CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 29 - Imines and Enamines; Oxidation and Reduction
Chapter 1. Imines [00:00:00]
Professor J. Michael McBride: OK. So as we said, most of the lectures nowadays are about carbonyl chemistry. And in particular today, we’re going to talk about imines, the nitrogen analog of a carbonyl group, and enamines, which you’ll see are closely related to those, and then get onto the general topic of oxidation reduction, and how it applies in organic chemistry and in its simplest case, where it involves electron transfer. In the unlikely event we have any time, we’ll review NMR a little bit at the end.
Bear in mind, this isn’t on the exam Friday, but it will be on the final.
OK. We talked last time about a ketone going to an imine. So you have an amine. Its unshared pair is a nucleophile to attack π*. Put a proton then, take a proton off the nitrogen onto the oxygen minus, you get this carbinolamine. And then the unshared pair allows you to lose hydroxide, right? So you get this cation, an aminium ion, protonated imine, and hydroxide. But the hydroxide can take the proton away.
So that’s how you make imines. And we mentioned that last time. The imine is the nitrogen analog of a carbonyl group.
The equilibrium constant for this, however, is unfavorable. It lies toward the starting ketone. But there are some cases where the equilibrium lies the other way. For example, with OH as the R group, it lies toward the right. That’s called an oxime. Or when this complicated thing is used as the imine, the equilibrium lies to the right. And that’s called a 2,4-dinitrophenylhydrazone, or a 2,4-DNP derivative.
And these things were really important before NMR came along, and IR, to allow you to characterize compounds, because you had these ketones, liquids typically, that were very difficult to deal with. But if you treated them with hydroxylamine at the top, or 2,4-dinitrophenylhydrazine at the bottom, they created the oxime, or the 2,4-DNP derivatives, which were easy to crystallize and gave nice, sharp melting points.
So there were tables of all different ketones, and what their oximes melted at, and what their 2,4-DNP derivatives melted at. So you made a couple of these derivatives, measure their melting point, and then you knew what ketone you had, if you didn’t have NMR to help you out.
So these were very important. And you’ll see that such compounds, this kind of imine, is important in biology, as well.
Now, why do these equilibria lie to the right, when with an alkyl group it lies to the left? So let’s think about it. We have an unshared pair here on the group that’s attached to nitrogen, where we don’t have an unshared pair here.
How might that help the equilibrium lie to the right? How might it stabilize the products? The fact that you have an imine which has a heteroatom with an unshared pair next door. What special stability might you get on the right from that that you wouldn’t get on the left? Anybody got an idea? What’s good about this compound having an unshared pair here attached to the imine? Matt, you have an idea?
Student: You have resonance stabilization of the compound.
Professor J. Michael McBride: Sure. You got the π* here, and it can overlap with that. So you get special stability on the right. It’s allylic. The unshared pair is adjacent to the double bond. So you got special stability on the right.
Now, how about on the left? Does having the unshared pair on the left make any difference, that would help the reaction go to the right? Is there any reason it should be unstable to have that unshared pair adjacent to the nitrogen on the left? Bear in mind that the nitrogen itself has an unshared pair. What if they overlap?
If you have two occupied orbitals come together, they mix up and down, but up more than down. The unshared pairs repel one another, right? So it’s unstable on the left, because of the overlap of unshared pairs.
So both these factors shift the equilibrium toward being stable as imines, as oxime or 2,4-dinitrophenylhydrazone.
Now, there are other cases where imines are very important. For example, in the olden days, when it was difficult to separate compounds– it still could be difficult to separate compounds, but there are a lot more techniques nowadays, with fancy chromatographic techniques, for example.
But Girard developed this reagent. Which is, you see, one of these things that has an unshared pair next to the nitrogen, so it’ll make a nice, strong imine, stable imine? And it could be reacted, for example, with a ketone in cortisone, a steroid hormone, right? To make it soluble in water.
So without Girard’s reagent, it would dissolve, when you do an extraction, it would be in the organic layer. But if you add Girard’s reagent, it goes into the water layer. So you can separate it from other things that would go into the one or the other of those layers.
How does that work? Why does Gigard’s reagent make the ketone soluble in water? Wojtek?
Student: Does it change polarity?
