In Williamson's synthesis, methoxyethane is prepared by

Options

A. passing ethanol over heated `Al_(2)O_(3)`
B. heating sodium ethoxide with methyl bromide
C. treating ethyl alcohol with excess of `H_(2)SO_(4)` at 440 K
D. heating ethanol with dry `Ag_(2)O.`

Explanation

To prepare methoxyethane using Williamson's synthesis, we follow these steps:

### Step 1: Identify the reactants
In Williamson's synthesis, we need a sodium alkoxide and a primary alkyl halide. For methoxyethane (also known as ethyl methyl ether), we will use sodium ethoxide as the alkoxide and methyl bromide as the alkyl halide.

### Step 2: Write the structures of the reactants
- **Sodium Ethoxide (C2H5O^-Na^+)**: This is derived from ethanol (C2H5OH) by deprotonating it with sodium (Na).
- **Methyl Bromide (CH3Br)**: This is a primary alkyl halide where bromine is the leaving group.

### Step 3: Set up the reaction
The reaction can be represented as follows:
\[ \text{C2H5O}^- \text{Na}^+ + \text{CH3Br} \rightarrow \text{C2H5OCH3} + \text{NaBr} \]

### Step 4: Mechanism of the reaction
1. The sodium ethoxide (C2H5O^-) acts as a nucleophile and attacks the electrophilic carbon of the methyl bromide (CH3Br).
2. The bond between the carbon and bromine breaks, leading to the formation of methoxyethane (C2H5OCH3) and sodium bromide (NaBr).

### Step 5: Conclusion
Thus, methoxyethane is prepared by the reaction of sodium ethoxide with methyl bromide.

### Final Answer
Methoxyethane is prepared by heating sodium ethoxide with methyl bromide.

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