# Can a system of 2 consecutive biphenyl systems be optically active , even if any adjacent biphenyl systems aren't optically active?

This question came in an online test, where they asked whether or not 22,26-dibromo-12,32-dichloro-23,25-diiodo-16,36-dimethyl-11,21:24,31-terphenyl was optically active.

The answer is that, it is optically active.

According to Solomon and Fryhle:

"Biphenyl is an asymmetric compound so it must not have a plane of symmetry or center of symmetry to be optically active". Hence compounds which have 2 same adjacent groups on a single phenyl in any biphenyl system cant be considered optically active .

Applying that concept above as same groups (Br and I) are attached to the middle phenyl system, the individual biphenyl systems on both sides are optically inactive, hence the compound should be optically inactive.

Please tell me where I am going wrong in my interpretation.

The key point you have to remember is that optical activity isn't decided by a single section of the molecule but rather it is a property of the entire molecule - and while not being able to find a mirror plane is almost a guarantee, the nail in the coffin to decide that a molecule is optically active is if you cannot find an improper rotation axis (reflect and rotate).

There is no mirror plane in the tri-phenyl molecule you show (cut it into left and right the Br and I don't overlap; cut it up and down the methyl and chloride won't overlap; and so on and so forth). Hence, it is optically active.

• This is not entirely true, the condition for chirality is that the molecule must not possess an improper rotation axis (of which mirror planes are a subset of) Jul 8 '16 at 6:52
• The presented compounds are typical examples of atropisomerism. Jul 8 '16 at 7:18

Here is the correct representation of possible forms of this compound.

Let central ring be in the plain. Then other two rings are twisted. The darker line on a diagram indicates a group pointing towards us. It is important to remember that the angle between the planes of the rings is neither 0$^o$ nor 90$^o$ (otherwise your question would be valid).

Note that -CClBrI radical is not optically active. However, combining two such radicals you can have an optically active molecule IBrClC-CClBrI.

• why wouldnt the angle between the planes of the ring be 90 degrees ? due to steric hindrance isn't that the best configuration for them to minimize repulsion ? Jul 21 '16 at 17:02
• @IshitaGupta You would guess it is, but it is not a case. In the case of unsubstituted biphenyl, the equilibrium torsional angle is 44.4° (from wiki about biphenyl). Jul 21 '16 at 18:42