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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2026 · Paper II

Q6(a). Chemoselective reductions

  • 10 marks
  • Answered
Question

Write the structures of the products formed in the following reactions:
(i) [structure] → (NaBH₄/CeCl₃, MeOH) (ii) [structure] → (Na/NH₃ (l))
(iii) [structure] → ((i) B₂H₆, (ii) H⁺) (iv) [structure] → ((i) LiAlH₄, (ii) H⁺)
(v) [structure] → (B₂H₆, THF)

Answer in brief

(i) Cyclohex-2-en-1-ol; (ii) cyclohexanone; (iii) and (iv) 4-methyl-δ-valerolactone (borane reduces the acid, LiAlH₄ the ester); (v) methyl 1-benzylpyrrolidine-3-carboxylate.

Key points

  • (iii) B₂H₆ is electrophilic: it reacts fastest with the electron-rich, acidic CO₂H (forming an acyloxyborane) and leaves the ester alone. The acid is reduced to CH₂OH: ethyl 5-hydroxy-3-methylpentanoate. On acidification the…
  • (iv) LiAlH₄ first deprotonates the CO₂H; the carboxylate is negatively charged and is reduced only slowly. The nucleophilic hydride reduces the ester to CH₂OH: 5-hydroxy-3-methylpentanoic acid, which lactonises on acidification.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Introduction. Each reagent reduces only one of the groups present. The choice depends on whether the reducing agent is a nucleophilic hydride (NaBH₄, LiAlH₄), an electrophilic, Lewis-acidic hydride (borane) or an electron donor (sodium in ammonia).

(i) Luche reduction

CeCl₃ is a hard Lewis acid; in methanol it converts NaBH₄ into NaBH(OMe)₃-type hard hydrides and activates the C=O. Hard hydride attacks the hard carbonyl carbon (1,2-addition) instead of the soft β-carbon. Product: cyclohex-2-en-1-ol (allylic alcohol), C=C kept.

(ii) Dissolving-metal reduction

Sodium gives an electron to the enone (radical anion); protonation at the β-carbon and a second electron give the enolate, which is protonated on work-up. The C=C is reduced (conjugate reduction) and the C=O survives as the enolate. Product: cyclohexanone.

Scheme: 1,2-reduction (Luche) and conjugate reduction (Na/NH₃) of the same enone.

(iii) and (iv) Borane and LiAlH₄ on the half-ester

  • (iii) B₂H₆ is electrophilic: it reacts fastest with the electron-rich, acidic CO₂H (forming an acyloxyborane) and leaves the ester alone. The acid is reduced to CH₂OH: ethyl 5-hydroxy-3-methylpentanoate. On acidification the alcohol attacks the ester intramolecularly and closes the δ-lactone.
  • (iv) LiAlH₄ first deprotonates the CO₂H; the carboxylate is negatively charged and is reduced only slowly. The nucleophilic hydride reduces the ester to CH₂OH: 5-hydroxy-3-methylpentanoic acid, which lactonises on acidification.

Scheme: (iii) via the hydroxy ester, (iv) via the hydroxy acid: the same δ-lactone.

Both give 4-methyltetrahydro-2H-pyran-2-one; the opposite ends of the chain are reduced. (If the half-ester were a single enantiomer, the two routes would give the two enantiomers of the lactone.)

(v) Borane with a lactam ester

Borane reduces amides (here the lactam C=O) quickly, because the nitrogen lone pair makes the oxygen basic enough to coordinate boron; esters react only slowly. The lactam C=O becomes CH₂, and the methyl ester is kept.

Scheme: Amide reduced to amine; ester untouched.

Conclusion. (i) Cyclohex-2-en-1-ol; (ii) cyclohexanone; (iii) and (iv) 4-methyl-δ-valerolactone (borane reduces the acid, LiAlH₄ the ester); (v) methyl 1-benzylpyrrolidine-3-carboxylate.

Application. Luche and borane chemoselectivity are routine in drug synthesis (prostaglandin allylic alcohols; amine drugs from amides without touching esters).

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Model answer to Q6(a), 2026 Paper II: solution page 17
Solution page 17 of 26