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Which among these hydrocarbons is more acidic?

Here, I thought about hyperconjugation. For this, I counted the alpha H i.e. the number of hydrogens attached to the carbon attached to a sp2 hybridized carbon. What I get is, (C) has most no. of those Hydrogens. So, my assumption is that it should be more acidic. But that's not the case. Where am I going wrong?

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    $\begingroup$ Interestingly, why do you think more hyperconjugation means the compound is more acidic? Where did you even pick that thought from? I earnestly suggest you drop that thought immediately, it has more problems than we can discuss. Acidity is always directly proportional to stability of the conjugate base (barring some very rare cases). $\endgroup$ – Gaurang Tandon Mar 23 '18 at 2:45
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Acidity is directly proportional to the stability of conjugate base. Aromatic compounds are very stable.

In essence only one compound after deprotonating is aromatic - which is the conjugate base of (A).

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The conjugate base of compound A becomes aromatic post deprotonation:

This is essentially a cyclopentadienyl anion derivative, that has 6 $\pi$ electrons that are delocalised over the ring. Thus this compound will be by far the most acidic.

As for the other cyclopentadiene derivative, it cannot be deprotonated to make it aromatic: it is localised on an $sp^2$ hybridised carbon atom, and the electron pair will thus be in the plane of the ring itself.

A similar logic can be extended to the methylated tropylium derivative. As for the tropylium derivative that can be deprotonated to give a delocalised electron system, it will have $8\pi$ electrons and thus will be antiaromatic. The molecule will thus prefer to be non-planar instead of being destabilised by the antiaromatic effect.

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