Aldehydes, Ketones and Carboxylic Acids Part 2

$m$-Chlorobenzaldehyde does not possess any $\alpha$-hydrogen atom, so it cannot undergo aldol-type condensation.In the presence of a strong base such as $50\%$ $\mathrm{KOH}$, it undergoes the Cannizzaro reaction, i.e. a base-induced disproportionation (self-oxidation–reduction).In this reaction one molecule of the aldehyde is oxidised to the $m$-chlorobenzoate ion, while a second molecule of the same aldehyde is reduced to $m$-chlorobenzyl alcohol.$$2\,\mathrm{ClC_6H_4CHO} + \mathrm{KOH} \xrightarrow{50\%} \mathrm{ClC_6H_4COO^-K^+} + \mathrm{ClC_6H_4CH_2OH}$$Mechanism: $\mathrm{OH^-}$ adds to the carbonyl carbon of one aldehyde molecule to give a tetrahedral intermediate; hydride transfer from this intermediate to the carbonyl carbon of a second aldehyde molecule gives the carboxylate (oxidised product) and an alkoxide, which is protonated to give the alcohol (reduced product).Thus the products formed are potassium $m$-chlorobenzoate ($m$-chlorobenzoate ion) and $m$-chlorobenzyl alcohol.