 Okay, let's jump back into electrophilic addition. So currently we're talking about using alkenes as nucleophiles, really, is the topic that we're, and we're talking about the kinds of reactions that you should have already covered at some point back in your undergraduate organic chemistry courses. So addition of halogens, hydrohalogen acids, things like that. So we left off talking about the reaction of alkenes with things like bromine and chlorine. And we talked about the way we're going to push arrows for this. There's initially some pi complexation. There's initially formation of an actual complex in which bromine or chlorine is literally coordinated to the pi bond, but you haven't formed any sigma bonds yet. And we said we're not going to draw that. We said we're not going to draw these pi complexes with halogens sort of sticking to double bonds. We're just simply going to draw it like this and pretend there is no initial pi complex formation. And we have this weird set of arrows to show that bromide is getting pushed out as a leaving group. But at the same time that the double bond is attacking this bromine, the lone pairs on bromine are coming back to make sure there's no carbocation that gets formed. And that's what leads to this cyclic bromonium ion. If you take most olefins and this under typical conditions and you throw in iodine as a reagent, you won't get any of this product. And so why is that? Why is it that this reaction works with chlorine and bromine? It turns out that iodine and iodine bonds are weak. Bromine-bromine bonds are weak. Chlorine-chlorine bonds are relatively weak. There's a strong thermodynamic driving force in order to generate a dibromide, all right, at the dibromide product here, or to generate a dichloride. But because the bond dissociation energy for a carbon iodine bond is only about 51 k-cals per mole, this is just barely anthropically favorable. And if I rewrite this as an equation, in other words, if this is only 2 k-cals per mole favorable and thoughtfully, the entropic cost counteracts that. Overall, in terms of delta G under typical conditions, you can't get iodination to work. The equilibrium actually favors the starting material side. So you're not going to see very many cases where people add iodine to simple double bonds. There's special conditions. Obviously, you could use potentially Le Chatelier's principle, dump in an excess iodine, and help drive that equilibrium. But it's not going to be a good situation. So you can take advantage of this. I'll show you an olefination procedure that's based on this instability. And so why is it that, so in other words, if you put iodine with cyclohexene, what's the important point here? The important point is that it's not that there's no reaction. That would be a misstatement if we said there's no reaction. There is reaction. You do iodinate the double bond. It's that the equilibrium is poor. So in other words, it's not that you're not forming this iodide. It's that when you do, it kicks the iodine right back out, exactly the same mechanism you drew for forming it, it undergoes the reverse. So this is an equilibrium problem. So how do I draw this out? The equilibrium here favors the starting material side because of entropy. Okay, so if you take any kind of a diol and you don't want to diol, and I'm going to give you a very specific example here, the substituents on this system don't matter. I'm just giving you a literature example. Now you don't need to know the mechanism for this replacement reaction, but there's a reaction called the appell reaction. We'll cover that when we get to phosphorus. Where you take phosphine, triphenolphosphine is typical. You take halogens like bromine or chlorine and iodine, and you can replace these OHs with halogens. This is much better than using acid like HI or HBR. This special combination of conditions allows you to replace hydroxyl groups with halogens. But in this procedure, you don't get the halogen in place of the OHs. You get a double bond. And so what's happening here is you're undergoing something called an appell reaction. Let me try to redraw this starting material here. And again, we'll talk about this later. I think it's called appell instead of apple, but you get the idea. And so essentially, you get the equivalent of replacement of those OHs with iodines. And then of course, as I mentioned before, those are unstable. Iodine is a good leaving group. The lone pairs on iodine are good at pushing the other one out, and so that goes right back to the olefin. So it spits out iodine as a byproduct in that reaction. Okay, so bromination and chlorination of double bonds is thermodynamically favorable. Iodination is kinetically a good reaction. It's just thermodynamically unfavorable. Okay, so you don't always have to add the same halogen that acts as the electrophile as the nucleophile. So here's a reaction that's related to halo-hydrin formation where you form a cyclic halonium ion like this bromonium ion here. You don't always have to have bromine act as your nucleophile. If you do these reactions in water, hopefully you know this already, water will come in and attack the bromonium ion. And so you get an OH group anti to the bromide. And it doesn't have to be an OH group. So here's an example where if you have a nucleophile that's poised nearby like this carbonyl lone pair and you form an Iodonium, a