The product formed in the reaction
`PhCH = CH_(2) underset(H_(2)O_(2)//OH^(-))overset(B_(2)H_(6))rarr is`

Options

A. `PhCH_(2)CHO`
B. `PhCH_(2)CH_(2)OH`
C. `PhCHOHCH_(3)`
D. `PhCOCH_(3)`

Explanation

To solve the problem, we need to analyze the reaction step by step, focusing on the hydroboration-oxidation mechanism.

### Step-by-Step Solution:

1. **Identify the Reactant**:
The reactant given is PhCH=CH2, which is styrene (C6H5CH=CH2).

2. **Recognize the Reagents**:
The reagents mentioned are B2H6 (diborane) and H2O2 in the presence of OH⁻. This indicates that we will be performing a hydroboration-oxidation reaction.

3. **Hydroboration Step**:
In the hydroboration step, diborane (B2H6) reacts with the alkene (styrene). The boron atom adds to the less substituted carbon of the double bond (anti-Markovnikov addition).
- The product of this step will be a trialkyl borane:
\[ \text{C6H5-CH2-CH2-B} \text{ (trialkyl borane)} \]

4. **Oxidation Step**:
In the oxidation step, hydrogen peroxide (H2O2) in the presence of a base (OH⁻) will convert the trialkyl borane to an alcohol. The OH group will add to the same carbon where boron was added during hydroboration.
- The final product will be:
\[ \text{C6H5-CH2-CH2-OH} \]
This is phenethyl alcohol.

5. **Final Product**:
Therefore, the product formed from the reaction of PhCH=CH2 with B2H6 followed by H2O2/OH⁻ is phenethyl alcohol (C6H5-CH2-CH2-OH).

### Conclusion:
The product formed in the reaction is phenethyl alcohol.