CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 37 - Proving the Configuration of Glucose and Synthesizing Two Unnatural Products
Chapter 1. Glucose Structure by IR, NMR, and X-Ray [00:00:00]
Professor J. Michael McBride: OK. I realize it’s reading period, so we’re supposed to be having fun. So the structure of glucose is really fun.
I originally had this lecture mostly about the synthesis of two unnatural products. These last two lectures were going to be about synthesis. One about unnatural products and one about natural products. But glucose is such a great story that I couldn’t miss telling it. So we might not get to the synthesis of two unnatural products, or I might just mention it at the end. It’s also Nobel Prize work, but so is the structure of glucose. And there are Nobel Prizes and Nobel Prizes, but the structure of glucose is a really great Nobel Prize.
So we saw last time that there are carbohydrates, and I found who used the word first. It was a guy named Schmidt in 1844. He said, “I may be allowed to denote compounds [with] carbon plus hydrogen and oxygen in the same ratio that obtains in water…as carbohydrates.” That is, it doesn’t have to be one to one, one carbon to one water.
So there were examples known. Like grape sugar was isolated pure in 1792. And it was given by Dumas the name glucose, which comes from the Greek word gleukos, which means sweetness. So that’s where glucose came from. But even more important, the -ose ending, which is just part of the word glucose, came to be the generic ending for all sugars. So when you see something -ose, is means it’s one of these carbohydrates.
And we remember that Couper, as soon as structure became something that you could talk about, because of the tetravalence of carbon, he proposed that there was another water added to give the hydrate of glucose, which is just fine. That would still be a carbohydrate. And he was right.
So you can have aldoses like glucose where you take two hydrogens from the top carbon, put them on either end so you have an aldehyde. Or you can do the next carbon or the next carbon and ketones, so you can have ketoses. For example, fructose is a ketose, and glucose is an aldose.
Now, how do you know this? Of course, nowadays we know the structure of things by spectroscopy. For example, here’s the IR spectrum of glucose, this aldehyde. What do you notice specially about this spectrum? Anything? What would you look for if you were looking for a sugar? You’d look for OH groups– sure enough, big O-H peak. And you’d look for what other group? Antonia?
Student: The aldehyde?
Professor J. Michael McBride: The aldehyde carbonyl. And where would you look for it, do you remember?
Student: The C=O stretch.
Professor J. Michael McBride: You look for it around 1,700, and it’s not there.
Or you could do the NMR. So here’s the proton NMR of crystalline glucose. You dissolve it in D2O, so now you don’t have the OH protons there. And there’s the proton NMR spectrum. And again, there’s no way downfield peak. No aldehyde. Or if you look at the carbon NMR spectrum, no downfield peak for aldehyde, or ketone, either. They would both come down there in the carbon spectrum. Of course, there wouldn’t be a hydrogen on the ketone.
So let me draw it, to try to explain this, draw it in a funny conformation. Which could be in equilibrium with a hemiketal. Remember? With either acid or base catalysis in water, here deuterated water, you could make a hemiketal down here at the bottom right. So all the Os are ODs, but here you have Hs.
Now let’s blow up that part of the NMR spectrum that you can see there. What peak stands out? One far downfield. What’s it due to? Which of the hydrogens in this compound should come the furthest downfield? Leen?
Student: The one that’s attached to the carbon attached to oxygen.
Professor J. Michael McBride: That’s attached to what?
Student: Attach the carbons to O.
Professor J. Michael McBride: They all have an O.
Student: Well yeah, but attached to one without the alcohol.
Professor J. Michael McBride: The one here, you mean.
Student: Yeah, over there.
Professor J. Michael McBride: Now, that one is on a carbon that has two carbons attached to it and an oxygen, and oxygens withdraw, tend to shift things downfield. Can you do better than that? There’s one oxygen to withdraw from that carbon, right? Megan?
Student: The equatorial hydrogen.
Professor J. Michael McBride: The equatorial one, the one at the far right, has two oxygens pulling on it, the one that’s on a hemiketal. So it’s way downfield.
Now notice that there’s a little peak here, a little doublet. What’s that due to? Well, it could be just impurity, but it’s not. Because if you let it sit a day, it grows, and the other one shrinks. It’s also downfield, so it looks like it’s got more oxygens on it. Anybody got an idea?
What did you call this one, Megan?
Professor J. Michael McBride: Suppose it had been axial. Then if you had a ring current, as in benzene– you don’t here– this one would be downfield, and the one where it was axial would be upfield. More over the top of the ring. Can you get it there? Yeah, you can go back here and attack the carbonyl from the other side, and get that one.
Now, how do you know which is which? How do you know which one was axial and which one was equatorial? Well, you might use this argument about ring currents. People talk about that. I’m not sure whether it means anything. There are other kinds of arguments you could make that might have to do with that, too.
