 Good morning. Good morning. So I'd like to finish up Chapter 25 today on the chemistry of amines and as I said before, Chapter 25 is a little bit of a hodgepodge because we've already seen a lot of chemistry of amines but there's some really beautiful and some really useful chemistry that we're going to see and some of it because we're on the quarter system harkens back to some of the aromatic chemistry that you learned at the very end of the 51B course so that very last part of the 51B course and so we're going to review a little chemistry from that and put it all together. So when we were talking last time we were talking about alkylation of amines, in other words treatment of amines with alkyl halides and I said that the basic thing about amines is it's not easy to control their alkylation reactions. We know amines are nucleophiles and we know alkyl halides are electrophiles and that primary alkyl halides and to a lesser extent secondary alkyl halides participate in SN2 displacement reactions but at the same time what we would think would be an obvious reaction of mixing an amine and an alkyl halide doesn't lead to controlled monoalkylation. You get proton exchange and you get a mixture of monodye and trialkylated compounds. It's kind of sort of possible to take an alkyl halide and treat it with a huge excess of ammonia and get a primary amine although often in practice chemists use other procedures because you do get over alkylation but in general taking a primary amine and treating it with an alkyl halide gives a mixture of products. Now on the other hand we can control this reaction if we want heavy alkylation so just by way of repeating what I was just saying if we take benzilamine and we treat it with methyl iodide right methyl iodide is a great alkylating agent. Primary alkyl halides are very good at SN2 displacement reactions but methyl halides methyl bromide methyl chloride methyl iodide are especially good also other sorts of compounds like benzylic halides are very very good alilyl halides as well and so if we treat benzilamine with methyl iodide we end up with a mixture and this was what I was talking about last time a mixture of monodye and trialkylated products. So in other words we don't just get the just for the sake of clarity I'll say we don't don't just get not just the monoalkylated product where I've drawn say a single methyl group and technically we would have it as the ammonium iodide salt NH2I minus. Now one of the things about organic chemists is we really like to control what's going on. A lot of organic chemistry focuses on synthesis. It focuses on making stuff, taking a reaction and figuring out how to take charge of it, how to make it useful and so one of the things that one can do is to drive this reaction by use of an excess of methyl iodide and I'll show you where we're headed with this in just a second but we can drive this reaction to with an excess of methyl iodide and you need a base to pull off a proton because every alkylation gives you an ammonium salt so I'll just write out this equation and then we're going to do some play with molecules of this sort. So instead of getting a mixture if I take benzolamine and I treat it with methyl iodide so I'll say excess later on you'll see me write it as E as excess because chemists like shorthand. Excess methyl iodide and you really need a base in there. I think your textbook kind of neglects it. It's not super important until you go into the lab and actually try to make the chemistry work but you need a base to pull off the proton so typically you would do this under heterogeneous conditions with say solid potassium carbonate in there and this would give you the trialkylated product, benzol trimethylammonium iodide so iodide is the counter ion. If you ever look at your eye drops or nose drops or other sorts of products that are liquids that you put in your body particularly in your eyes or contact lens solution you'll often see on the back of it, benzol trialkyl ammonium salts. These types of salts have antibacterial properties. They're surfactants. They're soap like molecules that disrupt the membranes of bacteria and so this helps keep nasty things from growing in your contact lens solution or whatever. So this is part of a broad class of what we would call quaternary ammonium salts. Quaternary just means remember we talked about amines and we said they're primary, secondary and tertiary amines depending on the number of substituents on nitrogen. Quaternary means there are four substituents and of course it's a salt because by that point you'll have a charge on the nitrogen. Let's see a question. Ah, okay, another good question here. Would it be possible for methyl iodide to add on the benzene ring? So one of the distinctions I've made, we think about nucleophiles and we think about electrophiles and one of the distinctions I've tried to make in the class is the distinction between things that are slightly nucleophilic and things that are very nucleophilic and then things that are slightly electrophilic and very electrophilic. So methyl iodide is slightly electrophilic. You need a nucleophile with a relatively available lone pair of electrons to attack it. We often use pKa as a surrogate for thinking about nucleophilicity when you're comparing elements of the same type like nitrogen over a series like an aniline nitrogen pKa of the conjugate acid 5 and say this nitrogen pKa of the conjugate acid 10, it's a relatively good comparison to say benzolamine is much more nucleophilic than aniline. When you're comparing atoms of different types, the comparison is a little bit more loose and yet still, you can use this general idea that benzolamine has a pKa of the conjugate acid of 10, it's relatively nucleophilic. Benzene, of course, is not at all basic. It'll protonate in sulfuric acid but it's not a very strong nucleophile at all. So most of the electrophiles that react with benzene are cations or else they're very good electrophiles like halogens and we'll talk more about this at the end. So methyl iodide is in no way a good enough electrophile to do electrophilic aromatic substitution on benzene. Another example that's perfect is say formaldehyde. Formaldehyde is not a good enough electrophile. We talked about aldehydes and protonated aldehydes. Formaldehyde is not a