Lecture 26. Electrophilic Aromatic Substitution, Part 3
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Vranken, David Van
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. This video is part of a 26-lecture undergraduate-level course titled "Organic Chemistry" taught at UC Irvine by Professor David Van Vranken. 
 I just saw a reference to a news article in Google about some new kind of compound or actually a polymer that could be used to remove toxic chemicals, not as a source of environmental cleanup but from the air that you breathe. The article refers to something called POPs, that's clearly an acronym. So I wanted to find out more about what is this POPs business and what does it have to do with cleaning up the air that you breathe. This is work that was developed by Son Bin Nguyen at Northwestern University. The basis for this idea POPs, it's based on a cool reaction that you've never seen before. It's a cobalt catalyzed cyclotrimerization. So you've never learned about any cool ways to make benzene rings. What we've spent our time in this chapter doing is talking about ways to substitute benzene rings. In fact, benzene is very cheap. You typically buy benzene and then add substituents to it. But here's a reaction where you take alkynes and you can ignore these green dash lines for the moment here. And when you throw in a cobalt catalyst and you heat this, if you've chosen the right cobalt catalyst, the cobalt atoms will grab the alkynes and start forming carbon-carbon bonds. So a cobalt atom will grab two alkynes and make a carbon-carbon bond here. And then it will maneuver over and make a new carbon-carbon bond here. And then it will grab another alkyne and make a new carbon-carbon bond there. And if you've designed your alkynes very cleverly, and this is what the shtick here, if you've designed them very cleverly, if you have an extra alkyne on the other end, now that one can engage in this cyclotrimerization reaction. And if there's another alkyn on this end of this other substrate, that can now engage in cyclotrimerization reactions. And if over on this piece here you've got three more alkynes dangling off, that can engage in three more alkynes in a cyclotrimerization reaction. So you end up with these benzene rings that serve as rigid spacers. And at the ends of those you have sites for polymerization. You get this massive polymeric network composed of these rigid benzene rings, and it has lots of space inside. It can't possibly pack very closely. So if you take these kinds of monomers, here's a monomer with two alkynes, and just ignore that functional group for now. And here's a monomer that has four alkynes. It's shaped like a tetrahedron. It's sort of because it's based on this sp3 hybridized center. And if you apply this cobalt catalyzed polymerization reaction making benzene rings, all the benzene rings are highlighted in red. And what this group did is that they incorporated these two alcohol moieties in a protected form and they hydrolyzed off this group. That's chemistry you'll see in Chem 51c. And it left behind two oxygens ready to grip onto metal atoms, and in particular copper and zinc metal atoms. So what that means is this very porous polymer now has metal atoms, and copper is exceptional at binding to amines. So if you're breathing in some sort of toxic vapor that has bad stuff, this will help absorb those vapors, and the copper atoms specifically will latch onto amines and suck them out of the air very quickly. So for example, Vx nerve gas has amine groups on there, and wouldn't it be nice to have some polymer that selectively binds to those kinds of amines? OK, so this is a very cool way to make benzene rings using transition metal catalysis. And it really gets closer to the kinds of reactions that are being developed now in the field of organic synthesis. So let's switch back to chapter 18 stuff in the Gorginski-Smith textbook, and we'll talk about electrophilic aromatic substitution. We left off with one of the ideas that I introduced to you during our last lecture was that if you do electrophilic aromatic substitution on a benzene ring that already has a substrate there like this R group up here, that R group will now control where the second addition of a functional group will go. Groups that are very activating and speed up electrophilic aromatic substitution tend to be ortho-pair directors. They tend to cause the next substituent to add to the ortho-in-pair positions, and usually that's a resonance effect with lone pairs, but alkyl groups can also do that. And then if you look at the groups that are deactivating, which tend not to have lone pairs but have electronegative atoms, those tend to slow down electrophilic aromatic substitution and also lead, they slow down, in particular they slow down addition to ortho-in-pair positions, and so it leaves you with meta-substitution. Those are meta-directors. So I showed you this chart, and I expect you to know this chart, and you'll see in just a little while why you really need to memorize this chart. There's some things in here like this, the position of halogens. That's a deactivator, but it's still ortho-pair, and you just need to memorize that. There's no simple principles you could use to have guessed that. Now when we left off, I showed you these special rules for Friedel-Kraftz reactions. I said Friedel-Kraftz reactions have all these exceptions you have to memorize, and so it is cool to form carbon-carbon bonds. It's undeniably cool, but the problem with Friedel-Kraftz reactions is that acyliam ions are nowhere near as reactive as nitronium ions or those other electrophiles we showed you. Carbocations are nowhere near as reactive as these other kinds of electrophiles that I showed you. So anything below this line here, and let me draw that line, anything below this line, any functional group below this line deactivates the benzene ring so much that you can no longer do Friedel-Kraftz reactions with acyliam ions and carbocations. And that line there just happens to correspond to this line between meta-directors and ortho-pair directors. It's just a happenstance. So we've got this simple rule that any benzene ring that has a meta-director on it is simply too deactivated to work with Friedel-Kraftz acylations or alkylations. Anything above the line you can do it on, but then there's this additional problem, and I don't want to go into the details of this, but that is that if you have amino groups on benzene rings with two alkyl groups or Hs on the amine, that's too activated. It's too activated to work in Friedel-Kraftz reactions, acylations and alkylations. So you can do other types of electrophilic aromatic substitution, but not CC bond formation. So those are all those exceptions or some of the exceptions with Friedel-Kraftz alkylations and acylations. All those horrible exceptions. Okay, so let's continue to talk about exceptions. What are some more of the exceptions that we have to remember that's going to make it difficult? And let's focus again on this over-activated end of that chart, the top end of the chart where things are just too active and get too hard to control. And let's talk about anilines and phenols, and I'll start off by talking about phenols. Boy, that lone pair on oxygen, if you have a hydroxy group on a benzene ring, that really donates into this benzene ring and activates that benzene ring to the point where once you add a single chlorine atom, and it is true that a chlorine atom is slightly deactivating, but this ring is still so activated that you cannot stop a second chlorination. So it adds again. And you still cannot stop another chlorination, so it adds again. Even if you add only one equivalent of Cl2, chlorine, you'll still get trisubstituted products coming out of this. You just can't stop this. You'll get this ugly mixture. So you can't stop over chlorination when you're using this super active recipe with chlorine and ferric chloride. But there's a simple thing that you can do that will allow you to get monosubstituted products, and that is to leave out the ferric halide catalyst, to leave out the ferric bromide, FEBR3, or to leave out the ferric chloride, FeCl3. If you have a phenol and you want to add just one halogen, and invariably you only want to add, nobody wants to add three halogens. If you just want to add one halogen, leave out the FEBR3. It's that simple. That's how you stop. Now you're still going to get a mixture of ortho and para. That's okay if we get a mixture of ortho and para. At least we didn't add all those extra halogens on here. So those hydroxy groups activate benzene rings so much that we get over halogenation unless we leave out that ferric chloride or ferric bromide catalyst. So hydroxy groups are very, very activating. Amino groups are more activating than hydroxy groups. And it turns out there is nothing you can do if you've got a dialkyl amino group or just an NH2 group sticking off of a benzene ring. There's nothing you can do. You're going to get overbromination in these cases. So in this case, if it's, I'll do the case where it's two methyl groups there. If you've got a dialkyl amino group, you're going to get every single ortho and para substitution brominated. And it doesn't matter whether you include the ferric bromide, you may think, oh, well, I'll just leave the ferric, no, that's not going to help. It'll still overbrominate. Remember, these anilines are at the very top of our activated benzene rings list and you just can't stop overactivation. So that makes, so again, there's all these exceptions and you have to memorize these exceptions. I expect you to know those exceptions for electrophilic aromatic substitution. I expect you to know that if you want to brominate phenol that you have to leave out that ferric catalyst and just go with straight bromine. That's how activating that hydroxy group is. Yes? Is that all about the monobrominates? With amids, you can mono-brominate and we used to do that in the 51L labs and I don't know if we do that anymore. So in other words, if you had an amide and you're, we're going to cover amids in amazing and infinite detail next quarter. So if you have a carbonyl group on there, you can