The compound which undergoes fastest reaction with aq. NaOH solution is :

Options

A. `C_(6)H_(5)-underset(Cl)underset(|)(CH)-OMe`
B. `C_(6)H_(5)-underset(Cl)underset(|)(CH)-CH_(3)`
C. `C_(6)H_(5)-underset(Cl)underset(|)(CH)-CH_(2)CH_(3)`
D. `C_(6)H_(5)-underset(Cl)underset(|)(CH)-C_(6)H_(5)`

Answer

To determine which compound undergoes the fastest reaction with aqueous NaOH, we need to analyze the stability of the intermediates formed during the reaction. The reaction rate with NaOH is influenced by the leaving group and the stability of the resulting product.

### Step-by-Step Solution:

1. **Identify the Compounds**: We have four compounds that contain a chlorine atom as the leaving group. The compounds are:
- Compound 1: C6H5-CH(CH3)-Cl
- Compound 2: C6H5-CH(CH3)-Cl
- Compound 3: C6H5-CH(CH2CH3)-Cl
- Compound 4: C6H5-CH(C6H5)-Cl

2. **Analyze the Leaving Group**: All compounds have chlorine as the leaving group. Therefore, the rate of reaction will not depend on the leaving group but rather on the stability of the carbocation formed after the leaving group departs.

3. **Consider the Stability of Carbocations**: The stability of the carbocation formed after the departure of the leaving group is crucial. The more stable the carbocation, the faster the reaction will occur. The stability order of carbocations is:
- Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl

4. **Evaluate Each Compound**:
- **Compound 1**: C6H5-CH(CH3)-Cl (Secondary carbocation)
- **Compound 2**: C6H5-CH(CH3)-Cl (Secondary carbocation)
- **Compound 3**: C6H5-CH(CH2CH3)-Cl (Secondary carbocation)
- **Compound 4**: C6H5-CH(C6H5)-Cl (Tertiary carbocation)

5. **Determine the Fastest Reaction**: Among these compounds, Compound 4 has a tertiary carbocation, which is more stable than the secondary carbocations formed in the other compounds. Therefore, it will undergo the fastest reaction with aqueous NaOH.

### Conclusion:
The compound that undergoes the fastest reaction with aqueous NaOH is **Compound 4: C6H5-CH(C6H5)-Cl**.