The Williamson ether synthesis makes an ether (R–O–R′) in two steps: deprotonate an alcohol with a strong base such as NaH or sodium metal to form an alkoxide (RO⁻), then let that alkoxide attack an alkyl halide in an SN2 reaction.
Because the key step is SN2, the alkyl halide must be methyl or primary. When two disconnections are possible, always alkylate the less hindered carbon; secondary halides give poor yields and tertiary halides just eliminate.
The whole reaction in one line: ethoxide plus iodomethane gives ethyl methyl ether by SN2.
Ethers look simple, but you rarely find a good one-step route to them. The Williamson ether synthesis is the workhorse method precisely because it is so reliable: it stitches two carbon fragments together through an oxygen using a reaction you already know cold — the SN2. Master four ideas (make the alkoxide, run the SN2, choose the right partner, and watch for elimination) and you can build almost any ether, symmetric or mixed, open-chain or cyclic.
1. The alkoxide nucleophile forms first
An alcohol on its own is a weak nucleophile and its O–H proton is in the way. So step one is deprotonation. Treating the alcohol with a strong, non-nucleophilic base — sodium hydride (NaH) or sodium metal — removes the hydroxyl proton and leaves behind an alkoxide, RO⁻. The alkoxide is a far stronger nucleophile than the neutral alcohol, and with NaH the only by-product is harmless H2 gas, which bubbles off and drives the equilibrium to completion.
Ethanol is deprotonated by NaH to give ethoxide, releasing H2.
The same trick works on phenols. A phenol (pKa ≈ 10) is acidic enough that even a mild base gives the phenoxide ion, which then behaves as the nucleophile in the alkylation step. That is how you reach aryl alkyl ethers such as anisole.
2. The alkoxide does an SN2 on the alkyl halide
With the nucleophile prepared, the alkoxide oxygen attacks the electrophilic carbon of an alkyl halide from the back side, displacing the halide leaving group in a single concerted step. The new C–O bond is the ether linkage. Everything you learned about SN2 applies directly: rate depends on both the alkoxide and the halide, a good leaving group (I > Br > Cl, or a tosylate) helps, and a polar aprotic solvent speeds things up.
Methoxide attacks the primary carbon of bromoethane to give ethyl methyl ether.
Notice you can build the same ether, ethyl methyl ether, from either "half" as the alkoxide and the other half as the halide. That freedom to choose is exactly what the next two sections exploit.
3. The alkyl halide must be methyl or primary
Because the bond-forming event is an SN2, steric bulk at the electrophilic carbon is fatal. A methyl or primary halide is ideal — the back side is open and yields are high. A secondary halide reacts sluggishly and loses yield to competing elimination. A tertiary halide does not undergo SN2 at all. The alkoxide is both a strong nucleophile and a strong base, so the more crowded the carbon, the more the reaction tips away from substitution and toward E2.
The rule of thumb: keep the SN2 on the least-hindered carbon and let oxygen carry the branching. A bulky group can live on the alkoxide side all it likes — the alkoxide is the nucleophile, not the electrophile, so its bulk never blocks the reacting carbon.
4. Choose the disconnection that alkylates the less-hindered carbon
Most target ethers can be taken apart in two ways, but usually only one of them works. Consider anisole (methyl phenyl ether). The winning disconnection makes the aromatic oxygen the nucleophile — phenoxide — and puts the SN2 on methyl iodide.
Right disconnection: phenoxide + iodomethane cleanly gives anisole.
The mirror-image plan — methoxide plus a halobenzene — fails completely, because an aryl halide has no SN2 pathway (its carbon is sp² and locked in the ring). Whenever you plan a Williamson synthesis, list both disconnections and pick the one that alkylates the more accessible, less-substituted carbon.
5. Tertiary halides eliminate instead of substituting
The classic trap is a target like tert-butyl methyl ether. The tempting-but-wrong route pairs methoxide with a tert-butyl halide. Instead of forming the ether, the alkoxide acts as a base, plucks off a β-hydrogen, and the tertiary halide undergoes E2 elimination to give an alkene.
Wrong disconnection: methoxide + tert-butyl bromide just eliminates to isobutylene — no ether.
The fix is to flip the roles. Deprotonate tert-butanol to the bulky tert-butoxide and let that be the nucleophile against methyl iodide. Now the SN2 happens on an unhindered methyl carbon, the branching sits safely on oxygen, and the ether forms in good yield. Same two fragments, opposite assignment of nucleophile and electrophile — that single choice decides success or failure.
6. The intramolecular version makes epoxides and cyclic ethers
When the alkoxide and the leaving group live on the same molecule, the SN2 closes a ring. A halohydrin (a molecule with an –OH and a C–X a few atoms apart) is deprotonated to an alkoxide, which then attacks its own carbon bearing the halide. This is the standard route to epoxides: treat a 1,2-halohydrin such as 2-chloroethanol with base and the intramolecular Williamson closure gives a three-membered ring.
Intramolecular Williamson: 2-chloroethanol closes onto its own C–Cl to give an epoxide.
The same logic builds larger cyclic ethers (oxetanes, tetrahydrofurans, tetrahydropyrans) whenever the geometry lets the oxygen reach the back side of the carbon–halide bond. Three- and five-membered rings form fastest because the alkoxide can line up for a clean backside attack.
7. Summary
The Williamson ether synthesis is deprotonate-then-alkylate. Make an alkoxide (RO⁻) from an alcohol or phenol with NaH or Na, then run an SN2 on a methyl or primary alkyl halide to forge the C–O bond. Because the mechanism is SN2, always alkylate the less hindered carbon and keep branching on the alkoxide side; secondary halides underperform and tertiary halides eliminate to alkenes. Run the alkoxide and leaving group on one molecule and the reaction closes a ring — the go-to synthesis of epoxides and other cyclic ethers.
Quiz yourself
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NaH is a strong, non-nucleophilic base that deprotonates the alcohol essentially irreversibly, releasing only H2 gas. NaOH would set up an equilibrium (its conjugate acid, water, has a similar pKa to the alcohol) and the hydroxide could compete as a nucleophile. NaH drives the reaction fully to the alkoxide.
Use tert-butoxide as the nucleophile and methyl iodide (CH3I) as the electrophile. This puts the SN2 on the unhindered methyl carbon. The reverse — methoxide plus a tert-butyl halide — fails because the tertiary halide undergoes E2 elimination to isobutylene instead of substitution.
Chlorobenzene is an aryl halide: its carbon is sp² and part of the aromatic ring, so it cannot undergo SN2. The correct route reverses the roles — phenoxide is the nucleophile and iodomethane is the electrophile, so the SN2 occurs on the methyl carbon.
Start from a 1,2-halohydrin (e.g., 2-chloroethanol). Base deprotonates the –OH to an alkoxide, which then performs an intramolecular SN2 on the adjacent carbon bearing the halide. The oxygen closes onto that carbon, expelling the halide and forming the three-membered epoxide ring.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.