By a proper choice of reagents, both symmetrical and unsymmetrical ethers can be prepared by Williamson synthesis which involves the reaction between an alkyl halide and an alkoxide ion. The reverse process involving the cleavage of ethers to give back the original alkyl halide and the alcohol can be carried out by heating the ether with HI at 373K.

Benzyl ether reacts with HI to form:

Benzyl alcohol and ethyl iodide

The correct answer is option 2. Benzyl alcohol and ethyl iodide.

The given reaction refers to a molecule with the general structure \(R-CH_2-O-R'\), where \(R\) is a phenyl group \((C_6H_5)\) and \(R'\) is an alkyl group (in this case, ethyl \(-CH_3CH_2-\)). Hydrogen iodide \((HI)\) acts as a cleaving agent for ethers. It breaks the \(C-O\) bond in the ether molecule.

In the reaction between benzyl ether and \(HI\), the \(C-O\) bond breaks. The benzyl group \((C_6H_5CH_2-)\) attached to the oxygen gets converted to benzyl alcohol \((C_6H_5CH_2OH)\) by acquiring a hydroxyl group \((OH)\) from \(HI\). The ethyl group \((CH_3CH_2)\) attached to the oxygen gains an iodine atom (I) from \(HI\), forming ethyl iodide \((CH_3CH_2I)\).

Therefore, the reaction between benzyl ether and HI results in the formation of benzyl alcohol and ethyl iodide.