CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 33 - Green Chemistry; Acids and Acid Derivatives
Chapter 1. Mitsunobu Inversion and Atom Efficiency [00:00:00]
Professor J. Michael McBride: OK, So now we’ll come back down to earth after Professor Ellison’s talk.
So we were talking about green chemistry. Today we’ll talk about one aspect, the Mitsunobu reaction, and then get on to acids and acid derivatives. And then we’ll do that again next lecture, more about acid derivatives, condensations, and some stuff about carbohydrates, and then the last two lectures.
Now, remember when we were talking about green chemistry, there were some new processes that were desired. There were old processes that need improvement, and new processes that need to be developed, according to the wishes of the pharmaceutical industry.
Remember, there were six votes for aromatic cross-coupling I mentioned right at the end last time. That’s how to put two benzene rings or other aromatic rings together, without using haloaromatics. We haven’t talked about those reactions, but that’s their top desire.
Then how to use aldehydes or ketones plus ammonia and reduction to give a chiral amine. Now, notice how many of these things have to do with getting a single enantiomer, because that’s so important in drugs, to have a single enantiomer. And this, notice what they want here is to do what nature does. We talked before about making glutamic acid by what’s called reductive amination, where you have a reducing agent, NADH. You make the imine and then reduce it. So four of the six pharmaceutical companies said that was something they would really like to be able to do. Or asymmetric hydrogenation of olefins and enamines or imines. And notice, this is hydrogenation of an imine, this very reaction here. But what they want is asymmetric. That is, the ability to make a single enantiomer, so you don’t have to throw half the stuff away and do a separation.
Four voted for a greener fluorination method. A lot of drugs have fluorine in them, and the reagents that put those in are often pretty vigorous ones.
Nitrogen chemistry, avoiding azides, and three, avoiding hydrazine, which is very poisonous.
Asymmetric hydramination. Again, asymmetric.
Greener electrophilic nitration. We talked about electrophilic aromatic substitution, nitration. But what they want is a greener way to do it.
And the final one is asymmetric addition of HCN.
Notice, incidentally, that four of these eight– half of them– have to do with asymmetric synthesis, making chiral things. And notice that six of them, six of the eight, have to do with nitrogen.
Now at the end last time, we were talking about the other table, the one of processes that need improving. And in particular, something that seems surprising to many chemists is O activation for nucleophilic substitution. We’ve talked about that so long, getting a halogen or a tosylate that can be substituted, but they want something that’s better than that.
And noticed that three voted for a safer and more environmental Mitsunobu reaction. So let’s see what the Mitsunobu reaction is. And you’ll see that it is a type of that second case, activation of OH for nucleophilic substitution.
So there’s a picture of Mitsunobu. And in the 1960s, he invented this reaction with a great leaving group. The OP+Ph3 leaving group. And so if you bring in a nucleophile, bingo! It’s a very clean substitution reaction, that is clean in the sense that it gives a high yield. And it’s very general. Any nucleophile will work, it seems, if it has a pKa, if it’s more acidic, than 15. For example, a carboxylate can get an R put on it, or the phosphoric acid, or the imide, or N3-, or most significantly, active methylene compounds.
So let’s see what that means, active methylene compound. So here’s a specific example taken from Mitsunobu’s paper in the journal Synthesis to make (S)-(-)-methylsuccinic acid. So it’s an asymmetric synthesis to make it, and it starts with a chiral starting material, (S)-(-)-ethyl lactate. So the idea is to make the OH into a leaving group, so let’s read through his thing and see how he does it.
He uses the reagents 1 and 2. Now, 1 is triphenylphosphine, and 2 is a compound whose acronym is DEAD, and we’ll see later what– it’s diethyl azodicarboxylate. So we’ll see how that works. But at any rate– those two in the next slide– but for now, let’s just take that that makes that leaving group, puts the triphenyl phosphorus on the oxygen, with a plus charge.
So now we’ve got a good leaving group, and we want a nucleophile to displace it. And the nucleophile they use in this example is ethyl cyanoacetate. So it’s the ethyl ester of acetic acid, but the carbon has a cyano on it. And that makes it an active methylene compound. And the reason is it has a pKa of 13. It’s pretty acidic for a CH bond.
