Freshman Organic Chemistry II

CHEM 125b - Lecture 14 - Epoxide Opening, Dipolar Cycloaddition, and Ozonolysis

Chapter 1. Regiospecificity in Epoxide Opening: Interpreting Experimental Data [00:00:00] 

Professor McBride: OK, more on addition to alkenes. In particular today we’re talking about cycloaddition, addition to make rings. So we’ll talk more about epoxides. Then ozonolysis and how acetals play a role in ozonolysis. And then we’ll get on, I hope, to transition metal catalysis, hydrogenation and polymerization.

So we’re talking– and remember reaction with alkenes, the reagents are called electrophilic, but they’re nucleophilic as well. And we’ve talked about these, and at the end last time we were talking about epoxidation, which we’ll continue with now. And then about ozonolysis and about things attached to metals– hydrogen attached to metals, and alkyl groups attached to metals.

So this is where we were last time. Remember there were a whole bunch of arrows we drew happening all at the same time, although not in perfect synchrony. Some were a little more established at the transition state than others.

Now after class I was asked wouldn’t it have been simpler to abbreviate arrows, as in textbooks, so you don’t draw quite as many arrows. I didn’t point this out, but there was a reason I drew so many arrows. You might imagine that you could abbreviate it this way, which shows the electrons here forming a bond, the electrons here going to here to make a double bond. This double bond coming down here to make a single bond. And this bond coming over here.

That accounts for all the electron pairs, but it doesn’t account for them properly, and the reason it doesn’t is some of them are p and some of them are s. So it can’t be the same electron pair that’s being both p in the plane of the ring and s in the plane of the ring, and p out of the plane of the ring. So I accounted for the electrons differently in drawing those arrows as to which kind was doing what.

So at the very end last time we were talking about ethylene oxide, how it was a big item of commerce, 20 million tons a year and $20 billion, because of the importance of ethylene glycol. Then we were going to see how you get ethylene glycol from either base or acid catalysis with water on ethylene oxide. So let’s start with base catalysis. Where’s it going to attack? Po-Yi, what do you say? Where will OH^– attack ethylene oxide? What kind of reaction’s going to happen?

Student: It will attach onto hydrogen, CH molecule.

Professor McBride: OK, so you want to take off a proton, is that right, to attack a hydrogen so as to remove a proton? Attack σ*_CH, is that the idea?

Student: Yeah.

Professor McBride: Now the problem with that is that the anion you get is not at all a good anion. It’s a carbon anion. The hybridization makes it better than most carbon anions, but not that much better. My guess is its pK_a is some place in the 40s, whereas the pK_a OH^–, or of water to give OH^– is 16, right– 15.5. 

So that there’s whatever that is, a 10²⁵ difference. So it’s going to be very hard to pull off a proton, even though hydroxide, you’re right, is a strong base– better as a base than what you would expect for its nucleophilicity. So how about another possibility? If it’s not going to be CH s*, what will attack?

Now of course, ethylene oxide is an ether. And ethers don’t usually undergo displacement, because it’s a bad leaving group. O^–is not a great leaving group, RO^–. But this is a special case. Can you see why? Anybody else got an idea? Arvind? 

Student: Is it because of the strain of the– 

Professor McBride: High strain, right, will make that bond weak. So, in fact, it is a backside attack, an S_N2 reaction on the CO s*, which opens the ring because of the ring strain to give that. And now that, of course, picks up a proton from water, the pK_a’s are essentially the same for an alcohol and water and generates ethylene glycol, the item of commerce we were interested in making.

Now suppose we’re going to do with H⁺ catalysis. Who’s got an idea for that one? Rahul, you’ve got an idea? How is H⁺ going to get into the act?

Student: It’s a low LUMO so it’s going to be attacked by a high HOMO. So the oxygen.

Professor McBride: What about oxygen?

Student: The lone pairs.

Professor McBride: Right, the lone pair on the oxygen. So it’ll come down, attack the lone pair on the oxygen to get this. And now what do you expect for the reactivity of this thing? Rahul, keep us going. 

Student: There’s still ring strain, right? 

Professor McBride: There’s still a ring strain, and you’ve got a much better leaving group now– O⁺, rather than going to become O^–. So now even water can do it. It doesn’t have to be hydroxide. So we’ve got a great leaving group, water comes in and does the trick, and again, we get it. So either acid or base catalysis with water will convert ethylene oxide into ethylene glycol.

