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The given compound is optically active. Initially I had reasoned that the compound has center of symmetry( two xylene groups across each other). Why doesn't it have a center of symmetry?

enter image description here

Source: MS Chouhan, Advanced Problems In Organic Chemistry, Page 116 (Chapter Isomerism Level 2 q6a)

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  • $\begingroup$ It probably racemises pretty quickly. $\endgroup$
    – Mithoron
    Dec 30 '17 at 14:54
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    $\begingroup$ @Mithoron I don't think so. $\endgroup$ Dec 30 '17 at 17:47
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You assume too much, starting with the words "The given compound". There is no given compound. The picture corresponds to quite a few different compounds, and here is why.

As I told you before on a remotely similar question, the biphenyl fragment totally can't be planar. The xylene substituents must be rotated so their ortho methyl group points either up or down relative to the plane of the central ring. Since there are two of them, we have three options: "both up", "both down", and "one up, one down". The first two compounds are indeed chiral, and enantiomers of each other. The third one is centrosymmetric, just as you've reasoned.

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  • $\begingroup$ It was a question ; the book had provided that picture and asked to state whether it was chiral or achiral. How do I figure out which of the three isomeric states would it be in? (the question book specifies whether the bonds are in or out of the plane{through wedges and dashes} ). So I was taking the compound to be planar (which is not possible as you did tell me earlier). $\endgroup$ Dec 30 '17 at 10:18
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    $\begingroup$ If you have wedges and dashes, then you have one specific isomer. If you don't, then you have all three at once. In that case any answer shorter than mine would be incomplete. $\endgroup$ Dec 30 '17 at 13:34
  • $\begingroup$ @IvanNeretin "one up, one down" would that stereoisomer's m-xylene rings be exactly perpendicular to the central benzene ring? Because if they are not exactly perpendicular, and rather inclined at some angle, then the centrosymmetry would be lost. $\endgroup$ Feb 24 '18 at 15:47
  • $\begingroup$ The central symmetry would still be there if they are inclined at the same angle in the opposite direction. $\endgroup$ Feb 25 '18 at 3:29
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enter image description here Look at the picture, although the two methyl groups are located equidistant from the centre but they aren't in opposite direction. Therefore, no centre of symmetry and so it's optically active.

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