Ethers: Preparation and Properties

Ethers are the quiet member of the alcohols-phenols-ethers trio — no acidic hydrogen, no easy oxidation, no colour tests to memorise. That very unreactivity is the point: it is exactly why diethyl ether was the classic laboratory and anaesthetic solvent for over a century, dissolving organic compounds without reacting with them. But "unreactive" is not "inert." Ethers have a bent C–O–C skeleton built the same way an alcohol's C–O–H is built, and that oxygen lone pair leaves two doors open — a controlled way to build the ether in the first place (Williamson synthesis) and a controlled way to break it apart again (cleavage by hot concentrated HI). Both doors turn on the same idea, carbon stability, and both are exam favourites precisely because a student who has not internalised that idea will guess wrong on which carbon reacts.

The core insight

Ethers are made by an $S_N2$ attack of an alkoxide on a primary alkyl halide (Williamson synthesis) and are broken by hot concentrated HI along whichever C–O bond gives the more stable carbocation or the less hindered $S_N2$ pathway.

The core idea

Shape and physical behaviour. An ether $\text{R–O–R}'$ has its oxygen roughly $sp^3$ hybridised, so the C–O–C angle sits a little above the tetrahedral $109.5°$ , widened by steric repulsion between the two R groups — geometrically it looks like a bent water molecule with both hydrogens replaced by carbon chains. Because there is no O–H bond, ether molecules cannot hydrogen-bond to one another, so they boil far below isomeric alcohols of the same molecular formula: diethyl ether boils near $308\ \text{K}$ ( $35\ °\text{C}$ ) while butan-1-ol, its structural isomer $\text{C}_4\text{H}_{10}\text{O}$ , boils at $391\ \text{K}$ ( $118\ °\text{C}$ ). Yet the lone pairs on that same oxygen can accept a hydrogen bond from water, so lower ethers stay reasonably water-soluble, roughly matching alcohols of similar chain length despite the huge gap in boiling point.

Making the bond — Williamson synthesis. The syllabus-prescribed route reacts a sodium alkoxide (or sodium phenoxide, for an alkyl-aryl ether) with an alkyl halide:

\text{R–ONa} + \text{R}'\text{–X} \rightarrow \text{R–O–R}' + \text{NaX}

where: R–ONa is the alkoxide nucleophile, R'–X the alkyl halide electrophile (X = Cl, Br or I), and NaX the salt by-product. The mechanism and why the halide must be primary are worked in full below.

Breaking the bond — cleavage by HI. Hot excess concentrated HI cleaves an ether at its C–O bond, ultimately converting both halves to alkyl iodides:

\text{R–O–R}' + \text{HI} \rightarrow \text{R–OH} + \text{R}'\text{I} \xrightarrow[\text{excess HI}]{} \text{R–I} + \text{R}'\text{I} + \text{H}_2\text{O}

where: the first equivalent of HI protonates the oxygen and cleaves one C–O bond, giving an alcohol plus an alkyl iodide; a second equivalent of HI then converts that alcohol to a second alkyl iodide. Which C–O bond breaks first is the crux of this reaction and is worked out fully in the worked example below.

Figure

Backside attack of an alkoxide ion on a primary alkyl bromide showing the leaving bromide and the new ether bond forming in one step

Derivation AWilliamson Ether Synthesis and Why the Halide Must Be Primary

The goal is to show, step by step, why treating a sodium alkoxide with a primary or methyl alkyl halide gives a clean ether by $S_N2$ substitution, while swapping in a secondary or tertiary halide instead throws away most of the yield to elimination — a distinction the syllabus expects a student to apply, not just quote, when picking reagents for a target ether.

Step 1. Identify the nucleophile and electrophile. The alkoxide ion $\text{R–O}^-$ (from sodium alkoxide, R–ONa) carries a full negative charge concentrated on oxygen and is a strong nucleophile. The alkyl halide $\text{R}'\text{–X}$ presents an electrophilic carbon, polarised $\delta^+$ because the halogen X pulls electron density toward itself; this carbon is the site of attack.

