
How would the alkaline \[KMn{O_4}\] act in the given reaction?
$C{H_3}C{H_2}OH\xrightarrow{{Alkaline\;KMn{O_4} + Heat}}C{H_3}COOH$
A) Reducing agent
B) Oxidising agent
C) Catalyst
D) Dehydrating agent
Answer
552.3k+ views
Hint: As we know that the alcohol reacts with various reagents where these reagents basically help in breaking the bonds between the alkyl or aryl group and the oxygen atom and results into the formation of an aldehyde compound which on further reaction with various oxidising and reducing agents results into acids or alcohol respectively.
Complete answer:
We know very well that when alcohols reacts with alkaline potassium permanganate, it first oxidises to an aldehyde and this aldehydic compound on further oxidation results into the formation of a carboxylic acid because potassium permanganate is a strong oxidising agent. So, we can draw a basic concept of oxidation of alcohol here as:
$R - C{H_2} - OH\xrightarrow{{[O]}}R - CHO\xrightarrow{{[O]}}R - COOH$
Now we are aware with alkaline potassium permanganate being the oxidising agent because the oxidation state of permanganate increases in the presence of an acidic medium and possess a $ + 7$ oxidation in anionic form which makes it highly electronegative and thus it can oxidise the alcoholic compounds to ketonic compounds. Also potassium permanganate can donate its four oxygen atoms in acidic medium to attain an oxidation state of $ + 2$ which is more stable thereby acting as an oxidising agent.
Hence, from the above explanation we can say that alkaline \[KMn{O_4}\] can act as an oxidising agent in the given reaction and would convert ethanol to ethanoic acid.
$C{H_3}C{H_2}OH\xrightarrow{{Alkaline\;KMn{O_4} + Heat}}C{H_3}COOH$
And hence, the correct answer is Option B.
Note: Remember that potassium permanganate is a strong oxidising agent and in acidic medium as it is electron deficient medium, potassium permanganate reduces itself to $M{n^{2 + }}$ and in basic or neutral medium, it gets reduced to $Mn{O_2}$. In acidic medium, it becomes easy for the electron exchange because water formation can takes place to combine with the cleaved oxygen atoms during the course of reaction and the colour of potassium permanganate changes from dark purple to colourless to brown while conversion of alcohols to ketones.
Complete answer:
We know very well that when alcohols reacts with alkaline potassium permanganate, it first oxidises to an aldehyde and this aldehydic compound on further oxidation results into the formation of a carboxylic acid because potassium permanganate is a strong oxidising agent. So, we can draw a basic concept of oxidation of alcohol here as:
$R - C{H_2} - OH\xrightarrow{{[O]}}R - CHO\xrightarrow{{[O]}}R - COOH$
Now we are aware with alkaline potassium permanganate being the oxidising agent because the oxidation state of permanganate increases in the presence of an acidic medium and possess a $ + 7$ oxidation in anionic form which makes it highly electronegative and thus it can oxidise the alcoholic compounds to ketonic compounds. Also potassium permanganate can donate its four oxygen atoms in acidic medium to attain an oxidation state of $ + 2$ which is more stable thereby acting as an oxidising agent.
Hence, from the above explanation we can say that alkaline \[KMn{O_4}\] can act as an oxidising agent in the given reaction and would convert ethanol to ethanoic acid.
$C{H_3}C{H_2}OH\xrightarrow{{Alkaline\;KMn{O_4} + Heat}}C{H_3}COOH$
And hence, the correct answer is Option B.
Note: Remember that potassium permanganate is a strong oxidising agent and in acidic medium as it is electron deficient medium, potassium permanganate reduces itself to $M{n^{2 + }}$ and in basic or neutral medium, it gets reduced to $Mn{O_2}$. In acidic medium, it becomes easy for the electron exchange because water formation can takes place to combine with the cleaved oxygen atoms during the course of reaction and the colour of potassium permanganate changes from dark purple to colourless to brown while conversion of alcohols to ketones.
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