Aniline reacts with `Br_(2)` water, `NaNO_(2)//HCL` gives respectively :

Options

A. p-Bromo aniline, p-chloro aniline
B. 2, 4, 6 tri bromo aniline, p-chloro aniline
C. 2, 4, 6 tri bromo aniline, Benzene diazonium chloride
D. p-bromo, aniline, Benzene diazonium chloride

Answer

To solve the problem of what products are formed when aniline reacts with Br₂ water and NaNO₂/HCl, we can break it down into two steps:

### Step 1: Reaction of Aniline with Br₂ Water
Aniline (C₆H₅NH₂) reacts with bromine water (Br₂ in water) through an electrophilic substitution reaction. The amino group (-NH₂) is an activating group and directs the incoming bromine to the ortho and para positions of the benzene ring.

1. **Identify the structure of aniline**:
- Aniline has a benzene ring with an amino group attached to it.

2. **Electrophilic substitution**:
- When Br₂ water is added to aniline, bromine atoms will substitute hydrogen atoms at the ortho and para positions.
- This results in the formation of 2,4,6-tribromoaniline.

### Step 2: Reaction of 2,4,6-Tribromoaniline with NaNO₂/HCl
Next, we consider the reaction of 2,4,6-tribromoaniline with sodium nitrite (NaNO₂) in the presence of hydrochloric acid (HCl).

1. **Formation of diazonium salt**:
- Under these conditions (0 to 5 degrees Celsius), the amino group (-NH₂) of 2,4,6-tribromoaniline is converted into a diazonium salt.
- The product formed is benzene diazonium chloride (C₆H₅N₂Cl).

### Final Products
Thus, the overall reaction sequence gives us:
- From aniline and Br₂ water: **2,4,6-Tribromoaniline**
- From 2,4,6-Tribromoaniline and NaNO₂/HCl: **Benzene diazonium chloride**

### Conclusion
The final answer is:
- **2,4,6-Tribromoaniline and Benzene diazonium chloride**.