Cyclopentadiene can react with itself in a Diels–Alder reactio...

Question

Cyclopentadiene can react with itself in a Diels–Alder reaction. Draw the endo and exo products.

Accepted Answer

All right. Hello everyone. So this question says, and it deals all the reactions cyclohexane can react with itself to form a bicyclic compound illustrate both the endo and exo products. So here on the left side of the screen, I went ahead and pre drew one molecule of cyclo Xa 13 diet. So we have a six membered ring with double bonds in between carbons, one and two and in between carbons three and four. So before we go ahead and talk about the deals, all the reaction in general, let me just go ahead and draw another molecule of cyclo hea 1 3d. So the difference is that for this second molecule of cyclohexane 13 diane, I went ahead and made sure that one of the double bonds inside of the ring was oriented towards the first molecule or the first ring that I drew earlier. And to understand why let's go through deals all their reactions. Now recall first and foremost that a deals all the reaction can otherwise be described as a four and two CYO addition. And so our starting materials are a conjugated dien and a dienno. So in this context, a Dienno describes an all keen or an all kind that has a high affinity for the conjugated diane. Oftentimes you'll see Diano files with electronic drawing groups. So on that note, the reason why it's called four and two Cyclo edition is because four carbons from the conjugated dyane combine with two from the Diop to create a six member green. So to illustrate this on the first molecule of cyclohexane 103 diane, I'm going to label the first four carbons that is the carbons involved in the conjugated dying one through four. So now on the other molecule that I have towards the right side, I'm going to label just one of the double bonds because our second molecule of cyclohexane 13 dien provides our dienno. So after selecting one of the double bonds, I would label its carbons five and six. Now this is not Iupac naming, of course, the idea of this exercise is to illustrate what carbons are taking part in the formation of the ring in the final product. All right. So speaking of a ring, we are going to end up with a characteristic six membered ring in our final product. Now, characteristically, right, if we preserve the numbering scheme that we did earlier for the starting materials, there should be a double bond between carbons, two and three of the ring itself of our product. So let's showcase how this happens. Now, recall that deals alder is concerted which means that everything I'm going to describe right now is happening all at the same time and it's done in the presence of heat. OK. So as mentioned before, right, at the same time that the pi bond between carbons five and six establishes a bond between carbons one and six, the pi bond between carbons one and two moves so that it's between carbons two and three from there. The car, excuse me, the pi bond between carbons three and four helps to establish a bond between carbons four and five. And so that's how we get the six membered ring that we observe in the product of deals all the reactions. But we still have to account for the rest of the starting materials, right? The rest of their structure. And the reason I bring this up is because recall that both our Dyne and Dienno were part of a larger structure. So for example, in the conjugated dyne, we have two more carbons from the ring that did not participate in the reaction. So those two carbons would actually form a bridge on top of the newly formed ring. I'm going to edit this, right. So we have the remaining two carbons from our starting material as a bridge above the product itself. So here in addition to this, because carbons five and six were carbons from the Diop that are themselves part of a different ring, that ring would be associated to the one that was just formed, right? And then the ring from the Diop has its own remaining double bond. So at this point, we have a sense of what the product looks like, but not quite right, because the question is specifically asking us to distinguish between the endo and exo products. Now recall the terms endo and exo describe the relative stereo chemistry of any substituents relative to the newly formed ring. Now, this is especially relevant in bicyclic compounds like the one that we just created. So the endo product, the endo product would have this ring towards the right side from the Diop below the one that we just formed from deals alder. So how do we show that? Well, here, I'm going to start by drawing out our product ring, right, the ring that was formed from deals alder. So before I forget, I'm also going to go ahead and add the two carbon bridge. So from the product ring, the second ring from the Dienno would be depicted beneath the first. And so these hydrogens that are attached to carbons five and six would therefore be pointing upwards. So that's the endo product just so we're clear. Now, the exo product would simply flip the stereo chemistry of the second ring and the hydrogens, right. So for the exo product, I would start by drawing out our newly formed ring. But this time around the ring from the Diop would mean that excuse me, now that I've gone ahead and drawn our deals, all the ring, our substituent that is our ring from the Diop would be pointing upwards this time. And if that ring is pointing upwards, then our hydrogens should be pointing downwards and there you have it. So now here we have both the ando and exo deals with all their products. And so if you watch this video all the way through, thank you so very much for doing so. And I hope you found this helpful.

Cyclopentadiene can react with itself in a Diels–Alder reaction. Draw the endo and exo products.

Key Concepts

Diels–Alder Reaction

Endo and Exo Products

Stereochemistry

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