- Drawing enantiomers
- Cahn-Ingold-Prelog system for naming enantiomers
- R,S system
- R,S (Cahn-Ingold-Prelog) naming system example 2
- R,S system practice
- More R,S practice
- Fischer projection introduction
- Fischer projection practice
- Optical activity
- Optical activity calculations
How to assign the configuration of a chirality center using the R,S system.
Want to join the conversation?
- how do we know if it's R or S if there are 2 asymmetric center?(7 votes)
- Both asymetric centers will have their own R/S configuration. You evaluate them independently.(26 votes)
- how should we name them if they are not drawn in the dash-wedge system?
how can we know what is coming out of the page and into the page?(7 votes)
- Sometimes you just can't tell, in which case you have to leave off the R/S.(10 votes)
- Between COOH and OH, which one should be given higher priority?(2 votes)
- The OH because O has a higher atomic number than C.(13 votes)
- Does the trick in the end always work?(2 votes)
- Yes, if the lowest priority group is in front, the "trick" will always work.(10 votes)
- Just for clarification, on the (R)-2-butanol molecule, I notice that in using Cahn-Ingold-prelog system that you wrote out the ethyl and methyl before ranking them according to priority. Is it possible at that stage to just say that ethyl will having a higher priority instead of writing out CH2 CH3?(2 votes)
- It is possible, it’s better to be thorough though. Lots of people mess up priorities when it comes to alkyl chains.
The best way to do it is to compare the immediately bonded atom, if there’s a tie (most often is with carbon chains) look at the other three bonds that the tied groups have and rank them from highest atomic number to lowest. The first point of different gives you the higher priority group.(4 votes)
- Does the centre carbon have to be chiral for the R/S naming system to apply?(1 vote)
- You can't have the R/S system in place unless the carbon is chiral. If it is achiral the RS system doesn't apply.(5 votes)
- In both examples the Br is highest priority and oriented coming towards you. Is that part of the rules? Orient lowest priority going away and highest priority coming out?(1 vote)
- It'll be very helpful for you to get a 3D model or practice your imagination. Whenever the lowest priority is going away, the atoms in the front will always be oriented counter clockwise or clockwise. No matter how you move the molecule - highest priority to the front or not, or as in a Newman projection - the rotation of position of 1, 2 and 3 will always be in the same orientation.(2 votes)
- At9:30he says that we could say it looks R and the H is projected at us: therefore it actually is an S. What would happen if H was on the same parallel and looked like R/S? What would happen then?(1 vote)
- If the C-H bond were in the plane of the paper, we would have to rotate the molecule so that the H was behind (or in front of) the paper.(2 votes)
- why do we start with the highest priority and then go towards the lower priority. what property makes it so(1 vote)
- That is what the Cahn-Ingol-Prelog rules say you do, they’re used to determine priorities for R/S and E/Z(2 votes)
- At8:07, I don't get why those are representing the same molecule. When I tried to rotate the molecule on the right, so that the H goes behind the page, I got the OH group in the plane of the page, and methyl group coming out at us (wedge) Hmm, I guess I have to think about this a bit more.(1 vote)
- Don't rotate around the bonds, rotate the entire molecule 180 degrees. Pretend you grabbed it and then flipped your hand over.
The methyl group that was on the right is now on the left
The ethyl group that was on the left is now on the right
The OH that was going away from us is now coming towards us
The hydrogen that was coming towards us is now going away from us
And all of this is exactly the same as the molecule on the left!(2 votes)
the RS system is used to describe the configuration of a chirality center. Down here we have a pair of enantiomers. On the left we have one compound, on the right we have its mirror image. We're going to assign an R orS to each of our enantiomers so let's start with step one. In step one we prioritize the four groups attached to our chiral center, and we do that according to atomic number. Let's start with the enantiomer on the left. We know that this carbon is our chiral center and we have four different things attached to this carbon. Over here to the right, to a very shortened version of our periodic table, and we can see that bromine has the highest atomic number out of these four atoms, so bromine gets highest priority. We're going to give the bromine a number one. Chlorine has the next highest atomic number at seventeen so chlorine gets second highest priority. Fluorine has the next highest with a nine, so fluorine gets a 3, and finally hydrogen is the lowest priority group, of the lowest atomic number of one, so hydrogen gets a four. Step one is done. Step two - orient the groups so the lowest priority group is projecting, or pointing away from you. Our lowest priority group is hydrogen, and it's already going away from us. The hydrogen is on a dash here, so step two is pretty much done. Step 3 - determine if the sequence 1, 2, 3, is clockwise or counterclockwise. We're going to ignore our hydrogen here, so I'll just kind of rub it out here, and look at what's happening with one two and three. So one two and three are going around in this direction that is clockwise, therefore this is the R enantiomer, so i'm going to write R-bromochlorofluoromethane. Let's do the same thing for its mirror image. So this compound on the right, we know that this carbon is our chiral center. We already know how to assign priority: Bromine gets a number one, Chlorine that's number two, Fluorine gets number three, and Hydrogen gets a number four, so step one is done. Step 2: point the lowest priority group away from you, that's already happening here, so step two is done. Step 3: determine if the sequence 1, 2, 3 is clockwise or counterclockwise. Let's ignore this group going away from us, the lowest priority group. Let's go around one, two, and three, so going around in a circle one two and three in this direction, we know that is counterclockwise. So this must be the S enantiomer, so this would be S-bromochlorofluoromethane. So step three is done. So that's how you