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CHEM 125b: Freshman Organic Chemistry II
- Triphenylmethyl and an Introduction to Carbonyl Chemistry
Painstaking studies of his “hexaphenylethane” and its reactivity convinced Gomberg that he had prepared the first trivalent carbon compound, triphenylmethyl radical, the discovery of which marked the emergence of fundamental organic chemistry in America. Isotopic labeling could decide whether protonated cyclopropane plays a role in Friedel-Crafts alkylation. C-13 NMR spectra of aldehydes and ketones show how characteristic chemical shifts are established empirically. The carbonyl group is thermodynamically stable but kinetically reactive. Its acid- and base-catalyzed reactions often involve loss of α-proton to form an enol or enolate intermediate. Carboxylic acids display four fundamentally different reaction patterns. Acid-catalyzed hydrolysis of acetals illustrates a multistep reaction mechanism involving the carbonyl group.
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Freshman Organic Chemistry II
CHEM 125b - Lecture 27 - Triphenylmethyl and an Introduction to Carbonyl Chemistry
Chapter 1. Triphenylmethyl: Chemistry Comes to America [00:00:00]
Professor J. Michael McBride: We introduced the idea last time about triphenylmethyl, so I thought you might like to see a space-filling model of triphenylmethyl, and feel it, how it works. Remember, it was very difficult to find a way of making tetraphenylmethane, which means you have to take this phenyl with its little hook there and get it into there. So I’ll pass this around and you can try twisting it and see if you can make tetraphenylmethane better than the people before Gomberg did.
And then there’s a question of whether you can make hexaphenylethane. So here are two triphenylmethyls, one has the hook in it, and one doesn’t. So the question is, can you put those two together? So Debbie, you’re the first. So whoever gets it should bring it up and show it to us.
Now, the projector is warming up here. So we’re speaking about triphenylmethyl and then I have a little second thought about Friedel-Crafts reactions that I’ll give to you. And then we’re going to go on to the next subject, which is carbonyl compounds, once the center projector gets warmed up here.
So we talked last time about how steric hindrance– how you doing with your steric hindrance there, Debbie?
Student: Pretty poorly.
Professor J. Michael McBride: Pretty poorly, OK. It’s really tough to get these things together, when you have these three propellers that all have to twist because of the way they run into one another. I still don’t have my projector, I don’t know how it got turned off. I guess I did it somehow. So it makes– as Debbie’s finding out–
Professor J. Michael McBride: It makes tetraphenylethane– did you try hexaphenylethane?
Student: That’s even worse.
Professor J. Michael McBride: It’s even worse? OK. So let us see. It looks like it’s on up there. How are you guys doing? Helen, Antonia? Can you get them together? Maybe you could get a picture of them trying to– hold it up so they can get a picture there, just turn it between you. But work on it as you’re doing it. Don’t hold your hand in the way of the camera. Yes? I can’t, pardon me – Quiet a minute, yes?
Student: Shouldn’t the structures of the models be able to be bent a little?
Professor J. Michael McBride: Yeah, it’s true. It’s true that the models have fixed angles and don’t distort easily, these models. If you made it with Fisher’s models, the ones that had rubber tubing, you’d have no problem making it, because they’re not space-filling either. And there’s the question of how big the atoms are, they’re a little bit crushable, right. These models, obviously, are more or less rigid. And in fact, you can make tetraphenylmethane, but as we’ll see shortly, once we get a projector going for us, you can’t make hexaphenylethane. Any luck? I don’t know why this is taking so long. I turned off the side one, and somehow the center one went off too. I going to turn on one of the other projectors, just in case this one doesn’t come on. In fact, I’ll turn them all on. We’ll see where we get any luck. Ah, there we go.
OK. So here you see what we were just saying. Steric hindrance in triphenylmethyl causes twists that reduce overlap by 25% from what they would be in diphenylmethyl, and makes tetraphenyl[correction:methane] very difficult to prepare, not to mention hexaphenylethane. So, but Gomberg was able to make tetraphenylmethane using this trick that he’d learned in Munich of knocking nitrogen out and having the two radicals come together. And he tried, as we mentioned last time, to prove that that’s what he had. The analysis was correct. And he was good at analysis, remember. And he got the right molecular weight, almost. So then he went back. He spent just one year in Germany. Not like Friedel who spent four years first, and then 17 years in Paris.
