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JEE Advanced 2026 · Paper 1, Chemistry question 4

Considering LiBHX4\ce{LiBH4} reduces an ester group to the corresponding alcohol and does not reduce a carboxylic acid group, the correct statement about the major products P, Q, R and S is

Top: a cyclobutane whose C1 carries CH₂CO₂Et (wedge) and CH₂CO₂H (hash) and whose neighbouring C2 carries CH₃ (wedge), cis to the ester arm. With (1) LiBH₄, (2) H⁺ it gives P; with (1) BH₃, (2) H⁺ it gives Q.Bottom: a cyclobutane whose C1 carries CH₂CO₂Et (wedge) and CH₂CO₂H (hash) and whose opposite carbon C3 carries CH₃ (wedge), cis to the ester arm. With (1) LiBH₄, (2) H⁺ it gives R; with (1) BH₃, (2) H⁺ it gives S.
Choose one option
Answer and worked solution

Official answer: C

The two reductions.

  • LiBHX4\ce{LiBH4} reduces the ester arm to CHX2CHX2OH\ce{CH2CH2OH}. On acidification, this OH closes a six-membered spiro lactone with the acid arm, so the lactone C=O comes from the arm drawn hashed.
  • BHX3\ce{BH3} reduces the acid arm instead. The new OH lactonizes with the ester arm, so the C=O comes from the arm drawn wedged.

So in each pair the products have the same lactone ring, but the C=O side and the CHX2CHX2O\ce{CH2CH2O} side are swapped on the spiro carbon.

P and Q. The methyl is on the neighbouring carbon C2. The spiro carbon is a stereocentre and C2 is fixed, so swapping two groups on the spiro carbon inverts it alone: diastereomers.

R and S. The methyl is at C3, across the ring. Swapping the arms turns the CHX2C(=O)\ce{CH2C(=O)} group from trans to the methyl (R) into cis (S): cis/trans diastereomers.

Topics:Optical isomerism and chiralityAcid derivativesGeometrical isomerism
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