R,S (Cahn-Ingold-Prelog) naming system example 2

Let's see if we can name this molecule using the-- sometimes called the R-S system, or the Cahn-Ingold-Prelog system. And the first thing to do is just to see if there are any chiral centers in this molecule. If there aren't, then we don't even have to use the R-S system. We can just use our standard nomenclature rules and we'd be done. So if we look here, this carbon is attached to three hydrogens, so it's definitely not attached to four different groups. Same thing about this carbon right here. This carbon right here is attached to a fluorine, but then it's attached to two methyl groups. So it's the same group, so this is also not a chiral carbon or an asymmetric carbon. This carbon right here is attached to a hydrogen and three other carbons, but each of these three carbons look like different groups. This carbon is attached to two methyls and a fluorine. This carbon is attached to two hydrogens and a bromine. This carbon is just a methyl group. So this right here does look like a chiral center. It does look like a chiral carbon, and the other ones don't. This is just a methyl group. It has three hydrogens, so definitely not attached to four different groups. And this is attached to two hydrogens, and those are obviously the same group, so this is also not a chiral center. So we have one chiral center, so the R-S naming system will apply. But a good starting point will just be naming it using our standard nomenclature rules. And to do that we look for the longest carbon chain here. Let's see, if we start over here, and I don't know what direction I'm going to name it from yet, but I just want to identify the longest chain. If we went from here, we have one, two, three. We can either go to four or to four there, so we definitely have four carbons, Four carbon, longest chain. So that tells us that we will be using the prefix but-, or it will be a butane, because they're all single bonds here, so it is a butane. But to decide whether we branch off, it doesn't matter whether we use this CH3 or this CH3, they're the same group. But to decide whether we use this part of the longest chain or we use that, we think about the rule that the core chain to use should have as many simple groups attached to it as possible, as opposed to as few complex groups. So if we used this carbon as part of our longest chain, then this will be a group that's attached to it, which would be a bromomethyl group, which is not as simple as maybe it could be. But if we use this carbon in our longest chain, we'll have two groups. We'll have a bromo attached, and we'll also have a methyl group. And that's what we want. We want more simple groups attached to the longest chain. So what we're going to do is we're going to use this carbon, this carbon, this carbon, and that carbon as our longest chain. And we want to start from the end that is closest to something being attached to it, and that bromine is right there. So there's going to be our number one carbon, our number two carbon, our number three carbon, and our number four carbon. And then we can label the different groups and then figure out what order they should be listed in. So this is a 1-bromo and then this will be a 2-methyl right here. And then just a hydrogen. Then three we have a fluoro, so on a carbon three, we have a fluoro, and then on carbon three, we also have a methyl group right here, so we also have a 3-methyl. So when we name it, we put in alphabetical order. Bromo comes first, so this thing right here is 1-bromo. Then alphabetically, fluoro comes next, 1-bromo-3-fluoro. We have two methyls, so it's going to be 2 comma 3-dimethyl. And remember, the D doesn't count in alphabetical order. Dimethylbutane, because we have the longest chain is four carbons. Dimethylbutane. So that's just the standard nomenclature rules. We still haven't used the R-S system. Now we can do that. Now to think about that, we already said that this is our chiral center, so we just have to essentially rank the groups attached to it in order of atomic number and then use the Cahn-Ingold-Prelog rules, and we'll do all that in this example. So let's look at the different groups attached to it. So when you look at it, this guy has three carbons and a hydrogen. Carbon is definitely higher in atomic number on the Periodic Table. It has an atomic number 6. Hydrogen is 1. You probably know that already. So hydrogen is definitely going to be number four. So let me put number four there next to the hydrogen. And let me find a nice color. I'll do it in white. So hydrogen is definitely the number four group. We have to differentiate between this carbon group, that carbon group, and that carbon group. And the way you do it, if there's a tie on the three carbons, you then look at what is attached to those carbons, and you compare the highest thing attached to each of those carbons to the highest things attached to the other carbons, and then you do the same ranking. And if that's a tie, then you keep going on and on and on. So on this carbon right here, we have a bromine. Bromine has an atomic number of 35, which is higher than carbon. So this guy has a bromine attached to it. This guy only has hydrogen attached to it. This guy has a fluorine attached to it. That's the highest thing. So this is going to be the third lowest, or I should say the second to lowest, because it only has hydrogens attached to it, so that is number three. The one has the bromine attached to it is going to be number one, and the one that has the fluorine attached to it is number two. And just a reminder, we were tied with the carbon, so we have to look at the next highest constituent, and even if this had three fluorines attached to it, the bromine would still trump it. You compare the highest to the highest. So now that we've done that, let me redraw this molecule so it's a little bit easier to visualize. So I'll draw our chiral carbon in the middle. And I'm just doing this for visualization purposes. And right here we have our number one group. I'll literally just call that our number one group. So right there that is our number one group. It's in the plane of the screen. So I'll just call that our number one group. Over here, also in the plane of the screen, I have our number two group. So let me do it like that. So then you have your number two group, just like that. And then you have your number three group behind the molecule right now the way it's drawn. I'll do that in magenta. So then you have your number three group. It's behind the molecule, so I'll draw it like this. This is our number three group. And then we have our number four group, which is the hydrogen pointing out right now. And I'll just do that in a yellow. We have our number four group pointing out in front right now. So that is number four, just like that. Actually, let me draw it a little bit clearer, so it looks a little bit more like the tripod structure that it's supposed to be. So let me redraw the number three group. The number three group should look like-- so this is our number three group. Let me draw it a little bit more like this. The number three group is behind us. And then finally, you have your number four group in yellow, which is just a hydrogen that's coming straight out. So that is coming straight out of-- well, not straight out, but at an angle out of the page. So that's our number four group, I'll just label it number four. It really is just a hydrogen, so I really didn't have to simplify it much there. Now by the R-S system, or by the Cahn-Ingold-Prelog system, we want our number four group to be the one furthest back. So we really want it where the number three position is. And so the easiest way I can think of doing that is you can imagine this is a tripod that's leaning upside down. Or another way to view it is you can view it as an umbrella, where this is the handle of the umbrella and that's the top of the umbrella that would block the rain, I guess. But the easiest way to get the number four group that's actually a hydrogen in the number three position would be to rotate it. You could imagine, rotate it around the axis defined by the number one group. So the number one group is just going to stay where it is. The number four is going to rotate to the number three group. Number three is going to rotate around to the number two group, and then the number two group is going to rotate to where the number four group is right now. So if we were to redraw that, let's redraw our chiral carbon. So let me scroll over a little bit. So we have our chiral carbon. I put the little asterisk there to say that that's our chiral carbon. The number four group is now behind. I'll do it with the circles. It makes it look a little bit more like atoms. So the number four group is now behind where the number three group used to be, so number four is now there. Number one hasn't changed. That's kind of the axis that we rotated around. So the number one group has not changed. Number one is still there. Number two is now where number four used to be, so number two is now jutting out of the page. And then we have number three is now where number two was. So number three is there. And now that we've put our fourth group behind the molecule, we literally just figure out whether we have to go clockwise or counterclockwise to go from one, two to three. And that's pretty straightforward. To go from one to two to three, we have to go counterclockwise. Or another way to think of it, we're going to the left, counterclockwise. At least on the top of the clock, we're going to the left. And so, since we're going to left, this is S or sinister. This is S, which stands for sinister, which is Latin for left. So we're done. We've named it using the R-S system. This molecule is (S)-- sinister-- 1-bromo-3-fluoro-2,3-di--