Why are some substituted biphenyl compounds optically active and what kind of substituted biphenyls are optically active . I read this in a book but no satisfactory reasons were given and no examples were given thanks in advance .
$\begingroup$ Related: chemistry.stackexchange.com/questions/5489/… $\endgroup$– Klaus-Dieter WarzechaJul 20, 2016 at 4:58
In general, compounds without improper axes of rotations are chiral. More concretely (and less correctly), compounds with less symmetry are more likely to be chiral.
Substituted biphenyl compounds generally have little symmetry. Why is this? Steric hindrance causes the two phenyl rings to be tilted relative to each other and serves as a barrier to rotation. This introduces dimensionality to the compound and breaks symmetry in the plane of the ring. If, in addition, there are substituents located asymmetrically on the rings, then the two rings can be differentiated from one another and thus lead to chirality. A similar situation is observed for spiro compounds.
BINAP is an example:
Remark. Most textbooks mention chiral centers as a sufficient (but not necessary) condition for chirality, and fail to emphasize other possibilities. Substituted biphenyl compounds, for example, may be chiral even without chiral centers.
$\begingroup$ "If, in addition, there are substituents located asymmetrically on the rings, then the two rings can be differentiated from one another and thus lead to chirality. " - doesn't the symmetry breaking due to strain suffice to enforce chirality? Are substituents necessary for that? $\endgroup$– harryApr 25, 2021 at 4:11
$\begingroup$ @HarryHolmes, the activation energy required for rotation in an unsubstituted biphenyl is low enough to be overcome quite readily, so the two stereoisomers (more specifically, atropisomers) can interconvert. Substituents induce chirality by increasing steric hindrance and hence the activation energy for rotation. I think the substituents do not have to be asymmetrically distributed on the rings, but this is a five-year-old answer so I’m not certain what exactly I meant. $\endgroup$ Apr 26, 2021 at 5:39