Past exam of the chemistry B course of the University of Cambridge / 2021 / Paper 1 / B3 / b / Solution

Both reactions are E2 eliminations requiring an axial chloride and an axial beta hydrogen in an antiperiplanar conformation relative to the C–Cl bond. The reacting C–H bond supplies the alkene $\pi$ bond as the C–Cl bond breaks; hydroxide accepts the proton.

For the trans-1,2 substrate, the low-energy chair has methyl and chlorine both equatorial. ring inversion is needed to place chlorine axial, but this also places methyl axial. The population of the reactive chair is consequently small, explaining the slow reaction. At the methyl-bearing neighboring carbon the axial position is occupied by methyl, not hydrogen. Only the other beta carbon has the needed anti axial H, so elimination gives 3-methylcyclohexene rather than the more substituted alkene.

For the cis substrate, a favorable chair has methyl equatorial and chlorine axial. It can react without paying the axial-methyl penalty, explaining the faster rate. Both adjacent carbons now supply an anti axial hydrogen. Removing the H on the methyl-bearing carbon gives 1-methylcyclohexene, the major product; removing the other gives 3-methylcyclohexene, the minor product. The former has the more substituted double bond.

Thus alkene stability selects between accessible pathways in the cis case, but it cannot create the missing antiperiplanar conformation H in the trans case.