2
$\begingroup$

In school they ask us to count the number of resonance structures in different organic compounds to determine it's stability. The one with more resonance structures is more stable. Is this notion correct?

$\endgroup$
  • 1
    $\begingroup$ Hello and welcome to Chemistry.SE. If you have any questions about the site I suggest starting off with the short tour. For information on homework-type of questions in particular, you can read through this discussion. Regarding your question, do you understand why the presence of resonance structures might lead to increased stability of a molecule? $\endgroup$ – airhuff Feb 26 '17 at 5:03
  • 3
    $\begingroup$ It's generally not that simple. See for example Why does having equivalent resonance structures give more stability? $\endgroup$ – Mithoron Feb 26 '17 at 14:57
  • 2
    $\begingroup$ Here is a good overview about resonance: What is resonance, and are resonance structures real? $\endgroup$ – Martin - マーチン Feb 27 '17 at 10:24
  • $\begingroup$ Isn't stability in molecules is relative; relative to atoms, bonds, connectivity angles in it? $\endgroup$ – bonCodigo Mar 19 '17 at 7:40
5
$\begingroup$

Stability is a funny word. It gets tossed around all the time, but a molecule's stability really depends on the context you are considering it in. For example, benzene is totally unreactive towards electrophilic addition of $\ce{Br2}$ like most alkenes, but $\Delta_\mathrm{f}H^\circ(\text{benzene}) = 49.0\ \mathrm{kJ/mol}$, while $\Delta_\mathrm{f}H^\circ(\text{cyclohexane}) = -37.8\ \mathrm{kJ/mol}$.

So when we talk about resonance and stability, what are we really talking about? What this really refers to the stabilization energy associated with conjugated systems. For every p orbital that is added to $\pi$-system, it mixes in phase to create a new molecular orbital lower that is lower in energy than the two starting orbitals, mixes out of phase to create and orbital that is higher in energy than the two starting orbitals, and in some instances doesn't mix at all. This is illustrated here:

You can't directly compare the relative energies of the molecular orbitals of two unlike molecules simply based on their number of resonance structures, though. Different atoms and substituents will cause deviations and make comparison meaningless, so it is really only useful to compare homologous compounds, like butane, but-1-ene, and buta-1,3-diene.

$\endgroup$
4
$\begingroup$

You cannot use a simple concept such as ‘counting resonance structures’ to determine stability. This has multiple reasons.

  1. Stability has much more aspects to it than just counting resonance structures. For example, I can draw six canonic resonance structures of the benzyl radical. I can only draw two canonic resonance structures for toluene — the same molecular structure only with an additional hydrogen on the side chain. Yet undoubtedly, toluene is more stable than any radical.

  2. You cannot really count resonance structures as that would imply some finite number. What one generally does is merely count the most probable resonance structures. You can draw many, many more less favourable ones; e.g. charge separation, homolytic bond cleavage and all that follows from them.

    Thankfully, though, the canonic resonance structures which are typically taken to be the most probable can be counted somewhat.

  3. Resonance as a concept is based on localised bonds. Molecular orbital theory tells us that better descriptions of molecular structures typically do not have localised bonds — although there are often mathematically equivalent descriptions that would.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.