CHEM 125b: Freshman Organic Chemistry II

Lecture 12

 - Nucleophilic Participation During Electrophilic Addition to Alkenes: Halogen, Carbene, and Borane


When electrophilic addition involves a localized carbocation intermediate, skeletal rearrangement sometimes occurs, but it can be avoided when both alkene carbons are involved in an unsymmetrical 3-center-2-electron bond, as in Markovnikov hydration via alkoxymercuration followed by reduction. Similarly a reagent that attacks both alkene carbons simultaneously by providing a nucleophilic component during electrophilic attack can avoid rearrangement, as in reactions that proceed via three-membered-ring halonium intermediates. Simultaneity in making two bonds during formation of cyclopropanes from carbenes can be demonstrated using stereochemistry. Anti-Markovnikov hydration can be achieved via hydroboration followed by oxidation with hydroperoxide. Rearrangement of the borane hydroperoxide intermediate with frontside C-O bond formation shows close orbital analogy to backside attack during SN2 substitution. Again syn-addition shows that nucleophilic attack occurs simultaneously with electrophilic attack on the alkene.

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Freshman Organic Chemistry II

CHEM 125b - Lecture 12 - Nucleophilic Participation During Electrophilic Addition to Alkenes: Halogen, Carbene, and Borane

Chapter 1. Forming Unrearranged Alcohols via Hydroxymercuration [00:00:00] 

Professor McBride: So what is it that does the attacking in SN1 and SN2 reactions? What kind of reagent? Pardon me?

Student: A nucleophile?

Professor McBride: A nucleophile. Why do we have nucleophiles attacking things with leaving groups on them? What’s the orbital that makes those things reactive, the substrate reactive?

Student: A high HOMO.

Professor McBride: Not the nucleophile, the nucleophile’s a high HOMO. It’s a low LUMO. So when you attack alkyl halides, it’s because of the low LUMO. When you attack alkenes, the characteristic functionality of the double bond is mostly remarkable because it’s a high HOMO. So we talk about electrophilic addition to double bonds. So it’s low LUMOs that are attacking double bonds, like H+

But many of the reagents that do this, that are classed as electrophilic addition reagents, also have high HOMOs. So they also function as nucleophiles. So the subject of the lecture today is electrophilic addition to double bonds, but stressing that there’s also a nucleophile that may be participating at the same time, that could react with the LUMO of the alkene, that is, the p*.

OK, first I wanted to suggest a problem for you to work on. The pinacol rearrangement is a neat rearrangement. Notice that the way the carbons are connected together in the product, pinacolone, is different from the way they’re connected in the starting material called pinacol. The origin of that name is interesting. It comes from an ancient Greek word for tablet. And it was coined in 1859 as a name for this compound because its crystals were plates. So it was called pinacol, because of the pretty plates of the hydrate crystal. 

But anyhow, see if this will be good exercise for you, to draw nice curved arrows that start where they should start, and end where they should end. To show how you do this, it requires several steps. So that’ll be fun for you. 

OK, now last time, we talked about rearrangement. In particular, that when you had acid reacting with the alkene, you got rearrangement of the product. So for example, you add the proton to get the more stable cation, the secondary instead of the primary. But there’s a more stable one still, which is the tertiary. So it involves a methide shift that we talked about last time. And the product, then, when you add water and then lose a proton from that group, is this rearranged alcohol. 

So this could be a problem, if you wanted to get the unrearranged alcohol. You couldn’t do it by acid-catalyzed hydration. But there’s an alternative, called oxymercuration reduction. So the oxymercuration adds Hg and OH to the double bond. But the electrophile is the Hg+2 cation, so one of the acetates comes off, and you add the Hg. And then there’s a second reaction in this sequence. First, the oxymercuration, then a reduction, which avoids rearrangement. So it gives the alcohol that you would have expected if you hadn’t been sophisticated enough to suspect rearrangement of the secondary to the tertiary cation.

Now let’s think how this works, and why it works here, when you get a rearrangement, when you have a proton. The proton is small, and isn’t able to overlap well with both carbons at the same time. We talked about the fact that it gives a double minimum. It can rearrange. Hydrogen, with its electrons, hydride, can shift the same way methide shifted in the rearrangement we just talked about on the previous slide.

