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CHEM 125b: Freshman Organic Chemistry II
- Metals and Catalysis in Alkene Oxidation, Hydrogenation, Metathesis, and Polymerization
Alkenes may be oxidized to diols by permanganate or by OsO4 catalysis. Metal catalysts provide orbitals that allow simultaneous formation of two bonds from metal to alkene or H2. Coupling such oxidative additions to reductive eliminations, provides a low-energy catalytic path for addition of H2 to an alkene. Such catalytic hydrogenation is often said to involve syn stereochemistry, but the primary literature shows that addition can be anti when allylic rearrangement occurs on the catalyst. Similar oxidative/reductive cycles operate in olefin metathesis and metal-catalyzed polymerization. Careful catalyst design allows control over polymer stereochemistry (tacticity). Polymerizations catalyzed by free-radicals or acids typically lack stereochemical control, but there are ways to control regiochemistry and chain length. Latex, a natural polymer, coagulates to form a rubber ball.
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Freshman Organic Chemistry II
CHEM 125b - Lecture 15 - Metals and Catalysis in Alkene Oxidation, Hydrogenation, Metathesis, and Polymerization
Chapter 1. Alkene Dihydroxylation [00:00:00]
Professor McBride: OK, we’re still talking about addition to alkenes. And we’ve looked at these various electrophile-cum-nucleophiles adding to alkenes. And now, at the beginning last time, we introduced the idea of osmium tetroxide and we’ll go on to permanganate as reagents for dihydroxylation of alkenes. Then we’ll go onto things involving transition metals, catalytic hydrogenation, and also metathesis and catalyzed polymerization.
So last time, we looked at the end at OsO4, which looked a little bit like O3 and could do the same trick O3 could do, to give this osmate ester, which is like an acetal. So it’s possible to add water and get the osmium equivalent of a carbonyl group, and the two hydroxyl groups added to the termini of the alkene.
So the problem with this is OsO4 is a very efficient reagent for that, but it’s poisonous, and it’s also very expensive. So you have to figure out some way not to use so much of it. And the way you could do it is to use just a little bit of it as a catalyst, and put hydrogen peroxide in there, in order to convert the osmium equivalent of trioxide back to the tetroxide again, so it can cycle around and keep doing the reaction.
In fact, it was improved in 1976, by using a different oxidizing agent, this N-oxide, the morpholine oxide. And in fact, this is called the Upjohn process, which is relevant because the guy whose picture is over here, whose estate gave the money for redoing this room awhile back, was the research director at Upjohn Laboratories.
So anyhow, that was a good way to do it. But then, we should notice something about the stereochemistry of this. We showed here a trans-2-butene. And of course, if it’s like ozone, the two oxygens will add at the same time, therefore to the same face. So the configuration you get at the two carbons, where the OH’s add, are related to one another. They could be S, S, as shown here.
Of course, the osmium tetroxide could equally well have come in from the bottom, or the 2-butene could have been turned over, and then you would have gotten R, R. So of course you don’t get chirality without starting with chirality. That is, a single chirality. However, you can start with it in a ligand that attaches to the to the osmium. So if you use a chiral amine ligand, then, you only get one of the two enantiomers of the dihydroxylated alkene.
And this, you won’t be surprised to hear, was also invented by Professor Sharpless, whom we talked about last semester with respect to adding oxygen. And so this is called the Sharpless asymmetric dihydroxylation. He invented it in 1988, and I think this room was the first place he talked about it in public. He was standing right there, and in fact, he did a demonstration on this table, as I’m going to do in the end.
But I’m going to do a different demonstration. Because osmium tetroxide which he used, is very dangerous and poisonous. And he actually spilled it on this table. I think it’s now long gone, since that was 1988. But at least what I’m going to do just uses vinegar, so you don’t have to worry about that.
OK, so anyhow, here’s a specific example of using Sharpless asymmetric dihydroxylation with that diene. It gives a 97% enantiomer excess of the indicated configuration here. Now, the permanganate is much cheaper, and looks very much like osmium tetroxide, so it won’t surprise you that it also can do this trick. And here, you get 85% yield with cyclohexene. And again, it’s a syn addition, the two OH’s come in on the same face. But that’s not one where there’s a catalyst that’ll make it give only one enantiomer.
