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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.

4-methylcyclopent-1-ene

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: stereoisomers of 1,2-dibromo-4-methylcyclopentane

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:

stereoisomers of puckered 1,2-dibromo-4-methylcyclopentane

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.

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  • $\begingroup$ So propene or pentene? $\endgroup$ – Mithoron Jul 24 '19 at 0:05
  • $\begingroup$ In either case there's only 4 isomers, because of plane of symmetry. $\endgroup$ – Mithoron Jul 24 '19 at 0:09
  • $\begingroup$ Thank you Mithoron for your comments: pentene is what I meant, I've edited my question. I will also attempt to upload pictures of what I exactly mean. $\endgroup$ – Tim Jul 24 '19 at 0:32
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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.

enter image description here

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  • $\begingroup$ Thanks for this info, really informative and contains the terminology and information I was looking for! $\endgroup$ – Tim Jul 24 '19 at 17:55
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Nice pictures and analysis!

There are four different stereoisomers: one pair of enantiomers (the trans structures) and two achiral isomers (the cis structures).

In a mixture of all four, there will be 3 sets of boiling points, peaks in chromatographic separation (with achiral columns), etc.

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

This argument works only if both bromine-bound carbons are (R), or both are (S). Otherwise, the methyl-bound carbon has two different substituents, and could be counted as chiral as well.

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