Chapter 7 · Alcohols, Phenols and Ethers

7.3 Ethers

Ethers are made by dehydrating alcohols or, more generally, by the Williamson synthesis, where the choice of alkyl halide decides between substitution and elimination. This section covers their physical properties, their cleavage by hydrogen halides (which bond breaks, and why), and electrophilic substitution in anisole.

On 16 October 1846, in a Boston operating theatre, a dentist held a glass globe soaked in diethyl ether to a patient's face, and a surgeon removed a tumour from his neck while he slept. Within months ether anaesthesia had spread around the world. Ethoxyethane is no longer used for that purpose, but ethers are everywhere in the laboratory and in industry: as solvents that dissolve almost anything organic yet do not react with most reagents, as the fragrant methoxybenzene (anisole) of perfumes, and, for a time, as the petrol additive MTBE, 2-methoxy-2-methylpropane.

An ether has no $\mathrm{O{-}H}$ group, so it cannot act as an acid or form hydrogen bonds with itself, and its C–O bonds are hard to break. That chemical calm is exactly what makes ethers good solvents. In this section we see how ethers are made, why their physical properties resemble those of alkanes more than those of alcohols, how the C–O bond can be cleaved by hot concentrated hydrogen iodide, and how the $\mathrm{{-}OCH_{3}}$ group of anisole steers electrophilic substitution.

Preparing Ethers

📐 Rule 7.15 — Ethers by dehydration of alcohols

Primary alcohols heated with concentrated sulphuric acid at about $413\ \mathrm{K}$, with the alcohol in excess, give symmetrical ethers:

$$\mathrm{2C_{2}H_{5}OH\xrightarrow[413\ K]{conc.\ H_{2}SO_{4}}C_{2}H_{5}OC_{2}H_{5}+H_{2}O}$$

Mechanism (a nucleophilic substitution, $\mathrm{S_{N}2}$). Step 1: the alcohol is protonated, $\mathrm{CH_{3}CH_{2}OH+H^{+}\rightleftharpoons CH_{3}CH_{2}\overset{+}{O}H_{2}}$. Step 2: the oxygen of a second alcohol molecule attacks the $\mathrm{CH_{2}}$ carbon of the protonated alcohol, pushing out water in a single step:

$$\mathrm{C_{2}H_{5}OH+CH_{3}CH_{2}\overset{+}{O}H_{2}}$$ $$\mathrm{\rightarrow C_{2}H_{5}\overset{+}{O}(H)CH_{2}CH_{3}+H_{2}O}$$

Step 3: the protonated ether loses $\mathrm{H^{+}}$.

Limits. (i) The temperature must be controlled: at $443\ \mathrm{K}$ elimination wins and ethanol gives ethene (Section 7.2). (ii) The method suits primary alcohols only: secondary and tertiary alcohols, which form carbocations easily, give alkenes instead. (iii) It makes symmetrical ethers; a mixture of two alcohols gives a mixture of three ethers.

📘 Definition 7.6 — Williamson synthesis

In the Williamson synthesis an alkyl halide reacts with a sodium alkoxide (or phenoxide); the alkoxide ion is the nucleophile and displaces the halide ion in an $\mathrm{S_{N}2}$ reaction:

$$\mathrm{R{-}X+R'{-}O^{-}Na^{+}\rightarrow R{-}O{-}R'+Na^{+}X^{-}}$$

It makes symmetrical and unsymmetrical ethers, including ethers with a tertiary alkyl group. The rules for choosing the pair are:

  • The alkyl halide should be methyl or primary (benzylic and allylic primary halides work very well). The alkoxide may be primary, secondary or tertiary.
  • With a secondary or tertiary halide the alkoxide, which is also a strong base, removes a β-hydrogen instead and elimination gives an alkene (Chapter 6): $$\mathrm{(CH_{3})_{3}CBr+CH_{3}ONa}$$ $$\mathrm{\rightarrow(CH_{3})_{2}C{=}CH_{2}+CH_{3}OH+NaBr}$$
  • Alkyl aryl ethers are made from a phenoxide and an alkyl halide, $\mathrm{C_{6}H_{5}ONa+CH_{3}I\rightarrow C_{6}H_{5}OCH_{3}+NaI}$, never from an aryl halide and an alkoxide: aryl halides do not undergo $\mathrm{S_{N}2}$ substitution.
🧭 Problem-solving strategy — planning a Williamson synthesis
  1. Draw the target and look at the two C–O bonds. Each can be the bond made in the synthesis.
  2. For each choice, the carbon group on the "cut" side must come from the halide; the oxygen stays with the other group, as the alkoxide.
  3. Classify the carbon that would carry the halogen: methyl or 1° ✓; 2° poor (much alkene); 3° ✗ (alkene only); aryl or vinylic ✗ (no reaction).
  4. Choose the route whose halide is methyl or primary. If both are, either works.
✏️ Example 1 — Choosing the halide and the alkoxide
Suggest the best Williamson synthesis of (a) 1-ethoxypropane; (b) 2-methoxy-2-methylpropane; (c) 2-methoxypropane; (d) methoxybenzene. (e) Why can diphenyl ether, $\mathrm{C_{6}H_{5}OC_{6}H_{5}}$, not be made this way?
(a) Both groups are primary, so either route works: $\mathrm{CH_{3}CH_{2}CH_{2}Br+C_{2}H_{5}ONa}$, or $\mathrm{C_{2}H_{5}Br+CH_{3}CH_{2}CH_{2}ONa}$.
(b) The tertiary group must be the alkoxide: $$\mathrm{(CH_{3})_{3}CONa+CH_{3}Br\rightarrow(CH_{3})_{3}COCH_{3}+NaBr}$$ The other pair, $\mathrm{(CH_{3})_{3}CBr+CH_{3}ONa}$, gives 2-methylpropene by elimination.
(c) Use sodium propan-2-olate and iodomethane: $$\mathrm{(CH_{3})_{2}CHONa+CH_{3}I\rightarrow(CH_{3})_{2}CHOCH_{3}+NaI}$$ With 2-bromopropane and sodium methoxide, much of the secondary halide would be eliminated to propene.
(d) $\mathrm{C_{6}H_{5}ONa+CH_{3}I\rightarrow C_{6}H_{5}OCH_{3}+NaI}$ (sodium phenoxide is made from phenol and $\mathrm{NaOH}$). Bromobenzene with sodium methoxide does not react.
(e) Either route needs an aryl halide as the substrate, and aryl halides do not undergo $\mathrm{S_{N}2}$ reactions (the C–X bond has partial double-bond character and the back of the carbon is shielded by the ring).
Interactive 1 — Williamson Planner: Which Bond Should You Make?