good enough electrophile to react with an aromatic ring. I'll take phenol as an example, hydroxybenzene which we'll see later on but if you have acid present, the formaldehyde can protonate. You can get an oxacarbenium ion and that's a good enough electrophile and that in fact is a basis for the synthesis of polymers particularly Bakelite that can be done either with acid or base. So this having this notion of things that react with benzene generally very hot electrophiles, carbocation, tert-butyl carbocation undergoing a Friedl-Crafts reaction, tert-butyl chloride and aluminum chloride or an acylium ion and acid chloride and aluminum chloride, those are the sorts of electrophiles that attack the benzene ring and we'll get to talk about some electrophilic aromatic substitution in a second. All right, I want to show you an example and I think this is going to bring us back to some concepts of stereochemistry and conformational analysis. I want to introduce us now to the Hoffman degradation. Your textbook introduces it at this point. This reaction used to be fantastically important back before techniques like NMR spectroscopy. So when I say that chemists took 50 years to determine the structure of morphine before NMR spectroscopy or IR spectroscopy, they were carrying out chemical degradation reactions and combustion analyses to figure out what molecules they were producing. And hydrogenation and measuring how much hydrogen was up taken. So the Hoffman degradation, which I'll show you now, used to be a very important reaction. The reason I'd like to teach it now is for concepts of mechanism, stereochemistry and conformational analysis, the latter of which is going to tie into our discussion of sugars. So if we take an amine and I'm just taking sort of the simplest stupidest amine I can think of here, I'm taking ethylamine and if we treat it with an excess of methyl iodide, so here I'm writing it as excess, just a little abbreviation. Empotassium carbonate, some sort of base to keep pulling off protons after the SN2 alkylation so you can go on smoothly to the next step and the next step until you completely alkylated the amine. You would get ethyl trimethylamonium iodide. This is what we would call an exhaustive methylation, just like I've written here. As part of the Hoffman degradation, it happens to be called the Hoffman exhaustive methylation. If we now take our, I'll show you the overall sequence and then I'll talk about how it works. If we now take our ammonium iodide salt, our ethyl trimethylamonium iodide and treat it with silver oxide in water, AG2O is silver oxide, we replace the iodine with a hydroxide. We replace the iodide with a hydroxide. I'll tell you how that occurs in just a second. And of course, hydroxide is a good base and now we have ethyl trimethylamonium hydroxide and if we heat this material, so I'll just write an arrow with a delta here so now you go ahead, you boil up your solution. This compound's going to break down to three components. Ethylene, I'll write it like this. If you're not comfortable writing it like that, just write it like this here. Ethylene plus trimethylamine. And organic chemists hate to balance equations so I'll put this in parenthesis water as the product. I want to show us how we get to this point. So first let's take a look at the silver because silver oxide is a reagent you've seen before as an oxidizing agent, a mild oxidizing agent that can allow oxidation of aldehydes up to carboxylic acid. So let's look at what else silver oxide can do. So in water, a lot of metal oxides are basically in equilibrium or equivalent to the hydroxide so if you think about it, you can think of a hydroxide as of an oxide as a hydroxide without water. What do I mean? You can think of it as silver oxide plus H2O is equivalent to or is in equilibrium with two molecules of silver hydroxide. Now the main thing about silver, silver absolutely loves the halogens. Silver precipitates out with chloride, bromide and iodide to form silver chloride, silver bromide, silver iodide. It's a very pretty reaction, particularly iodide because silver iodide is a pale yellow solid. So just like the iodiform reaction I talked to you about before, we can generate pale yellow solid but the real gist of this is when we treat our alkyl ammonium iodide, our ethyl trimethylammonium iodide with silver oxide in water, I'll write it as silver hydroxide so I'll write this sort of in quotes as silver hydroxide. You end up precipitating out silver iodide and getting the trimethylammonium hydroxide salt. That's just a sort of if you think back to it, it's sort of your standard inorganic reaction back from G chem days where you would mix two different species, two different salts together and you would end up with a precipitate of one salt in Le Chatelier's principle driving the reaction. Okay, but the main thing that's occurring in the real organic chemistry that's occurring in this process now is the last step, the elimination step that sometimes referred to as the Hoffman elimination. So the Hoffman elimination involves just an E2 elimination. Hydroxide is pulling off a beta proton and kicking out the trimethylammonium group. So I'm writing this with ethylene as our example as ethyl as our example for starters because it's nice to see the geometry here and it's nice to see how this can occur. So there's our ethyl trimethylammonium iodide and you can think of your hydroxide here. I'll write it as OH and I'll be a good person and put in all my lone pairs of electrons and so forth. You can think of hydroxide as acting like a base, pulling off the beta proton, pushing the electrons into the bond between carbon and carbon and spitting out trimethylamine and when all is said and done here at the end, I'll just write an arrow, we go to HOH water of course plus ethylene which I'll write like this plus trimethylamine. So this is the basis of an E2 elimination reaction. I guess technically my word and E2 are semi-redundant because E2 means elimination bimolecular but the big thing that's important in the E2 elimination is that you have an anti-periplanar relationship. In other words, as you look down the bonds, you go in a zigzag relationship or if you do a Newman projection, think back to Newman projections in your first quarter, you have a relationship where the