Now the reason, of course, is what you get, is an enolate, because there’s an α carbonyl. It’s α to a carbonyl, but it’s also α to the CN triple bond. So it’s doubly activated by low LUMOs next door. So it easily makes that anion.
So this then can be a nucleophile that will displace on the carbon. So it’s going to come across and invert the carbon and make the new bond. And that’s the first product they want: diethyl 2-cyano-3-methylsuccinate.
Now, notice they say the yield is pretty good, 61%, right? And notice especially that the conditions are very mild. These reactions started at minus 20°, and then just warmed to room temperature. It’s not something that needs a lot of heating. On the other hand, they had to isolate it by preparative layer chromatography. So that’s not an easy thing to do on a large-ish scale. So that’s certainly a painful thing to do, to try to get– it tends to be insoluble so you can filter it off. But to get the last bits of it out, you have to do this layer chromatography. So that’s certainly not so great.
Notice that it says, ultimately the optical purity is greater than 99%. Now, this reaction is indeed very, very clean inversion. So that’s one of the hallmarks of the Mitsunobu reaction. But, in fact, it’s not proved by that 99% purity, because you notice, at the end it was recrystallized. The product is recrystallized. So even if there was a little bit of the wrong enantiomer in there, when you crystallize it, you’d get only the right enantiomer, in purifying it that way. So this doesn’t prove it, but it’s true that it gives very, very clean inversion, and that’s one of its main advantages.
OK. But notice, incidentally, that this isn’t the product they wanted. This is an intermediate. And notice that there are two chiral centers. This one, which was 99% inverted, and this one, which came from the anion that did the attacking, the active methylene compound. And there’s no reason that that should be only one, right? It could have inverted at this stage of the anion. That is, the anion would have been planar because of the conjugation with the double bonds next door.
So wouldn’t that lower the yield? No, because of the product they’re finally going to. And you’ll see how that is. They hydrolyze this intermediate to (S)-(-)-methylsuccinic acid, which you notice is what it is they’re trying to make up at the top. So they were able to convert the ester to an acid, the other ester to an acid. That’s the reverse of a Fischer esterification, acid and water. And the CN triple bond also hydrolyzes to an acid, although it takes more work. It was heated under reflux for 16 hours, but that converted that one to an acid. And now you notice this carbon isn’t chiral anymore, because it’s got two acid groups. And even more than that, it turns out that when you have a situation like that, it decarboxylates. It loses the CO2, and the H comes on here. So again, it’s not a chiral carbon.
So this is the product, and the ultimate yield was 29% overall. So that’s not 100% reaction, but it’s very, very clean, in terms of the product being stereochemically pure. And the purity, the stereochemical purity, the inversion, is sometimes used by synthetic chemists when they design a complicated synthesis and they make an alcohol, but it turns out not to be the one they want, right? It’s the wrong enantiomer.
So then they use a version of this reaction called the Mitsunobu Inversion. So for example, suppose they made this (S)-configured alcohol, but what they really wanted was the ()R. So what you need to do is make that OH into a leaving group and bring an oxygen in from the backside to make the other one. So they can do this, to correct the synthetic “mistake”, by using these reagents triphenylphosphine and DEAD and acetic acids. So the acetic acid is something that’s more acidic than pKa 15. It’s got pKa 5, so it’ll do the trick. So you mix these things together, do the reaction, and it makes the acetate in the back. And now that’s the ester of the alcohol you want, so all you have to do is treat it with acid [correction: base] and get the inverted alcohol.
Now, how does this activation work? How do you do this in order to get the great leaving group?
Well, you react triphenylphosphine with HOR to do that. But they won’t interact directly. That’s, notice, -3, and in the product it’s -1, so you’re going to need an oxidizing agent in order to go there. And that oxidizing agent is this compound number 1 up here. Diethyl– so it’s an ethyl ester on both ends, notice it’s symmetrical– azodicarboxylate, DEAD.
