Question

Write the reaction products of $2 - $pentene when oxidized with acidified potassium permanganate at ${100^ \circ }C$.

Answer 1

Hint: Dilute solutions of $KMn{O_4}$ convert alkenes into acids. This behaviour is also used as a qualitative test for the presence of double or triple bonds in a molecule, since the reaction decolorizes the initially purple permanganate solution and generates a brown precipitate. It is sometimes called Baeyer’s reagent .

Complete answer: Baeyer's reagent is an alkaline solution of cold potassium permanganate of violet colour solution. It is a strong oxidizing agent. As this solution interacts with a double bond compound the colour disappears and becomes colourless.
Baeyer's reagent is a strong oxidizing reagent which is used to identify the presence of double or triple bonds in a hydrocarbon.
In this reaction, $2 - $pentene gets oxidized with acidified potassium permanganate and gives propanoic acid and ethanoic acid.
$C{H_2}C{H_2}CH = C{H_2} + 4(O) + KMn{O_4} \to C{H_3}C{H_2}COOH + C{H_3}COOH$

Additional information:
Potassium permanganate is delivered industrially from manganese dioxide, which additionally happens as the mineral pyrolusite. In $2000,$overall production was assessed at $30,000$ tonnes. The$Mn{O_2}$ is melded with potassium hydroxide and warmed in air or with another wellspring of oxygen, similar to potassium nitrate or potassium chlorate. This interaction gives potassium manganate:
$2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O$
The potassium manganate is then converted into permanganate by electrolytic oxidation in alkaline media:
$2KMn{O_4} + 2{H_2}O \to 2KMn{O_4} + 2KOH + {H_2}$

Note:
Potassium permanganate does not dissolve in many organic solvents. It is soluble in acetone, water, pyridine, methanol and acetic acid. It is also readily soluble in inorganic solvents. It has a rich purple colour in concentrated solution and pink colour in diluted solution.