Organic Chemistry Conjugated Dienes

Welcome back to Educator.com.0000

Today, we are going to be talking about conjugated dienes--what they look like and how they behave and what sort of reactions they undergo.0002

A diene has two π bonds; a conjugated system has those π bonds in an alternating fashion--so we have a double bond, and then a single bond, and then a double bond.0009

OK, so what we have is a p orbital on each carbon for the first π bond, and then a p orbital on each carbon for the second π bond; so what we end up with is aligned sets of p orbitals, where those π electrons can delocalize over multiple aligned p orbitals.0022

What we get is resonance, and that makes these systems very special and allows them to undergo some unique reactions.0042

Now, there are two orientations--two conformations that a diene can have.0050

This one is described as an s-trans conformation; so you can see that these groups are trans to each other (on opposite sides), but we call it s-trans because it is trans about a single bond, or a σ bond.0057

In other words, you can rotate around this; there is nothing stopping this from rotating.0073

And if it did so, it would have this conformation: we call this conformation the s-cis conformation, because it's cis about a single bond.0078

And, although these two conformations exist, the equilibrium is favored in the reverse here; the trans is more stable, because the s-cis has hydrogens that are on these end carbons, and they are forced to be in a coplanar arrangement here.0090

And so, we have some steric interactions with those two hydrogens.0109

So, s-cis is less stable; we are going to see that this conformation is going to be necessary for certain reactions to take place.0112

Now, let's talk about this nature of being conjugated, and what kind of stabilization this resonance imparts, and how we can get some evidence for that.0122

Let's take a look at three different types of dienes: each of these has five carbons, so these are all pentadiene examples.0134

Here, the two double bonds are right on top of each other--we call this a cumulated diene, where the p orbitals are perpendicular--the ones on this π bond are perpendicular to the ones for the second π bond.0143

Here, we have π bonds that are unrelated--they are isolated; we have...the p orbitals on these two carbons have no relationship at all to the p orbitals on the other two carbons.0157

And then, finally, we have the conjugated system, where we are saying we can have some interaction here, and some delocalization, because of resonance.0171

We can take a look at the heats of hydrogenation for each of these dienes to see if we can observe some differences in their energies.0181

When you catalytic-hydrogenate each of these, they give the same product (they give pentane)--all of those have the same energy, so when we measure the ΔH, that gives us a relationship that tells us something about the initial energy of our starting materials.0190

Let's take a look at the isolated π bonds, initially: the energy released there is -60.7 kilocalories per mole--that is essentially the heat of hydrogenation for an alkene, times 2.0204

So, because these double bonds are separate from each other and have no interaction, you simply get twice as much energy out as if you had just one π bond.0216

However, when you force those two π bonds to be cumulated to one another, we actually get out more energy from the reduction of this diene, meaning it started out higher energy.0227

This is an unstable arrangement of double bonds, and when you allow them to be conjugated with one another, you have less energy being released, which means your diene must have started at a lower energy.0237

So again, if we do a quick little energy diagram comparing these three, they are all going to the same product--the pentane product--but A must be starting at a higher energy than B, and C must be starting at a lower energy than B.0251

C is the most stable diene, because it has the smallest amount of energy to release upon hydrogenation.0268

OK, and it because of this resonance energy, the energy that we get from that stabilization.0277

And one thing I forgot to point out on the past slides--some physical evidence for this resonance interaction--is that this σ bond is shorter than a normal single bond.0283

OK, because you have these p orbitals on each carbon that are drawing them closer together because of that overlap, it gives us some hint that those electrons are being delocalized over all 4 carbon atoms.0302

OK, related to these conjugated systems are allylic systems, like an allylic carbocation and allylic radicals; let's take a look at how those look.0316

Allylic means that you are next to a π bond; so this is an allylic carbocation; an allylic carbocation is uniquely stable, because it has resonance; you can delocalize that positive charge to another carbon.0327

One way that you can look at an allylic carbocation is to say, "Well, we have a π bond here, which means a p orbital on each of these carbons sharing two electrons, and a carbocation is also sp² hybridized, so it has an empty p orbital."0340

And so, when you look at this orbital picture, you can see how very easily those two electrons can be shared among these aligned p orbitals, and it would be very easy to delocalize the charge and delocalize those two electrons over the p orbitals.0359

That is an allylic carbocation, allylic radicals, the nearly identical picture...we have two electrons in a π bond, and then we have a radical in a third p orbital.0378

And each of these discrete Lewis structures has a localized π bond and then a radical on the third p orbital.0391

But again, of course, because this resonance exists, what does the true structure look like?--it is some kind of blend of both of these resonance forms--something in between, where we have partial π bond character, and those three electrons are being distributed among all three carbons.0399

Now, let's take a look at some reactions we have seen in the past for alkenes, and see how that is going to apply or how they differ when it comes to dienes.0422

For an alkene, we saw some addition reactions--electrophilic additions; for example, if we react an alkene with HX (HBr, HCl, something like that), we could break the π bond, add a hydrogen and a halogen, and we do so with a regiochemistry called Markovnikov addition, where the hydrogen with more hydrogens is the one that gets the hydrogen.0432

OK, we could also do electrophilic additions with halogens like chlorine and bromine, and we could form the bromonium ion, and the second bromine can come in and open it up, and so we can add two chlorines or two bromines across a π bond, as well.0454

We have also seen some reactions of conjugated systems; if you recall, an alpha, beta unsaturated ketone (or aldehyde, in this case)--an alpha, beta unsaturated ketone is electrophilic in two positions because of resonance delocalization.0469

And, when a nucleophile adds, it can either add to this end carbon (we call that 1,4 addition), or it can add to the carbonyl carbon (we call that 1,2 addition).0487

We have seen some evidence in the past where, when we have a conjugated π system, those are going to interact, and we can get a mixture of products based on that interaction.0498

When we take a look at a conjugated diene, and we add an electrophile to it (like HX or X₂), we are going to get two possible products out.0510

In one product, we are going to break just the first π bond here and add H and X; we are going to do it with Markovnikov regiochemistry, just like we have seen before.0523

So, we'll add the hydrogen to the end carbon and the halogen to the middle carbon.0534