But there’s another possibility for a nucleophile in the system, which is this O^–. So you could imagine it coming down there and doing the trick, and now you get another O^– that can attack again, another O^– that could attack again.

So you can get a whole bunch of these CH₂-CH₂-O-CH₂-CH₂-O’s strung together. So you can get a polyether. There can be as few as three to call it poly, although often you’d call that oligo, meaning a few. But up to 20,000 units or more can be strung together that way. And these things are big items in commerce as solvents among other things. They’re also used in toothpaste, in paintball and lots of different applications.

Here’s another application. So you can take the trimer– that’s three of the CH₂O’s put together. And it’s been treated with potassium hydroxide to make the salt. And you could treat another one of the alcohols with one of the agents to convert OH into Cl, like thionyl chloride and get a dichloride. Can you see what I’m after here? What kind of reaction do I want to do? Ellen?

Student: S_N2?

Professor McBride: Right, S_N2. It’s a Williamson ether synthesis, RO^– reacting with RCl to make an O-C bond. And what’s the product going to be?

Student: A crown ether.

Professor McBride: Right, that’s how 18-crown-6 is made. And in fact, this is what’s called an Org. Syn. prep. Org. Syn. is Organic Syntheses. Every year they put out one of these volumes– I’ll pass around so you can look at them– which have various interesting preparations that have been proposed by competent chemists and tested by them to make sure they work.

Here’s another one, you can send it back the other way.

So you can see each of these has the order of 20 really good, careful descriptions with lots of information about exactly how you do it, where you get the starting materials, and lots and lots of footnotes. So textbooks don’t have time to do this. And original publications in the paper often haven’t been carefully checked.

But if you want to see how a reaction is really carried out, this is a very good source, and it tells you exactly what faithful yields are, not ones that have been exaggerated. And one of the Org. Syn. preps is of 18-crown-6. It’s interesting, and you might think about that, that they used the potassium salt, the potassium hydroxide was the one they made the dianion with. You might think why that might be. Well actually, incidentally, if you click on that you will get this, which is the description of how to make 18-crown-6 on the web from Org. Syn.

So that’s how we open ethylene glycol.

Now, notice that if the epoxide is unsymmetrical– substituted on one side and not the other– then there are two different ways of attacking. For example, in the textbook it shows that you could attack the less substituted or the more substituted. And it says “in acid the nucleophile generally becomes attached to the more substituted carbon. In base the nucleophile generally becomes attached to the less substituted carbon.”

So this is obviously a generality. And one would be interested in seeing precise data to see just how selective it is. So here’s an example from the book where they used sodium amide [correction: sodium azide]. So N₃^– is the nucleophile. And indeed here, with the anion attacking, it attacks the less substituted, as you would expect in an S_N2, and gives a 41% yield of the product from attacking the less substituted carbon.

Now, do you find that then convincing in terms of in acid you attack one and in base you attack the other? Noah, what do you say?

Student: No. 41% yield really isn’t very much yield.

Professor McBride: Yeah. Now of course, you’ve been in lab, and I wouldn’t be surprised that from time to time you’ve been happy with a 41% yield. Right? So if you really do get 41%– remember we’ve talked about one synthesis where 5% was plenty. So from a point of preparative– being able to make the stuff– 41% sounds pretty good. But if you’re interested in does it really support that generalization, that with base catalysis that is, anion attacking, you attack the less substituted one, you’d want to know what the other 59% is. If the other 59% were attacking the other carbon that would be bad news for this generalization.

So one has to be careful in reading textbooks, because often references aren’t given about that 41%. So here’s a generalization that says that– here’s the other example with acid catalysis. So it says when you attack with acid, indeed, the chloride goes more on the more substituted carbon. That’s what the generalization was. But notice that the product ratio is 55% to 45%. It’s not 99:1. It then explains it by saying if the charge is on the more substituted carbon, that is when you protonate the O, the bridge is unsymmetrical– we’ve talked about that– with more charge on the more substituted carbon so that’s the one that gets attacked. “That’s where the nucleophile adds,” as it says.

But notice that this is a tough thing to generalize, not a very reliable example to generalize with, the 55:45. Because that ratio is 10^0.09, which means that the difference in activation energy for going one way and going the other at room temperature is 0.12 kilocalories per mole. And you’re sophisticated enough by now to know that that’s nothing. That’s only 1/25 as big as a hydrogen bond. So it’s true that it gives more of the more substituted one in this particular case, but you wouldn’t feel very confident in interpreting that.

