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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:
- o-ethylaniline, N-ethylaniline, p-ethylaniline
- o-ethylaniline, p-ethyltoluene, m-ethylaniline
- o-ethylaniline, ethylbenzene, p-ethylaniline
- 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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