In Williamson's synthesis, ethoxy ethane is prepard by

Options

A. a. heating sodium ethoxide with ethyl bromide
B. b. passing ethanol over heated alumina
C. c. treating ethyl alcohol with excess of cond. `H_SO_4" at " 430-440 K`
D. d. heating ethanol wit dry `AG_2O`

Answer

To solve the question regarding the preparation of ethoxy ethane in Williamson's synthesis, let's break down the steps:

### Step-by-Step Solution:

1. **Understanding Williamson's Synthesis**:
- Williamson's synthesis is a method used to prepare ethers. It involves the reaction of an alkoxide ion with a primary alkyl halide.

2. **Identifying the Product**:
- The product we want to prepare is ethoxy ethane, which has the chemical formula C2H5-O-C2H5.

3. **Determining the Alkoxide Ion**:
- In this case, the alkoxide ion needed is derived from ethanol (C2H5OH). When ethanol loses a hydrogen ion (H+), it forms the ethoxide ion (C2H5O^-).

4. **Identifying the Alkyl Halide**:
- The alkyl halide required for this reaction is ethyl bromide (C2H5Br). This is a primary alkyl halide, which is necessary for the Williamson synthesis.

5. **Setting Up the Reaction**:
- The reaction can be represented as:
\[ \text{C2H5O}^- + \text{C2H5Br} \rightarrow \text{C2H5-O-C2H5} + \text{Br}^- \]
- Here, the ethoxide ion attacks the ethyl bromide, displacing the bromide ion (Br^-) and forming ethoxy ethane.

6. **Conclusion**:
- Therefore, the reactants required to prepare ethoxy ethane in Williamson's synthesis are sodium ethoxide (which provides the alkoxide ion) and ethyl bromide (the primary alkyl halide).

### Final Answer:
Ethoxy ethane is prepared by the reaction of sodium ethoxide with ethyl bromide.