In the addition of $\ce{NaHSO3}$ to aldehydes and methyl ketones (higher ketones do not respond well to this reaction), crystalline addition products are formed. An $\ce{-SO3H}$ and an $\ce{-OH}$ group are attached to the same carbon.

$$\ce{R2C=O + NaHSO3 -> R2C(OH)(SO3^{-}Na^+) <=> R2C(O^{-}Na^+)(SO3H)}$$

The line I read online, bothering me is:-

"The proton transfer equilibrium lies to the right for most ketones and to the left for most aldehydes."

What is the reason for this (assuming that it is correct in the first place)? Since the sulphonic acid group is more acidic than the hydroxyl group, should not the proton tranfer equilibrium lie towards the left for both the cases?

  • $\begingroup$ Could you provide a reference to the source where you read that claim? $\endgroup$ – Philipp Mar 6 '14 at 19:21
  • $\begingroup$ @Philipp Can't seem to find my original source. But found another one here pg 359. This appears to be comparatively more credible than the original source (some sort of notes), but still am not sure whether this is true. $\endgroup$ – stochastic13 Mar 7 '14 at 4:55
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    $\begingroup$ Hmm, your linked book says it is so "for steric reasons". That is rather vague especially because they don't explain in detail which steric interactions get relieved. Another problem is that they don't say whether their statement is true only for the crystalline compound or for the compound in solution too. If I had to guess, I'd say that maybe in the crystalline structure it might be true (although I don't think anyone could have predicted that without knowing the crystal structure) but not for the compound in solution. $\endgroup$ – Philipp Mar 7 '14 at 15:03
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    $\begingroup$ A little bit of searching gives no indication that the structure on the right is formed. See here, here, here and here $\endgroup$ – Nicolau Saker Neto Mar 7 '14 at 22:29
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    $\begingroup$ I tried to calculate some simple model, apparently (in gasphase) $\ce{MeHC(OH)(SO3^{-}Na+)}$ may exist, while $\ce{MeHC(O^{-}Na+)(SO3H)}$ is blown into pieces. I have not yet tried a ketone. (Solvent might also have to be considered. While I have some time for that I might be coming back on that one.) I honestly cannot think of a plausible reason why the right structure should exist (esp. in solution). In what solvent would this reaction be carried out? $\endgroup$ – Martin - マーチン May 16 '14 at 12:57

I think your confusion arose because of an error in the translation of the text you linked (the discussion on p. 359) and that this text was unclear about which equilibrium it was referring to. In that figure, the equilibrium that the authors are presumably referring to is the (planar) carbonyl- (tetrahedral) hemiacetal equilibrium (as described on page 358). So the equilibrium that the text refers to isn't the "proton transfer equilibrium," but the planar-tetrahedral equilibrium.

enter image description here

In contrast, the scheme you posted with your question assumed that the formation of a hemiacetal (or hemiketal) is irreversible, as indicated by a unidirectional arrow. This reaction should be treated with an bidirectional equilibrium arrow as shown in p. 359 of the document that you linked.

In this context, the steric argument is reasonable and the reaction as a whole analogous to the more well-known relative hydration equilibria of aldehydes and ketones (which is related the general notion of aldehydes being more electrophilic than ketones).


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