 All right, so I am thinking about where we're going in the course here, and I'm starting to put together the discussion section problems for next week as well as thinking about what's important. And there are a whole bunch of cool reaction mechanisms involving the chemistry of carbonyl compounds. And these are really on my mind, and I think we're going to be working on them in discussion. And I'm going to want you to participate a little more in discussion. We're going to get up to the blackboard and put our thoughts and really work together on thinking about these problems. So we're in the midst of the chemistry of ketones and aldehydes, and we did a little bit of basic introduction of their chemistry last time. We introduced how we name them. We introduced some ideas about reactivity at the carbonyl group. We talked a little bit about their spectroscopy. And then I introduced the idea of how I like to think about reactions of carbonyl compounds with nucleophiles. And I like to think about there being sort of three types of nucleophiles, broad types, the very strongly basic nucleophiles like hydride and alko groups, things who the pKa groups for which the pKa of their conjugate acid is a very large number. In other words, things that although the conjugate acids we call a very weak acid would not to any measurable extent dissociate in water. An alkane like methane doesn't dissociate to methyl anion even though it is a technically a very, very weak acid. If you work out an equilibrium for a mole of a compound with pKa of 50 in water and try to figure out the concentration of the conjugate base and a proton, the concentration is less than one molecule, well under one molecule. So basically things like methane and hydrogen don't dissociate in water. We then moved on to a category that I roughly grouped as moderately basic nucleophiles and that fits a lot of things where you'd think of their conjugate acids as weak acids. Hydrogen cyanide for example pKa 9.4 dissociates in water a little bit so you have cyanide anion in water which actually acts as the nucleophile. Alcohols and water are sort of at the lower edge of that category as very weakly basic nucleophiles. And then finally I mentioned very weak nucleophiles, things where the conjugate acid really is a strong acid. Things like an alcohol where the conjugate acid, ROH2 plus is a strong acid. We're sort of off in the negative numbers there. And in our third lecture we'll be talking about that class but here I want to continue our discussion of the category of moderately basic nucleophiles. And I just want to lay two examples from the previous lecture to talk about what they taught us because we're going to go on now to looking at the reaction of amines with carbonyl compounds with ketones and aldehydes. Amines fall in that moderately basic category. The pKa of the conjugate acid, the pKa of an ammonium ion is about 10 or 11. And we saw what set the stage for these reactions in two different reactions. We saw the reaction of hydrogen cyanide with a generic ketone or aldehyde. I'm not meaning this to mean acetone per se but any sort of ketone or aldehyde. And the reaction was to form a cyanohydrin. And then the other reaction we talked about last time was the reaction of a vitigre agent. And again I'll just sort of write a semi-generic vitigre agent. I'll write ROH2, PPH3 double bonded. And remember we talked about there being two resonance structures. So we talked about there being a second resonance structure of the vitigre agent that had an illid form. In other words, had a separated charge. And only because phosphorous can take an expanded octet. Only because phosphorous can take 10 electrons around it. Could we write this other resonance form here? And then the product of that, again, generic ketone or aldehyde with this sort of semi-generic vitigre agent is an alkene. Now what these reactions set the stage for is the idea that nucleophiles like to add to the carbonyl group. We see that hydrogen cyanide we can think of it as dissociating to H plus and CN minus or in water H3O plus and CN minus. CN minus adding to the carbonyl, the resulting alkoxide anion then picking up a proton from another molecule of hydrogen cyanide or from hydronium ion. In the case of the vitigre action, the start of this theme was very similar. In other words, we start our mechanism by electrons flowing from the nucleophilic species. In this case, the basic carbon, the carbonyl like carbon to the oxygen. But then at the same time, electrons flow on to the phosphorus now from the oxygen. Now phosphorus really, really loves oxygen and here's where the theme for today's lecture gets set in. Phosphorus binds on to the oxygen and then pulls off the oxygen. And so the oxygen is now taken out of the molecule. And this is the same theme, although not with the pole from the same atom, not with the pole from phosphorus, that we're going to