On boiling concentrated \(HBr\), phenyl ethyl ether will give:

Phenol and Ethyl bromide

The correct answer is option 1. Phenol and Ethyl bromide.

When phenyl ethyl ether (phenetole) reacts with concentrated $HBr$, the ether linkage ($C - O$ bond) is cleaved.

1. Protonation: The oxygen of the ether is protonated by $HBr$ to form an oxonium ion.
2. Bond Cleavage: The cleavage occurs at the Alkyl-Oxygen bond rather than the Aryl-Oxygen bond.
• The $O - C_6H_5$ (Aryl-Oxygen) bond has a partial double bond character due to the resonance of the lone pair on oxygen with the benzene ring, making it much stronger and harder to break.
• The $O - C_2H_5$ (Alkyl-Oxygen) bond is a standard single bond and is more susceptible to nucleophilic attack.
3. Nucleophilic Attack: The bromide ion ($Br^-$) attacks the smaller alkyl group (ethyl) via an $S_N2$ mechanism.

Correct Products

The reaction yields Phenol and Ethyl bromide:

$C_6H_5OCH_2CH_3 + HBr \overset{{\Delta}}{\longrightarrow} C_6H_5OH + CH_3CH_2Br$