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I was wondering if it is possible to draw the Newman projections of the (E,E) and (E,Z) isomers of 1,4-diphenyl-1,3-butadiene in the following manner that I have? I wasn't quite sure about how to present the rings in such projections.

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I don't know what tempts you to use the Newman projection for these compounds. Trying to draw them, you might have missed the existence of the actual conformers. (Not speaking of the confusing 120° angle representation of the planar system.)

Thanks to the totally conjugated system, the molecules tend to maintain a planar shape. The two conformations, ‘cisoid’ and ‘transoid’ (deprecated terms; recommended: s-cis, s-trans; discussed also here) resemble double bond configurations because of the stability of the planar conformation (sterical hindrance has to be taken into account as well, it can be significant for the s-cis E,Z).

For example, the s-trans and s-cis conformations of the E,E isomer can be represented as follows.

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Trying to show the conformation at all the three single bonds at once wound be difficult, but could be done. And there's more missing info, so the two Newman projections would represent the (E,Z)'s conformers as well.

I am aware that excluding the hydrogens in the Newman projection is unorthodox. With or without H's, I don't think that the Newman projection is useful here, as for someone unaware of the planarity tendency, might consider e.g. torsional angle 90° much more favorable than the 0° one.

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