cyclic Iodonium intermediate, you can get oxygen atoms to act as nucleophiles in these reactions. And so this would be an example where you form initially a three-membered ring Iodonium intermediate, iodine adds here. But before an I- can come back in and attack, this carbonyl snaps shut to make a five-membered ring. And I'm not going to show the whole mechanism for that. But this turns out to be stereoselective. There's two non-equal faces to this Iodonium ion. You can form an Iodonium on one face or the other face, one face or the other face reversibly. But then eventually this will cyclize preferably to make this ring system stereoselectively. Okay, so let's talk about another disturbing type of electrophilic addition that relates to this idea that there's this other, there's this pi complex that I haven't been showing you or that we don't draw when we draw arrow-pushing mechanisms. And I don't think I have such a problem thinking about this pi complex. Let me draw it out in case you forgot this from the last lecture. I don't want us to draw this when we do arrow-pushing, but I just want you to remember that when electrophiles come close to double bonds, they can form complexes. And those complexes, you know, forming them has a transition state, disrupting them has a transition state. But because we can't push arrows with these dative-type bonds, I would prefer that we don't draw them, but I don't want you to forget those types of dative bonds. So now let's come to what seems like a simple reaction, but maybe is not as simple as we think. And that is the addition of acids across double bonds, hydrochloric acid, hydrobromic acid, even hydronium ions. And I'm going to remind you of an empirical rule for these types of additions called Markonikov's rule. So Markonikov's rule is really just an empirical rule. I'd prefer that you understood the underlying mechanistic basis for this rule that allows you to predict the regiochemistry for HX addition. And so the idea is that whenever you add electrophiles across double bonds, that you end up getting the product with the heteroatom nucleophile, and I'll draw the H that initially got added. See that H there? I'll just circle it so you can see the H that was part of our electrophile. So Markonikov's rule says that the heteroatom, the nucleophilic heteroatom, ends up attached to the more substituted position. So I'll just write this here on more substituted carbon. And the H ends up added to the less. So I hope you remember that rule. But more better than that, I would prefer that you understood why this rule works. The rule works because when you initially protonate this pi bond with either HBR or HCl or some sort of acid source, you would prefer to protonate so that you get the more stable carbocation. That's supposed to be the basis for this rule. And that is the basis for this rule. You protonate so that you leave behind the more stable carbocation. And then the nucleophile comes in and attacks and I won't show you that arrow pushing mechanism. That's the idea behind Markonikov's rule. But let me show you some, the results of some isotopic reaction studies. Or instead of adding HBR or HCl, you add the deuterium analog. And that allows you to track where the proton is going. And in particular the stereochemistry. So let me start off by drawing a substrate. And I'll draw this sort of cis-olefin substrate. And I'm going to have it go in two different directions. So we can track two types of reactions. So let's try tracking a reaction in this direction where we're adding deuterium bromide instead of HBR. And so the deuterium is going to add to one side and the bromine is going to add to other. So if R is a phenyl group that stabilizes a carbocation, our prediction should be that if the deuterium adds from, for example, the bottom face, you'll get a carbocation next to the phenyl ring. You'll end up with a benzylic carbocation here. And then the bromide can attack from either face of that carbocation. That would be your expectation. That you would protonate the double bond to leave a stabilized benzylic carbocation. And then there's two faces to that carbocation. Bromide doesn't care which face it's adding to. But if R is not a good carbocation stabilizing group, and here's something that may seem a little bit bothersome. So I'll just imagine when R is equal to methyl. If you protonate from the bottom side, it's equally probable to protonate from top or bottom. I'm just going to draw the product of protonating on the bottom. It turns out that you get mostly anti-addition. If you're postulating that there's some free carbocation there and rotation around this bond, then something's wrong here. Right? Something has to be wrong here. Let me give you another piece of information. So I'll just write down that the observation is that under most conditions, you get stereospecific anti-addition. It's not 100% anti, but it's mostly anti. Addition of the D and the Br. So here is another observation. I'll just draw a line here. So you take some sort of an alkene and you add HCl to this. And this in particular was done in nitromethane. So I'm not going to, I won't claim that these kinetic results are the same under every set of conditions. Nitromethane is very poor at hydrogen bonding, but it's very polar. And I want to draw the product out for you. It undergoes Markovnikov addition. So the H and the Cl are