How to predict the outcome of a crossed Cannizzaro reaction?

Question

My question is, how do we predict the products of a cross-Cannizzaro disproportionation in presence of a concentrated base? i.e, How to judge which of the aldehydes (both don't have any alpha hydrogen) is reduced and which one is oxidized?

If we are asked

$$\ce{CH3-O-C6H4-CHO + HCHO ->[OH^- (conc)] ?}$$

Case: if we see acc to RDS, which is hydride transfer, we must see which compound is better hydride donor. Then it will be $\ce{CH3-O-C6H4-CHO}$. This will get oxidized, and $\ce{HCHO}$ reduced.

Case: if we see which is a better nucleophilic substrate, $\ce{HCHO}$ will get oxidised, and $\ce{CH3-O-C6H4-CHO}$, reduced.

Now my problem is, which of the above cases is correct? i.e, do we see which one of the substrates is better $\ce{H-}$ donor, or which one of them is a better nucleophilic substrate?

Please advise which case is correct, and how to approach such problems.

Answer 1

To answer this, first let us look at the mechanism of the Cannizzaro reaction in general. It is depicted in scheme 1 as a cyclic transition state. The cycle itself isn’t as important, the realisation that one aldehyde is present as a (deprotonated) hydrate after the action of $\ce{OH-}$ is.

Scheme 1: General mechanism of a Cannizzaro reaction.

In your reaction, you are reacting formaldehyde with a (substituted) benzaldehyde. The question boils down to ‘can we predict, which aldehyde will form a hydrate mor easily?’ And the answer is: we can!

$$\begin{array}{cc}\hline \text{carbonyl} & \text{hydrate ratio} \\ \hline \ce{H2CO} & 99~\% \\ \ce{MeCHO} & 60~\% \\ \ce{Me2CO} & < 1~\% \\ \hline\end{array}$$

We have a clear tendency for formaldehyde to almost exclusively form hydrates under standard conditions. Thus, we can conclude that formaldehyde will alwys be the hydrate partner, end up oxidised and reduce its reaction partner as shown in scheme 2.

Scheme 2: Cannizzaro reduction of an aldehyde with formaldehyde as sacrificial aldehyde.

Answer 2

Canizzaro Reaction

$$\ce{A +B->[\text{conc. OH}^-]A' +B'}$$

If A has more partial positive charge, equivalently better hydride donor or more reactive, it will reduce and B will oxidise. Exception is formaldehyde where it will always get oxidised to formate as it's hydrate is more stable in aqueous medium.

Answer 3

It has been found that a crossed Cannizzaro reaction occurs on mixtures of aromatic aldehydes and formaldehyde, which leads to the formation of formic acid and the corresponding aromatic alcohol (Ref.1) as precise prediction of Jan's answer: $$\ce{R-CHO + HCHO + H2O -> R-CH2OH + HCOOH}$$ This method offers a particularly convenient method of preparing certain aromatic alcohols of interest of which the corresponding aldehydes are readily available. In this work, the mole ratios of aromatic aldehyde to formalin to $\ce{NaOH}$ was $1:1.3:3$ and yield of corresponding aromatic alcohol was reported to be 85-90%. The reference also reported that about 2-5% of corresponding aromatic carboxylic acid was also discovered.

In addition, it is also possible that crossed Cannizzaro Reaction with with two aromatic aldehydes (Ref.2). For example, when a mixture of 4-chlorobenzaldehyde and 4-methoxybenzaldehyde (p-anisaldehyde) was allowed to react with alkali, a definite disproportionation has taken place (chloro acid to methoxy acid: $2.3:1$ and chloro alcohol to methoxy alcohol : $1:2.5$).

Interestingly, total of seven reactions have been studied in this manner where one member of the aldehyde pair was halogen substituted: 4-chlorobenzaldehyde and 4-methoxybenzaldehyde; 4-chlorobenzaldehyde and benzaldehyde; 4-chlorobenzaldehyde and 3,4-methylenedioxybenzaldehyde; 4-bromobenzaldehyde and 4-methoxybenzaldehyde; 4-bromobenzaldehyde and benzaldehyde; 3-bromobenzaldehyde and 4-methoxybenzaldehyde; and 3-bromobenzaldehyde and benzaldehyde. In all seven systems, a crossed Cannizzaro reaction has taken place between corresponding aromatic aldehydes and in every case, the halogen substituted molecule showed a tendency to be oxidized to the acid at the expense of the other aldehyde. Also note worthy that when the second aldehyde is benzaldehyde, proportionating ratio has been reduced (halo acid to benzoic acid: ~$1.4:1$ and halo alcohol to benzyl alcohol : ~$1:1.4$). Yet, 3-bromobenzaldehyde and benzaldehyde mixture gave halo acid to benzoic acid ratio of ~$2.4:1$.

References:

1. David Davidson and Marston Taylor Bogert, "The Preparation of Aromatic Alcohols by the Crossed Cannizzaro Reaction with Formaldehyde," J. Am. Chem. Soc. 1935, 57(5), 905-905 (DOI: https://doi.org/10.1021/ja01308a036).
2. John C. Bailar, Jr., Allan J. Barney, and R. F. Miller, "The Action of Alkalies on Mixtures of Aromatic Aldehydes," J. Am. Chem. Soc. 1936, 58(11), 2110–2111 (DOI: https://doi.org/10.1021/ja01302a009).

Answer 4

The Cannizzaro reaction, is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the alpha position. The oxidation product is a salt of a carboxylic acid and the reduction product is an alcohol.

I believe that your fist compound is an isomer of methoxybenzaldehyde, but the other compound, "hcho" is... Formaldehyde?

Methoxybenzaldehyde would disproportionate in their corresponding carboxylic acid and alcohol if a base were present, which is not, so it will not undergo Canizzaro disproportion.

I'm not sure if formaldehyde can undergo Canizzaro reaction to yield methanol plus formic acid, but a base would be surely needed.

If you talk about a crossed Cannizzaro reaction, in which an aldehyde is used in combination with a more valuable chemical. In this variation, the reductant is formaldehyde, which is oxidized to sodium formate, and the other aldehyde is transformed into the correspondent alcohol. But you also need base.