hydrogen is opposite the leaving group. So I'll write anti-periplanar relationship and that's really one of the requirements for an E2 elimination is this geometrical relationship. Question, why couldn't, ah, you could use NaOA. So the question is why do we have to use potassium carbonate here? And I think the answer is you really could. The nice thing about the conditions, you're trying to minimize SN2 displacement on methyl iodide. Hydroxide at that point is a little bit better nucleophile so you'll kill some of your methyl iodide. But if you were to try this reaction, say with potassium hydroxide and methyl iodide under the right conditions of solvent, you could probably probably get it to work. It might not work quite as smoothly. One of the beauties of a lot of this really ancient chemistry is a lot of it works really cleanly and easily meaning you just mix some things together by a general, really a recipe here and the chemistry works. And that's cool because when you're trying to figure out what's going on, knowing this chemistry goes in the same way, whether I have an ethyl group, whether I have a propane group, whether I have morphine or whether I have cocaine is really, really cool and really valuable. Now here's the cool thing about this E2 elimination and this actually brings us to some stuff I wanted to talk about. All right. One of the cool things about this chemistry is that you get regiochemical control and you get regiochemistry that's different than other eliminations you've seen before. You know that controlling regiochemistry is important. When you take one butene, you learn back in either 51a or 51b that when you take one butene, you could go ahead and control the regiochemistry of addition of water across the double bond depending on whether you use an oxymercuration reduction system to give reaction sequence to give the so-called Markovnikov addition or a hydroboration oxidation sequence to give the anti-Markovnikov addition. You also learn that in general, E2 eliminations if possible prefer the more substituted product. So for example, if I have one methyl, one iota, one methyl cyclohexane and I treat it with potassium hydroxide and I heat it up, we get as our major product the more substituted alkene and our minor product, we get the less substituted alkene and that's sort of a hallmark of atypical E2 elimination. In the transition state for atypical E2 elimination, you start to feel the greater stability of the more substituted alkene and so in general, you end up with the more substituted product being favored over the less substituted product. But what's cool about the Hoffman degradation sequence is if we take the same basic molecule except instead of our leaving group being an iodide, it's going to become a trimethylammonium, we get the other regiochemical preference. In other words, if I subject this molecule, this one amino, one methyl cyclohexane, first to excess methyl iodide and potassium carbonate, then to silver oxide and water and then to heat, now our main product is the less substituted alkene and our more substituted alkene is very, very minor. The selectivity is huge on this. We get a relative ratio of 99 to 1, 99 percent. This is our major product and that's really, really cool and really significant. It's the same element of control that you were getting emphasized in the hydroboration, oxidation sequence that you were getting that I've emphasized in talking about chemists in synthesis. So I'm going to just sort of summarize this and then we'll talk why. So I'll say that conventional E2 elimination and I'll say generally because I can always find some type of counter example, so I'll say generally gives the more substituted alkene and then I'll say the Hoffman elimination has, I'll say special steric and stereochemical requirements and I'll say that these requirements often lead to, I'll say often leading to the less substituted alkene. All right, let me give you, before we go on into other examples, I want to show you my thinking on the rationale for the difference we're seeing on the upper right-hand blackboard there. Here's our iodide, the first example from the blackboard. You have a methyl group here. You have the iodide in the axial position. You require the hydrogens to go ahead and have an anti-periplanar relationship and if possible you favor the more substituted double bond. So here's hydroxide, we're using potassium hydroxide. I'll just draw in one lone pair here and so we push our arrows like so that's going to lead to E2 elimination to give the more substituted alkene. Now, by comparison, here's our example with the cyclohexyl trimethylammonium iodide. Now, trimethylammonium is isostructural with the tert-butyl group. Remember the tert-butyl group back from 51A? Remember how big the tert-butyl group is? The tert-butyl group wants to keep out of the way of everything. If it can avoid it, it doesn't want to be axial on a cyclohexane ring. If it can avoid it, it doesn't want to be ghost of things. It wants to keep out of their way. So it's not great to put a methyl group axial on a cyclohexane ring but it's a lot worse to put a tert-butyl group axial and by the same dent, it's a lot worse to put a trimethylammonium group axial. So our trimethylammonium group prefers a conformation where it's equatorial. Now, let me just do one thing here and I'm going to redraw this molecule and draw it in the hydrogen and now think about our elimination. Actually, I've given it its own blackboard because I want to have a little bit of room here to make my point. All right, so here's our methyl group. We have a hydrogen here and two more hydrogens on it. I'll just write it as CH2. Here's our trimethylammonium group like so. Here are our hydrogens like so and like so axial and our equatorial hydrogens. Which hydrogen is antiperiplanar? Which one has a zigzag relationship to that trimethylammonium group? Which one? The one on the CH2, yes. So here's our zigzag relationship. So here we have a ghost relationship, not an anti-relationship and here we have an antiperiplanar relationship and so our hydroxide can come in like so, kick, kick, kick and do our E2 elimination. Now the reason I hedged, the reason I qualified my statement over here often leading is I was going to give you a counter example and it involves some