Alcohols, Epoxides and Ethers
The Williamson Ether Synthesis
Last updated: November 4th, 2022 |
All About The Williamson Ether Synthesis
In the last post, we discussed the acid-base properties of alcohols. Two posts ago, we said that acid-base reactions are often used to “set up” substitution and elimination reactions of alcohols. In this post, we’ll talk about what is probably the best example of this last point – the Williamson Ether Synthesis.
Spoiler: It’s basically just an SN2 reaction between RO(-) and an alkyl halide, but there are lots of little wrinkles. Read on!
Table of Contents
- What Is The Williamson Ether Synthesis?
- Why Do We Use R–O(–) and not R–OH As The Nucleophile?
- Which Alkyl Halides Work Well In The Williamson Ether Synthesis?
- Starting With An Alcohol: Good and Bad Choices of Base
- How Do We Choose The Solvent In The Williamson Ether Synthesis?
- Summary: The Williamson Ether Synthesis
- (Advanced) References and Further Reading
1. What Is The Williamson Ether Synthesis?
The Williamson Ether Synthesis is an old reaction, dating back to 1851, but hasn’t been surpassed. There just isn’t a simpler way out there to make an ether.
It’s a type of reaction we’ve already seen many times before – an SN2 reaction between a deprotonated alcohol [“alkoxide”] and an alkyl halide that forms an ether. (See article: The SN2 Mechanism)
Note that we’re forming and breaking a bond on carbon here – the textbook sign of a substitution reaction.
That’s the standard reaction. It works. But just by looking at a Williamson that “works”, we only get half the picture. In the rest of this post we’re going to look at some ways the Williamson can go wrong, and answer the following questions:
- Why RO- and not ROH as the nucleophile?
- Which alkyl halides work well, and which don’t?
- If we start with an alcohol (ROH) and add a base to get RO- , how do we choose a good base to use?
- How do we choose our solvent?
Let’s dig in.
2. Why Do We Use RO– and Not ROH As The Nucleophile?
The first thing you might notice about the Williamson is the fact that we use the alkoxide (RO–) in addition to the alcohol (ROH) in the reaction. You might ask, why bother?
Won’t ROH react with an alkyl halide the same way RO–does and still give us an ether?
The answer comes back to what we talked about two posts ago: the conjugate base is always a better nucleophile. The reaction of RO– with an alkyl halide is always going to be much faster than the reaction of ROH because of the higher electron density on the nucleophile (oxygen). That’s why we use RO– . (See post: What Makes a Good Nucleophile?)
The most common way to present the Williamson is to show the alkoxide added in the presence of the alcohol.
It’s also possible to start with the alcohol, add a base to give RO– , and then add the alkyl halide [note the shaded section]. We’ll talk about that below.
3. Which Alkyl Halides Work Well In The Williamson Ether Synthesis? (and which don’t)
Here’s where we come back to concepts from midway through Org 1 that resurface in this chapter on alcohols.
The Williamson Ether synthesis is an SN2 reaction. Remember that since the SN2 reaction proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the “big barrier” for the SN2 reaction is steric hindrance.
The rate of the SN2 reaction was highest for methyl halides, then primary, then secondary, then tertiary (which essentially don’t happen at all). The same pattern exists for the Williamson Ether reaction.
Methyl and primary alkyl halides are excellent substrates for the Williamson.
[one exception: the very hindered tert-butoxide anion (t-BuO-) is slower to perform the SN2 reaction than its other alkoxide counterparts. This alkoxide, also being strongly basic, may instead start to produce elimination (E2) byproducts when primary alkyl halides are used, especially if heated].
Because alkoxides are strong bases (recall the pKa of alcohols is in the range 16-18), competition with elimination [E2] pathways becomes a concern once the alkyl halide becomes more sterically hindered.
For this reason trying to perform a Williamson on a secondary alkyl halide is a bit more problematic than it is for a primary alkyl halide.
One way to attempt to get the SN2 to be favoured over the E2 is to use a polar aprotic solvent (such as acetonitrile or DMSO) that will increase the nucleophilicity of the alkoxide. If heat is applied, however, the E2 will most likely dominate.
One substrate that fails completely with the Williamson is tertiary alkyl halides. This should be no surprise, since a backside attack on a tertiary alkyl halide encounters tremendous steric hindrance. Instead of substitution, elimination reactions occur instead, via the E2 mechanism (See post: The E2 Mechanism)
4. Starting With An Alcohol And Adding Base – What Works?
As mentioned above, the most common way to present the Williamson is to show the alkoxide base being added to the alkyl halide in the presence of its conjugate acid as solvent.
However, it’s also possible to start with the alcohol, add base (generating the alkoxide) and then add the alkyl halide.
