Monday, May 22, 2006

Lecture 018: Chirality part 3

Lecture 018: Chirality part 3

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Jean-Claude Bradley: I had a request for a question that I will start with.

[Background noise]

Jean-Claude Bradley: This is one of the ones that we skipped over. That I said I would be happy to revisit if there was a question about it. This is where we need to determine the different number of hydrogens and therefore the different number of products we would get from a free radical halogenation reaction.

Jean-Claude Bradley: So there are a lot of different hydrogens here. In fact, I hope I have enough colors for all of them. First of all the methyl groups, all the three Hs on each methyl group are obviously going to be the same, but the three methyl groups that are listed, they are all different. OK? Because the two methyl groups that are on the ring, one of them is going to be pointing up, one of them is going to be pointing down, and the ring is not symmetrical horizontally so all the three hydrogens, all of the nine hydrogens, are going to different.

Jean-Claude Bradley: In the red I have one population, and you have another population here. Then another population here. OK, so also we don't have a symmetry. This is drawn at an angle so you are kind of looking at the molecule on the side because of the two methyl groups that I just circled that are on the ring, that basically breaks the symmetry and so we have the hydrogens in the front being different than the hydrogens in the back. In green here I would draw a different hydrogen that would be different from the one in yellow.

Jean-Claude Bradley: And the same thing for the other two hydrogens that are on the ring. One of them is pointing up, the other one is pointing down. So those two will be different from each other, and the other four hydrogens will also be different from each other for the same reason.

Jean-Claude Bradley: We would have, and one more I forgot here, pointing out. OK, so we would have one, two, three, four, five, six seven, eight, nine, ten, eleven, twelve. Twelve different hydrogens. The consequence of that would be twelve different products, and let's see, how many major products would we have for this reaction? Let's say you did the reaction with bromine and the presence of light. We have twelve total products, how many are major and how many are minor? Yeah?

Student: One major?

Jean-Claude Bradley: One major, I can see more than that. You can see three? Well let's start pointing them out. So which one, which color would be one of the replacements?

Student: For major?

Jean-Claude Bradley: For major.

Student : [Unintelligible]

Jean-Claude Bradley: The yellow one?

Student : Yes.

Jean-Claude Bradley: OK, so let's see. I guess I almost have no colors here. So, this one is tertiary and so it will be major. So if you were to put a bromine there it would be one of the major products. Where else?

Student : [Unintelligible]

Jean-Claude Bradley: Which color?

Student : Right next to the green one.

Jean-Claude Bradley: Yeah. OK. So this one here?

Student : Yeah.

Jean-Claude Bradley: Yup. That one is also tertiary. And what else?

Student : The green one is also tertiary.

Jean-Claude Bradley: And the green one, OK. And everything else would be either secondary or primary, and so we would have twelve products, three major products and nine minor products. That's what happens when you start to break the symmetry, you get lots and lots of different products. Were there any other questions on that?

Jean-Claude Bradley: These three destructions are a little bit harder to do because you have to take into account perspective but it's really the same principle as doing the straight chain.

Student : [Unintelligible]

Jean-Claude Bradley: yea.

Student: The ones that you do with the straight box, because one is the axial and the one is the equatorial, is that why there are two different types like the ones that you drew over there.

Jean-Claude Bradley: Well it's a little bit hard to talk about equatorial and axial because this is not a chair confirmation.

Student : Ok, so the one pointing down [unintelligible].

Jean-Claude Bradley: What you are trying to draw, is that you have SP-3 hybridized centers at each point, so I'm putting a wedge to kind of make it obvious that it's coming out towards you. Because when you draw these on a two dimensional surface, you can have the optical illusion, some people may actually see it from a different angle and so I'm just drawing the wedge to show you that I mean that that one is actually coming out. In case you did not see the perspective the way that I was trying to draw it.

Student : So, the other ones, can you make it clearer for us, we could draw them as wedges with dashes? [Unintelligible]

Jean-Claude Bradley: Well the other ones are pretty close to the plane I would say, expect for the one that I drew the dashed wedge which is definitely going in the back. But otherwise, no, I don't think any other one, it would really?" I don't think there are any other ones that are clearly going in the back and clearly in the front. In the way that I drew this.