be seeing in the reaction of amine nucleophiles with ketones and aldehydes. So by way of diving into things, let me start with a generic reaction here. If we have some compound, organic chemists often use R groups to indicate unspecified alcohol groups or sometimes even alcohol and hydrogen. So I'll say if we have some ketone or aldehyde, this is a very general reaction. And we treat it with a primary amine and I'll use double prime to indicate just that we're talking about some R group here, not necessarily the same R group. A primary amine, of course, is an amine with one alcohol group on the nitrogen. Remember, it's a little confusing. And alcohols, of course, the only alcohols are alcohols with one R group on the oxygen. So a primary alcohol is an alcohol where you have an RCH2 connected to the alcohol where you have only one other carbon. A secondary alcohol is an alcohol where you have two carbons connected to that carbon like isopropanol and a tertiary alcohol is an alcohol where you have three carbons connected to that carbon like tertiary butanol. But in the case of amines, because you can have multiple R groups on the nitrogen, we call a primary amine and amine with one alcohol group on nitrogen. We call a compound with two alcohol groups on nitrogen a secondary amine and a compound and amine with three alcohol groups on nitrogen, a tertiary amine. So with a primary amine and a ketone or aldehyde, they react. Your textbook indicates catalytic acid and we will invoke this in the mechanism. And I'm going to write a little caveat on this sometimes because ketones and amines can react without the presence of catalytic acid. Remember, a catalyst ends up facilitating the pathway but that doesn't necessarily mean the reaction can't occur without it plus there are various sources of H plus. As you go through the reaction, you're going to generate protic species. So the product of this reaction is an amine where the nitrogens R group is derived from the primary amine and the amine has this structure. And again, organic chemists are terrible about writing balanced equations but the other product of the reaction is water and that actually gets important in driving the reaction. So I'll write out the name for this. I'll just say an amine and maybe I'll give you a concrete example of this reaction. So for example, if I take benzaldehyde, remember benzaldehyde is one of those aldehydes whose common name is so widely used that we would probably never call it anything else. And let's take benzalamine as our other partner in this reaction. Usually it takes some heat to drive this chemistry but really what this reaction takes, the reaction's an equilibrium. In other words, and I'll write it later as such, right now I'm writing this more in a synthetic fashion. But the equilibrium water and the amine exist in equilibrium with the ketone or aldehyde and the primary amine. And the real problem is if you're trying to make this chemistry go and remember chemists often try to control their chemistry and drive the reaction to a product. In other words, get rid of all the starting material and form as much product as they can as possible. You do this in your chem lab class when you're monitoring by TLC and you're saying as my reaction done yet, is it done yet, can I work it up, have I used up all my reactants and the problem in an equilibrium is that unless you employ Le Chatelier's principle, unless you push it, you will often end up with your reaction never quote being done because done is in the middle. So the way that chemists often drive this reaction is by doing it in a solvent that water doesn't dissolve in such as benzene, sometimes they use toluene at reflux, reflux means boiling. Now benzene and toluene have a very special property, benzene and water, toluene and water co-distilled to form what's called an azeotrope. In other words, the benzene drives the water up in the vapor phase. And if you go ahead and then have a receiving vessel, the water is heavier than benzene. If you have a little receiving vessel under the distillation apparatus underneath the condenser, the water and benzene drip back in but once the receiving vessel fills up with benzene, the benzene keeps sluicing back into the reaction flask and continuing to distill while the water goes to the bottom. This apparatus is called the Dean Stark Trap. After its inventors and by carrying out the reaction under these conditions, you can get your imine. Your textbook writes, quote, mild acid, this particular reaction proceeds just fine without an acid catalyst although an acid catalyst will make it go faster and so here's our imine product and again at least for now I'll be good about balancing my equation in writing water and sometimes you'll hear an imine derived from an aldehyde referred to as an aldimine so I'll write that out and sometimes you'll hear of an imine derived from a ketone as being referred