ChemistryCBSE Class 12 · Alcohols, Phenols and Ethers

Ethers: Williamson Synthesis and Cleavage by HI

Updated 15 October 2026 · Fact-checked

In the Williamson synthesis an alkyl halide reacts with a sodium alkoxide or phenoxide by SN2 to give an ether; the halide must be primary, because tertiary halides undergo elimination instead. Ethers are cleaved by hot HI: the iodide attacks the smaller alkyl group (SN2), but a tertiary group forms a tertiary iodide (SN1); anisole gives phenol and methyl iodide.

To make t-butyl methyl ether, put the bulky tertiary group on the alkoxide and use a primary halide; a tertiary halide gives an alkene.

Illustration drawn for QuestionBuddy.

Understand Ethers: Williamson Synthesis and Cleavage

Ethers ( $R - O - R^{'}$ ) have an oxygen bonded to two carbon groups. They are less reactive than alcohols because there is no O–H, so their key reactions are how they are made and how the C–O bond is broken.

1. Preparation by dehydration of alcohols. Primary alcohols heated with concentrated $H_{2} S O_{4}$ at 413 K give symmetrical ethers: $2 C_{2} H_{5} O H H_{2} S O_{4}, 413 K C_{2} H_{5} O C_{2} H_{5} + H_{2} O$ . At 443 K the same mixture gives ethene, so temperature control is essential. The mechanism is SN2: one alcohol is protonated, a second alcohol molecule attacks the carbon and pushes out water, and the protonated ether loses $H^{+}$ . This method suits only primary alcohols: secondary and tertiary alcohols form stable carbocations and give alkenes.

2. Williamson synthesis. This is the important laboratory method for both symmetrical and unsymmetrical ethers. An alkyl halide reacts with a sodium alkoxide: $R - X + R^{'} - O^{-} N a^{+} \to R - O - R^{'} + N a X$ . The alkoxide ion is the nucleophile and attacks the alkyl halide by SN2. So the best results come from primary alkyl halides. With a tertiary halide the alkoxide, which is also a strong base, removes a β-hydrogen and an alkene forms instead. To make tert-butyl methyl ether, therefore, use sodium tert-butoxide with bromomethane, not 2-bromo-2-methylpropane with sodium methoxide (which gives 2-methylpropene). Phenols are converted into ethers the same way, after turning them into phenoxide with NaOH: $C_{6} H_{5} O^{-} N a^{+} + C H_{3} I \to C_{6} H_{5} O C H_{3}$ (anisole) $+ N a I$ . The halide part must be alkyl; aryl halides do not undergo this SN2 reaction.

Physical properties. The C–O–C bond angle is a little more than tetrahedral (about 111.7° in methoxymethane) because of repulsion between the bulky alkyl groups. Ethers cannot hydrogen-bond with each other, so their boiling points are close to those of alkanes of similar mass and much lower than isomeric alcohols. They can accept hydrogen bonds from water, so small ethers are about as soluble as alcohols of similar size.

3. Cleavage of the C–O bond. Ethers are cleaved only under drastic conditions with excess hydrogen halide. The order of reactivity is $H I > H B r > H C l$ (iodide is the best nucleophile). The ether oxygen is first protonated. Then:

If both groups are primary or secondary, $I^{-}$ attacks the less hindered (smaller) carbon by SN2: $C H_{3} O C H_{2} C H_{3} + H I \to C H_{3} I + C H_{3} C H_{2} O H$ . With excess HI the alcohol is also converted into an iodide.
If one group is tertiary, the protonated ether ionises to the stable tertiary carbocation (SN1), so the tertiary halide forms: $(C H_{3})_{3} C - O - C H_{3} + H I \to (C H_{3})_{3} C - I + C H_{3} O H$ .
Alkyl aryl ethers always break at the alkyl–oxygen bond: $C_{6} H_{5} O C H_{3} + H I \to C_{6} H_{5} O H + C H_{3} I$ . The O–aryl bond has partial double-bond character, and SN2 attack on a benzene carbon is impossible, so phenol (not iodobenzene) forms.

4. Electrophilic substitution in aryl ethers. The –OR group activates the ring (+R) and directs to ortho and para, like –OH but a little less strongly. Anisole undergoes bromination with $B r_{2}$ in ethanoic acid without a catalyst (p-bromoanisole major), Friedel–Crafts alkylation ( $C H_{3} C l$ / $A l C l_{3}$ → 2- and 4-methoxytoluene), acylation ( $C H_{3} C O C l$ / $A l C l_{3}$ → 2- and 4-methoxyacetophenone) and nitration (→ 2- and 4-nitroanisole).

