[narrator] Let’s take a moment to review halogen additions to alkynes.
What do we know about this reaction?
Well, first of all, we know that with alkynes versus alkenes we have two pi bonds to work
with, double the reactivity.
We know that halogens add, in an anti-fashion, and we can stop at one addition or keep going.
Let’s see what the product is after one halogen addition.
We would end up with this double bond with an anti Br-Br relationship.
Even on internal alkyne versus a terminal, so we’ve got a terminal here--end of the
line, with a hydrogen on this end.
We could easily do this with an internal and have the same effect.
If we add a second halogen, again this works with bromine and chlorine, we would end up
losing the stereochemistry here because we would lose our double bond and not form a
chiral center, either, because we add two bromines to each position.
A couple things to look at.
Let’s look at the mechanism.
It’s very similar to what we’ve seen before.
We’re going to form a cyclic bromonium, as we did before, invoking a lone pair on
bromine.
Pi bond grabs bromine, kicks off a bromide.
Bromine with a +1 formal charge.
Not part of the mechanism but I’ll show, to illustrate markovnikov’s rule, a bigger
delta plus forms on the more substituted carbon which attracts the nucleophile in step two,
Br- here, kicking up the bromine.
This adds in an anti-fashion.
We also need to check for a nucleophilic solvent because, at this point, if we do have methanol,
or ethanol, or a nucleophilic solvent that can add it will because it’s in solution.
The bromide ions are very rare compared to the solvent molecule.
There are proximity effects that give a bromine a little bit of an advantage but the methanol
and solvent could easily add, at this point, so check for that in your given information.
What do we get?
Well, we get the anti-relationship vicinal dihalide; 1,2-dihalide.
A vinyl dihalide, even.
Then it’s just the same as our alkene material so we can do this process again with another
bromine molecule
forming a cyclic bromonium.
I’ll show the hydrogen just for clarity and then, again, we attack at the markovnikov
position, although it will not affect the final stereochemistry
because there is no final stereochemistry.