Ethers

Methoxypropane, \(\mathrm{CH_3OCH_2CH_2CH_3}\), is prepared by Williamson’s synthesis from sodium methoxide and a primary propyl halide (1-bromopropane) in dry ether:

Sodium propoxide with iodomethane also works. Isopropyl bromide is not used, because a secondary halide gives propene.

Sodium ethoxide is heated with chloroethane (or bromoethane or iodoethane) in dry ether. The ethoxide ion replaces the halide by an SN2 reaction:

Requirements: sodium alkoxide, a primary alkyl halide, dry ether and heat.

Phenol is first converted into sodium phenoxide with NaOH, and the phenoxide is then heated with iodomethane (or bromomethane) to give methoxybenzene (anisole), an aromatic ether:

Diethyl ether is heated with excess concentrated hydroiodic acid. Both C-O bonds break; the ethanol formed first is converted into the second molecule of iodoethane:

Two molecules of ethanol lose one molecule of water (intermolecular dehydration) and ethoxyethane (diethyl ether) is formed. Ethanol must be in excess and the temperature must stay near 413 K; at 443 K ethene is formed.

The ether is cleaved and only iodoethane is obtained (the alcohol formed first reacts with more HI):

On standing in air and light, ethoxyethane slowly takes up oxygen and forms an ether peroxide, 1-ethoxyethyl hydroperoxide. Peroxides are explosive on heating or distilling. Their presence is shown by the blood-red colour of \(\mathrm{Fe(CNS)_3}\) with acidified \(\mathrm{FeSO_4}\) and KCNS, and they are prevented by storing the ether over iron or copper wire.

In bright sunlight, chlorine replaces all ten hydrogen atoms (free-radical substitution) and perchlorodiethyl ether is formed with hydrogen chloride:

In the dark only the hydrogens next to oxygen are replaced, giving alpha,alpha’-dichlorodiethyl ether: \(\mathrm{CH_3CH_2OCH_2CH_3+2Cl_2\xrightarrow{dark}CH_3CHClOCHClCH_3+2HCl}\).

Williamson synthesis takes place; ethoxyethane and sodium chloride are formed:

Reasoning (working backwards). The only symmetrical ether with 4 carbon atoms is ethoxyethane, so S is ethoxyethane. It is made from sodium ethoxide and an ethyl halide, so Q, made with HI, is iodoethane. Iodoethane is obtained from an ether by excess HI, and the ether obtained from P at 140 °C with conc. \(\mathrm{H_2SO_4}\) is a symmetrical ether, so R is ethoxyethane and P is ethanol. (P → R is intermolecular dehydration of ethanol.)

R and S are the same compound, made in two different ways (dehydration of an alcohol and Williamson synthesis).

Reasoning. The second member of the alcohol series is ethanol, so P = \(\mathrm{CH_3CH_2OH}\). It fits every clue: it releases hydrogen with sodium (Q = sodium ethoxide), it contains the \(\mathrm{CH_3CH(OH)-}\) group and so gives yellow iodoform with iodine and NaOH (R), and at 140 °C it forms an ether with the formula \(\mathrm{C_4H_{10}O}\), which is ethoxyethane (S). Conc. HI converts S into iodoethane (T), and Q + T give S back by Williamson synthesis.

Ethers slowly react with oxygen of the air to form explosive peroxides. Iron (or copper) has a greater tendency to combine with oxygen than the ether has, so the metal reacts with the peroxide and with traces of oxygen as they form, and the peroxide cannot build up in the bottle.

An old sample has formed ether peroxides. The peroxides are less volatile than the ether, so when the ether is boiled or distilled they stay behind and become concentrated in the flask. On further heating the concentrated peroxide decomposes violently and explodes. A sample is therefore tested with acidified \(\mathrm{FeSO_4}\) and KCNS (blood-red colour) before it is heated.

Both have the same general formula \(\mathrm{C}_n\mathrm{H}_{2n+2}\mathrm{O}\) but different functional groups: -OH in an alcohol and -O- in an ether.

a. Methoxybenzene + HI. The O-CH₃ bond breaks, giving phenol and iodomethane:

Iodobenzene and methanol are not formed, because the C₆H₅-O bond has partial double bond character by resonance and is too strong to break.

b. Ethoxyethane + Cl₂ in sunlight.

The product is perchlorodiethyl ether.

(A) Williamson synthesis of methoxyethane: \(\mathrm{CH_3ONa+CH_3CH_2Cl\longrightarrow CH_3OCH_2CH_3+NaCl}\). Sodium ethoxide with chloromethane (or iodomethane) is also correct, because both halides are primary.

(B) \(\mathrm{2CH_3CH_2OH\xrightarrow{conc.\ H_2SO_4,\ 413\ K}CH_3CH_2OCH_2CH_3+H_2O}\)

(C + D) Excess HI breaks both C-O bonds: \(\mathrm{CH_3OCH_2CH_3+2HI\longrightarrow CH_3I+CH_3CH_2I+H_2O}\).

(E) Ethoxyethane has ten hydrogens, all replaced by chlorine: \(\mathrm{CH_3CH_2OCH_2CH_3+10Cl_2\longrightarrow CCl_3CCl_2OCCl_2CCl_3+10HCl}\).

Step 1: what each clue shows.

Step 2: identify. Methanol has one carbon and cannot give the iodoform test. Ethanol has two carbons and gives the test (the only primary alcohol that does), so P is ethanol. S is ethoxyethane, and excess HI changes it to iodoethane (T). Check: sodium ethoxide (Q) + iodoethane (T) gives ethoxyethane (S) by Williamson synthesis, as the question states.

Step 3: reactions.

1. Ethoxyethane with oxygen. In air and light, ethoxyethane is slowly oxidised at the carbon next to oxygen and forms an ether peroxide (1-ethoxyethyl hydroperoxide), which is explosive:

2. Methoxymethane with excess HI. Both C-O bonds are broken, so Y is iodomethane:

3. Y with sodium phenoxide. The phenoxide ion attacks the carbon of iodomethane (SN2, Williamson synthesis) and Z is methoxybenzene (anisole), an aromatic ether:

The ratio is C : H : O = 3.94 : 9.45 : 1, so there are 4 carbon atoms and 1 oxygen atom. The percentages are rounded, so the hydrogen figure is not exactly a whole number. A saturated ether has the formula \(\mathrm{C}_n\mathrm{H}_{2n+2}\mathrm{O}\), so for n = 4 there are 10 hydrogen atoms, and the formula is \(\mathrm{C_4H_{10}O}\).

Check: M(\(\mathrm{C_4H_{10}O}\)) = 4(12) + 10(1) + 16 = 74. Percentages: C = 48/74 = 64.9% (65), H = 10/74 = 13.5% (about 13), O = 16/74 = 21.6% (22). The data agree.

The ether is symmetric, so the 4 carbons are divided into two ethyl groups: \(\mathrm{CH_3CH_2OCH_2CH_3}\), IUPAC name ethoxyethane (diethyl ether).

The oxygen atom of ethoxyethane has lone pairs, so the ether acts as a base and accepts a proton from cold conc. HCl. An oxonium salt, diethyloxonium chloride, is formed and the ether is not cleaved:

Adding water decomposes the salt and gives back the ether.

Ether must be protected from oxygen, light and heat, because these form explosive peroxides in a bottle left for a long time:

Even with these precautions, no storage can make ether completely safe for such a long time, so in practice old samples are tested regularly and discarded safely if peroxide appears.