-1
$\begingroup$

I am having a hard time understanding the more acidic hydrogen in the following two compounds.

enter image description here

I attempted it by looking at two things, finding the possible resonance and the EN of the Sulfur and Oxygen. All these point to me that h1 is more acidic as the carbon anion formed after removal of H shall be more stable than the h2. But, h2 is more acidic. How is it so? What am I missing?

$\endgroup$
  • 1
    $\begingroup$ There is a possibility of back bonding of lone pair generated by Proton removal in second case in the vacant 3d orbitals of Sulphur, which is not possible in case of Oxygen. So, in second case, conjugate base will be more stabilised. $\endgroup$ – Soumik Das Dec 8 '18 at 6:31
  • $\begingroup$ @SoumikDas No. $\endgroup$ – Mithoron Dec 8 '18 at 19:20
  • $\begingroup$ @Mithoron Why not ? $\endgroup$ – Soumik Das Dec 9 '18 at 3:47
  • 1
    $\begingroup$ Because outer $d$ orbitals are not really valence orbitals. Can we construct molecular orbitals that favor the sulfur containing anion? $\endgroup$ – Oscar Lanzi Dec 9 '18 at 11:21
0
$\begingroup$

When comparing acidic strength, compare the stability of conjugate base. One with more stable conjugate base will be a stronger acid.

When $H+$ will leave, a negative charge will be left on the carbon. And remember:

Lone pair and negative charge can't go in resonance due to high repulsion between electrons of both

And

Stabilization due to resonance is greater than that due to inductive effect

While drawing resonance structures, both of them will have resonance with the +ve charge on the $\beta$-ketone, so the only difference will be due to S and O. Sulphur and oxygen both have lone pair, but it can't participate in resonance. But S has empty d-orbitals which can do back-bonding, and thus participate in response while oxygen cannot do so. And because resonance trumps over inductive effect, it even if Oxygen is more EN, will not stabilize as much as by Sulphur's back-bonding.

Thus $H_2$ is more acidic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.