to as a ketimine. Let me give you an example of a ketimine forming. The examples I've chosen are real examples from real experiments just to show you typical conditions that one employs for this chemistry. So the example I'll give you is methylcyclohexanone and I'll give you phenylethylamine. It turns out that there's a lot of really beautiful chemistry of imines and enamines that we're not going to talk extensively about today and your textbook doesn't present extensively but they can be really important in carbon-carbon bond forming reactions and I told you early on and we've talked about synthesis how important it is to build up molecules, build up greater degrees of complexity this is how we make new medicines for example this is how you make chemically synthesized steroids and so forth and drugs and so if one uses a chiral amine and I'm not going to draw a particular enantiomer here one can go ahead and form a valuable imine from this. It turns out imines have stereochemistry to them about the carbon-nitrogen double bond and not surprisingly if you have a choice you want to form the least sterically hindered product. Immine formation is thermodynamic. I guess I was writing pH for benzene here so I will continue to do so. Immine formation is thermodynamic and so you form the thermodynamic less stable compound and here the experimental conditions are very similar. The reactions done in toluene, benzene boils at a little bit lower temperature than toluene. Toluene boils at about 110 benzene at about 80 degrees and refluxing in toluene, oops, toluene again with a dean's stark apparatus hives the reaction and of course you get water as a byproduct of the reaction which you collect in the dean's stark trap. If I had my choice personally between using benzene or toluene I prefer to use toluene. Benzene is carcinogenic and so in general people don't like to work with it as much it's not really carcinogenic. I mean it is really carcinogenic. It's not super carcinogenic. We use it in an experiment in the Chem 160 lab course that I use. You just work with it carefully like you do when you work with other toxic chemicals. Question? Yeah. Thank you. Yes, it is the primary amine NH2. Thank you very much and once again thank you. You have led in to the next point. So let's talk about the mechanism of amine formation and so now we come to the stage that's been set by the Vittig reaction. We come to the stage that's been set by cyanohydrin formation and by the Vittig reaction where nucleophiles like to add to the carbonyl compound. The carbonyl of the ketone or aldehyde is electrophilic and the oxygen can under the right circumstances go away. It can leave and obviously it becomes water because I've written water several times. So let's see how that happens. So as I said the reaction is catalyzed by acid. There's an intermediate that forms half way along the way called the hemiaminal or an amino carbonyl depending on whether you prefer more common name or the name the book prefers for it and that intermediate is slow to break down. That's where the water is about to leave the reaction. Acid catalyzes that step. It makes it occur more quickly. So what I'm going to write out for you is a mechanism of acid catalyzed imine formation and for the most part we can think our way through the mechanism. So we know that an amine is reasonably nucleophilic. I told you the pKa is 10 or 11 for the conjugate acid. In other words, it is nucleophilic enough to attack a carbonyl compound. So here's that generic ketone or aldehyde that I started with and I'm going to be a good person and try to remember to write in all of my charges, all of my lone pairs of electrons and to use curved arrows to always show the flow of electrons during this reaction. So here's our amine. The amine has electrons, the carbonyl wants electrons. Electrons can flow from the nucleophile to the electrophile. Of course, we can't have 10 electrons around oxygen. So just as in cyanohydrin formation, electrons flow up onto oxygen to give an oxyanion species. All of these steps and indeed all of imine formation is in equilibrium. In other words, every step that can occur forward can occur in reverse. In fact, under identical conditions, if you go through the reverse process, the hydrolysis of imines, it has to occur through the exact same pathway that the imine formation occurs because that's the lowest energy pathway. So we add amine to the carbonyl and now we have a species where we have a negative charge and a positive charge and a proton is going to get transferred. I'm going to be a little bit vague and loose with how the proton transfer occurs. You could think of it as occurring in one step. Your textbook implies this and I think for where you're at now in terms of your thinking and particularly for carrying out this reaction in the presence of, let me get my r's and r primes here. I think for carrying out this