Professor J. Michael McBride: What’s the nature of the polarity? It’s more than polarity. Polarity is having something that’s dipolar, plus here and minus here, right? It’s much more so than that. This reagent has a charge that you can’t get rid of, right? Because that nitrogen has four alkyl groups on it. So it can’t lose a proton and get rid of its charge. So once this is hooked onto here, the molecule is permanently a cation, right? So then it dissolves in water. It’s a cationic imine. A hydrazone. That is, hydrazone is the things that have nitrogen-nitrogen double bond, that carbon.
OK. So there’s another example of the importance of imine.
Now, you remember this slide that you’ve been reviewing for the test about chromophores. But let’s just now focus on this part down at the bottom. Notice that it’s the imine that holds the chromophore, the thing that’s going to be colored, that conjugated system, onto a protein to make rhodopsin. So that’s what holds retinaldehyde. This was originally an aldehyde with an oxygen on here. But it formed an imine, and got held to the protein. So it functions the way it’s supposed to do in your eye.
Chapter 2. Amino Acid Synthesis [00:07:16]
Imines are also important in the synthesis of amino acids. Particularly and originally in the technique developed by Adolf Strecker. Strecker was a student and an assistant later in Liebig’s lab in Giessen that we talked about last time. But in 1854, he figured out how to make the amino acid alanine by starting with acetaldehyde, reacting it with ammonium chloride and potassium cyanide.
Now let’s figure out how that happens. First you react the ammonia of ammonium chloride, which is in equilibrium, of course, with ammonium chloride, with the aldehyde to make an imine. OK? Then what’s going to happen next, you think, given the reagents you have there?
Another reagent comes in. The reagent that comes in is cyanide. What will cyanide do with an imine? Cyanide is an anion, a nucleophile. It’s a special nucleophile, because it’s nucleophilic at carbon. So a carbon has a high HOMO on it in cyanide. So where’s it going to attack? Cassie?
Student: It’ll attack the π*.
Professor J. Michael McBride: The π* of the imine. It would also attack the ketone, if you had the ketone there. OK, so cyanide comes down, attacks the imine. That makes a negative charge on the nitrogen when you formed a bond. But you could put a proton on there, right? So you get this compound. OK now that’s the first stage. Then you add strong acid and heat it up– hydrochloric acid and water, and heat it up. Now you protonate the nitrogen– the unshared pair on the nitrogen– with the acid. And now water can attack the cation to give this and lose a proton.
But now notice you have another sort of imine-like thing– carbon nitrogen double bond. And we already know that could react with water, right, to liberate ammonia. We look– the very first step was the same reaction going the opposite direction, taking the carbonyl off, putting the nitrogen on to make an imine. Now we put an oxygen on, pull the nitrogen off, right? And that’s alanine.
So he was able to make that in 50% to 60% yield this way. So such a high yield from such very simple compounds suggests to a lot of people that this could be involved in the origin of life. There are people who get into discussions and debates about how life might have begun. One thing you need is amino acids. This could have provided amino acids without any extraordinary requirements in terms of starting materials and complicated mechanisms.
And notice in passing here that this is an important thing– that RCN—triple, nitrile, carbon nitrogen triple bond– gives RCOOH. So you could make carboxylic acids by first using carbon as a nucleophile because the cyanide is an anion, and so it can displace a halide, do these other things. And then you can change it into a carboxylic acid group. So that’s a handy thing in synthesis, which is exemplified here.
Now it’s not only in the laboratory that this kind of thing happens where you go through an imine. Here’s an imine being used in a biosynthetic scheme to make the amino acid glutamic acid. So it’s the same deal as before. You have ammonia and you have NADH which we talked about last time. We talked about it in terms of nucleophilic aromatic substitution. And of course there are enzymes that control this whole thing. L-glutamate dehydrogenase is the one that does it.
OK. So first the ammonia reacts and gives the imine. And then you protonate the nitrogen, so you have a carbon cation. And now the NADH gets into the act. Because you remember, NADH was NAD+ to which H- has been added. If you go back and review that, you’ll see it. So it can go the other way, too. It can give up H- and make NAD+. So now the H minus can come down and attack that carbon, right? And you have glutamic acid.
And this is not, for mammals, an essential amino acid. They don’t have to get it in their diet, because they run this scheme themselves, right? So they make it inside themselves, using L-glutamate dehydrogenase.
