The problem asks for the structure(s) of the final product X in a multi-step reaction sequence starting from phenol.

Let's analyze the reaction sequence step by step:

Step 1: Reaction of Phenol with Sodium Methoxide ($CH_3ONa$) Phenol ($C_6H_5OH$) is acidic and reacts with a strong base like sodium methoxide ($CH_3ONa$) in an acid-base reaction. The methoxide ion ($CH_3O^−$) deprotonates phenol to form sodium phenoxide ($C_6H_5O^-Na^+$).

$C_6H_5OH + CH_3ONa \rightarrow C_6H_5O^-Na^+ + CH_3OH$

Step 2: Reaction of Sodium Phenoxide with Methyl Iodide ($CH_3I$) The sodium phenoxide formed is a good nucleophile. It undergoes a Williamson ether synthesis reaction with methyl iodide ($CH_3I$), which is an $S_N2$ reaction, to form anisole (methoxybenzene).

$C_6H_5O^-Na^+ + CH_3I \rightarrow C_6H_5OCH_3 + NaI$

Step 3: Reaction of Anisole with $Br_2$ in $CS_2$ This is an electrophilic aromatic substitution (bromination) of anisole. The methoxy group ($-OCH_3$) is a strongly activating and ortho, para-directing group due to its +R (resonance) effect. The reaction is carried out in carbon disulfide ($CS_2$), a non-polar solvent. In such solvents, the para-substituted product is strongly favored over the ortho-substituted product due to steric hindrance at the ortho positions. Therefore, the major product is p-bromoanisole (4-bromoanisole), and the minor product is o-bromoanisole (2-bromoanisole).

$C_6H_5OCH_3 \xrightarrow{Br_2, CS_2} p\text{-Br-}C_6H_4OCH_3 \text{ (Major)} + o\text{-Br-}C_6H_4OCH_3 \text{ (Minor)}$

Step 4: Reaction with Anhydrous $AlCl_3$ The mixture of bromoanisoles is treated with anhydrous aluminum chloride ($AlCl_3$), a strong Lewis acid. This reagent is used for the cleavage of aryl ethers (demethylation). The $AlCl_3$ coordinates with the ether oxygen, weakening the $O-CH_3$ bond, which then cleaves. After aqueous workup (implicit), the phenoxide intermediate is protonated to give a phenol. This reaction converts the bromoanisoles to bromophenols.

$\text{Br-}C_6H_4OCH_3 \xrightarrow{\text{1. Anhydrous } AlCl_3 \quad \text{2. } H_2O} \text{Br-}C_6H_4OH$

Based on the products from Step 3, we would expect to get a mixture of p-bromophenol (from the major product) and o-bromophenol (from the minor product).

However, strong Lewis acids like $AlCl_3$ can also catalyze the isomerization of substituents on an aromatic ring to the most thermodynamically stable position. The para isomer is generally more stable than the ortho isomer due to reduced steric strain. Therefore, under the reaction conditions with anhydrous $AlCl_3$, the minor o-bromoanisole isomer can rearrange to the more stable p-bromoanisole isomer. Consequently, the final product after demethylation is almost exclusively p-bromophenol.

Conclusion The final product X is p-bromophenol (4-bromophenol).

Let's evaluate the given options:

• A: 3-bromophenol (m-bromophenol). Incorrect, as the $-OCH_3$ group is an o,p-director.
• B: 4-bromophenol (p-bromophenol). This is the major and thermodynamically most stable product. Correct.
• C: 2,4-dibromophenol. Incorrect. The reaction conditions ($Br_2/CS_2$) favor monobromination.
• D: 2-bromophenol (o-bromophenol). This is a minor product of bromination, but it is likely to isomerize to the para-isomer in the final step. So it's not the final major product.

Therefore, the correct structure of X is p-bromophenol.