The acidity of a compound is determined by the stability of its conjugate base and the major structure of enolate has a negative charge on the Oxygen atom. Moreover, the negative charge on the O-atom of ethoxide is increased by the +I effect of $\ce{-CH2CH3}$.

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Image source: chem.libretext.org

So, how is ethoxide more stable than enolate?

  • 2
    $\begingroup$ If enol was the major tautomer, then it could be stronger, but it isn't. $\endgroup$
    – Mithoron
    Commented Jun 4, 2021 at 13:15
  • 2
    $\begingroup$ @Neha if you got your answer, please feel free to self-answer it. Self-answering is always appreciated and it will help future readers. $\endgroup$ Commented Jun 4, 2021 at 13:33
  • $\begingroup$ @NilayGhosh Well... I think only good self answers are really appreciated. Usually, if more experienced user then one that asks, answers, then result is better. $\endgroup$
    – Mithoron
    Commented Jun 4, 2021 at 14:18

1 Answer 1


Taking the data from this answer by ron, the equilibrium constant for the keto-enol tautomerism seen in acetone is

$$K_\mathrm{eq} = \frac{\text{[enol]}}{\text{[carbonyl]}}$$

$$ \begin{array}{lc} \hline \text{compound} & K_\mathrm{eq} \\ \hline \text{acetaldehyde} & 6 \times 10^{-7} \\ \text{acetone} & 5 \times 10^{-9} \\ \hline \end{array} $$

This means that acetone exists predominantly (if not wholly) in its keto form. For every one molecule of the enol form observed, we can expect $\pu{2E8}$ molecules of the keto form.

Because of this, any deprotonation that takes place in acetone will have to be via breaking a $\ce{C-H}$ bond, and that is not a simple thing to do when you compare it to deprotonating an $\ce{O-H}$ bond in ethanol. Thus ethanol is more acidic than acetone.


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