VKC ScienceChemistry for written exams Telegram

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

Q7(d). 1H NMR of ethyl (E)-but-2-enoate

  • 10 marks
  • Answered
Question

Assign the observed chemical shift values of 1.22, 1.88, 4.13, 5.81 and 6.95 for the following compound:

Answer in brief

1.22 OCH₂CH₃; 1.88 CH₃-C=; 4.13 OCH₂; 5.81 H-α; 6.95 H-β. The β-H is most deshielded by conjugation with C=O, and the large J (15–18 Hz range) shows the trans alkene.

Key points

  • 4.13 against 1.22: the ethyl group of the ester; OCH₂ is next to electronegative O and appears at 4.13 (quartet), its CH₃ at 1.22 (triplet).
  • 6.95 against 5.81: in an α,β-unsaturated carbonyl compound the π electrons are pulled towards C=O (C=C-C=O <→ ⁺C-C=C-O⁻), leaving a partial positive charge on the β-carbon. H-3 (β) is therefore far more deshielded (6.95) than…
  • 1.88: the CH₃ on C-3 is allylic; it is coupled to H-3 (about 7 Hz) and, weakly, to H-2 (four-bond allylic coupling).
  • The large vinylic coupling, J_(H2,H3) in the 15–18 Hz range, confirms the trans (E) geometry (cis vinylic couplings are about 10 Hz).

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 compound is ethyl (E)-but-2-enoate (ethyl crotonate): an α,β-unsaturated ester with H and H trans. Five kinds of proton give five signals. Shifts follow the electronic environment; multiplicities follow the n+1 rule.

Figure: Proton sets: a OCH₂CH₃, b OCH₂, c allylic CH₃, d H-2 (α), e H-3 (β).

δHmultiplicityreason
1.22a, OCH₂CH₃ (3H)triplet (J≈7 Hz)ordinary CH₃, next to CH₂
1.88c, CH₃-CH= (3H)doublet of doublets (7 Hz and a small allylic J)allylic CH₃, slightly deshielded by C=C
4.13b, OCH₂ (2H)quartet (J≈7 Hz)on the ester oxygen: strongly deshielded
5.81d, H-2, =CH-CO (1H)doublet of quartets (large trans J, small allylic J)vinylic, α to C=O
6.95e, H-3, CH₃-CH= (1H)doublet of quartets (large trans J, 7 Hz)vinylic, β to C=O: most deshielded
  1. 4.13 against 1.22: the ethyl group of the ester; OCH₂ is next to electronegative O and appears at 4.13 (quartet), its CH₃ at 1.22 (triplet).
  2. 6.95 against 5.81: in an α,β-unsaturated carbonyl compound the π electrons are pulled towards C=O (C=C-C=O <→ ⁺C-C=C-O⁻), leaving a partial positive charge on the β-carbon. H-3 (β) is therefore far more deshielded (6.95) than H-2 (α, 5.81). H-3 is also split by the CH₃ (quartet).
  3. 1.88: the CH₃ on C-3 is allylic; it is coupled to H-3 (about 7 Hz) and, weakly, to H-2 (four-bond allylic coupling).
  4. The large vinylic coupling, J_(H2,H3) in the 15–18 Hz range, confirms the trans (E) geometry (cis vinylic couplings are about 10 Hz).

Conclusion. 1.22 OCH₂CH₃; 1.88 CH₃-C=; 4.13 OCH₂; 5.81 H-α; 6.95 H-β. The β-H is most deshielded by conjugation with C=O, and the large J (15–18 Hz range) shows the trans alkene.

Application. The α/β vinyl-proton pattern and the size of J are used routinely to assign E/Z geometry of enones and acrylates, for example in checking the products of Wittig and Knoevenagel reactions.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q7(d), 2016 Paper II: book edition page 33
Book edition, page 33
Model answer to Q7(d), 2016 Paper II: book edition page 34
Book edition, page 34