Alcohols, Ethers & Related Chemistry Section 68 of 123

Ether chemistry

the why on every wrong step

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Diethyl ether, CH₃CH₂–O–CH₂CH₃, has an oxygen bonded to two carbons and nothing else. Any compound built that way, R–O–R′ with carbon groups R and R′, is an ether.

Ethers are best known for what they do not do: they are almost inert, and that inertness makes them valuable as solvents.

Naming them, and what they are like to handle

Common names list the two groups on the oxygen in alphabetical order and add the word ether: CH₃–O–CH₂CH₃ is ethyl methyl ether. IUPAC names make the larger group, or the ring, the parent and treat the other half as a substituent. That substituent, RO–, is an alkoxy group, named by changing -yl to -oxy (methyl becomes methoxy). So the same compound is methoxyethane, and cyclohexyl methyl ether is methoxycyclohexane. Exams use both systems, so learn to read both.

methoxyethaneethyl methyl ethermethoxycyclohexanecyclohexyl methyl etherethoxyethanediethyl etheroxolaneTHF (tetrahydrofuran)OOOO
Under each structure, the upper name is the IUPAC name and the green one is the common name.

Physically, an ether behaves like an alcohol that has lost its O–H hydrogen. The C–O–C unit is bent and the molecule still has a dipole, but no hydrogen sits on the oxygen, so ether molecules cannot hydrogen-bond to each other. Diethyl ether boils at 35 °C, while butan-1-ol, with the same formula, C₄H₁₀O, boils at 118 °C.

The oxygen's two lone pairs are still there, though, so an ether can accept a hydrogen bond from a molecule that has an O–H, even though it cannot donate one. That is why about 6 g of diethyl ether dissolves in 100 mL of water, far more than pentane does. It is also why ethers dissolve polar and nonpolar compounds alike.

BUTAN-1-OL · bp 118 °COHOHmolecules held by O–H···O bondsDIETHYL ETHER · bp 35 °COOsame formula, C₄H₁₀Ono O–H: nothing to donateDIETHYL ETHER + WATEROOHHthe ether accepts a hydrogen bond
Dashed lines are hydrogen bonds. Each one runs from an O–H hydrogen to a lone pair on another oxygen.

Why ethers are so unreactive

Compare ethanol with diethyl ether. A strong base such as NaH pulls the O–H proton off ethanol, as Alcohol reactions showed. Diethyl ether has no O–H, so a base finds no acidic proton to remove.

A nucleophile fares no better. To replace the oxygen at a carbon, it would have to push out an alkoxide ion, RO⁻. An alkoxide is the conjugate base of an alcohol (pKa about 16), so it is a strong base, and it is as poor a leaving group as hydroxide, the conjugate base of water (pKa 15.7). An ether therefore has nothing to give a base and nothing to give a nucleophile.

So ethers stand up to bases, nucleophiles, reducing agents and mild oxidizing agents. That is why diethyl ether and THF (tetrahydrofuran, the ring ether drawn above) are the usual solvents for very reactive reagents such as Grignard reagents and LiAlH₄, which would attack almost any other functional group in the flask. A Grignard reagent, RMgX, is a carbon nucleophile made from an alkyl halide and magnesium (Organometallic Bonding & Grignard Reagents, after this chapter, covers it). LiAlH₄ is a powerful source of hydride, H⁻, taught in Oxidation & Reduction. Here, all that matters is that both are violently reactive and need a solvent that will not react back.

The solvent does more than sit there. Each ether oxygen gives one lone pair to the magnesium of a Grignard reagent, and two ether molecules usually bind each magnesium. That coordination keeps the reagent dissolved and reactive. THF binds more strongly than diethyl ether: its ring ties the carbons back, which leaves the oxygen less crowded, and it is the stronger Lewis base. So THF is the solvent of choice when a Grignard reagent is hard to form.

Williamson ether synthesis

Methoxyethane can be made from ethanol and iodomethane in two steps. First, sodium hydride, NaH, removes the O–H proton from ethanol and gives sodium ethoxide. The by-product, hydrogen gas, bubbles out of the flask.

Second, the ethoxide oxygen attacks the carbon of iodomethane from the side opposite the iodine, and iodide leaves. That step is an ordinary SN2 reaction, and the new C–O bond it forms completes the ether.

STEP 1 · NaH TAKES THE O–H PROTONOHH−Na⁺hydride, H⁻O−+ H–H + Na⁺ethoxide ionH₂ gas bubbles offSTEP 2 · SN2 AT THE CH₃ OF CH₃–IO−CH₃Iattacked from the backO+ I⁻ + Na⁺methoxyethane
Step 1 makes the nucleophile. Step 2 forms the new C–O bond, highlighted in the product.

