Alkene `R-CH=CH_2` reacts readily with `B_2H_6` and the product on oxidation with alkaline hydrogen peroxides produces

Options

A. `R-CH_2-CHO`
B. `R-CH_2-CH_2-OH`
C. `R-undersetunderset(O)(||)C-CH_3`
D. `R-undersetunderset(OH)|CH-undersetunderset(OH)|CH_3`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Alkene**:
The given alkene is \( R-CH=CH_2 \). This structure indicates that it is a terminal alkene where \( R \) can be any alkyl group.

2. **Reaction with Diborane**:
When alkenes react with diborane (\( B_2H_6 \)), they undergo hydroboration. In this reaction, diborane adds across the double bond of the alkene. The boron atom attaches to the less substituted carbon (the terminal carbon in this case), while hydrogen adds to the more substituted carbon.

The reaction can be represented as:
\[ R-CH=CH_2 + B_2H_6 \rightarrow R-CH_2-CH_2-BH_2 \]

3. **Formation of Trialkylborane**:
The product of the hydroboration is a trialkylborane. In this case, the product is \( R-CH_2-CH_2-BH_2 \).

4. **Oxidation with Alkaline Hydrogen Peroxide**:
The trialkylborane undergoes oxidation when treated with alkaline hydrogen peroxide (\( H_2O_2 \) in a basic medium). This process converts the boron atom into a hydroxyl group. The reaction can be summarized as:
\[ R-CH_2-CH_2-BH_2 + H_2O_2 \rightarrow R-CH_2-CH_2-OH \]

5. **Final Product**:
The final product after oxidation is an alcohol. Specifically, the product is \( R-CH_2-CH_2-OH \), which is a primary alcohol.

### Conclusion:
The product formed after the reaction of the alkene \( R-CH=CH_2 \) with \( B_2H_6 \) followed by oxidation with alkaline hydrogen peroxide is a primary alcohol \( R-CH_2-CH_2-OH \).