00:04
The first part of this problem asks us to draw the mechanism of the reaction of two butyne with one equivalent of br2.
00:12
So here's our two butyne.
00:14
So the first thing that will happen is let's make br2 like that.
00:18
So the first thing that will happen is one of these pie bonds will come and attack the bromine.
00:22
And then the bromine, one of the lumb pairs on the bromine, will actually come and attack one of the carbons as well.
00:28
And we'll kick off that second bromine.
00:30
So we get a transition state that looks like this.
00:33
Now we only have two bonds there because we used one of the pie bonds to attack the bromine.
00:38
It's attached to both carbons and it's positive because it has two bonds.
00:43
We also have this br minus left over.
00:45
Now the br minus is going to come in and attack one side.
00:50
But because this br pyramid transition state, or not transition state, intermediate is so big, it has to come in from the opposite side.
01:00
It cannot attack from the same side...