But it goes over a barrier, whereas mercury is a much larger ion, and is big enough to overlap with both carbons at the same time. So it’s a single minimum. Instead of rearranging over a bump from one to the other side, it lives near the middle. So as shown there, now it’s a Y-bond. The electrons that were in the p bond, on the carbon-carbon double bond, are now shared with the vacant orbital on the mercury cation. So it’s one of those three-center, two-electron bonds that we talked about before. OK, so now, since the mercury is bonded to both carbons, you can’t get the rearrangement.

Now here, we draw resonance structures to denote the bonding that we also can denote by that Y, upside-down Y, in this case. But notice that I haven’t drawn it symmetrical. The mercury atom is not in the middle, halfway between the two carbons. We can have the three resonance structures on the right, with Hg+ on the left with the primary C+, and on the far left, with the secondary C+. But that secondary cation would be thought to be more stable than the primary one. 

So it won’t surprise you that the best position for the mercury is not halfway between the carbons, but nearer the one on the left, so that it’s the most important resonance structure. And what that means is that the cation is mostly on the secondary carbon, and that bond is weaker. The bond on the left, the red bond shown here, is stronger. So when the nucleophile comes in, in the second stage of the reaction when the water comes in, it attacks that cation, not the other one. If it were symmetrically bridged, you would think that it could attack both of them. Maybe fifty-fifty. But in fact, it attacks that one.

So it gives this product, that’s the hydroxymercuration product, the hydroxymercuration addition to the double bond. That, then, nobody wants that, but you could remove the mercury by treating with a reducing agent, sodium borohydride. And it’s replaced then by hydrogen. 

So this subsequent reduction step by the BH4 anion completes Markovnikov hydration. Notice that it’s the same orientation of alcohol that you would have gotten with H+. It’s the more stable of the two cations that you could get as a primary cation or secondary cation. The nucleophile, OH, went to the more substituted carbon. So it’s Markovnikov, it’s the same product you get. But you suppressed the rearrangement by the fact that the mercury was bonded to both carbons at once, and didn’t allow the methide to shift. OK?

Chapter 2. Electrophilic Addition to Alkenes with Nucleophilic Participation: Halonium Ions [00:07:36]

Now, this idea of bridging and unsymmetrical bridging occurs also in the addition of halogens to the double bonds, chlorine and bromine. These are examples from the lecture that Professor Siegel gave last year, so they show you the way you would write this quickly. And notice that the addition of chlorine adds Cl and OH. Cl goes, in this case, to the primary carbon, and OH to the more substituted carbon, analogous to the way in the previous slide, mercury went to the less substituted carbon and OH, ultimately, to the more substituted carbon. So it’s Markovnikov addition. 

Now what was the electrophile? What was the thing that added to give a carbon cation intermediate in this case here? OH came in as a nucleophile in the second step, and then lost a proton. Water came in, lost a proton to give OH. But chlorine must’ve been the electrophile. That meant it was something like Cl+, not Cl, which you usually think about.

So let’s contemplate this reaction in a little more detail. So here’s ethylene, viewed edge on. And an electrophile will attack its HOMO, which looks like that. Now, the electrophile is going to be chlorine. Now, what orbital is the LUMO of chlorine? Cl2. Arvind, you got an idea?

Student: s*.

Professor McBride: Right, s* is it. So let’s draw the s* orbital. LUMO is s*, so it looks like that. Does that look good to you, Arvind? First, is the energy match good? Is it an unusually low LUMO, s* of Cl2?

Student: It’s probably unusually low, but it doesn’t look like the overlap is–

Professor McBride: Ah, it is unusually low, because it’s chlorine, with that big nuclear charge, so low energy. But as you say, the overlap is bad. It looks almost orthogonal. So what could you do about that? How could you increase the overlap? Do you have any idea?

If you were a nucleophile, a high HOMO, and wanted to attack that LUMO, the s* of Cl-Cl, where would you approach from? Would you approach the way the ethylene is doing here? Where would you come from?

Student: You’d probably approach from the top-right.

