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CHEM 125b: Freshman Organic Chemistry II
- Mechanism and Equilibrium of Carbonyl Reactions
This lecture aims at developing facility with devising plausible mechanisms for acid- and base-catalyzed reactions of carbonyl compounds, carboxylic acids, and their derivatives. When steric hindrance inhibits the A/D mechanism of Fischer esterification, an acid-catalyzed D/A mechanism can still occur. Substituent influence on the equilibrium constants for carbonyl hydration demonstrates four effects: bond strength, steric, electron withdrawal, and conjugation. Cyclic acetals play an important role in protecting the carbonyl groups of sugars, but acetals also can be used to protect alcohols, as can silyl ethers. Using amines instead of alcohols allows converting carbonyl compounds to imines via carbinolamines.
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Freshman Organic Chemistry II
CHEM 125b - Lecture 28 - Mechanism and Equilibrium of Carbonyl Reactions
Chapter 1. Fischer Esterification and Steric Hindrance [00:00:00]
Professor J. Michael McBride: So we’ve started talking about the mechanism about carbonyl chemistry. We’ll talk today about addition to carbonyl, the mechanism, and the equilibrium constant. About protecting groups, a little bit of NMR review at the end if we have time. And if we really have time, a little bit about the next topic, which is oxidation and reduction as it involves carbonyl groups and electron transfers.
So last time we took an overview of basic carbonyl reactivity. That you can have addition. You can have addition where the nucleophile attacks, that’s the old Burgi-Dunitz story. And then the acid A comes on to oxygen and you complete it. Or you can do it the other direction. You can start with the acid protonating the oxygen first. Or whatever acid it is. And then to give an oxygen cation that’s stabilized by the double bond. You can draw a resonance structure with plus on carbon and the nucleophile comes in. So there are two ways of doing it. Initially with a nucleophile, initially with an acid.
And we saw that if you have an α hydrogen, then it’s possible to pull it off with base to make an enolate and put it on– back on the oxygen. So that’s– you can do an allylic rearrangement to an enol. It can be base-catalyzed, by the enolate. It also could be acid-catalyzed. You could first put the proton on the oxygen and take it off the α carbon. You can also, because of having that functionality, enolate an anion at the α carbon, or the double bond of the enol, you can also do α substitution. So we’ll see examples of that.
And if you have a leaving group, the thing is called an acid derivative, right? And you can replace it by a nucleophile, by– in the two-step reaction, first add to the carbonyl then remove the nucleophile. So nucleophilic substitution of acid derivatives.
And in the case of carboxylic acids you can have addition, of course. just as we did before. You can have substitution at carbon, where you make a new carbon-oxygen bond. Or you can have substitution at oxygen, where you make a new O-R bond. And again, if you have an α carbon it’s possible to do substitution at the α carbon. And we’ll see examples of that.
Now, one particular example is what’s called Fischer esterification which was invented, you’ll not be surprised to hear, by Emil Fischer. And it’s substitution at carbon of a carboxylic acid to convert it into an ester. And it’s acid-catalyzed. So let’s think about how this works. Where’s the acid going to come in and protonate this? Karl?
Student: At the oxygen.
Professor J. Michael McBride: There are two oxygens.
Student: Oh. The oxygen that’s connected by the single bond.
Professor J. Michael McBride: How do you know? Why should it do that one? why not the other?
Student: Because water’s a good leaving group, and so that O just put the H… or you put the proton on–
Professor J. Michael McBride: I’m not asking what’s a good leaving group. I’m asking where the best place to protonate is.
Student: Um, from– that O has– that O has the unshared pairs.
Professor J. Michael McBride: How about the oxygen on top?
Student: That also has unshared pairs, so–
Professor J. Michael McBride: Let’s look at what would happen, if you brought the proton in on top. You’d get that cation. Anything good about that one that you can’t have if you protonate the bottom one? Anybody see? Debbie?
