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Ether synthesis — SN2

Williamson ether synthesis

How the Williamson ether synthesis works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.

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In short. A 3-step mechanism. Typical reagents: R–OH + NaH (or Na, KOH); R′–X (1° or methyl). Conditions: THF or DMF, 25–80 °C. Bonds formed: C–O ether bond; bonds broken: O–H; C–X. Stereochemistry: Inversion at the alkyl halide carbon. Rate and selectivity: SN2; 2° / 3° halides give E2 alkene instead.

At a glance

ReagentsR–OH + NaH (or Na, KOH); R′–X (1° or methyl)
Solvent / conditionsTHF or DMF, 25–80 °C
Bonds formedC–O ether bond
Bonds brokenO–H; C–X
StereochemistryInversion at the alkyl halide carbon
Rate / selectivitySN2; 2° / 3° halides give E2 alkene instead

Step 1: Base forms the alkoxide

+−H₂OHHO1 · BASE FORMS THE ALKOXIDEethanolC₂H₆OSUBSTRATEhydride (from NaH)H⁻BASEethoxideC₂H₅O⁻NUCLEOPHILERO⁻ is a far stronger nucleophile than ROH.

Base deprotonates an alcohol to form an alkoxide — a much stronger, cleaner nucleophile than the neutral alcohol.

Step 2: Backside SN2 on a primary halide

+OI2 · BACKSIDE SN2 ON A PRIMARY HALIDEethoxideC₂H₅O⁻NUCLEOPHILEiodomethaneCH₃IELECTROPHILEbackside attackNucleophile approaches 180° from the leaving group.

The alkoxide performs a standard backside SN2 attack on a primary alkyl halide, exactly like the mechanism you learned first — no new chemistry, just a new nucleophile.

Step 3: Unsymmetrical ether

+OI3 · UNSYMMETRICAL ETHERethyl methyl etherC₃H₈OPRODUCTiodideI⁻LEAVING GROUPUse the less hindered halide partner; a 2°/3° halide would eliminate (E2).

The product is an unsymmetrical ether. Because it's a clean SN2, this only works well with unhindered (primary, methyl) alkyl halides — a secondary or tertiary halide would give elimination instead.

Abbreviations and terms

R
R group. Any carbon substituent (alkyl, aryl…) whose identity does not matter for the mechanism.
X
Halogen / leaving group. Cl, Br or I (sometimes any leaving group).
THF
Tetrahydrofuran. Cyclic ether solvent; its oxygen lone pairs solvate Li⁺/Mg²⁺ so organometallics stay dissolved and reactive.
DMF
N,N-Dimethylformamide. Polar aprotic solvent – dissolves ionic nucleophiles but does not H-bond to them, speeding SN2 and SNAr.
NaH
Sodium hydride. Insoluble, irreversible base: removes H⁺ and bubbles off H₂.
KOH
Potassium hydroxide. Strong aqueous/alcoholic base.
H2
Hydrogen gas. Reductant for catalytic hydrogenation.
SN2
Bimolecular nucleophilic substitution. One concerted step with backside attack and inversion; rate = k[RX][Nu].
E2
Bimolecular elimination. Concerted base removal of β-H and loss of the leaving group (anti-periplanar).

How it compares with related reactions

Williamson ether synthesisAza-Wittig reactionBaeyer-Drewson indigo synthesis
ReagentsR–OH + NaH (or Na, KOH); R′–X (1° or methyl)R–N=PPh₃ (from R–N₃ + PPh₃, Staudinger), aldehyde or ketoneo-Nitrobenzaldehyde, acetone, aq. NaOH
ConditionsTHF or DMF, 25–80 °CTHF or toluene, 25–110 °C, anhydrousH₂O/acetone, 20–25 °C, then air
StereochemistryInversion at the alkyl halide carbonImine E/Z set at ring collapse; no stereocentre neededProduct is the E-indigo, planar and strongly H-bonded
Rate / selectivitySN2; 2° / 3° halides give E2 alkene insteadCycloreversion is fast; driven by strong P=O bond (≈ 130 kcal/mol)Aldol step fast; indigo precipitates as deep blue solid

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Williamson ether synthesis?

R–OH + NaH (or Na, KOH); R′–X (1° or methyl). Conditions: THF or DMF, 25–80 °C.

Which bonds are formed and which are broken?

Formed: C–O ether bond. Broken: O–H; C–X.

What is the stereochemical outcome?

Inversion at the alkyl halide carbon

What controls the rate or the selectivity?

SN2; 2° / 3° halides give E2 alkene instead

Study guides that help with this reaction

Related mechanisms

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