The alkene `R-CH=CH_(2)` reacts readily with `B_(2)H_(6)` and formed the product `B` which on oxidation with alkaline hydrogen peroxide produces

Options

A. `R-CH_(2)-CHO`
B. `R-CH_(2)-CH_(2)-OH`
C. `R-underset(CH_(3))underset(|)(C )=O`
D. `R-underset(OH)underset(|)(C )H-underset(OH)underset(|)(C )H_(2)`

Explanation

To solve the problem step by step, we will follow the reaction of the alkene \( R-CH=CH_2 \) with diborane \( B_2H_6 \) and then the subsequent oxidation with alkaline hydrogen peroxide.

### Step 1: Identify the Alkene
The given alkene is \( R-CH=CH_2 \). This is a simple alkene where \( R \) is an alkyl group.

### Step 2: Reaction with Diborane
When \( R-CH=CH_2 \) reacts with diborane \( B_2H_6 \), it undergoes hydroboration. In this reaction, diborane adds across the double bond of the alkene.

- The boron atom from diborane will bond with the less substituted carbon atom (the terminal carbon in this case), while the hydrogen atoms will add to the more substituted carbon atom.

The reaction can be represented as:
\[ R-CH=CH_2 + B_2H_6 \rightarrow R-CH_2-CH_2B \]

### Step 3: Formation of Product B
The product formed after the hydroboration is \( R-CH_2-CH_2B \), where \( B \) is a boron atom attached to the second carbon.

### Step 4: Oxidation with Alkaline Hydrogen Peroxide
Next, we oxidize the boron compound \( R-CH_2-CH_2B \) using alkaline hydrogen peroxide (H2O2 in the presence of NaOH). This oxidation converts the boron group into a hydroxyl group (-OH).

The reaction can be represented as:
\[ R-CH_2-CH_2B + H_2O_2/NaOH \rightarrow R-CH_2-CH_2OH \]

### Step 5: Final Product
The final product after oxidation is \( R-CH_2-CH_2OH \), which is an alcohol.

### Summary of the Reaction
1. **Starting Alkene**: \( R-CH=CH_2 \)
2. **Reacts with**: \( B_2H_6 \)
3. **Forms**: \( R-CH_2-CH_2B \)
4. **Oxidized by**: Alkaline hydrogen peroxide
5. **Final Product**: \( R-CH_2-CH_2OH \) (an alcohol)