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.

Hint:

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.

Answer 1

Step 1: Resonance in anisole:
In anisole the oxygen lone pair overlaps with the benzene ring.
This overlap gives the $\mathrm{C_6H_5{-}O}$ bond a partial double bond nature.
So the bond between the ring carbon and oxygen is shorter and stronger than the $\mathrm{CH_3{-}O}$ bond.

Step 2: Cleavage with HI:
HI protonates oxygen. The weaker bond is then the one that breaks.
Iodide ion attacks the methyl group, which is small and unhindered ($\mathrm{S_N2}$).
\[ \mathrm{C_6H_5OCH_3 \xrightarrow{HI} [C_6H_5O^+HCH_3]\,I^- \longrightarrow C_6H_5OH + CH_3I} \]

Step 3: Why the other path fails:
For iodobenzene to form, iodide must attack the sp2 carbon of the ring.
Backside attack at a ring carbon is blocked by the ring itself, and the aryl cation is unstable.
Methanol and iodobenzene therefore do not form.

Final Answer:
The $\mathrm{C_6H_5{-}O}$ bond is stronger than the $\mathrm{CH_3{-}O}$ bond, so HI cleaves the methyl-oxygen bond. The products are phenol and methyl iodide.