Question

P-hydroxybenzaldehyde/cannizzaro reaction

Answer 1

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.[4] 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.