Aryl ether (Williamson with phenoxide): practice and mechanism
Phenols are more acidic than alcohols, so NaOH turns them into the phenoxide. It attacks CH₃I at the oxygen (SN2): an aryl methyl ether forms (e.g. anisole).
Example
Phenols are much more acidic than alcohols (pKa ≈ 10), so NaOH already deprotonates them to the phenoxide. The phenoxide attacks methyl iodide via SN2 at the oxygen, giving the aryl methyl ether.
Mechanism step by step
- 1
NaOH fully deprotonates the phenol (pKa ≈ 10) to the phenoxide ion.
- 2
The phenoxide attacks CH₃I from the back with its oxygen (SN2).
- 3
Iodide leaves: the aryl methyl ether forms (e.g. anisole from phenol).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: Esterification with acetic anhydride · Friedel–Crafts alkylation
Frequently asked questions
Why does NaOH suffice here but not for alcohols?
Thanks to resonance in the phenoxide, phenol is about a million times more acidic than ethanol. Alcohols (pKa ≈ 16) need NaH or Na.
Does it work the other way round with bromobenzene and methoxide?
No. SN2 is impossible at an aromatic carbon. The arene must be the nucleophile (phenoxide) and the alkyl halide the electrophile.