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- 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--