Alcohols, Phenols and Ethers · Lesson 10 of 12
Preparing Ethers
“Design ether syntheses by choosing which carbon is attacked and distinguish substitution from competing elimination.”
• Explain acid-catalysed intermolecular dehydration of primary alcohols. • Choose an alkoxide and alkyl halide for a Williamson synthesis. • Predict when steric hindrance and basicity favour elimination. • Prepare aryl alkyl ethers through phenoxide alkylation.
Join carbon groups through oxygen without confusing an ester
An ether has R–O–R′ connectivity and no carbonyl within that linkage. Its two carbon groups may be identical or different. Forming an ether therefore means constructing a C–O bond between an oxygen nucleophile and an appropriate carbon electrophile. In acid dehydration the oxygen nucleophile is an alcohol; in Williamson synthesis it is an alkoxide or phenoxide. These routes have different conditions because a strong acid would destroy an alkoxide by protonation.
Intermolecular dehydration uses two alcohol molecules
First protonate one ethanol molecule to make CH₃CH₂OH₂⁺. The oxygen lone pair of a second, neutral ethanol attacks its primary carbon from the backside while the C–O bond to the leaving water breaks. This SN2 step gives protonated ethoxyethane, (C₂H₅)₂OH⁺. Deprotonation by water or another suitable base yields neutral ether and regenerates acid. The nucleophile is an ethanol oxygen, not an ethoxide ion surviving in concentrated acid.
The two participating alcohol molecules lose one net molecule of water. In contrast, alkene formation loses water from one molecule and creates a C=C bond. A sterically accessible primary carbon is important for the substitution step. With secondary or tertiary alcohols, elimination is more competitive, making acid dehydration an unreliable general preparation of the corresponding ethers. “Add acid” without temperature and substrate identity does not select a unique product.
Problem
Show the route and mechanism from propan-1-ol to 1-propoxypropane.
- 1.Use controlled acid-catalysed intermolecular dehydration of the primary alcohol. The net equation is 2CH₃CH₂CH₂OH ⇌ CH₃CH₂CH₂OCH₂CH₂CH₃ + H₂O.
- 2.Protonate one OH, then another propan-1-ol oxygen attacks its primary CH₂ carbon as water leaves in an SN2 event. The resulting oxonium ion loses an O-bound proton to give the ether.
- 3.Choose conditions favouring intermolecular substitution rather than high-temperature alkene formation. The 413 K ethanol comparison is a useful example, not an exact universal optimum for every alcohol.
Mixing methanol and ethanol under acid dehydration can produce methoxymethane, ethoxyethane and methoxyethane. Two different oxygen nucleophiles can meet two different protonated alcohols, so the unsymmetrical product is not obtained selectively by simply mixing the two liquids. A planned Williamson disconnection is usually more useful when the groups differ.
Williamson synthesis: choose the attacked carbon wisely
The new bond forms when oxygen attacks the carbon bearing X and X leaves with the old C–X bonding pair. It is that halide carbon, not the carbon already attached to oxygen in the alkoxide, that must be accessible. A tertiary alkoxide can therefore be one partner in an ether synthesis if the other partner is suitably unhindered. Conversely, a small alkoxide does not make a tertiary alkyl halide suitable for SN2.
To plan an unsymmetrical ether, disconnect either C–O bond on paper. Compare the two possible halide partners. Choose methyl or the less hindered primary group as the electrophile when possible, leaving the more substituted group on oxygen. Bromides and iodides are commonly convenient leaving groups; chlorides may react less readily. Nucleophilicity, basicity, solvent and temperature also influence the outcome, so a valid disconnection is not a guarantee of quantitative isolated yield.
| Target ether | Conventional useful pairing | Pairing to avoid for ordinary SN2 |
|---|---|---|
| CH₃OC₂H₅ | CH₃ONa + C₂H₅Br, or C₂H₅ONa + CH₃I | Acid mixed dehydration if a selective single product is needed |
| (CH₃)₃COCH₃ | (CH₃)₃CONa + CH₃I | CH₃ONa + (CH₃)₃CBr |
| (CH₃)₃COC₂H₅ | (CH₃)₃CONa + C₂H₅Br | C₂H₅ONa + (CH₃)₃CCl |
| C₆H₅OC₂H₅ | C₆H₅ONa + C₂H₅Br | C₆H₅Br + C₂H₅ONa as an ordinary Williamson SN2 route |
An alkoxide is also a base
A strongly basic alkoxide can remove a β hydrogen from an alkyl halide while the C–H pair forms C=C and the C–X bond breaks. This E2 elimination competes increasingly as the halide becomes hindered. A tertiary halide is blocked for ordinary SN2, so strong-base elimination generally dominates. Secondary halides often give mixtures. Even primary halides can eliminate with bulky bases under suitable conditions; the standard primary preference is a design guide, not permission to ignore every competing pathway.
Problem
Why is ethoxide plus tert-butyl chloride unsuitable, and what alternative is conventionally used?
