ChemistryISC Class 12 · Alcohols, Phenols and Ethers

Ethers: Williamson Synthesis and Cleavage (ISC Class 12)

Updated 15 October 2026 · Fact-checked

Ethers are best made by Williamson's synthesis: a sodium alkoxide or phenoxide reacts with a primary alkyl halide by SN2, R–ONa + R′–X → R–O–R′ + NaX. Ethers are fairly unreactive, but hot HI cleaves the C–O bond; with a methyl or primary group the iodide forms at the smaller group.

Keep the bulky tertiary group on the alkoxide and let the halide be primary, or elimination to an alkene beats SN2.

Illustration drawn for QuestionBuddy.

Understand Ethers: Williamson Synthesis and Cleavage

Structure and properties. In an ether, R–O–R′, oxygen is sp³ hybridised with two lone pairs; the C–O–C angle (about 111.7° in methoxymethane) is a little more than tetrahedral because of the bulky alkyl groups. Ethers have no O–H bond, so they cannot hydrogen-bond with one another: their boiling points are close to those of alkanes of similar mass and far below those of isomeric alcohols. They can accept hydrogen bonds from water, so lower ethers are about as soluble in water as alcohols of the same mass. Diethyl ether is a volatile, highly flammable liquid used as a solvent, for example for Grignard reagents; it was once used as an inhalation anaesthetic.

Williamson's synthesis. A sodium alkoxide (made from the alcohol and sodium) reacts with an alkyl halide by an SN2 mechanism: $C_{2} H_{5} O N a + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5} + N a B r$ . Both symmetrical and unsymmetrical ethers can be made. Choice of partners matters: the alkyl halide must be primary (or methyl). With a tertiary halide the strongly basic alkoxide causes elimination instead, giving an alkene. So to make tert-butyl ethyl ether, use sodium tert-butoxide with bromoethane, $(C H_{3})_{3} C O N a + C_{2} H_{5} B r \to (C H_{3})_{3} C O C_{2} H_{5} + N a B r$ , not 2-bromo-2-methylpropane with sodium ethoxide. Aryl alkyl ethers are made from a phenoxide and an alkyl halide, $C_{6} H_{5} O N a + C H_{3} I \to C_{6} H_{5} O C H_{3} + N a I$ (anisole); an aryl halide and an alkoxide do not work, because aryl halides resist nucleophilic substitution. Simple symmetrical ethers can also be made by dehydrating alcohols with conc. H₂SO₄ at 413 K.

Reactions of aliphatic ethers.
• No reaction with sodium: there is no H on oxygen, so no hydrogen is evolved (sodium is used to dry ether).
• With HI (cleavage): the ether oxygen is protonated, and I⁻ attacks a carbon. With one mole of HI in the cold, an alcohol and an alkyl iodide form: $C_{2} H_{5} O C_{2} H_{5} + H I \to C_{2} H_{5} I + C_{2} H_{5} O H$ . With excess hot HI, both groups become iodides: $C_{2} H_{5} O C_{2} H_{5} + 2 H I \to 2 C_{2} H_{5} I + H_{2} O$ . For mixed ethers with primary or methyl groups, I⁻ attacks the less hindered carbon by SN2, so the smaller group becomes the iodide: methoxypropane gives iodomethane and propan-1-ol. If one group is tertiary, the C–O bond to it breaks by SN1 (stable carbocation), so the tertiary group becomes the iodide: $(C H_{3})_{3} C O C H_{3} + H I \to (C H_{3})_{3} C I + C H_{3} O H$ . HI is more effective than HBr, and HCl hardly reacts, because I⁻ is the best nucleophile.
• With PCl₅ on warming: $C_{2} H_{5} O C_{2} H_{5} + P C l_{5} \to 2 C_{2} H_{5} C l + P O C l_{3}$ .
• With chlorine: in the dark, substitution at the carbon next to oxygen (α-carbon) gives α-chloroethers, $C_{2} H_{5} O C_{2} H_{5} + C l_{2} \to C H_{3} C H C l O C_{2} H_{5} + H C l$ ; with excess chlorine in light, all H atoms are replaced, giving perchlorodiethyl ether, $C_{2} H_{5} O C_{2} H_{5} + 10 C l_{2} \to C_{2} C l_{5} O C_{2} C l_{5} + 10 H C l$ .
• Oxidation (peroxide formation): on long exposure to air and light, ethers slowly form peroxides, which are explosive when concentrated; ether should never be distilled to dryness, and is stored in dark bottles.

