What is the major product formed when ${\text{CH}}_3\text{CH}(\text{Br}){\text{CH}}_2{\text{CH}}_3$ is treated with alcoholic $\text{KOH}$?

Solution

1. Treatment with alcoholic $\text{KOH}$ induces a dehydrohalogenation reaction via an $\text{E}2$ mechanism.
2. According to Saytzeff rule, the more substituted alkene is the major product.
3. But-2-ene is more substituted than but-1-ene.
4. Hence the answer is (B).