Two Routes, One Functional Group
An ether is two carbon fragments bridged by an oxygen, $\ce{R-O-R'}$. To assemble that linkage you must form a new carbon–oxygen bond, and there are exactly two NCERT-level ways to do it: persuade one alcohol oxygen to attack a second carbon (dehydration), or pre-form an oxygen nucleophile and feed it a carbon electrophile (Williamson synthesis). Both are nucleophilic substitutions on carbon, but they differ sharply in scope.
Dehydration is the industrial method: cheap reagents, a single feedstock, but a narrow product range. Williamson synthesis is the laboratory method: a little more setup, but it can deliver almost any ether you draw, including unsymmetrical and aryl alkyl ethers. The strategic skill NEET tests is matching the target ether to the route that can actually make it.
| Feature | Dehydration of alcohols | Williamson synthesis |
|---|---|---|
| Reagents | Alcohol + protic acid ($\ce{H2SO4}$, $\ce{H3PO4}$) | Sodium alkoxide/phenoxide + alkyl halide |
| Bond-forming step | $\ce{SN2}$ (1° alcohols) | $\ce{SN2}$ |
| Symmetrical ethers | Yes (best use) | Yes |
| Unsymmetrical ethers | No (gives mixtures) | Yes |
| Aryl alkyl ethers | No | Yes |
| Main rival reaction | Alkene formation | Alkene formation (elimination) |
Dehydration of Alcohols
Alcohols undergo dehydration in the presence of protic acids such as sulphuric or phosphoric acid. Whether the product is an alkene or an ether depends entirely on the temperature. With ethanol and concentrated sulphuric acid, the high-temperature path at 443 K gives ethene by intramolecular dehydration, while the lower-temperature path at 413 K gives ethoxyethane (diethyl ether) by intermolecular dehydration.
$$\ce{2 C2H5OH ->[\text{H2SO4}][413\ \text{K}] C2H5-O-C2H5 + H2O}$$
NCERT classifies the ether-forming step as a nucleophilic bimolecular reaction ($\ce{SN2}$): a neutral alcohol molecule attacks a protonated alcohol. The acid first protonates one alcohol to give an oxonium ion, $\ce{R-OH2+}$, converting the poor leaving group $\ce{-OH}$ into water. A second alcohol molecule, acting as the nucleophile through its oxygen lone pair, then displaces that water in a single concerted step, and loss of a proton unmasks the ether.
$$\ce{C2H5OH + H+ -> C2H5-\overset{+}{O}H2}$$ $$\ce{C2H5OH + C2H5-\overset{+}{O}H2 -> C2H5-\overset{+}{O}H-C2H5 + H2O}$$ $$\ce{C2H5-\overset{+}{O}H-C2H5 -> C2H5-O-C2H5 + H+}$$
Because the slow step is a backside $\ce{SN2}$ attack on a carbon, the carbon must be unhindered. NCERT states the rule plainly: the method is suitable for ethers having primary alkyl groups only, the alkyl group should be unhindered, and the temperature should be kept low. Raise the temperature and elimination takes over to give the alkene instead.
Why Dehydration Fails for 2°, 3° and Unsymmetrical Ethers
Two independent limitations confine dehydration to a single product class. The first is the substrate. For secondary and tertiary alcohols the reaction switches to an $\ce{SN1}$-type pathway through a carbocation, and at that point elimination competes so strongly that the alcohol simply dehydrates to the alkene. NCERT is explicit: the dehydration of secondary and tertiary alcohols to give the corresponding ethers is unsuccessful, because elimination outcompetes substitution and alkenes form readily.
Dehydration cannot build an unsymmetrical ether
A single alcohol gives only the symmetrical ether $\ce{R-O-R}$. If you heat a mixture of two different alcohols hoping for $\ce{R-O-R'}$, every oxonium ion can be attacked by either alcohol, so you obtain a statistical mixture of all three possible ethers, which cannot be separated cleanly. This is precisely why bimolecular dehydration cannot be used to prepare ethyl methyl ether.
Rule: dehydration → symmetrical 1° ethers only. For anything unsymmetrical or aryl-containing, switch to Williamson synthesis.