Then he went back to Ann Arbor and got back to work, stayed there the rest of his life. And he confirmed the molecular weight by doing an even better measurement, he got 318 versus 320. So he was really confident that he had tetraphenylmethane. But there was one thing that was funny. He nitrated it, did electrophilic aromatic substitution on the four phenyls, got the tetranitro compound, and he got it at 99.5% yield. But there was a problem. He wrote, “Unlike trinitrotriphenlmethane…it does not dissolve in sodium ethylate,” that’s ethoxide anion strong base, “nor does it give any coloration…” So these nitro compounds with a lot of phenyls seemed to dissolve in strong base and to give colors, but his compound didn’t.
Now why does this dissolve in strong base? Any ideas? What does base do with it, strong base? Wojtek?
Student: It takes off the H+.
Professor J. Michael McBride: Yeah, it takes off H+. This is a very good anion. It’s triply benzylic, plus it has the electron-withdrawing nitro group, so it’s easy to pull off the proton and get the anion, which then dissolves and is colored. It has an unusually high HOMO and a low LUMO nearby, so the electrons can move, absorbing light that’s in the colored region. OK, so it’s possible to get that. Of course, he didn’t know this theory that we know. So for him, it’s was just funny that it didn’t dissolve in base the way other nitrophenyl compounds did.
But he wondered, how about hexaphenylethane? If you nitrate it, will it be colored? Will it dissolve in base? So he set about trying to make hexaphenylethane. Now how did he try? He tried using the techniques that failed to make tetraphenylmethane. It’s very much like Wurtz coupling, right? You put a metal in, pull out two chlorides, the things come together is the idea. Here, it’s plausible because it’s easy to pull that chloride off, as we’ll see later.
So in fact he did get– he thought– hexaphenylethane. So he did analyses. So he got the total of carbon plus hydrogen– that is CO2 and water that he burned it and got– was 93.97 by one method. This is a method, not a run. He did many, many analyses using the same method. Their average was 93.97. He tried another method: 94.20, 94.00, 94.57. Four different methods, maybe 20 analyses he did. But he could never get it near 100%. And he was confident of his ability as an analytical chemist. Remember all that time he spent in lab. Ultimately he used more than 17 different methods to analyze it, and none of them would give 100%.
So he concluded that there’s something else there besides carbon and hydrogen. And that something else is oxygen. Because if you have two oxygens between the triphenylmethyl groups, then the sum of C and H is 93.82%, which is what he got within a few parts per thousand.
So there must be O2 in between. So he prepared the authentic peroxide. You can easily do an SN2 [correction: SN1] reaction, because the chloride will come away to give the triphenylmethyl cation, and then O2-2 will go between. So you can get the authentic peroxide, and it was the same compound.
So then he knew how to prepare the hydrocarbon. You do it by preparing it without oxygen present. It reacts easily with oxygen. So he used a special apparatus he said, “with ground glass joints.” Then they used corks and sometimes rubber stoppers, but never or rarely, ground glass joints. That was really high tech. But he did it, and he got the compound that analyzed correctly. But he found it was very funny because this substance not only reacted with O2, it reacted easily with Cl2, with Br2, and even with I2, which you remember doesn’t add to alkenes. So this is a weird, unsaturated compound. So unsaturated it reacts spontaneously even with I2. And certainly hexaphenylethane shouldn’t do that. Tetraphenylmethane didn’t do anything like that.
