 Okay. When we left off on Friday, we've been talking about alkali, metal, organometallics, alkylithiums, potentially alkyl-sodiums or alkyl-potasiums. You won't see those. But alkylithiums, you'll see a lot of. And we brought up this problem that alkali, metal, organometallics aggregate. They always aggregate in solution. They don't exist as monomers. Even in THF, which is a coordinating solvent, this demand of lithium for extra electrons causes it to do this weird stuff. So if you put butylithium in THF, it adopts these cube-shaped structures. That's the structure of butylithium in THF. And it's an equilibrium in THF with this dimeric form. So not a cube but a square. So each butylithium, each carbon atom here has not just four bonds but five bonds to the carbon atom. As long as it doesn't violate the octet rule, that's okay. So you can see what's happening because of this demand, the high nucleophilicity of this carbon-lithium bond and the demand of lithium to satisfy the octet rule. You get these weird aggregates. So this is not the exception. This is the rule. And so we're sort of stuck in this situation that we want to be truthful about our representations. And so what are we going to do? And so let me show you some heartening stuff for us as arrow pushers. I'll bring you back to a study of Andy Streitweiser that deals with the chemistry of enolates. So we're going to very soon get into the chemistry of pi bonds, of enolates, of azillion things that are very similar. And let me just point out that when you put a lithium enolate in THF, so you make lithium enolates with strong bases like lithium to isopropyl amide, you deprotonate at the alpha position. And how are we going to represent this kind of lithium enolate? Over 99% of this lithium enolate exists as a dimer in solution, kind of like the one I'm showing you over there. And so here's what is true. If you take this and you treat this with benzyl bromide, just to do an SN2 reaction where you displace the bromide using the lithium enolate. So I guess maybe I should draw out some sort of an arrow pushing mechanism, although that's not that important. You could draw out something like this, some sort of an SN2 mechanism. How are we going to represent this species, this lithium enolate, when in fact it exists over 99% in solution like this? So this is, if you want to really describe what this enolate looks like, you're going to end up stuck doing something like this. And I'm not going to draw the THF molecules on those lithiums. This is not a Lewis structure. These are dative bonds because right of oxygen has three bonds that have a positive charge. So these are dative bonds in here. I'm ignoring the charges. So this is what lithium enolates look like in THF. And these striatwizers showed that even though this is 99% of your lithium enolate in solution, that small amount of monomer that's floating around is actually what's doing the chemical reaction. In other words, if you want to draw the correct arrow pushing mechanism, you ought to be drawing the monomer in there. Even though that's not most of what this lithium enolate looks like, we're going to draw all of our organometallic reactions. Alkyl lithiums, lithium enolates, drawing the monomer. So never forget that these aggregates are present in your solution, but when you draw arrow pushing mechanisms, just draw the monomer and do the same kind of arrow pushing that you've been doing since sophomore organic chemistry. And that's going to be our rule for this class. And I hate to use this word ignore because I don't want you to ignore aggregation, but I'll say it. For arrow pushing, ignore aggregation. Just acknowledge that it's there mentally, but when you push your arrows, just draw the monomers. Draw monomers of you to lithium, draw monomers of your Grignard reagents, draw monomers of your lithium enolates. This is the way we're going to do our arrow pushing in this class because, and again, we find some justification from this important study of Schreitweiser showing that it's the monomers that are doing the business work there. Okay, so that's our rule for alkyl lithiums, for enolates. Just draw the monomer. Okay, so let's go ahead and cover the last little piece of our puzzle here for, and this is really not an alkali or organometallic, but I want to contrast alkyl lithiums and alkyl sodiums with one class of organometallic reagents. And that's not having lithium or sodium on the carbon atom, it's having copper. And so let's talk about two types of organo-copper reagents. So the first type we call just simple organo-copper compound, and it looks like this. So this would be a copper one with some alkyl group attached to it. Could be butyl, could be phenyl. So if it's phenyl, I'd call it phenyl-copper. And one characteristic of simple organo-copper reagents is they don't do much. You'd have to have something very reactive to get it to be worth your while to make this