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UPSC Combined Geo-Scientist

Chemist

  • 3 descriptive papers
  • 200 marks each
  • 3 h per paper

Mains Paper III covers analytical and organic chemistry.

UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper III

Q10(b). Norbornenone from cyclopentadiene; hydroboration with 9-BBN

  • 10 marks
  • Topic: Synthetic methods
  • Answered
Question

(i) Suggest a suitable method for the synthesis of the compound (Z), bicyclo[2.2.1]hept-5-en-2-one, from cyclopentadiene. (ii) Identify the major products (X–Z) formed in the following reaction sequence: cycloocta-1,5-diene + BH₃ → X; X + α-methylstyrene → Y; H₂O₂/NaOH → Z.

Answer in brief

(i) Diels–Alder with 2-chloroacrylonitrile (a ketene equivalent), then hydrolysis of the chloronitrile with KOH to the ketone. (ii) X = 9-BBN; Y = the alkylborane PhCH(CH₃)CH₂-B(9-BBN); Z = 2-phenylpropan-1-ol (anti-Markovnikov).

Key points

  • Ketene itself cannot be used as a dienophile (it gives cyclobutanones by [2+2]). 2-Chloroacrylonitrile is the ketene equivalent.
  • Diels–Alder gives 2-chloro-2-cyanonorborn-5-ene (endo/exo mixture; both isomers are used).
  • hydrolysis of the cyanohydrin-like chloronitrile gives CN⁻ and the ketone; Z = bicyclo[2.2.1]hept-5-en-2-one.
  • BH₃ adds to both C=C of 1,5-cyclooctadiene (two hydroborations, syn, one intramolecular) to give 9-borabicyclo[3.3.1]nonane (9-BBN).

The full model answer

Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.

(i) Synthesis of norbornenone (5 marks)

  1. Ketene itself cannot be used as a dienophile (it gives cyclobutanones by [2+2]). 2-Chloroacrylonitrile is the ketene equivalent.
  2. Cyclopentadiene + CH₂=C(Cl)CN, 0°C: Diels–Alder gives 2-chloro-2-cyanonorborn-5-ene (endo/exo mixture; both isomers are used).
  3. KOH in aq. DMSO (or aqueous KOH, then H⁺): hydrolysis of the cyanohydrin-like chloronitrile gives CN⁻ and the ketone; Z = bicyclo[2.2.1]hept-5-en-2-one.

(ii) 9-BBN sequence (5 marks)

  1. X: BH₃ adds to both C=C of 1,5-cyclooctadiene (two hydroborations, syn, one intramolecular) to give 9-borabicyclo[3.3.1]nonane (9-BBN).
  2. Y: the B–H of 9-BBN adds to α-methylstyrene, boron going to the less substituted terminal CH₂, H to the benzylic carbon: PhCH(CH₃)CH₂-BR₂ (anti-Markovnikov, regioselectivity >99% because 9-BBN is bulky).
  3. Z: H₂O₂/NaOH oxidises C–B to C–OH with retention of configuration: PhCH(CH₃)CH₂OH, 2-phenylpropan-1-ol; the by-product is cis-1,5-cyclooctanediol.
Model answer to Q10(b), 2026 Mains Paper III: solution page 28
Solution page 28 of 30
Read the whole 2026 solution2026 question paper

CoversDiels–Alder, hydroborationNorbornenone from cyclopentadienehydroboration with 9-BBN

All parts of Question 10