- 1.In C₂H₅ONa + (CH₃)₃CCl, oxygen would need to attack a tertiary carbon for the ether route. Ordinary SN2 is sterically blocked, while ethoxide can remove a β hydrogen.
- 2.The major elimination product is 2-methylpropene: (CH₃)₃CCl + NaOC₂H₅ → (CH₃)₂C=CH₂ + C₂H₅OH + NaCl.
- 3.Reverse the disconnection: use sodium tert-butoxide with an ethyl halide, for example C₂H₅Br, so substitution occurs at primary ethyl carbon. Elimination can still compete because tert-butoxide is bulky and basic; appropriate conditions and product verification matter. Methyl electrophiles are still more favourable when the target permits them.
Problem
Use ethanol and 3-methylpentan-2-ol as starting materials.
- 1.Convert ethanol to an ethyl halide, for example C₂H₅OH + HBr → C₂H₅Br + H₂O. Separately convert the secondary alcohol to its sodium alkoxide with a suitable reagent under dry conditions.
- 2.The alkoxide is CH₃CH(O⁻Na⁺)CH(CH₃)CH₂CH₃. React it with C₂H₅Br to create the O–ethyl bond at a primary electrophilic carbon.
- 3.The product is CH₃CH(OC₂H₅)CH(CH₃)CH₂CH₃. Making the secondary fragment into the halide instead would increase substitution/elimination competition; the two apparent disconnections are not equally favourable.
Use phenoxide on the oxygen side
Phenol reacts with NaOH to form phenoxide; then a suitable methyl or primary halide can alkylate oxygen. C₆H₅ONa + CH₃Br → C₆H₅OCH₃ + NaBr makes anisole, and C₆H₅ONa + C₂H₅Br → C₆H₅OC₂H₅ + NaBr makes ethoxybenzene. The benzene ring stays on the oxygen nucleophile side. An ordinary unactivated aryl halide cannot be substituted by the Williamson SN2 pathway at its sp² carbon.
For 1-methoxy-4-nitrobenzene, sodium 4-nitrophenoxide plus methyl bromide is the direct Williamson pairing: methyl carbon is accessible to SN2. The alternative 4-bromonitrobenzene plus methoxide is not an ordinary Williamson SN2 route. An ortho/para nitro group can activate an aryl halide for a separate addition–elimination nucleophilic aromatic substitution under suitable conditions. That special chemistry should not be denied, but it must not be confused with the route being taught.
After synthesis, ether oxygen remains a hydrogen-bond acceptor, but the product has no OH donor. This explains why replacing a phenolic H by an alkyl group changes acid–base behaviour and self hydrogen bonding. Anisole does not behave as a phenol with a removable OH proton. The structural change explains both the synthesis and its consequences.
Quick Quiz
Quiz
Which carbon is attacked in sodium tert-butoxide plus methyl iodide?
Explanation: The electrophile is CH₃I; the tert-butyl group stays on the nucleophilic oxygen side.
Why does methoxide with tert-butyl bromide mainly form an alkene?
Explanation: A tertiary electrophile and strong base favour E2 rather than the desired ether substitution.
What is the nucleophile in acid-catalysed ether formation from ethanol?
Explanation: Neutral alcohol attacks a protonated alcohol, then the product loses a proton.
Which is the direct Williamson pairing for anisole?
Explanation: The phenoxide oxygen attacks the accessible methyl carbon.
Why is mixed methanol/ethanol dehydration poorly selective for methoxyethane?
Explanation: Symmetrical and unsymmetrical ethers can form in the mixed reaction.
Practice Problems
- Write a Williamson synthesis for 1-propoxypropane and for methoxyethane.
- Prepare 2-methoxy-2-methylpropane and explain why the reversed pairing is poor.
- Give the three mechanistic stages for acid dehydration of propan-1-ol to its symmetrical ether.
- Plan 2-ethoxy-3-methylpentane from ethanol and 3-methylpentan-2-ol.
- Choose a Williamson route to 1-methoxy-4-nitrobenzene and qualify the aryl-halide alternative.
- Why are secondary/tertiary alcohols poor general substrates for acid dehydration to ethers?
Use CH₃CH₂CH₂ONa + CH₃CH₂CH₂Br → CH₃CH₂CH₂OCH₂CH₂CH₃ + NaBr for the symmetrical ether. For methoxyethane, use CH₃ONa + C₂H₅Br → CH₃OC₂H₅ + NaBr, or ethoxide plus methyl iodide. Both use accessible primary/methyl electrophiles.
Key Takeaways
• Acid ether synthesis uses neutral alcohol attack on a protonated alcohol. • Temperature and steric accessibility decide whether substitution or alkene formation is favoured. • Mixed alcohol dehydration can give several ethers. • Williamson synthesis makes C–O by SN2 attack of alkoxide or phenoxide. • The carbon bearing halide must be accessible; methyl and primary electrophiles are preferred. • Alkoxides are also strong bases and can promote E2 elimination. • Aryl alkyl ethers are conveniently made from phenoxide and an alkyl halide. • Activated nucleophilic aromatic substitution is distinct from ordinary Williamson SN2.