really cool confirmation stuff but I decided all right, I'm not going to. It's a beautiful example where you've got an antiperiplanar relationship that leads to the more substituted alkene for your purposes and your textbook shows a few additional examples, you can think, yeah, generally less substituted alkene but just keep in mind the steric and stereochemical requirement. I'll show you one more specific example for comparison and then I'd like to move on to some aromatic chemistry. All right, just kind of a simple example. Again, harkening back to E2 elimination that you learned back at 51A or 51B. So we'll take 2-bromobutane by example here. We'll treat it with a strong base. Remember our favorite bases we often like sodium ethoxide and ethanol because you can generate it by throwing sodium metal into ethanol. They react to give sodium ethoxide and hydrogen. Secondary alcohol halides and strong bases tend to give elimination. You don't get a Williamson ether synthesis and so you get the 2-butene and the 1-butene. So I'll write 2-butene and I'll write parenthesis mixture of cis and trans. I'm not going to break down what, which amount you have of which and then the 1-butene. So you get the more substituted alkene and the less substituted alkene and in general reactions like this favor the more substituted alkene. It's not 100% preference but it's a good preference and this particular example we have an 81% to 19% ratio. In other words it's four parts of 2-butene to one part of 1-butene. Now by comparison we take 2-amino-butene and we subject it to the same types of conditions. In other words we go ahead and we take 2-amino-butene. We treat it first with excess methyl iodide and potassium carbonate then we treat it with silver oxide and water and then we heat it up. So it's our Hoffman whoops it's supposed to be a delta to indicate we're heating it and we get vastly, vastly predominant the 1-butene. So our 2-butene which again I will write sort of with a squiggly line is 5% and our 1-butene now is 95%. So a very, very big preference for the less substituted alkene. So in a way this is sort of the antithesis, the opposite of your oxymercuration reduction, hydroboration, oxidation control in that now you're going the opposite way and you're exerting regiochemical control. Thoughts or questions at this point? Oh, okay, why do I draw it like this? This means stereochemistry. Squiggly line means stereochemistry unspecified. So this is equivalent to trans plus cis. And we often use a squiggly line. So for example, if I'm going ahead and saying let's say pentane diol and I don't want to specify the stereochemistry of it, 1, 2 pentane diol, that could mean either cisisomer, transisomer, a mixture of isomers. It could mean a mixture of the trans as the RR and SS. And anti-emers, so it basically is sort of all encompassing. Okay, if you're going to remember good question because this is confusing for students and also as I said, the Hoffman elimination is not that important. If you're going to remember one thing from the Hoffman elimination, I'd remember two things. One of the things I'd remember is anti-periplanar attack. Always wants anti-periplanar attack. That's the stereoelectronics. That's how your orbitals line up. Then the simple rule is generally it's going to give you the less substituted alkene. But because of the reasons for it, you can, I could come up with an example where you had the more substituted alkene and your textbook doesn't even touch on it. So it's kind of a little bit advanced. Other questions? All right, I want to take us back to aromatic chemistry because the aromatic chemistry of amines is super important. It's really the basis of all commercial organic chemistry and it's something that you didn't get in chapter 18. So by way of refreshing, electrophilic aromatic substitution is a great way to introduce substituents onto benzene. You can get simple benzene compounds, benzene and toluene and naphthalene and xylene, which is dimethyl benzene. Toluene is methyl benzene. You can get them all from petroleum. They're present in crude petroleum. And so petroleum is a great source, a great feedstock. So if we take toluene, the other chemical that's a great feedstock for all of chemistry is, well, two chemicals, is nitric acid and sulfuric acid. These are readily, readily available earliest sorts of chemical industry in the U.S. And so taking a benzene compound and treating it with nitric acid, sometimes with sulfuric acid, gives rise to nitrobenzene, electrophilic aromatic nitration. You learned about that before. And then if you reduce the nitro group, for example, by catalytic hydrogenation with hydrogen and palladium on carbon or by zinc, by tin and HCl reduction or by iron and HCl reduction, there's a zillion ways to do it. You can get the amine para, it's para-amino toluene but it's called toluene or p-toluene if you want to know the trivial name. I'm not going to write it down because that's one that's sort of, aniline you should know, toluene you should know, phenol you should know, hydroxybenzene, p-toluene is kind of one for people in the know. Okay, here I want to show you now though some cool chemistry of this and this will get us into some new places both mechanistically and synthetically. So let's take our p-toluene and we're going to treat it with sodium nitrite NaNO2 and an acid. Your textbook always uses HCl and that's largely fine. I'll make some distinction about using HCl or HBr, H2SO4 or fluoroboric acid and give you a little bit of subtlety and you want to keep this cold because the product that we're getting is not very stable and it can blow up. And what you get out of this is an araldiazonium salt. The diazonium salt of course is the nitri, is the triple bond to nitrogen and the reason this isn't very stable is this is one step away from releasing molecular nitrogen and molecular nitrogen into is the world's greatest leaving group. Now I'm going to get into some chemistry that shows us what happens when the nitrogen leaves and it's largely SN1 type of chemistry, nucleophilic displacement, unimolecular carbocation chemistry but I want to show you first this diazotization reaction so I'll write this out as diazotization. All right there's a lot of chemistry going on here and a lot of it's just simple acid based