reaction in benzene or in toluene, I think it's reasonable to think of it that way. So I'll write this as proton transfer. I think you could also think of it as occurring with, remember you have some catalytic acid, of course you have other molecules around. You have water around eventually. I think you could also think of it as occurring by picking up a proton and then losing a proton. You don't really have water in this reaction or large quantities but you may have a little bit of catalytic acid. Anyway, however you wish to think about this, I think is fine whether you think about this oxygen pulling off a proton or you think about BH plus some acid that's present transferring a proton and then pulling off a proton or even the reverse I think is fine in other words first pulling off a proton with a base like amine and then putting on a proton. So I will write all of these ways that I think are valid for thinking about this. So at this point what we have done is we have gotten to a species that looks like it should be a neutral molecule. This species is an intermediate. It's not in general a stable species. In general you can't put this species in a bottle. We call this a hemiaminal. Hemi means half. You're going to see or if you've completed the chapter reading you'll see the term hemiasatal and hemiketal. Hemiaminal means half of an aminal. An aminal is another unstable species in which you have two nitrogens on one oxygen. A hemiaminal is one in which you have one on the nitrogen, one nitrogen and one OH group. Your textbook also refers to this as an amino carbonyl. Carbonyl is a very old term for alcohol. So aminocarbonol is a fancy way of saying an aminocalcahol which is a fancy way of saying a molecule that has an aminogroup and an alcohol group that happen to be attached to the same carbon. Anyway, in general this is not a stable intermediate. So I'm going to write in parenthesis unstable. In other words, it's going to break down and that's going to be the remaining part of our mechanism. Another question, you mean over here? You mean first, you have hit an absolutely beautiful question. So the question that's being asked from the front row is a really good one and it caused me a little bit of thought and a little bit of consternation in putting together today's lecture. And the reason for it, so the question that's being asked which is an extremely valid one is why don't we protonate the carbonyl oxygen first by acid? And what causes me real consternation, the mechanism I'm giving you is the best one for this reaction catalyzed by mild acid. However, and it's the one that most closely accords to experiment for the rate determining step being the breakdown of the hemiaminal which we will see in just a moment. However, if you for example look at other sources or you look at wikipedia or more not wikipedia, the organic chemistry wiki text from the University of Michigan, you will indeed see that mechanism and it doesn't accord quite as well with experiment even though it represents very sound mechanistic thinking. So coming back to the question of why I'm writing the mechanism this way because this is important because your thinking is sound and mechanistic and specifically not wrong for writing the mechanism even though experiment is different. Why am I harping on this? Why am I now taking the time of 200 people? Because you're going to worry about this later on when we start to come into our last segment and you worry why is there a difference between the mechanism I wrote here and the mechanism that we're going to write with very weak nucleophiles. Here's the reason why I consider this mechanism that I've written to be a more realistic, more valid mechanism, a better mechanism. So we're doing this reaction with catalytic amounts of acid. In other words, this reaction is being done with a little bit of toxic acid which I will show you later, for example, or some acetic acid or just with the amount of inherent acid that's present. You have one mole of amine present. Amines are basic. So the strongest acid, if I mix a strong acid with the amine, if I mix a little bit of HCl, remember less than one equivalent, a catalytic amount of acid or a little bit of toxic acid which I'll show you later with the amine, the strongest acid that's present will be the ammonium ion, RNH3 plus. RNH3 plus is a very weak acid or is a weak acid to keep consistent with my terminology. PKA, about 10 or 11, a carbonyl, that carbonyl is a relatively weak base. It's weaker than water. Water has a PKA of roughly for the conjugate acid for a hydronium ion of negative 1.7. A carbonyl group is even less basic. It's sp2. The electrons are held more tightly to the oxygen atom. So the PKA of the conjugate acid to protonate that carbonyl is well into the negatives, negative 3, negative 4, somewhere around there. In general, an equilibrium where you have 10 orders of magnitude or greater