Chemical properties of ethers

An ether $R\text{–O–}R'$ has two C–O single bonds and two oxygen lone pairs, but no O–H bond and no carbonyl. That single structural fact explains most of its chemistry: bases, mild nucleophiles, and mild oxidising or reducing agents leave it alone, while strong acids and radical conditions can still attack the C–O bond or an alpha C–H bond.

Halogenation. In the dark, chlorine substitutes hydrogen at the alpha-carbon (the carbon next to oxygen), because the adjacent oxygen lone pair stabilises the radical intermediate formed during the substitution.

where: the product is the alpha-chloro ether; in diffused sunlight the reaction does not stop there, and every hydrogen is eventually replaced to give the fully chlorinated $C_2Cl_5\text{–O–}C_2Cl_5$.

Autoxidation to peroxides. Left standing in air, especially as diethyl ether, the alpha C–H is slowly abstracted by atmospheric oxygen to build up an unstable, shock-sensitive peroxide.

where: the peroxide is far less volatile than the ether itself, so distilling an old bottle of ether to dryness concentrates the peroxide in the residue — a genuine explosion hazard, which is why old ether stock is tested for peroxides (or discarded) before use.

Cleavage by HI (or HBr). Ethers react with hydrohalic acids because the oxygen lone pair is basic enough to be protonated, converting a poor leaving group (alkoxide/ether oxygen) into a good one (a neutral alcohol or water).

where: cold, dilute HI stops after one C–O bond breaks (alcohol plus alkyl iodide); excess hot, concentrated HI pushes the alcohol on to a second substitution, so the net product is two molecules of alkyl iodide.

Reaction with $PCl_5$. Phosphorus pentachloride cleaves both C–O bonds of the ether in one step, giving two alkyl chlorides and phosphorus oxychloride.

Reactivity in ethers hinges on two ideas: what oxygen can offer and what carbon can stabilise. Oxygen offers only lone pairs, never a hydrogen to lose, so ethers cannot act as acids the way alcohols weakly can, and there is no O–H bond for a base to remove. That rules out most of the reactions alcohols undergo (esterification with acids under mild conditions, reaction with active metals to liberate hydrogen, and so on). What survives is exactly the chemistry that only needs a lone pair or an alpha C–H: protonation by a strong acid, radical abstraction of an alpha hydrogen by $Cl_2$ or by atmospheric $O_2$, and nucleophilic attack at carbon once that lone pair has been protonated to make the oxygen a workable leaving group. Once the ether oxygen is protonated (by HI, for instance), the molecule effectively becomes an "oxonium" species where one of the two C–O bonds can break to release a neutral alcohol — a far better leaving group than a bare alkoxide. Which C–O bond breaks, and by which pathway, is decided by how well each attached carbon can support the transition state. A primary or secondary carbon cannot stabilise a positive charge, so iodide ion is forced to attack it directly from the back side in a single concerted step ($S_N2$) — the leaving group departs exactly as the nucleophile bonds, and the attacked carbon is inverted. A tertiary carbon, by contrast, can stabilise a genuine carbocation through hyperconjugation and inductive donation from three alkyl groups, so the C–O bond to that carbon breaks first, on its own, before iodide ever gets involved ($S_N1$). The practical rule that follows directly from this stabilisation argument: with a symmetrical primary/secondary ether, iodide attacks the less hindered carbon; with an ether bearing one tertiary alkyl group, cleavage occurs at the tertiary carbon instead, because forming the stable carbocation is faster than forcing a hindered back-side attack.

Problem. Diethyl ether is treated first with cold, dilute HI, and separately with excess hot, concentrated HI. Then methyl tert-butyl ether, $(CH_3)_3C\text{–O–CH}_3$, is treated with cold HI. Give the products in each case and explain why the regiochemistry differs between the two ethers.

Step 1. Cold, dilute HI on diethyl ether. The ether oxygen is protonated, then iodide performs a single back-side ($S_N2$) attack on one of the two equivalent, primary ethyl carbons. Because both C–O bonds are equivalent in a symmetrical ether, cleavage stops after one bond breaks:

Step 2. Excess hot, concentrated HI on diethyl ether. The ethanol formed in Step 1 does not survive the forcing conditions — it is itself converted to ethyl iodide by a second $S_N2$ substitution:

so the overall, net transformation under forcing conditions is complete conversion to two molecules of alkyl iodide:

Step 3. Cold HI on methyl tert-butyl ether. Protonation gives the same oxonium intermediate, but now the two carbons attached to oxygen are not equivalent: one is a methyl carbon, the other a tertiary carbon from the tert-butyl group. The tertiary carbon can form a stabilised carbocation, so the mechanism switches to $S_N1$, and the C–O bond to the tertiary carbon breaks first:

Step 4. Compare the two outcomes: with the symmetrical, all-primary diethyl ether, the mechanism is forced to be $S_N2$ and iodide simply attacks either equivalent carbon; with the unsymmetrical ether, the presence of one tertiary carbon flips the mechanism to $S_N1$, so the iodide ends up on the bulkier carbon instead of the less hindered one. Final answer: cold dilute HI gives $C_2H_5OH + C_2H_5I$ from diethyl ether, excess hot HI gives $2\,C_2H_5I + H_2O$, and cold HI on methyl tert-butyl ether gives $(CH_3)_3C\text{–I} + CH_3OH$, with the tertiary carbon cleaved because it stabilises a carbocation.

• Calling ethers "completely unreactive" — they lack an O–H bond and so resist mild bases and nucleophiles, but they still react with strong acids (HI, HBr), with $Cl_2$, and undergo slow autoxidation to peroxides.
• Forgetting that regiochemistry flips with mechanism: iodide attacks the primary carbon under $S_N2$ (symmetrical or all-primary/secondary ethers) but the tertiary carbon under $S_N1$ (once a tertiary alkyl group is present).
• Writing ether plus $PCl_5$ as giving only one alkyl chloride — both C–O bonds break, so the products are two alkyl chlorides plus $POCl_3$, not one chloride and an unreacted fragment.
• Ignoring the peroxide-safety point — distilling old ether to dryness concentrates the low-volatility peroxide in the residue, a classic "why" question in exams.

Ask the tutor: if you are unsure whether an unsymmetrical ether cleaves by $S_N1$ or $S_N2$ with HI, ask the AI tutor to check which carbon can stabilise a carbocation before you commit to an answer.