But one thing that’s clear in the spectrum is that there’s a doublet splitting here which is about 8 Hz, and a doublet splitting here which isn’t even half as big. So one is a big splitting, and one is a small splitting. Now, what is it that’s splitting this hydrogen? Karl?
Student: Sometimes that D is replaced with an H?
Professor J. Michael McBride: So there has to be another H on the adjacent carbon that’s splitting that one, right? So this H here. And in this case, this H here would be doing it. But one of them is a big splitting, and the other is a small splitting. Sebastian, you got an idea?
Student: In the top one there, anti.
Professor J. Michael McBride: Right. The top one is anti, and the other one is a gauche. So there’s much better overlap, anti than gauche. So this one is the one that’s anti. That’s the one with axial hydrogen. And this is the one that came from the crystal. So the crystal was all the same. You dissolve it and it begins to equilibrate, and ultimately has more of the equatorial OD than of the axial OD.
And in fact, if you look at the original material, just after you dissolve it, it has a rotation of 112°, optically active. But after a day in D2O, it’s only 53°, because the other isomer is 19°. So people used that, before there was spectroscopy, to try to figure out what the ratio of the two was at equilibrium.
Or if you really want to do it, you do it by X-ray diffraction, and you get Cartesian coordinates for all the atoms. That’s what you do nowadays.
Chapter 2. Glucose Constitution from van’t Hoff to the Kiliani-Fischer Synthesis [00:08:12]
But the interesting thing, and what got a Nobel Prize, was not this. It was, how did they know? How could they figure out what the structure of glucose was?
Well, a lot of sugars were known in the last half of the 19th century. When van ‘t Hoff wrote the German version of his book in 1877, he mentioned a lot of sugars as optically active, for example, here, he cites glucose. He also cites levulose. And in fact, those both come from sucrose. So sucrose is a more complex sugar. It’s a ketal that has two of them together. But if you take them apart, one of them is glucose, and the other is levulose, which is also called fructose.
Do you know why it’s called levulose? What does levo mean? Actually, the English word has the same root. Left. Levo is left, dextro is right in Latin. So this rotates white to the left. This rotates light to the right, and is called dextrose, also, or dextrin down here.
He also mentioned lactose. And it was found in 1876 that lactose treated with acid gives also glucose, but another one called galactose, which comes from the Greek word for milk. So it’s called milk sugar in German. So lactose, obviously, came from milk. 5% of milk, I think, by weight is lactose sugar.
Or there’s stuff called mannitol, which came from manna. They would slash the bark of flowering ash trees around the Mediterranean, and this stuff would ooze out which they called manna. And from that, they got mannitol. And mannitol turns out to have these kind of CHOH groups that van ‘t Hoff was interested in because they made things chiral. But notice in mannitol, the two ends are both OHs. The aldehyde on the end has been reduced, in this case.
Or both ends can be oxided as in what’s called sugar acid. So you could have both ends be an acid.
Or you can have arabinose, which was isolated in 1873 from gum arabic. A gum that came from Arabia. You know, if you’re an actor, you put your mustache on with gum arabic. I think you used to, anyhow. I don’t know.
And in fact, it was possible to take mannitol and oxidize it back to an aldehyde. That must have been a tough thing to do, because aldehydes are– very hard to stop an aldehyde. But somehow, in 1888, they were able to get mannose, maybe just a teeny, teeny bit of mannose, doing that. So aldehyde.
So there were lots and lots of these known, and people fiddled around with them, and found out what you’d get if you oxidized, what you’d get if you reduced, and so on.
And on comes Heinrich Kiliani, the hydrogen cyanide man. Because in this paper in 1855, which fortunately I’ve translated here, he says, “While a large number of compounds are very easily formed upon oxidation of dextrose”– remember, dextrose is glucose, it rotates light to the right– “by dilute nitric acid or by halogens, these molecules retain six carbons bound together in a chain. Under the same conditions, levulose”– the other one that came from sucrose– “yields substances containing chains with a smaller number of carbons (like glycolic acid, or inactive tartaric acid).”– that’s meso-tartaric acid, OK? “Here oxidation causes immediate splitting of the molecule,”– it breaks the carbon chain– “a fact which means that levulose is a ketone.”
So that glucose, or dextrose, has an aldehyde at the end of the chain, you oxidize that to an acid. But if you have a ketone in the middle of the chain and you oxidize it, you break the chain. So it must be a ketose of some sort.
