What is the major product formed when $\text{1-bromo-2-methylcyclohexane}$ is treated with $\text{KOEt}$ in $\text{EtOH}$?

Solution

1. The reaction of a secondary alkyl halide with a strong base like ${\text{EtO}}^{-}$ proceeds via an $\text{E}2$ mechanism.
2. The $\text{E}2$ mechanism requires an anti-periplanar orientation of the leaving group and the beta-hydrogen.
3. Elimination leads to the formation of the more substituted alkene, which is $\text{1-methylcyclohexene}$.
4. This follows the Zaitsev rule for alkene stability.
5. Hence the answer is (A).