# Why is ethanol more acidic than acetone?

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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So, how is ethoxide more stable than enolate?

• If enol was the major tautomer, then it could be stronger, but it isn't. Jun 4, 2021 at 13:15
$$K_\mathrm{eq} = \frac{\text{[enol]}}{\text{[carbonyl]}}$$
$$\begin{array}{lc} \hline \text{compound} & K_\mathrm{eq} \\ \hline \text{acetaldehyde} & 6 \times 10^{-7} \\ \text{acetone} & 5 \times 10^{-9} \\ \hline \end{array}$$
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.