Professor McBride: You’d approach from one of the ends, where you get the overlap without hitting the wrong orbital. So what do you need to do to the picture?

Student: Turn it around. 

Professor McBride: Right. You need to rotate the chlorine, because you have poor overlap here. But if you rotate the chlorine, it looks just fine. So now you can take the electrons in the HOMO and mix them with this p* orbital. Which means, remember, there’s antibonding, it’s an antibonding orbital of Cl2. So we can draw arrows like this. Break the Cl-Cl bond. So it looks like a nucleophilic substitution reaction. And what atom is being attacked if this is a nucleophilic substitution?

Student: Chlorine?

Professor McBride: The chlorine. It’s a nucleophilic substitution at chlorine. OK, so that’s what we would get, where the Cl+ has attacked the C-C double bond and formed this three-membered ring, still a cation. And Cl has gone away. It took the electrons that were in the s orbital. OK? So there’s what’s going on.

So if you consider this product, the two different species in the product as one, that’s the HOMO. It’s where the electrons of this chlorine-chlorine bond went. That top arrow, the curved arrow. And this is the HOMO – 2. And that, you can see, is distorted. But it’s the p orbital of the C-C double bond, mixed with the p orbital on the chlorine, or some hybrid orbital on the chlorine, to form the bond in the blue region there. That’s where the p electrons went.

But notice this is HOMO – 2. Where is the HOMO, and where is HOMO – 1 of this product pair of things? So let’s look at it again. Notice that there are two chlorine-chlorine bonds drawn in this three-membered ring. So there must be some other pair of electrons that’s involved, if we can draw two bonds. Where did those other two electrons come from? That other pair?

Well, we had the LUMO on ethylene, p*, and the HOMO on chlorine. Now, what do you think the HOMO on Cl2 would be? Ruoyi?

Student: A p orbital chlorine.

Professor McBride: So it’s a p orbital on one of the chlorines, an unshared pair. In fact, we now want a chlorine nucleophile, so it’s that unshared pair. In fact, we want to draw the molecular orbital that has it, the high HOMO. It turns out to be the analog of p*. It’s not only the p orbital on the bottom chlorine, which would be relatively high in energy, but it’s a mixture with the p orbital of the other chlorine overlapping with it, antibonding to raise the energy. So it’s still higher in energy than it would be if it were just a single p orbital. So it’s this orbital here. You see, it’s precisely set up to form good overlap, so its electrons on the chlorine will be stabilized by the LUMO of the ethylene.

At the same time that the s* orbital of the chlorine is stabilizing the p electrons. So it mixes both ways. The chlorine is both an electrophile and a nucleophile. Two different pairs of electrons are involved, and that gives the HOMO – 1. Which you see. So now we have two pairs of electrons that are forming those bonds, those new bonds in the three-membered ring. So we can denote that this way.

Now, where is the HOMO? That’s just an unshared pair on the oxygen [correction: chlorine] that wasn’t involved in this, because it’s orthogonal to the other bonds, it’s in and out of the plane. So that’s the HOMO. So we’ve seen the HOMO, the HOMO – 1, and the HOMO – 2. There’s the LUMO, which is the antibonding combination of the chlorine and the p.

OK, so we’ve traced this through, but the crucial lesson is that there were two reactions going on. The chlorine acting as a nucleophile as well as an electrophile. So although you call it electrophilic addition, and sort of imagine that it’s Cl+ attacking, in fact, both things are happening at the same time, and two bonds are being formed.

Now, what would happen next? There’s the LUMO, and we have the HOMO of the Cl ion that was formed at the same time. So if those two reacted in the way shown here, it would be the exact reverse. You would form a new bond between chlorine and chlorine, you would put electrons into this LUMO, this bond would break, and you’d go back to the starting material. But there’s a different arrangement that will allow you to go on from here toward different products, which is to move that chloride, an independent ion, after all, down to the bottom. So now, we have that chloride unshared pair HOMO and this LUMO, and they can interact with one another like this.

What kind of reaction does that remind you of? Did you ever see a reaction like that before?

Student: An epoxide.

