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- [Lecturer] In the last
video, we looked at a Fischer
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projection of a compound that
had only one chirality center.
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This molecule has two, and
our goal is to determine
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the configuration of
each chirality center,
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so at the intersection of
these lines here, we know that
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this is a chiral center
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and then we have another
one down here, so those are
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the two chirality centers.
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Let's focus in on the top
one here, and let's assign
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a configuration to that carbon.
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Let me go ahead and draw
the carbon on the right.
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We know that in a Fischer
projection, the horizontal lines
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represent a bond that's
coming out of the page at us,
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so over here on the left,
the bond to this hydrogen is
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coming out of the page and
so is the bond to this OH,
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so let's draw them on the right.
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We have a bond to a hydrogen
coming out of the page
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on the left, and then we
have a bond to OH coming out
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of the page on the right, so those,
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those are represented by a wedge.
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We know that the vertical
lines in a Fischer projection
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represent a bond that's
going away from us in space,
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so this line right here is
a bond that's going away
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from us, so that would be a
dash, so let's draw that in,
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and that's going to a
carboxylic acid which we know
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is a carbon double bonded to
an oxygen, bonded to an OH.
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Let's look at the other vertical line,
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so this one down here,
so that represents a bond
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going away from us in space,
so we draw that one in,
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so we're essentially staring
down at our chiral center.
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And this carbon down here is
directly bonded to an oxygen,
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a carbon, and a hydrogen,
and those are the only atoms
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that we need to assign a
configuration so those are
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the only ones that I'm going to draw in.
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Let's go back to our chiral
center, so this carbon right
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here, we know to assign a
configuration, we look at
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the atoms that are directly
bonded to that carbon.
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There is a hydrogen, an oxygen
and then our two carbons.
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Out of those atoms, oxygen
has the highest atomic number
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so the OH group gets the highest
priority and we say that is
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a number one.
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The hydrogen has the lowest
atomic number so that's
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the lowest priority, so we
say that's a number four.
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Next, we have carbon versus
carbon, so this carbon
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versus this carbon and we
know that carbon has the same
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atomic number, and so we've
seen how to break our tie
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in earlier videos.
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We look at the atoms that
are directly bonded to those
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carbons.
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This bottom carbon here is
directly bonded to an oxygen,
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a carbon, and a hydrogen,
so we put those in order
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of decreasing atomic number,
so that's oxygen, carbon,
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hydrogen.
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This top carbon here, the
one in our carboxylic acid,
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has a double bond to an oxygen,
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and so we treat that like it's
two bonds to two different
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oxygens, even though it's
really only one double bond
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to one oxygen, and then we
have another bond to an oxygen
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on the right here so we
could say, for the purposes
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of assigning a configuration,
that would be oxygen,
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oxygen, oxygen.
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Next, we compare so we have
an oxygen versus an oxygen,
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so that's a tie.
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We go on to our next atom
which is oxygen versus carbon,
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and obviously, oxygen wins.
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It has the higher atomic
number, so the carboxylic acid
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group gets the higher priority
so this must be group two,
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and this group down here
must be group three.
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Once we've done this, our next
goal is to put the hydrogen
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going away from us in space,
but I'm going to use the trick
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that I talked about in the
last video because it means
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you don't have to move the
molecule around in your mind.
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You can just use this little trick.
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And the trick was to ignore
the hydrogen for the time being
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and just look at one, two, and three.
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Here's one, here's two, and
here's three, so if we go
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around in a circle from one
to two to three, we're going
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around in this direction.
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That is counterclockwise, and
counterclockwise we know is S.
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So this looks S, but the
hydrogen is coming out at us
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in space and so we know
all we have to do is take
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the opposite so even though
it looks S, since the hydrogen
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is coming out at us, we
know it's actually R,
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so this is actually, actually R,
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so if you actually made
this molecule and rotated it
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and put the hydrogen going
away from you in space,
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then you would see that it's
R directly, but this trick
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allows you not to worry about
that, so I find this to be
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the easiest way to assign a configuration.