on different carbons and the chlorine is attached to the more substituted carbon. That's exactly what you expect. And if you were to draw out some sort of a mechanism, let me draw that bond there. You draw some, I would draw something like this, right? That makes total sense. You have a carbocation, a stabilized tertiary carbocation. And then the chloride comes in and attacks that from the opposite face. And as with most reactions that make carbocations, it's formation of the carbocation that's the rate determining slow step. It's easy to attack a tertiary carbocation, but if this reaction is faster or slower, it's because of that first step. But somebody, some crazy person went and measured the rate dependence. And here's what they found. Let me just write equals here. The rate depends on the concentration of the olefin. 10 times more olefin, you form product 10 times faster. Well, that makes sense. 10 times more olefin, I'll form the product 10 times more faster. There's also a dependence on the acid here, on the HCl concentration. 10 times more HCl, it turns out the reaction's 100 times faster. Now that does not make sense. There's a second order dependence on the concentration of HCl. And if this is the mechanism, then that's not the rate equation, right? My equation here says if I have 10 times more, or my mechanism says if I have 10 times more hydrochloric acid, then it ought to collide 10 times faster and form this 10 times faster. Why would it be 100 times faster? Something is wrong with this mechanism. We instantly know this mechanism can't be correct. You need to have some sort of an explanation for why this thing might be depend on with second order dependence on the HCl concentration. So in order to write out a mechanism for this, you have to have two molecules of HCl in here. Now I'm going to draw a kind of picture that's most consistent with what's going on in this reaction mechanism. And I'm going to have one of these. I'm trying to draw this sort of dashed pi bond here. Remember one of the things we said about proton transfers, the eigenmechanism for proton transfers starts off with a hydrogen bond. That's what eigen showed. And if you want to show what's going on here and follow that eigenmechanism, then you need to draw some hydrogen bond or something we can't push arrows with. And have it bonded to the hydrogen. So if you want to have some sort of an explanation here for what's going on, you need to draw something that has two HCl's in it. And you can have this chloride still bonded to an H, doesn't matter. Now that's the kind of mechanism that will explain second order dependence. It's like if you cut the concentration of HCl by a factor of 10, it hurts you both ways. It hurts you because there's less nucleophile and it hurts you because there's less proton on the top face. This is probably a better description in nitromethane of what's going on in this mechanism. Okay, but when we draw mechanisms, so I've made all this hullabaloo about the actual mechanism, I'd like you to continue to draw the simple mechanism, okay? But someday when somebody says, well, Van Branken said, okay, just remember that and be ready to fling it out when you're talking over coffee, but draw your arrow pushing like this. But never forget this sort of picture of pi-complexation that we saw with halogens and even for HCl. Okay, so let me bring you back to a series of substrates and let's talk about reactivity. You're going to see a very similar pattern of reactivity when we look at various types of substitutions on olefins. So how do the substitution patterns on this affect the rate at which you add HX? It's very similar to the kind of pattern that we saw for addition of halogens like bromine and chlorine, Br2 and Cl2. So let's go ahead and start off by asking, what would happen if I put a single substituent on here, a single CH3? There's CH bonds on this methyl. I should have drawn it with CH bonds. And those CH bonds are donating into this pi system and making it more nucleophilic. Or if you like this sort of sophomore level picture, you could say, well, it stabilizes the carbocation and by Hammond's posh, whatever. The bottom line is just a single methyl with those CHs accelerates rates of HX addition. So if you look at the relative rates for addition of, and I don't even know what acid I, sorry, I don't have the, I don't know which acid this is for, whether it's HCl or HBr. So when you're adding HCl or HBr across these double bonds, it's 10 to the 7th faster, 10 million times faster to have that methyl group on the double bond. You may look at that and think, wow. When you have a double bond at the end of an olefin, that's so reactive. Well, it's not that reactive. So if you put two methyl groups, well, it depends on how you put the two methyl groups. And it doesn't matter whether this is cis or trans. If I put a second methyl group, not on the same side but on the other side, this is about the same rate. It's about 60% faster. But the problem is that if I think about how this CH is donating, when this CH donates, it's going to make it more nucleophilic here and more basic at that carbon, right? If you draw the carbocation, you would generate by protonating here, it would be next door to this. But the problem is these CHs are not activating the same carbon, and so that's why you don't get a significant increase when the two methyl groups