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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 · 2020 · Paper II

Q2(a). Cyclobutene ring opening and closing, then Diels–Alder

  • 15 marks
  • Answered
Question

Complete the following conversion writing missing intermediate compounds A, B and C:
[structure] → (Δ) A → (hν) B → (Δ) C → ( [structure], Δ) [structure]

Answer in brief

A = (2E,4E)-hexa-2,4-diene (thermal conrotatory opening); B = cis-3,4-dimethylcyclobutene (photochemical disrotatory closure); C = (2E,4Z)-hexa-2,4-diene (thermal conrotatory opening), whose endo Diels–Alder adduct with maleic anhydride has trans methyl groups.

Key points

  • A. The thermal HOMO of the opening butadiene is ψ₂; its end lobes have opposite phase on the same face, so the C3–C4 bond breaks conrotatorily. For the trans isomer, conrotation can turn both methyls outward or both inward;…
  • B. On irradiation the HOMO is ψ₃, whose end lobes have the same phase on the same face: closure is disrotatory. Both outward methyls of the (E,E)-diene turn the same way and finish on the same face: B =…
  • C. Heating B opens it again conrotatorily. In the cis isomer conrotation must turn one methyl outward and one inward: C = (2E,4Z)-hexa-2,4-diene. The overall result is a geometrical isomerisation of the diene through the…
  • Diels–Alder. C in its s-cis form adds to maleic anhydride, suprafacial on both components and endo (secondary orbital overlap). The outside (E) methyl and the inside (Z) methyl end up on opposite faces of the new ring, so the…

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. The starting material is trans-3,4-dimethylcyclobutene (one methyl above the ring, one below). Each step is pericyclic and stereospecific: thermal 4π ring opening is conrotatory, photochemical 4π ring closure is disrotatory, and the Diels–Alder reaction is a suprafacial [4πₛ+2πₛ] cycloaddition. The printed adduct has its two methyls trans, which fixes the geometry of C.

Scheme: trans-Cyclobutene → (E,E)-diene → cis-cyclobutene → (E,Z)-diene; con = conrotatory, dis = disrotatory.

  1. A. The thermal HOMO of the opening butadiene is ψ₂; its end lobes have opposite phase on the same face, so the C3–C4 bond breaks conrotatorily. For the trans isomer, conrotation can turn both methyls outward or both inward; both-outward avoids steric clash, giving A = (2E,4E)-hexa-2,4-diene.
  2. B. On irradiation the HOMO is ψ₃, whose end lobes have the same phase on the same face: closure is disrotatory. Both outward methyls of the (E,E)-diene turn the same way and finish on the same face: B = cis-3,4-dimethylcyclobutene. The cyclobutene absorbs below the irradiation wavelength, so it accumulates.
  3. C. Heating B opens it again conrotatorily. In the cis isomer conrotation must turn one methyl outward and one inward: C = (2E,4Z)-hexa-2,4-diene. The overall result is a geometrical isomerisation of the diene through the cyclobutene.
  4. Diels–Alder. C in its s-cis form adds to maleic anhydride, suprafacial on both components and endo (secondary orbital overlap). The outside (E) methyl and the inside (Z) methyl end up on opposite faces of the new ring, so the adduct is trans-3,6-dimethylcyclohex-4-ene-cis-1,2-dicarboxylic anhydride, exactly as printed. An (E,E)-diene would have given the cis-3,6-dimethyl adduct.

Conclusion. A = (2E,4E)-hexa-2,4-diene (thermal conrotatory opening); B = cis-3,4-dimethylcyclobutene (photochemical disrotatory closure); C = (2E,4Z)-hexa-2,4-diene (thermal conrotatory opening), whose endo Diels–Alder adduct with maleic anhydride has trans methyl groups.

Application. Thermal ring opening of cyclobutenes (and of benzocyclobutenes to o-xylylenes) is used to generate dienes of defined geometry for Diels–Alder steps in steroid and alkaloid synthesis.

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Model answer to Q2(a), 2020 Paper II: solution page 6
Solution page 6 of 21

All parts of Question 2