Question

Name the process 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 1

Hint: Alkaline solution of potassium permanganate ($KMn{{O}_{4}}$) acts as a very strong oxidizing agent. Primary alcohols are converted to corresponding aldehydes by controlled oxidation. Using strong oxidizing agents primary alcohols are readily oxidized to carboxylic acids.

Complete step by step solution:
Propanol ($C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH$) is primary alcohol as the hydroxy ($-OH$) group is attached to the carbon-containing one-carbon chain.
Propanol can be oxidized to give propanal ($C{{H}_{3}}C{{H}_{2}}CHO$). Propanal, in turn, can be oxidized to propanoic acid ($C{{H}_{3}}C{{H}_{2}}COOH$).
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow{\left[ O \right]}C{{H}_{3}}C{{H}_{3}}CHO\xrightarrow{\left[ O \right]}C{{H}_{3}}C{{H}_{2}}COOH\]
$KMn{{O}_{4}}$ acts as an oxidizing agent in neutral, acidic as well as alkaline solution. Since alkaline $KMn{{O}_{4}}$ is a strong oxidizing agent and therefore when it reacts with propanol, the reaction does not stop at aldehyde, i.e. propanal but forms propanoic acid as the end product. \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow{alkaline\,KMn{{O}_{4}}}C{{H}_{3}}C{{H}_{2}}COOH\]

Therefore, propanol is converted to propanoic acid by the process of oxidation using alkaline $KMn{{O}_{4}}$.
Hence, the correct option will be (C).

Additional information:
Dehydration (removal of a water molecule) of propanol is carried out in the presence of protic acids like sulphuric acid (${{H}_{2}}S{{O}_{4}}$) to give propene.
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH\xrightarrow{conc.\text{ }{{H}_{2}}S{{O}_{4}}}C{{H}_{3}}CH=C{{H}_{2}}+{{H}_{2}}O\]

Note: Note here that oxidation is the addition of oxygen or removal of hydrogen. Conversion of propanol to propanoic acid by alkaline $KMn{{O}_{4}}$ involves the addition of one oxygen with the removal of two hydrogen atoms, thus, it is an oxidation process.