And that same reaction can happen with chlorine or bromine, where we add a halogen to each of the first two carbons.0537

OK, so we'll get that product: in other words, the diene can act just like an ordinary alkene; but there is a second product that can be formed, in which case, the hydrogen adds to the first carbon and the halogen adds to the last carbon, and the double bond has moved to be between the middle two carbons.0545

This can happen with the addition of HX; it can also happen with the addition of X₂ (chlorine or bromine), and we describe these products--we call these products 1,2 addition, because they have added 1,2; we call these 1,4 addition because they have added 1,4.0564

We will take a look at the mechanism; we will see in which cases one product is favored over the other and so on.0580

OK, now, before we get into the mechanism of the diene, let's make sure we review the mechanism of the alkene, and we understand what was going on there.0587

If you had propene and reacted with HCl, the very first step--it's a two-step mechanism; the first step is going to be to protonate the alkene with the strong acid.0595

And this is the step where you decide the regiochemistry--is the proton going to go to the first carbon or the second carbon?0606

Now, one carbon is going to get the proton; the other carbon just lost this π bond, so it is going to have a positive charge; we are going to form a carbocation intermediate.0615

And so, the way we decide which is the major product: we compare these two intermediates; we see that one is a secondary carbocation, and the other is a primary carbocation.0625

What do we know about carbocation stability--which is the more stable one?--the more carbon groups, the more stable, because remember: carbon groups are electron-donating, so that is going to be a good thing for the positive charge.0641

The secondary carbocation is more stable, so the primary carbocation is less stable, and it is not formed.0653

This was the rationale behind what we described as Markovnikov's rule.0663

The hydrogen goes to the carbon with more hydrogens so that the carbocation ends up on the carbon with more carbons; that is going to be a more stable carbocation.0668

What is our second step?--we form the carbocation; now, the Cl^- we formed in the first step is going to attack; so we protonate, and then we attack, and we are done.0679

We have our addition reaction.0690

OK, so reviewing that for an alkene--let's see how that is going to vary when we start with a diene instead of an alkene starting material.0693

Well, it is going to begin the same way: if you take a diene and react it with HCl, you are going to protonate the π bond, and we are always going to do it in a Markovnikov fashion so that we always get the more stable carbocation.0702

We are always going to form the most stable intermediate possible, which means we are going to add the proton to one of these end carbons to give a carbocation (and this second double bond is still here).0714

And when we look at the carbocation we just formed, we see that it is a special kind of carbocation, because it is next to a π bond: what do we call that?--it is an allylic carbocation.0733

What is special about it?--well, it has resonance: I can move this π bond over, and that will locate the positive charge at two possible positions.0746

Hopefully, now you can see how it is that we can get to a 1,2 product or a 1,4 product; that comes from having the chloride attack at the first carbon or having the chloride attack at the last carbon.0763

Because both of those carbons have partial positive character, the chloride could attack either way; and we get these two different products out.0781

Now, let's talk about these two products a little bit more: this first one, the 1,2 product, is described as the kinetic product; this is the product that can be favored if we have cold reaction conditions.0789

OK, it is called the kinetic product because this has the lower transition state energy; we will look at that transition state in just a moment to see why that is.0799

It has the lower transition state energy, and therefore it is formed the fastest.0810

And remember, kinetics has to do with the rate of the reaction, so the kinetic product is going to be the one that is formed the fastest.0818

OK, a little note here: it is not about having the more stable carbocation intermediate; this is a common misconception, or a common misstatement, when trying to explain this reactivity.0825

It is not acceptable to say that this came from the better carbocation--why not?--I mean, it looks good: this is secondary; this is primary; but why can't you say that this kinetic product came from a better carbocation?0845

Are these two different carbocations that were competing, like in Markovnikov's rule--if we protonate here, or protonate there, we get two different carbocations?0858

No, there is one carbocation: there are just two ways to draw it.0868

Remember, the actual structure is a blend of these two; so they both--both of these products--come from the same carbocation; we will just see that the one leading to the 1,2 product goes through a lower-energy transition state; that is what makes it faster.0872

OK, the 1,4 product is described as the thermodynamic product, because this is what is favored when we have hot reaction conditions, and it is called the thermodynamic product because this is the more stable product.0887

Now, if you compare these two products (the 1,2 and the 1,4), what do you see about the structures that would help you explain why this is more stable?0902

Is it about where the chlorine wants to be?--we have never heard anything about chlorine's preferences.0914

But how about the double bond--how would you describe these double bonds?0920

This one is terminal; it is monosubstituted; this one is internal--it's disubstituted, right?0924

The more substituted a double bond, the more stable it is; so it's because we have an internal π bond that it is the more stable product.0933

OK, so kinetics has to do with the rate of a reaction; thermodynamics has to do with the product stability.0943

If we take a look at an energy versus progress of reaction diagram, we will be able to see both of these things competing.0951

We are starting at some preliminary energy for our butadiene (in this case, 1,3 butadiene and HCl).0959

OK, the first step of the reaction, remember, is protonation to give a carbocation; so our carbocation is going to be some high-energy intermediate.0966

We could draw both resonance forms here; but remember, that is the same carbocation, and so, to go here and protonate, we have to overcome some barrier, some transition state, to get here.0980

OK, but what is key is: whether you are trying to make the 1,2 product or the 1,4 product, they both start the same way: they always both start by protonating the diene in Markovnikov regiochemistry to give an allylic carbocation--a resonance-stabilized carbocation.0993

OK, and when we are done, we expect our products to be lower in energy than our starting material, because we are doing an addition reaction: we are breaking a weak π bond and forming a stronger σ bond.1011

OK, and what we said was that, of our two products, we said that the 1,4 product is the more stable product; so that is where the double bond was in between.1023

The 1,2 product was not as stable, because our double bond was terminal (or actually...the wrong way...double bond over here, and chlorine over here).1036

OK, so we know where we are going; and what we just pointed out in the last slide (and will talk about next) is that the 1,2 transition state is lower energy than the 1,4 transition state.1048

The path to go to the 1,2 product goes through a lower-energy transition state and ends up at this higher-energy product.1066

