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JEE Main 2018 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · medium

4-Methylbenzoic acid is converted by the sequence SOCl2; NH3; Br2/NaOH; NaNO2/HCl at 273 K; then warming with water. The final organic product is

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4-Methylbenzoic acid is converted by the sequence SOCl2\mathrm{SOCl_2}; NH3\mathrm{NH_3}; Br2/NaOH\mathrm{Br_2/NaOH}; NaNO2/HCl\mathrm{NaNO_2/HCl} at 273273 K; then warming with water. The final organic product is
  1. A. 4-methylphenol
  2. B. 3-methylphenol
  3. C. (4-methylphenyl)methanol
  4. D. 4-methylbenzoic acid
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Answer: (A) 4-methylphenol

SOCl2\mathrm{SOCl_2} gives the acid chloride and NH3\mathrm{NH_3} the amide, p-CH3C6H4CONH2p\text{-}\mathrm{CH_3C_6H_4CONH_2}. Hofmann bromamide degradation (Br2/NaOH\mathrm{Br_2/NaOH}) removes the carbonyl carbon, giving pp-toluidine. NaNO2/HCl\mathrm{NaNO_2/HCl} at 00-5∘5^\circC gives the diazonium salt, which on warming with water is hydrolysed to the phenol, with the methyl group still para: 4-methylphenol (pp-cresol). No rearrangement to the meta isomer occurs.
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