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JEE Main 2016 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · easy

A compound C7H9N gives an alkali-soluble product with benzenesulfonyl chloride, forms an orange dye when diazotised at 273 K and added to alkaline…

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A compound C7H9N\mathrm{C_7H_9N} gives an alkali-soluble product with benzenesulfonyl chloride, forms an orange dye when diazotised at 273 K273\ \mathrm{K} and added to alkaline β\beta-naphthol, and shows only two doublets in the aromatic region of its 1^1H NMR spectrum. The compound is
  1. A. 4-Methylaniline
  2. B. Benzylamine
  3. C. 2-Methylaniline
  4. D. 3-Methylaniline
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Answer: (A) 4-Methylaniline

Alkali-soluble sulfonamide means a primary amine (Hinsberg test). Benzylamine is primary but its diazonium salt, being aliphatic, decomposes with N2\mathrm{N_2} loss to the alcohol and cannot couple, so the dye test excludes it. Of the toluidines, only the para isomer has a symmetric 1,4-disubstituted ring, which gives the AA′BB′\mathrm{AA'BB'} pattern of two doublets; the ortho and meta isomers show four distinct aromatic protons. Hence 4-methylaniline (p-toluidine).
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