Diels-Alder Reaction

Concept: Concept: General Features

10m
Video Transcript

In this video, we're going to discuss the most famous pericyclic reaction in Organic Chemistry, and that's called the diels alder reaction.
The diels alder reaction is a heat-catalyzed reversible pericyclic reaction between two different molecules that we’re going to go into more depth on. The one thing in common between all diels alder reactions is that they're always going to yield a six-membered ring as their product. You always know that you’re going to create one new ring through the formation of a diels alder reaction.
We need two components plus heat to make this happen. We're going to need one, a 1,3 diene. Two, we're going to need a dienophile. I’ve recognized that these are terms that you’re probably not that familiar with. Let’s really just dive into what that is.
First of all, a 1,3 diene is pretty simple. It sounds like exactly what you’re thinking. It’s a diene that is at the 1 and the 3 position. Basically, another way to say it is that it just has to be a conjugated diene because if it’s not 1,3, let's say that we used a 1,4 diene. Then that would no longer be a conjugated diene. We would actually call that an isolated diene because now the double bonds wouldn’t be next to each other. They would have a space in between. You can't use a 1,4 diene. That’s isolated. You need to use a conjugated diene.
Let’s look at some examples. This first one is pretty easy. That looks like a typical diene. You can add any other R-groups. The important thing is that you at least have that 1,3 diene. Here you see that we actually have a cyclic in the middle. We have a cyclic 1,3 diene because one of them starts at the 1 and one of them starts at the 3.
You might be wondering, ‘Johnny, why are you using those specific atoms to count as 1 and 3?’ It doesn't matter where you start as long as you have diene starting, basically two carbons away from each other, at the 1 and at the 3. You have two diene, two double bonds starting. That's another diene. Here we have another example of a 1,3 diene. I’m just trying to show you guys how 1,3 dienes can come in all shapes and sizes. We're just caring about the fact that they're conjugated to each other.
What’s a dienophile? By definition, the word phile means lover. A dienophile would be a molecule that loves dienes. Dienophiles are actually really easy. All they are is that they are alkenes or alkynes. That's it. It's really that easy. A dienophile could just be a simple cyclohexene, just having that double bond there because it’s a dienophile.
Notice that this next molecule here has two double bonds. Which part of it do you think is the dienophile? I said in the definition it has to be an alkene or an alkyne. This is actually the dienophile, nothing else. The carbonyl doesn't count.
Check this out. That’s weird. Did I make a mistake? Did I drag the wrong molecule at the same molecule to this box? No, I didn’t because it turns out that dienes have alkenes in them. That means both of these double bonds can act as dienophiles. That means you can sometimes see dimerization taking place with these reactions where one molecule reacts as the diene and the other reacts as the dienophile. They react together to form a dimer or something that there's two of them now attached to each other. That's something we need to be aware of.
Here's our last example. Triple bonds have pi bonds in them. This can also be a dienophile. Pretty simple so far. We know that we need a diene, a 1,3 diene. We need a dienophile. We need heat because I told you it’s a heat-catalyzed reaction. But it turns out there's a few more technicalities we have to go over before you’re ready to draw these. One is that your 1,3 diene has to be in a certain shape. You can’t just use any 1,3 diene. For the mechanism to work, you’re going to need to have your 1,3 diene rotated into what's called the s-cis or sigma-sis conformation.
Remember from organic chemistry one that sigma bonds are able to freely rotate as much as they want, meaning that just because it's in that position now doesn't mean it will always stay there. But you have to make sure that at least it’s able to rotate into the s-cis conformation momentarily. Let me show you what I mean.
This diene is not rotated into s-cis. This is what we would call s-trans. Why? Because if you were to draw a dotted line along the single bond or the sigma bond, what you would find is that your R-groups are in opposite sides. That’s what we call s-trans because your sigma bond is rotated in such a way that they’re in opposite sides.
If we were to rotate that sigma bond, what we would find is that now when we draw that same line, now they're on the same side. This is what we would call s-cis. This is the way we need it to be rotated. This would be a big no no. You can’t start off like that. In order to begin your diels alders reaction, you must rotate it first into the s-cis and then you can proceed. Not that bad.
Now let’s look at the general mechanism. The general mechanism is going to be an s-cis diene, specifically s-cis 1,3 diene with a dienophile. Remember I told you guys that a dienophile can be any alkene or alkyne. This molecule right here is a perfect dienophile because it’s just got that double bond.
The cyclization reaction is a 3-membered or 3-arrow reaction where you would get the dienophile initiating because remember it’s like the lover of the dienes. It just wants to attack it. I would go ahead and I would attack one of the edges. But if I make a bond, I have to break a bond because I'm in violation of an octet if I don’t. then this double bond is going to make a new double bond here. Once again, make a bond, break a bond. I’m going to need to break that last diene. This one comes over and attaches to the other side of the double bond.
This is going to form two new single bonds. This forms a new single bond here and this forms a new single bond here. Then this arrow that’s going in between the dienes forms a new pi bond here. Our final product has two new bonds and a double bond. As you can see, I now have a six-membered product, a six-membered cyclic product. Cool so far? That's the general mechanism.
You could get a problem that's just that easy. But diels alder can get a little more complicated as I’ll show you guys. I’m going to start layering on the complications. The first of which being it’s pretty straightforward that the stereochemistry of all substituents must be retained. You have to identify the stereochemistry of your dienophile and your diene so that when you react this together and make a ring, that the stereochemistry is preserved.
Check out this first diels alder. We have a 1,3 diene and a dienophile. But notice that my R-groups on this dienophile are in a cis position. This would be a cis alkene. Remember, double bonds don't twist. It’s always going to be cis. It can't be a trans.
The way we can tell it’s cis is if we were to draw that dotted line or fence that I like to use. They’re both on one side of it. When we react this product, we’re going to draw our arrows. When we react this product, you need to make sure that your R-groups remain cis to each other. They have to remain on the same side of the ring.
Did I have to face them up? No. I could have also faced them down. The important part is that they’re both facing the same exact direction. It wouldn't make sense if I put one R up and one R down because that would like trans. That's not what it began with.
Awesome. That’s pretty straightforward. In the same way if I begin with a trans double bond, as you can see this one’s different. Then I’m going to get a pair of enantiomers because as you can see, now I'm going to get trans products but there's two different ways that those trans products could orient each other. I could get R1 in the front, but I could also get R2 in the front just depending on how the attack works.
In this case, I would have to draw a pair of enantiomers because I have two different trans products that are possible. It’s just really basic stuff that you have to make sure you get right and make sure you pay attention to the stereochemistry in order to get the full credit for this question.
Pretty easy so far, but that's not it. There's a few more complications with diels alder. Let's go ahead and move on to a few more concepts. 

