I am in the David Klein textbook in Chapter 3 going over rule 2 of finding the more acidic proton: if the electron pair in the conjugated base is delocalized, then the conjugate base is more stable, ergo, the proton at that location is more acidic. Here's the practice problem I am stuck on:enter image description here

I generated two different conjugated bases. The book indicates the red proton is the more acidic proton. Is this the proper resonance structure for the red proton's conjugate base? enter image description here If so, then this explains why the red proton is more acidic, because it has a delocalized pair of electrons, while the blue proton does not have a delocalized pair of electrons.

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    $\begingroup$ Yes, your reasoning sounds right. $\endgroup$ Dec 25, 2018 at 1:37
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    $\begingroup$ What is your question? You seem to have answered it yourself. $\endgroup$ Dec 25, 2018 at 1:58
  • $\begingroup$ @orthocresol, I apologize if I have hidden the question with adding too much detail. I am trying to confirm with someone if I properly drew the resonance structure that would explain why the top proton is more acidic. Thanks Tan Yong Boon! I feel much better with you thinking my reasoning is correct. I suppose that will answer my question. $\endgroup$ Dec 25, 2018 at 2:45
  • $\begingroup$ Just an additional information, as soon as you made the conjugate base, the compound became aromatic(Huckel's rule), so the red proton is definitely very acidic. Had you got a conjugate base which was antiaromatic, then inspite of conjugation, the acidity of the proton would have been quite less. So while checking acid base equilibrium, Gibbs free energy as a whole is to be considered $\endgroup$ Dec 25, 2018 at 3:19

2 Answers 2


More is involved here than just resonance structure. When you have pi electrons delocalized around a single ring there is the $4n+2$ rule where the most stable configuration of pi electrons in a ring has $4n+2$ delocalized electrons. A planar, conjugated ring with the proper $4n+2$ number of pi electrons is aromatic and will be strongly stabilized. If instead, you have $4n$ delocalized electrons it doesn't work, and especially for smaller rings you may end up with something highly unstable.

When you draw the resonance structures by removing the red proton, you count six delocalized pi electrons -- $4n+2$. So you get that highly stabilized aromatic ring. That's why the red proton is a relatively strong acid.

Bonus questions: 1) Would it work if you used cyclopropene instead of cyclopentadiene? 2) What if you had 3-chlorocyclopropene instead and you used a Lewis acid to remove a chloride ion(see slides 30 and 31)?


Figure c does not look like a real compound. If it represents the conjugate base of cyclopentadiene, it is lacking a negative sign, and all five protons around the ring are equivalent, so the question of whether the red or blue proton is more acidic makes no sense.

If Figure c represents cyclopentadiene, the acid, it does not clearly show that the red position really has 2 hydrogens. However, the double bond structure shown is consistent with 2 hydrogens in the red position, and the question becomes very simple: in cyclopentadiene, is a methylene proton or a vinylic proton more acidic? Removing a vinylic proton puts a negative charge in the localized sigma system - no aromaticity develops. However, removing one of the methylene protons causes that carbon to go from sp3 to sp2p, and the negative charge goes into the pi system where it can be delocalized over the 5-membered ring. And aromaticity is a characteristic feature of cyclopentadienide ion.


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