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Williamson Ether Synthesis: Planning

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). [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.]

Here’s the bottom line for today’s post. 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:

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?

Keeping these factors in mind, Williamson Ether syntheses involving a tertiary alkyl halide or alkenyl halide will fail. 

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 (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 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.

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.

The Bottom Line

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!

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