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**Anisole (CH₃-O-C₆H₅)** with HI gives:

APhenol + methyl iodide
BAnisol's methyl side leaves as CH₃-I, leaving phenol — because phenoxide carbon (sp² aromatic) is not susceptible to S$_N$2
CBoth A and B describe the same outcome
DCyclohexanol
Answer & Solution
Correct answer: C. Both A and B describe the same outcome
Mechanism: protonation of ether O, then iodide attacks the methyl group (sp³ C, S$_N$2). The aryl C is not attacked (sp² and stabilised). Products: C₆H₅-OH (phenol) + CH₃-I. Phenol does NOT react with HI further (aryl C-O is strong, sp² C resists S$_N$2). Classic textbook example.
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