00:01
We have the following structure, and we would like to know what would the resulting product be if we performed an e2 elimination reaction on it.
00:08
So an important thing to remember with e2 elimination reactions is that we need the beta hydrogen to the alkaliahe and the alkali halide to be antiparriplanner to each other.
00:18
But in this situation here, we see that our hydrogen and our alkali are on the same side, and they do not exhibit anti -paraparoplaneer geometry.
00:26
But a cool thing that we can do here is that we can rotate this carbon -carbon bond, such that it will make them again.
00:31
Exhibit anti -paraparraline or geometry, and it will look like this.
00:38
It's the same structure just with the carbon -carbon bond flipped.
00:47
So these bonds here are going to remain the same, while the ethanol, or the ethyl group rather is going to be in the plane of the screen.
00:59
Our hydrogen is now coming towards us, and the methyl group is going away from us.
01:08
And here we see that the reacting beta hydrogen and the chlorine are now on opposite size to each other and they exhibit anti -parapraplanar geometry...