- 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
R,S system practice
How to assign a configuration to a chirality center using the R,S system.
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- In one of your videos before you teached an easier way to identify if its R ou S simply by ignoring the fact that the molecule with the lowest atomic number was not oriented back and then if you concluded that it was S it was because it would be R, why doesn't this work here?(22 votes)
- The lowest priority group is in the plane of the paper.
It is not in the front, so you can't use the easier way for a group in the front.(28 votes)
- Are the name of the molecules (R)-1-methyl-cyclohexene and (S)-3-bromo-3-chloro-pentane?(9 votes)
- The name of the first molecule is (R)-4-methylcyclohexene.
The double bond gets priority in the naming, and the numbers must include both alkene carbon atoms.
So C-1 is at the bottom, C-2 is at the 4-o'clock position, and the methyl group is on C-4.
The name of the second molecule is (S)-3-bromo-3-chloro-2-methylpentane.
You forgot the methyl group.(20 votes)
- At8:38, how is that molecule in the S formation? Since the chlorine is pointing away from us, why didn't we flip the molecule so that it would be in the R formation?(7 votes)
- but how do we vision holding the molecule and doing the rotation in a test where we only have pen and paper? is there any better way to help see it?(3 votes)
- Why wouldn't the second problem be R? I put S at first but then saw that the lowest priority was not going away in the figure the chlorine was so then I just chose the opposite configuration which was R.(4 votes)
- You have to view the molecule so the lowest priority group (ethyl) is going either directly away from you.
The ethyl group must be either a wedge or a dash. You can't leave it in the plane of the paper.(5 votes)
- How is this a chiral molecule? I thought that a carbon was a chiral center if it had 4 different groups attached.(2 votes)
- Every chiral carbon in this video has four different groups attached.(5 votes)
- So...if 1 is to the left of 2 - regardless of where 3 and 4 are - is it ever not counter-clockwise? It seems to me that the only information you need to determine R or S is 1's and 2's positions.(2 votes)
- Since we're always moving from group 1 to group 2, that initial direction always determines the configuration of the molecule. The third and fourth groups are good to label so that we know for sure we have the correct first and second group, but the rotation direction really depends on that initial group 1 to group 2 placement.
Hope that helps.(2 votes)
- when counting the carbons at3:04why don't you include all the carbons attached to the 3rd carbon on the right (blue?) so instead of CCH it would be CCCH and the third carbon on the left(red) it would be CCHH instead of CHH.(1 vote)
- We count only the atoms that are further out from the chiral centre.
We don't count backwards.
The atoms that are directly attached to the blue carbon and are further out from the chiral carbon are CHH.(3 votes)
- how the 1st molecule has a chiral centre because the carbon is not attached to 4 different groups??(1 vote)
- Because that top carbon is bonded to an H, a CH3, and the left side of the ring isn’t the same as the right side of the ring due to the double bond.(3 votes)
- at1:57why it is HHH ? instead of that it should be CHHH(1 vote)
- We don't count going back to bonds we have already been to when making that list, it saves us time as it's always going to be there and it's always going to be a C.(2 votes)
- How would you do this for a ring structure without a double bond?(1 vote)
- Do a video search for "determining r and s practice". You should get some useful results.(2 votes)
- [Instructor] Let's get some more practice with the R, S system. So we'll start with this compound right here. We already know from earlier videos that this carbon is a chiral center. So let me go ahead and redraw everything because it's gonna help us assign a configuration. So that's that carbon. We have a methyl group coming out at us. I'm gonna draw in the carbon with the hydrogens here. We have a hydrogen going away from us. And going to the right around the ring, we hit a CH2, so I'm gonna draw in a carbon with two hydrogens here. And then we hit a CH. So this carbon bonded to a hydrogen, that's this carbon on the ring. Notice that this carbon that I just marked is double bonded to this carbon. And for the purposes of R and S system, we're going to pretend like this carbon is bonded to two different carbons, even though it's really one. So that's how to handle a double bond. Going this way around the ring, we hit a CH2, so I'm gonna draw that in. So here's our CH2. And then we hit another CH2 right here. So CH2. And then this carbon is bonded to this carbon. So I'm just gonna draw a line in there like that. All right, let's think about priority. So this is our chiral center, let's look at the four groups attached to the chiral centers. So this is step one. Prioritize the four groups using atomic number. So what's directly attached to this carbon? There's a hydrogen, there's a carbon, there's a carbon and there's a carbon. So carbon beats hydrogen in terms of atomic number. So hydrogen is the lowest priority group. So we assign that a group number four. All right. Now we have a tie, we have three carbons. We have three carbons. So we need to see what those carbons are directly bonded to. Let's start with this top carbon here. This carbon is bonded to hydrogen, hydrogen, hydrogen. So let me write that down. So hydrogen, a hydrogen, a hydrogen. Let's move this carbon on the right. This one's bonded to carbon, hydrogen and hydrogen. So carbon, hydrogen, hydrogen. We're thinking about the atoms directly bonded to it, and we're going in decreasing atomic number, which is why I put the carbon first. Now let's look at this carbon. So this carbon is bonded to a carbon, a hydrogen and a hydrogen. So CHH. All right, let's compare now. Well, first point of difference. We look at the