The Johnson–Claisen reaction is a variant of the Claisen react...

Question

The Johnson–Claisen reaction is a variant of the Claisen reaction studied in this chapter. (a) Suggest a mechanism for the first step. (b) Predict the product (B) that would result from heating intermediate A.

Chemical reaction diagram showing reactants, intermediate A, and product B with conditions for the Johnson–Claisen reaction.

Accepted Answer

All right. Hi, everyone. So this question says that the following is an example of a Johnson Klein reaction, which is a Klein reaction variant provide the possible mechanism or the first step in the reaction. So here based on the reaction scheme provided in the text of the question, right, the Johnson Clyn reaction involves taking an AIC alcohol with a tri alkyl Ortho acetate. And in this case, specifically a tr methyl Ortho acetate. So we combine these two starting materials together in the presence of slight acid to yield a gamma sigma unsaturated era at the end. So in the first step of the reaction, I'm going to go ahead and actually redraw the tri methyl Ortho acetate to go ahead and get started, right. So here are my three metal groups and so to start off the mechanism, right, one of the oxy groups is going to first get pronated with the small amount of acid that is in solution, right? So one of my oxy oxygens, specifically meth oxy oxygens goes ahead and takes a proton to therefore proton name one of the Aloxi groups or alcoy groups. And the first thing or another thing I want to point out here is that this reaction is reversible, right? So I'm going to be using equilibrium arrows or all the steps in this reaction mechanism unless otherwise specified, right. So here in the next step, one of the metox groups of the tral Ortho acetate has since been proton it. And so what that does as a result after I go ahead and adjust this a little bit, this goes ahead and facilitates the formation of an Oxonian cion. And the reason for this is because one of the two remaining meth oxy groups that is neutral is going to donate a pair of electrons in between itself and the carbon it is attached to, thereby displacing the prote needed me Foxy group as one equivalent of methanol. And so once that is completed, we get the following Oxon cion in which oxygen has a positive charge and one of our metox groups has since been removed. So now the aic alcohol can go ahead and interact with the Oxonian cion. So here is the alcohol making sure I'm not missing any carbons like. So, right, so here the aic alcohol is going to go ahead and interact with the oxonian cion. Specifically, it's going to go ahead and attack the carbon that is double bonded to the oxygen in the ox sodium cion, which therefore displaces the pair of electrons of the pi bond towards that oxygen. So that as a result, now our two starting materials have since combined together like so right now, the oxygen of the alcohol has since combined two, the carbon of the Oxonian intermediate. So now a proton transfer is going to occur between the proton of the alcohol and one of our alcoy oxygens. So here one of the oxy oxygens is going to go ahead and take a proton from our alcohol, thereby restoring a neutral charge on the alcoholic oxygen. So here is the metox group then now has a positive charge due to the presence of that proton. So now our alcohol or the oxygen that came from our alcohol no longer has a proton and therefore has a neutral charge. Once again, just checking to make sure I have all my carbons, which I do. OK. So now the remaining meth oxy group, right, specifically that one that used to have a double bond between itself and carbon during the ox sodium intermediate, that one is going to reform a double bond and therefore restore the Oxonian cion. In other words, bring it back because it's going to donate a pair of electrons in between itself and carbon, which is therefore going to once again displace the meth boxy group that has been proton in the prior step, right? So here we get a second intermediate or a second Oxonian cion, I should say here is the product of that, right. So now oxygen has a positive charge. Once again, here is the R group of my original alcohol with my double bond and my methyl groups. So now, right, what I want to point out one more time is that the idea here is to produce a dying intermediate, right, in which there is a carbon, carbon double bond. So the last step of the mechanism requires a dep protonation of that aceto carbon or that methyl carbon. Right, recall that we do in fact have methanol floating in solution. So one equivalent of methanol is going to appear, right? And it is going to remove one of our methyl protons to facilitate the formation of a diet. Now, the pi bond of the Oxonian cion is going to go ahead and move towards oxygen once again to go ahead and yield our final product. Let me scroll down one more time. This is the R group of the alcohol. Here is the oxygen and the only remaining meth oxy group from our trio Ortho acetate as well as our second all keen to go ahead and create our dying intermediate and there you have it right. So at this point, we have formed the first intermediate, which therefore, or by then undergoes a sigma tropic rearrangement to go ahead and produce the final unsaturated Esther product and there you have it. So with that being said, if you stuck around to the end, thank you so very much for watching and I hope you found this helpful

The Johnson–Claisen reaction is a variant of the Claisen reaction studied in this chapter. (a) Suggest a mechanism for the first step. (b) Predict the product (B) that would result from heating intermediate A.

Key Concepts

Johnson–Claisen Reaction

Mechanism of the Johnson–Claisen Reaction

Sigmatropic Rearrangement

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