Oxidation of Aldehydes and Ketones: Tollens' and Fehling's Tests

Benzaldehyde, $\text{C}_6\text{H}_5\text{CHO}$, has no $\alpha$-hydrogen at all -- its carbonyl carbon is attached directly to the aromatic ring on one side and to hydrogen on the other, and neither of these is a carbon bearing hydrogens the way an ordinary alkyl $\alpha$-carbon would be (the ring carbons are part of the aromatic system and their hydrogens are not acidified by the adjacent carbonyl in the same enolisable way). Without an $\alpha$-hydrogen, benzaldehyde cannot form an enolate at all, and so it cannot undergo self-aldol condensation (§8.8), which absolutely requires an enolate as the attacking nucleophile.

The Cannizzaro reaction as the alternative pathway. When such an $\alpha$-hydrogen-free aldehyde is instead treated with concentrated ($\approx 50\%$) $\text{NaOH}$, it undergoes a completely different reaction: the Cannizzaro reaction, an intermolecular self-oxidation-reduction (disproportionation) between two molecules of the same aldehyde. The mechanism proceeds in two steps. First, hydroxide ion adds directly to the carbonyl carbon of one benzaldehyde molecule -- an ordinary nucleophilic addition exactly as in §8.5 -- giving a tetrahedral intermediate that carries two oxygen substituents on the same carbon (a gem-diolate anion), $\text{C}_6\text{H}_5\text{CH(O}^-\text{)(OH)}$. Second, this electron-rich tetrahedral intermediate transfers a hydride ion ($\text{H}^-$) directly from its own carbon to the carbonyl carbon of a second molecule of benzaldehyde. The molecule that donated the hydride is thereby left as a carboxylate anion (benzoate, $\text{C}_6\text{H}_5\text{COO}^-$, effectively oxidised), while the molecule that accepted the hydride becomes an alkoxide, protonated on work-up to give benzyl alcohol, $\text{C}_6\text{H}_5\text{CH}_2\text{OH}$ (effectively reduced).

$2\ \text{C}_6\text{H}_5\text{CHO} + \text{NaOH} \longrightarrow \text{C}_6\text{H}_5\text{COONa} + \text{C}_6\text{H}_5\text{CH}_2\text{OH}$ …