Cleaving ethers with hydrogen iodide

Why ethoxyethane boils like pentane

Ethoxyethane, $\text{C}_{2}\text{H}_{5}\text{OC}_{2}\text{H}_{5}$, has a molar mass of $74\ \text{g/mol}$ and boils at $35\ ^\circ\text{C}$. Pentane, with a molar mass of $72\ \text{g/mol}$, boils at $36\ ^\circ\text{C}$. Butan-1-ol, with a molar mass of $74\ \text{g/mol}$, boils at $118\ ^\circ\text{C}$.

An ether has a bent C-O-C shape, so it has a small dipole and some attraction between molecules. It has no O-H bond, so one ether molecule cannot give a hydrogen bond to another. The boiling points of ethers are close to those of alkanes of similar molar mass, and far below those of alcohols. The ether oxygen can accept a hydrogen bond from water. For that reason the solubility of an ether in water is close to that of an alcohol of the same molar mass.

Ethers are unreactive, with one big exception

An ether has no acidic hydrogen and no polar double bond. Sodium does not react with it. Dilute acids and bases leave it alone. This is why ethers are used as solvents for Grignard reagents and for hydroboration. The oxygen lone pairs can bind to the magnesium of a Grignard reagent, and that helps it dissolve.

The strong acid hydrogen iodide cuts the C-O bond.

Cleavage by hydrogen halides

An ether reacts with HI (or HBr, or hot concentrated HCl) to give an alkyl halide and an alcohol.

The order of the acids is HI, then HBr, then HCl. Three steps explain the products.

1. Protonation. The oxygen takes a proton from HI and becomes an oxonium ion, $\text{R-O}^+\text{(H)-R'}$. Now the leaving group is an alcohol, which is far better than $\text{RO}^-$.
2. Attack by iodide. An iodide ion attacks one of the two carbons and pushes the alcohol out.
3. Excess HI. The alcohol that comes off then reacts with more HI and becomes an iodide as well.

Which carbon is attacked? The rule depends on the groups.

• Two primary or one methyl and one primary. Iodide attacks the less crowded carbon, by $S_N2$. Methoxyethane gives iodomethane and ethanol.
• One tertiary group. The tertiary carbon leaves as a carbocation, by $S_N1$, and the iodide joins it. 2-Methoxy-2-methylpropane gives 2-iodo-2-methylpropane and methanol.
• An aryl-oxygen bond. It is not cleaved. The aryl carbon is $sp^2$, so $S_N2$ cannot happen there, and the C-O bond has some double bond character from resonance. Anisole gives phenol and iodomethane. Phenol does not go on to iodobenzene.

With excess HI, an alkyl ether gives two alkyl iodides: $\text{C}_{2}\text{H}_{5}\text{OC}_{2}\text{H}_{5} + 2 \text{HI} \to 2 \text{C}_{2}\text{H}_{5}\text{I} + \text{H}_{2}\text{O}$.

Both products come from the same moles of anisole. The iodomethane curve is steeper, because iodomethane has the larger molar mass.

Practise with new numbers

Try it in Python

The program applies the three rules to four ethers. It prints which group becomes the iodide and what the other group becomes.

Common mistakes

• Making iodobenzene from anisole. The aryl C-O bond stays. The products are phenol and iodomethane.
• Always attacking the smaller group. A tertiary group is attacked even when it is the larger.
• Forgetting the second HI. With excess HI an alcohol product becomes an iodide as well. Phenol is the exception.
• Saying an ether can give hydrogen bonds to itself. It has no O-H bond.
• Expecting dilute acid to cleave an ether. Strong, hot HI is needed.

Remember these

• Ethers boil near alkanes of the same molar mass, and far below alcohols.
• Ethers are poor to react, which suits solvents. HI cleaves them: $S_N2$ at the less crowded carbon, $S_N1$ at a tertiary carbon.
• Aryl alkyl ethers give a phenol and an alkyl iodide.
• Excess HI turns an alkyl ether into two alkyl iodides.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What are the two organic products when $\text{CH}_{3}\text{OC}_{2}\text{H}_{5}$ is heated with one equivalent of HI?

2. What happens to $\text{CH}_{3}\text{OC}(\text{CH}_{3})_{3}$ with HI?

3. Why does phenyl methyl ether give phenol and not iodobenzene with HI?

4. Why is the boiling point of ethoxyethane much lower than that of butan-1-ol?

References

• NCERT Chemistry Part II (Class 12), Chapter 7, the sections on ethers.
• Morrison and Boyd, Organic Chemistry, the chapter on ethers.
• Lide (ed.), CRC Handbook of Chemistry and Physics, physical constants of organic compounds (boiling points).