Chapter 2. Catalytic Hydrogenation of Alkenes: Oxidative Addition, Reductive Elimination [00:04:28]
Now on to catalytic hydrogenation. So we looked before at the idea of drawing two curved arrows and making H2 add to an alkene. And saw that the orbitals weren’t set up for that. HOMO wouldn’t match HOMO, LUMO with LUMO, so there’s no profit in that. However, we said, even though that doesn’t work, that it does work if you have a metal catalyst, like palladium or platinum, or many other transition metals.
So let’s look a little bit at how it is that palladium does this trick. It does it by providing orbital variety. Rather than just having the orbitals that we showed before, the s and s* of the H-H bond, and the p and p* of the alkene, you have lots of different shapes of orbitals on the metal. For example, in ethylene, we have the p, and the p*, but in the metal, palladium has 4d10, 5s0, and 5p0. So the occupied orbitals of palladium, then, are this dz2. Now, is that set up to overlap well with p*? The HOMO below and the LUMO above? What do you think, is that going to be a good reaction? Who’s got an opinion, Karl?
Student: I don’t think so because how you have the red on one side of the blue and on the other side of the blue you might be able to get one side of the red to overlap?
Professor McBride: Maybe on one side, but not as shown there. Right? That’s what you’re saying there? And you’re right, they’re orthogonal as drawn. But there are other d orbitals. There are five of these d orbitals, so we could check out others. How about the dyz, is that going to do the trick? Are you going to get overlap if you bring that up? Kate, what do you say? If you turned it, you could…
Professor McBride: but as shown, it’s orthogonal. Right? OK, how about that one? Is that going to do the trick? No? Chris is shaking his head. That won’t do it. How about that one? No? It’s still that one of them is symmetrical right to left, the other is antisymmetric right to left, so they’d have no overlap. How about that one? Ah, just right.
OK, so that one can come up and overlap, and stabilize the electrons that were on the metal. And that new orbital shown there is, in fact, the HOMO – 4. It’s gotten quite stable, right? In fact, part of the reason it got stable is the nature of the orbitals changes as atoms come up, and the geometry changes. So although it looks very much like those two original orbitals, and in a sense, came from them, by the time it gets to this point at the complex, it’s 40% of that dxy orbital on the bottom. 47%, actually, on the top is sCH’s, but 13% is still the p*. So that’s just to tell you the ugly truth. But the fact is that the reaction began, and went through a transition state because of the mixing of those orbitals.
So that’s two electrons. But how about the other way around? How about if we use the HOMO on the top? Now, of course, we have all these d orbitals on the bottom, and few of them would overlap, and even that one which does overlap you don’t care about, because it’s HOMO with HOMO. We want unoccupied orbitals on the bottom, which means that 5s and 5p. So there’s the 5p. That’s an unoccupied orbital, that’s going to overlap well, and there’s the 5s. It’ll overlap. So you use some mixture of those two. You’ll use a hybrid orbital. An sp hybrid of some sort on the palladium, which will overlap with that HOMO and stabilize another pair of electrons. And there it is in the complex. You see, it’s a hybrid orbital below, and the p orbital above, forming that new bond, stabilizing those electrons.
In fact, if you look at the ugly truth on this one, by the time you get to the complex, it’s changed. It’s 67% of p, and it’s 11% of an sp hybrid. But it’s also 15% of that d orbital that we showed at the beginning, so things have gotten mixed up. But that’s the way it happened, because you had these vacant orbitals that could overlap from below on the metal, overlap with p, you had the filled orbital, the particular one, the orbital that could overlap with p*. So you stabilize two pairs of electrons. That means you make two new bonds to the alkene.
So the palladium, or platinum, or many of these transition metals can do this trick of adding two at the same time to the alkene. Now you can also do it to s bonds. So here’s palladium reacting with H2. It’s just s and s* now. But if that’s the energy of the starting material in a pretty crummy calculation, so don’t hang your hat on this, these are successive stages as you go left to right. First, you begin to associate the H2 with the palladium, and bond to it. Then the H2 gets even closer, and as it gets closer and closer, the hydrogens spread apart. And the reason I’ve drawn them up and down a little bit is to show energy minima and transition states.
So first, you get the H2 associated with palladium, without very close contact. Then it gets closer, and goes over a transition state, and becomes this one, which is closer bonded. And then, the two hydrogens come apart, and get quite far apart.
