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Diels-Alder reaction
How to draw the products for a Diels-Alder reaction.
Want to join the conversation?
- What's the significance of Diels-Alder reactions? When and where do they occur in nature? Thanks(3 votes)
- The Diels-Alder reaction is most useful for synthesizing molecules in the lab.
But scientists believe that specific enzymes catalyze Diels-Alder reactions in some organisms.
Examples are the formation of lovastatin, a cholesterol-lowering drug found in oyster mushrooms, and Spinosyn A, a natural insecticide produced by a certain bacterium.(11 votes)
- At5:29, how is it that the diene can simply be rotated to complete the reaction? I thought an example such as this was stuck in the trans conformation due to the presence of the double bond.(4 votes)
- Generally,as a rule of thumb, everything can be rotated about a single bond!(6 votes)
- What do you mean as EWG? Oxygen, X?(3 votes)
- In Diels-Alder reactions, an EWG (electron withdrawing group) is usually something that contains a carbonyl group, such as an aldehyde, carboxylic acid, ester, or anhydride.(3 votes)
- How do Dienes form and interact prior to dies alder reactions?(3 votes)
- What is the exact defenition of diels alder reaction(2 votes)
- The Diels-Alder reaction is a cycloaddition of a 4 pi + 2 pi (diene + dienophile) system that forms a more stable product due to the fact that the sigma bonds created are more stable than the pi bonds destroyed.(3 votes)
- what makes a good dienophile(2 votes)
- A good dienophile usually has an electron withdrawing group (EWG) attached to one or both alkene carbons.
Good EWGs include ketone, aldehyde , nitrile, nitro, and trifluoromethyl groups.
Maleic anhydride is an excellent dienophile.(2 votes)
- @3:58he says "we can just do that in our head"... Not really. Don't you need to provide some sort of input of energy (roughly 15kcal/mol, usually in the form of heat) in order to go from the s-trans to s-cis conformation?
Otherwise, the reaction wouldn't proceed. The s-trans conformation is more stable (due to sterics), and therefore, we would need to destabilize it and force it into the s-cis conformation in order for the reaction to proceed.
Right?(2 votes)- You're right that s-trans is more stable and that some energy is required for it to become s-cis, but it is very low.
While the s-cis is higher in energy, the amount of energy needed for the transformation is low enough that the molecule can usually accomplish it as long as it can rotate about the single bond. For 1,3-butadiene, the energy required is only 2.8kcal/mol (so for larger molecules, it may take more energy). The only obstacle that I learned of was if there is steric hindrance that prevents the molecule from going from s-trans to s-cis, otherwise, it can generally accomplish that transformation.(2 votes)
- Does this reaction require any specific condition or catalyst?(2 votes)
- At1:26, Does the cyclohexene exhibit resonance? Why or why not?(2 votes)
- No, because there is only one pi bond and nowhere else for those electrons to go.(2 votes)
- At the reaction at5:20, could the diene be added twice, i.e. once at each side of the dienophile to create just a single bond in the middle?(2 votes)
Video transcript
- [Narrator] Diels–Alder
reaction is a very important reaction because it's used a lot in synthesis to make
complicated molecules. On the left we have our diene, so we have two double
bonds in that molecule. On the right is our dienophile. Let's take a look at that word. We know that phile means love so the dienophile loves the diene, and the dienophile usually has at least one electron withdrawing group, which withdraws electron
density from this double bond. The dienophile is
relatively electron poor. The diene, on the other hand, is relatively electron rich. We have four pi electrons, so you can think about the electrons flowing from the diene to the dienophile, and this is what's called
a pericyclic reaction. This is a one step reaction that proceeds through a cyclic transition state, and if we think about electron density flowing from the diene to the dienophile, we could start with these pi
electrons moving into here, so we form a bond between
these two carbons. Next, these pi electrons
would move into here to form a bond between these
two carbons, and then finally, these pi electrons
would move over to here. Think about all of those six pi electrons moving at the same time. And that would give us our product on the right over here, which
is a cyclohexene ring. If we follow our pi electrons, we'll start with these
pi electrons in red. So these pi electrons formed this bond. Next let's follow these pi
electrons on the dienophile. They start on the dienophile and they end up forming this bond
between those two carbons. And then finally, the electrons
in magenta right here, on the diene, move down
to here to form the double bond and to give
us our cyclohexene ring. But all this happens at once. Now I drew my electrons going around in a counterclockwise fashion. I showed my electrons going
