Preparation of Amines
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King, Susan
This is the third (and final) 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. 
 All right, so we left off last time and we're predicting the acidity and basisity of imidazole. All right, so do we have any acidic protons on imidazole? Anybody? Yeah, we have this. This is an acidic proton. If we remove that proton, it does not affect acidity or base. It doesn't affect aromaticity. So yeah, we can remove a proton. We can't protonate this nitrogen here on the bottom, but we can deprotonate it. It does not affect aromaticity. Still aromatic. So, pKa of that, and here's some pKa of several nitrogen heterocycles. You do not need to memorize these numbers. But this is a nice place to have all these pKa's because they're sometimes hard to find. And so here we have all the pKa's just for comparison. All right, questions on acidity and basisity. Anybody? Let's talk about preparation of amines and then we're going to do reactions of amines. So, preparations of amines. You might remember that when you were doing chapter 7 problems, and you had ammonia attacking and alkyl halide that you got a protonated product as your product, and that might have thrown you off a little bit. And it turns out that it's a little more complicated than we'd like you to believe in chapter 7. So let's get the real story here. All right, so that would be a straight chapter 7 problem. And this would have been the answer you would find in your answer key for a chapter 7 substitution. Maybe showing the counter ion. Okay, so you would get a salt. But this is actually going to further react. So if you look here, if you have ammonia and we have methyl amine, our product, we could get an acid-base reaction. We're going to predict the direction of equilibrium here. So I'm just going to draw a box here. All right, so we want to predict which way is the equilibrium favored. Just by not looking at pKas, but just by knowing what we know about structure. So let's compare the two bases and see which one's more basic. So here is our, this should be protonated here. Here's our base on this side. And here's our base. So we can compare acids on both sides, or we can also compare bases. If we're just using structure, then we can compare those two things directly. We can compare the bases. So which one of those bases is stronger? Ammonia or methyl amine? So we have this lone pair here. Is the methyl electron donating or electron withdrawing group? Electron donating. So it's going to be feeding some, a little bit of electron density, more electron density into the nitrogen than ammonia. So this is a stronger base. So the equilibrium is actually favored to the left. But you can get a little bit of that happening. So remember, the equilibrium always goes from stronger acid, stronger base to weaker acid, weaker base. Here's the problem that we didn't tell you about in, back in Chapter 7, and that's over-alkylation. So as soon as we form some of this compound here, and we can form this compound by this deprotonation that can happen here, even though it's not favored, it is still going back and forth. And when we form this compound here, it can actually also attack another molecule of methyl iodide. So that's what that would look like. And in fact, if you're comparing nucleophile strength, compare methyl amine versus ammonia, which one's more nucleophilic? Anybody have an idea of which one's more nucleophilic? We're looking at the concentration of charge on nitrogen. Methyl is slightly electron donating. And so this one has some electron donating methyl, making that more electron rich. So this is actually a stronger nucleophile than ammonia. And that's why we get over-alkylation, because this is more reactive than that. So as soon as we form some of this, and how we form it is we do a substitution, and then we deprotonate, now we have this. It's going to be more likely to attack methyl iodide than ammonia. And every single time we put another methyl on there, it gets more reactive still. So what you actually get in this reaction, and why I really never asked you this on a test, is you get this, you're going to also get, and it doesn't really matter how much you use, you're going to get a combination of all these things. If you completely methylate nitrogen, that's called exhaustive methylation. Alright, so even if you limit the methyl iodide, and the mean to one equivalent of each, you're going to get a combination of all of these products. So you're going to get one methyl, two methyl, three methyl, four methyl combination. And so that's, of course, all accounting for the different nucleophilicity, why over-alkylation? That's because once your product is more reactive than your starting material. So now we'll draw this out what I was talking about. Methyl is an electron-donated group. So it's donating electron density to nitrogen. That's making it more electron-wrench. So it's going to add more electron density on nitrogen. Therefore it's more nucleophilic. And every time you methylate this, you add one more methyl group, which is going to add more electron density to nitrogen. So every single time you add a methyl group, you make it more nucleophilic. Alright, so if we want just one methyl, if we want to take ammonia, and we want to just form methyl amine, one way around this is to use a very large excess of ammonia. Use a very large excess of ammonia, so that ammonia is the most likely thing to collide with methyl iodide rather than once you methylate ammonia and you get methyl amine, there's less collisions by using a very large excess of ammonia. And so that's one solution. And that's fine if you're just using ammonia, if you're using something that is not expendable like ammonia, that maybe you've made in amine and you want to use a large excess, you don't want to do that. If you synthesize something, you don't want to use a large excess and throw it out. So I'm going to show you some alternative methods below that will work a lot better for you than using a sesame. Alright, so let's look at some alternatives. Alternative number one, reduction of azides, another route to primary means. And so what you do is you do a substitution, you guys probably remember sodium azide from 51A, that was in chapter 7, and so what you do, you have azide ion, negatively charged nitrogen, and you just do an SN2 reaction, works really well, attack the carbon-bearing leaving group, that leaves you with benzoyl azide, and then you can reduce the azide to an amine. So two ways to do that, with the aluminum hydride, step one, followed by water, so we're familiar with that reagent, or you can use H2 and palladium, hydrogen and palladium, so you can do a hydrogenation, and that will take you to the amine. Alright, so what do you remember about azide from chapter 7? I used that one a lot as a nucleophile, why did I use that as a nucleophile a lot? Do you remember? I preferred azide over ammonia, and now you can see why I prefer azide over ammonia. What was the big reason though? What did we have to worry about with