But notice, at least in this calculation, and in some others, as you’ll see, it goes up hill a little bit. So that would seem to be unfavorable. You could do it, you could split the hydrogen apart on the metal, but in terms of enthalpy, kilocalories per mole, it’s uphill. However, once it gets up there, these hydrogens can move from one metal to the next, because we’re talking a solid metal, in this case. So in fact, it dissociates on the palladium surface, and then the hydrides move. So you get an advantage from entropy, from the hydrogens getting apart, even though it’s uphill a little bit in energy. It’s very favorable to move them apart in entropy, rather than having to have them right next to one another. So it’s possible to dissociate hydrogen on palladium.
Now here’s a calculation certainly better than the previous one. It was published in the Journal of Physical Chemistry in 2004. And it shows the same thing, palladium approaching hydrogen above, palladium approaching methane below. And again, I don’t think that these are great numbers, probably, but it gives you the idea of what I showed on the previous slide, and some specific geometries.
OK, now, catalytic hydrogenation puts these two reactions together that we just looked at. You have what inorganic chemists call “oxidative addition.” That’s where we do this, which we just saw is possible, two electrons can get stabilized because of the orbitals that are available, to form something like a three-member ring here. And that’s called oxidative addition. The reason it’s called oxidative is the way inorganic chemists count electrons. They say that if you have a bond to metal, those electrons belong to the ligand, not to the metal. So electrons, formally, have been taken away from the metal. We’ll talk more about oxidation and reduction before too long. But anyhow, that’s called oxidative addition.
But of course, you could do the reverse, as well. You could do reductive elimination, and pull the alkene back off again. OK, now you do the same thing with hydrogen, as we just saw. So you could have oxidative addition, those two arrows move like that, to give that, and then the blue reverse, the reductive elimination, and you’re back where you started. But now, suppose you do both of these things. Let’s suppose that we first do the oxidative addition, the insertion of palladium into the H-H bond, and one of the H’s, let’s say, moves away. And now we have room on the surface of the palladium, there, to bring in the alkene and do its trick.
So now, we have this form. Now of course, you could do reductive elimination again. Now incidentally, in that form, the experts discuss just how much the different bonds are formed. Sometimes they draw it this way, and just say the alkene is complexed. But that begs the question of what is it that’s holding the alkene there, and it’s those orbital interactions that we looked at. The electrons going one way, the electrons going the other. So we’ll for our purposes of our lecture today, we’ll say that the bonds are completely formed, even though they may not be.
OK, so you can do reductive elimination, it could come off again. But suppose you did the other reductive elimination? Suppose you did that one? Now you’re over here. And you’ve put hydrogen on one carbon, and palladium on the other. Now suppose we have that hydrogen that’s wandering around the surface, suppose we bring it back? Can you see what I’m going to do now? Chris, do you have an idea of what step is next? Look at the title of the slide.
Student: We might hydrogenate?
Professor McBride: Yeah, we want to put a hydrogen on each of the carbons, how are we going to do that? We’ve got a hydrogen on the right carbon already, how do we get a hydrogen on the left carbon?
Student: We have a hydrogen nearby, that’s bonded to the palladium.
Professor McBride: Yeah, and the carbon is bonded to palladium. And what will we call the process that gets the hydrogen bonded to carbon, instead of the carbon and the hydrogen bonded to palladium?
Professor McBride: Nope. It’s that. Anybody got a name for that? That’s the reductive elimination, the blue thing we’ve been talking about, all these, right? They come on, they come off, they come on, they come off. But if you put them on one way, and take them off a different way, you’ve done the trick. Got it? OK.
Chapter 3. Catalytic Hydrogenation of Alkenes: Stereochemistry [00:15:08]
Now, notice, that that addition of the palladium, the oxidative addition of alkene to palladium, happened two at the same time, so they come in on the same face. It’s a syn addition. And then, when the hydrogens replace palladium, they come in on the front side. It’s not a backside attack. When we curve those two arrows, it’s not attacking backside. So, they come in from the same side as well, so the overall product is syn addition. The H’s came in on the same– here, the bottom face of the double bond. And of course, if you put methyl groups on there, and it were 2-butene, then you’d expect to get that particular isomer of the 2-butene, where the two hydrogens operated, added on same side. Of course, you couldn’t tell it, because the other two things would be hydrogens as well. How could you tell? How could you tell that the two came in on the same side? You wouldn’t use hydrogen, what would you use?