Pick a target ether, then pick which C–O bond to make (the red cut). The group on the cut side must come from the alkyl halide; the oxygen stays with the other group as the sodium alkoxide. The bars show, qualitatively, how the alkoxide shares its effort between substitution ($\mathrm{S_{N}2}$, giving the ether) and elimination (giving an alkene) for that kind of halide. Play runs through every target and route.

sodium alkoxide—
alkyl / aryl halide—
halide carbon is—
main product—
✏️ Example 2 — Dehydration or Williamson?
Which of these ethers can be made in good yield by heating the corresponding alcohol(s) with concentrated $\mathrm{H_{2}SO_{4}}$, and which need the Williamson synthesis? Explain. (a) ethoxyethane; (b) 1-methoxypropane; (c) 2-(tert-butoxy)-2-methylpropane, $\mathrm{(CH_{3})_{3}COC(CH_{3})_{3}}$; (d) dipropyl ether, $\mathrm{(CH_{3}CH_{2}CH_{2})_{2}O}$.
(a) Dehydration of ethanol at $413\ \mathrm{K}$ works: a symmetrical ether from a primary alcohol.
(b) Unsymmetrical: heating methanol with propan-1-ol gives a mixture of methoxymethane, 1-methoxypropane and 1-propoxypropane. Use Williamson: $\mathrm{CH_{3}CH_{2}CH_{2}ONa+CH_{3}I}$ (or $\mathrm{CH_{3}ONa+CH_{3}CH_{2}CH_{2}Br}$).
(c) Neither works well. 2-Methylpropan-2-ol with acid gives 2-methylpropene; in a Williamson synthesis one partner would have to be a tertiary halide, which eliminates. (This very crowded ether is made by other methods.)
(d) Dehydration of propan-1-ol (a primary alcohol) gives 1-propoxypropane, with careful temperature control; Williamson ($\mathrm{CH_{3}CH_{2}CH_{2}ONa+CH_{3}CH_{2}CH_{2}Br}$) also works.

Physical Properties of Ethers

📐 Rule 7.16 — Physical properties of ethers
  • Polarity. The C–O bonds are polar and the molecule is bent ($\mathrm{C{-}O{-}C}$ about $110^{\circ}$), so ethers have a small net dipole moment; methoxymethane is about $1.3\ \mathrm{D}$.
  • Boiling points are close to those of alkanes of similar molar mass and far below those of alcohols, because ether molecules cannot hydrogen-bond to one another: pentane about $309\ \mathrm{K}$, ethoxyethane about $308\ \mathrm{K}$, butan-1-ol about $391\ \mathrm{K}$ (all $M\approx72$–$74$).
  • Solubility in water is similar to that of alcohols of the same molar mass, because the ether oxygen can accept a hydrogen bond from water: ethoxyethane and butan-1-ol both dissolve to about $7$–$9\ \mathrm{g}$ per $100\ \mathrm{g}$ of water. Lower ethers are fairly soluble; solubility falls as the carbon chains grow.
  • As solvents, ethers dissolve most organic compounds and are unreactive towards bases, dilute acids, oxidising and reducing agents; dry ethoxyethane is the usual solvent for Grignard reagents. They are very volatile and highly flammable, and on standing in air they slowly form explosive peroxides.

Chemical Reactions of Ethers

Ethers are among the least reactive organic compounds. Their C–O bond is broken only under drastic conditions, by heating with a concentrated hydrogen halide, and the question of which C–O bond breaks has a neat answer.

