This reaction involves E2 elimination of a tertiary alkyl halide with a strong base (ethoxide). The major product follows Zaitsev's rule, favoring the more substituted, stable alkene. The substrate is 2-bromo-2-methylbutane. Removal of a beta-hydrogen from a methyl group (less substituted carbon) leads to the trisubstituted alkene, 2-methylbut-2-ene, which is the major product.

The major product is the trisubstituted alkene: ${\mathrm{CH}}_3-C\left({\mathrm{CH}}_3\right)={\mathrm{CH}}_2$

Final Answer: The first option (top image)