substitution? Elimination, and azide is a much weaker base than ammonia, so it means you get less elimination, so this has a bunch of things going for it. You don't get any over-alkylation, and you get much less elimination. So azide is a great nucleophile, and a weak base. How do we know it's a weak base? We look at the pKa of the conjugate acid, pKa of hydrogen, and hydrogen azide is 4.3. Compare with ammonia. Conjugate acid is ammonium ion, with the pKa of about 10. So this is a much weaker acid. It's about six times weaker. So that means we're going to get a whole lot less elimination when we use it. So there's two advantages here. Those are two advantages that we don't get as much elimination and we don't get over-alkylation. But the big deal here, in this part right here, is that we're competing. Each two eliminations are lessened. So that means that this is great for primary and secondary substrates. So, advantage is using azide over ammonia, as we already mentioned, no over-alkylation. And of course, let's just say it one more time, azide ion is a strong nucleophile and a weak base. So less E2, less E2 than with ammonia. Alright, so let's give an example here. And all of those beta-phenethylamines, the two types, we're going to make those, synthesize those various different ways throughout this chapter. You can use this method to synthesize the beta-phenethylamine that has a hydroxyl in the alpha position. So we would just have azide come in here. This is a good example. Attack on the least substituted side. After protonation, we get this compound. And H2-palladium gives us the type of beta-phenethylamine that has a hydroxyl in the alpha position. So let's label that. And let's just remind ourselves, attack at the least substituted side. That goes all the way back to chapter 9. Alright, questions, anybody on using azide instead of ammonia for substitution? Alright, let's see another alternative here. To using azide. I mean to using ammonia for substitution. This alternative is called the Gabriel synthesis. And it uses a compound that's like an anhydride, except it has a nitrogen instead. So that's what we see here. So if this was an oxygen, this would be an anhydride. Instead, it's a nitrogen. And so let's go through the Gabriel synthesis. This is also a really good alternative. Hydroxide ion. The hydroxide ion is going to deprotonate nitrogen. Then we add in our alkyl halide and we do a substitution. And so that nitrogen is not going to be nearly as basic as ammonia for, also because it has electrons that can be delocalized onto two carbonyls, right? So that's going to be less basic than ammonia. So then we do substitution. Now the nitrogen is bonded to the R group, whatever that R group may be. And then, and now we're ready to take this, this other part of the ring off. That was only there sort of as a temporary measure. And so there's actually two ways to do this. You can use HClH2O and heat, what's that going to do? So we have sort of something that looks like a diamond. It's going to hydrolyze that. So we're actually going to get a carboxylic acid. So you may want to try doing the mechanism for that. Dicarboxylic acid. And our NH3 plus. So if we're forming an amine, an acid is going to be protonated. And then we would use hydroxide in a second step to deprotonate. So notice this is a great alternative to using ammonia. The other way that you can do this, I'll do this in a different color. The other way that you can do this is to use hydrazine. That will give you this compound. And after I draw it, you can kind of think about how we would make this compound. Plus RNH2. So can you see what's happening there? So what's going to happen is the hydrogen on the hydrogen is going to attack the carbonyl. Kick electrons up onto oxygen. Right here. Kick electrons up on oxygen. Electrons of oxygen come down and kick this nitrogen off. And now this is going to attack the other side also. So it's almost like transesterification, but with an amine. So those are two alternatives. I believe there's a sapping problem that has this. And you can also do it this way on the test. Alright, so it only gives one product and therefore there's a great way to synthesize primary amines. So that's alternative number two. We have some more alternatives. Questions so far though. Stop me now if you have any questions so far. The other way is to make amines is reducing carbon nitrogen double and triple bonds. This doesn't necessarily give you primary amines, but it can give you primary amines. So that if you treat a nitrile with lithium, aluminum, hydride, and water, we already talked about this reaction, right? So even though I'm talking about this today, this is fair game for the test, right? Didn't we mention that one? And what was that? Chapter 23. Chapter 22. End of chapter 22. Okay, so we can reduce the nitrile. That will give us a primary amine. That primary amine happens to be one carbon longer than the previous one that we did. But maybe that's what you want. The other thing is to reduce an amide with lithium, aluminum, hydride. So another reaction where I'm talking about it today, but it is not a new reaction. So this is chapter 21, right? So you know that reaction. There are two more ways to make amines. Reducing amines and nitriles. Reducing nitriles provides a route to two different types of beta-phenethylamines. So let's look at this one here. So we have cyanide. We're just going to do an SN2 reaction. Backside attack. Kickoff bromide as a leaving group. SN2, right? That gives us a cyanide group. And then we can reduce that with lithium, aluminum, hydride. Followed by water to give us the other type of beta-phenethylamine, where we don't have a hydroxyl in the alpha position. So this would be your simplest beta-phenethylamine that you can make. But you can also, with the nitrile, you can make the other type also. The beta-phenethylamine that has the hydroxyl in the beta position. So the cyanide group is going to attack the aldehyde. Remember what that does? I'm just trying to draw right on top of this. After acid, that gives you a cyanohydrin. Remember that? That is also a review reaction. Which chapter was that in? That's chapter 21. We'll label this after I finish this. So now we can lithium, aluminum, hydride, right? So here's a good bet. There's a good bet that on the final you'll be asked to synthesize both types of beta-phenethylamine. So I'm going to show you several ways to do that. So you'll have choices. Okay, so that's lithium, aluminum, hydride reduces the nitrile to a CH2NH2. And that gives you the second type of beta-phenethylamine. All right, so we got lots of choices here. So let's just remind you, let's label this. This is a cyanohydrin. And that is chapter 21. And the lithium, aluminum, hydride plus a nitrile, that is chapter 23. Or chapter 22. Chapter 22. Okay? Right? So none of that's actually new, except the fact that now we're recognizing what we made as a beta-phenethylamine. All right, so that's more ways to make amines. We've got a lot of fun ways to make amines. And then another way to make amines is called a reductive amination. This