- 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
More R,S practice
How to assign configurations to a molecule with two chirality centers.
Want to join the conversation?
- I have a hard time figuring out how to rotate so that the molecule/atom with the least priority is going "in to the paper". With the second chiral center in this example could one have just kept the pentanol group still and rotated the other three groups around the carbon so the hydrogen ends up in the back?(7 votes)
- Yes, you could, and you would still get the same answer.
But there's an even easier way.
If the H is in front, don't bother to rotate. Just determine the priorities in the usual way.
If the 123 direction is clockwise (R), just change it to (S).
It works every time.(20 votes)
- Why is it always necessary that the lowest priority group into the space??(4 votes)
- You mean away from the chiral centre, right?
I guess it is so that we could determine the whether it is R or S. All the high priority groups being in the same plane helps in determine whether it is clockwise or anticlockwise rotation. As the structure is 3-D it is kind of like neglecting the lowest priority group by placing it in the different plane from other groups and there by making it easier to see the rotation.
I hope this helps.(6 votes)
- I am unable to understand the orientation of the lowest priority group away from our vision. pls help(3 votes)
- The lowest priority group (the group with the lowest atomic number) should be projecting away from you ("into the page"). So, in our example, Hydrogen is the lowest priority so it needs to be pointing away from you in order to assign a R or S chirality. Then you can use the other three groups (generally 1, 2, 3) to determine which direction your circle goes in: clockwise-R, counterclockwise-S.(1 vote)
- Is there a trick when your 4th priority is at the sheet plane? Just like in the previous video.(2 votes)
- Unfortunately, no. Because your wedges and dashes will be in arbitrary locations.(2 votes)
- I still don't know how to visualize rotations of the molecules (like when the lowest priority group is in the linear position, and I have to make it go away from me). Help me, please!(2 votes)
- Yes, that is not easy. Firstly, like Sal did in the last video, it helps to simplify the diagram and just draw #1, #2, #3, #4 instead of drawing bulky groups. Also, use your fingers, and make a tripod! assign groups to each of your fingers, and try! Have a nice day );(2 votes)
- What does R and S stand for?(1 vote)
- actually its vice versa R for rectus and S for sinister(4 votes)
- It's really difficult for me to rotate the molecule whenever we use skeletal structures. If the 4th group is on the back, then it's easy but in problems like these, I just can't do it without the model kit. I've noticed, though, that if you get R configuration whenever the 4th group is NOT in the back, you can change it to S and that will be the right answer without having to mentally rotate the molecule. Would this work with all molecules? I'm not sure if I explained myself correctly.(2 votes)
- I can understand where the "S" comes from but where does the "3" and "2R" come from?(2 votes)
- The 2R and 3 are basically the position of the chiral carbon on the Carbon chain.(1 vote)
- I'm having trouble understand how Jay arrived at an S configuration for the second chiral center in the pentanol example. If you rotate the molecule holding the carbon bonded to the alcohol in place, the hydrogen pointed towards us would end up facing away, the methyl group facing away would end up in the plane of the molecule, and the C2H5 that was in the plane of the molecule would end up pointing towards us. That would make it go 1, 2, 3, left, top, right. Which is R configuration.(1 vote)
- Remember it is not “top” it’s towards us in space. It will be an anticlockwise rotation (left->towards->right) so it is S.
Honestly though rather than worrying about rotating the molecule, there’s a really neat trick to working these that will save time. If you swap any two groups around you change a chiral centre from R to S or S to R.
Using this idea, if you swap the hydrogen and methyl so the hydrogen is going away while the methyl is coming towards us, you should be able to see that is R (left->right->towards=clockwise), so if that’s R then the stereocentre of this chiral centre will be S.
This video introduced the trick to me: https://youtu.be/Z10oC7BF4ig(2 votes)
- From3:11to3:27and7:04to7:32, after finding out how the priority goes , the compound is named in as (2R, 3S) 3-methyl-2-pentanol. How did that happen? Please clarify this doubt ASAP.(1 vote)
- What exactly aren't you sure about here?
I just watched the video and it seemed pretty straight forward.