chemistry. Sodium nitrate, nitrite rather is the sodium salt of nitrous acid and I'll take us all the way back to freshman chemistry and write a Lewis structure of the nitrite anion and I'll even remember to put in my charges and if we just protonate the nitrite anion not going to show you curved arrows but if we imagine H3O plus right H3O plus is the strongest acid that can exist in water in H2O. Your textbook writes HCl and write curved arrows probably not quite as correct. Anyway if you go ahead and write this out now you know that you'll have an equilibrium where you'll get nitrous acid HONO this is again sort of freshman chemistry. Now if we imagine another proton going on to the molecule protons go on and off in strong acid. One of the places the proton can go is onto the OH group. You saw this in electrophilic aromatic substitution think all the way back to the mechanism of nitric acid and sulfuric acid to generate the NO2 plus the nitrosonium ion. So let's go ahead and we'll write a protonation H3O plus over the arrow to now protonate the oxygen. I'll be a good person remember all of my charges and lone pairs of electrons and now imagine water leaving. This is all inorganic chemistry right now. Imagine water leaving with its lone pair of electrons and so I'll write one last equilibrium here. I'll write this up and down just for the heck of it. And I'm going to go ahead and just take us to the more substituted, the more important resonance structure. If you don't like the resonance structure that I'm writing you can write the one in the textbook. I'm going to give everyone a complete octet here so we'll just push out the water. So this species here is called the nitrosonium ion. The main thing is even though this is an inorganic species, even though it looks very different than a lot of the species you're seeing in organic chemistry, basically it's the same gist of what we've seen of a lot of other positively charged compounds. They tend to be electrophilic. We can push arrows and form bonds that give rise to new bonds to them. So let's do this and I'm going to use the abbreviation for arrow and write AR. Arrow is sort of generic aromatic and so I'll write AR and H2 in this case to represent toluidine but of course it could represent aniline. It could represent any other amino group on a benzene ring. So here's our amino group. So just as water left and I guess I can even balance my equation here and say H2O, just as water left we can have the amine attack and so I will write my nitrosonium ion like so and now we just imagine what really is the reverse of that reaction there except with nitrogen attacking. In other words a lone pair comes in. We push up electrons onto the oxygen and you can think of this as a series of equilibria if you like. I think I like to. Acid. In acid protons go on, protons come off everything with a lone pair of electrons. So what's going to happen at this point is we're going to swap some protons around. We're going to tautomerize. Your textbook probably writes chloride as a base. I think it's probably a little bit more reasonable to write water as a base. So I'll just write H2O and I can give you a curved arrow here. I'm not going to give you curved arrows for absolutely everything. So I'll give you a curved arrow. We'll put our electrons onto the nitrogen. Now this compound here, it's not stable under the conditions. We're going to tautomerize it. The very end I'll show you another compound of this class that's stable. We call this compound a nitrosamine. Just an amine with a nitroso group on it, hence nitrosamine. Nitrosamines when you have a hydrogen on them, when they're from primary amines, let me write this more clearly. When they're from primary amines, they're not stable and they tautomerize. But on the other hand, secondary amines give stable ones and those are nasty compounds. They're nasty carcinogens. Question? Ah, this one. That's the nitrosamine, so that's our neutral compound. All right, we're in acid and we're going to continue to tautomerize and again, one of the beautiful things about really seeing things from a point of view of curved arrows is you can start to see in many ways this is like a ketototomer here. In other words, we have, well, it's not a carbonyl. It's a nitroso group and here we have an alpha hydrogen and so we can imagine swapping protons around. So in this particular mechanism, here's our nitrosamine. I'll start with this again and we're going to tautomerize. Protons go on, protons go off, all different positions. So just as we've tautomerized, just as we've tautomerized carbonyl ketones and aldehydes, we can tautomerize here. First step in the tautomerization in acid is going to be putting a proton on. Protons go on and off all your lone pairs. So I will go ahead and put on a proton here like so. And just like we, well, not just like, but I will show you, there's a second resonance structure that I can write here. If you just imagine pushing in this lone pair of electrons and pushing up electrons onto oxygen. And by this point, you should remember that different resonance structures aren't different molecules, they're just different representations of the same thing. I could write one, I could write the other, but it's really easy to see where we go from here if we look at the next one. Just imagine at this point losing H plus. Maybe for the sake of balance, I think, since we're just doing tautomerization, I'm actually going to write plus H plus here. I'm giving you a little bit of a shorthand of the mechanism. Of course, water is supplying H3O, hydronium ion, is supplying H plus. I'll write minus H plus. Of course, water is acting as a base for taking off that proton. And so now we just imagine that proton coming off like so. And now we're going to continue our chemistry. Was there a question? 