What is an ether? Naming and structure
An ether has an oxygen atom joined to two carbon groups: R–O–R′. If both groups are the same it is symmetrical (C₂H₅OC₂H₅); if different, unsymmetrical or mixed (CH₃OC₂H₅).
Common name: name both groups alphabetically + 'ether' – ethyl methyl ether. IUPAC name: the smaller group with O becomes the alkoxy prefix on the larger alkane – methoxyethane, ethoxyethane, methoxybenzene (anisole).
Structure: O has two bonds and two lone pairs. The C–O–C angle is slightly more than 109°28′ (about 111.7° in methoxymethane) because the two carbon groups push each other apart.
Preparation of ethers
1. Dehydration of alcohols
2 C₂H₅OH → (conc. H₂SO₄, 413 K) C₂H₅OC₂H₅ + H₂O. Steps: (i) one alcohol takes H⁺ and becomes R–OH₂⁺, (ii) a second alcohol's O attacks that carbon and water leaves (SN2), (iii) H⁺ is lost. Temperature matters: at 443 K ethene forms instead. Works for primary alcohols and gives symmetrical ethers; 2° and 3° alcohols mostly give alkenes.
2. Williamson synthesis
R–X + R′–O⁻Na⁺ → R–O–R′ + NaX. The alkoxide ion attacks the carbon holding the halogen from the back (SN2). This works for both symmetrical and unsymmetrical ethers, and for aryl ethers (phenoxide + alkyl halide → anisole type).
Choose the pair wisely: the alkyl halide must be primary. With a tertiary halide the strong base pulls off a H and an alkene forms. So for (CH₃)₃C–O–CH₃ use (CH₃)₃CO⁻Na⁺ + CH₃Br, not (CH₃)₃CBr + CH₃O⁻Na⁺. Aryl halides cannot be used as the halide part.
Physical properties of ethers
The C–O bonds are polar and the bent shape gives ethers a small dipole moment. But ether molecules have no H on O, so they cannot form hydrogen bonds with one another. So their boiling points are low – close to alkanes of similar mass and far below alcohols (ethoxyethane 308 K vs butan-1-ol 391 K).
Solubility: the O of an ether can accept an H-bond from water, so small ethers dissolve to about the same extent as alcohols of similar mass (ethoxyethane about 7.5 g per 100 g water). Bigger ethers are nearly insoluble.
Chemical properties of ethers
A. Cleavage of C–O bond by HX
Ethers are quite unreactive, but hot conc. HI (or HBr) breaks them: R–O–R + HX → R–X + R–OH; with excess HX the alcohol also becomes R–X.
- Step 1: H⁺ joins the ether O (protonation).
- Step 2: I⁻ attacks. With two primary/methyl groups it attacks the less crowded carbon (SN2): CH₃–O–C₂H₅ + HI → CH₃I + C₂H₅OH.
- If one group is tertiary, that C–O bond breaks by SN1 and the tertiary group becomes the iodide: (CH₃)₃C–O–CH₃ + HI → (CH₃)₃C–I + CH₃OH.
- Anisole + HI → phenol + CH₃I. The ring–O bond has partial double-bond character, so it does not break.
Reactivity of HX: HI > HBr > HCl.
B. Electrophilic substitution in aromatic ethers
–OCH₃ pushes electrons into the ring and directs to ortho and para. Bromination in ethanoic acid (no FeBr₃ needed) → mainly p-bromoanisole. Friedel–Crafts with CH₃Cl or CH₃COCl (AlCl₃) → o- and p- products. Nitration (conc. HNO₃ + H₂SO₄) → 2- and 4-nitroanisole.
C. Safety
On standing in air and light, ethers slowly form peroxides, which can explode on distillation. Store in dark bottles.
Key formulas and definitions
- General: R–O–R′; CₙH₂ₙ₊₂O (same as alcohols – they are functional isomers)
- 2 C₂H₅OH (conc. H₂SO₄, 413 K) → C₂H₅OC₂H₅ + H₂O
- Williamson: R–X + R′ONa → R–O–R′ + NaX (R–X primary)
- CH₃OC₂H₅ + HI → CH₃I + C₂H₅OH
- (CH₃)₃COCH₃ + HI → (CH₃)₃CI + CH₃OH
- C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I
- HX reactivity: HI > HBr > HCl
Worked examples
1. Give the IUPAC name of CH₃–O–CH₂CH₂CH₃.
The smaller group CH₃–O– becomes 'methoxy'; the bigger chain has 3 C (propane) and O is on C1. Name: 1-methoxypropane.
2. Write the best Williamson route to 2-methoxy-2-methylpropane, (CH₃)₃C–O–CH₃.
Two choices: (a) (CH₃)₃C–Br + CH₃ONa, (b) CH₃Br + (CH₃)₃CONa. In (a) the tertiary halide with a strong base gives 2-methylpropene (elimination). In (b) the halide is methyl, so SN2 works. Use (b).
3. Predict the products of C₂H₅–O–CH(CH₃)₂ with one mole of hot HI.
Both sides are not tertiary, so SN2 on the less crowded carbon. Ethyl (1°) is less crowded than isopropyl (2°). I⁻ attacks ethyl: products C₂H₅I + (CH₃)₂CHOH.
4. How many grams of ethoxyethane (M = 74 g/mol) can be made from 92 g of ethanol (M = 46 g/mol)?
2 C₂H₅OH → C₂H₅OC₂H₅ + H₂O. Moles of ethanol = 92/46 = 2 mol → 1 mol ether → 74 g.
5. Anisole (M = 108 g/mol, 10.8 g) is heated with excess HI. What mass of phenol (M = 94 g/mol) forms?
C₆H₅OCH₃ + HI → C₆H₅OH + CH₃I. Moles of anisole = 10.8/108 = 0.1 mol → 0.1 mol phenol = 9.4 g.
6. Why does ethoxyethane (M = 74) boil at 308 K while butan-1-ol (M = 74) boils at 391 K?
Same mass, so the difference comes from forces between molecules. Butan-1-ol has O–H and forms hydrogen bonds between molecules; ethoxyethane has no O–H, so it has only weak dipole and dispersion forces. Less energy is needed to separate ether molecules.
Common mistakes
- Using a tertiary alkyl halide in Williamson synthesis – it gives an alkene, not an ether.
- Thinking anisole + HI gives iodobenzene. The ring–O bond does not break; products are phenol and CH₃I.
- Saying ethers have high boiling points like alcohols. They cannot H-bond to each other, so boiling points are low.
- Heating ethanol with conc. H₂SO₄ at 443 K and calling the product ether – at 443 K it is ethene; ether forms at 413 K.