mono-brominate that. And again, they used to do that, I'm not sure if they still do that in the 51L labs in 51LC but you may do that. So here you could mono-brominate and what's particular about that is that the carbonyl group is electron withdrawing and these lone pairs spend a lot of time donating into the carbonyl instead of into the benzene ring. But you don't know how to make amids yet. You don't know how to hydrolyze those back off. Wait till Chem 51C and we'll show you that and they will be asking you about this amid case in Chem 51C. They're going to check to see if you remembered this over halogenation of amids. Okay, so that's more exceptions and more exceptions. Let's stop talking about the exceptions and let's start talking about some powerful tools to help you further figure out how to selectively add substituents to benzene rings. Now I talked about this case. What happens if you have a substituent on a benzene ring? You need to know which groups are ortho-paradirectors. You need to know which groups are metadirectors so that you can make a prediction where the second substituent is going to go. But what happens if you have two substituents on a benzene ring? How do you predict where the third substituent is going to go? Which of these groups do we pay attention to when we look at a ring? And we have a very, very simple and powerful rule to help us predict which groups will control the regiochemistry. And our rule says the strongest activating group on the benzene ring controls the regiochemistry, controls this issue of ortho-para versus meta. So if I take this top example here, I think this is called cresol, this para-methylphenol here. I've got two groups there. How do I know whether the brominate, and here's one of these examples where I'm going to brominate and I see the hydroxy group. So I'm smart enough not to add that ferric chloride catalyst because I don't want to over brominate it. I just want to brominate it once. And so which of these two groups controls the regiochemistry? According to the rule I gave you, I go to the highest group on this list and it's over here. It's a hydroxy group. So in other words, the hydroxy group controls the regiochemistry, not the methyl group, and hydroxy groups are ortho-para directors. Let me just write that here. They direct ortho-para. They're alkyl, that's not as high as hydroxy in our list here. So the product of this is going to have a halogen wherever the hydroxy group makes the halogen go. And so I'm going to put that, I can't put it here at the para position, there's already a methyl group there. So really the only product you'll get out of this in high yield is going to be that monobromination product ortho to the hydroxy group. It's the hydroxy group that controls the regiochemistry. Okay, let's go down to this next example. Here I've got a carboxylic acid substituent and again I have a methyl substituent on there. And so when I think about where that third substitution is going to occur, I have to decide whether it's the carboxylic acid or the methyl group that's going to control the regiochemistry. And so when I come over here and I look at my list, well carbonyl groups are down here, solphonic acids, ketones, aldehyamides, any carbonyl compound. But higher than that are alkyl groups like methyl. And that is once again, that's an ortho para director. And you need to know that. So when I look at the two groups, the methyl group is controlling the regiochemistry and the para group, the group para to methyl is already substituted. And so in this case the new substituent, the solphonic acid substituent is going to occur ortho or para, but para is blocked ortho to the methyl group. So that's how you use that rule. It's a pretty simple rule but it requires that you have memorized this list. And you just straight up have to memorize that list of activating, deactivating and ortho-paradirecting. So know that list and know how to use it. So that's the simple idea there. So let me just circle again the groups that are controlling the regiochemistry. This is the group that controls the regiochemistry here. It's ortho-paradirecting. And down here it's the methyl group which is controlling the regiochemistry. And that's ortho-paradirecting. If you have two meta-directing substituents, it's the most activating or maybe least deactivating meta-substituent. You just find the highest group on the list. Okay so and there are problems and many, many problems in the back of the chapter that you should work that will give you practice with that. There are problems on the online sapling homework that will give you practice with that. So practice that. Work all those problems and keep working. Okay, here's an observation that's based, you know, sterics is really overblown as a source of control in organic chemistry. Usually resonance effects are much more important in organic chemistry. What's symbolously referred to as electronic effects that involve molecular orbitals, explanations. But here's a very simple steric effect and the idea is simple. If you have two groups that are meta to each other, like this, if you have a choice, for example here, this group is an ortho-paradirector, so both of these are equally activating. So if this group is an ortho-paradirector, it's going to direct the new substituents to go ortho or para. And this methyl group just so happens because of its positioning will direct to the same positions. So now I have three positions and really these positions here, this position here is different from these two here. Position two and two are both identical. There's a symmetry element in this molecule. You couldn't tell if I substituted at the top position two or the bottom position two. And so how do you decide? And our simple rule is that if there's a choice, you don't substitute between meta-substituents, right here in between it's too sterically hindered. It will be faster to substitute not between those two methyl groups. So here, even though there's all these positions where you could substitute, you only get one product as the major product and in good yield. You'll get very little of the product where you substitute between these two methyl groups, even though our ortho-paradoles predict that this middle position is activated. So one product would have the carbonyl group down here and the other product would have the carbonyl group over here at this other position two, but that's the same. That's the same product as I just drew. So simple rule, it's based on sterics and I hope, maybe that's kind of obvious that it's a little more hindered between the methyl groups, but you wouldn't have known that that would be useful enough to actually really dissuade groups to go in between. Okay, so again, if there's a choice, don't add between the two meta substituents and that's another thing that you have to look out for and it will help you to do synthesis and to answer questions in the back of the chapter and in the sapling problem sets. Okay let me show you a useful reaction. We covered all this radical chemistry back in Chapter 15. It was all kinds of radical substitution stuff. I'm glad you guys are seeing this too. I thought this was just me. Let me see if I can stop this. That's driving me insane. Let me go back to PC here. Okay, let me, let's see what we can do here. This looks like it's on the PC end of the, let me try turning this thing off. Just give me a minute here. We'll sort this out. Okay. This is going to kill us. Okay, this seems to be a doc cam problem. So let me just check out what we've got here. I may not have this lecture on my, I don't know how to fix this doc cam so we may have to make a choice here between chalk or, okay chalk, let's do it. Okay people in the back. Yeah well, I'd like to be able to do that. I apologize for the doc cam here. I don't know how to fix that. Maybe I can give it one more shot here just to see if we can make this work out. Yeah, the last time I taught Chem 51B it was chalk and people in the back complained that they, even, and I write with the big chalk, they complained they couldn't see my writing and I thought I wrote big. So okay, let's see if we have any chance here of not spazzing out. If it spazzes out again I'm going to go right to the chalk and I'll write big to help you. So far so good? Okay, let's cross our fingers here. Okay, so all of that radical stuff and you're never going to see radicals again? Yeah. Okay. I'm considering getting it a couple times. Okay, let's go to chalk. I'm sorry for the, you guys in back I'll try to make an attempt to make sure that we can see what's going on. Okay, I want to remind you of chapter 15 stuff. You learned all of this stuff in section 15.10 about allillic bromination and in cases like this when we treated allillamines with NBS and heat or peroxides you don't use BR2 on here because that will add bromine across the double bonds. We use this N-bromosexinimate compound. The problem with this is you get a radical where the, an allill radical where the BR group can substitute on both ends of that allill radical and so you almost always got these hideous mixtures when you do allillic bromination. And who wants a mixture? And there's also on top of that this issue that you generate new stereogenic centers. That's not such a big deal. But nobody likes to get mixtures of compounds as products in organic chemistry. And so what I'll show you now is if we do the same reaction and instead of doing it on this kind of an allill substrate we simply switch to a benzylic substrate it does the same stuff. So let me redraw that same substrate here and all I'm going to do is make this part of a benzene ring. And I do the same conditions N-bromosexinimate and light. But this time I don't get mixtures of products. You'll get selective bromination of one of the two H's that are at these benzylic positions. And that makes this way useful whereas allillic bromination is not way useful. I would say way useless would be a better description of allillic bromination. You have to have super symmetrical substrates for allillic bromination to give you clean product mixtures. So look at this. I can replace one of the two H's with this same reaction that you learned in