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ExamsJEE MainChemistry

Three isomers A, B and C (mol. formula C8H11N) give the following results: A and C — Azotization → P + Q (i) Hydrolysis (ii) Oxidation (KMnO4/H+) → R (product of A) → S (product of C) R has lower boiling point than S B — C6H5SO2Cl → alkali-insoluble product A, B and C, respectively are:

  1. o-ethylaniline, N-ethylaniline, p-ethylaniline
  2. o-ethylaniline, p-ethyltoluene, m-ethylaniline
  3. o-ethylaniline, ethylbenzene, p-ethylaniline
  4. o-ethylaniline, p-ethylaniline, m-ethylaniline

Correct answer: o-ethylaniline, N-ethylaniline, p-ethylaniline

Solution

A and C undergo diazotization, so they are primary aromatic amines; B gives an alkali-insoluble sulfonamide with PhSO2Cl, so it is a secondary amine (N-ethylaniline). The ortho product (from A) has intramolecular H-bonding and lower boiling point than the para product (from C). Thus A = o-ethylaniline, B = N-ethylaniline, C = p-ethylaniline.

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