00:01
Hi, in this question we are asked to draw the fisher projection of s2 bromobutane.
00:06
We have butane as a parent chain which contain four carbon atom.
00:11
We can draw in this manner.
00:13
At second position br group is attached but we need to make sure that the configuration of second carbon should be s.
00:21
Suppose we are taking br present below the plane and hydrogen present above the plane in this manner.
00:33
This is the chiral center.
00:35
According to cip rule, we will give numbering to the each group which is attached to this carbon atom in order to find out the configuration.
00:44
According to cip rule, we have to give numbering from 1 to 4 and highest numbering will be given as 1 and it will be given to that atom which has highest atomic number.
00:55
Here we can see bromine is having the highest atomic number.
00:59
Therefore we have have given one numbering to it.
01:02
And least numbering will be provided to hydrogen atom as four.
01:07
Now we have to decide between two and three.
01:10
Here at this position we can see we are having a carbon atom which is further attached with three hydrogen atom.
01:17
And here we are having a carbon atom which is further attached with two hydrogen atom and a carbon atom with three hydrogen atom.
01:26
Therefore, we can say the second numbering will be provided to this carbon atom and third will be to this ch3 group...