Now, this is sort of a complicated scheme. And it’s balanced just right so the correct nucleophile will attack at each step. There are three nucleophiles that are involved. There’s the phosphorus unshared pair. There’s the unshared pair on an oxygen of the alcohol. And there’s X minus that was formed from the acid that you put in, which is ultimately the thing that’s going to do the displacing. That’s the nucleophile that comes in. So you don’t want the wrong nucleophile to be operating at any given time.
OK. So first, the unshared pair on phosphorus attacks the double bond on nitrogen, which will put a charge on this nitrogen here, when these electrons go on there.
What’s good about having the charge on that nitrogen? Why would that be better than adding to any old N=N
Professor J. Michael McBride: Yeah. It’s got the resonance with the carbonyl, so it’s going to make this allylic system here. OK. So that comes in.
And now this HX is one of these things that has a pKa less than 15. So it’s able, then, to protonate this– Remember, this is a fairly stable anion, but it’s not as stable as X-. It doesn’t have that low a pKa. So this one’s able to protonate here.
Now, notice, if X attacked– X is the anion of a pretty good acid. You could imagine that attacking the phosphorus to make a fifth bond to phosphorus. Right? Phosphorus has vacant d-orbitals. But if it did, it would come off again, because it’s a much better anion than the anion formed if you broke the phosphorus-nitrogen bond. So this is an example of it being tuned just right.
So if the X came on it, would just come off again. But if the unshared pair of oxygen goes on, then you would get the bond formed like this. And now that would come off again. The protonated oxygen is a stronger acid than this would be. But if it loses the proton, then this one is now a better leaving group, especially if it gets protonated.
So now we’ve made this thing here. We’ve made the oxygen into a leaving group. And notice, at the same time, the nitrogen-nitrogen, which originally was a double bond, now has two hydrogens on it. So it got reduced, right? That is, it was the oxidizing agent that allowed this to happen. And now X- is able to do the nucleophilic substitution, and that’s how the Mitsunobu activation happens.
And notice that this last one here, then, is irreversible. Once you put the X on, this is not a good nucleophile. It’s not going to come back on. So this is a high-yield process, when you do that one.
But the problem is that complete separation required chromatography. That was bad about it. And at least this aspect of it has been improved. And it’s an interesting way in which it’s been improved. Instead of using triphenyl phosphine, the benzene rings of the triphenyl phosphine, some of them, are attached to a polymer, right, like that. So that means if you have a polymer bead that holds this reagent, at the end, you don’t have to do the chromatography in order to separate your product from the triphenylphosphine oxide. You just have to filter it, because it’s attached to a solid polymer.
But what disturbs the green chemists about this otherwise handy reaction is that your goal is to eliminate water between HX and ROH. But in order to do that, you generate byproducts that have a molecular weight of 450. So this is what’s called atom inefficient. You generate lots of waste for the way to the molecule you generate.
Chapter 2. Green Oxidation of Aldehydes and Alcohols [00:15:59]
Now, we’re going to be talking about oxidation of aldehydes and alcohols. We have talked about some of them already. But it’s nice to do those in a green way, too. And in fact, we already talked about such a reaction last semester. It was one of the first oxidations of an aldehyde that was studied, which was the oil of bitter almonds. You remember what was special about this sample?
When you turned it over, it turned out that this aldehyde was in fact mostly solid. It had converted spontaneously to benzoic acid without adding any chromium, manganese, or anything like that. You got benzoic acid just by reacting it with oxygen. So there could hardly be a cleaner, greener process.
So the air oxidation of benzaldehyde is a process that I think we’ve mentioned before, but I’ll talk about it again. So it’s a free-radical chain reaction. You have some radical which pulls the hydrogen away from the aldehyde, which generates this acyl radical. But the acyl radical can now attack oxygen to form a new bond, and a single bond to oxygen, which itself is now a radical, which is the X. So X comes back and takes the hydrogen from the other, and the product, then, is not benzoic acid, but peroxybenzoic acid. So that’s what you would expect to get at first.
But in fact, it might be fun for you just to review free-radical chain reactions by writing this thing in one of those cyclic diagrams as to what comes in, where the radicals are, what comes out. That would be a good exercise.