So you might do a calculation to see what the minimum energy geometry is for such a protonated epoxide. This is not a really high quality calculation. And it’s in the gas phase, and, of course, where a solvent is can make a lot more difference than 0.12 kilocalories per mole. But anyhow, you could look and see where positive things would like to be or negative things would like to be on the van der Waals surface of this molecule.

So if you look up here at the top, remarkably enough, when you drew this structure you draw it O⁺– it’s a trivalent oxygen, the O of the three-membered ring has been protonated. So you draw O⁺. But notice that this is the energy a proton would have on the surface of that. It’s high near the oxygen, but it’s 120 kilocalories higher near the hydrogen. That’s the worst place to put a positive charge. So it’s not what you would draw if you just put a plus charge on the oxygen. So that’s the worst place for H⁺. It would be the best place for a negatively charged nucleophile. So the best thing would be to pull the proton back off again.

On the other hand, you could, if you turn the molecule upside-down from the top right and look at the bottom of it, you see the two carbons that could be attacked. And at that place, the energy for the proton would be +140; there it would be +141.5. So it’s true that it’s a little bit more positive near the central part, near the more substituted carbon, but not very much. And of course, it’s sterically hindered there. So it seems to be very touchy to try to interpret this. But still, there are many examples where it is more dramatic than that. 

Here, for example, are cases from the textbook where a carbon nucleophile does the attack. Now you notice that when the phenyl attached to copper attacks, so that the polarization is such that the phenyl is mostly minus, so it’s the nucleophile, although it’s really more complicated because the copper’s around at the same time. But notice it attacks backside and it gives a 98% yield. So that’s the kind of thing when you see it in the book that you can say ah-ha, that’s really something. 55:45 is not the basis for confident interpretation. 98% is more like it.

Or in this other example where it’s 85:8 attack at one position versus the other with base catalysis, attacking, as the generalization had said, at the less substituted carbon to give the 85% product. So there are plenty of examples that support this generalization, but it’s not always reliable. It’s important to be critical in looking at the evidence that people provide to support their generalizations. So that one shows stereospecificity. The other one shows regiospecificity, and much more impressive than that 55:45 example; 85:8 is much more practical for synthesis purposes and much more convincing for interpretation purposes. So that much for the epoxides.

Chapter 2. Ozonolysis and 1,3-Dipolar Cycloaddition [00:16:02]

Now onto ozonolysis, which takes us into other examples of cycloaddition. Of course, forming the epoxide is a three-membered ring, so it is a cycloaddition that forms a ring. But ozone is more interesting because it’s three atoms. So we can imagine this shift of electron pairs, perhaps happening all at the same time, to make a five-membered ring. And on the left is the arrangement of those atoms in the transition state, as calculated by quantum mechanics.

So we could look at the motion along the reaction coordinate, both face on here, the side view, as I said, and also the end view. Notice that O₃ is bent so that the central oxygen is not as near the carbons being attacked as the terminal oxygens are.

So now watch, and we’ll watch it move. So the oxygens move down, the carbons move up and form the bond. We’ll reverse it and go back and forth several times. So you can see it’s another one of these cases where it looks like the attacking group is rotating down toward the carbon as the carbon moves up. So notice everything is happening at the same time. It’s not that you form one bond and then the other bond.

So let’s look at the orbitals that are involved in making two new bonds at the same time, so there must be two electron pairs that are involved. So there we have the LUMO on O₃, and the HOMO on ethylene. Remember that that central oxygen up on top is not getting into the act, it’s back behind the screen.

So you see the two reds with the LUMO on top are in position to make good overlap. And then give this orbital at the transition state, which is the HOMO– those overlapping with one another– to be bonding, and that’s one electron pair. So that would account for one bond.

But by the same token you have the LUMO of the alkene on the bottom right interacting with the HOMO of O₃ up on top. And those overlap well and give the HOMO – 1. So another pair of electrons gets stabilized as these atoms come together. So two electron pairs get stabilized, two bonds are formed.

Now this is cycloaddition of allylic 1,3-dipole to alkenes. There are other examples besides ozone. Now a 1,3-dipole is a thing like ozone. You may remember that in the third lecture, way back at the beginning of the fall, we talked about drawing valence structures for ozone, and the fact that the lowest energy structure is not a three-membered ring, but an open form.