To do the 1,4 product, we go to a higher-energy transition state, but then we end up at a lower-energy product for the 1,4.1077

OK, now let's see if we can explain why we have this energy difference between the transition states.1088

What is happening in our transition state (the 1,4 transition state) is: this is our second step--what we are doing is: our chloride (we have a chloride here--sorry--as our other intermediate in this step)...now, that chloride is attacking the carbocation.1095

It is attacking either at this position or this position; so in the transition state, the nucleophile and electrophile are coming together; you are starting to form a bond.1112

So, we have a partial bond between the carbon and the chlorine; we have partial charges: as the charge on chlorine starts to dissipate, the charge on the carbocation starts to dissipate; so this is what our transition state looks like.1120

We use a double dagger to indicate a transition state.1131

That is the 1,4 transition state; and here is the 1,2 transition state, with the chlorine attacking the second carbon instead.1135

Now, when you compare these two structures of the transition states, what do you see about them that might be different in their stabilities?1143

Now, the double bond is in different places; we saw that was an issue to determine the product stability; but for the intermediate, it is going to be the charges that you have that are the biggest source of instability.1153

So, how would you describe the two charges that you have?1166

In the 1,4 transition state, I have a primary partial positive; and in the 1,2 transition state, I have a secondary partial positive.1169

And, just like a secondary carbocation would be more stable than a primary, a secondary partial positive will be more stable, as well.1182

OK, so what we could say is that for the 1,2 transition state, we have a better partial positive; it is secondary; it is also allylic (they are both allylic--we could say this is primary allyic versus secondary allylic).1190

OK, that means that the 1,2 transition state is lower in energy (which we have shown here--we have drawn it at a lower-energy spot), and if the transition state is lower in energy, that means the 1,2 energy of activation is lower.1210

If the transition state is lower, that means the energy of activation (the energy of activation is right here)--the energy required to get to that transition state--is lower; we are comparing that, of course, to the energy of activation for the 1,4.1232

So, if the energy of activation is lower, that means it is a faster reaction.1249

That is why we describe it as being the kinetic product.1257

The faster reaction is the one that goes through the lower-energy transition state: so the 1,2 is called the kinetic product because it is formed faster; the 1,4 is called the thermodynamic product because it is a more stable product.1263

So, even though it took more energy to get there (it is a slower reaction--it is more difficult to do this reaction), it goes to a better place; so we will see different reaction conditions favor each one.1276

If you have cold reaction conditions, then you are limited in the amount of energy that you have; so therefore, whatever products form fastest (the 1,2 in this case)--that is formed immediately, and that is going to be your major product.1292

OK, but how is it that adding heat to a reaction is going to somehow cause this 1,4 product to be the major product?--that is what we want to talk about next.1307

OK, we will call that kinetic versus thermodynamic control; and so, here are two examples.1318

If we take butadiene and we react it with chlorine (just one equivalent of chlorine in each case), and we do one at -15 degrees (that looks like cold reaction conditions), and then we have a little Δ here to symbolize heat--this looks like hot reaction conditions.1322

OK, what do we expect to have as our major product?--well, we can form both the 1,2 and the 1,4; this is not an all-or-nothing.1340

You will usually get a mixture out; OK, but in cold reaction conditions, the major product is going to be the 1,2; so we get something like 60% and 40%; so the 1,2 is major and the 1,4 is minor.1351

OK, but if we do this under hot reaction conditions, we still get a mixture of 1,2 and 1,4, but it moves: we get about 30-70, where now the 1,4 is the major product.1373

Kinetic control will favor the cold conditions, while hot conditions will have thermodynamic control.1387

But what is very interesting is: if you take this product mixture that you got initially, and you heated it, it will redistribute to not have the 1,2 as a major product anymore, but have the 1,4 as the major product.1395

So, in other words, there is no longer any butadiene or chlorine left; but just these products can reorganize themselves to be different products--meaning somehow, this 1,2 product is disappearing, and more 1,4 product is reappearing.1414

How is that happening--what is going on here?1430

Well, the key is that the reaction that we are looking at is reversible.1432

Even though you can form 1,2 product quickly initially, it is possible for that reaction to reverse and go to 1,4 product.1437

Take a look at our 1,2 product; remember, our 1,2 product is less stable.1446

Even though it is formed the fastest (because it had that secondary partial positive), because it is less stable and higher-energy as a product, it is more likely to do the reverse reaction.1453

So, if it was fast to form the product, that means it also is fast (or a little faster, compared to the other) to do the reverse reaction.1473

That reverse reaction is our leaving group leaving to give back the carbocation; and once you are at that allylic carbocation--once you are here--now when the Cl^- adds back in, it can add to either position again.1481

It is going to preferentially want to do that to get to this more stable product.1502

OK, so this is how when you add heat--when you give it enough energy--you give an opportunity for things to equilibrate.1507

OK, so let's say some things about this 1,4 product: remember, this is the more stable product, because it has the internal π bond; and because it is more stable, it is less likely to do the reverse reaction.1516

So, once you make the 1,4 product, you are more likely to stay as the 1,4 product.1531

A reversible reaction--the key is having a reversible reaction, and with the addition of heat (so if you have a reversible reaction and you provide it with enough energy to do that reversible reaction), it allows products to equilibrate.1537

That is the key here: thermodynamic control assumes that you have equilibration; and when you allow a reaction to go forward and back and forward and back and forward and back, it eventually is going to build up in concentration of whatever is the most stable product.1564

OK, we call that the thermodynamic sink--the notion that our product mixture is going to continue until you get to the most stable place that you can be.1580

OK, now let's shift gears and take a look at a different reaction that dienes can undergo.1595

What we just looked at were electrophilic addition reactions and the idea that an electrophile can either add 1,2 or 1,4; another reaction that is unique to dienes is a reaction called the Diels Alder reaction.1600

Now, what happens in this is: we need a diene, as shown here; it needs to be a conjugated diene (something like 1,3 butadiene), and the thing we are going to react it with is something that loves to react with diene, so we call it a dienophile.1613

We are going to react a diene and a dienophile; we are going to have, as our only reaction conditions, heat (and sometimes pressure--you could do that too).1630

