`R-CH=CH_(2)` reacts with `B_(2) H_(6) ` in presence of `H_(2) O_(2)` to give :

Options

A. `RCOCH_(3)`
B. `RCHOH CH_(2) OH`
C. `RCH_(2) CH_(2)OH`
D. `RCH_(2) CHO`

Answer

To solve the question regarding the reaction of the alkene \( R-CH=CH_2 \) with \( B_2H_6 \) in the presence of \( H_2O_2 \), we can follow these steps:

### Step 1: Identify the Reaction Type
The reaction of an alkene with diborane (\( B_2H_6 \)) followed by hydrogen peroxide (\( H_2O_2 \)) is known as the Hydroboration-Oxidation reaction. This reaction is characterized by the addition of water across the double bond.

**Hint:** Remember that Hydroboration-Oxidation involves the addition of boron and then oxidation to form an alcohol.

### Step 2: Hydroboration Step
In the hydroboration step, diborane (\( B_2H_6 \)) adds across the double bond of the alkene. The boron atom attaches to the less substituted carbon (the terminal carbon in this case), while a hydrogen atom attaches to the more substituted carbon. This follows the anti-Markovnikov rule.

For the alkene \( R-CH=CH_2 \):
- The boron atom (\( B \)) will attach to the \( CH_2 \) end, and a hydrogen atom will attach to the \( CH \) end.

The intermediate formed will be:
\[ R-CH_2-CH_2-BH_2 \]

**Hint:** Focus on the anti-Markovnikov addition where boron attaches to the less substituted carbon.

### Step 3: Oxidation Step
In the oxidation step, the trialkyl borane formed in the previous step reacts with hydrogen peroxide (\( H_2O_2 \)) in the presence of a base (like \( NaOH \) or \( KOH \)). This leads to the replacement of the boron atom with a hydroxyl group (\( OH \)).

The reaction can be summarized as:
\[ R-CH_2-CH_2-BH_2 + H_2O_2 \rightarrow R-CH_2-CH_2-OH + B(OH)_2 \]

**Hint:** Remember that the hydroxyl group will attach to the carbon that has more hydrogen atoms, following the anti-Markovnikov rule.

### Step 4: Final Product
The final product of this reaction will be a primary alcohol:
\[ R-CH_2-CH_2-OH \]

This indicates that the \( OH \) group is attached to the carbon that was originally part of the double bond and had more hydrogen atoms.

**Hint:** The product is a primary alcohol due to the addition of the hydroxyl group to the less substituted carbon.

### Conclusion
Thus, the product formed when \( R-CH=CH_2 \) reacts with \( B_2H_6 \) in the presence of \( H_2O_2 \) is:
\[ R-CH_2-CH_2-OH \] (a primary alcohol).