Elimination of bromine from 2-bromobutane results in the formation of

Elimination of bromine from 2-bromobutane takes place in the presence of an alcoholic KOH . In this case, two products can be formed. But 2-butene is the major product. According to the Saytzeff rule, the formation of a more alkylated double bond takes place. 3 2 3 3 3 3 2 2 CH –CH –CHBr–CH +alc.KOH CH –CH=CH–CH (major)+CH –CH –CH=CH (minor) ->