One mole of 4 - nitrocatechol (4 -nitro - 1, 2 - dihydroxybenene) on treatment with an excess of `NaH` followed by one mole of methyl iodide and `H_(2)O` respectively gives

Options

A. 4- nitro -1, 2- diamethoxybenzene
B. 4 - nitro - 5methyl -1, 2- dimethyloxybenzene
C. 2 - methoxy - 5 nitrophenol
D. 2-methoxy - 4nitrophenol

Answer

To solve the problem, we need to analyze the reaction of 4-nitro-1,2-dihydroxybenzene with sodium hydride (NaH), followed by methyl iodide (CH3I) and water (H2O).

### Step-by-Step Solution:

1. **Identify the Structure of 4-Nitro-1,2-Dihydroxybenzene**:
- The compound has a benzene ring with two hydroxyl (–OH) groups at positions 1 and 2, and a nitro (–NO2) group at position 4.
- Structure:
```
OH
|
OH-C6H3-NO2
|
```

2. **Reaction with Sodium Hydride (NaH)**:
- Sodium hydride acts as a strong base and will deprotonate the hydroxyl groups (–OH) on the benzene ring, generating phenoxide ions (–O⁻).
- Since we have two hydroxyl groups, both will be deprotonated:
```
O⁻
|
O⁻-C6H3-NO2
|
```

3. **Nucleophilic Attack on Methyl Iodide (CH3I)**:
- The negatively charged oxygen (phenoxide ion) is a good nucleophile and will attack the electrophilic carbon in methyl iodide (CH3I).
- Since we have two phenoxide ions, both can react with methyl iodide. However, the more stable negative charge will dictate which oxygen attacks first.
- The oxygen at position 2 (ortho) is less stable due to the electron-withdrawing effect of the nitro group at position 4, making it more reactive than the oxygen at position 1 (para).

4. **Formation of Methoxy Groups**:
- The reaction will lead to the formation of two methoxy groups (–OCH3) on the benzene ring:
```
OCH3
|
OCH3-C6H3-NO2
|
```

5. **Hydrolysis with Water (H2O)**:
- The final step involves hydrolysis, where the remaining negative charges on the oxygens will be neutralized by water, converting them into hydroxyl groups.
- The final product will be 2-methoxy-4-nitrophenol.

### Final Product:
The final product of the reaction is **2-methoxy-4-nitrophenol**.