Step 2. Approach and bond formation. Because this is an $S_N2$ (bimolecular nucleophilic substitution) process, the alkoxide oxygen must approach the electrophilic carbon from the side directly opposite the C–X bond — a backside attack — since approaching from the same side as X would be blocked by X's own electron cloud. As the new C–O bond forms, the C–X bond simultaneously weakens and breaks in one concerted step; there is no intermediate carbocation.

Step 3. Why the halide component must be primary or methyl. A backside attack needs open space behind the carbon bearing the leaving group. On a primary (or methyl) carbon, only hydrogens and one R group crowd that space, so the alkoxide reaches it easily. On a secondary halide the approach is already hindered, and on a tertiary halide it is blocked almost completely by three bulky groups — so instead of substituting, the strongly basic alkoxide ion pulls off a $\beta$ -hydrogen and triggers E2 elimination, forming an alkene and regenerating R–OH, wasting the reagents without making any ether.

Step 4. Assign the roles correctly. Given a target ether $\text{R–O–R}'$ with R bulkier than R', the rule that follows from Steps 2–3 is: make the bulkier group R the alkoxide half (R–ONa) and keep the less hindered group R' as the alkyl halide half (R'–X), since the alkoxide's steric bulk does not block its own approach the way it would if it were sitting on the electrophile. Reversing this — putting the bulky group on the halide — is the single most common way this synthesis is set up wrong on an exam. Verified: the product of $\text{CH}_3\text{ONa} + \text{CH}_3\text{CH}_2\text{Br}$ is $\text{CH}_3\text{OCH}_2\text{CH}_3$ (ethyl methyl ether) + NaBr, matching the mass balance of the reactants exactly — one Na and one Br relocate into NaBr, and all carbons and the ether oxygen are conserved in the product.

Worked example

Prepare ethyl methyl ether ( $\text{CH}_3–\text{O}–\text{CH}_2\text{CH}_3$ ) by Williamson synthesis, choosing the better of two reagent combinations, then predict what happens if that ether is later heated with excess concentrated HI.

Step 1. List the two possible pairings. Option A: sodium methoxide ( $\text{CH}_3\text{ONa}$ ) plus bromoethane ( $\text{CH}_3\text{CH}_2\text{Br}$ ). Option B: sodium ethoxide ( $\text{CH}_3\text{CH}_2\text{ONa}$ ) plus bromomethane ( $\text{CH}_3\text{Br}$ ). Both give the same target, $\text{CH}_3\text{OCH}_2\text{CH}_3 + \text{NaBr}$ , since the product ether does not "remember" which oxygen came from which reagent.

Step 2. Apply the rule from the derivation. Both options use a primary or methyl halide as the electrophile (bromoethane is primary, bromomethane is methyl), so neither risks E2 elimination — both are mechanistically sound $S_N2$ routes here, because ethyl methyl ether happens to have no bulky group on either side. Either combination is an acceptable answer; the general habit to state is that whichever group is more hindered should always be placed on the alkoxide, never on the halide.

Step 3. Now cleave the finished ether. Heating $\text{CH}_3\text{OCH}_2\text{CH}_3$ with excess concentrated HI protonates the ether oxygen, and since both attached groups (methyl and ethyl) are primary or methyl, cleavage proceeds by $S_N2$ : iodide attacks the less hindered carbon, which for a methyl group is essentially barrier-free. This first gives methanol + iodoethane (or ethanol + iodomethane, attack being fast at whichever carbon iodide reaches); with excess HI present, the alcohol formed is itself converted to a second alkyl iodide.

Step 4. State the final products. Ethyl methyl ether reacts completely with excess HI to give iodomethane ( $\text{CH}_3\text{I}$ ) + iodoethane ( $\text{CH}_3\text{CH}_2\text{I}$ ) + $\text{H}_2\text{O}$ , the two alkyl fragments of the original ether now separated as two different alkyl iodides.

