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JEE Advanced 2025 Paper 1 · Chemistry · General Organic Chemistry and Isomerism · medium

Match each reaction in List-I with the description of its major organic product in List-II. List-I (P) 2,3-Dimethylbutane-2,3-diol heated with dilute…

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Match each reaction in List-I with the description of its major organic product in List-II.

List-I
(P) 2,3-Dimethylbutane-2,3-diol heated with dilute H2SO4\mathrm{H_2SO_4}
(Q) Benzamide treated with Br2\mathrm{Br_2} and aqueous NaOH\mathrm{NaOH}, then warmed
(R) Benzaldehyde and formaldehyde treated with concentrated NaOH\mathrm{NaOH} (organic product other than formate)
(S) Benzil (PhCOCOPh\mathrm{PhCOCOPh}) boiled with aqueous KOH\mathrm{KOH} followed by acidification

List-II
(1) A cyclic amide
(2) A primary aromatic amine
(3) A ketone containing a quaternary carbon atom
(4) An α\alpha-hydroxy carboxylic acid
(5) A primary alcohol in which the carbinol carbon is attached to a phenyl ring
  1. A. P→3, Q→2, R→5, S→4
  2. B. P→3, Q→2, R→4, S→5
  3. C. P→1, Q→2, R→5, S→4
  4. D. P→3, Q→1, R→5, S→2
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Answer: (A) P→3, Q→2, R→5, S→4

P: the pinacol rearrangement. Protonation and loss of water gives a tertiary cation, then a methyl shift gives pinacolone, (CH3)3C ⁣− ⁣CO ⁣− ⁣CH3\mathrm{(CH_3)_3C\!-\!CO\!-\!CH_3}, a ketone with a quaternary carbon (3). Q: the Hofmann bromamide degradation converts PhCONH2\mathrm{PhCONH_2} into PhNH2\mathrm{PhNH_2} with loss of the carbonyl carbon as carbonate, a primary aromatic amine (2). R: crossed Cannizzaro reaction. Formaldehyde is the better hydride donor and is oxidised to formate, while benzaldehyde is reduced to benzyl alcohol, PhCH2OH\mathrm{PhCH_2OH}, a primary alcohol with a benzylic carbinol carbon (5). S: benzilic acid rearrangement of a non-enolisable 1,2-diketone gives Ph2C(OH)COOH\mathrm{Ph_2C(OH)COOH} after acidification, an α\alpha-hydroxy acid (4). Cyclic amide (1) belongs to a Beckmann product and is not used. The correct code is P→3, Q→2, R→5, S→4.
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