[narrator] Just want to take a minute to review and revisit the stereochemistry of alkene
reactions beginning with our anti additions.
You remember the anti additions look like this in our
Lewis structures.
We form a cyclic bromonium intermediate and bromination followed by addition of a bromide
anion in anti fashion, kicking up the bromine.
One of our products, then, looks like this using the stereochemistry formulation that
I have shown with this particular mechanism.
Now, let’s look at what that looks like with a model structure.
The full mechanism with 3D models would indicate that, as we see here that, bromine
must add from the top or the bottom.
It can only add from one of the two places at any given time.
We see the double bond here of our alkene, grabs a bromine, lone pair from the bromine,
reattach the alkene at the other carbon, and then kicks off a bromide ion.
As we’ve drawn--we’ve drawn the cyclic bromonium forming on top and so then the bromide
ion that’s formed must come in from the bottom and attack either one of two positions.
It can attack the right most carbon that we’ve shown, and again this is cyclopentene that
we’re dealing with, and so we end up with the formation that we see here.
Our product, then, looks like this.
As we’ve drawn.
But, we also could attack the other position.
I’m kind of combining mechanistic steps.
This is more for finding the stereochemistry than being a complete mechanism, so please
don’t try to draw all these arrows at once.
These are two separate processes that form these from two separate intermediates.
I could also form this product.
It’s up to you to be able to tell that these are mirror images and that they are not superimposable
with one another.
We get enantiomers.
Two chiral carbons in each and if we draw the mirror images and then try to superimpose
them we cannot do it.
We get to enantiomers, two chiral centers each.
That’s, again, from deciding which way to add--which carbon to add from on these compounds.
Well, what if we don’t have a symmetrical alkene?
So, we start with an alkene that looks like this instead?
Well, we form a cyclic bromonium but we can form that cyclic bromonium on the top or the
bottom of our alkene.
I’m going to draw both of them here and you should be able to draw--I’m trying to
keep the perspective similar.
That’s what’s already there, as drawn.
Now I’m going to draw the other one.
The other bromonium that adds from the bottom.
Cause remember alkenes are planar.
Here we would force the CH3 up.
Our bromide that formed from the first step of the mechanism, in this case it’s not
symmetrical.
We added the more substituted carbon and we do that again, except this time anti addition
means adding from the top.
Cyclic bromonium is on the bottom and we end up with two different products based on this
addition.
Methyl goes up, bromine down.
Then the other bromine goes up and in the bottom example where the cyclic bromonium
is on the bottom anti addition means adding from the top which pushes
the methyl down, the bromine is up, and then the other bromine was already on the bottom
of the ring.
Remember there was a hydrogen here and so it’s now down in this position.
We get two different enantiomers based on whether the bromonium formed on the top or
the bottom, equally likely to occur.
For syn additions like hydroboration let’s take the same compound.
A methyl, 1-methyl-cyclopentene.
Treat it with BH3 and THF.
Well, if we look here we’ve got an orientation in syn fashion where we’ve got our double
bond here and that double bond fills the empty orbital on the brom--can’t draw oval there--got
the empty orbital on the boron so that alkene can fill in.
Then the bond from hydrogen fills the orbital that’s beginning to empty
in a syn fashion.
So we’re the same side.
The BH2 and the hydrogen, both are adding from the bottom so that’s going to push
the methyl up and push the hydrogen up on the other side leaving the BH2 on the bottom.
Again, remember hydroboration is anti markovnikov.
If we finish this process wherever the boron ends up that’s where the alcohol is going
to be.
We end up with opposite orientation of the methyl and the OH, because of the syn addition,
but if we’d added from the top you end up with the enantiomer product
because the additions are reversed.
Still a syn addition but the additions are reversed.