I have two compounds (1R,2S)-3,3-dimethyl-2-bromocyclohexanol and (1R,2R)-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 (1R,2R)-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 (1R,2S)-3,3-dimethyl-2-bromocyclohexanol is going to have a E2 reaction due steric hindrance but that will end with an alcohol anyway.