Anisole (methoxybenzene), an aryl ether. A colourless liquid insoluble in water. The –OCH₃ group donates electrons into the ring by resonance, so anisole is activated and ortho-para directing in electrophilic substitution (less strongly than phenol):
• Halogenation: Br₂ in ethanoic acid (no Lewis acid needed) gives mainly 4-bromoanisole.
• Nitration: conc. HNO₃ and H₂SO₄ give 2- and 4-nitroanisole (para major): $C_{6} H_{5} O C H_{3} + H N O_{3} \to C_{6} H_{4} (N O_{2}) O C H_{3} + H_{2} O$ .
• Friedel–Crafts reactions with CH₃Cl or CH₃COCl and anhydrous AlCl₃ give 2- and 4-methylanisole or 2- and 4-methoxyacetophenone.
• With HI, anisole gives phenol and iodomethane, never iodobenzene, because the O–aryl bond has partial double-bond character and an sp² carbon resists attack.

Ethers: Williamson Synthesis and Cleavage: formulas and key facts

Williamson's synthesis: $R - O N a + R^{'} - X \to R - O - R^{'} + N a X$; SN2; R′X must be methyl or primary.
Cleavage by HI: $C_{2} H_{5} O C_{2} H_{5} + H I \to C_{2} H_{5} I + C_{2} H_{5} O H; C_{2} H_{5} O C_{2} H_{5} + 2 H I heat 2 C_{2} H_{5} I + H_{2} O$; Mixed ether: smaller group → iodide (SN2); tertiary group → iodide (SN1).
With PCl₅: $C_{2} H_{5} O C_{2} H_{5} + P C l_{5} \to 2 C_{2} H_{5} C l + P O C l_{3}$; On warming.
Preparation of anisole: $C_{6} H_{5} O N a + C H_{3} I \to C_{6} H_{5} O C H_{3} + N a I$; Phenoxide + alkyl halide (never aryl halide + alkoxide).
Anisole with HI: $C_{6} H_{5} O C H_{3} + H I \to C_{6} H_{5} O H + C H_{3} I$; The O–aryl bond is not broken.

How to solve Ethers: Williamson Synthesis and Cleavage questions

ISC asks ether conversions, the products of HI cleavage, reasons (no reaction with Na, choice of halide) and anisole substitutions. Treat the ether oxygen as a weak base and the alkyl groups as SN1 or SN2 targets.

1For a synthesis, split the target ether at either C–O bond; pick the split in which the halide part is methyl or primary and the alkoxide (or phenoxide) carries the bulky or aryl group.
2Write the alkoxide formation (ROH + Na) if the question starts from an alcohol, then the Williamson step.
3For HI cleavage, protonate the oxygen first; then decide: methyl/primary group → I⁻ attacks it (SN2); tertiary group → it leaves as a carbocation (SN1) and becomes the iodide.
4With excess hot HI, convert the alcohol formed into a second iodide.
5For anisole, remember: ring reactions at ortho and para (para major); HI breaks only the O–CH₃ bond.
6Balance all equations and name products.

Ethers: Williamson Synthesis and Cleavage in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Williamson synthesis in practice

Suppose a teaching laboratory prepares diethyl ether for a Grignard experiment by adding 10.9 g of bromoethane (M = 109) to an excess of sodium ethoxide in ethanol. After washing and distilling, 5.55 g of dry ether (M = 74) is collected.

Write the equation, calculate the percentage yield, and explain why ethanol alone, without sodium, would not work.

Try it first, then show the solutionSolution

$C_{2} H_{5} O N a + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5} + N a B r$ .
Moles of bromoethane $= \frac{10.9}{109} = 0.100$ mol, so the theoretical mass of ether $= 0.100 \times 74 = 7.40$ g.
Percentage yield $= \frac{5.55}{7.40} \times 100 = 75%$ .
Ethanol is a weak nucleophile; the ethoxide ion, formed by reacting ethanol with sodium, is a much stronger one and attacks the C–Br carbon readily (SN2).

Answer: Yield about 75%. An alkoxide ion, not the neutral alcohol, is needed as the nucleophile.

Memory hook: Williamson needs the alkoxide: alcohol plus sodium first, then the primary halide.