The interesting thing about it is when you draw Lewis structures you can’t draw a structure that doesn’t have charges or free radicals on the atoms. So those examples are shown there. So these things are called 1,3-dipoles. There are three atoms in a row and all the structures you can draw are dipolar. But that’s not the only example.

Here on an exam from several years ago, I asked the question saying, “Having learned that the allylic p system of O₃“– remember, allylic means three p orbitals overlapping in a row, that’s allylic and that’s, obviously, the case with ozone. So, “Having learned that allylic p system of ozone forms two bonds at once to an alkene, as shown” bottom left there, “one might think to try the same thing with the apparently analogous boron compound.”

I’m not actually confident that that molecule exists, but it might. “Explain in terms of the orbitals involved why this might not” work. It looks like the same thing. You have a 1,3-dipole on the right, and we draw curved arrows to make the two new bonds, we have a 1,3-dipole on the left and we draw curved arrows to make the new bonds. But why wouldn’t the one on the right work?

Let’s try to do a theoretical analysis of that.

So here’s ozone viewed from the top. Now if we rotate this central oxygen at the top back to see the p orbitals we see that there are three p orbitals in a row, an allylic system. The lowest-energy p orbital then will have them all overlapping favorably with one another. That will be Y₁, the lowest-energy p orbital. It has no antibonding node. And the middle atomic orbital will be largest. We can choose the size of the p orbitals of the three.

There’s an advantage to making the one– the sum of their squares has to be one. What do you call that condition, you remember?

Student: Nomalization.

Professor McBride: Normalization, right. That the electron has to be somewhere with the total probability of one. So the sum of the squares has to be one, but they don’t have to be the same size. Symmetry would suggest that the two ends should be the same size, but the one in the middle could be smaller or bigger. And I say that it’s a little bit bigger. Why should it be bigger?

Why do you get advantage by having the one in the middle big compared to the ones at the end, if the sum of their squares is fixed? What you want to get is bonding, overlap. So the bigger the orbitals the more overlap. What’s special about the one in the middle? Why do you want to make it bigger? Lauren?

Student: Because it’s bonded to two of them?

Professor McBride: It overlaps twice. Good.

So then the next one has a node in the middle– that doesn’t surprise you. And the highest one has two nodes. Now, I’m confused by this because starting with three atomic orbitals and making three molecular orbitals, you have to use up the three atomic orbitals. If you have more in one place there has to be less in some place else. Obviously, there’s none on the middle oxygen in the one that has no node.

That means that the others have to have more of that and less of what’s on the end. It doesn’t appear that way in this picture. It doesn’t appear that the central one is big. But, in fact, it has to be, and I think this is some artifact of the particular contour that’s been chosen, and I can’t really explain it.

But in general, the favorable one, Y₁, is as favorable as it can be, big in the middle. The second one with the node in the middle is obviously 50:50 on the two ends. And the top one will be big in the middle to be as antibonding as it can be. When they were good they were very, very good. And when they were bad they were horrid. That’s the way it works out.

So those are the three molecular orbitals we’re going to be dealing with.

Now if you take another allylic system like CH₂BHCH₂, the one we were talking about, we can look at its p orbitals and there’s Y₁ and there’s Y₂ and there’s Y₃, and this one does, indeed, show biggest in the middle, so that’s satisfying. Of course, boron is less electronegative than carbon, less nuclear charge. So you would expect the lowest orbital not to have so much of it, and the highest orbital to have a lot of the less favorable orbital.

Now if we look at a third example, carbonyloxide, which you notice is a carbonyl compound, H₂CO, formaldehyde, with an extra oxygen on it. And that turns out also to be a 1,3-dipole to be bent in the structure shown. And now it’s not symmetrical end to end. So we see the same orbitals, but they’re distorted.

This one looks mostly like the C double bond O, p, but a little bit of the oxygen on the end. That one has a node in it, and that one is mostly the antibonding p* of carbonyl with a little bit of the oxygen on the end.

Now why should it be that way? Well notice that the carbon is much closer to the central oxygen than the central oxygen is to the terminal oxygen. So the big interaction will be between carbon and the central oxygen. So we could first look at the interaction between carbon and the central oxygen, and then look at the interaction of that, whatever we get from that with the terminal oxygen, because we’ll look at the strongest interaction first and then at the weaker one. So that says what I just said, and it will make p and p* out of that.