But all we need to do is heat this up; and the reaction that occurs is as follows.1637

One end of the diene reacts with one end of the dienophile, and we form a bond; and the other end of the diene reacts with the other end of the dienophile, and we form a bond.1642

The mechanism is actually just a single-step mechanism--one step, all occurring at once: the diene attacks the dienophile, the dienophile kicks its π bonds up and attacks the diene again, and the double bond moves over here.1653

6 electrons moving in a ring--we describe this as a paracyclic reaction, where all of our cyclic transition states and all of our electrons move in a concerted fashion--and what happens is: we have 1, 2, 3, 4 carbons in the diene and 5, 6--two more carbons in the dienophile.1670

We end up forming a 6-membered ring: 1, 2, 3, 4, 5, 6.1689

We get a 6-membered ring...and something else is missing here: notice that these double bonds are gone, but we just moved a double bond here to be between carbons 2 and 3.1696

So, when we track all of those 6 electrons and where they go, we are going to get a cyclohexene product.1705

As a Diels Alder reaction, it always gives a 6-membered ring, and it always has a double bond here.1714

It also forms two new carbon-carbon bonds; so this is a really cool technique for organic synthesis, because it is a way of building new carbon structures and building 6-membered rings.1721

This is described as a cyclo addition, because we have an addition product and we form a ring; it is called a 4+2, because one component has 4 π electrons, and the other component has 2 π electrons; so it is described as a 4+2 cyclo addition.1735

It is one subset of a larger class of reactions, called pericyclic reactions; the Diels Alder is just kind of the most famous one of those, so that is the one we will study as an introduction to this class of reactions.1751

OK, so let's study, one by one, the different components: let's look at a dienophile in the Diels Alder reaction.1764

In most cases, the dienophile acts as an electrophile; it can be either an alkene or an alkyne, and it is usually best--it is going to be a good Diels Alder reaction if it's electron poor.1770

A lot of times, what is done is: an electron withdrawing group is added; so we have a double bond, and we add one or two or three or even four electron withdrawing groups to that to make it readily undergo Diels Alders.1782

Our EWGs (our electron withdrawing groups) are the same groups you have seen before: carbonyls, like an ester or a ketone or an aldehyde; cyano...right?--those are the groups that we have seen as electron withdrawing groups.1796

OK, and why do they make them good electrophiles?--well, what all electron withdrawing groups do is: they withdraw electron density out of the π bond.1810

So, if we look at the resonance form for this ester, we know that this resonance form exists: and what does it do here?1820

It makes that electron poor.1831

That makes it a good electrophile: it puts some partial positive character here and pulls electron density out of the double bond, and that makes it behave as a good dienophile.1835

Let's see an example where our dienophile is an alkyne--does this look like it would be a good dienophile?1850

I have 2 ester groups, 2 electron withdrawing groups; so this would probably be a very good Diels Alder.1856

We have our dienophile; we have our diene; and we are always going to line it up like this, so that the two double bonds in the diene are kind of pointing toward the dienophile.1862

We are going to do that so we can very easily see the two bonds that are being made.1875

And then, our mechanism, because it is just a one-step mechanism--it's a concerted reaction, one-step mechanism; it's so easy to draw; it's helpful to do that.1880

It also doesn't hurt to number your carbons: we will always have four carbons in our diene--at least those four carbons, and then there will be two more (5, 6): 1, 2, 3, 4, 5, 6.1889

And then, what is missing here?--we will always have this double bond between carbons 2 and 3; there is our cyclohexene; we have our ester groups on carbons 5 and 6 (ethyl ester, ethyl ester).1902

And what else is missing?--between 5 and 6, we had two π bonds; only one of them was involved in the Diels Alder reaction, so the other one stays behind.1918

If we were to use an alkyne as our starting material, we would get a cyclohexadiene product out.1926

We would always have these two double bonds opposite each other after using an alkyne.1936

OK, what does a diene look like?--now again, the diene, typically, in most Diels Alder reactions, acts as the nucleophilic component.1943

OK, so it is best if it is electron rich; it often has electron donating groups, like an alkyl group or an ether group.1951

OK, it is not possible to have an OH: an OH can also be an electron donating group, but we can't have that on the diene; and what is the problem there?1960

Let's imagine trying that: what happens if we put an OH group on a diene--what functional group does that give us here?1968

When you have an alkene and an alcohol on the same carbon, this is an enol; OK, and an enol is not a stable functional group.1976

What does it do?--it tautomerizes to the ketone...in this case, the aldehyde.1984

OK, so it would be impossible to make an OH on a diene, but you can have an O-R; you could have just an alkyl group; and so remember, alkyl groups donate electron density inductively, so that makes this more electron rich.1991

Something like a methoxy group would donate electron density by resonance; this lone pair is allylic, so we can add it in.2005

And when we take a look at that resonance form, what do we see?--this is now electron rich.2017

That is what makes an electron donating group; it makes an electron rich diene, which makes it even hotter for doing a Diels Alder reaction.2023

OK, now another thing that we have to have with the s-cis: I mentioned how that we will always want those two double bonds being pointed towards the dienophile--it turns out that that is a requirement for the Diels Alder.2033

If it is going to be a concerted mechanism, and it happens all at once in the 6-membered transition state, those have to be able to line up with the dienophile so that we can form the 6-membered ring.2041

OK, but now, sometimes we are going to see our diene presented in an s-trans configuration; that is OK, and that is usually the way we draw it, with a zigzag here, because it is the most stable.2051

But what we have to recognize is: before we do the Diels Alder, we must rotate it into the required s-cis conformation; now, we can line it up and do the Diels Alder.2061

That is part of what the heat is here for--that is part of the requirements of the Diels Alder: part of that energy that is going in is to rotate it to the less stable s-cis conformation.2074

Now that it is here, now it is able to do the Diels Alder.2085

2, 4, 6 electrons...and so what does our product look like?--we have a double bond up here, and our two methyl groups; and on this carbon...this carbon is right here; it has the two cyano groups.2089

OK, so a lot of times, to do our Diels Alder, we have to flip this over; we are just taking that double bond and flipping it up--in this case, flipping it down.2107

