Identify the product formed when $2\text{-bromobutane}$ is treated with potassium ethoxide in ethanol.

Solution

1. Potassium ethoxide is a strong base, favoring the E2 elimination mechanism.
2. Elimination can occur at ${\text{C}}_1$ or ${\text{C}}_3$ to form 1-butene or 2-butene.
3. Zaitsev's rule states that the more substituted alkene is the major product.
4. Among the 2-butene isomers, the trans isomer is more stable due to reduced steric hindrance.
5. Hence the answer is (C).