To solve the problem of what product is formed when aniline is treated with bromine water, we can follow these steps:
### Step 1: Identify the Reactants
Aniline (C6H5NH2) is an aromatic amine, which means it has an amino group (-NH2) attached to a benzene ring.
### Step 2: Understand the Reaction Type
When aniline reacts with bromine water, it undergoes an electrophilic aromatic substitution reaction. The amino group is an electron-donating group, which increases the electron density on the benzene ring, particularly at the ortho (2-position) and para (4-position) positions relative to the amino group.
### Step 3: Analyze the Electrophile
In this reaction, bromine (Br2) acts as the electrophile. The presence of water helps facilitate the reaction, and the bromine can be generated in situ from the bromine water.
### Step 4: Predict the Substitution Positions
Since the amino group is activating the ring, bromine will preferentially attack the ortho and para positions:
- Ortho position (2-position)
- Para position (4-position)
### Step 5: Determine the Product
When three moles of bromine react with aniline, bromination occurs at all three positions (ortho, para, and meta). However, due to steric hindrance and the activating nature of the amino group, the major product will be 2,4,6-tribromoaniline, where bromine is substituted at all three positions.
### Step 6: Write the Balanced Equation
The balanced equation for the reaction can be represented as:
\[ \text{C}_6\text{H}_5\text{NH}_2 + 3 \text{Br}_2 + 3 \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{NH}_2 + 3 \text{HBr} \]
### Conclusion
The major product formed when aniline is treated with bromine water is **2,4,6-tribromoaniline**.
### Final Answer
The correct option is **4. 2,4,6-tribromoaniline**.
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