Q4(d). E2 of menthyl and neomenthyl chlorides: anti-periplanar requirement

Introduction. (i) is menthyl chloride (Cl, isopropyl and methyl all equatorial); (ii) is neomenthyl chloride (Cl axial, the other two equatorial). E2 needs the H and Cl anti-periplanar; in a cyclohexane this means both axial (trans-diaxial).

1. (ii) Neomenthyl chloride. In its stable chair Cl is already axial. Both neighbouring carbons have an axial H anti to Cl: H on C-2 (the carbon bearing isopropyl) and H on C-6 (the CH₂). Elimination is fast. Removal of the C-2 H gives the more substituted 3-menthene (Zaitsev, 75%); removal of the C-6 H gives 2-menthene (25%).
2. (i) Menthyl chloride. With Cl equatorial no anti H is available. The ring must flip to the conformer with all three groups axial, which is present only in very small amount. In that conformer the C-2 H is now equatorial (the isopropyl is axial), so the only anti-periplanar H is on C-6. Product: 2-menthene only (100%, anti-Zaitsev), and the reaction is about 200 times slower than for neomenthyl chloride.
3. The difference shows that geometry (anti-periplanar H) overrides product stability (Zaitsev rule) in E2 reactions of rings.

Figure: Axial bonds (vertical) anti to the axial C–Cl bond; red = H that can be removed.

Scheme: The two possible alkenes.

Conclusion. (i) Menthyl chloride gives only 2-menthene, slowly; (ii) neomenthyl chloride gives 3-menthene (75%) + 2-menthene (25%), fast. E2 in cyclohexanes requires a trans-diaxial H–C–C–Cl arrangement.

Application. The trans-diaxial rule is used to predict eliminations in steroids and sugars and explains why some ring halides resist E2 altogether.