5.5 Fischer Projections

So here, we want to take a couple minutes out and look at what are called Fischer projections here. So and, in a Fischer projection, it's just another perspective in looking at a tetrahedral atom. So in this case, these horizontal bonds on a Fischer projection correspond to the wedged bonds here, in the horizontal. So and, the vertical positions on a Fischer projection, correspond to dashed bonds. So and, it's kind of a useful perspective in looking at a tetrahedral atom, because you'll never have the fourth part of group in the plane. It will always be a wedge or a dash. And that makes it much easier to assign R and S. In this case, you need to know a couple things about it that it's only an accurate portrayal in looking at one chiral center at a time. Nut it's not a true 3-dimensional portrayal of the molecule. So it turns out you're not allowed to rotate them any anyway you want to willy-nilly. So it turns out the only rotation you're allowed to do is to rotate them a hundred and eighty degrees in the plane. So at a time. So in this case, if we put R4 here and then R3 here, R2 here, and R1 here, turns out this is the same thing as the first Fischer projection. But you're not allowed to rotate it 90 degrees. So let's just say we started with R1 here, and then R4 here and then R3 here, and R2 here. So this is not equivalent to any of the two structures we've drawn here. Not allowed to just rotate it 90 degrees. You're also not allowed to rotate it out of the plane 180 degrees either. So it's rotating in the plane 180 degrees. So if we try to rotate it out of the plane 180 degrees, and maybe put R3 here, and R1 here. And so left R4 and R2 right where they were. So this would actually give you its enantiomer, and again this would not be equivalent to our original Fischer projection. So and again, this is not true three-dimensional portrayal of a molecule. So you got to be a little careful in what rotations are possible. So with real three-dimensional portrayals of a molecule, you can rotate them any way you want to and it's the same molecule. Not true with a Fischer projection. So you have got to be a little careful with them. So let's take a look at the structures we've got here and let's turn them into Fischer projections. So in this case it's customary for a Fischer projection to draw them vertically when you have many tetrahedral atoms. So in this case we've got three tetrahedral atoms. So we've got methyl group at the top, methyl group at the bottom. And so in this case it doesn't really matter what we put at the top or bottom. But normally whichever you find your most oxidized carbon, you put them as close to the top as you can. It's kind of the the general rule in doing this. Now in this case, the question is where did the bromine and the hydrogen go and stuff like that. The middle carbon here has got two hydrogens he's easy. So but if we get the bromine wrong here, so that's going to lead us to some problems. Let's just choose a way here. I'm going to put a bromine on the right. Hydrogen on the left. I might have done it exactly backwards and if I did I'll have those two groups trade places and get us right back to where we were. So and typically what I like doing is just assigning R and s here. So here's carbon one, carbon two, carbon three- and this is in the R configuration. If I did the same thing for this carbon here. So here would be priority one, priority two, priority three- and this thing looks R and it's really S. And so it turns out I got this exactly backwards. So let's go back and erase these and get them correct here. So in this case, we've got then, properly, the bromine on the left and the hydrogen on the right. So and now this would also be in the R configuration. So again the hydrogen here is on the wedged position being, on the horizontal. And so it looks like now, that it's going to be S but it's really R. So in fact let's go to the last one here. Down at the bottom. So again if we assign chirality, here's the miso version. So this is S. So right off the bat. And in this case to get this bottom one S- So one, if I know he's the miso, I know that there should be an internal plane of symmetry, so I'll do it that right- right off the bat and kind of cheat. But how we got it wrong we'd have the hydrogen and bromine trade places. But in this case here, let's do this in blue to contrast, so here's priority one, priority two, priority three- and as we go around the circle we can see it's a right-handed turn, which would correspond too R. But again with our lowest priority group hydrogen on the horizontal, that's a wedged position. So this thing is really the opposite, and therefore is really cool. And so there's our two chiral centers. One R, one S. And again this is the miso. And you can so see that internal mirror plane verifying that it's achiral. And then again with two chiral centers, being achiral and having chiral centers, it is, again, our miso compound that we've already looked at. So if we kind of look at the other two. Having this first one drawn will make these two really easy. So again, methyl group on top. Methyl group on bottom. Three tetrahedral atoms. Middle one's got two hydrogens. So in this case, I can see that this carbon right here is in the same configuration as this one right here, and so I'll draw it in the same configuration. So, and it's still going to be R. So but this carbon right here is in the opposite one of the previous molecule, and so I'll put it in the opposite configuration. I'll put the bromine on the right. I'll put the hydrogen on the left. And what's nice about these Fischer projections, as well, is if there's an internal mirror plane, you're likely to see it. And there's not one here. And it'd be easier to see that this thing's definitely chiral. So in the kind of line angle formula here, I can put that in a whole bunch of different rotational confirmations to try and hide that internal mirror plane from you. But it's going to be much more difficult for me to hide it in a Fischer projection. So, and finally if we do the last one here. Again middle carbon has got the two hydrogens. So and, I see that this is perfectly the enantiomer here, of that middle one here. So both chiral centers get flipped. So here I put the bromine on the right and the hydrogen on the left; opposite of what we did right here. So on the bottom chiral center I'll put the bromine on the left, and the hydrogen on the right. Again opposite of what we did right here. Now that's going to make these configurations S and the other one S, where in the previous molecule they were both R. Cool. And there's our Fischer projections. So I highly recommend if you're going from line angle to Fischer or Fischer to line angle, that you just kind of assign R and S and make them match. So there can be a little bit of method to madness, but it's prone with errors. Especially for undergraduates. So I again highly recommend just assign R and S, and going from one to the other.