Professor J. Michael McBride: You can get resonance stabilization of that one. Why protonate this one instead of that one? Well notice in the starting– that this unshared pair is stabilized by that. By the low LUMO next door, so that makes it a little harder to protonate, right? On the other hand, the one that’s up here is– it could be a lower-energy one, if it were in those rabbit-ear sp2 orbitals. So that would suggest that this one might be better, right? But it’s killed by that. And when you protonate– but notice the stability you get there gets even more so when you protonate on the top. Whatever stabilization you had of this unshared pair before you protonated it, you have more now, because you have a lower LUMO to stabilize it. So you get that special stabilization, which you can draw by a resonance structure, that resonance structure. So you in fact protonate the top, not the bottom. OK?
So once you’ve done that then you can bring in an alcohol and attack the carbon. It’s going to be easy to attack now, since a very low LUMO. And then lose– so you get a tetrahedral intermediate. You started with a trigonal carbon, three atoms attached to it. Now there are four. It’s a tetrahedral intermediate. So notice that as opposed to SN2 substitution, this one first does an addition, and then loses the leaving group. They’re not both happening at the same time.
In that sense it’s analogous the nucleophilic aromatic substitution. Here’s a nucleophile that comes in, forms a tetracoordinated carbon, and then this one is going to leave. Here the amine came in, formed a tetracoordinated carbon, from the one that had only three things when the fluorine was down here. And now this is strictly analogous to that. What we do next is lose a proton from here, and then lose the flouride. Here we’re going to lose a proton from here and then lose hydroxide. So we lose the proton– now, of course there’s an easier way than losing hydroxide directly. Now you can do what you want to do, Karl.
Student: Bring on the proton to that oxygen.
Professor J. Michael McBride: To that– to your favorite oxygen there. Now you’ve got the leaving group. But the first protonation was up here and you went through that tetrahedral intermediate. It wasn’t the case of making this a leaving group, and then doing something like an SN2. You first did an addition to the carbonyl and then eliminate the water, and of course the proton then comes off again.
So that’s the mechanism of Fischer esterification. And of course you can run it either direction. You can run it forwards or backwards. It’s a way of replacing OR on the acid carbon, the carbonyl carbon. And you can do it with where R is H, as well as where R is some alkyl group. So you can go either way, trading them back and forth. So the way you decide to get ester, rather than getting acid, which way are you going to go, is to have a lot more alcohol than water there. So you drive the equilibrium one way. If you turn it around and have a lot more water than alcohol, you’ll drive it the other way. OK. So that’s Fischer esterification.
Now Victor Meyer, we saw last time, was the short-term supervisor of Moses Gomberg. And remember what Gomberg did in his lab? He made tetraphenylmethane, which was hard to make because it was so sterically hindered, as we found out in trying to make it with models last time. But in fact it was Victor Myer who introduced the idea of steric hindrance in 1894. He wrote, “the source of this behavior is stereochemical, the space filling of the neighboring groups.” So this was something else that was added to the organic model in the 19th century that we didn’t really discuss last time. We discussed configuration– things involving arrangement in space– which was van’t Hoff in 1874. We talked about strain, ring strain. That was Baeyer, 1885. And we talked about conformation which Sachse tried, unsuccessfully, to introduce into the discussion in 1891.
But Meyer had the idea of steric hindrance. And he said the source of this behavior. What was this behavior? What was it, the basis of his proposing steric hindrance? It was Fischer esterification in the situation where it doesn’t work. So it doesn’t work for 2,6-dimethylbenzoic acid. When you have a methyl group here and a methyl group here adjacent to that, then you can’t do Fischer esterification. Now why not? You can clearly start out here by bringing in a proton and putting it on there. No problem. Where do you get into trouble, do you see? Karl, try again.
Student: When you’re going to bring in the OR group, because it’s going to have to attack from that way.
Professor J. Michael McBride: It’s going to have to come in from here, where this thing is. And you’re going to try to make that into a tetrahedral carbon with three extra substituants on it. So it’s going to to have the– it’s going to have whatever just came in, right? Some OR group. Plus, this one protonated and this one protonated. So you’ve got to get all these three things here, where that’s not an H. That’s whatever just came in. Adjacent to these two things, and it’s quite crowded as you can see here, that there’s going to be strain involved in these things running into one another. So you can’t get to the tetrahedral intermediate of the Fischer esterification. And this was actually the basis of the first suggestion that there was such a thing as steric hindrance.
