CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 13 - Addition to Form Three-Membered Rings: Carbenoids and Epoxidation
Chapter 1. The Pinacol Rearrangement Mechanism [00:00:00]
Professor McBride: OK, so we’ve been talking about electrophilic addition and the nucleophilic aspect of it. Today we’ll extend that to talk about cycloaddition, and also there’ll be some practical applications with respect to epoxides and acetals and how they work. So first, about that problem I gave you last time, about drawing nice curved arrows to show the pinacol rearrangement. So, how does it start out? What’s the first thing that happens? Wells, do you have an idea for acid catalysis here? What’s the very first thing that happens? Too early in the morning? Chris, what do you say?
Student: The proton protonates.
Professor McBride: Protonates what?
Student: Probably OH.
Professor McBride: OK. So we’ll protonate the OH, and then water can leave. Why is it good for water to leave, Chris?
Student: The pKa of hydronium is low, negative.
Professor McBride: Yeah, O3+ [correction: H3O+] is good at losing things, how about the other aspect of it? How about the thing that’s leaving, other than the water?
Student: It’s charged?
Professor McBride: Yeah, how about where the charge goes?
Professor McBride: What’s it going to be when the water leaves? The water will be neutral. Pardon me.
Student: It will go from a positive charge to a neutral charge in this case.
Professor McBride: The water will, but we care about the organic thing more than about the water. What will be left?
Student: Negative charge.
Professor McBride: No, no. The thing is a cation, if you lose neutral water, the charge will stay. Where will it be?
Student: On the carbon.
Professor McBride: Is that a good place for it to be?
Professor McBride: Why? Why is that good?
Student: It’s tertiary.
Professor McBride: It’s tertiary. All right, so we get a nice tertiary cation. What did anybody think about the next step here? Yeah, Kate.
Student: I thought that maybe there was a methide shift.
Professor McBride: A methide shift, good. So the methide could shift over, a rearrangement, and now you have this cation. Now, Kate, you’re going to tell me about this cation. We started already with a very good cation, a tertiary cation, right, that’s as good as they get with– unless you have double bonds next door. OK. But, what’s the driving force? Why should it go this direction? What’s good about being on the right?
Student: It has to do with the fact the CO bond is a low LUMO.
Professor McBride: No, I don’t think you could say the CO, it’s true– the CO bond is a lower LUMO. But to the extent that the oxygen, the reason it’s low is because of the oxygen, right. And, that oxygen is withdrawing charge from the carbon, withdrawing electrons from the carbon, tending to make the carbon positive anyhow. So, if a carbon is already positive, it’s not such a great place to put more positive charge. So, to the extent that that σ bond is important, you’d think it’d go the other direction. The oxygen would be electron withdrawing and destabilize the cation. Can anybody see why it could be good to have the oxygen next door? It’s certainly not that σ bond.
Student: Isn’t the unshared pair the reason?
Professor McBride: What about the unshared pair?
Student: They stabilize the positive charge.
Professor McBride: You know actually, I would say the other way around. The electrons are something real, right, and they get stabilized by mixing with the low vacant orbital that’s associated with the positive charge. It’s the electrons that get lower in energy. OK, but you’re absolutely right, so you can draw a resonance structure like that, and that’s why it’s good to go to the right. OK, now, how do you get from there to the product? Jack?
Student: Hydrogen can go off.
Professor McBride: Yeah, you just lose a proton, right, and you’ve got the product. So, that’s the mechanism of the pinacol rearrangement. It’s all steps that you know well, but you have to think and put them together, so that’s just an exercise in that.
Chapter 2. Carbenoids and Simmons-Smith Cyclopropanation [00:04:36]
OK, now back to these simultaneous reactions that are involved in addition to alkenes. Simultaneous in the sense that there’s both electrophile, which is what they’re typically called, electrophilic additions, but, also a nucleophile that’s participating at the same time. We saw this last time with dichlorocarbene which you remember came on edge on, and then rotated in order to make the two new bonds to give us cyclopropane. We saw it in hydroboration, where the vacant orbital on the B attacks, but at the same time the high HOMO B-H, mixes with the π* orbital.
