Reactions of differents halohydrins with tert-butoxide

Question

I have two compounds (1​R,2​S)-3,3-dimethyl-2-bromocyclohexanol and (1​R,2​R)-3,3-dimethyl-2-bromocyclohexanol, both of them are treated with $\ce{KO^tBu}$ (a strong but bulky base) to form two different isomers of $\ce{C8H14O}$ which is not an alcohol.

I though that (1​R,2​R)-3,3-dimethyl-2-bromocyclohexanol is going to form an epoxide due the configuration of the halogen and the hydroxyl (although I don't know if it is required an specific configuration) and (1​R,2​S)-3,3-dimethyl-2-bromocyclohexanol is going to have a E2 reaction due steric hindrance but that will end with an alcohol anyway.

Answer 1

Chair conformation RR-1 (shown as its alkoxides) is in equilibrium with conformation RR-2, which is the unfavorable partner owing to the strong 1,3-diaxial Me/alkoxide interaction. It is this conformation that leads to the epoxide 1. This route is possible if the rate of formation is much greater than the rate constant for RR-1 --> aldehyde 2. RR-1 has concentration on its side and anti-periplanar alignment of bonds for ring contraction. I prefer ring contraction but this does not preclude epoxide formation.

The RS isomer has two chair conformations, RS-1 and RS-2. Again, RS-2 is an unfavorable conformation (vide supra) but has proper bond alignment for ring contraction. Conformation RS-1 has the necessary bond alignment for E2 elimination to form cyclohexanone 4 via the enol 3.

A priori, it is not easy to predict with certainty what the product distribution will be. The epoxide can only arise from the RR isomer while aldehyde 2 can arise from either bromohydrin. Organic chemistry is an experimental science.