Q5(e). Methyl groups of 7,7-dimethylnorbornene: anisotropy of C=C

Introduction. The compound is 7,7-dimethylnorborn-2-ene. Both methyls sit on the one-carbon bridge C-7. Methyl A lies over the C2=C3 double bond (syn); methyl B lies over the saturated CH₂-CH₂ bridge (anti).

Figure: Anisotropy of the C=C: the region above the π cloud (along the p orbitals) is shielded; methyl A lies in it.

1. (i) Methyl B has the higher chemical shift (resonates downfield); methyl A is upfield.
2. (ii) The circulating π electrons of the C=C set up a local field. Above and below the plane of the double bond (the face of the π cloud) this field opposes B₀: a shielding cone. In the rigid norbornene skeleton methyl A is held over the face of the C2=C3 π bond, inside this cone, so it is shielded. Methyl B points over the saturated C5–C6 bridge, outside the cone, and feels no such shielding.
3. The two methyls are therefore non-equivalent and appear as two separate 3H singlets; in 7,7-dimethylnorbornane (no C=C) they are much closer in shift.
4. (iii) Approximate values: methyl A (syn to C=C) about δ 0.7–0.8; methyl B (anti) about δ 0.9–1.0. Both are near the normal value for a methyl on a quaternary sp³ carbon (δ≈ 0.9).

Conclusion. Methyl B (anti, over the CH₂CH₂ bridge) has the higher δ, about 0.9–1.0; methyl A, held over the π face of the C=C, is shielded by the double-bond anisotropy and appears near δ 0.7–0.8.

Application. The same anisotropy shifts the 7-syn and 7-anti protons of norbornenes apart and is used to assign the configuration of bridged bicyclic terpenes (camphor and pinene derivatives) from their NMR spectra.