# Reverse Cannizzaro reaction to make aldehydes?

Is it possible (and practicable) to reverse Cannizzaro reaction (disproportionation of aldehydes to carboxylic acids and alcohols) to produce aldehydes from carboxylic acid and alcohol without need to use neither chromate or $$\ce{LiAlH4}$$? For example: 1,4 butanediol + 1,4 butanodioic acid - > 1,4 butanedial

• What reagents you were suggesting to be able to undergo reverse Cannizzaro reaction? – Mathew Mahindaratne May 14 at 21:11
• In case it makes a difference I'll edit the question to include an example – Francis L. May 14 at 21:19
• That'd be ideal.:-) – Mathew Mahindaratne May 14 at 21:22
• I think it's impossible. Somehow alkoxide would need to be strong enough donor of H- that carboxylate would accept it - definitely not the case you mention. – Mithoron May 14 at 21:51
• @Mithoron: I agree with you. It need a $\ce{H-}$ source to start with and that would react with alcohol first as acid-base reaction. Thus, no reverse Cannizzaro reaction is possible. – Mathew Mahindaratne May 14 at 22:32

No, this is not practicable because aldehydes are higher in free energy than either alcohols or acids. Thus, the equilibrium constant of the Cannizaro reaction lies strongly against the aldehydes.

The case of formaldehyde may be even more extreme than the example reaction you gave, but it is what I could find data for.

The eQuilibrator database lists the equilibrium constant for the reaction of

$$\ce{2 HCHO + H2O <=> CH3OH + HCOOH}$$

as $$K'_{eq} = 3.4 × 10^{13}$$

This is for each reactant at 1 mM in concentration, at pH 7 and an ionic strength of 0.1 mol / L. Thus, you'd need a teramole or two of formaldehyde to get one mole of acid/alcohol. The thermodynamics for other aldehydes may be slightly more "favorable" but you aren't going to get 13 orders of magnitude of favorability for any "regular" pair of aldehyde and acid/alcohol.

• Maybe under scrupulously anhydrous conditions with continuous removal of trace water you could get the backwards reaction to go a tiny tiny bit, since lack of water would shift the equilibrium constant. But probably still not to a practicable degree. – Curt F. May 15 at 21:55