Now, that’s an interesting name for the enzyme that does this. Because notice what we’ve done, is that the enzyme, using the NADH, actually hydrogenates this. It puts H2 on. H here and H here, right? The first H came in as a proton, and then the key one came in from up here as H-.
But you might call this α-Ketoglutaric acid hydrogenase. Right? The thing that does this. Why is it called dehydrogenase? Because a catalyst makes the reaction go faster in either direction, and it was first studied in the direction where it goes from bottom-left to top-right. Right? So it’s called dehydrogenase, but the catalyst just makes the transformation happen in either direction.
Now once you have an amino acid like glutamic acid that we got last time, then you could make other amino acids by what’s called “transamination.” So you have a compound like this that has a ketone in it that reacts with that amino group to form an imine. And then this thing ultimately comes apart to go back to the α-ketoglutaric acid. So you sacrifice having made glutamic acid, having made this amino acid, but you get back alanine, the one that’s CH3 here, CH, NH2. Right?
Now, what was the key step? Once you form this — form this imine – what had to happen in order to make it unzip to make alanine and go back to the ketoglutaric acid?
Notice that in this first step, we made this double bond. Obviously, if we had the double bond here and a single bond here, then if we undid the double bond by adding water, we’d put the ketone at the top carbon, up here, right? And we’d have the amino acid at the bottom.
So what’s key is to move the double bond from here to here, and the single bond from top to bottom, right? What kind of rearrangement is that? Can you think of a name for a rearrangement like that? Amy, did you say? Or somebody said. Ayesha?
Professor J. Michael McBride: It’s an allylic rearrangement, right? Where you say, take off this proton, which is adjacent to the double bond, then put it on down here. So that’s called transamination.
So again, many other amino acids are not essential. You don’t have to get them in your diet, because you have the enzymes that help make it by transamination. OK?
Now, does anybody recognize what this fragment is? It’s a fragment of a much bigger molecule. Do you know what the bigger molecule is? Pardon me? I heard someone say something.
Professor J. Michael McBride: It’s a chain of DNA. Here’s a stick diagram of it. And it’s a bunch of Cs in a row.
But let’s just look at that bit in the middle there, and let’s break the chain here, these phosphates, and so we’ll make these Os into OHs, here. So there’s the key thing that’s in the chain.
Now, there’s an interesting functional group there. That’s called hemiaminal, that functional group. So it has a nitrogen and an oxygen on the carbon.
Does that remind you of any other functional groups? What would the oxygen analog be, if you had, instead of the nitrogen there, an oxygen? What would it be?
Student: An acetal.
Professor J. Michael McBride: It would be an acetal, right? And you can see that you could take it apart the way you take an acetal apart. You have an OH, a carbonyl and here an NH, when you would have an alcohol, an OH, if it were the oxygen analog.
So you have a sugar and the base, as it’s called, right? But if that base were an alcohol instead, and you then reversed the reaction, you’d make a ketal.
So the DNA is a whole bunch of, essentially, ketals that are called hemiaminals, because an aminal would be two nitrogens. A ketal is two oxygens. A hemiaminal is a nitrogen and an oxygen. So you have a whole bunch of these essentially ketals.
So that’s a protecting group for a sugar, right? So in a sense, DNA is a protecting group for sugars, although it has other uses, as well.
Chapter 3. Enamine Alkylation and Acylation [00:17:14]
Now, you can also get what’s called [ge]alpha-substitution via a relative of the imine called enamine. So here we saw before, you take the imine, you put it together to get the carbinolamine. You have the unshared pair. You can lose hydroxide.
And you get here– and remember what happened before? The OH- took off a proton to make an imine. Now you can’t do that. Can you see why you can’t make an imine? Why can’t you make an imine here? Chris?
The imine is the C=N double bond, but without the charge, because you lost the proton. But there’s no proton on the nitrogen, right? Because we’ve started with a secondary amine, one that has two R groups on it. So you don’t have the proton to take off from the iminium ion, right?
But you do have that proton on the carbon. And notice, that’s in the α position, adjacent to the double bond. And this is very much like a ketone. In fact, it’s very reactive, because the N+ makes that double bond even a lower LUMO than the π*. So It’s something that’s going to be able to stabilize an electron pair, if you pull that proton off.
