The major product obtained from the following reaction is:

The substrate contains an $$-OCH_3$$ group directly attached to the benzene ring. This $$-OCH_3$$ group has an unshared pair of electrons on the oxygen. Because of resonance donation, it pushes electron density into the ring, making the ring more nucleophilic and activating it toward electrophilic aromatic substitution.

Resonance structures show that the additional electron density appears predominantly at the ortho and para positions. Hence, when an electrophile approaches, substitution is directed to those positions (the group is ortho-, para- directing).

The reagent provided in the question is molecular bromine, $$Br_2$$, in a polar protic medium (water or acetic acid) and no Lewis-acid catalyst such as $$FeBr_3$$ is present. In such a medium, the ring is already strongly activated by $$-OCH_3$$; therefore bromination proceeds rapidly without a catalyst, and usually stops after introduction of a single bromine atom.

Because steric hindrance at the ortho positions is higher (the incoming $$Br^+$$ must squeeze past the bulky $$-OCH_3$$ group), the para product is favoured over the two equivalent ortho products. Thus the major product is para-bromoanisole.

Therefore, among the options, the structure that shows bromine para to the $$-OCH_3$$ group (Option A) is the major product.

Hence the correct answer is:
Option A which is: para-bromoanisole.