Question: Find out the number of products of monochlorination of 2-methylbutane including stereoisomers.
The given answer is 6, but why can't it be 7? I am getting 7 products, the one which I am getting but not in the answer is given below:
Question: Find out the number of products of monochlorination of 2-methylbutane including stereoisomers.
The given answer is 6, but why can't it be 7? I am getting 7 products, the one which I am getting but not in the answer is given below:
Let us take the base compound, 2-methylbutane, as below.
I shall list down each compound, and explain my reasoning also.
This question is one of the more confusing ones as such, because people can easily leave out / count extra structures which might be same.
You must note that the below compound is also the same - the name will be the same.
Note that the 2nd carbon is chiral in nature (marked by the $\color{orange}{\text{orange letter A}}$) and hence we need to count this as 2 isomers - the d-isomer and the l-isomer.
Note that this is achiral and thus is counted as 1.
So, we've got 3.
Note that the carbon marked 'A' is chiral and thus, this isomer is counted as 2.
That makes it 5!
That makes it six, and we have come to the end of the list of monochloro derivatives from the reaction.
The reason why you might have got 7 is that (refer point 1) you have counted the 1st compound as 2, but in fact those 2 structures point to the same compound.