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Diels-Alder: intramolecular

How to analyze the product of an intramolecular Diels-Alder reaction.

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  • piceratops seed style avatar for user einkim
    At , the end product shows the stereochemistry of carboxylic acid as dashed- wouldn't it be a wedge since it is on the right side of the dienophile?
    (3 votes)
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  • blobby green style avatar for user jesse.caron41
    At the carboxylic acid group is on the right of the dienophile. However, at the carboxylic acid group is on the left side of the dienophile. I understand we switched the orientation, but why are we allowed to do this? How do we know how to look at it to determine what is right and what is left?
    (4 votes)
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    • leafers seed style avatar for user kberd003
      He was explaining that we could not obtain an endo product from the first reaction, and basically wanted us to work backwards as in, "Which steriochemistry will give us the endo product?" He rotated the dienophile portion of the molecule so that the carbonyl groups were inward in order to obtain the endo product. Otherwise, it would have been an exo product.
      (1 vote)
  • duskpin ultimate style avatar for user Aritra Mitra
    At , they have the wrong product as the oxygen of ester is replaced by a -CH2 group
    (3 votes)
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  • leafers tree style avatar for user samuel
    I still don't understand WHY atoms on the right side rotate a certain way while the atoms on the other side rotate a different way (wedge vs. dash). Is there a way to know this WITHOUT memorization (maybe conceptually). I don't understand this for the Diene and the Dienophile.
    (2 votes)
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    • starky ultimate style avatar for user Yellow
      Samuel, first of all we must consider that this drawings are representations of the molecules and do not are the molecules. Therefore, the real molecules are three dimensional and our drawings are two dimensional. So, we need some form of representation in the 2D paper of the 3D molecules. This is the wedge and dashes. The 'body' of the molecule is represented as in the plane of the paper, the dashes are going behind this plane, and the wedges its coming at you. That said, when he says that the part right of the molecule relative to the line is going in a certain direction, that is another standard of the model, but is not exactly what really happens in the real word. IS a model of visualization for us. However, some reactions are more favored than others, but this is another kind of problem. Hope that helped!
      (2 votes)
  • blobby green style avatar for user saad.a.khan
    So basically in this molecule, we can just change whether the COOH group is on the left or the right because one of the sigma bonds (attached to the ether Oxygen) can rotate. The endo product is naturally favored, so the sigma bond will rotate to accommodate that since sigma bonds aren't normally fixed in space?
    I know I put that as a statement; but I'm really I'm just asking to make sure I'm not just entirely off.
    (2 votes)
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  • blobby green style avatar for user alpasalp220
    Why do inside substituents on the diene go up? What even determines "up"? Would they be going down if I looked from the opposite direction? So, why this direction and why up?
    Furthermore, what is with the line through the dienophile, determining the same thing?
    (1 vote)
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  • blobby green style avatar for user Henry Kay
    Does adding pressure change bond lengths?
    (1 vote)
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  • female robot grace style avatar for user tyersome
    Is there is a second possible product (the enantiomer of the one shown) for this reaction?

    I think we can get a second product by having the dienophile portion of the molecule approach the diene portion from above (rather than below as Jay shows ).
    (1 vote)
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Video transcript