Competitive stretch

tert-Butyl methyl ether, $(\text{CH}_3)_3\text{C–O–CH}_3$ , behaves differently: the tertiary side heterolyses to the stable carbocation $(\text{CH}_3)_3\text{C}^+$ ( $S_N1$ -type), while the methyl–oxygen bond stays intact. Final products: 2-iodo-2-methylpropane, $(\text{CH}_3)_3\text{CI}$ , + methanol. Lesson: all-primary ethers cleave $S_N2$ (iodide picks the less hindered carbon); one secondary/tertiary group flips it to $S_N1$ (iodide lands on the more substituted carbon).

Common traps

Choosing a secondary or tertiary alkyl halide as the electrophilic partner in Williamson synthesis — this favours E2 elimination over substitution and is a classic "why did this route fail" question.
Assuming ether cleavage by HI always attacks the carbon that "seems more reactive," without first checking whether both R groups are primary ( $S_N2$ , attack the less hindered carbon) or one is secondary/tertiary ( $S_N1$ , attack via the more stable carbocation).
Forgetting that ethers cannot hydrogen-bond to each other (no O–H bond), so they boil far below isomeric alcohols even while showing comparable water solubility.
Treating ethers as inert everywhere — they resist oxidation, bases and most acids under mild conditions, but hot concentrated HI (or HBr) cleaves them, and this is the one reaction the syllabus tests in depth.

Ask the tutor: Not sure which side of an unsymmetrical ether gets the iodine when it is cleaved by HI, or want a reagent-pairing check for a Williamson synthesis target? Talk it through with the AI tutor.

Why can we use either alkoxide-halide pairing for ethyl methyl ether but not for a bulkier ether like tert-butyl ethyl ether?

Ethyl methyl ether has no bulky group on either side, so putting either fragment on the alkoxide or the halide leaves the electrophilic carbon equally open to backside attack — both pairings are $S_N2$ -clean. For tert-butyl ethyl ether, though, one fragment is tertiary. If you make the tert-butyl group the halide (tert-butyl bromide plus sodium ethoxide), the strongly basic ethoxide cannot reach the crowded tertiary carbon and instead pulls off a hydrogen, giving isobutylene by E2 elimination instead of the ether. The safe pairing is sodium tert-butoxide plus bromoethane, where the bulky group sits on the alkoxide and the halide is the unhindered primary ethyl group.

If ethers resist most acids, why does concentrated HI cleave them so readily?

Ethers are basic at the oxygen lone pairs just like alcohols, so they do get protonated by a strong acid; the difference is what happens next. Dilute or weak acids protonate the oxygen but there is no good nucleophile around to attack the resulting oxonium ion, so nothing further happens and the ether is recovered unchanged on workup. Concentrated HI supplies both a strong protonating acid and, crucially, iodide, an excellent nucleophile that readily displaces one alkyl group from the protonated ether by $S_N2$ or $S_N1$ . It is the combination of a good leaving ability (protonated ether oxygen is a fine leaving group) with a good nucleophile that makes hot concentrated HI uniquely effective at ether cleavage.

How do I know in advance which carbon iodide will attack when cleaving an unsymmetrical ether?

Look at what carbocation, if any, could form. If both alkyl groups attached to the ether oxygen are primary or methyl, no carbocation is stable enough to form, so the mechanism is $S_N2$ and iodide attacks the less hindered (usually the smaller) carbon directly. If one group is secondary or, especially, tertiary, the C–O bond on that side breaks first to generate the more stable carbocation, an $S_N1$ pathway, and iodide then captures that carbocation — so the halogen ends up on the more substituted carbon while the other, less substituted group leaves as the alcohol. The single question to ask is "can a reasonably stable carbocation form on either side," and if yes, that is the side that gets iodine.

Ethers: Preparation and Properties

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads