Major product 'B' of the following reaction sequence is

In the first step, the alkene reacts with bromine in the presence of methanol.

$$\mathrm{Br_2/CH_3OH}$$

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

Since methanol is present in excess, it acts as the nucleophile instead of $$\mathrm{Br^-}$$.

Methanol attacks the more substituted carbon because it carries greater partial positive character.

After deprotonation, intermediate $$\mathrm{A}$$ is formed.

$$\mathrm{Alkene \xrightarrow{Br_2/CH_3OH} Bromomethoxy\ Compound\ (A)}$$

Intermediate $$\mathrm{A}$$ contains:

1. A methoxy group $$\mathrm{(-OCH_3)}$$ on the tertiary carbon

2. A bromine atom on the adjacent secondary carbon

In the second step, compound $$\mathrm{A}$$ reacts with hydroiodic acid.

$$\mathrm{HI}$$ protonates the ether oxygen, converting it into a good leaving group.

$$\mathrm{R-OCH_3 + H^+ \longrightarrow R-O^+HCH_3}$$

Since the oxygen is attached to a tertiary carbon, cleavage occurs through an $$\mathrm{S_N1}$$ mechanism.

Methanol leaves, generating a stable tertiary carbocation.

$$\mathrm{R-O^+HCH_3 \longrightarrow R^+ + CH_3OH}$$

The iodide ion then attacks the tertiary carbocation.

$$\mathrm{R^+ + I^- \longrightarrow R-I}$$

The bromine atom on the neighbouring carbon remains unchanged throughout the reaction.

Thus, in the final product $$\mathrm{B}$$:

1. The bromine atom remains on the secondary carbon

2. The methoxy group is replaced by iodine

Therefore, the correct structure is:

$$\boxed{\mathrm{Option\ containing\ Br\ and\ I\ on\ adjacent\ carbons}}$$