Professor McBride: You’d use deuterium, as you say. Yeah, so then you could tell. And here’s an example, from the book, where they use deuterium and platinum oxide, in this case, and the two D’s came in on the same side. That was not necessary, in this case, to tell, because you also have the cyclohexane ring there. So, it says here that deuterium is added in a syn fashion. It’s a syn addition. And here’s a typical passage from another textbook, the Loudon textbook that we used several years ago, about stereochemistry of this addition reaction, catalytic hydrogenation. And it says, “stereospecific syn-addition” happens, it says, for “most alkenes.” And it gives this example. But I’m actually doing this little bit to show you how to read textbooks.
So it says, most alkenes do this, and give stereospecific syn addition, and it gives an example. But notice, it doesn’t say what the yield is. It could be one of these things– remember last time, we said there was a 55:45, that was said to be specific, or to favor one over the other. So it doesn’t say what the yield is. And it doesn’t give any reference to the literature where you could look up and see what they yield was, or how they did it, or how much of the other isomer there might have been there.
Now this is very typical for textbooks, and I have no doubt that this happens with high specificity, in this case. But on the other hand, you shouldn’t be satisfied with that. So you would look in another book, like this book from 1972 by House, Modern Synthetic Chemistry. Now this is a graduate-level text. So it shows an example, here, using palladium on carbon as the catalyst. Palladium on carbon means you have palladium metal, but you don’t want to have a big chunk of metal, because there’s not much surface. So you want it to be very finely divided, but you want to be able to handle it. You don’t have very much, so you put it on charcoal. It’s absorbed on the surface of the charcoal. So that’s what palladium on carbon is.
So it’s done in acetic acid, with hydrogen gas at one atmosphere, and you have that double bond. And what do you would notice about the stereochemistry of the product? Is it a syn addition? Which is the product of syn addition? Sebastian? The one on the left or the right?
Student: The one on the left.
Professor McBride: The one on the left. The two came in on the same side. On the right, one came in on the top, one on the bottom. And 90% of the product is the wrong one, the one the book tells you, or almost any book will tell you, is a syn addition. And it often is. Now, look at this example from 1948. What’s going on here? This is Barton in 1948. Anybody else remember what Barton was talking about in 1948 or ‘49, from last semester? That was the first conformational analysis. And it was molecules very like this. So this is what he was working on, these steroids.
But notice these two compounds are allylic isomers. They only differ in the position of that double bond. There are three carbons in a row, and a double bond between two of them. That’s an allylic system.
So these are isomers where you move a hydrogen from the top to the bottom, and now you have a double bond here and a single bond here. So those are allylic isomers. And now if you look at the details of this primary literature, it says that there’s a rearrangement of one of these compounds to the other, right? And it says that it happens in 80% yield. 80% of the product has gone from here to here. Now, this is what you would use for catalytic hydrogenation. But instead of hydrogenating, it first went by allylic rearrangement.
Now, how did that happen? So let’s look back again at the mechanism of the catalytic hydrogenation. First, you have the oxidative addition of the double bond, and then you have the reductive elimination, on the right, to put the first hydrogen on. And then you put the other hydrogen on the left, and continue. Of course that reaction would be reversible. You could have taken this back off again, in the wrong way, and instead of going there, pardon me, you could go there.
But the other thing you could have done is, if there’s an allylic hydrogen in the compound you’re talking about, so not only the alkene, but the next door carbon has a hydrogen on it, you’re here, you’re here, you’re here, you’re here, you get to there. OK? And now, one possibility is to reverse that way. That’s the reductive elimination, or pardon me, oxidative addition that takes you back here, that undoes those blue arrows, and would take you back. But what other option is there? Can anybody see another option? Liang?
Student: The other side.
Professor McBride: Aha, that compound is symmetric, if you have an allylic hydrogen. So it could unzip that way, too. And now, instead of going back to this, you go back to this. And if that does the reductive elimination, you come back here. And now, you’ll notice that has isomerized the alkene. The double bond moved next door.
So it’s all these oxidative addition reductive eliminations, but if you go through that symmetric intermediate, then you can take the wrong one off as you go back toward starting material. So the starting material is actually isomerized product. So that’s presumably how it happens. So catalytic hydrogenation can lead to allylic rearrangement.
And now let’s go back and look at that paper. And we see that if we had that rearrangement, this could go to this, in fact, it could go to four different isomers. It could be the double bond here, or here, or here, or here. So we could come back to this. And now, once you’re here, if you add hydrogen, even if you add them from the same side, those two hydrogens, they could be the opposite side from where the first hydrogen was. So now you could expect to get this, and you’d expect more of this than this, because that form is more stable. It’s conformationally a better structure. OK, so 90% there.