OK. So this tetrahedral intermediate is hard to make if R is really big and crowded. OK? But there’s a way to make the ester, nonetheless. And it was invented by a guy who sat in some of your seats. I don’t know which seats he sat in when he was a student here, because it was a long time ago. So we can’t do that. We can’t do Fischer esterification that way. But Melvin Newman, whose picture we showed last semester, figured out in 1941 how to do this. And he did it by putting the acid into concentrated sulfuric acid. Now what happens when you put it into concentrated sulfuric acid? Karl? So you put it into concentrated acid, it’s going to protonate it. Where will it protonate it?
Student: On the top.
Professor J. Michael McBride: On the top, right? Wrong again. No, actually you’re right of course. You do protonate it. You put it in 100% sulfuric acid, you do protonate it on the top, right? For the same reason we went through before. But once in a while– you were right originally. Once in a while, it’ll protonate on the bottom. The one you wanted. And what’s good about that?
Student: It’s a leaving group.
Professor J. Michael McBride: It’s a leaving group. So in fact, that can now go away. Now of course it could come back again. But, in concentrated sulfuric acid, water is all protonated. And as long as it’s protonated, it can’t come back. It doesn’t have the high HOMO. So in fact, you can make this thing, which is called in an acylium ion. And now, what he does is take this concentrated sulfuric acid and pour it, carefully, into water– or pardon me, into alcohol. So as he pours it into alcohol it’s no longer concentrated sulfuric acid. So the acid– so you won’t be able to make these things anymore. But alcohol rather than water is the thing that will get the carbon. So then you’ll– that inhibits the reversal, as we just said. And you pour it into ROH and now the OR traps it, and you’ll lose a proton, and you’ve made the ester. But notice the difference. Here you made it by dissociation followed by association, when it’s crowded. The normal Fishcer esterification is association and then dissociation. But by doing it in real strong sulfuric acid, Newman figured out how to do this in the very, very hindered cases.
Chapter 2. Carbonyl Hydration [00:05:38]
OK. So now, hydration of carbonyl groups, which can be done either acid or base. And that was a note to myself which I forgot to take out. So let’s talk about acid-catalyzed hydration of formaldehyde. So I want somebody to help me out on here. Here, down at the bottom, we have formaldehyde and water. And what I want to do is add to get the hydrate, as it’s called, of the aldehyde right? The diol. That’s a hydrate or a gem-diol, the two OH’s on the same carbon. So how do we do that? Here we have– it’s acid-catalyzed, it’s hydration, so we have water, and it’s formaldehyde. How are we going to begin here? Amy what do you say?
Professor J. Michael McBride: It’s acid-catalyzed, so what’s going to happen?
Student: You’re going to protonate one of the–
Professor J. Michael McBride: So we’re going to protonate something. So far, so good. What are you going to protonate?
Student: We’re starting on that side, right?
Professor J. Michael McBride: Pardon me?
Student: We’re starting on that side, right?
Professor J. Michael McBride: We’re going to protonate. Where are we going to protonate it?
Student: The carbonyl group?
Professor J. Michael McBride: Where on the carbonyl group? Which atom?
Student: The O.
Professor J. Michael McBride: OK, good. So bring it down, we bring that on, and now we have that, which is a nice stable cation, right? It’s resonance stabilized. OK. What’s going to happen next? You want to keep us going?
Student: Then the O on the H2O’s going to come in and attack the cation.
Professor J. Michael McBride: OK, so that will come in, attack the cation. So now we got that. And finally? We’re getting pretty close to the product, what do we have to do?
Student: We have to deprotonate the OH.
Professor J. Michael McBride: Right, so you just have to lose the proton. And it doesn’t just fly off. An alcohol will take it or something. OK. Or water. OK. So we’ve done it. So that’s acid-catalyzed hydration of formaldehyde.
Now how about base-catalyzed hydration of formaldehyde? Need someone to help us here. Lauren, how about you? So what’s going to attack what?
Professor J. Michael McBride: We’ve got base for a catalyst. Notice, we got the H+ back again here. It was catalytic, right? Came in at first, then came out again. So OH- is going to come in at first, and come out again– it’s base-catalyzed– and we’ve got water, and we’ve got an aldehyde. What are we going to attack with the OH?