10 electron, oops, thank you, thank you. 10 electrons on nitrogen. All right, so let's continue with our species here and imagine putting on another proton. In other words, we just have H3O plus transfer proton on to here, protons go on, protons come off every lone pair of electrons. And now the final step in this whole sequence, remember, we're headed toward the diazonium ion. The final step in this sequence is just that the nitrogen pushes in its lone pair of electrons, kicks out water, and so we go AR, N triple bond, N with a positive charge on nitrogen and a lone pair of electrons plus H2O. And so we have our diazonium ion over here. In a way, in a way what we've done is just review a lot of carbonyl chemistry because basically everything that you've seen here, it's a steppy mechanism, but everything you've seen here is carbonyl chemistry with heteroatoms with nitrogen and oxygen just pushing some protons around and kicking out leaving groups. Where I want to go, I don't want to dwell on the mechanism. I think it's worth seeing, I think it's worth thinking through and I think it's worth taking a few moments on your own after class to think through it. But where I want to go right now is what we do with these diazonium salts because it's a gateway to all of aromatic chemistry, to the earliest drug chemistry, to dichemistry before that and in some cases, dichemistry is drug chemistry. All right, so here we are at our diazonium ion derived from taluodine, the one that we got from nitrating benzene, reducing the nitro compound and diasatizing with sodium nitrite and HCl. And I said you want to keep that cold because it can explode. All right, the main thing is that if we heat it up, we end up losing nitrogen and we can get a carbocation. And I want to talk about that carbocation. We can get a carbocation. Carbocations are not stable species with very few exceptions. They're not things that you could put in a bottle. They react right away. So I'll say an aryl carbocation. I don't think you've seen one before. I'll tell you a little more about it in a second. But the main thing, it doesn't stick around very long and any sort of nucleophile NU reacts with it to form a bond. I'll write this sort of as generic NU plus. We'll go through specific details in a second. But the point is that it reacts with nucleophiles. It does not want to stick around. The thing is, remember, of course, what you have here before, think about valence. You have a hydrogen on this carbon. You have a hydrogen on this carbon. All you've got on this is this nitrogen and it takes away its lone pair, it takes away its pair of bonding electrons and leaves that poor carbon without anything on it. It leaves it trivalent with a broken heart and an empty sp2 orbital. Now, it's pretty bad to have an empty sp3 orbital. Sp3 orbitals, though, are higher in energy. You have more p character. It's really, really bad to have an empty sp2 orbital. Because you just want to have electrons in there getting stabilized. So what reacts as a nucleophile? Well, anything, anything that's around can react as a nucleophile. This is SN1 chemistry. In the case, in some cases you'll add some copper and it's a little bit more complicated then. But basically, I'll write this as SN1 reaction where nitrogen leaves and it's replaced by a nucleophile. So if we do this reaction in water, and I'm going to go ahead and make a slight tweak to the conditions, you don't absolutely have to do this. But for the most part, acids are interchangeable in the basic reaction they undergo. I'm going to go ahead and do the same diasodization chemistry with sodium nitrite, but I'll use sulfuric acid. Because sulfuric acid's a little bit less nucleophilic for its counter ion than hydrochloric acid. So I'll write H2SO4. And I'm going to remind us that you always do this reaction initially in an ice bath if you want to keep both of your hands. So we're going to write 5 degrees Celsius, you know, whatever temperature your ice bath will keep it. And then we're going to heat it up. Of course, this is in water, so I'll write H2O. Basically, you do this in dilute aqueous acid. The product of this reaction is creosol. It's P-creosol. That's just paramethylphenol if you don't want a trivial old name for it. What's happening is the water is being your nucleophile. You get OH2 plus, and of course in water, protons go on. Protons go off. Point is that we've taken a benzene ring that had nothing on it. We nitrated it, and now we've gone ahead after the reduction, the diazotization, and the breaking down of the diazocompound. We've introduced a hydroxy group into the benzene. We've accessed phenols. And this is really the start of this wonderful, wonderful gateway of electrophilic aromatic chemistry that you get into here. The whole family of chemistry, and a lot of this is borrowed right from your textbook. So with a few subtle tweaks, you can substitute just about anything on there. I'm going to do this all with our Ptolyudine example. So go ahead, and again, your textbook uses HCl for everything, and I think that's fine. If Johnny or Kim were to go into the lab, I'd probably say use sulfuric acid here. Use hydrochloric acid here. Use hydrochloric acid. So I'll show you how we do it in the laboratory. All right, to introduce a chlorine to the ring, right where the nitrogen is, you go ahead and you do this with copper chloride. Now, you'd say, well, why don't I just let the chloride react or add some more chloride? You probably could, but chloride's a little bit less nucleophilic than some of the other ions. It doesn't trap the carbocation as well. You get mixtures of phenol and the chloride. Copper chloride mediates the reaction. You can think of it just as SN1 chemistry, but there's really some involvement of organometallic intermediates. Anyway, we're not going to fuss with the mechanism. I want to show you about five different beautiful reactions. So going ahead and using cuprous chloride introduces the chlorine here when you allow them to react. They call this the Sandmeyer reaction, and again, a lot of this is really, really old chemistry. Of course, organic chemistry works at any time, any place, and Mars on the moon, on Earth, all the same way. It works in the 19th century. It works in the 20th century. So I want to show you some more variants of this. A lot of this is just straight from your textbook. It's just some cookbook. All right, same conditions, another variant of the Sandmeyer reaction. Sodium nitrite, and in this particular example, I'll use HCl as well. Chloride just isn't that good a nucleophile. Sodium nitrite, HCl, 5 degrees. Of course, this is all aqueous. Copper cyanide, cuprous cyanide, Cucn is a way of mediating the introduction of a nitrile group, a cyanogroup, like so. That's cool. We formed a carbon-carbon