Diels-Alder: stereochemistry of diene

- [Instructor] This video we're gonna focus on the stereochemistry of the diene. On the left is our diene, on the right is our dienophile. So first let's draw our product. This Diels-Alder reaction without worrying about stereochemistry. So we know that a Diels-Alder reaction, involves a concerted movement of six pi electrons. So if these pi electrons move into here before a bond between these two carbons, these pi electrons move into here. We form a bond between these two carbons and then these pi electrons would move down. So let's draw the product, again ignoring stereochemistry for the time being. So let's put in, let's put in these bonds, we have these methyls right here. And then I'm going to abbreviate these esters as CO2Me. So CO2Me down here. The reason that we chose this alkyne as our dienophile is because we don't have to worry about stereochemistry at these two carbons. We do need to worry about stereochemistry at these two carbons and those came from our diene. So let's look in more detail at the structure of the diene. So this is a trans-trans diene. So this double bond is trans and so is this one. Let's focus in on this inside substituents first so these two hydrogens and let's look at them in the picture. So those two hydrogens are on the inside and when we form a bond between these two carbons and between these two carbons. Notice what happens to those hydrogens. On the left here those hydrogens are on sp2 hybridized carbon so this carbon is sp2 hybridized and so is this one. But on the right, notice how those two carbons are now sp3 hybridized. So those two hydrogens, those inside substituents go up in our product. So if we're looking down at our ring this way, those two hydrogens are coming out at us in space. So let's go ahead and draw that. So we have our ring and we put in our double bonds here, and we have our esters, let me go ahead and put that in. So CO2Me and CO2Me and let's draw in those hydrogens coming out at us in space. So those two carbons that I marked in red, those are these two carbons. So we should have a hydrogen coming out at us at both of those carbons. So here's a hydrogen coming out at us on a wedge and then here is a hydrogen coming out at us on a wedge. All right next, let's look at the outside substituents. So let me use blue for this. So our outside substituents are these two. So in blue, these are the outside. So we have two methyl groups that are outside and I made the methyl groups yellow in the model set here. So here are the two methyl groups and when these two carbons, when these two carbons go from sp2 to sp3 hybridized notice what happens to the methyl groups. If the inside substituents go up, the outside substituents go down. So these are a methyl groups and they go down in space. So looking down at our ring both methyl groups will be going away from us in space. So let's go ahead and draw those in on a dash. So that methyl group is going away from us and so is this one. For this next problem we have the same dienophile but we have a different diene. This is cyclopentadiene. You can see the double bonds are both cis. So when we think about our Diels-Alder reaction, we move our six pi electrons. So these pi electrons move in here to form a bond between these two carbons. These pi electrons move into here so we form a bond between these two carbons and this pi electrons move over to here. So when we draw our products, we go ahead and draw in this ring here. And let's highlight some of those electrons. So we're not done drawing the product yet obviously. These electrons in red move into here and then we had electrons in blue so these pi electrons in blue move into here. And then finally our high electrons in magenta move over to here. And then, but we still have a bridging CH2 so this bridging CH2 we need to draw that in on our product. So let me just sketch that in. So this is actually a top view of our product. Then we still have our esters, so CO2Me coming off of this carbon and CO2Me coming off of this carbon. So let me draw those in here. So this is one way to draw your product but if you wanna draw a little more three dimensional standpoint, let's look at the picture down here. So our bridging is CH2 would be this carbon. So we have these bonds right here going to the bridging CH2 and those are these bonds. Those are inside substituents and we know inside substituents go up. So when we form our bonds, so we formed our bond in red over here so this one, that's the bond between this carbon and this carbon. So we form a bond between these two carbons which is this bond. And then we also formed our bond in blue over here so that's the bond between this carbon and this one. So we form a bond between these two and that's this bond right here. Those inside substituents, this bridging CH2 goes up in space. So if this goes up and we can sketch in our product from a different point of view so I'm gonna redraw what we see here in the picture. So this is a form of drawing a product that you see a lot. So let me sketch this in, so this takes some practice. All right and let's draw in our double bonds and then we have our CO2Me and our CO2Me. So our bridging CH2 is right here so that went up in space. For our last problem we have our same dienophile. But notice this time for our diene we have cis at this double bond, but this double bond is trans. So let's first move our six pi electrons. So this pi electrons move into here, so we form a bond between these two carbons. These pi electrons move into here so we form a bond between those two carbons. And these pi electrons move down. So let's draw, let's start to draw our product so we have our ring and then we have two double bonds in our ring. Let's go ahead and put in our esters here. So we have CO2Me and then CO2Me. Next, let's think about our substituents. So look at the inside ones first. So we have this hydrogen and this methyl group. And we go down here to the picture here is that hydrogen and here is that methyl group. So when we form our bonds between these two carbons and these two carbons are inside substituents go up. So here is that hydrogen and here is that methyl group so for staring down this way at our product those two will be coming out at us in space. So let's go ahead and put in those groups so the methyl group right here is on this carbon and that's this one. So we have a methyl group coming out at us in space at this carbon so let me draw that in. So we would have on a wedge CH3 so let me fill in that wedge. And then for our other carbon, let me mark it right here so this carbon which is this one would have a hydrogen coming out at us in space. So let me draw in that hydrogen coming out at us right here. Next let's think about our outside substituents. So I'll make those in blue. So our outside substituents are this hydrogen and this methyl group. So let's find those on our picture. So here is that hydrogen and here is our methyl group. And we can see where they end up so this hydrogen goes down in space. So we need to put that on a dash, so let's draw in our hydrogen on a dash at this carbon. So the outside substituents go down. And this methyl group right here, this one goes right here. So it's down relative to the hydrogen so let's go ahead and put that in. So let's put in our methyl group, our CH3 down at this carbon. So what would happen if you took your diene here and you looked at it from a different perspective. So let's go to the video so you see what I mean. So here's our diene with the yellow representing the methyl groups. And if I just rotate it this way, now we can see we have this methyl group on an inside substituent and this methyl group would be an outside substituent. As we saw in the video this is the same diene as before. It's the same as this one. It's just we viewed in a different perspective. So we move our six pi electrons around. So we form a bond between these two carbons, we move these pi electrons down and let's draw in our ring. So we have our ring, we have two double bonds in the ring and then we have our esters coming off so CO2Me and CO2Me. Next we think about our inside substituents so we have a hyrdrogen and we have our methyl group. The inside substituents go up so the hydrogen and the methyl group are going to be on a wedge. So this hydrogen is on a wedge at this carbon. So we put that in. And then this methyl group is on a wedge at this carbon so filling that wedge here and write CH3. Next, we think about our outside substituents so we have a methyl group and a hydrogen. Outside substituents go down so we could draw in that methyl group is going away from us in space so there's our CH3. And this hydrogen is also going away from us in space so it's on a dash. So now we have this as our product. Before, when we looked at our diene in this way, we got this as our product. What is the relationship between this molecule and this molecule? They are enantiomers of each other. So really this reaction would give us a pair of enantiomers.