Reactions of differents halohydrins with tert-butoxide

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.




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. 


[![enter image description here](https://i.stack.imgur.com/2Be3e.jpg)](https://i.stack.imgur.com/2Be3e.jpg)


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