So this is the initial product. But this itself as an oxidizing agent. It’s got this weak oxygen-oxygen bond, a low LUMO, which is a little bit like using a halogen-halogen bond. So it also is like an alcohol. So it can form a hemiketal by attacking the carbonyl. So here’s the group that attacked, here’s the new bond, and we have OH. So it’s like a hemiketal, except it’s peroxy.
And now notice that it’s converted what would have been an OH into an O with a leaving group. If this had just been a regular old hydrate of the aldehyde, there would be OH here. But now we have O with a leaving group. So a base can pull off the proton, the electrons come in to make a double bond, these leave, and now you have benzoic acid and benzoate, which is a base that can do that reaction next time.
So what could be a more efficient reaction? All you do is expose benzaldehyde to oxygen, and it becomes benzoic acid. So there’s a very green oxidation reaction. You can see Section 18.12, which talks about this.
But last year when I was talking about this subject, we had a seminar from a professor at Weizmann Institute, David Milstein, who talked about devising catalysts for greener operations. And I thought this one was a very interesting one.
So this is a ruthenium catalyst, and it has on it carbon monoxide and hydrogen, and then this funny ligand. It’s a pyridine, benzene with the nitrogen in it, with two groups, CH2 here, CH2, and it has a nitrogen, diethylamine, on this end, and di-t-butylphosphine on the other end. So it has three things with unshared pairs, phosphorus, nitrogen, and nitrogen, that can mix with the orbitals of the ruthenium to make this complex.
Now, I’ve colored two of the hydrogens here green, because they can come off. You could lose H2, having lost those two hydrogens, notice, or you can add H2 and come back again.
Now, how does that happen? Well, notice that this stuff here, the nitrogen plus charge with an extra bond here, going to this kind of ring, is a little bit reminiscent of what we saw before with NAD+ and NADH. That this is a nucleophilic aromatic substitution, the start of it. So the H- adds here. These go on to the nitrogen. So you have a ring like that one, and H attached. But the H- can come off again, and be given to something else.
So this is closely balanced, right? This one has the advantage of being aromatic. This one has the advantage of having a new bond, and not being charged. So being closely balanced, it can go back and forth. So the same way that NADH in nature can function as both an oxidizing and a reducing agent, going one way or another, you could imagine that a thing like this can go back and forth. So lose the H+ and go back, as far as the stability in this part of it goes.
But there’s a problem. The H could be thought of as H-, because it’s bonded to a metal. So the electrons are mostly on the more electronegative hydrogen. But it can’t reach to that position, so you can’t do that kind of thing. But what it can reach is the hydrogen here, if we change the conformation of the thing a little bit.
So you can imagine that the hydride attacks this hydrogen, which loses these electrons to make that bond. And that’s helped by the fact that these electrons then get stabilized by moving on to the positive nitrogen. Right? So this type of effect is what allows this to go back and forth in either direction.
But we’re most interested in the direction going up. So let’s redraw that thing and look at what happens if there’s an alcohol. So an alcohol comes in, and its unshared pair makes a bond to the ruthenium, and now it’s O+.
Now we can have the hydride– notice it’s H– Wait a second, have I done the right thing here? Oh no. What we’re doing is using the unshared pair of nitrogen. This is not like hydride. I said the wrong thing. This is a proton. The electrons are going away from this hydrogen. So the proton is able to go on here, to generate the N+ again.
And now notice, at the same time, that that alcohol that came on had two H’s that I drew explicitly. The one down here is the one I’m interested in, because it’s near the ruthenium.
And now if this were O-– and notice, in a sense, it is O-, because it’s bonded to a metal– then the electrons of the O-, these electrons, can make a double bond here, giving hydride. A hydride can attack the ruthenium. Like that.
So notice what the product of this is, then. At the top, we get a carbonyl group, and the thing we have at the bottom is this, which can start the whole process over again by giving off H2.
So what has this whole catalytic cycle accomplished? You’ve brought in an alcohol, and you removed from it these two hydrogens as H2. So it’s an oxidation of the alcohol, not by bringing in oxygen and making water, but by giving off H2.