Now we’re going to mix p_O, the terminal oxygen, with p and p*. And we see it’s closer to the p than to the p* so we get three new ones that look like that. There’s less mixing at the top, more mixing at the bottom, a better energy match. But Y₁ still looks mostly like the lower energy one, p_CO. The p* looks mostly like p*_CO. And this one looks mostly like the terminal oxygen. That’s just to show that the orbitals are reasonable. So that’s big on p and a little bit of the terminal oxygen. That’s that one. Then you have one that’s mostly the terminal oxygen. Then you have one that’s mostly p*. Good.

Now let’s look at how many p electrons we put in to these systems. So we have to count electrons. So here we have three oxygens, there are the s bonds, but also unpaired electrons that are s, in the plane of the three O’s. And there are also then, if we count up, electrons on the terminal oxygens– there’s one more electron on each of the terminal oxygens, and two on the central oxygen. So there are four electrons in the p system. So in these p orbitals we have to put in four electrons. So that’s two pairs, and they go into the lowest two orbitals, of course.

So the HOMO is the middle orbital, the one with the node. And the LUMO is the one with two nodes. Now suppose we’re going to take ozone and react it with an alkene. So we bring an alkene up, and remember it touches the two oxygens on the end, the ones in front, not the one that’s behind the screen. So that’ll make a nice bond, nice bonding interaction. So the symmetry matches and we get stabilization of one pair of electrons.

But if we take the HOMO of the alkene, it mixes with the LUMO of the ozone. So another pair of electrons gets stabilized, and we make two bonds at the same time and make the five-membered ring. So that’s fine. And as I said here, don’t worry about the oxygen in the center, it doesn’t overlap because it’s behind the screen.

Now let’s go over to look at the carbonyl oxide and count its electrons. So that’s the s electrons, and in the p there’s one from carbon, one from oxygen, and two from the terminal oxygen. So again, there are four electrons, which means we’re going to occupy those two. And again, the HOMO is going to be the one with one node, and the LUMO the one with two nodes. And when we bring the alkene up it’s going to do its trick. So that’s fine, too. That’ll form a five-membered ring.

Now let’s look at the other one, the one that has boron in the center. So that one made two bonds. Now the one with boron in the center has just those two p electrons– the boron’s not contributing anything. So we have only two p electrons and we occupy only the lowest of these orbitals. So it’s the HOMO and the middle one is the LUMO. Now you see when you bring alkene in, it doesn’t do the same thing, because the HOMO of the alkene looks just like this and they’re repulsive with one another. And the LUMO of the alkene looks pretty much like this. So it’s HOMO with HOMO, LUMO with LUMO to give overlap, and nothing gets stabilized, it’s repulsive.

So that one won’t do the trick. There’s no alkene-HOMO match with that LUMO, and no alkene-LUMO match with that HOMO. So the ones on the end will do cycloadditions and give a five-membered ring, but that one with boron in the middle won’t because of how many p electrons it has. So you can’t make two bonds simultaneously for cycloaddition to the alkene of that hypothetical compound.

Are there any questions about that? Does it seem unutterably complicated? Then review it.

So let’s look at what happens then in the process of ozonolysis. Incidentally, that name, the lysis part means a loosening or a breaking apart. It seems a funny name, doesn’t it, for a cycloaddition, which is clearly bringing things together. But you’ll see in a second what happens, it’s pretty neat.

So we bring alkene in and make those two new bonds by shifting the orbitals, as we just were describing. So we get this five-membered ring. But this isn’t so stable. Why is it not stable? Wojciech, you got an idea?

Student: All HOMOs.

Professor McBride: Yeah, two oxygens in a row of peroxide is bad. Three oxygens in a row must be even worse. Remember what makes two oxygens in a row bad?

Student: Lone pairs.

Professor McBride: The lone pairs repelling. And in this case, we’ve got– this compound is called a molozonide, if you have to give it a name. But it’s unstable because of the mixing of these HOMOs with one another, the unshared pairs. So that’s really bad. So it breaks apart. Of course, it could break apart just the way it formed.

But in fact, it breaks apart a different way. It draws these curved arrows, which is the same kind of thing we formed it with, but we’re breaking different bonds from the one we formed. We formed this bond and this bond, and now we’re breaking this bond and this bond.

