Aniline when treated with `K_2 Cr_2 O_7` in conc `H_2 SO_4` gives :

Options

A.
B.
C.
D.

Answer

To solve the question of what happens when aniline is treated with \( K_2Cr_2O_7 \) in concentrated \( H_2SO_4 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have aniline (\( C_6H_5NH_2 \)) and potassium dichromate (\( K_2Cr_2O_7 \)) in the presence of concentrated sulfuric acid (\( H_2SO_4 \)).

2. **Understand the Role of \( K_2Cr_2O_7 \)**:
- Potassium dichromate is a strong oxidizing agent. In acidic medium, it can oxidize various organic compounds.

3. **Determine the Type of Reaction**:
- The reaction involves oxidation. Aniline, being a primary amine, can be oxidized to form an imine or further to a carboxylic acid depending on the conditions.

4. **Oxidation of Aniline**:
- When aniline is oxidized by the nascent oxygen produced from \( K_2Cr_2O_7 \) in \( H_2SO_4 \), it can be converted to a product known as benzoic acid (\( C_6H_5COOH \)).
- The reaction can be summarized as:
\[ C_6H_5NH_2 + 3 \, O \rightarrow C_6H_5COOH + NH_3 \]
- Here, aniline is oxidized to benzoic acid, and ammonia is released as a byproduct.

5. **Final Products**:
- The main product of the reaction is benzoic acid (\( C_6H_5COOH \)).

### Conclusion:
When aniline is treated with \( K_2Cr_2O_7 \) in concentrated \( H_2SO_4 \), the product formed is benzoic acid (\( C_6H_5COOH \)).

---