OK, so, now we’re going to go on to a different kind of a way of making a cyclopropane with what’s called the Simmons-Smith reagent, and then into epoxidation, adding just an oxygen to a double bond to give us a 3-membered oxirane ring, and then ozonolysis which is a more dramatic kind of cycloaddition to give a ring, and then we probably won’t get today to catalytic hydrogenation, and to polymerization and metathesis, where metals are involved. So, we’ll do those others today.
First, the so-called Simmons-Smith reagent or a carbenoid. We talked about CCl2, a carbene, last time, that’s a free species, it’s pretty reactive, but it floats free. It has an existence. These are called carbenoids, because they don’t really give a free carbene, but they give a product as if it were a carbene, that is the CH2 in a 3-membered ring. The reagents as you can see, are methylene iodide, CH2I2, and the thing called zinc-copper couple, and I don’t know where the name couple came from except that there two metals there. But it turned out that having a little copper with the zinc made it work better. So, I don’t know, anyhow, zinc is the operative group here, but copper clearly has something to do with it. This was developed at DuPont Central Research in Wilmington, Delaware, where Howard Simmons was the Director of Research.
First, a word about metals and alkyl halides. So, you have a metal and an alkyl halide. Now, metals tend to be over on the left of the periodic table, what does that mean about their electrons? Are the highest valence electrons held tightly or are they able to give up electrons? Compared to most things. Do they have a high HOMO? Things that are on the left of the periodic table don’t have such a big nuclear charge, given the row they are in, so they’re relatively good at giving up electrons. How about RX? What makes it reactive, a halide on R? What’s its characteristic reactivity? Ayesha.
Student: Low LUMO, σ*.
Professor McBride: Low LUMO, σ*. So, we have something that can give electrons and something that can take it, so it won’t surprise you that you can get an electronic transfer from the metal, and it’s just a metallic metal, it’s not an ion in solution, it’s just a metal. But, at the surface of the metal an electron could be transferred to RX. Now, where does that electron go in the RX? Ayesha, you got us on to this, tell us again, what orbital does it go into?
Professor McBride: And what does star mean?
Professor McBride: That it’s antibonding. So, what’s, going to happen when you put an electron and get RX–? It’s also a radical, it’s got an odd number of electrons. What’s the role of that last electron? What does the star mean?
Professor McBride: Antibonding. What happens? We put electrons in the antibonding orbital, what does it do?
Professor McBride: Right, breaks the bond. So, a pair will go there, and now you got X– and the R•. But, the R• is generated right next to the metal, so you can get a making of a bond between that radical and an atom in the metal. OK. So, you get an alkyl-metal bond and X–. This is shown as if the metal is divalent, like zinc, so it makes a bond to R, and has a positive charge, and then it’s associated with the X–. If it had been a monovalent metal like lithium, then it wouldn’t have had the charge, it would be RLi, right. Two lithium atoms would have been involved. Another one would be Li+ that goes with X–, but at any rate, this is how you make alkyl-metal bonds in the first place.
The next three slides suggest a plausible mechanism for this formation of the cyclopropane, but as we’ll see, it’s probably not correct. But, let’s just take this as an opportunity to practice figuring out how mechanisms might work. So, we could react zinc with methyl chloride, and we get this reaction that was shown on the bottom of the previous slide, a metal-zinc bond and chloride, and I’ve shown it covalent between the chloride and the zinc. We showed +/– last time.
This we’re going to use as a model for the actual reaction, which had CH2I2 as the starting material with zinc. So, it gave I-Zn-CH2I, so we’ve simplified it here so we can draw orbitals that are a little simpler. That’s the LUMO of that model compound and mostly it’s a 4s orbital on the zinc, which is antibonding to the things on either side, which take away part of it, because you have antibonding nodes there, of course. So, that’s the LUMO. The HOMO is mostly a p orbital on the zinc, actually the LUMO+1.
Now, if you bend the zinc, you’ll mix those two, they’ll hybridize. So, if you bend it, that’s the 4p on zinc, if you bend it, as it’s approaching the transition state, where the zinc is going to be approaching the C-C double bond, then you hybridize that central one and you’ve got an orbital that looks like this. And, that’s an sp hybrid, sp something or other on the zinc. Notice that that’s the LUMO, so what’s it well set up to do with respect to the alkene? What could it overlap with? Amy?
Student: The σ bond HOMO.
Professor McBride: σ is down in the middle of the carbons.
Student: Wouldn’t the π be better?
