CH_3-CH_2-CH_2-Br Alc. KOH Final product is

When haloalkane containing $\beta$-hydrogen atom is heated with alc. KOH solution, then alkene is formed $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \xrightarrow{\text{ alc. } \mathrm{KOH}} \mathrm{CH}_3 - \underset{\text{ Propene }}{\mathrm{CH}}=\mathrm{CH}_2$ In this reaction, hydrogen is elimated from $\beta$-carbon and the halogen is lost from $\alpha$-carbon atom. As a result, an alkene is formed as a product. Due to the involvement of elimination of $\beta$ -hydrogen atom, the process is often called $\beta$-elimination.