This is the question:

Which one is more acidic?

enter image description here

According to me, cyclopentadiene should have been the answer, as losing a proton would make it aromatic; whereas the first option doesn't need deprotonation to be aromatic, it already is. However the first option is given as the correct one.

I have come up with the following reasonings as to why the first option should be more acidic:

  1. Deprotonation might impart aromaticity to the pentagonal rings too which would act as the driving force to lose the proton.
  2. The negative charge on central carbon (after deprotonation) would be in conjugation with the $\pi$ electrons of the three rings which would make the entire molecule planar and hence reduce the strain on the three bonds that are directly linking the benzene rings. This could be another driving force for deprotonation.

The second reasoning seems much convincing to me but I fail to understand how can I compare that factor with the aromaticity factor involved in cyclopentadiene.

  • 2
    $\begingroup$ related chemistry.stackexchange.com/questions/124356/… $\endgroup$ – Mithoron Mar 28 '20 at 20:42
  • $\begingroup$ "...as losing a proton would make it aromatic..." - cyclopentadiene would become anti-aromatic instead (note the 4n electrons, and planar molecule) $\endgroup$ – Rahul Verma Mar 29 '20 at 17:49
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    $\begingroup$ @RahulVerma It's 6 electrons in conjugation (4n+2). $\endgroup$ – Aumkaar Pranav Mar 29 '20 at 18:28
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    $\begingroup$ @AumkaarPranav: Oh yeah, you're right! Late night, brain farts. $\endgroup$ – Rahul Verma Mar 30 '20 at 2:09

The method presented by Yusuf is a simplified model, which works best for well-behaved molecules. Unfortunately in this case it presents the 'correct' result based on incorrect assumptions. (Even though I did a calculation, I am not 100% convinced this is actually the correct result in the first place.)

In order for the resonance to stabilise a negative charge, the π-systems must be aligned, which is the case for the deprotonation of cyclopentadiene, but it certainly is not true for the other molecule F (Loong said it's called 3a2H-benzo[3,4]pentaleno[2,1,6,5-jklm]fluorene). Here is the calculated structure at the ri-BP86/def2-SVP level of theory:

structure of 3a2H-benzo[3,4]pentaleno[2,1,6,5-jklm]fluorene

As you can see, it is domed, and there are three (almost) independent aromatic systems. Through deprotonation this will not change much.

A reason for its relatively high acidity can be found in applying Coulson's Theorem, where the bond angle between two $\mathrm{sp}^x$ hybrid orbitals can be calculated from their angle $\omega_{ij}$ and their hybridisation indexes $\lambda_x$: $$\cos(\omega_{ij}) = -\frac{1}{\sqrt{\lambda_i\lambda_j}}$$ Since the molecule is C3v symmetric, therefore $\lambda_i = \lambda_j \mathop{:=} x$, this simplifies to: $$ x = -\sec(x)$$

The inner angle is $\approx\pu{102^\circ}$, therefore $x \approx 5$. The term for hybrid orbitals is actually just an abbreviation: $$\mathrm{sp}^{x} = \mathrm{s}^{\frac{1}{x+1}}\mathrm{p}^{\frac{x}{x+1}}$$

Given that we just form linear combinations, we can see: \begin{align} 1\times\mathrm{s}, 3\times\mathrm{p} &\leadsto 3\times\mathrm{sp}^5, 1\times\mathrm{sp} \end{align} Given the rounding I have done, the remaining orbital is very roughly speaking an $\mathrm{sp}$ orbital; in the same ballpark as acetylene, from which we know that it has a very acidic proton.

The reason for the high acidity is therefore not the stabilisation of the conjugate base, but the destabilisation of the acid. This is the completely reverse reasoning as cor cyclopentadiene, in which the corresponding base is massively stabilised by resonance.

Given the above, how are you supposed to know, and how are you supposed to judge these compounds?

In my opinion, the exercise is something more on the level of quantum chemistry course and belongs into the canon of a masters student. While simply models will accidentally produce the correct result, it is important to see, that they completely fail for F. Structure-property relationships cannot be inferred from a 2D-graph of the molecule. To be honest, the reasoning above was also rather post-hoc, after I have calculated the molecule and analysed its structure.

I believe, (as long as you are not a human calculator), that there is no simple way to tell which of those molecules is more acidic; there is no easy way to weigh the stabilising effect of resonance against the destabilising effect of pyramidalization.

On the ri-BP86/def2-SVP level of theory (which is rather crude, but I don't have a supercomputer at home) it really comes out that F is more acidic. The following isodesmic reaction yields a difference in electronic energy of about $\Delta E_\mathrm{el} = \pu{-92.8 kJ mol-1}$.

