Overall reaction
2 C₆H₅CHO + NaOH (conc.) → C₆H₅CH₂OH + C₆H₅COO⁻Na⁺
An aldehyde with no α-H disproportionates in strong base: one is oxidised, one reduced.
In the Cannizzaro reaction an aldehyde without α-hydrogens (HCHO, benzaldehyde, pivalaldehyde) disproportionates in concentrated base: one molecule is oxidised to the carboxylate and another is reduced to the alcohol. Hydroxide adds to one aldehyde, and the tetrahedral anion transfers a hydride to a second aldehyde.
2 C₆H₅CHO + NaOH (conc.) → C₆H₅CH₂OH + C₆H₅COO⁻Na⁺
An aldehyde with no α-H disproportionates in strong base: one is oxidised, one reduced.
Each step is checked electron by electron: the curly arrows give exactly the structures drawn.
Base removes the α-hydrogen first, so they undergo aldol addition/condensation instead.
A mixture of formaldehyde and another non-enolisable aldehyde: formaldehyde (the better hydride donor after OH⁻ addition) becomes formate, and the other aldehyde becomes the alcohol.
No. It moves directly from the carbon of the tetrahedral anion to the carbonyl carbon of the second aldehyde.
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