Professor McBride: The π is sticking up where you need it to overlap. So that can overlap with the HOMO, and you’re going to get a pair of electrons that’s bonding between these two things, between the zinc and the two carbons. But, this is again one of those cases, so we have an electrophile on top, a LUMO, attacking an alkene, but we can also look at the HOMO on top, that’s the electrons shifting into the bonding region. We can also look at the HOMO on top, and what does that look well set up to do? Matt, what do you say, with respect to the alkene? What orbital of the alkene could that react with?
Student: The double bond, I guess.
Professor McBride: What about the double bond? Which specific orbital of the double bond?
Student: The π*.
Professor McBride: π*, right, which is the vacant orbital, which is what you want. This a HOMO, the LUMO downstairs is what we want there, and you see those are well set up to react. We can do electron donation in the other direction, top to bottom, right, so were taking an electron pair from top to bottom, also previous slide, bottom to top, so we’re forming two new bonds. We can draw curved arrows, like this and like that, and what we’ve done now is form this product where zinc is added at one side and methyl at the other.
Now, remember this we just drew so the orbitals would look simpler. If we had been using that thing that had two iodines instead of one chlorine, then the product would have been this, within the same reaction, but that would have been the product. And now, what makes this reactive? Can you see where there might be a low LUMO in this? Any ideas? Cassie?
Student: The σ* orbital of the C-I.
Professor McBride: σ* in the carbon-halogen bond. That would be a low LUMO. How about the HOMO? Where would be an unusually high HOMO? Carbon with a halogen is unusually low, what’s the analog that’s unusually high? Carbon with what?
Student: The zinc atom.
Professor McBride: With a metal is unusually high. So, the carbon-zinc bond would be unusually high. σZnC, and the LUMO is σ*CCl. I left one of the C’s out. Can you see what can happen? So, you have a high HOMO attacking a C-X bond, where have we seen that before? What reaction has a high HOMO attack σ*CX, where X is a halogen?
Professor McBride: Pardon me?
Professor McBride: I can’t hear you.
Professor McBride: Hydroboration is such a case and I’d have to think about why this is an example. I had it even simpler example in mind.
Professor McBride: SN2 right, a backside attack by the high HOMO. We can have the HOMO attack the carbon, the iodide is the leaving group. That then would form the three-membered ring. So that’s a very plausible mechanism, it has two steps. First, you have the addition of zinc halide and the carbon group, and then in a second step, you do the SN2-type reaction to form the second bond. The first reaction formed the bond on the right between the two CH2’s and the second reaction formed this one. But that probably is not the way it goes. It’s probably not two transition states. We just guessed that.
There was a quantum mechanical calculation of what the transition state for this should look like and they found something different. Although that mechanism is plausible, it probably occurs in a single step according to this calculation, with this bent transition state. This is the transition state where everything is happening at once, and let’s look at what that is. First, we could look at how the thing moves through the transition state. You see, the carbons going closer, the iodine is going away from the carbon, and the zinc going away. Let’s see what happens. Notice incidentally that the CH2 group is like this, and is rocking in. Where have you seen that before? Where the CH2 group approaches sideways and then rocks in? Seen that before? Ellen?
Student: The carbene with chlorines.
Professor McBride: A true carbene, CCl2 did exactly that in order to get the orbitals to match up. Let’s look at the orbitals in the transition state for this on the way to making the cyclopropane. This is at the transition-state geometry. Here’s the LUMO. What is that LUMO of the zinc reagent? What would you call that orbital?
Professor McBride: σ* between zinc and the iodine. That’s the LUMO, and you can see it’s well set up to mix with the π HOMO. At the same time, this is the HOMO – 2. It’s very near the HOMO, so a very high-energy orbital. What’s that well set up to overlap with? Matt, you’re our expert on this? What about the alkene is well set up there?
Student: The π*.
Professor McBride: π*, right. Blue on the right, red on the left. We can do both of them at once and make the two new bonds and break the others.
Chapter 3. Epoxidation by Peroxycarboxylic Acids [00:17:56]
Here’s the next reaction we want to talk about, which is an analogous reaction in the sense of forming two bonds at once with a reaction between an alkene and a percarboxylic acid. If we didn’t have the red oxygen there, if that hydrogen were directly attached to the oxygen below, that group would be a carboxylic acid. This is called the peracid, per- means two oxygens linked together, which is of course not a very stable bond. Do you remember why it’s not so stable to have two oxygens in a row? Natalie?
