In alcoholic $KOH$ alkoxide ions $(R{O^ - })$ are present which is a strong base. They abstract proton from ?-carbon of alkyl halide and favours elimination reaction

$\mathop {ROH}\limits_{{\rm{Alcohol}}} + KOH \to \mathop {ROK + {H_2}O}\limits_{{\rm{Potassium \,alkoxide}}} $

$ROK \to \mathop {R{O^ - }}\limits_{{\rm{Alkoxide \,ion}}} + {K^ + }$

$R{O^ - } + H - \mathop {C{H_2}}\limits^\beta - \mathop {C{H_2}}\limits^\alpha - Br \to ROH + C{H_2} = C{H_2} + Br$