So this is from his paper in 1900, just at the turn of the century. He said, “Triphenylmethyl…The experimental evidence presented above forces me to the conclusion that we have to deal, here, with a free radical…”
Remember where radicals came from? Remember, it was the nomenclature in the 1830s, the organic radical? At first they thought they were real things, Kolbe thought he had made methyl, but in fact he had made the dimer, ethane. And then from then on, radical was just a name. It wasn’t something that was real, it wasn’t the organic element. Here, he tried to make the dimer hexaphenylethane but concluded that he had the monomer, triphenylmethyl free radicals. Not just names. “On this assumption alone do the results described above become intelligible and receive an adequate explanation. The action of zinc results, as it seems to me, in the mere abstraction of the halogen, leaving the free radical. The radical so formed is apparently stable, for it can be kept both in solution and in the dry crystalline state for weeks.” So he had prepared a free radical.
Now of course, when you do a reaction, you either have to go association/dissociation or dissociation and then association. And when you dissociate from carbon, you get something that’s trivalent. And this was the first example of an authentic trivalent carbon compound. So this is the watershed between the 19th century, when it was just making compounds and things being mysterious, and the 20th century when you could begin to understand mechanisms in terms of actually knowing something about these trivalent carbon compounds. And this was the first.
So this launched the American century of chemistry, because this experiment was done in America, in Ann Arbor. And here in October 1900, he wrote a paper, published it in America, received October 4th. “An Instance of Trivalent Carbon: Triphenylmethyl.” But at the same time, he published it in Germany. A preliminary paper, “Triphenylmethyl: a Case of Trivalent Carbon,” received the 1st of October, 1900. So he published it both places to cover his bases. Now you remember, we talked about American organic chemistry and about James Mason Crafts and that essentially everything he did, he did in Paris. And many people thought he was French. He did essentially nothing in the U.S. There were people doing more than that, but not a lot more than that. All the center of gravity was in Europe, especially in Germany, at the end of the 19th century. But Gomberg comes along, born 25 years later.
Oh, I should have said that there’s an exception, which is Yale. Because Gibbs really made at complete international splash with his “Equilibrium of Heterogeneous Substances,” which was published the same period as the Friedel-Crafts reaction. But that was in France, and this was in America. So in physical chemistry, America had done something, and it was Gibbs. But in organic chemistry, Gomberg was the first one to really do something. And these are Gomberg’s publications. And contrast that with the previous one.
So here, he went for just one year to Munich and Heidelberg, where he did this work and prepared this compound, the tetraphenylmethane, which was itself notable. But much more drastic was the preparation of the radical triphenylmethyl, which he did, obviously, after he got back to America. So he published these things in German, because that’s where the big readership was. But within 10 years or so, he was doing all his publishing in the United States, and a lot of publishing.
So in that 25 year period between Friedel and Gomberg, things started really happening in this country. And then after the Second World War, of course, Europe had such difficulty that things really took off here, and it was a long time for them to catch up again. So this launched the American chemical century.
And incidentally, one other thing about it, is the big German journal for organic chemistry was Berichte der Deutschen Chemischen Gesellschaft, “The Reports of the German Chemical Society.” And in 1900 there were 4,290 pages, and I leafed through them to see how many graphs there were, how many quantitative things were being discussed. This was mainly a journal of organic chemistry and there were very, very few graphs. In fact, in two of the papers from this period here, Gomberg’s papers had more graphs than all those things in all of 1900 in the German main literature. So it showed not only were things coming to America, but physical-organic chemistry, the quantitative approach to organic chemistry was rising as well.
Chapter 2. Protonated Cyclopropane in Friedel-Crafts Alkylation? [00:15:39]
Now I said I had second thoughts on the Friedel-Crafts reaction, of what I told you last time. So I’m going to take just a minute to tell you about those. So we talked about rearrangement happening in Friedel-Crafts alkylation, and the idea that you can get a chloride being complexed with triphenylmethyl chloride, which makes it a good leaving group. Which means that the phenyl can come in and displace that, it’s essentially an SN1 [correction: SN2] reaction. Here’s the substrate with its good leaving group, and the high HOMO here comes in displaces, and then you lose a hydrogen from the same place and get the product. You get the normal propyl product, the first one there. But if it dissociates a little bit and rearranges with a hydride shift, then you get the other product. This is what we talked about.