So it undergoes a reverse of the previous process to give that. So that’s an aldehyde, plus this carbonyl oxide that we saw on just the previous slide, a 1,3-dipole. But the carbonyl oxide can do a cycloaddition. So it could reverse this, but it wouldn’t because the product would be unstable.

Do you see how to reverse it and make the product more stable where you don’t have three oxygens in a row? Anybody got an idea? How could you make the cycloaddition go, of the 1,3-dipole to the double bond, and not have three oxygens in a row? Elisa? 

Student: You’d use the thing you just formed…

Professor McBride: Can’t hear you. 

Student: Use what you just formed instead of the three O’s?

Professor McBride: Use what you just formed instead of the three O’s, I’m not quite sure what you mean. 

Student: You react to the CH₂-OO to make the ring. 

Professor McBride: Aha. React the CH₂ with O. In other words, you could turn that molecule over and now do the cycloaddition and you’d get this form, which only has two oxygens in a row. That’s the product that you actually get by reacting O₃ with the alkene.

Chapter 3. Acetal Hydrolysis and the Completion of Ozonolysis [00:32:59] 

Now this ozonide, as it’s called– remember that first one with three oxygens in a row was called a molozonide, this one’s called an ozonide. That’s not so fantastically important, but it’s what they call them. This I call here a double acetal. Now an acetal is a carbon that has two oxygens on it, two ether oxygens. So you’ve got that one, but you also have that one, so it’s a double acetal. You can hydrolyze acetals with acid. So we’re going to go through and figure out the mechanism of that as another practice in using curved arrows.

So Natalie, I want you to start us. It’s acid catalyzed. So what’s going to be the very first step? 

Student: To introduce your proton, and your proton is the LUMO, and so and your lone pair on the oxygen–

Professor McBride: OK. So the proton will attack an unshared pair on the oxygen. Good.

Now what’s our goal? They’ll be lots of different things we can do as we go along, but as we try to formulate a mechanism, it’s important to remember what kind of thing we’re trying to do. And the kind of thing we’re aiming toward is to remove RO and replace it with HO. That’s the goal of the first step. So you got us started well. We take a proton, come down and attack the unshared pair and we get that. Now what’s the next step? Natalie, can you help us some more? 

Student: You have your hold the positive charge on the oxygen is your HOMO.

Professor McBride: Now the positive charge, remember, makes a HOMO, makes orbitals be high [correction: low] in energy. But the positive charge is not an orbital, it’s just a charge. What particular electrons or what particular vacant orbital would that be? We want a low vacant orbital. The plus charge will lower orbitals. What’s the low orbital that we’re going for?

Student: π*.

Professor McBride: There aren’t any p electrons in this, no double bonds.

Student: σ*. 

Professor McBride: σ*. So the CO σ*. So we can do this. Have you ever seen a reaction like that before? Where a leaving group leaves and leaves a cation? Anybody seen a reaction like that? Alex?

Student: S_N1.

Professor McBride: Right. That’s an S_N1 reaction where the leaving group leaves– we made it a good leaving group by putting the positive charge on there.

Now that’s a primary cation. That doesn’t look like a very good candidate for S_N1. Alex, do you see any reason why S_N1 might be feasible? Why that cation might be better than you would imagine? I could tell you a reason why it’s bad. Oxygen is electron withdrawing– right, high nuclear charge. So it tends not to let cations form. It pulls electrons. Why is that a good cation? Anybody else got a suggestion? Debby?

Student: The electrons on the O are stabilized.

Professor McBride: The electrons of the adjacent oxygen get stabilized. The unshared pair on the oxygen get stabilized. We can draw that resonance structure. So it is an unusually stable cation, not because C⁺ is stable, but because that low vacant orbital stabilizes the oxygen electrons next door. So we get that cation. And now we have– it’s hydrolysis. So water is going to get into the act, form a new bond, and we have this. And we’ve done what we wanted to do here. Remove RO and replace it with OH. Of course, we replaced it with H₂O, but that’s no problem, we could lose the proton to another water molecule.

So we’ve done the first step here. We have what’s called a half acetal, a hemiacetal. Remember an acetal had two RO groups attached to CH₂. Now we only have one, the other is OH, a hemiacetal. Now this whole process here, if you had to give it a name, a mechanistic name, what would you call it Alex? S_N1. OK. So it’s an S_N1 reaction. 