chapter 15. And why is that important? Because that's a leaving group. We can go from something that has no functional groups here really to something that has a leaving group at that position. And now you can do SN2 reactions, elimination reactions. So that's very powerful. This idea that we can add leaving groups wherever you want to have those. Because there's no easily bromonatable double bonds in a benzene ring, you don't have to worry about using N-bromosexinimate. So in other words, we can take this same kind and you can have other substituent on the benzene ring. They don't get affected in that reaction. I don't know if this is a distraction that we erased this. And because we don't have bromonatable double bonds where you add Br2 across, you can use Br2 in this reaction if you want. In fact, that's more typical. And the book shows you other types of ways to do this. Or you could simply heat the reaction so hot that you generate small amounts of radicals. So you can't do allillic bromination with Br2, but you can do benzylic bromination with molecular bromine. But it works just fine. And look how powerful that is that we can add a leaving group anywhere that we want to have a leaving group. So you should know that reaction and be ready to use that reaction. Because again, that's very powerful to add a leaving group. OK, I'm going to show you two more useful transformations of side chains. So really what we're talking about now is what are the useful things you can do once you've added a side chain to a benzene ring? How do we turn that into a useful functional group? And let me show you two powerful reductions of carbonyl groups. When there's an arrow group on one side, these tend to be very efficient reduction reactions. So I'll show you a way to take a carbonyl group attached to a benzene ring and burn that down to a CH2 group. And so here's one version of that reaction. This is called a Wolf-Kieschner reduction. And you don't need to know the name of this reaction. It's not important. This is a reagent called hydrazine. There's an NN bond, H2NNH2. And you have to do this under strongly basic conditions, sodium hydroxide. And there's a third thing for you to remember, and that is you have to heat these reactions. These are not mild reaction conditions. So I'm going to write a big delta under here. So hydrazine, H2NNH2, sodium hydroxide, and you boil it. And when you do that, the product of this reaction is now missing that carbonyl. But it's not so much that it's missing the carbonyl. It's that you've now replaced that carbonyl with a CH2 group. And I'll draw that CH2 group for you so you can see. It's not that there's nothing there. There's H's there. And you don't need to know anything about the mechanism for this Wolf-Kieschner reaction. But I think you can see that's useful. And the reason why this is useful is because you can't make that product through a Friedel-Krafts alkylation. So I want you to recall that Friedel-Krafts alkylations on primary alkohalides suffer from this problem of migration. So I want you to recall this problem here. Well, actually let me cover that in just a second. Let me first talk about an alternative set of conditions. Suppose you can't remember hydrazine for some reason, and it's just completely escaping you. There's an alternative set of conditions we can do. This is a reaction called a Cleminson reduction. And it involves taking, you buy this stuff called zinc-mercury amalgam. Here's the way we represent zinc-mercury amalgam. We write it like this. You take zinc powder and you mix it with mercury, and the mercury absorbs onto the surface. So you put the mercury in parentheses. That's called zinc-mercury amalgam. And in this case, it's not strongly basic conditions. It's strongly acidic conditions. And again, these conditions are not mild. You have to boil your ketone with zinc-mercury amalgam and hydrochloric acid. So I'll draw a delta symbol here. That means you're boiling the heck out of this to get it to work. And this will do the same thing as that Wolf-Kieschner reduction with hydrazine. Boy, as soon as I see the CH2 group, I want to brominate it within Bromo6-Inimate. You can see why it's cool to introduce a CH2 group. OK, so this is a way to take carbonyl groups and reduce them down to CH2 groups. You're not going to see anything like that in other chapters. It's a, well, maybe there's a different type of reaction with lithium-aluminum hydrochloric acid. Never mind. In this chapter, that's the only kind of reaction that will take a carbonyl group and reduce it down. So let me make one important point to you, because you're going to tend to get this confused in chapter 20, that you cannot burn down the groups of esters or amids. There's a different side reaction I'm not going to show you. It has to do with hydrolysis. And so you're not going to be able to ever take something like a methyl ester or an ethyl ester and convert those into CH2 groups using this Wolf-Kieschner reduction. We may have some other chemistry to do something like that. But there's, but not exactly like that. So don't do this with esters, not with amids, only for ketones and aldehydes. That's the only place where this is useful. And you'll have a tendency to want to do that. Now, you might ask, which of these two should I memorize? Which of these two is more useful? Typically, the base version is more useful. The hydrazine and sodium hydroxide, because there are so many functional groups that are acid-sensitive, you're going to find that boiling things with acid is usually going to screw up some other part of your molecule. So I usually prefer this Wolf-Kieschner version, this hydrazine and sodium hydroxide. If you want to have, you know, if there's one thing you want to really commit to memory the most, that would be the one that would be most useful to you this quarter. Okay, so let me remind you of why this is useful. It's because of this limitation in the Friedel-Kraftz reaction. Remember this problem with primary alkohalides is that they will rearrange. And so it really limits when you do Friedel-Kraftz reactions. So this is just a continuation. Why would you want to do this Clemensen reduction or a Wolf-Kieschner reduction? If you wanted to take this benzene ring and do a Friedel-Kraftz reaction with this primary alkohalide, if you wanted to make that kind of product, look at this CH2 group here. You might see that and try to write this out as a reaction. But there's a problem with this. You wouldn't get any of the product I drew over there. What you would get would be a T-butyl group substituted on there. And the reason is when you draw out the primary carbocation like this, is that that will do anything so that it's not a primary carbocation. And it will do it quickly. The primary carbocation intermediate that you generate, and I'll draw the, let me draw that H2C group here, will very quickly rearrange to give you a tertiary carbocation. And that's what's going to attack the benzene ring. It's going to be the tertiary carbocation that attacks the benzene ring and does this Friedel-Kraftz alkylation reaction. So if you want to add a substituent like this over here, you have to do it in two steps. First you do a Friedel-Kraftz acylation. And then you use one of these reductions to reduce the carbonyl down to a CH2 group. So that's the way to do things. Now there is an exception to this. I said don't try to do primary alkohalides. But you can do, there are two alkohalides that you can do that kind of look like they're primary. And the book seems to love these and the online homework seems to love these. And I don't want you to worry about the mechanism. You can do it with this because methyl is not susceptible to any kind of a hydride migration. There's no 1, 2 migration possible for a methyl because there's not even a second carbon. The other type of starting material that you'll see used in Friedel-Kraftz alkylations, mainly in the book, not in real life, is ethyl chloride. And why is that not a problem with hydride migrations or alkyl migrations? It's because if you migrate a hydride from one side of an ethyl cation to the other, you get back an ethyl cation. So those are the only two instances of things that look primary where you don't suffer from this problem. In general, if you see a CH2 group attached to a benzene ring, you should make that through Friedel-Kraftz's isolation and then reduce the carbonyl group down to a CH2 group. That's the general way to attach CH2 groups to benzene rings. And that's the way I would recommend to you. So lots of stuff you can do with these carbon substituents. You can brominate them. You can reduce down carbonyls to CH2 groups. That's powerful stuff. And let's talk about carbon nitrogen bonds or nitrogen substituents on benzene rings. And I apologize for those of you in back. The problem is these erasers don't do the best job of erasing usually. And so you're trying to see my writing through this haze of chalk on the board. So let's talk about converting nitro groups, which are pretty useful. You can't make medicines that have nitro groups on them. Those cause problems with toxicity when they're metabolized. So if you want to make useful drugs that have nitrogen on benzene rings, typically invariably you're reducing those nitro groups down and doing something else with those. So here's an example of ortho-nitrotoluene to nitrotoluene. And it's very, very simple to reduce these down to give you something to give you something that's useful to work with. And it's the same conditions that you use for hydrogenation. And that's H2 and palladium on carbon. I usually write only solvents underneath the arrow. That's why I was hesitating there. So palladium on carbon is a catalyst. It's the same thing that you were doing with hydrogenation of double bonds. Typically we don't show the stoichiometry. In fact, it takes three equivalents of hydrogen. Generally I want you to write the stoichiometry. But for hydrogenations, you don't chemically do not draw out the stoichiometry for hydrogen. So I think you can see you're going to lose two H2O's at a minimum here. And then another set of H2's go on to the nitrogen atom. So we don't show stoichiometry