So we said that this is what gives the primary product, it’s essentially an SN1 [correction: SN2] reaction. But this SN2 [correction: SN1] reaction after rearrangment, when you get the more stable secondary cation, then it can dissociate and then you get the so-called cumene, isopropylbenzene.
Now that may be true, but it may be wrong, too. I got thinking about this. OK, so suppose it rotated so that methyl was on the backside. Then what might you expect to happen? What happened over here to get the secondary cation?
Student: Hydride shift.
Professor J. Michael McBride: It was a hydride shift, right? What might you get now?
Student: Methide shift.
Professor J. Michael McBride: A methide shift. And methides and hydrides rearrange at about the same rate. So why not do this, why not do methide shifts? So what do you get if you do a methide shift? You get that. But it’s the same thing you had before, it’s a degenerate rearrangment. It’s still the primary compound and then you could do the SN1 and get the product, so it doesn’t make any difference that you do a methide shift.
But suppose the methide shift gets stuck halfway across. Suppose it stops there. Because another way to draw that is as protonated cyclopropane. It’s cyclopropane that’s taken on an extra proton. And that substance has been studied. And it’s known that its stability is between the primary cation and the secondary cation. So suppose you got that, what would that do? Well, you could have the nucleophile attack that, I mean the methyl+ is a great leaving group. So you could do that, and again, it would still give that same product, if you had the protonated cyclopropane instead of this thing doing the alkylation.
Now in fact, Norman Deno at Penn State University in 1968, unbeknownst to me, had actually done that reaction. He took benzene and cyclopropane and reacted it with acid, and got these products. And he used deuterium here. So there’s deuterium in this normal propylbenzene, just one deuterium. Now what I want you to tell me is where you expect it to be. On which carbon should that deuterium be, which of the three carbons? Let’s suppose that it happens by making protonated cyclopropane. Chris, what do you think?
Student: The terminal one?
Professor J. Michael McBride: It should be the terminal one, because if you put it on here then you displace– that’ll be the methyl group and the deuterium should be there, one deuterium. Here’s what he found. Where is it? 3/7 of it is where Chris said, 2/7 here and 2/7 here. It’s completely scrambled, right? There are three hydrogens here, two here, and two here. And that one deuterium is completely scrambled among those positions.
How did it get scrambled? Well, this corner-protonated cyclopropane very easily rearranges to edge-protonated cyclopropane. And then that can become protonated at this corner, and then this edge, and then this corner, and then this edge, and then back to where you started. And when it’s that CH3 it doesn’t have to be the deuterium you added that moves, it could have been the hydrogen. So everything gets scrambled very quickly going around there. And in fact, this was shown in a key experiment by a man named Richard Baird, who was an assistant professor here in this department in 1964. And it was on the basis of that that Deno tried this experiment.
So protonated cyclopropane does give this compound. So now we have a question, which of these is the thing that gives the Friedel-Crafts product? Is it an SN2 reaction with this, or an SN1 reaction that involves the protonated cyclopropane? And it could be either. And no one knows, as far as I am aware. I’ve just been going through the literature in the last two days on this. I don’t think anybody knows this. I tried to call Professor Olah, and wasn’t able to get him, at University of Southern California, who got the Nobel Prize for Friedel-Crafts things, and I haven’t been able to get him yet. But Professor Saunders, who knows a lot about this, doesn’t think he knows. We’ll see. But anyhow, I think it’s not known. So what I want you guys to do is to try to think up an experiment that can tell you whether it’s the SN2 or the SN1 mechanism. And I’ll give you a hint, I think the good experiments involve isotopes. So here’s a problem for you. So maybe you can become famous and determine the mechanism of the Friedel-Crafts reaction after 143 years.
Chapter 3. Carbonyl Compounds: Energy and Spectroscopy [00:22:05]
OK, now onto the next subject, which is carbonyl compounds. I didn’t bring the Jones textbook along, but carbonyl compounds are discussed in Chapter 16 to 19, which is 268 pages. It’s about a quarter of the text book, I think, and that’s not atypical. You have many other texts and you can check, but there’s a lot of stuff in there about carbonyl compounds. The good news is that almost all of it is review, because we’ve talked a lot about the reactivity of carbonyl compounds. So Chapter 16, for example, is aldehydes and ketones. So for the rest of this lecture, I’m just going to go over the big view of this, and then in the next lectures we’ll take a more detailed view of some of the lore.
