When we were first taught resonance it was with the example of benzene and I myself justified it by thinking that the double bond oscillates because no two carbon atoms are special and hence there is no reason there will be a double bond between any two and not any other two. I now know that's not true. But just to avoid any latent confusion, can you verify my reasoning is wrong by answering this question? By my method, all three bonds will have same length and bond order 4/3

  • $\begingroup$ For the purpose of this question, it might be good to assume for a while that what you thought earlier about benzene is true. 2-methylpropene, however, is not like that. Its carbons are special. Some have 3 hydrogens, some only have 2. $\endgroup$ – Ivan Neretin Mar 26 '19 at 7:19
  • $\begingroup$ You're correct. I didn't see that. Guess I'll have to find another example to attempt to disprove my theory. Thanks a bunch! $\endgroup$ – LoktarOgar Mar 26 '19 at 7:29
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    $\begingroup$ It is not clear " what isn't true and now you know it" but independent of the question resonance is not a kind of oscillation. $\endgroup$ – Alchimista Mar 26 '19 at 8:08
  • $\begingroup$ @Alchimista I understand that it is not an oscillation, just that's the terms all us amateurs(students around me) use without thinking about it and I too just went with the flow $\endgroup$ – LoktarOgar Mar 26 '19 at 8:50

The cases of benzene and 2-methyl propene are different.

In benzene, all carbon atoms are equivalent. One can also imagine the electron resonance.

In case of $\ce{CH3-C(CH3)=CH2}$ , the electrons have no place to move to create a delocalization cloud.

Additionally, it would have to switch a hydrogen atoms between 3 atoms, or create highly unstable $\ce{C=CH3}$

Such molecule would be highly reactive.

  • $\begingroup$ I completely ignored that the hydrogen would have to be displaced for it to happen. Thanks! $\endgroup$ – LoktarOgar Mar 26 '19 at 7:22

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