are on a different position. If you come over and you put a benzylic position on here by having a benzene ring right next to this carbon, your rates of HX addition will become significantly faster. So this is now a benzene ring. A benzene ring makes that 100 times faster than just a simple alkyl group like methyl. So you sort of know that carbocation forms more quickly when they're benzylic. And neither of these is as good, and now I'm going to draw these CHs on here, as when I have two alkyl groups. This goes back to the idea that a tertiary carbocation is more stable than a benzylic carbocation. It's related. This is now a thousand times faster to have these two methyl groups on the same carbon as the benzene ring. And it's 10 to the 5th faster than when the two methyls are on the same carbon. So again, if you want to understand why this is true and you want to sort of come back to principles that you learned on sophomore organic chemistry, draw the carbocation you would generate and you can see that, yeah, that's a good carbocation. It matters where those methyl groups are on the double bond. So if you ever want to do selective additions to double bonds, if you've got some big substrate molecule with five double bonds in there, you can get significant selectivity just based on the substitution pattern of your double bond. And of course, if you put a second methyl group on here, it doesn't really have that much of a difference in terms of rates. These are pretty similar to each other. Putting that second methyl group doesn't buy you a lot. Okay, so HX addition, you know, all of these sort of start with this pi complexation under typical conditions. You don't have to draw that. Just draw regular arrow pushing mechanisms that generate carbocations and we'll be able to explain a lot. Okay, so let's switch off of protons as electrophiles. And we're going to start off by looking at peroxidations with peroxy acids. So I'm going to try to draw a double bond substrate as best as I can edge on here. So if we were looking at the edge of some sort of an olefin and I've got this particular olefin is trans substituted, so E configuration. If we come in with a peroxy acid, so a peroxy acid is not like a regular carboxylic acid, immediately you can see that there's an OO bond in there, a weak OO bond. There's going to be some sort of OO bond cleavage in this reaction. I can't think of that many reactions that involve peroxides where the peroxide bond doesn't break. Okay, so when you look at reactions, epoxidation reactions, they're stereospecific. So when you form epoxides, those reactions are stereospecific when you use peroxy acids or peroxy type substrates. So if the two R groups start off anti to each other in the starting material, they end up anti to each other in the product. And I hope that that's not new information. Okay, here's what we're going to do for the, let me give you the, one of the most common reagents, very cheap. It's metachloroperoxybenzoic acid. They used to sell this in relatively pure form until they found that in very rare cases the stuff can go off. I've seen jars of MCPBA with blue flames shooting out the top, you don't want to see that. So now they ship this only 35% pure. The other 65% is just the carboxylic acid. So the reagent you buy is not pure and it's got all this acid in there. Whenever you do these reactions, it's important to remember the conditions are acidic unless you've pre-purified the MCPBA. So when we draw these out, we're going to draw a transition state for this. And once again, I'm trying to draw an olefin edge on. How would I envision the transition state for delivery of a single oxygen atom? And it's going to look really bizarre and weird. Here's the transition state, looks something like this. And so we're going to struggle to figure out how do we draw this with an arrow pushing mechanism? And I need to draw partial bonds for the bonds that are breaking. So here's a set of bonds that are involved in the transition state. The MCPBA comes in and delivers not the oxygen that's closest to the carbonyl, that stays on there. It's the oxygen that starts off with the H on it that gets delivered. And you can tell that some nucleophile is going to come in and break that oxygen-oxygen bond in this transition state. And we'll show you a way to draw this that doesn't drive us insane. Because if you try to draw too many arrows in one step, it'll drive you crazy. I want to start off by looking at the rates for epoxidation. They're not exactly like this reaction. So in these types of reactions, we tend to draw carbocation intermediates. There's no carbocation intermediate in an epoxidation reaction. There is no carbocation. And so we should expect this order, or at least the quantitative order of reactivity won't be exactly the same, but there will be similarities. So for reasons I don't know, I'm going to follow my notes. For reasons I don't understand, I drew this in the opposite order with the most reactive olefin first. So I'm going to show you relative rates for epoxidation with metachloroperoxybenzoic acid. And as before, the trisubstituted olefins are most reactive. Even though there's no carbocation intermediate, trisubstituted olefins are more reactive than disubstituted, which are more reactive than monosubstituted. So MCPBA is an electrophile. Trisubstituted olefins are more reactive. Why is this