonyl chloride (Hinsberg reagent) to give sulfonamides. The sulfonamide from a primary amine is soluble in NaOH. Both being primary amines, this test does NOT differentiate them. Step 2 - Evaluate option (b) Isocyanide test (carbylamine test): Both aniline and cyclohexylamine are primary amines. Both react with CHCl3 and KOH to give isocyanides (carbylamines) with a foul smell. This test distinguishes primary amines from secondary/tertiary amines, but NOT aromatic primary from aliphatic primary amines. Step 3 - Evaluate option (c) NaNO2, HCl (diazotization), then beta-Naphthol (azo coupling): Aniline, being an aromatic primary amine, reacts with NaNO2 and HCl at 0-5°C to form a stable diazonium salt (benzenediazonium chloride). This diazonium salt couples with beta-Naphthol in alkaline medium to give a brilliant orange-red azo dye. Cyclohexylamine, being an aliphatic primary amine, reacts with NaNO2 and HCl to form an unstable aliphatic diazonium salt that immediately decomposes (losing N2 gas) and does NOT form a stable diazonium salt, so no azo dye is formed with beta-Naphthol. Thus, aniline gives a bright orange-red color while cyclohexylamine does not — this clearly differentiates the two compounds. Step 4 - Evaluate option (d) NaOH: Aniline does not react with NaOH (it is a base, not an acid). Cyclohexylamine also does not react with NaOH in a distinguishing manner. This test does not differentiate them. Conclusion: Only option (c) NaNO2, HCl followed by beta-Naphthol (diazotization and azo coupling) differentiates aniline (gives orange-red azo dye) from cyclohexylamine (no stable diazonium salt, no dye formed). Therefore, the correct answer is C.