Student: Repulsion between the electrons.
Professor McBride: You have two pairs of electrons that are repelling one another across the bond. So, meta-chloroperbenzoic acid is a commonly used reagent for this, and the product, you see, gives the more stable carboxylic acid, and oxygen attached to the double bond. Our question is what the mechanism is. It’s also interesting to wonder why you use meta-chloroperoxybenzoic acid. Why meta? Why not para? It just seems a weird choice. It’s interesting why it’s chosen. It’s because it crystallizes very easily, and the crystals of it are very stable, so it could be stored for a very long time without any decomposition, even though essentially it’s a rather unstable compound. It’s that practical application that means that this particular one is the one that’s often chosen, although others could be used as well, other R groups on the peroxy acid functionality.
OK, so you see in this particular case, which I took as an example from the Jones book, it was done at 25 degrees in benzene for five hours and gave an 81% yield when R is n-hexyl, so it’s a pretty good reaction. So, to look at what the orbitals are involved, let’s just make that R group instead of meta-chlorobenzene, make it just the hydrogen, so this is peroxyformic acid, where the extra oxygen is obviously here. Had that hydrogen been here, it would have been formic acid. Let’s just distort that to the geometry it has at the transition state. So you see the oxygen-oxygen length increased a little bit, and the hydrogen bent up a bit. Now we’ll rotate it, so to get the idea of how we’re going to look at it and rotate it back a bit. Now we’re going to look at the orbitals it has, it has a bunch of occupied orbitals, of course. And, it has a bunch of unoccupied orbitals. But, we’re particularly interested in its LUMO. What’s the LUMO going to be? Elisa, what do you think? There’re several possibilities, so come up with anything. Anybody else got an idea, Karl?
Professor McBride: σ*OO. Remember, that’s an oxygen, big nuclear charge. Can you see any other thing that might be a low LUMO? Karl?
Professor McBride: σ*CO is also low. So in principle, either of those might be the reaction. So, that’s really the problem often in choosing mechanisms, that there’s several possibilities, and you have to consider them all, and if they’re two possibilities, and then at the next step two more, and two more, it gets to be a pretty big number. But at any rate, those are the two that you might consider at the beginning and once you’ve tried these several paths, then you can see which one’s going to give you the product that you know by experiment is actually the product, and it turns out to be this one, σ*OO. How about the LUMO of this thing? Pardon me, that was the LUMO, here’s the HOMO, actually it’s not the HOMO, it’s the HOMO – 3, but do you see what it’s made of? What’s this big thing here?
Student: Unshared pairs.
Professor McBride: That’s the p orbital on oxygen, the unshared pair, and it’s mixed with this up here, which notice is three p orbitals in a row. So that’s a p orbital on this oxygen and the π electrons of the C-O double bond, mixing together in a bonding way here but antibonding here. That’s why had this been red and that been blue, then this would have been a lower energy orbital, so this is the antibonding combination of those two, and it happens to be HOMO – 3. So that’s the HOMO that we’re going to be interested in. So that’s mostly, as we said, the π on that oxygen, the p orbital, and also π-allyllic. Remember allyllic is three p orbitals in a row all pointing in the same direction so they overlap side to side. And there we’ve turned it a little further just so you can see that it’s those p orbitals.
Now we’re going to look at it as it interacts with the carbon-carbon double bond. The first interaction is going to be that LUMO of the oxygen-oxygen bond, σ*, that’s the electrophile, that’s the LUMO, and the high occupied orbital is going to be our old favorite, σ of the double bond. We’ll do this. What does that reaction remind you of a little bit, when something comes in from one side. and the group leaves with its electrons from the other side? SN2 of course, so it’s the same kind of reaction. So that’s an SN2 attacking oxygen, not carbon of course. So, that’s going to form a new bond, and the leaving group, notice, is a fairly good anion, carboxylate, better than RO–. Now, what makes this reactive? What could happen next? Elisa?
Student: You have a positive charge on the carbon.
Professor McBride: We’ve got a positive charge on carbon, so that’s going to be a very low energy orbital there. Where could be a HOMO that would react with that? A high occupied orbital. Liang?
Student: The lone pair on oxygen.
Professor McBride: OK, it could be. That wasn’t what I expected you to say, that’s going to be what it is. What other possibility? Liang? [LAUGHTER] Liang, what other possibility is there when you look at this picture for a high HOMO?
