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 question regarding the reaction \( C_2H_5ONa + IC_2H_5 \rightarrow C_2H_5OC_2H_5 + NaI \), we can break it down step by step.

### Step 1: Identify the Reactants
The reactants in the given reaction are:
- Sodium ethoxide (\( C_2H_5ONa \))
- Iodine ethane (\( IC_2H_5 \))

### Step 2: Understand the Reaction Mechanism
This reaction is a classic example of ether formation through a process known as Williamson's Synthesis. In this reaction, sodium ethoxide acts as a nucleophile.

### Step 3: Nucleophilic Attack
In the reaction, the ethoxide ion (\( C_2H_5O^- \)) from sodium ethoxide attacks the ethyl iodide (\( IC_2H_5 \)). The oxygen atom in the ethoxide ion has a negative charge, making it a strong nucleophile.

### Step 4: Formation of Ether
The nucleophilic attack leads to the formation of diethyl ether (\( C_2H_5OC_2H_5 \)). During this process, the iodide ion (\( I^- \)) is released as a byproduct.

### Step 5: Formation of Sodium Iodide
As a result of the reaction, sodium iodide (\( NaI \)) is formed. The sodium ion (\( Na^+ \)) pairs with the iodide ion (\( I^- \)).

### Final Reaction
Putting it all together, the overall reaction can be summarized as:
\[ C_2H_5ONa + IC_2H_5 \rightarrow C_2H_5OC_2H_5 + NaI \]

### Conclusion
This reaction is known as Williamson's Synthesis, which is a method for preparing ethers.

### Answer
The answer to the question is **Williamson's Synthesis**.

---