How to predict the outcome of a crossed Cannizzaro reaction? 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 

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

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.
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!

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

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.