around in this direction, but it doesn't matter, you could've drawn your electrons going around in a clockwise fashion. What matters is thinking
about moving your six pi electrons to give you
your product on the right. Let's get some practice with some simple Diels–Alder reactions. And we won't worry about
stereochemistry in this video. First you need to recognize
the diene and the dienophile. On the left is our diene. On the right is our dienophile. If we think about electron density flowing from the diene to the dienophile, I can move these electrons into here so we form a bond between these two carbons, and these electrons move into here to form a bond between these two carbons, and then these electrons down to give us our cyclohexene ring. So we could draw our product right away. We know we get a cyclohexene ring here and then we would have our aldehyde coming off of that carbon. So, following our electrons, I'll be consistent with the
colors that we used before, so these pi electrons are red and those electrons move over
here to form this bond. Next, these pi electrons in blue moved into here to form this bond, and then finally the pi electrons in magenta moved into here to form this bond. All six pie electrons move at the same time in this one-step reaction. Let's do the next problem. Down here on the left, this is our diene, and on the right is our dienophile. This on the left, this is a diene, but notice that it has an
interesting confirmation. Up here, we had our diene and what's called the s-cis confirmation. S refers to this single,
or sigma, bond here. But here we have the s-trans confirmation. We have our double bonds
trans about this single bond, so we have to rotate about this single bond here to go from the s-trans confirmation to the s-cis confirmation. You have to do that. If you've got this problem on a test, you can just do that in your head, and now you have your diene
in the s-cis confirmation. It needs to be in this confirmation in order to undergo a
Diels–Alder reaction. Now we're ready for our reactions. We think about our six pi electrons. We're going to move these electrons into here to form a bond
between these two carbons. Notice this time we're
dealing with a triple bond. Up here we only had a double bond, but alkynes can act as dienophiles too. We're going to take these pi electrons and move them into here, so there's a bond that forms
between these two carbons. And then finally move
these electrons into here. Let's draw our product. This time we have two double
bonds in the ring like that, and then we would have this
group coming off of this carbon, which is this one right here. So let's draw that in.
So we have our esters. We've put that in. And then the same thing down here. Let's follow our electrons again. We'll start with the electrons in red. These electrons moved into
here to form this bond. Next, let's look at these electrons right here on our alkyne, These pi electrons move
into here to form this bond. And then finally our electrons
in magenta move in to here. So this is our product. All right, let's do another one. Let's go down here and let's look at this Diels–Alder reaction. On the left we have our diene. On the right we have our dienophile. And we can start moving
our electrons around because we already have
an s-cis confirmation. If we move these electrons into here, then we form a bond right here. Then we move these pie electrons into here to form a bond here. Then move these electrons. We can draw our product. We would have this ring on the left, and then on the right we would have this, and we'll draw in this portion. Draw in our carbonyls here,
and this is our product. Following our electrons as usual, electrons in red moved into here, our pi electrons in blue moved into here, and our electrons in
magenta moved over to here. Six pi electrons moving at the same time. What if you were given the product and asked to come up with the necessary diene and dienophile? Let's do a problem like that. So, thinking backwards. So let's say you were
given this on the right, and asked what combination of diene and dienophile do you need. Well, think about moving
those electrons in reverse, so let's look at our product here. Let's start with these electrons. I'll make them the same
colors we've been using. Move them in the reverse order this time, so these electrons
would move over to here, and then these blue electrons in this bond would move over to here, and finally, these electrons in red would move over to here, so let's go ahead and draw
our diene and our dienophile. On the left, our diene, we would have our double
bonds looking like that, and then on the right, for our dienophile, let me go ahead and draw our ring here and put in the carbonyls. Let's follow our electrons along here. The electrons in magenta
moved over to here, the electrons in blue were over to here, and our electrons in
red moved over to here. To check yourself on a problem like this, you can just take the
diene and the dienophile that you drew and double
check and make sure they give you the product on the right. Thinking about the Diels–Alder reaction in reverse gets you your starting materials.