for hydrogenation type reactions. Look how easy this. These are the best reactions ever. If ever you want to have a feel good reaction in the lab where you get 100% yield and no byproducts and all you do is filter your reaction, this is the way to go. Now the book gives you two other recipes for reducing nitro groups to amino groups. And this only works when they're attached to benzene rings by the way. You can't do this on alkyl nitro groups. Let's take a look at another set of conditions. Sorry, I don't know how to push this past so you've got some light here to see. So here's the other set of conditions that the book shows. The book shows tin and HCl. And then some books, I can't remember whether it's this one, shows SNCl2 and HCl. And that's actually more typical conditions, stanis chloride. But the book shows tin and HCl or iron powder and HCl. So tin powder, tin is SN, and hydrochloric acid. And the product of this reaction that they show in the book is the same thing that you get with hydrogenation. And so why does the book give you all of these different recipes? Why don't they just? I don't know. But why does the book show you in Chem 51, see all these recipes? Here's why any chemist would want to do this. It turns out that if you have a halogen up above here, those hydrogenation conditions will cleave the carbon halogen bond. Whereas the tin HCl conditions do not cleave the carbon halogen bond. And you're never going to run across that in the Chem 51 series. I'm not going to ask you that. The book is careful not to ask you that. So that's why people might not want to use the hydrogenation is because maybe they don't want to lose that bromine atom. There's a second little trick that I expect you to add that they didn't add here in this book. And that is, if you're stirring around an amine in hydrochloric acid, it's not the amine anymore. The product of this reaction is actually an ammonium salt. And I want you, even if the book doesn't, I want you and expect you to draw out that if you want the free amine NH2, that you have to treat this with sodium hydroxide to deprotonate that. And that's free-basing. Like when you free-base coke this weekend, this is what you're doing. You're deprotonating an amine. And I want you to show that free-basing step. If you do Stannous chloride, that's NH2 I'm drawing here. So don't just draw this one step tin in hydrochloric acid. I want you to do a second step with sodium hydroxide. And usually we just enumerate those. Step one, tin hydrochloric acid, step two, sodium hydroxide. So generally I would use hydrogenation if I didn't have some other reaction I was trying to avoid. Yeah? No, the byproducts here, you get 2H2O's as a byproduct. So you see two of the molecules of hydrogen combined with the oxygens give you two waters. The third equivalent of hydrogen ends up, don't worry about the mechanism. You don't need to know that. You just need to know that that's a fantastic reaction that works every time. Okay, last, let's talk about multi-step synthesis. You've learned a lot of reactions in Chapter 18, actually not that many. Five electrophilic aromatic substitution reactions. And then I just now showed you some fancy ways to dress up those side chains that you put on through reduction, et cetera. So for multi-step synthesis, when you do that involves electrophilic aromatic substitution, the main admonition that I can give you is to strategize regio control. If you want to put three groups onto a benzene ring with a super particular arrangement, you have to know which group you add first because each group is either an ortho-pera director or a meta director. And so you need to strategize regio control. So let's take an example where we want to do some sort of a synthesis. Suppose that you're asked by somebody to take benzene. And then through some number of steps show reagents and steps that would allow you to put on these kinds of groups here. And so you have to think about what do I add first and what do I add second? And what are all those exceptions that I have that I have to worry about? So let's go ahead and talk about this. How would we, typically when you're given a synthesis that you have to do, you have a choice to either work forward or work backwards? I usually start by working backwards. I usually start by saying, what are all the different ways that I could make this? And so what are some of the ways that you know for how to make that? Well, maybe one possibility is we could start off with nitro-benzene and then do some sort of an alkylation reaction. But you already know that doesn't work. Nitro groups are meta directors. Nitro-benzene is too deactivated for you to do any kind of Friedel-Krafts alkylations or acylations. So just forget that. That's out. Really your only choice is to somehow have some kind of a group here. But there is an alternative here. And that is maybe my last step in my synthesis was reduction of one of these carbonyl groups using that Wolf-Kieschner reduction or the Clemenson reduction. Maybe the last step in my synthesis was reduction of a carbonyl. This has to do with me being so clever that I could see that CH2 group. And I remembered those carbonyl reduction reactions that I just taught you. Maybe that was the last step. Or maybe the other possibility that we have here is that the last step involved in nitration. Now that actually looks pretty promising. And that looks promising because alkyl groups are ortho-paradirectors. So we could have made this by first putting some sort of an alkyl group here on the benzene ring and then doing a nitration reaction. Let's remind ourselves of what those conditions are for nitration. It's nitric acid and sulfuric acid. So of these choices, to me what's looking pretty promising here is this nitration. So maybe the last step in this synthesis is involved in nitration. But I think we can probably figure out how to put a propyl group on a benzene ring. And so how would we do that? How would we add a propyl group to a benzene ring? Well, this comes back to another exception that you can't directly add propyl groups to benzene rings. Let me just write that problem out for you here. What was the problem that we just finished talking about with trying to do this in one step? The problem that we talked about is that if you take chloropropane and you try to do a freol-crafts alkylation, that you'll end up with all these hydride migrations. That primary carbocation would be unstable. And if you made something that looked anything like a primary carbocation, I'm not saying that that's what would happen here, but if you started to generate something that looked like a primary carbocation, these hydrides would migrate over. And so before it does freol-crafts. And so in that case, you wouldn't have n-propyl groups attached to your benzene ring. You'd have isopropyl groups attached to your benzene ring. Right? So you can't do just straight freol-crafts to make this. How do we make this? Well, this is where those carbonyl reductions come into play. The best way to put that CH2 group directly attached to a benzene ring is to do one of these reductions. And I'm going to choose the Wolf-Kieschner reduction, that hydrazine one. There's an NN bond there. You don't have to draw that NN bond. And there's two other parts to this recipe. Sodium hydroxide and heat. And if you wanted to do that zinc-mercury thing with HCl and heat, that's perfectly fine, too. So now it's starting to look like we've got a root here. We could start here with benzene and do freol-crafts, acylation, and aluminum trichloride. And then for step two, we could use some sort of a reduction, hydrazine and sodium hydroxide. And then that's the molecule we have over here. And then our third step, we could nitrate. Now there's one little stickly point here that I'm leaving out. And this is the last thing that I'll tell you in this chapter. And that is in cases when you're doing synthesis problems, in cases where I ask you to efficiently make some product, I'm going to allow you to claim that separation of ortho- and para-isomers constitutes an efficient chemical step. And I want to add that in here. Ortho- in other words, if you take something that looks like this and you brominate this, you will get a mixture of ortho- and para-isomers out of this. And there's no way that you can get around that. So as one of your chemical steps in any synthesis, I will allow you to do this. And separate ortho-para. And then just draw whichever one that you want. I admit, yeah, you only got a 50% yield of each one. And that's not really that efficient. But we're just going to pretend that this overall process of making both ortho- and para, and separating out the one you want. Let me just draw one of those. Suppose I want the ortho here. We'll just pretend that that's an efficient combination of steps. So what I need to do here is acknowledge the fact that nitration is going to give me a mixture of ortho- and para. And I need to add an additional step here. So what we need to do here is write step. And I don't know how to raise this board here. So step one is going to be nitrate. And let me go back a little bit here. Let me do it like this. I'll start back over here with my propyl benzene. And I'll write step one, nitric acid and sulfuric acid. And then step two, just like I just told you, separate ortho- and para. And that's the way you deal with that. When you generate ortho-para mixtures, but you only want one of those. I would like you to do that. Okay, that's it. That's the end of what we've got to cover. And I have one last thing to tell you. Our final exam is on Monday morning at 8 a.m. Seating charts are posted and I'd like you to be in your seats at 10 before so we can start passing out the exams. And this weekend, drive thy business. And when you finish this class and you are going on into 51C, drive your business and invest in your career by learning organic chemistry. And so thank you. Have a great time with you this morning.