more reactive? Well, there's CH bonds here that are donating into the pi system and making it more reactive. And the more neighboring bonds you have, next to that pi system, the more they're going to up the energy of the HOMO. Okay, if I have a benzene ring or a diene, it doesn't have to be a benzene ring. Any sort of pi system that I put next door that can donate into that double bond will increase the HOMO energy of that double bond. And so that makes that more reactive than just a simple disubstituted olefin. If I come back here to just a simple internal alkene, again, there's no carbocation intermediates in epoxidation reactions. And then, of course, this is vast. I don't have a number for a terminal olefin, but it wouldn't be as good. Just simple ethylene is 50 times slower. This just gives you a sense for the importance of having these bonds nearby. These CH bonds or C-alkyl bonds, simple ethylene that has nothing on here that can donate into that pi system and activate it, means that that's not a very reactive olefin. And we can deactivate olefins. So in other words, let's try to, what could I do to this one to make it less reactive? What I could do is I could put an acceptor here, a pi acceptor, like a carbonyl group. So in other words, instead of reacting like this simple styrene derivative, synamic acid is now even less reactive than ethylene itself. So in other words, if you have a substrate that has an enone and just regular alkenes, you will have no problem selectively epoxidizing a regular alkene and leave the enone alone. Enones are not very reactive. Yes, because the carbonyl is electron withdrawing. I mean, let me draw this out in terms of, before I go on, because I don't want this to be vague in any way. So usually when I'm drawing arrow-pushing mechanisms, I'm not drawing MO diagrams. The MO diagrams are just something you could fall back on if you're looking to explain things. So how do you explain the reactivity of some sort of a pi-CC double bond, right? It's got a pair of electrons, CC double bonds have a pair of electrons. The higher an energy that is, the more nucleophilic the olefin is. The lower an energy that is, the less nucleophilic the olefin is. So if I have some unfilled orbital next door, so let's suppose I have some molecule and I conjugate on a carbonyl, well, what I'm saying really is that if there's a carbonyl next door, if there's this empty orbital, a pi star CO, then instead of my double bonds attacking the nucleophile, they're going to be attacking that carbonyl. And if I draw that new, if I draw this new molecular orbital out that shows what happens when I mix these orbitals together, you can graphically see what's the effect of the carbonyl, right? That's sort of the graphical depiction that allows us to see the energetic consequences of having the double bond attacking the carbonyl instead of attacking your electrophile. That's a less nucleophilic, so the HOMO for this overall system now, I'm not sure how I would describe this, I'll get it this long pi C double bond CC double bond O. So now the HOMO for this system ends up lower in energy, it's just a less nucleophilic, and you don't have to draw this MO diagram to explain that. Hopefully, when you push arrows, you just, you can see the carbonyl there and have an intuition that the carbonyl is kind of sucking the life out of that olefin, making it less nucleophilic. Okay, so how are we going to push arrows for this very complex reaction here? And here's the way I would like you to do it. I'm going to redraw the peroxy acid like this. Here's a picture of a peroxy acid, and I'm going to draw it hovering right above the double bond that gets epoxidized, and I don't want you to just start drawing arrows because you end up with 15,000 arrows on there, and it'll look so strange and crazy that you don't know what's going on. In peroxy acids, there is an internal hydrogen bond in here. Peroxy acids have an internal hydrogen bond. That means this H is going, it's vibrating back and forth between these two oxygen atoms. And so I'm going to start this mechanism by simply drawing the H, transfer it over. There's a fast vibration there where the H goes back and forth, and so I'm simply going to transfer the proton there. And as soon as I do that, now it becomes much easier to see how I can deliver a single oxygen atom in a concerted reaction. Now that I've broken that H off, now it's much easier for me to see how I'm going to deliver one oxygen atom. The thing that's going to look crazy to you is my electrophilic oxygen in this particular structure now has a negative charge, and that's going to throw you off. So let me go ahead and draw this initial H atom, the proton transfer. Okay, so don't worry about the O minus here. That's still the electrophilic oxygen. Worry about the fact that there's this pathetically weak oxygen oxygen bond that is going to break. So I have a nucleophile, that's my olefin. So hopefully this won't be weird for you to think that, oh gee, my olefin can attack and break that oxygen-oxygen bond. So you may not like that O minus there, but look at this leaving group now. Look how much better is a leaving group that is with that positive charge. And how satisfying this is. It's just like bromination or chlorination at this point. You've got this leaving group, it leaves, but you don't want to leave a carbocation here. In fact, let me