Concept: Concept: Bridged-Products

12m
Video Transcript

Now, let's discuss a few added complexities of the diels-alder reaction. So it turns out that sometimes when you run a diels-alder reaction you get a bicyclic bridged molecule as your products, as shown in this diagram right here, so, how does that happen and why would we get a bicyclic as our product? Well, it turns out that by cyclic bridge products are obtained, when you're s-cis diene cyclic, okay? Remember, that I told you guys that one three dienes could be regular straight chains but they could also be found within rings. So, if your diene is found within a ring it's going to produce, we call a bridge product, okay? So, let's just look at these two different examples, if you're starting off with a normal acyclic diene as, we have here, okay? Then you're just going to wind up getting a six membered ring as your product and we're used to seeing that, however, if you start off with a cyclic diene. Notice that we already have one ring, that one ring is here. So, when we go to react, let me just make that a little more clear one, okay? So, when we go to react this and form another six round ring on top of it, we're actually going to get a bycyclic product, there's going to be two rings present not just one because we started with the ring to begin with. So, now we're layering another ring on top of that okay? Well, let's just go through this really quick, I know that you guys already know how to draw the cyclic product for an acyclic diene but let's see why you get a bridge in the cyclic sign, because notice that the cyclic sign is going to have a diene portion of the molecule and it's also going to have a non diene portion as identified, that's a little star there, there's some portion of this molecule that's in the ring that is not a part of the diene, it's outside of the diene. So, when the dienophile goes to attack with its three mechanistic arrows, you've got the one, two, three, what we find is that one or more of these atoms get pushed out of the way and get pushed above the entire reaction, you can almost imagine it like the dienophile is a diene lover, right? So, imagine, that the diene file and the diene are trying to like hook up or something and there's like some third wheel, you're that awkward third wheel in the middle and you're just like, hey I need to get out of the way because these guys are just going at it okay? Well, that is exactly how this red carbon is feeling right now, he's feeling super awkward. So, instead of getting involved in the mix he's going to go ahead and stay out of it and just move right on top of the Ring. So, as, we see you end up still getting the six membered ring, there's nothing you can do about that but now we're going to get a bridge on top of it because this carbon really wanted nothing to do with what was going on. So, we went ahead and stayed above the whole situation. Now, the difference between these two molecules here is that they're just represented differently, this is the planar representation and this is the 3d representation and you should be able to understand both of them, I know they look a little bit weird, but this is essentially the same carbon and then you have your six membered ring below, okay? So, that's what we call a bycyclic bridge product and that happens when your diene is a ring. So, let's look at this example here, I'm actually going to draw this one, we're just going to do this as a worked example since I think that it's still a little too hard for you guys to do this. So, how would, we draw this product? Well, as you can see, this is going to be a dimerization, that means that we have the same molecule acting as the diene and acting as the dienophile, okay? So, how would this happen? Well, first of all is this diene in the right conformation to even react. Remember, that we stated how your one, three diene always has to be in the S cis conformation, is it in the right conformation? yes it is because if you were to draw a line between you would notice that both of the R groups are faced in the same side. So, we want to do is want to rotate that so that it's going to be facing opposite to the dienophile. So, in order to line this up correctly I would actually flip my cyclo pentadiene over to the right so that now, it's going to be able to correctly face the dienophile. Now, the dienophile, I told you guys for cyclopentadiene, either one of these double bonds could be the dienophile so it doesn't really matter which one we pick I'm just going to line one of them up next to it, okay? So, there we go. Now, it's time to draw our arrows and we're going to see that we get one two three with. Notice, we've got a bridge because this carbon here is not so happy about what's going on, it doesn't want a part of this, just doesn't want to be there. So, we're going to go ahead and draw this product it's going to be a new six membered ring. Now, let me just show you how