The reaction given below is known as `C_2H_5ONa+IC_2H_5rarrC_2H_5OC_2H_5+ NaI`

Options

A. Kolbe's synthesis
B. Wurtz's synthesis
C. Williamson's synthesis
D. Grignard's synthesis

Answer

To solve the problem, we need to identify the type of reaction represented by the equation:

\[ C_2H_5ONa + IC_2H_5 \rightarrow C_2H_5OC_2H_5 + NaI \]

### Step-by-Step Solution:

1. **Identify the Reactants:**
- The reactants in the equation are sodium ethoxide (\(C_2H_5ONa\)) and ethyl iodide (\(IC_2H_5\)).
- Sodium ethoxide is an alkoxide, and ethyl iodide is a haloalkane.

2. **Identify the Products:**
- The products formed are diethyl ether (\(C_2H_5OC_2H_5\)) and sodium iodide (\(NaI\)).
- Diethyl ether is an ether, which is formed from the reaction of two alkyl groups with an oxygen atom.

3. **Understand the Mechanism:**
- In this reaction, the alkoxide ion (\(C_2H_5O^-\)) acts as a nucleophile and attacks the electrophilic carbon in the ethyl iodide.
- The iodine atom acts as a leaving group, resulting in the formation of the ether.

4. **Classify the Reaction:**
- This type of reaction, where an alkoxide reacts with a haloalkane to form an ether, is known as Williamson Ether Synthesis.
- The Williamson synthesis is characterized by the nucleophilic substitution mechanism where an alkoxide ion attacks a haloalkane.

5. **Conclusion:**
- Therefore, the reaction given is classified as Williamson Ether Synthesis.

### Final Answer:
The reaction \(C_2H_5ONa + IC_2H_5 \rightarrow C_2H_5OC_2H_5 + NaI\) is known as **Williamson Ether Synthesis**.

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