Q.n-propyl bromide on reaction with alcoholic KOH gives:

When a base removes a beta-hydrogen as the halogen leaves the alpha-carbon, a new C=C double bond forms between them -- beta-elimination (dehydrohalogenation), the substitution reaction's direct competitor, typically run with a strong, poorly-nucleophilic base in a poorly-ionising solvent: alcoholic (not aqueous) KOH. Where more than one type of beta-hydrogen is available, Saytzeff's rule predicts the major product is the MORE substituted alkene (more alkyl groups on the double-bond carbons), since a more substituted double bond is thermodynamically more stable -- 2-bromobutane + alcoholic KOH gives mainly 2-butene, with 1-butene as the minor product. Two mechanisms, mirroring SN2/SN1: E2 (bimolecular): Rate = k[RX][base], second order, ONE step -- the base removes the beta-H and the halide leaves the alpha-carbon simultaneously (concerted, no intermediate); favoured by primary substrates with a strong base. E1 (unimolecular): first order, TWO steps -- (1) slow heterolysis to a carbocation (identical first step to SN1), then (2) fast loss of a beta-proton to form the double bond; favoured by tertiary substrates, and because the intermediate is the SAME carbocation as in SN1, a tertiary substrate under basic, ion …