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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 · 2016 · Paper II

Q5(b). Cope elimination of an amine oxide

  • 10 marks
  • Answered
Question

Write the major product (X) in the reaction given below and explain its formation:
[structure] → (Δ) (X)

Answer in brief

X = 1-methylcyclopentene, formed by Cope (Ei) syn elimination through a planar five-membered transition state; methylenecyclopentane is the minor alkene and Me₂NOH the by-product.

Key points

  • The substrate is (1-methylcyclopentyl)dimethylamine oxide. The β-hydrogens are on the ring CH₂ groups (C-2, C-5; only the one H on each that is cis to the C–N bond can be used) and on the methyl group (three H, any of which…
  • In the transition state the O⁻ pulls a β-H while the C–N bond breaks; the five atoms (C_α, C_β, H, O, N) must lie nearly in one plane, so the H and the N-O group must be syn-periplanar.
  • In a cyclopentane ring a ring C–H next to C–N can be eclipsed with it (the ring is nearly planar and flexible), so a ring H can reach the planar transition state as easily as a methyl H. When more than one β-H can reach it,…
  • Removing a methyl H gives methylenecyclopentane (disubstituted, exocyclic), the minor product. (In the six-membered analogue the ring H cannot be syn-periplanar to an equatorial or axial C–N in the chair, and…

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. A tertiary amine oxide with a β-hydrogen decomposes on heating (100–150°C) into an alkene and N,N-dimethylhydroxylamine. This is the Cope elimination: an intramolecular syn elimination (Ei) through a five-membered cyclic transition state, with no base or ion intermediates.

Scheme: The O⁻ removes a β-H on the same side as the C–N bond.

Figure: Sketch of the Cope transition state: O⁻ takes the syn β-H, the C–H pair becomes the C=C π bond, and the C–N pair stays on N.

  1. The substrate is (1-methylcyclopentyl)dimethylamine oxide. The β-hydrogens are on the ring CH₂ groups (C-2, C-5; only the one H on each that is cis to the C–N bond can be used) and on the methyl group (three H, any of which can rotate into place). Counted alone, the H atoms favour the methyl side: the Cope elimination, like the Hofmann elimination, usually removes H from the β-carbon with the most H.
  2. In the transition state the O⁻ pulls a β-H while the C–N bond breaks; the five atoms (C_α, C_β, H, O, N) must lie nearly in one plane, so the H and the N-O group must be syn-periplanar.
  3. In a cyclopentane ring a ring C–H next to C–N can be eclipsed with it (the ring is nearly planar and flexible), so a ring H can reach the planar transition state as easily as a methyl H. When more than one β-H can reach it, the product ratio follows the stability of the competing transition states; the one leading to the trisubstituted endocyclic alkene is lower: X = 1-methylcyclopent-1-ene is expected as the major product (the selectivity is not high).
  4. Removing a methyl H gives methylenecyclopentane (disubstituted, exocyclic), the minor product. (In the six-membered analogue the ring H cannot be syn-periplanar to an equatorial or axial C–N in the chair, and methylenecyclohexane becomes the major product.)
  5. By-product: (CH₃)₂N-OH. The elimination is concerted; no rearrangement of the carbon skeleton occurs.

Conclusion. X = 1-methylcyclopentene, formed by Cope (Ei) syn elimination through a planar five-membered transition state; methylenecyclopentane is the minor alkene and Me₂NOH the by-product.

Application. The mild Cope elimination, which gives no skeletal rearrangement, is used to make alkenes from amines (via H₂O₂ oxidation) and to make trans-cyclooctene from N,N-dimethylcyclooctylamine oxide.

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Model answer to Q5(b), 2016 Paper II: solution page 16
Solution page 16 of 25