 I just saw a reference to a news article in Google about some new kind of compound or  actually a polymer that could be used to remove toxic chemicals, not as a source of environmental  cleanup but from the air that you breathe.  The article refers to something called POPs, that's clearly an acronym.  So I wanted to find out more about what is this POPs business and what does it have to  do with cleaning up the air that you breathe.  This is work that was developed by Son Bin Nguyen at Northwestern University.  The basis for this idea POPs, it's based on a cool reaction that you've never seen before.  It's a cobalt catalyzed cyclotrimerization.  So you've never learned about any cool ways to make benzene rings.  What we've spent our time in this chapter doing is talking about ways to substitute benzene  rings.  In fact, benzene is very cheap.  You typically buy benzene and then add substituents to it.  But here's a reaction where you take alkynes and you can ignore these green dash lines  for the moment here.  And when you throw in a cobalt catalyst and you heat this, if you've chosen the right  cobalt catalyst, the cobalt atoms will grab the alkynes and start forming carbon-carbon  bonds.  So a cobalt atom will grab two alkynes and make a carbon-carbon bond here.  And then it will maneuver over and make a new carbon-carbon bond here.  And then it will grab another alkyne and make a new carbon-carbon bond there.  And if you've designed your alkynes very cleverly, and this is what the shtick here, if you've  designed them very cleverly, if you have an extra alkyne on the other end, now that one  can engage in this cyclotrimerization reaction.  And if there's another alkyn on this end of this other substrate, that can now engage  in cyclotrimerization reactions.  And if over on this piece here you've got three more alkynes dangling off, that can engage  in three more alkynes in a cyclotrimerization reaction.  So you end up with these benzene rings that serve as rigid spacers.  And at the ends of those you have sites for polymerization.  You get this massive polymeric network composed of these rigid benzene rings, and it has lots  of space inside.  It can't possibly pack very closely.  So if you take these kinds of monomers, here's a monomer with two alkynes, and just ignore  that functional group for now.  And here's a monomer that has four alkynes.  It's shaped like a tetrahedron.  It's sort of because it's based on this sp3 hybridized center.  And if you apply this cobalt catalyzed polymerization reaction making benzene rings, all the benzene  rings are highlighted in red.  And what this group did is that they incorporated these two alcohol moieties in a protected  form and they hydrolyzed off this group.  That's chemistry you'll see in Chem 51c.  And it left behind two oxygens ready to grip onto metal atoms, and in particular copper  and zinc metal atoms.  So what that means is this very porous polymer now has metal atoms, and copper is exceptional  at binding to amines.  So if you're breathing in some sort of toxic vapor that has bad stuff, this will help absorb  those vapors, and the copper atoms specifically will latch onto amines and suck them out of  the air very quickly.  So for example, Vx nerve gas has amine groups on there, and wouldn't it be nice to have  some polymer that selectively binds to those kinds of amines?  OK, so this is a very cool way to make benzene rings using transition metal catalysis.  And it really gets closer to the kinds of reactions that are being developed now in the  field of organic synthesis.  So let's switch back to chapter 18 stuff in the Gorginski-Smith textbook, and we'll talk  about electrophilic aromatic substitution.  We left off with one of the ideas that I introduced to you during our last lecture was that if  you do electrophilic aromatic substitution on a benzene ring that already has a substrate  there like this R group up here, that R group will now control where the second addition  of a functional group will go.  Groups that are very activating and speed up electrophilic aromatic substitution tend  to be ortho-pair directors.  They tend to cause the next substituent to add to the ortho-in-pair positions, and usually  that's a resonance effect with lone pairs, but alkyl groups can also do that.  And then if you look at the groups that are deactivating, which tend not to have lone  pairs but have electronegative atoms, those tend to slow down electrophilic aromatic substitution  and also lead, they slow down, in particular they slow down addition to ortho-in-pair positions,  and so it leaves you with meta-substitution.  Those are meta-directors.  So I showed you this chart, and I expect you to know this chart, and you'll see in just  a little while why you really need to memorize this chart.  There's some things in here like this, the position of halogens.  That's a deactivator, but it's still ortho-pair, and you just need to memorize that.  There's no simple principles you could use to have guessed that.  Now when we left off, I showed you these special rules for Friedel-Kraftz reactions.  I said Friedel-Kraftz reactions have all these exceptions you have to memorize, and  so it is cool to form carbon-carbon bonds.  It's undeniably cool, but the problem with Friedel-Kraftz reactions is that acyliam ions  are nowhere near as reactive as nitronium ions or those other electrophiles we showed  you.  Carbocations are nowhere near as reactive as these other kinds of electrophiles that  I showed you.  So anything below this line here, and let me draw that line, anything below this line,  any functional group below this line deactivates the benzene ring so much that you can no  longer do Friedel-Kraftz reactions with acyliam ions and carbocations.  And that line there just happens to correspond to this line between meta-directors and ortho-pair  directors.  It's just a happenstance.  So we've got this simple rule that any benzene ring that has a meta-director on it is simply  too deactivated to work with Friedel-Kraftz acylations or alkylations.  Anything above the line you can do it on, but then there's this additional problem,  and I don't want to go into the details of this, but that is that if you have amino groups  on benzene rings with two alkyl groups or Hs on the amine, that's too activated.  It's too activated to work in Friedel-Kraftz reactions, acylations and alkylations.  So you can do other types of electrophilic aromatic substitution, but not CC bond formation.  So those are all those exceptions or some of the exceptions with Friedel-Kraftz alkylations  and acylations.  All those horrible exceptions.  Okay, so let's continue to talk about exceptions.  What are some more of the exceptions that we have to remember that's going to make it difficult?  And let's focus again on this over-activated end of that chart, the top end of the chart  where things are just too active and get too hard to control.  And let's talk about anilines and phenols, and I'll start off by talking about phenols.  Boy, that lone pair on oxygen, if you have a hydroxy group on a benzene ring, that really  donates into this benzene ring and activates that benzene ring to the point where once  you add a single chlorine atom, and it is true that a chlorine atom is slightly deactivating,  but this ring is still so activated that you cannot stop a second chlorination.  So it adds again.  And you still cannot stop another chlorination, so it adds again.  Even if you add only one equivalent of Cl2, chlorine, you'll still get trisubstituted products  coming out of this.  You just can't stop this.  You'll get this ugly mixture.  So you can't stop over chlorination when you're using this super active recipe with chlorine  and ferric chloride.  But there's a simple thing that you can do that will allow you to get monosubstituted  products, and that is to leave out the ferric halide catalyst, to leave out the ferric bromide,  FEBR3, or to leave out the ferric chloride, FeCl3.  If you have a phenol and you want to add just one halogen, and invariably you only want  to add, nobody wants to add three halogens.  If you just want to add one halogen, leave out the FEBR3.  It's that simple.  That's how you stop.  Now you're still going to get a mixture of ortho and para.  That's okay if we get a mixture of ortho and para.  At least we didn't add all those extra halogens on here.  So those hydroxy groups activate benzene rings so much that we get over halogenation unless  we leave out that ferric chloride or ferric bromide catalyst.  So hydroxy groups are very, very activating.  Amino groups are more activating than hydroxy groups.  And it turns out there is nothing you can do if you've got a dialkyl amino group or  just an NH2 group sticking off of a benzene ring.  There's nothing you can do.  You're going to get overbromination in these cases.  So in this case, if it's, I'll do the case where it's two methyl groups there.  If you've got a dialkyl amino group, you're going to get every single ortho and para substitution  brominated.  And it doesn't matter whether you include the ferric bromide, you may think, oh, well,  I'll just leave the ferric, no, that's not going to help.  It'll still overbrominate.  Remember, these anilines are at the very top of our activated benzene rings list and you  just can't stop overactivation.  So that makes, so again, there's all these exceptions and you have to memorize these exceptions.  I expect you to know those exceptions for electrophilic aromatic substitution.  I expect you to know that if you want to brominate phenol that you have to leave out that ferric  catalyst and just go with straight bromine.  That's how activating that hydroxy group is.  Yes?  Is that all about the monobrominates?  With amids, you can mono-brominate and we used to do that in the 51L labs and I don't  know if we do that anymore.  So in other words, if you had an amide and you're, we're going to cover amids in amazing  and infinite detail next quarter.  So if you have a carbonyl group on there, you can mono-brominate that.  And again, they used to do that, I'm not sure if they still do that in the 51L labs in 51LC  but you may do that.  So here you could mono-brominate and what's particular about that is that the carbonyl  group is electron withdrawing and these lone pairs spend a lot of time donating into the  carbonyl instead of into the benzene ring.  But you don't know how to make amids yet.  