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For me, it's definitely the fastest.
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Let's move on to our
other chirality centers,
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so this carbon down here,
so let me draw that one in,
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and let's look at our
horizontal lines, so we have
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a bond to an OH on the left,
and a bond to a hydrogen
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on the right and those are
coming out at us, since those
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are horizontal lines so
that OH is on the left,
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and we have a hydrogen on the right.
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Let's look at our vertical
lines, so this line right here
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means a bond going away from
us in space, so we're staring
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down at our chiral center,
and that's a bond to a carbon
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and that carbon is directly
bonded to an oxygen,
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a carbon, and a hydrogen,
so let's draw those in.
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Again, we don't need all of
the atoms because this is all
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we need to assign a
configuration so that's all
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I'm drawing in, and then let's
look at this other vertical
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line, so this one down here,
so that's a bond going away
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from us in space to this
carbon, so let's draw that in
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so we have a dash over here on
the right, going to a carbon
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and this carbon is directly
bonded to two hydrogens
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and an oxygen, so let's
draw that in, so we have our
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two hydrogens and an oxygen.
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Going back to our chiral centers,
so this carbon right here,
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and we look at our atoms that
are directly bonded to it
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and we have an oxygen, a
hydrogen, and two carbons,
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so we know that the oxygen
has the highest priority
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so once again, the OH group
is the highest priority group
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and the hydrogen is the lowest
priority group and we have
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a tie again between the carbons,
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so the carbon versus carbon,
the top carbon is directly
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bonded to oxygen, carbon,
hydrogen, so in order
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of decreasing atomic number,
oxygen, carbon, hydrogen,
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the bottom carbon would be
oxygen, hydrogen, hydrogen,
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so oxygen, hydrogen, hydrogen.
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We look for the first point
of difference so we have
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oxygen versus oxygen, so that's a tie.
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Next, we have carbon versus
hydrogen and we know that
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carbon has the higher atomic
number so this top group here
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gets higher priority so
this should be a number two
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and this group down here
should be a number three.
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Let's use our trick again,
so we ignore the hydrogen
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for the time being and we
look at one, two, and three,
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so one to two to three is
going around in this direction
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which we know is clockwise, so
we can say that this looks R
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because clockwise is
R, but because this is
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a Fischer projection, the
hydrogen is coming out at us
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in space, so we know our trick
is just to take the opposite
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of how it looks, so if it
looks R, we can say that it's
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actually S, so we write
down here actually,
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actually S.
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All right, so we've done it.
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We've determined the configuration
of each chirality center
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so this chiral center, let
me use a different color,
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let me use a red over here,
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so this chiral center,
we said that one was R,
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so it's R at this carbon and
then at this carbon down here
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it is S.
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Let's also practice drawing the
enantiomer of this compound.
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We saw how to do that in the last video.
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We put a mirror right here
and we reflected our groups
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in the mirror so this OH here,
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I like to draw a dashed line,
so we just reflect that,
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and therefore we would have
this, and then on this side,
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we would have a hydrogen so
we're reflecting that hydrogen
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and then let's go to
this hydrogen down here,
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so we just reflect this in our mirror
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and then we draw our horizontal
line and then we would
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have an OH, so we draw in an OH here.
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We can put in our vertical
line like that and then we have
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a carboxylic acid, but this
carbon is not a chiral center
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so we don't really need to worry about it.
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We can just write in
COOH here, and then same
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with this carbon, not a chiral
center so we can just write
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CH2OH.
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The important part is if
we have an OH on the right,
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it'll be on the left in the mirror image.
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If we have an OH on the left
in a Fischer projection,
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it'll be on the right in a mirror image,
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so now we have a pair of
enantiomers, so these two
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are mirror images, right?
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The one on the right is
a mirror image of the one
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on the left, but the one on
the right is not superimposable
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on the one on the left.