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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

Q8(c). Bullatenone: correcting a structure from IR, NMR and MS

  • 20 marks
  • Answered
Question

An organic compound X was erroneously assigned the following structure:
[X]
The spectral data of the organic compound are given below. On the basis of the spectral data, justify why X is definitely a wrong structure and write the correct structure compatible with the spectral data:
tabular@ll@
IR (cm⁻¹): & 1705, 1604
¹H NMR: & δ_H (ppm) 1.43 (s, 6H), 5.82 (s, 1H), 7.35 (m, 3H), 7.68 (m, 2H)
MS: & m/z 188 (10%) (high resolution confirms C₁₂H₁₂O₂), 105 (20%), 102 (100%), 77 (20%)
tabular

Answer in brief

X cannot be right: it would show a 3H doublet, a CH multiplet and an OCH₂ signal and a C=O band below 1700 cm⁻¹. The data fit 2,2-dimethyl-5-phenylfuran-3(2H)-one (bullatenone): 6H singlet for C(CH₃)₂, vinyl singlet at 5.82, five-ring enone C=O at 1705, and the fragments 102 (PhC#CH⁺), 105 (PhCO⁺) and 77.

Key points

  • The 6H singlet at δ 1.43. X has only one methyl, which would be a 3H doublet (coupled to the ring CH). A 6H singlet requires two equivalent methyl groups on a carbon with no hydrogen, i.e. a C(CH₃)₂ unit next to oxygen…
  • Missing signals. X would show a 1H multiplet for CH-CH₃ (near δ 2.5–2.7) and a 2H O-CH₂ signal (two double doublets near δ 4.1–4.4). None appears; the spectrum has only the vinyl singlet and the aromatic protons besides the 6H…
  • IR. The cyclic vinylogous ester of X (a six-membered O-C=C-C=O ring) would absorb well below 1700 cm⁻¹ (near 1660–1670). The observed 1705 cm⁻¹ fits the same conjugated system in a five-membered ring, where ring strain raises…
  • (CH₃)₂C-O at C-2. δ 5.82 (s, 1H): H-4, the vinyl proton between C=O and the phenyl-bearing C-5, with no neighbours (singlet). δ 7.68 (m, 2H) and 7.35 (m, 3H): ortho and meta + para protons of a phenyl conjugated with C=C–C=O…

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. This is the natural product bullatenone, first assigned the dihydropyranone structure X and later shown by synthesis to be a furanone. The formula C₁₂H₁₂O₂ has seven degrees of unsaturation: a phenyl ring (four), a C=O, a C=C and one more ring.

Scheme: Same formula; the correct structure has two equivalent methyls on a quaternary carbon and no CH₂ or CH.

Why X is wrong

  1. The 6H singlet at δ 1.43. X has only one methyl, which would be a 3H doublet (coupled to the ring CH). A 6H singlet requires two equivalent methyl groups on a carbon with no hydrogen, i.e. a C(CH₃)₂ unit next to oxygen (shifted downfield to 1.43).
  2. Missing signals. X would show a 1H multiplet for CH-CH₃ (near δ 2.5–2.7) and a 2H O-CH₂ signal (two double doublets near δ 4.1–4.4). None appears; the spectrum has only the vinyl singlet and the aromatic protons besides the 6H singlet.
  3. IR. The cyclic vinylogous ester of X (a six-membered O-C=C-C=O ring) would absorb well below 1700 cm⁻¹ (near 1660–1670). The observed 1705 cm⁻¹ fits the same conjugated system in a five-membered ring, where ring strain raises the C=O wavenumber. 1604 cm⁻¹ is the conjugated C=C and aromatic ring stretch.

Correct structure: 2,2-dimethyl-5-phenylfuran-3(2H)-one

  1. NMR assignment. δ 1.43 (s, 6H): (CH₃)₂C-O at C-2. δ 5.82 (s, 1H): H-4, the vinyl proton between C=O and the phenyl-bearing C-5, with no neighbours (singlet). δ 7.68 (m, 2H) and 7.35 (m, 3H): ortho and meta + para protons of a phenyl conjugated with C=C–C=O (the ortho H are deshielded most).
  2. Mass spectrum. m/z 188 = M⁺ (C₁₂H₁₂O₂). 105 = PhC#O⁺ (benzoyl), from C-5 with its ring oxygen. 77 = C₆H₅⁺ (105 - CO). 102 (base peak) = PhC#CH⁺: the ring loses acetone (58) and CO (28) together (188 - 86 = 102), leaving C-4 and C-5 with the phenyl. This fragmentation is natural for the furanone and impossible to explain for X.

Scheme: Key fragments of bullatenone: phenylacetylene ion (base peak), benzoyl and phenyl cations.

Conclusion. X cannot be right: it would show a 3H doublet, a CH multiplet and an OCH₂ signal and a C=O band below 1700 cm⁻¹. The data fit 2,2-dimethyl-5-phenylfuran-3(2H)-one (bullatenone): 6H singlet for C(CH₃)₂, vinyl singlet at 5.82, five-ring enone C=O at 1705, and the fragments 102 (PhC#CH⁺), 105 (PhCO⁺) and 77.

Application. Bullatenone is a classic case of a structure corrected by spectroscopy and synthesis; 3(2H)-furanones of this type occur as flavour compounds (e.g. furaneol in strawberries).

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Model answer to Q8(c), 2020 Paper II: book edition page 30
Book edition, page 30
Model answer to Q8(c), 2020 Paper II: book edition page 31
Book edition, page 31

All parts of Question 8