The question here is, what base should we use?
First of all, it goes without saying that the base must be strong enough to actually deprotonate the alcohol. Using something like Cl- or RCO2– (acetate) is not going to do the job. Ideally, we’d like something at least as strong a base as alkoxide, or stronger.
Secondly, we need to worry about side reactions. It might help to reflect on how these reactions are run. We typically start with a flask of our alcohol solvent, add base, and then add our alkyl halide. Then, when the reaction is complete, we isolate the product. That means that after the base does its deprotonation, its conjugate acid is still swimming around in solution, and therefore has the potential to react with our alkyl halide (screwing things up).
Here’s an example of a bad choice of base:
NaNH2 is certainly a strong enough choice of base to deprotonate the alcohol. However, after that’s done, we have NH3 in solution, and that’s a good enough nucleophile to react with the alkyl halide, giving us amine byproducts in our reaction. Not ideal!
How can we do this the right way? It’s possible to make the alkoxide directly from the alcohol, simply by adding sodium or potassium metal, which liberates hydrogen. Hydrogen is a perfectly innocuous byproduct as far as the alkyl halide is concerned – it will not act as a competing nucleophile, and being a gas, simply bubbles out of solution. After alkoxide formation we can then add our alkyl halide.
A different (but more common) way to do this is to add sodium or potassium hydride (e.g. NaH or KH). This has the same effect as adding sodium or potassium metal – forms the alkoxide and also H2 – and has the extra bonus of not being strongly reducing, a potential concern if we’re dealing with a complicated starting material that is easily reduced.
5. How Do We Choose Our Solvent In The Williamson Ether Synthesis?
As mentioned above, our normal choice of solvent is the conjugate acid of the alkoxide. [There are exceptions – we might choose to try a polar aprotic solvent if competition with E2 is a concern].
The question is, “why” ? Have you figured it out?
Imagine we were to decide to add sodium ethoxide to propanol, and then add our alkyl halide. What might happen?
This will set up an equilibrium! Our “sodium ethoxide” won’t stay that way for long – it can deprotonate propanol to give sodium propoxide, along with ethanol. We can theoretically have a mixture of sodium ethoxide and sodium propoxide in solution, which could lead to a mixture of ether products. Again, not ideal.
Why give ourselves this headache? It’s pointless. For that reason, we greatly simplify matters if we just use the alcohol solvent that is the conjugate acid of the alkoxide.
6. Summary: The Williamson Ether Synthesis
That’s all there is to say about the Williamson for today. In the next post, however, we’re going to think about this reaction backwards. Starting with a given ether, how can we plan to synthesize it by using a Williamson reaction?
Next Post – The Williamson Ether Synthesis: Planning
(Advanced) References And Further Reading:
- XLV. Theory of ætherification.
Alexander Williamson (1850) , The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 37:251, 350-356,
The original Williamson paper.
- Equilenin 3-Benzyl Ether
M. Hoehn, Clifford R. Dorn, and Bernard A. Nelson
The Journal of Organic Chemistry 1965 30 (1), 316-316
DOI: DOI: 10.1021/jo01012a520
One of the reactions in this paper is a classic Williamson reaction – protection of the alcohol in dehydroestrone as a benzyl ether, using benzyl chloride.
- Total Synthesis of (+)-7-Deoxypancratistatin: A Radical Cyclization Approach
Gary E. Keck, Stanton F. McHardy, and Jerry A. Murry
Journal of the American Chemical Society 1995 117 (27), 7289-7290
In modern organic synthesis, the Williamson reaction is used for the protection of reactive alcohols in a substrate. Common protecting groups include methoxymethyl (MOM) and 2-methoxyethoxymethyl (MEM). MOM protection is employed in this total synthesis by Prof. Keck and coworkers.
31 thoughts on “The Williamson Ether Synthesis”
I had a similar question to the one above–bulky bases are supposed to be an example of a poor nucleophile but a really strong base, so cause E2 reactions, with Hoffmann elimination when applicable for steric reasons. But the textbook and this site also say that using a tert-butoxide ion is a better bet for making an ether by SN2 reaction than say, a tertiary alkyl halide in which case only elimination product comes.
But then, I guess you could make an ether by SN1 reaction on a tertiary alkyl halide? As you said in a response to another comment. The textbook says the reaction of CH3ONa with (CH3)3C-Br gives exclusively 2-methylpropene. Guess that’s an experimental result.
NCERT chemistry class 12
Will reaction of sodium tert-butoxide and chloroethane give ethene (due to E2) or t-butyl ethyl ether (due to SN2) as major product?