So we talked about before that the carbonyl compound is stable, but it’s reactive. Those things we’ve said before seem contradictory. You have the average bond energies we’ve looked at before for the single and double bond, and for the single and double bonds to oxygen. And we’ve seen that the second bond in carbon is very much weaker than the first bond. But the second bond for oxygen is stronger than the first bond. And it’s not all due to the π bond, because it has also to do with a change and hybridization of the carbon, the σ bond changes when you form a double bond. So it’s very stable compared to carbon-carbon double bonds. But it’s also reactive. because it has a very low LUMO the π*.
Now just to give you something to practice with about NMR, to see how people figured out chemical shifts in NMR. They did it by taking a compound, a known compound like acetaldehyde where n is 3 here, you’ll see why I used n. And they found it had two peaks, 199.6 and 31.2 in the CMR spectrum, carbon magnetic resonance. And then they put another carbon on and found peaks at those positions, and then another one on and found peaks at those positions. And then they did a ketone, acetone, and found peaks at those positions. And with one more carbon, 2-butanone, they found that. And with 2-pentanone they found that, and with 3-pentanone they found that. Now with that empirical evidence behind you, I want you to figure out what this CMR spectrum is of and which peaks are which. And that’s taken from that Chem 220 problem site that we’ve mentioned before.
So there are the values of the chemical shifts in this proton-decoupled CMR spectrum, and use the table to identify what each of those carbons is. And in the process, you’ll figure out what regularities there are within this table, and how you go about figuring out new things. And just to give you an extra hint, here’s the proton magnetic resonance spectrum of the same compound. Notice this is 400 MHz for protons. It was a 100 MHz for carbon. That’s the same magnet, because the proton is four times as strong as carbon. And just to help you out with this, we can blow up each of those peaks and see what the spin-spin splitting is. And there they are with the values of their chemical shifts, and here’s the scale for the detail, that’s 20 Hz. So this splitting here is 1.8 Hz and the one over here is 6.8 Hz, just for reference. So there are two problems for you. Make your name by devising an experiment to prove the mechanism of the Friedel-Crafts reaction, and figure out what this compound is.
Chapter 4. Carbonyl Compounds: Reactivity Patterns [00:26:06]
OK now, so carbonyl is very stable, but it’s also, paradoxically, reactive. And we know that’s because of π*. We’ll see other ways as well. So the nucleophile comes in, we’ve done this is a zillion times, and we have nucleophilic addition. And last semester, we talked about the Buergi-Dunitz angle, and all this is discussed in Chapter 16. And you remember, this contrasts with C=C, where the addition reactions are electrophilic, not nucleophilic. So then you bring a proton on and you get an alcohol. And especially interesting alcohols, that will be discussed in Chapter 16, are ones where you form a new bond to R, where the nucleophile is an alkyl group. For example, methyllithium has that property, and we talked about that some last semester. Or where the nucleophile is H-, as in lithium aluminum hydride. So here are some sections that talk about that. And again, we’ll talk more about this.
Now, a different kind of reactivity is when one of the substituents on the carbonyl can be a leaving group in an SN2 reaction, so it’s an OK anion. Because when you do that, there’s something else you can do at this stage. Of course, you can reverse it, you can bring these electrons in to reform the double bond and lose the nucleophile and go back. But as in SN2 reactions, you can also lose the leaving group if it’s a reasonable anion. So you can go like this, and now what you’ve done is not an addition. What kind of reaction is this, instead of an addition? What’s overall? Linda?
Professor J. Michael McBride: It’s a substitution instead of an addition. So nucleophilic substitution of things that are called acid derivatives– these things when L is a leaving group, when that thing is a leaving group are called acid derivatives. So it’s association/dissociation. It’s not like SN2 where you go through a pentavalent transition state. This has an actual intermediate, this has a finite lifetime. So it’s association and then dissocation, as in aromatic substitution. Remember, electrophilic aromatic substitution goes through a tetravalent carbon, then the proton gets lost and you have the electrophile left on there. And that’s what’s discussed in Chapter 18.
