Cahn-Ingold-Prelog System for Naming Enantiomers

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 call them.