organo-copper reagent. So organo-coppers are not very interesting. And let me contrast that with a cuprate, which is very interesting and very reactive. Here's a cuprate, an organo-cuprate typically looks like this, and guess how you make the organo-cuprate? You make it the same way you make the organo-copper reagent. So an organo-copper reagent, you take copper bromide and alkyl lithium, you mix them together. But if you add a second equivalent of the alkyl lithium to your organo-copper reagent, now you can imagine how this alkyl can again add to copper. And so now you have two alkyl groups on copper. And these are very nucleophilic and not very basic. And if you were as interested in CC bond formation as I am, then that's an interesting set of properties, things that are very nucleophilic and not very basic. Okay, so let's just go ahead and talk about how to think about these. And I gave you a peek when I showed you that little slide show, the PowerPoint presentation on Friday. I gave you a little peek of what organo-copper reagents look like they're aggregates, worse even than these organo-lithium species. But we're going to draw these like this if you ever need to. So this little suffix ending here, 8, cuprate, tells me there's a negative charge on copper. That tells me that the bonds to copper are nucleophilic. So it doesn't actually look like this. It exists as an aggregate. But you can just pretend it looks something like this, where I've added a second alkyl group to my organo-copper reagents and now there's two alkyl groups on there. And so those bonds are nucleophilic on that organo-copper reagent. So we've seen other eight complexes like hydrido-boreate, hydrido-illuminate, siliconate, it's the same idea. You get this nucleophilicity out of the bond. So one property of cuprates, and I think I can sum this up in a simple statement, all the things that you wish organo-lithiums would do, this is what organo-copperates do. They do substitution reactions with alkyl halides. They efficiently open epoxides to make CC bonds, which alkylithiums do not do. They do 1, 4 conjugate addition reactions. They do SN2 prime reactions. All the arrow pushing mistakes that you make with organo-lithiums, these are the things that organo-copperates are good for. Let's take some examples of some of the things that you can do with cuprates. So let's suppose you wanted to make a carbon-carbon bond there. We have lots of techniques now that involve transition metals and palladium catalysis. But in the days before palladium catalysis, if you wanted to make a carbon-carbon bond here, you can't just do SN2 substitution at that center. Even with copper, you can't. But at least when you mix lithium-dithenyl-copperate in here, you get very efficient substitution. Don't worry about the mechanism. It's not SN2, but you get very efficient substitution of that bromide. And it doesn't have to be an SP2 center. It can be an alkyl halide if you wanted. So I'm just using the vinyl group here to emphasize it. That's something you couldn't do with other reagents, that kind of a substitution process. SN2 prime reactions. Cuprates are good for that. SN2 prime reactions are very unusual. But if you take cuprate reagents, so here I have an allylic bromide. So how do I form a bond to this end carbon and not directly here to this secondary position? If you add some sort of an alkoxide nucleophile, you're not going to get SN2 prime reactions out of that. They'll still add here or else do eliminations. So here, if you wanted to substitute this through an SN2 prime reaction where your nucleophile adds to the end, then you could just add a cuprate reagent. They prefer SN2 prime. And there's my butyl group adding to the end. You know, if I drew some sort of a mechanism for this, this would be completely fictional. But if you wanted to have some way to think about this with arrow pushing, I wouldn't count you wrong if you did an arrow pushing reaction, even if it's completely unrealistic. This will start off with some copper bound to the olefin. But if you did this, I wouldn't complain. Here's what I'm thinking when I say SN2 prime in case you've forgotten. What is SN2 prime? It's when you attack the double bond of an allylic leaving group system. Okay, so those are two examples of reactions that cuprates do. They open epoxides. And I'm not going to draw that, but this is one of those reactions they show you in sophomore organic chemistry when they're trying to teach you that SN2 prime is the best reaction in the world, cuprates are great at opening epoxides and forming C-C bonds and other types of organometallics or not. Okay, the last major class of reactions that cuprates are great at is one four addition reactions. So let me just help you contrast some patterns of reactivity here. If you take a simple enone and you add an alkyl lithium or even a very pure green yard reagent that has no contaminating copper in there. So