0:00
So first reaction.
00:01
The first reaction represent an example of is the example of e2 elimination in which hydrogen or deuterium should be at anti to the halogen, in which hydrogen or deuterium should be at antiposition, in which hydrogen or deuterium should be at antiposition.
00:41
To halogen.
00:43
So here anti -edition, sorry anti -elimination occurs.
00:47
So the reactant here given is this above the position, above the plane, hydrogen, below the plane this ch3 group, above the plane br atom, above the plane br atom, below the plane h, above the plane here hydrogen and below the plane d is present.
01:18
So above the plane or the below the plane, two below the plane or two above the plane, this represent what, seen position, seen positions.
01:28
While anti -position means they are anti to each other, they are their positions will be opposite to each other like one should be above the plane and another should be below the plane.
01:38
So this represent anti -position.
01:40
So here this deuterium and this bromine are at anti -position.
01:45
So here the elimination occur from d, this deuterium, okay? so therefore elimination will occur.
01:55
Elimination will occur from deuterium.
02:06
Hence the reaction will be like this.
02:09
This is the reactant, hydrogen below the plane, ch3, above the plane dr, below the plane hydrogen.
02:19
Here, deuterium is below the plane, and hydrogen is above the plane.
02:24
The reactant is ch3ona, that is sodium alcoxide.
02:30
This sodium alcoxide out of which ch3o minus is the nucleophile which abstracts this deuterium and the bond between deuterium and carbon will shift towards the ring and the product will be an alken hydrogen and this bromine will leave the group as a leaving group and below the plane ch3 along with byproducts nabr plus ch3 o d as the alcohol.
02:59
So this is the reaction...