Diels-Alder: endo rule
How to draw the endo product for a Diels-Alder reaction.
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- May you Please tell that when will the Exo Product favoured,
I have seen one example of furan(Diene) to give more stable Exo product, not endo(5 votes)
- The Diels-Alder reaction is a reversible reaction.
Th formation of exo vs endo is a case of kinetic vs. thermodynamic control.
The exo product is more stable, but the activation energy for endo is lower, so the less stable endo product is formed faster.
At lower temperatures, kinetic control prevails and the less stable endo isomer is the main product.
At higher temperatures and after long reaction times, the chemical equilibrium can take over and the thermodynamically more stable exo isomer is formed.(23 votes)
- At6:40; Is the product formed have another enantiomer ?(4 votes)
- Nice catch! It does have an enantiomer because a chiral center exists.(4 votes)
- Wouldn't steric hinderance make the Endo product less likely to happen.(3 votes)
- Yes steric hindrance is higher in the formation of the endo product, but the activation energy of the endo product is lower than the exo product. This means that at high temperatures the exo product will form and at low temperatures the endo product will form.(5 votes)
- From5:20to6:20, endo product is preferred over Exo product in reality. Scientists think this phenomenon is due to stabilizing effect from the reaction between carbonyl group and newly formed pi bond in diene. However, would not carbonyl group cause greater steric hindrance in endo product than in exo product, thereby supposedly making endo product a less favored product instead?(2 votes)
- Endo is the thermodynamic product, because of the stabilization, but exo is the more stable product because of steric considerations. If you heat the endo product, it slowly converts to the more stable exo form.(4 votes)
- How do dienes form and react prior to interacting with each other?(2 votes)
- One example of forming dienes is through electrocyclic reactions, in which a pi bond within a cycloalkene is converted into a sigma bond (or vice-versa).
Here is the wikipedia page on electrocyclic reactions:
Additionally, here is the wikipedia page for dienes:
Good luck!(2 votes)
- I know the due to the endo rule the endo product is favored, but at9:06, is that the only product that is formed in this reaction is conditions are held stable (i.e. no change in temp)? Or is there just a majority of the endo product and trace amounts of the exo product?(2 votes)
- Is the endo product only preferred with carbonyl groups or any bulky R group?(1 vote)
- Usually, the EWGs are carbonyl or other groups on the dienophile that cause the stereospecificity.
In the endo transition states, the p orbitals on the carbonyl groups (or other EWGs) can overlap with the p orbitals on C2 and C3 of the dienophile and provide extra stabilization.(3 votes)
- how are the undo and 'endo' and 'exo' products biciclic?(1 vote)
- At2:32why do you put two extra hydrogens? As i can see at3:49the final product has 2 cho and two extra hydrogens.(1 vote)
- The two hydrogens were always there in the butenedial.
He just chose to draw them explicitly at2:32and at3:49.(1 vote)
- at 8.22 why are the methyl groupgoing down ?(1 vote)
- The "inner" hydrogens have to go up (because if they went down they would need to pass through the bonds forming between the diene and the dienophile). Consequently the "outer" methyl groups must go down. Using a molecular modeling kit to see this is likely to be very helpful.(1 vote)
- [Narrator] On the left is our diene, this is cyclopentadiene, and we have this bridging CH2 to think about. On the right is our dienophile, and we have these two aldehydes that are cis to each other. So in this video, we have stereochemistry for the diene, and the dienophile, and we're also going to talk about the endo and exo products. But first, let's just think about our Diels-Alder Reaction without any stereochemistry. So we're going to move our six pi electrons around. Take these pi electrons, and we're going to move them into here, to form a bond between these two carbons. These pi electrons are going to form a bond between these two carbons, and finally, these pi electrons are going to move down. So we draw the product, again not worrying about stereochemistry, we put our double bond in here, and these aldehydes, I'm going to abbreviate as CHO. So we have CHO, and then another aldehyde down here, so CHO. We can put in our bridging CH2, so I'll just draw that in, and let's follow our pi electrons. So our pi electrons in red formed a bond here. Our pi electrons in blue formed this bond, and our pi electrons in magenta formed this bond. But notice for our product, we haven't really shown any stereochemistry. So where are these aldehydes in space? How do they relate to this bridging CH2? Let's say the diene and the dienophile approach each other in this way. So we know that a bond forms here, so this bond in red, so that's a bond that forms between this carbon and this carbon, so that bond forms in here. And for the product, here you can see that bond. And then we also have this bond in blue, which forms between this carbon and this carbon. So it's the bond that forms in here. And here is that bond on our product. We need to think about the stereochemistry of the diene first. So we look at our cyclopentadiene, and we have this bridging CH2. We know that inside substituents go up, so on our models set, here's that CH2, and here are the bonds that are leading to it, and when we form our product, that CH2 goes up in space. Next we need to think about the stereochem of our dienophile. So we have a double bond, we have hydrogens on the left side, and if we draw a line here, we saw how to deal with this in earlier videos. The stuff on the left side goes down, so these hydrogens, if we follow them along, so here are the hydrogens, we can see them in our final product. If we're staring at the final product in this direction, those hydrogens are going away from us in space, so the hydrogens would be down. And these aldehydes here, so these aldehydes I've made red, in the model set. So here you can see the