I'll start off by adding an alkyl lithium. That carbon atom and alkyl lithium has a very significant amount of anionic of negative charge. If you look at the formal charge, there's a partial negative on the methyl carbon right there. And that makes it want to add here to the carbon. You get preferential addition for most simple organometallics to the carbonyl carbon. And so after the, well, I guess I don't need to say after the workup, I'll just draw it as the alkoxide. That's the preferred side of addition. Don't expect to see any one four. We call this one two addition because normally we number the carbons in this enone system. Let me get a red pen here just so we can see the numbering and we can remind you. So one term for this process is we call this one two addition to this enone system. We're changing the bonding to atoms one and two. We also call this conjugate addition. This is the other phrase for this. In contrast, if we add a cuprate to this, lithium dimethyl cuprate, now you see only the conjugate addition reaction here. This is a function of the fact that there isn't a, even though we draw it like this, there isn't a lot of negative charge on these carbons of the methyl group. They're very, those methyl carbon, sorry, methyl copper bonds are very nucleophilic, but they're not very basic. And I don't know why I'm doing this. That looks awful to me. So in this case, those methyl carbon bonds attack in a conjugate addition faction to give you this copper enolate. And the other copper stuff is, so what's the remainder that's floating around before I reconnect things? I've got methyl copper. Remember I said that's not very reactive. That's one of the byproducts. And I have this lithium floating around until I end up with recordination of this O minus to copper and lithium species that are floating around in solution. Okay, so fantastic reaction. It, I could draw an arrow pushing reaction where I simply make the methyl copper bond attack at carbon four. It's not really like that. In fact, the copper starts off bonded, pi bonded to the enone pi bond, the C-C pi bond. But you don't need to worry about that. Just if you want to do arrow pushing where you just take the methyl copper bond and attack, I'll accept that. Okay, so let's talk about one way to make those reactions even better. And this is more of a recipe issue than an arrow pushing strategy for you. And that is that trimethylsiloclide is known to accelerate these additions by copper reagents, specifically by cuprates. And so just to give you a sense for the effect here, it's normally very hard to do conjugate additions when you have this kind of disubstituted beta carbon here. But having those two substituents really slows down additions here. It's very hard. But if you add, take your standard cuprate addition, lithium-dibutyl cuprate, and you add in some TMS chloride to that, it phenomenally accelerates an otherwise slow reaction. So you can imagine that the silicon is pre-bonding to the carbonyl oxygen, making the system more electrophilic. And again, the precise mechanism, I don't want to go through that in any detail simply to say that generally when you see people doing, conjugate additions with any sort of steric challenges in there with cuprate reagents, there's usually TMS chloride in there in order to help you accelerate that. And the product is not the lithium enolate, you get something called a silil enol ether. And we'll talk more about silil enol ethers later, which have their own benefits. Okay, so the last sort of dramatic example of the effect of this combination of TMS chloride and cuprates is even enols, aldehydes are so reactive. We talked about the rates of addition of aldehydes versus ketones. We talked about the thermodynamic ease with which you can pyramidalize at aldehydes. And so it is very hard to keep nucleos files from attacking aldehydes and to get them to attack that beta position on an enone. And if you use this combination of TMS chloride and cuprate, you favor, you still favor addition to the beta carbon over addition to the aldehyde carbon. So that's phenomenal. It's phenomenal that you can get that kind of selectivity and not attack that aldehyde. Okay, I've got one last sort of not alkali organometallic here in this lecture that's called alkali organometallic. Just to get you, so I hope when you see a copper attached to carbon that you don't mistake that for the same reactivity as an alkyl lithium. That's been the point here. Don't mistake dialkyl cuprates as organolithiums. The reactivity is very different. And here's one last type of organometallic and that zinc that has none of these basicity issues of an organolithium reagent. And this particular example will convince you that the alkyl zinx are very different. They must be very different from alkyl lithiums. So I'm going to start off by showing you, it's very easy to make acetals of cyclopropanone. Cyclopropanone hates