Alcohols, Epoxides and Ethers
The Williamson Ether Synthesis
Last updated: June 4th, 2019 |
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
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. 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 don’t 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.
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– .
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 in section 3, below.
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 E2].
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.
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.
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?