Explanation: Propene yields two products, however only one predominates as per Markovnikov's rule i.e., 2-bromopropane which on heating with aq. $\mathrm{KOH}$ gives secondary alcohol. Aq. $\mathrm{KOH}$ is alkaline in nature so it gives hydroxide ion which is a nucleophile. It replaces halide(bromide in this case) ion and form alcohols. $\underset{\text{Propene}}{\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2} + \mathrm{HBr} \rightarrow \underset{\text{2-bromopropane}}{\mathrm{CH}_3-\overset{\mathrm{Br}}{\overset{\vert}{\mathrm{CH}}}-\mathrm{CH}_3}$ $\xrightarrow{\text{aq. KOH}} \mathrm{CH}_3-\overset{\mathrm{OH}}{\overset{\vert}{\mathrm{CH}}}-\mathrm{CH}_3$