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Checking whether a terphenyl is 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 . enter image description here

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.

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2 Answers 2

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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.

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    $\begingroup$ 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) $\endgroup$
    – orthocresol
    Commented Jul 8, 2016 at 6:52
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    $\begingroup$ The presented compounds are typical examples of atropisomerism. $\endgroup$
    – vapid
    Commented Jul 8, 2016 at 7:18
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Here is the correct representation of possible forms of this compound.

enter image description here

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.

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  • $\begingroup$ 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 ? $\endgroup$
    – Ishita Gupta
    Commented Jul 21, 2016 at 17:02
  • $\begingroup$ @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). $\endgroup$
    – sixtytrees
    Commented Jul 21, 2016 at 18:42

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