So now in this particular reaction, there’s another oxidation removing H2 from a different alcohol. Notice the product is an aldehyde, but in water it will form a diol. So it could be the alcohol that comes in, and you remove another H2 from it, which gives then a carboxylic acid. Plus you’re oxidizing another alcohol, and have CO coupling, and you get–
Have I done this right? I think I counted wrong here. I think I wrote this wrong. I don’t think you oxidize that alcohol. But at any rate, it couples to give the ester. I think that should be two H2s there. I’ll correct that.
But this happens in the presence of this catalyst with no other activation. So what comes in is the alcohol, what comes out is two moles of hydrogen and the ester. Right? A very clean process.
So here’s a table taken from Milstein’s paper, and you can see, with several different alcohols he was able to get high conversions, higher than 90, up to 99%, very high yields of the ester, and only a little bit of the aldehyde. Most of it got oxidized.
Let’s look at the thermochemistry of this. And again, I think that should be two H2s.
So if you look at the heat of formation of ethanol and the heat of formation of the ester that’s the product from that, and hydrogen has a free energy of zero, so the two hydrogens don’t count. But if we add it together, this reaction is endothermic by seventeen kilocalories per mole. So it’s way uphill. You’d think the equilibrium constant would be very unfavorable.
What makes it possible for this to happen catalytically? We’re not putting in something that’s a strong reagent that makes one of these very reactive, and then it can react with the other things. We’re just putting the alcohol in. And to have the alcohol come in, and ester come out, and everything stay the same otherwise with the catalyst, you have to go uphill in energy by 17 kilocalories per mole, which sounds very tough.
But there are two things that make this good. You start with two molecules, and you form three molecules. Two H2 and the ester. Again, I counted wrong here.
So by making more molecules, you make more entropy available. And furthermore, this is especially helped because you make those two hydrogen gas molecules, and you can remove those from the solution.
So notice how this was done. It was done, in one case, under a flow of argon to sweep the hydrogen gas out. And in the other case, the toluene solution was refluxed, which again, the solvent pushes the other gas out. So at low pressure of hydrogen, it’s possible to do this spontaneously. So that’s an important type of analysis to do on a catalytic cycle that you’ve proposed, as to whether, in fact, it could be thermodynamically allowed.
They also were able to make amines that way, starting with alcohols make an imine by reacting with ammonia. So alcohol plus ammonia gives an amine. And also, incidentally, an imine.
So probably, it’s the same kind of thing as before. The alcohol gets oxidized to an aldehyde, but in the presence of the nitrogen, it forms the imine, the CN double bond, and then that gets reduced by the hydrogen that’s there in order to give the amine.
But again, these are very high yield reactions. When you look at a synthetic paper and a table in it, it’s important to look at two different kinds of yields. The normal yields given here are analyzed by gas chromatography to figure out how much was in there. But what you’re really interested in is what you can put in a bottle as a pure form, right? So the yields in parentheses are isolated products. So sometimes they’re very good, compared to the gas chromatography yield. But in this particular case, it’s hard to separate, and you only get half as much out to sell to your neighbor as you had analyzed by GC.
So you can also make imines and amides and so on with these catalysts. So that’s an interesting development which seems to address, if not to solve perhaps, some of these problems that the pharmaceutical industry identified.
Chapter 3. Understanding the Acidity of Carboxylic Acids [00:29:23]
Now, we’ve been talking in these last examples of green chemistry about acids and acid derivatives. So let’s talk about them a little more systematically.
First, the acidity of our CO2H, of a carboxylic acid. You remember that the pKa of a normal carboxylic acid is between 4 and 5, like butyric acid is 4.8.
Now let’s look at the effect of putting a chlorine in there. So if you put the chlorine in on the terminal carbon, it hardly changes the pKa at all, just by 0.3 units. If you move it a little bit closer to the carboxylic acid, it’s 4.1, and if you move it as close as possible to the carboxylic acid, now it makes a real difference. 2.8.
What is it that the chlorine is doing?
Student: It’s electron withdrawing?
Professor J. Michael McBride: So it’s electron withdrawing. And how does that help?