Tricks to remember Ethers: Williamson Synthesis and Cleavage

Short memory aids. Say each one aloud once, then use it on a question the same day.

Bulky on the oxygen, small on the halide
In Williamson's synthesis put the big or aryl group on the O⁻ and the small (methyl or primary) group on the halide.
Use it when: Choosing reagents for unsymmetrical ethers.
HI picks on the little one, unless there's a tertiary
HI turns the smaller group into the iodide (SN2), except that a tertiary group always leaves as the iodide (SN1). The aryl group never does.
Use it when: Predicting cleavage products.
No O–H, no H₂
Ether has no hydrogen on oxygen, so sodium has nothing to replace: no hydrogen gas.
Use it when: Distinguishing ethers from alcohols.

Ethers: Williamson Synthesis and Cleavage: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Ethers on standing · 1 mark · MCQBoard level

A sample of diethyl ether stored for a long time in a clear bottle exposed to air should not be distilled to dryness because it may contain:
(a) ethanol
(b) explosive peroxides
(c) ethanal
(d) chloroethane

Try it first, then show the solutionSolution

Ethers slowly react with oxygen in air and light to form peroxides.
Peroxides concentrate in the residue on distillation and can explode.

Answer: (b) explosive peroxides.

Example 2: Ethyl bromide to diethyl ether · 2 marks · Short answerBoard level

How will you convert ethyl bromide into diethyl ether? Give the equation and name the reaction.

Try it first, then show the solutionSolution

Heat ethyl bromide with sodium ethoxide (made from ethanol and sodium): Williamson's synthesis (1 mark).
$C_{2} H_{5} O N a + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5} + N a B r$ (1 mark).

Answer: Williamson's synthesis with sodium ethoxide.

Example 3: Which group becomes the iodide? · 3 marks · Short answerHarder board level

Predict the products, with equations, when each ether is treated with one mole of HI, and name the mechanism of the C–O bond cleavage: (a) 1-methoxypropane, CH₃OCH₂CH₂CH₃; (b) 2-methoxy-2-methylpropane, (CH₃)₃COCH₃.

Try it first, then show the solutionSolution

In both, H⁺ first protonates the ether oxygen (½ mark).
(a) I⁻ attacks the less hindered methyl carbon (SN2): $C H_{3} O C H_{2} C H_{2} C H_{3} + H I \to C H_{3} I + C H_{3} C H_{2} C H_{2} O H$ (1 mark).
(b) The protonated ether ionises to the stable tert-butyl carbocation (SN1), which takes I⁻: $(C H_{3})_{3} C O C H_{3} + H I \to (C H_{3})_{3} C I + C H_{3} O H$ (1 mark).
So the methyl group becomes the iodide in (a), but the tertiary group in (b) (½ mark).

Answer: (a) Iodomethane + propan-1-ol (SN2). (b) 2-Iodo-2-methylpropane + methanol (SN1).

Example 4: Substitution in anisole · 3 marks · Short answerHarder board level

Write equations for (a) nitration and (b) Friedel–Crafts acetylation of anisole, naming the major products. Why is anisole more reactive than benzene in these reactions?

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} O C H_{3} + H N O_{3} conc. H_{2} S O_{4} C_{6} H_{4} (N O_{2}) O C H_{3} + H_{2} O$ : 4-nitroanisole (major) and 2-nitroanisole (1 mark).
(b) $C_{6} H_{5} O C H_{3} + C H_{3} C O C l anhyd. AlCl_{3} C_{6} H_{4} (C O C H_{3}) O C H_{3} + H C l$ : 4-methoxyacetophenone (major) and 2-methoxyacetophenone (1 mark).
The –OCH₃ group donates a lone pair into the ring (+R effect), increasing electron density, especially at ortho and para positions (1 mark).

Answer: (a) 4-Nitroanisole. (b) 4-Methoxyacetophenone. –OCH₃ activates the ring by +R.