redraw this so I can obey the three-arrow rule. Let me draw a different resonance structure. I'll make my resonance structure look like this. Uh-oh, sorry, I hate it. I know you hate it when I, I'm going to draw another resonance structure of this that looks like this, and now maybe it's even easier to see what's going on. So here's another resonance structure we could draw where now it's even more satisfying to see that leaving group leave. So I'll do the same arrow pushing with this resonance structure. I'll have my double bond acts as a nucleophile. It breaks that weak, weak oxygen-oxygen bond. I'll give the electrons there. And now if I draw a lone pair on this O minus, I can see how it can simultaneously come in and attack any carbocation before it gets a chance to form. And so that's how you get sort of simultaneous bonding. And so you can see now, readily, from either of these, I'm worried about this mechanism here because I violated the three-arrow rule. I told you I wanted you to, so I'd like you to draw this resonance structure. And so now you can see how you get an epoxide and how your leaving group is going to be a carboxylic acid. So by the end of the reaction, even if you start with pure MCPBA with no carboxylic acid, you're generating carboxylic acid as the reaction proceeds. So unless you add some sort of a base, like bicarbonate is common, you're going to end up with acid in your reaction conditions. Okay, so how could you make that more reactive? How could you make MCPBA more reactive? Only when it's epoxidizing. So this is a long-known observation, known probably before we had a clear understanding of what's happening with peroxy acids. So if you have a lillic alcohols, and in this case I'll draw a stereochemistry here. When you epoxidize a lillic alcohols with peroxy acids like metachloroperoxybenzoic acid, you end up with syn-apoxidation, syn to the hydroxyl group. And that's important for stereoselective synthesis. So in this particular case, you get about 24 to 1 syn to anti, and the reason why this is true, I'll draw this, this is called the henbest transition state for, I'm going to draw this, this lillic alcohol. In other words, how do I think about what's the effect of an OH group? It could be that this oxygen lone pair or the two oxygen lone pairs, one of those two, is somehow hydrogen bonding with the H. Or it could be the H on the alcohol is somehow hydrogen bonding with one of the oxygens in the MCPBA. And it's this effect here that causes you to attack or prefer to attack from the same face as the alcohol. It turns out that what makes this react selectively from the top face is that when MCPBA hydrogen bonds to the peroxy acid reagent, you can imagine what will happen if you put any more partial positive charge on that oxygen that is the leaving group. At the same time that you're making this more delta plus, you're also making this more delta minus by having the hydrogen spend any time bonded to that other oxygen. And if this can act as a nucleophile and attack bonds here, it'll make this system more nucleophilic, it'll make the electrophile reagent more electrophilic, you get an acceleration from that effect. That hydrogen bond makes the electrophile more electrophilic and makes the nucleophile more nucleophilic. So the transition state, I'll call this the, both of these transition states that I drew a transition state somewhere, these are called the henvest transition state where you deliver this oxygen here to the double bond. So as you might expect, other hydrogen bonding groups like amides, NHs can also have this effect of directing epoxidation from the same face. You'll see the same sort of influence. Okay, so let's switch away from epoxidation. So that's all about epoxidation. Let's turn to another classical reaction of double bonds acting as nucleophiles. Also a concerted reaction like epoxidation. So if you add borane reagents and it doesn't matter what R is, these go with reasonably good regioselectivity. I'll discuss that in just a little bit. But what you see is the opposite of Markovnikoff additions. That's not bromine, that's a borane, that's a boron atom. So in the product, Br2 is not bromine. Okay, so when you add boranes, the H adds to the most substituted carbon and the heteroatom adds to the least substituted carbon. And so we call this anti-Markovnikoff addition. And I don't want to really dwell on that Markovnikoff sort of nomenclature. I just want to go straight to talking about the transition state. I don't know a simple way to introduce this, so I'm just going to start off straight with the complex picture here. So I'm going to draw a borane molecule and this is not perfect, so borane is planar. But borane, just like HBr, just like Br2, just like transition metals, can complex with pi bonds. So I hope that's kind of obvious that borane would desperately love to have an extra pair of electrons. Boranes is second row atom, it wants eight electrons. So whenever you mix boranes in with pi bonds, you're going to end up seeing some sort of a pi complex forming. But the problem with this pi complex, even though it's a real intermediate, is that we can't push arrows with that pi bond. So when I ask you to show me an arrow pushing mechanism, you don't have to show this pi complex. Let me draw out a transition state for this hydroboration reaction. In the transition state, let's just ignore