Ethers: Williamson Synthesis and Cleavage: formulas and key facts

Williamson synthesis: $R - X + R^{'} - O^{-} N a^{+} \to R - O - R^{'} + N a X$; SN2; use a primary alkyl halide; the alkoxide may be primary, secondary or tertiary.
Dehydration route: $2 C_{2} H_{5} O H conc. H_{2} S O_{4}, 413 K C_{2} H_{5} O C_{2} H_{5} + H_{2} O$; At 443 K ethene forms instead; only for primary alcohols.
Cleavage, SN2 (small group gets I): $C H_{3} - O - C_{2} H_{5} + H I \to C H_{3} I + C_{2} H_{5} O H$; Iodide attacks the less hindered carbon.
Cleavage, SN1 (tertiary group gets I): $(C H_{3})_{3} C - O - C H_{3} + H I \to (C H_{3})_{3} C - I + C H_{3} O H$; Stable tertiary carbocation forms.
Cleavage of anisole: $C_{6} H_{5} - O - C H_{3} + H I \to C_{6} H_{5} O H + C H_{3} I$; Aryl–O bond is not broken.
Reactivity of hydrogen halides: $H I > H B r > H C l$; Iodide is the best nucleophile.

How to solve Ethers: Williamson Synthesis and Cleavage questions

Questions ask you to choose reagents for a Williamson synthesis, predict products of cleavage, write a mechanism, or complete substitution reactions of anisole. Decide which bond is made or broken and what kind of carbon is involved.

1For a target ether $R - O - R^{'}$ , split it into two possible halide + alkoxide pairs.
2Choose the pair in which the halide carbon is primary (or methyl); put the bulky or aryl group on the alkoxide/phenoxide side.
3Reject a pair with a tertiary halide (elimination gives an alkene) or an aryl halide (no SN2).
4For cleavage, identify the two groups on oxygen: primary/secondary → iodide goes to the smaller group (SN2); tertiary → iodide goes to the tertiary group (SN1); aryl → phenol + alkyl iodide.
5With excess HI, convert the alcohol formed into a second alkyl iodide (but phenol stays as phenol).
6For anisole substitution, place the new group ortho and para to –OCH₃, with para as the major product.
7Write mechanisms in steps: protonation, nucleophilic attack (or ionisation), and the 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.

Planning a Williamson synthesis

Suppose a perfume chemist needs anisole (methoxybenzene), which has a pleasant smell, and has phenol, sodium hydroxide, iodomethane and iodobenzene on the shelf. A trainee suggests heating iodobenzene with sodium methoxide.

Explain why the trainee's route fails and give a route that works, with equations.

Try it first, then show the solutionSolution

Williamson synthesis is SN2 at the carbon bearing the halogen. In iodobenzene that carbon is part of the ring: the C–I bond has partial double-bond character and backside attack is impossible, so no ether forms.
Instead, put the aryl group on the nucleophile: $C_{6} H_{5} O H + N a O H \to C_{6} H_{5} O^{-} N a^{+} + H_{2} O$ .
Then react sodium phenoxide with iodomethane (a methyl halide, ideal for SN2): $C_{6} H_{5} O^{-} N a^{+} + C H_{3} I \to C_{6} H_{5} O C H_{3} + N a I$ .
The product is anisole.

Answer: Aryl halides do not undergo SN2; make sodium phenoxide from phenol and NaOH, then react it with $C H_{3} I$ to get anisole.

Memory hook: In Williamson, the halide must be small and simple; the big or aromatic part brings the oxygen.

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.

Halide small, alkoxide big
Put the methyl or primary group on the halide and the bulky (3°) or aryl group on the oxygen.
Use it when: Choosing Williamson reagents.
Small gets I, unless there's a tertiary
HI cleavage: iodine goes to the smaller group (SN2), except a tertiary group grabs it (SN1). Aryl never takes it.
Use it when: Predicting cleavage products.
413 ether, 443 ethene
Thirteen makes the ether (two molecules join); forty-three makes ethene.
Use it when: Dehydration temperature questions.

