Number of stereoisomers of trisubstituted cyclopentane When 4-methylcyclopent-1-ene reacts with bromine, then 1,2-dibromo-4-methylcyclopentane forms. And I am stuck on determining the amount of stereoisomers that can form as a result.



1,2-dibromo-4-methylcyclopentane has two chiral carbon atoms, those of bromine. And an internal mirror plane, we should have 3 optical isomers.

But I am confused upon drawing the following:


1 and 4 appear to be the same, 2 and 3 appear to be the same.

5 and 7 appear to be the same, 6 and 8 appear to be the same.

This gives me 4 isomers. So I draw them different again:



The puckered structure leaves me doubting myself: while I think that the internal mirror plane causes there to be only 2 different cis structures instead of 4 (or instead of 1 if it had been 1,2-dibromo-4,4-dimethylcyclopentane), I am not sure if the trans structures are actually the same.
Almost everything that needs to be said has been said. But I do have a few comments about configurations. In enantiomeric, trans-dibromides 1 and 2, the methyl groups are chirotopic and non-stereogenic. Inversion of the stereochemistry of the methyl group in either 1 or 2 returns the same structure, respectively. Thus, there is no designation for stereochemistry at the C4 methyl center. On the other hand, inversion of the stereochemistry of the methyl group in 3 gives 4 and the same operation on 4 produces 3. Therefore, the methyl center in achiral isomers 3 and 4 is achirotopic and stereogenic. Accordingly, the C4 centers have lowercase descriptors, r and s, respectively. This site might be of help.