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Williamson synthesis looks like the easiest reaction in the ethers section: alkoxide plus alkyl halide gives ether. The marks are in one decision. An unsymmetrical ether R–O–R′ can be made from R–X and R′O⁻, or from R′–X and RO⁻, and usually only one of those works. This post gives a three-question method for choosing, works through four ethers, and finishes with a yield calculation.
Is Williamson synthesis still in the syllabus?
It depends on the exam. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both open their oxygen-compounds unit with "general methods of preparation, properties, reactions and uses", but under ethers they list only "Ethers: Structure". The JEE (Advanced) 2026 brochure is explicit: "Preparation by Williamson's synthesis; C-O bond cleavage reactions." The NCERT Class 12 chapter Alcohols, Phenols and Ethers (2026-27 reprint) still teaches the method in Section 7.6.1, with Example 7.6 and Intext Question 7.10, and the CBSE 2026-27 curriculum lists ethers in Unit 7. So it is fully examinable for CBSE boards and JEE Advanced; for JEE Main and NEET, treat it as preparation knowledge that may appear inside a reaction question. Check the current year's documents.
What happens in the reaction?
NCERT describes it as an alkyl halide reacting with a sodium alkoxide:
R–X + R′–O⁻Na⁺ → R–O–R′ + NaX
The alkoxide attacks the carbon bearing the halogen from the back, pushing out the halide in one step: an SN2 reaction. The method is named after Alexander Williamson, whose paper Theory of ætherification appeared in the Philosophical Magazine in 1850.
The catch, as NCERT explains, is that alkoxides are strong bases as well as nucleophiles. With secondary halides, elimination competes with substitution. With a tertiary halide, an alkene is the only product. NCERT's example: CH₃ONa with (CH₃)₃C–Br gives exclusively 2-methylpropene.
How do you choose the halide and the alkoxide?
Ask three questions in order.
Is either group aryl (a benzene ring on the oxygen)? Then that side must be the phenoxide. NCERT's haloalkanes and haloarenes chapter notes that aryl halides are extremely less reactive towards nucleophilic substitution, so a halogen on a ring will not be displaced.
Which group is more crowded at the carbon next to oxygen? Put that group on the alkoxide.
Is the halide now methyl or primary? If yes, go ahead. If it is secondary, expect a mixture; if tertiary, rethink.
Four worked choices
Example 1: 2-ethoxypropane, (CH₃)₂CH–O–CH₂CH₃.
Route A: sodium propan-2-olate, (CH₃)₂CHO⁻Na⁺, with bromoethane. The halide is primary.
Route B: sodium ethoxide with 2-bromopropane. The halide is secondary, so elimination to propene competes.
Choose Route A.
Example 2: 2-ethoxy-3-methylpentane (NCERT Intext Question 7.10, starting from ethanol and 3-methylpentan-2-ol). The ethyl group is primary; the other group joins oxygen through a secondary carbon. So convert ethanol to bromoethane, convert 3-methylpentan-2-ol to its sodium alkoxide with sodium, and combine them.
Example 3: ethoxybenzene, C₆H₅–O–CH₂CH₃. Question 1 settles it. Use sodium phenoxide with bromoethane. Bromobenzene with sodium ethoxide gives no ether.
Example 4: benzyl methyl ether, C₆H₅CH₂–O–CH₃. Here both routes can work: benzyl bromide with sodium methoxide, or sodium benzyloxide with iodomethane. Neither halide can undergo the usual elimination: a methyl halide has no neighbouring carbon, and in benzyl bromide the neighbouring carbon is part of the benzene ring. Questions like this test whether you understand the rule rather than memorising "always the smaller halide".
Worked example: a yield calculation
These masses are sample numbers. A student converts 4.70 g of phenol to sodium phenoxide, reacts it with excess bromoethane and isolates 4.90 g of ethoxybenzene.
Moles of phenol: 4.70 ÷ 94.11 g/mol ≈ 0.0499 mol.
One mole of phenol gives one mole of ether, so the theoretical yield is 0.0499 × 122.17 g/mol ≈ 6.10 g.
Percentage yield: 4.90 ÷ 6.10 × 100 ≈ 80%.
Which mistakes should you avoid?
Putting the tertiary group on the halide. You get an alkene. NCERT's Example 7.6 on tert-butyl ethyl ether tests exactly this.
Using an aryl halide. The ring must come from phenoxide.
Using the alcohol instead of the alkoxide. The neutral alcohol is a poor nucleophile; make the alkoxide first.
Assuming "smaller halide" always. The real rule is "uncrowded carbon attacked".
To see the same ethers broken apart again, read our guide to ether cleavage with HI. For why primary and tertiary carbons behave so differently, see SN1 and SN2 in haloalkanes.
A practice routine
Write both routes for five unsymmetrical ethers and choose one, using the three questions.
Answer NCERT Intext Question 7.10 and Exercise 7.25 on the limitations of the method.
For one ether, write the competing elimination product of the wrong route.
Do two yield calculations with your own sample masses.
If you keep picking the wrong route, compare your reasoning with a worked solution: the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Try it now: which reactants would you use to make 2-methoxy-2-methylbutane?