to exist as the ketone. You can't buy cyclopropanone, but you can buy acetals. They're very cheap and so you can even buy the monoethoxy adduct of cyclopropanone and it's very easy to sililate that OH. This is probably commercially available as well. Whenever I see a cyclopropane and I hope you're thinking is the same, I think gee, those bonds have so much p-character. I want to start using those bonds as nucleophiles. They're nucleophilic. There's lots of p-character in those bonds. And when I see lone pairs here on these oxygens, I'm thinking gee, those are donating here into this antibonding orbital. Let me make it end precisely on this antibond. Those lone pairs on these oxygen atoms are souping up the nucleophilicity of these two C-C bonds. So if I thought a cyclopropane bond was nucleophilic and I see an oxygen lone pair nearby, I ought to think that's super-duper nucleophilic. So when you treat this with zinc chloride, just buy some zinc chloride. You typically get it as a hydrate, dry it off, put it under vacuum and heat it. What you'll find is when you mix zinc chloride in there is that bond, that C-C bond that's strained in nucleophilic will attack the zinc chloride. And so you get this very weird-looking structure. And I'll start off by drawing it like drawing the initial intermediate. But this doesn't just sit there like this. One of the chloride pops off. Imagine the chloride popping off and then coming back in and displacing silicon. That's not a one-step process. It's three steps. Don't worry about the number of steps. Ultimately what you get is this very strange-looking reagent that at first glance looks okay. It's got this nice sort of chelate structure in there. You can imagine what I'm going to do with those carbonyls in just a second. But what's amazing about this structure is it looks kind of like an enolate, but it's not an enolate. There's an ACH2 group here. Let me just draw those H's so we can clearly see them. In other words, I've got nucleophilic bonds and I've got electrophilic carbonyls built into the same reagent. What is amazing is that these bonds are not so nucleophilic that they go around attacking other carbonyls. Alkyl lithiums attack esters and carbonyls. These carbon zinc bonds are so unreactive that they don't go around attacking the carbonyls in other molecules that are floating around. Right? You have more than one of these in solution. These carbon zinc bonds don't go around attacking the ester bonds in your carbonyl. You can isolate these types of reagents. So what are these alkyl zinc reagents great for? If you have superduper activated aldehydes, they'll attack them. If you have palladium complexes floating around, these bonds will attack the palladium complexes. And you can transfer those alkyl groups to palladium. That's where they tend to be useful. So this is just to convince you that these types of bonds are not at all nucleophilic and basic like an alkyl lithium. So much, much, much attenuated reactivity. It's extraordinary that you can have esters in the same pot as those alkyl zinc bonds. Okay. So that's all we're going to say about alkyl sodium, alkyl potassium, alkyl lithium. Dialkyl cuprates, alkyl zinc species. And you should be getting lots of practice using those as nucleophiles to attack things. So we're going to have a major switch in years now and start talking about pi bonds as nucleophiles. We've spent all kinds of time talking now about lone pairs as nucleophiles or bonds to lithium as nucleophiles. And so let's change what we're doing. So we're going to talk about adding electrophiles to carbon-carbon pi bonds or conversely, like I said, we're going to talk about pi bonds as nucleophiles. And these are the most important carbon-carbon bond forming reactions or the most important class of bonding reactions in the field of organic chemistry. Or maybe I should say organic synthesis. Okay. So we're just going to talk about over and over again, over and over and over. We're going to talk about that. That's what we're going to talk about. So I want to start off by helping us establish a picture, an MO picture of the pi bond or various types of pi bonds. And what I want to do is I want to contrast these types of pi systems. There's three different pi bonds. And we're going to see all kinds of reactions where these act potentially as nucleophiles. Here's what we're going to see a lot of. If I take some sort of an electrophile, like a carbocation, and I think about that addition, that explains electrophilic aromatic substitution and all those other, there's so many reactions that that explains. But what I want you to do is I want you to avoid doing this. And it doesn't matter what the electrophile is. It could be a proton, like an acid. It could be a carbocation. It could be aluminum. Right? That's not the most nucleophilic part and basic part of that molecule. We've said this over and over. When you react with carbonyls, I