It is definitely worth redrawing it before you go to predict a product, because then you are less likely to make mistakes in drawing the product.2118

OK, let's see another example--how about this dienophile reacting with this diene?2126

Well, the problem here is...now again, this is s-trans (right?--we have one double bond going in this direction, and another going in this direction), but how about if we tried to rotate, to flip one over so that it would be s-cis--is that possible in this case?2132

The presence of those rings locks them in this position; we could described this as being locked in the s-trans conformation; so no matter how much we heat this or how much we try, there is going to be no Diels Alder reaction.2147

It must get into that s-cis conformation in order for the Diels Alder to occur.2161

Now, a lot of times, the kind of diene we use is a cyclic diene, like this one: cyclopentadiene is very good at doing a Diels Alder.2169

What is great about cyclopentadiene is: because it is in a ring, these double bonds are held in an s-cis conformation; they are locked in that conformation, which means they are just always ready to go (to do the Diels Alder) in an instant.2176

In fact, cyclopentadiene is so good at doing the Diels Alder, it reacts with itself and does a self Diels Alder, where one equivalent acts as the diene; the other acts as the dienophile; and you get a dimer out.2190

Any time you want to use cyclopentadiene, you have to distill it fresh, and you add heat to cause the reverse Diels Alder, so that it breaks up and gets back to the cyclopentadiene; and then you can distill that and use it fresh for your Diels Alder reaction.2206

Now, the stereochemistry we are going to see gives something that we are going to describe as an endo product: and we will see what that looks like in just a moment.2224

First, let's draw our Diels Alder product.2231

Again, it is important to identify the terminal carbons of the diene, because those are the carbons that are going to react with the dienophile.2234

OK, again, a good dienophile--it has some cyano groups on there, so that looks good.2246

Let's number our carbons: 1, 2, 3, 4, 5, 6.2250

OK, and we are going to be forming a 6-membered ring: 1, 2, 3, 4, 5, 6.2256

Our double bond...I'm sorry, let's go ahead and do our mechanism.2265

This π bond comes up; this π bond comes up; this π bond comes over; so there are our 2, 4, 6 electrons.2268

We'll have a double bond here between 2 and 3, and we have our cyanos at 5 and 6; but notice that, right here, connecting carbons 1 and 4, we have this CH₂ group.2275

This is now going to act as a bridge over the cyclohexane ring that we just formed.2289

We can draw it as a wedge coming up out of the page; we can draw our 1-carbon bridge, our methylene bridge.2296

Now, if this bridge is drawn up, as shown, it turns out that these cyano groups end up pointing away from the bridge: that is what we call the endo stereochemistry.2306

Now, it is possible to draw it like this; but usually, we draw it in a three-dimensional perspective that looks more like the actual structure, the actual shape of the molecule, where we kind of look at it from the side.2318

So, kind of like we use the chair conformation (we draw the line drawings of that, where we are kind of looking at it from the side, and then we get rid of our dashes and wedges), we are doing the same thing here.2334

OK, so usually we draw it like this; this is good to practice this; and what this is saying is that, when this carbon chain is up, then the cyano groups are going to be pointing down.2345

We have a hydrogen here, and we have a hydrogen here; this is called the endo product.2359

The endo product is preferred; this is going to be the major product.2367

OK, just for comparison, what is not happening is: we don't get the bicyclic product where the cyano groups are pointing up in the same direction as the bridge.2373

This product we call the exo product, and this is not formed.2388

OK, so this is just a discussion of stereochemistry: we are asking about the groups that are on the dienophile: what is their relationship to the bridge that we have?2395

Now, I am not going to get into why the endo is preferred; it has to do with when these cyano groups line up with the dienophile...with the diene...2406

The electron withdrawing groups can either line up so that they go underneath the diene, or they can line up the other way, so that they point away from the diene.2418

They prefer to line up underneath the diene, because then you have additional orbital overlap between the electron withdrawing groups and the diene p orbitals.2426

OK, so because that orientation is preferred, we end up with the cyano groups, or whatever electron withdrawing groups we have, pointing in the down position.2436

OK, let me just talk a little bit more about this endo versus exo, so we can get this terminology down--especially because this is the first time we have seen these kind of bicyclic systems.2445

This is called norbornane as the common name, but the IUPAC name--how would you name a bicyclic compound?2456

Well, it is called a bicyclic compound because you would have to cut two bonds in order to become an acyclic product; so you would have to cut one bond, and you would still have a ring; you would cut a second bond, and then you would have no more rings left.2464

That is why it is described as a bicyclic compound.2478

Now, overall, this is still a 7-atom ring, a 7-carbon molecule; so it's a heptane.2481

If there were just one ring, we would call that cycloheptane; because there are two rings, we call that bicyloheptane.2490

And then, what we do is: in these brackets, we list the bridge sizes in decreasing order.2497

OK, so if you look at this norbornane, we describe these positions as the bridgehead carbons.2503

Those are the carbons that are shared with both rings; and if you look at it a certain way, you can see that those two carbons are connected by three bridges: here is a 1-carbon bridge; here is a 2-carbon bridge; and here is a 2-carbon bridge.2517

The way we would name this compound is: we would call this bicyclo[2.2.1]heptane.2534

We list the sizes of the bridges in decreasing order, with periods in between: bicyclo[2.2.1]heptane.2541

Really briefly, let's look at another bicyclic example, and see if we can do the name for that.2549

This is another bicycloheptane: there are two rings, and there are 7 carbons total, so this is bicycloheptane.2554

Find the bridgehead carbons (right here and right here), and then ask how many carbons are in all of the bridges connecting those two bridgehead carbons.2564

Here, starting with our biggest one: our biggest carbon bridge is right here; it has 1, 2, 3, 4 carbons.2577

The next biggest bridge is right here: it has 1 carbon; and the last bridge is right here--how many carbons are there in that bridge?--there are 0 carbons; the two bridgehead carbons are directly connected.2585

So, we call that a bicyclo[4.1.0]heptane.2596

OK, so just a little brief introduction to that IUPAC: so, when we look at norbornane, we have this as our smallest bridge; we look at the positions here.2601

Knowing these are tetrahedral...we have 2 positions that are pointing down and 2 positions that are pointing up.2614

