In Williamson's synthesis, ethoxyethane is pprepared by

Options

A. Heating sodium ethoxide with ethyl bromide
B. Passing ethanol over heated alumina
C. treating ethyl alcohol with excess of conc. `H_(2)SO_(4)` at 430-440K.
D. heating ethanol with dry `Ag_(2)O`.

Answer

**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 an alkyl halide through an SN2 mechanism.

2. **Formation of Alkoxide Ion**:
In this synthesis, we start with ethanol (C2H5OH). When sodium (Na) is added to dry ether, it reacts with ethanol to form sodium ethoxide (C2H5O^-Na^+) and hydrogen gas (H2). The reaction can be represented as:
\[ 2 \text{C}_2\text{H}_5\text{OH} + 2 \text{Na} \rightarrow 2 \text{C}_2\text{H}_5\text{O}^- \text{Na}^+ + \text{H}_2 \]

3. **Reaction with Alkyl Halide**:
The alkoxide ion (C2H5O^-) then acts as a nucleophile and attacks an alkyl halide, such as ethyl bromide (C2H5Br), through an SN2 mechanism. This results in the formation of ethoxyethane (C2H5OC2H5) and the release of bromide ion (Br^-):
\[ \text{C}_2\text{H}_5\text{O}^- + \text{C}_2\text{H}_5\text{Br} \rightarrow \text{C}_2\text{H}_5\text{O}\text{C}_2\text{H}_5 + \text{Br}^- \]

4. **Final Product**:
The final product of this reaction is ethoxyethane (also known as diethyl ether), which is represented as:
\[ \text{C}_2\text{H}_5\text{O}\text{C}_2\text{H}_5 \]

5. **Conclusion**:
Therefore, in Williamson's synthesis, ethoxyethane is prepared by heating sodium ethoxide with ethyl bromide.

**Final Answer**: Ethoxyethane is prepared by heating sodium ethoxide with ethyl bromide in Williamson's synthesis.

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