📐 Rule 7.17 — Cleavage of the C–O bond by hydrogen halides

Dialkyl ethers heated with concentrated $\mathrm{HI}$ (or $\mathrm{HBr}$) give an alkyl halide and an alcohol; with excess acid the alcohol is also converted into the halide:

$$\mathrm{R{-}O{-}R+HX\rightarrow R{-}X+R{-}OH}$$ $$\mathrm{R{-}OH+HX\rightarrow R{-}X+H_{2}O}$$

Reactivity: $\mathrm{HI\gt HBr\gt HCl}$ (the order of acid strength and of nucleophilic power of $\mathrm{X^{-}}$). Mechanism. The ether oxygen is first protonated, $\mathrm{R{-}\overset{+}{O}(H){-}R'}$, which turns $\mathrm{R'OH}$ into a good leaving group. Then:

  • if both groups are methyl, primary or secondary, $\mathrm{I^{-}}$ attacks the less hindered carbon by $\mathrm{S_{N}2}$: $$\mathrm{CH_{3}OCH_{2}CH_{3}+HI\rightarrow CH_{3}I+CH_{3}CH_{2}OH}$$
  • if one group is tertiary, the protonated ether ionises by $\mathrm{S_{N}1}$ to the stable tertiary carbocation, and the halide ends up on the tertiary carbon: $\mathrm{(CH_{3})_{3}COCH_{3}+HI\rightarrow(CH_{3})_{3}CI+CH_{3}OH}$;
  • in an alkyl aryl ether the aryl–O bond is never broken (partial double-bond character, $sp^{2}$ carbon, no aryl $\mathrm{S_{N}}$ reaction), so the products are a phenol and an alkyl halide: $\mathrm{C_{6}H_{5}OCH_{3}+HI\rightarrow C_{6}H_{5}OH+CH_{3}I}$. Excess $\mathrm{HI}$ does not convert the phenol into iodobenzene.
✏️ Example 3 — Which bond breaks?
Give the products of heating each ether with one mole of concentrated $\mathrm{HI}$: (a) 1-methoxybutane; (b) 2-methoxy-2-methylpropane; (c) ethoxybenzene; (d) ethoxyethane. (e) What does (d) give with excess $\mathrm{HI}$?
(a) Methyl is the less hindered group; $\mathrm{S_{N}2}$ at $\mathrm{CH_{3}}$: iodomethane + butan-1-ol.
(b) One group is tertiary; $\mathrm{S_{N}1}$ through $\mathrm{(CH_{3})_{3}C^{+}}$: 2-iodo-2-methylpropane + methanol.
(c) The aryl–O bond survives: phenol + iodoethane.
(d) Symmetrical: iodoethane + ethanol.
(e) The ethanol reacts too: $\mathrm{C_{2}H_{5}OC_{2}H_{5}+2HI\rightarrow2C_{2}H_{5}I+H_{2}O}$.
Interactive 2 — Ether Splitter: Where Does the Iodide Go?

Choose an ether and the amount of $\mathrm{HI}$. The lab first protonates the oxygen, then shows which C–O bond breaks (red) and by which mechanism: $\mathrm{S_{N}2}$ attack by $\mathrm{I^{-}}$ on the less hindered carbon, or $\mathrm{S_{N}1}$ through a tertiary carbocation. The aryl–O bond (green) never breaks. Press Play to cycle through protonation, cleavage and products.

stage—
bond broken—
mechanism—
products—

Aromatic ethers. In anisole the $\mathrm{{-}OCH_{3}}$ group, like $\mathrm{{-}OH}$, releases electrons into the ring by resonance (+R), so it activates the ring and directs incoming electrophiles to the ortho and para positions. The para product is usually the main one, because the ortho positions are crowded by the methoxy group.

📐 Rule 7.18 — Electrophilic substitution in anisole

(a) Halogenation. Bromine in ethanoic acid, even without an iron(III) bromide catalyst, gives mainly 4-bromoanisole (1-bromo-4-methoxybenzene) with some 2-bromoanisole.

(b) Friedel–Crafts reactions with anhydrous $\mathrm{AlCl_{3}}$: alkylation with $\mathrm{CH_{3}Cl}$ gives 4-methoxytoluene (1-methoxy-4-methylbenzene, main) and 2-methoxytoluene; acylation with $\mathrm{CH_{3}COCl}$ gives 4-methoxyacetophenone, $\mathrm{CH_{3}OC_{6}H_{4}COCH_{3}}$, i.e. 1-(4-methoxyphenyl)ethan-1-one (main), and 2-methoxyacetophenone.

(c) Nitration with a mixture of concentrated $\mathrm{H_{2}SO_{4}}$ and $\mathrm{HNO_{3}}$ gives 4-nitroanisole (1-methoxy-4-nitrobenzene, main) and 2-nitroanisole.

