Source: Chapter 9 Lecture Notes
Tags: ethers, Williamson ether synthesis, acidic ether synthesis, ether cleavage, protecting groups, intramolecular reactions, ring size, SN2, SN1, E2, THF, diethyl ether, epoxides, organic chemistry
Difficulty: Intermediate Prerequisites: Part 1 of these Chapter 9 notes (alkoxides, alkyloxoniums, alcohols to alkyl halides). You need to be comfortable with SN2 conditions and the concept of alkoxides as nucleophiles.
Once you know how to make alkoxides (Part 1), you can use them to build ethers. Ethers are among the most common solvents in organic chemistry, and understanding how to make them and how to break them apart is essential. This section covers the two main synthetic routes (Williamson and acid-catalysed), the limitations of each, how ethers react under forcing conditions, and the idea of protecting groups, which is a critical strategy in multi-step synthesis. If you skipped Part 1, go back and cover alkoxides and leaving groups first.
Ethers are made by Williamson synthesis (SN2: alkoxide + primary alkyl halide) or by acid-catalysed condensation of alcohols. Ethers are normally inert but can be cleaved by strong acids such as HBr. Protecting groups temporarily mask an –OH as an ether so that other reactions can be carried out elsewhere on the molecule without interference.
Ether
A compound with the general structure R–O–R'. Ethers are common solvents (diethyl ether, THF, 1,4-dioxane). They have lower boiling points than alcohols of similar molecular weight because they cannot hydrogen-bond with themselves. In simple terms, an ether is two carbon groups joined through an oxygen atom.
Williamson ether synthesis
An SN2 reaction between an alkoxide nucleophile and a primary alkyl halide electrophile to form an ether. The classic method for making unsymmetrical (mixed) ethers. Think of it as "alkoxide attacks an unhindered carbon to make an ether."
Acidic ether synthesis
An acid-catalysed (usually H₂SO₄) condensation of two alcohols to form an ether plus water. Works well for symmetrical ethers from primary alcohols (SN2-like). For secondary and tertiary alcohols, the mechanism is SN1-like and produces symmetrical ethers only.
Ether cleavage
The breaking of a C–O bond in an ether using a strong acid (typically HBr or HI) and a nucleophilic halide. Ethers are usually inert, so harsh conditions are required.
Protecting group
A temporary modification to a functional group (commonly –OH) that renders it unreactive during a synthetic step. The protecting group is removed afterwards to restore the original functionality. In simple terms, it is a "mask" you put on a group so it does not interfere with what you are trying to do elsewhere on the molecule.
Intramolecular reaction
A reaction where the nucleophile and electrophile are in the same molecule, forming a ring. Ring size controls the rate, with 5-membered and 3-membered rings forming fastest.
General properties
General formula: CₙH₂ₙ₊₂O (same as alcohols for saturated examples)
Common solvents: diethyl ether (BP 35 °C), THF (BP 66 °C), 1,4-dioxane (BP 101 °C)
Lower boiling points than comparable alcohols
The reason: no O–H bond means no hydrogen bonding between ether molecules
Inert and unreactive under most conditions, which is precisely why they make good solvents
The reaction
An alkoxide (RO⁻) reacts with a primary alkyl halide (R'–X) in an SN2 mechanism
Solvent is typically DMSO or THF
Product: R–O–R' (an ether)
Critical limitations
The alkyl halide must be primary (1° only)
The alkoxide must be non-bulky
Alkoxides are strong bases as well as strong nucleophiles. If the alkyl halide is secondary or tertiary, E2 elimination dominates over SN2 substitution.
In other words: if you pair an alkoxide with a 2° or 3° alkyl halide, you get an alkene (elimination product), not an ether
Planning a Williamson synthesis
When you need to make a mixed ether (R–O–R'), you have to decide which fragment becomes the alkoxide and which becomes the alkyl halide. The rule: the less substituted side carries the halide (so it is primary), and the other side is the alkoxide.