- [Narrator] If a diene and a dienophile are both contained in the same molecule, that molecule can undergo an intramolecular Diels-Alder reaction. This molecule on the left undergoes an intramolecular Diels-Alder reaction to form the product on the right. And notice how we form two rings for our product, so this is a pretty cool reaction. The first thing we need to do is find our diene and our dienophile. Our diene is over here on the left, so let me highlight the pi electrons. These pi electrons in red are part of the diene, and so are these pi electrons in magenta. Our dienophile is over here on the right. These pi electrons, let me make them blue, these pi electrons are part of our dienophile. First let's show how those two rings form, and we won't worry about stereochemistry right now. Let's just think about moving those six pi electrons to form our two rings. Our pi electrons in red move into here to form a bond between these two carbons. Our pi electrons in blue move into here to form a bond between these two carbons. Obviously they'd have to be a lot closer together in space in reality, and our pi electrons in magenta move into here. Let's draw our cyclohexene ring. Here's our cyclohexene ring. Our electrons in red formed this bond, our electrons in blue formed this bond, and the electrons in magenta formed this bond. Alright, let's look at what's attached to those carbons. This carbon right here is this one in our product, and so there must be a CH2 and then an oxygen. So there is a CH2 and then an oxygen coming off of that carbon. Next let's look at this carbon. Maybe I should change colors here. Let me make this blue. This carbon right here is this one. And we know we have a carbonyl coming off of that carbon, and then we hit this oxygen. That's the same oxygen as before. We have a carbonyl coming off of that carbon, and then that goes straight to the oxygen. Already we see those two rings. The next carbon is this one. I'll make it green. And we can see we have a methyl group and a carboxylic acid coming off of that carbon. Again, I'll just draw these in without any stereochemistry. There's our methyl group, there is our carboxylic acid, CO2H. And then finally, our last carbon, and I'll just make this yellow here, this carbon has a methyl group coming off of it. There's our methyl group. Now we can see that the formation of these two rings. But the next thing we need to do is to account for the stereochemistry that we see in our product here. And notice we're told that this is the endo product. The endo product needs to have an endo approach, and we've talked about this in earlier videos. If this is our dienophile, these carbonyls on the dienophile need to point towards the diene. Let's go to a video so we can see how to use the model set to better visualize the endo approach. Here's our molecule, and I'm going to rotate about this bond, just so we can make our diene and our dienophile approach each other easier. Now we have our carbonyls pointing towards the diene, so this is the endo approach, and now if I hold it like this, you can see the diene and the dienophile. On the left is a picture of what we saw in the video, and we know that a bond forms between this carbon and this one, so I'll draw in a dotted line here. On the right is a picture of our product, so that bond that forms is between this carbon and this one. And let me put it in on the drawing too, so we're talking about this bond. We know that another bond forms between this carbon and this one, so I'll draw in a dotted line here. And that's a bond between this carbon and this one, so I'll draw in that bond. And then on this drawing we're talking about this bond in blue. Next let's think about the stereochemistry of the diene. Here is our diene, and let's put in these two hydrogens. These two hydrogens are inside substituents. We know that inside substituents go up. If we find those two hydrogens on our picture, here they are, and for our product, those two hydrogens are going up in space. If we are staring down at the molecule from this direction, those two hydrogens are coming out at us. In our drawing, here's one of the hydrogens coming out at us on a wedge, and at this carbon would be the other one. Let me go ahead and draw in this wedge. Here's our other hydrogen coming out at us in space. The outside substituents go down, so let me use blue for this. This methyl group is an outside substituent. So is this CH2. On our picture, here is our methyl group and here is that CH2. And for our product, those two are going away from us. It's a little bit hard to see, but this methyl group is going away from us in space, and so is this CH2, so for our product, here is our methyl group going away from us and here is our CH2 going away from us. Next let's think about the stereochemistry of the dienophile. I like to draw a line right here, and we analyze everything on both sides of that double bond. First let's look at what's on the left side. We have a carboxylic acid on the left side of that line and we have this carbonyl. It's hard to see the carbonyl, but it's right back here, and then this carboxylic acid I've represented with this red atom here just to make it easier to work with in the model set. Those two are gonna end up on the same side, so these two are gonna end up on the same side, and they're to the left of this line that I drew. We know those two are going to end up down. If we find them in our product, here is the carboxylic acid. It's actually going down, away from us, and then this bond to that carbonyl is going away from us too. Let's find those on the drawing of our product. Well, here is the carbonyl, and you can see it's going away from us, it's on a dash for our drawing, and here is the carboxylic acid. It's also going away from us on a dash. So those two end up on the same side of the ring, in this case, down in space. Let's look at what's on the right side of our double bond. We know there's a hydrogen here, so let me change colors so we can see these things. There's a hydrogen and we have a methyl group on the right side of our double bond. And here is the methyl group, which I've used as a yellow atom, and here is the hydrogen. The stuff on the right ends up on the same side, and it goes up for our final product. If we look at our product here, this hydrogen is actually going up in space, and it's really hard to see for this methyl group, but it's actually going up if you reorient this molecule. So these two are going up in space. Here is that hydrogen and here is that methyl group. So the stuff to the right side of the double bond goes up, and the stuff to the left side, and again, this is the endo approach, ends up going down.