00:01
In this problem, we want to start with s2 bromobutane.
00:10
So, butane is four carbons.
00:14
Two bromal.
00:15
We have bromine here.
00:17
And then we have priority one, two, and three.
00:21
Here we're going clockwise, which is not what we want.
00:25
We want the opposite.
00:26
So we put bromine behind the plane.
00:31
So that's going to be our starting material.
00:35
And we're told that this is an sn2 reaction.
00:39
With iodide.
00:44
So with sn2, it's a backside attack.
00:48
So we calm attack here and kick out bromine.
00:52
So it's a concerted mechanism, it all happens in one step.
00:56
So that backside attack leads to an inversion of stereochemistry.
01:02
And if we needed a transition state here we look something like that, and we could put that in our transition state brackets just to be clear.
01:23
And to name the product, now this will be r, right, because we inverted the stereochemistry.
01:30
And then two iota butane.
01:37
And for an energy diagram for this, we have our reaction progress, energy.
01:47
We'll have our reactants is r.
01:53
This will go up and down.
01:57
This is our transition state.
01:59
And then we have our products at the end.
02:14
And then the next one, we have r3 -r -3 -romo, three methyl hexane...