Student: Is it going to attack the double bond?
Professor J. Michael McBride: Where will it attack?
Student: The lone pair will attack the π*.
Professor J. Michael McBride: On which atom, O or C?
Student: On the C.
Professor J. Michael McBride: Pardon me?
Student: On the C.
Professor J. Michael McBride: On the C, right. This is our old favorite attacking π*. So that will come down there. And now we got that. But what we need is the hydrate, the gem-diol. So what’s going to happen next? What do we need in order to get there? To get from here to here, what atom do we need?
Student: A proton.
Professor J. Michael McBride: A proton. Where do we get it?
Student: From water.
Professor J. Michael McBride: From water. It’s in base, right? So we’re not going to get it from H+. We’re going to get it from water. So the water will come down, do an SN2-type reaction at hydrogen– substitution at hydrogen– and we’ve got the product and hydroxide, which comes back and catalyzes another one. Lives to fight another day. OK. So either acid- or base-catalyzed schemes will work for that.
Now, how about, independent of which catalyst you use, there is going to be an equilibrium constant between water plus the aldehyde, and the diol, the gem-diol, the hydrate. And as luck would have it, for acetaldehyde, the one which has CH3 and H on the carbonyl, the equilibrium constant is one. And now we’re going to see how that equilibrium constant changes, as we change what kind of aldehyde or ketone we’re dealing with. OK.
So suppose we have formaldehyde. I want someone to predict for me whether this equilibrium constant is going to be greater than one or less than one. Can anybody see any factors that will be relevant? Alex, can you see a difference– we want differences between this and this, or between this and this, that it’s going to make one or the other more stable, and therefore change the equilibrium constant.
Student: Well I guess the top-right one might be crowded by the methyl.
Professor J. Michael McBride: Aha. That’s a good point. This angle gets smaller here, right? So this CH3 could run into the H. But also, the CH3 gets more crowded because we’ve got two O’s on here now. So that’s a good point. But it’s the one that I’ve animated second. OK. Which way is that going to change it? Is it going to make the equilibrium constant greater or less than one?
Student: It’ll make the bottom one higher than one.
Professor J. Michael McBride: It’ll make this one less energetic than that was. Lower in energy, so that’ll tend to shift it to the right. OK? Any other factors? How about anything on the left? Anybody see anything there? Between this and this. Notice that here we have an sp2 bond to carbon and here we have an sp2 bond to hydrogen. Karl?
Student: Would the methyl be more electron withdrawing?
Professor J. Michael McBride: Generally electrons consider– carbon is considered donating, but that’s not the factor. The strength of bond, remember the strength of the carbon-carbon bond depends on hybridization. The strength of the carbon-hydrogen bond depends on hybridization. But the carbon-carbon bond is more sensitive to hybridization, gets stronger when you go sp2 , than the carbon-hydrogen does. So this one was a stronger bond than that one, with respect to the sp2 hybridization. So that made this one low. And this one was high originally because of sterics, as we said. Now we’ve got– this one is now higher, and this one is lower. So it should shift it to the right, and indeed, both the bond strength that we just talked about, and the steric hindrance both suggest that it should shift to the right. And it does. It’s 20,000, the equilibrium constant.
Now let’s try this way. Let’s look at acetone. Two CH3s. Now what do you think? How about on the left? Is this going to be better or worse than before, than the ones above? Po-Yi.
Student: As for bond strengths
Professor J. Michael McBride: It’s got two of these carbon-carbon bonds that are especially strong. So this is good on the left. How about on the right?
Student: There is more steric hindrance.
Professor J. Michael McBride: Right, there’s even more steric hindrance. So that one should go to the left from one. And in fact it goes the other way, right? By a factor of a thousand roughly. OK. Now let’s look at this. Let’s put chlorine on the methyl group on top. Now, does anybody have any ideas about that? So the fact of the chlorine will change this bond. It’ll also change that bond. Because it’s electron-withdrawing. So where can it withdraw electrons better from? Can it withdraw electrons better from an sp2 or from an sp3 bond? Where does it withdraw better? Cassie, you got an idea? Which is the more likely to donate electrons, sp3 or sp2? Which one is going to hold the electrons tighter?
