JEE Advanced 2023 Paper 1 Q02 Chemistry Organic Chemistry Isomerism & Stereochemistry Hard

JEE Advanced 2023 Paper 1 · Q02 · Isomerism & Stereochemistry

In the following reactions, $P$, $Q$, $R$, and $S$ are the major products. ($P$: $CH_3CH_2CH(CH_3)CH_2CN$ with (i) PhMgBr then $H_3O^+$, (ii) PhMgBr then $H_2O$; $Q$: $C_6H_6$ + $CH_3COCl$ with anhyd $AlCl_3$, then PhMgBr/$H_2O$; $R$: $CH_3CH_2COCl$ with $(PhCH_2)_2Cd$ then PhMgBr/$H_2O$; $S$: $PhCH_2CHO$ with PhMgBr/$H_2O$, then $CrO_3$/dil $H_2SO_4$, then HCN, then $H_2SO_4/\Delta$.) The correct statement(s) about $P$, $Q$, $R$, and $S$ is(are)

  1. A. Both $P$ and $Q$ have asymmetric carbon(s).
  2. B. Both $Q$ and $R$ have asymmetric carbon(s).
  3. C. Both $P$ and $R$ have asymmetric carbon(s).
  4. D. $P$ has asymmetric carbon(s), $S$ does not have any asymmetric carbon.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:C, D

Solution

$P$: PhMgBr adds to nitrile, hydrolysis gives a ketone, second PhMgBr addition gives a tertiary alcohol $CH_3CH_2C^*H(CH_3)CH_2C(OH)(Ph)(CH_3)$ — chiral carbon present. $Q$: Friedel–Crafts acylation gives acetophenone; PhMgBr addition gives $PhC(OH)(CH_3)Ph$ (1,1-diphenylethanol) — no asymmetric carbon. $R$: $(PhCH_2)_2Cd$ converts acyl chloride to benzyl ethyl ketone, PhMgBr addition gives $CH_3CH_2C^*(OH)(CH_2Ph)(Ph)$ — chiral. $S$: terminal sequence ends as $PhCH=C(Ph)COOH$ via dehydration — planar, no chirality. Therefore (C) and (D) are correct.

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