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Arrange the following in decreasing order of acidity

enter image description here

I think it's S>P>R>Q

(Q) forms an anti aromatic conjugate base. So its the least acidic compound here.

(R) forms forms a conjugate base which is only stabilized by electron withdrawing (inductive) effect of $\ce{S}$.

(P) forms an aromatic conjugate base and (S) forms am ion stabilised by resonance (two resonance structures). But negative charge in $\ce{O}$ is more stable than negative charge on $\ce{C}$. So (S)>(P).

Is this correct?

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    $\begingroup$ Looks right to me. $\endgroup$ Oct 13, 2015 at 13:36
  • $\begingroup$ I would say P>S>R>Q as I think aromaticity of P would take preference over non-aromatic anion of S. $\endgroup$ Oct 13, 2015 at 13:50
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    $\begingroup$ Let's look at some data. pKa of cyclopentadiene is about 16, while that of acetylacetone is about 9. Now who's stronger? $\endgroup$ Oct 13, 2015 at 14:00
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    $\begingroup$ Normally a person would know that resonance stabilization is not an absolute weapon. It can be measured and thus compared to other things; some are weaker than it, and some are stronger. $\endgroup$ Oct 13, 2015 at 14:12
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    $\begingroup$ 3d-orbitals of S are now commonly thought to be less important than we believed before. (Maybe unimportant at all.) $\endgroup$ Mar 11 at 15:14

1 Answer 1

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(Q) is the least acidic compound because the conjugate base formed is anti aromatic and hence unstable.

(R) is more acidic than (Q) because the conjugate base is stabilized by the electron withdrawing inductive effect of two sulphur atoms.

Conjugate bases of both (P) and (S) are stable due to resonance. Among the two, conjugate base of (S) is more stable because a negative charge on Oxygen atom is more stable than a negative charge on carbon atom. Here are the resonance structures: enter image description here

This is verified by @Ivan Neretin's comment on the OP.

Let's look at some data. pKa of cyclopentadiene is about 16, while that of acetylacetone is about 9.

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