Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: gpavanb on April 18, 2009, 12:46:32 PM
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Look at the diagram accompanying the question.
Which compound is the most acidic?
The compound C forms a carbanion which is a massive 14-electron aromatic system. Thus it is the most acidic.
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molecule A would form a 6 electron aromatic system upon proton abstraction, so it is basic according to the 4N+2 rule. Molecule B would form an antiaromatic system with 10pi electrons. molecule C would provide a 14 pi electron system, which is also aromatic as you said...molecule D would again form an antiaromatic system with 10 pi electrons. in comparing A and C i would say C is more acidic because that carbanion can be resonated all throughout the things, while molecule A only provides a couple positions
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Molecule B would form an antiaromatic system with 10pi electrons.
no it wouldn't. the system is non-aromatic when deprotonated.
as mentioned, the choices are between A and C.
fluorene (http://en.wikipedia.org/wiki/Fluorene)
and cyclopentadiene (http://en.wikipedia.org/wiki/Cyclopentadiene)
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just to be clear:
an antiaromatic system with 10pi electrons.
no it wouldn't. the system is non-aromatic when deprotonated.
a. 4(2) + 2 = 10 (10 is a Huckel number)
b. Compound B is aromatic from the start...it won't become non-aromatic when it's deprotonated.
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B cannot be antiaromatic, because it does not have a complete set of conjugated pi electrons. Try drawing resonance throughout the right ring, you can't.
I'm sure you wouldn't say that deprotonation of 1,3-Cyclohexadiene (http://en.wikipedia.org/wiki/1,3-Cyclohexadiene) gives you an aromatic system would you?
The most that B offers is that the hydrogen is benzylic.
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Just out of curiosity, why is cyclopentadiene so much more acidic than the other system that becomes aromatic upon deprotonation?
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Forget that you know the pKa's for a minute and consider this situation: you have a solution of the conjugate base of fluorene (Fl-) and treat it with cyclopentadiene (Cp-H)
Fl- + Cp-H ::equil:: Fl-H + Cp-
arom. non-arom arom. arom.
It's common practice to compare the relative stabilities of the charged species of an acid-base equilibrium, but that doesn't give the whole picture. Here, you also have to compare the stability of Cp-H to Fl-H
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If I had been asked by a professor on the spot, I would have said Fluorene. It has more resonance forms, and should make the Fl-H bond weaker, so more acidic. However, what about considering any possible electron-donating character of the benzene rings? Enough to destabilize the anion of the Fl- to increase the pKa enough?
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Squirmy - what you're saying makes sense to me, thanks.
Macman - I'm not sure I get quite what you're saying. The negative charge in that system would be spread over numerous (maybe all, not drawing out the resonance structures =P) carbons, right?
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Right. If you deprotonate fluorene, then the negative charge can resonante throughout the whole molecule. Much more delocalization than in cyclopentadiene. This will help stabilize the negative charge, so you would think that fluroene should be more acidic.
I am wondering if the inductive effects of the benzene rings are enough to cause the decrease in acidity, or what else may be causing that.
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just to be clear:
Quote
an antiaromatic system with 10pi electrons.
Quote
no it wouldn't. the system is non-aromatic when deprotonated.
a. 4(2) + 2 = 10 (10 is a Huckel number)
b. Compound B is aromatic from the start...it won't become non-aromatic when it's deprotonated.
yes, quite right. it's aromatic going it, it's aromatic coming out. deprotonation does nothing special and does not have any bearing on this question. i stand corrected :)