Student: The negative charge on the–
Professor McBride: Right, we got a negative charge up here. This oxygen looks to me like it has a higher HOMO than this one does. What advantages does yours have over the one that’s up there?
Student: It’s closer.
Professor McBride: Pardon me?
Student: It’s closer.
Professor McBride: It’s closer, right. The electrophile is going to be what Elisa said, the p orbital on the C+. But, the p orbital on the oxygen is going to be the nucleophile, and it has the advantage of being nearby. So that means we can do this and make a second bond, and of course the plus charge will move when we share the electrons. Now, what do we have for a LUMO here? If we’re going to do another reaction? It’s something to do with a positive charge. What particular orbital would you point to, if we had to choose a particular orbital? The charge is not an orbital. The charge makes an orbital low, makes the electrons in an orbital, low in energy nearby. But, what orbital will we look at? Jack, you got an idea?
Professor McBride: Pardon me.
Professor McBride: I couldn’t hear.
Professor McBride: Which π?
Student: The one with oxygen.
Professor McBride: A p orbital on oxygen? The p orbitals on oxygen are all filled with electrons. We’re looking for a vacant orbital, right? It’s true that the unshared pair on oxygen is lower in energy than it would normally be, but that doesn’t make it reactive. For a pair of electrons to be reactive they have to be high in energy, not low. We want some vacant orbital that will be unusually low in energy. Chris?
Student: d orbital?
Professor McBride: The d orbital? No, you know, if you get to sulfur or something, if you get down in the next row of the periodic table, then there are d orbitals that are vacant in the valence shell, but on oxygen d is way up in the next shell.
Professor McBride: σ*.
Student: of C-O?
Professor McBride: It could be σ*CO. Any other possibilities? It could be σ*CO. Any other possibilities?
Professor McBride: σ*OH. Right. That’s the kind of thing you’re thinking of, because when you have an O+ that’s protonated, three bonds, one of them to a proton, what do you think of happening? Losing a proton, right, and that’s that σ*OH. So that’s going to be our electrophile, that’s where the electrons are going to go, into σ*OH and break that bond. Where they going to come from? Where’s the high HOMO? Liang, let’s go back to you now.
Student: Negative charge.
Professor McBride: Now you’ve got the negative charge here. But there’s something interesting about this negative charge. The reason that was a good place to put a negative charge, that carboxylate was a good leaving group, was that there was a carbonyl group, a C-O double bond next door. We can denote that– so we’re going to have an SN2 at H. We’re going to attack, and notice where we’re going to attack from, backside. These attacks are always backside to break a σ*. Now, the trouble with this O– is that it’s not backside, it’s out in the front. Can you see how we could get it into the back? How could we get this charge back near here?
One way that Karl suggests is move the anion, it’s not bonded after all. But, there’s a cleverer way to do it. Not that you’re not clever, Karl. Watch this. Does that suggest anything? That’s why carboxylate is a good leaving group, because it’s resonance-stabilized. But, what does that suggest? Sebastian? You don’t need to use this anion, you can use this anion, because the charge is both places here. So we can draw that, put the charge over there, and a double bond here in front.
So now we’ve got it where it needs to be. Now we can do the backside attack, move the proton over. We have those products, the carboxylic acid without it’s oxygen, that extra oxygen that the peracid had, and oxygen added to the carbon-carbon double bond to give a three-membered ring, an oxirane as it’s called, or an epoxide. The interesting thing about this– Karl had the great idea, let’s just move it to where it needs to be. But there’s a reason I didn’t want to move things. The reason is, that in fact, all those steps are not steps. They all happen at the same time. So in fact when you start with that material, and bring it up to the carbon-carbon double bond, all those things happen at once. It made sense looking at them one at a time, but in fact they can all happen at the same time.
So you have minimal atomic displacement if you go directly from this to transfer the oxygen atom and get that. But, they don’t happen strictly in parallel, just because they’re all happening at the same time, doesn’t mean that they’re all half way done at the same time. Some are a little leading, some are a little trailing, but they’re all happening together with minimal atomic motion. How do I know that? Because of this paper that was published in JACS in 1991. Now that’s 20 years ago, and you can do much better calculations I’m sure now. But, this is when it was done and published and it’s believable.
