[narrator] Let’s review halogen addition reactions of alkenes.
First, let’s recap a few things we know about the reaction.
Number one, in this example we’ll form a cyclic bromonium intermediate.
Of course, a chloronium would also be possible, if we were using chlorine instead of bromine
addition.
Cyclic bromonium; this is an anti addition.
We will also get a markovnikov addition of the nucleophile and we will--which means attacking
the more substituted carbon when the nucleophile adds in the second step of the mechanism.
Also, we know that we only need to get, or predict--only need to predict
one stereoisomer to get the others.
We don’t have to go through the whole process, both times.
Let’s just go ahead and do the mechanism because that dictates the stereochemistry
of the reaction.
What I’m going to do is I’m never going to rearrange my carbon skeleton from how it’s
drawn for me and I’m going to add the hydrogen in for clarity.
I’m just going to pick that the bromonium forms on top, as we showed all we need to
do is to get to one of the products and we can get the other if chiral centers are formed.
For step one of the mechanism, I’m going to show the bromonium formation invoking a
lone pair on the bromine.
We’re going to attack the carbon of the double bond.
The double bond will then grab the bromine and will eliminate a bromide from the bromine
molecule, and we’ll form our cyclic bromonium.
For the sake of simplicity we’ll just say that it adds to the top, as we are looking
at the carbon skeleton, because we know that if we can at least predict one then we can
predict the other.
I’m going to, in the second step of the mechanism, I’m going to attack the more
substituted carbon with the bromide that I just formed.
It’s an anti addition so it attacks on the back side, kicks up the bromine to the other
carbon, and I get--as we’ve drawn it--this structure.
Again, I’m not moving anything relative to my view to make things easier for us, for
visualization.
Here, this isn’t a true perspective diagram because the way I’ve drawn it I’m looking
flat down over the carbon skeleton and, technically, I can pick either one of these to come forward
as long as I keep everything in the same orientation relative to one another.
I’m going to pick the methyl group to have the wedge and I’m going to pick the hydrogen
on the other side to have the dash.
At this point, I have one of my products.
I can rewrite it if I want to in a more common method where I keep the parent chain in the
plane, six carbons.
If you see here; down, up, down, down, up, down.
This stays the same.
But, I’ve got to rotate the other around to be the up position and when I do that it’s
going to put the bromine to the back and the hydrogen to the front, which is not drawn
in a skeletal structure.
Then I know that to get the other compound that’s formed, since I formed two chiral
centers--one here and one here, I must flip both chiral centers to get the other product
of the reaction.
This product would have originated from the cyclic bromonium forming on the bottom of
the original alkene.