2-Bromopropane is heated with alcoholic $\mathrm{KOH}$. The major product is

Alcoholic $\mathrm{KOH}$ supplies the strong base $\mathrm{RO}{}^{-}$/$\mathrm{OH}{}^{-}$ in a poorly solvating medium, favouring E2 elimination (dehydrohalogenation). Loss of $\mathrm{HBr}$ from 2-bromopropane gives propene.