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There are numerous websites out there which claim that the pKa of benzaldehyde – C6H5CHO – is 14.90. (Just do a Google search for benzaldehyde pka to see what I mean.) This doesn't make sense chemically, as there are no protons in benzaldehyde which would be so acidic.

However, it doesn't seem to have been made up. The CRC Handbook of Chemistry and Physics lists the same value of 14.90 for the pKa of benzaldehyde, at $\pu{25 ^\circ C}$ in aqueous solution. But I can't find this information in the primary literature.

If this value refers to the acidity of the hydrate, does it include the equilibrium constant for hydrate formation? i.e. is this Ka the equilibrium constant for the reaction:

$$\ce{PhCHO + 2H2O <=> H3O+ + PhCH(OH)O-}$$

or is it just

$$\ce{PhCH(OH)2 + H2O <=> H3O+ + PhCH(OH)O-}?$$

I would appreciate some kind of definitive evidence. Using chemical reasoning and intuition is great, but if this value is to be of any use to anybody, then we need to know exactly what it describes.

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    $\begingroup$ Is it possible this is a value for the hydrate? $\endgroup$
    – Dennis Cao
    Commented Jan 7, 2019 at 19:57
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    $\begingroup$ @DennisCao, indeed, that's my primary suspicion: it might be analogous to the "pKa" of CO2. $\endgroup$
    – orthocresol
    Commented Jan 7, 2019 at 19:59
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    $\begingroup$ I think this must be for the deprotonation of the hydrate. $\endgroup$
    – Waylander
    Commented Jan 7, 2019 at 22:05
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    $\begingroup$ Not that it's the definitive source, but the book "Dictionary of Food Compounds" (see Google Books preview) suggests it is indeed the hydrate. $\endgroup$
    – SendersReagent
    Commented Jan 7, 2019 at 22:07
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    $\begingroup$ Just for some additional context, it does appear to be possible to abstract the aldehyde proton from benzaldehyde directly (as opposed to the hydrate), though it is indeed a much weaker acid. This reference indicates that LiTMP in toluene/hexane at 0 °C performs the deprotonation, but the resulting conjugate base understandably immediately condenses with additional benzaldehyde, eventually forming benzyl benzoate on aqueous workup. $\endgroup$
    – Nicolau Saker Neto
    Commented Sep 7, 2021 at 1:20

1 Answer 1

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Significant amount of geminal diol of benzaldehyde exists in an aqueous solution of benzaldehyde at 25 °C because $\mathrm{p}K_{\text{hyd}} = 2$ (Ref. 1)

Benzaldehyde Hydration

The $\mathrm{p}K_{\mathrm a}$ of benzyl alcohol is listed as 15.40 (Wikipedia). Thus, one can reasonably assume that the given value of $\mathrm{p}K_{\mathrm a}$ 14.9 represents a composite equilibrium constant for the hydration of benzaldehyde and dissociation of the geminal diol thus formed.

Late addition to the answer:

In his paper, "Acidity constants of benzimidazolium ketone and pyridinium aldehyde hydrates" (Ref.2), Terence C. Owen states that:

It is known that the acidity constants of gem-diols typically are about 2.5 units lower than those of the corresponding monohydric alcohols.

When do some extensive literature survey, one can find few examples to backup that statement. For example, $\mathrm{p}K_{\mathrm a}$ of methanol is reported as 15.7 while that of formaldehyde hydrate is 13.3, between which $\Delta\mathrm{p}K_{\mathrm a} = 2.5$ (Ref.2). Interestingly, $\mathrm{p}K_{\mathrm a}$ of 1,1,1,3,3,3-hexafluoropropan-2-ol is reported as 9.22 while that of hexafluoroacetone hydrate is 6.45 where $\Delta\mathrm{p}K_{\mathrm a} = 2.77$ (Ref.3). However, $\mathrm{p}K_{\mathrm a}$ of 2,2,2-trifluoroethanol is reported as 12.37 (Ref.4) while that of 2,2,2-trifluoroethanal hydrate is 10.05 (Ref.3) where the difference is < 2.5 ($\Delta\mathrm{p}K_{\mathrm a} = 2.33$).

Thus, we can conclude that the $\mathrm{p}K_{\mathrm a}$ of benzaldehyde is derived from its hydrate (gem-diol).

Also see: Yoshiro Ogata and Atsushi Kawasaki, In The Chemistry of Carbonyl Group, Volume 2; Jacob Zabicky, Ed.; John Wiley & Sons Ltd.: New York, NY, 1970, Chapter 1: Equilibrium additions to carbonyl compounds, pp 1–69 (https://doi.org/10.1002/9780470771228.ch1).

Late Edit:

It is also noteworthy that $\mathrm{p}K_{\mathrm a1}$ of the cyclic hydrate of phthalaldehyde is reported to be $11.50 \pm 0.10$ at $\pu{25 ^\circ C}$ (Ref. 5):

Phthalaldehyde hydrate

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    $\begingroup$ A bit of further research I did backs this up (several sources also mention the hydration equilibrium constant, e.g. nrcresearchpress.com/doi/pdf/10.1139/v79-084), and I think it is plausible that the gem-diol is more acidic than benzyl alcohol (inductive withdrawal). However, I'm still holding out for a more detailed confirmation of what exactly this 14.9 entails, specifically whether it includes the hydration equilibrium or not. $\endgroup$
    – orthocresol
    Commented Jan 7, 2019 at 23:09
  • $\begingroup$ This is a good reference. Accordingly, $pK_a$ of hydrated Phthalaldehyde is suggested as 11.6. $\endgroup$
    – Mathew Mahindaratne
    Commented Jan 7, 2019 at 23:33
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    $\begingroup$ Note that if you combine the 2 orders of magnitude from the hydration equilibrium and the 2.5 orders of magnitude from the gem-diol expected acidity constant, you would expect that the pKa is about 0.5 units smaller, which is actually spot on. $\endgroup$
    – Zhe
    Commented Jan 14, 2019 at 15:30
  • $\begingroup$ I like that you're still working on this answer. ;) Another paper further suggests that the pKa of benzaldehyde hydrate is 12.0 (observed) or 12.43 (calculated). Together with the hydration equilibrium constant we arrive at a pKa of ~14 to 14.5 for benzaldehyde. That is probably convincing enough for me so I will accept (I upvoted long ago of course!)... I am guessing that this 14.9 number was obtained just by direct pH / conductivity / etc. measurements without any regard for the underlying equilibria... $\endgroup$
    – orthocresol
    Commented Jul 7, 2020 at 7:12
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    $\begingroup$ In a way, we don't have to suppose the existence of a geninal diol. We get the same effect if we assume that the aldehyde acts similarly to boric acid, simply adding a small amount of hydroxide ions from water/a basic solution to its own molecule. $\endgroup$
    – Oscar Lanzi
    Commented Jul 7, 2020 at 9:56

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