Student : So [unintelligible] symmetry. [Unintelligible]

Jean-Claude Bradley: When you have symmetry, right? If I didn't have the methyl groups then basically the green hydrogens would be the same as the yellow hydrogen. Because I could take the molecule, flip it over, and I couldn't tell the difference. So, anything else? Any other questions, actually, on chapters one to four?

Student: [Unintelligible]

Jean-Claude Bradley: Could I have quiet please? It's hard for me to listen to the question.

Student: Can you kind of explain how it's tertiary? I know it has to do with the carbon...

Jean-Claude Bradley: OK, tertiary. What I'm looking at when I am saying that this hydrogen is tertiary. Let's say I'm looking at a yellow one, I'm looking at the carbon where that hydrogen is connected and I'm counting how many carbons are directly connected to that carbon. Let me see. I'll use another color here. If I'm looking at the yellow one, I'm looking at that carbon and I count one, two, three carbons.

Student: [Unintelligible]

Jean-Claude Bradley: Yes. That's the definition of primary tertiary, secondary tertiary. Anything else?

Jean-Claude Bradley: All right. We had left off with, let's see, looking at compounds that have two chiral centers in them and we had talked about diastereomers and entromers, and I think the one thing that we didn't mention is how you would actually name these things. So let me actually draw back a Fisher projection. And, just to be different, we'll make a slightly different molecule.

Jean-Claude Bradley: So this is a Fisher projection, so once again I'm going to draw three lines, which means equivalency. So I am going to re-draw this in a way, because we just started to use Fisher projections, kind of to remind you what they mean. All of the horizontal bonds are coming out towards you. And the vertical bonds are in the back of the board when possible.

Jean-Claude Bradley: Clearly I can't put the central carbon-carbon bond between carbons two and three, both behind the board, because it is the same bond. So when you are actually looking at a Fisher projection that has multiple chiral centers in a row, what you're doing is it's actually curved. You can think about it that way. So I can't really draw it exactly the way that it would appear if you were to make a model of it. But when you analyze the Fisher projection for R and S, you would definitely think of the horizontal bonds as being behind the board, I'm sorry, the vertical bonds being behind the board and the horizontal ones being in the front. Think of this equivalency when you have two chiral centers.

Jean-Claude Bradley: If you wanted to name this molecule, the first thing we have to do is to figure out how many chiral centers we have. And we have two carbons that have four different groups onto them, so we have two chiral centers to name. When you name them, let's say I'm naming the first one. In fact let me put a different color for these. So I've got a yellow chiral center, and I have an orange one. So if I'm trying to figure out if the yellow one is "RS??, I'm not concerned that the other one is chiral or not. Ok? When I'm looking at the yellow one, I'm simply asking, "OK I've got four different groups??, I'm going to have a certain order or priority in those four groups. The lowest priority group is the hydrogen and the highest priority group between those three would be what?

Student : Bromine.

Jean-Claude Bradley: Bromine. That one is easy because it's bromine versus carbon versus carbon. So we're going to put a one next to the bromine. Now when we go to the next groups we have a CH-3, and the other group that we're looking at is actually this entire thing here, the left, this whole fragment that is the remainder of a molecule. So if I look at the shells for CH-3, I have a C followed by H, H, H. Whereas for the bottom group I have C. That C corresponds to that chiral center, and then that carbon will be connected to three other things: an H, a CL, and a carbon. Does everybody see that?

Jean-Claude Bradley: When I'm now comparing the other two groups, the two carbons from the first shell cancel out and what I'm comparing next is the highest atom from the second shells. The highest atom from the CH-3 would be an H, and the highest atom in the bottom fragment would be CL. So I'm comparing the CL versus the hydrogen, and, for that reason, the bottom group would be number two in priority, and the methyl would be third. As I go one, two, three, I find that I am going clockwise. Clockwise is normally R, but the lowest priority group is in the front, so this is actually S. So I have determined that carbon two has an S configuration. When you guys do this on a piece of paper what I very strongly recommend you do is to copy the molecule to analyze your second chiral center, because otherwise you're going to end up with a mess, especially if you don't have different colors, it's going to be truly a mess to try to figure out what's happening.

