The alkene shown is treated first with $\mathrm{HBr}$ and the major bromide so formed is heated with alcoholic $\mathrm{KOH}$ (Zaitsev elimination). Which alkene is the major final product?

The alkene is 3,3-dimethylbut-1-ene. Methyl shift in the cation gives 2-bromo-2,3-dimethylbutane as the major bromide. Dehydrohalogenation with alcoholic $\mathrm{KOH}$ follows Zaitsev's rule and gives the more substituted, tetrasubstituted alkene, 2,3-dimethylbut-2-ene.