When Williamson does not work
If both possible disconnections give a secondary or tertiary alkyl halide, Williamson synthesis is not a viable route
E2 elimination will dominate under the basic conditions of the alkoxide
You need an alternative method (e.g. acidic ether synthesis for symmetrical ethers, or a different synthetic strategy)
Ring formation
When the nucleophile and electrophile are in the same molecule, an intramolecular SN2 reaction can form a cyclic ether
Ring size controls the rate of reaction
Ring-size preference (fastest to slowest): 3 ≈ 5 > 6 > 4 > 7 > 8
3-membered and 5-membered rings form fastest
Why 3? Despite ring strain, the two reacting atoms are very close (proximity effect)
Why 5? Good balance of proximity and low strain
3-membered rings form via an SN2 backside attack mechanism
Why proximity and entropy matter
The nucleophile and electrophile are already tethered in the same molecule
The activation energy for organisation of the reactants is already "built in" to the system, so no extra entropic cost is paid
Larger rings require the chain to adopt an unlikely conformation, which slows the reaction
Example: epoxide formation
A halohydrin (alcohol with a halide on an adjacent carbon) is treated with NaH in THF
NaH deprotonates the –OH to give an alkoxide
The alkoxide attacks the carbon bearing the halide intramolecularly (SN2, backside attack)
Product: a 3-membered ring epoxide with inversion at the carbon that was attacked
Counting carbons in intramolecular reactions
Count the number of atoms from the nucleophile to the electrophile (inclusive) to determine ring size
The product is the ring that forms fastest, not necessarily the largest possible ring
Primary alcohols → SN2-like
Two equivalents of a primary alcohol are heated with H₂SO₄
The acid protonates one alcohol to make it a good leaving group (alkyloxonium)
A second molecule of alcohol acts as the nucleophile and attacks via SN2
Water is lost; the product is a symmetrical ether
This works because primary substrates favour SN2
Secondary and tertiary alcohols → SN1-like
Protonation gives an alkyloxonium, water departs, and a carbocation forms
A second alcohol molecule attacks the carbocation
This produces symmetrical ethers only (because both equivalents of alcohol are the same)
Mixed ethers are difficult to make this way since you cannot control which alcohol acts as nucleophile vs electrophile
The mechanism varies with substrate class
Limitations
Best for making symmetrical ethers from primary alcohols
With 2°/3° substrates, you get symmetrical ethers but also risk rearrangement and elimination side products
Not suitable for mixed ethers
General principle: ethers are usually inert
Ether cleavage requires harsh conditions: strong acids
These reactions need a nucleophilic acid (e.g. HBr, HI) where the halide ion (X⁻) acts as the nucleophile
Cleavage mechanism
Depending on conditions: SN1, SN2, or E1 pathways are all possible
E2 is not observed here because E2 requires a strong base, and the conditions are acidic
The acid protonates the ether oxygen, converting one of the C–O bonds into a good leaving group
The halide (Br⁻ or I⁻) then attacks
Step-by-step (with HBr, 1 equivalent)
Step 1: HBr protonates the ether oxygen
Step 2: Br⁻ attacks the less hindered carbon (SN2) to cleave one C–O bond
Products: one alkyl bromide + one alcohol
A second equivalent of HBr can then convert the alcohol to a second alkyl bromide
Mechanism matters
For unsymmetrical ethers, the halide attacks the less substituted carbon (SN2 pathway)
If one side is tertiary, that C–O bond breaks heterolytically to give a carbocation, which the halide then attacks (SN1)
The regiochemistry of cleavage depends on the mechanism
The problem
In multi-step synthesis, you sometimes need to react at one functional group while leaving another one alone
Example: a molecule with both an aldehyde (C=O) and an –OH group. If you want to do a Grignard reaction on the aldehyde, the Grignard reagent (PhMgBr) will also deprotonate or react with the free –OH. The –OH is "in the way."
The solution: protect, react, deprotect
Step 1: Protect the –OH group by converting it to an ether (e.g. using acid-catalysed conditions with another alcohol)
Step 2: Carry out the desired reaction (e.g. Grignard addition to the aldehyde) without interference from the now-masked –OH
Step 3: Remove the protecting group (deprotect) to regenerate the free –OH
Requirements for a good protecting group
Easy to install (mild conditions, high yield)
Stable under the conditions of the main reaction
Easy to remove afterwards without damaging the rest of the molecule
Worked example from the notes
Goal: add a phenyl group (via PhMgBr) to an aldehyde in a molecule that also has a free –OH.
Bad approach: run the Grignard directly. The reagent deprotonates the –OH (acid-base chemistry happens first, or ABC: acid-base chemistry before nucleophilic addition), consuming the Grignard before it reaches the aldehyde.
Correct approach:
Protect the –OH as an ether (treat the diol with H₂SO₄ to form a cyclic ether)
Run the Grignard reaction (PhMgBr, 1 eq, THF) on the aldehyde
Deprotect (H₃O⁺ / H₂SO₄ aqueous workup) to remove the ether and restore the –OH
Product: the desired alcohol with a new C–Ph bond and the original –OH intact
Williamson ether synthesis: RO⁻ + R'–X → R–O–R' + X⁻ (SN2, requires 1° R'–X)
Acidic ether synthesis: 2 R–OH + H₂SO₄ (cat.) → R–O–R + H₂O
Ether cleavage: R–O–R' + HBr (excess) → R–Br + R'–Br (via sequential cleavage and substitution)
Ring-size formation rate: 3 ≈ 5 > 6 > 4 > 7 > 8
THF and diethyl ether are two of the most widely used solvents in organic and organometallic chemistry, precisely because ethers are inert under most reaction conditions. The concept of protecting groups is fundamental in pharmaceutical synthesis, where complex molecules routinely have multiple reactive functional groups that must be addressed one at a time.