And the ones that are pointing up in the same direction as that bridge, we describe as the exo position; and the ones that are pointing away from the bridge, and kind of in this cave on the inside of that cup shape, we call the endo.2621

So, the exo and the endo...and so, what we find is: when we get a bicyclic product like this in a Diels Alder reaction, we choose to put the electron withdrawing groups in the endo position, away from this bridge.2639

OK, let's see if we can do an example.2656

Here is a cyclic diene: there are the two carbons of our diene that are going to react; here are the two carbons of our dienophile that are going to react.2660

Those are the two new bonds that we are going to form; you can always find your 6 carbons that are involved: 1, 2, 3, 4, 5, 6.2671

There will always be 6 carbons involved; so we are going to form a 6-membered ring, and now we are going to have a 2-carbon bridge connecting those bridgehead positions.2679

The shape of the molecule is going to look like this: it's going to go up, and we are going to have 2 carbons up top.2691

This is the shape of our product.2702

Now, let's number our carbons: right here is the 6-membered ring that was formed: 1, 2, 3, 4--so carbons 1 and 4...here is the 2-carbon bridge above that got pushed up, and then 1 and 4 are now connected to 5 and 6.2704

On 6, we have our two positions; and where are we going to put that aldehyde group?--are we going to put it on the top or the bottom?2724

We are going to put it down here, because that is the endo position.2732

Make sure you always draw in the hydrogen, because otherwise, if you kind of draw it all out in the side, it might be ambiguous whether it's exo or endo.2739

So, always draw in the hydrogen; kind of like when we did axial and equatorial on a cyclohexane, it is always a good idea to draw in the hydrogen, not just the substituent.2745

OK, what else is missing on this?--something else is missing.2753

Remember, our mechanism always has this π bond moving; so between the middle two carbons of the diene will always be a remaining π bond.2756

This is our product: now, what is interesting about this product--one last thing about it--is: we have just drawn a chiral product.2769

It is impossible to draw a chiral product and just have a single chiral product after starting with achiral starting materials.2775

What we always want to remember is that you will always have an enantiomer formed with a chiral product (so in other words, a racemic mixture, a racemate).2786

Now, what does the enantiomer look like for this?--well, if we drew our same product with that same orientation, and we still need to have the endo electron withdrawing group, but if we chose to put it on this carbon instead, that would also be the right answer: it is actually the enantiomer of this.2798

Now, if we rotate them around, hopefully you can see that mirror-image relationship that we know exists for enantiomers.2826

OK, but I just want you to keep that in mind: when maybe you draw one product, and you check your answer, and you see a different product drawn, it might be both the right answer; they are just different enantiomers that are formed.2832

It doesn't matter if we just flip this over; that could also react, and that would just form the enantiomer; and both of those are just as equally likely to happen, so we are going to get a 1:1 mixture, a racemate.2844

Let's talk, next, about the stereochemistry of the Diels Alder reaction.2861

Now, we already talked a little bit about stereochemistry for those bicyclic cases; we saw that endo is the major product.2864

OK, the other issue about stereochemistry is that the stereochemistry of your dienophile is retained.2871

We also talked about how, if you form a chiral product, it is going to be formed as the racemate.2878

OK, but let's see an example where the dienophile has some stereochemistry; and this is an example...where we have this ester and this cyano, we can see that those two groups are cis to one another in the starting material.2883

So, when we draw our product, let's draw our two bonds that we are forming; number our carbons: 1, 2, 3, 4, 5, 6; go ahead and do our mechanism, since it's so quick and easy--just concerted mechanism.2897

OK, 1, 2, 3, 4, 5, 6; we have a double bond between 2 and 3, and we have an ester and a cyano to put on 6 and 5.2915

OK, but what we have to show is: they started out cis to each other; they are still going to be cis in the product.2926

So, how do we draw a cis when we are looking at a cyclic product--how do we show substituents being cis to one another?2932

They are going to be pointing at the same side of the ring: so they are both pointing up, or they are pointing down; so, in other words, you could draw them both as wedges, and that is a way of showing that they are still cis.2940

So, because it is a concerted mechanism and it happens all in one step, the spatial arrangement--the stereochemistry of the substituents remain fixed.2955

They start out cis, and they end up cis.2965

Now, I have chosen them both to be wedges; they, of course, could be dashes instead; let's take a look at that product to see if that is the same thing.2968

What do you think?--is that the same thing--are these two the same molecule or not?2979

I can't get them to superimpose; and in fact, if I flip this over, I can see their mirror-image relationship; these are, in fact, the enantiomers.2984

This is another example where, because this product is a chiral product that I drew, I know it can't be the only product; the enantiomer also must be formed.2994

Now, that doesn't mean you have to draw both every time, but if you draw one chiral product, you have to say "plus enantiomer," or you have to say "racemic."3004

You have to indicate the fact that this is not the only product that is formed.3012

OK, so cis means they are both wedges or they are both dashes.3016

Let's see an example of that: now we have a cyclopentadiene reacting with an ester; here, our groups are trans to each other, so we have to keep that in mind.3023

We know it is these end carbons of the diene that are going to react; so this looks like one of those bicyclic systems that are going to be formed.3037

Let's number our carbons: 1, 2, 3, 4, 5, 6.3048

And we are going to have a 1-carbon bridge, so we're going to have a norbornene-type backbone.3054

Double bond between the middle carbons; 1, 2, 3, 4, 5, 6; so, in these bridge systems, in these bicyclic systems, the 6-membered ring that we are forming, we are drawing as a boat down at the bottom, and then we are having a bridge (a 1-carbon bridge, in this case) connecting these opposite carbons.3064

OK, so there is our product; now, how do we draw these ester groups being trans to one another?3084

They are on 5 and 6; let's draw in our positions of what a tetrahedral carbon looks like--we can draw them trans by putting one in the down position and one in the up position.3092

Don't forget, hydrogen is in the opposite position; so we could draw one up and one down.3107

Now remember, we said we like these electron withdrawing groups to be where?--we want them to be in the endo position, the down position, but in this case, we can't put them both in the down position, because they have to stay trans.3111