Mechanism for Williamson synthesis
You are right that it isn’t in there. Need to fix that. It’s just an SN2.
Thanks for sharing with such useful details.
If the product is water insoluble, you can get rid of excess base simply by pouring the reaction in water; however, my product is highly water soluble, do you have any suggestions how to separate it from the excess base? (I use K2CO3)
Is your product soluble in any organic solvents at all? One way to do it would be to quench the base with saturated NH4Cl solution, and then add equal volumes of brine and n-BuOH. Perform 3 extractions with n-BuOH and your organic molecule should persist in that layer while all the salts will be in the aqueous layer.
Is it possible to use NaOH instead of NaH as a base in the formation reaction of alcohol to Alkoxide?
Sure! There will be an equilibrium between alkoxide and alcohol but will still get the job done. It’s best when the solvent is the conjugate acid of the alkoxide (e.g. EtO- / EtOH).
hello and thank you very much for your notes, I found them really helpful.
I have a question: in my project I had to form a benzylic ether to protect a phenol. I did it using NaI, K2CO3 and BnBr in DMF at 80oC as literature suggested. Is this a Williamson synthesis and does the ether form through an Sn2? If yes then why NaI and K2CO3 and not a stronger base?
The NaI makes benzyl iodide from benzyl bromide. Benzyl iodide is too unstable to isolate.
The pKa of phenol is 10. The pKa of carbonic acid is about 6. The pKa difference is 4. A good rule of thumb is that a pKa difference of 8 or less will be sufficient to get your conjugate base to participate. So although the acid-base reaction will like far on the carbonate side, there will be enough phenoxide to react with your benzyl iodide.
Can we please tell me if SN1 is possible in Williamson ether synthesis?..Because our Teacher today said it can… And i have learned otherwise 😕..Iam a 12th standard student
*ether formation* is possible via SN1. For example, take a tertiary alkyl halide like t-butyl bromide and dissolve in methanol; you’ll get a new ether, t butyl methyl ether.
However that’s not technically a Williamson; a Williamson involves deprotonating an alcohol to give an alkoxide, and then having that react with an alkyl halide to give a new ether. This proceeds through an SN2 reaction.
Ultimately the name “Williamson” isn’t very important; what’s more important is to realize that ethers can form both through SN1 and SN2 reactions.
Hii, I have a dought. Is that possible of good yield with a primary alcohol and a secondary halide having steric hindered groups ?. I tried with a strong base in polar solvent but yield was very less 6-7% with unreacted SM.
The SN2 with secondary alkyl halides, particularly hindered secondary alkyl halides, is quite poor. Why don’t you switch it around and use a secondary alkoxide with a primary alkyl halide? That would work much better.
According to you, the following sequence :
is a SN1 or SN2 mechanism ?
Thanks for your answer
I don’t see the molecules you’re working on, but I assume that NaH is to deprotonate an alcohol, and the BnBr reacts with the resulting alkoxide.
Hi, I learned a lot in this page. However, could you give some more examples about the conjugate acid of the alkoxide while choosing solvent.
Hello! I have a question: why heptanol can not be deprtonated by NaOH?
It can, it’s just that it will be highly reversible. you might have some solubility issues as well as heptanol is on the greasier side of things. Try a phase transfer catalyst. Tetraammonium hydroxide
Found your web pages while helping my Daughter find good sources of info for her Organic course.
Concise but not Terse…
Pros, Cons, Comparisons, Rationale for choosing Reactant-Substrates with Reagents to yield preferred products along with related ” Be Aware Of This” notations are on point.
Your presentations perfectly full fill my Golidlocks criteria for selecting a Professors and Teachers.
Not To Little
Not Too Much,
Tom Jones, D.D.S.
Can we use K2CO3 as a base to make the alkoxide ?
Not a great base to use because it’s quite weak. Equilibrium greatly favours the alcohol, not the alkoxide.
It depends also on the irreversibility, the extrusion of CO2 may help; maybe you need then the alkyliodide ….
Excellent stuff! World needs authors like you in chemistry! This is how an information should be conveyed. Short & covered all essential points..
Using DMSO(aprotic solvent) will favour E2 and not Sn2.Instead of Aprotic,protic solvent should be used.
This is not correct. Aprotic solvents favor SN2 over E2, and DMSO is a useful solvent for an SN2 reaction. Think of it this way. A polar PROTIC solvent would hydrogen bond to the nucleophile and hinder its backside attack.
“… after the base does its deprotonation, its conjugate base is still swimming around in solution…”
“it is non-nucloephilic”
“Easily avoided if we we just…”
Please make the appropriate corrections.
But a really great and useful post, actually (well, as usual).
Glad the mistakes are minor this time. Thank you, as always.