00:02
We have cs3, ch, ch, ch, ch, ch, cs3, cs3, cs3.
00:15
Here, there are no chiral center present.
00:27
Because here we see that at this carbon, the groups attached are same, and also at this carbon, the groups attached are same.
00:35
Also there is pos present so the number of stereo centers is zero so number of stereo isomers equal to two key power n this will be equal to two raised to power zero and this will come out to be one so correct option is a now we have another molecule which is cs3 ch3 ch, ch 3, br, oh.
01:15
Here we have two chiral centers.
01:17
First is present here and second is present here.
01:21
Because you see that at this carbon the group attached are different four groups.
01:26
Also similarly at this carbon we have different four groups attached.
01:30
So the number of chiral center is two.
01:33
So, number of stereo isomers equal to 2 raise to power n, this will be equal to 2 raise to power 2, this will be equal to 4.
01:47
So the correct option is option d.
01:51
Now we have here we have this molecule which is cs3 and cs3.
02:05
We will draw the chair transformation...