Example 5: Identify the C₄H₁₀O ethers · 5 marks · Long answerHOTS / competency

Compound A, C₄H₁₀O, does not react with sodium metal. Heated with excess HI it gives a single alkyl iodide B, which with aqueous KOH gives C; C gives a yellow precipitate with I₂ and NaOH. An isomer D of A, which also does not react with sodium, gives iodomethane and 2-iodopropane with excess hot HI.
(a) Identify A, B, C and D. [2]
(b) Write equations for A with excess hot HI and for the conversion of B into C. [2]
(c) Write the equation for the reaction of D with excess hot HI. [1]

Try it first, then show the solutionSolution

No reaction with Na: A and D are ethers. A gives a single iodide, so it is symmetrical: A = ethoxyethane; B = iodoethane; C = ethanol (iodoform positive). D gives CH₃I and (CH₃)₂CHI: D = 2-methoxypropane (2 marks).
(b) $C_{2} H_{5} O C_{2} H_{5} + 2 H I \to 2 C_{2} H_{5} I + H_{2} O$ (1 mark); $C_{2} H_{5} I + K O H \to C_{2} H_{5} O H + K I$ (1 mark).
(c) $C H_{3} O C H (C H_{3})_{2} + 2 H I \to C H_{3} I + (C H_{3})_{2} C H I + H_{2} O$ (1 mark).

Answer: A ethoxyethane, B iodoethane, C ethanol, D 2-methoxypropane.

Where students lose marks in Ethers: Williamson Synthesis and Cleavage

The slips examiners see most often, and what to write instead.

Mistake: Using an alcohol instead of an alkoxide in Williamson's synthesis
ISC examiners reported ethyl alcohol used instead of sodium ethoxide (2024, 2025).
Wrong working
$C_{2} H_{5} O H + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5} + H B r$
Right working
$C_{2} H_{5} O N a + C_{2} H_{5} B r \to C_{2} H_{5} O C_{2} H_{5} + N a B r$
Do this: Write the sodium alkoxide (made from the alcohol and Na) as the nucleophile.
Usually costs: 1 mark.
Mistake: Not giving the reason for no reaction with sodium
Examiners noted that candidates did not state that ethanol has an active hydrogen while ether has none (2024).
Do this: Say: ethanol has an H on oxygen that Na replaces (H₂ evolved); ether has no such H.
Usually costs: 1 mark.
Mistake: Choosing a tertiary halide in Williamson's synthesis
Students write sodium methoxide with 2-bromo-2-methylpropane.
Do this: Tertiary halides undergo elimination with alkoxides; use the tertiary alkoxide with the methyl or primary halide.
Usually costs: 1 to 2 marks.
Mistake: Writing iodobenzene from anisole and HI
Students treat the aryl–O bond like an alkyl–O bond.
Do this: Anisole + HI gives phenol and iodomethane.
Usually costs: 1 mark.

Exam tips

When asked to name a reaction, write 'Williamson's synthesis' and show the alkoxide or phenoxide.
For HI questions, say whether one mole (cold) or excess (hot) HI is used; the products differ.
In anisole substitution questions, give both ortho and para products and state that para is major.
Learn the four aliphatic ether reactions in the CISCE list: chlorine, oxidation (peroxides), HI and PCl₅.
Use IUPAC names (methoxyethane, ethoxyethane, methoxybenzene) alongside common names.