So, we get one that is endo and one that is exo, and that is just the way it is; no problem.3123

The only product is that chiral product that just formed: this actually is a chiral product that has no plane of symmetry, so we want to say "racemic."3128

Can you envision what the enantiomer would look like?--how could you draw the enantiomer instead (draw a product that is right, but looks different from this one)?3138

They still need to be trans, but instead of having the front carbon down and the back carbon up, you could have the back carbon down and the front carbon up; that would still be trans, and that would simply be the enantiomer of what we have already drawn.3149

OK, and finally, how about the regiochemistry of the Diels Alder?3164

Regiochemistry is when we are looking at which site of reactivity reacts with another site--which site do we react with?3167

And this is what we are going to have when we have a dienophile that is not symmetrically substituted--a diene that is not symmetrically substituted reacting with a dienophile that is also not symmetrically substituted.3176

We are going to get two possible products: if we take this diene and this dienophile, and we line them up as drawn, we would get this product.3188

But if we flip the dienophile over, and we lined up the opposite ends, we would get this product.3196

This is a question of regiochemistry.3202

OK, and the rule is that 1,2 product is favored over 1,3; so here, these two groups are 1,2, so this is the preferred product; this is the 1,3, and it is not formed.3205

1,2 is favored over 1,3.3224

Notice that I haven't shown the stereochemistry here; and that is kind of an advanced topic--typically we don't worry about it.3227

It is possible to predict the stereochemical relationship between the group that was on the diene and the group that was on the dienophile, but we usually don't get into that at this level.3234

OK, so we will just leave it as straight lines now, and assume that all diastereomers are formed.3246

OK, so one rule is that 1,2 is preferred over 1,3; if we move our substituent here to be at this position, now when we look at the two possible alignments, we get that the two groups can be 1,3 or 1,4 to each other.3252

The rule there is that, again, 1,3 is disfavored; this is not formed--the 1,4 is favored instead; this is the major.3270

The 1,2 product and the 1,4 product are the major products; now sometimes, some books call it ortho-like and para-like; I don't like using those terms, because we are not dealing with benzene rings.3283

OK, but the two groups will either be 1,4 to each other or 1,2 to each other; that is going to be better than 1,3.3294

Now, why is that--what is going on here?3303

Again, this is a pretty advanced topic, and so a complete discussion of this and an explanation of why we get this observed regiochemistry has to do with molecular orbital theory.3306

So, you really need to look at the orbitals that are involved--the orbitals from the diene that are interacting with the orbitals of the dienophile--and you need to have them line up just right, and have their symmetries conserved, and so on.3317

OK, so without going into a complete, lengthy discussion on that, I'll just kind of refer to that slightly.3331

The diene is the part that is your nucleophile, that has...you look at the molecular orbitals, and the highest occupied molecular orbital (we call that the HOMO), and that is going to be interacting with the lowest unoccupied orbital (called the LUMO) on the electrophile (the dienophile).3341

That is the interaction that you want to make as good as possible.3365

OK, on certain cases, we can predict that, and we can explain it, even without looking at the MO theory; and those are cases that involve resonance.3371

So, let's do that for a few examples.3380

Here is an example: if you look at a methoxy substituted diene, and you consider the resonance...we know that this is an electron donation group, so let's go to this resonance form.3382

We know that this is a contributing resonance form to this diene; and, when we look at this dienophile with an aldehyde here (an electron withdrawing group), we know that it has resonance; and this is a contributing resonance form.3399

When you consider lining these two up, it is not random how they are going to line up.3414

On the previous slide, it kind of looked like, "Well, we could put it this way, or we could put it this way"--it seems kind of 50/50; but when you really look at the reactivity of the substrates, you see that what you want is the diene with its partial minus...and right here, you want that lining up with the dienophile and its electron withdrawing group, so that you have the best nucleophile, the most nucleophilic site, joining up with the best electrophile, the most electrophilic site.3420

It is not random how they line up; and when you look at the molecular orbital diagrams, it is not random how they line up.3455

When you look at the molecular orbitals themselves, it is not random.3462

This is the product we are going to get, which would lead, in this case, to a 1,2 substituted product; and that is what we predicted to be major.3465

Our predictions work very well; and there are a few ways you explain it, but when it comes to resonance, you can use resonance theory to explain it quite nicely in certain situations.3477

OK, let's take a look at several examples: how about this one?3488

Let's consider both the regiochemistry of the reaction (in other words, do we line them up as shown, or do we flip them over and line up the other ends?) and the stereochemistry of the problem.3491

There is a lot to be involved with here.3503

Well, here is the interaction that we have; and again, don't be tempted to just connect them as drawn; we might need to flip one of them over before we connect them.3508

OK, but right away, I see something that is confusing, because I know that, for the regiochemistry, I want 1,2 product or 1,4 product, but I have a group on my diene--but I have two groups on my dienophile!3519

What am I comparing the position of this methyl group...what am I comparing it against when I decide if it is 1,2 or 1,4?3538

OK, so now I want you to look at those two groups that are on the dienophile: you have a methyl and you have a cyano.3545

Which of those two groups do you think is going to be more influential in guiding the regiochemistry of the problem?3550

Which one will have a bigger impact, and more of an interaction with this π bond?3557

Certainly it is the electron withdrawing group that has the resonance and has the bigger impact; so the electron withdrawing group controls the regiochemistry.3563

If I were to line it up as drawn, what kind of regiochemistry would I get?--I would get 1, 2, 3--I would get 1,3 as my product.3576

So, what I am going to do is: I'm going to flip this over; I'm going to put the cyano down here and the methyl up here.3586

And when I flip it, I want to make sure that I keep my groups trans to each other; I need to keep them trans.3594

It doesn't matter whether you have it this way or this way; it doesn't matter whether the cyano is on the left or right; but they still have to be trans, because that is going to be important.3600

Now, I'm going to bring these two together: 1, 2, 3, 4, 5, 6; 1, 2, 3, 4, 5, 6; numbering the carbons is so important, so that you can track all of your substituents properly.3608

Double bond between 2 and 3; I have a methyl group on carbon 2--that is going to be planar, because it's on a double bond; and then, on 5 and 6, I have a cyano and a methyl--they started out trans; they are still going to be trans.3624