Professor McBride: It’s like opening epoxide, which is in an even more general sense, an SN2  reaction. A nucleophile attacking a LUMO, which is antibonding here, and the leaving group leaves, it’s an attack on carbon. So that, then, would open up the ring, and give an addition of two chlorines to the carbon-carbon double bond.

So you could think of it first as addition of Cl+, and then addition of Cl. It’s a two-step process. But the interesting thing is that that first step involved both a nucleophile and an electrophile activity of the chlorine. So we could draw it again here. That first step, it makes two bonds in the three-membered ring while it breaks the chlorine-chlorine bond. So the red is what you normally think of as an electrophile attacking the double bond. But the blue is a nucleophile attacking. So it’s an electrophilic addition of Cl2 to an alkene that’s both electrophilic and nucleophilic simultaneously, which generates this so-called chloronium ion, this three-membered ring intermediate, which then undergoes an SN2 reaction to open up and give the product.

Now, remember, since it’s two steps, the nucleophile in the second step doesn’t have to be the chloride. It could be some other nucleophile. For example, you could do this reaction, generate that intermediate, but then have water come in and be the nucleophile. In which case the chloride would go away, and you’d lose the proton, so the product is HCl, the product not shown is HCl. But, you have the compound that’s called a halohydrin.

Now, let’s look at the bromonium ion, which you get when you add Br2 and look at the regiochemistry that’s involved. So you have this three-membered ring, but it’s not symmetrical. The bromine I’ve drawn is more on the right than on the left. Why would I draw it that way? Connor, you have any idea? Why not put it in the middle?

Student: Because the carbon on the left is the more substituted one.

Professor McBride: Right. The more substituted carbon on the left is a better place to have a cation. So you have a stronger bond on the right, a longer, weaker bond on the left. So it’s an unsymmetrical bridged bromonium ion. And then if the bromine comes in, backside at the carbon to do its SN2 reaction, you get the product that’s just like the one we’ve showed before with chlorine, except it’s two bromines that have added in this case. But that’s only formed a 24% yield, in this case. What do you think the dominant product is? Antonia, can you see any alternative?

Student: Can you add to the other side?

Professor McBride: Well, you could add to the other side, but we’ve just been talking about– you mean on the right, rather than on the left. You could, but that would be not preferred. You might say maybe steric hindrance would help it attack from the right. But in fact, it doesn’t attack from the right, because of the unsymmetrical nature, the bond between bromine and carbon is stronger on the right. The positive charge is bigger on the left, so that’s where the bromide attacks. So it’s not that it attacks on the right. But there’s another possibility. Anybody got an idea?

Student: The water attacks?

Professor McBride: Pardon me?

Student: The water attacks?

Professor McBride: Aha, the water can attack. So in this case, instead of the bromide, the water can come backside to the carbon, and do its trick. 60% yield, in this case, is the bromohydrin. So this is a competition between forming the halohydrin and forming the dihalide. And notice that it shows Markovnikov regiochemistry for the reason we’ve just been detailing. That the electrophile that added first was bromine, gave the more stable cation, or unsymmetrically the thing that had more positive charge here, and that’s where the water, the nucleophile ends up, on the more substituted carbon.

But how do we know the ion is bridged? The same thing would happen if we hadn’t formed that second bond– right, if it had just been like H+, if the bromine had just attacked to get the more stable cation rather than a bridged one. How could you know that it’s bridged? What effect was there from bridging in the case of mercury? Yeah, Arvind?

Student: There’d be no rearrangement?

Professor McBride: There’d be no rearrangement. That’s one thing you could do, but there’s another as well, another way of showing that. And this is the key to it here, that it’s stereospecific. If you start with the trans- or the cis-2-butene you get different stereochemical dibromides. Stereochemically different dibromides. But we’ll look at them in another case here.

Suppose you have a ring compound like cyclohexene? And you add bromine, you get the bromonium ion, obviously both bonds in the three-membered ring have to be on the same face of the six-membered ring. And then, when you have backside attack by the water, it must form that compound. It can’t form the OH on the same face as the bromine in the SN2 process. It has to be a backside, not a frontside attack.

