What happens when phenol reacts with ethyl iodide in the presence of an alkali?

Phenetole is formed

The correct answer is option 2. Phenetole is formed.

When phenol reacts with ethyl iodide \((C_2H_5I)\) in the presence of an alkali (such as sodium hydroxide, \(NaOH\)), phenetole \((C_6H_5OC_2H_5)\) is formed.

This reaction is an example of a Williamson ether synthesis. Here's a breakdown of the steps involved:

Alkali deprotonation: The alkali \((NaOH)\) acts as a strong base and deprotonates the hydroxyl group \((OH)\) of phenol, forming a phenoxide ion \((C_6H_5O^-)\) and water \((H_2O)\).

Nucleophilic attack: The phenoxide ion acts as a nucleophile due to the negative charge on the oxygen atom. This nucleophile attacks the carbon atom of the ethyl iodide, which is partially positive due to the inductive effect of the halogen atom (iodine).

Bond formation and leaving group: This nucleophilic attack leads to the formation of a new C-O bond between the phenoxide ion and the ethyl group \((C_2H_5)\) of the ethyl iodide. The iodide ion0 \((I^-)\) from the ethyl iodide acts as a leaving group.

Overall reaction:

Therefore, the product formed is phenetole, an aromatic ether. The reaction replaces the hydroxyl group in phenol with an ethoxy group \((C_2H_5O^-)\).