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Cannizzaro reaction

Organic chemical reaction

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Cannizzaro first accomplished this transformation in 1863, when he obtained benzyl alcohol and potassium benzoate from the treatment of benzaldehyde with potash (potassium carbonate). More typically, the reaction would be conducted with sodium hydroxide or potassium hydroxide, giving the sodium or potassium carboxylate salt of the carboxylic-acid product:

2 C6H5CHO + KOH → C6H5CH2OH + C6H5COOK

The process is a redox reaction involving transfer of a hydride from one substrate molecule to the other: one aldehyde is oxidized to form the acid, the other is reduced to form the alcohol.

Mechanism

The reaction involves a nucleophilic acyl substitution on an aldehyde, with the leaving group concurrently attacking another aldehyde in the second step. First, hydroxide attacks a carbonyl. The resulting tetrahedral intermediate then collapses, re-forming the carbonyl and transferring hydride to attack another carbonyl. In the final step of the reaction, the acid and alkoxide ions formed exchange a proton. In the presence of a very high concentration of base, the aldehyde first forms a doubly charged anion from which a hydride ion is transferred to the second molecule of aldehyde to form carboxylate and alkoxide ions. Subsequently, the alkoxide ion acquires a proton from the solvent.

Overall, the reaction follows third-order kinetics. It is second order in aldehyde and first order in base:

rate = k[RCHO]2[OH−]

At very high base a second path (k') becomes important that is second order in base:

rate = k[RCHO]2[OH−] + k'[RCHO]2[OH−]2

The k' pathway implicates a reaction between the doubly charged anion (RCHO22−) and the aldehyde. The direct transfer of hydride ion is evident from the observation that the recovered alcohol does not contain any deuterium attached to the α-carbon when the reaction is performed in the presence of D2O.

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