Amongst the following, the major product of the given chemical reaction is

The reaction involves electrophilic addition of bromine to an enol ether in methanol.

The $$\mathrm{\pi}$$ electrons of the double bond attack $$\mathrm{Br_2}$$ to form a cyclic bromonium ion intermediate.

The bromonium ion is unsymmetrical because the carbon adjacent to oxygen develops greater positive character due to resonance stabilisation by the oxygen lone pair.

Methanol $$\mathrm{(CH_3OH)}$$ acts as the nucleophile.

Since methanol is the solvent and present in large excess, it attacks the bromonium ion preferentially instead of $$\mathrm{Br^-}$$.

The nucleophilic attack occurs at the carbon adjacent to the oxygen atom because it is the more electrophilic centre.

Ring opening occurs through backside attack, giving:

$$\mathrm{Anti\ Addition}$$

Hence, the methoxy group $$\mathrm{(-OCH_3)}$$ and bromine atom become trans to each other in the final product.

The major product contains the methoxy group on the carbon adjacent to oxygen and bromine on the neighbouring carbon in an anti stereochemical arrangement.

Thus, the right option is A