Compound P, →(HCl, Δ), Filter, {→ Residue Q; → Filtrate; (M.F. C₁₄H₁₃ON), ↓ NaOH; M.F. = MOLECULAR FORMULA, Oily Liquid R.
Compound P is neutral, Q gives effervescence with NaHCO₃ while R reacts with Hinsbergs reagent to give solid soluble in NaOH. Compound P is

Solution

Chemistry ma’am · 3 steps

Neutral P with nitrogen and oxygen is an amide. Hydrolyse it and the NaOH test on the amine tells which one.

1. Read the clues. P is neutral, so it is an amide, not an amine. Boiling with HCl splits it into a carboxylic acid and an amine. The acid is the solid residue Q, which fizzes with NaHCO₃. The amine stays in the filtrate as its hydrochloride salt.
2. Free the amine. NaOH releases the free amine R as an oily liquid. Its Hinsberg product dissolves in NaOH, so the sulphonamide still has an N–H. That makes R a primary amine.

   Option	What HCl does	Amine released by NaOH	Hinsberg product in NaOH
   (a) C₆H₅CON(CH₃)C₆H₅	benzoic acid + C₆H₅NH(CH₃)	N-methylaniline, secondary	insoluble
   (b) p-CH₃C₆H₄CONHC₆H₅	p-toluic acid + C₆H₅NH₃⁺Cl⁻	aniline, primary, oily	soluble
   (c) C₆H₅COC₆H₄NHCH₃	no amide to hydrolyse, basic amine dissolves	no acid residue forms	secondary, so insoluble
   (d) ArCONH₂	acid + NH₄Cl	NH₃ gas, not an oil	does not fit

3. So. Only (b) fits every clue: Q is p-toluic acid and R is aniline, whose benzenesulphonamide C₆H₅SO₂NHC₆H₅ dissolves in NaOH. P is 4-methyl-N-phenylbenzamide, option (b).

The idea to keep. In a Hinsberg puzzle, a product soluble in NaOH means a primary amine, because the sulphonamide keeps an acidic N–H. A secondary amine gives an insoluble product and a tertiary amine gives none.

Answer from the NTA final answer key. Solution by PrepWiser.