You don't know how to hydrolyze those back off.  Wait till Chem 51C and we'll show you that and they will be asking you about this amid  case in Chem 51C.  They're going to check to see if you remembered this over halogenation of amids.  Okay, so that's more exceptions and more exceptions.  Let's stop talking about the exceptions and let's start talking about some powerful tools  to help you further figure out how to selectively add substituents to benzene rings.  Now I talked about this case.  What happens if you have a substituent on a benzene ring?  You need to know which groups are ortho-paradirectors.  You need to know which groups are metadirectors so that you can make a prediction where the  second substituent is going to go.  But what happens if you have two substituents on a benzene ring?  How do you predict where the third substituent is going to go?  Which of these groups do we pay attention to when we look at a ring?  And we have a very, very simple and powerful rule to help us predict which groups will  control the regiochemistry.  And our rule says the strongest activating group on the benzene ring controls the regiochemistry,  controls this issue of ortho-para versus meta.  So if I take this top example here, I think this is called cresol, this para-methylphenol  here.  I've got two groups there.  How do I know whether the brominate, and here's one of these examples where I'm going  to brominate and I see the hydroxy group.  So I'm smart enough not to add that ferric chloride catalyst because I don't want to  over brominate it.  I just want to brominate it once.  And so which of these two groups controls the regiochemistry?  According to the rule I gave you, I go to the highest group on this list and it's over  here.  It's a hydroxy group.  So in other words, the hydroxy group controls the regiochemistry, not the methyl group, and  hydroxy groups are ortho-para directors.  Let me just write that here.  They direct ortho-para.  They're alkyl, that's not as high as hydroxy in our list here.  So the product of this is going to have a halogen wherever the hydroxy group makes the  halogen go.  And so I'm going to put that, I can't put it here at the para position, there's already  a methyl group there.  So really the only product you'll get out of this in high yield is going to be that monobromination  product ortho to the hydroxy group.  It's the hydroxy group that controls the regiochemistry.  Okay, let's go down to this next example.  Here I've got a carboxylic acid substituent and again I have a methyl substituent on there.  And so when I think about where that third substitution is going to occur, I have to  decide whether it's the carboxylic acid or the methyl group that's going to control the  regiochemistry.  And so when I come over here and I look at my list, well carbonyl groups are down here,  solphonic acids, ketones, aldehyamides, any carbonyl compound.  But higher than that are alkyl groups like methyl.  And that is once again, that's an ortho para director.  And you need to know that.  So when I look at the two groups, the methyl group is controlling the regiochemistry and  the para group, the group para to methyl is already substituted.  And so in this case the new substituent, the solphonic acid substituent is going to occur  ortho or para, but para is blocked ortho to the methyl group.  So that's how you use that rule.  It's a pretty simple rule but it requires that you have memorized this list.  And you just straight up have to memorize that list of activating, deactivating and  ortho-paradirecting.  So know that list and know how to use it.  So that's the simple idea there.  So let me just circle again the groups that are controlling the regiochemistry.  This is the group that controls the regiochemistry here.  It's ortho-paradirecting.  And down here it's the methyl group which is controlling the regiochemistry.  And that's ortho-paradirecting.  If you have two meta-directing substituents, it's the most activating or maybe least deactivating  meta-substituent.  You just find the highest group on the list.  Okay so and there are problems and many, many problems in the back of the chapter that you  should work that will give you practice with that.  There are problems on the online sapling homework that will give you practice with that.  So practice that.  Work all those problems and keep working.  Okay, here's an observation that's based, you know, sterics is really overblown as a  source of control in organic chemistry.  Usually resonance effects are much more important in organic chemistry.  What's symbolously referred to as electronic effects that involve molecular orbitals, explanations.  But here's a very simple steric effect and the idea is simple.  If you have two groups that are meta to each other, like this, if you have a choice, for  example here, this group is an ortho-paradirector, so both of these are equally activating.  So if this group is an ortho-paradirector, it's going to direct the new substituents  to go ortho or para.  And this methyl group just so happens because of its positioning will direct to the same  positions.  So now I have three positions and really these positions here, this position here is different  from these two here.  Position two and two are both identical.  There's a symmetry element in this molecule.  You couldn't tell if I substituted at the top position two or the bottom position two.  And so how do you decide?  And our simple rule is that if there's a choice, you don't substitute between meta-substituents,  right here in between it's too sterically hindered.  It will be faster to substitute not between those two methyl groups.  So here, even though there's all these positions where you could substitute, you only get one  product as the major product and in good yield.  You'll get very little of the product where you substitute between these two methyl groups,  even though our ortho-paradoles predict that this middle position is activated.  So one product would have the carbonyl group down here and the other product would have  the carbonyl group over here at this other position two, but that's the same.  That's the same product as I just drew.  So simple rule, it's based on sterics and I hope, maybe that's kind of obvious that  it's a little more hindered between the methyl groups, but you wouldn't have known that that  would be useful enough to actually really dissuade groups to go in between.  Okay, so again, if there's a choice, don't add between the two meta substituents and  that's another thing that you have to look out for and it will help you to do synthesis  and to answer questions in the back of the chapter and in the sapling problem sets.  Okay let me show you a useful reaction.  We covered all this radical chemistry back in Chapter 15.  It was all kinds of radical substitution stuff.  I'm glad you guys are seeing this too.  I thought this was just me.  Let me see if I can stop this.  That's driving me insane.  Let me go back to PC here.  Okay, let me, let's see what we can do here.  This looks like it's on the PC end of the, let me try turning this thing off.  Just give me a minute here.  We'll sort this out.  Okay.  This is going to kill us.  Okay, this seems to be a doc cam problem.  So let me just check out what we've got here.  I may not have this lecture on my, I don't know how to fix this doc cam so we may have  to make a choice here between chalk or, okay chalk, let's do it.  Okay people in the back.  Yeah well, I'd like to be able to do that.  I apologize for the doc cam here.  I don't know how to fix that.  Maybe I can give it one more shot here just to see if we can make this work out.  Yeah, the last time I taught Chem 51B it was chalk and people in the back complained that  they, even, and I write with the big chalk, they complained they couldn't see my writing  and I thought I wrote big.  So okay, let's see if we have any chance here of not spazzing out.  If it spazzes out again I'm going to go right to the chalk and I'll write big to help you.  So far so good?  Okay, let's cross our fingers here.  Okay, so all of that radical stuff and you're never going to see radicals again?  Yeah.  Okay.  I'm considering getting it a couple times.  Okay, let's go to chalk.  I'm sorry for the, you guys in back I'll try to make an attempt to make sure that we can  see what's going on.  Okay, I want to remind you of chapter 15 stuff.  You learned all of this stuff in section 15.10 about allillic bromination and in cases like  this when we treated allillamines with NBS and heat or peroxides you don't use BR2 on  here because that will add bromine across the double bonds.  We use this N-bromosexinimate compound.  The problem with this is you get a radical where the, an allill radical where the BR  group can substitute on both ends of that allill radical and so you almost always got  these hideous mixtures when you do allillic bromination.  And who wants a mixture?  And there's also on top of that this issue that you generate new stereogenic centers.  That's not such a big deal.  But nobody likes to get mixtures of compounds as products in organic chemistry.  And so what I'll show you now is if we do the same reaction and instead of doing it  on this kind of an allill substrate we simply switch to a benzylic substrate it does the  same stuff.  So let me redraw that same substrate here and all I'm going to do is make this part  of a benzene ring.  And I do the same conditions N-bromosexinimate and light.  But this time I don't get mixtures of products.  You'll get selective bromination of one of the two H's that are at these benzylic positions.  And that makes this way useful whereas allillic bromination is not way useful.  I would say way useless would be a better description of allillic bromination.  You have to have super symmetrical substrates for allillic bromination to give you clean  product mixtures.  So look at this.  I can replace one of the two H's with this same reaction that you learned in chapter  15.  And why is that important?  Because that's a leaving group.  We can go from something that has no functional groups here really to something that has a  leaving group at that position.  