Williamson ether synthesis

Every ether on your syllabus that isn't a symmetrical one made by simple dehydration was almost certainly built by Williamson ether synthesis — the reaction of an alkoxide (or phenoxide) with an alkyl halide. It looks like a small reaction, one oxygen lone pair displacing one halide, but it is the reaction ISC examiners lean on to test whether you truly understand SN2S_N2: which partner can be bulky, which partner must stay small, and why choosing the wrong combination quietly swaps your ether for an alkene. Master the selection rule here and you can build any unsymmetrical ether, aliphatic or aryl, on demand.

The core insight

An alkoxide or phenoxide ion performs a bimolecular nucleophilic (SN2S_N2) backside attack on a primary (or methyl) alkyl halide, displacing the halide and forming the ether — so the halide must always stay small while the bulky fragment is built into the alkoxide.

The core idea

Williamson synthesis has exactly one moving part: a good oxygen nucleophile attacking a good SN2S_N2 electrophile.

The nucleophile. Sodium (or potassium) alkoxide or phenoxide, generated in situ from an alcohol or phenol plus sodium metal or a strong base. Because oxygen carries a full negative charge, it is a far stronger nucleophile than the neutral alcohol it came from.

The electrophile. An alkyl halide, R′–XR'\text{–X}. The carbon bearing the halogen is what the oxygen lone pair attacks, and the whole reaction lives or dies on how accessible that carbon is.

R–O−Na++R′–X⟶R–O–R′+NaXR\text{–O}^-Na^+ + R'\text{–X} \longrightarrow R\text{–O–}R' + NaX

where: R–O−Na+R\text{–O}^-Na^+ is the sodium alkoxide (nucleophile), R′–XR'\text{–X} is the alkyl halide (electrophile, X = Cl, Br or I), and the product R–O–R′R\text{–O–}R' is the ether. The aryl-ether version is identical in spirit, with a phenoxide standing in for the alkoxide:

Ar–O−Na++R–X⟶Ar–O–R+NaXAr\text{–O}^-Na^+ + R\text{–X} \longrightarrow Ar\text{–O–}R + NaX

Why this is one-directional. Because the mechanism is a single-step, concerted SN2S_N2 — the nucleophile attacks from directly behind the leaving halide, in the same motion that the C–X bond breaks — the alkyl halide component must be unhindered: primary or, best of all, methyl. A secondary or tertiary halide blocks that backside approach with its own alkyl groups, and the alkoxide, being a strong base as well as a nucleophile, instead rips off a β\beta-hydrogen in a competing E2E2 elimination, turning your ether synthesis into an alkene synthesis. The alkoxide/phenoxide side has no such restriction — it never undergoes backside attack itself, so it can be built from any alcohol or phenol, however bulky.

Rate=k [RO−][R’X]\text{Rate} = k\,[\text{RO}^-][\text{R'X}]

where: kk is the second-order rate constant, [RO−][\text{RO}^-] is the alkoxide concentration and [R’X][\text{R'X}] is the alkyl halide concentration — both appear to the first power because both molecules are present, colliding, in the single transition state of the one-step mechanism.

Figure
Backside attack of an alkoxide oxygen on an alkyl halide carbon giving inversion and ether product with sodium halide
Williamson synthesis — backside SN2 attack with inversion R–O⁻ nucleophile C X⁻ leaving group transition state — inversion at carbon R–O–R' ether + NaX
Derivation AThe second-order rate law confirms bimolecular attack

The goal is to show, from the rate law of Williamson synthesis, why doubling either reactant's concentration doubles the initial rate — the kinetic signature that distinguishes a genuine one-step SN2S_N2 displacement (both partners colliding in the same transition state) from an SN1S_N1-type process that would depend only on the halide.

Step 1. Write the rate law for the elementary bimolecular step. Since alkoxide and alkyl halide react in a single concerted step with no intermediate, the rate law mirrors the stoichiometry directly: Rate=k[RO−][R’X]\text{Rate} = k[\text{RO}^-][\text{R'X}], first order in each reactant, second order overall.

Step 2. Take a reference case. Suppose k=5.0×10−4 dm3 mol−1 s−1k = 5.0\times10^{-4}\ \text{dm}^3\,\text{mol}^{-1}\,\text{s}^{-1}, [RO−]=0.20 mol dm−3[\text{RO}^-] = 0.20\ \text{mol dm}^{-3} and [R’X]=0.10 mol dm−3[\text{R'X}] = 0.10\ \text{mol dm}^{-3}. The initial rate is Rate1=(5.0×10−4)(0.20)(0.10)=1.0×10−5 mol dm−3 s−1\text{Rate}_1 = (5.0\times10^{-4})(0.20)(0.10) = 1.0\times10^{-5}\ \text{mol dm}^{-3}\,\text{s}^{-1}.

Step 3. Double the alkyl halide concentration only, to 0.20 mol dm−30.20\ \text{mol dm}^{-3}, holding alkoxide fixed. Rate2=(5.0×10−4)(0.20)(0.20)=2.0×10−5 mol dm−3 s−1\text{Rate}_2 = (5.0\times10^{-4})(0.20)(0.20) = 2.0\times10^{-5}\ \text{mol dm}^{-3}\,\text{s}^{-1} — exactly double Rate1\text{Rate}_1. Doubling the alkoxide concentration instead, with halide back at 0.100.10, gives the identical doubled value, 2.0×10−52.0\times10^{-5}.

