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JEE Main 2017 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · medium

4-Methylbenzamide is treated successively with (i) Br2/KOH, (ii) NaNO2/HCl at 273-278 K, (iii) CuCN, and (iv) H3O^+, heat.

Single correct option

4-Methylbenzamide is treated successively with (i) Br2/KOH\mathrm{Br_2/KOH}, (ii) NaNO2/HCl\mathrm{NaNO_2/HCl} at 273273-278 K278\ \mathrm{K}, (iii) CuCN\mathrm{CuCN}, and (iv) H3O+\mathrm{H_3O^+}, heat. The final product is
  1. A. 4-methylphenol
  2. B. 4-methylaniline
  3. C. 4-methylbenzamide
  4. D. 4-methylbenzoic acid
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Answer: (D) 4-methylbenzoic acid

(i) Hofmann bromamide degradation removes the carbonyl carbon: ArCONH2→ArNH2\mathrm{ArCONH_2\to ArNH_2} (4-methylaniline, one carbon fewer). (ii) Diazotisation gives the diazonium chloride. (iii) Sandmeyer reaction with CuCN gives 4-methylbenzonitrile, restoring a carbon on the ring. (iv) Acid hydrolysis converts the nitrile to the carboxylic acid: 4-methylbenzoic acid. Note that the product has the same carbon count as the starting amide but is the acid, not the amide.
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