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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 · 2017 · Paper II

Q6(d). Photodecarbonylation of norcamphor and bicyclo[3.2.0]heptan-3-one

  • 10 marks
  • Answered
Question

Compounds A and B undergo photodecarbonylation resulting in the formation of I and II (From A) and III and IV (From B). Give their structures.
[A] → (hν) CO + I + II [B] → (hν) CO + III + IV

Answer in brief

I = bicyclo[2.1.1]hexane, II = hexa-1,5-diene; III = bicyclo[2.2.0]hexane, IV = hexa-1,5-diene.

Key points

  • C1–C2 cleavage and loss of CO leave radicals at C-1 and C-3 (C₆H₁₀). Bonding C-1 to C-3 gives I = bicyclo[2.1.1]hexane; cleavage of the C-4–C-7 bond, which links the two radical-bearing arms, gives II = hexa-1,5-diene.
  • loss of CO leaves radicals at C-2 and C-4, attached to the fused carbons C-1 and C-5. Closure (C-2–C-4) gives III = bicyclo[2.2.0]hexane; cleavage of C-1–C-5 (also releasing cyclobutane strain) gives IV = hexa-1,5-diene.
  • All four products are C₆H₁₀ isomers; the ring-closed products are strained bicyclic alkanes that are hard to make by thermal methods.

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. Photodecarbonylation of a cyclic ketone is a Norrish type I process: α-cleavage of the excited C=O gives an acyl–alkyl biradical, which loses CO. The remaining 1,4-biradical either closes (a ring smaller by one carbon) or fragments its central C–C bond to two C=C.

Scheme: From norcamphor: the two radical centres (C-1 and C-3) are 1,4-related through C-7–C-4.

Scheme: From bicyclo[3.2.0]heptan-3-one: radical centres on C-2 and C-4, linked through the C-1–C-5 bond.

  1. A (norcamphor, C₇H₁₀O): C1–C2 cleavage and loss of CO leave radicals at C-1 and C-3 (C₆H₁₀). Bonding C-1 to C-3 gives I = bicyclo[2.1.1]hexane; cleavage of the C-4–C-7 bond, which links the two radical-bearing arms, gives II = hexa-1,5-diene.
  2. B (bicyclo[3.2.0]heptan-3-one): loss of CO leaves radicals at C-2 and C-4, attached to the fused carbons C-1 and C-5. Closure (C-2–C-4) gives III = bicyclo[2.2.0]hexane; cleavage of C-1–C-5 (also releasing cyclobutane strain) gives IV = hexa-1,5-diene.
  3. All four products are C₆H₁₀ isomers; the ring-closed products are strained bicyclic alkanes that are hard to make by thermal methods.

Conclusion. I = bicyclo[2.1.1]hexane, II = hexa-1,5-diene; III = bicyclo[2.2.0]hexane, IV = hexa-1,5-diene.

Application. Photodecarbonylation was the first route to bicyclo[2.1.1]hexane and is still used to make strained small rings (cyclobutanes, bicyclobutanes) from easily made ketones.

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Model answer to Q6(d), 2017 Paper II: solution page 21
Solution page 21 of 23