Step 4. Interpret. Both reactants independently double the rate when doubled, confirming that both are present in the rate-determining transition state — the hallmark of a single concerted SN2S_N2 step. This is also why bulky alkyl halides slow the reaction so sharply: steric hindrance lowers the effective kk for backside approach, and once kk falls low enough, the competing E2E2 elimination pathway (which does not need clean backside access) takes over as the major product-forming route.

Worked example

Problem. You react 9.6 g9.6\ \text{g} of sodium tert-butoxide, (CH3)3CONa(CH_3)_3CONa (molar mass 96 g mol−196\ \text{g mol}^{-1}), with excess methyl iodide by Williamson's synthesis to prepare 2-methoxy-2-methylpropane, (CH3)3C–O–CH3(CH_3)_3C\text{–O–}CH_3 (molar mass 88 g mol−188\ \text{g mol}^{-1}). Calculate the theoretical mass of ether obtainable, and the percentage yield if the actual mass recovered is 7.48 g7.48\ \text{g}.

Step 1. Confirm the route is correct before computing anything. The alkyl halide here is methyl iodide — methyl, the best possible SN2S_N2 substrate — while the bulky tertiary group sits in the alkoxide, which never undergoes backside attack. This is exactly the safe combination identified earlier: bulky group as nucleophile, small group as halide.

Step 2. Find moles of the limiting reagent, sodium tert-butoxide (methyl iodide is in excess, so it cannot limit the yield). n=9.6 g96 g mol−1=0.100 moln = \dfrac{9.6\ \text{g}}{96\ \text{g mol}^{-1}} = 0.100\ \text{mol}.

Step 3. Apply 1:1 stoichiometry from (CH3)3CONa+CH3I→(CH3)3C–O–CH3+NaI(CH_3)_3CONa + CH_3I \rightarrow (CH_3)_3C\text{–O–}CH_3 + NaI: 0.100 mol0.100\ \text{mol} of alkoxide gives 0.100 mol0.100\ \text{mol} of ether. Theoretical mass =0.100×88=8.8 g= 0.100 \times 88 = 8.8\ \text{g}.

Step 4. Percentage yield =actualtheoretical×100=7.488.8×100=85%= \dfrac{\text{actual}}{\text{theoretical}} \times 100 = \dfrac{7.48}{8.8}\times 100 = 85\%. The theoretical yield is 8.8 g of 2-methoxy-2-methylpropane, and the actual percentage yield is 85%.

Competitive stretch

If you had instead (wrongly) reacted sodium methoxide with tert-butyl iodide to try the same product, the tertiary halide would undergo E2E2 elimination with the strongly basic methoxide, and the major isolated product would be isobutylene (2-methylpropene), not the ether — so no yield calculation on the ether would even apply. Always check which fragment is the halide before doing stoichiometry.

Common traps

  • Assigning the bulkier fragment to the alkyl halide instead of the alkoxide — always keep the halide primary/methyl and push steric bulk into the nucleophile side.
  • Forgetting that percentage-yield questions still require you to first verify the reaction actually proceeds by SN2S_N2; a tertiary halide gives elimination, not a lower-yield ether.
  • Writing the mechanism as SN1S_N1 (via a carbocation) — Williamson synthesis is valued precisely because it is concerted, stereospecific SN2S_N2, with no carbocation intermediate and no rearrangement risk.
  • Trying to make phenol act as the alkyl-halide partner — phenol must first be converted to sodium phenoxide (the nucleophile); aryl halides never undergo this kind of SN2S_N2, so an aryl ether always needs the alkyl halide on the other side.

Ask the tutor: Not sure which alcohol should become the alkoxide and which the halide for your specific unsymmetrical ether? Ask the AI tutor to walk through the choice with your target molecule.

Why can't I just use aryl halide plus alcohol to make an aryl ether?

Aryl halides do not undergo SN2S_N2 substitution: the halogen sits on an sp2sp^2 ring carbon, backside attack is geometrically blocked by the ring, and the C–X bond has partial double-bond character from conjugation with the ring, making it far too strong to break this way. That is why aryl ethers are always made with the aryl group on the nucleophile side, as sodium phenoxide, reacting with an ordinary alkyl halide as the electrophile.

If I need an unsymmetrical ether, how do I decide which alcohol becomes the alkoxide and which becomes the halide?

Look at both possible alcohol fragments and identify which one is more hindered (secondary, tertiary, or otherwise bulky at the carbon that would bear the halogen). Convert that bulky fragment into the alkoxide, since the alkoxide never has to squeeze through a backside attack. Convert the less hindered, ideally primary or methyl, fragment into the alkyl halide, since that is the carbon under direct nucleophilic attack. Getting this backwards invites E2 elimination instead of substitution.

Does the percentage-yield idea in the worked example apply to any Williamson synthesis, or just this one?

It applies generally, but only once you have confirmed the reaction is a genuine SN2S_N2 substitution. Percentage yield compares actual product mass to the theoretical mass from stoichiometry, and that calculation is only meaningful if the ether is actually the major product. If you accidentally chose a tertiary or bulky secondary halide, elimination competes and dominates, so a naive stoichiometric yield calculation on the ether would overstate what you can really isolate.

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