# Which biphenyl is optically active?

Which biphenyl is optically active?

I know that it can never be 1. I don't think it will be 4 either as I read that it should be ortho-substituted biphenyl.

So, when I look at 2 and 3, I can't make out which one should be the correct answer.

Should it always be two ortho substituents on each phenyl group for it to be optically active?

Also, what if the two groups on the same phenyl are different, will it always be optically active in that case?

Biphenyl 3 is the only optically active compound here. These stereoisomers are due to the hindered rotation about the 1,1'-single bond of the compound (Ref.1). Biphenyl 2 is noty optically active, because partially allowed rotation about the 1,1'-single bond of the compound (rotation is only partially restricted). To illustrate this phenomenon, I depicted the following diagram:

Note that compound 3 can rotate through two simultaneous $$\ce{I}$$ and $$\ce{H}$$ atoms allowing last $$180 ^\circ$$ rotation, which is well illustrated in the diagram posted by Karsten Theis.

References:

1. Paul Newman, Philip RutkinKurt Mislow, "The Configurational Correlation of Optically Active Biphenyls with Centrally Asymmetric Compounds. The Absolute Configuration of 6,6'-Dinitro-2,2'-diphenic Acid," J. Am. Chem. Soc. 1958, 80(2), 465-473 (https://doi.org/10.1021/ja01535a054).
• So only biphenyl 3 in your diagram is optically active? Why is that so? They all look pretty much the same, 2 different atoms at the same positions in each case? – studious Nov 21 '19 at 1:44
• I think your third diagram has an error in the lower left corner-- it looks like you are saying the hydrogens in Biphenyl 3 will turn into bromine when you rotate it certain directions. – Brian Rogers Nov 21 '19 at 16:11

(3) is not chiral, so maybe it is (2).

• So should it have 2 different groups at ortho positions of each biphenyl? What if there are 2 groups on one biphenyl and different 2 groups on the other biphenyl? – studious Nov 21 '19 at 1:48
• @studious To restrict rotation, you need two large groups in the ortho position of one phenyl, and at least one large group in the ortho position of the second phenyl. If you have identical groups on one phenyl (2-A, 6-A) and identical groups on the other (2'-B, 6'-B) none of the conformations are chiral. – Karsten Theis Nov 21 '19 at 2:08
• ok thank you got it – studious Nov 21 '19 at 2:22