00:01
Hello students, in this question we have to carry out the reaction of this given substrate with sodium methoxide base.
00:09
So, this will give och3 negative which will act as a base and cause elimination reaction from here like this.
00:18
Hence, we will get an alkene with phenyl groups and methyl group.
00:24
Now, i am drawing these distorted lines because we have to draw this product by correct stereochemistry.
00:32
So, for that we will have to place the leaving groups antiperiplanar to each other.
00:39
Now, hydrogen and bromine are the two leaving groups.
00:43
So, let's place them antiperiplanar.
00:45
To place them antiperiplanar, we will have to rotate this sigma bond.
00:50
We have to rotate the sigma bond in such a way that hydrogen come on the plane here.
00:56
Now, we will have to rotate this like this.
01:00
Hence, phenyl will now come above the plane and methyl will stay below the plane.
01:12
To place bromine atom antiperiplanar to hydrogen, we will have to rotate this carbon atom in such a way that bromine comes here.
01:23
Therefore, if we rotate it like this, this phenyl group will come above the plane like this and the hydrogen atom will remain below the plane.
01:35
So, now this hydrogen atom and this bromine atom bonds are both antiperiplanar to each other.
01:42
Now, when these bonds are antiperiplanar to each other, then sigma star of cbr bond comes parallel to the ch bond.
01:52
Now, this elimination can take place.
01:55
So, in this elimination, och3 negative will abstract this hydrogen and electrons will be transferred from sigma bond of ch to sigma star of cbr...