The Claisen Condensation

by James

I’ve talked about how important enolates are as nucleophiles, and how they’re used in the Aldol reaction (among others).

Let’s talk about a cousin of the Aldol reaction today. Whereas the Aldol reaction is the addition of enolates to aldehydes or ketones, this reaction is the addition of enolates to esters.

It’s called the Claisen condensation.

It follows the same 2 first steps that the Aldol reaction does.

First, we use a strong base – in this case, EtO(-) , to form the enolate. Next, the enolate adds to another equivalent of the ester (addition reaction).

Now comes the third step, which doesn’t happen in the aldol. Since we’ve got an O(-) and an OR group on the same atom, we can have an elimination reaction occur. This expels RO(-), giving us a ketone. So the RO(-) is regenerated, which can then go and repeat step 1 on another equivalent of ester.

So 3 steps are required: 1) deprotonation 2) addition 3) elimination

It’s one of the most important reactions of esters you’ll encounter. But you can see that it’s built out of the same mechanistic steps we’ve been talking about for the past few weeks. Addition. Elimination. Deprotonation.

thanks for reading! James

P.S.. Common question. Notice that the base here (NaOEt) is the same as the OR on the ester? (OEt). Why is that? Well, if we used a different base such as NaOMe, we’d see “transesterifiation” – that is, replacement of OEt by OCH3. Here, with NaOEt, if transesterification occurs, it’s “invisible”.

P.P.S. The version of the Claisen where we form a ring is called the “Dieckmann condensation”. Same reaction in all respects. Confusing, I know.

P.P.P.S. Reaction Guide – the Claisen condensation


{ 1 comment… read it below or add one }


As always, another well written reaction! Thank you for the simplification!


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