✏️ Example 4 — Reactions of anisole
Give the main organic product, with its IUPAC name, of anisole with: (a) bromine in ethanoic acid; (b) chloromethane and anhydrous $\mathrm{AlCl_{3}}$; (c) ethanoyl chloride and anhydrous $\mathrm{AlCl_{3}}$; (d) the nitrating mixture; (e) hot concentrated $\mathrm{HI}$. Why is no Lewis acid needed in (a)?
(a) 1-bromo-4-methoxybenzene (4-bromoanisole), with some 2-bromoanisole.
(b) 1-methoxy-4-methylbenzene (4-methoxytoluene), with some 1-methoxy-2-methylbenzene.
(c) 1-(4-methoxyphenyl)ethan-1-one (4-methoxyacetophenone), with some of the 2-isomer.
(d) 1-methoxy-4-nitrobenzene (4-nitroanisole), with some 1-methoxy-2-nitrobenzene.
(e) Not a ring substitution: the $\mathrm{CH_{3}{-}O}$ bond is cleaved, giving phenol and iodomethane.
Why The methoxy group makes the ring so electron-rich that $\mathrm{Br_{2}}$, slightly polarised by the polar solvent, is a strong enough electrophile; benzene itself needs $\mathrm{FeBr_{3}}$.
✏️ Example 5 — Identify the ether
(a) An ether $\mathrm{C_{4}H_{10}O}$ gives only one organic product, an alkyl iodide, when heated with excess $\mathrm{HI}$. Identify it. (b) An isomeric ether gives iodomethane and a secondary alcohol with one mole of $\mathrm{HI}$. Identify it. (c) An ether $\mathrm{C_{7}H_{8}O}$ is insoluble in aqueous $\mathrm{NaOH}$; heated with $\mathrm{HI}$ it gives a compound that is soluble in $\mathrm{NaOH}$ and gives a white precipitate with bromine water. Identify the ether and both products.
(a) The three ethers of formula $\mathrm{C_{4}H_{10}O}$ are 1-methoxypropane, 2-methoxypropane and ethoxyethane. Only a symmetrical ether gives a single iodide: ethoxyethane, giving iodoethane.
(b) The secondary alcohol must be propan-2-ol, so the ether is 2-methoxypropane, $\mathrm{CH_{3}OCH(CH_{3})_{2}}$: $\mathrm{I^{-}}$ attacks the methyl carbon by $\mathrm{S_{N}2}$.
(c) $\mathrm{C_{7}H_{8}O}$, not acidic, cleaved to a phenol: methoxybenzene (anisole). Products: phenol (dissolves in $\mathrm{NaOH}$, gives 2,4,6-tribromophenol with bromine water) and iodomethane.
✏️ Example 6 — How much product?
(a) $1.08\ \mathrm{g}$ of anisole is heated with excess concentrated $\mathrm{HI}$. What masses of phenol and iodomethane form if the reaction is complete? (b) In a Williamson synthesis, $2.30\ \mathrm{g}$ of sodium is dissolved in excess ethanol and the sodium ethoxide formed is treated with excess bromoethane. What is the maximum mass of ethoxyethane? ($M$: C 12, H 1, O 16, Na 23, I 127.)
(a) $M(\mathrm{C_{7}H_{8}O})=108$: $0.0100\ \mathrm{mol}$. Phenol ($94$): $0.94\ \mathrm{g}$; iodomethane, $\mathrm{CH_{3}I}$ ($142$): $1.42\ \mathrm{g}$.
(b) Moles $2.30\ \mathrm{g}$ of $\mathrm{Na}$ is $0.100\ \mathrm{mol}$; $\mathrm{2C_{2}H_{5}OH+2Na\rightarrow2C_{2}H_{5}ONa+H_{2}}$ gives $0.100\ \mathrm{mol}$ of ethoxide.
(b) Mass $\mathrm{C_{2}H_{5}ONa+C_{2}H_{5}Br\rightarrow C_{2}H_{5}OC_{2}H_{5}+NaBr}$: $0.100\ \mathrm{mol}\times74\ \mathrm{g\,mol^{-1}}=7.4\ \mathrm{g}$.
Check Mass balance in (a): $1.08+0.0100\times128\ (\mathrm{HI})$ $=2.36\ \mathrm{g}$ $=0.94+1.42$ ✓.
⚠️ Common mistakes with ethers

1. Williamson with the wrong halide. $\mathrm{(CH_{3})_{3}CBr+CH_{3}ONa}$ gives 2-methylpropene, not the ether. Put the bulky group on the alkoxide, the small group on the halide.

2. Aryl halides in Williamson. $\mathrm{C_{6}H_{5}Br+CH_{3}ONa}$ does not give anisole; use $\mathrm{C_{6}H_{5}ONa+CH_{3}I}$.

3. HI cleavage of anisole. The products are phenol and iodomethane, never iodobenzene and methanol.

4. Mixed methyl ethers. With $\mathrm{HI}$, methyl primary/secondary ethers give $\mathrm{CH_{3}I}$ ($\mathrm{S_{N}2}$ at the less hindered carbon), but methyl tertiary ethers give the tertiary iodide and methanol ($\mathrm{S_{N}1}$).

5. "Ethers are insoluble in water because they cannot hydrogen-bond." They cannot hydrogen-bond to each other, but their oxygen accepts hydrogen bonds from water, so the lower ethers are fairly soluble.

🚀 Beyond the board syllabus (enrichment only) — epoxides: ethers under strain

Enrichment only: no 2026 syllabus names epoxides (JEE Advanced lists C–O cleavage of ethers in general); not in the CBSE syllabus, JEE Main or NEET.

When the oxygen of an ether is part of a three-membered ring, as in oxirane (ethylene oxide, $\mathrm{C_{2}H_{4}O}$), the ring angles are about $60^{\circ}$ instead of $109.5^{\circ}$. This ring strain makes epoxides far more reactive than ordinary ethers: they are opened by water with a trace of acid, giving ethane-1,2-diol (the antifreeze of Section 7.1, made this way industrially), and by nucleophiles such as alkoxides, amines and Grignard reagents. Grignard reagents with oxirane give primary alcohols with two extra carbons, $\mathrm{RMgX+C_{2}H_{4}O\rightarrow RCH_{2}CH_{2}OMgX}$, which hydrolysis turns into $\mathrm{RCH_{2}CH_{2}OH}$. Five- and six-membered cyclic ethers (oxolane, also called tetrahydrofuran, and 1,4-dioxane) have no strain and are used as unreactive solvents.

