# Crossed Cannizzaro products

I have understood the basic mechanism of cannizzaro reaction. But I am unable to understand the products formed in case of a crossed Cannizzaro reaction.

Suppose Benzaldehyde and Formaldehyde are heated in presence of $\ce{NaOH}$ to carry out Cannizzaro transformation. I have read that it is Benzyl Alcohol and Formic Acid formed in the greatest proportion and not Benzoic acid and Methyl alcohol.

Why is this so? How can we understand this in terms of the stability of Cannnizzaro intermediates?

The first step of the Cannizzaro reaction is nucleophilic attack of hydroxide on an aldehyde (generically $\ce{RCHO}$) to form a tetrahedral intermediate.

\begin{aligned} &\ \ \ \ \ \ce{O} && \ \ \ \ \ \ce{O-} \\ &\ \ \ \ \ \parallel && \ \ \ \ \ \mid \\ &\ce{R-C-H +OH- ->} && \ce{R-C-H} \\ & \ && \ \ \ \ \ \mid \\ & \ && \ \ \ \ \ \ce{OH} \end{aligned}

Formaldehyde $(\ce{R}=\ce{H})$ is more reactive that benzaldehyde $(\ce{R}=\ce{Ph})$ for both steric and electronic reasons:

• Sterics: The Ph group is larger than hydrogen, so nucleophiles will have more trouble getting to the aldehyde in benzaldehyde
• Electronics: The Ph group is electron rich (all those $\pi$ electrons), which repels the electron rich nucleophile.

If formaldehyde reacts faster than benzaldehyde with hydroxide anion, then more of formaldehyde's tetrahedral intermediate will form, which means there will be more benzaldehyde left for the second step - hydride transfer.

• The rate determining step in this case is the hydride transfer step. Surely it depends on which of the two reactants is more reactive towards the hydride? In that case it is formaldehyde. Even in the first step, formaldehyde should be attacked by the hydroxyl - in other words, formic acid and methanol should be the major products. This is a bit like cross aldol condensation. Why doesn't this happen? – Charles Aug 25 '15 at 11:40