Isodesmic reaction of **F** with cyclopentadiene

Important: Even though the above sound reasonable, it still neglects plenty of things and has other shortcomings; to name the most important: thermal corrections, solvent effects, level of theory.


Upon deprotonation the structure does not change very much (see larger image), it remains very domed, see the overlay of the structures below.

overlay of the structures of **F** with its deprotonated form

Judging from the three highest occupied molecular orbitals (HOMO, HOMO -1, HOMO -2) we can still presume some delocalisation of the negative charge into the adjacent aromatic rings.

HOMOS of deprotonated **F**

I'm talking more about hybridisation, Bent's rule, and Coulson's theorem in the following: Are the lone pairs in water equivalent?

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    $\begingroup$ Interesting take. One thing that comes to mind: Does the polycyclic structure become more planar when we remove the proton? If so, would that mean we can still invoke the aromaticity/resonance effect? $\endgroup$ – Oscar Lanzi Mar 31 '20 at 10:09
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    $\begingroup$ @OscarLanzi not significantly, it remains domed. If I don't forget, I can post a picture of the anion, and I could have a look at the orbitals. $\endgroup$ – Martin - マーチン Mar 31 '20 at 23:22
  • $\begingroup$ Thanks. Something else: You render three central orbitals as $sp^5$ meaning each has 1/6 of the $s$ orbital. If so, the orbital with the proton has the remaining half of the $s$ orbital and so is all the way to $sp$. May want to check this. $\endgroup$ – Oscar Lanzi Apr 1 '20 at 0:09
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    $\begingroup$ @Oscar I've added the images. Thanks again for catching my calculation mishap. $\endgroup$ – Martin - マーチン Apr 1 '20 at 18:36

You can take both factors (1) and (2) combined as the reasoning for the greater acidity of the former. When you are comparing stability (in thermodynamic terms) of two conjugate bases, you are basically trying to position them with respect to each other on the energy axis of the energy profile (or reaction coordinate diagram).

In this case, the reaction coordinate (RC) for both the intermediates remains fixed (as a similar reaction, namely, deprotonation has occurred for both the reactants), while you try to use qualitative arguments to position them relative to each other on the energy axis.

In that spirit, one basic idea which we can use to check is that by seeing that:

More is the energy released by the system in form of resonance, more will be the stability of the negative charge.

Note that if you look at the reactants themselves, the first species is releasing roughly 3 times the resonance energy of benzene more as compared to the second one due to the presence of three benzene rings. Hence, we can estimate that the former should have had a lower energy positioning than the latter to start with.

After proton abstraction from both, like you said, 3 more five-membered aromatic rings will be generated in the former system, plus the negative charge can be delocalised over the three benzene rings. So now, if you see, the latter species is stabilising by an extent equivalent to the resonance energy of one cyclopentadienyl anion. For the former, the system is releasing energies equivalent to (resonance energy of 3 cyclopentadienyl anions) + (resonance energy of 3 benzyl anions).

So, by a qualitative estimate, the (1) species should be energetically more stable than the (2).

  • $\begingroup$ Thank you for the answer. Can you also please explain (or maybe provide a link to do so) the nature of aromaticity in the first case: do all the three cyclopentadienyl rings show aromaticity along with the benzene rings at the same time, or is there some kind of partial aromaticity? I am unable to wrap my head around aromaticity of the conjugate base of the first molecule (and aslo, do you know what it's called?). $\endgroup$ – Aumkaar Pranav Mar 29 '20 at 7:09
  • $\begingroup$ I'm sorry, but I have to disagree with this simplified reasoning. $\endgroup$ – Martin - マーチン Mar 29 '20 at 20:24
  • $\begingroup$ @Martin - マーチン♦Well,I am not a master's student(undergrad presently) so I had no knowledge of the calculations you have presented in your answer,I simply used what I have been taught by my instructors till now. However,the 3-D structure was quite enlightening about the how the overall aromaticity of the system was functioning,I didn't think about looking at that. Thank you for your answer :) $\endgroup$ – Yusuf Hasan Mar 29 '20 at 23:14
  • $\begingroup$ Yes, you are at no fault at all. I believe many of these exercises are hastily given out, without the proper reasoning beforehand. I have noticed that manytimes such exercises heavily rely on these inaccurate models. That is mainly a problem within the educational system, but I do not wish to diverge in politics. I hope that 3D-structure considerations will play more important roles in your future career ;) $\endgroup$ – Martin - マーチン Mar 30 '20 at 9:40

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