generally doesn't get invoked in a mechanism. It's just too far to the left for an ammonium ion, PKA 10 or 11, to be protonating a carbonyl. When we come to weak acids, we are going to see a different mechanism because now you've got the strongest, now you don't have an amine present, you don't have that base and so we're going to have weak acids being acid catalyzed where yes, you protonate. Now we take the flip side in favor of this mechanism. The flip side in favor of this mechanism is your nucleophile, is PKA for the conjugate acid of about 10. That's like cyanide. That's like other nucleophiles that we see happily add in an unacid catalyzed step to the carbonyl. So this first step occurs without protonation of the carbonyl. However, the thinking is not long. It's not bad thinking to think about it in the other way. So back to my consternation at putting this lecture together. So what gave me some consternation is I wanted to follow the order of the textbook here and this is a nice reaction because it introduces the principle of a nucleophile that's not phosphorous pulling the water out and I think they made a good choice. But mechanistically, today's mechanism is the oddball mechanism because you precisely have the conundrum that you've cited. So thank you for raising that. It really is good. And coming to your point as students about should I be confused about this mechanism or if I'm feeling confused or the parallel, oh do I need to know this? If I'm feeling confused, yes, this is the one of all the mechanisms that we're going to see. Today's mechanisms are the ones that will cause you the most consternation. Next Tuesday's mechanism should end up showing more elucidation and so my consternation in presenting this was do we want to go with the most confusing one first and I decided to follow your textbook. But yes, if you're scratching your head, you ought to a little bit. All right, at this point our hemiaminal undergoes acid catalyzed breakdown. In very, very special circumstances, certain hemiaminals can be stable. But in general, they're not. And the rate determining step, the slowest step is the breakdown of the hemiaminal. Acid and it can be as simple as some of the other water that's present or other adventitious sources of acid or it can be a little bit of added acid. Which in the end, although again you're going to find some mechanisms writing as H3O plus really would end up being a protonated ammonium ion. It would really end up being a protonated amine. So some source of acid is going to transfer a proton. You can think of electrons as flowing from the oxygen to the acid back onto the base. Maybe I'll skip the curved arrows at that point just because we're very good at that. Whoops, onto the oxygen. So both of these atoms are basic and to get over the hump. So if we're headed back, basically reversibly you can protonate the nitrogen, but if we're headed back, if we're headed forward now, we have to get over this hump, the nitrogen has to push out a molecule of water. So protons come on and off every Lewis basic acid in the molecule, but sometimes they go on to this oxygen and when they do we're now set up to overcome the rate determining step. In this step, electrons push from the nitrogen and push out a water molecule. And I guess since I've been, I'm going to actually write this in a separate line because things are getting crowded here. So here we have our hemiaminol protonated on the oxygen atom and now in the final, the penultimate, the rate determining step. Now we're going to break this guy down. So I'm just rewriting the species and electrons are going to flow from the nitrogen and push out water. And again, every step in this reaction mechanism is an equilibrium. Whoops, now we're going to get a double bond to nitrogen. All right, now we're ready to finish up our mechanism. And in the final step of the mechanism, so now we have our aminium ion. And as I said, this is the rate determining step. And now in the final step of this process, we're going to simply take a proton off. Electrons flow from the base to the proton. We pull the proton off. We end up with our amin. And you notice in these three steps I've written here, the first step where the base is protonating the oxygen, the second step where our protonated intermediate is breaking down, and in the third step where we're getting our base back, you see how the acid is acting as a catalyst. A little bit of acid gets invoked at the beginning, but it comes back at the end. It's regenerated at the end. So if you had a catalytic amount of an ammonium ion present, now that ammonium ion would be given up at the beginning, but by the end of this catalytic cycle would be regenerated. Okay. So remember, and I thank you for my lone pair. So remember, this nitrogen is sp2 hybridized. This carbon is sp2 hybridized. And so you have 120 degree angle here and 120 degree angle here. Another