Question

What happens when a few drops of alkaline ${\text{KMn}}{{\text{O}}_4}$ are added to ethanol?

Answer 1

Hint: Alkaline ${\text{KMn}}{{\text{O}}_4}$ is known to be a strong oxidising agent and hence will first oxidise ethanol. This might cause the reaction to occur in a different way than expected.

Complete step by step solution:
When ${\text{KMn}}{{\text{O}}_4}$ is added to a solution of ethanol, and heated, the colour of the solution disappears. This is because, ${\text{KMn}}{{\text{O}}_4}$ causes oxidation of ethanol to ethanoic acid by donating nascent oxygen.

Following reaction occurs -
${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH + 2[O] }}\xrightarrow[{{\text{Heat}}}]{{{\text{Alkaline KMn}}{{\text{O}}_4}}}{\text{ C}}{{\text{H}}_3}{\text{COOH + }}{{\text{H}}_2}{\text{O}}$
Due to the acidic environment of ethanoic acid, the dark purple colour of ${\text{KMn}}{{\text{O}}_4}$disappears.
Hence, when alkaline${\text{KMn}}{{\text{O}}_4}$is added to a solution of ethanol and warmed a little, the colour of potassium permanganate disappears.

Additional information: ${\text{KMn}}{{\text{O}}_4}$is dark purple in colour originally. It changes its colour to brown when used in a redox titration. This is because, in ${\text{KMn}}{{\text{O}}_4}$ the oxidation state of ${\text{Mn}}$is ${\text{M}}{{\text{n}}^{ + 7}}$which changes to ${\text{M}}{{\text{n}}^{ + 2}}$when it is reduced. Hence it is known as a self-indicator. It is also called ‘Bayer’s reagent’.

In acidic solutions, ${\text{KMn}}{{\text{O}}_4}$with is reduced to pale pink with ${\text{M}}{{\text{n}}^{ + 2}}$oxidation state of manganese ion. The reaction occurs is –
$8{{\text{H}}^ + }{\text{ + Mn}}{{\text{O}}^ - }_4{\text{ + 5}}{{\text{e}}^ - }{\text{ }} \to {\text{ M}}{{\text{n}}^{ + 2}}{\text{ + 4}}{{\text{H}}_2}{\text{O}}$
In basic solutions, ${\text{KMn}}{{\text{O}}_4}$with ${\text{M}}{{\text{n}}^{ + 7}}$is reduced to green with ${\text{M}}{{\text{n}}^{ + 6}}$ oxidation state of manganese. The reaction that occurs is –
${\text{Mn}}{{\text{O}}^ - }_4{\text{ + }}{{\text{e}}^ - }{\text{ }} \to {\text{ Mn}}{{\text{O}}^{ - 2}}_4$
In neutral solutions, it gets reduced to brown colour with ${\text{M}}{{\text{n}}^{ + 4}}$oxidation state of manganese ion. The reaction that occurs is –
${\text{2}}{{\text{H}}_2}{\text{O + Mn}}{{\text{O}}_4}{\text{ + 3}}{{\text{e}}^ - }{\text{ }} \to {\text{ Mn}}{{\text{O}}_2}{\text{ + 4O}}{{\text{H}}^ - }$

Note: We must keep in mind that if the environment becomes extremely acidic, the colour of potassium permanganate disappears completely.