red, which is just standing in for our aldehydes, and those aldehydes end up on the same side, and if we're staring down this way, here are the aldehydes, those are actually coming out at us. So when we draw in our product, let's go ahead and start drawing this in here, we have a bicyclic compound, so we sketch that in, and we have our bridging and CH2, so let's put that in. We have our double bond here, and then we have our aldehydes. Our aldehydes are up. So at this carbon, we need to draw in an aldehyde up, so CHO. We have a hydrogen that's down, so let's put in our hydrogen going down at this carbon. And then we have, at this carbon an aldehyde going up, so CHO, and then a hydrogen going down. So this is the exo product. So this is said to be the exo product, because we have our aldehydes up, our bridging CH2 is up, and then here is our double bond. So that's one of the possible products. What about if our dienophile approaches in this direction? So if we have our hydrogens here, on our double bond, and we draw a line, notice stuff on the right side of the double bond, ends up. So here are the two hydrogens. And when those two bonds form right in here, these hydrogens are up, if you're staring down this way. The stuff on the left side of the double bond, now our aldehydes are on the left side. So here are the aldehydes, and when our product is formed, we're looking this way. The aldehydes are going away from us in space. The aldehydes are down. Let's draw in this product. So we have our bicyclic, let me go ahead and draw that in here. So here is our bicyclic with our bridging CH2. Our double bond is in the back. And now when we're drawing this in, the hydrogens are up. So at this carbon, there's a hydrogen up, and then there's an aldehyde going down. So here is our CHO going down. Same thing at this carbon. So a hydrogen's going up, and an aldehyde is going down. This is called the endo product. So let me write that in here. So this is endo. Our aldehydes are going down, our bridging CH2 is up, and here is our double bond. It turns out, the endo product is the preferred product most of the time. And this is called the Endo Rule, or the Alder Endo Rule. And it's thought to be due to an interaction between the developing pi bond, which occurs between these two carbons, and the carbonyl groups. The carbonyl groups for our aldehyde here. And even though I didn't put in the carbonyls, we know that these red spears here symbolize those aldehydes. And so there's an interaction between the developing pi bond and those carbonyls, which stabilizes the dienophile approaching in this way. And that gives us our aldehydes down in space. That gives us our endo product. So the endo product is preferred. With the exo product, these carbonyls, these aldehydes would be way out here, that's too far away from the developing pi bonds. There's no interaction. So the exo product is not preferred. Let's do one more problem. On the left is our diene; on the right is our dienophile. And let's draw the part without stereochemistry again. So we move our six pi electrons. So we form a bond between those two carbons, we form a bond between these two carbons, and these pi electrons move down. So if we draw in our products, again not worrying about stereochemistry, we have our aldehyde coming off of this carbon, we have a methyl group here, and a methyl group here. Following our electrons, electrons in red, these pi electrons, formed this bond. Our pi electrons in blue formed this bond, and our pi electrons in magenta moved down to here. So let's highlight those bonds on the model set, on the picture. So this bond in red, formed between this carbon and this carbon, so formed in here. There's that bond on our final product. The bond in blue formed between this carbon and this carbon, so here's that bond on our final product. But notice, if we draw the dienophile approaching this way, there is no stabilizing interaction between the developing pi bond and the carbonyl groups. So down here, I flipped over the dienophile, so it has an endo approach, to the diene. And now we think about our stereochemistry. So let's worry about the stereochemistry of our diene first. We know that inside substituents go up. So these two hydrogens, here they are on the diene. And then our product, those two hydrogens, are going up in space, if we're looking down in this direction. So those methyl groups, let me change colors here, so these two methyl groups, which I've used in yellow, in the model set, those two methyl groups go down. So down relative to those hydrogens going up. So those methyl groups are going down, if we're staring down this direction. Let's draw that in for our products. We have our ring, so let's put in our double bond, and those methyl groups are going away from us. So we put those on a dash. So there's one methyl group, and there's the other one. Next, let's think about the dienophiles. So here is our dienophile, and if I draw a line here, we know this stuff to the left ends up down, so this aldehyde is going to be down in space. So let's follow that aldehyde group. So here is the aldehyde. We know that this endo approach gives us some stabilization between the developing pi bond and our carbonyl. So for our product, here is the aldehyde. And again, if we're looking down, this aldehyde is going away from us in space. We'd have to put that aldehyde on a dash. So here is the aldehyde going away from us in space. What if the dienophile approached this way? Well this is still an endo approach. Because here would be our aldehyde, and our developing pi bond would be back here. So there's a stabilizing interaction. So here's the aldehyde in our product, and if we're staring down this way, this aldehyde is going away from us in space. So let's draw this product. We would have our ring, we would have our double bonds, and the aldehydes going away from us at this carbon. So we put that aldehyde on a dash. And our methyl groups are still going away from us. So this one's going down, and this one's going down. So we draw in our methyl groups. What is the relationship between these two molecules? Well, they are enantiomers of each other. So you should get a pair of enantiomers.