question. Yeah, you could use a water. And the only reason I'm not using water and not using H3O for this, well, actually, the main reason is in the presence of amine, and we have amine present, in the presence of amine, you get only a minuscule amount of H3O plus. Remember, H3O plus pKa negative 1.7 ammonium ion, pKa roughly 10 or 11. So you'll have very little auto dissociation to hydronium ion, but yes, and as I said, for so many reasons, today's mechanism is the oddball mechanism for so many reasons, today's mechanism is hard to think about. And this is one of them. Yes, we will be using H3O plus and H2O in, or ROH in Tuesday's mechanisms, but here realistically speaking, most of the acid we're going to have around is going to be the ammonium ion. Now, coming back to this proton here, because if you're thinking and on your toes, you're probably saying, well, why is this going to protonate? This is even more basic than a typical alcohol. So we will protonate this. And you're only generating a little bit of protonated species. This is the toughest part, the breakdown of that intermediate. Other question, another great question. Is that lone pair of electrons taking up a little more space back to VSEPR theory and geometry of molecules? Yes, a little more space. Remember, methane ammonia water, the hydrogens in methane, 109.5 degree angles, hydrogens in ammonia about 107. I don't remember, I think it's 107.3 degree angle. They get pushed ever so slightly, 2 degrees closer together. Hydrogens in water, 104, I think it's 104.5 degrees. So about 2 degrees closer. So yes, technically this bond angle is probably closer to 118 degrees rather than 120 degrees, but 120 or 118 is a hell of a way away from 109.5 or 107. So it is sp2 hybridized. I saw another question. All right, one last question. Will a racemic mixture of both E and Z, oops, bad question. Remember, alkenes, no chirality to the double bond, diastereomers. So the question is racemic means handed. So the answer, will we yes, we will if we had let's say a methyl group and an ethyl group here form a mixture where it would be very close to equal. But if we have one group that's big like a methyl group and one group that's relatively big like a methyl group and one group that's small like a hydrogen as in an aldehyde imine and aldimine, then indeed you will have exclusively the trans. Or if you have one group that's very big like the methyl on the cyclohexanone that I showed and the other that's less big. But if they're roughly the same, yep, you will get two of them. All right, so we've walked through, we've, another question. Will this reaction, oh, I love it. I love the question and this is every consternation I was thinking about and putting together today's lecture. So brilliant, brilliant, brilliant question. What do we think? How many think it will? How many think it won't? This is good. And how many think I'm asking the wrong question? Yay. It turns out in general, in general, imines of this sort are unstable. In general they go on to form other products. In general, whereas you can put most imines in a bottle, remember by stable I mean things you can put in a bottle, you can isolate, the aminocarbonol, the hemiaminal intermediate, in general you can't put in a bottle. Although I can give you examples where you can. In general, you can't. In general, you can put an aldamine or a ketamine in a bottle. I showed you two at the start. In general, you can't put this one in a bottle. Now, just to check at the beginning of putting this lecture together, I was curious, I went to one of my favorite sources which is organic syntheses and indeed they have a preparation of one very, very special such imine with trifluoromethyl groups which are very electron withdrawing and very special here. But in general, this ammonia imine, the oddball imine, which isn't primary, it isn't secondary, in general that is unstable and undergoes further reaction. So, yes, you would form it transiently as an intermediate and then it would undergo further reaction. All right, now, for coming back down to earth here because this is good, we have had a really important discussion of some reasonably heady topics. We have wrestled with mechanism and this is good and this coming week's discussion section from Tuesday on, I'm going to ask you to wrestle with mechanism. So this is good. We have been engaging and thinking about it. We have asked questions that are deliberately outside the box and outside the textbook. This is exactly what you should be doing. So I want to come back to earth with a few basic principles here to ground us because the other thing that comes with heady discussions is this is confusing. This is scary. Will this be on the exam? So let's come back to some general principles that are less scary. All right, one of the principles I want to mention is that imine formation is an equilibrium. In other words, in writing these