The difference in the reaction of phenol with bromine in chloroform and bromine in water medium is due to

We need to determine the reason for the difference in bromination products of phenol in chloroform versus water medium.

When phenol is treated with $$Br_2$$ in chloroform ($$CHCl_3$$), we get monobromination—specifically a mixture of ortho-bromophenol and para-bromophenol because in a non-polar solvent, $$Br_2$$ remains as a molecular species and acts as a weaker electrophile. The reaction proceeds through a typical electrophilic aromatic substitution requiring a Lewis acid catalyst, and only one hydrogen is replaced.

When phenol is treated with $$Br_2$$ in water ($$Br_2$$/aq), we get 2,4,6-tribromophenol as a white precipitate. In water, bromine partially dissociates as follows:

The polar solvent facilitates the generation of a more reactive electrophile ($$Br^+$$ or its equivalent from $$HOBr$$), and the phenoxide ion ($$C_6H_5O^-$$) formed partially in water is even more reactive than phenol. This leads to tribromination at all available ortho and para positions.

The key difference arises due to the polarity of the solvent:

- In a polar solvent (water), $$Br_2$$ generates a stronger electrophilic species, and phenol is more activated, leading to tribromination.

- In a non-polar solvent (chloroform), $$Br_2$$ remains molecular and is a weaker electrophile, leading to only monobromination.

Option A: Hyperconjugation — Not relevant here.

Option B: Polarity of solvent — Correct. The solvent polarity determines the strength of the electrophile generated.

Option C: Free radical formation — The reaction is electrophilic, not free radical.

Option D: Electromeric effect of the substrate — The electromeric effect of phenol is the same regardless of solvent; it is the solvent that changes the outcome.

Hence, the correct answer is Option B: Polarity of solvent.