So whether it’s the OH, and you’re forming the halohydrin, or whether it’s Br, and you attack to form the dibromide, in either case, it has to be a trans product. The bromine and the nucleophile have to be on opposite sides of the ring. So an unbridged C+, the nucleophile could have attacked from the top or the bottom if the bromine weren’t bridging. So the bridging is evidenced not only by a lack of rearrangement, but also by the stereochemistry of the product. It’s an anti addition. The reverse, you’ll notice, of anti elimination, that we talked about in E2 reactions.

Now how about other kinds of reagents that can be both an electrophile and a nucleophile at the same time? That do two things. There are a lot of them. There are things called carbenes, there’s the reagent involved in hydroboration, which– things that have a B-H bond in them. There’s a thing called a carbenoid, which involves these reagents. There’s epoxidation, which forms the three-membered ring with oxygen by a peroxy acid. There’s ozonolysis, O3 to form a five-membered ring, initially.

And there’s this thing we referred to last time, where, remember, H2 doesn’t have the proper symmetry of its orbitals to add directly to a C-C double bond. But with the help of a metal, it can. And that’s another case where you have simultaneous nucleophile and electrophile behavior. So we’re going to go through these, we won’t get through all of them today, to see how these reagents work. And finally, there’s polymerization, which is another topic we’ll get to then.

Chapter 3. Electrophilic Addition to Alkenes with Nucleophilic Participation: Carbenes [00:24:56]

OK, so first, carbenes. There was a fellow named Jack Hine, who got his PhD in 1950, and went to be a post-doctorate with Professor Bartlett, your chemical grandfather. And when Bartlett spoke about this sometime, he mentioned that the project that Hine decided to work on, all on his own, I believe, without much guidance from Bartlett as a post-doctoral, was the reaction of chloromethanes with hydroxide.

Now, by 1950, this was pretty old hat. Because the reaction had been known right from the time of Williamson ether synthesis 100 years earlier. And all this stuff about SN1 and SN2, E1, E2,  had all been done in the 1920s, ’30s. So this was pretty old. Bartlett described this, he said it was “about as earthy a research project as could be appropriate for a post-doctoral.” But Hine knew what he was doing. He had an interesting idea he was following up on. He found that the reaction with hydroxide with chloroform is very fast. With methylene chloride is– wait a second, I’ve got the first one is supposed to be CH3Cl I’m sorry, I put it in there wrong.

So with methyl chloride, it’s fast. With methylene chloride, it’s slow. With carbon tetrachloride, it’s very slow.

Now, if you were thinking of the halogens as roughly like a methyl group, which they’re about the same size, then this looks like the same steric effect that was well known already in the rates of SN2 reactions. That methyl is faster than ethyl is faster than t-butyl. So that much is sensible. But what was interesting was this: that chloroform is fast. What does that suggest to you?

We saw this before, so just in general, you saw before that when we were doing SN2 reactions, you had methyl, ethyl, isopropyl, and suddenly, t-butyl was very fast. What did that suggest?

Student: A different mechanism.

Professor McBride: It suggested, yeah, that there is a different mechanism being involved in that reaction, somehow, since it doesn’t fall in line. Now, what’s special about that? Well, he also could study the reaction with this sulfur anion, which is a nucleophile. Hydroxide, smaller, better overlap with hydrogen, is a good base. But phenyl sulfide is better as a nucleophile. Better at bonding to carbon. So it’s the thing that would be doing the SN2 kind of reaction. And that is fast, slow, and very slow, which makes sense for chloroform. So it’s something about OH being a base that makes chloroform unusual, Hine discovered.

So what it is, is when you have a very strong base like the base from t-butyl alcohol, t-butoxide, it turns out that you can extract H+. Because the electronegative three chlorines make the carbon anion sufficiently stable that it can lose H+. It’s not a strong acid to lose H+, but it’s much better than having just two or only one chlorine.

So you can do this SN2 reaction at hydrogen and get the CCl3. But then, it has a good leaving group on it, chloride. So you get elimination of HCl from a single carbon. It’s a-elimination. We talked about b-elimination before in the E2 or E1 reaction, where you take a hydrogen, the proton, off from one, chloride, or whatever leaving group, off of the adjacent one.

