How would you draw stereoisomers of 1,2,3-trimethylcyclopentane?

1 Answer
Dec 30, 2015

I would draw all the possibilities and then eliminate meso compounds and any other identical structures

There are three chiral centres, so the maximum number of stereoisomers is #2^3=8#.

Here are the four mirror image pairs. Let's say that up is U and down is D and call the members of each pair A and B.

1. UUU


2. UUD


3. UDU


4. DUU



1A and 1B are the same meso compound.

2A and 2B are a pair of enantiomers.

3A and 3B are the same meso compound

4A and 4B are a pair of enantiomers, but they are identical with 2B and 2A,so we can ignore them.

So, of the eight possibilities, there are only four stereoisomers: two meso compounds and a pair of enantiomers.

Here they are again.