Monday, May 22, 2006

Lecture 017: Chirality part 2

Lecture 017: Chirality part 2

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Jean-Claude Bradley: We determined last time how to determine if a carbon center is chiral or not. It needed to have four different groups, if it has four different groups; we know we definitely have a chiral center. One thing we did not talk about was how to actually name these things, so that's what we're going to start off with.

When you're drawing a chiral center, you don't have to do it this way, but it really helps a lot if you use this kind of projection, where the horizontal bonds are coming out towards you, and the vertical bonds are in the back. I'm going to draw a chiral molecule at random here. The only thing I have to do is draw four different groups, and I'm guaranteed to have chiral molecule. I'll put an H here, I'll put an F, Cl, and a Br. We know immediately that this is definitely going to be a chiral molecule; it has a chiral center, and when we talk about chiral centers, generally put a little star, if you look through the chapter you'll see that that's the way that they're referred to.

If we wanted to have this compound and do something with it, we'd have to find it in a lab, we'd have to order it, we'd have to have some way of actually naming it, so what we're going to do is find a way to name all these chiral molecules so that you can talk about them. What you'll notice is that because we have four different groups, we can differentiate these atoms depending on what different groups they have and the order in which they appear. What we're going to do is actually prioritize the groups based on their molecular weight, so we're looking at the molecular weight of the atom that is directly connected to the chiral center; not the whole group, but just the atom. This is the simplest example, all I have is atoms, but later on we're going to have actually groups that have multiple atoms, and we'll have to deal with those a different way, but the first thing we're going to be concerned with is the molecular weight of each of these atoms. What you generally do is look for the lowest priority group or the lowest atomic weight group, or actually atomic number in this case, and put a box around it. So hydrogen is easy, because nothing is going to have a lower weight than hydrogen, so that's going to be your lowest priority. For the others, we're going to look at the highest priority and go down, so the highest priority between bromine, chlorine, and fluorine, is what? Bromine, and then chlorine. So we just go down our little periodic table that I asked you to memorize. Bromine first, then chlorine, then fluorine. You'll find that you will always be able to draw a circle that is going either clockwise, or counterclockwise, and in this particular case, we're going counterclockwise.

The definition to name these is that when you're going counterclockwise and the lowest priority group is in the back, this will be called S. 'In the back' means that it has the bond with the dotted lines that tells you it's in the back of the board. Something you'll notice about the way that we draw this sp3 hybridized carbon is that there's no bonds that are in the plane of the board, they're either in the back, or they're in the front. That's part of what makes this useful. When you have a group that's neither in the back or in the front, you have to start to do some things so that' you're able to put the lowest priority group in the back. If you draw your chiral groups like this, you're never going to have that problem. Counterclockwise is S, and when the lowest priority group is in the front, we're going to call that R, so let me draw the mirror image of this molecule...

This is just the diastereomer of the previous molecule. Lowest priority group is in the back, I put a square around it, you start up at bromine, then chlorine, then fluorine. We're going clockwise. That's the definition that you'll find to be consistent through all the organic textbooks that you're going to pick up. Any questions on this first part?

Let's do an example where the lowest priority group is in the front, and let's see if we can handle that. Let's say that I redraw this molecule, and this time I put the H in the front, the lowest priority group. So if you proceed to analyze this, there are two ways to solve this at this point. What you can do is, you can rotate the molecule so that the lowest priority group is in the back, and that's a little bit difficult for some people to do. Some people are really good at rotating three-dimensional objects, some of you don't. The method that I will give you will work for everybody. What you do is still determine everything the same way you did the other, put a square on top of the lowest priority group, and still order the other groups, so bromine is one, fluorine is second, chlorine is third... We are going counter clockwise, and the way that we talk about this is, counter-clockwise is normally S, but the lowest priority group is in the front, therefore the molecule is R, and the reason that I do that particular logic, is that again, if you look in any organic chemistry textbook, you'll see that the counter-clockwise always corresponds to S, and I want that association to be held constant so that you don't get confused. So this is counter-clockwise, so the logic is counter-clockwise is normally S, but the lowest priority group is in the front, so it is R.