And now you can do SN2 reactions, elimination reactions.  So that's very powerful.  This idea that we can add leaving groups wherever you want to have those.  Because there's no easily bromonatable double bonds in a benzene ring, you don't have to  worry about using N-bromosexinimate.  So in other words, we can take this same kind and you can have other substituent on the  benzene ring.  They don't get affected in that reaction.  I don't know if this is a distraction that we erased this.  And because we don't have bromonatable double bonds where you add Br2 across, you can use  Br2 in this reaction if you want.  In fact, that's more typical.  And the book shows you other types of ways to do this.  Or you could simply heat the reaction so hot that you generate small amounts of radicals.  So you can't do allillic bromination with Br2, but you can do benzylic bromination with  molecular bromine.  But it works just fine.  And look how powerful that is that we can add a leaving group anywhere that we want to  have a leaving group.  So you should know that reaction and be ready to use that reaction.  Because again, that's very powerful to add a leaving group.  OK, I'm going to show you two more useful transformations of side chains.  So really what we're talking about now is what are the useful things you can do once  you've added a side chain to a benzene ring?  How do we turn that into a useful functional group?  And let me show you two powerful reductions of carbonyl groups.  When there's an arrow group on one side, these tend to be very efficient reduction reactions.  So I'll show you a way to take a carbonyl group attached to a benzene ring and burn  that down to a CH2 group.  And so here's one version of that reaction.  This is called a Wolf-Kieschner reduction.  And you don't need to know the name of this reaction.  It's not important.  This is a reagent called hydrazine.  There's an NN bond, H2NNH2.  And you have to do this under strongly basic conditions, sodium hydroxide.  And there's a third thing for you to remember, and that is you have to heat these reactions.  These are not mild reaction conditions.  So I'm going to write a big delta under here.  So hydrazine, H2NNH2, sodium hydroxide, and you boil it.  And when you do that, the product of this reaction is now missing that carbonyl.  But it's not so much that it's missing the carbonyl.  It's that you've now replaced that carbonyl with a CH2 group.  And I'll draw that CH2 group for you so you can see.  It's not that there's nothing there.  There's H's there.  And you don't need to know anything about the mechanism for this Wolf-Kieschner reaction.  But I think you can see that's useful.  And the reason why this is useful is because you can't make that product through a Friedel-Krafts  alkylation.  So I want you to recall that Friedel-Krafts alkylations on primary alkohalides suffer  from this problem of migration.  So I want you to recall this problem here.  Well, actually let me cover that in just a second.  Let me first talk about an alternative set of conditions.  Suppose you can't remember hydrazine for some reason, and it's just completely escaping  you.  There's an alternative set of conditions we can do.  This is a reaction called a Cleminson reduction.  And it involves taking, you buy this stuff called zinc-mercury amalgam.  Here's the way we represent zinc-mercury amalgam.  We write it like this.  You take zinc powder and you mix it with mercury, and the mercury absorbs onto the surface.  So you put the mercury in parentheses.  That's called zinc-mercury amalgam.  And in this case, it's not strongly basic conditions.  It's strongly acidic conditions.  And again, these conditions are not mild.  You have to boil your ketone with zinc-mercury amalgam and hydrochloric acid.  So I'll draw a delta symbol here.  That means you're boiling the heck out of this to get it to work.  And this will do the same thing as that Wolf-Kieschner reduction with hydrazine.  Boy, as soon as I see the CH2 group, I want to brominate it within Bromo6-Inimate.  You can see why it's cool to introduce a CH2 group.  OK, so this is a way to take carbonyl groups and reduce them down to CH2 groups.  You're not going to see anything like that in other chapters.  It's a, well, maybe there's a different type of reaction with lithium-aluminum hydrochloric acid.  Never mind.  In this chapter, that's the only kind of reaction that will take a carbonyl group and reduce  it down.  So let me make one important point to you, because you're going to tend to get this confused  in chapter 20, that you cannot burn down the groups of esters or amids.  There's a different side reaction I'm not going to show you.  It has to do with hydrolysis.  And so you're not going to be able to ever take something like a methyl ester or an ethyl  ester and convert those into CH2 groups using this Wolf-Kieschner reduction.  We may have some other chemistry to do something like that.  But there's, but not exactly like that.  So don't do this with esters, not with amids, only for ketones and aldehydes.  That's the only place where this is useful.  And you'll have a tendency to want to do that.  Now, you might ask, which of these two should I memorize?  Which of these two is more useful?  Typically, the base version is more useful.  The hydrazine and sodium hydroxide, because there are so many functional groups that are  acid-sensitive, you're going to find that boiling things with acid is usually going to  screw up some other part of your molecule.  So I usually prefer this Wolf-Kieschner version, this hydrazine and sodium hydroxide.  If you want to have, you know, if there's one thing you want to really commit to memory  the most, that would be the one that would be most useful to you this quarter.  Okay, so let me remind you of why this is useful.  It's because of this limitation in the Friedel-Kraftz reaction.  Remember this problem with primary alkohalides is that they will rearrange.  And so it really limits when you do Friedel-Kraftz reactions.  So this is just a continuation.  Why would you want to do this Clemensen reduction or a Wolf-Kieschner reduction?  If you wanted to take this benzene ring and do a Friedel-Kraftz reaction with this primary  alkohalide, if you wanted to make that kind of product, look at this CH2 group here.  You might see that and try to write this out as a reaction.  But there's a problem with this.  You wouldn't get any of the product I drew over there.  What you would get would be a T-butyl group substituted on there.  And the reason is when you draw out the primary carbocation like this, is that that will do  anything so that it's not a primary carbocation.  And it will do it quickly.  The primary carbocation intermediate that you generate, and I'll draw the, let me draw  that H2C group here, will very quickly rearrange to give you a tertiary carbocation.  And that's what's going to attack the benzene ring.  It's going to be the tertiary carbocation that attacks the benzene ring and does this  Friedel-Kraftz alkylation reaction.  So if you want to add a substituent like this over here, you have to do it in two steps.  First you do a Friedel-Kraftz acylation.  And then you use one of these reductions to reduce the carbonyl down to a CH2 group.  So that's the way to do things.  Now there is an exception to this.  I said don't try to do primary alkohalides.  But you can do, there are two alkohalides that you can do that kind of look like they're  primary.  And the book seems to love these and the online homework seems to love these.  And I don't want you to worry about the mechanism.  You can do it with this because methyl is not susceptible to any kind of a hydride migration.  There's no 1, 2 migration possible for a methyl because there's not even a second carbon.  The other type of starting material that you'll see used in Friedel-Kraftz alkylations, mainly  in the book, not in real life, is ethyl chloride.  And why is that not a problem with hydride migrations or alkyl migrations?  It's because if you migrate a hydride from one side of an ethyl cation to the other,  you get back an ethyl cation.  So those are the only two instances of things that look primary where you don't suffer  from this problem.  In general, if you see a CH2 group attached to a benzene ring, you should make that through  Friedel-Kraftz's isolation and then reduce the carbonyl group down to a CH2 group.  That's the general way to attach CH2 groups to benzene rings.  And that's the way I would recommend to you.  So lots of stuff you can do with these carbon substituents.  You can brominate them.  You can reduce down carbonyls to CH2 groups.  That's powerful stuff.  And let's talk about carbon nitrogen bonds or nitrogen substituents on benzene rings.  And I apologize for those of you in back.  The problem is these erasers don't do the best job of erasing usually.  And so you're trying to see my writing through this haze of chalk on the board.  So let's talk about converting nitro groups, which are pretty useful.  You can't make medicines that have nitro groups on them.  Those cause problems with toxicity when they're metabolized.  So if you want to make useful drugs that have nitrogen on benzene rings, typically invariably  you're reducing those nitro groups down and doing something else with those.  So here's an example of ortho-nitrotoluene to nitrotoluene.  And it's very, very simple to reduce these down to give you something  to give you something that's useful to work with.  And it's the same conditions that you use for hydrogenation.  And that's H2 and palladium on carbon.  I usually write only solvents underneath the arrow.  That's why I was hesitating there.  So palladium on carbon is a catalyst.  It's the same thing that you were doing with hydrogenation of double bonds.  Typically we don't show the stoichiometry.  In fact, it takes three equivalents of hydrogen.  Generally I want you to write the stoichiometry.  But for hydrogenations, you don't chemically do not draw out the stoichiometry for hydrogen.  So I think you can see you're going to lose two H2O's at a minimum here.  And then another set of H2's go on to the nitrogen atom.  So we don't show stoichiometry for hydrogenation type reactions.  