This is what they calculated for the geometry at the transition state, and they give in every case what the length of that bond, and what the bond angles are. Notice that that O-O is strongly stretched at the transition state, so it’s really broken a lot, putting the electrons in the anti-bond. It’s started to break, it’s 1.87, where normally it was 1.5, roughly. Notice that the O-H bond, remember that H is going to transfer from O1 to O3, at the transition state it is hardly stretched at all. It started at about 1 Å and it’s 1.01 here at the transition state. It actually is going to do most of its motion after the transition state. I put kH/kD is about one, that is there’s not a kinetic isotope effect on this. Does that surprise you that there’s not a kinetic isotope effect? That deuterium and hydrogen are transferred at the same rate? What does it tell you, when there’s a hydrogen isotope effect, a hydrogen-deuterium isotope effect? What does that tell you about the mechanism, about the transition state?
Student: That it’s an intermediate.
Professor McBride: No, it doesn’t tell you it’s an intermediate.
Student: That the deprotonation is the limiting step.
Professor McBride: The rate-limiting step, the hydrogen is being transferred, not before, not after, but at the rate-limiting step. But, in this case, the bond isn’t much stretched at the transition state, so there’s not an isotope effect, because most of the transfer of the hydrogen is happening later. This shows the motion, successive motions. Here’s the transition state, then P1, P2 are subsequent stages on the reaction pathway. If you look at the motion of each atom, you see the hydrogen does most of its motion from this oxygen to this oxygen after the transition state.
So, just because all these things are coordinated, you don’t have distinct intermediates, you have only one transition state, but they don’t all have to be 50% of the way from starting material to product at exactly the same instant. You can see how the others are moving as well. So, notice that after the transition state, the hydrogens are moving down a little bit, the carbons are moving up, the oxygen is moving in here to make the new bond, and so on, and the hydrogen is moving away. This thing, the carboxylate at the top is rocking away, breaking the oxygen-oxygen bond here, and forming the OH bond here.
There’s only one transition state, this is said to be concerted in the sense things are happening at the same time, not several distinct intermediates. But it’s not synchronous, it’s not all happening exactly in parallel. The name of this transition state, they called it spiro transition state. Spiro means two perpendicular rings sharing a common atom. Here it’s O1, here’s this five-membered ring here, and a three-membered ring here, but they’re perpendicular to one another, they share an atom. That’s what spiro means, to have two rings that are spiro, they share an atom and are perpendicular to one another.
But, in fact, a very similar mechanism, in fact arguably the same mechanism, was proposed by Professor Bartlett, your grandfather, in 1950. This was before people thought so much about spiro and about the arrangement in space of orbitals as they do now. In fact, they didn’t think about orbitals at all really in 1950. So, in a paper in 1950, he drew this structure, which shows the peroxycarboxylic acid twisted around, so the hydrogen can make it from here to here, and the oxygen then be transferred to that. So this picture is taken from that publication in 1950. You’ll notice that the arrows were not at all carefully drawn, and it’s not clear what they mean. This hydrogen in fact started attached to this oxygen, and is now moved over to this oxygen in the product, so I have no idea really what that arrow meant.
But, that was old days, right, before people talked about orbitals and so on, 60 years ago. The problem is, how about now, how carefully do people draw such things now? Well, we could look at a modern textbook that has a drawing of this particular reaction, and draws it in that way. What I’d like you to think of for a problem is, compare the arrows in this textbook illustration, with the ones that we developed in the previous frame to show where HOMOs and LUMOs were interacting, and how electron pairs were shifting, and see if you can draw your own diagram that’s more accurate than this textbook one is.
Stereospecificity of this epoxidation, if the oxygen is really being transferred this way, all at once, the two new bonds have to be on the same face of the double bond. So, if you started with trans-2-butene and used this meta-chloroperbenzoic acid, and this is a specific example done in this solvent, dioxane, at 0 degrees for 10 hours. Notice that these two methyl groups are on opposite sides of the 3-membered ring, just as they were on opposite sides of the π bond here. This happens with 52% to 60% percent yield, but more significant, is that it’s greater than 99.5% trans. This is an actual 52% to 60% yield, that’s what they actually got of pure stuff and put it in the bottle. And you know, that if you have to distill stuff at the end and so on, you don’t always get 100%. If all they had said, was that a 60% yield, you’d worry, what’s that other 40%? Could it be the cis isomer?
