Question:

Ethers are prepared by the dehydration of alcohols in presence of protic acids at 413 K. Symmetrical and unsymmetrical ethers can also be prepared by Williamson synthesis. This reaction involves \( \mathrm{S_N2} \) attack of an alkoxide ion on primary alkyl halide. If a tertiary alkyl halide is used, elimination reaction occurs and an alkene is formed, no ether is formed. C - O bond in ethers are cleaved under drastic conditions with excess of HI. When unsymmetrical ethers react with HI, the alkyl halide is formed from smaller alkyl group. If one of the alkyl group is tertiary, the alkyl halide is formed from the tertiary alkyl group because tertiary alkyl carbocation is more stable than the primary carbocation. Cleavage of alkyl aryl ethers takes place at the alkyl-oxygen bond due to more stable aryl-oxygen bond. The order of reactivity of hydrogen halides is HI > HBr > HCl. Aromatic ethers undergo electrophilic substitution reactions. The alkoxy group attached to the aromatic ring activates the ring towards electrophilic substitution and directs the incoming group to ortho- and para-positions.

Answer the following questions :

(b) Anisole on reaction with HI gives phenol and \( \mathrm{CH_3-I} \) and not methanol and iodobenzene. Justify the statement.

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The aryl-oxygen bond has partial double bond character because of resonance, so it is stronger. Iodide attacks the methyl group by an SN2 path.
Updated On: Oct 22, 2026
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Solution and Explanation

Step 1: Write the reaction:
Anisole is \( \mathrm{C_6H_5{-}O{-}CH_3} \). With HI it gives phenol and methyl iodide.
\[ \mathrm{C_6H_5OCH_3 + HI \longrightarrow C_6H_5OH + CH_3I} \]

Step 2: Protonation of the ether:
HI first protonates the oxygen atom. This gives the oxonium ion \( \mathrm{C_6H_5{-}O^+(H){-}CH_3} \).
Now the iodide ion must attack one of the two carbons joined to oxygen.

Step 3: Why iodide attacks the methyl carbon:
The methyl carbon is sp3 and easy to reach. \( \mathrm{I^-} \) attacks it by an \( \mathrm{S_N2} \) path. The C-O bond of the methyl group breaks.
The phenyl carbon is sp2. The lone pair on oxygen is in resonance with the ring, so the C(aryl)-O bond has partial double bond character. This bond is stronger and hard to break.
An \( \mathrm{S_N2} \) attack at an sp2 ring carbon is also not possible.

Step 4: Why no more reaction:
Phenol does not react further with HI to give iodobenzene, because its C-O bond is also strong. So iodobenzene and methanol are not formed.

Final Answer:
The aryl-oxygen bond is stronger (partial double bond by resonance) and the sp2 carbon cannot undergo \( \mathrm{S_N2} \), so only the methyl-oxygen bond breaks. This gives phenol and \( \mathrm{CH_3I} \).
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