Step 1: Understanding the Concept:
The OH group of phenol activates the benzene ring strongly towards electrophilic substitution at the ortho and para positions.
Step 2: Reaction:
With bromine water (no catalyst needed), all three positions are substituted at once. The product is 2,4,6-tribromophenol:
\[ C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr \]
Step 3: Observation:
2,4,6-Tribromophenol is insoluble in water and appears as a white precipitate, and the brown colour of bromine water disappears. So the answer is (C).
Step 4: Why the others are wrong:
No gas is released (HBr dissolves in water), and no pink colour forms. A brown liquid would mean no reaction.
Final Answer:
Phenol gives a white precipitate of 2,4,6-tribromophenol.
\[ \boxed{C:\ \text{White precipitate}} \]