Provide a mechanism of the Diels–Alder reaction shown and pred...

Question

Provide a mechanism of the Diels–Alder reaction shown and predict the regioisomer that will form.

Chemical reaction showing Diels-Alder mechanism with maleic anhydride and benzoic acid forming two possible regioisomers.

Accepted Answer

All right. Hi, everyone. So this question is asking us to show the mechanism of the given deals all their reaction which radio isomer will form the 13 or the 14 product. Now recall first and foremost that the deals alter reaction is also referred to as a four plus two cyclo edition involving a conjugated dien as well as either an alkene or an all kind referred to as a dienno file. Now, in this case, the diane is two ey beta 13 diane, whereas the Diop is acrylic acid. And so here in a deal, all the reaction, four carbons from the conjugated diane combine with two from the dienno to create a six R ring. This reaction follows a concerted mechanism in which everything or all steps are occurring at the same time in the presence of heat as catalysts. No, the reason for this question is because our dying and our dienno happened to be asymmetrical, which means that two reg isomers are possible such as the 13 and the 14 product. Now, even though two reg isomers are possible, meaning the 13 or the 14 product are possible only one of them is going to be favored. So the first thing I want to point out here is that the dienno, in this case, acrylic acid, its substituent is a carboxylic acid group which is electron withdrawing, right? It is an G or electron withdrawing group. By contrast, the epoxy group of the dying is electron donating, otherwise known as an eg. So I also want to point out here the importance of the positionings of either the EDG and ewg. So let's say I scroll down and give myself some room here. Let's say that I have a hypothetical conjugated, dying as well as another Diop oops. And let's say that the Diop has only one substituent, which happens to be electron withdrawing denoted by our prime. Now are dying. Our hypothetical diane has one substituent which is electron donating that I have abbreviated as are now the ache that I drew on the screen here is known as one substituted because notice how the electron donating substituent or R is on carbon one of the dying itself. Now a deals holder reaction with a one substituted, dying is going to favor the 12 product. So were going to form a six memberm ring. We're going to observe a double bond in between carbons two and three. And the electron donating and electron withdrawing substituents are going to be on adjacent carbons in the ring itself. Hence the term 12 product. And the reason for this is because the idea is to ensure that all partial charges are in alignment with each other when the ring actually forms, right, like I said previously or indicates an electron donating substitute, which means that the carbon it is attached to. In this case, carbon one is going to have somewhat of a partial negative charge because the electron donating group increases its electron density. Now, by contrast, right, the electron withdrawing substituent is going to remove electron density away from the carbon, it is attached to which means that the carbon it is attached to. In this case, carbon six of my diagram is going to have a partial positive charge, it's going to be electron deficient. So the reason why in this case, the 12 product is going to form is because my partial negative carbon of the dien aligns well with the partial positive carbon of the dienno when creating the ring. However, this is not the scenario that's being described in the actual question because in my dying, I do have an electron donating group, right. However, it's on carbon two instead of carbon one, meaning that it is a too substituted diet. And so when a two substituted diane combines with a dienno file in the presence of heat, which I didn't draw. Previously, we get the 14 product instead we'll see the same six member ring but instead, right, or and or prime are going to be on opposite sides of the ring with two carbons in between them. Because by changing the positioning of the electron donating and electron withdrawing substituents, the partial charges on our carbons are going to be arrange differently and therefore need to align differently when the actual product is being formed. So if I scroll up here one more time, it's worth noting that the 13 product doesn't actually form in either case. Right. So here our 14 product is going to be formed. So let's go ahead and proceed with the mechanism. If I scroll down some more. First, I'll start by drawing my diane, my two substituted diane in which the Foxy group is on carbon too. And now I'm actually going to draw my Diano file a little bit differently. Why? Because by changing the orientation of my Diano file, I'm better showcasing the 14 product. Wait. Oh no, let's go ahead and proceed with the actual mechanism and just to recap this is a simultaneous process, everything is happening at the same time. So at the same time that the pi bond between carbons five and six of the Diop interacts with the diane, the pi bond in between carbons, one and two is going to move in between carbons two and three. Instead, therefore, forcing the Pi bond in between carbons three and four to interact with the dienno once more. Now that create the signature six member ring in which there is going to be a double bond in between carbons two and three. So in this case, carbons two and five had substituents. Carbon two has the oxy group, whereas carbon five has the carboxylic acid and there you have it, here is our 14 product in our deals, all the reaction. So if you stuck around to the end, I thank you so very much for watching and I hope you found this helpful.

Provide a mechanism of the Diels–Alder reaction shown and predict the regioisomer that will form.

Key Concepts

Diels–Alder Reaction

Regioselectivity

Mechanism of Cycloaddition

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