I do that, you've always got your four carbons from the diene, right? And you've got your fifth and sixth carbon from the diamond file, right? So, that's going to make your six membered ring. Now, we know that in between, this is one, two, three, four, five, six, we know that the double bond should be in between, which carbons 2 and 3 because 2 and 3 are the ones that received an arrow and we know that we're going to need the rest of the Ring attached to carbons 5 and 6. So, just hold for a second 5 and 6, we're going to get a five carbon ring coming off of that and we should have a double bond here since that's the double bond really nothing happens to it from the original cycle penta diene, right? But, we're still missing something, what are we missing? we're missing the bridge. Notice that the bridge was attached to which atoms? it's attached to atoms, let me just highlight this, attached to atoms 1 and 4, okay? So, I'm going to go ahead and draw a bridge between 1 and 4 and that's going to represent the fact, that's going to represent the carbon that didn't want to be there in the first place, that is not participating in the reaction and it's now above the ring as a bridge bicyclic, okay? So, hopefully that made sense so far. Now, it turns out we're not done because bridge compounds add an extra complication to the product of the diels-alder, it turns out that anytime that a bridge product is made, you have to be aware of stereochemistry that we call EXO and endo stereochemistry. Now, what the heck do those words mean? those are words that specifically relate to the diels-alder products and it turns out that this ring actually could have faced two different ways. Now, if you guys just notice this molecule here is the same one that I drew up there. So, it's just drawn more professionally, and what you notice is that there's actually two different orientations that ring could have taken, either it could have been faced towards the top or it could have been faced towards the bottom, right? Is one of them preferred, is one of them not preferred, okay? Well, it turns out that yes one of them is highly preferred and one of them is highly not favored and that would be the direction that is away from the bridge, so that means that basically we've got two different options, let's look at 3D to see if we can figure this out, we've got one ring that is really close to that bridge. Now, this bridge has hydrogen sticking off of it so this is going to do something called a flagpole interaction, where you actually have hydrogens that are too close to each other, okay? So, you've got this hydrogen, I know it looks like it's kind of far but still, there's going to be some interaction there, they're not going to be very happy about being so close together, okay? Now, we've got those same hydrogen's here, but now notice that in this situation I face my ring down. So, there's a lot less, you know there's a lot more room for these hydrogen's to exist and this ring is way happier being on the bottom side of the entire molecule. So, when the rings are close together, we call that EXO, when the Rings are far apart, we call that endo, Well, when a bridge product is made you're always going to face the in dough direction because you want your ring or your substituents to be away from the bridge, so it doesn't just apply to rings it could apply to any substituents that we're facing off of the dienophile, if it's a bulky substituent, even if it's just a methyl it's going to want to be down, away from the bridge so it can avoid that flagpole interaction, okay? So, that means that for the product that I was drawing above, this is actually not drawn correctly yet, because I have to be specific about, is this going to be endo or is it going to be EXO, and your professor isn't going to be satisfied until you draw that correctly. So, how could we change this product so that it's correct? what I would want to do is I would want to erase those bonds and put them on dashes, okay? That shows that my ring or my substituents is now away from the bridge, okay? Another thing you want to do that you're going to see very often, many times, pretty much all the time, we ignore hydrogen's on hydrocarbons, we don't draw them, they're implied you never have to draw them, unless stereochemistry is involved and then they can be helpful. So, what you'll see many times and what I recommend to do is not only put the ring facing down, put the hydrogen's here facing up, draw those in explicitly, why? Because that's going to show your professor that you know what you're doing and that you agree that the H's should go up closer to the bridge because they're smaller and the ring should face away, this basically shows that you understand what an endo position is and that you know how to draw it, alright? So, remember guys, when do you use endo and EXO only when you have what? A bridge, when you have a bridge, when you have a cyclic diene.

