Diels-Alder: stereochemistry of dienophile
How to analyze the stereochemistry of the dienophile in a Diels-Alder reaction.
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- Will we always care about stereo chemistry? or are there time where we don't have to worry about it?(3 votes)
- You'll always have to worry about stereochemistry. It's important for drug manufacturing. One enantiomer might alleviate morning sickness during pregnancy, the other might cause terrible deformations in the child.
In fact, that was documented in Germany, it's fairly well known among organic chemistry teachers to drive home why stereochemistry is important. Look up stereochemistry on wikipedia and read about the thalidomide example.(9 votes)
- The choice has been made to always draw de groups on the left side pointing down in the final product. I assume this choice is arbitrary? The diene and the dienophile could react in a way that the groups on the left end up pointing up, couldn't it?
In the next videos (stereochemistry of diene; endo rule) the same choice has been made.(3 votes)
- no, it's not determined by left or right -ness
if that's the case, we can't get the enantiomer for the last example in video
to be specific, it depends on which one of the double bonds in a dienophile would attend DA reaction. this means you can get a wedge or dash r group as a product from both of a leftward and rightward r group as a reactant(1 vote)
- at the very end, instead of placing "+ enantiomer" couldn't you put a (+/-) in front of the product?(2 votes)
- No. That's never used in notation. "+ enantiomer" is how a mixture of enatiomers is always noted.(2 votes)
- Is there any way of shortcut to remember which r is going into the plane and which one is coming out of the plane...(2 votes)
- you don't need to remember any shortcut
cause there's no shortcut
cause any r can be coming out or going away from the plane, depending on which one of the double bonds would attend DA reaction
thus we get 1 pair of enantiomers in the last example even though we start with one specific direction of r-groups (upper-rightward and lower-leftward)(1 vote)
- I thought there wouldn't be stereochem at5:27since neither C is truly a chiral center...is the path around the ring not the same for both?(1 vote)
- You are generating two stereocentres from achiral reactants.
If you start with achiral reactants, the product must be achiral.
In this case, the product is a meso compound, because it has an internal plane of symmetry.(4 votes)
- I am confused, my text shows the reaction of a cyclopentadiene with a dienophile containing the R group on the right side of the double bond, but the product shows the R group on dashes instead of wedges....(1 vote)
- Enantiomers because the dienophile can be facing the diene in two different orientations.(2 votes)
- Since both are planar, do they react one on top of another or next to each other?(1 vote)
- What if there are multiple double bonds between the carbons? Then how would we decide across which double to draw the cyclohexane ring?
- Isn't there one conformation that's more favorable than the other (i.e. endo versus exo products)?(1 vote)
- At2:27, Jay talks about the -R (EWG) group(s) of the dienophile going out of the plane of the cyclohexene, if they are away from the conjugated system, and later he says that the -R(EWG) group(s) will go into of the plane of the ring if they are oriented opposite to each other. But, why should this happen? In any approach, both the possibilities are equally likely.(1 vote)
- [Narrator] In this video, we're going to look at the stereochemistry of the dienophile. But first, a quick review of the Diels-Alder Reaction. On the left we have our diene; on the right is our dienophile. We know that our Diels-Alder Reaction involves a concerted movement of six pi electrons. So these pi electrons move into here to form a bond, these pi electrons move into here to form a bond, and these pi electrons move down. So that gives us our product. So the electrons in red move in to here to form this bond. The pi electrons in blue form this bond, and finally, the pi electrons in magenta are these electrons in our product. So let's think about the stereochemistry of these dienophiles. So here is our dienophile, and we have our two R groups cis to each other, so they're on the same side. So what does that do for the product, in terms of the stereochemistry? How do those two R groups end up? Well down here, let's look at the picture of the diene and dienophile. So up top here is the diene, which you could think about as being one plane, and down here is the dienophile, which would be another plane. So the two R groups I made red right here. And the planes approach each other. So the diene and the dienophile approach each other, and we know a bond forms between this carbon and this carbon, so think about a bond forming here. And a bond forms between this carbon and this carbon. So a bond forms here. And notice what happens when we go to our product, over here on the right. We're thinking about our dienophile. This carbon goes from being sp2 hybridized, to being sp3 hybridized. And the same thing with this carbon. From sp2 to sp3 hybridized. So we need to think about those R groups. Since we have a concerted movement of electrons, these two R groups end up on the same side. And if you're staring this way, down at your cyclohexene ring, these hydrogens will be going away from you in space. So these R groups are actually coming out at you in space. So we're drawing in our stereochemistry, we draw in our cyclohexene ring, and we put in our R groups coming out at us in space. So let's go back to the drawing over here on the left, and let's think about stereochemistry. So I like to draw in the hydrogens over here sometimes, and think about the groups on the left and the right of this line. So if I'm drawing the dienophile approaching the diene in this