2,4,6-tibromo aniline is a product of

Options

A. Electrophilic addition on `C_(6)H_(5)NH_(2)`
B. Electrophilic substitution on `C_(6)H_(5)NH_(2)`
C. Nucleophilic addition on `C_(6)H_(5)NH_(2)`
D. Nucleophilic substition on `C_(6)H_(5)NH_(2)`

Answer

To determine the product 2,4,6-tribromoaniline, we need to analyze the reaction of aniline (C6H5NH2) with bromine. Here’s a step-by-step solution:

### Step 1: Identify the structure of aniline
Aniline is a compound that consists of a benzene ring (C6H5) with an amino group (NH2) attached to it. The structure can be represented as:
\[ \text{C}_6\text{H}_5\text{NH}_2 \]

### Step 2: Understand the reactivity of aniline
The amino group (NH2) is an electron-donating group, which makes the benzene ring more reactive towards electrophilic substitution reactions. It is a strongly activating group and directs electrophiles to the ortho and para positions on the benzene ring.

### Step 3: Identify the electrophile
In the bromination reaction, bromine (Br2) acts as the electrophile. The reaction involves the substitution of hydrogen atoms on the benzene ring with bromine atoms.

### Step 4: Determine the positions of substitution
Since the amino group directs substitution to the ortho (2 and 6 positions) and para (4 position) locations, we expect bromination to occur at these positions. Thus, the bromination of aniline will lead to the substitution of hydrogen atoms at the 2nd, 4th, and 6th positions of the benzene ring.

### Step 5: Write the product structure
The product formed from the bromination of aniline will be 2,4,6-tribromoaniline, which can be represented as:
\[ \text{Br} \text{C}_6\text{H}_2(\text{Br})_3\text{NH}_2 \]
This indicates that bromine atoms are present at the 2nd, 4th, and 6th positions of the benzene ring.

### Step 6: Conclusion
The reaction that produces 2,4,6-tribromoaniline is an electrophilic substitution reaction. Therefore, the correct answer is:
**Electrophilic substitution on C6H5NH2.**

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