Jean-Claude Bradley: So I am going to do that to determine the bottom one here. Well in fact I can use the one on the right here, it's not a problem. If I'm analyzing the one on the right then the lowest priority group would be H, and, using the same analysis, chlorine would be number one, followed by this whole group here that has the bromine, and then the ethyl. I am going counter clockwise, counter clockwise is normally S, but the lowest priority group is in the front, so this is R, this one. When I want to name this compound, what I'm going to do is to specify, so this is one, two, three, four, five, this is pentane. I'm going to specify that the two position is an S. I would write 2S, 3R, and then I would name the compound. So that would be 2-bromo, 3-chloro, pentane. So that's how you can unambiguously name chiral compounds.

Student : 2-S or 3-R what determines what it is?

Jean-Claude Bradley: Well you have to number from the side that will give you the lowest number. If I number from the top methyl, that would be one, two, three. If I started to number from the other end it would be three, four, which you can't do.

Student : ok.

Jean-Claude Bradley: Ok. That's basically all there is to naming these things and we'll do some more examples in the problem set. Now we're going to look at reactions involving chiral centers.

Jean-Claude Bradley: The first one we will look at is creation of a chiral center. The reaction of interest here is this. What happens when I take butane and react it with light and bromine? What's the major product from that? We saw from the last chapter that the major product would be a substitution of the secondary position. But now what we didn't pay attention to in the last chapter is that when we do that reaction we create a chiral center, because now that carbon has four different groups. It didn't in butane. Butane is not chiral, it has no chiral centers, but that product has a chiral center. Knowing what we know about the mechanism of that reaction and knowing what we know about chirality now, can we predict what the products are going to be? Are we going to get the R isomer, the S isomer, are we going to get a mixture of the R and S? Or is it not possible to predict what's going to happen? When we try to predict consequences of reactions, we're going to use the mechanism we know. We're going to step through it and see what actually comes out of that.

Jean-Claude Bradley: To work out what happens in this reaction we are going to go through the steps. The first step is always the same. It is the reaction of light --the reaction of bromine with light-- to generate two bromine radicals. The second step, the first propagation step, we will have the reaction of the bromine radical with butane. I'm only drawing the hydrogens here that I'm going to be doing a bond cleavage with. The first thing that we need to think about is that this bromine radical is approaching the butane, we already know that it's going to prefer to react with the secondary hydrogens, but now we can see that there's actually a choice of two hydrogens that he can actually extract. Is it going to make a difference which hydrogen that it extracts at this point?

Student : [Unintelligible]

Jean-Claude Bradley: Let me draw the radical and we'll take a look at the question a little bit more carefully. When I draw this radical, what is the proper way of drawing it? In other words, what is the hybridization of that radical? Of that carbon? What do you think? We never actually asked that question.

Jean-Claude Bradley: If the single electron is defined as a group of electrons it would be SP-3. But I never said that a single electron is considered a group of electrons. So, in fact, based on what we know, there's no way for you to know what the hybridization of that center is. It turns out that you don't count a single electron. Now that's something new that you know and that's going to have consequences. So that single electron is actually in the leftover P-orbital of that carbon. The radial is SP-2 hybridized, which is why I drew it this way. It's flat, 120 degrees, and the radical is in the leftover P-orbital that's on the top and the bottom of that flat structure. If you have an SP-2 center you can't have chirality at that point. In other words, I can't draw this. I can't draw the groups in any order that when I take their mirror image I can't superimpose them. And the reason is that I could take it, I could just flip it over, go from one flat structure to another flat structure, and re-order the group in any way that I want. This is important because that has a very significant consequence with respect to predicting what's going to happen here. So we know therefore, from this first step, that it makes no difference which one of those two hydrogens gets taken out. We're going to get the same radical, and, furthermore, we know that that radical is not chiral. There's only one possible radical from that reaction.

Jean-Claude Bradley: In the next step, the second propagation step. So I start with that radical. Let me start labeling this stuff here. This is SP-2, make sure we're clear about that, and we are going to react it with the bromine molecule.

