The reaction of `p-HOC_(6)H_(4)COOH` with excess `Br_(2)` forms

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Answer

To solve the problem regarding the reaction of para-hydroxybenzoic acid (`p-HOC6H4COOH`) with excess `Br2`, we will follow these steps:

### Step 1: Identify the Structure of the Reactant
The reactant is para-hydroxybenzoic acid, which can be represented as follows:
- It has a hydroxyl group (-OH) and a carboxylic acid group (-COOH) attached to a benzene ring at the para position.

### Step 2: Understand the Reaction with Bromine
When para-hydroxybenzoic acid reacts with excess bromine (`Br2`), the bromine will undergo electrophilic aromatic substitution. The hydroxyl group (-OH) is an activating group and directs the bromination to the ortho and para positions relative to itself.

### Step 3: Draw the Reaction Mechanism
1. The bromine molecule (Br2) will dissociate to form bromine radicals or bromonium ions.
2. The bromine will then attack the ortho and para positions of the benzene ring, leading to the substitution of hydrogen atoms with bromine atoms.

### Step 4: Identify the Final Product
In the case of excess bromine, both ortho and para positions will be brominated. Therefore, the final product will have:
- Two bromine atoms attached to the benzene ring: one at the ortho position and one at the para position relative to the hydroxyl group.

### Step 5: Write the Final Structure
The final product can be represented as:
- A benzene ring with one hydroxyl group (-OH), one carboxylic acid group (-COOH), one bromine at the ortho position, and one bromine at the para position.

### Final Product Structure:
The final product is 2,4-dibromo-5-hydroxybenzoic acid.

### Conclusion
Thus, the reaction of para-hydroxybenzoic acid with excess `Br2` forms 2,4-dibromo-5-hydroxybenzoic acid.

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