is a brand new reaction. And it's a really, really widely used in organic synthesis. Guaranteed I will put a reductive amination product on the, a problem on the final. If, one, if not two, it's a very important reaction. So here's what it looks like. And you guys love amines and ammonium ions, okay? So that's what we're going to do here. We're going to take an amine, and it could be primary or secondary. And we want catalytic acid, or we want, you could just write pH 5. So you're familiar with that reaction. So we're going to take a chapter 21 reaction. I'll show a reaction with a primary amine. So we make an amine. You can actually isolate the amine, or you can throw everything together. We'll, I'll show you both ways to do it. So amine is often not isolated, so let's label that. That's why I have it in brackets. You can isolate it. Often not isolated. If you want to break this into two steps, you certainly can. But you don't have to break it into two steps. Then we do a reduction. And what I usually use, H2 palladium, or another way to do this is a reagent that's very similar to sodium borohydride. Only it has one of the hydrogens replaced by a cyanide group. Sodium cyanoborohydride. And that's another way to do this. And that's the most common way to do it here, I would say. Let's label this. Most common. And so what you're doing here, let me show you what we're doing before we draw the product. It's easier to see from the hydrogenation. So we're just going to add a hydrogen to nitrogen. It's just like we're hydrogenating a double bond. We're going to have a hydrogen to nitrogen, or we're going to have a hydrogen to carbon. And so we're hydrogenating that double bond, even though it's an alkene, we certainly can do that. So let me, I'm going to draw this out. Well, let me draw it in red, and then I'm going to draw the hydrogens that just came on in a different color. So you can see what's happening here. Little hard to visualize. So I'm just going to draw here. And now let's add hydrogen here. Hydrogen to nitrogen, hydrogen to carbon. Okay. So what that means is that we are, and this double bond goes away, of course. So what we're going to make is this. All right. So the nitrogen goes where the carbonyl oxygen was, and then you hydrogenate everything. Great way to make a means. So that's one example. So you can also throw everything together all in the same pot. And that includes whether you use hydrogen and palladium or platinum. This one's using platinum. I almost always use palladium just to have less for you to worry about. But you can, you can combine everything here. Can combine everything. We still want to have some catalytic acid here though. Catalytic H3O plus, or you can write pH 5. So now notice that's a secondary amine. So we're not going to make an amine. We're going to make an iminium ion. So what's going to happen as your intermediate, and I'll draw the intermediate in brackets, you're going to make an iminium ion because you can't make an amine here. So that's formed in the reaction mixture, and as soon as that's formed, it's going to be hydrogenated. So that what you isolate is this amine here. Alright, so that works also really well. Here's reductive amination to make a well speed, and amphetamine. This would be, is this amphetamine or methamphetamine? So what you do is you make the, let's add our pH 5 here, or you can add H3O plus catalytic. In this case we're not throwing everything together. We're doing it stepwise. We're making an amine, and then H2 palladium. So again, the type of beta phenethylamine where you have no hydroxyl in the beta position. Questions on reductive amination? Yes? Excuse me? Enamine formation? Even if you do form the enamine, the enamine can be reduced by the hydrogen and palladium. Okay? Yeah. Alright, let's talk about reduction of nitro compounds. Another way to make amines. We've got a bunch of different ways. Are you sensing the amount of synthesis in this chapter? There's a lot already. And we're not even, you know, it's just one lecture. Okay, reduction of nitro compounds. So this is what it looks like. We have a nitro compound. We can use H2 palladium, or H2 nickel, or H2 platinum. And we can reduce the nitro group to an amine. The one I usually use is the palladium. This is primarily, well, not only, but it's primarily used to synthesize aromatic amines. That is also not a new reaction. Didn't we learn that in Chapter 18? So you can take benzene. Remember, you nitrated this. So we're just in a lot of these. The reductive amination is new, but a lot of these are not new reactions. This is Chapter 18. Was there another way to do this? We do some nitro group? Do you remember the other way to do it? Claminson. Claminson, yeah. Or zinc mercury amalgam and HCl. That also works, too. A great way to make beta-phenethyl amines is the Henry reaction. We've got another name reaction. Similar to the aldol, we've seen examples like this. This is an aldol-like condensation. All right, so here we go. Here. We're going to go through the mechanism here. Of course, the mechanism here, for everything, we're not going to show the hydrogenation mechanism, but the mechanism here, this is not a new mechanism. This is review, right? So I could put this mechanism on the test, right? This is just aldol, right? You agree with that? Sure, it's aldol. Okay, so let's write that down. Not a new mechanism. This is Chapter 24, and Chapter 24 is on the test, so who knows? I don't even know. I haven't written the test yet. I did start it, though, so I'm thinking about it. It's so tough because I have so many mechanisms, so many mechanisms, so little time, right? To have you guys take the test. If we could have a machine grade the mechanisms, I would have way more mechanisms than I do. All right, so everybody see what's going on? We deprotect this. What is this? What is this functional group right here? So now we have acetal, so we can deprotect that. Not deprotonate. Deprotect. If we deprotect that, we're going to get a diol and formaldehyde, and that's dopamine. So that's a way to make dopamine. How about that? But just to remind us that this is a review mechanism, let's go through the mechanism of the Henry. I'm not going to show the hydrogenation, but I do want to go through the mechanism for the Henry. All right, so we have CH3. We know that hydrogen's alpha to a nitro group are acidic. In fact, what does this have? A pKa of 9? It's really high. I mean, it's really a low number. Pretty acidic. We can go ahead and push electrons on carbon, or we could push them onto the nitro. And that's why this guy is so acidic. We actually have good resonance stabilization for this molecule. So we can go here. We can push electrons onto this oxygen here. We can de-localize electrons, definitely. And so I think a good idea for the final, if you're thinking about the final at all, I think a good idea for the final is knowing how to draw a nitro group. Because I feel like if you guys have had O-chem for the whole year, you should know how to draw a nitro group without having five bonds to nitrogen. And I see that a lot