Three isomers A, B and C (mol. formula \(\mathrm{C_8H_{11}N}\)) give the following results : R has lower boiling point than S A, B and C, respectively are :
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Hinsberg test: only B gives an alkali-insoluble sulfonamide, so B is a secondary amine; A and C are primary aromatic amines that diazotize and finally give hydroxybenzoic acids, whose boiling points depend on ortho (intramolecular H-bond) versus para (intermolecular H-bond).
A and C undergo diazotization, so both are primary aromatic amines (ArNH2). Their diazonium salts P and Q on hydrolysis (phenols) and then KMnO4/H+ oxidation (alkyl to -COOH) give hydroxybenzoic acids R (from A) and S (from C). B reacts with benzenesulfonyl chloride to give an alkali-insoluble product, i.e. B is a secondary amine (the sulfonamide has no N-H). This rejects option (1), where B (3-methylbenzylamine) is primary. R has a lower boiling point than S: o-hydroxybenzoic acid (salicylic acid) has intramolecular H-bonding (lower b.p.), while p-hydroxybenzoic acid has intermolecular H-bonding (higher b.p.). So A must be the ortho isomer (2-ethylaniline) and C the para isomer (4-ethylaniline). Option (2) has A = 3-ethylaniline (meta acid) and C = 2-ethylaniline (ortho acid), so R would boil higher: rejected. Option (3) has A para and C ortho: R boils higher: rejected. Option (4): A = 2-ethylaniline, B = N-methylbenzylamine (secondary), C = 4-ethylaniline satisfies everything. Answer: (4) A = 2-ethylaniline, B = N-methylbenzylamine, C = 4-ethylaniline
