Correct Answer - C
`CH_3-CH_2-Br+underset"alcoholic""KOH"overset"dehydrohal ogenation"to CH_2=CH_2+KBr +H_2O`
In alcoholic KOH alkoxide ions `(RO^-)` are present which is a strong base. They abstract proton from `beta`-carbon of alkyl halide and favours elimination reaction.
`underset"Alcohol""ROH"+KOHto underset"Potassium alkoxide"(ROK+H_2O)`
`ROK to underset"Alkoxide ion"(RO^-) + K^+`
`RO^(-) + H -oversetbeta(CH_2)-oversetalpha(CH_2)-Br to ROH + CH_2 =CH_2 +KBr`