Professor J. Michael McBride: Right, the one with more s will hold tighter and not be willing to give them up. So if chlorine wants to get them, it’s going to get them better from here than from here. So it should shift the equilibrium in that direction. So the electron donation– and it shifts it, not nearly by factors of a thousand, but by a factor of 37, compared to when I had an H instead of a chlorine.
Now, suppose we have three chlorines. Obviously the same direction should happen. But how big should it be? Sebastian do you have an idea? If one chlorine did 37, a factor of 37– so it changed the free-energy difference enough to make a factor of 37, what should two more do? What should one more– suppose we just put one more in. By what factor should we increase it, do you think? Bear in mind that if you make the same energy change, that translats… because it’s in an exponent for an equilibrium, the energy, it’s a certain factor, because if you add things in exponents it’s like multiplying things, right? So if one changes it by 37, the second one, if things are simple, would change it by a factor of 37. And the third would change it by another factor of 37. So if we go all the way from one to three, by what factor should we increase the equilibrium constant?
Student: 37 squared.
Professor J. Michael McBride: 37 squared. Is everybody me on that? OK. Now, 37 squared is a little bigger than 1,000, right? And look at that it’s 28,000. That’s better than we had any reason to expect. But it makes sense. And finally, what if we have benzene as the substituant. So benzaldehyde. Is that going to shift it right or left, what do you think? Arvind, do you have an idea? Where will the phenyl be stabilizing things more, on the right or on the left?
Student: Probably on the left.
Professor J. Michael McBride: Why?
Student: Because there’ll be more resonance at the–
Professor J. Michael McBride: Right. Because you’re going to have HOMO/LUMO interaction between the carbonyl and the π system of benzene ring. So it should shift things to the left. Conjugation should stabilize the aldehyde, and in fact it does. It’s a factor of 100, right? Instead of one, it’s 0.01. So it shifted it to the left. So all these make sense in terms of what we understand of what controls molecular energy.
Chapter 3. Protecting Carbonyls and Sugars [00:18:49]
OK now, acetal formation happens only with acids. And I think that this is the one. Oh no, OK. Yeah, OK. Now, so let’s look at acid-catalyzed hemiketal formation from formaldehyde. This is closely related to a slide that I just showed you. So what we have is, we want to get from the aldehyde to this, right? a hemiketal. And so what we– it’s acid-catalyzed, ROH is going to add to this. And how’s it going to happen?
Well let’s not mess around here. Let’s just go there, OK? So we protonate just as we did last time, when we were adding water. Alcohol comes in, forms the bond, bingo, we lose the proton, it’s catalytic, and we have the product. OK, nobody’s surprised by that. I’m doing the– I’m belaboring this, I realize. We did the same thing last time essentially. Or we did the reaction going the other direction, now I’m going back. But it’s really important that you get this.
Now, base-catalyzed. We can do this just like with water, right? We bring that in, attack the π*, get that. It takes a proton this time from alcohol, instead of from water. And we have the hemiketal. So fine. OK.
But suppose what we wanted to do was get the ketal, not the hemiketal, OK? Now, notice that that proton can attack either oxygen. It could either protonate OR, in which case what would happen if you then broke a bond? You see which way– where you’re going, if you protonate OR and then break the bond between O and C. Have alcohol leave. Amy?
Student: Would you go back?
Professor J. Michael McBride: You’re going right back where you started. But you can protonate this one as well as that one. So if you protonate this one and lose water, then instead of reversing, you can continue. So you lose water and you get that stable cation. Just like this stable cation, except it has an R on it instead of the H. And then you bring another alcohol in, and do your trick, and you’ve got the full ketal. Fine. We talked last time we went– spent a lot of time going from the ketal back to the carbonyl. Now we’re just going the other direction, reversing all of them.
Now let’s try to do the same thing with base catalysis, is it going to work? What’s the first step going backwards for base catalysis? If we want to start from here and go backwards, what’s the first thing that happens? Linda?
Student: It deprotonates.