In fact, that’s not what it was, because they tested and found that there was no cis there, it was more than 99.5, the limits of their detection, was trans, so it’s a concerted syn addition, both new bonds from the same face of the alkene. And, you’d worry about this also, that this is the more stable isomer, where the methyls aren’t running into one another. Maybe it wasn’t specific, but you just got the one that was most stable. So, of course what they did is do the cis one as well, and that’s also greater than 99.5%, cis now, not trans. It’s clear that the reaction is stereospecific.
Chapter 4. Other Routes to Epoxides [00:38:40]
They did that in order to prepare those epoxides for another purpose. It’d been prepared before, in 1936, by an alternative mechanism. It was two steps to do it, so it was a harder way to make it. They started with the reagent HOCl, hypochlorous acid. How do you think HOCl will react? Any ideas? What other reagent that we’ve talked about reacting does that remind you of? This is a chlorine, with an electronegative bond to it, bond to oxygen. What’s the LUMO of that, do you think? Cassie?
Student: The O-Cl σ*.
Professor McBride: Right, O-Cl σ*, the same as Cl-Cl σ*. What kind of thing happens with Cl-Cl σ* remember, if you react it with an alkene? It forms a halonium ion, like that. The only difference is that in that case, it was Cl– that’s leaving, in this case it’s OH– that’s leaving. But, that’s drawn in brackets because it’s just an intermediate, it’s not something that you isolate, it’s very reactive. Remember you made hydroxide, so what’s going to be the next stage of the reaction. How will this react with hydroxide? How does the one that had a chlorine on it react with chlorine?
It’s an SN2 kind of reaction. In fact, you have water here that can also do it. It gives a 55% yield after distillation of this stuff, where you notice what happened was the oxygen attacked backside, opened this ring. So this methyl group is back, and this methyl group is back, since we started with cis. We’ve got correlated configurations at these two carbons. Notice it would have been just as easy for the oxygen to attack this carbon, it attacked this one, it could have attacked this one. So it came this way, it could have come this way. What’s the relationship between the two products you would get if you did those two reactions? Ellen?
Professor McBride: They’re enantiomers. So you get not just this compound, but also it’s enantiomer. But, you don’t get ones where this methyl would be in front of the hydrogen in back, as shown there. There are four diastereomers, you only get two of them, you get the two enantiomers. So that’s the first reaction, they did it, they got a 55% yield after distillation. Then they reacted it with KOH, notice it’s 20 M KOH, pretty strong solution at 90 degrees, that’s vigorous conditions, in water for two hours. How do you think OH– will react with this stuff? I want several possibilities. What could OH– attack in this molecule? Chris?
Student: Deprotonate the hydrogen.
Professor McBride: Deprotonate the hydrogen. What would have been another possibility?
Student: Attack the chlorine.
Professor McBride: It could’ve attacked– done an SN2 here on the other one, but it’s a little bit hindered. Generally proton transfers are pretty fast, so you were right the first time. The KOH takes off that and generates that anion. Now again, I’ve drawn that in brackets because it’s just an intermediate. What does it do? What does this O– do? Chris, you just told me. What can it attack?
Student: The lone pair on the oxygen.
Professor McBride: It is the lone pair on the oxygen, that’s the high HOMO, what’s the low LUMO?
Student: The σ* carbon…
Professor McBride: Yeah, σ* carbon-chlorine. What would you call that kind of reaction, ever see a reaction like that before, where O– attacks carbon and chloride leaves? Noelle, did you ever see a reaction like that?
Professor McBride: Can’t hear.
Professor McBride: It’s SN2, right, and it’s helped out because this is held very close to where it needs to be to do the reaction. OK, so we do that. Now, we form the epoxide. Notice the stereochemistry of this is interesting. The two carbons are on the same side of the ring, the same way they were on the same side of the π bond here. But, the stereochemistry happened in sort of an interesting way. This is a 90% yield that they got in this in the second step. But, since they had to do two reactions from the starting material, the overall yield was only 45% from the original starting material. Do you remember what it was when they used meta-chloroperoxybenzoic acid in the previous example to make the same substance from the same starting material? You remember what the yield was?
