Physical & Chemical Properties of Ethers

The C–O–C framework: a quick recap

In an ether the oxygen atom is bonded to two carbon atoms through two sigma bonds, and it still carries two lone pairs. These four electron pairs — two bond pairs and two lone pairs — adopt an approximately tetrahedral arrangement. The C–O–C bond angle is therefore close to the tetrahedral value but is opened slightly wider, because the two bulky alkyl groups repel each other more strongly than two hydrogen atoms would in water. The C–O bond length in ethers is about 141 pm, almost identical to the C–O length in alcohols.

Two structural facts drive everything that follows. First, oxygen is more electronegative than carbon, so each C–O bond is polar and the bent geometry prevents the two bond dipoles from cancelling — an ether has a small but real net dipole moment and behaves as a weakly polar molecule. Second, and decisively, there is no hydrogen atom attached to the oxygen. That single absence separates the physical behaviour of ethers from that of their isomeric alcohols.

Boiling points: why ethers behave like alkanes

The C–O bonds in ethers are polar and the molecule has a net dipole moment, but this weak polarity does not appreciably raise the boiling point. The reason is that ether molecules cannot form intermolecular hydrogen bonds with one another: hydrogen bonding requires an H atom bonded to a strongly electronegative atom such as O, and an ether has none on its oxygen. The forces holding ether molecules together in the liquid are therefore essentially the same van der Waals forces that hold alkanes together.

The consequence is a striking pair of comparisons drawn directly from NCERT. The boiling point of an ether is close to that of an alkane of comparable molecular mass, and far below that of the isomeric alcohol of the same molecular mass. The large gap between ether and alcohol is due entirely to the hydrogen bonding present in the alcohol and absent in the ether.

Read the table as a single sentence: pentane and ethoxyethane, both unable to self-hydrogen-bond, boil within two kelvin of each other, while butan-1-ol — same molecular mass as the ether but able to hydrogen-bond — boils roughly 80 K higher. An ether is, thermally, an alkane with an oxygen quietly tucked inside the chain.

Solubility in water: as good as alcohols

Here lies the most elegant point in the physical properties of ethers, and a favourite of question-setters. Despite boiling like an alkane, an ether dissolves in water almost as well as the isomeric alcohol. The miscibility of ethers with water resembles that of alcohols of the same molecular mass, while a comparable alkane is essentially immiscible.

The resolution is to distinguish hydrogen-bond donating from hydrogen-bond accepting. An ether cannot donate a hydrogen bond, since it has no O–H. But its oxygen still carries two lone pairs, and those lone pairs readily accept hydrogen bonds from the O–H bonds of water. Water, in effect, hydrogen-bonds to the ether even though the ether cannot hydrogen-bond to itself. An alkane offers neither donor nor acceptor and is left out in the cold.

As with alcohols, solubility falls as the alkyl groups grow larger, because the hydrophobic hydrocarbon portion increasingly outweighs the small polar oxygen. The lower ethers are appreciably water-soluble; the higher homologues are not.

Chemical reactions: the least reactive group

Ethers are described in NCERT as the least reactive of the common functional groups, and this inertness is precisely why diethyl ether and tetrahydrofuran are prized as solvents. The C–O–C linkage has no easily attacked centre under ordinary conditions. Only three behaviours are examinable: cleavage of the C–O bond by hydrogen halides under forcing conditions, electrophilic aromatic substitution when an aryl ring is attached to the oxygen, and slow peroxide formation on storage. The first two carry essentially all the marks.

