Question:
Identify the compounds formed when anisole is heated with concentrated HI.
Identify the compounds formed when anisole is heated with concentrated HI.
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The ether oxygen is protonated, iodide attacks the methyl carbon and the aryl C-O bond stays intact.
Updated On: Nov 12, 2026
- Iodobenzene
- Iodobenzene + methanol
- Phenol + methanol
- Phenol + Iodomethane
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Anisole is methyl phenyl ether, \(\text{C}_{6}\text{H}_{5}\text{OCH}_{3}\). Ethers are cleaved by hot concentrated HI.
Step 2: Mechanism
HI first protonates the ether oxygen.
The iodide ion then attacks the less hindered methyl carbon by an \(S_{N}2\) path and breaks the \(\text{O-CH}_{3}\) bond.
Step 3: Why the ring bond stays
The aryl \(\text{C-O}\) bond has partial double bond character and the sp2 ring carbon resists substitution.
So the products are \(\text{C}_{6}\text{H}_{5}\text{OH}\) and \(\text{CH}_{3}\text{I}\).
Final Answer:
Anisole gives phenol and iodomethane with conc. HI. \[ \boxed{\text{(D) }\text{Phenol + Iodomethane}} \]
Anisole is methyl phenyl ether, \(\text{C}_{6}\text{H}_{5}\text{OCH}_{3}\). Ethers are cleaved by hot concentrated HI.
Step 2: Mechanism
HI first protonates the ether oxygen.
The iodide ion then attacks the less hindered methyl carbon by an \(S_{N}2\) path and breaks the \(\text{O-CH}_{3}\) bond.
Step 3: Why the ring bond stays
The aryl \(\text{C-O}\) bond has partial double bond character and the sp2 ring carbon resists substitution.
So the products are \(\text{C}_{6}\text{H}_{5}\text{OH}\) and \(\text{CH}_{3}\text{I}\).
Final Answer:
Anisole gives phenol and iodomethane with conc. HI. \[ \boxed{\text{(D) }\text{Phenol + Iodomethane}} \]
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