in 51C, so I may be testing you on that. I may be asking you that on the final. All right, so you get the idea here. Highly resonance stabilized. But it's an enolate-like compound. So I'm going to redraw it so we don't get confused here. So we're going to redraw our aldehyde. Now what I usually do is I say, you guys already know how to do this. I'm not going to draw this mechanism, but I'm showing it. I'm showing you because we're in good shape here time-wise. Let's redraw. I'm just going to redraw this second resonance structure here. I'm going to leave the electrons on carbon and use that. All right, so we attack the carbon-U. Kick electrons up onto oxygen. We get an alkoxide. Since we're attacking a type I carbon-U, we don't have a leaving group. So our choices are we have two choices. We can make a beta hydroxy carbonyl. Or we can do an elimination and get an alpha-beta unsaturated carbonyl. What's going to happen here? Because of this benzene ring here, we're not going to be able to stop at the beta hydroxy carbonyl. But I'm still going to draw that because we do need to form that first. And then we're going to eliminate. So just remember if I do ask you Aldol condensation, that when you do the elimination, that you do a two-step elimination. You make the enolate first and then you eliminate. Don't do it in one step, you lose four points. So if you've decided you're not going to do nomenclature, you're just going to miss the four-point nomenclature question. And then you do a one-step elimination. There's eight already. Points gone. So you want to make sure you make the enolate first. So now I'm going to draw the nitro group again because I need to remove this. And I can leave it on the nitro group or I can leave it on carbon. Or I can put it onto the nitro group. Let's just leave it on carbon. You don't have to draw the nitro group. So we make the enolate first. And then after we make the enolate, we kick off the leaving group. So just a condensation. But helpful leading up to the test to see this mechanism again. So make the enolate and then we kick electrons down and kick off hydroxyl in a second step. So when we reduce with the hydrogen and the palladium, it's going to reduce the double bond. So it won't be alpha-v8 unsaturated any longer. It will reduce the double bond and it will reduce the nitro group. So the reduction is going to, with H2 and palladium, which we're not going to share, the mechanism for is going to lead to protected dopamine. Questions? Anybody? Anybody have questions on that one? Yes. How is the first step resonance? Yeah, you know what? That should be resonance, huh? This should be reversible arrows, huh? Thank you for pointing that out. I didn't realize I did that. Yeah, thank you. More questions? All right, so that's the aldol condensation. We know that really well. So now we talked about a bunch of ways to make amines. Now, what can we do when we have amines? So this is reactions of amines. Once we already have one, what are the things we can do with it? We already know some of this. We could do one for addition to carbonyls. We could do with acid chlorides. We can make amides. We know a lot about amines. With ketones, we can make amines. If it's a primary, if it's a secondary amine, we can make enamines. With esters, we can do aminolysis to make amines. With anhydrides, we can also make amines. And then, of course, there's all of the SN2 reactions that we did way back in 51A, the end of 51A. As in one substitution with alkyl halides is problematic due to overalkylation. We already talked about that. So we have some alternatives. Amines are also strong races, so E2 competes with SN2. So I want to always keep that in mind. There's a new reaction that we're going to talk about. A reaction of primary amines with nitrous acid. And so nitrous acid in each NO2 is a weak, unstable acid generated in situ by treating NaNO2 with sodium nitrite. That's the same sodium nitrite that you find in hot dogs. And with hydrochloric acid, which is in your stomach, right? So this is some potential reactivity that's happening in your stomach when you eat cured meats that have nitrite in them, right? In the presence of acids, nitrous acid decomposes to a nitrosonium ion, which is a nasty, carcinogenic, nasty compound. A powerful electrophile, which reacts with primary amines to form disonium salts. And so this is what it looks like, how we form it in the first place. We're going to reprotonate oxygen, just like that. And actually, we're going to protonate that oxygen twice. This is not a mechanism I usually ask on exams, by the way. So just showing you what's happening here. Hydrochloric acid, super strong acid, is going to protonate that oxygen twice. And then once it does that, as you can kind of guess, that's going to, we're going to lose that hydroxyl as water. Now that we've protonated it twice, it's going to come off as water. And so as you can imagine, once that leaves, once this leaves to make water, we form a compound that's super, super unstable. Notice we do not have an octet on nitrogen. So this is a very reactive intermediate. So we pretty much avoided ever having nitrogen without an octet. So this is a reactive intermediate. It has a very, very short lifetime. So if you form this intermediate in the presence of a primary amine, what's going to happen is it's going to, anything with lone pairs, including DNA, okay, is going to attack that. So it would look like this after that happens. Again, don't worry about mechanism here. I'm just trying to show you what's happening here. You form this, and then deprotonate with chloride ion. And then we're going to skip a bunch of steps. And whenever we skip a bunch of steps, it means I'm not going to ask you this. But what you get is something. After you deprotonate with the chlorine. Oops. Do that wrong. This tablet's going quickly here. You get something that looks like that. And this is N-nitrosamine. Which is a potent carcinogen. This gets converted in many steps, which I'm not going to show you, to something that you probably haven't seen before. But what it looks like is, this is called a diazonium salt. And it turns out that this group here, this leaves as a leaving group. It is the world's best leaving group. Right here, nitrogen. So this is an electrophile with a fantastic leaving group. So it does all sorts of things. All the things that we talked about in 51A, 51B, SN1, E1, SN2, E2. Not very synthetically useful. So why did I show you this in the first place? Because there actually is some very good synthetic utility if you use aromatic amines to do this reaction. So, but let's just label this. Because I'm going to show you a good thing that we can do with this reaction. But for this, it's not synthetically useful. All right. We will stop right there. And we will continue this. On Monday, I'll see you on Friday for the exam.
