Concept:

The use of a polar solvent (like water) enhances the ionization of both the phenol (into a highly reactive phenoxide ion) and the electrophile, drastically increasing reaction speed and substitution extent.

Formula:

$$ \text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \xrightarrow{\text{H}_2\text{O}} \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr} $$

Solution:

• Because the ring is immensely activated in aqueous conditions, polyhalogenation cannot be stopped.



• Bromine attacks all available ortho and para positions simultaneously.



• The final product is a white precipitate of 2,4,6-Tribromophenol.

Why other options are incorrect:

To get mono-substituted ortho/para bromophenol (Option A), you must use a non-polar solvent like \(\text{CS}_2\) at low temperature. The -OH group is ortho/para directing, so meta-substitution (Options B and D) is impossible via direct bromination.