
Which of the following reagents cannot be used to oxidize primary alcohols to aldehydes?
A.Pyridinium chlorochromate
B.Heating in the presence of Cu at 573K
C.$KmnO_4$ in acidic medium
D.$CrO_3$ in the anhydrous medium.
Answer
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Hint:The oxidation of alcohol is known as a dehydrogenation reaction since this involves the loss of hydrogen from the alcohol molecule and can be carried out with a variety of reagents such as neutral, acidic, or alkaline.
Complete step by step answer:
In the case of option A, i.e. pyridinium chlorochromate (PCC) having chemical formula $Cr{O_3}.{C_5}{H_5}N.HCl{\text{ }}or{\text{ }}{{\text{C}}_5}{H_5}N{H^ + }Cr{O_3}C{l^ - }$ is also known as Corey's reagent. It is a milder version of chromic acid and it oxidizes the primary alcohol into aldehyde only.
In the case of option B i.e. when the vapour of primary alcohol is passed over heated copper at 573K, dehydrogenation takes place, and formation of aldehyde takes place.
In the case of option C, acidic \[KMn{O_4}\] is a strong oxidizing agent. It can easily oxidize primary alcohol first to aldehyde then to acids, both containing the same number of carbon atoms as original alcohol.
In the case of option D i.e. \[Cr{O_3}\] in the anhydrous medium primary alcohol oxidized to aldehyde.
Hence we can say the reagent which is not used to oxidize primary alcohol to aldehyde is \[KMn{O_4}\] in acidic medium.
Hence the correct answer is Option C.
Note:
The oxidation of alcohol is an important organic reaction and involves the formation of a carbon-oxygen double bond with the cleavage of O-H and C-H bonds. The ease of oxidation and nature of the product of oxidation depends upon the type of alcohol used whether it is primary secondary and tertiary. It is also an important step in the degradation of fat in human metabolism.
Complete step by step answer:
In the case of option A, i.e. pyridinium chlorochromate (PCC) having chemical formula $Cr{O_3}.{C_5}{H_5}N.HCl{\text{ }}or{\text{ }}{{\text{C}}_5}{H_5}N{H^ + }Cr{O_3}C{l^ - }$ is also known as Corey's reagent. It is a milder version of chromic acid and it oxidizes the primary alcohol into aldehyde only.
In the case of option B i.e. when the vapour of primary alcohol is passed over heated copper at 573K, dehydrogenation takes place, and formation of aldehyde takes place.
In the case of option C, acidic \[KMn{O_4}\] is a strong oxidizing agent. It can easily oxidize primary alcohol first to aldehyde then to acids, both containing the same number of carbon atoms as original alcohol.
In the case of option D i.e. \[Cr{O_3}\] in the anhydrous medium primary alcohol oxidized to aldehyde.
Hence we can say the reagent which is not used to oxidize primary alcohol to aldehyde is \[KMn{O_4}\] in acidic medium.
Hence the correct answer is Option C.
Note:
The oxidation of alcohol is an important organic reaction and involves the formation of a carbon-oxygen double bond with the cleavage of O-H and C-H bonds. The ease of oxidation and nature of the product of oxidation depends upon the type of alcohol used whether it is primary secondary and tertiary. It is also an important step in the degradation of fat in human metabolism.
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