Student: It makes the H come off.
Professor J. Michael McBride: And how does it make the H come off?
Student: By reducing the bond strength.
Professor J. Michael McBride: Why does it reduce the bond strength? You know, actually, I bet that’s not true. I bet it doesn’t reduce the bond strength. The bond strength is the kind of thing that Professor Ellison talked about last time. Breaking a bond to give two– people laugh already. I just mentioned his name, and they laugh. OK? You break the body and you get two free radicals.
I don’t know the results, but my strong suspicion is that chlorine hardly affects that at all. What it affects is dissociation as an acid. So what is it the chlorine is stabilizing?
Student: The LUMO?
Professor J. Michael McBride: It affects the anion, right? So the electron withdrawal by the chlorine makes the nearby region of the molecule a better place to put more electrons, as in the anion rather than the neutral molecule.
So this inductive effect of the chlorine works, but only in the vicinity of the chlorine. It dies away as you get further and further from the chlorine. So this is so-called inductive effect. And it’s not very big if you’re far away, but if you get pretty close, it can be big.
Now let’s just see this inductive effect, if it could be quantitatively reasonable. If the withdrawal by the chlorine stabilizes the negative charge by a certain amount of energy, then the equilibrium becomes easier by that amount of energy, and that energy difference would appear in the exponent for the equilibrium constant. Remember, 102/3 [gr]deltaH.
So it would make a certain energy effect in the exponent, and if you put two of them in, you’d make twice that change in the exponent, if it were simple. And that means results would be multiplicative, because when you add things in an exponent, that means you multiply the numbers.
OK. So let’s look at acetic acid, chloroacetic acid, dichloro- and trichloroacetic acid, and see whether this inductive effect of the chlorine is additive in energy, therefore multiplicative in equilibrium constant. Or if you do the log of the equilibrium constant, then it would be additive, because it’s up in the exponent.
So notice, here it’s 4.8. Same as here. Here it’s 2.9, same as here. So it’s the same kind of effect.
But let’s look at the values here, not at the values themselves, but at the difference. How much difference did that first chlorine make? It made a difference of 1.9 pKa units. Now, if the second one had the same effect on the energy of an anion, that would also be 1.9. It’s actually 1.6. Not quite as big. And the third one hardly makes any difference at all. It’s only 0.6. I mean, it’s substantial, but it’s not anything like the first ones.
And the reason is that once the anion is already stabilized, it gets harder and harder to withdraw subsequent electrons. One chlorine withdraws a certain amount. The next chlorine doesn’t have as much to withdraw, so it can’t do quite as good a job. The third one does a much poorer job.
But fluorine is better than chlorine, and trifluoroacetic acid is quite a strong acid. It’s got a negative pKa. So that’s the inductive effect.
And now we might wonder, what is it that makes the carboxylic acid acidic altogether, ignoring these additional inductive effects?
So Paul Rablen from Swarthmore College published a paper in the Journal of the American Chemical Society in 2000 that had an interesting analysis of this. And I put it on because Professor Ellison is such a big advocate of resonance structures, as you saw last time, rather than molecular orbital theory.
So this is an interesting analysis, in terms of resonance structures. Now, what he did is do very good calculations, but calculations in the gas phase, not in a solvent, of what the equilibrium constant should be, or what the energy involved should be in transferring a proton from an alcohol to a carboxylic acid. So you get the anion of the alcohol, and the protonated carboxylic acid.
And now, the carboxylic acids are called acids because they give their protons to bases. So this reaction lies to the left. And to go to the right is uphill by almost 28 kilocalories per mole in the gas phase.
Now, the difference in pKa… the pKa of an alcohol is about 16. The pKa of a carboxylic acid is about 5. So the difference in pKa is about 11. Therefore, you would expect the equilibrium constant at room temperature to be about 4/3 of that. So it should be about 15 kilocalories per mole.
But notice that pKa s are defined in water, so it must be that the solvent makes an enormous effect, right? It reduces this to about half of its value, from close to 30 to 15, by making it easier to go to the right. Only 15 kilocalories uphill, instead of 30 kilocalories uphill. Why? It could be because this one has a much more concentrated negative charge, so it gets more effectively solvated by the water, compared to this one, which helps drive the reaction on the left.
