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.

  • $\begingroup$ Would be real nice if you upgraded to good English ;) $\endgroup$
    – Mithoron
    Jan 29, 2018 at 0:05

1 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.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.