We have the name. We just need to determine R or S for both the stereocentres.(2 votes)
- [Lecturer] Now, let's look at a compound with two chiral centers, so our goal is to finish naming this molecule, so if I number my carbons, this would be carbon one, two, three, four, five. You can see there's an OH at carbon two and there are five carbons so that's where we get the 2-pentanol, and there's a methyl group at carbon three, so that's 3-methyl-2-pentanol, but we need to put in stereochemistry because we know from an earlier video that carbon two is a chiral center and carbon three is a chiral center, so we need to use the R-S system to finish our name. So let's focus in at carbon two, and this drawing on the right will determine the configuration at this chiral center, so here's our carbon, and if there's an OH coming out at us in space, we know there's a hydrogen going away from us in space, and our chiral center is directly connected to a carbon on the left and this carbon is bonded to three hydrogens so I'll draw those in there, and our chiral center is directly bonded to a carbon on the right, and this carbon on the right is directly bonded to another carbon, another carbon, and there must be a hydrogen coming out at us in space here, so let's assign priority to our four groups. Remember, that was our first step when we're trying to determine the configuration of a chirality center, so here's our chiral center. We look at the atoms directly bonded to that carbon. There's an oxygen directly bonded to that carbon. There's a hydrogen directly bonded to it. There's a carbon on the left and there's a carbon on the right, so we assign priority based on atomic number. Out of those atoms, oxygen has the highest atomic number so oxygen gets highest priority so we call that group a number one, so the OH group gets a number one. Hydrogen has the lowest atomic number, so hydrogen gets lowest priority and we call that group four here, so hydrogen is a four, so now we have a tie. We have these two carbons. Let me go ahead and circle them, so carbon obviously has the same atomic number, so to break this tie, we look at the atoms directly bonded to these carbons. The carbon on the left is directly bonded to three hydrogens, so we write down here three hydrogens and the carbon on the right is directly bonded to two carbons and one hydrogen, so we write carbon, carbon, hydrogen. Next, we look for the first point of difference and that's the first atom here, so carbon versus hydrogen. Carbon has the higher atomic number, so this group wins, so this group on the right is higher priority which means this must be number two for our groups, and the methyl group gets a number three. Now that we've assigned priority to our four groups, the next step was to orient the molecules so that the lowest priority group is projecting away from us, and that's what we have here because the hydrogen is the lowest priority group so we can ignore this and we can focus in on one, two, and three, so let me label this on the right, the OH was the highest priority and this group to the right was the second highest priority and the methyl group was the third highest priority. Next we draw a circle and determine whether we're going around clockwise or counterclockwise, so we draw a circle from one to two to three, and obviously, we're going around in a clockwise direction and clockwise is R, so we are R at carbon two, so I write here 2R. Now, let's focus on carbon three, so we know that carbon three is a chiral center, so what is directly bonded to this carbon? Well, there's a carbon attached to three hydrogens, so there's a methyl group, there's a hydrogen that must be coming out at us in space and on the right, there's a carbon bonded to two hydrogens and directly bonded to another carbon. On the left, there's a carbon directly bonded to an oxygen. There must be a hydrogen going away from us, and this carbon is directly bonded to another carbon, so let's assign priority to our four groups, so here is our chiral center. We look at the atoms directly bonded to that carbon. There are three carbons, one, two, three, and a hydrogen. We know the hydrogen has the lowest atomic number, so hydrogen is the lowest priority and we call that group four. Now we have three carbons. We have a tie because carbon, of course, has the same atomic number. To break a tie, we look at the atoms that are bonded to those carbons, so let's start with the carbon on the right here, so the carbon on the right is directly bonded to one carbon and then two hydrogens, so we write carbon, hydrogen, hydrogen, so in decreasing atomic number, in order of decreasing atomic number. This carbon down here is directly bonded to three hydrogens so one, two, and three. The carbon on the left is directly bonded to an oxygen, a carbon and a hydrogen, so we put these in order of decreasing atomic number, so we put oxygen first, then carbon, then hydrogen. Let's look at the first point of difference here so the first atom, we're comparing a carbon to a hydrogen to an oxygen. Out of those atoms, oxygen has the highest atomic number which means that this group gets highest priority so the group with the OH has the highest priority. This gets a number one. Next is carbon 'cause carbon has a higher atomic number than hydrogen so that wins it for this group so this group on the right, this ethyl group, gets second highest priority and finally, the methyl group would get third highest priority so now that we've assigned priority, step two is to orient the molecule so that the lowest priority group is pointing away from you in space, and here are hydrogens coming out at us in space, so that's not what we want. We want the hydrogen to point away from us in space, so one thing we could do is to think about an axis through our chiral center and we could rotate the molecule about that axis and that would put the hydrogen going away from us in space, so let's go to the video so we can visualize what happens. So here's our compound. You can see at carbon two, there's an OH coming out at us in space and at carbon three, there's a methyl group going away from us in space. If we rotate about an axis through this carbon, let's go ahead and do that, we can see what happens to those two groups. Now, now at carbon two, the OH is going away from us in space and at carbon three, the methyl group is coming out at us in space. Let's draw our compound with the hydrogen going away from us, so first we put in our carbon chain, and we know at carbon two our OH is gong away from us so here is our OH, and then at carbon three, our methyl group is coming out at us so let me go ahead and put that in, so there's a CH3 coming out, and there's a hydrogen going away from us in space. We can't really see it very well in the picture but we know it's there, so we know that the group with the OH had the highest priority so this was number one. We know the ethyl group had the second highest priority so we say that's a number two. The methyl group was a number three and the hydrogen was a number four, so with our lowest priority group going away from us, we can now ignore it and see that we're going around in this direction so we're going around counterclockwise and that would be S so we are S at carbon three, so I could finish off the name by writing it in here, 3S, so it's 2R, 3S, 3-methyl-2-pentanol. Now, you didn't have to do this whole trick with rotating about an axis. I showed you another way to do it in an earlier video so let's go back to our original drawing over here on the left. Let me use dark blue. There is a trick that you can use, so if your hydrogen's coming out at you in space, you can just ignore it for the time being and look at one, two, and three, and one, two, and three are going around this way, they're going around clockwise, so it looks R, it looks like it's R but since the hydrogen is coming out at you in space, you can just reverse it. You can take the opposite. If it looks R with the hydrogen coming out at you, it must be S, so that's a nice trick and it means you don't have to rotate the molecule in your head.