But this is losing them both from the same atom. And chloroform is uniquely set up to do that. Obviously, carbon tetrachloride can’t do it, because it doesn’t have the proton to be lost. So now we have this divalent carbon, which is called a carbene. And it can come together with the alkene to form a cyclopropane. So we’re going to try to analyze this in terms of the HOMOs and LUMOs that are involved in mixing the orbitals of the carbene, the dichlorocarbene with ethylene.

OK, so here are our reagents, we know the orbitals of ethylene. So here’s the HOMO of dichlorocarbene, which turns out to be a bent molecule. And here’s its LUMO, the vacant p orbital. Now does that look well set up to react with ethylene? To mix HOMO with LUMO and LUMO with HOMO? Noelle, what’s the shape of the HOMO of ethylene? That’s the thing that electrophiles usually attack. What is it? It’s the p. So it would be positive here, and positive here. Is that well set up to make sure that this LUMO– No, because the LUMO, the p orbital would be plus on one side and minus on the other. They’d be orthogonal.

And the same is true, the LUMO of the ethylene is– would be plus here, minus here. And this one is symmetrically between it, and they’d both be orthogonal. So both the HOMO-LUMO and the LUMO-HOMO pairs are orthogonal. So it doesn’t look like dichlorocarbene should be able to do this, to form these two bonds at once. But there’s a way to do it. And the person who can tell us how to do it is Arvind.

Student: We could rotate it again?

Professor McBride: Aha. We’ll rotate it. So this is actually the three frames through the transition state. This is just before the transition state. But notice the dichlorocarbene is coming in sideways rather than head on. So that bond is being formed, and the two bonds are being formed, the p are going here, those in that hybrid orbital here are going into this. So we’ll watch the motion here. Back up, come forward again.

So what happens actually as you go through the transition state is it’s rotating. Because it starts getting overlapped with the thing, and then as the new bonds form, it goes into the lower energy geometry of the three-membered ring, which of course has the chlorines pointing up. So then, this is the LUMO of the– I used difluorocarbene, to have fewer electrons, in order to do the calculation. But then, you see that the LUMO is the p orbital on the dichlorocarbene, as we suspected. The HOMO is the p of the alkene, and they overlap well. And then HOMO is that sp-hybrid on the carbon, and that’s well set up to overlap with the LUMO. So again, you can do both electrophilic and nucleophilic attack at the same time. But it requires coming in sideways and rotating.

OK, now how would we know that both bonds form at once? Because an alternative mechanism would be for this thing to act as if it were a radical. For one electron to come in here, one electron of the p bond to go here, the other to go on here, to give this diradical intermediate, which then, in the second step, could close the bond. How do we know whether there is an intermediate like this in the reaction, or whether we just have a transition state, that we just drew, where both of the bonds are being formed at once? What experimental test can one perform to see whether both bonds are being formed at once? Anybody got an idea? Chris, you have any idea of how you can tell if two bonds are being formed at once? Pardon me? Wells, you have an idea? Have you ever seen an example before where you could tell that two bonds are being formed at once, and not one bond at a time? Daisuke, you got an idea?

Student: It’s analogous to SN1 and SN2, so you could figure out the rate.

Professor McBride: Rates are one way one tries to do things like that. Ruoyi?

Student: Stereochemistry?

Professor McBride: Stereochemistry of the product. Because if you have this intermediate, if it lasts longer than about 10–10 seconds, that’s how long it takes a single bond to rotate. Something like that. So you could rotate– this was a double bond before, so these two methyls were cis to one another, both going back into the board. But if you get this thing with a single bond, then it can rotate in about 10–10 seconds, to have this methyl coming out in front, the other methyl still in back. And now, you should do that if you can, because this is less strained. Can anybody see why this would be less strained than that one? Lauren?

Student: The methyl groups aren’t running into each other.

Professor McBride: Right. In this one, this methyl, is between this CBr2 and this CH3. It’s gauche to both of them, or something like that, close to both of them.

