Question

In the elimination reaction of 1-bromo-2-ethylcyclohexane in sodium hydroxide in ethanol solution A) is faster for the trans isomer since the bromine and ethyl group are of different sides of the ring. B) is faster for the trans isomer since the bromine and the ethyl groups are both axial. C) is faster for the trans isomer since the bromine and the ethyl groups are both equitorial. D) is faster for the trans isomer since there is a hydrogen anti-periplanar to the bromine in the preferred chair conformer of the compound. E) None of the above provides a correct response.

Answer 1

Answer:

D

Explanation:

Option D is correct

Cycloalkanes(1-bromo 2-ehtyl cyclohexane is this case) mainly preferred chair conformers while undergoing elimination reaction and further products are not those that predicted by zaitsev's rule.

Hence, the correct answer is faster for the trans isomer since there is a hydrogen anti-periplanar to the bromine in the preferred chair conformer of the compound.