Look how easy this.  These are the best reactions ever.  If ever you want to have a feel good reaction in the lab where you get 100% yield and no  byproducts and all you do is filter your reaction, this is the way to go.  Now the book gives you two other recipes for reducing nitro groups to amino groups.  And this only works when they're attached to benzene rings by the way.  You can't do this on alkyl nitro groups.  Let's take a look at another set of conditions.  Sorry, I don't know how to push this past so you've got some light here to see.  So here's the other set of conditions that the book shows.  The book shows tin and HCl.  And then some books, I can't remember whether it's this one, shows SNCl2 and HCl.  And that's actually more typical conditions, stanis chloride.  But the book shows tin and HCl or iron powder and HCl.  So tin powder, tin is SN, and hydrochloric acid.  And the product of this reaction that they show in the book is the same thing that you  get with hydrogenation.  And so why does the book give you all of these different recipes?  Why don't they just?  I don't know.  But why does the book show you in Chem 51, see all these recipes?  Here's why any chemist would want to do this.  It turns out that if you have a halogen up above here, those hydrogenation conditions  will cleave the carbon halogen bond.  Whereas the tin HCl conditions do not cleave the carbon halogen bond.  And you're never going to run across that in the Chem 51 series.  I'm not going to ask you that.  The book is careful not to ask you that.  So that's why people might not want to use the hydrogenation is because maybe they don't  want to lose that bromine atom.  There's a second little trick that I expect you to add that they didn't add here in this  book.  And that is, if you're stirring around an amine in hydrochloric acid, it's not the amine  anymore.  The product of this reaction is actually an ammonium salt.  And I want you, even if the book doesn't, I want you and expect you to draw out that  if you want the free amine NH2, that you have to treat this with sodium hydroxide to deprotonate  that.  And that's free-basing.  Like when you free-base coke this weekend, this is what you're doing.  You're deprotonating an amine.  And I want you to show that free-basing step.  If you do Stannous chloride, that's NH2 I'm drawing here.  So don't just draw this one step tin in hydrochloric acid.  I want you to do a second step with sodium hydroxide.  And usually we just enumerate those.  Step one, tin hydrochloric acid, step two, sodium hydroxide.  So generally I would use hydrogenation if I didn't have some other reaction I was trying  to avoid.  Yeah?  No, the byproducts here, you get 2H2O's as a byproduct.  So you see two of the molecules of hydrogen combined with the oxygens give you two waters.  The third equivalent of hydrogen ends up, don't worry about the mechanism.  You don't need to know that.  You just need to know that that's a fantastic reaction that works every time.  Okay, last, let's talk about multi-step synthesis.  You've learned a lot of reactions in Chapter 18, actually not that many.  Five electrophilic aromatic substitution reactions.  And then I just now showed you some fancy ways to dress up those side chains that you  put on through reduction, et cetera.  So for multi-step synthesis, when you do that involves electrophilic aromatic substitution,  the main admonition that I can give you is to strategize regio control.  If you want to put three groups onto a benzene ring with a super particular arrangement,  you have to know which group you add first because each group is either an ortho-pera  director or a meta director.  And so you need to strategize regio control.  So let's take an example where we want to do some sort of a synthesis.  Suppose that you're asked by somebody to take benzene.  And then through some number of steps show reagents and steps that would allow you to  put on these kinds of groups here.  And so you have to think about what do I add first and what do I add second?  And what are all those exceptions that I have that I have to worry about?  So let's go ahead and talk about this.  How would we, typically when you're given a synthesis that you have to do, you have  a choice to either work forward or work backwards?  I usually start by working backwards.  I usually start by saying, what are all the different ways that I could make this?  And so what are some of the ways that you know for how to make that?  Well, maybe one possibility is we could start off with nitro-benzene and then do some sort  of an alkylation reaction.  But you already know that doesn't work.  Nitro groups are meta directors.  Nitro-benzene is too deactivated for you to do any kind of Friedel-Krafts alkylations  or acylations.  So just forget that.  That's out.  Really your only choice is to somehow have some kind of a group here.  But there is an alternative here.  And that is maybe my last step in my synthesis was reduction of one of these carbonyl groups  using that Wolf-Kieschner reduction or the Clemenson reduction.  Maybe the last step in my synthesis was reduction of a carbonyl.  This has to do with me being so clever that I could see that CH2 group.  And I remembered those carbonyl reduction reactions that I just taught you.  Maybe that was the last step.  Or maybe the other possibility that we have here is that the last step involved in nitration.  Now that actually looks pretty promising.  And that looks promising because alkyl groups are ortho-paradirectors.  So we could have made this by first putting some sort of an alkyl group here on the benzene  ring and then doing a nitration reaction.  Let's remind ourselves of what those conditions are for nitration.  It's nitric acid and sulfuric acid.  So of these choices, to me what's looking pretty promising here is this nitration.  So maybe the last step in this synthesis is involved in nitration.  But I think we can probably figure out how to put a propyl group on a benzene ring.  And so how would we do that?  How would we add a propyl group to a benzene ring?  Well, this comes back to another exception that you can't directly add propyl groups  to benzene rings.  Let me just write that problem out for you here.  What was the problem that we just finished talking about with trying to do this in one  step?  The problem that we talked about is that if you take chloropropane and you try to do a  freol-crafts alkylation, that you'll end up with all these hydride migrations.  That primary carbocation would be unstable.  And if you made something that looked anything like a primary carbocation, I'm not saying  that that's what would happen here, but if you started to generate something that looked  like a primary carbocation, these hydrides would migrate over.  And so before it does freol-crafts.  And so in that case, you wouldn't have n-propyl groups attached to your benzene ring.  You'd have isopropyl groups attached to your benzene ring.  Right?  So you can't do just straight freol-crafts to make this.  How do we make this?  Well, this is where those carbonyl reductions come into play.  The best way to put that CH2 group directly attached to a benzene ring is to do one of  these reductions.  And I'm going to choose the Wolf-Kieschner reduction, that hydrazine one.  There's an NN bond there.  You don't have to draw that NN bond.  And there's two other parts to this recipe.  Sodium hydroxide and heat.  And if you wanted to do that zinc-mercury thing with HCl and heat, that's perfectly fine,  too.  So now it's starting to look like we've got a root here.  We could start here with benzene and do freol-crafts, acylation, and aluminum trichloride.  And then for step two, we could use some sort of a reduction, hydrazine and sodium hydroxide.  And then that's the molecule we have over here.  And then our third step, we could nitrate.  Now there's one little stickly point here that I'm leaving out.  And this is the last thing that I'll tell you in this chapter.  And that is in cases when you're doing synthesis problems, in cases where I ask you to efficiently  make some product, I'm going to allow you to claim that separation of ortho- and para-isomers  constitutes an efficient chemical step.  And I want to add that in here.  Ortho- in other words, if you take something that looks like this and you brominate this,  you will get a mixture of ortho- and para-isomers out of this.  And there's no way that you can get around that.  So as one of your chemical steps in any synthesis, I will allow you to do this.  And separate ortho-para.  And then just draw whichever one that you want.  I admit, yeah, you only got a 50% yield of each one.  And that's not really that efficient.  But we're just going to pretend that this overall process of making both ortho- and para, and  separating out the one you want.  Let me just draw one of those.  Suppose I want the ortho here.  We'll just pretend that that's an efficient combination of steps.  So what I need to do here is acknowledge the fact that nitration is going to give me a mixture  of ortho- and para.  And I need to add an additional step here.  So what we need to do here is write step.  And I don't know how to raise this board here.  So step one is going to be nitrate.  And let me go back a little bit here.  Let me do it like this.  I'll start back over here with my propyl benzene.  And I'll write step one, nitric acid and sulfuric acid.  And then step two, just like I just told you, separate ortho- and para.  And that's the way you deal with that.  When you generate ortho-para mixtures, but you only want one of those.  I would like you to do that.  Okay, that's it.  That's the end of what we've got to cover.  And I have one last thing to tell you.  Our final exam is on Monday morning at 8 a.m.  Seating charts are posted and I'd like you to be in your seats at 10 before so we can  start passing out the exams.  And this weekend, drive thy business.  And when you finish this class and you are going on into 51C, drive your business and  invest in your career by learning organic chemistry.  And so thank you.  Have a great time with you this morning.
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