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Explain optical activity in Biphenyl compounds
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+ Stereochemistry
+ Isomers
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Karen Simon
Explain optical activity in Biphenyl compounds
In biphenyls, the two ortho substituents in each aromatic ring must be different and bulky (e.g. COOH and NO2).
Also the two aromatic rings must not be co-planar (i.e. must not be in the same plane) but perpendicular.
In biphenyls, the two ortho substituents in each aromatic ring must be different and bulky (e.g. COOH and NO2).Also the two aromatic rings must not be co-planar (i.e. must not be in the same plane) but perpendicular.
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.
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.
In biphenyls, the two ortho substituents in each aromatic ring must be different and bulky (e.g. COOH and NO2). Also the two aromatic rings must not be co-planar (i.e. must not be in the same plane) but perpendicular.
In biphenyls, the two ortho substituents in each aromatic ring must be different and bulky (e.g. COOH and NO2).Also the two aromatic rings must not be co-planar (i.e. must not be in the same plane) but perpendicular.
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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.
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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