This route is the Williamson ether synthesis: turn an alcohol into its alkoxide, then let the alkoxide displace a leaving group from a methyl or primary carbon. The leaving group can be a halide or a tosylate, the sulfonate leaving group made from an alcohol in Alcohol reactions.

Because the key step is SN2, the rules from Substitution & Elimination decide what works. An alkoxide is a good nucleophile and a strong base. A methyl or primary electrophile gives clean substitution. A secondary electrophile gives poor yields, because E2 competes and often wins. A tertiary one gives no ether at all: the alkoxide removes a β hydrogen instead, and E2 gives an alkene.

Choose which half becomes the alkoxide. tert-Butyl methyl ether could be made from either of two pairs of reagents. Methoxide with tert-butyl bromide fails: the halide is tertiary and the base is strong, so E2 gives 2-methylpropene. tert-Butoxide with iodomethane works well. Iodomethane is the ideal SN2 electrophile, and a methyl group has no β hydrogen, so E2 cannot compete. The bulk sits on the nucleophile, where it does far less harm. Put the crowded group on the alkoxide and the unhindered group on the halide.
WORKS: tert-BUTOXIDE + CH₃–IO−CH₃IOclean SN2CH₃ has no β-HFAILS: CH₃O⁻ + tert-BUTYL BROMIDEO−HCH₂CBrCH₃CH₃E2: 2-methylpropene + CH₃OH + Br⁻no ether forms
Both routes aim at the same ether. Only the top one gives the alkoxide a carbon it can reach.
Worked example — planning an unsymmetrical ether

Target: ethoxycyclohexane (cyclohexyl ethyl ether).

Route A: NaH turns cyclohexanol into cyclohexoxide, which then reacts with bromoethane. The electrophile is primary, so SN2 runs cleanly. ✓

Route B: NaH turns ethanol into ethoxide, which then reacts with bromocyclohexane. The electrophile is secondary and sits on a ring, so E2 wins and the main product is cyclohexene. ✗

Same target, same two pieces, and only one order works.

ROUTE A · the ring is the alkoxideO−+Brprimary halideSN2Oethoxycyclohexanethe target ✓ROUTE B · the ring carries the halideO−+Brsecondary halideE2 winsmostly cyclohexenelittle ether ✗
Compare the carbon each alkoxide has to attack: primary in route A, secondary in route B.

Alkoxymercuration: adding an alcohol across an alkene

Treat 2-methylpropene with mercuric acetate, Hg(OAc)₂, in methanol, then add sodium borohydride, NaBH₄. The product is tert-butyl methyl ether, made without an alkoxide or an alkyl halide. The methanol oxygen ends up on the more substituted carbon of the old double bond.

This reaction is alkoxymercuration–demercuration. It is the oxymercuration from Addition reactions with an alcohol in place of water, and it runs through the same steps.

First, mercury bonds to both alkene carbons at once, as a three-membered ring called a mercurinium ion. The alcohol then opens that ring at the carbon better able to carry positive charge, which is the more substituted one. So the OR group goes to the more substituted carbon: Markovnikov placement.

The alcohol attacks from the face opposite the mercury, so the OR group and the mercury add anti. NaBH₄ then replaces the mercury with hydrogen. As in oxymercuration, that step goes through radicals and scrambles whatever stereochemistry the first step set, so the overall reaction is not stereospecific.

THE MERCURINIUM IONCCHgOAcHHCH₃CH₃+C1C2δ+OCH₃Htop faceCH₃OH attacks C2 from the bottom face−H⁺ANTI ADDITIONCCHgOAcOCH₃HHCH₃CH₃HgOAc top face, OCH₃ bottom faceNaBH₄THE ETHEROtert-butylmethyl etherHg replaced by H
Wedges point toward you and hashes away, so the alkene plane stands edge-on and the top and bottom faces are the two π faces. This product has no stereocenter, so the anti relationship cannot be seen in it.

The bridge never opens into a free carbocation, so nothing rearranges. A crowded carbon is no obstacle either. The bridge has already put most of the positive charge on the more substituted carbon, so the alcohol bonds there. In a Williamson synthesis the new C–O bond always forms at the halide's carbon, so that carbon must be methyl or primary. In alkoxymercuration it forms at the more substituted alkene carbon, which can be tertiary.

Ether cleavage: only strong acid gets through

Heat methyl propyl ether, CH₃–O–CH₂CH₂CH₃, with concentrated HI, and it splits into iodomethane and propan-1-ol. Hot, concentrated HBr works the same way. The cleavage takes two steps.

First, the acid protonates the ether oxygen. Before protonation, the group that would have to leave is RO⁻. After it, the leaving group is a neutral alcohol, ROH, which is a far weaker base than RO⁻: its conjugate acid, ROH₂⁺, has pKa about −2. A weaker base is a better leaving group, as Leaving groups showed. This is the same fix that activates an alcohol.