But there are other ways to go at it too, because the oxygen has unshared pairs, so acids can attack it. So you have an acid come in, for example a proton to protonate it and get that cation, and then you can have electrophilic addition. So a nucleophile then can come in on a second step. In the previous slide, we had the nucleophile come in first, and then brought a proton in later. Now we’re bringing the proton in first on the oxygen, and bringing a nucleophile in later. So we get compounds that have OH and a nucleophile on there. Now what nucleophiles can you use? Notice that this is much easier than for C=C. So the nucleophile can be OH, in which case this thing over here is called a hydrate. It’s H20 added to the carbonyl. Or it’s sometimes called a gem diol– gem for geminal. Gemini is twins, right, so it’s twin OHs on there.
Or it can be OR, and we’ve already talked about that, hemiacetal, or on the way to an acetal. Or it could be a nitrogen, and then the equivalent of a hemiacetal for nitrogen is called a carbinolamine. And you can unzip this, of course, and go back to the carbonyl. But if you have two groups on there, you could unzip, losing the other group and keeping the nucleophile. And then you have a nucleophile double bond. That’s what the imine is, a carbon-nitrogen double bond. Or you can have a bisulfite addition product, or CN. That one’s special, because you form a new carbon-carbon bond. You’re always on the lookout for things that will form new carbon-carbon bonds, not just change the functional group. So that’s called a cyanohydrin. So there are sections 16.6 to 16.11 \ about that in the Jones book, but all these things will be covered in all these textbooks.
Now, the reason this was stable, one reason it was easy to protonate that was you have this resonance structure. But notice that if you have a proton here, on the carbon adjacent to the carbonyl– that’s the α position. So if you have an α proton, then a nucleophile can act as a base. So a base can come in and take the proton. So instead of attacking here, it can attack here and generate this double bond. And you remember what those things are called? Enols. So that this is an allylic rearrangement, the fourth pattern of reactivity is allylic rearrangement, ketone to enol. As I’ve been going along, I’ve been numbering all the different ways carbonyls react.
So now you could a bring a low LUMO in here and have the electrons of the double bond attack it. So this is a double bond. So this is like electrophilic addition to a double bond, except especially good because you have the unshared pair on the oxygen that can help out and make this easy to do that. So it’s better than other carbon-carbon double bonds. So the enol is essentially a carbon nucleophile. It can attack low LUMOs, especially easy. So you get something like this. And notice that after you lost the proton then, what you have – what you’ve done is to do a substitution at the α position, an electrophilic substitution at the α position of a ketone or an aldehyde. So that’s another pattern of reactivity, and that’s Chapter 19, these reactions of enols.
So enols, enolates, and enolization. A little bit from Chapter 19, here. So here we have this α hydrogen, we bring the base in, as we said last time. This time, we’re doing it without having first put something on the O and having a double– so what we’ve made now is and anion. Its pKa is 19. Pretty acidic. But what’s it normally for alkanes, if you didn’t have the carbonyl there? You told me last time, Chris.
Student: A little over 50.
Professor J. Michael McBride: A little over 50, right. So here we’ve got 31 powers of 10 advantage. That’s how easy it is to make the enolate, the anion. Once you have the enolate with a double bond here and a charge on the oxygen, then you can put the proton on the oxygen and you have an enol. So another number that’s interesting is the equilibrium constant between here and here. So this is for forming the anion, this is for the allylic rearrangement of the proton. So it’s 5 times 10-9. So you’re going to have very little enol compared to the ketone. But notice at equilibrium, they’ll be in equilibrium with the same anion. Because if you lose the proton from here, you get the same enolate anion that you get from losing the proton here.