But anyhow, that’s a secondary consideration, because we’re just looking, as Rablen did, at the molecules in the gas phase.
OK. So it’s 28 kilocalories uphill to do that.
Now, what’s the reason? Why does it want to go that way? Well, we’re tempted to say resonance. That’s certainly what Ellison would have said. You have two resonance structures of this anion, and only one resonance structure that’s reasonable on the right.
Can you think of any other reason that it might be good to go to the left? Anything else that makes that a better anion compared to this anion, other than resonance? Any idea?
These are Hs’s. Incidentally. CH3. And here’s HCOO-.
If resonance weren’t a factor, would you expect these to be the same? Would you expect the equilibrium constant to be one? Amy?
Student: Well, no, because the π* is helping out.
Professor J. Michael McBride: Yeah. The π* helping out is the molecular orbital way of saying there’s resonance. That’s how they do it. So that, in fact, is the same thing. But is there anything else that would tend to make it go that way?
Student: Having the other oxygen up there?
Professor J. Michael McBride: And how would the oxygen help?
Student: Inductive effect?
Professor J. Michael McBride: So it could be an inductive effect. But notice this oxygen is closer than the chlorine that was making the difference before. The chlorine was out on the next carbon. So you could also have an inductive effect due to the oxygen.
Now, notice that it’s a double bond. So it’s an interesting question whether you count the oxygen once or count it twice. So for purposes of this paper, Rablen hypothesized that you count the oxygen twice.
So now we have resonance and inductive effects, and we can do some numerology about these, about how important resonance structures are.
OK. So notice that this 27.0 kilocalories is due to loss of resonance and to loss of two inductive effects for oxygen, if you’re supposed to count it twice with the double bond.
Now, let’s look at this one, where instead of using alcohol, we used the ketone and a protonated acid here. So instead of the anion, it’s protonated now. And now we transfer the proton from here to the ketone– or it’s the aldehyde, actually, these are hydrogens. So we make the protonated ketone and the carboxylic acid.
Now, here the reaction lies to the right. It’s exothermic by 6.2 kilocalories per mole in the gas phase.
Now, which side would resonance favor? Debbie, can you help us out? Would you draw resonance structures of this?
Student: You could.
Professor J. Michael McBride: You could. You could use the unshared pair on the OH, and it would be stabilized by the π*, but that would require separation of charge, and Ellison told us we don’t want to do that.
How about on the left? Any resonance structures there?
Student: Yes. Because of the positive charge?
Professor J. Michael McBride: Speak up.
Student: Because of the positive charge.
Professor J. Michael McBride: Now you could draw a double bond here and a single bond there and put the charge on the bottom. And you’re not separating charge, you’re just moving the charge.
So here you have good resonance. Here on the right you don’t have reasonable resonance. On this one you had resonance on the left and not on the right. So now– or no, pardon. In both cases, we had resonance on the left.
OK. So resonance is going to favor the left and cause you to go uphill, in either case.
Now, how about the inductive effect? How about the oxygen? Let’s try somebody else here. Chris, which side is favored by the inductive effect? In the top, the inductive effect favored the left, because we had two bonds to oxygen withdrawing electrons, stabilizing the anion, right? Now how about in this case, when we have two bonds here?
Student: Favors the right.
Professor J. Michael McBride: If it stabilizes the anion, the argument is, it destabilizes the cation. So it’s the same deal. Except now… the same resonance that would cause it to go uphill to the right, but the oxygen now makes it go down [correction: up] to the left.
Now, how many oxygens did we have? This time the difference between this and this is just this bond. So it’s not a question of counting one or two. There’s only one bond there. So that’s one inductive effect of oxygen.
Now we have two equations and two unknowns. We have the resonance effect, which is the same direction in both cases, and the inductive effect, which is in opposite directions and twice as strong, in one case.
So if we take the difference of these two, minus this one, we get plus 34.1 is three of the oxygen inductive effect. So the oxygen inductive effect is 11.4 kilocalories per mole. And once we have that, we can plug back in to see how big the resonance effect is. It’s 4.8.
