Which of the following is produced when anisole reacts with HI?

Question

Accepted Answer

1. **Reactants:** Anisole is methyl phenyl ether ($C_6H_5-O-CH_3$). HI is Hydrogen Iodide.
2. **Mechanism:** The reaction involves the cleavage of the ether linkage. In alkyl aryl ethers, the oxygen atom is attached to a benzene ring. The lone pair on the oxygen enters into resonance with the benzene ring, imparting a **partial double bond character** to the $C_{phenyl}-O$ bond. This makes the $C_{phenyl}-O$ bond much stronger and difficult to break compared to the $C_{methyl}-O$ bond.
3. **Cleavage:** Consequently, the weaker $O-CH_3$ bond breaks. The proton ($H^+$) from HI protonates the oxygen.
4. **Nucleophilic Attack:** The iodide ion ($I^-$) attacks the smaller alkyl group (methyl group) via an $S_N2$ mechanism.
5. **Products:** This results in the formation of **Phenol** ($C_6H_5OH$) and **Methyl Iodide** ($CH_3I$).
6. **Conclusion:** The products are Phenol and methyl iodide.

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