So if this is ever so much more energetic, higher in energy, than this, then this one is going to be unusually acidic. It’s going to be easy for it to lose that proton. Since the ketone is about 11 kilocalories below the enol, the pKa of this is about 11. Very acidic. Remember, normal alcohols are like 16 and 17 pKa. So this is very acidic. Now let’s go to – if one’s good, more is better. Put two carbonyls adjacent to this α thing, so now both of them can stabilize it. And now the pKa 9, and the equilibrium constant is now 3. It’s better to have this than to have this. Whereas here, it was 10-8 of this compared to this. Now it’s three times as much this as this at equilibrium.
Notice, incidentally, this would be hydrogen-bonded. And you could imagine that the hydrogen could shift from here to here– this double bond to here, this double bond to here– the hydrogen could bounce back between those two things. In fact, it’s even conceivable it would be halfway in between. And Professor Vaccaro in this lab is doing a lot of work on that kind of thing, compounds of that sort. So here, the ketone is approximately equal to the enol, because of the help from conjugation and from hydrogen bonding.
Now if you have these double bonds in there, now this compound is favored by a factor of 1013 . This enol is 1013 favored at equilibrium over this. so you got none of this, it’s all here. Why? Because this one is– you’ve gained the aromaticity, that 30 kcal/mole that you didn’t have over here. So the ketone is now 17 kilocalories above the enol. Here, the enol was 11 kilocalories above the ketone. It’s a complete range of ketone to enol. The pKa of this thing here is 10. It’s the same as this. The anion is only about 13 kcal/mole above phenol. So it’s very easy to lose that. And phenol, if you ever get it on your skin, burns like fire. It’s just terrible because as an organic compound, it dissolves in your lipids and then it’s a strong acid. So if you ever get it, you’ll know. Did you ever get into nettles? Stinging nettles? That’s phenols for you. And the pKa in this case is -3. So this really easily loses a proton.
Now, how about reactions of carboxylic acids? We went through, I think it was five patterns of reactivity of carbonyl groups. Now we’re going to look at the carboxylic acids, which are in Chapter 17, and see what patterns of reactivity we’re going to expect. Well first, you can get addition, just as if it were a carbonyl group. So you can imagine doing that kind of thing and there, we’ll see examples.
Another possibility is to do substitution at carbon. So this is like the OH became a leaving group somehow. And then OR came in. So it’s just like the ketones. We said if there was a leaving group here, here the leaving group is OH. Or we make it into a better leaving group somehow, maybe by protonating it, something like that. So we can get substitution at carbon.
But there’s another possibility in the acid. You could start here, and you could do the substitution at oxygen. Get the same product, R on R, O, C=O, get the same thing, but make this bond, rather than making that bond. So you can have substitution at carbon, you can have substitution at oxygen.
And there’s another possibility, of course, substitution at the α carbon. So if you have an RH here, an α hydrogen, then, as we saw with aldehydes and ketones, you can do a substitution there. And the good news in all this, is almost all of this is review. In fact, I should have said one thing on the previous slide. That when you have the OH and put the R on, making the R-O bond, that’s an α substitution also. It’s just that it’s an oxygen rather than a carbon that has the α hydrogen. So you can have that, or on the carbon side. And the good news is, very much of this is review. For example, if you look at pages 785 to 787, sections 16.9 to 16.10 in the Jones book, you’ll find that it’s an old favorite of ours. So let’s just conclude the hour, here, by going through that again.
It’s the mechanism for acid-catalyzed hydrolysis of acetal. We’ve already done this, but it doesn’t hurt to do it again. And it’s this part of the text. So we have this acetal, and the idea is to hydrolyze it, to go back to an aldehyde. And how do we do it, do you remember acid catalysis? Where do we start in? Kate, are you going to help us out, here? Where does an acid attack this compound, H+?
Student: Something to do with one of the oxygens?
Professor J. Michael McBride: What is it about an oxygen that makes the H+ want to go there?
Student: The lone pair.
Professor J. Michael McBride: The lone pair. So the goal is going to be– it’s two steps. First we’re going to remove OR and replace it by OH, then we’re going to remove the other OR and replace it by OH. So the way we do it is with H+, as Kate tells us, and we have the unshared pair, and we do that and we get here. This is exactly one we went over before, but the old ones are the best ones, right? OK, what’s going to happen next? John, what do you say? Remember what the goal is? What are we trying to do in the first stage?
