Monocyclic compounds P, Q, R and S are the major products formed in the reaction sequences given below.
The product having the highest number of unsaturated carbon atom(s) is
← JEE Advanced 2025 · Paper 2![Four reactions. (1) 3-Phenylpropanoic acid, PhCH2CH2COOH → [(i) Br2/red phosphorus; (ii) H2O] → P. (2) Benzaldehyde + acetaldehyde → [aq. NaOH, 293 K] → Q. (3) Phenylacetylene → [(i) NaNH2, allyl bromide (CH2=CHCH2Br); (ii) Hg2+, H3O+] → R. (4) 2-Methylindene (indene with a methyl on C2) → [(i) O3, Zn-H2O; (ii) CH3MgBr (2 equiv.); (iii) H+, Δ] → S.](/figures/ja-2025-p2/ja-2025-p2-che-q03-f1.png)
ChemistrySingle correct+3 / -1Question 35 of 48
JEE Advanced 2025 · Paper 2, Chemistry question 3
Answer and worked solution
Official answer: D
Count the unsaturated (sp² or sp) carbons in each product.
- P: HVZ bromination gives PhCH₂CHBrCOOH. The ring has 6 and the carboxyl has 1: 7.
- Q: Claisen-Schmidt condensation gives cinnamaldehyde, PhCH=CHCHO. Ring 6, alkene 2, CHO 1: 9.
- R: Alkylating the acetylide gives PhC≡CCH₂CH=CH₂. Hydration then gives PhCOCH₂CH₂CH=CH₂. Ring 6, C=O 1, alkene 2: 9.
- S:
- Ozonolysis of 2-methylindene gives o-(2-oxopropyl)benzaldehyde.
- Two equivalents of CH₃MgBr turn both carbonyls into alcohols.
- Acid and heat dehydrate both alcohols, giving a benzene ring carrying a vinyl group and a dimethylvinyl (or isopropenyl-type) group.
- Ring 6 plus two C=C (4): 10.
S has the most.