So we'll do several more examples to see how this works. If you change the chlorine and the fluorine, then you would get clockwise, and then it would be S. Actually, if you ask a question like 'What happens if you switch the chlorine and the fluorine?' Any two groups that you switch, you will automatically create the mirror image, and so if you've already determined that this is R, then if you simply switch two groups, automatically the molecule that you draw will be S, you don't need to redetermine it. That will always be the case. And we will use that to speed things up when we do analysis.

All you want to do is switch the chlorine and fluorine. So now we're going one, two, three, now we're going clockwise. The lowest priority group is the same in each case. The lowest priority group you can think of is four, and you don't include that in the circle; the clockwise and the counter-clockwise only refers to one, two, and three. When I draw these, they're going to overlap the square, but I don't use the square to draw the circle.

So what happens when we have multiple atoms? First thing is you have a molecule here, you have 1sp3 center, and you have four different groups, so we can immediately say that this is a chiral center. Now we want to determine if it's R or S. We are going to put a square around the lowest priority group, and then we're going to compare the other three groups to each other. Remember that it is not the groups we are concerned with; it is the atoms that are directly connected to the center. Now in the case that it is F vs. C vs. C, it is clear that F is going to be higher priority than the other two groups, so F will be number one. Now, between the C and the C, those are equivalent at this stage, so what I need to do at that point is to use this concept of shells, which is I'm going to keep going outward into the groups until I find a difference between them, and at that point I'm going to make an assignment as to which is a higher priority, and one way to do that is to draw brackets, and to list the shells separated by commas. In the first shell for methyl, I have a carbon, comma, and then I ask myself 'What are the three things that are connected to that carbon?', and the answer is H, H, and another H, for methyl, so now I'm going to do the same thing for this CH2I group. The first shell is the atom directly connected to the chiral center; that will be a carbon, comma, three things are directly connected to that carbon, in this case it's H, H, and I. Okay, so when I compare these again, I'm not looking for averages or sums, I'm actually looking for the highest priority atom in each one of those collections. So in the first shell, carbon vs. Carbon, those two cancel out; in the second shell, the highest atom in H, H, H is an H. In H, H, I, the highest atom is an I, so the only thing I'm comparing is an I vs. An H, and so basically I and H, the I will win out. That will make the CH2I the second priority group, and the CH3 will be the third. So as I go around, I find that I am going counter-clockwise, and the lowest priority group is in the back, so that means it's going to be S.

That's basically all there is to it. Sometimes you've got to go 3, four shells until you find that difference, that just means you need to be careful and check what's going on, but certainly the largest group is not the highest priority group. Like in this case, the CH2I is much bigger than the fluorine, but it's not a higher priority. It even has an iodine in it; that doesn't make a difference because the iodine is in the second shell and not the first. That way all the chemists can agree on a standard, and now when you name things, all you need to do is put an R or an S in front of it, and everyone will understand what you're talking about.

One of the reasons that I used the representation that I did with the horizontal bonds coming out toward you and the vertical bonds being in the back, not only is it convenient for finding R and S, it's also an easier way to make the other projection that we call a Fischer projection. You remember the Newman projection? It's got a circle; it's got three bonds in the front, three bonds in the back. If you're not taught how to read a Newman projection, it doesn't make any sense; it's the same thing with a Fischer projection. Fischer projection is always going to be in the form of a cross, or a multiple of crosses on a vertical line, like this. In a Fischer projection, you never put a carbon in the middle, so it's not like a skeletal formula, you don't have a choice of putting a carbon in the middle, you always have to put it as an X, and you can't tilt it, so if you only have a skeletal formula, there's no rule that says you can't draw it at any angle you want. In a Fischer projection, you must draw it exactly like this, vertical and horizontal, and a Fischer projection, it is implied that the vertical bonds are behind the board, and the horizontal ones are coming out towards you. If we draw this molecule, and we say it's a Fischer projection, then it's really the equivalent of drawing it like this.

