Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Shadow on April 05, 2013, 10:45:36 AM
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Is there a plane of symmetry in (1R,2S)-1-bromine-2-methylcyclohexane? I think that there is, so it isn't optically active. It should be meso, but I need a confirmation from someone else.
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Is there a plane of symmetry in (1R,2S)-1-bromine-2-methylcyclohexane? I think that there is, so it isn't optically active. It should be meso, but I need a confirmation from someone else.
Can you draw a picture showing where your proposed plane of symmetry is?
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Surely meso compounds have the same substituents on adjacent carbons?
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So it is not meso. Then I have the (1S,2R) isomer, too?
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Yes, I would expect as well.
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There are four compounds then: (1S,2R), (1R,2S), (1R,2R), (1S,2S) and not two are superimposable, right?
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Yes that's correct, 4 possible diastereoisomers, two enantiomeric pairs.
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Have to ask, if the substituents were the same, these would be meso?
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Have to ask, if the substituents were the same, these would be meso?
Look here:
http://en.wikipedia.org/wiki/Meso_compound (http://en.wikipedia.org/wiki/Meso_compound)