Now, the nice thing is you don’t have to even look at the secondary literature, you go back to the primary literature, because it gives a reference here. So we can look at that reference, and see here where they give the table about studying these things. And down here, in row 7, you see they’re talking about compound VII, ∆9,10-octalin So the ring is numbered in a certain way, and that double bond is between 9 and 10.
Now, if they started with that material, you see they got 90% of the trans product. That’s this one, that’s what shown here: 90%, 10%. But if they did only partial hydrogenation, they still got 90% of this, but they had 100%– after doing 80% of the hydrogenation. It says down here they did it 80%. After doing that, they don’t have any of this stuff. It’s all this stuff. So if these things can go back and forth, it must be that the equilibrium lies in this direction. And occasionally, when you get over here, you go on to the product. And that’s the explanation of it.
But it’s interesting to look even a little closer, to look at what happened if they started with VIII. So suppose they start with VIII. If you start at VII, and you can go through VIII to get down here, you get 90% of the product. That’s their explanation. Suppose you start with pure VIII. What should you get? In terms of the product, if you didn’t have VII, if you start with VIII? Debby, you got an idea?
Student: Wouldn’t you get the one below?
Professor McBride: You should get at least this much, maybe more. Right? If the way you get here– if some of it goes this way, and some of it goes this way, and you start here and get 10 to 90, if you start here, you should get even more than 90 to 10. Because you don’t have the chance of going here initially. Got it? OK, what does it say? You get 80%. So there’s something rotten here, but no one’s ever talked about it. It could be just that the experiments weren’t done really carefully, or maybe the catalyst was a little bit different in one case than the other. I don’t know what the answer is. But it’s when you go back to dig in to where the real results are that you find things that are interesting.
So evidently, this isn’t 100% understood. But I have no doubt that this allylic rearrangement is a big source of getting what appears to be the wrong addition–that it really is syn addition, but it happens after you’ve rearranged things. So the product isn’t what you would have expected, naively.
Chapter 4. Olefin Metathesis, Polymerization, and Tacticity [00:25:50]
OK, now, a process called metathesis of an alkene. This was reported when I first came here to Yale, the first year I was here, and it was amazing. People were amazed by this reaction. And the mechanism was quite speculative. Nobody had any idea what happened at the beginning. But in the subsequent whatever number of years, 45 years, it’s been figured out. And what is involved is a double bond from the metal to carbon. Of course, there are there two other things on the carbon. But this is called an alkylidene complex. And in fact, this process was the basis for the Nobel Prize in 2005, which went to Grubbs.
So this is a metal alkylidene complex. We can come in with an alkene, and do the oxidative addition, to form that, and a reductive elimination. So it gives a metallacyclobutane, a cyclobutane that has a metal in it. Just by the same processes we talked about on the previous slide. Now, those processes are reversible. And what do you notice about reversing it? Liang? What about the intermediate here?
Student: It’s symmetrical.
Professor McBride: It’s symmetrical. So you can reverse the other direction, to go to there, and then that can come apart. And you’ve put the central carbon, the one that’s here, changed it from being attached to this carbon to being attached to this carbon. So you can take, effectively, two alkenes, and mix up their carbons, which is double-bonded to which.
Now, you’ve heard of Professor Ziegler, who last semester taught the other organic course. I don’t know if you’ve seen him around in the lab, maybe. He’s unusually tall, he’s 6’ 6”, or something like that. And I point out, here, that he’s not Professor Karl Ziegler. We’ll talk about Professor Karl Ziegler a few slides later. But he was visiting Japan, in 1986, and this delegation of junior high girls thought he was so tall, they should get their picture taken with him, which is here. But he wasn’t the only tall guy visiting Japan at that time. Here he is, touring together with Professor Grubbs, the guy that got the Nobel Prize in 2005, who’s equally tall. In fact, Professor Grubbs’s daughter played basketball at Yale. She’s quite tall. I think she holds the rebounding record for women’s basketball at Yale. So this’ll help you remember Professor Grubbs, I hope. In fact, they took another picture together with their host, Professor Murahashi. So again, I underline that they’re unusually tall people.
