Aktualny czas:0:00Całkowity czas trwania:7:22

Transkrypcja filmu video

- [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.