Practice questions on Ethers: Williamson Synthesis and Cleavage

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The IUPAC name of CH₃OCH₂CH₃ is:
(a) ethyl methyl ether
(b) methoxyethane
(c) ethoxymethane
(d) propan-1-ol
Hint
The larger group is the parent alkane.
Answer
(b) methoxyethane (ethyl methyl ether is the common name).
Q2 · 2 marks · Short answerBoard level
Why is diethyl ether stored in dark bottles? Give one use of diethyl ether.
Hint
Air and light.
Answer
In air and light it slowly forms explosive peroxides; dark bottles (well stoppered) reduce this. Use: solvent for fats, oils and Grignard reagents (or formerly as an anaesthetic).
Q3 · 2 marks · Short answerHarder board level
What happens when diethyl ether reacts with chlorine (a) in the dark and (b) in excess in bright light? Give equations.
Hint
Substitution at the α-carbon first.
Answer
(a) $C_{2} H_{5} O C_{2} H_{5} + C l_{2} \to C H_{3} C H C l O C_{2} H_{5} + H C l$ (α-chloro ether; further substitution can follow). (b) $C_{2} H_{5} O C_{2} H_{5} + 10 C l_{2} \to C_{2} C l_{5} O C_{2} C l_{5} + 10 H C l$ (perchlorodiethyl ether).
Q4 · 3 marks · Short answerHarder board level
Convert phenol into 4-methoxyacetophenone in two steps, giving equations and conditions.
Hint
Make anisole first.
Answer
Step 1: $C_{6} H_{5} O H + N a O H \to C_{6} H_{5} O N a + H_{2} O$ , then $C_{6} H_{5} O N a + C H_{3} I \to C_{6} H_{5} O C H_{3} + N a I$ (Williamson). Step 2: $C_{6} H_{5} O C H_{3} + C H_{3} C O C l anhyd. AlCl_{3} C_{6} H_{4} (C O C H_{3}) O C H_{3} + H C l$ ; separate the para isomer.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Ethoxyethane does not liberate hydrogen with sodium metal, but ethanol does.
Reason (R): Ethoxyethane has no hydrogen atom attached to oxygen.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which hydrogen does sodium replace?
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Diphenyl ether cannot be made by heating chlorobenzene with sodium phenoxide under ordinary laboratory conditions, but anisole is easily made from sodium phenoxide and iodomethane. Explain.
Hint
Compare the halides.
Answer
Williamson's synthesis needs SN2 attack on the halide carbon. Iodomethane is an unhindered methyl halide and reacts easily. Chlorobenzene's C–Cl bond has partial double-bond character, the carbon is sp², and backside attack is blocked by the ring, so nucleophilic substitution does not occur under ordinary conditions.
Q7 · 3 marks · Short answerHOTS / competency
Write the mechanism of the reaction of ethoxyethane with one mole of HI, and explain why HI is more effective than HCl for cleaving ethers.
Hint
Protonation, then nucleophilic attack.
Answer
(1) $C_{2} H_{5} O C_{2} H_{5} + H I \to C_{2} H_{5} O^{+} (H) C_{2} H_{5} + I^{-}$ (protonation makes a good leaving group). (2) I⁻ attacks a CH₂ carbon by SN2, the C–O bond breaks: $C_{2} H_{5} I + C_{2} H_{5} O H$ . HI is a stronger acid (protonates the ether well) and I⁻ is a much better nucleophile than Cl⁻, so HI reacts readily while HCl hardly reacts.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student prepares a Grignard reagent, ethylmagnesium bromide, in dry diethyl ether. A friend suggests using ethanol as the solvent instead, because it dissolves many substances and is cheaper.
(i) Why would ethanol destroy the Grignard reagent? Write the equation. [1]
(ii) Why is diethyl ether a good solvent for Grignard reagents? [1]
(iii) Why must the ether be dry and free from peroxides? [1]
Hint
Grignard reagents are strong bases.
Answer
(i) Ethanol has an acidic O–H, which protonates the Grignard reagent: $C_{2} H_{5} M g B r + C_{2} H_{5} O H \to C_{2} H_{6} + C_{2} H_{5} O M g B r$ . (ii) Ether has no acidic hydrogen and its oxygen lone pairs coordinate to magnesium, stabilising the reagent. (iii) Water also destroys the reagent (giving ethane), and peroxides react with it and are an explosion hazard.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Mechanism of Williamson's synthesis?Show answer
SN2.
2.Products of ethoxyethane with PCl₅?Show answer
Chloroethane (2 mol) and POCl₃.
3.Products of anisole with HI?Show answer
Phenol and iodomethane.
4.Why do ethers have low boiling points?Show answer
They cannot form hydrogen bonds between their own molecules.
5.What forms when ethers stand in air and light?Show answer
Explosive peroxides.
6.Which group becomes the iodide when tert-butyl methyl ether reacts with HI?Show answer
The tert-butyl group (SN1).

Ethers: Williamson Synthesis and Cleavage: frequently asked questions

What is Williamson's ether synthesis?

The reaction of a sodium alkoxide or phenoxide with a primary alkyl halide, by SN2, to give an ether and a sodium halide.

Why can't tertiary alkyl halides be used in Williamson's synthesis?

The alkoxide is a strong base; with a tertiary halide it removes a proton and causes elimination, giving an alkene instead of an ether.

What happens when an ether reacts with HI?

The C–O bond breaks: with one mole of cold HI an alcohol and an alkyl iodide form; with excess hot HI two alkyl iodides form.

Why does anisole give phenol, not iodobenzene, with HI?

The O–aryl bond is strengthened by resonance and the sp² carbon resists nucleophilic attack, so I⁻ attacks the methyl carbon.

Why is anisole ortho-para directing?

The methoxy group donates electrons into the ring by resonance, increasing electron density at the ortho and para positions.