Fischer projection practice

- [Lecturer] In the last video, we looked at a Fischer projection of a compound that had only one chirality center. This molecule has two, and our goal is to determine the configuration of each chirality center, so at the intersection of these lines here, we know that this is a chiral center and then we have another one down here, so those are the two chirality centers. Let's focus in on the top one here, and let's assign a configuration to that carbon. Let me go ahead and draw the carbon on the right. We know that in a Fischer projection, the horizontal lines represent a bond that's coming out of the page at us, so over here on the left, the bond to this hydrogen is coming out of the page and so is the bond to this OH, so let's draw them on the right. We have a bond to a hydrogen coming out of the page on the left, and then we have a bond to OH coming out of the page on the right, so those, those are represented by a wedge. We know that the vertical lines in a Fischer projection represent a bond that's going away from us in space, so this line right here is a bond that's going away from us, so that would be a dash, so let's draw that in, and that's going to a carboxylic acid which we know is a carbon double bonded to an oxygen, bonded to an OH. Let's look at the other vertical line, so this one down here, so that represents a bond going away from us in space, so we draw that one in, so we're essentially staring down at our chiral center. And this carbon down here is directly bonded to an oxygen, a carbon, and a hydrogen, and those are the only atoms that we need to assign a configuration so those are the only ones that I'm going to draw in. Let's go back to our chiral center, so this carbon right here, we know to assign a configuration, we look at the atoms that are directly bonded to that carbon. There is a hydrogen, an oxygen and then our two carbons. Out of those atoms, oxygen has the highest atomic number so the OH group gets the highest priority and we say that is a number one. The hydrogen has the lowest atomic number so that's the lowest priority, so we say that's a number four. Next, we have carbon versus carbon, so this carbon versus this carbon and we know that carbon has the same atomic number, and so we've seen how to break our tie in earlier videos. We look at the atoms that are directly bonded to those carbons. This bottom carbon here is directly bonded to an oxygen, a carbon, and a hydrogen, so we put those in order of decreasing atomic number, so that's oxygen, carbon, hydrogen. This top carbon here, the one in our carboxylic acid, has a double bond to an oxygen, and so we treat that like it's two bonds to two different oxygens, even though it's really only one double bond to one oxygen, and then we have another bond to an oxygen on the right here so we could say, for the purposes of assigning a configuration, that would be oxygen, oxygen, oxygen. Next, we compare so we have an oxygen versus an oxygen, so that's a tie. We go on to our next atom which is oxygen versus carbon, and obviously, oxygen wins. It has the higher atomic number, so the carboxylic acid group gets the higher priority so this must be group two, and this group down here must be group three. Once we've done this, our next goal is to put the hydrogen going away from us in space, but I'm going to use the trick that I talked about in the last video because it means you don't have to move the molecule around in your mind. You can just use this little trick. And the trick was to ignore the hydrogen for the time being and just look at one, two, and three. Here's one, here's two, and here's three, so if we go around in a circle from one to two to three, we're going around in this direction. That is counterclockwise, and counterclockwise we know is S. So this looks S, but the hydrogen is coming out at us in space and so we know all we have to do is take the opposite so even though it looks S, since the hydrogen is coming out at us, we know it's actually R, so this is actually, actually R, so if you actually made this molecule and rotated it and put the hydrogen going away from you in space, then you would see that it's R directly, but this trick allows you not to worry about that, so I find this to be the easiest way to assign a configuration. For me, it's definitely the fastest. Let's move on to our other chirality centers, so this carbon down here, so let me draw that one in, and let's look at our horizontal lines, so we have a bond to an OH on the left, and a bond to a hydrogen on the right and those are coming out at us, since those are horizontal lines so that OH is on the left, and we have a hydrogen on the right. Let's look at our vertical lines, so this line right here means a bond going away from us in space, so we're staring down at our chiral center, and that's a bond to a carbon and that carbon is directly bonded to an oxygen, a carbon, and a hydrogen, so let's draw those in. Again, we don't need all of the atoms because this is all we need to assign a configuration so that's all I'm drawing in, and then let's look at this other vertical line, so this one down here, so that's a bond going away from us in space to this carbon, so let's draw that in so we have a dash over here on the right, going to a carbon and this carbon is directly bonded to two hydrogens and an oxygen, so let's draw that in, so we have our two hydrogens and an oxygen. Going back to our chiral centers, so this carbon right here, and we look at our atoms that are directly bonded to it and we have an oxygen, a hydrogen, and two carbons, so we know that the oxygen has the highest priority so once again, the OH group is the highest priority group and the hydrogen is the lowest priority group and we have a tie again between the carbons, so the carbon versus carbon, the top carbon is directly bonded to oxygen, carbon, hydrogen, so in order of decreasing atomic number, oxygen, carbon, hydrogen, the bottom carbon would be oxygen, hydrogen, hydrogen, so oxygen, hydrogen, hydrogen. We look for the first point of difference so we have oxygen versus oxygen, so that's a tie. Next, we have carbon versus hydrogen and we know that carbon has the higher atomic number so this top group here gets higher priority so this should be a number two and this group down here should be a number three. Let's use our trick again, so we ignore the hydrogen for the time being and we look at one, two, and three, so one to two to three is going around in this direction which we know is clockwise, so we can say that this looks R because clockwise is R, but because this is a Fischer projection, the hydrogen is coming out at us in space, so we know our trick is just to take the opposite of how it looks, so if it looks R, we can say that it's actually S, so we write down here actually, actually S. All right, so we've done it. We've determined the configuration of each chirality center so this chiral center, let me use a different color, let me use a red over here, so this chiral center, we said that one was R, so it's R at this carbon and then at this carbon down here it is S. Let's also practice drawing the enantiomer of this compound. We saw how to do that in the last video. We put a mirror right here and we reflected our groups in the mirror so this OH here, I like to draw a dashed line, so we just reflect that, and therefore we would have this, and then on this side, we would have a hydrogen so we're reflecting that hydrogen and then let's go to this hydrogen down here, so we just reflect this in our mirror and then we draw our horizontal line and then we would have an OH, so we draw in an OH here. We can put in our vertical line like that and then we have a carboxylic acid, but this carbon is not a chiral center so we don't really need to worry about it. We can just write in COOH here, and then same with this carbon, not a chiral center so we can just write CH2OH. The important part is if we have an OH on the right, it'll be on the left in the mirror image. If we have an OH on the left in a Fischer projection, it'll be on the right in a mirror image, so now we have a pair of enantiomers, so these two are mirror images, right? The one on the right is a mirror image of the one on the left, but the one on the right is not superimposable on the one on the left.