Ethers: Williamson Synthesis and Cleavage: solved examples

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

Example 1: Anisole with HI · 1 mark · MCQBoard level

The products of the reaction of anisole with hot concentrated HI are:
(a) iodobenzene and methanol
(b) phenol and iodomethane
(c) benzene and methanol
(d) iodobenzene and iodomethane

Try it first, then show the solutionSolution

The aryl–O bond has partial double-bond character and SN2 cannot occur at the aryl carbon; iodide attacks the methyl carbon.

Answer: (b) Phenol and iodomethane.

Example 2: The Williamson synthesis · 2 marks · Very short answerBoard level

What is the Williamson synthesis? Write an equation for the preparation of ethoxyethane by this method, and state why primary alkyl halides are preferred.

Try it first, then show the solutionSolution

An alkyl halide reacts with a sodium alkoxide to form an ether: $C_{2} H_{5} B r + C_{2} H_{5} O N a \to C_{2} H_{5} O C_{2} H_{5} + N a B r$ (1 mark).
The reaction is SN2; primary halides are least hindered, and with tertiary halides the strongly basic alkoxide causes elimination to an alkene (1 mark).

Answer: $C_{2} H_{5} B r + C_{2} H_{5} O N a \to C_{2} H_{5} O C_{2} H_{5} + N a B r$ ; SN2 works best with primary halides, tertiary halides eliminate.

Example 3: Choosing reagents for tert-butyl methyl ether · 3 marks · Short answerHarder board level

2-Methoxy-2-methylpropane (tert-butyl methyl ether) is to be prepared by the Williamson synthesis. Two pairs of reagents are possible: (i) $(C H_{3})_{3} C B r + C H_{3} O N a$ and (ii) $(C H_{3})_{3} C O N a + C H_{3} B r$ . Which pair should be used? Write the reaction for each pair.

Try it first, then show the solutionSolution

Pair (ii): $(C H_{3})_{3} C O N a + C H_{3} B r \to (C H_{3})_{3} C - O - C H_{3} + N a B r$ . The halide is methyl, ideal for SN2 (1 mark).
Pair (i): the tertiary halide is too hindered for SN2; methoxide acts as a base and removes a β-hydrogen: $(C H_{3})_{3} C B r + C H_{3} O N a \to (C H_{3})_{2} C = C H_{2} + C H_{3} O H + N a B r$ (1 mark).
So pair (ii) must be used (1 mark).

Answer: Use sodium tert-butoxide and bromomethane; the other pair gives 2-methylpropene.

Example 4: Mechanism of cleavage by HI · 3 marks · Short answerHarder board level

Write the mechanism of the reaction of methoxyethane ( $C H_{3} O C_{2} H_{5}$ ) with HI and explain why iodomethane rather than iodoethane is formed.

Try it first, then show the solutionSolution

Step 1: protonation of the ether oxygen: $C H_{3} - O - C_{2} H_{5} + H I ⇌ C H_{3} - O^{+} (H) - C_{2} H_{5} + I^{-}$ (1 mark).
Step 2: $I^{-}$ attacks the methyl carbon by SN2, and ethanol leaves: $I^{-} + C H_{3} - O^{+} (H) C_{2} H_{5} \to C H_{3} I + C_{2} H_{5} O H$ (1 mark).
Iodide attacks the smaller, less hindered methyl carbon; SN2 is fastest there (1 mark).

Answer: Protonation, then SN2 attack of $I^{-}$ on the methyl carbon; products $C H_{3} I$ and $C_{2} H_{5} O H$ .

Example 5: Predicting cleavage and substitution products · 5 marks · Long answerHOTS / competency

Give the main products of: (a) $(C H_{3})_{3} C - O - C H_{3}$ with HI; (b) ethoxyethane with excess HI; (c) anisole with $B r_{2}$ in ethanoic acid; (d) anisole with $C H_{3} C O C l$ /anhydrous $A l C l_{3}$ . Explain (a) with the mechanism type.

