Whilst trying to answer this question I came across an interesting situation regarding the relationship between resonance structures and molecular orbitals. There are two places that you can deprotonate this compound. The two different deprotonated forms are shown below along with their resonance structures.

Both compounds have a continuous system of p orbitals which can align in a plane and overlap to form a $\pi$ system. However, in the first example it is not possible to draw resonance structures that delocalise the negative charge over all five atoms that have the available p orbitals. Does this mean that the negative charge is not delocalised over all five atoms? @Martin said in his comment on this answer that valence bond theory and molecular orbital theory ultimately give the same result so this should also apply here. In the molecular orbital description, what is the reason why the negative charge is not delocalised over all five atoms? What do the molecular orbitals for these anions look like?

enter image description here

  • $\begingroup$ Perhaps it would be simpler to start with the basic allyl anion system. In that system you can only delocalise the negative charge onto C-1 and C-3; this is consistent with the HOMO having no coefficient on C-2. That means that if you were to react the allyl anion with an electrophile you would get no substitution on C-2. chemtube3d.com/orbitalsallylanion.htm $\endgroup$ May 27, 2016 at 20:22
  • $\begingroup$ The fact that you cannot delocalise negative charge onto every single carbon is significant and I wouldn't call it a "limitation". Another example is the o,p-directing effect of the phenoxide anion, which can be somewhat explained by showing that the negative charge can only be delocalised onto the ortho and para carbons (a more rigorous explanation would talk about the intermediate or transition state instead of the starting material, but the argument is the same). $\endgroup$ May 27, 2016 at 20:25
  • $\begingroup$ @orthocresol I think I phrased that badly. What I meant was that, is it actually true that the charge is not fully delocalised or does resonance give the wrong answer (i.e. it has a limitation)? $\endgroup$
    – bon
    May 27, 2016 at 20:29
  • $\begingroup$ If I deprotonated in the kinetic direction (first set of resonance structures, i.e. BuLi or LDA), I would expect only alkylation on that carbon or the oxygen possible. If I deprotonated in the thermodynamic direction (a Lewis acid and triethylamine), I expect either central alkylation or vinylogous alkylation (or oxygen, but that is harder with a Lewis acid sitting ther). So yes, I think there is a strong physical implication there. $\endgroup$
    – Jan
    May 27, 2016 at 20:39

1 Answer 1


If there are adjacent, parallel, overlapping p-orbitals in a molecule, the electrons will be delocalized over all such atoms. While the negative charge isn't delocalized over every atom in the conjugated system, it's more than you have shown. What you are failing to consider is:


What's important to note about these resonances structures is that the net charge remains constant. These structures are consistent with both the reactivities of the enolate and $\alpha$,$\beta$-unsaturated carbonyl compounds. There negative charge is delocalized between the carbonyl oxygen and its $\alpha$-carbon, and the positive charge is delocalized between the carbonyl carbon (giving rise to 1,2-addition) and its $\beta$-carbon (giving rise to 1,4-addition).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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