The following reaction gives :
4-nitrotoluene `underset(H_(2)S_(4))overset(K_(2)Cr_(2)O_(7)) to `?

Options

A. 4-nitrobenzaldehyde
B. 4-nitrobenzyl alcohol
C. 4-aminotoluene
D. 4-nitrobenzoic acid

Explanation

To solve the question regarding the reaction of 4-nitrotoluene with potassium dichromate (K2Cr2O7) in the presence of sulfuric acid (H2SO4), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in the reaction are:
- 4-nitrotoluene (C6H4(NO2)(CH3))
- Potassium dichromate (K2Cr2O7)
- Sulfuric acid (H2SO4)

### Step 2: Understand the Role of the Oxidizing Agent
Potassium dichromate (K2Cr2O7) is a strong oxidizing agent. In the presence of sulfuric acid, it can oxidize the alkyl group (methyl group in this case) of 4-nitrotoluene.

### Step 3: Predict the Oxidation Reaction
When 4-nitrotoluene is oxidized:
- The methyl group (–CH3) is oxidized to a carboxylic acid group (–COOH).
- The nitro group (–NO2) remains unchanged.

### Step 4: Write the Product Structure
The product formed from the oxidation of 4-nitrotoluene will be:
- 4-nitrobenzoic acid (C6H4(NO2)(COOH))

### Step 5: Conclusion
Thus, the final product of the reaction between 4-nitrotoluene and potassium dichromate in the presence of sulfuric acid is **4-nitrobenzoic acid**.

### Final Answer
The product is **4-nitrobenzoic acid**.

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