See image — Amines Chemistry Question

💡 Solution & Explanation

Concept: The Hinsberg test uses benzenesulfonyl chloride to distinguish primary, secondary, and tertiary amines. Primary amines react to give sulfonamides that are soluble in alkali (because the N-H is acidic). Secondary amines react to give sulfonamides that are INSOLUBLE in alkali (no N-H to ionize). Tertiary amines do not react. Step 1: Identify what type of amine gives a product insoluble in alkali. When a secondary amine reacts with benzenesulfonyl chloride, the product is a disubstituted sulfonamide (no N-H bond). Without an acidic N-H, the sulfonamide cannot form a sodium salt in alkali, so it remains insoluble in alkali. Step 2: Check the molecular formula of X (C7H9N). For option (a): N-methylaniline = C6H5-NH-CH3 = C7H9N. This is a secondary amine. Degree of unsaturation = (2×7 + 2 - 9 + 1)/2 = (14+2-9+1)/2 = 8/2 = 4, consistent with a benzene ring. Step 3: Verify the product formula Y (C13H13NO2S). N-methylaniline (C7H9N) reacts with benzenesulfonyl chloride (C6H5SO2Cl) to give C6H5-SO2-N(CH3)-C6H5. Formula: C13H13NO2S. Check: C6H5 (6C,5H) + SO2 (S,2O) + N(CH3)(C6H5) = 6+1+6=13 C; 5+3+5=13 H; 1 N; 2 O; 1 S = C13H13NO2S. This matches Y exactly. Step 4: Confirm insolubility in alkali. The product C6H5-SO2-N(CH3)-C6H5 has no N-H bond, so it cannot be deprotonated by alkali. It is insoluble in alkali. This is consistent with the problem statement. Step 5: Eliminate other options. (b) is not a standard stable aromatic amine and its formula doesn't correspond to a simple secondary amine matching these criteria. (c) 2-methylaniline (o-toluidine) is a PRIMARY amine (C7H9N), which would give a sulfonamide soluble in alkali. (d) 4-methylaniline (p-toluidine) is also a PRIMARY amine (C7H9N), which would give a sulfonamide soluble in alkali. Only option (a), N-methylaniline (a secondary amine), reacts with benzenesulfonyl chloride to give a product insoluble in alkali with formula C13H13NO2S. Therefore, the correct answer is A.