And now, from this point of view, if we look in the top case of that 30 kilocalories per mole going uphill, only 5 of it was due to resonance. By far and away, the dominant effect is the inductive effect. Only 20% of the special acidity of the carboxylic acid is due to resonance from this viewpoint.
Now, you may buy this resonance argument, and you may not. I’m personally skeptical. But one way or the other, it would be good to check it.
So Rablen checked it. He did this one with carbonic acid, the anion of carbonic acid. And now that’s plus 37.3.
Now, what does it involve? Here’s the anion, which is stabilized by resonance, as this one was. And there’s no resonance involved in this one, because it has the hydrogen on it. So it has resonance, the same as the others.
How about inductive effect? Compared to this, how many oxygens are stabilizing this O minus? Matt?
Student: I can count three.
Professor J. Michael McBride: Count three. If you count double bond twice, which is the rules he set, then you have three. So that’s O inductive times three.
Now let’s put that together. So we have O inductive times 3. That’s 34.2. And the resonance is 4.8, is 39. Not bad.
So we can try another one. Here we do the analog of what we did here, put a proton on. And now we have no resonance on the right. We have resonance on the left. How many resonance structures in addition to the original structure?
Professor J. Michael McBride: Chris? You can draw the double bond here. That’s what we did in that case. But you can also draw the double bond over here. So there should be resonance times two. And that should favor the left, make it uphill. But also, you’re going to have oxygens destabilizing this cation, two oxygens destabilizing that cation. And if you put those together, it’s minus 13.2, and observed is 11.
So there are people who think this, right? I personally don’t like this kind of argument so much, because I don’t think it comes down to really the physics of the thing. It comes to whether you count a double bond as two bonds and things like that. Which, it works out. So there are people who talk about this thing. And from that point of view, resonance isn’t as important to making acids acidic as inductive effect is. So the inductive effect is certainly important.
Chapter 4. Preparing Carboxylic Acids by Oxidation and Reduction [00:46:08]
Now, making acids by oxidation and reduction. And we’ve talked about these things. Oxidizing an alcohol with chromium, oxidizing– and remember, on the way, you get an aldehyde, so you can also oxidize aldehydes. And you can do it with chromium. I put potassium permanganate in here. We also talked about oxidizing double bonds with ozone and then hydrogen peroxide to get carboxylic acids.
There’s one different type of oxidation that we haven’t talked about yet which is mentioned in the Jones textbook here, which is alkyl groups attached to a benzene ring can be oxidized by permanganate at 100° to make carboxylic acids, as well.
Now, you can also make carboxylic acids by reduction. Of course, the carbon in a carboxylic acid is pretty much oxidized, right? But there are more oxidized things, like CO2.
So if you had CO2, what would you react it with in order to form a carbon-carbon bond to make a carboxylic acid? What kind of reagent? What would attack the carbon of CO2? High HOMO or low LUMO?
What would make CO2 double bond, double bond? What would make it reactive? Suppose it were just a single double bond. What makes it reactive? The π*, right? The low LUMO. So you want a high HOMO. It’s like the things we talked about with alcohols, right, where you could have a ketone, react it with R-, react it with H- to go on, and in this case that would be a reduction of the central carbon and would form a carboxylic acid. So you can do that kind of thing, too.
You can also react carboxylic acids with alkyl lithium, or hydride compounds. So for example, you could react a generic carboxylic acid here with alkyl lithium. What do you think the product is?
Alkyl lithium is like R-, right? It’s a high HOMO on the R, the R-lithium bond. So what will it attack? Mary?
Student: Is it addition at the carbon?
Professor J. Michael McBride: Addition to what?
Student: To the carbony π*.
Professor J. Michael McBride: To the carbonyl, right. That’s what I was trying to trick you into saying. In fact it reacts in another way. What else does a high HOMO do, besides make something a nucleophile? It makes it a base. And here we have an acid. So in fact the easy thing to react with is the proton. That’s easier to access than the carbonyl.
Oops! I’m sorry I’ve talked too long. Thanks for your patience.
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