my dative bond here, this dashed interaction bond. And imagine what would happen if I had this double bond attack the boron atom. If I take the electrons away from one of these carbons, like this one here, and I attack the boron, the boron will start to become more nucleophilic. And you'll start to develop a carbocation here at the end. And that means at the same time, one of these Hs will come back in and attack. A hydridogroup will come back in and attack. I'm going to draw the transition state here so you can see there's a four-membered ring in the transition state. This is the transition state for a hydroboration reaction. It's a four-membered ring transition state. There is no carbocation intermediate in the hydroboration reaction. It's a concerted transition state. And I'll put the double dagger there just so you know that that's not an intermediate. It's in transition state. So the prediction is, if this is true, and it is, is that you'll always get sin addition. And that is true. You don't get anti-addition like with bromination or addition of Hx double bonds. So how do we draw the arrow pushing for this? Because that's what I'd like you to do in this course. I'd like you to draw arrow pushing. We're going to just not draw the pi complex. And so this is the way I would like you to draw arrow pushing for these types of reactions. Just attack the boron. And at the same time that you're attacking the boron, don't leave a carbocation there. Use this Hb bond to come back in and attack one of the two carbon atoms. So the pi bond attacks the boron. And then this BH bond at the same time will simultaneously come back in and attack. Here's what's weird about this. It's not going to necessarily bother you until the next quarter. Your next quarter in Chem 202, if you take that course, you're going to cover paricyclic reactions like the Diels-Alder reaction, Ene reactions, sigmatropic rearrangements. And one of the first things you'll learn is that this is not possible. That this concerted 2 plus 2 reaction violates fundamental rules of orbital symmetry. And yet here's a reaction that is concerted. The key here is that this is concerted, but it doesn't have a particular property. And this won't mean anything to you now, but I'll give it to you to hold on to for later. This is not what we call a paricyclic reaction. It doesn't violate properties of orbital symmetry because it's not paricyclic. There are many concerted reactions that are not paricyclic. So later in Chem 202, I should put Xs through here. So later you're going to learn that these types of reactions are not possible as paricyclic reactions. And the reason why this can be concerted is because it's not paricyclic. Okay, so let's come back in and try to think about how do we think about this hydroboration reaction. And unfortunately, the best way to think about a hydroboration reaction is to draw it as though it's stepwise. So let me help you think about a stepwise hydroboration reaction. If I were to think about any sort of electrophilic addition to an alkene, I might want to draw a resonance structure. Maybe this will help us. And you might, it might offend you to draw this resonance structure because you look at that carbocation and you think, ah, one of those carbons doesn't have an octet. That's a crummy resonance structure. But just draw it for now. Not when you draw the arrow pushing, but just to help you think about the regiochemistry. Okay, so it sucks because one of the carbons doesn't have an octet. That's a sucky resonance structure. But now let's draw our hydroborating reagent. And it doesn't matter what X is. It can be an oxygen substituent. In this particular substrate, boron does not have an octet of electrons. And it wants an octet of electrons. Moreover, if I draw a periodic table here, boron, carbon, nitrogen, oxygen, and fluorine, does anybody know the electronegativity for carbon? It's the only one I know. I don't know any others. Well, maybe I know a couple others. But I can figure these others out. The electronegativity for nitrogen is three. Does anybody know what it is for oxygen? Oh, wait, what did I do here? 3.5 oxygen, 3.5. Does anybody know the electronegativity for fluorine? Four. It's monotonic. The only thing I'm changing, I'm not changing the size here. I'm just changing the number of protons in the nucleus. So when you come back to boron, carbon is more electronegative than boron. And a proton, that's the other electronegativity that I know, is 2.2. So when I look at this HB bond here, when I look at the HB bond, hydrogen is more electronegative than boron. So if I knew nothing but charge, even if I didn't understand the octet rule, I would tend to want to make this carbon atom attack the boron. So if we drew this out stepwise, it's not stepwise. If we drew it out stepwise and acknowledged that it's not stepwise, here's what we would see. As we come in and we attack this boron that doesn't have an octet, we'd end up with this intermediate where this nucleophilic borate, minus. And in that intermediate, all the bonds to boron are more nucleophilic. So it makes sense that if I start to develop this hydridoborate character, that this hydridoborate bond would come back in and attack the carbocation. So again, this is pure fiction. It's not stepwise like this, so remember it's not stepwise. But you can totally see why you get this regiochemistry. When