-1
$\begingroup$

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?

$\endgroup$
1

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.