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I have read about it and i found that the alcohol is more acidic, but isn't the conjugate base produced from losing the alpha hydrogen of the ketone more stable ?!

The negative charge in this conjugate base is spread over the oxygen atom and the alpha carbon atom by resonance ( charge is delocalized ) as shown in the attached image.

resonance structures of the conjugate base produced from ketone

while in the conjugate base produced from the alcohol we see that the negative charge is carried only by the oxygen atom and no resonance structures which means that ketone should be more acidic since its conjugate base is more stable and it will easy be formed.

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    $\begingroup$ Resonance is only one factor when it comes to anion stability. For a more extreme example, consider HCl and acetic acid... $\endgroup$
    – orthocresol
    Feb 9, 2022 at 12:25
  • $\begingroup$ No, I want this example what makes a ketone less acidic than alcohol although the conjugate base of a ketone is more stable than the conjugate base of the alcohol by resonance . $\endgroup$
    – Ahmed Hossam
    Feb 9, 2022 at 14:38
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    $\begingroup$ It isn’t more stable. If the pKa is larger, then it’s a weaker acid, and the anion is less stable. No exceptions. $\endgroup$
    – orthocresol
    Feb 9, 2022 at 15:20
  • $\begingroup$ What makes an apple taste different from an orange? $\endgroup$
    – user55119
    Feb 9, 2022 at 15:59

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You are considering two different Hs. I mean: one is a $\ce{O-H}$ bond; the other is an $\ce{C-H}$ bond.
You can't compare these two bonds.
Yes, $pK_a$ of the alcohol is around 16/17 and $pK_a$ of the ketone is around 25/26, but are two bonds completely different.

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  • $\begingroup$ no it is not about bonds at all H2O is more acidic than NH3 and here we are comparing OH bond with NH bond . $\endgroup$
    – Ahmed Hossam
    Feb 9, 2022 at 14:30
  • $\begingroup$ @AhmedHossam If not for enol / ketone tautomerism, and hydration equilibrium, you could perhaps simply compare them with alcohols, but their properties aren't nearly as simple as both you and Andrea seem to think. $\endgroup$
    – Mithoron
    Feb 11, 2022 at 1:24

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