Professor J. Michael McBride: The base will pull off the proton, right? Here the acid could protonate either oxygen. But here, when the base has to pull off a proton, there’s only one to pull off. It can’t pull it off from the R. And you can’t do an SN2– pulling the R off would be like doing an SN2 reaction, where hydroxide attacks the R, and this O leaves. But that’s a terrible leaving group. That’s not a good one.
So it can only take– so the ether– ethers are inert to base. They don’t under go SN2 reactions. So it can only reverse, it can’t go on. So you can get a ketal by acid catalysis. But you can only get a hemiketal with base catalysis. And furthermore, not only is there no way to get the ketal, but you can’t attack the ketal with base either. Once you’ve got the ketal, it won’t attack with base, because you can’t attack ethers with a base. So it’s the old joke, “You can’t get there from here.” Base will not do it with ketals, either making them or taking them apart again.
OK. Now, this relates to protecting groups. And I’m going to do this on the board. If I were talking about pedagogy, I would say it’s good for me occasionally to draw these things, just so you can see a human draw them, rather than have them all animated there. But in fact I didn’t have time to do this last night. So there we go.
So let’s take this up a bit.
So we’re going to look at ketals as protecting groups. Just a couple examples. There are plenty of examples in the book. Now, a particularly interesting compound is glucose, which is an aldehyde, and an alcohol. In fact it has many alcohol groups. It’s got OH here, OH here, OH here, OH here. Now remember, the ketal, or hemiketal, is formed by reaction of an alcohol with the aldehyde. But now we don’t have to add an alcohol. We’ve got alcohols in there already to react with the aldehyde. In fact, we’ve got plenty of them. And the question is, which one?
Now if we were to react this oxygen to make a ring with this carbon, then how many atoms would be in the ring? How big a ring would it be? It would have the oxygen– one, two, three, four, five, six, seven. It would be a 7-membered ring. 7-Membered rings are not so easy to form because they pucker up and things, and opposite sides run into one another. So that’s not so good.
Obviously if you use this one to attack the carbon, it would be a 3-membered ring. That’s bad. That’s strain. This one would be a 4-membered ring. But this one would be a 5-membered ring, that’s not so bad. This one would be a 6-membered ring and that’s good.
So let’s see how we can do that. If we made a hemiketal, we could have C, OH, H, COH, COH. I’ll just draw it that way. COH I’ll draw it that way. C, and now– and down here is CH2OH. And now, this oxygen has to go to there. So I could draw it this way. But nobody thinks I’m trying to draw what it actually looks like. In fact, it’s a 6-membered ring. It would look like a cyclohexane. But, just for purposes of showing what’s bonded to what, it’s often drawn that way. OK. So that would be a hemiketal. And it’s 6-membered ring, 6-ring, and that thing is called a pyranose, when it’s a 6-membered ring.
Now if it were this carbon, then we would have C, H, OH, and we can come down here, and we’d have CH2OH, and we’d have OH here, and we’d have CH, O coming out, O here. And we’d have OH and OH. And this would be one, two, three, four, 5-membered ring. And that’s called a furanose. OK. So it could go either way, and they’re reversible, too. Because we know that with either acid- or base-catalysis you can go back and forth between hemiketals and the carbonyl group.
OK. Now how could you find out which ring size it is? Well the way it’s actually done nowadays is by NMR. But, in the old days, when they didn’t have NMR, they had to do chemical transformations to figure out what was what. So chemical degradation.
So one thing they could do would be to treat it with acid and alcohol, say methanol. So acid can protonate an OH, and then lose water to leave a cation. But it turns out that only one of the– that one of the OH’s reacts much more rapidly than the others. Which one is easy to protonate and lose water, so as to generate a stable cation?
OK. If we – say we protonated this one, for example. We’d generate a secondary cation. Protonate this one, secondary cation, this one, secondary cation, this one, primary cation– not so good. This one– what would happen if we protonated that one, and lost water? I’ll write it down below here. So we had OH, H, this O down here, et cetera. Suppose I protonated this one and lost water. Here is the cation I would get. And then I could bring methanol in to put the oxygen on there, lose the proton from methanol, and I would have made what functional group? If I now have O– instead of OH up here I have OCH3. Everyone sees how I did that, don’t I? It’s the same thing we’ve been doing all the time. You protonate, lose water, put the alcohol on, lose proton. That replaces OH with OR.