And now the second goal is to remove the second RO, then a proton from OH, and then we’re going to generate a double bond here. We’re going to get CO double bond. So again, we have acid catalysis. Where are we going to attack for this purpose? Ruoyi?

Student: Attack the oxygen π*.

Professor McBride: Right. That’s the one we want to lose, so we protonate there. Next step, Ruoyi?

Student: Lose the alcohol.

Professor McBride: Right. So it’s another S_N1 kind of thing where we make the cation. And now what’s the last step if we want to lose– remember, we want to lose H, too, from here. We just do it, right? And what would you call that process, the one that went from here to here, if you had to give it a mechanistic name? Ruoyi?

Student: E1. 

Professor McBride: That’s E1, right, formation of the cation. So the process is E1. OK, so that did it. What we’ve done, the overall transformation is to take water plus an acetal, acid catalysis, and get a carbonyl plus two ROH’s– the one here and the one up there. So we started with the acetal and water, and the water comes across and we get two alcohols and a carbonyl, just by the S_N1 and E1 mechanisms that we just worked through. So that’s how you hydrolyze an acetal.

And here’s an acetal. The CH₂ has two O’s in it. So we can hydrolyze it. And we can hydrolyze that one. We can hydrolyze both– the one on the right and the one on the left. And you see then what we’ve done– now you see the lysis part, the breaking apart. What was the starting material? Where were these carbons in the starting material, the very early, before we added ozone? It was an alkene, a C-C double bond. What we’ve done is break the C-C double bond and make two C-O double bonds. So that’s the lysis part, the breaking apart. We change a C-C double bond into two C-O double bonds. 

But notice that one of the byproducts is water, the other one is hydrogen peroxide, which is an oxidizing agent. So it gives two carbonyl compounds and hydrogen peroxide. But it turns out that hydrogen peroxide oxidizes aldehydes to carboxylic acids. So this is a problem. You don’t just get the two carbonyls, you also have something that can react with the aldehyde, the carbonyl. 

So now, as the book points out, and any of your books will give this, there are two ways you can cope with this problem. One is if you want to make the aldehyde, put something else in that will react with the hydrogen peroxide to get rid of it. So you use a reducing agent, and we’ll talk shortly about oxidizing and reducing agents. But things like dimethyl sulfide or zinc, which destroys the HOOH. So the book now gives examples where you see you could do this, first treat it with ozone to get the ozonide. Then treat it with water, but with zinc there in order to reduce the hydrogen peroxide. And now you get the aldehyde in 93%. That’s the kind of yield that’s really convincing.

Or here’s another example where this one is first ozonized, and then when it’s hydrolized it’s treated with the sulfur to get rid of the hydrogen peroxide, and it gives a 75% yield of this aldehyde, and also the other one, CH₂O from the other side of the double bond.

The other possibility is just go with the flow. If it’s going to oxidize, fine, put in more oxidizing agent. So there we use the dimethyl sulfide and there zinc as reducing agents. But if you add more hydrogen peroxide, as is done here, then you can make the carboxylic acid in reasonably good yield, like this one, for example, 85% yield.

So those are the two possibilities with ozonolysis. You break the CC double bond, but the carbons could become either carbonyls or carboxylic acids.

Now how does that happen? How does the aldehyde react with hydrogen peroxide to give carboxylic acid? So let’s try another mechanism here. Derek, you’re going to start us in. What’s going to react with what here? What’s the HOMO, and what’s the LUMO? 

Student: Well, the minus on the O is a HOMO.

Professor McBride: The minus on the O is a HOMO.

Student: Would this be a double, like two at the same time?

Professor McBride: Let’s suppose we aren’t that sophisticated.

Student: Then C double bond O is the LUMO.

Professor McBride: C double bond O π* is a good example. Where would it attack, the C or the O? 

Student:  It would attack the C.

Professor McBride: Attack the C, OK. So there we go. That’s going to give this.

Now, can you think of a second reaction? Matt, let’s go to you.

Student: All right. Well, the– 

Professor McBride: Where do we have a HOMO? 

Student: You have a HOMO on the unshared pair on the O. 

Professor McBride: So here’s the HOMO, where’s the LUMO? 

Student: LUMO would be the OO σ*.

Professor McBride: σ*_OO. So we can do an S_N2 kind of reaction, right? Like that. That what you had in mind? 

