Question

What is the product formed when Benzaldehyde and formaldehyde are heated with an aqueous $NaOH$ solution?

Answer 1

Hint: We know that Cannizzaro response is a synthetic response named after Stanislao Cannizzaro that includes the base-actuated disproportionation of two atoms of a non-enolizable aldehyde to yield primary alcohol and carboxylic acid.

Complete answer:
Cannizzaro previously refined this change in \[1853\] , when he got benzyl alcohol and potassium benzoate from the treatment of Benzaldehyde with potash (potassium carbonate). All the more commonly, the response would be led with sodium hydroxide or potassium hydroxide, giving the sodium or potassium carboxylate salt of the carboxylic-acid item:
\[2{C_6}{H_5}CHO + KOH \to {C_6}{H_5}C{H_2}OH + {C_6}{H_5}COOK\]
The cycle is a redox response including the move of a hydride from one substrate particle to the next: one aldehyde is oxidized to shape the acid; the other is diminished to give alcohol.
A combination of Benzaldehyde and formaldehyde on warming with fluid $NaOH$ arrangement gives the item benzyl alcohol and sodium formate. It's anything but an illustration of the crossed Cannizzaro response.
\[{C_6}{H_5}CHO + HCHO \to {C_6}{H_5}C{H_2}OH + HCOONa\]

Note:
Now we can see about variation in cannizzaro reaction as,
In the Tischenko response, the base utilized is an alkoxide as opposed to hydroxide, and the item is an ester instead of the different liquor and carboxylate gatherings. After the nucleophilic base assaults an aldehyde, the subsequent new oxygen anion assaults another aldehyde to give a hemiacetal linkage between two of the earlier aldehyde-containing reactants as opposed to going through tetrahedral breakdown. At last tetrahedral breakdown happens, giving the steady ester item. Certain ketones can go through a Cannizzaro-type response, moving one of their two carbons bunches as opposed to the hydride that would be available on an aldehyde.