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--