Second, iodide attacks one of the two carbons bonded to the oxygen, and the C–O bond breaks. Iodide and bromide suit this step: both are good nucleophiles and very weak bases, and iodide is the better of the two. In methyl propyl ether neither carbon can form a stable carbocation, so the step is SN2. SN2 goes to the less hindered carbon, the methyl.

1 · HI PROTONATES THE OXYGENH₃COHIH₃COH++ I⁻2 · I⁻ ATTACKS THE CH₃ FROM THE BACKIH₃COH−+CH₃–I + HO–CH₂CH₂CH₃iodomethane + propan-1-ol
Step 1 makes the leaving group. In step 2 the highlighted C–O bond breaks as the new C–I bond forms.

That attack comes from the back, so the carbon it hits is inverted. The carbon on the other side of the oxygen keeps all of its bonds and so keeps its configuration.

Now switch to tert-butyl methyl ether. Once the oxygen is protonated, the tertiary carbon does not wait for iodide. Its C–O bond breaks on its own and gives a tertiary carbocation plus methanol. Iodide then bonds to the cation, so the mechanism is SN1, and the products are 2-iodo-2-methylpropane (tert-butyl iodide) and methanol. With excess HI, the methanol then becomes iodomethane by SN2. A benzylic carbon, one bonded to a benzene ring, reacts the same way, because its cation is stabilized by the ring (Carbocations).

So the iodide ends up on the more crowded carbon, not because it is easy to reach, but because it can hold a positive charge.

1 · THE C–O BOND BREAKS ON ITS OWN(the oxygen is already protonated)CCH₃CH₃CH₃OHCH₃+CCH₃CH₃CH₃++ CH₃OHmethanol2 · I⁻ CAPTURES THE CATIONICCH₃CH₃CH₃−+(CH₃)₃C–I2-iodo-2-methylpropane + methanoliodide ends up on the more crowded carbon
Watch the central carbon: four bonds before the C–O bond breaks, three and flat as the cation, four again once iodide bonds.

A secondary carbon ionizes only slowly. Paired with a methyl or primary carbon, it loses out to SN2 at the smaller carbon. Only when both carbons are secondary do the acid, the temperature and the halide decide the mix of SN1 and SN2.

To predict where the halide goes in an unsymmetrical ether, decide the mechanism first. If one carbon is tertiary or benzylic, it ionizes, and the halide ends up there (SN1). Otherwise the halide attacks the less hindered carbon (SN2), even when the other carbon is secondary; only two secondary carbons give a mix.

With excess HI, the alcohol formed in the first cleavage reacts as well: it is protonated, and iodide replaces the OH. So an ether ends up as two alkyl halides. HCl does not work well, because chloride is too weak a nucleophile for the SN2 case.

Worked example — which carbon, and what happens to the stereocenter?

(R)-2-Methoxybutane is heated with one equivalent of HI (one mole of HI per mole of ether).

Step 1 — protonate. The oxygen takes a proton, and now a neutral alcohol can leave.

Step 2 — pick the mechanism. The two carbons on the oxygen are a methyl carbon and C2, a secondary carbon. Neither is tertiary or benzylic, and the secondary carbon ionizes too slowly to compete, so the step is SN2 at the less hindered carbon.

Step 3 — write the products. Iodide attacks the methyl carbon, giving iodomethane and butan-2-ol. The C2–O bond never breaks, so the four groups on C2 stay where they were.

Step 4 — check the descriptor. The oxygen on C2 still outranks ethyl, methyl and H, so the priority order is unchanged, and the alcohol is (R)-butan-2-ol.

(R)-2-methoxybutaneOCH₃C2this bond breaksC2–O bond:untouchedone CH₃ and one secondary carbon on OHI (one equiv.)SN2 at the CH₃(R)-butan-2-ol + CH₃–IOHC2+ CH₃–Isame arrangement at C2, so still R
Compare the wedge at C2 before and after the reaction.

Epoxides: the reactive exception

Everything above is about open-chain ethers. An epoxide is a cyclic ether with a three-membered ring of two carbons and one oxygen. Its ring angles are about 60°, and the ring is strained in the same way as cyclopropane (Cyclohexanes). Opening the ring releases that strain, which turns the least reactive functional group in this chapter into one of the most useful. The next section, Epoxides, covers them.

DIETHYL ETHEROC–O–C about 110°no strainOXIRANE (AN EPOXIDE)Oring angles about 60°strained
Both are ethers. Only the three-membered ring forces its angles far below the usual value.

A safety note worth knowing

A bottle of diethyl ether left open to air for months slowly builds up peroxides, compounds with an O–O single bond, and these can explode. They form at one particular carbon: the α carbon, the carbon bonded to the oxygen.

