Question

Two organic compounds \(A\) and B have sp \(^{2}\) hybridised carbon atoms. \(\Lambda\) can decolourise alkaline \(\mathrm{KMnO}_{4}\) while B camot. \(\Lambda\) and \(\mathrm{B}\) could be (1) Ethylene and acetylene (2) Propylene and acetylene (3) Benzene and acetylene (4) Ethylene and benzene

Short answer

Option (4): Ethylene and Benzene.

Step-by-step solution

Understand the Concept of Decolourisation

To decolourise alkaline \(\text{KMnO}_4\), a compound must have a carbon-carbon double bond (C=C) or triple bond (C≡C). This is because alkaline \(\text{KMnO}_4\) is an oxidizing agent that reacts with these bonds.

Identify \(\text{sp}^2\) Hybridised Compounds

Both Ethylene (\(\text{C}_2\text{H}_4\)) and Benzene (\(\text{C}_6\text{H}_6\)) contain \(\text{sp}^2\)-hybridised carbon atoms because they have C=C bonds.

Determine Which Compound Decolourises Alkaline \(\text{KMnO}_4\)

Ethylene (\(\text{C}_2\text{H}_4\)) can decolourise alkaline \(\text{KMnO}_4\) because it has a single double bond. Benzene (\(\text{C}_6\text{H}_6\)) does not typically decolourise \(\text{KMnO}_4\) due to its stable aromatic ring structure.

Eliminate the Incorrect Options

Given the properties, eliminate options where both compounds can decolourise \(\text{KMnO}_4\) or neither can. Acetylene (\(\text{C}_2\text{H}_2\)) has a triple bond and is \(\text{sp}\)-hybridised, not \(\text{sp}^2\) hybridised.

Select the Correct Option

The correct compounds are Ethylene (\(\text{C}_2\text{H}_4\)) and Benzene (\(\text{C}_6\text{H}_6\)). Ethylene can decolourise alkaline \(\text{KMnO}_4\), while Benzene cannot.

Key concepts

decolourisation with KMnO4

When dealing with decolourisation by alkaline potassium permanganate (KMnO4), it’s important to understand its role as an oxidizing agent. It reacts with compounds that have carbon-carbon double or triple bonds. During this reaction, the purple color of KMnO4 disappears or decolourises. This is a useful test to identify unsaturated compounds which contain these bonds. For example, ethylene (C2H4) has a carbon-carbon double bond and can decolourise KMnO4, turning the solution from purple to colorless or brown.

carbon-carbon double bond

A carbon-carbon double bond (C=C) is a functional group in organic chemistry where two carbon atoms are bonded by sharing two pairs of electrons. This double bond creates a region of high electron density which can interact with oxidizing agents like KMnO4. In such reactions, the double bond can be broken, and additional atoms can be added to each carbon, leading to compounds with more bonds to oxygen atoms. This double bond feature is present in ethylene (C2H4), making it reactive in decolourisation tests.

aromatic compound stability

Aromatic compounds, such as benzene (C6H6), possess a unique stability due to the delocalized electrons over a conjugated pi system. In benzene, the six p-orbitals overlap to form a stable electron cloud above and below the plane of the molecule, providing exceptional stability. Because of this stability, benzene does not typically react with KMnO4, making it resistant to decolourisation. This stability contrasts sharply with simpler molecules like ethylene, which do not have this stabilizing effect and thus react more readily with oxidizing agents.