I know that a group exerts an inductive effect on another one if there is a difference in electronegativity. In the case of styrene (1) all the atoms are sp2-hybridized. So, how could the phenyl ring show a negative inductive effect on the ethenyl group?


  • $\begingroup$ Negative inductive effect, no. Negative mesomeric effect, yes. $\endgroup$ Nov 9, 2020 at 10:11
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    $\begingroup$ It may be meant as a -I effect, compared to an ethene hydrogen formally replaced by the phenyl. Similarly as methyl has +I effect on benzene, formally replacing a benzene hydrogen. By everyday life analogy, if 2 persons stand on the same stair, one of them is higher than the person that originally stood 1 stair lower. $\endgroup$
    – Poutnik
    Nov 9, 2020 at 10:14
  • $\begingroup$ ... because the phenyl ring has 6 sp2 atoms, and the ethylene group only two? $\endgroup$
    – Karl
    Nov 9, 2020 at 12:03

1 Answer 1


This can be explained by looking at the molecular orbitals. The molecular LUMO of benzene (which accepts electrons) is closer to the HOMO of ethylene (which given electrons) as compared to the large gap between HOMO of benzene and LUMO of ethylene.

HOMO-LUMO schematic of benzene and ethylene

Surely, the HOMO, LUMO don't exist separately and only exist as a molecular orbital. But the separation makes for easy transfer of electrons from ethylene to benzene unlike the opposite.

Because I have involved pi orbitals into the solution, I believe mesomeric effect is more apt here. To talk about inductive effect we have to involve a more complex idea (group electronegativities). Resources for such data are not easily accessible. I would appreciate if someone can find the group electronegativities of benzene and ethylene and post them here for the analysis of inductive effect as well.

Surely, if present, a soft nucleophile will it will attack the 2-carbon on ethylene side-group.

  • $\begingroup$ I have tried to make as good a picture as I could from what I know. Any correction to the energy levels is appreciated here. I wish I could do this over a software but I don't have access at the moment. $\endgroup$ Nov 9, 2020 at 12:04
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    $\begingroup$ I'm rather curious as to how you come up with these energy levels. From Hückel theory, both benzene and ethylene have the same frontier orbital energies: $\alpha + \beta$ for the HOMO and $\alpha - \beta$ for the LUMO. I'm sure joining them together, as in styrene, will make a difference, but how do you come up with this particular arrangement? $\endgroup$ Nov 9, 2020 at 12:21
  • $\begingroup$ @orthocresol I will look for an explanation for this. Even I have to verify this once before I go on telling you about the correct way to do it xD $\endgroup$ Nov 9, 2020 at 12:49
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    $\begingroup$ You didn't just make it up, did you? $\endgroup$ Nov 9, 2020 at 13:04
  • $\begingroup$ Lol, no. This is accurate to the best of my knowledge. Once again, I will verify this and get back. The statement I made regarding a nucleophile attacking the 2-carbon of ethylene side-group ascertains my hypothesis. There is no other place a nucleophile would attack this molecule as the pi bond must shift towards benzene ring in that case. $\endgroup$ Nov 9, 2020 at 13:18

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