Professor McBride: Yeah, it was 60%. So nearly half again as much, at least a third again as much, and in just one reaction, you didn’t have to do this distillation and so on, so it was much more convenient. That one, the oxygen just went straight on, and it was a syn addition. This one happens in an interesting way. First, it’s a syn addition of the Cl+, both from the same side, but then there’s this SN2 kind of reaction, and it’s an inversion at this carbon. And, then here, there’s an inversion at that carbon. So overall, it’s still a syn addition, but by two inversions. So, that’s interesting.
This is somewhat reminiscent of what we talked about last semester, which is the Sharpless asymmetric epoxidation. I’ll just run through this quickly, we’ve done it several times before. So, you lose the RO–, then a peroxy group comes up to the titanium. We get that intermediate and react it with allyl alcohol. The OH of the allyl alcohol replaces the RO, it replaces there, to get this. And, now this double bond is being held near this oxygen. The LUMO is the σ* up there, the same thing we’ve been talking about, and the HOMO is π. So, we’ve made a bond. In fact, this pair of electrons probably then makes a bond between oxygen and this titanium, rather than the RO going away.
But, at the same time, there’s also the p on the oxygen attacking π*CC. This is exactly the same thing we just saw a few slides ago with peroxybenzoic acid. We make two bonds, and put the oxygen on from the same face. This particular arrangement makes the R configuration at this tetrahedral carbon. If we wanted to get the other one, we’d have to rotate this bond to be in front, and this to be in back, so that we could attack the other face, rotate it back like that. You’ll notice that if you do that, those two groups in front are big, and they would run into one another. So you don’t do that, you do the other one, and it gives us a specific enantiomer of the epoxide.
Chapter 5. Practical Utility of Epoxides [00:46:44]
But the relevance at this stage– that we talked about last semester, the relevance at this stage is that it’s essentially the same mechanism that’s involved in the peroxybenzoic acid of the electrons forming the two new bonds. That was a chiral oxidizing agent. Now, this is a big time– making epoxides from alkenes is really a big time operation, as is shown here. This is done with silver catalysis at 15 atmospheres of pressure and 250 degrees Celsius. Notice that the source of oxygen is O2 in this case, that’s a really cheap source of oxygen. In fact, they do use O2. There are other people who do the same kind of thing who use air, but it turns out to be worthwhile to use O2 rather than air, because you get a higher yield.
So what you get is an oxygen that’s transferred like that, that’s called ethylene oxide. That process generates 20 million tons a year, worth 20 billion dollars, of making ethylene oxide. Just to give you an idea, here’s an aerial view of New Haven, and down here is where most of you live on the old campus. If you had a bucket that would hold that much ethylene oxide as a liquid, it’s a gas, but if you condensed it, it would be a liquid. That’s how big the bucket would be. [LAUGHTER] I think it’s 15 times as high as Harkness Tower and covering the entire old campus. So, that’s a lot of material. Now that reaction gives 84% yield, the rest oxidizes, the rest of the original ethylene oxidizes to CO2 and H2O. Suppose you could adjust the conditions to make the yield higher. Suppose instead of 84%, you could increase it by 5%. If you could raise the yield by 5%, that would be worth more than a billion dollars a year. So, here’s a way for you to make your fortune. Although, I will caution you that people have worked on this a lot to try to get every last percentage out of it.
In fact, why make 20 million tons of ethylene oxide? Only 0.05% of it is used as ethylene oxide. It’s a disinfectant that’s used in some applications, it’s a gas, and you can put it in to kill microbes or something. But, very, very little of it is used for. What do they use it for? 2/3 of it is used to make that compound, just add water in the reaction that we were just talking about, to make that compound.
Does anybody know what that compound is called? You could call it dihydroxyethane. But, it has a more common name. It’s called ethylene glycol. It’s antifreeze. So, a lot of it is used for antifreeze, but it’s also used it for solvent. Also it’s incorporated in polymers, the polyethylene terephthalate, the stuff that makes soft drink bottles, for example. It’s used in making that. Glycol is an interesting word. The gly- is from Greek root that means sweet, like glucose, it’s the same root. It’s because ethylene glycol tastes sweet, although we don’t taste it because it’s poisonous, but the original people did, so that’s why it’s called glycol.
That reaction, to change ethylene oxide into ethylene glycol, occurs either with base catalysis or acid catalysis, so since the focus of this lecture has been– woops, this lecture is over. Sorry, I got carried away. We’ll talk about the mechanisms next time. Thanks.
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