Erythro and Threo Stereoisomers
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Erythro and Threo Stereoisomers

March 5, 2020


[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.

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