OK, but Grubbs invented a process called ROMP: ring-opening metathesis polymerization. Now here’s a double bond where both of the carbons in the double bond are part of the same molecule, I mean, the same framework. So if you cleave that double bond, and exchange partners with something else, you still have the two linked together through this other ring. So if you bring in this alkylidene compound, and do the trick of metathesis, you get that. And now the product is still an alkylidene ruthenium here, so you can do that with another one of these molecules. In fact, you can do with the n of the molecules, and they’ll all link together in a chain like this, to give a polymer. So this ring opening metathesis was a very important part of this Nobel Prize work to generate a polymer.
OK, now catalytic hydrogenation, we saw, looked like this. But Ziegler-Natta polymerization, this is the other Ziegler, Karl Ziegler, a German, looks very similar. So it has titanium, with an R group on it. An alkene comes in, oxidative addition, to give this. Reductive elimination on the other side gives that. But now, that we could just call a different R group. It’s not the original R group, it’s the R group with two carbons in between, that we just brought in.
Now what can you do? You can bring it back to start over again, right? And that’ll be the R, and then you bring it in, and that’s the R. And then you do it again, and again, and again, and you can make a very long chain of ethylene molecules, for example. Now, Ziegler found this– I contacted Grubbs, yesterday, in order to get permission to show his picture to you, and Murahashi as well, and Ziegler, incidentally. But he told me an interesting story of how Ziegler discovered this. It was during the Second World War, and in Germany they were short of lubricating oil. So he was trying to figure out how to use triethyl aluminum to add to alkenes, and then polymerize on the basis of that to get things that could be lubricating oils. And after the war, he continued that work, but sometimes they found different products, rather than these oils they were trying to make. And they traced it to the presence of metals in the reactor.
So then they tried all different salts, and when they used titanium tetrachloride, they found that they had made a heterogeneous catalyst for polymerization, the thing that I’ve shown here. Now I underline that this is heterogeneous. You mix these things together, it’s a witch’s brew, and nobody knows, really, exactly what’s in there. And the reason you don’t know is because it’s a solid. It’s not a pure crystalline compound, it’s not something in solution that you could do spectroscopy on to figure out exactly what it is.
So it’s very hard to study mechanism with these heterogeneous catalysts. But that doesn’t mean they’re not useful. For example, high density polyethylene, the number two of your recycling thing, with ns of from 800 to a quarter million chain lengths, was made in 2004 to the tune of 25 times 106 tons, 25 million tons. Or polypropylene, with up to 105 units in a chain, again, 45 million tons in 2007. So this Ziegler-Natta catalyst was a really revolutionary thing for making polymers. But it was hard to do the mechanisms.
Now, one thing that we’ll see later, that free radicals also can polymerize things, can polymerize alkenes. But a difference here is that these are isotactic, that you get from Ziegler-Natta polymerization. So I should tell you what tacticity is. So if you have a polymer chain, and the alkenes had, say, methyl groups on them, propylene instead of ethylene, then you have a bunch of stereocenters. Now, notice that every alkene went in head to tail, that is, CH2 then CH, CH2, CHCH3, CH2, CHCH3. CH2, CHCH3. So it’s head to tail, but it’s stereoregular. All the methyls are coming out that you. And that’s called isotactic.
An alternative would be all head to tail but syndiotactic: out, back, out, back, out, back, every other one. And of course it could be random. Out and back. So there are these three different fundamental types of polymers: isotactic, syndiotactic, and atactic. And they have different properties, and we’ll talk about polymer properties in the next lecture, and why they should be different. So for some purposes you’d want one, for some another.
Now there are two questions that we’ll address here. One is, how do you know which is which? How do you know whether you have an isotactic or a syndiotactic or atactic. And for that, you’ve got to wait until we get to NMR, it’s coming very soon. It’s a really neat way to figure out which is which. But a more fundamental question is how do you control what you make? So there are these catalysts, instead of using titanium, use these zirconium catalysts, which are homogeneous. That means they’re dissolved; they’re not some solid that’s in there. So it’s possible to do spectroscopy on these things. And Kaminsky, the name associated with this catalysis, has to do with his accidental discovery that adding MAO makes these very reactive. MAO is methylaluminoxane, which is a sophisticated name for something that nobody knows what it is. You mix stuff together and you get that, and it’s not always the same. But if you add it to the zirconium dichloride complexes, you get a very active catalyst. So there have been, I think, since this was discovered in 1980, there have been something like 7,000 papers written on it, or something like that, and 2,000 patents.