Student: Yes.

Professor McBride: And it’s going to form a three-membered ring with two oxygens next to one another. That’s not so great, because the oxygen-oxygen’s not a strong bond and the three-membered ring is going to be strained. So that was a reasonable idea, that’s why I wrote it that way. But if I had written it a different way, the OO bond is weak and the three-membered ring is even worse, so forget that. 

So I’ll write a different conformation. And now this is a neat reaction. Again we’re going to use s*_OO with OH the leaving group. It’s a weak bond, OO. But now we’re doing something else. We’re taking the electrons from CH to be the attacking group. So it’s a rearrangement. It’s a hydride shift. And what pushes the hydride away, what allows it to do that, is the fact that these electrons can go in and form a very strong CO double bond at the same time.

And this will remind you, if you think about it a little bit, of another reaction we looked at. This hydride shift reminds you of this alkide shift here. Remember when you had B^– it was possible for these electrons to displace the OO. Remember that’s the second step of the hydroboration/oxidation process. So it’s the same kind of reaction.

So it’s a rearrangement, and that gives the carboxylic acid, but of course, it also generates hydroxide, which reacts with the carboxylic acid. So that’s the stable product is the carboxylate anion. So that’s why when you have hydrogen peroxide there the aldehyde becomes a carboxylic acid.

Now here’s a problem for you to work on. Try doing the same thing– here we used base catalysis. Suppose it were acidic conditions and you had to do it with acidic catalysis. Could you draw a mechanism that gets to the same place. Of course, it would be protonated, if it’s acidic rather than basic. But see if you can put together a mechanism that’s acid catalyzed that does the same thing.

Chapter 4. Electrophilic Participation in Nucleophilic Attack on C=O [00:46:02]

Now we’ve been talking about electrophilic additions to carbon-carbon double bonds. We talked already last semester about nucleophilic addition to CO double bonds. So nucleophile versus electrophile. But remember that the addition to the carbon-carbon double bond was both electrophilic and nucleophilic. So it wouldn’t be surprising if the addition to the CO double bond is both nucleophilic and electrophilic. So let’s just compare it.

So the nucleophilic addition of methyllithium to a carbonyl group– and we’ll discuss this more when we’re talking about the chemistry of carbonyl groups– is formally quite different because you add electrophiles to alkenes. But this shows you that there’s really quite a similarity. 

So here on top, this is the transition state for the addition of methyllithium, so the yellow or orange thing is lithium, and this CH₃, of course, is methyl. So that’s it coming down on top of formaldehyde. And that’s the transition state. 

So what’s happening is we’re changing partners here with these electron pairs, and going toward that product. So we start with this, methyllithium and the carbonyl, and there we have a HOMO, which is the lithium-carbon bond, mostly on carbon– that doesn’t surprise you. But notice that it’s well set up to interact with the p* of the carbonyl. And by the same token here’s LUMO + 1, a p orbital on the lithium, a vacant orbital, which is well set up to overlap with the red part of a HOMO. So again, there are two electrons that do the trick– HOMO with LUMO, LUMO with HOMO. And you get both things happening at once. And we’ll come back and talk about that more, but I just wanted to show you that it’s not a completely different thing from the addition to alkenes.

Chapter 5. Cycloaddition for Dihydroxylation [00:48:02]

Now we’re ready to talk about metals. Let me just do a little bit at the beginning and then we’ll go on next time to talk about more. So osmium textroxide, or permanganate, is Os, or Mn minus, in the center of four oxygens. And notice that that looks like ozone, these three atoms in a row above the two of the alkene. So there’s its HOMO, and it’s well set up to interact with the alkene LUMO. And there’s its LUMO, which is well set up to act with that alkene HOMO, although I seem to have messed up the animation a little bit.

So here we can see it happening. It gives us osmate ester. Now notice it forms two bonds at the same time. It’s like this 1,3-dipole or cycloaddition. And that’s the osmium analog of that cyclic acetal that we looked at before. So we can imagine water coming in and breaking it apart to give a double bond and two OH groups. So the product then from that osmate ester of hydrolysis is the diol. So you could make a carbon-carbon double bond, break this π bond and put two OH’s on. So osmium tetroxide is a very good reagent for giving that in high yield.

In other respects, it’s not such a great reagent, and we’ll talk about that next time.

[end of transcript] 

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