Exercises

I
Preparing ethers.
  • 1 Write the equation and conditions for the preparation of ethoxyethane from ethanol. What forms instead at $443\ \mathrm{K}$?
  • 2 Write the mechanism of the formation of ethoxyethane from ethanol and concentrated sulphuric acid.
  • 3 Why is acid-catalysed dehydration unsuitable for making (a) unsymmetrical ethers; (b) ethers from secondary or tertiary alcohols?
  • 4 Give the best Williamson synthesis for: (a) 1-methoxybutane; (b) 2-ethoxy-2-methylpropane; (c) ethoxybenzene; (d) (ethoxymethyl)benzene, $\mathrm{C_{6}H_{5}CH_{2}OC_{2}H_{5}}$.
  • 5 Predict the main organic product, and name the reaction type, for: (a) $\mathrm{CH_{3}CH_{2}Br+(CH_{3})_{3}CONa}$; (b) $\mathrm{(CH_{3})_{3}CBr+CH_{3}CH_{2}ONa}$.
  • 6 2-Bromopropane and sodium ethoxide give some 2-ethoxypropane but mostly another product. Name it, and suggest a better Williamson route to 2-ethoxypropane.
  • 7 How would you make methoxybenzene from phenol? Write the equations.
  • 8 Explain why the Williamson synthesis cannot give diphenyl ether, $\mathrm{(C_{6}H_{5})_{2}O}$.
1 $\mathrm{2C_{2}H_{5}OH\xrightarrow[413\ K]{conc.\ H_{2}SO_{4}}C_{2}H_{5}OC_{2}H_{5}+H_{2}O}$ (excess ethanol). At $443\ \mathrm{K}$ elimination gives ethene.
2 (1) $\mathrm{C_{2}H_{5}OH+H^{+}\rightleftharpoons C_{2}H_{5}\overset{+}{O}H_{2}}$; (2) a second $\mathrm{C_{2}H_{5}OH}$ attacks the $\mathrm{CH_{2}}$ carbon ($\mathrm{S_{N}2}$) and water leaves, giving $\mathrm{C_{2}H_{5}\overset{+}{O}(H)C_{2}H_{5}}$; (3) loss of $\mathrm{H^{+}}$ gives ethoxyethane.
3 (a) Two different alcohols give a mixture of three ethers (two symmetrical, one mixed). (b) Secondary and tertiary alcohols form carbocations readily under these conditions and eliminate to alkenes.
4 (a) $\mathrm{CH_{3}(CH_{2})_{3}ONa+CH_{3}I}$ (or $\mathrm{CH_{3}ONa+CH_{3}(CH_{2})_{3}Br}$); (b) $\mathrm{(CH_{3})_{3}CONa+C_{2}H_{5}Br}$; (c) $\mathrm{C_{6}H_{5}ONa+C_{2}H_{5}Br}$; (d) $\mathrm{C_{6}H_{5}CH_{2}Br+C_{2}H_{5}ONa}$ or $\mathrm{C_{6}H_{5}CH_{2}ONa+C_{2}H_{5}Br}$ (both halides are primary).
5 (a) Substitution ($\mathrm{S_{N}2}$) on a primary halide: 2-ethoxy-2-methylpropane. (b) Elimination on a tertiary halide: 2-methylpropene (with ethanol and $\mathrm{NaBr}$).
6 Propene, by elimination (the secondary halide is crowded and the ethoxide is a strong base). Better: sodium propan-2-olate with bromoethane, $\mathrm{(CH_{3})_{2}CHONa+C_{2}H_{5}Br}$.
7 $\mathrm{C_{6}H_{5}OH+NaOH\rightarrow C_{6}H_{5}ONa+H_{2}O}$; then $\mathrm{C_{6}H_{5}ONa+CH_{3}I\rightarrow C_{6}H_{5}OCH_{3}+NaI}$.
8 One phenyl group would have to come from an aryl halide such as bromobenzene, and aryl halides do not undergo $\mathrm{S_{N}2}$ substitution.
II
Physical properties. (Approximate boiling points: pentane $309\ \mathrm{K}$, ethoxyethane $308\ \mathrm{K}$, butan-1-ol $391\ \mathrm{K}$, methoxymethane $249\ \mathrm{K}$, ethanol $351\ \mathrm{K}$.)
  • 1 Explain why ethoxyethane and pentane have almost the same boiling point, but butan-1-ol boils about $80\ \mathrm{K}$ higher.
  • 2 Methoxymethane and ethanol are isomers. Explain the difference of about $100\ \mathrm{K}$ in their boiling points.
  • 3 Explain why ethoxyethane is about as soluble in water as butan-1-ol, although it has no $\mathrm{O{-}H}$ group.
  • 4 Ethers have a dipole moment although they contain no very polar bond such as $\mathrm{O{-}H}$. Explain.
  • 5 Give two reasons why ethoxyethane is a good solvent for Grignard reagents and for extracting organic compounds from water.
  • 6 Safety. Give two hazards of working with ethoxyethane in the laboratory.
1 Ether molecules, like alkane molecules, are held only by dispersion (and weak dipole) forces of similar size; butan-1-ol molecules are also hydrogen-bonded to one another, which costs extra energy to break.
2 Ethanol molecules form $\mathrm{O{-}H\cdots O}$ hydrogen bonds; methoxymethane has no H on oxygen and its molecules attract only weakly.
3 Its oxygen has lone pairs that accept hydrogen bonds from water molecules, and its hydrocarbon part is about the same size as butan-1-ol's.
4 The C–O bonds are polar, and the molecule is bent at oxygen (about $110^{\circ}$), so the two bond dipoles do not cancel.
5 Ethers dissolve most organic compounds but are unreactive towards bases, Grignard reagents and most reagents; the oxygen lone pairs also stabilise the magnesium of a Grignard reagent. Ethoxyethane is immiscible enough with water to form a separate layer and is easy to remove by evaporation (bp about $308\ \mathrm{K}$).
6 It is very volatile and highly flammable (its vapour is heavier than air and can travel to a flame); old samples can contain explosive peroxides. (It is also an anaesthetic when inhaled.)
III
Cleavage by hydrogen halides.
  • 1 Give the products of heating with one mole of concentrated $\mathrm{HI}$: (a) methoxymethane; (b) 1-methoxypropane; (c) 2-methoxybutane.
  • 2 Give the products of 2-ethoxy-2-methylpropane with $\mathrm{HI}$, and the mechanism.