Now one thing that’s pretty much agreed on is that what this methylaluminoxane takes the chlorides off, puts an R group like methyl on the zirconium, and leaves it as a cation. Now, what happens next? The alkenes approach, but they approach from alternate faces. Let’s see what that means. So here we take propylene, and bring it into the front to make one of these complexes. And then the R group shifts across, and you make a bond. Have you ever seen a reaction like that, where you have a metal with a group attached to it, alkene comes up, and you form a bond here and this thing comes over? The BH, hydroboration, is exactly that same kind of thing.
So we’ll call that new group R’. The original one was R. But now the next alkene comes in and associates with the zirconium, but what’s different from the first one I drew? It’s now in the back. The first alkene came in in the front. So it goes in the back, now the R’ moves back there, and then the next alkene comes in in front. And it comes over there. And so on, and so on. And you build a long chain.
But notice, every other one is reacting in front, in back, in front, in back. Now let’s compare the front and back of this zirconium complex. Notice it has an axis of symmetry there. If you rotate 180º about that axis of symmetry, you get the same thing again. So the front and the back are superimposable. They look just the same.
So whatever happens in front with respect to stereochemistry will happen the same way in back. So the two faces, front and back, are homotopic, they’re the same, so you’ll get the same chemistry, the same stereochemistry, at either one. Now how is it different in this case? That instead of having an axis of symmetry has a mirror plane of symmetry. What does that mean about the front and the back? Sebastian? What do you say? What’s the relationship between the front and the back? Are they superimposable?
Professor McBride: So what would you call them?
Professor McBride: Right, so those are enantiotopic. So whatever the one’s face does, the other one will do the opposite. And as you bounce back and forth, what product are you going to get now?
Professor McBride: It’ll be syndiotactic. And now suppose you take this one, which has a horizontal plane, which means the top and the bottom are both accessible. On the front, you can go either to the top or to the bottom, equally. So there’s no selection. So what do you think you’re going to get now? If it doesn’t make any difference which way you go. Now the faces are achiral, and you get an atactic product. So you can control which one you get by choosing the symmetry of the ligand that’s attached to the zirconium.
Chapter 5. Radical Polymerization [00:39:00]
Now as I said, as I promised, you can also do free-radical polymerization. I said we’d discuss that. This was the earliest kind of polymerization studied in laboratories. So alkene comes in, bingo. Another one, bingo. Bingo. Bingo. Bingo. So we’re growing a long chain. But the difference is that in this case, you can get rotation around this bond to get that form. And when you do, that radical is close to this hydrogen. And you can go from a primary radical to a secondary radical. That’s downhill. And now you can add an alkene to that.
So if you do radical polymerization, as opposed to doing these metal-catalyzed polymerizations, you can get branches, you get occasional butyl chains, C4 chains, stuck on your long chain. And that means that they don’t pack together as efficiently, so it’s not as crystalline. So since it doesn’t pack as well, it’s not as dense. So that gives low density polyethylene, the kind of stuff that you get in dry cleaner bags, the flimsy stuff, as opposed to what you would get in a supermarket to carry groceries in, which is tougher, and opaque.
Now let’s look at it again. We start polymerization going, fine, free-radical polymerization. But, we may not want it to keep going, because properties of the polymer, like its viscosity, its melting point, depend on how long the chain is. We may want to control how long the chain is. How could we control the length of the chain? One way to control the polymer chain length is to add a molecule, like carbon tetrachloride. There are many such reagents, not just carbon tetrachloride. OK, now, the radical can attack the chlorine, like this. And now you have a trichloromethyl radical, and your chain is done. But, you have the trichloromethyl radical. So it can add to another alkene. And then that one can grow a new chain.
So you see, what you’ve done by this is used what’s called a chain-transfer agent. It doesn’t destroy the kinetic chain. The radical still keeps working, but on a new molecule. So you shorten the polymer molecules without terminating the chain reaction. So if you add more of the chain-transfer agent, you get shorter chains.
So if you measure the rate of the transfer, the reaction with the carbon tetrachloride, versus the rate constant for polymerization–in the particular case of styrene and carbon tetrachloride, that ratio was 0.01. So, if you don’t have to worry about radicals finding radicals, they’re rather dilute, then the molecular length will be how likely it is to react with styrene compared to how likely it is to react with carbon tetrachloride and terminate the chain. But that depends on the concentration of styrene, and the concentration.
