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JEE Advanced 2025 Paper 2 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · medium

Which of the following statements is/are correct?

One or more options correct

Which of the following statements is/are correct?
  1. A. Benzaldehyde, treated with excess formaldehyde in concentrated NaOH\mathrm{NaOH}, gives benzyl alcohol and sodium formate
  2. B. Acetophenone, with excess I2/NaOH\mathrm{I_2/NaOH} followed by acidification, gives iodoform and phenylacetic acid
  3. C. terttert-Butyl chloride with NaOEt\mathrm{NaOEt} in ethanol gives mainly ethyl terttert-butyl ether by an SN2S_N2 pathway
  4. D. Propanamide on Hofmann bromamide degradation (Br2/NaOH\mathrm{Br_2/NaOH}) gives propan-1-amine
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Answer: A

(A) Correct: crossed Cannizzaro reaction. Neither aldehyde has an α\alpha-hydrogen; the more reactive formaldehyde is attacked by hydroxide first and acts as the hydride donor, so it is oxidised to formate while benzaldehyde accepts the hydride and is reduced to benzyl alcohol. (B) Incorrect: the haloform reaction cleaves the C−COCH3\mathrm{C{-}COCH_3} bond, so the acid formed is benzoic acid (PhCOOH\mathrm{PhCOOH}), not phenylacetic acid. (C) Incorrect: a tertiary halide with a strong base such as ethoxide undergoes mainly E2 elimination to give isobutene; SN2S_N2 at a tertiary carbon is not possible. (D) Incorrect: Hofmann degradation removes the carbonyl carbon, so CH3CH2CONH2\mathrm{CH_3CH_2CONH_2} gives ethanamine (CH3CH2NH2\mathrm{CH_3CH_2NH_2}), one carbon fewer.
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