Williamson ether synthesis — mechanism and scope

Question

Write the mechanism of Williamson ether synthesis for the preparation of ethyl methyl ether ( $\text{CH}_3\text{OC}_2\text{H}_5$ ). Why does this method fail when we try to prepare di-tert-butyl ether? What type of mechanism is involved?

(NCERT Class 12, commonly asked in CBSE and JEE Main)

Solution — Step by Step

First, we treat an alcohol with metallic sodium or $\text{NaH}$ to form the sodium alkoxide:

\text{CH}_3\text{OH} + \text{Na} \rightarrow \text{CH}_3\text{O}^-\text{Na}^+ + \frac{1}{2}\text{H}_2

Sodium methoxide ( $\text{CH}_3\text{O}^-\text{Na}^+$ ) is our nucleophile.

The alkoxide attacks the electrophilic carbon of a primary alkyl halide (bromoethane):

\text{CH}_3\text{O}^-\text{Na}^+ + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{OC}_2\text{H}_5 + \text{NaBr}

This is a classic $\text{S}_\text{N}2$ reaction — backside attack by the alkoxide on the carbon bearing the leaving group, with inversion of configuration.

To make di-tert-butyl ether, we’d need tert-butoxide attacking a tert-butyl halide. But:

Tert-butyl halides are tertiary — they can’t undergo $\text{S}_\text{N}2$ (too much steric hindrance around the carbon)
Tert-butoxide is a bulky, strong base
Instead of substitution, elimination (E2) dominates — we get an alkene, not an ether

That’s why Williamson synthesis works best when the alkyl halide is primary (or methyl).

Why This Works

The Williamson synthesis exploits the nucleophilicity of alkoxide ions. Alkoxides are strong nucleophiles because the negative charge on oxygen makes them eager to attack electrophilic carbons.

The reaction works through $\text{S}_\text{N}2$ , which requires a relatively unhindered carbon. Primary alkyl halides have open access to the back side of the C–halide bond. Secondary halides work sometimes but give mixed results. Tertiary halides? The three bulky groups block the nucleophile entirely, and elimination takes over.

This is why the “which alkoxide + which halide” choice matters so much. To make an unsymmetrical ether, always pair the more complex part as the alkoxide and the simpler part as the halide.

Alternative Method — Choosing the Right Pair

For an unsymmetrical ether like $\text{C}_6\text{H}_5\text{OCH}_3$ (anisole), you have two possible routes:

Route A: Phenoxide + $\text{CH}_3\text{Br}$ → Works perfectly ( $\text{S}_\text{N}2$ on methyl halide)

Route B: Methoxide + $\text{C}_6\text{H}_5\text{Br}$ → Fails (aryl halides don’t undergo $\text{S}_\text{N}2$ — the $\text{sp}^2$ carbon is not accessible)

For CBSE and JEE: always make the alkyl halide the simpler, primary component. The bulkier part should be the alkoxide. This gives maximum yield and avoids elimination side products.

Common Mistake

The most common error: students reverse the pair. They write tert-butoxide + tert-butyl bromide and expect an ether. In reality, you get isobutylene (elimination product). For any ether synthesis question, ask yourself — is the halide primary or methyl? If not, expect trouble.

Also watch out: some students confuse Williamson synthesis with acid-catalysed dehydration of alcohols. Dehydration works only for symmetrical ethers (e.g., diethyl ether from ethanol + $\text{H}_2\text{SO}_4$ ). For unsymmetrical ethers, Williamson is the only reliable method.

Question

Solution — Step by Step

Why This Works

Alternative Method — Choosing the Right Pair

Common Mistake

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