Try it first, then show the solutionSolution

(a) After protonation, the ether ionises to the stable tertiary carbocation, which captures $I^{-}$ (SN1): products 2-iodo-2-methylpropane and methanol (1½ marks).
(b) $C_{2} H_{5} O C_{2} H_{5} + H I \to C_{2} H_{5} I + C_{2} H_{5} O H$ , then the ethanol reacts too: overall $C_{2} H_{5} O C_{2} H_{5} + 2 H I \to 2 C_{2} H_{5} I + H_{2} O$ (1 mark).
(c) –OCH₃ activates the ring, so no catalyst is needed: 4-bromoanisole (major) and 2-bromoanisole (1 mark).
(d) Friedel–Crafts acylation: 4-methoxyacetophenone (major) and 2-methoxyacetophenone (1 mark).
Explanation of (a): SN2 at the tertiary carbon is blocked, but the tertiary carbocation is stable, so the C–O bond to the tertiary group breaks (½ mark).

Answer: (a) $(C H_{3})_{3} C I$ + $C H_{3} O H$ (SN1). (b) 2 $C_{2} H_{5} I$ + $H_{2} O$ . (c) p-Bromoanisole (major) + o-isomer. (d) p-Methoxyacetophenone (major) + o-isomer.

Where students lose marks in Ethers: Williamson Synthesis and Cleavage

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

Mistake: Using a tertiary halide in Williamson synthesis
Wrong working
$(C H_{3})_{3} C B r + C H_{3} O N a \to (C H_{3})_{3} C O C H_{3}$
Right working
$(C H_{3})_{3} C O N a + C H_{3} B r \to (C H_{3})_{3} C O C H_{3} + N a B r$
Do this: Keep the halide primary or methyl.
Usually costs: 1-2 marks.
Mistake: Writing iodobenzene from anisole
Do this: Alkyl aryl ethers give phenol and alkyl iodide.
Usually costs: 1 mark.
Mistake: Iodide to the larger group in SN2 cleavage
Wrong working
$C H_{3} O C_{2} H_{5} + H I \to C_{2} H_{5} I + C H_{3} O H$
Right working
$C H_{3} O C_{2} H_{5} + H I \to C H_{3} I + C_{2} H_{5} O H$
Do this: SN2: iodide attacks the less hindered carbon; only a tertiary group takes the iodide (SN1).
Usually costs: 1 mark.
Mistake: Wrong temperature for ether formation
Do this: 413 K for the ether, 443 K for ethene.
Usually costs: ½ mark.

Exam tips

For Williamson questions, always show both possible reagent pairs and say why one fails.
In cleavage questions, state the mechanism (SN1 or SN2) that decides the products.
Give the mechanism of ether formation from alcohols in three steps with $H^{+}$ regenerated.
Name ortho and para products for anisole reactions, with para major.
Mention 'excess HI' explicitly when both groups become iodides.