 All right, so we left off last time and we're predicting the acidity and basisity of imidazole.  All right, so do we have any acidic protons on imidazole? Anybody? Yeah, we have this. This is an acidic proton.  If we remove that proton, it does not affect acidity or base. It doesn't affect aromaticity.  So yeah, we can remove a proton. We can't protonate this nitrogen here on the bottom, but we can deprotonate it.  It does not affect aromaticity. Still aromatic.  So, pKa of that, and here's some pKa of several nitrogen heterocycles. You do not need to memorize these numbers.  But this is a nice place to have all these pKa's because they're sometimes hard to find.  And so here we have all the pKa's just for comparison.  All right, questions on acidity and basisity. Anybody?  Let's talk about preparation of amines and then we're going to do reactions of amines.  So, preparations of amines. You might remember that when you were doing chapter 7 problems,  and you had ammonia attacking and alkyl halide that you got a protonated product as your product,  and that might have thrown you off a little bit.  And it turns out that it's a little more complicated than we'd like you to believe in chapter 7.  So let's get the real story here.  All right, so that would be a straight chapter 7 problem.  And this would have been the answer you would find in your answer key for a chapter 7 substitution.  Maybe showing the counter ion.  Okay, so you would get a salt.  But this is actually going to further react.  So if you look here, if you have ammonia and we have methyl amine, our product,  we could get an acid-base reaction.  We're going to predict the direction of equilibrium here.  So I'm just going to draw a box here.  All right, so we want to predict which way is the equilibrium favored.  Just by not looking at pKas, but just by knowing what we know about structure.  So let's compare the two bases and see which one's more basic.  So here is our, this should be protonated here.  Here's our base on this side.  And here's our base.  So we can compare acids on both sides, or we can also compare bases.  If we're just using structure, then we can compare those two things directly.  We can compare the bases.  So which one of those bases is stronger?  Ammonia or methyl amine?  So we have this lone pair here.  Is the methyl electron donating or electron withdrawing group?  Electron donating.  So it's going to be feeding some, a little bit of electron density,  more electron density into the nitrogen than ammonia.  So this is a stronger base.  So the equilibrium is actually favored to the left.  But you can get a little bit of that happening.  So remember, the equilibrium always goes from stronger acid,  stronger base to weaker acid, weaker base.  Here's the problem that we didn't tell you about in, back in Chapter 7,  and that's over-alkylation.  So as soon as we form some of this compound here,  and we can form this compound by this deprotonation that can happen here,  even though it's not favored, it is still going back and forth.  And when we form this compound here,  it can actually also attack another molecule of methyl iodide.  So that's what that would look like.  And in fact, if you're comparing nucleophile strength,  compare methyl amine versus ammonia, which one's more nucleophilic?  Anybody have an idea of which one's more nucleophilic?  We're looking at the concentration of charge on nitrogen.  Methyl is slightly electron donating.  And so this one has some electron donating methyl,  making that more electron rich.  So this is actually a stronger nucleophile than ammonia.  And that's why we get over-alkylation,  because this is more reactive than that.  So as soon as we form some of this,  and how we form it is we do a substitution,  and then we deprotonate, now we have this.  It's going to be more likely to attack methyl iodide than ammonia.  And every single time we put another methyl on there,  it gets more reactive still.  So what you actually get in this reaction,  and why I really never asked you this on a test,  is you get this, you're going to also get,  and it doesn't really matter how much you use,  you're going to get a combination of all these things.  If you completely methylate nitrogen, that's called exhaustive methylation.  Alright, so even if you limit the methyl iodide,  and the mean to one equivalent of each,  you're going to get a combination of all of these products.  So you're going to get one methyl, two methyl,  three methyl, four methyl combination.  And so that's, of course, all accounting for the different nucleophilicity,  why over-alkylation?  That's because once your product is more reactive  than your starting material.  So now we'll draw this out what I was talking about.  Methyl is an electron-donated group.  So it's donating electron density to nitrogen.  That's making it more electron-wrench.  So it's going to add more electron density on nitrogen.  Therefore it's more nucleophilic.  And every time you methylate this,  you add one more methyl group,  which is going to add more electron density to nitrogen.  So every single time you add a methyl group,  you make it more nucleophilic.  Alright, so if we want just one methyl,  if we want to take ammonia,  and we want to just form methyl amine,  one way around this is to use a very large excess of ammonia.  Use a very large excess of ammonia,  so that ammonia is the most likely thing to collide with methyl iodide  rather than once you methylate ammonia and you get methyl amine,  there's less collisions by using a very large excess of ammonia.  And so that's one solution.  And that's fine if you're just using ammonia,  if you're using something that is not expendable like ammonia,  that maybe you've made in amine  and you want to use a large excess, you don't want to do that.  If you synthesize something,  you don't want to use a large excess and throw it out.  So I'm going to show you some alternative methods below  that will work a lot better for you than using a sesame.  Alright, so let's look at some alternatives.  Alternative number one, reduction of azides,  another route to primary means.  And so what you do is you do a substitution,  you guys probably remember sodium azide from 51A,  that was in chapter 7,  and so what you do, you have azide ion,  negatively charged nitrogen,  and you just do an SN2 reaction, works really well,  attack the carbon-bearing leaving group,  that leaves you with benzoyl azide,  and then you can reduce the azide to an amine.  So two ways to do that,  with the aluminum hydride,  step one, followed by water,  so we're familiar with that reagent,  or you can use H2 and palladium,  hydrogen and palladium,  so you can do a hydrogenation,  and that will take you to the amine.  Alright, so what do you remember about azide from chapter 7?  I used that one a lot as a nucleophile,  why did I use that as a nucleophile a lot?  Do you remember?  I preferred azide over ammonia,  and now you can see why I prefer azide over ammonia.  What was the big reason though?  What did we have to worry about with substitution?  Elimination, and azide is a much weaker base than ammonia,  so it means you get less elimination,  so this has a bunch of things going for it.  You don't get any over-alkylation,  and you get much less elimination.  So azide is a great nucleophile,  and a weak base.  