So if you know these two, you can then work out– since the transfer reaction with this process here is only 1% as fast. So if you had equal amounts of styrene and carbon tetrachloride, you’d tend to go about 100 molecules before you got a chain transfer. So you get, on average, about 100 molecular– 100 units in a chain. If you had more carbon tetrachloride, you’d get shorter chains. If you had less, you’d get longer chains. So you can control it, but not precisely.
There’s a property called dispersity, which is they’re not all exactly the same length, they’re average that, but some are longer, some are shorter. And how wide is this determines something about, obviously, the properties of the material you get. OK, we won’t go into that any further.
Chapter 6. Electrophilic Oligomerization and Polymerization and Rubber [00:43:16]
Now alkene and diene oligomerization and polymerization, using carbon electrophiles as the active reagents. Now oligo- means a few, so oligomerization is putting just a few such units together. And so we’re going to use either s* of R with a leaving group, or R cations as the reagent to do the polymerization.
So if you have R+ as an electrophile, you do a really interesting reaction. Notice you can take isobutane and isobutylene, react them with H2SO4, and add the elements of the isobutane, and I’ve colored them because you can do isotopic labeling to see what’s what. Notice that what added to the alkene was this C-H s bond. How do you like that, Noelle? Can you see any problem with doing a reaction like that? What’s the HOMO and LUMO of the purple molecule here?
Student: There doesn’t seem to be any.
Professor McBride: There’s no functional group there. How can you do it? Well, tell me what would happen first, if you have a sulfuric acid and react with this mixture. What you think would happen first? Where is there a functional group?
Student: The π.
Professor McBride: The π bond. So proton would attack the p electrons. And which end would it attack? Would you put the proton here, or would you put the proton on the more substituted carbon?
Student: It would be Markovnikov.
Professor McBride: It would be Markovnikov. You’d want to get a t-butyl cation. OK, so we’ll do that. And incidentally, this product was not irrelevant, because that’s what they mean when they say octane. That compound is defined as 100 octane, for the engine performance. So this was being worked on, during the Second World War, to try to make high-quality fuel for airplanes.
OK, so we’ll follow what Noelle says, and make the t-butyl cation. And now the t-butyl cation can react with the alkene, to give this. And now you’d say, why don’t we just keep going? Well, it depends on what the situation is, and you’ll see it can do that. But another thing it could do is react with this. Watch this.
This is a tertiary cation. But it’s rather hindered. This one, if you transfer this with its electrons, curved arrow there, that’s a hydride shift. We’ve seen hydride shifts before. The difference is this is happening between two molecules, rather than for one carbon to the next in the same molecule. But it was shown by Bartlett in 1944 that this process happens, using isotopic labels. So it gives isooctane, and what’s the other product? What’s the purple product? The H went– the hydride, not proton, hydride, the H with its electrons went up here. What was left behind? The t-butyl cation. But that’s what you need here. So this is a chain, it can go round and round. So you add isobutane to isobutylene.
Now, if you don’t have the isobutane there, then it turns out that it could do this polymerization, add another one and another one, et cetera, et cetera. And that gives polyisobutylene, which is called butyl rubber. And it’s not a very strong material, but it has the virtue of being airtight. So that’s what’s made to seal tires. When they made inner tubes in those days, they were made out of butyl rubber, to hold the air in.
OK, so now these electrophiles in terpene and steroid biogenesis. Now I’m going to get on to this next time, but since I brought stuff here, I’m going to entertain you with show-and-tell. It comes after this set of slides. But since I’ve got it here, I’m going to do it. So we’ll turn on the light here.
And what I have here is water and vinegar, dilute acetic acid, and this stuff that I bought down at the art supply store called mold builder. But if I read here, it says, it’s ammonia, 100% natural latex, that means it came right out of the tree, and water. So all that’s in there is tree sap, and water, and ammonia. So I’m going to pour some into the water here. I’m going to take my jacket off and tuck my tie in actually. I that think might be prudent. OK. So we’re going to open it up here, and we’ll pour some in there. OK, now the ammonia keeps the molecules– we’ll talk about this next time – apart from one another of this latex. But if we add some vinegar, it’ll get rid of the ammonia. Obviously I didn’t measure it very carefully. Oh, there, we’re doing it. So now the molecules are coming together. So if I reach in here and grab it, get rid of that water, dry it a few times here. So we’ve made a rubber ball. Good. OK, if anybody wants to make a rubber ball, you can come and do it. Thank you.
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