When 1-bromopropane react with alcoholic $\mathrm{KOH}$, it will form prop-1-ene $(X)$. Alcoholic $\mathrm{KOH}$ is used to give elimination reaction. In next step,
$\mathrm{HBr}$ (II) is used to give Markownikoff's addition to the alkene and produces 2-bromopropane.