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JEE Advanced levelSingle correct optionDifficulty 3 of 3o1203

Methyl 4-(chlorocarbonyl)benzoate is treated with 1 equiv of aniline and pyridine at 0\,^\circ\mathrm{C} in dry . The product is

  1. AN1,N4-diphenylbenzene-1,4-dicarboxamide
  2. B4-(methoxycarbonyl)benzoic acid
  3. Cphenyl 4-(chlorocarbonyl)benzoate
  4. Dmethyl 4-(phenylcarbamoyl)benzoate

Answer: D

The acid chloride is far more reactive than the methyl ester (better leaving group). Aniline attacks it selectively at low temperature with one equivalent, giving the amide; pyridine takes up HCl. The ester survives, so the product is methyl 4-(phenylcarbamoyl)benzoate.
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