Cleavage of the C–O bond by HX

The defining reaction of ethers is cleavage of the C–O bond by excess hydrogen halide under drastic conditions. A simple dialkyl ether on heating with concentrated HI or HBr is split, and with excess reagent both halves can ultimately be converted to alkyl halides:

$$\ce{R-O-R' + HX -> R-X + R'-OH}$$

The order of reactivity of the hydrogen halides mirrors their acidity and the nucleophilicity of the halide:

$$\text{Reactivity:}\quad \ce{HI > HBr > HCl}$$

HI is the reagent of choice; HCl is too weak to do the job. The mechanism, set out in NCERT, has three logical steps. The ether is first protonated at oxygen by the strong acid to give an oxonium ion, converting the poor C–O leaving group into a far better one (a neutral alcohol). The halide ion, a good nucleophile, then attacks a carbon and displaces the alcohol.

$$\ce{R-\overset{\displaystyle ..}{\underset{\displaystyle ..}{O}}-R' + HI -> [R-\overset{+}{O}(H)-R'] + I^-}$$

$$\ce{I^- + R-\overset{+}{O}(H)-R' ->[\text{S}_\text{N}2] R-I + R'-OH}$$

When HI is present in excess and the temperature is high, the alcohol fragment is itself converted to a second alkyl iodide by reaction with another molecule of HI.

Dialkyl ethers: which fragment becomes the halide

For a mixed dialkyl ether the marks lie entirely in predicting which group leaves as the alkyl iodide. When both alkyl groups are primary or secondary, the iodide attacks the carbon by an S_N2 pathway, and S_N2 prefers the less hindered carbon. The rule is therefore:

With primary or secondary alkyl groups, the smaller (less hindered) group becomes the alkyl iodide, and the larger group is released as the alcohol.

The decisive exception is a tertiary alkyl group. Here the protonated ether does not wait for backside attack; instead the C–O bond breaks first to release the highly stable tertiary carbocation, which is then captured by iodide. This is an S_N1 process, so the tertiary group becomes the halide and the smaller group leaves as the alcohol:

$$\ce{(CH3)3C-O-CH3 + HI -> (CH3)3C-I + CH3OH}$$

Alkyl aryl ethers: anisole gives phenol

An alkyl aryl ether such as anisole (methoxybenzene) is always cleaved at the alkyl–oxygen bond, never at the aryl–oxygen bond. The products are therefore phenol and an alkyl halide, not an aryl halide and an alcohol.

$$\ce{C6H5-O-CH3 + HI -> C6H5-OH + CH3-I}$$

The reasoning is twofold. First, after protonation gives the methylphenyl oxonium ion, the O–CH₃ bond is weaker than the O–C₆H₅ bond, because the phenyl carbon is sp² hybridised and the aryl C–O bond has partial double-bond character from resonance. Second, even if one imagined attacking the aryl carbon, an sp² aromatic carbon cannot undergo the nucleophilic substitution required to install a halide. So iodide attacks only the methyl carbon, the O–CH₃ bond breaks to give CH₃I, and the aryl–oxygen fragment survives as phenol, which does not react further.

Electrophilic substitution in anisole

Once an aromatic ring is attached to the ether oxygen, the ring itself becomes reactive. The alkoxy group (–OR) is activating and ortho/para-directing, behaving exactly like the –OH group of phenol. The oxygen lone pairs are donated into the ring by resonance, building up electron density at the ortho and para carbons and making them the sites of attack by electrophiles.

NCERT lists three electrophilic substitutions of anisole, all directed to the ortho and para positions:

The bromination detail is worth fixing in memory: anisole is brominated even in the absence of an iron(III) bromide catalyst, because the methoxy group activates the ring strongly enough on its own. The para isomer dominates at about 90% yield, the ortho position being partly blocked by the bulk of the –OCH₃ group.

Peroxide formation on storage

One safety-related reaction completes the picture. On standing in contact with air, ethers slowly react with oxygen at the carbon adjacent to oxygen to form hydroperoxides and then peroxides. These peroxides have a strong tendency to explode, particularly when an old sample is concentrated or distilled to dryness. NCERT and the NIOS supplement both stress that ethers stored for some time must be handled with care and tested for peroxides before use.

$$\ce{R-O-CH2-R' + O2 ->[\text{air, slow}] R-O-CH(OOH)-R'}$$