The α C–H bond is the weakest C–H bond in the molecule. Removing that hydrogen as a hydrogen atom, H·, leaves a carbon radical next to the oxygen. The radical's half-filled p orbital lies beside a lone pair on the oxygen, and the two overlap, which spreads the unpaired electron over both atoms. It is the same kind of delocalization that stabilizes allylic and benzylic radicals (Conjugated systems).

A trace of some other radical, often made by light, removes the first α hydrogen. After that, a chain reaction takes over, like the one in Radical halogenation. O₂ bonds to the carbon radical to give a peroxyl radical, C–O–O·. The peroxyl radical then takes an α hydrogen from another ether molecule. This gives a hydroperoxide, C–O–O–H, plus a new α radical that carries the chain on.

WHY THE α C–H IS WEAKCCH₃HOone electronO lone pairthe orbitals overlap side by side,so the odd electron spreads over C and Oα carbon = the carbon bonded to OTHE CHAIN THAT MAKES PEROXIDESE–H = the ether; E· = its α radicalSTEP 1E· + O₂ → E–O–O·O₂ bonds to the carbon radicalSTEP 2E–O–O· + H–E → E–O–O–H + E·takes an α H from another ethermolecule: a hydroperoxide formsthe new E· starts step 1 again,so peroxide builds up while air is present
Left: the orbital picture behind the weak α C–H bond. Right: the two steps that repeat once the first radical has formed.

That is why ether bottles carry the date they were opened, and why old ether must never be distilled to dryness: the peroxides boil at a higher temperature than the ether, so they concentrate in the flask as the ether boils off. It is also why THF is usually sold with an added inhibitor, a compound that stops the radical chain.

Ethers as Lewis bases, and crown ethers

Boron trifluoride, BF₃, is a gas, so chemists usually buy it as BF₃·OEt₂, "BF₃ etherate". In this liquid, the oxygen of diethyl ether gives a lone pair to boron's empty p orbital. The ether is acting as a Lewis base, an electron-pair donor (Lewis acids/bases). The complex is stable and easy to pour, yet it hands the BF₃ over to a stronger donor when one is present.

Lewis base + Lewis acidOBFFFOBFFF+−BF₃·OEt₂
The new O–B bond, highlighted, is made from an oxygen lone pair.

The bond between an ether and the magnesium of a Grignard reagent, described above, is the same kind of Lewis acid–base bond.

Now put six ether oxygens into one ring. The compound 18-crown-6 has an 18-atom ring made of six oxygens, each separated from the next by two CH₂ groups. Rings like this are called crown ethers, and the name gives the ring size first and the number of oxygens second.

The hole in the middle of 18-crown-6 is about 2.7 Å across. A potassium ion, 2.66 Å across, fits snugly and is held by one lone pair from each of the six oxygens at once. A sodium ion, 1.9 Å across, is too small to reach all six oxygens, so it binds far more weakly. The binding is size-selective.

That selectivity has a use. Potassium fluoride does not dissolve in benzene. Add 18-crown-6, and the ring wraps the K⁺ in a shell whose outside is made of CH₂ groups, so the whole ion pair dissolves. The fluoride comes along, with no solvent shell around it and its K⁺ locked inside the ring. A naked anion like that is a far stronger nucleophile than the same anion in water, where hydrogen bonds tie up its lone pairs. Nucleophiles mentioned an additive that wraps up the K⁺ of potassium fluoride. A crown ether like this one is that additive.

18-CROWN-6, WITH K⁺ IN THE MIDDLEOOOOOOK⁺each corner between two O atoms is a CH₂one lone pair on each O points at K⁺18 ring atoms, 6 of them oxygenWHY IT PICKS K⁺ AND NOT Na⁺cavityabout 2.7 ÅK⁺2.66 Å: fitsNa⁺1.9 Å: too smallDiameters, drawn to the same scale.WHAT THAT MAKES POSSIBLEKF does not dissolve in benzene.With 18-crown-6 around the K⁺,the whole ion pair dissolves, and the F⁻is neither solvated nor held tightly by K⁺.a naked anion: a far better nucleophile
Left: the ring holds K⁺ the way a shell of solvent molecules would. Top right: compare the three circles by size.

What carries forward

The Williamson synthesis is a standard step in synthesis problems, and a good test of SN2 judgment. Ethers are also used to protect alcohols: an ether is put on to keep an OH out of the way, then taken off later. The ethers most often used for that job are silyl ethers, which come off with fluoride rather than hot HI. They are covered with the other protecting groups in Synthesis & Retrosynthesis.

Finally, the protonation that lets acid cleave an ether is the same move that lets an alcohol react at its C–O bond. The general lesson: acid can often fix a bad leaving group.