“Bearing in mind further the fact that levulose is transformed into mannitol by nascent hydrogen”– nascent means at the time it’s being born. They had this interesting idea in those days that when they dissolved sodium in alcohol, it generated hydrogen, but this hydrogen was more reactive than most hydrogen and could reduce things that normal hydrogen gas couldn’t do. So they said it was nascent hydrogen that could do things. In fact, of course, it was the metal actually transferring the electrons to the compound to be reduced.
But anyhow, he said, if you do that with levulose, you get mannitol. Which, remember, was the thing that has OH on both ends of the chain. So if it was a ketose, a ketone in the middle chain, it already had alcohols on the end. But if you reduce the ketone to an alcohol, then you got alcohols all the way, much like mannitol, which they got from that.
So “one comes to the conclusion that levulose must be adjudged to have one of the following two constitutional formulae.” Either C=O– they can’t be the first carbon. That’s the aldehyde. That’s the one where you reduce it or oxidize it, it doesn’t break the chain. You just get an acid there.
If it’s a ketone here, you break the chain. If it’s a ketone here, you break the chain. What if it’s a ketone on the next carbon down? Why didn’t he consider that? Matt?
Student: It’s the same as that one.
Professor J. Michael McBride: It’s the same as the one above, right? There are only two.
OK. So is it I, or is it II? So he goes on in the paper to say, “One could hope to distinguish definitively between one and the other formula by succeeding in adding HCN to the ketone radical of that levulose and transforming the cyanhydrin into the corresponding carboxylic acid.”
So you add CN, hydrolyze it so it’s COOH. It adds to the carbonyl carbon. So from compound one, you would get this. Add in CN here and turn it into carboxylic acid. From compound II, the ketone on the next position down, you’d add here and get that carboxylic acid.
So the question is, if it has this structure, then you add HCN and HCl. You get this– and it’s very concentrated HCl he used, incidentally– or if it’s the next one down, you get this. OK. So you do it and decide. All you have to know is which product you get. Get it pure, see what it is.
The problem is, these things, as almost all things are in sugar chemistry, are syrups. It’s very, very, very difficult to get crystals from them. So it was really hard, in those days, to purify things. Nowadays you to do chromatography, maybe, or something like that. But then it was essentially impossible. So they had to try to make things simpler so that they had a chance of crystallization.
He says, “It must upon exhaustive reduction by concentrated HI yield methylbutylacetic acid.”– so this one– the one on the top– would give methylbutylacetic acid, reduced with HI and phosphorus—“and on the contrary, the carboxylic acid from compound II…would give ethylpropylacetic acid.” So all you have to do is know which of these you’ve got.
And he said, this is fine. But when he did the reaction and got the product, which should be either this or this, he says, “[It] does not agree with the description that Hecht gives of the Ca salt of methylbutylacetic acid.” So Hecht, he said, had done a lot of good work on this compound, including making the calcium salt. He was able to crystallize the calcium salt of the product he got. And it didn’t agree with Hecht.
So he says, “which means that my heptanoic acid is identical with ethylpropylacetic acid.” So this is the one he’s got. But that compound wasn’t known, so he couldn’t be sure about it. But anyhow, he concluded that it must be this, since it’s not Hecht’s compound.
Now the next year, in 1886, he reacted ethyl acetoacetate. So this is acetyl group on acetic ester. What’s special about those hydrogens? Active methylene adjacent to two carbonyls. You make the anion, you can put an R group on. You can make the anion again, put another R group on.
So he put on ethyl. Put on propyl. And then, what I told you last time is that when you hydrolyze this, you get the acid, and then it decarboxylates. So from the acetoacetic ester synthesis, you lose CO2 and get a ketone.
But he did it differently in those days. He used really, really strong KOH, which doesn’t give a ketone. It gives acetate. That is, the hydroxide attacks here. Acetic acid comes off, generating that anion, which then picks up a proton from the acetic acid. So you have acetate and an acid anion. Which you then put a proton on, and you’ve got this.
So he made authentic stuff, but it didn’t agree with what he got from levulose. So now from Hecht, he knows it wasn’t this, and from his own work, he knows it’s not that.
So now he began to wonder about Hecht and just how good Hecht was. So he went on and did the same thing, putting on methyl and butyl, to make this. And he compared it with the product from this, and found that the calcium, barium, strontium, and lead salts all agreed in their properties. So Hecht was wrong. He said at the end of the paper that Hecht is going to go back and look at this a little more.
But at any rate, he was then able to prove that levulose or fructose has that structure. The carbonyl is in the second position. So he now knows the constitution of fructose.
This paper from 1887 is on the composition and constitution of arabaric acid and arabinose. So now he’s going to look at a different sugar. And what he says is, “In a report dated 27 November 1886, I showed on the one hand that arabaric acid formed by oxidation of arabinose has the formula [shown there], but on the other hand described several derivatives of arabinose carboxylic acid with the formula [C7].”
