Alcohols, Phenols and Ethers · Lesson 11 of 12
Cleavage and Aromatic Reactions of Ethers
“Predict which ether bond breaks with hydrogen halides and distinguish that chemistry from substitution on an alkoxy-activated aromatic ring.”
• Explain protonation and SN2 cleavage of ordinary dialkyl ethers. • Identify tertiary, benzylic and aryl exceptions to simple steric predictions. • Distinguish limited-acid products from further reaction with excess hot acid. • Predict anisole bromination, nitration, alkylation and acylation.
Relative inertness is useful, but not absolute
Ethers are often useful solvents because they resist many reagents under ordinary conditions. Their oxygen lone pairs still accept protons and coordinate to Lewis acids. Strong concentrated HI or HBr with heating can activate and cleave a C–O bond. Ordinary HCl is much less effective for this purpose. Within the conventional comparison HI > HBr > HCl, both acid activation and the halide’s ability to attack matter; acidity alone does not explain every rate difference.
First protonate oxygen, then select the reaction centre
For an ordinary methyl/primary dialkyl ether, iodide attacks the less hindered carbon of the protonated ether from the backside. As a C–I bond forms, the original C–O bond pair returns to oxygen and a neutral alcohol departs. This is SN2 displacement at carbon, not substitution at the oxygen atom. The product is an alkyl iodide from the attacked group and an alcohol from the group that remains attached to oxygen.
Problem
Predict the first cleavage products of CH₃OCH₂CH₂CH₃ with HBr and of CH₃CH₂OCH(CH₃)₂ with HI.
- 1.For methyl propyl ether, Br⁻ attacks methyl carbon after protonation. Products are CH₃Br and CH₃CH₂CH₂OH.
- 2.For ethyl isopropyl ether, I⁻ attacks the primary ethyl carbon preferentially rather than the secondary isopropyl carbon. Products are C₂H₅I and propan-2-ol.
- 3.The principle is accessibility and mechanism, not a universal claim that whichever group has fewer total carbons must always form the halide. Resonance and carbocation pathways can change the selection.
Excess acid can change the final isolated products
Methoxymethane illustrates the mechanism without regioselectivity: protonate CH₃OCH₃, then I⁻ displaces methanol by SN2 attack at either equivalent methyl carbon. With one HI the products are CH₃I and CH₃OH; with excess hot HI the methanol can also become CH₃I, giving two methyl iodides and water overall. State the extent of reaction before deciding whether an alcohol should remain in the answer.
Stable carbocations can override the ordinary less-hindered choice
A protonated ether containing a tertiary carbon can lose neutral alcohol to form a tertiary carbocation. Capture by iodide gives the tertiary iodide in the standard cleavage representation. For (CH₃)₃COCH₃, the first products are (CH₃)₃CI and CH₃OH. This SN1 pathway differs from attacking the methyl end of an ordinary primary ether. Elimination from the cation can compete, particularly with heating, so a real product distribution depends on the conditions.
A benzylic ether can likewise react at its resonance-stabilised benzyl position. Benzyl ethyl ether, C₆H₅CH₂OC₂H₅, commonly gives C₆H₅CH₂I and ethanol in its initial HI cleavage. Here the benzyl group has more carbons than ethyl yet forms the halide. This is why the phrase “the smaller group becomes the halide” is inadequate without identifying the substrate and reaction pathway.
Problem
Predict first-stage HI cleavage of ethyl 2-methylbutyl ether, propyl 3-methylpentan-3-yl ether and benzyl phenyl ether.
- 1.Ethyl 2-methylbutyl ether is C₂H₅OCH₂CH(CH₃)CH₂CH₃. Both oxygen-bound carbons are primary, but ethyl is less hindered than the branched-side primary group; the conventional products are C₂H₅I and HOCH₂CH(CH₃)CH₂CH₃.
- 2.In CH₃CH₂CH₂O–C(CH₃)(CH₂CH₃)₂, the oxygen-bound carbon on the right is tertiary. Carbocation cleavage gives 3-iodo-3-methylpentane and propan-1-ol in the standard substitution outcome.
- 3.Benzyl phenyl ether, C₆H₅CH₂OC₆H₅, cleaves at benzyl–O to give benzyl iodide and phenol. The aryl–O bond is retained; distinguish benzyl CH₂ from a ring carbon.
An aryl C–O bond is not an ordinary SN2 site
Anisole is C₆H₅OCH₃. After protonation, iodide attacks methyl carbon; it does not displace oxygen from the aromatic sp² carbon. The aryl–O bond also has conjugative partial double-bond character. Products are phenol and methyl iodide. Phenol does not then undergo ordinary HI substitution to iodobenzene, so even excess acid does not justify writing two iodides by the dialkyl formula. Diaryl ethers are correspondingly resistant to this ordinary cleavage route.