OK. So you do that. The proton gets pulled off by the hydroxide. And you have this, which is now called an enamine. It’s like an enol, right?
Now, you can make that in a 90% yield by driving the equilibrium to the right by removing the water, which was the product you just made. So you distill out the water, then you can get a very high yield of this enamine, which was shown to be a very useful synthetic intermediate. And the reason it’s useful is the unshared pair can get involved in making that carbon nucleophilic.
Remember, we just said that cyanide is an important synthetic intermediate, because it has a carbon that’s a nucleophile that can attack another carbon and form a carbon-carbon bond. In the Strecker synthesis of the amino acid, we saw that.
But here’s another one that can function as a nucleophile to make new carbon-carbon bonds. For example, it can react with an acid chloride. So it adds, association, and then dissociation of the chloride, and puts an acyl group on the α position.
Now you can bring water in, and reverse the formation of the immonium ion, and you get– what you see there is a β-diketone. It’s two ketone groups, not adjacent to one another, not α, but next adjacent, with the carbon in between. A β-diketone.
So if you want to synthesize a β-diketone, you can use this reaction. And in fact, when R is a C5 chain, this was done in 70% percent yield.
That process is called Stork enamine acylation. Acyl, remember, is RC=O. so this put RC=O on the α position of a ketone. So if you want a carbonyl group here, you can make the enamine, and then react with an acyl chloride.
But that’s not the only thing you can attack with that carbon nucleophile. You could attack RCl. Then that puts an R-group on the α position. Bring in water and hydrolyze it, and you get that. And when R is phenyl-CH2, benzyl, there’s a 55% yield of that purified compound was obtained.
But that’s not an acylation anymore. You’re putting R on, not RC=O. This is alkylation. So this is called the Stork enamine alkylation.
And there’s a much fancier kind of alkylation that was done. And all these are in the same paper. If you have an α,β-unsaturated ketone, so a conjugated ketone, now you can attack not just the carbon that has the double bond O on it, but you can attack adjacent to it, right? To the terminal double bond, there, and then form this.
So it would form, first, that enolate. Then, since the enolate could be on the α position, proton come in, and you’ve got that compound. So you can put that fancier alkyl group on, bring in water, get rid of the helping group there, and you have this diketone.
Now these are called Stork. The reason they’re called Stork is not because of the bird, but because of Gilbert Stork, shown on the left there, who was the PhD advisor of my neighbor, Fred Ziegler, who teaches Chem 220. And in fact, Ziegler was a graduate student in Stork’s lab just at the time this work was being done. So when I went to talk to him about getting permission to use this picture in the slide show, he told me some lore about it that I wasn’t aware of that I think is interesting for you to know.
So here, remember that there’s an enormous amount of lore that you just have to be around the way Ziegler was around, or to have used these things, in order to know.
So we showed this 70% yield of the Stork acylation. Right? But in fact, what Ziegler pointed out to me is that to do this, you need two equivalents of the enamine, not just one. Now, why do you need two equivalents, rather than just one mole per mole of the acyl halide?
Because the first product, remember, before you got rid of this helping group from the enamine, was that. And that’s a super carbonyl group, which means that that hydrogen is α both to this one and to this carbonyl group. So what property should that hydrogen have, unusual property?
Student: It should be very acidic.
Professor J. Michael McBride: It’s very acidic. The anion will be very strongly stabilized. So it’ll be easy for it to disassociate to give H+.
Now can you think of anything that would be a base in here? Notice, we identified the enamine as a nucleophile. But nucleophiles are also basic. So this product transfers a proton to a mole of enamine. So that uses a second enamine up. So you need two of them, if you want to get the reaction to go. Otherwise, the first half of the product chews up the rest of the starting material, right?
But there’s a cheaper way to do it, which is to put in some other base that can eat up the proton. So in fact, they use – triethylamine is added in to soak up the H+.
And that wasn’t the only bit of lore Ziegler told me about. If you go to Wikipedia and look up enamine alkylation, you’ll get this as your example of what the product is. But in fact, that’s not the product. The product is this, in 71% yield, from doing this reaction. So where in the world did that come from?
Well, remember, before we hydrolyzed that, it was in that form. A super carbonyl type of group. And now here, that carbon is α to the carbonyl, so you can make an enolate there.