first atom here, this is a hydrogen and this is a carbon. So carbon beats hydrogen. And then over here we have a carbon. So this one, this one doesn't win. This one must be third in terms of priority. So I put a three here for this methyl group. Now we continue on. We have hydrogen versus hydrogen. So that's a tie. Another hydrogen, another hydrogen. So we have another tie. So we need to go to the next atom to break this tie. So we go to our next carbon. So this one right here. What atom is this carbon directly bonded to? Well, it's bonded to carbon, carbon, hydrogen. So carbon, carbon, hydrogen. This carbon is directly bonded to carbon, hydrogen, hydrogen. So carbon, hydrogen, hydrogen. We look for the first point of difference. This is a carbon versus a carbon. This is a carbon versus a hydrogen. So the carbon wins. And that means that this, this way around the ring, this is the higher priority path around the ring. So this gets a number one, and then this path around the ring going this way gets a number two. So now we've assigned a priority to our four groups. So now we're ready for step two. Orient the groups so that the lowest priority group is projecting away from us in space. So let's go back to our original dot structure here. We said that this way around the ring was the highest priority. So this got a number one. Next, going this way around the ring was second highest priority, so a number two. Our methyl group was a number three, and our hydrogen was a number four. Our lowest priority group is going away from us. Our hydrogen, our hydrogen is on a dash. So that's going away from us in space. So now we have, finally onto step three. And let me change colors again because it's getting a little busy here. So step three, determine if the sequence one, two, three is clockwise or counterclockwise. So if I look one, two, three and I go around the circle, here's one, here's two and here's three. So going around one, two, three in a circle is this way, all right? That is clockwise, right? You can see that's clockwise right here. And so clockwise is R. So the configuration of this chiral center is R. Now let's look at this compound. So we have only one chiral center to worry about, it's this one right here. Let's think about the atoms that are directly bonded to our chiral center. Well, there's a bromine directly bonded to it, a chlorine, and over here would be a carbon; and that we have another carbon. So we prioritize our groups in terms of atomic number. Bromine has the highest atomic number out of those atoms so we give bromine a number one. Next would be chlorine with atomic number of 17. So that's number two. Then we have a tie for our carbons. So we need to see what is directly bonded to those carbons. So for the carbon on the right here, this carbon is directly bonded to a carbon here, a carbon here, and then of course a hydrogen. So we write that in as carbon, carbon, hydrogen. And then for the carbon on the left, so this carbon, that carbon's bonded to a carbon and two hydrogens. So we write in carbon, hydrogen, hydrogen. We look for the first point of difference. So we have carbon versus carbon. So that's a tie. So we keep going and we get carbon versus hydrogen. So the carbon wins, and this group gets the higher priority. So this isopropyl group is a higher priority than this ethyl group. So that means the isopropyl group is gonna get a number three. So this is three. And the ethyl group is the lowest priority. It gets a number four. Now that we've assigned priority to our groups, we need to orient the molecules so the lowest priority group is pointing away from us. And the lowest priority group is group number four. So I'm going to a video in a second, and in the video I'm gonna show you two different ways to think about putting this group going away from you in space. So one way would be to just think about an axis through this carbon here and then rotate, and then rotate around this axis until your lowest priority group is pointing away from you. Another way to think about it is like a Newman projection. If you stare down, if you stare down this bond, let me go ahead and change colors here, if you look down at this carbon, carbon bond here, so if you put your eye along this axis, so here's your eye, that would mean your lowest priority group, your ethyl group, would be going away from you. So that's another way to think about looking at the model or the molecule, I should say, in a way where the lowest priority group points away from you. So here's our compound. Let's say that red represents bromine, so there's bromine; and yellow represents chlorine. So our goal is to rotate this to put our lowest priority group going away from us. So we think about an axis through our chiral center, we'll rotate it so the ethyl group is pointing away. And now we can see that our red bromine is to the left, our yellow chlorine is to the right, and the isopropyl group is up in space. We go back to where we where. This time let's think about a Newman projections. So we're staring down, staring down this carbon, carbon bond. So let's rotate the molecule. And it's a little bit different perspective but we're still able to see a red bromine to the left, a yellow chlorine to the right and an isopropyl group up. So here's what we saw when we stared down our carbon, carbon bond. So this carbon is our chiral center. So that's this one right here. And then I used red for bromine and then I used yellow for chlorine. So we can see that our bromine, our bromine is to the left here, and this was the highest priority group. The chlorine is to the right. That was the second highest priority. And then we have our isopropyl group here up. So that's one, two, three, with our lowest priority group pointing away from us. So all we have to do now is go around in a circle and see what we get. So we're going from one to two to three in a circle. That means we're going this way around our circle, and that of course is counterclockwise. So counterclockwise is S. So the configuration of this chiral center is S.