bond reaction. The other way you had learned to do this was Friedelcraft's alkylation and Friedelcraft's acylation. There's lots and lots of other things you can do with this as well. All right, introduce a bromide. To introduce a bromide, same thing. Sodium bromide's a reasonably good nucleophile, but copper bromide's even better. You don't want to mix up your bromide and chloride. So again, your textbook I think says HCl, but I would use HBr here. They're both available. You basically want to keep your ion separate. So copper bromide, these are all examples. Cuprous bromide, CUBr, these are all examples of what generally fall under the broad heading of the sandmire reaction. And it's a great way to get into some aromatic chemistry. All right, let's put the emergency handkerchief to the test and I'll put it to the laundry. So a couple of more reactions on benzene chemistry. Basically, you can put almost anything you want, almost any sort of atom you want, except alkyl groups in this way. So give you another example. I'm not sure if this one's in your textbook, but it's the same basic gist. And then I want to show you one that's really important we'll use. I want to put in fluoride. A lot of drug molecules contain fluorine in place of hydrogen. It gives them better biological properties. Chemistry that introduces this pretty well is fluoroboric acid, HBF4. So again, we're just taking strong acids in sodium nitrite. HBF4 and sodium nitrite. And again, we'll do this at 5 degrees Celsius and then we're just going to heat it up. This basically serves as a source of fluoride ion. So I'll write fluoroboric acid. It's a strong acid that serves as a source of F minus source of fluoride ion. And the overall result is that you get the fluorobenzene out of here. All right, you've put in absolutely anything here. Now I want to show you how to get rid of it and why we'll see in a second. All right, same thing. I'm doing all of these examples with toluidine but you could do them with benzene. You could do them with aminonaphthalene. You could do them with trimethylbenzene. All right, that carbocation can be reduced. It can take up hydride and a reagent that allows this to occur is phosphorous acid, H3PO2. And so if we go ahead, we do the same chemistry. Sodium nitrite, diazadization with acid. Your textbook uses HCl. That's fine. I would probably use H2SO4 here because I want a non-nucleophilic counter ion. Again, 5 degrees Celsius. H3PO2 ends up being like a source of hydride. It's kind of, it's something that kind of like we saw in the Canizaro reaction. Remember the Canizaro reaction from discussion? Where a hydride gets pushed off an atom by an oxygen pushing electrons down. It's kind of the same gist here with phosphorous acid. And here's the cool thing. I'm not writing anything. We've replaced the nitro group with hydrogen. We've reduced it, so we've done diazatized and reduced. Okay. I want to show you why all of this chemistry has me and your textbook so excited and why it's the gateway to so much good aromatic chemistry. I want to come back to something straight out of last quarter. And I'll write what really is a very trivial problem. No, it's the wrong type, Johnny. Thanks. It still works, though. Does it? I will try it. Maybe. Thank you. Thank you. We will try. One way, I wave the white flag of surrender. All right. One way or another, we'll model through. We always do. All right. All right. The first honest-to-goodness truth is a ton of simple aromatic compounds are available. Benzene, toluene, these are things that come from petroleum. We learned how to get to aniline. Phenol is easy to get to. So I'm going to do a typical sort of problem I do. Synthesize this molecule from benzene or simple, probably the sort of thing I'd put on the final. We're simple monosubstituted compounds, simple monosubstituted aromatics. This is a great time to review Chapter 18. Remember all those good concepts of resonance and stabilization of carbocations and electrophilic aromatic substitution and reagents to carry this out. And if you think back, you'll remember OH is an ortho-paradirector. If you think back a little more, you'll remember in electrophilic aromatic substitution, you go through a carbocation and that it is, the oxygen is stabilizing to that carbocation. So OH is an ortho-paradirector and it's activating. You'll also remember that electrophiles like chlorine undergo electrophilic aromatic substitution. So if I take phenol and I treat it with chlorine, sometimes in your textbook we'll use iron trichloride, ferric chloride as a catalyst. To tell you the truth, the choice of whether you use a catalyst is a little bit dependent on how good a nucleophile your aromatic is. If you have something that's activating, you may not need a catalyst. If you have something that's deactivating, you might need a catalyst. I'll put ferric chloride in brackets here to say I might not use it if I don't need it. I think your textbook or sapling, actually sapling I think doesn't and your textbook does in activating cases. Because as I said, if you don't need a catalyst, you don't need it. Point is that's a relatively simple operation. Phenol is a readily available organic compound. You've done electrophilic aromatic chlorination. Phenol is an ortho-paradirector. In general, para predominates over ortho. So in general, you can get mostly the para compound and that's good. All right, we will try Johnny's eraser. Let's see which one is better. All right, but remember, chemists are control freaks. We want to be able to do what we want to do and we want to do it the way we want to do it. Because different compounds are useful. You may make a drug with a parachlorophenol compound as a building block, but you may want the metachlorophenol building block and again, synthesize this molecule from simple compounds, same criteria there. All right, so now we have a problem. No matter how we slice it, so the problem is we already know that the OH is an ortho-paradirector and it's activating. Chlorine is also an ortho-paradirector and so we're kind of stuck if you look at this. You say, well, I can't go ahead and just directly chlorinate it. But now you think, wait a second, wait a second. We know a group that's a meta-director. What's a meta-director? NO2, the nitro group is a meta-director. And