If you have one chiral center, it's not really that big of an advantage to use a Fischer projection. Where the Fischer projection comes in really handy is when you have things like this. When you have multiple chiral centers in a molecule, Fischer projections are very useful because it'd be very difficult to draw a molecule like this showing all wedges going in, going out, and this is used for a very specific application in biochemistry. What kind of molecules have lots of chiral centers? Not proteins, sugars, yes, carbohydrates. Carbohydrates are really a pain to draw unless you have something like a Fischer projection, where all you need to do is draw a line, put a couple of horizontal lines, and you're done, pretty much. So that's the kind of projection we're going to use usually if we have more than one chiral center. Any questions on that?

The definition of Fischer projection is that all the horizontal ones are in the front, and that's it. That's why we have to agree on what it means, otherwise I couldn't draw a molecule like I did with the bottom and have anyone know what it is I'm referring to.

In this case, there's no Sys or Trans because there's no alkene, and there's no cyclic alkene, so sys and trans has no meaning in this particular projection. The Fischer projection is just a convenience, it's not like the cyclohexane conformation, that's actually the way that the molecule looks. In a Fischer projection, the molecule will look very very different from this, actually.

Let's start to use these Fischer projections in a practical way. Let's pick a molecule that has two chiral centers in it. Actually the molecule I'm going to draw all the isomers for is 2, 3, dicholorabutane. If you take a look at the quiz, you'll see that there's some questions that are related to this kind of processing. You have 2, three dichlorabutane, the way that I drew it a the bottom is I didn't draw it with any regard to chirality. You can't tell anything, basically they're drawn flat, it's just to see the connection of the atoms. At the top left, I drew a Fischer projection. That tells you at no uncertain terms what the chirality is of those two chiral centers. I've got two chiral centers in that molecule, because I have two sp3 hybridized carbons that have four different groups attached to them; two chiral centers.

One of the things that you'll need to determine is how many different isomers can you have like 2, three dicholrabutane? Fischer projections are a great way to do that efficiently. What you do is you draw a Fischer projection, and you only draw a cross where you have a chiral center. I've got two crossing points. Then, you switch left and right, the two groups that are on the horizontal bonds, until you generate all of the possibilities. I've got two groups, two times two is going to be four, there will be four different ways in which I could draw this. I could draw the two chlorines on the right, or I could draw the two chlorines on the left, and when you start to do this with Fischer projections, you can't change what's on the vertical axis, so if you decide that you're going to put the methyl's on the top and the bottom, and the chlorines on the horizontal bonds, you can't change your mind after and now put the methyl's on the horizontal bonds, because you will not be able to analyze the problem properly. Initially, there's nothing really that tells you that you must put the methyl groups vertically. Convention says that you probably put the alkyl groups on the top and bottom, but that's certainly not an absolute. It wouldn't put a wrong Fischer projection had I put the chlorine on the top and the methyl on the horizontal ones, but once I start, like in the Newman projection; I'm committed to finish the problem that way. I have the two chlorines on the right and the two chlorines on the left, the other two possibilities are as follows. I could put one chlorine on the left, one chlorine on the right, and then the opposite. If you look at the number of combinations we can make here, that's it, there's no other ways that I can mix up those two chiral centers. The question now is how many different molecules do I actually have on this page? How many molecules here are mirror images on each other? There's two pairs, I drew them in such a way to make it easy for you to see that. That's another really really nice thing about Fischer projections; it makes it really easy to see if you have mirror images or not because the vertical groups don't change.

