Aniline on oxidation with `Na_(2)Cr_(2)O_(7)` and `H_(2)SO_(4)` gives

Options

A. benzoic acid
B. m-amino benzoic acid
C. Schiff's base
D. p-benzoquinone.

Explanation

To solve the question regarding the oxidation of aniline with `Na2Cr2O7` and `H2SO4`, we will follow these steps:

### Step 1: Identify the Reactants
The reactant in this case is aniline, which has the molecular formula C6H5NH2. It is an aromatic amine.

### Step 2: Understand the Oxidizing Agent
The oxidizing agent used is sodium dichromate (Na2Cr2O7) in the presence of sulfuric acid (H2SO4). This combination is known to oxidize amines.

### Step 3: Reaction Mechanism
When aniline is treated with sodium dichromate and sulfuric acid, the amine group (-NH2) is oxidized. The oxidation process involves the conversion of the amine group into a carbonyl compound.

### Step 4: Identify the Product
The oxidation of aniline leads to the formation of a ketone. In this case, the product formed is para-aminobenzoic acid or para-benzocunone (C6H4(CO)C6H5) depending on the conditions of the reaction. However, under strong oxidizing conditions, it typically leads to the formation of benzoic acid (C6H5COOH).

### Step 5: Conclusion
Thus, the final product of the oxidation of aniline with sodium dichromate and sulfuric acid is **benzoic acid**.

### Final Answer:
Aniline on oxidation with `Na2Cr2O7` and `H2SO4` gives **benzoic acid**.

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