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(C) The reaction of methoxybenzene $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond.Since the $C_6H_5-O$ bond has partial double bond

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Methyl iodide + phenol

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(C) The reaction of methoxybenzene $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond.Since the $C_6H_5-O$ bond has partial double bond character due to resonance,it is stronger and harder to break than the $CH_3-O$ bond.Therefore,the $HI$ attacks the $CH_3$ group,resulting in the formation of methyl iodide $(CH_3I)$ and phenol $(C_6H_5OH)$.The reaction is: $C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I$.

What will be the products of the reaction if methoxybenzene reacts with $HI$?

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