Q.Benzaldehyde, C6H5CHO, has no alpha-hydrogen atom. Predict the product(s) formed when it is treated with concentrated aqueous KOH solution.

The Cannizzaro Reaction: An Intuition

Imagine two identical molecules of an aldehyde meeting in a strongly basic solution. Normally, aldehydes with a hydrogen on the carbon next to the carbonyl (the $\alpha$-carbon) undergo aldol condensation. But what if that $\alpha$-carbon has no hydrogen at all? The molecule cannot do the usual reaction. Instead, something remarkable happens: one aldehyde molecule gets reduced to an alcohol, while the other gets oxidized to a carboxylic acid (which, in base, exists as its salt). One molecule gives away electrons; the other accepts them. This is a disproportionation — a single species (the aldehyde) acts as both the oxidising and the reducing agent.

The Precise Statement

The Cannizzaro reaction is the base-catalysed disproportionation of an aldehyde that lacks an $\alpha$-hydrogen atom (i.e., the carbon adjacent to the $-\text{CHO}$ group has no hydrogen attached). In the presence of concentrated aqueous or alcoholic alkali, two molecules of such an aldehyde yield one molecule of a primary alcohol and one molecule of the salt of a carboxylic acid.

The general equation (using benzaldehyde as the classic example):

Benzaldehyde gives benzyl alcohol and sodium benzoate.

Why "No $\alpha$-Hydrogen" Matters

The key is the mechanism. The first step is the attack of hydroxide ion ($\text{OH}^-$) on the carbonyl carbon. This forms a tetrahedral intermediate. If an $\alpha$-hydrogen were present, this intermediate would lose water and form an enolate — leading to aldol condensation. Without that hydrogen, the intermediate cannot do that. Instead, it transfers a hydride ion ($\text{H}^-$) to a second molecule of aldehyde. That second molecule gets reduced to the alkoxide (which later picks up a proton to become the alcohol), while the first molecule becomes the carboxylate.

Crossed Cannizzaro Reaction …