Write a note on self oxidation- reduction reaction of aldehyde with suitable example.
Answer
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Hint: As we know that oxidation-reduction reaction basically involves both oxidation as well as reduction of the compound and we also know that aldehyde that do not have $ \alpha $ -hydrogen atoms undergo disproportionation reaction.
Complete answer:
As we are aware with the fact that a redox reaction is basically a reaction which involves both phenomenon oxidation as well as reduction and we also know that aldehydes that do not have any $ \alpha $ -hydrogen atom such as formaldehyde $ (HCHO) $ and benzaldehyde $ ({C_6}{H_5}CHO) $ etc undergo self-oxidation and self-reduction that is disproportionation on treatment with concentrated alkali to form equal amounts of products.
In this reaction, one molecule is reduced to alcohol and the other is oxidised to salt of carboxylic acid. This reaction is commonly known as Cannizzaro reaction.
Reaction of formaldehyde can be represented as a chemical equation given as:
$ HCHO + HCHO\xrightarrow{{conc.\;KOH}}C{H_3}OH + HCOOK $
Here, one formaldehyde is reduced to methanol and one formaldehyde is oxidised to potassium salt of formic acid. Similarly benzaldehyde reaction will give:
$ 2{C_6}{H_5}CHO\xrightarrow{{conc.\;NaOH}}{C_6}{H_5}C{H_2}OH + {C_6}{H_5}C{O_2}Na $
Here also, one benzaldehyde is reduced to benzyl alcohol and one benzaldehyde is oxidised to potassium salt of benzoic acid.
Cross cannizzaro reaction can also take place between two different aldehydes but the only condition is both should not have $ \alpha $ -hydrogen atoms. For example:
$ C{H_3}O - {C_6}{H_5} - CHO + HCHO\xrightarrow{{50\% \;NaOH}}C{H_3}O - {C_6}{H_5} - C{H_2}OH + HCOONa $
Here also we can see that formaldehyde gets oxidised to sodium formate or sodium salt of formaldehyde and other gets reduced to alcohol.
Note:
Always remember that hydride ion transfer is the rate determining step in cannizzaro reaction. Cross-cannizzaro reaction is more preferable as the atomic economy is low in this reaction but the production or yield of valuable compounds is high. Some ketones may also undergo a cannizzaro reaction where they transfer one of their two carbon groups rather than the hydride on aldehyde.
Complete answer:
As we are aware with the fact that a redox reaction is basically a reaction which involves both phenomenon oxidation as well as reduction and we also know that aldehydes that do not have any $ \alpha $ -hydrogen atom such as formaldehyde $ (HCHO) $ and benzaldehyde $ ({C_6}{H_5}CHO) $ etc undergo self-oxidation and self-reduction that is disproportionation on treatment with concentrated alkali to form equal amounts of products.
In this reaction, one molecule is reduced to alcohol and the other is oxidised to salt of carboxylic acid. This reaction is commonly known as Cannizzaro reaction.
Reaction of formaldehyde can be represented as a chemical equation given as:
$ HCHO + HCHO\xrightarrow{{conc.\;KOH}}C{H_3}OH + HCOOK $
Here, one formaldehyde is reduced to methanol and one formaldehyde is oxidised to potassium salt of formic acid. Similarly benzaldehyde reaction will give:
$ 2{C_6}{H_5}CHO\xrightarrow{{conc.\;NaOH}}{C_6}{H_5}C{H_2}OH + {C_6}{H_5}C{O_2}Na $
Here also, one benzaldehyde is reduced to benzyl alcohol and one benzaldehyde is oxidised to potassium salt of benzoic acid.
Cross cannizzaro reaction can also take place between two different aldehydes but the only condition is both should not have $ \alpha $ -hydrogen atoms. For example:
$ C{H_3}O - {C_6}{H_5} - CHO + HCHO\xrightarrow{{50\% \;NaOH}}C{H_3}O - {C_6}{H_5} - C{H_2}OH + HCOONa $
Here also we can see that formaldehyde gets oxidised to sodium formate or sodium salt of formaldehyde and other gets reduced to alcohol.
Note:
Always remember that hydride ion transfer is the rate determining step in cannizzaro reaction. Cross-cannizzaro reaction is more preferable as the atomic economy is low in this reaction but the production or yield of valuable compounds is high. Some ketones may also undergo a cannizzaro reaction where they transfer one of their two carbon groups rather than the hydride on aldehyde.
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