Step 1: Understanding the Question:
We need to determine the major organic product of the chemical reaction between phenol (\(C_6H_5OH\)) and an excess of bromine water (\(Br_2\) in \(H_2O\)).
Step 2: Key Formula or Approach:
This reaction is an example of electrophilic aromatic substitution. The hydroxyl (\(-OH\)) group attached to the benzene ring in phenol is a very strong activating group. Activating groups make the benzene ring more reactive towards electrophiles and direct the substitution to the ortho- and para-positions.
Step 3: Detailed Explanation:
1. Activation of the Ring: The lone pair of electrons on the oxygen atom of the \(-OH\) group is delocalized into the benzene ring, increasing the electron density, especially at the ortho (positions 2 and 6) and para (position 4) positions.
2. Electrophile: In bromine water, the \(Br_2\) molecule acts as the electrophile. The high electron density in the phenol ring polarizes the \(Br-Br\) bond, allowing for the attack.
3. Polysubstitution: Due to the strong activating effect of the \(-OH\) group, the reaction is very fast and does not require a Lewis acid catalyst (like \(FeBr_3\)). The reaction proceeds to completion, with bromine atoms substituting at all available activated positions (both ortho positions and the para position).
The overall reaction is:
\[ C_6H_5OH + 3Br_2(aq) \rightarrow C_6H_2Br_3OH(s) + 3HBr(aq) \]
The product, 2,4,6-tribromophenol, is a white solid that precipitates out of the solution.
Step 4: Final Answer:
The product formed is 2,4,6-Tribromophenol.