Q1(c). trans-1,2-Dibromo-1-methylcyclohexane from methylcyclohexane

Introduction. The target is the anti addition product of Br₂ and 1-methylcyclohexene. The alkene is reached from methylcyclohexane in two steps: selective radical bromination at the tertiary C–H and E2 elimination.

Scheme: Three steps: radical bromination, Zaitsev E2, anti addition of bromine through a bromonium ion.

1. Radical bromination. Br₂/hν (or heat). Bromine radicals abstract H selectively: the tertiary C–H gives the most stable (tertiary) radical, and bromination is far more selective than chlorination (late, product-like transition state). Product: 1-bromo-1-methylcyclohexane.
2. Elimination. NaOEt in ethanol (E2). The anti-periplanar ring H is removed and the Zaitsev (more substituted) alkene forms: 1-methylcyclohexene. Some methylenecyclohexane is the minor product; heating a tertiary bromide in ethanol (E1) gives the same major alkene.
3. Bromination of the alkene. Br₂ in CCl₄ forms a cyclic bromonium ion on one face. Br⁻ opens it from the opposite face, at the tertiary carbon, which carries more positive charge, and the ring opens trans-diaxially.
4. The product is trans-1,2-dibromo-1-methylcyclohexane, formed as a racemate (attack on the other face of the alkene gives the enantiomer). The diaxial product first formed flips to the diequatorial chair; both chairs are the same trans compound.

Conclusion. Brominate methylcyclohexane at the tertiary carbon (Br₂, hν), eliminate HBr with NaOEt to give 1-methylcyclohexene, and add Br₂ in CCl₄. Anti opening of the bromonium ion gives ()-trans-1,2-dibromo-1-methylcyclohexane.

Application. Selective tertiary bromination followed by elimination is the usual laboratory way to introduce a double bond into an unreactive hydrocarbon skeleton.