JEE Advanced 2025 Paper 2 · Q03 · Named Reactions & Reagents

Question

Monocyclic compounds P, Q, R and S are the major products formed in the reaction sequences given below.

Sequence 1 (-> P): 3-phenylpropanoic acid ($Ph-CH_2-CH_2-COOH$) treated with (i) $Br_2$/red phosphorus, then (ii) $H_2O$. (Hell-Volhard-Zelinsky alpha-bromination, gives $Ph-CH_2-CHBr-COOH$.)

Sequence 2 (-> Q): benzaldehyde ($PhCHO$) and acetaldehyde ($CH_3CHO$) in aqueous $NaOH$, 293 K. (Crossed aldol followed by dehydration; gives cinnamaldehyde, $Ph-CH=CH-CHO$.)

Sequence 3 (-> R): a vinyl bromide / haloalkene treated with (i) $NaNH_2$, (ii) $Hg^{2+}/H_3O^+$. (Generates an alkyne via elimination, then Markovnikov hydration to a methyl ketone.)

Sequence 4 (-> S): cyclohexene-derived aldehyde treated with (i) $O_3/Zn-H_2O$, (ii) $CH_3MgBr$ (2 equiv), (iii) $H^+$/heat. (Ozonolysis to a dialdehyde, double Grignard, then acid-catalysed dehydration to a phenyl-substituted cyclic conjugated diene.)

Among P, Q, R, S, the product having the highest number of unsaturated carbon atom(s) is:

SolveKar AI · Accepted Answer

Counting unsaturated ($sp$ / $sp^2$) carbons (aromatic carbons + $C=O$ + $C=C$ + $C$-triple-bond):
$P = Ph-CH_2-CHBr-COOH$: 6 (aromatic) + 1 (carboxyl $C=O$) = 7.
$Q = Ph-CH=CH-CHO$: 6 (aromatic) + 2 (alkene) + 1 (aldehyde $C=O$) = 9.
$R$ = a methyl cyclic ketone with one $C=O$: ~5-6 unsaturated carbons.
$S$ = a phenyl-substituted cyclic 1,3-diene: 6 (aromatic) + 4 (two $C=C$ of conjugated diene) = 10.
S has the most unsaturated carbons. Answer: (D) S.

JEE Advanced 2025 Paper 2 · Q03 · Named Reactions & Reagents

Solution