The correct sequence of reaction to be performed to convert benzene into m-bromoaniline is:

Options

A. Nitration, reduction, bromination
B. Bromination, nitration, reduction
C. Nitration, bromination, reduction
D. Reduction, nitration, bromination

Explanation

To convert benzene into m-bromoaniline, we need to follow a specific sequence of reactions. Here is a step-by-step solution:

### Step 1: Nitration of Benzene
- **Reaction**: Treat benzene (C6H6) with a mixture of concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4).
- **Product**: This will introduce a nitro group (-NO2) to the benzene ring, resulting in nitrobenzene (C6H5NO2).

### Step 2: Bromination of Nitrobenzene
- **Reaction**: Next, we perform bromination on nitrobenzene using bromine (Br2) in the presence of a catalyst like iron (Fe) or aluminum chloride (AlCl3).
- **Product**: The nitro group is a meta-directing group, so bromination will occur at the meta position relative to the nitro group. This will yield m-bromonitrobenzene (C6H4BrNO2).

### Step 3: Reduction of m-Bromonitrobenzene
- **Reaction**: Finally, we reduce m-bromonitrobenzene using a reducing agent such as tin (Sn) in hydrochloric acid (HCl) or iron (Fe) in HCl.
- **Product**: This reduction will convert the nitro group (-NO2) into an amino group (-NH2), resulting in m-bromoaniline (C6H4BrNH2).

### Summary of Steps
1. **Nitration**: Benzene + HNO3/H2SO4 → Nitrobenzene
2. **Bromination**: Nitrobenzene + Br2 → m-Bromonitrobenzene
3. **Reduction**: m-Bromonitrobenzene + Sn/HCl → m-Bromoaniline

### Final Product
The final product of this sequence is m-bromoaniline.

---