we pretend, just momentarily that it's stepwise, why boron always attacks the less substituted position on simple alkenes. Okay, so let's go ahead and take a look at the regiochemistry for hydroboration reactions. Now that I've made all this big deal about this Markonikoff and regiochemistry. And I want to start off by drawing out, here's one source of borane that you might use very commonly. It's a stable reagent. It's borane, THF complex. That's the Lewis structure. It's very stable. Alternatively, you might use diborane. Here's the structure of diborane. If you put borane in some solvent like pentane, that boron is just desperate for another pair of electrons. And it doesn't get it from the alkane solvent. It'll get it from another boron molecule. In other words, you'll have this weird structure where this boron starts to steal electron density from other HB bonds. How do we draw this? We draw it like this. And it just looks crazy. If you wanted to draw some charges, you'd end up writing minus here, minus here, and plus, but don't worry about the charges. That's diborane. It dissociates to get boron. As soon as you put it into solution, it's in equilibrium dissociating. And it's the same with borane THF complex. It's a more stable form of borane because the boron is spending most of its time hugging the lone pairs of the THF. But for either of these in solution, there's an equilibrium where you're releasing free borane. And what I want to do is I want to draw a table here where I look at various different substrates. And I look at what happens when I hydroborate a double bond with borane derived from these reagents. If you squirt in borane THF complex or if you squirt in diborane, you're really just adding this is the real reagent that's doing the hydroboration. And I'm going to give you a series of different substrates so we can see how good is the regioselectivity. So what do I have here? I have a simple terminal olefin. One end is more substituted with an end butyl. The other end is less substituted. And then over here I have a styro group. In other words, I've got a benzene ring that's conjugated next to this double bond. So it's sort of a conjugation effect on top of a steric effect. And then I have just two alkyl groups on opposite sides. One's bulkier, one's less bulky. Isopropyl versus methyl. And so what do I see for regioselection when I use borane? Remember our rule said that the boron is going to end up attached to the less substituted carbon. And that's basically what you see for a simple terminal alkene. It's 94 percent, the products have, 94 percent of the products have boron on the less substituted carbon. It's not 100 percent. There's still 6 percent of this other regioisomer where boron is on the more substituted carbon. But that's still pretty good. 20 to 1 selectivity approximately. When I look at a styro group where there's an extra electronic effect there, it's not quite as good. Now it's 80, 20. You can get a significant amount of the boron attached next to the phenyl. And then when I come back and try to use sterics as my basis, really the electronics here are not that different for each of those individual carbons. It's almost one to one. Boron on one side versus boron on the other. Boron is not a very big reagent. So there's not a big difference. There's not a huge difference. There's some difference between the isopropyl and the methyl as far as boron is concerned. If you want high levels of regioselectivity, you can buy cheaply reagents that have a huge steric bias. And I'm going to draw one of them for you. This is called 9BBNH. I meant to draw this up here. So at the same level. So I'll erase that. Sorry when I erase. I just want to draw these here from my table. So this is called 9BBNH. 9 borobicyclononane. If I replace the boron here with carbon, let me try to draw. I'm not doing a good job of drawing my chair. But I'm going to draw two axial bonds here on my chair. There's one and there's the other. And so if this were all carbons, it would be nonane. There would be nine carbons. A bicyclic system with nine carbons. Bicyclononane. But it's got a boron in there. So it's borobicyclononane. But you could see, I mean, what's the deal with this? It's got a BH bond that's reactive and it's very sterically hindered. There's a huge steric, there's not an electronic change here. It's a steric change. This is why you typically see people use BBNH. 9BBN is what you usually refer to that as. Right? If you want good regioselectivity, you don't go get a bottle of borane. You get a bottle of 9BBN. It's greater than 99%. Better than 99 to 1 regioselectivity. It's 99 to 1 in the case of that styrene. And in a case of just an isopropyl versus a methyl group, it's again better than 99 to 1 selectivity. That's just a steric effect. So sterics can be, in this particular case, the steric effects are very powerful for distinguishing that isopropyl. Okay, I've run out of time here. I've got just two last things to cover when we come back on Wednesday that relate to boronates and stuff we've already covered before. But we're pretty close to done. When we come back on Wednesday, I'll just finish up some last things talking about these boron intermediates. And then we can start talking about using double bonds, CC double bonds to form carbon, carbon bonds. And that's 90% of the field of organic chemistry is that reaction.