But here, it’s rotated to be far from the methyl. So it’s much less strained here, when it’s not between the two groups, here, it’s between the CBr2 and the hydrogen, although I’ve drawn it as if it were eclipsed. So if this is cis and it doesn’t rotate, this product will be cis. The two methyls will be on the same side of the ring, both going back into the board. But if this happens, the product will be trans. So you do the reaction, and you find that cis-starting material gives only cis product. So there wasn’t the opportunity for rotation. Therefore, it must have formed both bonds either at the same time, or within a really, really short period of time.

So fundamentally, what you did here is time how fast this could have been, how fast this closure could have been, by knowing something about how rapidly things rotate. You know this collapse must be faster than rotation. So it doesn’t really prove that there’s not an intermediate, it proves that it doesn’t last longer, if there is one, that it doesn’t last longer than 10–10 seconds, which is roughly the rate at which you would rotate.

Chapter 4. Anti-Markovnikov Hydration via Hydroboration and Oxidation [00:36:55]

OK, now the second example was hydroboration/oxidation. So remember we said that if you had a B-H bond, you could add it across a double bond, and then, treat that in a second oxidation step by reacting with hydrogen peroxide and hydroxide, and that replaces BR2 which people don’t want, with OH, which is a kind of compound that people do want.

OK, now, so what this process, the two steps, hydroboration and then oxidation, achieves addition of water to the C-C double bond. Do you know any other ways to add water to a C-C double bond? H+ and water. That’s what we added, HOH to a double bond. Oxymercuration and reduction, which we just talked about, also forms the alcohol. Why do you need another way to make an alcohol? Sebastian?

Student: It’s anti-Markovnikov.

Professor McBride: Ah, it’s anti-Markovnikov. Let’s look at how this happens. OK, so here’s the BH3, which is an electrophile, because of the vacant orbital on boron. And this is the motion that will happen. These electrons that are in the double bond go in to bond to boron. But at the same time, these electrons, the B-H bond, an unusually high HOMO. Why is that unusually high, the B-H bond? The overlap looks good to me. But boron doesn’t have as big a nuclear charge as carbon. So it’s a high energy orbital for the same reason that bonds in chlorine are low in energy. There’s not much nuclear charge in this one. So this nucleophile is attacking p* at the same time. So we can again watch this, watch them stretch and shrink as it goes toward the transition state.

So we’re breaking this bond, forming the C-H bond, and forming the B-C bond.

So, here we’re looking at the orbitals. So the LUMO on the BH3 is the p orbital, the HOMO of the alkene is this p, the HOMO on the BH3 is that B-H bond, and the LUMO of the alkene is this thing, which looks ugly. But the reason is that I’ve distorted it, to be in the shape it would have at the transition state, not where it started. And so that was the transition state, and you can see that it mixes HOMOs and LUMOs, LUMOs and HOMOs.

OK, so then, this then establishes a B-C bond. Now, the particular one– it has an R group on the B. Here I’ve drawn methyl. If this had been formed by addition of BH to the double bond, there would be another carbon here. There would have been a double bond, boron added to one end, hydrogen added to the other. But I’ve made it, for simplicity, made it a methyl group, just so we wouldn’t have as much to deal with. And now, I’m going to look at the second step, the oxidation. So what’s the LUMO that’s going to be involved in this reaction? Luke, what do you say?

Student: The B-C s.

Professor McBride: Though if it’s going to be vacant, it’s not going to be the s, it could be s*. But that’s not the LUMO, there’s a lower orbital than that of the boron compound.

Student: It’s, it’s–

Professor McBride: Derek?

Student: It’s bonded three times.  Boron has a vacant orbital.

Professor McBride: Ah, boron. BX3 from last semester. The vacant orbital on boron. So it has an atomic orbital that hasn’t mixed with anything. So the lowest orbital is the p orbital on boron. What’s the high HOMO? What’s going to react? Amy, you got an idea? Can’t hear very well.

Student: The negative charge.

Professor McBride: Right, the oxygen that has a negative charge on it. So we can just mix those two. OK, so we’ve got this, and now we draw a negative charge on boron, because it gained half interest in the pair of electrons that was on oxygen.

And now we have an interesting rearrangement that looks a lot like SN2. Where is there a low LUMO here in this molecule? Linda, you have an idea? Can’t hear very well.

Student: The O-O s*.