So it was uncertain whether arabinose was a five-carbon sugar or a six-carbon sugar. If it was a five-carbon sugar and he oxidized it, he’d get five carbons. If it was a six-carbon sugar, and he added HCN to add another carbon, he would get C7.
So here says, he admits that he had said contradictory things. One that implied that it was C5, another that implied it was C6.
“Since at that time I had no basis in truth to dispute the generally accepted formula for arabinose,” which is that it’s C6. The analyses are very similar to one another, whether it’s one or the other.
“and my analytical results did not contradict either assumption.”
So he couldn’t be sure whether arabinose, the stuff that came from gum arabic, was C5 or C6.
So he took arabinose. He added HCN He made the arabinose carboxylic acid, the stuff that he talked about up there, but he said then it was C7. Here he made it with one, two, three, four, five, six. Here he made the six– assuming that it starts with five, right? And he was able to reduce that with sodium with mercury to get mannitol, which was known to be C6. So it was clear that arabinose wasn’t a C6 sugar. It was a C5 sugar. So he had resolved this.
But it was pointed out that– One of the reasons it was difficult to work with this was this original compound was a mixture. Why would it be a mixture? Because the cyanide could add to either face of this carbon to make right- or left-handed carbon. And these other carbons are, of course, chiral, too. So you get diastereomers, depending on which side you add from.
This was pointed out by someone who was very sensitive to not just the constitution, the sequence of bonds, but to the configuration of these sugars, which is going to make a lot of difference. With so many chiral centers, there are going to be lots of isomers. So the person who pointed that out was an Emil Fischer, who was three years older than Kiliani, and died much sooner, for reasons you’ll see.
So one thing he did with Kiliani’s acid– this is the acid you get by adding CN to the five-carbon sugar, and then hydrolyzing it. So Kiliani treated it with sodium and mercury, and got mannitol.
But Fischer did something different. Bear in mind, this is a reduction. So you take the acid all the way to the alcohol. It’s very difficult to stop at the aldehyde. But Fischer found a way to stop at the aldehyde, which was to first heat it to make this lactone, a cyclic ester, and then reduce that with sodium and mercury, and now he got the aldehyde here.
Now, why was that handy? Because it gave a way to convert one sugar into another sugar. Now you could make new sugars synthetically, because it was possible to stop at the aldehyde. So a Kiliani-Fischer synthesis on arabinose gave this compound. So Kiliani-Fischer synthesis elongates an aldose, makes it one carbon longer.
Chapter 3. Fischer’s Osazones, Fischer’s Projection, and Fischer’s Evidence [00:23:28]
OK. Now, that wasn’t the reason Fischer got the Nobel prize. A contributing factor was phenylhydrazine. When he was a graduate student, he synthesized phenylhydrazine, and he loved to make derivatives with it. Remember, we talked about how you could make derivatives of carbonyl compounds with N-N things to make imine bonds, and they’re very strong, if you have two Ns together like that.
So he could do it with sugars. You could react with a sugar, and make an imine. It’s called a hydrazone, right? Now, this hydrogen here is allylic, so it could shift up to here and make the double bond here. But now this hydrogen is allylic, and it could shift to there, so you can get this.
Now you’ve got another ketone. So what’s going to happen? You can make another hydrazone. So now you’ve used the second mole of phenylhydrazine.
So now that one’s allylic. You can shift it over here. Here we do a cyclic rearrangement. This hydrogen goes to that nitrogen. This bond breaks. These electrons go here to make a double bond, becomes a single, double. So it breaks apart into two molecules, one of which leaves.
And now we have a C=N double bond, which is very like a C=O double bond, so you can get a third molecule of phenylhydrazine to react and make a new C=N double bond, losing ammonia, and you get this compound, that has two of his favorite molecules stitched in there. It’s called an osazone. It’s a hydrazone that’s made from an ose, from a sugar. By treating this with his favorite compound, you could make this.
Now, it was his favorite compound, but it’s also deadly poisonous. And he poisoned himself very badly for almost a decade he was in very bad health because of poisoning himself with this. And in fact, in 1919, he committed suicide. He had cancer at the time, and who knows whether it had to do with exposure to his favorite molecule.
But if what was so dangerous, and he knew he was being poisoned, why did he like it so much? It was because ozasones are crystalline. Remember, these sugars are always syrups, and you can’t do anything with them. But with crystals, you could purify and tell what things were, tell if one thing is the same as something else or different. So that was the beauty, for him, of phenylhydrazine. That you could make an osazone and identify things in sugar chemistry.
So he was interested in identifying the structures of these things. So let’s try to be systematic about seeing what the configurations could be of different sugars.