Students often try to use Williamson synthesis with a secondary or tertiary alkyl halide. This gives elimination (E2), not the ether. Always use a primary alkyl halide.
A frequent mistake is assuming acidic ether synthesis can make mixed ethers. It cannot do so reliably, because you have no control over which alcohol acts as nucleophile and which acts as electrophile.
Students sometimes forget that ethers are inert under normal conditions and try to cleave them with weak acids or nucleophiles. You need a strong, nucleophilic acid like HBr or HI.
With protecting groups, students sometimes skip the deprotection step or forget that the protecting group itself must survive the conditions of the main reaction. Both points are tested.
⚠️ When planning a Williamson synthesis, always assign the primary fragment as the alkyl halide and the other fragment as the alkoxide. If you get this backwards, you will predict an elimination product on the exam.
⚠️ Intramolecular ring-closure problems commonly ask you to predict which ring size forms. Remember: 3 and 5 are fastest.
⚠️ Protecting groups are a multi-step synthesis staple. Be ready for problems that ask you to identify where protection is needed, what protecting group to use, and how to remove it.
⚠️ Ether cleavage regiochemistry: the halide attacks the less hindered carbon (SN2 side) in an unsymmetrical ether. If one side is 3°, expect SN1 cleavage on that side.
True or False: Williamson ether synthesis works well with tertiary alkyl halides.
Fill in the blank: Ethers have lower boiling points than comparable alcohols because they lack _______.
True or False: Acidic ether synthesis with H₂SO₄ is the best method for making mixed (unsymmetrical) ethers.
Fill in the blank: The two ring sizes that form fastest in intramolecular reactions are _______ and _______.
True or False: A protecting group must be stable under the conditions of the main reaction it is protecting through.
Answers: 1. False (E2 dominates). 2. Hydrogen bonding (between molecules). 3. False (it works only for symmetrical ethers). 4. 3-membered and 5-membered. 5. True.
Q: You need to synthesise methyl tert-butyl ether (MTBE) using Williamson ether synthesis. Which fragment should be the alkoxide and which should be the alkyl halide? Explain.
A: The tert-butyl group must be the alkoxide (potassium tert-butoxide), and the methyl group must be the alkyl halide (methyl iodide or methyl bromide). Methyl halides are primary (unhindered), so SN2 proceeds smoothly. If you tried it the other way, tert-butyl bromide with methoxide, the strong base would cause E2 elimination to give isobutylene.
Q: Why can't Williamson ether synthesis be used if both possible disconnections give secondary or tertiary alkyl halides?
A: Alkoxides are strong bases. When paired with 2° or 3° alkyl halides, E2 elimination is faster than SN2 substitution. You would get alkene products instead of the ether.
Q: An unsymmetrical ether (ethyl tert-butyl ether) is treated with one equivalent of HBr. Predict the products and explain the regiochemistry.
A: The ether oxygen is protonated first. The tert-butyl C–O bond cleaves via SN1 (forming a stable 3° carbocation), which is then attacked by Br⁻. The products are tert-butyl bromide and ethanol. The halide ends up on the more substituted side because SN1 is favoured at the 3° carbon.
Q: Explain why a protecting group is needed when performing a Grignard reaction on an aldehyde that also contains a free –OH group.
A: Grignard reagents are extremely strong bases. They will deprotonate the free –OH (acid-base reaction) before they have a chance to add to the aldehyde (nucleophilic addition). This wastes the Grignard reagent. Protecting the –OH as an ether removes the acidic proton, allowing the Grignard to react selectively with the aldehyde.
Q: In an intramolecular reaction, a bromoalkoxide can cyclise to form either a 4-membered ring or a 6-membered ring. Which product predominates?
A: The 6-membered ring predominates. In the ring-size preference order (3 ≈ 5 > 6 > 4 > 7 > 8), 6-membered rings form faster than 4-membered rings due to better balance of proximity and low ring strain.
Williamson ether synthesis connects directly back to SN2 from earlier chapters, and the limitations (E2 with hindered substrates) reinforce the competition between substitution and elimination. Protecting groups will reappear throughout the rest of the organic chemistry course, particularly in carbonyl chemistry and multi-step synthesis problems. Ether cleavage ties back to the acid-catalysed reactions of Part 1, reinforcing that protonation of oxygen makes it a better leaving group.
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