How do we know it's a weak base?  We look at the pKa of the conjugate acid, pKa of hydrogen,  and hydrogen azide is 4.3.  Compare with ammonia.  Conjugate acid is ammonium ion,  with the pKa of about 10.  So this is a much weaker acid.  It's about six times weaker.  So that means we're going to get a whole lot less elimination  when we use it.  So there's two advantages here.  Those are two advantages that we don't get  as much elimination and we don't get over-alkylation.  But the big deal here, in this part right here,  is that we're competing.  Each two eliminations are lessened.  So that means that this is great for primary and secondary substrates.  So, advantage is using azide over ammonia,  as we already mentioned, no over-alkylation.  And of course, let's just say it one more time,  azide ion is a strong nucleophile and a weak base.  So less E2, less E2 than with ammonia.  Alright, so let's give an example here.  And all of those beta-phenethylamines, the two types,  we're going to make those, synthesize those various different ways  throughout this chapter.  You can use this method to synthesize the beta-phenethylamine  that has a hydroxyl in the alpha position.  So we would just have azide come in here.  This is a good example.  Attack on the least substituted side.  After protonation, we get this compound.  And H2-palladium gives us the type of beta-phenethylamine  that has a hydroxyl in the alpha position.  So let's label that.  And let's just remind ourselves, attack at the least substituted side.  That goes all the way back to chapter 9.  Alright, questions, anybody on using azide instead of ammonia for substitution?  Alright, let's see another alternative here.  To using azide.  I mean to using ammonia for substitution.  This alternative is called the Gabriel synthesis.  And it uses a compound that's like an anhydride, except it has a nitrogen instead.  So that's what we see here.  So if this was an oxygen, this would be an anhydride.  Instead, it's a nitrogen.  And so let's go through the Gabriel synthesis.  This is also a really good alternative.  Hydroxide ion.  The hydroxide ion is going to deprotonate nitrogen.  Then we add in our alkyl halide and we do a substitution.  And so that nitrogen is not going to be nearly as basic as ammonia for,  also because it has electrons that can be delocalized onto two carbonyls, right?  So that's going to be less basic than ammonia.  So then we do substitution.  Now the nitrogen is bonded to the R group, whatever that R group may be.  And then, and now we're ready to take this, this other part of the ring off.  That was only there sort of as a temporary measure.  And so there's actually two ways to do this.  You can use HClH2O and heat, what's that going to do?  So we have sort of something that looks like a diamond.  It's going to hydrolyze that.  So we're actually going to get a carboxylic acid.  So you may want to try doing the mechanism for that.  Dicarboxylic acid.  And our NH3 plus.  So if we're forming an amine, an acid is going to be protonated.  And then we would use hydroxide in a second step to deprotonate.  So notice this is a great alternative to using ammonia.  The other way that you can do this, I'll do this in a different color.  The other way that you can do this is to use hydrazine.  That will give you this compound.  And after I draw it, you can kind of think about how we would make this compound.  Plus RNH2.  So can you see what's happening there?  So what's going to happen is the hydrogen on the hydrogen is going to attack the carbonyl.  Kick electrons up onto oxygen.  Right here.  Kick electrons up on oxygen.  Electrons of oxygen come down and kick this nitrogen off.  And now this is going to attack the other side also.  So it's almost like transesterification, but with an amine.  So those are two alternatives.  I believe there's a sapping problem that has this.  And you can also do it this way on the test.  Alright, so it only gives one product and therefore there's a great way to synthesize primary amines.  So that's alternative number two.  We have some more alternatives.  Questions so far though.  Stop me now if you have any questions so far.  The other way is to make amines is reducing carbon nitrogen double and triple bonds.  This doesn't necessarily give you primary amines, but it can give you primary amines.  So that if you treat a nitrile with lithium, aluminum, hydride, and water,  we already talked about this reaction, right?  So even though I'm talking about this today, this is fair game for the test, right?  Didn't we mention that one?  And what was that?  Chapter 23.  Chapter 22.  End of chapter 22.  Okay, so we can reduce the nitrile.  That will give us a primary amine.  That primary amine happens to be one carbon longer than the previous one that we did.  But maybe that's what you want.  The other thing is to reduce an amide with lithium, aluminum, hydride.  So another reaction where I'm talking about it today, but it is not a new reaction.  So this is chapter 21, right?  So you know that reaction.  There are two more ways to make amines.  Reducing amines and nitriles.  Reducing nitriles provides a route to two different types of beta-phenethylamines.  So let's look at this one here.  So we have cyanide.  We're just going to do an SN2 reaction.  Backside attack.  Kickoff bromide as a leaving group.  SN2, right?  That gives us a cyanide group.  And then we can reduce that with lithium, aluminum, hydride.  Followed by water to give us the other type of beta-phenethylamine,  where we don't have a hydroxyl in the alpha position.  So this would be your simplest beta-phenethylamine that you can make.  But you can also, with the nitrile, you can make the other type also.  The beta-phenethylamine that has the hydroxyl in the beta position.  So the cyanide group is going to attack the aldehyde.  Remember what that does?  I'm just trying to draw right on top of this.  After acid, that gives you a cyanohydrin.  Remember that?  That is also a review reaction.  Which chapter was that in?  That's chapter 21.  We'll label this after I finish this.  So now we can lithium, aluminum, hydride, right?  So here's a good bet.  There's a good bet that on the final you'll be asked to synthesize both types of beta-phenethylamine.  So I'm going to show you several ways to do that.  So you'll have choices.  Okay, so that's lithium, aluminum, hydride reduces the nitrile to a CH2NH2.  And that gives you the second type of beta-phenethylamine.  All right, so we got lots of choices here.  So let's just remind you, let's label this.  This is a cyanohydrin.  And that is chapter 21.  And the lithium, aluminum, hydride plus a nitrile, that is chapter 23.  Or chapter 22.  Chapter 22.  Okay?  Right?  So none of that's actually new, except the fact that now we're recognizing what we made as a beta-phenethylamine.  All right, so that's more ways to make amines.  We've got a lot of fun ways to make amines.  And then another way to make amines is called a reductive amination.  This is a brand new reaction.  And it's a really, really widely used in organic synthesis.  