Student: Remove RO?
Professor J. Michael McBride: We want to remove RO, how are we going to do it?
Student: I’m not sure.
Professor J. Michael McBride: Well, we could just have it leave now, right? Remove ROH. What do we leave behind when we pull off ROH from this thing? Where do the electrons go– it’s this bond we’re going to break– where are the electrons going go, are they going to go on the O, or are they going to go on the C?
Student: On the O.
Professor J. Michael McBride: On the O, right? Because it’s O+, so it’s going to gain full ownership of that pair of electrons, so it becomes ROH neutral. So what’s left behind on the carbon? Nothing, right? The electrons went away. So what do we draw, then? How do we draw that? We have ROCH2 left, what do we draw on the CH2?
Student: A plus charge
Professor J. Michael McBride: Right, OK. So it’s going to do that, and we’ll get that. If this weren’t RO, if it were just H, say, that would be a crummy reaction. Why does it work here? What’s good about that cation?
Student: The lone pairs on the oxygen get stable.
Professor J. Michael McBride: The vacant orbital is able to stabilize those electrons, so the thing gets unusually stable. So you have that, and bingo. So we’ve got the first RO off. Now we have to replace it by OH. Cassie, how are we going to do that? How are we going to put OH on here after we took RO off?
Student: It can attack, the O on the π*
Professor J. Michael McBride: Speak up, please?
Student: The O on the π*?
Professor J. Michael McBride: The O of the what?
Professor J. Michael McBride: OK, so we want to put OH on that carbon. How do we do it? Say it again, please?
Student: Can you attack the π*?
[correction: Cassie is at least partly right. The instructor is thinking of the resonance structure on the left]
Professor J. Michael McBride: I’m not sure what π* you’re talking about. There’s a π here, in this resonance structure, but the important thing that we have is a vacant orbital on carbon.
Student: There are lone pairs on oxygen.
Professor J. Michael McBride: What oxygen? O2?
Student: Of the alcohol.
Professor J. Michael McBride: So suppose it attacks the unshared pair on the alcohol, you form a bond here. Now you’re here. That’s going backwards. So we’re going to need something else.
Student: You have a water group.
Professor J. Michael McBride: Yigit, what would you use?
Professor J. Michael McBride: Water. So if you put water in… That cation is usually stable, as John told us. So we bring water in, now we make that bond and now we have OH there, but we also have H. Is that a problem? Yigit? We want OH here, not H2O.
No, we just lose the proton. So we’ve accomplished our first step up here, right? We have a hemiacetal and we can check that one off.
Now we have to take the second RO off and replace it with OH. How are we going to do that? Nathan?
Student: Same way
Professor J. Michael McBride: Tell me what we’re going to do. What’s the first step?
Student: An H+
Professor J. Michael McBride: And where do we want to put the H+? Here or here?
Student: On the top one.
Professor J. Michael McBride: On the top one, OK. So we remove the second one by bringing H+ in and put it on the top. Next?
Student: Lose the water.
Professor J. Michael McBride: Lose water. Pardon me, lose alcohol to get this, and now?
Student: Add water.
Professor J. Michael McBride: Add water. No, we don’t need to add water. If we wanted to put the second one in, true. I misled you. But really where you go is here. So you just lose a proton. So you’ve taken the ketal to a carbonyl by way of a hemiacetal or a hemiketal. So the overall transformation, then, is water plus acetal, acid catalyzed to give carbonyl plus two ROHs. And we wrote it this way before. They come together and pull it apart, and so the H2O enters all three molecules. OK, that’s how much we wanted to do today. We’ve made it. We made it to the end, we’re caught up, actually, with where we were last year. Plus you have the chance to become famous by proving the mechanism of the Friedel-Crafts reaction. Thank you.
Oh, you did it. [A student had succeeded in assembing the space-filling model of tetraphenylmethane.]
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