In most academic courses you'd learn that aldehydes are more acidic (lower $\text{p}K_\text{a}$) than ketones due to the lower electron donating effect of the proton compared to the alkyl group of the ketone.

However, when talking about condensations, for example, it's not a weird choice to form a ketone enol over an aldehyde enol on the basis of "thermodynamic stability due to a more substituted alkene under thermodynamic control". Yet, $\text{p}K_\text{a}$ is the measurement of such stability, and thermodynamically the aldehyde enol is more stable. Adding to this, a following argument is that "aldehydes react much quicker towards nucleophilic attack than ketones". I agree with this, but weren't we just discussing thermodynamic control?

The questions are then:

  1. Is it valid to use the "more substituted alkene" to argue a step on a mechanism under claims of thermodynamic control?
  2. Can this type of argument be coupled with a kinetic one like the example provided as well? I.e. is it valid to mix thermodynamic and kinetic arguments one in the same? (e.g. this intermediate forms because it's more thermodynamically favored, and the next one forms because it is kinetically favored).

(I realize question 2 may be broader than this, but I'm trying to keep this in the sense of mixed aldol condensations)

  • $\begingroup$ I am probably wrong, but does the word "condensation" refer to simply aldol condensations (self and cross) or some other reactions as well? $\endgroup$ Mar 22, 2018 at 8:21
  • 1
    $\begingroup$ Since the whole post is about aldehydes, ketones and their formation of enolates, I think it is safe to assume I'm talking about aldol condensations. $\endgroup$
    – ralk912
    Mar 22, 2018 at 8:27

1 Answer 1


I would like to first comment on the reasoning given as to why aldehydes are more acidic than ketones. The usually provided reasoning is, as stated here:

  • H atoms are regarded as having no electronic effect : they don't withdraw or donate electrons.
  • Alkyl groups are weakly electron donating, they tend to destabilise anions (you should recall that they stabilise carbocations). This is because they will be "pushing" electrons towards a negative system which is unfavourable electrostatically. Hence, the anion of a ketone, where there are extra alkyl groups is less stable than that of an aldehyde, and so, a ketone is less acidic.

We need to be clear of what we are measuring when we measure the acidity of an aqueous solution of an aldehyde. It is important that we recall what is mentioned in this post, on the acidity of benzaldehyde, as it relates to the acidity of aldehydes in aqueous solution. Most aldehydes exist in aqueous solution with a significant proportion of their hydrated forms, which are geminal diols. Hence, it is very likely that the $\mathrm pK_\mathrm{a}$ values we have obtained are for the $\ce {-OH}$ protons of the hydrate and not for the $\ce {C-H}$ protons at the $\alpha$ position of the aldehyde.

Let's take a look at acetaldehyde. Ethanol has $\mathrm pK_\mathrm{a}$ value of $15.9$ and the hydration equilibrium constant for acetaldehyde is $\ce {1}$, as mentioned here. Based on the guideline that geminal diols have $\mathrm pK_\mathrm{a}$ $2.5$ units less than that of the monohydric alcohol, the predicted $\mathrm pK_\mathrm{a}$ value for acetaldehyde is $13.4$. The experimental measurement gives $13.57$, as seen from here, which is very close to the value determined via calculation with the guidelines. This closeness of the values suggests that the hydrate is indeed formed and the acidity that we are measuring for the aqueous aldehyde is indeed that of the hydrate.

If this is what we are measuring, then clearly the reasoning highlighted in the box may not even be correct. In fact, we can't even be sure that the $\ce {\alpha}$ $\ce {C-H}$ proton of the aldehyde is more acidic than that of a ketone since we won't be able to measure their acidities, at least not in water.

With the clarification I have made, I hope that there is no longer any inconsistency in the issues you have mentioned. In particular, it is correct to say that under thermodynamic control, the intermediate with greater stability (e.g. the enolate with the more substituted alkene) is formed. Regarding your 2nd question, I am unsure of what you mean but I will just say that kinetic and thermodynamic control refer to different reaction conditions resulting in the reaction undergoing different pathways to thus form different products.

As an extension, we can also consider if the $\mathrm pK_\mathrm{a}$ values given for the ketones are representative of the acidity of the $\alpha$ protons or that of the $\ce {-OH}$ protons of the hydrate. Let us consider acetone, with $\mathrm pK_\mathrm{a}$ given as $\ce {19.16}$ here. The hydration constant for acetone is $1.4 \times 10^{-3}$ and the $\mathrm pK_\mathrm{a}$ of 2-propanol is $16.5$ (as seen here). Based on the same sequence of calculations and the same guideline, the predicted $\mathrm pK_\mathrm{a}$ value is $16.9$. The significant difference between the predicted value and the experimentally measured value suggests that the acidity of the acetone is not just due to that of the hydrate $\ce {-OH}$ protons and that there may be significant deprotonation of the $\alpha$ protons taking place.

An important question that we would like to know the answer to is what then is the $\mathrm pK_\mathrm{a}$ value of an $\alpha$ proton of an aldehyde. Well... It seems like due to hydration of the aldehyde, we would not be able to directly measure the acidity of this proton. However, one would reasonably estimate that it is close to that of the ketone $\alpha$ hydrogens. Again, that is something we may not be able to measure directly for the same reason. An educated guess would be in the range of $18-22$ as it needs to be more lower than the $\mathrm pK_\mathrm{a}$ of esters but higher than that of alcohols.


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