The aromatic ring of anisole remains activated
Anisole has no OH proton to lose, yet its oxygen can donate a lone pair to the aromatic ring. The methoxy group is therefore activating and ortho/para directing, like OH in the relevant ring substitutions. Resonance contributors place positive charge on oxygen and enhanced electron density at ortho/para ring positions. Meta attack lacks the corresponding stabilising oxygen-donation contributor in the substitution intermediate.
Which part reacts depends on reagent. HI targets the ether linkage after oxygen protonation. Bromine in ethanoic acid targets aromatic C–H positions. Concentrated HNO₃/H₂SO₄ nitrates the ring under the stated conditions. A prediction that always cleaves the ether whenever an acid appears ignores the specific reagent system and reaction pathway.
| Reaction | Reagents/conditions | Products and retained connectivity |
|---|---|---|
| Bromination | Br₂ in ethanoic acid, without FeBr₃ | 2- and 4-bromoanisole; para generally major; OCH₃ retained |
| Nitration | Concentrated HNO₃/H₂SO₄ | 2- and 4-nitroanisole; para usually major in the stated comparison |
| Friedel–Crafts alkylation | CH₃Cl, anhydrous AlCl₃ in an appropriate dry medium, conventionally CS₂ | 2- and 4-methylanisole, para generally major |
| Friedel–Crafts acetylation | CH₃COCl, anhydrous AlCl₃ | 2- and 4-methoxyacetophenone; para generally major |
In Friedel–Crafts alkylation, AlCl₃ activates the alkyl halide toward electrophilic attack by the ring. In acylation, activation of the acyl chloride gives an acyl electrophile, conveniently represented as an acylium ion. Ring attack forms a sigma complex and loss of a ring proton restores aromaticity. The introduced acetyl group is bonded directly through its carbonyl carbon, Ar–C(=O)CH₃; the original methoxy group remains Ar–OCH₃. Do not confuse ring acetylation of anisole with oxygen acetylation of phenol.
Problem
Give the principal para-connectivity products from anisole bromination, methylation and acetylation.
- 1.Bromination gives 1-bromo-4-methoxybenzene, commonly 4-bromoanisole. The new bond is ring C–Br.
- 2.Methylation gives 1-methoxy-4-methylbenzene, commonly 4-methylanisole. The new bond is ring C–CH₃.
- 3.Acetylation gives 1-(4-methoxyphenyl)ethan-1-one, commonly 4-methoxyacetophenone. The new bond is ring C–C(=O)CH₃, and the methoxy oxygen is not the acylation site in this named ring reaction.
- 4.Para preference is commonly explained by less crowding than ortho, while electronic activation permits both. An exact percentage such as a reported 90% para yield belongs to particular bromination conditions, not every substitution.
Quick Quiz
Quiz
First-stage HI cleavage of anisole gives which pair?
Explanation: Iodide attacks methyl carbon while the aryl–O bond remains.
What can excess hot HI do after dialkyl ether cleavage?
Explanation: The initial alcohol can undergo a separate substitution; aryl phenol is an important limit.
Which group commonly forms the halide in initial tert-butyl methyl ether cleavage?
Explanation: Tertiary carbocation formation enables cleavage on the tertiary side.
What is the role of anhydrous AlCl₃ in anisole acetylation?
Explanation: It is a Lewis acid used to generate/activate the acyl electrophile.
Which new bond forms in Friedel–Crafts acetylation of anisole?
Explanation: The introduced group is –C(=O)CH₃ attached directly to the aromatic carbon.
Practice Problems
- Write the initial and excess-HI products for methoxymethane, including the mechanism.
- Compare initial HI cleavage of ethoxybenzene and benzyl ethyl ether.
- Predict first-stage HBr cleavage of methyl propyl ether and qualify further reaction.
- Predict initial HI cleavage of tert-butyl ethyl ether and mention a competing pathway.
- Write reagents and para products for anisole bromination, nitration and acetylation.
- Explain why anisole is ortho/para directing despite having no OH proton.
- A student predicts iodobenzene from anisole with excess HI. Explain the error.
Protonation gives CH₃–O⁺(H)–CH₃. I⁻ attacks methyl by SN2 and expels methanol: CH₃OCH₃ + HI → CH₃I + CH₃OH. Excess hot HI converts methanol to another CH₃I plus water. Net CH₃OCH₃ + 2HI → 2CH₃I + H₂O. The first products and final forcing-condition products should not be conflated.
Key Takeaways
• Strong concentrated HI/HBr activates ethers by oxygen protonation. • Ordinary dialkyl cleavage commonly uses SN2 attack at a less hindered carbon. • Tertiary and benzylic groups can alter the pathway through stabilised cationic centres. • One equivalent and excess hot acid can give different final product sets. • Aryl alkyl ethers cleave on the alkyl side and leave phenol. • Methoxy activates anisole toward ortho/para aromatic substitution. • Bromination, nitration and Friedel–Crafts reactions retain the ether linkage under their stated conditions. • Ring acetylation makes Ar–C(=O)CH₃ and differs from oxygen acetylation.