Consider the following reaction.

Statement I : In the above reaction, product formed will be a mixture of benzyl alcohol and iodobenzene.

Statement II : In the above reaction, the $$-O-CH_2-$$ bond is cleaved to give the product.

In the light of the above statements, choose the correct answer from the options given below :

The ether undergoes cleavage in the presence of HI. Benzyl phenyl ether contains two different C-O bonds:

1. Aryl-oxygen bond: $$\mathrm{Ph-O}$$
2. Alkyl-oxygen bond: $$\mathrm{PhCH_2-O}$$

The first step is protonation of the ether oxygen by HI, forming an oxonium ion.

The iodide ion then attacks the carbon atom attached to oxygen. The $$\mathrm{Ph-O}$$ bond is resistant to cleavage because it possesses partial double bond character due to resonance, and nucleophilic substitution at an $$sp^2$$ hybridized aromatic carbon is not feasible.

Therefore, the iodide ion attacks the benzyl carbon through an $$S_N2$$ mechanism, resulting in cleavage of the $$\mathrm{-O-CH_2-}$$ bond.

The products formed are:

$$\mathrm{Ph-OH\ +\ PhCH_2I}$$

Statement I: "The products formed are benzyl alcohol and iodobenzene."

This statement is incorrect because the actual products are phenol and benzyl iodide.

Hence, Statement I is false.

Statement II: "The $$\mathrm{-O-CH_2-}$$ bond is cleaved to give the product."

This statement is correct because cleavage occurs at the benzyl carbon.

Hence, Statement II is true.

Therefore, Statement I is false, but Statement II is true. Hence, the correct option is (D).