State the product formed during the reaction between sodium phenoxide and ethyl iodide on heating `:`

Options

A. Phenetole
B. Ethyl phenyl alcohol
C. Phenone
D. None of these

Explanation

To find the product formed during the reaction between sodium phenoxide and ethyl iodide upon heating, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are sodium phenoxide and ethyl iodide. Sodium phenoxide can be represented as C6H5O^-Na^+ and ethyl iodide as C2H5I.

### Step 2: Understand the Reaction Type
This reaction is an example of the Williamson ether synthesis, which involves the reaction between a sodium alkoxide (sodium phenoxide) and a primary alkyl halide (ethyl iodide) to form an ether.

### Step 3: Mechanism of the Reaction
In the Williamson synthesis:
- The negatively charged oxygen of sodium phenoxide (C6H5O^-) acts as a nucleophile.
- It attacks the electrophilic carbon of the ethyl iodide (C2H5I), which has a partial positive charge due to the presence of the iodine atom (a good leaving group).

### Step 4: Formation of the Ether
As the nucleophile attacks, the bond between the carbon and iodine in ethyl iodide breaks, leading to the release of iodide ion (I^-). The result is the formation of an ether:
- The product formed is ethyl phenyl ether, also known as phenetol (C6H5OCH2CH3).

### Step 5: Write the Balanced Reaction
The balanced chemical equation for the reaction can be written as:
\[ \text{C}_6\text{H}_5\text{O}^- \text{Na}^+ + \text{C}_2\text{H}_5\text{I} \rightarrow \text{C}_6\text{H}_5\text{O}\text{C}_2\text{H}_5 + \text{NaI} \]

### Conclusion
Thus, the product formed during the reaction between sodium phenoxide and ethyl iodide on heating is ethyl phenyl ether (or phenetol).

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