Alcohols, Epoxides and Ethers
Williamson Ether Synthesis: Planning
Last updated: November 4th, 2022 |
Planning A Williamson Ether Synthesis: Avoiding The Pitfalls
- The Williamson Ether synthesis proceeds through an SN2 mechanism
- When planning the synthesis of an ether through the Williamson, remember that SN2 reactions work well for primary and methyl alkyl halides and fail for tertiary alkyl halides.
- Also, SN2 reactions do not work at all with alkenyl or aryl halides.
- This means that in certain cases there are definitely “right” and “wrong” ways to plan the synthesis of ethers using the Williamson.
Table of Contents
- The Williamson Ether Synthesis: Thinking Backwards
- The Mechanism of The Williamson Ether Synthesis
- Planning A Williamson: Two Simple Examples
- Where Planning A Williamson Can Go Wrong
- Summary: Avoid Planning SN2 Reactions On Tertiary or Aryl Halides
1. The Williamson Ether Synthesis: Thinking Backwards
In the last post we introduced the Williamson ether synthesis, one of the most straightforward ways we know of to make an ether. If you’ve been in the trenches long enough, you would have also noticed that it’s actually nothing that new – the Williamson is just “rebranding” of a reaction we’ve seen before, the SN2 reaction between an alkoxide (RO- ) and an alkyl halide (R-X). (See post: The SN2 Mechanism)
[Side note: when Williamson reported the reaction in 1850 he didn’t know what an SN2 was – scientists didn’t even know what electrons were, for that matter – which again goes to show that the science of organic chemistry developed through a lot of empirical observations first, and the theory developed later.]
Often we’ll be in the position of having to plan the synthesis of an ether using the Williamson [this happens to be a very popular exam question]. Today we’re going to apply what we know about this reaction to think backwards. This will be good training for the many occasions later in the course where you’ll be asked to plan syntheses of increasingly complex molecules.
Here’s what we’re going to talk about today:
- Review the mechanism of the Williamson ether synthesis
- Show two easy examples of planning a Williamson
- Show two more difficult examples of planning a Williamson, and how to avoid the most common mistakes
2. The Mechanism of the Williamson Ether Synthesis
As we said above, the Williamson is just an SN2 reaction between an alkoxide and an alkyl halide. Here’s the mechanism.
“X” here is just a good leaving group, such as Cl, Br, I, OTs, or OMs. I like to show specific examples instead of “X”, so in the images below, I’ll be using Br, but keep in mind that they work perfectly well if you use any of those other leaving groups I just mentioned in the preceding sentence. Just don’t use F.
What are the most important points to recall about SN2 reactions?
- It’s a very simple reaction: a bond forms at carbon and a bond breaks on the same carbon.
- Because it proceeds through a backside attack of the nucleophile on the alkyl halide, the “big barrier” (h/t Steven) is “steric hindrance”. The rate decreases as the alkyl halide goes from methyl to primary to secondary to tertiary
- SN2 reactions don’t proceed at all with tertiary alkyl halides. They also don’t work with alkenyl or alkynyl halides (i.e. where the halogen is attached directly to a C-C double or triple bond).
Keeping these factors in mind,Williamson Ether syntheses involving a tertiary alkyl halide or alkenyl halide will fail.
3. Planning The Williamson: Two Simple Examples
Let’s take the first example (diethyl ether) and work backwards. How can we make this molecule through an SN2 reaction?
We need a nucleophile (alkoxide) and an electrophile (alkyl halide) to combine in a substitution reaction to form diethyl ether. So let’s “break” one of the C-O bonds of diethyl ether and tack a good leaving group (Br in this example) on the end.
Notice that it’s symmetrical. No matter what C-O bond you choose to form in the process of the SN2 reaction, you should end up with the same starting materials.
In this example we have an SN2 between ethoxide ion and an ethyl halide. This is the only possible way to use the Williamson to make this molecule. This is a perfectly good SN2 reaction because the electrophile is a primary alkyl halide.
Our next example is ethyl methyl ether (“methoxyethane”). Note here that unlike diethyl ether, this is not a symmetrical molecule.
Therefore there are two possible C-O bonds we could form in an SN2 reaction that generates the ether.
Let’s call them A (green) and B (blue). Imagine what starting materials would be necessary for the SN2 that forms bond A, and then think about what starting materials would be necessary for the SN2 that forms bond B. We call this “retrosynthetically” (reverse-synthesizing, if you will) breaking these bonds.
- Possibility A leads us to a reaction between methoxide (CH3O– ) and a primary alkyl halide (ethyl bromide in this example).
- Possibility B leads us to a reaction between ethoxide (CH3CH2O–) and a methyl halide (methyl bromide here).
Both of these SN2 reactions should work perfectly well. So there’s really no “wrong” way to make this molecule via the Williamson here.