  • 3 Write the steps of the mechanism for $\mathrm{CH_{3}OCH_{2}CH_{3}+HI\rightarrow CH_{3}I+CH_{3}CH_{2}OH}$.
  • 4 Why does anisole give phenol and iodomethane with $\mathrm{HI}$, rather than iodobenzene and methanol?
  • 5 What are the final products of ethoxyethane with excess hot concentrated $\mathrm{HI}$? Write the overall equation.
  • 6 Arrange $\mathrm{HCl}$, $\mathrm{HBr}$ and $\mathrm{HI}$ in order of their ability to cleave ethers, and explain.
  • 7 An ether $\mathrm{C_{5}H_{12}O}$ gives 2-iodo-2-methylpropane and methanol with $\mathrm{HI}$. Identify it.
  • 8 An ether gives iodoethane and phenol with $\mathrm{HI}$. Identify it and write the equation.
1 (a) iodomethane + methanol; (b) iodomethane + propan-1-ol ($\mathrm{S_{N}2}$ at the less hindered methyl carbon); (c) iodomethane + butan-2-ol.
2 One group is tertiary: protonation, then $\mathrm{S_{N}1}$ loss of ethanol to give $\mathrm{(CH_{3})_{3}C^{+}}$, which takes $\mathrm{I^{-}}$. Products: 2-iodo-2-methylpropane and ethanol.
3 (1) $$\mathrm{CH_{3}OC_{2}H_{5}+HI\rightleftharpoons CH_{3}\overset{+}{O}(H)C_{2}H_{5}+I^{-}}$$ (2) $\mathrm{I^{-}}$ attacks the methyl carbon from the back ($\mathrm{S_{N}2}$), and ethanol leaves: $\mathrm{CH_{3}I+CH_{3}CH_{2}OH}$.
4 The aryl–O bond has partial double-bond character (resonance of an O lone pair with the ring) and the ring carbon is $sp^{2}$; neither $\mathrm{S_{N}1}$ (no stable phenyl cation) nor $\mathrm{S_{N}2}$ (backside blocked by the ring) can occur there. The methyl–O bond breaks instead.
5 $\mathrm{C_{2}H_{5}OC_{2}H_{5}+2HI\rightarrow2C_{2}H_{5}I+H_{2}O}$: iodoethane only.
6 $\mathrm{HCl\lt HBr\lt HI}$. $\mathrm{HI}$ is the strongest acid (protonates the ether most) and $\mathrm{I^{-}}$ is the best nucleophile of the three in the cleavage step.
7 2-Methoxy-2-methylpropane, $\mathrm{(CH_{3})_{3}COCH_{3}}$ (MTBE).
8 Ethoxybenzene: $\mathrm{C_{6}H_{5}OC_{2}H_{5}+HI\rightarrow C_{6}H_{5}OH+C_{2}H_{5}I}$.
IV
Aromatic ethers and problems. ($M$: C 12, H 1, O 16, Na 23, Br 80, I 127.)
  • 1 Give the main product of anisole with (a) $\mathrm{Br_{2}}$ in ethanoic acid; (b) $\mathrm{CH_{3}Cl}$/anhydrous $\mathrm{AlCl_{3}}$; (c) $\mathrm{CH_{3}COCl}$/anhydrous $\mathrm{AlCl_{3}}$; (d) concentrated $\mathrm{HNO_{3}}$ + $\mathrm{H_{2}SO_{4}}$.
  • 2 Explain why the methoxy group directs electrophiles to the ortho and para positions, and why the para product is the major one.
  • 3 Convert phenol into 4-nitroanisole in two steps.
  • 4 Convert ethanol into ethoxyethane (a) in one step; (b) by a Williamson route.
  • 5 $3.24\ \mathrm{g}$ of anisole is nitrated, giving $70\%$ of the para product. What mass of 4-nitroanisole ($\mathrm{C_{7}H_{7}NO_{3}}$, $N=14$) forms?
  • 6 What mass of iodomethane is needed to convert $11.6\ \mathrm{g}$ of sodium phenoxide completely into anisole?
  • 7 An ether A, $\mathrm{C_{8}H_{10}O}$, contains a benzene ring and is not acidic. With $\mathrm{HI}$ it gives phenol and an alkyl iodide B. Identify A and B, and suggest a Williamson synthesis of A.
  • 8 Environment. MTBE leaking from fuel tanks contaminates groundwater, while hydrocarbons in petrol mostly float on it. Use the structure of MTBE to explain why.
1 (a) 1-bromo-4-methoxybenzene; (b) 1-methoxy-4-methylbenzene; (c) 1-(4-methoxyphenyl)ethan-1-one; (d) 1-methoxy-4-nitrobenzene (each with some ortho isomer).
2 A lone pair on oxygen is delocalised into the ring (+R), placing extra electron density at the ortho and para carbons and stabilising the intermediate formed by o/p attack. The bulky $\mathrm{{-}OCH_{3}}$ group hinders the neighbouring ortho positions, so para attack is favoured.
3 Phenol $\xrightarrow{\mathrm{NaOH},\ \mathrm{CH_{3}I}}$ anisole $\xrightarrow{\text{conc. }\mathrm{HNO_{3}}+\mathrm{H_{2}SO_{4}}}$ 4-nitroanisole (separated from the ortho isomer).
4 (a) Concentrated $\mathrm{H_{2}SO_{4}}$ at $413\ \mathrm{K}$ with excess ethanol. (b) Ethanol + $\mathrm{Na}$ $\rightarrow$ sodium ethoxide; ethanol + $\mathrm{PBr_{3}}$ (or $\mathrm{HBr}$) $\rightarrow$ bromoethane; then $\mathrm{C_{2}H_{5}ONa+C_{2}H_{5}Br\rightarrow C_{2}H_{5}OC_{2}H_{5}+NaBr}$.
5 $\dfrac{3.24}{108}=0.0300\ \mathrm{mol}$; $M(\mathrm{C_{7}H_{7}NO_{3}})=153$; $0.0300\times0.70\times153=3.2\ \mathrm{g}$.
6 $M(\mathrm{C_{6}H_{5}ONa})=116$, so $0.100\ \mathrm{mol}$; $0.100\times142=14.2\ \mathrm{g}$ of $\mathrm{CH_{3}I}$.
7 The products show an aryl–O–alkyl ether with a $\mathrm{C_{2}H_{5}}$ group: A is ethoxybenzene, $\mathrm{C_{6}H_{5}OC_{2}H_{5}}$; B is iodoethane. Synthesis: $\mathrm{C_{6}H_{5}ONa+C_{2}H_{5}Br}$.
8 MTBE's oxygen accepts hydrogen bonds from water, so it is far more soluble in water than hydrocarbons; and, as an ether with a tertiary group, its C–O bonds are hard for microbes to break, so it persists.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks What is the Williamson synthesis? Write the equation for preparing 2-ethoxy-2-methylpropane by this method, choosing the reactants correctly, and say why the other pair would fail.

