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JEE Main 2020 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · medium

Propanamide is treated with Br2/NaOH to give X. X on warming with CHCl3/alc. KOH gives Y. Complete acid hydrolysis of Y gives

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Propanamide is treated with Br2/NaOH\mathrm{Br_2/NaOH} to give X. X on warming with CHCl3/alc. KOH\mathrm{CHCl_3/alc.\ KOH} gives Y. Complete acid hydrolysis of Y gives
  1. A. C2H5NH3+\mathrm{C_2H_5NH_3^+} and HCOOH\mathrm{HCOOH}
  2. B. C2H5COOH\mathrm{C_2H_5COOH} and NH4+\mathrm{NH_4^+}
  3. C. C2H5NH2\mathrm{C_2H_5NH_2} and CH3COOH\mathrm{CH_3COOH}
  4. D. CH3COOH\mathrm{CH_3COOH} and NH4+\mathrm{NH_4^+}
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Answer: (A) C2H5NH3+\mathrm{C_2H_5NH_3^+} and HCOOH\mathrm{HCOOH}

Hofmann bromamide reaction removes the carbonyl carbon: C2H5CONH2→C2H5NH2\mathrm{C_2H_5CONH_2\to C_2H_5NH_2} (X, one carbon fewer). Carbylamine reaction converts the primary amine to ethyl isocyanide C2H5NC\mathrm{C_2H_5NC} (Y). Isocyanides hydrolyse in dilute acid to the primary amine salt and formic acid: C2H5NC+2H2O→C2H5NH2+HCOOH\mathrm{C_2H_5NC+2H_2O\to C_2H_5NH_2+HCOOH}.
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