[narrator] Let’s take a moment to clarify erythro and threo stereoisomers.
The term erythro and threo, basically, designates the difference between isomers where we place
the chiral centers in a Fischer projection on the horizontal bonds.
With a Fischer projection the horizontal bonds are coming at us and the vertical bonds are
going away from us.
By this designation erythro means that our group, say bromine for example, is on the
same side as its partner bromine.
Likewise we could have another erythro isomer that has the bromines on the other side.
We have two different erythro isomers.
With threo we would have them on opposite sides of the Fischer projection.
It all depends on whether the bonds are R-S and the designation R-S does not necessarily
predict whether we’ll be erythro or threo.
We could also have another threo isomer that is the enantiomer.
These are enantiomer pairs, both sets are enantiomer pairs with one another.
Erythro is on the left.
Threo--here are the two enantiomer threo isomers, are on the right.
The reason that I don’t particularly like erythro/threo designation for these types
of compounds is because as soon as you change your compound to where, say, you have a Br
on one side, you have methyl on the other, and maybe an ethyl here.
Suppose you have a methoxy followed by an ethyl and say you had a propyl here.
Well, I generally put the parent chain, the longer chain, on the vertical axis.
Now, the designation erythro and threo doesn’t mean very much because all the groups are
different.
Originally erythro and threo were designed for sugars, which all the way down their chain
they have OHs and then they have their designated R groups depending on the sugar.
Depending on the sugar the stereochemistry changes.
That’s where erythro and threo are most useful but if we do have to tell which is
which we can take, say, this compound and turn it into a Fischer projection.
I’m going to start with carbon one at the top.
One, two, three, four, five--as we would name it.
We’ll start with a methyl as carbon one.
We have two chiral centers with bromines and then we have an ethyl group that would be
part of the parent chain at the bottom.
Carbons four and five.
We’ve got to rotate everything so that it’s pointed away from view.
If we do that we have to be able to visualize the rotation of the compound that puts all
the groups away.
If we can’t do the visualization in our head, say I’m going to rotate to where this
bromine is on the left, and if I do that looking at the compound it puts the methyl away and
a hydrogen at us.
Bromine and hydrogen at us, methyl away.
In the next position I have to rotate this bond--well, really, I have to rotate this
bond around so that the ethyl group is pointed away from our field of vision.
If I do that it ends up putting this bromine on the same side as the other.
We get an erythro isomer for this.
How can we check our work?
Well, we can name these with R and S. If we know these are coming at us and then away.
At us.
Away.
As a Fischer projection we see with group number one bromine is heavier than carbon.
Carbon to bromine is the next heaviest.
Then carbon is heavier than hydrogen.
This looks like its S and because hydrogen is coming at us instead of away it’s actually
R in this position.
Do it again for the other carbon.
If we do that, that’s--bromine wins again.
Carbon to bromine, again.
Carbon to carbon, then hydrogen.
Looks like its R again--looks like its R because it’s clockwise.
One, two, to three.
But, because hydrogen’s coming at us it’s actually S instead of hydrogen going away,
like it should be for the naming convention.
We then have to show that carbon two and three are R and S, on our original compound.
One, two--I’ll erase the parent chain numbers for clarity there--so, two, three, and then
four.
Not drawn because it’s a skeletal structure.
I’ve got--looks like its S but it’s actually R because hydrogen is coming at us.
We’re right on the first carbon.
What about the second carbon?
Let’s renumber, again.
Bromine wins.
Carbon to bromine.
Next best, carbon to carbon.
Number three and then our hydrogen’s going away from us, good.
That’s four.
That looks like counter-clockwise, and it is because hydrogen’s going away so we have
R-S.
We did, indeed, transform it correctly into our Fischer projection.