Anisole on cleavage with HI gives : 1. Phenol (C ₆ H ₅ OH) + CH ₃ I 2. Iodobenzene (C ₆ H ₅ I) + CH ₃ OH 3. Phenol (C ₆ H ₅ OH) + C ₂ H ₅ I 4. Iodobenzene (C ₆ H ₅ I) + C ₂ H ₅ OH

Answer: 1. Phenol (C ₆ H ₅ OH) + CH ₃ I Aryl alkyl ethers are cleaved by HI at the alkyl-oxygen bond. The ether oxygen is protonated, then I^- attacks the smaller methyl group by S_N2 . The C₆H₅-O bond is stronger because of partial double bond character, and I^- cannot attack sp2 carbon of the ring. C₆H₅OCH₃+HI→ C₆H₅OH+CH₃I Answer: phenol + methyl iodide, option (1).

Topic: Ethers· Organic Compounds Containing Oxygen

Anisole on cleavage with HI gives :

four product pairs drawn as benzene-ring structures (OH or I) plus CH3I/CH3OH/C2H5I/C2H5OH

1. Phenol (C$_6$H$_5$OH) + CH$_3$I

2. Iodobenzene (C$_6$H$_5$I) + CH$_3$OH

3. Phenol (C$_6$H$_5$OH) + C$_2$H$_5$I

4. Iodobenzene (C$_6$H$_5$I) + C$_2$H$_5$OH

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Answer: 1. Phenol (C$_6$H$_5$OH) + CH$_3$I

Aryl alkyl ethers are cleaved by HI at the alkyl-oxygen bond. The ether oxygen is protonated, then $\mathrm{I^-}$ attacks the smaller methyl group by $\mathrm{S_N2}$. The $\mathrm{C_6H_5-O}$ bond is stronger because of partial double bond character, and $\mathrm{I^-}$ cannot attack sp2 carbon of the ring.

$$\mathrm{C_6H_5OCH_3+HI\rightarrow C_6H_5OH+CH_3I}$$

Answer: phenol + methyl iodide, option (1).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Anisole on cleavage with HI gives :

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