Professor McBride: Yeah, the s* of O-O, because of the high nuclear charge, like Cl2, or something like that. So low LUMO here. Where is there a high HOMO? Wells, you got an idea? When you look at this whole thing, here, what is there that tells you that there could be a high HOMO?

Student: Negative charge.

Professor McBride: Aha. The negative charge. But that negative charge is not an unshared pair of electrons. There aren’t any unshared electrons on the boron. So what is the orbital? It has to be one of the s bonds to boron. For example, this s bond here. So here we have a nucleophile, and here we have an electrophile. Here’s the HOMO, s* is the LUMO.

Now, here’s the geometry of the transition state. You see, what’s happening is that the CH3 is moving over, taking these electrons with it. It’s a methide shift, like we saw in the cation cases, except that in that case, there would be a cation here, and the CH3 with its electrons would shift, to give a more stable cation. In this case, it’s because you have the O-O bond. But it’s interesting to compare this transition state with the transition state for an SN2 reaction of, say, chloride attacking methyl chloride.

So you can see, here’s the nucleophile, the leaving group, here’s the nucleophile, the leaving group. And in fact– the s*CBr is being attacked here, in that reaction of the SN2, but in this case, it’s s*OO, and sBC is the nucleophile. So you see, it looks like the same reaction. And in fact, if we go to the transition state, you see the thing bouncing back and forth to the transition state, but if we look at the orbitals of the transition state, this is the HOMO for SN2, this is the HOMO for this rearrangement. You see, the orbitals look almost exactly the same. And if you look at the LUMO for the SN2, and the LUMO for this rearrangement, it’s the same reaction as SN2, this rearrangement is.

So, this is the question that Sebastian has already answered. Why do this procedure, if all it does is add H and OH to C double bond C? How about the regiochemistry? And how about the stereochemistry? So that initial addition adds BR2. BR2 is now the low LUMO in that initial addition, before the rearrangement. In the initial addition, B is the electrophile. So it adds to the less-substituted carbon, so that you get more of the carbon cation character here.

And the hydride is what shifts, remember? It’s the hydride that– B is the electrophile, and the B-H bond with the hydride shifting is the nucleophile. So it goes to the more substituted center. So whereas Markovnikov, of course, knew nothing about HOMOs and LUMOs, nucleophiles and electrophiles, all he knew was that the hydrogen went to the place that had the most hydrogens. And the other thing went where there were the most non-hydrogens.

But the other thing in this case is BR2, which turns out to be the electrophile, not the nucleophile. In the other cases, the other thing not hydrogen was the nucleophile. Now it’s the electrophile. So it gives the opposite orientation here from what Markovnikov probably would have said, but for the same reason that explains Markovnikov addition. And now, when we do the rearrangement, what we do is put the oxygen on boron, and then have this whole R group shift over. But that’s a frontside attack. It’s the s orbitals of this R-BR bond that are doing the attacking. So it’s frontside at that, it comes in and the OH replaces the BR2. It’s not a backside situation. So you get that.

So the alcohol product is syn. The two things that have added are on the same face, not on opposite faces. And it’s anti-Markovnikov. The OH now is on the opposite place from where it would go by these other mechanisms of adding the alcohol. So this is a great boon, then, if you’re in the synthesis business, and need to prepare an alcohol where the OH is on the less-substituted carbon. So it’s anti-Markovnikov. So you contrast that with acid-catalyzed hydration, which would give this cation, or pardon me, this alcohol, which is the Markovnikov product.

How about the stereochemistry in this case, is it cis, or was it syn addition or anti addition? Did H and OH come in on the same face, or opposite faces? Cassie? Can’t hear. You can’t tell, here, right? Because there are two H’s. So it’s a clear choice in this case, where it was anti-Markovnikov. When it’s Markovnikov, and the H goes where there are already H’s, there is going to be this ambiguity, unless you did what?

Student: Put a D in.

Professor McBride: Put a deuterium in, then you can tell that it is syn. But it appears random. Oh, no, pardon me. It not only appears random, if you did it with deuterium, you would find that it is random, because you generate a cation intermediate here when you add protons first. And that could be attacked from either side. So it’s syn here, but random down here, and that’s enough for today.

[end of transcript]

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