Jean-Claude Bradley: OK, so, I told you where the electron is; it's in the P-orbital, which is either on the top or the bottom. So, when the bromine actually comes and reacts with it, there's now a choice: the bromine can attack from the top, or it can attack from the bottom, and that choice is going to determine the chirality of the product because at that point we are going to make a chiral product. We are either going to make R or we are going to make S; so, what determines whether you make R or S is the approach of the bromine relative to your flat radical. That's where the decision happens.

Jean-Claude Bradley: So, I'm going to draw here two possibilities. And of course the - actually sorry, I made a mistake, it's HBr. Second step we get the bromine radical so, basically, at this point it attacks from the top or it attacks from the bottom and we're going to get those two chiral centers and I drew them like this out of convenience. The only thing I did is I switched the H and the Br. I could have switched any two groups on that tetrahedral center, however, this is the easiest way to tell that these are mirror images and andromers, so when I can I am always going to draw andromers like this actually this is. Unfortunately I do not have an eraser on this thing. Let's try that. Ah, very good.

Jean-Claude Bradley: All right, thank you. So I have Br on the left, H on the right. So there are my two entromers, and the question we have to ask ourselves is "is there a reason that the bromine would tend to attack from the top versus the bottom?" I have a flat structure, perfectly symmetrical; there is not any conceivable reason why the bromine would attack from the top or the bottom, so you would expect to get just random attacks. We would predict, based on our knowledge of all of this stuff, (the mechanism, the geometry) we would predict that we would get fifty percent - exactly fifty percent of one and the other, and that's exactly what we get.

Jean-Claude Bradley: Fifty percent of the R and fifty percent of the S, so this ratio of getting exactly the same amount of R and S happens very often in reactions, and so this ratio has a special definition, it is called a racemic mixture.

Jean-Claude Bradley: It just means fifty percent R, fifty percent'or, I shouldn't say that. I should say fifty percent of each enantiomer. One of the consequences of a racemic mixture is because one enantiomer will rotate and polarize light exactly the opposite direction as the other, the racemic mixture has no optical activity - ok, so there's one consequence if you have exactly the same amount of each, it will not rotate plain polarized light.

Jean-Claude Bradley: Ok, any questions about the creation of the chiral center? Ok, lets move on to the next one which is reaction at a chiral center.

Jean-Claude Bradley: So, here's a molecule that's chiral - it has one chiral center and lets say we start with this molecule and we react it with bromine in the presence of light, so, if we look at the major products, we see that there is only one tertiary center so we would expect the bromine to go on that center.

Jean-Claude Bradley: However, that's the chiral center so the question is: are we going to get a hundred percent of this or are we going to get a racemic mixture, or are we going to get inversion at that center? So, these are the options that we have. That's the question. So, we're going to step through the mechanism to see if we can predict what the answer would be if we were to do that reaction.

Jean-Claude Bradley: OK, so, again, we go through the mechanism, generate our bromine radicals. So now there is no question about which hydrogen will be attracted. In the last one we had a choice between two, now there's no choice, there's just one hydrogen so now the fact that we know the hybridization of the radical is of great importance, because that would actually change the answer, because we know that the radical is flat, we know the radical is not chiral. It doesn't matter which enantiomer we start with, we are going to destroy the chirality at this step right here.

Jean-Claude Bradley: So, left hand side is chiral, right hand side is not chiral. Now the problem gets solved in the same way as the previous one. Now again I have my flat radical. I got the bromine coming in. It has an equal probability of coming in from the bottom or the top, so again I would predict getting my racemic mixture.

Jean-Claude Bradley: The only product we would have - a little bromine radical you could generate here, so this has some interesting consequences, because what it tells you is that you start off with something that's chiral, and once the reaction is finished you have a racemic mixture that is not chiral. What makes this interesting is that you could actually monitor this reaction in a polarimeter. As the reaction proceeds, you would see that plain polarized light would initially be reflected. As the chiral material gets destroyed, it would go to zero so as you put together all these tools, that's how you start to design experiments and figure out what's going on and test out your mechanisms. We couldn't do that in the last example because there was nothing chiral to begin with or end with but here we do, we destroy the chirality. Any questions about this one? Number 2? All right, number three is a reaction next to a chiral center.