So, you can see all these things are kind of tied together to that entire idea of starting off with a cyclic Diene, if you start off with a cyclic sign you have to worry about EXO and endo, if you start off with a normal diene, like we did in past videos, skip all this explanation because, you don't need it, you only need it for the cyclic version, okay? Awesome. So, let's go ahead and move on to the next topic.

Concept: Example 1: Retrosynthesis

5m
Video Transcript

In this video I'm going to walk you guys through a technique that you might need to use for diels-alder problems. So, sometimes your professor, your textbook, your online homework is going to ask you to do a diels-alder retrosynthesis, that means that you're going to be given the final cyclisation product and then you're going to be asked which diene and, which dienophile were required to make this six membered ring in the first place. Now, it turns out this is a really easy type of question to answer, if you just have the right technique and that's what I'm here to show you. So, here's an example, which diene and dienophile would you use to synthesize the following compound, okay? It's really easy to get lost in these types of questions because as college students we hate doing stuff that we haven't been taught to do explicitly, right? And, this is a backwards question, you have to think backwards. So, I'm going to teach you guys how to think, I already tell you guys how to think forwards and I'm going to teach you guys how to think backwards as well. Well, the first thing that you guys want to notice with this product is that there's a certain like kind of like landmark that you're always going to be able to find on these products because you know about the mechanism and the mechanism. Remember, always makes a double bond between the second and the third carbon. Remember me showing you guys that? so that means the first thing you want to do is you want to find that landmark and you want to kind of orient yourself on that because that's going to provide the structure for the rest of the molecule, so the first thing I want to do is identify where was that new double bond, that new double bond is right here, okay? That means that this must have been my second and my third carbon, okay? Which means that that must have been where my diene was, okay? So, my first step is always to find the original diene and the way you do that is by identifying the double bond and then saying, well, this must have been the diene to begin with this was my 1, and and this one must have been my four, okay? Now, that I have that diene I understand the mechanism better, I know that the diene must have reacted with a dienophile and remember that the diene always creates two new to the dienophile, so the next step is going to be to cross out the new bonds because, I know that my diene must have made two bonds. So, here's my diene once again, okay? And it made two new bonds, I'm going to cross these out because I know what they didn't use to exist, okay? So, after I've crossed them out that means that whatever is attached on the other side must have been the dienophile. So, in my next step I isolate the dienophile and guess how you do that, really easy all you do is you take your X's, right? You still have your diene and you just cross a line, right through them. So, I'm going to just chop it off right there, right through the X's and what we see is that now we have a diene on one side and something on the other, the other side must have had a double bond in order to react in the first place, okay? So, basically, we're making those three double bonds that we started with so the answer for this question would be that I start off with a diene that looks like this a 1, 3 diene like this plus I started off with a dienophile that looked like this, okay? And now that we figured that out, isn't it crazy to see how actually it was a dimerization, this is a molecule that reacted with itself through diels-alder but it wouldn't have been obvious unless you use a system to figure it out, okay? So, I know this is a new skill, I know I might be making it look too easy. So, test out our skills on this example go ahead for the following two problems, try to figure them out, we'll do them one at a time, go ahead and do a right now and then I'll help you guys and show you the answer.

