In methyl alcohol solution, bromine reacts with ethylene to yield ${\mathrm{BrCH}}_2{\mathrm{CH}}_2{\mathrm{OCH}}_3$ in addition to 1, 2-dibromoethane because

When ethylene reacts with Br₂, then first there is the formation of a cyclic intermediate. At that instance, both Br^- and -CH₃O^- (from CH₃OH) attack on the cyclic intermediate. Therefore, 2 products are expected instead of one.