Why did it do it there and not do it at all these other places? What was specially good about this cation, that the others didn’t have? Chris? Let me draw the– I’ll draw the cation here, or here. So the cation has an O on it, an H on it, a carbon on it, +. What’s good about– and this one goes down then to the bottom. What’s good about that cation, Chris?
Student: Secondary cation.
Professor J. Michael McBride: Pardon me?
Student: Secondary cation.
Professor J. Michael McBride: It’s secondary if you want to consider an oxygen as if it were carbon, but so were all these others secondary. Right? If you protonated those and lost water. What’s especially good about this one? Matt?
Student: You have an electron-withdrawing oxygen.
Professor J. Michael McBride: Is that good to make a cation, if you have something that’s electron withdrawing? That means it’s already positive. It doesn’t want to get more positive.
Professor J. Michael McBride: So the electron withdrawal by the oxygen is going to be bad for this. But there’s something that makes it good. Brandon?
Student: The lone pair mixes with the carbon.
Professor J. Michael McBride: Right. The unshared pair on the oxygen is stabilized by this vacant orbital, right? You can draw a resonance structure that puts a double bond here. And + on the oxygen. OK. So this one is the one that’s reactive to make a ketal. This thing is a ketal. It’s got two OR groups on it. And that one we could make a ketal too, up here of course, and not the other places. So we can make a ketal, right?
Now, it’s possible under those circumstances to do reactions with alcohols to chew up the other alcohols, but not chew up this one. Because that alcohol is now in the form of OCH3. We’ll talk about reactions next class that involve reactions of alcohols. Oxidation for example. OK. So this is protecting that particular carbon from being reacted by having a full ketal there.
Now, then you can take it apart again by treating it with acid and water. Get it back and see what compound you have. Which of the oxygens in that chain got oxidized and which ones were not oxidized. Compare with known compounds, and you know what the size of the ring was. Because notice that this– not only did this group keep… this OH keep from getting oxidized, this OH, that is. this OH didn’t get oxidized on this side, right? And this OH didn’t get oxidized there. So the protection protects not only the aldehyde from being oxidized, it also protects the alcohol.
Chapter 4. Protecting Alcohols [00:29:26]
Professor Ziegler, who we’ve mentioned, who teaches Chem 220 often, is a synthetic chemist and always, when he teaches a class in synthesis he always points this out. That people have tunnel vision often. They think of ketals as something that protects a carbonyl. But it also protects an alcohol at the same time that’s tied up as an ether and won’t do the reactions that the alcohol would do. So you not only protect the carbonyl, you also protect the ROH. And a particular example of that is what’s called THP protection. So this compound can be reacted with ROH and acid catalysis. So how’s it going to start out? Derek, you tell us. Where are you going to protonate this?
Student: The O?
Professor J. Michael McBride: You could protonate this O, you could protonate this O, right? Either of them particularly good? There’s actually another place to protonate that’s very good. Can anybody see? Arvind?
Student: On the carbon-carbon double bond?
Professor J. Michael McBride: Where, top or bottom?
Student: Umm, bottom.
Professor J. Michael McBride: Nope.
Student: The top.
Professor J. Michael McBride: Top. OK, if we put the H+ here, why is that good? Got the H here, there are now two H’s there, of course. There was one there already. What’s good about that cation? Matt?
Student: Stabilized by the lone pair on the oxygen.
Professor J. Michael McBride: Yeah. You can say– that’s what everybody always says, that it’s stabilized by the lone pair on the oxygen. But actually what it does is stabilize the lone pair on the oxygen, which makes the molecule more stable. You’re right, that’s what everyone says, that it’s stabilized by the lone pair. But in truth, it stabilizes the lone pair. OK, so we got this.
Now we got ROH, what’s it going to do? John? What’s the ROH going to do now when you add that cation?
Student: It’ll– the lone pair on the oxygen will attack it.
Professor J. Michael McBride: And then what?
Student: And then it’ll lose H.
