`(CH_(3))_(3)-C-O-CH_(2)-C_(6)H_(5)` can be prepared from Williamson's synthesis, using:

Options

A. `(CH_(3))_(3)-C-Cl ` and `C_(6)H_(5)CH_(2)ONa`
B. `C_(6)H_(5)CH_(2)Cl` and `(CH_(3))_(3)C-ONa`
C. `(CH_(3))_(3)C-O-CH_(2)-Cl ` and `C_(6)H_(5)ONa`
D. All of these

Explanation

To prepare the compound `(CH₃)₃C-O-CH₂-C₆H₅` using Williamson's synthesis, we need to identify the appropriate reactants. Williamson's synthesis involves the reaction of an alkyl halide with a sodium alkoxide to form an ether.

### Step-by-Step Solution:

1. **Identify the Structure**: The given structure `(CH₃)₃C-O-CH₂-C₆H₅` is a tertiary butyl group `(CH₃)₃C-` connected to an ether bond `-O-` and then to a benzyl group `-CH₂-C₆H₅`.

2. **Determine the Alkoxide**: In Williamson's synthesis, we need a sodium alkoxide. The alkoxide corresponding to the tertiary butyl group is `(CH₃)₃C-O⁻Na⁺` (sodium tert-butoxide).

3. **Identify the Alkyl Halide**: We need an alkyl halide that can react with the sodium alkoxide to form the ether. The benzyl group `-CH₂-C₆H₅` suggests that we can use benzyl chloride `C₆H₅CH₂Cl` as the alkyl halide.

4. **Write the Reaction**: The reaction can be summarized as follows:
\[ (CH₃)₃C-O⁻Na⁺ + C₆H₅CH₂Cl \rightarrow (CH₃)₃C-O-CH₂-C₆H₅ + NaCl \]
Here, the sodium alkoxide `(CH₃)₃C-O⁻` acts as a nucleophile and attacks the benzyl chloride, resulting in the formation of the ether and sodium chloride as a byproduct.

5. **Conclusion**: Therefore, the correct reactants for the Williamson synthesis to prepare `(CH₃)₃C-O-CH₂-C₆H₅` are sodium tert-butoxide and benzyl chloride.

### Final Answer:
The compound `(CH₃)₃C-O-CH₂-C₆H₅` can be prepared from Williamson's synthesis using:
- **Sodium tert-butoxide** `(CH₃)₃C-O⁻Na⁺`
- **Benzyl chloride** `C₆H₅CH₂Cl`