We have two sets of mirror images, and we know we have to be careful about mirror images, because it doesn't tell you that you have enantiomers, it tells you either that you have enantiomers or the same molecule. Is there a case here where we might have the same molecule? How do you know that the first one is the same molecule? By definition, you can superimpose it, but how can you tell, just by looking at it? Are you doing a translation, a rotation? With a Fischer projection, another really nice property is that you can rotate it 180 degrees by keeping everything in the plane. The rotation that we're talking about here is this. If I rotate, follow that arrow here, and rotate 180 degrees, the top methyl group will end up at the bottom, the bottom methyl group will end up at the top, and the two H's on the right will end up on the left, the two chlorines on the left will end up on the right, and because you're not flipping through the plane, all the bonds in the back are staying in the back, and all the bonds that were in the front are staying in the front, you don't have to worry about that. The only kind of rotation you can do with a Fischer projection is the rotation that I'm drawing here. Don't take anything in and out of the plane, just rotate it, turn it upside-down basically. And if you do that you can see that I can superimpose the structure onto it's supposed mirror image. Everybody see that?

If you try to do the same thing to the other two, you've got a problem, because although the methyl groups match, the Cl group that is on the right here would end up on the bottom left, so if I rotated that fourth structure on the right, it would end up being the same thing. I could rotate it forever and never generate the third structure. That's a real advantage of using the Fischer projection.

What we know from drawing these, then is that we have actually three molecules. The first two possibilities that I drew are actually the same thing, that counts for one, and then we've got two, and then three, so one of the things that you need to be able to determine is how many different compounds do we have for 2, three dicholrabutane, the answer is three. There ought to be a couple of quiz questions to help you do that for some additional molecules.

The other thing that we have to do with this is name the relationship between the molecules that I drew. So the two molecules on the right that are mirrors of each other, but which are not the same molecule, are enantiomers. That's the definition of a enantiomers, that they are mirror images, so that definition doesn't change. Remember that the term diastereomer is a relationship, so a molecule is an diastereomer of another molecule, or I can say that those two molecules are enantiomers. So the one new things we have here is I have a molecule that is different chirality-wise from another one, but they're not enantiomers. The second structure I drew, and the third structure are not mirror images of each other, however, they have the same connectivity, so that's a new relationship, that's a new word, those two are diastereomers.

There'll be consequences of that fact. What we know from enantiomers is that enantiomers will rotate in polarized light, and they'll rotate it in opposite directions. We know that the third and the fourth compound have the same melting point. Sometimes I might ask you the question with a couple things that you have to integrate, like I can ask you do these compounds three and four have the same melting point, and to answer that, you have to understand that they're mirror images, that they're enantiomers, and one of the properties of enantiomers is that they have the same physical properties, so they have the same melting point. What we also know is that the first compounds and the third or the fourth compound, the diastereomers, those are not mirror images, and so they will not have the same physical properties; they won't have the same melting point, they won't have the same boiling point, they'll have different solubility's, they have different diapole movements... They're similar, they're pretty close, usually, but they're not going to be identical, and for that reason you can chemically separate diastereomers much more easily that you can enantiomers, so if you were to do a reaction and you had mixtures of those three compounds, it's generally much more easily to separate diastereomers from each other, because you could do things like crystallization or distillation, things like that, they would have different properties. You would have to use some other tricks to separate the enantiomers.

There's one more piece of nomenclature that we have to look out. The compound on the left, that I drew the mirror image, is that a chiral compound? It's got two chiral centers; however, its mirror image is super imposable upon itself, so although it has two chiral centers, it is not chiral. It's a little bit counter-intuitive at first, but if you think about it, this is how it's possible. Because this is a special molecule that has multiple chiral centers, yet it is not chiral itself, that's called a meso isomer. It's called a meso isomer, or meso compounds, but that's a special term you use whenever you have this kind of situation. Meso is not a relationship term, it's not like enantiomer, it's just a name, so I can say what is the meso isomer for this compound, and this is it.

Transcription by CastingWords


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