Professor J. Michael McBride: Yeah. Lose proton, right? So we got this. And what functional group is this that we now have? Does anybody recognize this functional group? It’s familiar. Leen? It’s a ketal, right? Two ethers on the same carbon. But we didn’t make it from a carbonyl, we made it this way.
OK, so now we could do reactions. Suppose this R had something else coming off that was going to react, maybe a double bond. And we wanted to react that, but we didn’t want to react the alcohol. But the ether keeps the alcohol safe. We do the reaction down here. Now we need to get the alcohol back again, having done what we wanted to do down here. How are we going to get it back? How can we remove this stuff? We can protonate. Protonate the O, lose this, generate this nice stable cation again, go back. We’d just go back and forth, right? With water. Go back with water. So we can make– so this is a protecting group for an alcohol, to keep it from reacting.
I’ll mention one other protecting group for alcohols here because they mention it in the text book that many of you have, which is called TBDMS. It’s t-butyldimethylsilyl. So t-butyl, dimethyl, silyl. And you start with the chloride. Now the chloride has a bond strength, this bond, of 90 kcal/mole. But you could react it with ROH, and the unshared pair will react to form a bond with silicon, right? So now the silicon has five things attached to it. But then the chloride can leave. It’s not an SN2 reaction. It’s an association, followed by dissociation. Because silicon down in the next row of the periodic table has vacant orbitals that can make a bond with that. So now we’ve got t-butyl, Si, methyl, methyl, O, R. And the strength of the SiO bond is 110 kcal/mole. So this is downhill in energy. OK.
Now, so now you can do what you want to do, other things in the R group, without touching– it’s not an alcohol anymore. It’s a little bit like an ether. But the neat thing is that there’s a special way to take it apart. You can take it apart by treating it with tetramethylammonium, methyl4, N+, a salt, but the business end is fluoride. Because fluoride can attack the silicon the same way the unshared pair on oxygen did here. And then you can protonate and lose the OR. So this is the nice thing– the product from the silicon is now S– which you don’t really care about– is the Si-F, and the Si-F bond is a fabulous bond. It’s 135 kcal/mole. Plus now your ROH, once you’ve protonated.
So the nice thing about this is there’s not many things that react with fluoride easily. It’s a terrible nucleophile generally. So now you have something that’s very selective, that will only do that and take this protecting group off and not mess up other things that’s in your R group. And this was invented in the 1970s by a professor at Harvard, who got the Nobel Prize in Synthetic Chemistry, Professor Corey.
Chapter 5. Imines [00:45:44]
So that’s those things. And now imines and their derivatives, and the role of electron-pair repulsion and allylic stabilization. So I’m going to just introduce this here, and we’ll finish it next time. So we have a carbonyl, and we’ve been reacting it with alcohols or water. But let’s consider reacting with an amine, which actually has certain similarity to water, of course. It’s got the unshared pair, and protons on the nitrogen the way alcohol has protons on the oxygen. So this can react, and of course that can lose a proton here, and it can put a proton on there. It’s probably not the same proton. And this is called a carbinolamine. What’s the analog when you have an oxygen instead of NH in there? What functional group is it? If this were just an O instead of an NH, what would you call this functional group? I see you’re all bored. It’s a hemiketal. That’s what we’ve just been talking about. But the carbinolamine is the amine analog.
Now you can do– because you have this hydrogen here, you can do something special. You can lose this OH, say protonate and lose it, helped out by this. And we get this, and water. Pardon me, it would be originally NH+, when we use the unshared pair, but then we can lose the H+. So now we have this functional group, which is called an imine, which is rather like a ketone. So we got rid of the oxygen and replaced with NR, and the oxygen became H2O, by this same kind of stuff we’ve been doing. So these imines turn out to be very important, not only in organic chemistry, but also in biology. And we’ll continue with that next time. Or actually– yeah, we’ll continue with that, but there will be no more on the exam on Friday than what we’ve done today. OK.
Student: Do we need to know about the amines?
Professor J. Michael McBride: Pardon me?
Student: We need to know about the amines?
Professor J. Michael McBride: Oh, forget the amines, OK? Just do the ketals and hemiketals. Imines will be on the final.
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