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 Williamson ether synthesis proceeds by:
(a) SN1 (b) SN2 (c) electrophilic addition (d) free-radical substitution
Hint
Alkoxide attacks a primary halide in one step.
Answer
(b) SN2.
Q2 · 2 marks · Very short answerBoard level
How is ethoxyethane prepared from ethanol? Give the conditions and explain why this method is not suitable for preparing ethers from tertiary alcohols.
Hint
Temperature matters.
Answer
Heat ethanol with conc. $H_{2} S O_{4}$ at 413 K: $2 C_{2} H_{5} O H \to C_{2} H_{5} O C_{2} H_{5} + H_{2} O$ . Tertiary alcohols form stable carbocations that lose a proton to give alkenes, so elimination dominates.
Q3 · 2 marks · Very short answerHarder board level
Complete the reactions and name the major products: (a) anisole + $C H_{3} C l$ (anhydrous $A l C l_{3}$ ) (b) anisole + conc. $H N O_{3}$ /conc. $H_{2} S O_{4}$
Hint
–OCH₃ is an o/p director.
Answer
(a) 4-Methoxytoluene (major) and 2-methoxytoluene. (b) 4-Nitroanisole (major) and 2-nitroanisole.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Anisole reacts with HI to give phenol and iodomethane, not iodobenzene and methanol.
Reason (R): The O–C₆H₅ bond has partial double-bond character, and SN2 attack on an aromatic carbon is not possible.
(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 C–O bond can iodide reach?
Answer
(a).
Q5 · 3 marks · Short answerHarder board level
Write the mechanism of the formation of ethoxyethane from ethanol in the presence of conc. $H_{2} S O_{4}$ at 413 K.
Hint
Protonation, SN2 attack, deprotonation.
Answer
(i) $C_{2} H_{5} O H + H^{+} \to C_{2} H_{5} O H_{2}^{+}$ . (ii) A second ethanol molecule attacks the carbon of the protonated alcohol by SN2, displacing water: $C_{2} H_{5} O H + C_{2} H_{5} O H_{2}^{+} \to C_{2} H_{5} - O^{+} (H) - C_{2} H_{5} + H_{2} O$ . (iii) Loss of a proton: $C_{2} H_{5} - O^{+} (H) - C_{2} H_{5} \to C_{2} H_{5} O C_{2} H_{5} + H^{+}$ .
Q6 · 2 marks · Very short answerHarder board level
Why do ethers have much lower boiling points than isomeric alcohols, yet similar solubility in water? Give an example pair.
Hint
Hydrogen bonding between molecules vs with water.
Answer
Ether molecules have no O–H, so they cannot hydrogen-bond with one another, giving low boiling points (ethoxyethane boils far below butan-1-ol). But the ether oxygen can accept hydrogen bonds from water, so small ethers dissolve to a similar extent as alcohols of similar size.
Q7 · 3 marks · Short answerHOTS / competency
An ether X ( $C_{5} H_{12} O$ ) reacts with one equivalent of hot HI to give methanol and a tertiary alkyl iodide $C_{4} H_{9} I$ . Identify X and the iodide, and explain why the iodide is tertiary. Suggest a Williamson synthesis of X.
Hint
Cleavage with a tertiary group is SN1.
Answer
X is 2-methoxy-2-methylpropane, $(C H_{3})_{3} C - O - C H_{3}$ ; the iodide is 2-iodo-2-methylpropane. The protonated ether ionises to the stable tertiary carbocation (SN1), which takes $I^{-}$ , while methanol leaves. Synthesis: $(C H_{3})_{3} C O N a + C H_{3} B r \to (C H_{3})_{3} C O C H_{3} + N a B r$ .
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student wants to prepare ethyl phenyl ether ( $C_{6} H_{5} O C_{2} H_{5}$ ). She considers two routes: Route 1: sodium phenoxide + bromoethane. Route 2: bromobenzene + sodium ethoxide.
(a) Which route works? Write its equation. (1 mark)
(b) Why does the other route fail? (1 mark)
(c) Predict the products when ethyl phenyl ether is heated with HI. (1 mark)
(d) Will the ring in ethyl phenyl ether be more or less reactive than benzene towards bromination? Give the reason. (1 mark)
Hint
SN2 needs an alkyl halide.
Answer
(a) Route 1: $C_{6} H_{5} O N a + C_{2} H_{5} B r \to C_{6} H_{5} O C_{2} H_{5} + N a B r$ . (b) Bromobenzene does not undergo SN2: the C–Br bond has partial double-bond character and backside attack on a ring carbon is impossible. (c) Phenol and iodoethane. (d) More reactive: the –OC₂H₅ group donates electrons to the ring (+R), activating it and directing to o/p.

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.Reagents in Williamson synthesis?Show answer
An alkyl halide and a sodium alkoxide (or phenoxide).
2.Temperature for making ethoxyethane from ethanol?Show answer
413 K with conc. $H_{2} S O_{4}$ .
3.Products of anisole with HI?Show answer
Phenol and iodomethane.
4.Order of reactivity of HX with ethers?Show answer
HI > HBr > HCl.
5.Product of tert-butyl methyl ether with HI?Show answer
2-Iodo-2-methylpropane and methanol.
6.Is –OCH₃ activating or deactivating?Show answer
Activating, ortho/para-directing.

Ethers: Williamson Synthesis and Cleavage: frequently asked questions

What is the Williamson synthesis in Class 12 Chemistry?

A method of preparing ethers in which an alkyl halide reacts with a sodium alkoxide or phenoxide by an SN2 reaction: R\text{–}X + R'ONa \to R\text{–}O\text{–}R' + NaX.

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

They are too hindered for SN2, and the strongly basic alkoxide removes a β-hydrogen, so an alkene forms instead of an ether.

Why does anisole give phenol with HI?

The bond between oxygen and the benzene ring has partial double-bond character and cannot undergo SN2, so iodide attacks the methyl group, giving phenol and iodomethane.

Which alkyl group becomes the iodide when a mixed ether is cleaved by HI?

The smaller (less hindered) group by SN2, unless one group is tertiary, in which case the tertiary iodide forms by SN1.

Why do ethers have low boiling points?

They cannot form hydrogen bonds between their own molecules because they have no O–H group.