Concept: Example 2: Retrosynthesis

2m
Video Transcript

So that this was again just a three-step process and you always start by finding the original diene, which must have been right here, because this double bond comes from the original diene, if that's the original diene that means the new bonds were these two right here, which means that, when I Isolate the dienophile, I'm going to chop, right through both of those X's, okay? Now, notice that this question did include stereochemistry so in order to get this question right you needed to include the stereochemistry.

So, let's go ahead and draw both of these components I would have a diene that looked like this, okay? With the two methyls coming off of it and I would have a dienophile that is a double bond, right? Because it must have been a double bond but notice that you can't really see that, sorry, must have been a double bond, right? But notice that my substituents are cis to each other on the ring versus, if they're both facing up that means that that original diene, dienophile must have had cis groups so this is your answer, you must have had a diene that looked like that with the two R groups and then a cis double bond dienophile, alright? Awesome, so go ahead and work on the next one and then I'll show you guys the answer.

Concept: Example 3: Retrosynthesis

3m
Video Transcript

Alright, so due to limited space once again I'm going to take myself out of the video, don't miss me too much home right here, okay? So, let's go ahead and do this one once again, it was pretty easy to identify the diene, it was right next to the double bond. So, we know that that is the diene but notice that I gave you guys a bridge compound this time. So, it's a little bit more challenging. So, you have to take out the new bond in this next step and I really hope you guys didn't cross out the bridge because that just doesn't make any sense you guys know that the bridge has to do with extra carbons that must have been on the diene to begin with. So, actually the bonds that we want to cross out are the ones that go to the rest of this ring okay, meaning that when we go to separate this we're separating the diene and the diene a file, that means that my original diene a file must have had a double bond here, but it means that my diene had extra carbons, right? So, how big of a molecule was the original diene, how big, it was a ring, right? Because, we know that rings make bridges, right? We kind of went over that extensively so the question is how big is that ring and you're just going to draw it exactly the way it is, we're going to draw, let me use red, we're going to show that the diene must have been a six membered ring because I've got double bond, double bond but then it's got these 1-2 extra carbons that didn't even want to be there that are now here, okay? So, we had a six membered ring that formed a two carbon bridge and then my dienophile is pretty straightforward guys, it's just going to be this ring, and really you don't have to worry about cis and trans here because a ring is always going to be cis and trans, wow that was really unclear, what I meant to say, let me just come back in, you don't have to worry about drawing my dienophile as like a sister trans ring because rings are always going to be on one side of the molecule they're automatically cis because they're small. So, don't worry about that, this is the right answer. So, if you just have this dienophile with this six membered diene you got the question right, okay? Awesome. So, I hope that that skill might help you on the exam, maybe get you a few more points, let's go ahead and move on to the next topic.