So now this carbon anion can attack π* at that carbon. So you form a bond from here to here.
Now of course, in order to reach, you have to flop this over 180°. So if you do that, then you get this, right? This is the new bond that was formed to the carbon adjacent to the nitrogen.
But notice with this, you can make an anion there, take a proton off. It’s an enolate. Where would the proton go? It would go on the nitrogen here.
And now we have a leaving group and a nucleophile, and that’s the product. Right? Elimination from that compound.
So that’s another bit of lore that you don’t get what Wikipedia gets, you get what Stork found in 1963.
OK. That’s what we want to say about imines. And now onto oxidation and reduction.
Chapter 4. Oxidation and Reduction as Electron Transfer [00:26:48]
Now, the most straightforward kind of oxidation-reduction– oxidation, where does the word come from? What does oxidation mean? Originally, what did it mean? Who thought up the term, oxidation or oxygen? Somebody? Yeah, Linda.
Student: Oxygen made things acidic.
Professor J. Michael McBride: Yeah, who had the idea that oxygen made things acidic?
Professor J. Michael McBride: Scheele didn’t have the idea. Scheele discovered some of those things. But the name came from France, came from Lavoisier and his colleagues. OK. So oxidation was reaction with oxygen, right, which made things acidic.
Where did the word reduction come from? What’s reduced? What does reduced mean? Duco, ducere in Latin means to lead. So to re-lead is to lead back again. So you’ve gone to oxidation. You’ve put oxygen onto something. And then you reduce it, you lead it back to what it was originally, take the oxygen off. When do you think that term originated? Where in technology would you be involved in removing oxygen from some material? It’s very old; thousands of years.
You take a metal oxide, treat it with a reducing agent, carbon, pull the oxygen off, make CO2, and you’re left with the metal. That’s how you get metal from metallic ores. Iron, for example. OK. So that’s what reduce is.
But nowadays, we understand it in terms of electron transfer. To oxidize something is to remove electrons from it. To reduce it is to put electrons on. And those actual reactions occur in organic chemistry. For example, metals are something that are often used as reducing agents, like solid magnesium. And an alkyl halide can take an electron.
Where would the electron go into methyl bromide if you took an electron out of magnesium? Where is the low LUMO? Antonia?
Student: C-Br bond?
Professor J. Michael McBride: Right. σ*C-Br. So we have reduction, σ*. So an electron comes out, we get it out of the metal. So the metal has lost an electron, and the CH3Br has gained one. How will that change the properties of methyl bromide, having an extra electron in σ*.? Antonia?
Student: It will make it more reactive.
Professor J. Michael McBride: In what sense, more reactive? What specific bond will change? What do you call σ*? What kind of orbital?
Student: I forget the word. Anti-bonding.
Professor J. Michael McBride: Anti-bonding. Why do you call it anti-bonding?
Student: Because it usually breaks, or it makes less attraction.
Professor J. Michael McBride: So what’s going to happen?
Student: So it will break the bond.
Professor J. Michael McBride: So the bond breaks. So you get CH3 dot. There were three electrons in that σ and σ*., two in σ, one in σ*.. So now those three electrons are a pair on bromine, bromide, and a single electron, a radical on methyl. Right?
Now you can get– the magnesium atom comes out and reacts with the methyl radical to give methyl magnesium bromide.
And that’s called the Grignard reagent. And it behaves a lot like you might expect CH3- to behave. It’s a carbon nucleophile, and it’s a strong base. Of course it’s not really CH3-. It’s actually the σ pair of electrons in here. But because the magnesium is so high in energy, its orbitals, this is an unusually high HOMO. And we’ve talked about it before. And lithium can do the same thing that magnesium did here.
Grignard got a very early Nobel prize, about 1908 or something like this, because it was so useful to have ways to make alkyl nucleophiles in synthesis. 1912, pardon me, he got the Nobel prize.
OK. Now another example is a ketone. So it can react with magnesium. Now where will the electron go? Rahul, what do you say? What’s the low LUMO in acetone?
Professor J. Michael McBride: So we’ll put an electron in π*. So now we can write that as a charge on oxygen and the radical on carbon. It doesn’t break the bond, because there was a double bond to begin with. But it went down to a single bond now.
OK. Now you can do that with another acetone on the surface of the magnesium and get that. And now you have Mg+2 between them.