so we get this idea in our head, if we could take nitrobenzene and chlorinate it, it would go meta and then we can draw upon all this other chemistry to get the control that we wanted. So imagine now you take your nitrobenzene, you chlorinate it. At this point, I probably would use ferric chloride. Again, this is something that maybe Johnny or Kim would worry about more in the lab when to use it. You might literally set up the reaction and monitor it by TLC and see whether you need the catalyst. So nitro is a deactivating meta-director and so you can meta-chlorinate it. And now you say, all right, I can see my way to the rest of this. I go ahead and I go ahead, I hydrogenate. We can do H2 and palladium on carbon. That gets us to the aniline. And then I can take the aniline and I can diazatize. And I'll use sulfuric acid. Again, my personal preference and I think the general experimental preference. I will diazatize to make the diazonium salt. And then I'll treat with phosphorus acid H3PO2. I'm getting ahead of myself here. That's the next example I want to give. I will then heat it up in water. I'll say delta water. Of course, you're running the whole reaction in water. So basically it's just heat the thing up. And so at this point, you get the phenol and the chlorine. Yeah, question. Okay, let's try one more example. I want to really give you a scope. This next example happens to be a problem in sapling so you'll enjoy paying attention to this. And again, starting with simple stuff, synthesize one, three, five tribromobenzene. All right, now you'll look at this and you say we really are in trouble. Bromine's an ortho-paradirector. I can start with benzene and brominate it, get bromobenzene. But if I dibrominate it, I'll get mainly parabromobenzene, very little orthobromobenzene. And I can't get one, remember, ortho is three, right? So ortho-dibromobenzene is with two bromines here. Paradibromobenzene would be with a bromine, two bromines here. But there's no way to get over to this. And so now you really need a spark of inspiration. So I'll just remind us, Br is ortho-paradirector and slightly deactivating. Now you need a flash of inspiration. The flash of inspiration is the amino group is a very good ortho-paradirector and it's activating. And so here's your flash of inspiration. Your flash of inspiration is to say, okay, we've been down the road a zillion times with aniline. And I can go ahead and I can brominate it. And one of the nice things about bromination and chlorination is you get a lot of control. That is, you can put in one bromine, you can put in two bromines, you can put in three bromines because bromination is deactivating. So in general, you have control. If you just use a little bromine and keep it cold, you won't go, you'll get a monobromo. If you put in more bromine, you'll get a dibromo. More, you'll get a tribromo. So we imagine treating this with excess bromine, maybe with iron-tribromide catalyst. Honestly, in the case of aniline, you probably don't need it. I think sapling doesn't use it. I think your textbook does. Johnny and Kim would go into the laboratory, add some bromine, monitor by TLC. If the reaction goes, they would probably not add a catalyst. If it's sluggish, they'd probably add some iron-tribromide as a catalyst or they'd heat it up. Point is, you can go ahead and brominate once, para, twice, or though, three times, or though again. And now you look, you say, all my bromines are where I want. And you say, oh, wait a second. I know how to get rid of that nitro, that amino group. We just diasotize and reduce it off with phosphorus acid. So if I go NaNO2, and you can use HCl or sulfuric acid, honestly it doesn't matter, 5 degrees Celsius, H3PO2, we get the 135-tribromo compound. We've achieved this concept of control. All right. I want to mention two things, one of which is we've been focusing on alcohol, on aromatic amines. So the big thing about aromatic amines is you can isolate the diazonium compounds. They're not very stable. You don't want to let them warm up to room temperature in general because they'll release nitrogen, which makes for volume. It'll release heat, which means they blow up. So you want to keep them ice cold. You want to use them quickly. But we get this carbocation that's really not very happy. Now, if you go ahead and do things with alcohol amines, instead you have a very different chemistry. Alcohol diazonium salts are unstable. So there's really not this beautiful controlled chemistry. You end up getting into carbocation chemistry in a very uncontrolled fashion. So if I take cyclo, oops, boy, I'm in a benzene mindset. If I take cyclohexalamine here and I subject it to these same conditions, NaNO2, sodium nitrite, HCl, even if I keep it cold. The problem is you never really have any lifetime of the diazonium ion. So I'm going to go ahead and say not stable. It forms transiently, but it doesn't get us, it gets us into carbocation chemistry. Basically, as soon as it forms, the nitrogen leaves with its lone pair of electrons. And you get all E1 and SN1 products. And so that kind of leads you nowhere as far as useful chemistry go. Secondary amines, of course, remember if you think back to the mechanism, secondary amines, you can't atomize. So secondary amines, secondary amines form nitrosocompounds. I'll say more specifically, nitrosamines. So let's take dimethylamine. I'll just take the simplest example. And we'll treat it with the same conditions, NaNO2 and HCl. This is a nitrosamine. I'll write out the resonance structure of it. Now, nitrosamines are really, really bad actors. They're carcinogens, they're DNA, they react with DNA, they mess up your DNA. And so this is why hot dogs and cured meats that have nitrites in them are often bad for you, particularly if you grill those cured meats because it makes nitrosamines that, well, the meats taste really good, but it messes up your DNA and gives you cancer. All right, the last thing I'll have you read about, there's some very nice dichemistry that's written about using electrophilic aromatic substitution on diazoniums that's very beautiful. Just take a look at that. It'll be really nice for you to read about it. We'll pick up next time talking about the following chapter. We'll pick up talking about, I think it's chapter 28 at this point.