Skip to content
dronacharya.ai
← JEE Advanced 2026 · Paper 1
ChemistryList match+4 / -1Question 47 of 48

JEE Advanced 2026 · Paper 1, Chemistry question 15

The List-II contains products obtained from the reaction of compounds in List-I with OX3\ce{O3}/Zn-HX2O\ce{H2O} followed by cyclization (via more stable enolate) in the presence of aqueous NaOH. Match each entry in List-I with appropriate entry in List-II and choose the correct option.

List-I

  • P
    Octahydronaphthalene with the C=C at the ring fusion; one CH₃ on a carbon of the left ring next to the upper fusion carbon, one on a carbon of the right ring next to the lower fusion carbon.
  • Q
    Bicyclo[5.3.0]decene with the C=C at the fusion; CH₃ on the seven-membered ring next to the upper fusion carbon and CH₃ on the five-membered ring next to the lower fusion carbon.
  • R
    Bicyclo[5.3.0]decene with the C=C at the fusion; CH₃ groups on the two seven-membered-ring carbons next to each fusion carbon.
  • S
    Octahydronaphthalene with the C=C at the fusion; CH₃ groups on the two left-ring carbons next to each fusion carbon.

List-II

  • 1
    Decalin: one fusion carbon bears OH, the other CH₃; a C=O in the right ring next to the CH₃-bearing fusion carbon; a CH₃ on the right-ring carbon next to the OH-bearing fusion carbon.
  • 2
    Bicyclo[5.3.0]decane: fusion carbons bear CH₃ and OH; C=O in the seven-membered ring next to the CH₃ fusion carbon; CH₃ on the seven-membered-ring carbon next to the OH fusion carbon.
  • 3
    Bicyclo[5.3.0]decane: fusion carbons bear CH₃ and OH; C=O in the seven-membered ring next to the CH₃ fusion carbon; CH₃ on the five-membered-ring carbon next to the OH fusion carbon.
  • 4
    Decalin: one fusion carbon bears OH, the other CH₃; C=O in the right ring next to the OH fusion carbon; CH₃ on the left-ring carbon next to the OH fusion carbon.
  • 5
    Decalin: one fusion carbon bears OH, the other CH₃; C=O in the left ring next to the CH₃ fusion carbon; CH₃ on the right-ring carbon next to the OH fusion carbon.
Choose one option
Answer and worked solution

Official answer: C

General course. Ozonolysis of the tetrasubstituted ring-fusion alkene opens both rings into one ten-membered ring with two ketones. Each ketone has a CH(CHX3)\ce{CH(CH3)} neighbour. The more stable (more substituted) enolate forms at that carbon and attacks the other ketone, giving a fused β-hydroxy ketone. The former CH(CHX3)\ce{CH(CH3)} carbon becomes a fusion carbon bearing CHX3\ce{CH3}; the attacked carbonyl becomes the other fusion carbon, bearing OH.

(P) Each ketone has its CH(CHX3)\ce{CH(CH3)} on opposite sides (3 carbons between one pair, 3 between the other). Bond formation closes a 5-ring and a 7-ring. The C=O stays in the 7-ring next to the CHX3\ce{CH3} fusion carbon, and the other CHX3\ce{CH3} is on the 7-ring next to the OH carbon → 2.

(Q) From the 7/5 system, the new bond closes two 6-rings. The C=O and the other CHX3\ce{CH3} end up in the same ring, next to the CHX3\ce{CH3} and OH fusion carbons respectively → 1.

(R) Both CHX3\ce{CH3} groups were on the 7-ring. The new bond again gives two 6-rings, but now the C=O is in one ring and the second CHX3\ce{CH3} is in the other, next to the OH carbon → 5.

(S) Both CHX3\ce{CH3} groups were on one 6-ring. The new bond closes a 5-ring (with the second CHX3\ce{CH3}, next to the OH carbon) and a 7-ring (with the C=O, next to the CHX3\ce{CH3} fusion carbon) → 3.

P→2, Q→1, R→5, S→3.

Topics:Aldol and related condensationsAlkenes
← Previous questionNext question →