The second limitation, then, is symmetry. Even when both fragments are primary, dehydration of a single alcohol can deliver only the symmetrical product. An unsymmetrical ether such as ethyl methyl ether, $\ce{CH3-O-C2H5}$, lies permanently out of reach of this route. These two constraints together are the reason dehydration is rarely the answer in a NEET synthesis question unless the target is a simple symmetrical primary ether like ethoxyethane.
The Williamson Ether Synthesis
The Williamson synthesis, named for Alexander William Williamson, is the most important laboratory method for preparing ethers, and crucially it makes both symmetrical and unsymmetrical ethers. An alkyl halide is allowed to react with a sodium alkoxide:
$$\ce{R-X + Na+ ^{-}O-R' -> R-O-R' + Na+X-}$$
The alkoxide is generated in advance by treating the parent alcohol with an active metal or a strong base — sodium metal, potassium metal, or sodium hydride — which deprotonates the $\ce{-OH}$ group:
$$\ce{R'-OH + Na -> R'-O^{-}Na+ + 1/2 H2 ^}$$ $$\ce{R'-OH + NaH -> R'-O^{-}Na+ + H2 ^}$$
Because the oxygen partner and the carbon partner are introduced as separate reagents, the two halves of the ether can be different by design. NCERT notes that ethers carrying substituted (secondary or tertiary) alkyl groups can also be prepared this way, provided the bulky group is supplied as the alkoxide and the partner halide is primary. The reaction is the $\ce{SN2}$ attack of an alkoxide ion on a primary alkyl halide.
Once the ether is built, how does it behave? See Properties of Ethers for cleavage by HI, low reactivity and boiling-point trends.
The SN2 Mechanism of Williamson Synthesis
Williamson synthesis is a textbook $\ce{SN2}$ displacement. The alkoxide oxygen, a strong nucleophile bearing a full negative charge, approaches the carbon of the alkyl halide from the side directly opposite the halogen. In a single concerted step the new $\ce{C-O}$ bond forms as the $\ce{C-X}$ bond breaks, the halide leaves, and the carbon configuration inverts. There is no intermediate.
Figure 1. The alkoxide ion (teal) attacks the primary carbon opposite the halogen; the halide (coral) departs as the new C–O bond forms in one concerted step.
Two consequences follow from this being $\ce{SN2}$. First, the rate is sensitive to steric crowding at the carbon under attack, so the halide carbon must be as open as possible — a primary carbon. Second, the alkoxide is not merely a nucleophile; it is also a strong base. Whenever it cannot reach the carbon easily, it will instead pluck off a $\beta$-hydrogen, triggering elimination to an alkene. This dual personality of the alkoxide is the single most examined idea in the entire subtopic.
Choosing the Right Reactant Pairing
Every unsymmetrical ether can in principle be sliced at the $\ce{C-O}$ bond in two ways, giving two possible alkoxide–halide pairings. Only one of them works. NCERT's rule is unambiguous: better results are obtained when the alkyl halide is primary. With a secondary halide elimination begins to compete; with a tertiary halide elimination is total, and an alkene is the only product. The classic NCERT illustration is that $\ce{CH3ONa}$ with $\ce{(CH3)3C-Br}$ gives exclusively 2-methylpropene, no ether at all.
The correct strategy therefore inverts the naive instinct: put the bulky group on the oxygen (as the alkoxide) and keep the carbon electrophile primary. The figure below contrasts the right and wrong pairings for tert-butyl methyl ether.
Figure 2. The same target ether, two retrosynthetic disconnections. Only the pairing with a primary halide survives; the tertiary halide is consumed entirely by elimination.
The same logic governs aryl alkyl ethers. An aromatic carbon never undergoes $\ce{SN2}$, so an aryl halide can never be the electrophile. The ring must instead be supplied as the phenoxide, with the alkyl group brought in as a primary halide. NCERT notes that phenols are converted to ethers by exactly this method, using phenol as the phenoxide moiety.
Worked Examples
Prepare anisole (methoxybenzene).