Definition Preparation of an ether by $\mathrm{S_{N}2}$ reaction of a sodium alkoxide (or phenoxide) with an alkyl halide: $\mathrm{RX+R'ONa\rightarrow ROR'+NaX}$. (1 mark)
Equation $$\mathrm{(CH_{3})_{3}CONa+C_{2}H_{5}Br}$$ $$\mathrm{\rightarrow(CH_{3})_{3}COC_{2}H_{5}+NaBr}$$ With $\mathrm{(CH_{3})_{3}CBr+C_{2}H_{5}ONa}$ the tertiary halide undergoes elimination to 2-methylpropene. (1 mark)

E2. JEE Main 2-Methoxy-2-methylpropane is heated with one mole of concentrated $\mathrm{HI}$. The products are (A) $\mathrm{CH_{3}I}$ and $\mathrm{(CH_{3})_{3}COH}$  (B) $\mathrm{(CH_{3})_{3}CI}$ and $\mathrm{CH_{3}OH}$  (C) $\mathrm{(CH_{3})_{2}C{=}CH_{2}}$ and $\mathrm{CH_{3}I}$  (D) $\mathrm{(CH_{3})_{3}CI}$ and $\mathrm{CH_{3}I}$

Idea After protonation, the bond to the tertiary carbon breaks by $\mathrm{S_{N}1}$ (stable $\mathrm{(CH_{3})_{3}C^{+}}$), and the iodide joins the tertiary carbon. (D) would need a second mole of $\mathrm{HI}$.
Answer (B).

E3. NEET Anisole is heated with hydrogen iodide. The products are (A) iodobenzene and methanol  (B) phenol and iodomethane  (C) 4-iodoanisole and water  (D) benzene and iodomethane

Idea The aryl–O bond has partial double-bond character and the ring carbon cannot undergo $\mathrm{S_{N}1}$ or $\mathrm{S_{N}2}$; $\mathrm{I^{-}}$ attacks the methyl carbon.
Answer (B).

E4. JEE Main Numerical value. Each ether is heated with one mole of concentrated $\mathrm{HI}$: methoxymethane, methoxyethane, 2-methoxypropane, 2-methoxy-2-methylpropane, methoxybenzene, ethoxyethane, ethoxybenzene, 1-methoxybutane. How many give iodomethane as a product?

Sort Methyl ethers whose other group is methyl, primary, secondary or aryl give $\mathrm{CH_{3}I}$: methoxymethane, methoxyethane, 2-methoxypropane, methoxybenzene, 1-methoxybutane. The methyl tertiary ether gives 2-iodo-2-methylpropane and methanol instead; the two ethyl ethers contain no methyl group.
Answer $\mathbf{5}$.

E5. JEE Advanced One or more correct. (A) The Williamson synthesis is an $\mathrm{S_{N}2}$ reaction of an alkoxide ion with an alkyl halide. (B) 2-Bromo-2-methylpropane with sodium methoxide gives mainly 2-methylpropene. (C) Diphenyl ether is made from bromobenzene and sodium phenoxide under Williamson conditions. (D) Sodium phenoxide and iodomethane give methoxybenzene.

(A), (B), (D) True: alkoxide displaces halide by $\mathrm{S_{N}2}$; a tertiary halide eliminates with the strongly basic methoxide; phenoxide + methyl halide is the standard route to anisole.
(C) False: bromobenzene does not undergo $\mathrm{S_{N}2}$ substitution.
Answer (A), (B), (D).

E6. Board · Case study Read the passage and answer. Ethoxyethane was the first widely used general anaesthetic. It is made from ethanol and concentrated sulphuric acid; if the mixture gets too hot, a gas is given off instead. Ethoxyethane is very volatile, boiling at about $308\ \mathrm{K}$, while ethanol boils at about $351\ \mathrm{K}$. (a) Write the equation for the preparation with conditions, and name the gas formed if the mixture gets too hot. (2 marks) (b) Why does ethoxyethane boil lower than ethanol, although its molar mass is higher? (1 mark) (c) What is formed when ethoxyethane is heated with excess concentrated $\mathrm{HI}$? (1 mark)

(a) $\mathrm{2C_{2}H_{5}OH\xrightarrow[413\ K]{conc.\ H_{2}SO_{4}}C_{2}H_{5}OC_{2}H_{5}+H_{2}O}$; at about $443\ \mathrm{K}$ ethene is formed.
(b) Ethanol molecules are linked by hydrogen bonds; ether molecules have no $\mathrm{O{-}H}$ and cannot hydrogen-bond to one another.
(c) Iodoethane (two moles) and water: $\mathrm{C_{2}H_{5}OC_{2}H_{5}+2HI\rightarrow2C_{2}H_{5}I+H_{2}O}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why did many countries stop adding the ether MTBE to petrol, and replace it with ethanol?

Why it matters. Oxygen-containing additives ("oxygenates") help petrol burn more completely and raise its octane rating. From the 1990s, regulations in some regions required cleaner-burning petrol to contain about $2\%$ oxygen by mass, and 2-methoxy-2-methylpropane (MTBE) was the cheapest way to provide it. Within a few years, MTBE leaking from underground fuel tanks was being found in drinking-water wells, where even small amounts give water an unpleasant taste and smell.

What to do.

  1. Calculate the percentage of oxygen by mass in MTBE, $\mathrm{C_{5}H_{12}O}$, and in ethanol, $\mathrm{C_{2}H_{5}OH}$ ($M$: C 12, H 1, O 16).
  2. Find the mass percentage of each additive needed to give petrol $2.0\%$ oxygen by mass (assume the rest of the petrol contains no oxygen).
  3. Approximate solubilities in water at room temperature: MTBE about $4$–$5\ \mathrm{g}$ per $100\ \mathrm{mL}$; benzene about $0.18\ \mathrm{g}$ per $100\ \mathrm{mL}$; octane less than $0.001\ \mathrm{g}$ per $100\ \mathrm{mL}$. Use hydrogen bonding (Rule 7.16) to explain this order.
  4. MTBE is broken down only very slowly by soil bacteria. Use what you know about the C–O bond of ethers (Rule 7.17) to suggest why, and find out how ethanol behaves in groundwater.

Tools: Calculator; the physical-property and cleavage rules of this section; environmental-agency reports on fuel oxygenates.

Hand in: Your two calculations and a one-page briefing for a city council comparing MTBE and ethanol as fuel additives (oxygen supplied, water solubility, persistence, source).

Going further. ETBE (2-ethoxy-2-methylpropane), made from bio-ethanol, is used in some countries instead of MTBE. Calculate its oxygen content and suggest how it would be made by a Williamson synthesis.

Instructor note (solubilities are rounded literature values): O in MTBE $=\tfrac{16}{88}=18.2\%$; in ethanol $\tfrac{16}{46}=34.8\%$. For $2.0\%$ O: MTBE $\tfrac{2.0}{0.182}=11.0\%$ by mass; ethanol $\tfrac{2.0}{0.348}=5.75\%$. MTBE's oxygen accepts hydrogen bonds from water; benzene and octane cannot (benzene's π cloud interacts weakly). The ether C–O bonds, one of them to a tertiary carbon, resist enzymatic cleavage, whereas ethanol is readily metabolised by microbes. ETBE: $\mathrm{C_{6}H_{14}O}$, $\tfrac{16}{102}=15.7\%$ O; Williamson: $\mathrm{(CH_{3})_{3}CONa+C_{2}H_{5}Br}$.

🧪 Self-check quiz

0/5 correct

Q1. The best reagents for making 2-methoxy-2-methylpropane by the Williamson synthesis are

Q2. Methoxybenzene heated with concentrated $\mathrm{HI}$ gives

Q3. Ethoxyethane boils far lower than its isomer butan-1-ol because

Q4. The main product of anisole with bromine in ethanoic acid is

Q5. Ethanol heated with concentrated $\mathrm{H_{2}SO_{4}}$ at $413\ \mathrm{K}$, with ethanol in excess, gives mainly

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