The reaction of phenylmethyl ether (anisole) with hydrogen iodide (HI) involves the cleavage of the C-O bond.

1. Bond Strength: The oxygen atom is bonded to a methyl group (alkyl) and a phenyl group (aryl). The bond between the oxygen and the benzene ring ($C_{sp^2}-O$) has a partial double bond character due to resonance, making it much stronger and difficult to break compared to the bond between oxygen and the methyl group ($C_{sp^3}-O$) .
2. Mechanism: The reaction proceeds via the protonation of the ether oxygen. The iodide ion ($I^-$), being a good nucleophile, attacks the less sterically hindered methyl carbon via an $S_N2$ mechanism.
3. Products: This results in the cleavage of the $O-CH_3$ bond, forming methyl iodide ($CH_3I$) and phenol ($C_6H_5OH$). Iodobenzene is not formed because the strong aryl-oxygen bond remains intact.

Reaction: $C_6H_5-O-CH_3 + HI \xrightarrow{\Delta} C_6H_5OH + CH_3I$