Jean-Claude Bradley: So we're using this molecule which is a Fisher projection. We want to react it with bromine and light and we want to look at the major products, so the reason we put a phenyl ring and a t-butyl group and a methyl group is we want to make sure that the major product is at the secondary center, up top here, so nothing else will be the major product except for replacing one of those two hydrogens. So the question is going to be, of course, what happens the possibilities are as follows.

Jean-Claude Bradley: ok there are those two secondary hydrogens and we can imagine replacing either one of those, and the question we want to answer is: do we get one completely and not the other? Do we get a mixture of them? If we get a mixture can we predict what the ratio is going to be between them? So, let's analyze it the same way. Yes?

Student: Is it [unintelligible]

Jean-Claude Bradley: Ah, is it?

Student: [Unintelligible]

Jean-Claude Bradley: So, it isn't, right? So you stole some of my thunder there, because that was one of the questions at the very end. You're right, they are not mirror images so they are not enantiomers, and that is actually going to be a very important consequence when we start to look at the ratios of these things. Let's do the reaction.

Jean-Claude Bradley: The first step is to generate the radicals. Now normally when you draw a Fisher projection you have an intersection, you have a chiral center. In this case you'll notice that the one with the two H's is not chiral, but I drew it that way because it is going to become chiral. So you rarely would use that unless you had a good reason to, and in this case we have a good reason because that is where our chiral center is going to be and there is only one chiral center in this molecule. Ok, so the bromine radical comes in.

Jean-Claude Bradley: It has a choice between those two hydrogens. Let's ask the question: does it matter which of those hydrogens it takes off? It doesn't, and the reason is because the radical will be the same.

Jean-Claude Bradley: Ok, so the bottom part here is the same. Now the top part, we have to be careful how we draw it because it's no longer SP3 hybridized, so it's no longer an intersection in the fisher projection; it is now going to become flat, so I have to draw it flat, and I got a radical, so plus HBr so, the next question is going to be for step number 3. I'm going to take that radical, so I'm going to have to use a new page here.

Jean-Claude Bradley: All right, so now we are going to attract this, and what we said in the previous two examples was the radical is flat. It's not chiral, so when the bromine comes in it has an equal probability of attacking from the bottom or the top. Is that still the case here? Is there an equal probability of the bromine coming in the top or the bottom? Anyone have an idea about that? It is not exactly the same scenario. When your radical is flat it looks absolutely identical from the bottom and from the top. It's not chiral, remember that, it's not chiral. This thing is chiral, the chiral center may not be at the radical, but it is chiral, so that means that when its going to be, when its going to twist to its lowest energy confirmation, that is going to look different from one side than the other. You got a big phenyl ring there, you got a big t-butyl group that's going to block the attack more from one side than the other so because this radical as a molecule as a radical is chiral, I am not going to get 50-50 mixtures here. I do know that I will get a mixture, so I will likely get a mixture but I can't predict based on what we know right now what the ratio is going to be between the two. The best I can say in this example is that we get a mixture of diastereomers.

Student: [Unintelligible]

Jean-Claude Bradley: Well, it's something that actually may be very difficult to draw on paper in order to actually answer the question, you'd have to make a molecule model and twist it around in such a way that you could see that the steric interactions are minimized. Remember, you've got a lot of single bonds here; the Fisher projection is a convenience: it doesn't tell you the way the molecule actually looks in 3-dimensional space. It's just a convenience. All it tells you is that you have a chiral center down here, you have a radical up here. I'm not trying to show you that one of these is easier than the other. What I'm trying to explain to you is that it looks different from the bottom and from the top because the molecule is not symmetrical about its mirror plane. You don't have to understand exactly how it's different, just that in principle it's different because its chiral, it doesn't look the same way one way as the other so, that's basically it, we are going to move on to the problems now. Next class, I want to make sure, the test starts on Friday make sure that you read the syllabus and the FAQ because I have been getting some questions that have been causing me some concern. There are not a lot of things to worry about, but definitely make sure you know what rooms you're going to, and when the test is. I won't accept as an excuse that you didn't read the FAQ so you didn't know where the rooms are, but if anything is unclear let me know, and I will be happy to answer any questions.

[Sounds of people leaving]

Transcription by CastingWords


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