Cahn-Ingold-Prelog System for Naming Enantiomers

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Cahn-Ingold-Prelog System for Naming Enantiomers

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Right now, based on what we know so far, if we wanted to
name this molecule, we would say, well, what's the longest
carbon chain here?
Well, we have a two-carbon chain, and there's all single
bonds, so we're dealing with an ethane.
Actually, I'll write it all at once.
And then we have on the one carbon, we can call this the
one carbon, and call this the two carbon, we have a bromine
and a fluorine.
So we could call this 1-bromo, and we're putting the bromo
instead of the fluoro because B comes before F
alphabetically.
1-bromo-1-fluoro, and then we're dealing with an ethane.
We have a two-carbon chain, all single bonds,
fluoroethane.
That's the name of that molecule there, just a review
of some of the earlier organic nomenclature
videos we had done.
Now, we know immediately, based on the last few videos,
that this is also a chiral carbon, and if we were to take
its mirror image, we would get another enantiomer of this
same molecule, or that they are enantiomers of each other.
So what is the mirror image of this
1-bromo-1-fluoroethane look like?
Well, you'd have the carbon right here.
I want to get all the colors right.
You would still have the bromine up above.
You would have this methyl group that's attached to the
carbon now pointing in the left direction, CH3.
The fluorine would now still be behind the carbon, and now
the hydrogen would still pop out of the page, but it would
now pop out and to the right.
That is the hydrogen.
Now, based on our naming so far, we would name this
1-bromo-1-fluoroethane, and we would also name this
1-bromo-1-fluoroethane, but these are fundamentally two
different molecules.
Even though they have the same molecules in them; they have
the same molecular formula; they have the same
constitution in that this carbon is connected to a
hydrogen, a fluorine, and a bromine; this carbon is
connected to the same things; this carbon is connected to a
carbon, three hydrogens; so is this one; these are
stereoisomers.
These are stereoisomers, and they're mirror images of each
other, so they're enantiomers.
And actually, they will, one, polarize light differently,
and they actually can often have very different chemical
properties, either in a chemical or biological system.
So it seems not good that we have the same
names for both of these.
So what we're going to focus on in this video is how do you
differentiate between the two?
So how do we differentiate between the two?
So the naming system we're going to use right here is
called the Cahn-Ingold-Prelog system, but it's a different
Cahn, it's not me.
It's C-A-H-N instead of K-H-A-N.
Cahn-Ingold-Prelog system, and it's a way of differentiating
between this enantiomer, which right now we would call
1-bromo-1-fluoroethane, and this enantiomer,
1-bromo-1-fluoroethane.
It's a pretty straightforward thing.
Really, the hardest part is to just visualize rotating the
molecules in the right way and figuring out in which
direction it's kind of-- whether it's kind of a
left-handed or right-handed molecule.
We're going to take it step by step.
So the first thing you do in the Cahn-Ingold-Prelog system
is to, one, identify your chiral molecule.
Here, it's pretty obvious.
It's this carbon right here.
We'll just focus on this left one, the one we started with
first. It's bonded to three different groups.
And then what you want to do is you want to rank the groups
by atomic number.
So if you go up here, out of bromine, hydrogen, fluorine,
and a carbon, this is what is bonded directly to this
carbon, which has the highest atomic number?
Bromine is over here-- let me do this in a darker color.
We have bromine at 35, we have fluorine at 9, we have carbon
at 6, and then we have hydrogen at 1.
So of all these, bromine is the largest. We'll just call
this number one.
Then after that, we have fluorine.
That is the number two.
Number three is the carbon.
And then hydrogen is the smallest, so
that is number four.
So now that we've numbered them, the next step is to
orient this molecule so that the smallest atomic number
group is sitting into the page.
It's sitting behind the molecule.
Right now, this hydrogen is the smallest of all of them.
Bromine's the largest, hydrogen is the smallest, so
we want to orient it behind the molecule.
The way it's drawn right now, it's oriented in
front of the molecule.
So to orient it behind the molecule, and this really is
the hardest part is just to visualize it properly.