Let’s look at the hexoses. We all we already have erythrose, arabinose, and xylose. Arabinose came from gum arabic, xylose came from wood. Erythrose comes from a lichen. Eryth means red in Greek, so erythrose was because it came from this red plant. But there were also hexoses glucose, mannose, galactose, and gulose. And all these were identified by Kiliani as being aldoses. Aldehydes, not ketones. We showed how he could tell the difference there.
So these must differ in their configuration. Right? Now, how many should there be? If you have six carbons in a chain that ends in an aldehyde, how many should there be? Well, for this purpose, we could write them all out like this. This is a particular one, using what projection?
Student: Fischer projection.
Professor J. Michael McBride: The Fischer projection. And one of the most amazing things about Fischer is that he figured this all out without having the Fischer projection. It’s easy for us to trace his thought now using the Fischer projection, but he used van ‘t Hoff’s notation, which was very, very complicated. And I have never figured it out. So that’s one of the real triumphs of Fischer, was to be able to think through this stuff without Fischer projections, which he published the next year.
But we could abbreviate that Fischer projection by writing it that way. Whoops, I wrote it wrong here. These lines show which way the OH points. So the bottom, you don’t care, it’s not chiral. The next one, the OH to right, OH to the right, that one should be OH to the left, OH to the right. So the code is clear.
Now we can just write out all the possibilities. So there are four chiral carbons, so there are going to be 16 possible isomers. But if we’re looking only at the D-sugars, those are the ones that are all to the right on the bottom.
We could draw the mirror image, too, the L-sugars. But let’s keep it simple just by showing the D-sugars, because as Fischer well knew, he had no way of telling D from L. You could measure the rotation, but not the absolute configuration. So let’s just assume that a given compound is D, in the D-series.
Now, how can you draw them all out? You can draw right, right, right, right, left, left, left, left. Then you can go right, right, left, left, right, right, left, left; and finally right, left, right, left, right, left, right, left.
So you’ve got eight isomers, and four of them must be glucose, mannose, galactose, and gulose. But not arabinose and xylose, because as we just saw, Kiliani showed that arabinose was a five -arbon sugar, not a six-carbon sugar. It’s a pentose. An erythrose is a tetrose, a four-carbon sugar. And finally, at the top, you have glyceraldehyde, the one that Fischer chose as the reference for what’s D.
So the question only is which is which? And at first glance, that looks impossible, until you come along with X-ray or NMR maybe, or something to figure it out. But Fischer figured it out.
So here was the evidence he used. You can look online on the website and see the translation of his paper, how he actually did it. Almost all textbooks nod and cover this a little bit, because it’s realized what a great contribution it was. I mean, it’s like Shakespeare or something like that. But they all mess it up. They use modern reagents, they say it should have worked this way, or something. This is what Fischer actually did, so you can read about it.
But anyhow, this is the evidence he used. He found that glucose could be oxidized by nitric acid to make the compound that is an acid on both ends of the chain. So it’s symmetrical with respect to acid. Not just the aldehyde, but also the primary alcohol got oxidized to an acid.
And then he did his heating to make a lactone reduce it with sodium, which made it back into a sugar again. And the sugar he got he called gulose. It’s different from what he started with. You’d think if you oxidize and then reduce back again, you got the same thing you started with. But he could get a different sugar this way, Why? How could it be different?
It could be the aldehyde was on the top, made two acids, reduced the top to an alcohol and the bottom to an aldehyde. So now it’s upside down, so you’d have to rotate it. So it could be a different one of these sugars that we drew on the last slide.
And he decided to call it gulose. Why? It’s a sort of an anagram. It reverses the U and the L. Glucose, gulose. Because he went end for end. So, you know, chemists sometimes have sort of geeky senses of humor, right? So gulose.
So that was the first evidence. And then subsidiary evidence for that is that the diacid he got on the way is chiral, and both enantiomers of it were known. That’s the first line of evidence.
The second line of evidence is that glucose and mannose give the same osazone. Remember, he made osazones to make crystals, but glucose and mannose gave the same one.
Also, arabinose, if you do a Kiliani reaction, gives two different acids. Gluconic and mannonic acid. We talked about that before, that it could add either side.
And fructose, when it’s reduced to give alcohols all along the chain, gives two different alcohols, glucitol and mannitol. And notice all of these are glucose-related and mannose-related. So there’s some special relationship between glucose and mannose, that all three of these things do that.
The second line of evidence is mannitol and mannonic acid are chiral. So if you have OHs on both ends or acids on both ends, it’s chiral.
Bear in mind that people were just fiddling around, seeing what they could make from what. But Fischer collected the key items of evidence that could be used to make a real logical argument.
So third, arabinose, when he did a Kiliani-Fischer synthesis, gave glucose. But also, it gave mannose. Right? Now we see this relationship between gluc- and mann-.
