E2 is concerted, so the β-hydrogen and leaving group must be anti-periplanar — on a cyclohexane chair that only happens when both are axial (trans-diaxial). A locked or equatorial leaving group slows E2 and lets geometry override Zaitsev.
Because the C–H and C–LG bonds break as the π bond forms, they must be lined up — the anti-periplanar requirement that makes E2 on rings a rich, exam-favorite topic.
Bromocyclohexane loses HBr to give cyclohexene.
1. E2 Requires the β-Hydrogen and Leaving Group to Be Anti-Periplanar
The H–C–C–LG dihedral must be about 180° so the developing p orbitals overlap into a clean π bond — in open-chain substrates free rotation reaches this easily.
2. On a Cyclohexane Ring, Anti-Periplanar Means Both Groups Must Be Trans-Diaxial
A chair locks each carbon's bonds into axial or equatorial, and only when both the leaving group and β-H are axial (trans-diaxial) do they reach the 180° dihedral; an equatorial leaving group blocks E2.
3. The Ring Must Flip to Place the Leaving Group Axial Before It Can Eliminate
Most cyclohexyl halides rest with the halogen equatorial, so the molecule must ring-flip into the axial chair to react — cheap for an unhindered ring, so the reactive conformer need not be the most populated one.
4. A Bulky tert-Butyl Group Locks the Chair and Can Shut E2 Down
A bulky tert-butyl group locks the chair with itself equatorial: the cis isomer is frozen with Br axial (fast E2), while the trans isomer freezes Br equatorial (E2 stalls, favoring substitution or E1).
5. Trans-Diaxial Geometry Can Override Zaitsev
On a locked ring you get whichever alkene the only available axial β-H can make: 1-bromo-2-methylcyclohexane can be forced to give the less-substituted 3-methylcyclohexene — the classic menthyl / neomenthyl chloride story where geometry, not stability, decides.
6. Summary
Anti-periplanar (dihedral ≈ 180°) · on a chair that means trans-diaxial · ring-flip to put the leaving group axial · tert-butyl locks the chair (cis fast, trans stalls) · one axial β-H can override Zaitsev · always draw the chair before predicting the product.
Quiz yourself
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They must be trans-diaxial — both axial, on opposite faces of the ring. Only that arrangement gives the H–C–C–LG dihedral of ~180° (anti-periplanar) needed for the concerted E2 transition state. If the leaving group is equatorial, no adjacent hydrogen is anti-periplanar and E2 is blocked.
The ring flips to the higher-energy chair that places Br axial. That conformer is less populated but reactive; because the flip is fast and low-cost for an unhindered ring, E2 proceeds through it even though the equatorial chair is more stable. The reactive conformer need not be the most abundant one.
The tert-butyl group locks the chair with itself equatorial. In the cis isomer that forces Br axial (E2 is fast); in the trans isomer it forces Br equatorial, where no β-H is anti-periplanar, so E2 is very slow and substitution or E1 competes instead.
If the only β-hydrogen that is trans-diaxial to the leaving group sits on the less-substituted side, that is the only H E2 can remove — so you get the less-substituted alkene regardless of Zaitsev. Geometry overrides thermodynamic preference, as in the menthyl/neomenthyl chloride case.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.