fashion, the two R groups are on the right side, and those end up as both wedges coming out at us in space. So the stuff on the right side of the line, ends up as a wedge, and for the cis, all right we have a cis alkene on the left, and those two R groups end up cis on our ring as well. The two groups are on the same side of the ring. The stuff on the left side of the line that I drew, in this case, two hydrogens, that ends up going down in our product. So here are those two hydrogens. Those would be dashes if we put them in on our product. What happens if we show our dienophile approaching our diene in a different way? In this case, if you think about our double bonds, the two R groups on the left side of the double bonds, let me go ahead and make those red, so these two R groups, on the model, you can see the R groups are right here. And on the right side of our double bond, in terms of how I've drawn it, we have two hydrogens, and here are the hydrogens on the model. You know that a bond forms between this carbon and this one, and a bond forms between this carbon and this one, so when we look at the product, here are the two bonds I just pointed out, and we can see that our two R groups are on the same side. And if we're staring down in this direction at our cyclohexene ring, let me go ahead and draw in our cyclohexene ring like that, these two R groups are going away from us in space. So those two R groups should go on dashes. So I'm going to go ahead and put these dashes in. And so the stuff on the left side of our double bond, let me go ahead and put that dienophile over here. Stuff on the left side, these two R groups, would end up going down. So I'm just very consistent in how I draw my dienophile, in thinking about my product. And I think a systematic way helps you, when you're doing Diels-Alder Reaction. The stuff on the right side, these hydrogens, these end up, so they're up relative to our plane, if we're staring down this direction, so we put those hydrogens as wedges, if we put them in for the products. Now if these R groups are the same, it doesn't matter how you would represent the product. So let's look at an example of what I mean. Let's draw the product for this Diels-Alder Reaction. We know these electrons move into here, so we form a bond between these two carbons. These electrons move into here, and these electrons move down. So we get a cyclohexene ring, so let me sketch that in first, and then we think about the stereochemistry of our dienophile. We know that the stereochemistry is cis. So we need to have those two groups on the same side of the ring. We could use either wedges or dashes here. So I'm just going to use two wedges, so let me put those wedges in. Then I'm going to draw in an ester up here. And we draw the same ester in down there. And there's our product of our Diels-Alder Reaction. What if we have our two R groups trans to each other? So let me go ahead and draw a line in. Look at the R group on the left, that's the one going towards the diene. So on the model, that's this R group, the one that's pointing towards the diene. The one to the right of the line, this R group is going away from the diene, so that's this one on the model. We know that a bond forms between these two carbons, and a bond forms between these two carbons, so we get the product on the right. So here are the bonds that I just pointed out. And if we're staring down at our cyclohexene ring, I'm going to draw that in, this R group is going away from us in space. So if we work backwards, that's this R group. That's the one on the left here. So that should be a dash, for our product. So we put that R group in as a dash. This R group is coming out at us in space. And that was this one, the one going away from our diene, that's the one on the right. But that's coming out at us in space, that should be a wedge. So let me put that R group in. What if our dienophile approaches our diene in a different way? So now when we look at this line here, this is the R group that's going towards the diene, the one on the left. So that's this one back here. And this R group is going away from the diene, so that must be this R group. So we draw in our bonds that we know are forming, a bond between those two carbons, and a bond between these two carbons. We look at our product on the right with those bonds in there, and also the double bond, which I'm just not really focusing on at the moment, and we look down at our cyclohexene ring, so I draw that in. Now when we look down, let's look at this R group first. That R group is going away from us in space. And that was this one. That's the one going towards the diene. That's this one right here. So to the left, the R group on the left, is going to get a dash. So that's this one. So I put in an R group on the dash, and then this R group is coming out at us, it's going up. And that was this one, the one going away from the diene. And that's this R group. So from the right, it gets a wedge. So this should be a wedge. Notice that, if we have R groups that are the same, these two are enantiomers of each other. So we would get a pair of enantiomers for our reaction. Let's say we were given this Diels-Alder Reaction on a test. On the left is our diene, on the right is our dienophile. We know a bond forms between these two carbons, and between these two carbons. So we can start by drawing in a cyclohexene ring. Let me put that in. Next, we think about the stereochemistry of our dienophile. If we draw a line like that, the stuff to the right of the line ends up as a wedge. So this ester is going to be on a wedge. So let me put that ester coming out at us in space. The stuff to the left of our line should be on a dash. It's going away from us in space. So we put that ester in here. And we know, from the pictures that we just saw, we're also going to get the enantiomer. So we get this compound plus the enantiomer, as our products.