00:01
Okay, we've got chlorium phenocle, and we've got to determine what the configuration of each of its asymmetric centers is.
00:08
So let me first draw a chloramphenicol, a little benzene ring here.
00:18
We go to wedge to oh h, wedge to oh h.
00:28
It's a nitrogen here.
00:35
A couple of chlorines here.
00:39
And on this side, we have a nitro.
00:47
So here's our chloram phenocle.
00:50
The first important thing we can do is find our asymmetric centers, right? so to be a stereogenic carbon or a stereocenter, it's necessary that the carbon have bonds to four different substituents.
01:05
First, we can rule out any sp2 hybridized carbons.
01:10
So that would be any of the carbons on the benzene ring.
01:12
Like these cannot be stereocenters.
01:15
One way you can tell is that, for example, this carbon here is bound to the same carbon twice.
01:20
And you need four unique bonds from a carbon for it to be a stereocenter.
01:24
Right.
01:24
Another one we can rule out is this one.
01:26
Once again, two bonds with the same oxygen can't be a stereo center.
01:30
In this carbon, you'll notice it's not as p2 hybridized, but there's two identical bonds.
01:37
Two bonds to chlorine with nothing else are identical.
01:40
Therefore, we can't have stereo isomerism coming from this carbon.
01:46
In fact, the only stereocenters on this molecule are right here and right here.
01:50
So we'll step through figuring out what the absolute configuration.
01:54
Of these stereocenters are the first thing that i like to do is number the substitution well actually the first thing i like to do is draw in any implied hydrogens right we know if we see a carbon with three bonds and there must also be a bond to hydrogen and because we've got two bonds in the plane in one wedge we know that that bond hydrogen must be in the back okay the next thing i'd like to do is number the carbons in the order of there or the number number the substituents in the con ingold prelog rules order of importance...