want you to use these. Right? That's what I want you to use. And those aren't the same. You might think they give you the same thing. Oh, they give me the same sort of resonance. But this pi bond is not as nucleophilic as those lone pairs on a carbonyl. So you use the lone pairs. And it's, if that were true for oxygen, it's doubly true for an amine. Right? That's the nucleophilic part of an amine. That's the basic part of an amine, not the pi bond. So use the lone pairs. Now, of course, with carbon, you don't have lone pairs on carbon, so with double bonds, you're going to use the pi bond. There's no choice with that. Okay, so just keep that in mind. Let's come back and talk about trajectory here. And I want to distinguish additions to carbonyls because we went into so much detail about the correct trajectories for addition to carbonyls. We said there was this thing called a Berge Dunnitz angle that honors the fact that pi star looks like this. Pi star has this sort of phasing picture that looks like this. And if I'm a nucleophile trying to overlap with this empty pi star orbital, if you're a nucleophile adding to some sort of pi star, and I'm going to arbitrarily phase my nucleophile to have some empty orbital like this that's unhashed, I want to minimize my interactions with these hashed lobes of pi star. Those are destructive bonding interactions. I want to tilt myself around so I don't have any interactions with these oppositely phased orbitals. And the only way to do that is to attack from this 109-degree angle, the Berge Dunnitz angle. But things are totally different when we add electrophiles to CC bonds. So if you're an electrophile, you're not adding to pi star, for an electrophile, it's opposite. This thing is the nucleophile. The CC bond is the nucleophile. And so if I draw this sort of orbital picture, I'll decompose this pi into the individual p orbitals. Now if I want to maximize the interaction of this pi bond, which is now the filled orbital in the nucleophile, and I'll have some sort of fictitious empty orbital here, now I do want to maximize the interactions between both of those lobes. The prediction is that when electrophiles attack pi bonds, they should approach from the middle. You simultaneously will get bonding from both of these carbon atoms to both of those carbon atoms. You can't avoid that. As close as you get to this carbon atom, you're also going to be somewhat close to that one. And both of those interactions are going to lead to bonding combinations. Okay, so there's a trajectory difference between adding electrophiles to pi bonds versus adding nucleophiles to pi bonds. So 109-degree angles here, whereas here you approach the middle of the double bond. That's the expectation. Okay, so let's skip for a brief moment back to an old idea. What I want to do is I want to contrast two different types of elimination geometries for you. And we'll take this sort of fictitious example here where I have, and it doesn't have to be a chloride leaving group. The leaving group really doesn't matter. It could be protonated water ready to leave. But if you contrast these two types of geometries for elimination of the chloride group, the one on the right-hand side, the one on the left-hand side, which one is better? Which one is going to eliminate chloride faster? If I think about where sigma star is located for this carbon-chlorine bond over here, I would expect this one over here on the left-hand side to be faster than this one over here on the right-hand side. So I hope you already know that anti-eliminations are favored over sin-type eliminations. So what we're really saying is that this has a lower energy transition state. The transition state in order to eliminate here is lower in energy than this other transition state. So this transition state is lower in energy to give you that. But what we're also saying, right, even though I could do in theory something like this and get to something that looks very similar or I guess identical, what I'm really actually saying is that if we think about the opposite reaction, if I think about nucleophiles adding to pi star, that transition state, I'm also saying that that transition state is lower in energy. Either the transition state is lower in energy or it's not. That means for both the forward and the backward reaction, this transition state is lower in energy than the one where I attack like this. In other words, when I attack with some sort of a nucleophile on a pi star, that lone pair that appears on the neighboring atom doesn't appear sin to the nucleophile. It appears anti to the nucleophile. It's the same transition state for pushing out a leaving group or for adding. And so here's the big point that I'm making right here. Here's the take home lesson for this that's important. Whenever I have some sort of a double bond and I have a nucleophile that's poised nearby and to make this interesting I'll make this a