But now since the two are attached to the same magnesium, the radicals can react with one another to make a bond. And if you then add water to put protons on the Os, you can get pinacol. And this is called the pinacol reduction. And we’ve seen before that if you treat pinacol with H+, you get pinacolone.
So here are two examples of reducing an organic compound, actually by electron transfer. But much more often in organic chemistry, we’re not talking about actual electron transfers. We’re just using it as a bookkeeping scheme.
Chapter 5. Oxidation and Reduction as Bookkeeping: Atomic Oxidation States [00:33:04]
Now, what do we want to keep books about? We arbitrarily assign shared electron pairs. We know from last semester the nature of a bond, and how electron pairs get shared. But for bookkeeping purposes, we pretend that if there’s a bond between two different atoms, it’s an ionic bond, and those two electrons– for bookkeeping purposes, even though they’re practically 50-50 shared, we assign them to the more electronegative of the two elements.
There’s nothing realistic about this. It’s just a scheme for bookkeeping. But it turns out to be handy, as you will see.
So let’s look at these two – the ketone and the alcohol here, and let’s figure out what the oxidation states of the various atoms are. So let’s start.
We’re just doing bookkeeping here, and I’ll tell you why. It’s because it will be very helpful in choosing reagents, or remembering reagents. Right? Because we’ve classified reagents as HOMOs and LUMOs and what they can react with. But there’s a completely different way of classifying reagents, not as having unusually high HOMOs, or usually low LUMOs, or both, or neither. But you can classify them as to whether they’re oxidizing or reducing agents.
So when you think, I need to make a transformation from here to here, I have to choose a reagent. It can be very helpful if you can exclude two-thirds of the reagents and think only about the others, or if you’re trying to design a new reagent, to design it sensibly, so that it could plausibly give this transformation, right?
So you can think about HOMOs and LUMOs, and we do do that when we try to think, for example, for a ketone, we’d think of having a high HOMO to attack the carbonyl, or a low LUMO, like a proton, to protonate the oxygen. That’s mechanistic thought. But we can throw out two-thirds of the candidates if we think about oxidation and reduction, so that’s why this bookkeeping scheme is handy.
So let’s look first at this carbon, and I’ll start you out here. OK. So this carbon has a bond to carbon. Now, that’s a carbon-carbon bond. So one electron came from the carbon we’re talking about, one came from the other carbon, but we now split it 50-50, since they’re the same electronegativity. So this carbon neither lost nor gained an electron by forming the carbon-carbon bond.
But it also formed three bonds to hydrogen. Now, carbon is more electronegative than hydrogen, right? So those three pairs, each of them has one electron that came from carbon, one electron that came from hydrogen. But for bookkeeping purposes, we assign them to carbon.
So now carbon has three electrons that it didn’t come in with. It came in with four, shared one with another carbon– bookkeeping keeps that one, but not the other one– made three bonds to hydrogen, get an electron from each of those. So that carbon has an oxidation state of -3. Everybody see how we did that?
OK. So if we see that, now tell me what the next carbon is. What should its oxidation state be? Ruoyi? This carbon here, the second carbon from the left. What bonds does it make?
Student: It has two carbon-carbon bonds.
Professor J. Michael McBride: How many carbon-carbon bonds?
Student: I guess four carbon-carbon bonds, so they’re all the same, so…
Professor J. Michael McBride: It makes four carbon-carbon bonds. Both of them split the electron, since it’s the same element. So what should its oxidation level be?
Professor J. Michael McBride: OK. How about the next one? Ruoyi, you’re on a roll here, so keep us going.
Student: There’s only the oxygen, so…
Professor J. Michael McBride: It has two bonds of carbon, but it has two bonds to oxygen. And those electrons go to who?
Student: The oxygen.
Professor J. Michael McBride: Those went to oxygen. Those two electrons of carbon went to oxygen. So what’s the carbon?
Professor J. Michael McBride: +2, OK. How about the next carbon?
Student: One bond to carbon and then three bonds to hydrogens, so that would be -3
Professor J. Michael McBride: So it’s -3 again, like the methyls on the left. OK. Now let’s look at the right. That one’s -3. How about the next one?
Professor J. Michael McBride: That’s 0. How do you answer so quickly?
Student: It’s the same as the other.
