When alkyl halide reacts with alcoholic KOH, then alkene forms with the removal of HBr.

${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Br}\stackrel{\mathrm{alc}. \mathrm{KOH}}{\to }\underset{\mathrm{A}}{{\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2}+\mathrm{HBr}$

Addition of HBr in alkene follows Markovnikov's rule, where Br attached to the carbon atom that has a fewer number of hydrogens.

${\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\to \underset{\mathrm{B}}{{\mathrm{CH}}_3-\mathrm{CH}\left(\mathrm{Br}\right)-{\mathrm{CH}}_3}$

In the presence of aqueous KOH, alkyl halide forms corresponding alcohol

$\mathrm{CH}{}_3-\mathrm{CH}\left(\mathrm{Br}\right)-{\mathrm{CH}}_3+\mathrm{aq}.\mathrm{KOH}\to \underset{\mathrm{C} (\mathrm{propan}-2-\mathrm{ol})}{{\mathrm{CH}}_3-\mathrm{CH}\left(\mathrm{OH}\right)-{\mathrm{CH}}_3}$