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JEE Main 2021 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · easy

Toluene is oxidised with alkaline KMnO4 followed by acidification to give A. A is heated with SOCl2 to give B, B reacts with excess NH3 to give C…

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Toluene is oxidised with alkaline KMnO4\mathrm{KMnO_4} followed by acidification to give A\mathrm{A}. A\mathrm{A} is heated with SOCl2\mathrm{SOCl_2} to give B\mathrm{B}, B\mathrm{B} reacts with excess NH3\mathrm{NH_3} to give C\mathrm{C}, and C\mathrm{C} on treatment with Br2\mathrm{Br_2} and KOH gives D\mathrm{D}. D\mathrm{D} is
  1. A. benzylamine
  2. B. aniline
  3. C. pp-toluidine
  4. D. NN-methylaniline
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Answer: (B) aniline

Toluene is oxidised to benzoic acid (A), which with SOCl2\mathrm{SOCl_2} gives benzoyl chloride (B). Ammonia converts it to benzamide (C). The last step is the Hofmann bromamide degradation: C6H5CONH2+Br2+4KOH→C6H5NH2+2KBr+K2CO3+2H2O\mathrm{C_6H_5CONH_2}+\mathrm{Br_2}+4\mathrm{KOH}\to\mathrm{C_6H_5NH_2}+2\mathrm{KBr}+\mathrm{K_2CO_3}+2\mathrm{H_2O}, in which the carbonyl carbon is lost, giving a primary amine with one carbon fewer. D is aniline, not benzylamine, because the amino group is attached directly to the ring.
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