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.
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.
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.
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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