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How is optical activity explained in biphenyl compounds using atropisomerism.

NOTE: I was taught that if vertical symmetry exists then optical activity is not shown by any substituted biphenyl compounds And I was also taught that 6,6'-dimethyl[1,1'-biphenyl]-2,2'-diamine shows optical activity. How is this possible because there exists a plane of symmetry right?

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In general, there's no reason why a simple substituted biphenyl is flat either. The key is that if the barrier to rotation is low enough, then the different stereoisomers can interconvert, so you end up with a racemic mixture.

For more substituted biphenyls, strain present in the planar form is high enough that it impedes free rotation around the bond that connects the two phenyl rings. This means that the axially chiral conformations cannot easily interconvert, and thus, you can persist any enantiomeric excess that you create.

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