Question

Which one of the following organic compounds decolourised by an alkaline $KMn{{O}_{4}}$
solution
A. $ C{{S}_{2}}$
B. $ {{C}_{3}}{{H}_{6}}$
C. $ {{C}_{3}}{{H}_{8}}$
D. $ C{{H}_{3}}OH$

Answer 1

Hint: Potassium permanganate, $KMn{{O}_{4}}$ decolourised unsaturated compounds (hydrocarbons) and consists of double or triple bonds between carbon and hydrogen atom ($C-H$). To solve this problem first we will have to find the unsaturated compounds among the given options.

Complete Step by Step Answer:
Alkenes are generally more reactive than alkanes due to the presence of unsaturated double bonds. When alkenes react with potassium permanganate, alkenes are oxidised to alcohols and one hydroxyl group ($-OH$ ) is attached to each carbon on either side of the carbon-carbon double bond, thereby producing $1,2-$glycols or vicinal. Potassium permanganate is further reduced to $Mn{{O}_{2}}$ a dark brown precipitate.

Alkenes are hydrocarbons, having the general formula ${{C}_{n}}{{H}_{2n}}$where n is the number of carbon atoms. In alkene, the number of hydrogen atoms is twice that of carbon atoms present in the alkene compound. Alkynes are also hydrocarbons with the general formula ${{C}_{n}}{{H}_{2n-2}}$.
Let us check the given four options one by one.

From (A) it is found that the given compound is an unsaturated compound but not a hydrocarbon. It is carbon disulfide. Hence it is not the correct option.

In the second option, the given compound ${{C}_{3}}{{H}_{6}}$ contains three carbon atoms hence the number of hydrogen atoms is six. It is a compound of alkene.
${{C}_{3}}{{H}_{2\times 3}}$ Or, ${{C}_{3}}{{H}_{6}}$ Here $n=3$
Thus, the compound ${{C}_{3}}{{H}_{6}}$is an unsaturated hydrocarbon or an alkene compound that reacts with alkaline potassium permanganate$KMn{{O}_{4}}$ and decolorizes the solution from a pink colour to dark brown.

The next given compound${{C}_{3}}{{H}_{8}}$ is neither alkene or alkyne compound, it is an alkane compound ( general formula ${{C}_{n}}{{H}_{2n+2}}$), it is also not a correct one. In the fourth option, it is an alcohol named methanol ($C{{H}_{3}}OH$). Hence (D) is also the wrong option.
Thus, option (B) is correct.

Note: Alkanes do not decolorize the alkaline potassium permanganate solution. As it contains only sigma bonds and we know sigma bonds are more strong than pi bonds. Hence they are stable and do not decolorize the alkaline $KMn{{O}_{4}}$ solution.