Question

Identify the product when Phenol reacts with Bromine water.

• A. \(o\)-Bromophenol
• B. \(m\)-Bromophenol
• C. \(2,4,6\text{-Tribromophenol}\)
• D. Bromobenzene

Hint:

Phenol reacts very rapidly with bromine water because the \(-OH\) group strongly activates the benzene ring. Unlike benzene, this reaction does not require a Lewis acid catalyst such as \(FeBr_3\). The reaction directly forms \(2,4,6\)-tribromophenol with a characteristic white precipitate.

Answer 1

The Correct Option is C

Concept: Phenol contains an \(-OH\) group attached to the benzene ring. The \(-OH\) group is a strong electron donating group due to resonance. It increases the electron density of the benzene ring and strongly activates the ortho and para positions toward electrophilic substitution reactions. When phenol reacts with bromine water, the ring becomes highly activated and substitution occurs rapidly at the ortho and para positions. As a result, multiple bromination occurs without the need for a catalyst.

Step 1: Understanding the directing effect of the \(-OH\) group. The \(-OH\) group directs incoming electrophiles to the:
• Ortho position
• Para position Because these positions become highly reactive, bromination takes place at: \[ 2,\,4,\ \text{and}\ 6 \text{ positions of the benzene ring.} \]

Step 2: Formation of the final product. Three bromine atoms substitute the hydrogen atoms at the ortho and para positions forming: \[ \text{Phenol} + 3Br_2 \rightarrow 2,4,6\text{-Tribromophenol} + 3HBr \] Thus, the major product formed is \[ \boxed{2,4,6\text{-Tribromophenol}} \] This compound appears as a white precipitate in bromine water.