reactions and thinking about synthesis, I've said, yes, we can drive this equilibrium. We're organic chemists. We can do anything. We can make anything. We can push compounds to go in almost any direction that we want them to go. But let's not forget little shadow. Yay. If you don't take out water, we come back to this reaction and it's an equilibrium which means it is just as valid to write the reverse reaction and if we don't take out water and we throw our imine together with water we can go ahead and drive the reaction in reverse. Again, Le Chatelier's principle. It is just as valid to write the hydrolysis. If instead of throwing a Dean Stark trap on there and instead of using benzene to azeotropically distill out water, I go ahead and I throw in lots of water, then I get the corresponding amine back. Further, I want to further generalize because again, organic chemistry can be scary and you say what general principles should I take away? Okay, in general, the reaction of moderately basic nucleophiles and weakly basic nucleophiles with ketones and aldehydes is reversible. We looked at cyanohydrin formation. That's an Le on the end by the way. We looked at cyanohydrin formation. That reaction is reversible. You can drive it in the reverse by distilling out or using acid to protonate the HCN and driving it away as a gas. So in general, the reaction of moderately basic nucleophiles like amines or hydrogen cyanide or weakly basic nucleophiles, we'll talk about alcohols next time. In general, that reaction is reversible and in general, the reaction of strongly basic nucleophiles is not. In other words, in general, when you add an alkyl nucleophile or a Grignard reagent or an organolithium compound or an acetylide to a ketone or an aldehyde, in general, it ain't coming back out. In general, if you add a hydride nucleophile to a ketone or an aldehyde, it's not coming back out. And the fun part are the exceptions and those exceptions go beyond the scope of the course but they end up being absolutely fascinating mechanistically. If you learned it in pyridine chemistry, you may have learned one weird example called the Chichibabin reaction where a hydride nucleophile comes out. It's a fun name. If you didn't learn it, Google it just for fun. All right, again, staying grounded in the stuff that people really fret about in sophomore organic chemistry, let's go ahead and just look at a couple of species that we haven't seen before. This particular species is called an oxime. And if you look at an oxime and you say, okay, I haven't seen this before but I know how to form it because I can think backwards retro synthetically. You can say, oh, that's just a funny looking amine called hydroxylamine reacting with an aldehyde. And so you can think backwards. You can say, ah, I'm smarter than I think, than he thinks I am. I know how to make that. I just react hydroxylamine with this aldehyde, maybe with a little bit of acid catalysis. But let's take another funny looking species called a phenylhydrazone. It used to be really, really important to make derivatives of carbonyl compounds because they were one of the ways people characterized them. And if you look at this, you can say, oh, again, I'm clever enough to figure out how to synthesize it. I just realize it's just a funny looking amine derived from acetone and phenylhydrazine. And so these principles that we're learning here for forming and taking apart molecules really ends up being much, much more general. Now, we've talked about primary amines. And we've talked about ammonia, surprisingly. And now I want to move on to secondary amines and show you the conundrum that they present. Tertiary amines really can't go anywhere. They can add reversibly, but you can't even pull off a proton and form a Hemi-Aminel. When people get tired of writing out long words like primary and secondary, they'll often go ahead and use a degree symbol. So two-naught amine is another, is a shorthand for secondary amine. And secondary amines with ketones or aldehydes go to a species called enamines. And I'll give you an example of this reaction first. Again, I've gone to one of my favorite sources for this Org-Syn, and so I've taken cyclohexanone. This turns out to be a reasonably important one. And unfortunately, your textbook doesn't do real justice to the beauty of this chemistry because it's useful in making steroids and chemical synthesis and asymmetric synthesis. And it was pioneered by a fellow named Gilbert Stork, really brought into very high utility by him. Maybe I'll put something on this week's discussion section. If you take a ketone or an aldehyde that has a hydrogen at the position adjacent to the carbon and you take a secondary amine and again you subject it to the same types of conditions that we used before. Benzene reflux removal of water with a Dean Stark trap. You might not even see that written. You might see acid catalysis. This