4. Where Planning A Williamson Can Go Wrong
In the preceding examples there was really no “wrong” way to plan the synthesis of our ethers via the Williamson synthesis. It’s not always so straightforward.
Let’s look at t-butyl ethyl ether, for example. As with methyl ethyl ether, there are two possible ways to “disconnect” our desired product into starting materials. Let’s call them Path A and Path B.
- Possibility A involves the SN2 between ethoxide (CH3CH2O–) and t-butyl bromide.
- Possibility B involves the SN2 between t-butoxide [(CH3)3CO– ]and ethyl bromide.
Do you notice something wrong here? One of these SN2 reactions is NOT going to work.
- Possibility A, the SN2 involving a tertiary alkyl halide is a no-go. In fact, what would happen instead would likely be an E2 reaction (elimination).
- On the other hand, Possibility B (anSN2 on the primary alkyl halide ) would work just fine.
So there’s definitely a “right way” and a “wrong way” to build this molecule using the Williamson. If you’re planning an SN2 that involves a tertiary alkyl halide, you’re doing it wrong.
Let’s look at another example, Phenyl methyl ether.
As before, let’s break it down into two possibilities.
- Possibility A involves the SN2 of methoxide (CH3O–) with phenyl halide.
- Possibility B involves the SN2 of phenoxide (PhO–) with methyl halide.
Which SN2 is going to work better?
If you said possibility B, you’re right. Again, Possibility A doesn’t work because we’d be trying to perform an SN2 on an sp2 hybridized carbon. These aren’t effective. The SN2 is a reaction that is only effective on sp3 hybridized carbon atoms.
5. Summary: Avoid Planning SN2 Reactions On Tertiary or Aryl Halides
The bottom line here is that you should plan the synthesis of an ether using a Williamson the same way you’d plan any SN2.
Choose to break down your ether in a way that allows you to employ, ideally, a methyl or primary alkyl halide. Avoid any syntheses that require employing a tertiary or alkenyl halide.
We didn’t really involve secondary alkyl halides here. They are borderline for a Williamson and it’s not as clear a choice.
This begs the question. What happens if we have to synthesize an ether like this one?
Clearly the Williamson is out. So what do we do?
The answer, as we’ll see in the next post, will involve more déja vu from Org 1.
Think about it for a little while. The oxygen is attached to tertiary carbons.
Can you think of a reaction which prefers tertiary carbons as a starting material (versus primary)?
Answer next time.
Next Post: Synthesis of Ethers (2) – Back To The Future!
19 thoughts on “Williamson Ether Synthesis: Planning”
In the synthesis of t-butyl ethyl ether, I think the product of B possibility is not correct. right?
There was a typo which has now been fixed. Thank you
How would K2CO3 act as a base in this reaction?
It it beneficial in some aspect?
K2CO3 could deprotonate the alcohol resulting in the alkoxide, which is more nucleophilic than the free alcohol and will perform the SN2 reaction much faster.
Synthesis of t-butyl ethyl ether, you have wrong the product B
Yes, thank you. Fixed.
Thank you so much for making this easy to understand! You are awesome.
Glad you find the website helpful Mellisa!
In that last example, possibility B, why would the negative charge on oxygen attack the alkyl halide? I think the nucleophilic attack should take place via carbon (negative charge on carbon via resonance).
To support this, I can provide two reasons which are although basically same.
1. (C-) is a better nucleophile than (O-)
2. (C-) is a softer base than (O-) and the site which is to be attacked is also a soft acidic site.
Please correct me if I am wrong.
No. You really need to go back and review nucleophilic substitution. We’re not dealing with a carbanion here. The carbon in CH3-Br is electrophilic.
I think your tert-butyl ethyl ether possibility B has the wrong product. Shouldn’t it be a tert-butyl ethyl ether instead of the methy ethyl ether?
Not for the third synthesis error in the picture? Is there a bad product !?
Yes, fixed. Thanks!
I’m a bit confused about the SN2 between t-butoxide and ethyl bromide. I thought that t-butoxide was a bulky base, so it would cause an E2 reaction. Why does it work for SN2 here?
Hi – with t-butoxide and alkyl halides, SN2 can work (although you might have some E2 byproducts, especially if you heat). It’s the secondary alkyl halides where you get a lot more elimination.
is a methyl or phenyl group a bigger hindrance? for example (ch3)2-ch-ch2-O-ch2-Ph
You’re comparing phenyl and isopropyl, not methyl. If you go by A-values as a proxy for steric hindrance, then isopropyl (A-value 2.15) is less sterically bulky than phenyl (A-value 3.0). https://en.wikipedia.org/wiki/A_value
“… alkenyl or alkynyl halides”:maybe you should include aryl halides as well, and use the term “aryl” when talking about phenyl methyl ether. Also, a quick proofreading of this post might be useful.
But otherwise great work, as usual