And xylose, which came from wood– that’s the, like, xylophone, right?– xylose, wood sugar, by the Kiliani-Fischer synthesis, added a carbon and made gulose. Remember, this one up here. And also another compound which I think is one called idose now. But these aren’t really relevant.
But then 3a, arabinose. If you make two OHs, or two carboxylic acids on each end, is optically active. But xylose gives inactive compounds with OH on both ends, or acid on both ends. In fact, that was tough. Because arabitol has no rotation. You put it in, you think it’s not chiral. But if you add boric acid, then it rotates light. Remember, it’s hard to know what causes rotation. And it just happens that arabitol has no rotation, or unmeasurably small rotation. But if you make a derivative of it, then you can see– and the acid is rotating. So it required a lot of work to know these things for sure.
So that’s the evidence, and therefore, he knows what glucose is and what all the others are.
Chapter 4. Fischer’s Proof of the Configuration of Glucose [00:35:25]
That’s pretty much of an accomplishment, right? Just to conceive that on the basis of this evidence you could figure it out is amazing. And this is only fifteen years after van ‘t Hoff proposed the tetrahedral carbon, and a lot of people still didn’t believe in it.
So here are the hexoses and pentoses. And we’re going to go through his lines of evidence and see how his logic worked.
Evidence 1 was that glucose gives a diacid, and on heating and reducing, it gives gulose. So when you turn it end for end, then you’ve got a different compound. So let’s go across and see. Would that be true of this?
If you turned it end for end, would it be a different compound? Remember, with a Fischer projection, the idea is things turn out at you. So can’t rotate 90°, but you can rotate 180°. So if we rotate this one 180°, do we get the same thing again? That is, what you do here with this is make a dot on the bottom and not on the top, so then you have to rotate it to bring it back into this series that we’ve drawn here. Right?
Obviously, those would all point to the left. It would be the mirror image of the one we started with. So we’ll put that on the back burner for a second.
How about the next one? If you turn that end for end, does it look the same, if you put the dot on the bottom and rotate it? Obviously different. Is everybody with me on its being different?
That one will be different.
This one, however, will be the same. If you rotate it 180°, it’s going to be left, left, right, right again. If you change top to bottom and rotate. So that one is not glucose, nor is it gulose, because if you rotate end for end it’s different.
How about the next one? Will that be different? Should we take it out? That’ll be different.
Will that be different? Ah, that’s going to be the same. Not different. So we can take that one out.
How about the next one? Will that be different? That’s going to be like the first one, where everything that’s right is going to be left, everything that’s left is going to be right. So we’ll put that in abeyance for a while.
And the last one also is different.
So from Item 1, we know for sure that it’s not that and not that, glucose.
Now, item 1a– and we’re not so confident about that, because it changes the hand in this only.
But the diacid is chiral. Both the enantiomers are known when the two ends are the same. So with this one, if the two ends were the same, would it be chiral? No. It would have a horizontal plane of symmetry through the middle. So it can’t be that one.
This one, would that have a plane of symmetry? No.
That one’s out already.
How about that one? No.
How about that one?
Ah. That one’s out. That wouldn’t be chiral.
The last one would be chiral.
OK. So from 1a, we confirm that it’s not that, and it’s not that. So we’re down to four candidates for what could be glucose.
Now we go to Item 2. Glucose and mannose give the same osazone.
Remember what the osazone did? It made a double bond at the top carbon and a double bond at the next carbon. So that means this next carbon here isn’t chiral anymore. Right?
And if you take arabinose, start with that being the carbonyl and add CN, it can add to either side. One would give that, one would give that. Depending on what arabinose is, one would give this, this; one would give that, that; one would give that, that. So again, it says they differ only at this second carbon.
And fructose, which has a ketone at the second carbon, as Kiliani showed, gives two alcohols. So it could give this alcohol and that alcohol; this alcohol, that alcohol; and so on. So anyhow, what that means is you disregard that.
So this Item 2 shows that glucose and mannose differ only in this carbon.
So now, when we look at Evidence 2, we know that if this is glucose, this is mannose. They differ only at that carbon, which this shows is not relevant here.
OK. So if this is glucose, this is mannose. If this is glucose, what’s mannose? This one, right? If this is glucose, this is mannose. And if this is glucose, this is mannose. That’s what Evidence 2 shows.
Now, Evidence 2a is that mannitol, OH on both ends, or acid on both ends, those molecules are chiral. So can this be mannose if the two ends are the same? Is it handed? No. So this can’t be glucose.
How about this? Could this be mannose? If the two ends are the same, will it be chiral? No [correction: Yes]. It doesn’t have a mirror in the middle. It’s a propeller. It’s not achiral. The one that’s right at the top and left at the bottom is the mirror image of this. Not identical to it. So it could be could still be that.
