Problem 34 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

When \(t\)-butanol and \(n\)-butanol are separately treated with a few drops of dilute \(\mathrm{KMnO}_{4}\), in one case only the purple colour disappears and a brown precipitate is formed. Which of the two alcohols gives the above reaction and what is the brown precipitate?

Step-by-Step Solution

Answer

\( n \)-butanol reacts with \( \mathrm{KMnO}_{4} \), forming MnO_2 as the brown precipitate.

Step 1: Understand the Reactants and Reagents

First, understand that \({t}\)-butanol and \({n}\)-butanol are two different isomers of butanol. Dilute \(\mathrm{KMnO}_{4}\) is often used as an oxidizing agent and will typically be used to test for the presence of oxidizable groups like alkenes or primary alcohols.

Step 2: Evaluate the Oxidation Possibilities

Identify that \({n}\)-butanol is a primary alcohol, which can be oxidized by \(\mathrm{KMnO}_{4}\) to create a carboxylic acid and can lead to a color change. \({t}\)-butanol, being a tertiary alcohol, typically does not get oxidized under such mild conditions.

Step 3: Chemical Reaction for \( n \)-butanol

When \(n\)-butanol is subjected to \(\mathrm{KMnO}_{4}\), it oxidizes to form butanoic acid. During this reaction, the purple color of \(\mathrm{KMnO}_{4}\) disappears as it is reduced to solid manganese dioxide (MnO_2), the brown precipitate.

Step 4: No Reaction for \( t \)-butanol

Because \({t}\)-butanol is a tertiary alcohol, it does not undergo oxidation readily with \(\mathrm{KMnO}_{4}\). Thus, the purple color remains unchanged when \({t}\)-butanol is tested.

Key Concepts

KMnO4 Oxidation

Potassium permanganate \(\mathrm{KMnO}_4\) is a powerful oxidizing agent used in chemistry to determine the presence of certain functional groups. \(\mathrm{KMnO}_4\) works by gaining electrons during reactions which causes its deep purple color to fade, often transforming into a different color. In organic chemistry, this reagent is commonly used to test for primary alcohols, alkenes, and other oxidizable compounds.
\(\mathrm{KMnO}_4\) oxidizes compounds by accepting electrons, altering its oxidation state from +7 to lower states. In the context of the textbook exercise, when \(n\)-butanol (a primary alcohol) is treated with \(\mathrm{KMnO}_4\), it gets oxidized leading to a visible color change as the purple color of \(\mathrm{KMnO}_4\) disappears. This transformation is a clear indication of the oxidation process occurring.

Primary vs Tertiary Alcohols

Primary and tertiary alcohols behave differently under oxidation by \(\mathrm{KMnO}_4\). This discrepancy stems from their structural differences.
A primary alcohol, like \(n\)-butanol, has the hydroxyl group (\(-OH\)) attached to a carbon that is connected to only one other carbon atom. This makes them more easily oxidizable, often turning into aldehydes or carboxylic acids. In this process, \(n\)-butanol is oxidized to butanoic acid when exposed to \(\mathrm{KMnO}_4\).
On the other hand, a tertiary alcohol, such as \(t\)-butanol, has its hydroxyl group attached to a carbon connected to three other carbon atoms. These alcohols do not easily undergo oxidation because they lack a hydrogen atom on the alcohol-bearing carbon, making oxidation through common agents like \(\mathrm{KMnO}_4\) improbable under mild conditions.

Manganese Dioxide Formation

One significant outcome of \(\mathrm{KMnO}_4\) oxidation reactions is the formation of manganese dioxide (\(\mathrm{MnO}_2\)), observed as a brown precipitate. This formation occurs when \(\mathrm{KMnO}_4\) acts as an oxidizing agent, reducing its manganese from an oxidation state of +7 to +4 in \(\mathrm{MnO}_2\).
For the reaction involving \(n\)-butanol, once it is oxidized, \(\mathrm{KMnO}_4\) is simultaneously reduced, forming this characteristic brown precipitate of \(\mathrm{MnO}_2\). This visible change not only confirms the occurrence of a redox reaction but also aids in distinguishing the results from reactions with other types of alcohols like \(t\)-butanol, which does not lead to such a precipitate because it does not oxidize under these conditions.