Name the method for propanol to propanoic acid conversion and reactants used.
A. Hydration of propanol using alkaline \[KMn{O_4}\]
B. dehydration of propanol using alkaline \[KMn{O_4}\]
C. oxidation of propanol using alkaline \[KMn{O_4}\]
D. reduction of propanol using alkaline \[KMn{O_4}\]
Answer
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Hint: We have to remember that an alkaline solution of permanganate potash\[(KMn{O_4})\] acts as a very strong oxidising agent. Primary alcohols are converted to corresponding aldehydes by controlled oxidation. Using strong oxidising agent’s primary alcohols are readily oxidised to carboxylic acids.
Complete step by step answer:
We have to remember that the propanol \[(C{H_3}C{H_2}C{H_2}OH)\] is primary alcohol because the hydroxyl –OH group is attached to the carbon containing one-carbon chain. Propanol is often oxidised to give propanal \[(C{H_3}C{H_2}CHO)\]. Propanal, in turn, are often oxidised to propanoic acid \[(C{H_3}C{H_2}COOH).\]. We can write the chemical equation for this formation of propionic acid from propanol as,
\[C{H_3}C{H_2}C{H_2}OH\xrightarrow[{Heat}]{{alk.KMn{O_4}}}C{H_3}C{H_2}COOH\]
We must know that the \[KMn{O_4}\] acts as an oxidizing agent in neutral, acidic also as an alkaline solution. Since alkaline \[KMn{O_4}\] is a strong oxidizing agent and thus when it reacts with propanol, the reaction doesn’t stop at aldehyde, that’s propanal but forms propionic acid because of the outcome product.
Propanoic acid is sorted by oxidation of propanol within the presence of alkaline permanganate potash.
So, the correct answer is Option C.
Note: Additional information:
We have to remember that the Dehydration (removal of a water molecule) of propanol is administered within the presence of protic acids like sulphuric acid to offer propene.
\[C{H_3}C{H_2}C{H_2}OH \to C{H_3}CH \doteq C{H_2} + {H_2}O\]
Note here that oxidation is the addition of oxygen or removal of hydrogen. Conversion of propanol to propanoic acid by alkaline permanganate potash involves the addition of one oxygen with removal of two hydrogen atoms. Thus, it is an oxidation process. We must know that the $KMn{O_4}$ is used in the medicinal field to cure a number of skin problems while the potassium permanganate is used as an indicator in titration.
Complete step by step answer:
We have to remember that the propanol \[(C{H_3}C{H_2}C{H_2}OH)\] is primary alcohol because the hydroxyl –OH group is attached to the carbon containing one-carbon chain. Propanol is often oxidised to give propanal \[(C{H_3}C{H_2}CHO)\]. Propanal, in turn, are often oxidised to propanoic acid \[(C{H_3}C{H_2}COOH).\]. We can write the chemical equation for this formation of propionic acid from propanol as,
\[C{H_3}C{H_2}C{H_2}OH\xrightarrow[{Heat}]{{alk.KMn{O_4}}}C{H_3}C{H_2}COOH\]
We must know that the \[KMn{O_4}\] acts as an oxidizing agent in neutral, acidic also as an alkaline solution. Since alkaline \[KMn{O_4}\] is a strong oxidizing agent and thus when it reacts with propanol, the reaction doesn’t stop at aldehyde, that’s propanal but forms propionic acid because of the outcome product.
Propanoic acid is sorted by oxidation of propanol within the presence of alkaline permanganate potash.
So, the correct answer is Option C.
Note: Additional information:
We have to remember that the Dehydration (removal of a water molecule) of propanol is administered within the presence of protic acids like sulphuric acid to offer propene.
\[C{H_3}C{H_2}C{H_2}OH \to C{H_3}CH \doteq C{H_2} + {H_2}O\]
Note here that oxidation is the addition of oxygen or removal of hydrogen. Conversion of propanol to propanoic acid by alkaline permanganate potash involves the addition of one oxygen with removal of two hydrogen atoms. Thus, it is an oxidation process. We must know that the $KMn{O_4}$ is used in the medicinal field to cure a number of skin problems while the potassium permanganate is used as an indicator in titration.
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