`CH_(3)CH=CH_(2)underset(H_(2)O_(2)//OH)overset(NaBD_(4))to` Product X, X is :

Options

A. `CH_(3)underset(OH)underset(|)(C)HCH_(2)D`
B. `CH_(3)underset(D)underset(|)(C)HCH_(2)OH`
C. `CH_(3)underset(OD)underset(|)(C)HCH_(3)`
D. none is correct

Explanation

To solve the question, we need to determine the product formed when 1-propene (CH3CH=CH2) reacts with NaBD4 in the presence of H2O2 and OH-.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is 1-propene (CH3CH=CH2). This is an alkene with a double bond between the second and third carbon atoms.

**Hint**: Look for the functional groups and the structure of the reactant.

2. **Identify the Reagents**:
The reagents are NaBD4 (sodium borodeuteride), H2O2 (hydrogen peroxide), and OH- (hydroxide ion). NaBD4 is a reducing agent that will facilitate the reduction of the alkene.

**Hint**: Recognize the role of each reagent in the reaction.

3. **Understand the Reaction Mechanism**:
The reaction involves the reduction of the alkene. NaBD4 will add deuterium (D) across the double bond. The presence of H2O2 and OH- suggests that the addition will follow the anti-Markovnikov rule, meaning that the -D will add to the less substituted carbon.

**Hint**: Recall the concept of Markovnikov's rule and its opposite, anti-Markovnikov's rule.

4. **Break the Double Bond**:
The double bond in 1-propene will break, and the -D will add to the terminal carbon (the less substituted carbon), while -OH will add to the more substituted carbon.

**Hint**: Visualize the breaking of the double bond and the addition of new groups.

5. **Draw the Product**:
After the addition, the structure of the product will be:
- The carbon that was part of the double bond will now have a -D and a -OH group attached.
- The product can be represented as CH3-CH(D)-CH2(OH).

**Hint**: Ensure that you correctly place the -D and -OH groups according to the anti-Markovnikov addition.

6. **Final Product Structure**:
The final product can be written as CH3-CH(D)-CH2OH. This indicates that we have a secondary alcohol with deuterium at the second carbon.

**Hint**: Confirm that the product reflects the correct addition pattern and that all valencies are satisfied.

### Conclusion:
The product X formed from the reaction of CH3CH=CH2 with NaBD4 in the presence of H2O2 and OH- is **CH3-CH(D)-CH2OH**.