See image — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

Concept: Compound (P) is 4-chloroanilinium chloride (p-chloroaniline hydrochloride), which contains an ionic chloride (Cl-) as a counter-ion to the ammonium group, as well as a covalent C-Cl bond on the ring. Compound (Q) is 1-chloro-4-nitrobenzene, which contains only a covalent C-Cl bond on the ring and no ionic chloride. Step 1 - Using ammoniacal AgNO3 (option a): AgNO3 reacts with ionic (free) chloride ions to give a white precipitate of AgCl. Compound (P) has an ionic Cl- (from the ammonium salt), so it will give a positive test (white precipitate) with amm. AgNO3. Compound (Q) has no free ionic chloride; its Cl is covalently bonded to the aromatic ring and will not precipitate with AgNO3 under normal conditions. Therefore, amm. AgNO3 differentiates (P) from (Q). Step 2 - Using NaOH (option b): NaOH is a base. Compound (P) is a salt of a weak base (aniline) and a strong acid (HCl); treatment with NaOH will liberate the free amine (4-chloroaniline), which may show a characteristic amine behavior or change in solubility/appearance. Compound (Q) is a neutral aromatic compound with no acidic or basic functional group that reacts with NaOH under normal conditions. Thus NaOH also differentiates (P) from (Q). Step 3 - Why FeCl3 (option c) does not differentiate: FeCl3 is used to distinguish phenols (giving violet/purple color) or certain enols. Neither compound (P) nor compound (Q) contains a free phenolic -OH group, so FeCl3 does not provide a clear differentiation between them. Step 4 - Conclusion: Both amm. AgNO3 and NaOH individually differentiate (P) from (Q), so the answer is 'Both (a) & (b)'. Therefore, the correct answer is D.