On reaction of haloalkane with methoxide ion, alkene is formed.
Mechanism
●
⊖
O
CH3 will act as a base due to its small size and high electron density and therefore, abstracts proton to form double bond which is in conjugation with aromatic ring. ● The
⊖
O
CH3 when acts as nucleophile undergoes nucleophilic substitution and replaces Br⊖ to form ether, which is a minor product.