Could this be mannose? Yes. That means this could be glucose.
Could this be mannose? No, because now it has a mirror in the middle. So that can’t be glucose.
So from 2a, then, we can exclude this, and we can exclude that.
So one of these must be glucose, and the other one is gulose, the one that’s turned end for end, top to bottom.
So how we going to decide that, or how did Fischer decide that? He went to Evidence 3. Which is, arabinose, in the Kiliani-Fischer synthesis, gives glucose, and xylose gives gulose. So one of them gives one of these, and the other one gives the other one of those.
So we know something now about arabinose and xylose. They can’t be this, and they can’t be this, because that, on a Kiliani-Fischer synthesis, couldn’t give either of these, and this couldn’t give either of those. So those are out. That one’s out and that one’s out for arabinose and xylose.
But what we know from Evidence 3 is that if this is glucose, that one had to be arabinose, the thing that you add another carbon here and make that. And if this was glucose, that would be arabinose. And vice versa. If it’s arabinose, it’s glucose.
So if we know arabinose, which is arabinose and which is xylose, then we know which is glucose and which is gulose, because of those connections.
So Evidence 3a is arabinose gives optically active arabitol and arabaric acid. When the two ends are the same, it’s chiral. But xylose is not chiral when the two ends are the same.
OK. Could this one be arabinose? Would it be chiral if the two ends are the same? How about that one? No. So that one’s got to be arabinose, and that one’s got to be xylose.
So we know that glucose can’t be this, because it comes from arabinose. So that one is glucose.
So he proved it!
So all these known compounds were then assigned. If you knew this was glucose, then you know that was gulose. You know that glucose and mannose differ only at this carbon. So you know which mannose is, you know which arabinose is, which xylose is. You also can figure out which one it erythrose is, by making arabinose from it.
So all these things now work. It’s like the end of a solitaire game, when everything begins to work, right?
So you now you get all the other sugars. In fact, of these sixteen sugars, the right- and left-handed ones in the bottom row, the hexose, of the sixteen Fischer made thirteen of them himself. And he made fifty artificial sugars overall in the 1880s and 1890s.
So he got others. Like he also got threose. So this was a new compound. How did he name it? Where did the name threose come from? It’s erythrose turned around, right? Sort of turned around.
Or he got ribose. Like ribonucleic acid, deoxyribonucleic acid. Where does the name ribose come from?
Professor J. Michael McBride: From arabinose. From gum arabic.
Or he got lyxose. That’s pretty obvious. Or you got talose from galactose. T A L, OK?
Or he got altrose. He was beginning to run out, so this just means “another one.” Alter, right?
Or he got idose. That’s idem, you know, the same as the one above in a list is idem. So it’s another one, right?
Or allose, which is the Greek version of altrose.
So he got all these, and he got the Nobel Prize. And he got the second Nobel Prize. The first Nobel Prize went to van ‘t Hoff. Not for chirality, but for his work in physical chemistry. But the second Nobel Prize went to chirality via Fischer, because everybody realized what a great contribution– you know, it’s not just doing it, but doing it without Fischer projections is a lot tougher. It’s to conceive that you can do it, that the evidence is there, that you apply this new theory that van ‘t Hoff came up with not so long ago, and follow it through with rigorous logic, and you’ve got it.
Rather just fiddling around, making this, making this, making this, making this, he puts it all together. He finds the crucial bits of evidence and puts them together to do this really beautiful proof. So I thought you would want to know about that.
Now, if you want to remember it, the guy who taught me Organic– I took Organic Chemistry the way you are, 50 years ago this year. And the guy that I took it from, Louis Fieser, made up this. “All altruists gladly make gum in gallon tanks.” And I remember that for fifty years. So I have no trouble doing allose, altrose, glucose, mannose, gulose, idose, galactose, and talose. It’s not a very fundamental contribution to science, but it’s very handy, if you want to teach about hexoses.
And if you want to do this one, it’s raxl. R A X L. And erythrose and threose, you just remember.
OK. So hooray for Emil Fischer.
Chapter 5. Synthesizing a Cyclobutadiene Precursor in a Designer Clamshell [00:47:16]
Now we’ve got two minutes just to show you– I’m not going to show you how this happened. It’s going to be a stream of consciousness thing. You’re not responsible for it on the exam. But I’ve got all the slides here, so what the hell, right? And it is a neat thing.
OK. So we talked about cyclobutadiene. Is it antiaromatic? Make it and see what its properties are. The problem is… So you can make it by this. Shine light on it. It closes up in a disrotatory process. And then with light, it loses CO2, and you get it. So study its properties.
