Williamson Ether Synthesis: Mechanism and Examples

The Williamson ether synthesis is the most reliable laboratory method for making ethers: an alkoxide ion (RO−) attacks an alkyl halide (R′–X) and displaces the halide, giving the ether R–O–R′. It is named after Alexander Williamson, who developed it in 1850.

It matters because it is one of the few general ways to build an unsymmetrical ether with full control over which two groups end up joined — you simply choose the alkoxide for one side and the halide for the other. The catch is that the reaction follows an SN2 pathway, so the choice of which side becomes the halide decides success or failure.

Key Concepts

• Alkoxide: the conjugate base of an alcohol (RO−Na+), made by reacting the alcohol with sodium metal or NaH; a strong nucleophile and base.
• SN2 mechanism: the alkoxide attacks carbon from the back side as the halide leaves, in one concerted step — fastest on unhindered (methyl, primary) halides.
• E2 elimination (the competing reaction): with secondary and especially tertiary alkyl halides, the basic alkoxide instead removes a β-hydrogen, giving an alkene rather than an ether.
• Alkyl halide reactivity: R–I > R–Br > R–Cl for the SN2 step.

How It Works

Step 1 — make the alkoxide. 2 ROH + 2 Na → 2 RO−Na+ + H2. The alcohol's acidic proton is removed; everything must be dry because water would destroy the alkoxide.

Step 2 — SN2 attack. RO− + R′–X → R–O–R′ + X−. Back-side attack inverts the configuration at the halide carbon (relevant for chiral halides).

Choosing sides — the golden rule. Put the more hindered group on the alkoxide side and the less hindered group on the halide side. To make tert-butyl methyl ether, use (CH3)3CO−Na+ + CH3I (works), never (CH3)3C–Br + CH3O− (the tertiary bromide eliminates to isobutene). Silver oxide (Ag2O) with an alkyl halide is an alternative for sensitive substrates.

Worked Example

Ethyl methyl ether is made from sodium ethoxide and methyl bromide. What masses react to give 60 g of product (complete conversion)?

1. Equation: C2H5O−Na+ + CH3Br → C2H5OCH3 + NaBr (1:1:1:1).
2. Molar mass of ethyl methyl ether (C3H8O) = 36 + 8 + 16 = 60 g/mol, so 60 g = 1.00 mol.
3. Sodium ethoxide (C2H5ONa): 24 + 5 + 16 + 23 = 68 g/mol → 68 g needed.
4. Methyl bromide (CH3Br): 12 + 3 + 80 = 95 g/mol → 95 g needed.
5. Mass check: 68 + 95 = 163 = 60 + 103 (NaBr = 103 g/mol) — balanced.

Common Misconceptions

• "Either side can be the halide." Wrong. A tertiary alkyl halide with a basic alkoxide gives elimination (alkene), not the ether. Always make the bulkier fragment the alkoxide.
• "Williamson synthesis follows SN1." Wrong. It is a classic SN2 reaction — one step, back-side attack, inversion of configuration, fastest with methyl and primary halides.
• "Phenols cannot be used." Wrong. Phenoxide ions work fine — the reaction of phenol with chloroacetic acid under this chemistry gives phenoxyacetic acid derivatives (herbicide precursors).

Real-World Applications

• MTBE: methyl tert-butyl ether, a petrol additive that raises octane number, is made by this logic (methanol + isobutene industrially).
• Anaesthetics and solvents: diethyl ether was the first general anaesthetic; ethers remain key laboratory solvents.
• Herbicides: 2,4-D (2,4-dichlorophenoxyacetic acid) is built by Williamson-type coupling of a phenoxide with chloroacetate.

To recap: deprotonate the alcohol to an alkoxide, then let it displace halide from an alkyl halide by SN2. The reaction is general and predictable — provided the halide is methyl or primary and the bulkier group is carried by the alkoxide, so elimination never gets a chance.