Professor J. Michael McBride: It’s the same as here, right? So that one’s 0. How about the next one?
Student: 0 as well.
Professor J. Michael McBride: 0. Because it was -1 with respect to the hydrogen, but +1 with respect to the oxygen, so it’s 0. And finally?
Professor J. Michael McBride: Finally -3, same as on the left. Now, notice that almost all of these were the same, left and right. So we were really wasting our time, except for getting some practice, in doing everything, except the ones that changed. So you look at the elements that change their bonds, right, to see what happened. And if we do that now, we see that only this one and this one changed, and it changed from +2 to 0. So it got reduced.
So that question mark now, we know that’s a reduction. So we had to choose a reducing agent. So now, if we had a way of classifying reagents as being oxidizing or reducing or neither, then we would know we would only have to look among reducing agents if we want to do this reaction. Unless the reaction might be a really complicated thing, where you do an oxidation and then two reductions or something like that. But for simple things, it should be a reducing agent.
OK. So let’s try some elements, or some compounds, and see whether we would expect them to be oxidizing, or reducing, or neither.
So let’s start with HCl. There’s a good reagent. Now, what’s the oxidation state of hydrogen in this? Derek? In HCl what’s the oxidation state of hydrogen?
Professor J. Michael McBride: It’s +1, because it’s shared its electron with chlorine. That means, for bookkeeping purposes, it gave it to chlorine. And so hydrogen is +1. How about chlorine, Chris?
Professor J. Michael McBride:-1. Now, what would those be if they were in organic compounds, if they were attached to carbon? What would hydrogen be if it were attached to carbon?
Student: It would also be +1.
Professor J. Michael McBride: It would be +1. What would chloride be if it were attached—chlorine if it were attached to carbon?
Professor J. Michael McBride: OK. So these have their normal oxidation levels. HCl has the same oxidation level for H and Cl as if they were attached to carbon. So it’s neither oxidizing nor reducing if it goes into carbon. Right? So that’s neither.
Now let’s take Br2. Po-Yi, what’s the oxidation level of bromine in Br2, in elemental bromine?
Student: Are they both 0?
Professor J. Michael McBride: I can’t hear.
Student: Are they both 0?
Professor J. Michael McBride: Yeah, they’re both 0. They shared a pair, but they get it 50-50 for bookkeeping, because they’re the same element. So they’re both 0. That one’s 0, that one’s 0.
OK. Now what if they were attached to carbon, what would they be?
Student: Negative one.
Professor J. Michael McBride: -1. They would take electrons from carbon, those atoms in bromine. Right? So are they oxidizing or reducing agents? Is Br2 oxidizing or reducing? If it reacts with carbon, it’ll take electrons from carbon for our bookkeeping purposes. Right? So it’s a reducing agent, it takes the electrons away, right? It… uh… uh… Oh, pardon me, I said the wrong thing. It’s an oxidizing agent. It takes electrons away. It makes the carbon become plus.
So Br2 is an oxidizing agent. And we’ve already seen, of course, Br2 act as an oxidizing agent. For exasmple, it reacts with ethylene. What’s the oxidation level of carbon in ethylene?
Well, what do you say? In CH2-CH2, what’s the oxidation level of the carbon atoms? There are two bonds, carbon to carbon, but those don’t take electrons away or give electrons, right, because it’s the same element. But there are two bonds to hydrogen on each carbon. And what do those do? Hydrogen does what, for our bookkeeping?
We consider the bond ionic for bookkeeping. H plus, C minus. So the carbon got an electron from each hydrogen. Right? So what should its oxidation level be of the carbon on the left side?
Professor J. Michael McBride: -2. And same on the right.
Now, we went through HOMOs, LUMOs and so on, and saw how it reacts with bromine. Remember, bromonium ion, bromide comes in, and so on. That’s the mechanism.
But for purposes of oxidation and reduction, they’re now -1, right? Because the bromine canceled one of the hydrogens. So we went from -2 on each carbon to -1. So if we want to go from ethylene to dibromo… to di-X-ethylene, we need an oxidizing agent, like Br2. Not Br-. NaBr isn’t going to do it. It has to be Br0 that’s going to do it. You need an oxidizer.
Now, how about CH4? What’s the oxidation state of carbon in CH4? Amy?