The problem is that it undergoes a really quick reaction with itself. A Diels-Alder reaction, which gives this. And then that opens up, because it’s very strained, of course, and gives cyclooctatetraene. So you get cyclooctatetraene, which is sort of evidence that you had cyclobutadiene, but it’s not like knowing its properties.
So the problem is that it reacts with itself. So it’s presumptive evidence of existence. Could you do spectroscopy? So you have to make it, somehow, under conditions that it won’t react with itself. So if you made it inside a cage, like a big clam shell, so only one molecule was in there, then it couldn’t react with itself.
So they made this compound with a guest inside. So it’s two big, like a clam shell. It holds the precursor inside. There’s a mouth that you can put things in and out from. And it was done by Donald Cram, who got the Nobel Prize, and Martin Tanner, the son of Dennis Tanner, who burned his eyebrows off with a Fourth of July demonstration? They grew back again, you know, he was not permanently marred. Thomas, I don’t know.
So these R groups were big things to make it soluble. And it was that, and you can make that by thinking backwards with an aldol reaction.
So first, they make the bottom half of the thing, and how to form this big ring with sixteen carbons in it. You take resorcinol and hydrocinnamaldehyde, and then– This is very active for electrophilic substituition. So you protonate that, you have an electrophile, it forms a bond, ortho-para, you’ve got the first bond made.
Put that in, dehydrate, you can react it was another one, so you’ve got another one. Now you just keep doing it to make all those all the way around. But you could imagine lots and lots of different ways of doing it. And it turns out that the electrophilic aromatic substitution is reversible, and ultimately, this compound, the one they wanted to make this, precipitates from the equilibrated mixture in a 69% yield. That is luck. So the fish jumped into this particular boat, right?
But now they need these bonds across the top of that bottom, too. The teeth for the clam. So what they did is they put bromine on. Then they made anions here with base, and reacted with CH2 with two leaving groups. So one leaving group, the other leaving group. And now do it on all the others, and you’ve got those carbons put in.
Now you have to join the top half to the bottom half.
Oh, they got this, incidentally. They didn’t get a 69% yield. That was 5% from the tetramer. And they had to isolate it by chromatography.
So now they had to join the two hemispheres, which they put a lithium on in place of the bromine. And then they put boron on, and then hydrogen peroxide, which we’ve seen before, to make those into alcohols. We’ve seen that before.
So you see, this was really to be review of stuff we’ve done.
So now you’ve got alcohols all the way up. So you’ve got the Os that are going to do the bridging. And now you just do that same thing you did before, to put a carbon in, this carbon the same way you did that one, by having a carbon with two halogens, so two leaving groups.
And now again, when those two things come together, if you have three things that can react here and three things that can react here that can go like that, which is what you want. It could go like that, it could go like that, it could go like that. There are lots of different ways.
But it turned out that they got 40% of it. 40% of this stuff went into that. Overall, from the stuff they started, and it’s a 1% yield. But they started with a lot of stuff. So they had the clam shell.
And now it turned out they could get the guest out. This is a crystal structure, so this stuff at the top is chloroform, and this is solvent. There are two different solvents in there. But in the middle of the cage is DMF. This is that clam shell here. And these things are just put on to make it soluble, so you can work with it.
Chapter 6. The Antiaromaticity of Cyclobutadiene [00:52:52]
OK. So they could get that stuff out, and they could replace it with this. So here’s that stuff on the cover of Angewandte Chemie inside that compound. And this is the NMR spectrum. So most of the peaks come from these hydrogens in the shell. But the ones marked in red, and one that’s under here, are these four hydrogens on the starting material.
And then they photolyzed it, and they got the purple peaks. It changed. And then they photolyzed it again, and they got just one peak. That’s the peak of cyclobutadiene. And it can’t react with any others, because there are no others around. And then if you get it out again, then it will react to give cyclooctatetraene.
So this is way upfield, antiaromatic. As we talked about before.
Now, is that what causes the upfield shift? Because you’ll notice something else. If you put benzene inside, its normal position would be here. But benzene inside that clam shell is there. It shifted way upfield. Why? Because it’s above all sorts of other benzene rings.
Remember, when you have something above but a benzene ring, it’s shifted upfield. If it’s out beside it, it’s shifted downfield. But this, the one that’s in the middle is above all. So some of that shift must be due to the same effect that makes benzene go up.
So most of the shift comes from the other rings. Still, it’s about one and a half ppm above benzene, so it is shifted upfield, and it is antiaromatic.
So I think that’s what I wanted to show you. That’s the end. But the big lesson is how Fischer did glucose. Thanks.
Oh, I should say that there will be a discussion section tonight. I haven’t checked my e-mail today. I’ll e-mail where the room is. The peer tutors are doing one tonight because they didn’t do it last Sunday.
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