metal. Let's make this metal carbon bond to the nucleophile. If the metal carbon bond is on top of this pi system and attacking it from the top, what that means is that this pi bond will be nucleophilic on the bottom face. That's a consequence of this stereoelectronic relationship that I showed you above. If I have nucleophiles attacking pi bonds from the top face, that nucleophilicity will be realized and this is really just a resonance structure. Well, that nucleophilicity will be realized on the bottom face. That's the important point. And it's all related to this iminium ion thing that I showed you up above. Okay, so let's take some examples of that. So how does that become useful in organic synthesis or of any consequence? So let me go ahead and give you an example. I'm going to try to draw out as best I can some sort of a cyclohexenol system edge on because I want to see both faces of that, the top and the bottom face of this cyclohexene. And of course, it doesn't make a difference between what is the top face of the bottom face unless I have some substituent there. So just to help me keep this straight, I'm going to put a methyl group on the top face of this cyclohexene. So now if a reaction occurs, I can tell, well, it occurred either on the same face as the methyl or the opposite face of the methyl. And of course, alkenes aren't very nucleophilic. So let me soup this up. Let me put some nucleophilic bond here that's long and nucleophilic. Let's make that tin. There's that long nucleophilic bond. You should now not be surprised after I went through this big hoopla over here about the fact that if I protonate that bond or react it with any kind of an electrophile, let's suppose I protonate that bond. Of course, I can't tell where I'm protonating if I use protons, so let me make this deuterium. So now if I protonate that bond, my expectation should be that if this is donating from the top face and I'm not going to actually move those electrons, but if this tin carbon bond is really nucleophilic and is really souping up, if it's really donating into that pi system and souping up the basicity and nucleophilicity of the pi bond, it ought to be happening more on the bottom face opposite this carbon tin bond than on the top face. And that's exactly true. When you look at deuteration of this exact substrate, what you find is that the deuterium gets picked up on the bottom face. And now I'm just going to draw a D instead of the sort of isotopic symbol. So methyl is on the top, deuterium is on the bottom, and you get this phenomenally stable beta stanyl carbocation with this long nucleophilic tin bond there, and then this acetate is floating around looking for something to attack. And gee, right there is that tin. Now it's not an SN2 substitution, but that acetate is going to attack that tin and pop it off. And so the final compound that you get out of this is that the deuterium is going to be on the opposite face of the methyl. This is really powerful, useful stuff. And we're going to see now a zillion reactions of allyl stananes and allylsilanes and allyl, any metal you want where there's this facial selectivity, the metals on one face, the electrophiles on the opposite face of the pi bond. So just keep that in mind. Okay, here's another example of exactly, so maybe you don't care about protonation reactions. Who cares about that? Protonation is a very interesting process. I would agree with that. So let's extrapolate this a little bit. Let's suppose we take an allylsilane and you had some magical way to make this optically pure. I mean it doesn't have to be magic. And now what you're really interested in is a chemist is C-C bond formation. And so here's my electrophile. You take an epoxide, a simple oxurane, and you coordinate to that titanium tetrachloride. So that now that oxurane is very electrophilic and very easy to attack. So when you see attack, your expectation should be is that the electrophile should come in from the bottom face opposite the silicon. And that's true. That's what you see. Here's the intermediate that you get. And here's this carbocation. I love that carbocation. It's got beta-silicon, silicon carbon bond next door, donating into the empty P orbital and stabilizing it. And I just made a new C-C bond on the bottom face of that system. It's the bottom face of that olefin that attacked the epoxide and opened it up. And I'm not going to go through all the steps here to exchange, but ultimately after you do the workup, what you get is this new carbon-carbon bond on the bottom face, opposite where the silicon was. Okay, so keep that feature in mind. When we have nucleophilic bonds next to pi systems, they're going to make those pi systems more nucleophilic than normal. And there's a facial selectivity to that. So keep that in mind. Okay, so now let's come back to some simple mechanisms that I kind of hope you already know, but maybe not in the detail that I'm going to show you. So at some point during about your fourth or fifth month in sophomore organic chemistry, you started learning about a zillion reactions where you add electrophiles to C-C bonds. Halogens like Br2 or Cl2 or HCl. And so let's come back and look at those reactions that I hope you already know. And so we'll start off by, I'm going to try to draw an alkene here. And I'm going to try to draw this alkene edge on so that we can clearly see that the alkene has two different faces associated with it. So I'm trying to envision an alkene edge on. So now there's a top face and a bottom face to that. And so if I add bromine to this, Br2, you'll end up making the dibromide, the visceral 1, 2 dibromide. And we know that these reactions involve bromonium ions. What does bromonium mean? Bromonium means a bromine that has two bonds. This one just happens to be a three-membered ring bromonium ion. You know, if you just had two bonds to bromine, it would be a bromonium ion. And this bromonium just happens to be strained in cyclic. So these reactions go through this cyclic bromonium ion, this strained cyclic bromonium ion. I'll just write bromonium here. Two bonds to bromine. So how would we push arrows for this? The way you push arrows for this is that we have to show that it's the pi bond that's attacking this BRBR bond. That's a weak bond. It's easy to attack. It's got a low-lying sigma star that makes it easy to attack. The challenge is that as you take these electrons away from one of these two carbon atoms, you're going to leave a carbocation. And that carbocation is not going to be happy. If you start taking away electrons from, say, this carbon, well, isn't there a lone pair that's right here? In fact, three lone pairs that are poised and ready to start adding into that carbocation as we develop it. So it shouldn't be surprising to you that as you're attacking this, you could come back in with these lone pairs and make a second bond. And so that's, this is the traditional way that we depict these, these bromonium ions. And so if we have some sort of a bromide here that's left over, the second part of this mechanism is to come back in and attack on either side. That would give you one enantiomer versus the other. If I attack this side, I'll get one enantiomer. If I attack the other side, I'll get the other enantiomer as long as those R groups are the same. But the bottom line is that bromination should give you an anti, an overall anti-addition of bromine where the two bromine atoms end up getting added from the opposite face. So the two bromides end up anti to each other. And of course, if the molecule can spin around that central bond, then it's, you know, it's, you know, it's only when you draw it like that that you can see the anti-relationship. Here's a disturbing piece of information that is actually disturbing. So this is not the correct mechanism in reality. But this is how I would like you to draw it. I'd like you to draw it the way you've been drawing it since day one. But in case anybody ever asks you, gee, didn't anybody ever teach you the correct mechanism for bromination? At least you can say you've seen it somewhere. It turns out that in reality, bromine forms a pi complex with olefins before there's any carbon-bromine bond formation. It turns out there's an initial complexation. And the problem with this is we don't have good Lewis structures to depict these. We have to draw things like dative bonds or sometimes people draw these dashed bonds. So you get this pi complex, sometimes, I'll say also known as a k a, a charge transfer complex. And so what is this charge transfer complex? A charge transfer complex basically means that the pi bond is donating into this bromine-bromine bond and it just sits there with this pi bond donating into that sigma star orbital for the bromine-bromine bond. And there's no, there's no dissociation of bromide initially. Initially they just sit here like this. And then this bromide gets pushed out. So there's a pre-coordination. And we're not going to draw that. We're going to continue drawing this exactly the way that you've drawn it before. Someday you're going to run into some reaction that you can't explain and I hope you'll remember this picture of bromination. So keep drawing halogenation reactions with Cl2 and Br2 in the same way that you've been drawing them. Just use these arrows here to push this. Don't draw it as two steps. But if you ever need to, you can fall back on that picture for bromination or chlorination of double bonds. So if you look at rates of bromination of alkenes and methanol, let's go ahead and talk about how the structure of the alkene influences the reactivity towards electrophiles. So I'm going to draw a series of alkenes here and I'll start with ethylene gas, pathetically unreactive olefin, awful, awful, on low reactivity olefin. There's no substituents on there. If I put even a single alkyl group on there, I got a 100-fold rate acceleration just by having a simple alkyl group. And if you want some way, I'm not going to go back through and explain this to you, but it's exactly the same as that carbon silo, well, it's the same type of idea as that carbon silicon bond or the carbon tin bond. These bonds, these proton carbon bonds are donating into pi star, they are souping up the nucleophilicity of that pi bond. So that's the effect of having these bonds nearby near that olefin is they soup up this pi bond nucleophilicity. And you don't have those bonds on ethylene. Of course, when I look at a terminal olefin, a C-C double bond at the end of an alkyl chain, I'm thinking that looks unreactive to me. So these are just relative rates of bromination. How fast do these pi bonds attack bromine? This terminal olefin looks unreactive to me. When I think of reactive alkenes, I think of internal olefins that have two alkyl substituents on them. That's a more typical type of substrate that you'll use. And this is another factor of 10, factor of 20 actually closer to a factor of 20 in reactivity. So having two alkyl groups is even more reactive. Just moving the double bond to the middle of butene makes it that much more reactive. And if I rearrange my butene, still four carbon atoms in the right way, now I get even higher reactivity. And you can have all kinds of explanations for this. The way to think about this arrangement, so these both have four carbons and look at the reactivity here, is that here if I think about what this H-carbon bond is doing is it's kind of souping up the reactivity over here. And there's a CH bond over, or three of them. There's CH bonds over here that are souping up the other carbon reactivity. But you'll get higher reactivity if both of these substituents are souping up the same carbon atom. Right? If you want to draw some kind of a resonance structure to help you see that both of these are cooperatively accelerating the nucleophilicity of this carbon atom at the end. And so you see this higher rate of addition. And then the fastest of all these would be a trisubstituted olefin. And the important point here is that all olefins are not equally reactive. And if you've got some complex substrate and you want to do a selective epoxidation of an internal olefin versus a terminal olefin, you will probably be able to get a high yield out of that. So different levels of nucleophilicity when you add electrophiles to olefins depending on the substituents that are on the olefin. Okay, let's go ahead and talk about the intermediates that are involved here. So this idea of brominating a double bond or. So when you draw some sort of a reaction like this, how do I know it's not like this? What's this cyclic bromonium business? Why can't I just make a free carbocation like that? How do I know whether it's going through this cyclic bromonium ion business or whether I'm simply attacking and having a free carbocation? Well, we can check on this. I'll show you how you check. What you do is you look at the percentage of anti. If it really is going through this kind of cyclic bromonium ion, the two bromines will always add anti to each other because the leaving group is on the opposite face of where the nucleophile is attacking. So I'm going to give you a series of olefins and we can see how these substituents affect whether you're going through the cyclic bromonium ion. So in this simple internal olefin butene, you get 100% anti product out of this, anti addition of the two bromides. That tells you that 100% of that was going through the cyclic bromonium ion. But if we put a carbocation stabilizing group, we can tip that balance, right? If we stabilize this enough, we can start to see some carbocation formation. Some of it goes through this cyclic bromonium ion. Some of it goes through the free carbocation. So remember, these are the two choices. Are these lone pairs coming back in and attacking the new carbocation as the bromine adds? Or is the double bond simply attacking the bromine and leaving a free carbocation? And so when we have a phenyl group here as one of our substituents, we see only 83% anti addition. In other words, some of this is going through the free carbocation like this. And if we really substitute, if we add a second substituent to really help stabilize this carbocation at that center, now we see only 68% anti addition. So 68% of it, or at least a significant fraction of this, is still going through the cyclic bromonium ion. Even though we could generate, in theory, a very stabilized carbocation, we're still getting a very significant amount of that cyclic bromonium ion formation. So cyclic bromonium ion formation is the rule. Even when you can form stabilized carbocations. Okay, I'm going to go ahead and stop right there. And we'll come back and say, I feel like I'm going to say almost all the same things when we add HBR and HCl and acids to double bonds.