Professor J. Michael McBride: That’s -4. And the hydrogens are all, Amy?
Professor J. Michael McBride: +1 for each of those. Now, should that be oxidized or reduced? Well, certainly it reduces O2. What reaction do you call it when you reduce O2 with methane? What do you call it? You call it fire. Right? You get CO2 and water. OK?
You can reduce Fe2O3 at great heat. That’s the original reduction. Getting a metal from the metal oxide. You could react it with Br2 hard in free-radical halogenation. But generally, methane is not very reactive, except under very special circumstances. It doesn’t have a HOMO/LUMO, but we could classify it as a reducing agent, sensibly.
How about LiAlH4? So this is a salt. Li+, right? So lithium is in what state? It’s obvious. +1, right?
Now aluminum. Aluminum is forming four bonds to hydrogen. Now how are we going to do that? OK. What we can regard this as is aluminum +3, and the hydrogens all being -1, right? Except that we also have a minus charge there. So the aluminum would be +4 with four hydrogens, because hydride is more electronegative than aluminum. So the aluminum would be +4, with respect to the four hydrogens. But then there’s an extra electron. It’s AlH4-. So that we put on the aluminum, and we have Al+3.
Now, what’s abnormal? Aluminum is a metal, not very electronegative, so it expects to be plus. trivalent, +3, that’s reasonable.
What element here is unnatural? That is, not what it would be if associated with carbon?
Professor J. Michael McBride: Hydrogen! It’s H-, not H+. So that’s going to make this a reducing agent, because of the H-.
How about sodium hydride? There’s an easier one. Karl, what’s the oxidation state of sodium in this?
Professor J. Michael McBride: +1. And hydrogen?
Professor J. Michael McBride: -1. What’s unnatural?
Professor J. Michael McBride: Hydrogen. So it’s a reducing agent, OK?
How about potassium metal? Nathan what do you say, in potassium metal, what’s the oxidation state of potassium?
Professor J. Michael McBride: 0. Is that what it is with respect to carbon, if you had a potassium-carbon bond? What would it be?
Student: It would be +1.
Professor J. Michael McBride: It would be +1, if it were with carbon. It’s less electronegative than carbon. So is it an oxidizing or reducing agent?
Professor J. Michael McBride: No, it’s able to give up an electron, right, to go from 0 to +1. So that’s another reducing agent.
How about KCl? Nathan help us with that one. What’s K?
Professor J. Michael McBride: +1. And chloride?
Professor J. Michael McBride: What are they with respect to carbon?
Student: The same.
Professor J. Michael McBride: So where should we put it? Which category? Oxidizing, reducing, or neither? Neither.
How about RSSR? What’s the oxidation state of sulfur? Lauren? Sulfur’s more electronegative, for these bookkeeping purposes, than carbon. So with respect to the sulfur-carbon bond, it’s more electronegative. It would be -1.
How about sulfur-sulfur bond?
Professor J. Michael McBride: 0. So what is it? The sulfur?
Professor J. Michael McBride: It’s -1. Same for the other one. So that’s an oxidizing agent, because if it were associated with two carbons, it would be -2. It would get an electron from carbon. So as compared to RSSR, right RSR, it’s an oxidizing agent. So it oxidizes things. Like two RSHs could be oxidized to this by this. So RSH is a reducing agent in that reaction.
In that reaction it’s a reducing agent. Did you ever see such a reaction?
Student: The sulfide bridges?
Professor J. Michael McBride: Yeah. The disulfide bridges are the permanent waves in hair, right? That’s exactly the reaction we talked about. We talked about it in HOMO/LUMO terms. But you know in order to make the disulfide bridge, you need an oxidizing agent. Something has to be the oxidizing agent, and it could be this.
CrO3 I think I just have two more here, let me finish it. That’s +6, that’s -2.
Now, there’s also a stable ion like this where it’s Cr +4. So Cr +6 can become Cr +4. That means it’s an oxidizing agent, CrO3, and we can put it over there.
Water, those are the normal oxidation level, +1, -2. So it’s neither, except–
Well, no. Actually, let’s quit it here. We can carry on next Monday.
So there’s a review session this evening, right? 8 o’clock– We said 8 to 10:30. So some people couldn’t come early, some couldn’t come late, so I’m going to do it for two and a half hours.
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