For a molecule with one chiral center
there are two different
enantiomeric possibilities, and we're
gonna want a way to differentiate
between them
in other words we want a way to assign
absolute configuration
to any chiral center. in order to do that
we're going to use something called that
cahn-ingold-prelog
convention. so what we're gonna do his
first we have to identify any chiral center
so here we can see that there is one chiral center on this molecule
because this carbon is connected to 4 different groups
it is connected to a hydroxyl group, the
implied hydrogen
which I'll go ahead and right now, a 2 carbon chain
and a 1 carbon chain. four different
groups, that is a chiral center
now what we're gonna need to do
is we're going to need to assign priority
to each of those four groups, and we're going to do that
by assessing atomic mass one atom at a time
in other words without looking at an extended chain
we're gonna see that this carbon atom
that represents the chiral center
is connected to an oxygen atom, not a
hydroxyl group but just simply
an oxygen atom, a carbon atom over here
a carbon atom over here, and a hydrogen
atom over there
so because we're going by atomic mass we know immediately that
the oxygen atom is going to be priority one
because that is the heaviest atom of
the atoms in question. likewise the hydrogen
is the lightest atom so that is certainly going to be
priority four.
it is these two carbon atoms that are equivalent
and if you hit an equivalency you're
going to have to extend
further down the chain until you reach a
point of difference
now this carbon atom
happens to be bound to the three
implied hydrogen atoms so that's hydrogen
hydrogen hydrogen... this
carbon atom is bound to to implied hydrogen atoms
and a carbon atom, so hydrogen hydrogen carbon
now one way to look at it is
as though these hydrogens cancel
these hydrogen cancel, then we're pitting
a carbon atom
versus a hydrogen atom, and because
carbon is heavier than hydrogen
that means that this carbon is going to get priority two
over this one which will now be priority three
now I just want to reiterate
what we have just said because this is
often a point of confusion
I want to stress the fact that as you
assign priority via this convention
you must go one atom at a time
so it is not the case that this is priority two
over that carbon because an ethyl group
is heavier than a methyl group
that is not the case, what is the case is
that they are identical carbon atoms
but then because we have to go one point
of difference further
we see that hydrogen hydrogen hydrogen
in totality weighs less than hydrogen, hydrogen
carbon, so that is why this gets priority
two over this
so on the basis of atomic mass we have
assigned priority to these four groups
we have 1, 2, 3, 4. in order to assign
absolute configuration you have to place yourself
or rotate the molecule in such a way so as to put
the lowest priority group away from you.
now fortunately in this case that is already
how the molecule is sitting because the
number four priority group, the lowest
priority groups is on this dash bond meaning it is
further away from us than the board so
just looking directly upon this molecule we are already
in the privileged perspective to be able to assign
absolute configuration. if that wasn't the case
and the number four group was in the
plane of the board or coming towards us
we would have to do something to either
redraw the molecule or rotate the
molecule in our minds or do some other
trick which we will discuss later.
but for the time being let's just be
happy that that's already away from us
and so now what we're going
to do is we're going to make a circle
from group 1 to 2 to 3, so here's one
here's two here's three, if the circle
went clockwise then that
is an R stereocenter.
if the circle went counterclockwise that would be
an S stereocenter. so these are the two
absolute configurations for any chiral
center because we have determined that
this is clockwise and therefore R.
that means that this is
(R)-2-butanol
so we already knew that this was 2-butanol based on
our knowledge of IUPAC nomenclature, however now
we are able to assign the extended
formality of the absolute configuration
because this is going to be either R or S
and in this particular case it is R
you're probably going to have to assign
absolute configuration on a number of
different kinds of chiral centers
and it may not always be the case that
the lowest priority substituent is
already away from you
so we're gonna have to be able to sign
absolute configuration
regardless of how the substituents are oriented
in space or on paper, so let's take a
look at a couple examples
for one, let's look at the opposite of
the example we just looked at
so this is a chiral center but in this case
the lowest priority group is towards us
so we are not in the privileged
perspective to be able to assign absolute configuration
so there's a couple things we can do, if
you are very good with
spatial orientation and that kind of
thing what you could do
is simply place yourself in a different position in space
mentally, so for example if I was on the
other side of this board
looking back this way towards you then I would be
in the appropriate orientation because
this hydrogen atom
that is projecting towards you right now
would then be away from
me and from that perspective I could see that bromine
ethyl, methyl, that that would be S
that would be counterclockwise
from my perspective behind the board. however,
a lot of people are maybe not as comfortable with that
and so there's other tricks that we can do. for example
we can arbitrarily switch the bromine and the hydrogen
and place the bromine on the wedge and
the hydrogen on the dash
now this is sort of a cheat because
we're redrawing a completely different molecule
but what we're doing is we're saying
that if I switch the bromine and hydrogen
it is going to be easier for me to
assign the absolute configuration
because now
I can just look at it the way we're
looking at the board right now
and we can go like this, that looks like
R, so but the thing is that we have
we have inverted the chiral center
to the opposite configuration so if we
do this kind of cheating method
what we're going to have to do is invert the answer we get
so we swapped them, the swapped molecule
looks like R, therefore the original
must have been S.
we can kind of swap things to make it
easier for ourselves
but then we must invert our answer
because it is the case
that if you switch the position of any two groups
on a chiral center you will invert
the stereochemistry of that chiral center.
so let's take a look at another example
but now it's the case that the lowest priority group,
the hydrogen, is in the plane of the board so
it's not gonna work switching the dash and the wedge bonds
is not going to help because that is not
going to be changing the positioning
of the lowest priority group. now again, if
you're very good with spatial reasoning
what you can do is orient yourself in a
position to imagine
that the lowest priority group is away
from you, so imagine I'm
sitting in the plane of the board and
i'm looking this way,
now the hydrogen is away from me
and I can see that the bromine is up here
the chlorine is down and to my left
because this wedge bond projects the chlorine
out this way, and then that fluorine is
down and to the right.
so I'm seeing bromine, chlorine, fluorine,
from my perspective
that looks like S. now once again not
everyone is comfortable with this kind of reasoning
so what we can do is apply the same rule that if you swap
any two groups on a chiral center you are going to
invert the stereochemistry of that
chiral center but
if we say for example switch the hydrogen
and the fluorine now the reason we would do that is because
if we place the lowest priority group
away from us, which would
be the case if it was on the dash bond that would be furthest away from us
so we can arbitrarily swap the hydrogen and the fluorine
like this and now simply by looking in
the direction towards the board
we are able to assess what's going on
and so we would see bromine, chlorine, fluorine
that would be clockwise
so that appears to be R, however once again
if we do this some kind of cheating thing like this
we have to then invert our answer so
in inverting the stereocenter we made
something that looks like R,
that means that the original must have been S.
so when you are assigning absolute
configuration for a chiral center
where the lowest priority group is not
already away from you
you have two options. you can change in
your mind where you are located and
the way you are looking at the molecule,
if you're very good with spatial reasoning
or if you are not you can swap any two groups
in that chiral center, but then you have to realize that
what it appears to be after you swap
them, you must then
invert to get the absolute configuration
reflective of the original molecule.
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