Guaranteed I will put a reductive amination product on the, a problem on the final.  If, one, if not two, it's a very important reaction.  So here's what it looks like.  And you guys love amines and ammonium ions, okay?  So that's what we're going to do here.  We're going to take an amine, and it could be primary or secondary.  And we want catalytic acid, or we want, you could just write pH 5.  So you're familiar with that reaction.  So we're going to take a chapter 21 reaction.  I'll show a reaction with a primary amine.  So we make an amine.  You can actually isolate the amine, or you can throw everything together.  We'll, I'll show you both ways to do it.  So amine is often not isolated, so let's label that.  That's why I have it in brackets.  You can isolate it.  Often not isolated.  If you want to break this into two steps, you certainly can.  But you don't have to break it into two steps.  Then we do a reduction.  And what I usually use, H2 palladium, or another way to do this is a reagent that's very similar to sodium borohydride.  Only it has one of the hydrogens replaced by a cyanide group.  Sodium cyanoborohydride.  And that's another way to do this.  And that's the most common way to do it here, I would say.  Let's label this.  Most common.  And so what you're doing here, let me show you what we're doing before we draw the product.  It's easier to see from the hydrogenation.  So we're just going to add a hydrogen to nitrogen.  It's just like we're hydrogenating a double bond.  We're going to have a hydrogen to nitrogen, or we're going to have a hydrogen to carbon.  And so we're hydrogenating that double bond, even though it's an alkene, we certainly can do that.  So let me, I'm going to draw this out.  Well, let me draw it in red, and then I'm going to draw the hydrogens that just came on in a different color.  So you can see what's happening here.  Little hard to visualize.  So I'm just going to draw here.  And now let's add hydrogen here.  Hydrogen to nitrogen, hydrogen to carbon.  Okay.  So what that means is that we are, and this double bond goes away, of course.  So what we're going to make is this.  All right.  So the nitrogen goes where the carbonyl oxygen was, and then you hydrogenate everything.  Great way to make a means.  So that's one example.  So you can also throw everything together all in the same pot.  And that includes whether you use hydrogen and palladium or platinum.  This one's using platinum.  I almost always use palladium just to have less for you to worry about.  But you can, you can combine everything here.  Can combine everything.  We still want to have some catalytic acid here though.  Catalytic H3O plus, or you can write pH 5.  So now notice that's a secondary amine.  So we're not going to make an amine.  We're going to make an iminium ion.  So what's going to happen as your intermediate, and I'll draw the intermediate in brackets,  you're going to make an iminium ion because you can't make an amine here.  So that's formed in the reaction mixture, and as soon as that's formed,  it's going to be hydrogenated.  So that what you isolate is this amine here.  Alright, so that works also really well.  Here's reductive amination to make a well speed, and amphetamine.  This would be, is this amphetamine or methamphetamine?  So what you do is you make the, let's add our pH 5 here,  or you can add H3O plus catalytic.  In this case we're not throwing everything together.  We're doing it stepwise.  We're making an amine, and then H2 palladium.  So again, the type of beta phenethylamine where you have no hydroxyl in the beta position.  Questions on reductive amination?  Yes?  Excuse me?  Enamine formation?  Even if you do form the enamine, the enamine can be reduced by the hydrogen and palladium.  Okay?  Yeah.  Alright, let's talk about reduction of nitro compounds.  Another way to make amines.  We've got a bunch of different ways.  Are you sensing the amount of synthesis in this chapter?  There's a lot already.  And we're not even, you know, it's just one lecture.  Okay, reduction of nitro compounds.  So this is what it looks like.  We have a nitro compound.  We can use H2 palladium, or H2 nickel, or H2 platinum.  And we can reduce the nitro group to an amine.  The one I usually use is the palladium.  This is primarily, well, not only, but it's primarily used to synthesize aromatic amines.  That is also not a new reaction.  Didn't we learn that in Chapter 18?  So you can take benzene.  Remember, you nitrated this.  So we're just in a lot of these.  The reductive amination is new, but a lot of these are not new reactions.  This is Chapter 18.  Was there another way to do this?  We do some nitro group?  Do you remember the other way to do it?  Claminson.  Claminson, yeah.  Or zinc mercury amalgam and HCl.  That also works, too.  A great way to make beta-phenethyl amines is the Henry reaction.  We've got another name reaction.  Similar to the aldol, we've seen examples like this.  This is an aldol-like condensation.  All right, so here we go.  Here.  We're going to go through the mechanism here.  Of course, the mechanism here, for everything, we're not going to show the hydrogenation mechanism,  but the mechanism here, this is not a new mechanism.  This is review, right?  So I could put this mechanism on the test, right?  This is just aldol, right?  You agree with that?  Sure, it's aldol.  Okay, so let's write that down.  Not a new mechanism.  This is Chapter 24, and Chapter 24 is on the test, so who knows?  I don't even know.  I haven't written the test yet.  I did start it, though, so I'm thinking about it.  It's so tough because I have so many mechanisms, so many mechanisms, so little time, right?  To have you guys take the test.  If we could have a machine grade the mechanisms, I would have way more mechanisms than I do.  All right, so everybody see what's going on?  We deprotect this.  What is this?  What is this functional group right here?  So now we have acetal, so we can deprotect that.  Not deprotonate.  Deprotect.  If we deprotect that, we're going to get a diol and formaldehyde, and that's dopamine.  So that's a way to make dopamine.  How about that?  But just to remind us that this is a review mechanism, let's go through the mechanism of the Henry.  I'm not going to show the hydrogenation, but I do want to go through the mechanism for the Henry.  All right, so we have CH3.  We know that hydrogen's alpha to a nitro group are acidic.  In fact, what does this have?  A pKa of 9?  It's really high.  I mean, it's really a low number.  Pretty acidic.  We can go ahead and push electrons on carbon, or we could push them onto the nitro.  And that's why this guy is so acidic.  We actually have good resonance stabilization for this molecule.  So we can go here.  We can push electrons onto this oxygen here.  We can de-localize electrons, definitely.  And so I think a good idea for the final, if you're thinking about the final at all, I think a good idea for the final is knowing how to draw a nitro group.  Because I feel like if you guys have had O-chem for the whole year, you should know how to draw a nitro group without having five bonds to nitrogen.  