Q8(b). 1H NMR of the three bromochloroethylenes
How will you distinguish various isomers of bromochloroethylene on the basis of 1H-NMR spectroscopy?
Answer in brief
All three isomers give two doublets; the coupling constant identifies them: geminal 2J about 0–3 Hz (CH₂=CBrCl, upfield), cis 3J about 6–8 Hz (Z-BrCH=CHCl) and trans 3J about 12–14 Hz (E-BrCH=CHCl).
Key points
- 1-Bromo-1-chloroethene, CH₂=CBrCl: the two H are on the same carbon (cis to Br and cis to Cl respectively). They couple by a very small geminal 2J (about 0–3 Hz) and appear at higher field, near δ 5.5–6, as two close doublets…
- (Z)-1-Bromo-2-chloroethene: H on adjacent carbons, cis 3J, about 6–8 Hz (halogens lower the usual 10–12 Hz).
- (E)-1-Bromo-2-chloroethene: trans 3J, about 12–14 Hz, the largest coupling.
- Why the couplings differ. Vicinal coupling through a C=C depends on the H–C=C–H dihedral angle: 180° (trans) gives the best orbital overlap and the largest 3J; 0° (cis) a smaller one. Geminal coupling on an sp² CH₂ is close…
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. C₂H₂BrCl has three isomers: 1-bromo-1-chloroethene (both halogens on one carbon), and the cis (Z) and trans (E) isomers of 1-bromo-2-chloroethene. Each has two non-equivalent vinyl protons that couple to each other, so each spectrum is an AB (or AX) pair of doublets. The isomers are told apart by the size of the H–H coupling constant and by the chemical shifts.
Scheme: Geminal, cis and trans H–H relationships.
- 1-Bromo-1-chloroethene, CH₂=CBrCl: the two H are on the same carbon (cis to Br and cis to Cl respectively). They couple by a very small geminal 2J (about 0–3 Hz) and appear at higher field, near δ 5.5–6, as two close doublets or an apparent singlet.
- (Z)-1-Bromo-2-chloroethene: H on adjacent carbons, cis 3J, about 6–8 Hz (halogens lower the usual 10–12 Hz).
- (E)-1-Bromo-2-chloroethene: trans 3J, about 12–14 Hz, the largest coupling.
Figure: Each isomer: two doublets; the splitting is set by the H–H relationship (schematic, not to scale).
- Why the couplings differ. Vicinal coupling through a C=C depends on the H–C=C–H dihedral angle: 180° (trans) gives the best orbital overlap and the largest 3J; 0° (cis) a smaller one. Geminal coupling on an sp² CH₂ is close to zero. Electronegative substituents (Cl, Br) reduce all of them, which is why the values are below the 10–12 (cis) and 14–18 Hz (trans) of simple alkenes.
- Chemical shifts. In the 1,2-isomers each H sits on a halogen-bearing carbon and is deshielded (about δ 6.3–6.8). In CH₂=CBrCl the CH₂ protons are on the unsubstituted carbon and appear further upfield.
- Order of identification. Smallest J and upfield signals: 1,1-isomer; medium J: Z-1,2; largest J: E-1,2.
Conclusion. All three isomers give two doublets; the coupling constant identifies them: geminal 2J about 0–3 Hz (CH₂=CBrCl, upfield), cis 3J about 6–8 Hz (Z-BrCH=CHCl) and trans 3J about 12–14 Hz (E-BrCH=CHCl).
Application. J(trans) > J(cis) > J(gem) is the routine NMR test for alkene geometry, used for example to assign E/Z pheromones and the products of Wittig reactions.
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All parts of Question 8
- 8(a) Structure of C₇H₅Cl₃O from its NMR spectrum15 marks
- 8(b) 1H NMR of the three bromochloroethylenes15 marks
- 8(c)(i) Monomers of PS, LDPE, PP, PMMA and Buna-S10 marks
- 8(c)(ii) Photodecarbonylation; trapping the o-xylylene10 marks