Question

When t- butanol and n- butanal are separately treated with a few drops of dilute $KMn{{O}_{4}}$ in one case only, the purple color disappears and a brown precipitate is formed. Which of the two alcohols gives the above reaction and which is the brown precipitate?

Answer 1

Hint: t- butanol is used as a name for tert butyl alcohol, where alcohol is tertiary alcohol. While n- butanal is used for 1- butanal, that has an aldehydic group on first carbon. Dilute $KMn{{O}_{4}}$ is used as an oxidizing agent.

Complete answer:
Dilute $KMn{{O}_{4}}$ is used for the oxidation of alcohols, and aldehydes. It converts the alcohols, and aldehydes into their respective carboxylic acids. This Dilute $KMn{{O}_{4}}$is only used in case of primary alcohols, and does not react to oxidize tertiary alcohols. So this reaction will be carried by the n- butanal. The reaction is as follows:
${{C}_{3}}{{H}_{7}}CHO+KMn{{O}_{4}}\to {{C}_{3}}{{H}_{7}}COOH+Mn{{O}_{2}}\downarrow $
The reaction will form butanoic acid and a precipitate of manganese dioxide. The precipitate is brown in color.
Hence, n- butanal gives the reaction with a brown precipitate $Mn{{O}_{2}}$.

Additional information:
Manganese dioxide $Mn{{O}_{2}}$ is a brown color organic compound. It is naturally found as a pyrolusite mineral. It is used as a catalyst in several reactions in chemistry.

Note:
Potassium permanganate $KMn{{O}_{4}}$is used for the oxidation of alcohols. It will oxidize primary alcohols into aldehyde which on further oxidation gives carboxylic acids. Secondary alcohols are converted into ketones, and tertiary alcohols do not react with$KMn{{O}_{4}}$, but under high temperatures they oxidize to form carboxylic acids.