assign a configuration at to a chiral center. Here we have another pair of enantiomers, so this alcohol on the left and its mirror image on the right. Let's start with the one on the left. we know from earlier videos that this is the chiral center, and our goal is to assign a configuration to this chiral center. Let's go ahead and redraw the molecule so that carbon is our chiral center. Attached to that carbon is our OH, so I'll draw the OH on a wedge, we know that hydrogen is there going away from us so that's a dash, so even though it's not drawn in we already know it's there. To the right we have a methyl group, so a CH3. So let's draw in a CH3, And then finally to the left we have an ethyl group, so that would be a CH2 and then CH3. All right let's go back to our chiral center, so here is our carbon that's our chiral center. Let's look at the atoms that are directly bonded to this carbon. So there's an oxygen directly bonded to the carbon, there's a hydrogen, and then we have two carbons. So let's assign priorities. If we look over here at our very shortened version of the periodic table we know that oxygen has the highest atomic number out of those atoms, so oxygen gets highest priority. The OH group gets the highest priority, so this gets a number one. Hydrogen has the lowest atomic number, so the hydrogen is the lowest priority group, so we say that's group number four. Finally we have two carbons, and two carbons would of course be a tie. Carbon has atomic number of six, so we have to find a way to break the tie. So the way to do a tiebreaker is to look at the atoms that are directly bonded to those carbons, so we'll start with the carbon on the right. The carbon on the right is directly bonded to three hydrogen's - one, two, three - so let's go and write that out - hydrogen, hydrogen, hydrogen. The carbon over here is directly bonded to a carbon, a hydrogen, and a hydrogen. We're going to put those atoms in order of decreasing atomic number, so this carbon has a higher atomic number than these hydrogens, so we write carbon, hydrogen, hydrogen. Next we compare these lists, so on the left we have carbon, hydrogen, hydrogen. On the right we have hydrogen, hydrogen, hydrogen. And we look for the first point of difference. Well, that's carbon versus this hydrogen here. Carbon has a higher atomic number than this hydrogen, so the carbon wins. So this group gets higher priority, so the ethyl group has a higher priority than the methyl group. So the ethyl group must be the second highest priority, so this gets a number two. And the methyl group must get a number three. So now we've assigned priority to all of our groups. Let's go down here and let's write it in. The OH got highest priority, so that gets a number one. The ethyl group got second highest priority, so that gets a number two. The methyl group got third highest priority, that's a number three. And the hydrogen, our lowest priority group, is pointing away from us, so that takes care of step two. Step three is to see what the sequence is doing. So if we go around the circle, 1, 2, 3, we're going around this way. All right, we're going around clockwise, and we know if we're going around clockwise that must be the R enantiomer. So this is R-2-butanol. If that's R-2-butanol, the mirror image must be S-2-butanol, so let's go ahead and let's double check and make sure that's true. So this is our chiral center. This is our chiral center. We know the OH is highest priority, so that gets a number one. Let me go ahead and change colors for this, so this one gets a number one. We know that our ethyl group gets a number two. We know our methyl group gets a number three, and we also know there's a hydrogen going away from us in space. So our lowest priority group is projecting away. So all we do now is look at what's happening with 1, 2, and 3. And 1, 2, 3 are going around this way, and that of course is counterclockwise, and counterclockwise is S, so this is S-2-butanol. We just saw that this is S-2-butanol, but what if you were given the dot structure on the right, and asked to assign a configuration to the chiral center? So here is the chiral center, and the OH this time is going away from us. That means that the hydrogen is coming out at us in space. So we've already seen how to assign priority: the OH gets a number one, the ethyl gets a number two, and the methyl gets a number three, and the hydrogen gets a number four. When you go to step two, step two says to orient the group so the lowest priority group is projecting away from you, but that's not what we have here. Here we have our lowest priority group coming out at us in space, because this is a wedge. So one thing you could do would be to take this compound, and in your head mentally rotate it so the hydrogen is pointing away from you in space. And when you do that, you'll see that it's the same as the one on the left. So this is just two different ways to represent the same enantiomer. So in the video on drawing enantiomers, I actually showed you the video where I rotated this compound to prove that these two drawings represent the same compound. So now your hydrogen is going away from you in space, and that's how we got, that's how we got S. Because this was number one, this is number two, and this was a number three, so we went around this way and we saw that it is counterclockwise, and so we got S. But what if you didn't want to rotate the molecule in your head? Sometimes it's pretty difficult to do. For this one it's not too bad, but it can get a little bit tricky, so there is a trick that you can use. There is a trick that you can do, you can just start with the drawing on the right. You can go ahead and number your your groups in terms of priority, and then you can go ahead and just ignore the hydrogen for the time being, even though the hydrogen is coming out at us in space here. If you ignore it and look at 1,2,3, so 1, 2,3 is going around this way, and that would be clockwise, and we know that clockwise is R so i'm going to write it looks R here. It looks R but since the hydrogen is coming out at us in space, we can take the opposite of how it looks. So it looks R and the hydrogen is coming out in space, you know it's actually the S enantiomer. So that's a little trick that you can do instead of rotating the molecule in your head, you can just assign your priorities, come up with R or S, and then take the opposite.