And I see that a lot in 51C, so I may be testing you on that.  I may be asking you that on the final.  All right, so you get the idea here.  Highly resonance stabilized.  But it's an enolate-like compound.  So I'm going to redraw it so we don't get confused here.  So we're going to redraw our aldehyde.  Now what I usually do is I say, you guys already know how to do this.  I'm not going to draw this mechanism, but I'm showing it.  I'm showing you because we're in good shape here time-wise.  Let's redraw.  I'm just going to redraw this second resonance structure here.  I'm going to leave the electrons on carbon and use that.  All right, so we attack the carbon-U.  Kick electrons up onto oxygen.  We get an alkoxide.  Since we're attacking a type I carbon-U, we don't have a leaving group.  So our choices are we have two choices.  We can make a beta hydroxy carbonyl.  Or we can do an elimination and get an alpha-beta unsaturated carbonyl.  What's going to happen here?  Because of this benzene ring here, we're not going to be able to stop  at the beta hydroxy carbonyl.  But I'm still going to draw that because we do need to form that first.  And then we're going to eliminate.  So just remember if I do ask you Aldol condensation,  that when you do the elimination, that you do a two-step elimination.  You make the enolate first and then you eliminate.  Don't do it in one step, you lose four points.  So if you've decided you're not going to do nomenclature,  you're just going to miss the four-point nomenclature question.  And then you do a one-step elimination.  There's eight already.  Points gone.  So you want to make sure you make the enolate first.  So now I'm going to draw the nitro group again because I need to remove this.  And I can leave it on the nitro group or I can leave it on carbon.  Or I can put it onto the nitro group.  Let's just leave it on carbon.  You don't have to draw the nitro group.  So we make the enolate first.  And then after we make the enolate, we kick off the leaving group.  So just a condensation.  But helpful leading up to the test to see this mechanism again.  So make the enolate and then we kick electrons down and kick off hydroxyl in a second step.  So when we reduce with the hydrogen and the palladium, it's going to reduce the double bond.  So it won't be alpha-v8 unsaturated any longer.  It will reduce the double bond and it will reduce the nitro group.  So the reduction is going to, with H2 and palladium, which we're not going to share,  the mechanism for is going to lead to protected dopamine.  Questions? Anybody?  Anybody have questions on that one?  Yes.  How is the first step resonance?  Yeah, you know what?  That should be resonance, huh?  This should be reversible arrows, huh?  Thank you for pointing that out.  I didn't realize I did that.  Yeah, thank you.  More questions?  All right, so that's the aldol condensation.  We know that really well.  So now we talked about a bunch of ways to make amines.  Now, what can we do when we have amines?  So this is reactions of amines.  Once we already have one, what are the things we can do with it?  We already know some of this.  We could do one for addition to carbonyls.  We could do with acid chlorides.  We can make amides.  We know a lot about amines.  With ketones, we can make amines.  If it's a primary, if it's a secondary amine, we can make enamines.  With esters, we can do aminolysis to make amines.  With anhydrides, we can also make amines.  And then, of course, there's all of the SN2 reactions that we did  way back in 51A, the end of 51A.  As in one substitution with alkyl halides is problematic  due to overalkylation.  We already talked about that.  So we have some alternatives.  Amines are also strong races, so E2 competes with SN2.  So I want to always keep that in mind.  There's a new reaction that we're going to talk about.  A reaction of primary amines with nitrous acid.  And so nitrous acid in each NO2 is a weak, unstable acid generated  in situ by treating NaNO2 with sodium nitrite.  That's the same sodium nitrite that you find in hot dogs.  And with hydrochloric acid, which is in your stomach, right?  So this is some potential reactivity that's happening in your stomach  when you eat cured meats that have nitrite in them, right?  In the presence of acids, nitrous acid decomposes to a nitrosonium ion,  which is a nasty, carcinogenic, nasty compound.  A powerful electrophile, which reacts with primary amines to form disonium salts.  And so this is what it looks like, how we form it in the first place.  We're going to reprotonate oxygen, just like that.  And actually, we're going to protonate that oxygen twice.  This is not a mechanism I usually ask on exams, by the way.  So just showing you what's happening here.  Hydrochloric acid, super strong acid, is going to protonate that oxygen twice.  And then once it does that, as you can kind of guess,  that's going to, we're going to lose that hydroxyl as water.  Now that we've protonated it twice, it's going to come off as water.  And so as you can imagine, once that leaves, once this leaves to make water,  we form a compound that's super, super unstable.  Notice we do not have an octet on nitrogen.  So this is a very reactive intermediate.  So we pretty much avoided ever having nitrogen without an octet.  So this is a reactive intermediate.  It has a very, very short lifetime.  So if you form this intermediate in the presence of a primary amine,  what's going to happen is it's going to, anything with lone pairs,  including DNA, okay, is going to attack that.  So it would look like this after that happens.  Again, don't worry about mechanism here.  I'm just trying to show you what's happening here.  You form this, and then deprotonate with chloride ion.  And then we're going to skip a bunch of steps.  And whenever we skip a bunch of steps, it means I'm not going to ask you this.  But what you get is something.  After you deprotonate with the chlorine.  Oops.  Do that wrong.  This tablet's going quickly here.  You get something that looks like that.  And this is N-nitrosamine.  Which is a potent carcinogen.  This gets converted in many steps, which I'm not going to show you,  to something that you probably haven't seen before.  But what it looks like is,  this is called a diazonium salt.  And it turns out that this group here,  this leaves as a leaving group.  It is the world's best leaving group.  Right here, nitrogen.  So this is an electrophile with a fantastic leaving group.  So it does all sorts of things.  All the things that we talked about in 51A, 51B, SN1, E1, SN2, E2.  Not very synthetically useful.  So why did I show you this in the first place?  Because there actually is some very good synthetic utility  if you use aromatic amines to do this reaction.  So, but let's just label this.  Because I'm going to show you a good thing that we can do with this reaction.  But for this, it's not synthetically useful.  All right. We will stop right there.  And we will continue this.  On Monday, I'll see you on Friday for the exam.
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