# What is the difference between resonance and π conjugation

I have seen electron delocalization mentioned frequently in 2 separate contexts: resonance (ozone, carbonate / nitrate...) and π-conjugation systems (aromatic compounds).

While the 2 initially sound the same, 2 different points lead me to believe otherwise:

1. π conjugation requires p- orbitals to be in the same plane, whereas many compounds said to have resonance ($$\ce{ClO4-}$$, $$\ce{SO4^{2-}}$$) are not planar so resonance can't be a subset of π conjugation
2. All molecules with (significant) resonance have multiple equivalent stable-lewis structures (nitrate has 3, sulfate has 4...) whereas many molecules with π-conjugation do not (haxa-1,3-diene has only one "most stable" lewis structure) so π-conjugation cant be a subset of resonance

NOTE ON POINT 2: I am aware that hexa-1,3-diene has minor contributing structures, but so do all possible molecules. However, the only molecules for which resonance hybrids are used are those with multiple major contributing lewis structures.

The only intersection between the 2 concepts I find is benzene; it has 3 equivalent major contributing structures implying resonance and 6 overlapping p-orbitals lying in the same plane, implying π conjugation

I feel like I'm missing some part of the bigger picture. What's actually happening?

• Resonance is one mechanism for checking if a system has π conjugation . – Zhe Feb 11 at 15:16

You might be confused but not missing anything big or fundamental.

A pi-conjugated system is one in which delocalisation of pi-electrons can take place. It is all. When effective, pi-conjugation is a subset, or a type, of delocalisation than can be described by resonance.

However:

to be strict, conjugation refers to bonds, and specifically to an alternating single and double bonds structure. It is just a positional nomenclature.

It turns out that in cases and in different extent the structure with such conjugated alternating single and double bonds does not represent the real molecule right because delocalisation takes place.

While the terms are linked, a not very accurate use of them leads to their treatment as if they are synonyms. That is what I have more or less done above, but just using a more nuanced term as pi-conjugation, thereby referring to electrons rather than bonds.

This is as to say that a sequence of alternating single and double bonds (as in at least one limiting form) is conjugated by definition. The conjugation (would be better saying delocalisation, here) can eventually/effectively be so true to the point of erasing the original meaning, i.e. in fact there are even not localized conjugated bonds.

Whit all the above in mind, a perfect organic metal with equalised bonds would be seen as "the king" of conjugated polymers.

Finally, within the context and dictionary of VB the above delocalisation is described by resonance, This again lead to the use of delocalisation and resonance as interchangeable terms.

• Oh ok so alternating single and double bonds are conjugated by definition and have delocalization only to a small extent (due to several but minor resonance structures). And conjugated molecules like benzene withe multiple major contributing structures have significant delocalization and are best described by a resonance hybrid. Is this understanding correct? – OVERWOOTCH Feb 12 at 11:43
• @OVERWOOTCH yes. Aromatic indeed are not really called or seen as conjugated moleculed, but you can somewhat say there is pi-conjugation to intend resonance /delocalisation bla bla bla. Case in between are conjugated polymers, they do have delocalisation but not the point of removing single and double bonds alternation. i.e. the actual hybrid has localized bonds, just for various reasons the weight of the opposite facing limiting forms isn't equal. I would say you got it. – Alchimista Feb 12 at 12:05