Animated Solution for Chemistry - Organic Chemistry: Major products of the following reaction are

Imagine you are looking at a chemical battlefield where two different aldehydes are placed in a highly basic environment. We have benzaldehyde ($PhCHO$) and formaldehyde ($HCHO$), reacting with a concentrated $50\%$ $NaOH$ solution, followed by an acidic workup ($H_3{O}^{+}$).

The first thing you must notice is the complete absence of $\alpha$-hydrogens in both molecules. When aldehydes lacking $\alpha$-hydrogens are subjected to a concentrated base, they cannot undergo the classic Aldol condensation. Instead, they are forced down the path of the Cannizzaro reaction. Because we are dealing with two distinct aldehydes, this is specifically a Crossed Cannizzaro reaction.

Who Gets Attacked?

In a Crossed Cannizzaro reaction, the two aldehydes undergo a disproportionation-like redox process: one is oxidized to a carboxylic acid salt, and the other is reduced to an alcohol. But how do we decide which one meets which fate?

The rule is simple: The aldehyde that is more reactive towards nucleophilic attack by the hydroxide ion ($O{H}^{-}$) will be the one that gets oxidized.

Let's compare our contenders. Formaldehyde ($HCHO$) is a tiny molecule with no bulky groups, making it sterically very accessible. Furthermore, its carbonyl carbon is highly electrophilic because it lacks any electron-donating groups. Benzaldehyde, on the other hand, possesses a bulky phenyl ring. This ring donates electron density to the carbonyl carbon via resonance ($+R$ effect), significantly reducing its electrophilicity.

Therefore, the $O{H}^{-}$ ion will preferentially and rapidly attack the highly electrophilic carbonyl carbon of formaldehyde, forming a rich, electron-dense hydroxyalkoxide intermediate.

Hydride Transfer

Now comes the most thrilling part of the mechanism. The negative charge on the oxygen of the hydroxyalkoxide intermediate pushes back down to reform the strong carbon-oxygen double bond. As this happens, the carbon atom must let go of something, and it kicks out a hydride ion ($H^{-}$).

This hydride ion is highly unstable and immediately attacks the carbonyl carbon of the waiting benzaldehyde molecule. This intermolecular hydride transfer is the slowest step of the entire process, making it the rate-determining step (RDS).

The Acidic Workup

As the hydride ion slams into benzaldehyde, it reduces it to a benzyl alkoxide ion ($PhC{H}_2{O}^{-}$). Simultaneously, the formaldehyde molecule that lost the hydride is oxidized to formic acid ($HCOOH$).

Because we are in a basic medium, a rapid proton exchange occurs instantly: the acidic proton of formic acid is snatched by the strongly basic benzyl alkoxide ion, yielding a formate ion ($HCO{O}^{-}$) and benzyl alcohol ($PhC{H}_{2}OH$).

Finally, the second step of our reaction introduces an acidic workup ($H_3{O}^{+}$). This simply protonates the formate ion back into formic acid.

Our final major products are benzyl alcohol and formic acid. A golden rule to remember for exams: In any Crossed Cannizzaro reaction involving formaldehyde, formaldehyde is exceptionally reactive and will almost always be the one oxidized to formate!