n-propyl bromide on treating with alcoholic $\mathrm{KOH}$ produces

$\underset{ }{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}+\mathrm{KOH}\to {\mathrm{CH}}_3-\mathrm{CH}}={\mathrm{CH}}_2+\mathrm{KBr}+{\mathrm{H}}_2\mathrm{O}$

This reaction removes a molecule of $\mathrm{HX}$,the reaction is called dehydrohalogenation. The hydrogen atom is eliminated from

 $\mathrm{\beta }$-carbon atom (carbon atom next to the carbon to which halogen is attached). Therefore the reaction is also called $\mathrm{\beta }$- elimination reaction.