The Eschenmoser–Claisen reaction is a variant of the Claisen r...

Question

The Eschenmoser–Claisen reaction is a variant of the Claisen reaction studied in this chapter. Based on your answers to Assessments 22.53–22.55, predict a product of this reaction.

Chemical reaction diagram showing reactants and conditions for the Eschenmoser–Claisen reaction.

Accepted Answer

All right. Hi Ron. So this question says that the Eschen Moser Klein reaction is a variation of the Klein reaction. What is the product of the following Eschen Moser Klein reaction? So recall first and foremost, right, that the Eschen Moser Klein reaction involves heating and aic alcohol. Like the one pictured on the left here in the presence of NN dimethyl aide dimethyl acetyl in order to produce a gamma sigma unsaturated a mite. So let's go ahead and showcase how this actually happens. But to do that, I'm actually going to move some stuff out of the way there we, so the first step of the reaction is going to involve or a mean intermediate because our amine here, our amino nitrogen is first going to create a double bond between itself and the carbon it's attached to, therefore creating an I minium ion after one meth oxy group is removed, right? So once that happens, we're only going to have one meth oxy group attached and there's going to be a double bond between carbon and nitrogen, therefore giving nitrogen a positive charge. So it's the I minium ion. Now that our alcohol can go ahead and attack. So here is the art group of my alcohol like. So, so here the hydroxy group of the alcohol is going to attack the carbon of the imi ion, therefore, restoring a neutral charge back to nitrogen. So now there's going to be a new carbon oxygen bond. However, because the oxygen of our alcohol still has hydrogen attached to it, right, it's going to have a positive charge in the set. However, right, we do have to recall the metox group that detach itself in the first step of this reaction. So that metox or meth ide ion, I should say is going to go ahead behave as a base and detonate the oxygen of our alcohol, which will therefore result in the same intermediate except now it's going to be neutral. I'm going to move this thing over because I'm seeing that it was a little bit crowded. So now at the next step at this point, in order to yield our aide in the beginning, we have to somehow facilitate the removal of our second meth oxy group in our intermediate. So for that, we're going to see another interaction between our meth oxy group and the molecule of methanol that formed in the prior step when meth ide took a proton, right. So heres methanol. So now our metox group is going to take the proton from methanol because by doing so, right, the metox group becomes a more suitable living group. So here is one of my R groups oxygen. So now our meth group is going to have an extra proton thereby giving it a positive charge. So then another I minium cion is going to be generated. So nitrogen is going to lend its pair of electrons in between itself and carbon. Therefore, displacing our proton in methylene of methanol, right. So once that happens, we get are nitrogen now with a double bond in between itself and carbon, thereby giving it a positive charge. Now, our second I minium kaion intermediate is going to eventually produce our dying once metox side actually detonates one of the protons on the carbon adjacent to the I minium itself, right? So this primary carbon here is going to be detonated and that's going to create a new pi bond between our two carbons, therefore, restoring a neutral charge on nitrogen. So that gives us our dying intermediate like. So, so now our dying intermediate can go ahead and and go through the cli in rearrangement, otherwise known as the sigma tropic rearrangement, write a 33 sigma tropic rearrangement. So for this step, I'd like you to imagine that the region here in our intermediate that involves both double bonds as well as the two atoms in between them. Wait, that's a total of six atoms. So those atoms can be labeled and they can be divided into two distinct three atom regions in which the two atoms in between both double bonds are each labeled number one. So now for the sigma tropic rearrangement recall that this is a concerted mechanism. So everythings going to happen at the same time. But first, what I'm going to draw first is that the bond between the carbon from our alcohol and our oxygen is going to go ahead and break and create a new pipe bond. So that by consequence is going to displace the pi bond already in between carbons two and three, therefore, creating a new sigma bond in between both carbons labeled number three. And then that is going to displace another pi bond, thereby creating a car bal. And with that, we get our final product with a new car bono instead. Now it's worth noting here also that because the involved double bond had a Z configuration that led to a sin rearrangement. So a rearrangement occurred on the same face of the molecule. And overall, that's going to give us our final product, which I'm going to go ahead and redraw at the very top of the screen here. Our final product which is a gamma sigma unsaturated amide is as follows showing stereo chemistry. And there you have it, our final product is going to be oops re three isopropyl nn six tri methyl hept four in amide. So with that being said, I know this was a lot, right? But if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

The Eschenmoser–Claisen reaction is a variant of the Claisen reaction studied in this chapter. Based on your answers to Assessments 22.53–22.55, predict a product of this reaction.

Key Concepts

Eschenmoser–Claisen Reaction

Claisen Condensation

Enolate Ion Formation

Watch next