00:01
Hello students, in this question we have to draw the products of some pericyclic reactions.
00:06
In the first reaction this substrate is undergoing a rearrangement.
00:11
So, we can see that it has pi bonds like this.
00:15
Hence, this will go under a 3 -3 sigmatropic rearrangement.
00:23
Now for the stereochemistry let's first draw it in its chair form like this.
00:30
So, this will be its chair form.
00:32
Now we can see that this ethyl group is present above the plane.
00:36
Let's show it above the plane like this.
00:39
This methyl group is present below the plane.
00:42
So, let's show this below the plane.
00:44
Now we have shown two alkenes here like this and these two alkenes are e alkenes.
00:53
So, to make this in e form we will have to add this methyl group here and this methyl group here.
01:00
Now we can show the rearrangement.
01:03
Rearrangement can be shown like this.
01:06
This pi bond will move here.
01:08
This bond will be broken and a new 3 -3 bond will be formed.
01:13
So, the new molecule that will be formed will have a bond between these two carbons like this.
01:20
Now here we have a double bond now and this will become a z alkene.
01:26
Similarly, here we will have a new double bond here which will have the ethyl group like this.
01:32
So, this will also become a z alkene like this.
01:36
Now we can see that this methyl group will be present at the equatorial position.
01:42
Similarly, this methyl group will be present at the equatorial position.
01:47
So, this will be the new structure.
01:49
Let's draw it in its open form.
01:52
So, we will have a new molecule like this in which two double bonds will be formed here and these two are z alkenes.
02:03
So, ethyl group will be present here.
02:06
Similarly, methyl group will be present over here.
02:10
Now these two will be present like this.
02:14
This will be present below the plane and this will be present above the plane.
02:21
So, this will be the correct structure.
02:23
Now let's move to the next question.
02:27
In second reaction we have to carry out the same 3 -3 rearrangement in this molecule.
02:33
So, let's first draw it in its chair form like this.
02:39
Now this contains two alkenes both are e alkenes.
02:45
So, they can be drawn like this.
02:47
This methyl group will be here.
02:49
Now this contains two hydrogen atoms above the plane.
02:54
Therefore, this three -membered ring will be present below the plane like this.
03:00
Now let's carry out the reaction.
03:02
We can move these bonds like this.
03:06
Hence, a new bond will be formed between these two carbon atoms.
03:11
Therefore, the new molecule will have a bond over here...