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JEE Advanced 2023 Paper 1 · Chemistry · General Organic Chemistry and Isomerism · medium

Match the transformation in List-I with the correct statement about its product in List-II. List-I: (P) Benzamide treated with Br2 and aqueous NaOH…

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Match the transformation in List-I with the correct statement about its product in List-II.

List-I: (P) Benzamide treated with Br2\mathrm{Br_2} and aqueous NaOH\mathrm{NaOH}, then warmed; (Q) the oxime of acetophenone treated with conc. H2SO4\mathrm{H_2SO_4}; (R) benzaldehyde treated with concentrated aqueous NaOH\mathrm{NaOH}; (S) phenol treated with CHCl3\mathrm{CHCl_3} and aqueous NaOH\mathrm{NaOH}, followed by acidification

List-II: (1) the product shows intramolecular hydrogen bonding; (2) a primary aromatic amine with one carbon atom fewer than the reactant is formed; (3) both an alcohol and a carboxylate salt are formed; (4) the product is NN-phenylacetamide; (5) a carboxylic acid with the same number of carbon atoms as the reactant is formed by oxidation with KMnO4\mathrm{KMnO_4}
  1. A. P→2, Q→4, R→3, S→1
  2. B. P→4, Q→2, R→3, S→1
  3. C. P→2, Q→4, R→5, S→1
  4. D. P→2, Q→1, R→3, S→4
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Answer: (A) P→2, Q→4, R→3, S→1

(P) Hofmann bromamide degradation: PhCONH2+Br2+4NaOH→PhNH2+2NaBr+Na2CO3+2H2O\mathrm{PhCONH_2+Br_2+4NaOH\to PhNH_2+2NaBr+Na_2CO_3+2H_2O}; aniline is a primary aromatic amine with one carbon fewer ⇒\Rightarrow 2. (Q) Beckmann rearrangement of acetophenone oxime: the group anti to OH\mathrm{OH} (phenyl) migrates to nitrogen, giving PhNHCOCH3\mathrm{PhNHCOCH_3}, NN-phenylacetamide (acetanilide), after tautomerisation of the imidic intermediate ⇒\Rightarrow 4. (R) Benzaldehyde has no α\alpha-hydrogen, so it undergoes the Cannizzaro reaction: 2PhCHO+NaOH→PhCH2OH+PhCOONa2\mathrm{PhCHO+NaOH\to PhCH_2OH+PhCOONa} ⇒\Rightarrow 3. (S) Reimer-Tiemann reaction gives salicylaldehyde via dichlorocarbene; the ortho CHO\mathrm{CHO} and OH\mathrm{OH} show intramolecular hydrogen bonding ⇒\Rightarrow 1. (5) is a distractor, since none of the reactions involves permanganate. Correct code: P→2, Q→4, R→3, S→1.
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Original question written in the JEE Advanced 2023 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake