Lecture 18: Radical Reactions
King, Susan
This is the second quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. 
 So what it's looking like, what it's looking like, we'll see how it goes. So everybody knows we have a midterm coming up. It looks like it's going to be through Chapter 15 for the midterm. This is going to be a really hard midterm. And so, you know, a lot of synthesis. So last year, we had spec plus this stuff. And so last year, we had spec plus this stuff. And so I couldn't ask as much synthesis because spec takes time, right? There's no spec on this test. So that means there's going to be more synthesis, right? So what you want to do is you want to do as much synthesis as you can. So do the sapling synthesis. Come straight at the end of each chapter. There's a synthesis section where it mixes reactions from all the chapters. You want to do that. Those are the things you want practice and synthesis. Okay? Questions? Anybody? Before we get started. All right. So Chapter 15, really different. So we had a little bit of terminology we were talking about. Feature of radicals, some terminology, bromide ion, bromine atom, bromine molecule. Let's talk about some general features of radical reactions. Because we're going to have some mechanisms in this chapter. Two. Two possible mechanisms from this chapter. And three possible reactions from this chapter. So there's three reactions we're going to take away and use in synthesis. There's two possible mechanisms from this chapter. All right. So radicals undergo two common reactions. They react with sigma bonds and they add to pi bonds. Here's what the arrow pushing looks like. You actually have seen this before. It was in one of the chapters in 51A. When we were introducing reactivity, we talked about homolytic bond cleavage, heterolytic bond cleavage. I didn't really emphasize it because I think this is a whole world unto itself, radical chemistry. The arrows are really different so I didn't really have students have to learn that. So the arrow pushing is completely different. So now remember when we're doing this reaction, we're breaking the carbon-hydrogen bond homolytically. That means one electron goes to this carbon. The other electron is going to combine with the unpaired electron on X and it looks like they're kind of combining midair. That's the way we draw the arrows for that. And then once you do that, you get a CH3 radical plus HX. So whenever we're breaking a bond in radical chemistry, we're going to break it homolytically. One electron goes to carbon, one electron goes to hydrogen. The electron that belongs to hydrogen keeps that electron and it combines with this unpaired electron to make a new HX bond. So once we have those two electrons, we have a new bond and that's the bond right here. So that's one type of arrow pushing. And then the second type is where a radical attacks a carbon-carbon double bond. And we're going to see both types. Alright so again, we're going to break the carbon-carbon-pied bond homolytically. So one electron is going to go to this carbon. The other electron from this carbon on the right is going to combine and it looks like it's combining midair to make a new sigma bond. So that's the arrow pushing and all of the mechanisms that we do in this chapter are going to be some combination of that arrow pushing. So the radical goes right here. So notice when we're doing these radical processes here, we start with a radical, we end with a radical. We start with a radical, we end with a radical. Alright so radical mechanisms have three parts. So very likely that I will put a radical mechanism, I'll give you more details on Monday. And you have to put things into the right spot. So there is an initiation, that can be one step or it can be multiple steps. So there's initiation, there's propagation and there's termination. And so if I give you a radical mechanism on this upcoming midterm, I will have this already written for you. Initiation, I will have it propagation and I will have it termination. And there's reactions and there's problems in sapling chapter 15 where they have you categorize reactions as initiation, propagation and termination. And I'll show you what to look for for that. But if you have the right reaction and you put it in the wrong spot, you don't get points for it. So you have to be able to recognize what type you have, what type of step you have. So initiation, reactive intermediates generated. So in initiation, we start with something that's not a radical and we break the bond homolytically and we make two radicals. That's the initiation. And sometimes that takes a couple steps to do that. Propagation, the reactive intermediate reacts with a stable molecule to form another reactive intermediate and the chain continues until the supply of reactants is exhausted or the reactive intermediate is destroyed. So these radical chain, so it's like every time the reaction, we get one molecule reacting, then it starts the chain again. And so every time it does the process, it's another link in the chain. Termination steps or side reactions that destroy reactive intermediates and tend to slow or stop the reaction. So we'll see examples of each of these. I first want to talk about halogenation and then we'll look at that mechanism. Chlorine or bromine will react with alkanes in the presence of lighter heat or added peroxide to give alkyl halides. So this is the first reaction that we want to talk about. And this is an important reaction because it allows us to take an alkane that is completely unreactive and once we put a chlorine on that alkane, now we have a functional group and now we can turn that into other functional groups in all ways that we've learned. So we have an alkyl halide. We can do reactions from chapter 7, chapter 8, chapter 9, chapter 10, chapter 11 and chapter 12, right? If we just have ethane, for example, there's nothing we can do with it. It doesn't have a functional group. So this allows us to put a functional group on an unreactive molecule. So here's an example here. This is methane plus chlorine and you get a big mixture here. This is high temperature. Starts this reaction. High temp. Light. Usually, a light of a specific frequency or an initiator. And so the initiators that we're going to be using in this class are usually peroxides. So this peroxide is a free radical initiator or just a radical initiator. All right, so you need the chlorine. You need high temperature, heat or light. Now what I normally do on my exams is I will show H new and that's a trigger to you to make you think, okay, this means it's a reaction from chapter 15. It's a radical reaction. That helps you out. Because I'm sure as of right now, all these reactions are swimming in your head right now. It's kind of hard to keep them straight in some ways. The second example uses bromine and this is light and temperature to get this reaction to go. So this is not necessarily easy to do. And this reaction is much more selective. We'll talk about why coming up. You actually only get one product here. So chlorination, not a selective. We'll see that coming up. Much more selective. All right, so this is a radical chain reaction. And I'm going to do the mechanism for example one and then we're going to do an abbreviated mechanism for example two. We'll do both of them. And I'm not going to show a formation of all of these products. I'm just going to show you a few here. Okay, sorry about that. I went the wrong way. Okay initiation is where the radical is formed to start the process going. In order for our free radical chain reaction to happen, you only need one radical to start. And if that radical sticks around a long time, I mean if it's there and it's going to create new radicals, that reaction will keep on going. So it's not, when you do a radical reaction, you're not hitting it with light and all the radicals are forming at the same time. It's one molecule at a time and every time you complete the reaction, you get another link in the chain. So for a radical formed, and this is going to always be where we break a weak bond. Alright so here's some examples in the top corner of weak bonds. So you notice bond association energies are all really universally low here. Fluorine fluorine bond 38, chlorine chlorine 59, bromine bromine 46, iodine iodine bond 36, peroxide 51, this is tert butyl peroxide, di-terp butyl peroxide 38. And what is something these all have in common? Besides the fact that they have small bond association energies, what do they all have in common? Well I think I heard it, both of these atoms have lone pairs and they're right next to each other. Okay so this has lone pairs, that has lone pairs and they're right next to each other. And so what happens with those lone pairs is they want to, they repel each other. And so they want to move further away, because there's electron-electron repulsion. And when they move further away that lengthens the bond and makes it weaker, right? So this goes all the way back to 51A. Strong bonds are short bonds, weak bonds are longer bonds. So the electron-electron repulsion from those lone pairs is going to want to make those things pull away from each other and it's going to weaken the bond. So that's what's going on. So let's throw some lone pairs on here so you can be reminded of that. I'm having too much fun drawing lone pairs here. So good point to make about weak bonds here. I like this because I can actually write these a little faster than I can with a pen for the transparencies. How about that? Okay, so you see a whole lot of lone pairs going on. So that makes a good point. So what we have is two heteroatoms. And the other thing that's really good is that when I have my hand, my hand is completely covering this and it's not covering it at all up there. So that's another thing I like about this tablet here. So two heteroatoms bonded together both with lone pairs. So that makes it for a weaker bond and a weaker bond is easier to break. So when we have chlorine, so chlorine, chlorine, that's on our weak bonds. So the radical is formed and so we're going to make a chlorine-chlorine bond or we're going to break it homolytically. So we have chlorine. So we start off in an initiation, we start off with a neutral molecule that's not a radical and we end up with two radicals. And that's what starts the process rolling. So how do we know if something's an initiation step? We start off with something that's not a radical and we create new radicals. That's how you know if it's initiation. Propagation. So what's going to happen here is we have methane, so I'll draw a methane and I'm going to draw out one of the bonds for methane. And the chlorine radical that we formed in the initiation. Now we're going to do the same arrow pushing we did on the previous page. The chlorine radical is going to abstract that hydrogen. Remember when we do radical reactions we're going to break each bond homolytically. So we break that carbon-hydrogen bond. Carbon gets one electron, hydrogen gets the one electron. The electron that hydrogen gets combines with the unpaired electron and chlorine to make a new hydrogen-chlorine bond. So we get hydrochloric acid as our side product. And then the methyl radical that we just formed, it's going to grab a chlorine. So we're going to break this chlorine-chlorine bond homolytically. The electron from the chlorine on the left is going to combine with the unpaired electron on the methyl radical and we're going to get our product. And that's the other thing that we're going to see in a propagation step. So that's one chain, that's one link in the chain. And as you can see, what we've done here is we've regenerated the radical that we started with. So this is going to come in and it's going to, there's a link in the chain and it's going to come in and it's going to do it again. It's going to keep cycling through. So we regenerate the radical we started with. You can, and I'm going to show that, that's a termination step. Okay, I'm going to show that in just a second. So we regenerate the radical that you started with. And the other thing I want you to notice is that we've always keeping the same number of radicals in propagation steps. So in this one, we start with a radical, we end with a radical. Here we start with a radical and in a neutral molecule we end with a radical, I mean, or a non-molecule with all electrons paired. So start with a radical and with a radical, start with a radical and with a radical. So when you're sorting those problems and sapling into boxes, that's what you're looking for. The number of radicals stays the same. One here, one here, one here, one here. In the first one we started with none and made two. That's what makes that an initiation step. So note, the number of radicals stays the same in propagation steps. So definitely want to make sure that you look for that. Now termination is really the opposite of initiation. In initiation you start with something that's not a radical and you break it into two radicals. In termination you combine two radicals and you make it into something that's not a non-radical. That's a termination step. So that's what you're going to look for when you're doing this ear sorting. So pretty much any radical that we have here can combine with any other radical. So for example we could have a chlorine radical and this is to answer your question combining with a methyl radical. So it looks like that. That would give you the product that we're going to get anyway. It's just made by a different pathway. That would give you methyl chloride. We could have two methyl radicals combining. That will give you ethane. So that will give you ethane. You can also have two chlorine radicals recombine. So these end up being some side products that you get. Well the first one's not a side product. The third one's not a side product but certainly ethane is not something that we're trying to make. Usually we don't get very much of these termination steps because let's imagine that we start off with one molecule of chlorine breaking. So let's imagine we have just one molecule do this. So we only have two chlorines here. Two chlorine radicals. What are the chances of those two chlorine radicals? They're going to be stirring in that reaction. They're going to be moving around doing this reaction over and over again in the propagation. The chance of those two actually interacting and colliding and recombining is very small. So there will be a lot more propagation than there will be termination. You still do get these side products. And so that's why it's not a very complete reaction because no one's going to want to have to separate all of these possible products here. So you see we have a lot. Here we have methyl chloride. Now you can imagine to form this one right here, a radical would come and remove another proton from methyl chloride. That's how we would get some of this. To get this, this is chloroform. We would have a radical come and remove another hydrogen from carbon here. And then that's how we would form this. You could do that four times. You could replace all four hydrogens with chlorine. And here, this product right here would be from the ethyl that we formed in our termination step. So you get a lot of, it's kind of a messy reaction here. So the competing propagation step here, I'll show formation of the dichloromethane here. So you can see what I'm talking about here. If this comes in and grabs a proton, instead of from methane, it grabs a proton from methyl chloride, then you form this radical. And now what's that radical going to do? That radical is going to come and grab a, it's going to come in right here. Where are we? It's going to come in here. It's going to grab a chlorine and that's how you get dichloromethane. Let's show that. So kind of a messy reaction here. Just my single headed arrows here. Everybody see how that's formed? So what did we see? We said about 80%, we get about 80% methyl chloride in this reaction and we get about 20% of these other little side products that have to be separated. One of the ways to get this reaction to go a little bit better and a little bit cleaner is to keep the concentration of these side products low. Keep concentration low by using a large excess of methane. Alright so since methane is a gas, you would use a large excess of methane and then that will increase the amount of methyl chloride you get and some of these other little side products will be much less. Question about that mechanism, anybody? Alright so as we say on the next page, this isn't a very good reaction because it's not very clean. It's not the best way to synthesize alkyl halides, there are better ways but it's still useful because it is the only way to convert an inert alkane into a reactive compound. It's the only way that we know to take an inert alkane like propane, ethane, methane and add a functional group on it. Once we put a chlorine or a bromine on there now we can take that and do all the other reactions we know that we can do with alkyl halides. But until we do that we're stuck. So this is, even though it's not a very good reaction it's the only thing we have. So another better way to synthesize alkyl chloride, we learned in chapter 10, right? If we do it this way we don't get a million little side products. So that's what we would do if we wanted to make an alkyl halide for alkyl bromides, we would use HBR. So there are better ways. That's the difference here though. Are we starting with a compound that doesn't have a functional group? No, we're starting with a compound that's an alkene that has a functional group. It has a functional group that can be converted into other functional groups. So this is different. It's a better way to make alkyl halides but there are starting material has a functional group right here. There's a functional group, that alkene is a functional group and so we're just converting it to another functional group. If we have an alkene we have no functional group. This has no functional groups. So we can't take base and remove that proton because it has a pKa of 50. So this is the only way that we have to put a functional group on there and then once we do that we can convert it into other things. Questions so far? Anybody? Let's look at the second mechanism. This one's a lot cleaner. So we take tertiary butane, bromine, light and heat and let's draw our product. Single product here, we don't get a lot of different products here. Alright so initiation, erratical form, we always break a weak bond. We got a weak bond here. Did I not leave you time? Sorry about that. Okay so we're going to break a weak bond. That's easy, break a weak bond homolytically. We start off with a non-ratical. We end up with two radicals. Now again I want to make the point that not all of the bromine, not all of the molecular bromine bonds are going to break. Just a few of them. Not all of them are going to break. We only really require one to break and that's enough to get all of our starting material converted into products. So propagation and I drew some of these already to make it a little easier. I should probably draw that bond a little longer. So we go here. So tert butyl radical plus HBr. And then the tert butyl radical comes in. It's really the same mechanism here. Grabs of bromine. So we still have molecular bromine. We still got this bromine in here that hasn't been broken apart. And this is going to come and grab a bromine from that. And when it does that it forms our product and we always want to look for this. It's going to regenerate the radical that we started with at propagation. So always the last step here is going to be regenerating the radical that we started with and definitely you want to look for that. So this one here. That's what we started with. And now it's going to cycle back through again. So most of the bromine is going to be broken apart this way, not this way. Questions I hear a lot of. Oh, what did I do? Oh I forgot. Yeah, yeah, yeah, thank you. I knew I did. Which means one of two things I've either scrolled too quickly or I made a mistake. How about that? That's looking too much like a dog. Let's try and fix that. How about that? Is that better? Okay. Regenerate the radical that we started with. All right. Termination. Just destroyed by recombining any two radicals here you fill in. There's three possible. And by the way on the test I will have an initiation section. I will have a propagation. I will have a termination. And what I always do on the test is I say show me one termination. That's all I want to see. You don't have to draw all three. I just want to see one. So whatever one you think you like the best. That's what you can include. All right. So possible mechanism. Radical halogenation. I will have some sort of alkane. I will have either bromine or chlorine and light and heat and all that. That's one possible mechanism for a midterm two. All right. I want to say something about selectivity. Very selective with bromine. It turns out that bromine is a much more selective reagent than chlorine and we're going to explain why that is. So let's look at this. We've got one tertiary hydrogen. Let's remind ourselves with a tertiary hydrogen as a tertiary hydrogen is a bonded to a tertiary carbon. That carbon is bonded to three other carbons so it's a tertiary carbon. And we have how many primary hydrogens? Nine equivalent tertiary hydrogens. Our primary hydrogens, nine primary hydrogens. So we've got nine to one ratio of primary tertiary hydrogens and we still get no primary hydrogens abstracted. We get no product that comes from our primary hydrogen being extracted. So statistically you are nine times as likely but you don't get any. You don't what do you not get any. You don't get any primary hydrogen abstracted so you don't get any of this product here. This product would result from a primary hydrogen being extracted. Any one of those nine would give you this product and you don't get any of that. So let's put a big X through that. None of that product is formed. So we need to explain why that is. There's two factors that we need to talk about to explain this. Number one is the relative stability of the radicals formed. So we have tertiary radical more stable than secondary radical, more stable than primary, more stable than methyl radical. So if we remove one of those primary hydrogens we're forming a primary radical. And so maybe not surprising that we don't get any of that formed at all because we know that primary radicals are very unstable. Similar to carbocation primaries, very unstable. The other thing that's not obvious is that we have to look at the ease at which the different hydrogens are removed and in order to do this you can look at the bond association energies and energy of activation. So we actually have to do some calculations. We're not going to do any calculations. So but I want to show you that with the relative rates of hydrogen abstraction by halogen atoms for bromine primary is 1 to 82 to 1640. So tertiary forming a tertiary radical is 1640 times more likely than forming a primary with a bromine radical. And since we only have a 9 to 1 ratio there's not a chance statistically that we're going to be removing one of those primary hydrogens. The selectivity for tertiary is way too great. Chlorine on the other hand we're going to just relative rates, we'll have 1 for primary, 2.5 and 4. So if we did this exact same reaction with chlorine we will get some of the primary product, and that's because there's only a 4 to 1 selectivity for tertiary over primary is 4 to 1 and we have 9 times the number of primary carbons so we will get primary here. So much more, much less selective. And you're beginning to see why I tend to favor bromine because it's a better leaving group, it works better in this reaction and it's better in a lot of different reactions so I tend to favor it. Alright so why is the bromine atom so much more selective? It seems like it just should have to do with carbocation stability but it's not. Remember we're doing other things in this reaction. So it turns out if you compare transition states and activation energies for the abstraction of a hydrogen atom by bromine radical versus a chlorine radical you will find that. Number one, abstraction of a hydrogen atom by bromine is endothermic while abstraction of a hydrogen atom by chlorine is exothermic. This sounds crazy difficult, it's really not that bad. And number two, the transition states for the endothermic, the bromination have a larger energy difference than those for the exothermic chlorination even though, and that shouldn't be on that page, but even though the energy difference in the products is the same in both reactions. Okay it's like whoa, wrap your head around that, right? So let's see if I can explain this graphically here for you. So what we said on the previous page is abstraction of a hydrogen atom by bromine is endothermic while abstraction of a hydrogen atom by chlorine is exothermic. So here is alkane and chlorine exothermic. So this radical here is down here. And then if we do bromine it's endothermic so this radical is higher in energy. Okay so let's label that. This is exothermic reaction and this is endothermic reaction. The rate of the reaction is going to have to do with the height of that energy barrier. How do we know about the structure of the transition state in an exothermic reaction? Does it look more like the reactant for that reaction or the product? It's closer in energy to this and we're starting off with one thing. We're starting off with one thing, we have a very exothermic reaction. The difference in the energy between these transition states is so close because it looks more like the alkane than it does the radical. Very very close in energy. So what that means is that when you do the chlorination you're going to be equally likely to take all three of these pathways. Yes, the red pathway down here is lower in energy but these are all very close by so they're all accessible. Over here we have an endothermic reaction by Hammett's postulate. We know that in an endothermic reaction the transition state looks more like the products of that reaction. And so since there's a big energy difference here between these three products there's also going to be a big energy difference here between the transition states. And so you can start to see now that with bromination this lower energy pathway is so low and these are much too high for it to overcome that energy barrier. So all the molecules are going to take that pathway because there's a big energy difference here. Here there's not, this is a small energy difference. So we definitely want to label that a little better here. Let's label that. So exothermic reaction therefore the transition state resembles. And we're only looking at one step here. So the transition state resembles reactants more than products. And an endothermic reaction the transition state resembles products more than reactants. And so if you look here this is a really small energy difference between these guys. These are two tiny little arrows. I don't know if you're going to be able to see those but very small energy difference. It is less than 1 kcal per mole. So three very accessible transition states and so you're going to get a big mixture here. And here these guys here this is a large energy difference. And what do we have here? This is about, I may be exaggerated a little bit more. This is 2 kcal per mole. And that's enough that you, so if you add 2 kcal per mole, if you've got tertiary versus primary that's 2 kcal per mole plus 2, we're talking about 4 kcal per mole. That's a really large energy difference for transition states. So we have the large energy difference here. This one has a much bigger energy difference. Therefore the rates of the reactions will be markedly different. And this is a really good example of the point that when you're talking about rates of reaction it does not have to do with the intermediates. These intermediates and these intermediates have the same energy difference. And if it was only because of that then we would have the same product outcome whether it's bromine and chlorine. It has to do with the height of the transition state here. That's what's key. Alright so here's an example here to show you what this means, difference wise in selectivity. So here we have bromination, here we have chlorination. With bromination we get one product 82%. With chlorination we get 26% of this product, 22% of this product, 22% of this product, 14% of this product and 17%. And you can actually predict these percentages by taking into account the difference in selectivity of chlorine versus the number of hydrogens that you have and you can actually calculate these differences. The textbook we had before this, had you do that, this textbook doesn't have it so I'm not going to have you learn that. You'll be glad about that. You're very glad. It's not that hard but you're glad. Alright questions on bromination of alkanes or chlorination of alkanes. That's reaction number one from this chapter. Radical substitution of benzylic and alec, that's reaction number two in this chapter. We are going to skip the mechanism here. But you need to know the reaction. Poor synthesis. Why am I having you skip the mechanism? It's a confusing mechanism. It has both single headed arrows and double headed arrows and it's just so hard to wrap your head around when you're just learning organic chemistry for the first time. How do I know that because I remember having to memorize this mechanism. And a lot of mechanisms that I didn't really need to memorize. So I'm having you, I'm selectively having you memorize certain things. As little as possible though. So we're skipping the mechanism and you don't need to know that. Recall that allyl and benzyl radicals are both more stable than tertiary radicals. We know the more stable the radical the faster it can be formed. So a hydrogen bonded to either benzylic or alelic carbon can be preferentially be substituted in a halogenation reaction. So again, these guys are more stable than tertiary. So that means that if you have a choice between removing a tertiary hydrogen or benzylic or alelic, benzylic or alelic will win. So here's some examples here. Br2, we can make this radical right here. We won't form a primary radical, we'll only form that one. And then if we carry that on, we will get benzyl bromide. A substituted benzyl bromide. So the bromine will go in the benzyl position, nowhere else. And you put that bromine. If we have allyl, then we can put, we have a choice between vinyl, hydrogen or allyl. So here's vinyl right here. Are we going to form any of that? Not when we've got allyl that we can form, right? So that's way better. So that means that we can form the allyl radical. Here we can convert that into alyl bromide. So who sees a competing reaction that can form with this, these two? Who sees another reaction that's competing with that? From chapter 10. Yeah, and you know, bromination of an alkene is an extremely fast reaction. You guys did this in the lab, didn't you? How fast was it? It was really fast. So how are we going to do this when we can do this just as fast? Okay, so I'm going to give you a different reagent for this. We're not going to use this reagent. We're going to give you a special reagent for this, then I want you to use instead. And that's because we have competing reactions here. With bromine, we can form a bromonium ion. Let's remember this. We don't want to forget this. This is midterm two material, right? Not to bring up an unpleasant subject, but you know, right? Nobody likes taking tests. In graduate school, in chemistry, you take one year of classes and then you work in a lab for the rest of the time while you earn your PhD. And I remember thinking at the time, I've only got one more year of classes in me. I cannot do this for two years. I just was so burned out on taking tests. So I haven't had to take a test since then, and it's a beautiful thing. Because nobody likes taking tests. All right, so remember this, okay? So that's a competing reaction. That's a competing, very fast reaction. So really, let's not use bromine for this. Even though Smith uses this a lot in synthesis, I want you to not use this reagent. And that's the reason why. We'll stop right there, and we will finish chapter 15 next time. So this is the reagent I want you to use instead.
 So what it's looking like, what it's looking like, we'll see how it goes.  So everybody knows we have a midterm coming up.  It looks like it's going to be through Chapter 15 for the midterm.  This is going to be a really hard midterm.  And so, you know, a lot of synthesis.  So last year, we had spec plus this stuff.  And so last year, we had spec plus this stuff.  And so I couldn't ask as much synthesis because spec takes time, right?  There's no spec on this test.  So that means there's going to be more synthesis, right?  So what you want to do is you want to do as much synthesis as you can.  So do the sapling synthesis.  Come straight at the end of each chapter.  There's a synthesis section where it mixes reactions from all the chapters.  You want to do that.  Those are the things you want practice and synthesis.  Okay?  Questions?  Anybody?  Before we get started.  All right.  So Chapter 15, really different.  So we had a little bit of terminology we were talking about.  Feature of radicals, some terminology, bromide ion, bromine atom, bromine molecule.  Let's talk about some general features of radical reactions.  Because we're going to have some mechanisms in this chapter.  Two.  Two possible mechanisms from this chapter.  And three possible reactions from this chapter.  So there's three reactions we're going to take away and use in synthesis.  There's two possible mechanisms from this chapter.  All right.  So radicals undergo two common reactions.  They react with sigma bonds and they add to pi bonds.  Here's what the arrow pushing looks like.  You actually have seen this before.  It was in one of the chapters in 51A.  When we were introducing reactivity, we talked about homolytic bond cleavage, heterolytic  bond cleavage.  I didn't really emphasize it because I think this is a whole world unto itself, radical  chemistry.  The arrows are really different so I didn't really have students have to learn that.  So the arrow pushing is completely different.  So now remember when we're doing this reaction, we're breaking the carbon-hydrogen bond homolytically.  That means one electron goes to this carbon.  The other electron is going to combine with the unpaired electron on X and it looks like  they're kind of combining midair.  That's the way we draw the arrows for that.  And then once you do that, you get a CH3 radical plus HX.  So whenever we're breaking a bond in radical chemistry, we're going to break it homolytically.  One electron goes to carbon, one electron goes to hydrogen.  The electron that belongs to hydrogen keeps that electron and it combines with this unpaired  electron to make a new HX bond.  So once we have those two electrons, we have a new bond and that's the bond right here.  So that's one type of arrow pushing.  And then the second type is where a radical attacks a carbon-carbon double bond.  And we're going to see both types.  Alright so again, we're going to break the carbon-carbon-pied bond homolytically.  So one electron is going to go to this carbon.  The other electron from this carbon on the right is going to combine and it looks like  it's combining midair to make a new sigma bond.  So that's the arrow pushing and all of the mechanisms that we do in this chapter are  going to be some combination of that arrow pushing.  So the radical goes right here.  So notice when we're doing these radical processes here, we start with a radical, we end with  a radical.  We start with a radical, we end with a radical.  Alright so radical mechanisms have three parts.  So very likely that I will put a radical mechanism, I'll give you more details on Monday.  And you have to put things into the right spot.  So there is an initiation, that can be one step or it can be multiple steps.  So there's initiation, there's propagation and there's termination.  And so if I give you a radical mechanism on this upcoming midterm, I will have this already  written for you.  Initiation, I will have it propagation and I will have it termination.  And there's reactions and there's problems in sapling chapter 15 where they have you  categorize reactions as initiation, propagation and termination.  And I'll show you what to look for for that.  But if you have the right reaction and you put it in the wrong spot, you don't get points  for it.  So you have to be able to recognize what type you have, what type of step you have.  So initiation, reactive intermediates generated.  So in initiation, we start with something that's not a radical and we break the bond  homolytically and we make two radicals.  That's the initiation.  And sometimes that takes a couple steps to do that.  Propagation, the reactive intermediate reacts with a stable molecule to form another reactive  intermediate and the chain continues until the supply of reactants is exhausted or the  reactive intermediate is destroyed.  So these radical chain, so it's like every time the reaction, we get one molecule reacting,  then it starts the chain again.  And so every time it does the process, it's another link in the chain.  Termination steps or side reactions that destroy reactive intermediates and tend to slow or  stop the reaction.  So we'll see examples of each of these.  I first want to talk about halogenation and then we'll look at that mechanism.  Chlorine or bromine will react with alkanes in the presence of lighter heat or added peroxide  to give alkyl halides.  So this is the first reaction that we want to talk about.  And this is an important reaction because it allows us to take an alkane that is completely  unreactive and once we put a chlorine on that alkane, now we have a functional group and  now we can turn that into other functional groups in all ways that we've learned.  So we have an alkyl halide.  We can do reactions from chapter 7, chapter 8, chapter 9, chapter 10, chapter 11 and chapter  12, right?  If we just have ethane, for example, there's nothing we can do with it.  It doesn't have a functional group.  So this allows us to put a functional group on an unreactive molecule.  So here's an example here.  This is methane plus chlorine and you get a big mixture here.  This is high temperature.  Starts this reaction.  High temp.  Light.  Usually, a light of a specific frequency or an initiator.  And so the initiators that we're going to be using in this class are usually peroxides.  So this peroxide is a free radical initiator or just a radical initiator.  All right, so you need the chlorine.  You need high temperature, heat or light.  Now what I normally do on my exams is I will show H new and that's a trigger to you to  make you think, okay, this means it's a reaction from chapter 15.  It's a radical reaction.  That helps you out.  Because I'm sure as of right now, all these reactions are swimming in your head right  now.  It's kind of hard to keep them straight in some ways.  The second example uses bromine and this is light and temperature to get this reaction  to go.  So this is not necessarily easy to do.  And this reaction is much more selective.  We'll talk about why coming up.  You actually only get one product here.  So chlorination, not a selective.  We'll see that coming up.  Much more selective.  All right, so this is a radical chain reaction.  And I'm going to do the mechanism for example one and then we're going to do an abbreviated  mechanism for example two.  We'll do both of them.  And I'm not going to show a formation of all of these products.  I'm just going to show you a few here.  Okay, sorry about that.  I went the wrong way.  Okay initiation is where the radical is formed to start the process going.  In order for our free radical chain reaction to happen, you only need one radical to start.  And if that radical sticks around a long time, I mean if it's there and it's going to create  new radicals, that reaction will keep on going.  So it's not, when you do a radical reaction, you're not hitting it with light and all the  radicals are forming at the same time.  It's one molecule at a time and every time you complete the reaction, you get another  link in the chain.  So for a radical formed, and this is going to always be where we break a weak bond.  Alright so here's some examples in the top corner of weak bonds.  So you notice bond association energies are all really universally low here.  Fluorine fluorine bond 38, chlorine chlorine 59, bromine bromine 46, iodine iodine bond  36, peroxide 51, this is tert butyl peroxide, di-terp butyl peroxide 38.  And what is something these all have in common?  Besides the fact that they have small bond association energies, what do they all have  in common?  Well I think I heard it, both of these atoms have lone pairs and they're right next to  each other.  Okay so this has lone pairs, that has lone pairs and they're right next to each other.  And so what happens with those lone pairs is they want to, they repel each other.  And so they want to move further away, because there's electron-electron repulsion.  And when they move further away that lengthens the bond and makes it weaker, right?  So this goes all the way back to 51A.  Strong bonds are short bonds, weak bonds are longer bonds.  So the electron-electron repulsion from those lone pairs is going to want to make those  things pull away from each other and it's going to weaken the bond.  So that's what's going on.  So let's throw some lone pairs on here so you can be reminded of that.  I'm having too much fun drawing lone pairs here.  So good point to make about weak bonds here.  I like this because I can actually write these a little faster than I can with a pen for  the transparencies.  How about that?  Okay, so you see a whole lot of lone pairs going on.  So that makes a good point.  So what we have is two heteroatoms.  And the other thing that's really good is that when I have my hand, my hand is completely  covering this and it's not covering it at all up there.  So that's another thing I like about this tablet here.  So two heteroatoms bonded together both with lone pairs.  So that makes it for a weaker bond and a weaker bond is easier to break.  So when we have chlorine, so chlorine, chlorine, that's on our weak bonds.  So the radical is formed and so we're going to make a chlorine-chlorine bond or we're  going to break it homolytically.  So we have chlorine.  So we start off in an initiation, we start off with a neutral molecule that's not a radical  and we end up with two radicals.  And that's what starts the process rolling.  So how do we know if something's an initiation step?  We start off with something that's not a radical and we create new radicals.  That's how you know if it's initiation.  Propagation.  So what's going to happen here is we have methane, so I'll draw a methane and I'm going  to draw out one of the bonds for methane.  And the chlorine radical that we formed in the initiation.  Now we're going to do the same arrow pushing we did on the previous page.  The chlorine radical is going to abstract that hydrogen.  Remember when we do radical reactions we're going to break each bond homolytically.  So we break that carbon-hydrogen bond.  Carbon gets one electron, hydrogen gets the one electron.  The electron that hydrogen gets combines with the unpaired electron and chlorine to make  a new hydrogen-chlorine bond.  So we get hydrochloric acid as our side product.  And then the methyl radical that we just formed, it's going to grab a chlorine.  So we're going to break this chlorine-chlorine bond homolytically.  The electron from the chlorine on the left is going to combine with the unpaired electron  on the methyl radical and we're going to get our product.  And that's the other thing that we're going to see in a propagation step.  So that's one chain, that's one link in the chain.  And as you can see, what we've done here is we've regenerated the radical that we started  with.  So this is going to come in and it's going to, there's a link in the chain and it's  going to come in and it's going to do it again.  It's going to keep cycling through.  So we regenerate the radical we started with.  You can, and I'm going to show that, that's a termination step.  Okay, I'm going to show that in just a second.  So we regenerate the radical that you started with.  And the other thing I want you to notice is that we've always keeping the same number  of radicals in propagation steps.  So in this one, we start with a radical, we end with a radical.  Here we start with a radical and in a neutral molecule we end with a radical, I mean, or  a non-molecule with all electrons paired.  So start with a radical and with a radical, start with a radical and with a radical.  So when you're sorting those problems and sapling into boxes, that's what you're looking  for.  The number of radicals stays the same.  One here, one here, one here, one here.  In the first one we started with none and made two.  That's what makes that an initiation step.  So note, the number of radicals stays the same in propagation steps.  So definitely want to make sure that you look for that.  Now termination is really the opposite of initiation.  In initiation you start with something that's not a radical and you break it into two radicals.  In termination you combine two radicals and you make it into something that's not a non-radical.  That's a termination step.  So that's what you're going to look for when you're doing this ear sorting.  So pretty much any radical that we have here can combine with any other radical.  So for example we could have a chlorine radical and this is to answer your question combining  with a methyl radical.  So it looks like that.  That would give you the product that we're going to get anyway.  It's just made by a different pathway.  That would give you methyl chloride.  We could have two methyl radicals combining.  That will give you ethane.  So that will give you ethane.  You can also have two chlorine radicals recombine.  So these end up being some side products that you get.  Well the first one's not a side product.  The third one's not a side product but certainly ethane is not something that we're trying  to make.  Usually we don't get very much of these termination steps because let's imagine that we start off  with one molecule of chlorine breaking.  So let's imagine we have just one molecule do this.  So we only have two chlorines here.  Two chlorine radicals.  What are the chances of those two chlorine radicals?  They're going to be stirring in that reaction.  They're going to be moving around doing this reaction over and over again in the propagation.  The chance of those two actually interacting and colliding and recombining is very small.  So there will be a lot more propagation than there will be termination.  You still do get these side products.  And so that's why it's not a very complete reaction because no one's going to want to  have to separate all of these possible products here.  So you see we have a lot.  Here we have methyl chloride.  Now you can imagine to form this one right here, a radical would come and remove another  proton from methyl chloride.  That's how we would get some of this.  To get this, this is chloroform.  We would have a radical come and remove another hydrogen from carbon here.  And then that's how we would form this.  You could do that four times.  You could replace all four hydrogens with chlorine.  And here, this product right here would be from the ethyl that we formed in our termination  step.  So you get a lot of, it's kind of a messy reaction here.  So the competing propagation step here, I'll show formation of the dichloromethane here.  So you can see what I'm talking about here.  If this comes in and grabs a proton, instead of from methane, it grabs a proton from methyl  chloride, then you form this radical.  And now what's that radical going to do?  That radical is going to come and grab a, it's going to come in right here.  Where are we?  It's going to come in here.  It's going to grab a chlorine and that's how you get dichloromethane.  Let's show that.  So kind of a messy reaction here.  Just my single headed arrows here.  Everybody see how that's formed?  So what did we see?  We said about 80%, we get about 80% methyl chloride in this reaction and we get about  20% of these other little side products that have to be separated.  One of the ways to get this reaction to go a little bit better and a little bit cleaner  is to keep the concentration of these side products low.  Keep concentration low by using a large excess of methane.  Alright so since methane is a gas, you would use a large excess of methane and then that  will increase the amount of methyl chloride you get and some of these other little side  products will be much less.  Question about that mechanism, anybody?  Alright so as we say on the next page, this isn't a very good reaction because it's not  very clean.  It's not the best way to synthesize alkyl halides, there are better ways but it's still  useful because it is the only way to convert an inert alkane into a reactive compound.  It's the only way that we know to take an inert alkane like propane, ethane, methane  and add a functional group on it.  Once we put a chlorine or a bromine on there now we can take that and do all the other  reactions we know that we can do with alkyl halides.  But until we do that we're stuck.  So this is, even though it's not a very good reaction it's the only thing we have.  So another better way to synthesize alkyl chloride, we learned in chapter 10, right?  If we do it this way we don't get a million little side products.  So that's what we would do if we wanted to make an alkyl halide for alkyl bromides, we  would use HBR.  So there are better ways.  That's the difference here though.  Are we starting with a compound that doesn't have a functional group?  No, we're starting with a compound that's an alkene that has a functional group.  It has a functional group that can be converted into other functional groups.  So this is different.  It's a better way to make alkyl halides but there are starting material has a functional  group right here.  There's a functional group, that alkene is a functional group and so we're just converting  it to another functional group.  If we have an alkene we have no functional group.  This has no functional groups.  So we can't take base and remove that proton because it has a pKa of 50.  So this is the only way that we have to put a functional group on there and then once  we do that we can convert it into other things.  Questions so far?  Anybody?  Let's look at the second mechanism.  This one's a lot cleaner.  So we take tertiary butane, bromine, light and heat and let's draw our product.  Single product here, we don't get a lot of different products here.  Alright so initiation, erratical form, we always break a weak bond.  We got a weak bond here.  Did I not leave you time?  Sorry about that.  Okay so we're going to break a weak bond.  That's easy, break a weak bond homolytically.  We start off with a non-ratical.  We end up with two radicals.  Now again I want to make the point that not all of the bromine, not all of the molecular  bromine bonds are going to break.  Just a few of them.  Not all of them are going to break.  We only really require one to break and that's enough to get all of our starting material  converted into products.  So propagation and I drew some of these already to make it a little easier.  I should probably draw that bond a little longer.  So we go here.  So tert butyl radical plus HBr.  And then the tert butyl radical comes in.  It's really the same mechanism here.  Grabs of bromine.  So we still have molecular bromine.  We still got this bromine in here that hasn't been broken apart.  And this is going to come and grab a bromine from that.  And when it does that it forms our product and we always want to look for this.  It's going to regenerate the radical that we started with at propagation.  So always the last step here is going to be regenerating the radical that we started with  and definitely you want to look for that.  So this one here.  That's what we started with.  And now it's going to cycle back through again.  So most of the bromine is going to be broken apart this way, not this way.  Questions I hear a lot of.  Oh, what did I do?  Oh I forgot.  Yeah, yeah, yeah, thank you.  I knew I did.  Which means one of two things I've either scrolled too quickly or I made a mistake.  How about that?  That's looking too much like a dog.  Let's try and fix that.  How about that?  Is that better?  Okay.  Regenerate the radical that we started with.  All right.  Termination.  Just destroyed by recombining any two radicals here you fill in.  There's three possible.  And by the way on the test I will have an initiation section.  I will have a propagation.  I will have a termination.  And what I always do on the test is I say show me one termination.  That's all I want to see.  You don't have to draw all three.  I just want to see one.  So whatever one you think you like the best.  That's what you can include.  All right.  So possible mechanism.  Radical halogenation.  I will have some sort of alkane.  I will have either bromine or chlorine and light and heat and all that.  That's one possible mechanism for a midterm two.  All right.  I want to say something about selectivity.  Very selective with bromine.  It turns out that bromine is a much more selective reagent than chlorine and we're going to explain  why that is.  So let's look at this.  We've got one tertiary hydrogen.  Let's remind ourselves with a tertiary hydrogen as a tertiary hydrogen is a bonded to a tertiary  carbon.  That carbon is bonded to three other carbons so it's a tertiary carbon.  And we have how many primary hydrogens?  Nine equivalent tertiary hydrogens.  Our primary hydrogens, nine primary hydrogens.  So we've got nine to one ratio of primary tertiary hydrogens and we still get no primary  hydrogens abstracted.  We get no product that comes from our primary hydrogen being extracted.  So statistically you are nine times as likely but you don't get any.  You don't what do you not get any.  You don't get any primary hydrogen abstracted so you don't get any of this product here.  This product would result from a primary hydrogen being extracted.  Any one of those nine would give you this product and you don't get any of that.  So let's put a big X through that.  None of that product is formed.  So we need to explain why that is.  There's two factors that we need to talk about to explain this.  Number one is the relative stability of the radicals formed.  So we have tertiary radical more stable than secondary radical, more stable than primary,  more stable than methyl radical.  So if we remove one of those primary hydrogens we're forming a primary radical.  And so maybe not surprising that we don't get any of that formed at all because we know  that primary radicals are very unstable.  Similar to carbocation primaries, very unstable.  The other thing that's not obvious is that we have to look at the ease at which the different  hydrogens are removed and in order to do this you can look at the bond association energies  and energy of activation.  So we actually have to do some calculations.  We're not going to do any calculations.  So but I want to show you that with the relative rates of hydrogen abstraction by halogen atoms  for bromine primary is 1 to 82 to 1640.  So tertiary forming a tertiary radical is 1640 times more likely than forming a primary  with a bromine radical.  And since we only have a 9 to 1 ratio there's not a chance statistically that we're going  to be removing one of those primary hydrogens.  The selectivity for tertiary is way too great.  Chlorine on the other hand we're going to just relative rates, we'll have 1 for primary,  2.5 and 4.  So if we did this exact same reaction with chlorine we will get some of the primary product,  and that's because there's only a 4 to 1 selectivity for tertiary over primary is 4  to 1 and we have 9 times the number of primary carbons so we will get primary here.  So much more, much less selective.  And you're beginning to see why I tend to favor bromine because it's a better leaving  group, it works better in this reaction and it's better in a lot of different reactions  so I tend to favor it.  Alright so why is the bromine atom so much more selective?  It seems like it just should have to do with carbocation stability but it's not.  Remember we're doing other things in this reaction.  So it turns out if you compare transition states and activation energies for the abstraction  of a hydrogen atom by bromine radical versus a chlorine radical you will find that.  Number one, abstraction of a hydrogen atom by bromine is endothermic while abstraction  of a hydrogen atom by chlorine is exothermic.  This sounds crazy difficult, it's really not that bad.  And number two, the transition states for the endothermic, the bromination have a larger  energy difference than those for the exothermic chlorination even though, and that shouldn't  be on that page, but even though the energy difference in the products is the same in  both reactions.  Okay it's like whoa, wrap your head around that, right?  So let's see if I can explain this graphically here for you.  So what we said on the previous page is abstraction of a hydrogen atom by bromine is endothermic  while abstraction of a hydrogen atom by chlorine is exothermic.  So here is alkane and chlorine exothermic.  So this radical here is down here.  And then if we do bromine it's endothermic so this radical is higher in energy.  Okay so let's label that.  This is exothermic reaction and this is endothermic reaction.  The rate of the reaction is going to have to do with the height of that energy barrier.  How do we know about the structure of the transition state in an exothermic reaction?  Does it look more like the reactant for that reaction or the product?  It's closer in energy to this and we're starting off with one thing.  We're starting off with one thing, we have a very exothermic reaction.  The difference in the energy between these transition states is so close because it looks  more like the alkane than it does the radical.  Very very close in energy.  So what that means is that when you do the chlorination you're going to be equally likely  to take all three of these pathways.  Yes, the red pathway down here is lower in energy but these are all very close by so  they're all accessible.  Over here we have an endothermic reaction by Hammett's postulate.  We know that in an endothermic reaction the transition state looks more like the products  of that reaction.  And so since there's a big energy difference here between these three products there's  also going to be a big energy difference here between the transition states.  And so you can start to see now that with bromination this lower energy pathway is so  low and these are much too high for it to overcome that energy barrier.  So all the molecules are going to take that pathway because there's a big energy difference  here.  Here there's not, this is a small energy difference.  So we definitely want to label that a little better here.  Let's label that.  So exothermic reaction therefore the transition state resembles.  And we're only looking at one step here.  So the transition state resembles reactants more than products.  And an endothermic reaction the transition state resembles products more than reactants.  And so if you look here this is a really small energy difference between these guys.  These are two tiny little arrows.  I don't know if you're going to be able to see those but very small energy difference.  It is less than 1 kcal per mole.  So three very accessible transition states and so you're going to get a big mixture  here.  And here these guys here this is a large energy difference.  And what do we have here?  This is about, I may be exaggerated a little bit more.  This is 2 kcal per mole.  And that's enough that you, so if you add 2 kcal per mole, if you've got tertiary versus  primary that's 2 kcal per mole plus 2, we're talking about 4 kcal per mole.  That's a really large energy difference for transition states.  So we have the large energy difference here.  This one has a much bigger energy difference.  Therefore the rates of the reactions will be markedly different.  And this is a really good example of the point that when you're talking about rates of reaction  it does not have to do with the intermediates.  These intermediates and these intermediates have the same energy difference.  And if it was only because of that then we would have the same product outcome whether  it's bromine and chlorine.  It has to do with the height of the transition state here.  That's what's key.  Alright so here's an example here to show you what this means, difference wise in selectivity.  So here we have bromination, here we have chlorination.  With bromination we get one product 82%.  With chlorination we get 26% of this product, 22% of this product, 22% of this product,  14% of this product and 17%.  And you can actually predict these percentages by taking into account the difference in selectivity  of chlorine versus the number of hydrogens that you have and you can actually calculate  these differences.  The textbook we had before this, had you do that, this textbook doesn't have it so I'm  not going to have you learn that.  You'll be glad about that.  You're very glad.  It's not that hard but you're glad.  Alright questions on bromination of alkanes or chlorination of alkanes.  That's reaction number one from this chapter.  Radical substitution of benzylic and alec, that's reaction number two in this chapter.  We are going to skip the mechanism here.  But you need to know the reaction.  Poor synthesis.  Why am I having you skip the mechanism?  It's a confusing mechanism.  It has both single headed arrows and double headed arrows and it's just so hard to wrap  your head around when you're just learning organic chemistry for the first time.  How do I know that because I remember having to memorize this mechanism.  And a lot of mechanisms that I didn't really need to memorize.  So I'm having you, I'm selectively having you memorize certain things.  As little as possible though.  So we're skipping the mechanism and you don't need to know that.  Recall that allyl and benzyl radicals are both more stable than tertiary radicals.  We know the more stable the radical the faster it can be formed.  So a hydrogen bonded to either benzylic or alelic carbon can be preferentially be substituted  in a halogenation reaction.  So again, these guys are more stable than tertiary.  So that means that if you have a choice between removing a tertiary hydrogen or benzylic or  alelic, benzylic or alelic will win.  So here's some examples here.  Br2, we can make this radical right here.  We won't form a primary radical, we'll only form that one.  And then if we carry that on, we will get benzyl bromide.  A substituted benzyl bromide.  So the bromine will go in the benzyl position, nowhere else.  And you put that bromine.  If we have allyl, then we can put, we have a choice between vinyl, hydrogen or allyl.  So here's vinyl right here.  Are we going to form any of that?  Not when we've got allyl that we can form, right?  So that's way better.  So that means that we can form the allyl radical.  Here we can convert that into alyl bromide.  So who sees a competing reaction that can form with this, these two?  Who sees another reaction that's competing with that?  From chapter 10.  Yeah, and you know, bromination of an alkene is an extremely fast reaction.  You guys did this in the lab, didn't you?  How fast was it?  It was really fast.  So how are we going to do this when we can do this just as fast?  Okay, so I'm going to give you a different reagent for this.  We're not going to use this reagent.  We're going to give you a special reagent for this, then I want you to use instead.  And that's because we have competing reactions here.  With bromine, we can form a bromonium ion.  Let's remember this.  We don't want to forget this.  This is midterm two material, right?  Not to bring up an unpleasant subject, but you know, right?  Nobody likes taking tests.  In graduate school, in chemistry, you take one year of classes and then you work in a  lab for the rest of the time while you earn your PhD.  And I remember thinking at the time, I've only got one more year of classes in me.  I cannot do this for two years.  I just was so burned out on taking tests.  So I haven't had to take a test since then, and it's a beautiful thing.  Because nobody likes taking tests.  All right, so remember this, okay?  So that's a competing reaction.  That's a competing, very fast reaction.  So really, let's not use bromine for this.  Even though Smith uses this a lot in synthesis, I want you to not use this reagent.  And that's the reason why.  We'll stop right there, and we will finish chapter 15 next time.  So this is the reagent I want you to use instead.
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So what it's looking like, what it's looking like, we'll see how it goes. So everybody knows we have a midterm coming up. It looks like it's going to be through Chapter 15 for the midterm. This is going to be a really hard midterm. And so, you know, a lot of synthesis. So last year, we had spec plus this stuff. And so last year, we had spec plus this stuff. And so I couldn't ask as much synthesis because spec takes time, right? There's no spec on this test. So that means there's going to be more synthesis, right? So what you want to do is you want to do as much synthesis as you can. So do the sapling synthesis. Come straight at the end of each chapter. There's a synthesis section where it mixes reactions from all the chapters. You want to do that. Those are the things you want practice and synthesis. Okay? Questions? Anybody? Before we get started. All right. So Chapter 15, really different. So we had a little bit of terminology we were talking about. Feature of radicals, some terminology, bromide ion, bromine atom, bromine molecule. Let's talk about some general features of radical reactions. Because we're going to have some mechanisms in this chapter. Two. Two possible mechanisms from this chapter. And three possible reactions from this chapter. So there's three reactions we're going to take away and use in synthesis. There's two possible mechanisms from this chapter. All right. So radicals undergo two common reactions. They react with sigma bonds and they add to pi bonds. Here's what the arrow pushing looks like. You actually have seen this before. It was in one of the chapters in 51A. When we were introducing reactivity, we talked about homolytic bond cleavage, heterolytic bond cleavage. I didn't really emphasize it because I think this is a whole world unto itself, radical chemistry. The arrows are really different so I didn't really have students have to learn that. So the arrow pushing is completely different. So now remember when we're doing this reaction, we're breaking the carbon-hydrogen bond homolytically. That means one electron goes to this carbon. The other electron is going to combine with the unpaired electron on X and it looks like they're kind of combining midair. That's the way we draw the arrows for that. And then once you do that, you get a CH3 radical plus HX. So whenever we're breaking a bond in radical chemistry, we're going to break it homolytically. One electron goes to carbon, one electron goes to hydrogen. The electron that belongs to hydrogen keeps that electron and it combines with this unpaired electron to make a new HX bond. So once we have those two electrons, we have a new bond and that's the bond right here. So that's one type of arrow pushing. And then the second type is where a radical attacks a carbon-carbon double bond. And we're going to see both types. Alright so again, we're going to break the carbon-carbon-pied bond homolytically. So one electron is going to go to this carbon. The other electron from this carbon on the right is going to combine and it looks like it's combining midair to make a new sigma bond. So that's the arrow pushing and all of the mechanisms that we do in this chapter are going to be some combination of that arrow pushing. So the radical goes right here. So notice when we're doing these radical processes here, we start with a radical, we end with a radical. We start with a radical, we end with a radical. Alright so radical mechanisms have three parts. So very likely that I will put a radical mechanism, I'll give you more details on Monday. And you have to put things into the right spot. So there is an initiation, that can be one step or it can be multiple steps. So there's initiation, there's propagation and there's termination. And so if I give you a radical mechanism on this upcoming midterm, I will have this already written for you. Initiation, I will have it propagation and I will have it termination. And there's reactions and there's problems in sapling chapter 15 where they have you categorize reactions as initiation, propagation and termination. And I'll show you what to look for for that. But if you have the right reaction and you put it in the wrong spot, you don't get points for it. So you have to be able to recognize what type you have, what type of step you have. So initiation, reactive intermediates generated. So in initiation, we start with something that's not a radical and we break the bond homolytically and we make two radicals. That's the initiation. And sometimes that takes a couple steps to do that. Propagation, the reactive intermediate reacts with a stable molecule to form another reactive intermediate and the chain continues until the supply of reactants is exhausted or the reactive intermediate is destroyed. So these radical chain, so it's like every time the reaction, we get one molecule reacting, then it starts the chain again. And so every time it does the process, it's another link in the chain. Termination steps or side reactions that destroy reactive intermediates and tend to slow or stop the reaction. So we'll see examples of each of these. I first want to talk about halogenation and then we'll look at that mechanism. Chlorine or bromine will react with alkanes in the presence of lighter heat or added peroxide to give alkyl halides. So this is the first reaction that we want to talk about. And this is an important reaction because it allows us to take an alkane that is completely unreactive and once we put a chlorine on that alkane, now we have a functional group and now we can turn that into other functional groups in all ways that we've learned. So we have an alkyl halide. We can do reactions from chapter 7, chapter 8, chapter 9, chapter 10, chapter 11 and chapter 12, right? If we just have ethane, for example, there's nothing we can do with it. It doesn't have a functional group. So this allows us to put a functional group on an unreactive molecule. So here's an example here. This is methane plus chlorine and you get a big mixture here. This is high temperature. Starts this reaction. High temp. Light. Usually, a light of a specific frequency or an initiator. And so the initiators that we're going to be using in this class are usually peroxides. So this peroxide is a free radical initiator or just a radical initiator. All right, so you need the chlorine. You need high temperature, heat or light. Now what I normally do on my exams is I will show H new and that's a trigger to you to make you think, okay, this means it's a reaction from chapter 15. It's a radical reaction. That helps you out. Because I'm sure as of right now, all these reactions are swimming in your head right now. It's kind of hard to keep them straight in some ways. The second example uses bromine and this is light and temperature to get this reaction to go. So this is not necessarily easy to do. And this reaction is much more selective. We'll talk about why coming up. You actually only get one product here. So chlorination, not a selective. We'll see that coming up. Much more selective. All right, so this is a radical chain reaction. And I'm going to do the mechanism for example one and then we're going to do an abbreviated mechanism for example two. We'll do both of them. And I'm not going to show a formation of all of these products. I'm just going to show you a few here. Okay, sorry about that. I went the wrong way. Okay initiation is where the radical is formed to start the process going. In order for our free radical chain reaction to happen, you only need one radical to start. And if that radical sticks around a long time, I mean if it's there and it's going to create new radicals, that reaction will keep on going. So it's not, when you do a radical reaction, you're not hitting it with light and all the radicals are forming at the same time. It's one molecule at a time and every time you complete the reaction, you get another link in the chain. So for a radical formed, and this is going to always be where we break a weak bond. Alright so here's some examples in the top corner of weak bonds. So you notice bond association energies are all really universally low here. Fluorine fluorine bond 38, chlorine chlorine 59, bromine bromine 46, iodine iodine bond 36, peroxide 51, this is tert butyl peroxide, di-terp butyl peroxide 38. And what is something these all have in common? Besides the fact that they have small bond association energies, what do they all have in common? Well I think I heard it, both of these atoms have lone pairs and they're right next to each other. Okay so this has lone pairs, that has lone pairs and they're right next to each other. And so what happens with those lone pairs is they want to, they repel each other. And so they want to move further away, because there's electron-electron repulsion. And when they move further away that lengthens the bond and makes it weaker, right? So this goes all the way back to 51A. Strong bonds are short bonds, weak bonds are longer bonds. So the electron-electron repulsion from those lone pairs is going to want to make those things pull away from each other and it's going to weaken the bond. So that's what's going on. So let's throw some lone pairs on here so you can be reminded of that. I'm having too much fun drawing lone pairs here. So good point to make about weak bonds here. I like this because I can actually write these a little faster than I can with a pen for the transparencies. How about that? Okay, so you see a whole lot of lone pairs going on. So that makes a good point. So what we have is two heteroatoms. And the other thing that's really good is that when I have my hand, my hand is completely covering this and it's not covering it at all up there. So that's another thing I like about this tablet here. So two heteroatoms bonded together both with lone pairs. So that makes it for a weaker bond and a weaker bond is easier to break. So when we have chlorine, so chlorine, chlorine, that's on our weak bonds. So the radical is formed and so we're going to make a chlorine-chlorine bond or we're going to break it homolytically. So we have chlorine. So we start off in an initiation, we start off with a neutral molecule that's not a radical and we end up with two radicals. And that's what starts the process rolling. So how do we know if something's an initiation step? We start off with something that's not a radical and we create new radicals. That's how you know if it's initiation. Propagation. So what's going to happen here is we have methane, so I'll draw a methane and I'm going to draw out one of the bonds for methane. And the chlorine radical that we formed in the initiation. Now we're going to do the same arrow pushing we did on the previous page. The chlorine radical is going to abstract that hydrogen. Remember when we do radical reactions we're going to break each bond homolytically. So we break that carbon-hydrogen bond. Carbon gets one electron, hydrogen gets the one electron. The electron that hydrogen gets combines with the unpaired electron and chlorine to make a new hydrogen-chlorine bond. So we get hydrochloric acid as our side product. And then the methyl radical that we just formed, it's going to grab a chlorine. So we're going to break this chlorine-chlorine bond homolytically. The electron from the chlorine on the left is going to combine with the unpaired electron on the methyl radical and we're going to get our product. And that's the other thing that we're going to see in a propagation step. So that's one chain, that's one link in the chain. And as you can see, what we've done here is we've regenerated the radical that we started with. So this is going to come in and it's going to, there's a link in the chain and it's going to come in and it's going to do it again. It's going to keep cycling through. So we regenerate the radical we started with. You can, and I'm going to show that, that's a termination step. Okay, I'm going to show that in just a second. So we regenerate the radical that you started with. And the other thing I want you to notice is that we've always keeping the same number of radicals in propagation steps. So in this one, we start with a radical, we end with a radical. Here we start with a radical and in a neutral molecule we end with a radical, I mean, or a non-molecule with all electrons paired. So start with a radical and with a radical, start with a radical and with a radical. So when you're sorting those problems and sapling into boxes, that's what you're looking for. The number of radicals stays the same. One here, one here, one here, one here. In the first one we started with none and made two. That's what makes that an initiation step. So note, the number of radicals stays the same in propagation steps. So definitely want to make sure that you look for that. Now termination is really the opposite of initiation. In initiation you start with something that's not a radical and you break it into two radicals. In termination you combine two radicals and you make it into something that's not a non-radical. That's a termination step. So that's what you're going to look for when you're doing this ear sorting. So pretty much any radical that we have here can combine with any other radical. So for example we could have a chlorine radical and this is to answer your question combining with a methyl radical. So it looks like that. That would give you the product that we're going to get anyway. It's just made by a different pathway. That would give you methyl chloride. We could have two methyl radicals combining. That will give you ethane. So that will give you ethane. You can also have two chlorine radicals recombine. So these end up being some side products that you get. Well the first one's not a side product. The third one's not a side product but certainly ethane is not something that we're trying to make. Usually we don't get very much of these termination steps because let's imagine that we start off with one molecule of chlorine breaking. So let's imagine we have just one molecule do this. So we only have two chlorines here. Two chlorine radicals. What are the chances of those two chlorine radicals? They're going to be stirring in that reaction. They're going to be moving around doing this reaction over and over again in the propagation. The chance of those two actually interacting and colliding and recombining is very small. So there will be a lot more propagation than there will be termination. You still do get these side products. And so that's why it's not a very complete reaction because no one's going to want to have to separate all of these possible products here. So you see we have a lot. Here we have methyl chloride. Now you can imagine to form this one right here, a radical would come and remove another proton from methyl chloride. That's how we would get some of this. To get this, this is chloroform. We would have a radical come and remove another hydrogen from carbon here. And then that's how we would form this. You could do that four times. You could replace all four hydrogens with chlorine. And here, this product right here would be from the ethyl that we formed in our termination step. So you get a lot of, it's kind of a messy reaction here. So the competing propagation step here, I'll show formation of the dichloromethane here. So you can see what I'm talking about here. If this comes in and grabs a proton, instead of from methane, it grabs a proton from methyl chloride, then you form this radical. And now what's that radical going to do? That radical is going to come and grab a, it's going to come in right here. Where are we? It's going to come in here. It's going to grab a chlorine and that's how you get dichloromethane. Let's show that. So kind of a messy reaction here. Just my single headed arrows here. Everybody see how that's formed? So what did we see? We said about 80%, we get about 80% methyl chloride in this reaction and we get about 20% of these other little side products that have to be separated. One of the ways to get this reaction to go a little bit better and a little bit cleaner is to keep the concentration of these side products low. Keep concentration low by using a large excess of methane. Alright so since methane is a gas, you would use a large excess of methane and then that will increase the amount of methyl chloride you get and some of these other little side products will be much less. Question about that mechanism, anybody? Alright so as we say on the next page, this isn't a very good reaction because it's not very clean. It's not the best way to synthesize alkyl halides, there are better ways but it's still useful because it is the only way to convert an inert alkane into a reactive compound. It's the only way that we know to take an inert alkane like propane, ethane, methane and add a functional group on it. Once we put a chlorine or a bromine on there now we can take that and do all the other reactions we know that we can do with alkyl halides. But until we do that we're stuck. So this is, even though it's not a very good reaction it's the only thing we have. So another better way to synthesize alkyl chloride, we learned in chapter 10, right? If we do it this way we don't get a million little side products. So that's what we would do if we wanted to make an alkyl halide for alkyl bromides, we would use HBR. So there are better ways. That's the difference here though. Are we starting with a compound that doesn't have a functional group? No, we're starting with a compound that's an alkene that has a functional group. It has a functional group that can be converted into other functional groups. So this is different. It's a better way to make alkyl halides but there are starting material has a functional group right here. There's a functional group, that alkene is a functional group and so we're just converting it to another functional group. If we have an alkene we have no functional group. This has no functional groups. So we can't take base and remove that proton because it has a pKa of 50. So this is the only way that we have to put a functional group on there and then once we do that we can convert it into other things. Questions so far? Anybody? Let's look at the second mechanism. This one's a lot cleaner. So we take tertiary butane, bromine, light and heat and let's draw our product. Single product here, we don't get a lot of different products here. Alright so initiation, erratical form, we always break a weak bond. We got a weak bond here. Did I not leave you time? Sorry about that. Okay so we're going to break a weak bond. That's easy, break a weak bond homolytically. We start off with a non-ratical. We end up with two radicals. Now again I want to make the point that not all of the bromine, not all of the molecular bromine bonds are going to break. Just a few of them. Not all of them are going to break. We only really require one to break and that's enough to get all of our starting material converted into products. So propagation and I drew some of these already to make it a little easier. I should probably draw that bond a little longer. So we go here. So tert butyl radical plus HBr. And then the tert butyl radical comes in. It's really the same mechanism here. Grabs of bromine. So we still have molecular bromine. We still got this bromine in here that hasn't been broken apart. And this is going to come and grab a bromine from that. And when it does that it forms our product and we always want to look for this. It's going to regenerate the radical that we started with at propagation. So always the last step here is going to be regenerating the radical that we started with and definitely you want to look for that. So this one here. That's what we started with. And now it's going to cycle back through again. So most of the bromine is going to be broken apart this way, not this way. Questions I hear a lot of. Oh, what did I do? Oh I forgot. Yeah, yeah, yeah, thank you. I knew I did. Which means one of two things I've either scrolled too quickly or I made a mistake. How about that? That's looking too much like a dog. Let's try and fix that. How about that? Is that better? Okay. Regenerate the radical that we started with. All right. Termination. Just destroyed by recombining any two radicals here you fill in. There's three possible. And by the way on the test I will have an initiation section. I will have a propagation. I will have a termination. And what I always do on the test is I say show me one termination. That's all I want to see. You don't have to draw all three. I just want to see one. So whatever one you think you like the best. That's what you can include. All right. So possible mechanism. Radical halogenation. I will have some sort of alkane. I will have either bromine or chlorine and light and heat and all that. That's one possible mechanism for a midterm two. All right. I want to say something about selectivity. Very selective with bromine. It turns out that bromine is a much more selective reagent than chlorine and we're going to explain why that is. So let's look at this. We've got one tertiary hydrogen. Let's remind ourselves with a tertiary hydrogen as a tertiary hydrogen is a bonded to a tertiary carbon. That carbon is bonded to three other carbons so it's a tertiary carbon. And we have how many primary hydrogens? Nine equivalent tertiary hydrogens. Our primary hydrogens, nine primary hydrogens. So we've got nine to one ratio of primary tertiary hydrogens and we still get no primary hydrogens abstracted. We get no product that comes from our primary hydrogen being extracted. So statistically you are nine times as likely but you don't get any. You don't what do you not get any. You don't get any primary hydrogen abstracted so you don't get any of this product here. This product would result from a primary hydrogen being extracted. Any one of those nine would give you this product and you don't get any of that. So let's put a big X through that. None of that product is formed. So we need to explain why that is. There's two factors that we need to talk about to explain this. Number one is the relative stability of the radicals formed. So we have tertiary radical more stable than secondary radical, more stable than primary, more stable than methyl radical. So if we remove one of those primary hydrogens we're forming a primary radical. And so maybe not surprising that we don't get any of that formed at all because we know that primary radicals are very unstable. Similar to carbocation primaries, very unstable. The other thing that's not obvious is that we have to look at the ease at which the different hydrogens are removed and in order to do this you can look at the bond association energies and energy of activation. So we actually have to do some calculations. We're not going to do any calculations. So but I want to show you that with the relative rates of hydrogen abstraction by halogen atoms for bromine primary is 1 to 82 to 1640. So tertiary forming a tertiary radical is 1640 times more likely than forming a primary with a bromine radical. And since we only have a 9 to 1 ratio there's not a chance statistically that we're going to be removing one of those primary hydrogens. The selectivity for tertiary is way too great. Chlorine on the other hand we're going to just relative rates, we'll have 1 for primary, 2.5 and 4. So if we did this exact same reaction with chlorine we will get some of the primary product, and that's because there's only a 4 to 1 selectivity for tertiary over primary is 4 to 1 and we have 9 times the number of primary carbons so we will get primary here. So much more, much less selective. And you're beginning to see why I tend to favor bromine because it's a better leaving group, it works better in this reaction and it's better in a lot of different reactions so I tend to favor it. Alright so why is the bromine atom so much more selective? It seems like it just should have to do with carbocation stability but it's not. Remember we're doing other things in this reaction. So it turns out if you compare transition states and activation energies for the abstraction of a hydrogen atom by bromine radical versus a chlorine radical you will find that. Number one, abstraction of a hydrogen atom by bromine is endothermic while abstraction of a hydrogen atom by chlorine is exothermic. This sounds crazy difficult, it's really not that bad. And number two, the transition states for the endothermic, the bromination have a larger energy difference than those for the exothermic chlorination even though, and that shouldn't be on that page, but even though the energy difference in the products is the same in both reactions. Okay it's like whoa, wrap your head around that, right? So let's see if I can explain this graphically here for you. So what we said on the previous page is abstraction of a hydrogen atom by bromine is endothermic while abstraction of a hydrogen atom by chlorine is exothermic. So here is alkane and chlorine exothermic. So this radical here is down here. And then if we do bromine it's endothermic so this radical is higher in energy. Okay so let's label that. This is exothermic reaction and this is endothermic reaction. The rate of the reaction is going to have to do with the height of that energy barrier. How do we know about the structure of the transition state in an exothermic reaction? Does it look more like the reactant for that reaction or the product? It's closer in energy to this and we're starting off with one thing. We're starting off with one thing, we have a very exothermic reaction. The difference in the energy between these transition states is so close because it looks more like the alkane than it does the radical. Very very close in energy. So what that means is that when you do the chlorination you're going to be equally likely to take all three of these pathways. Yes, the red pathway down here is lower in energy but these are all very close by so they're all accessible. Over here we have an endothermic reaction by Hammett's postulate. We know that in an endothermic reaction the transition state looks more like the products of that reaction. And so since there's a big energy difference here between these three products there's also going to be a big energy difference here between the transition states. And so you can start to see now that with bromination this lower energy pathway is so low and these are much too high for it to overcome that energy barrier. So all the molecules are going to take that pathway because there's a big energy difference here. Here there's not, this is a small energy difference. So we definitely want to label that a little better here. Let's label that. So exothermic reaction therefore the transition state resembles. And we're only looking at one step here. So the transition state resembles reactants more than products. And an endothermic reaction the transition state resembles products more than reactants. And so if you look here this is a really small energy difference between these guys. These are two tiny little arrows. I don't know if you're going to be able to see those but very small energy difference. It is less than 1 kcal per mole. So three very accessible transition states and so you're going to get a big mixture here. And here these guys here this is a large energy difference. And what do we have here? This is about, I may be exaggerated a little bit more. This is 2 kcal per mole. And that's enough that you, so if you add 2 kcal per mole, if you've got tertiary versus primary that's 2 kcal per mole plus 2, we're talking about 4 kcal per mole. That's a really large energy difference for transition states. So we have the large energy difference here. This one has a much bigger energy difference. Therefore the rates of the reactions will be markedly different. And this is a really good example of the point that when you're talking about rates of reaction it does not have to do with the intermediates. These intermediates and these intermediates have the same energy difference. And if it was only because of that then we would have the same product outcome whether it's bromine and chlorine. It has to do with the height of the transition state here. That's what's key. Alright so here's an example here to show you what this means, difference wise in selectivity. So here we have bromination, here we have chlorination. With bromination we get one product 82%. With chlorination we get 26% of this product, 22% of this product, 22% of this product, 14% of this product and 17%. And you can actually predict these percentages by taking into account the difference in selectivity of chlorine versus the number of hydrogens that you have and you can actually calculate these differences. The textbook we had before this, had you do that, this textbook doesn't have it so I'm not going to have you learn that. You'll be glad about that. You're very glad. It's not that hard but you're glad. Alright questions on bromination of alkanes or chlorination of alkanes. That's reaction number one from this chapter. Radical substitution of benzylic and alec, that's reaction number two in this chapter. We are going to skip the mechanism here. But you need to know the reaction. Poor synthesis. Why am I having you skip the mechanism? It's a confusing mechanism. It has both single headed arrows and double headed arrows and it's just so hard to wrap your head around when you're just learning organic chemistry for the first time. How do I know that because I remember having to memorize this mechanism. And a lot of mechanisms that I didn't really need to memorize. So I'm having you, I'm selectively having you memorize certain things. As little as possible though. So we're skipping the mechanism and you don't need to know that. Recall that allyl and benzyl radicals are both more stable than tertiary radicals. We know the more stable the radical the faster it can be formed. So a hydrogen bonded to either benzylic or alelic carbon can be preferentially be substituted in a halogenation reaction. So again, these guys are more stable than tertiary. So that means that if you have a choice between removing a tertiary hydrogen or benzylic or alelic, benzylic or alelic will win. So here's some examples here. Br2, we can make this radical right here. We won't form a primary radical, we'll only form that one. And then if we carry that on, we will get benzyl bromide. A substituted benzyl bromide. So the bromine will go in the benzyl position, nowhere else. And you put that bromine. If we have allyl, then we can put, we have a choice between vinyl, hydrogen or allyl. So here's vinyl right here. Are we going to form any of that? Not when we've got allyl that we can form, right? So that's way better. So that means that we can form the allyl radical. Here we can convert that into alyl bromide. So who sees a competing reaction that can form with this, these two? Who sees another reaction that's competing with that? From chapter 10. Yeah, and you know, bromination of an alkene is an extremely fast reaction. You guys did this in the lab, didn't you? How fast was it? It was really fast. So how are we going to do this when we can do this just as fast? Okay, so I'm going to give you a different reagent for this. We're not going to use this reagent. We're going to give you a special reagent for this, then I want you to use instead. And that's because we have competing reactions here. With bromine, we can form a bromonium ion. Let's remember this. We don't want to forget this. This is midterm two material, right? Not to bring up an unpleasant subject, but you know, right? Nobody likes taking tests. In graduate school, in chemistry, you take one year of classes and then you work in a lab for the rest of the time while you earn your PhD. And I remember thinking at the time, I've only got one more year of classes in me. I cannot do this for two years. I just was so burned out on taking tests. So I haven't had to take a test since then, and it's a beautiful thing. Because nobody likes taking tests. All right, so remember this, okay? So that's a competing reaction. That's a competing, very fast reaction. So really, let's not use bromine for this. Even though Smith uses this a lot in synthesis, I want you to not use this reagent. And that's the reason why. We'll stop right there, and we will finish chapter 15 next time. So this is the reagent I want you to use instead.
arrow pushing. And then the second type is where arrow pushing and all of the mechanisms that we to combine with the unpaired electron on the let's label that. This is exothermic reaction and