R and S of Fischer Projections - Video Tutorials & Practice Problems

R and S rule for Fischer Projections.

Now that we're pretty familiar with Fisher projections, it turns out that there's going to be a new way to determine R and S configurations for them. The reason is that converting these structures to bond line representations and then figuring out R and S would be alright? So, what we want to do is find an easy method to figure out R and S configurations, and there actually is one. There's a shortcut that we can use. So, I want to go ahead and show you guys that now. Alright? So, this is basically how it works. We want to determine the location of the lowest priority group. Okay? What I mean by location is, we want to figure out whether it is vertical or horizontal. Okay? If, and the lowest priority group is always going to be 4. So, basically, I just want to know where 4 is. Okay? If 4 is vertical, then the chirality is exactly as it looks. I'm just going to go from 1 to 3. I'm going to draw that arrow and that's going to be it, okay? Whereas, if 4 is horizontal, then the chirality is just going to be flipped. So, whatever you draw, you're just going to take the opposite sign. Alright? So, here I have two examples. This would be an example where h is vertical. So, notice that my 4th priority group is vertical. So, that means that it's going to be as it looks. So, I'm just going to say 1 to 2 to 3. It's going in an S direction. So that's actually going to be the final answer. The final answer for this chiral center would just be S. Does that make sense so far? Now let's look at this next one. Here's my chiral center. 4, as you can see, is now horizontal. So, that means it's going to be flipped. So, once again, I'm going to do 1 to 2, 2 to 3, 3 to 1. It looks like S, but it's actually going to be R because of the fact that it's horizontal. Now keep in mind, for Fischer projections, I'm never swapping groups. So, I'm not swapping out 1 or 4 or anything like that. All I'm doing is taking the chirality and flipping it if it happens to be horizontal or keeping it if it happens to be vertical. So, in some ways, this is actually easier than what we learned for the other compounds. And when you have a very big Fischer projection that has a lot of chiral centers, you're going to be thankful that you have this method, that you can just go really, really fast and figure it out. Okay? So, what I want you guys to do now is go ahead and determine the absolute configurations, meaning the R and the S, for this Fisher projection using this formula and using this format. And then when you're done, I'll go ahead and show you guys how to do it.

Determining Absolute Configurations

Alright, guys. So this wasn't an easy problem. This is actually a pretty complicated problem. If you got it right, power to you. Okay? But this is not easy. So first of all, here's my first chiral center and I need to recognize what these substituents are. I need to figure out priorities. So my first priority obviously has to be my oxygen because oxygen has the higher atomic weight. But then what I have here is carbon, and I'll say that's my blue carbon I mean, my red carbon and my blue carbon. Which one's going to get higher priority? Well, in order to understand this, I need to know what the CHO functional group is. Do you guys remember what that is? I hope you guys remember. That is an aldehyde. So what that actually looks like, I'm just going to scratch this out, is like this: double bond, O and H. That's what CHO means. I told you guys when I taught you all those functional groups that you have to memorize that. Okay? So there you go. That's one functional group. The other one is a carbon with a chlorine and oxygen. So if you can't immediately tell which one's going to win, you have to do the playoff system. And if you do the playoff system, what you would find is that the C that's in blue would be attached to an O, an O, and an H. Whereas, the C that's in red would be attached to a Cl, an O, and a C. Okay? So which one's going to win? Just the one that has the heaviest atom on it. So red has to win because red has a Cl on it. So this one's going to be 2, then this one's going to be 3, and then finally, this would be my 4. That was a little bit tricky. Okay? Now I'm going to say okay. Is my 4 vertical or horizontal? It is horizontal so that means whatever I draw has to get flipped. So I'm going to go ahead and go from 1 to 2, 2 to 3, 3 to 1. This looks counterclockwise, but really this is going to be clockwise. Okay? So that's my first one. See, that was a lot of work. Let's move on. Okay?

So let's go on to this next chiral center. For this next one, I have that this is going to be my 1 because chlorine is heavier than oxygen. This is my 2. Then I have 3 and 4, which is this blue carbon and this green carbon. Okay? So which one's going to win there? Once again, I mean, this one, you just have to do playoffs. So I'm going to do blue. That's C. What is that attached to? It's attached to an O. It's attached to a C, and it's attached to an H. Now let's go down to this green one. The green one is attached to an O, it's attached to a C, and it's attached to an H. So which one's going to win? I actually don't have a winner yet. They're both the same. So what happens if I get to the Illinois playoff and they're both the same? I have to keep going down the longest carbon chain. Alright? So that means that now I'm actually going to go to each carbon that it was attached to. So notice that this blue was attached to another carbon. That's this carbon. Now let us say what are the 3 things that that carbon is attached to? Well, we already did this one. It's attached to O, O, and H. The reason I'm counting O twice is because that is an aldehyde. So it has a double bond O. I'm going to do the same thing for this C. This C is attached to what? It's actually attached to just an O and an H and an H. Why is that? Because it's a CH2 group. Okay? So now do we have a winner? Yes, we do. Now we have a winner. This one has to be 3 and this one has to be 4. So it's going to be 3, and it's going to be 4. Like I said, this is just a really tricky question. So now is my 4th priority vertical or horizontal? It is vertical. Okay. So what that means is that I'm just going to draw this exactly how it looks. 1 to 2, 2 to 3, 3 to 1. It looks like an S and it is an S. Okay? So now let's go on to this last one. This last one would have this is my 1, This is my 4. And we already did this before in terms of we know the Cl is going to be better. So this one's my 2, and this one's my 3. Is my lowest priority horizontal or vertical? It's horizontal, so I'm going to flip whatever I get. 1 to 2, 2 to 3, 3 to 1. It looks like an S, but actually it's an R. Okay? And those are my 3 chiral centers. Now I know this seemed like it took forever, but it wasn't because of the rotation. The part that took a long time was that this just was a hard problem in terms of figuring out priorities. So no matter what strategy you use, this is going to take a while. But I think that what you're going to notice is that using the horizontal and vertical thing is going to save you guys a lot of time. And just so you know, there actually is no alternative. If you don't want to use this horizontal vertical thing, the only other alternative to get it right every time is to convert it completely to bond line and then do it from there. Alright? So let's go ahead and move on to the next topic.