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Is the compound in the picture chiral or achiral?

(2E)-3-(3-bromo-2,4,6-trimethylphenyl)-3-chloro-2-methylprop-2-enoic acid

I think it should be achiral for the following reasons:

  • There is free rotation. I know, but for every configuration produced by free rotation there is an exactly opposite configuration.

  • Also at certain point of time all the groups can come to the same plane.

This question is from my textbook exercise. And the answer says it is chiral. But I've no idea why. Also is it optically active or inactive?

EDIT: @K_P gave a very useful article related to these type of compounds:

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    $\begingroup$ Is there free rotation around the bond between the phenyl ring and the double bond? $\endgroup$
    – K_P
    Commented Oct 30, 2016 at 12:24
  • $\begingroup$ I think so @K_P tell me if I am wrong $\endgroup$
    – user14857
    Commented Oct 30, 2016 at 12:28
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    $\begingroup$ related chemistry.stackexchange.com/questions/37794/… $\endgroup$
    – Mithoron
    Commented Oct 30, 2016 at 12:51
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    $\begingroup$ Have you sudied atropisomerism? It is a very important topic and your exercise is an example of this $\endgroup$
    – K_P
    Commented Oct 30, 2016 at 12:54
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    $\begingroup$ also chemistry.stackexchange.com/questions/24901/… $\endgroup$
    – Mithoron
    Commented Oct 30, 2016 at 12:57

1 Answer 1

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Is the compound in the picture chiral or achiral?

The answer in practice is: it depends. You say that there would be free rotation along the phenyl–vinyl bond. However, every adjacent carbon atom — the 2- and 6-positions of the phenyl ring and the 1-position of the double bond — are substituted. For free rotation to occur, these substituents must pass each other. Since they are all bulky, though, free rotation is very effectively hindered. A planar rotamer with full π conjugation between the phenyl ring and the double bond is not populated at any extend which can be shown by the NMR shifts of the corresponding carbon atoms (and could also be shown if the methyl group on the double bond were a proton, because it’s shift would be vastly different from one in styryl conjugation).

Under low enough temperatures, the compound you have been given will indeed form to separable enantiomers due to the bromine atom which destroys the molecule’s symmetry. These are examples of axial chirality; if the chlorine atom is above the phenyl ring’s plane, it is (aR), if it is below, the compound is (aS). The two enantiomers can be separated, e.g. by chiral HPLC, and they will show optical activity.

At higher temperatures, thermal energy will be enough to overcome the rotation barrier and an optically active single enantiomer will racemise. This temperature is typically above room temperature for a system such as your’s but well below the temperature required for alkene (E/Z) isomerisation.

This concept is called atropisomerism and is also the reason why BINOL and BINAP are chiral.

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  • $\begingroup$ Do you know of any web resource which has listed down all known atropisomers ? $\endgroup$
    – user14857
    Commented Oct 31, 2016 at 5:04
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    $\begingroup$ @S007 All known is wa~y too broad; I don’t think that even all bisphenyl-based atropisomers are listed anywhere. Unless of course you’re talking about different conceptual building blocks but then again I still don’t know. (But at least the list would be manageable.) $\endgroup$
    – Jan
    Commented Nov 1, 2016 at 20:37