Disconnect at the $\ce{C-O}$ bond into a phenyl fragment and a methyl fragment. The aromatic carbon cannot be an $\ce{SN2}$ site, so the ring must be the alkoxide. Convert phenol to sodium phenoxide with $\ce{NaOH}$, then react it with methyl iodide:
$$\ce{C6H5OH + NaOH -> C6H5O^{-}Na+ + H2O}$$ $$\ce{C6H5O^{-}Na+ + CH3I -> C6H5-O-CH3 + NaI}$$
The phenoxide oxygen displaces iodide by $\ce{SN2}$ to give anisole.
Prepare tert-butyl methyl ether, $\ce{(CH3)3C-O-CH3}$.
The two fragments are a bulky tert-butyl group and a methyl group. The bulky group must become the alkoxide; the methyl supplies the primary halide. Make sodium tert-butoxide from tert-butyl alcohol, then add methyl iodide:
$$\ce{(CH3)3C-OH + Na -> (CH3)3C-O^{-}Na+ + 1/2 H2}$$ $$\ce{(CH3)3C-O^{-}Na+ + CH3I -> (CH3)3C-O-CH3 + NaI}$$
The reverse pairing — $\ce{CH3O^{-}Na+}$ with tert-butyl bromide — would give only 2-methylpropene, so it must be rejected.
Prepare 2-ethoxy-3-methylpentane.
The structure is $\ce{CH3-CH(OC2H5)-CH(CH3)-CH2-CH3}$. Disconnect at the $\ce{C-O}$ bond into an ethyl group and a secondary 3-methylpentan-2-yl group. The ethyl partner is the unhindered primary fragment, so it should be the halide; the secondary fragment becomes the alkoxide. Form sodium 3-methylpentan-2-oxide from 3-methylpentan-2-ol, then react it with ethyl bromide (or iodide):
$$\ce{CH3-CH(O^{-}Na+)-CH(CH3)-C2H5 + C2H5Br -> CH3-CH(OC2H5)-CH(CH3)-C2H5 + NaBr}$$
Pairing the secondary fragment as the halide with sodium ethoxide as the base would let elimination compete and lower the yield, so the secondary group is deliberately kept on the oxygen.
Aryl halides are never the Williamson electrophile
Diphenyl ether or anisole cannot be made by reacting an aryl halide with an alkoxide, because aromatic carbons do not undergo $\ce{SN2}$. The ring is always introduced as the phenoxide. Examiners often plant a tempting "$\ce{C6H5Br + CH3O^{-}}$" distractor — reject it.
Rule: ring → phenoxide; alkyl → primary halide. Never the reverse.
Dehydration vs Williamson — At a Glance
The decision tree is short. If the target is a symmetrical ether of a single unhindered primary alcohol, dehydration is acceptable and cheap. For everything else — unsymmetrical ethers, ethers with secondary or tertiary groups, and all aryl alkyl ethers — Williamson synthesis is the method, with the bulky fragment placed on oxygen and a primary halide supplying the carbon.
| Target ether | Best method | Key reagents |
|---|---|---|
| Ethoxyethane (symmetrical, 1°) | Dehydration | C2H5OH / H2SO4, 413 K |
| Ethyl methyl ether (unsymmetrical) | Williamson | C2H5O⁻Na⁺ + CH3I |
| Anisole (aryl alkyl) | Williamson | C6H5O⁻Na⁺ + CH3I |
| tert-Butyl methyl ether (3° group) | Williamson | (CH3)3CO⁻Na⁺ + CH3I |
Preparation of ethers in one screen
- Dehydration: protic acid + primary alcohol at low temperature gives a symmetrical 1° ether via $\ce{SN2}$ on a protonated alcohol; high temperature gives the alkene.
- Dehydration fails for 2°/3° alcohols (elimination wins) and for unsymmetrical ethers (gives mixtures).
- Williamson synthesis: sodium alkoxide/phenoxide + primary alkyl halide, an $\ce{SN2}$ displacement that makes symmetrical and unsymmetrical ethers.
- The alkyl halide must be primary; a 3° halide gives only the alkene because the alkoxide acts as a strong base.
- Pairing rule: put the bulky group on the oxygen (alkoxide), use a primary halide for the carbon; for aryl alkyl ethers the ring is always the phenoxide.