Remember, this fluorine is behind; this is right in the
plane of the paper; this is popping out of the paper.
We would want to rotate.
You could imagine we'd be rotating the molecule in this
direction so that-- let me redraw it.
We have the carbon here.
And now since we've rotated it like this, we've rotated it
roughly 1/3 around the circle, so it's about 120 degrees.
Now, this hydrogen is where the fluorine was.
So that's where the hydrogen is.
The fluorine is now where this methyl group is.
These dotted lines show that we're behind now.
This shows that we're in the plane.
And the methyl group is now where the hydrogen is.
It's now popping out of the page.
It's going to the left and out.
So this methyl group is now popping out of the page, out
and to the left.
That's where our methyl group is.
So all we've done is we've just rotated this around about
120 degrees.
We've just gotten this to go behind, and that's kind of the
first step after we've identified the chiral carbon
and ranked them by atomic number.
And.
Of course.
The bromine is still going to be on top.
Now, once you put the smallest atomic number molecule in the
back, then you want to look at the rankings
of one through three.
And we have four molecules here.
We look at the largest, which is bromine, number one.
Then number two is fluorine, number two, and then number
three is this methyl group.
That's the carbon that's bonded to this carbon, so it's
number three right there.
And in the Cahn-Ingold-Prelog system, we literally just
think about what would it take to go from number one to
number two to number three?
And in this case, we would go in this direction.
To go from number one to number two to number three, we
would go in the clockwise direction.
We're just ignoring the hydrogen right now.
That's just sitting behind it.
That was the first step, to orient it so it's sitting in
the back, the smallest molecule.
And then the three largest ones, you just say what
direction do we have to go to go from number one to number
two to number three?
In this case, we have to go clockwise.
And if we go clockwise now, then we call this a
right-handed molecule, or we use the Latin word for right,
which is rectus.
And so we would call this molecule right here not just
1-bromo-1-fluoroethane, this is R, R for rectus.
Or you could even think right, although we'll see left is
used as S, which is sinister, so the Latin is really where
the R comes from.
But this is (R)-1-bromo-1-fluoroethane
That's this one right here.
So you might guess, well, this must be the opposite, this
must be the counterclockwise version.
We can do it really fast.
So same idea.
We know the largest one.
Bromine is number one.
That's the largest in terms of atomic number.
Fluorine is number two.
Carbon is number three.
Hydrogen is number four.
What we want to do is put hydrogen in the back, so what
we're going to have to do is rotate it to the back to where
fluorine is right now.
So if we had to redraw this molecule right here, you'd
have your carbon still.
You still have your bromine sitting on top.
But we're going to put the hydrogen now to the back, so
the hydrogen is now where the fluorine used to be.
The hydrogen's there.
This methyl group, this carbon with the three hydrogens, is
going to be rotated to where the hydrogen used to be.
It's now going to pop out of the page, because we're
rotating it in that direction, so this is our methyl group
right there.
And then this fluorine is going to be moved where the
methyl group was, so this fluorine will go right here.
And now, using the Cahn-Ingold-Prelog system,
this is our number one, this is our number two, just by
atomic number, this is number three.
You go from number one through number two to number three.
You go in this direction.
You're going counterclockwise.
Or we are going to the left, or we use the Latin word for
it, which is sinister.
And the word sinister comes from the Latin word for left,
so I guess right is good, and people thought either
left-handed people were bad, or if you're not going to the
right, it's bad.
I don't know why sinister took on its sinister meaning now in
common language.
But it's now the sinister version of the molecule.
So we would call this version, this enantiomer of
1-bromo-1-fluoroethane, we would call this S, S for
sinister, or for left, or for counterclockwise:
(S)-1-bromo-1-fluoroethane.
So now we can differentiate the names.
We know that these are two different configurations.
And that's what the S and the R tell us, that if you have to
go from this to this, you would literally have to detach
and reattach different groups.
You'd actually have to break bonds.
You actually have to swap two of these groups in some way in
order to get from this enantiomer to this enantiomer.
They're different configurations, really
fundamentally different molecules, stereoisomers,
enantiomers, however you want to--