Aldehydes which do not have α-hydrogen atom react with concentrated sodium hydroxide (NaOH) or potassium hydroxide (KOH) in such a way that one molecule get oxidized to acid and the second molecule gets reduced to alcohol.
Note two molecules of aldehyde participates in the reaction.
This self oxidation-reduction under the influence of a base is known as the Cannizzaro's reaction.
Formaldehyde does not possess α-hydrogen atom and therefore undergoes Cannizzaro's reaction. Acetaldehyde does not give this reaction.
Formaldehyde (HCHO) two molecules + warm NaOH give Methanol (CH3OH) and Sodium Formate (CHOONa)
CH3OH is the reduction product HCHO becomes CH3OH (two hydrogen atoms are getting in).
CHOONa is the oxidation product. one 'H' has gone out and One 'O' came in along with Na.
1 comment:
can you tell me why a ketone cannot participate in cannizaro reaction?
aldehydes whch do not have a-hydrogen cn participate. i know that. But can a ketone which does not have a-hydrogen undergo cannizaro reaction? if no then why not? eg a ketone lyk "((CH3)3)C)2CO"
i.e. "2,2,4,4-Tetramethyl pentan-3-one" plz reply?
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