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JEE Main 2022 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · medium

P-Toluidine is treated successively with (i) NaNO2/HCl, 273-278 K; (ii) CuCN; (iii) H3O^+, heat; (iv) hot alkaline KMnO4 followed by acidification.

Single correct option

p-Toluidine is treated successively with (i) NaNO2/HCl\mathrm{NaNO_2/HCl}, 273273-278 K278\ \mathrm K; (ii) CuCN; (iii) H3O+\mathrm{H_3O^+}, heat; (iv) hot alkaline KMnO4\mathrm{KMnO_4} followed by acidification. The final product is
  1. A. Benzene-1,4-dicarboxylic acid
  2. B. 4-Methylbenzoic acid
  3. C. 4-Cyanobenzoic acid
  4. D. Benzoic acid
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Answer: (A) Benzene-1,4-dicarboxylic acid

(i) Diazotisation gives 4-methylbenzenediazonium chloride. (ii) The Sandmeyer cyanation replaces N2+\mathrm{N_2^+} by −CN\mathrm{-CN} to give 4-methylbenzonitrile. (iii) Acid hydrolysis converts −CN\mathrm{-CN} to −COOH\mathrm{-COOH}: 4-methylbenzoic acid. (iv) Alkaline KMnO4\mathrm{KMnO_4} oxidises the benzylic CH3\mathrm{CH_3} to COOH\mathrm{COOH}, so the final product is terephthalic acid (benzene-1,4-dicarboxylic acid).
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