Which one of the following cannot be prepared from `B_(2)H_(6)` ?

Options

A. `NaBH_(4)`
B. `B_(2)(CH_(3))_(4)H_(2)`
C. `B_(2)(CH_(3))_(6)`
D. `H_(3)BO_(3)`

Answer

To determine which compound cannot be prepared from \( B_2H_6 \), we will analyze the possible reactions of \( B_2H_6 \) and the compounds listed in the question.

### Step-by-Step Solution:

1. **Understanding the Structure of \( B_2H_6 \)**:
- \( B_2H_6 \) (Diborane) has a structure with two boron atoms and six hydrogen atoms. It contains two bridging hydrogen atoms and four terminal hydrogen atoms.

2. **Analyzing the First Compound: \( NaBH_4 \)**:
- \( NaBH_4 \) (Sodium Borohydride) can be prepared by reacting \( B_2H_6 \) with sodium hydride (\( NaH \)).
- The reaction is:
\[ B_2H_6 + 2NaH \rightarrow 2NaBH_4 \]
- **Conclusion**: \( NaBH_4 \) can be prepared from \( B_2H_6 \).

3. **Analyzing the Second Compound: \( B_2(CH_3)_4H_2 \)**:
- This compound can be synthesized by replacing the four terminal hydrogen atoms of \( B_2H_6 \) with methyl groups (\( CH_3 \)).
- The reaction is:
\[ B_2H_6 + 4CH_3OH \rightarrow B_2(CH_3)_4H_2 + 4H_2O \]
- **Conclusion**: \( B_2(CH_3)_4H_2 \) can be prepared from \( B_2H_6 \).

4. **Analyzing the Third Compound: \( B_2(CH_3)_6 \)**:
- This compound suggests that all six hydrogen atoms of \( B_2H_6 \) are replaced by methyl groups.
- However, only the terminal hydrogen atoms can be replaced, and since there are only four terminal hydrogens, it is impossible to replace all six with methyl groups.
- **Conclusion**: \( B_2(CH_3)_6 \) cannot be prepared from \( B_2H_6 \).

5. **Analyzing the Fourth Compound: \( H_3BO_3 \)**:
- \( H_3BO_3 \) (Boric Acid) can be prepared by oxidizing \( B_2H_6 \) with oxygen (\( O_2 \)).
- The reaction is:
\[ B_2H_6 + 3O_2 \rightarrow 2H_3BO_3 \]
- **Conclusion**: \( H_3BO_3 \) can be prepared from \( B_2H_6 \).

### Final Answer:
The compound that **cannot** be prepared from \( B_2H_6 \) is \( B_2(CH_3)_6 \).