If 2-pentanone is reacted with `NaBH_(4)` follwed by hydrolysis with `D_(2)O` the product will be

Options

A. `CH_(3)CH(OD)CH_(2)CH_(2)CH_(3)`
B. `CH_(3)CD(OH)CH_(2)CH_(2)CH_(3)`
C. `CH_(3)CH(OH)CH_(2)CH_(2)CH_(3)`
D. `CH_(3)CD(OD)CH_(2)CH_(2)CH_(3)`

Answer

To solve the problem of determining the product when 2-pentanone is reacted with sodium borohydride (NaBH4) followed by hydrolysis with deuterium oxide (D2O), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of 2-Pentanone**:
- The molecular formula of 2-pentanone is C5H10O. Its structure can be represented as:
```
CH3-CO-CH2-CH2-CH3
```
- Here, the carbonyl group (C=O) is located at the second carbon.

2. **Reaction with Sodium Borohydride (NaBH4)**:
- Sodium borohydride is a reducing agent that provides hydride ions (H-) to reduce the carbonyl group of the ketone.
- The reduction of 2-pentanone by NaBH4 will convert the carbonyl group (C=O) into a hydroxyl group (C-OH), resulting in the formation of 2-pentanol:
```
CH3-CHOH-CH2-CH2-CH3
```

3. **Hydrolysis with Deuterium Oxide (D2O)**:
- When 2-pentanol is subjected to hydrolysis using D2O, the hydroxyl group (-OH) will be replaced by a deuterated hydroxyl group (-OD).
- The reaction can be represented as follows:
```
CH3-CHOH-CH2-CH2-CH3 + D2O → CH3-CHOD-CH2-CH2-CH3
```

4. **Final Product**:
- The final product after the reaction with D2O will be:
```
CH3-CH(OD)-CH2-CH2-CH3
```
- This indicates that at the second carbon, there is a deuterium atom (D) instead of a hydrogen atom (H).

5. **Conclusion**:
- The correct product formed from the reaction is:
```
CH3-CH(OD)-CH2-CH2-CH3
```