Q5(d). Aromatic out-of-plane C–H wagging and adjacent hydrogens
Arrange the following compounds in increasing order of out of plane C–-H wagging vibrations, as shown, and give a suitable explanation to justify your answer:
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(A) 1,3-R₂C₆H₄, the H between the two R shown & (B) 1,2-R₂C₆H₄, the four ring H shown
(C) 1,3-R₂C₆H₄, the three adjacent H shown & (D) benzene, all six H shown
tabular
Answer in brief
Increasing wavenumber: D (6 adjacent H) < B (4) < C (3) < A (isolated H). The fewer adjacent hydrogens, the higher the out-of-plane wagging frequency.
Key points
- all six H wag together; the band is lowest, near 670–675 cm⁻¹.
- B, ortho-disubstituted (four adjacent H): 735–770 cm⁻¹.
- 750–810 cm⁻¹ (meta compounds also show a ring band at 690–710).
- A, the isolated H between the two R groups of a meta compound: it has no H neighbour to couple with, so it wags at the highest wavenumber, about 860–900 cm⁻¹.
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. In the out-of-plane wagging (γC–H) mode the ring hydrogens move above and below the ring plane. Adjacent hydrogens are coupled and move in phase, so the frequency depends on the number of adjacent ring hydrogens, not on the nature of R: the more adjacent H atoms, the lower the wavenumber.
Figure: Typical ranges of strong γC–H bands; each bar is labelled, colour is only a guide.
- D, benzene (six adjacent H): all six H wag together; the band is lowest, near 670–675 cm⁻¹.
- B, ortho-disubstituted (four adjacent H): 735–770 cm⁻¹.
- C, meta-disubstituted (three adjacent H): 750–810 cm⁻¹ (meta compounds also show a ring band at 690–710).
- A, the isolated H between the two R groups of a meta compound: it has no H neighbour to couple with, so it wags at the highest wavenumber, about 860–900 cm⁻¹.
- Reason. When several neighbouring C–H oscillators move in phase, the effective mass taking part grows and the restoring force per oscillator falls; the coupled mode therefore drops in frequency as the run of adjacent H atoms grows. Two adjacent H (para) appear at 810–840.
Conclusion. Increasing wavenumber: D (6 adjacent H) < B (4) < C (3) < A (isolated H). The fewer adjacent hydrogens, the higher the out-of-plane wagging frequency.
Application. The 650–900 cm⁻¹ pattern is the quickest IR test of the substitution pattern of a benzene ring, used for example to distinguish o-, m- and p-xylene in mixtures.
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All parts of Question 5
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- 5(c) Photoenolisation and intramolecular Diels–Alder: estrone skeleton10 marks
- 5(d) Aromatic out-of-plane C–H wagging and adjacent hydrogens10 marks
- 5(e) Methyl groups of 7,7-dimethylnorbornene: anisotropy of C=C10 marks