- [Instructor] A meso compound is a compound that has
chirality centers but is achiral.
We're gonna come back to this definition in a few minutes.
Right now, let's focus in on this drawing.
And our goal is to draw all possible stereoisomers
for this dot structure.
So we know from earlier videos that this carbon
is a chirality center, and so is this one.
And we would expect two to the n stereoisomers,
where n is the number of chirality centers.
And since we have two chiral centers here,
we would expect two to the second power,
or four stereoisomers.
So this is really just a maximum number,
so I'm gonna put a question mark right here.
So do we get four stereoisomers?
Let's draw out the four possibilities.
Our first stereoisomer could have both bromines
coming out at us in space.
So let me go ahead and draw that in.
So we could have both bromines coming out at us.
For the second possibility, we might have both bromines
going away from us in space.
So I'll draw that in there.
For the third possibility, we could have one bromine up
and one bromine down.
So I'll put those in.
And for the fourth, just reverse them.
Have the top bromine down and the bottom bromine up.
like that.
Well, all right, let's examine the relationships
between our stereoisomers.
And let's start with stereoisomer possibility
three and four.
So let's compare these and let's figure out
the relationship.
On the left is stereoisomer three.
We can see there's a bromine coming at us
and a bromine going away from us.
On the right is stereoisomer four.
Now we have a bromine going away from us at the top carbon
and a bromine coming out at us here.
So let's compare our two stereoisomers.
If I rotate the one on the right, we can see these are
mirror images of each other.
And if I try to superimpose one on top of the other,
here we get one pair of bromines to line up,
but the other pair doesn't match.
If we try to get the other pair of bromines to line up,
now the first pair doesn't match.
So these are non-superimposable mirror images of each other.
These are enantiomers.
So three and four are enantiomers.
They are non-superimposable mirror images.
And we could have guessed that by looking at the drawing,
because at this carbon we have bromine coming out
at us in space, and then now we have bromine
going away from us in space for the other stereoisomer.
So that's an opposite configuration at that carbon.
And at this carbon, we go from a dash to a wedge.
So that's an opposite configuration at this one too.
So since we have opposite configurations at all
chirality centers, we would expect these two
to be enantiomers of each other.
What about the relationship between one and two?
Well, at first we might say oh, those are enantiomers
because here we have a wedge, and then over here
we have a dash and here we have a wedge,
and here we have a dash.
So that should be the opposite configuration
at both chirality centers.
So those might be enantiomers.
But let's go to the video to see if that's true.
On the left is a model of drawing one,
with the two bromines coming out at us in space.
On the right is a model of drawing two,
with the two bromines going away from us in space.
And if I rotate the model on the right,
we can see that these are mirror images of each other.
But they are superimposable mirror images.
So if I put that one on top of the other,
you'll see that they are superimposable.
So these actually are two models of the same molecule.
This is a meso compound.
It's a compound that has chirality centers,
but it is achiral, the mirror image is superimposable.
So one and two really represent the same molecule.
This is a meso compound, a compound that has
chirality centers but is achiral.
The mirror image is superimposable on itself.
So we thought we would have four stereoisomers,
but really we only have three.
We have a pair of enantiomers and we have one meso compound.
So to look for a meso compound, one thing you could do
is what we did in the video.
We had the mirror image and we were able to superimpose
the mirror image on itself.
Another way to look for a meso compound is to look for
a plane of symmetry.
So if I draw a line here, think about this as being a plane,
and look for symmetry on either side.
So you can see, it's symmetrical.
I drew in the plane of symmetry with a dashed line here,
but it's hard to visualize it.
So up here is a better picture.
Here you can see the plane dividing the molecule in half.
And on the left side, we have our bonds here
and then we have our bromine going up
and our hydrogen going down.
The right side is symmetrical with the left side.
So look for symmetry on both sides of the plane.
Let's do another example.
This one's a little bit harder than the last one.
We know that we have two chiral centers.
So that's a chiral center and so is this one.
So we would expect two to the second stereoisomer.
So that's, of course, four.
So I'll put a question mark here again
because we're not sure if we actually will get
four stereoisomers.
That's a maximum number.
Down here I have the four possibilities.
So I've drawn them out just to save some time.
So we have one, two, three and four.
And let's examine the relationship between one
and two first.
On the left is stereoisomer one,
and I've left the hydrogens off the methyl groups
and the OH just so we can see the models better.
So here's our carbon chain, and we have both OHs
coming out at us in space.
And then for stereoisomer two, here's the carbon chain,
and we have both OHs going away from us.
So I'm gonna hold them in the way they are in the drawing
and I'm gonna rotate the one on the right.
And when I do that, we can see that these are mirror images
of each other.
But if I try to superimpose one on the other,
so I'll just rotate this one back here and flip it over,
you can see they don't match up.
So the atoms don't line up here.
So they're non-superimposable.
Doesn't matter how you do it.
I'll rotate it again, and we can see we can't
superimpose our atoms.
So these are non-superimposable mirror images of each other.
These are enantiomers.
And we could check for a plane of symmetry,
so I could take one of these and I can rotate it
so we have our hydrogens going away from us in space.
And I look for a plane of symmetry, but I don't see one.
So this is not a meso compound.
So one and two are enantiomers.
They're non-superimposable mirror images of each other.
And we could've guessed that because if we look at
our chiral centers here, so this one has an OH
coming out at us, and then that one has it going
away from us, this one has an OH coming out at us
and this one has it going away from us,
so we have opposite configurations at both
chirality centers.
Let's look at three and four next.
So what is the relationship between these two?
Well, first, we might think these could be enantiomers
because at this carbon, we have OH on a wedge,
and then here we have OH on a dash.
And then here we have OH on a dash,
and here we have it on a wedge.
So that might be your first guess.
But let's look at the video and let's look at
the model sets to help us out.
Remember, I'm leaving the hydrogens off the methyl groups
and the hydrogens off the oxygens in the video
just to help us see the molecule more clearly.
On the left, we have a model of drawing three.
So here's our carbon chain with an OH going away from us
in space, and an OH coming out at us in space.
On the right is a model of drawing four.
Here's our carbon chain with an OH coming out at us
in space and an OH going away from us in space.
So I'll hold the two models and we'll compare them.
First let's see if they are mirror images of each other.
So I'll take the one on the right and I'll rotate it
and I'll hold it up next to the one on the left.
And now we can see that these two are mirror images
of each other.
So next, let's see if one is superimposable on the other.
So I'll go back to the starting point and I'll rotate it
around like that, and let's see if we can superimpose
the one on the right, the mirror image,
on the molecule on the left.
And notice that we can.
All of the atoms line up.
So all of the hydrogens, carbons and oxygens
are in the same place.
So this is a compound that has chirality centers,
but it is achiral, the mirror image is superimposable
on itself.
So we should be able to find a plane of symmetry.
So I'll just pick one of these models,
doesn't matter which one because they represent
the same compound, and I'll rotate, I'll rotate it around
so we can see a plane of symmetry.
So right there is our plane of symmetry.
This is a meso compound.
So three and four actually represent the same compound.
So this is one meso compound.
So these two are the same, and we have one meso compound.
It's not really obvious looking at these
bond line structures that these represent the same molecule.
So definitely get a model set and try this out for yourself.
So we thought there might be four stereoisomers,
but actually there are only three.
We have a pair of enantiomers,
and we have one meso compound.