The major product of the following reaction is:

We need to determine the mechanism and major product for the reaction of sodium phenoxide with methyl iodide ($$\text{CH}_3\text{I}$$).

Reaction Mechanism: Williamson Ether Synthesis

1. Nucleophilic Attack via $$\text{S}_\text{N}2$$ Pathway:

   The reaction between sodium phenoxide ($$\text{Ph--O}^\ominus\text{Na}^\oplus$$) and methyl iodide ($$\text{CH}_3\text{I}$$) is a classic example of a Williamson ether synthesis. The strongly nucleophilic phenoxide oxygen atom attacks the relatively unhindered primary carbon atom of the methyl iodide electrophile:

   $$\text{Ph--O}^\ominus + \text{CH}_3\text{--I} \xrightarrow{\text{S}_\text{N}2} \text{Ph--O--CH}_3 + \text{I}^\ominus$$



2. Transition State and Displacement:

   As the nucleophilic oxygen atom forms a new covalent bond with the methyl group, the iodide ion simultaneously departs as an excellent leaving group ($$-\text{NaI}$$), yielding 2-methoxy-1-methylbenzene (or anisole depending on ring substitutions).

Conclusion:

The reaction cleanly proceeds via a bimolecular nucleophilic substitution ($$\text{S}_\text{N}2$$) process to give the corresponding alkyl aryl ether.

Answer: Option B (2-methoxy-1-methylbenzene formed via S$$_\text{N}$$2 reaction)