particular example doesn't require acid catalysis, but in general you might say sometimes acid catalysis is used or if you're concerned you could write acid catalysis. So what we've done here is we've combined our secondary amine and our ketone to form a species called an enamine. And again if I'm going to be good and write a balanced equation and remind us that we have an equilibrium that we're driving by our Dean Stark distillation, we drive that reaction removing water. All right, so I guess I will also write in parenthesis or maybe below here sometimes with catalytic acid. And where I want to close our discussion is to just talk about the mechanism for formation of the enamine and maybe a little bit more about it. So just as in the case of our primary amine, our amine is a nucleophile. The carbonyl is an electrophile. The amine is a moderately basic nucleophile. It's a good enough nucleophile to attack the carbonyl. It's like cyanide. It's reasonably basic. So just as we had done in our previous example, I am going to now assume that you can get yourself all the way to the hemiaminal, attack of the amine onto the carbonyl proton transfer. And at this point, again, we have some catalytic acid. This is probably the most confusing and worst one. And what was your name by the way? It was a good question you asked. Kyle. And Kyle asked this very good question about H3O plus. And so here I'm going to be a little bit vague, but honestly your acid is probably in this case the ammonium ion. So we have our equilibrium here. You have a small amount just as we have done before. You have a small amount protonating here. And now in our intermediate, and again I'll be a good person and try to fully balance my equation here. And maybe I'll even draw a curved arrow here just to complete things. All right. So now we come down to the crux of the matter, the breakdown of our protonated hemiaminal and nitrogen has a lone pair of electrons. It can give a good push. We push electrons in to form a double bond. We push out water. And up until now everything, everything, everything has been the same as we did in the amine formation. So here we have our aminium ion. But at this point we have ourselves in a conundrum. Because in the case of the aminium ion formation we could now pull a proton off, off of the nitrogen. And at this point we're stuck. We can't pull a proton off of the nitrogen. There's no proton to pull off the nitrogen. Aminium ions can be stable species under special conditions but in general they're intermediates and something else occurs. And let me show you what else occurs. So remember I said that we have to have beta protons. So here's our aminium ion and we can think about this two ways and they're both valid ways of thinking about this. So one way you can think about it is you have base present. The base among other things is the base that just got released over there. It can be a mean. You could think of it as being water. So we have various basic species around. One way you can think about it is with base pulls off this proton. And I'm perfectly happy with this view and we're going to be taking a view very much like this as we move into I believe it's chapter 23 and discuss the chemistry of enolates. We push our proton from the, push our electrons from the base to the proton. We pull off this proton. We push electrons in forming a new double bond and we push them up onto the nitrogen. All of these steps are reversible and so I will show you curved arrows. I will show you an equilibrium arrow rather and here we go. And that's a perfectly valid way of thinking about this. Now equally valid is recognizing the fact that we can think about there being two resonance structures of the aminium ion and I'll use a resonance arrow. We have a major resonance form in which everyone has a complete octet but the electronegative nitrogen bears a positive charge. We have a second resonance structure, a minor resonance structure in which the positive charge is on the carbon that's good carbon's less electronegative than nitrogen but it's bad in so far that the carbon doesn't have a complete octet. So we have a major resonance structure and a minor resonance structure and we can think of our base as pulling off a proton and if you're more comfortable with this type of way of doing it you can say all right the base pulls off our proton and we push electrons in over here. I guess if I want to keep track of my hydrogens I will write in a hydrogen there and now we have our protonated base and these are both valid ways of thinking about the same thing. Well I think that pretty much wraps up the chemistry that I would like to talk about. I will probably bring in some carbon-carbon bond forming mechanism with enamines into our discussion chemistry to show you a little bit about the beauty of their chemistry and a little bit about curved arrows with them. I will see you on Tuesday.