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Here, A has a greater atomic number than B, which has a greater atomic number than C. To assign E,Z designations you would divide the double bond into left and right portions and assign priorities. On the right side, A > C. However, on the left side, B = B. A (first priority on right side) is simultaneously on the same side and on opposite sides to B (first priority on left side). How would you class this compound then?

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  • $\begingroup$ How would you classify an ethylene molecule without any substituents? Is it E or Z? $\endgroup$ – Ivan Neretin Mar 13 '17 at 14:16
  • $\begingroup$ @IvanNeretin I take it that you mean E,Z nomenclature can permissibly fail to class a molecule? $\endgroup$ – lightweaver Mar 13 '17 at 14:17
  • $\begingroup$ How would you classify a methane molecule without any substituents and without any double bonds? Is it E or Z? $\endgroup$ – Ivan Neretin Mar 13 '17 at 14:18
  • $\begingroup$ @IvanNeretin I thought that E,Z nomenclature was used for compounds with double/triple bonds. $\endgroup$ – lightweaver Mar 13 '17 at 14:20
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    $\begingroup$ That's right (that is, if you throw away the triple bonds), but there's a catch. First and foremost, any nomenclature (not just E,Z) exists to tell apart different things. Your molecule is identical, no matter which way you orient those A and C. There is nothing to tell apart. $\endgroup$ – Ivan Neretin Mar 13 '17 at 14:25
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When both the substituents are identical on one side of the double bond, the compound does not show geometrical isomerism.

There is no point in using E-Z nomenclature in that case.

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