This cyano group on carbon 5--is it going to be a dash or a wedge?3638

It actually doesn't matter--you can pick one: let's make it a wedge; that is fine, but if I make it a wedge, that means my methyl group on carbon 6 has to be a dash; it has to be the opposite.3645

We have to have one up and one down; that is going to be trans.3654

So, of course, you can pick the cyano to be a wedge or a dash, because that would be drawing one enantiomer or the other.3659

Let me just draw both, so we can see what they look like--right?3666

If I drew this as a dash, then my methyl group would have to be a wedge, and so we still have trans.3670

We have a racemate here, as usual, because we have drawn a chiral product in this case.3678

OK, let's take a look at this interesting problem: it says, "Explain why no Diels Alder reaction takes place in this case."3688

OK, I have a diene...what are the components you need for a Diels Alder?--I have a diene; I have a conjugated diene, so that is good; I have something that could be a dienophile--does this look like a good dienophile?3697

Here is a double bond--that makes it a potential dienophile; and what does it have attached?--it has two carbonyls.3709

This is actually a great dienophile--so what is the problem?3714

If you went to try and do this reaction, what would you have to do?3719

Well, remember, we have always lined up the ends of the diene and the ends of the dienophile in order to predict the product.3725

So, what we would want to do first is: we would want to rotate this to get in the s-cis conformation.3730

OK, it has to be in the s-cis conformation to do the Diels Alder.3737

But here is the key: what does t-butyl stand for?--t-butyl stands for tert-butyl; and so, what happens is: once you rotate it, that brings those two tert-butyl groups toward each other and forces them to be coplanar and allows steric hindrance to interfere here.3740

OK, it is impossible for these to be coplanar, because the bulky tert-butyl groups keep twisting it out of the plane to avoid the steric interactions.3763

Sterics prohibit the s-cis conformation.3774

And, without an s-cis conformation, there is no Diels Alder.3783

OK, this is another interesting Diels Alder reaction, because it starts with, not an ordinary diene like cyclopentadiene or butadiene; it starts with anthracine.3790

Anthracine has, built in it, a diene--1, 2, 3, 4; and normally, we say, "Wait a minute; this is an aromatic compound; aromatic compounds do not behave as normal alkenes."3802

OK, but anthracine will do this reaction, and we will see why in just a second.3817

Let's try and draw this product: now again, we are going to have to imagine...let's keep this in the plane; let's bring in the dienophile from the top here.3821

I'm going to take that middle 6-membered ring and spread that out into the bottom.3833

This is going to be 1, 2, 3, 4, and we are adding 5 and 6 up here; so we are going to look at it in a slightly different point of view, just so we can keep these benzene rings down here.3841

On this carbon, we have a benzene ring; and on this one (let's bring this in here--let's see if we can do that without making a huge mess), we form this bond.3857

And when you do your arrows, you see that we move the π bond to be between carbons 2 and 3.3871

Look what it gives us: after it does the Diels Alder, we still have two aromatic rings left over.3884

We have lost very little aromatic resonance.3892

OK, we used to have three aromatic rings, but after forming these new carbon-carbon bonds, we still are left with two benzene rings, which is still very good.3903

And remember, we formed new carbon-carbon bonds, which is always a good thing; so it turns out that this is a favorable reaction.3914

Anthracine is a very interesting dienophile to use, and it is good to get some practice and see some examples there, and to understand why, in this case, it might be OK to do the Diels Alder, because we still are able to keep two of those benzene rings intact.3920

OK, and finally, let's look at a synthesis example that utilizes a Diels Alder reaction.3940

I'm thinking I might utilize a Diels Alder reaction, because my target molecule has one of these bicyclic systems that we have seen being made from Diels Alder reactions.3946

Our instructions are to synthesize this from starting materials with no more than 5 carbons; so it gives us some guidance as to how much we have to disconnect this molecule.3956

Now, first of all, I see an epoxide in the target molecule; so let's do a retrosynthesis asking what starting materials I need.3968

And what reaction have you seen that makes an epoxide--what functional group did you have initially that could be converted into an epoxide after?3980

How about a double bond?--if I had a double bond here, then I could convert that to epoxide--do an epoxidation reaction.3991

OK, and that is cool, because now this gives me my norbornene type structure--remember, that double bond is part of the Diels Alder product.4004

The question here is, "How do you do a retro Diels Alder?"4014

That is another skill that you can develop when you are learning about the Diels Alder reactions.4018

So remember that these two carbons--the bridgehead carbons--used to be the end carbons of the diene.4023

Here is the 5-membered ring that...this came from cyclopentadiene: 1, 2, 3, 4 carbons, plus this extra bridge.4030

Those reacted with these two carbons (5, 6); so we can break...these are the two bonds that were formed in the Diels Alder; these are the two bonds we can break to do the retro Diels Alder.4039

We can even use our arrows, starting with the π bond of the 6-membered ring.4052

We can even use our arrows (2, 4, 6) to do the retro Diels Alder, because that can help us track the electrons around.4056

See, we are breaking this bond now; we are breaking this bond.4063

And so, what do we end up with as our starting material?--we get cyclopentadiene, and we get a 2-carbon dienophile with an ester group.4067

That would be a very good dienophile; this would be a very good diene; this would be a great Diels Alder.4080

So, our synthesis is simply taking cyclopentadiene and reacting it with the dienophile and some heat; that will do the Diels Alder.4085

OK, this is racemic: notice, it says "racemic" here; so I'm not just getting the ester at this position--I'm also getting the ester at this position.4096

OK, and then, what do I have to do to get to my target molecule--how do I epoxidize a carbon-carbon double bond?4105

I need some kind of oxidizing agent: it is going to be mCPBA: you need some kind of a peroxide to do that oxidation.4113

So, that is what gives the target molecule.4124

Diels Alder: a very important reaction synthetically, and just the one of paracyclic reactions that we will be studying; and it is one of the examples of a type of reaction that the class of compounds of conjugated dienes can undergo.4128

That wraps it up for the conjugated dienes lesson.4145

I hope to see you again soon at Educator.com.4149