in this picture

identify the more acidic compound (and the answer is the III is more acidic than II ).

My reasoning: Consider II compound, we have oxygen attached to the carbon forming a sigma and a $\pi$ bond, this oxygen has two filled p orbitals. The oxygen in the ortho position to it (in II), also has two filled p orbitals. My theory is that as we have filled p orbitals, they can form a $\pi$ bond type interactions (*). After forming the two $\pi$ bonds, the compound is stable.

Now discussing III, applying the same idea as above, there is a larger distance for the $\pi$ overlap and hence I think $\pi$ overlap is less stable compared to the case in II.

Hence, we can say that III is more acidic (**) than II, as the pi overlap of the p orbitals of the two oxygen atoms are farther and long in III, and II is more stable (less acidic) as the pi overlap is over a small distance.

I reached the correct answer but I am not sure if my reasoning for it is correct because I haven't seen directly similar explanations in textbooks, however, I felt this idea was hinted at.

*: Not a $ \pi$ bond in the strong sense of the word

**: Less stable

  • $\begingroup$ Well for one thing, ketones are more acidic than esters (lactones). $\endgroup$
    – user55119
    Oct 15, 2020 at 19:08
  • $\begingroup$ I am not asking about which compound is more acidic or basic, its already known to us. I just want to verify if my reasoning for it is a correct one. @user55119 $\endgroup$
    – Floatoss
    Oct 15, 2020 at 19:28
  • $\begingroup$ It's not OK. In esters second oxygen donates electrons to the one in carbonyl group competing with similar donation from carbon in the conjugate base - you need to compare conjugate bases rather then acids. $\endgroup$
    – Mithoron
    Oct 15, 2020 at 20:56
  • $\begingroup$ So, you're saying the idea completely false/ inapplicable? I feel like it has some truth in it. For example consider how the p orbitals become a donut sort of shape in benzene, pretty sure there is some mixing and mashing of p's somewhere. $\endgroup$ Oct 16, 2020 at 5:43
  • $\begingroup$ Have you never heard of ester resonance? $\endgroup$
    – Jan
    Oct 16, 2020 at 13:39

1 Answer 1


First, let's get to the root of this problem, then I'll double back to your question. We are asked which of the two compounds is most acidic. As a side note there are various definitions of acidity, but note that broadly these theories describe acidity as linked to propensity to accept electron pairs in some form rather than general instability.

Considering neither II nor III have any immediately acidic functionalities (alcohol, carboxylic acid, sulfonic acid), we identify the alpha carbons of the carbonyl functionalities in each molecules as the most acidic hydrogens with pka ~20-25 (since all other hydrogens are alkyl (pKa ~50) and cannot form a stable conjugate base upon deprotonation).

With that in mind, the real difference between II and III is the identity of the carbonyl group. Comparing II to III we see II is an ester, with the carbonyl center bonded to an alkoxyl substituent and an alkyl substituent. Comparatively, III is a ketone, with its carbonyl center bound to to alkyl substituents. Recognizing that O lone pairs can hybridize to conjugate with the carbonyl center were C atoms cannot, we predict III to be more acidic.

The reasoning for this can be approached from a couple of angles, so here's two:

  1. O lone pairs present in II destabilize the enol formed by deprotonation, making it more likely to donate electrons (I.E has more basic character than III), thus making its conjugate acid form less acidic (since weaker acids have conjugate bases that are stronger bases and vice versa)

  2. O lone pairs can form an additional bond to the carbonyl carbon in a octet-satisfied resonance contributor, which greatly reduces the contribution of the analogous hyperconjugative resonance contributor with dissociation of H+. Note that this hyperconjugative contributor is a rather poor description in a thermodynamic sense, but it does play an effective stand in for a mechanism with a basic species.

Now to your actual question: You are correct with your analysis of the pi system of II, but I would note that your description is using atomic orbitals which can become very confusing in the world of ochem. In the hand-wavy hybridization terms, each oxygen would adopt an sp2 bonding orbital hybridization with unhybridized p orbitals participating in the pi-system. However, these pi-interactions can only occur if sequential atoms have unhybridized p orbitals.

This is due to the fact that the stability of pi systems arises from overlap of p orbitals. When two p orbitals are close enough to overlap, their wavefunctions can recombine into relatively lower energy (bonding) and relatively higher energy (antibonding) wavefunctions which become orbitals. This same recombination can occur for any number of p-orbitals, with the sole condition being that overlap actually occurs.

This can only have a fully stabilizing effect when atoms are ~1.20-1.5 angstroms apart as indicated by spectroscopic bond lengths. Overlap is feasible between single-bonded substituents (Like C-O in II), but impossible once the continuous system of unhybridized p orbitals is broken.

Thus while the O in II can directly conjugate its electrons with the carbonyl pi system by "rehybridizing" from the general expected sp3 hybridization to stabilize the system, in III there are no adjacent unhybridized p orbitals to interact with the carbonyl pi system. The only electronic interactions present between the substituent (cyclic ether) and the carbonyl system would be inductive/field effects.

While one could argue that a conformer of III could possibly involve delocalization of O p electrons into the carbonyl system, I would respond by noting that such a conformer

  1. Would have to have an unhybridized O p-orbital in order to be involved with pi interactions
  2. Would have to have a torsional angle Θ between the oxygen and the carbonyl group of Θ=/=0 in order for overlap to actually occur (Distance between orbitals r<~1.5).

However, this would mean that the oxygen's unhybridized p orbital and the carbonyl pi system would have different symmetry elements, meaning such a pi-interaction would be symmetry-forbidden

That last bit might be jumping the shark a little, but I hope this helps!

  • $\begingroup$ Could you give a reference for this re-hybridisation concept? $\endgroup$ Oct 16, 2020 at 7:25
  • $\begingroup$ "When two p orbitals are close enough to overlap, their wavefunctions can recombine into relatively lower energy (bonding) and relatively higher energy (antibonding) wavefunctions which become orbitals. " are you combining both molecular orbital theory and hybridisation here? $\endgroup$ Oct 16, 2020 at 7:27
  • $\begingroup$ " conformer of III" what kind of conformed are you talking about here? $\endgroup$ Oct 16, 2020 at 7:28
  • $\begingroup$ alright, your answer is really nice. But you are talking about pi interactions between the C and O, whereas I am talking about pi interactions being possible between the two Oxygen atoms themselves (as they have 2 filled p orbitals). Your reasoning for III being more acidic is right, but you never talked about my question which said "is pi interactions possible between the Oxygens themselves. And in the middle you say that the Oxygen is $sp^2$ hybridized, but isn't it $sp$ hybridized? Please do let me know what is wrong with my thinking. Thank you for your answer. @bigboss87 $\endgroup$
    – Floatoss
    Oct 16, 2020 at 7:42
  • 1
    $\begingroup$ @Buraian In organic, descriptions of molecular orbitals tend to be a little hand-wavy, so there can be a little mixing of theories that describe the same thing. Here, I'm discussing LCAO using hybridization to explain the presence of atomic orbitals on a molecular species. As I noted in my post, for favorable recombination to occur, the p-orbitals have to be able to interact through space, which is not possible if the probability density regions are not overlapping. This overlap becomes infinitesimal once you are >1 single bond away. $\endgroup$
    – bigboss87
    Oct 16, 2020 at 19:21

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