Ethene decolourises acidified potassium tetraoxomanganate(VII) solution. Which gas will decolourise bromine water?

Analyze the question: Ethene (an alkene) decolourises acidified KMnO₄ because it is unsaturated (it has a C=C double bond). The question asks which gas from the list will also decolourise bromine water, so we need another unsaturated hydrocarbon that is a gas.

Concept: Bromine water (orange/brown) is decolourised by compounds with C=C or C≡C bonds, through an addition reaction. Saturated hydrocarbons (alkanes) do not react with bromine water in the dark or at room temperature. Benzene has delocalised electrons and does not readily undergo addition, so it doesn't decolourise bromine water either.

Solution, checking each option:

• A. CH₄ (methane): alkane, saturated, no reaction.
• B. C₂H₆ (ethane): alkane, saturated, no reaction.
• C. C₃H₆ (propene): general formula CₙH₂ₙ, so it is an alkene with a C=C bond. It undergoes addition with bromine: CH₃CH=CH₂ + Br₂ → CH₃CHBrCH₂Br The orange colour disappears, so it decolourises bromine water. It is also a gas at room temperature.
• D. C₃H₈ (propane): alkane, saturated, no reaction.
• E. C₆H₆ (benzene): aromatic and a liquid, not a gas. It resists addition, so there is no decolourisation.

Answer: C. C₃H₆

Exam tip: Use the general formulas to spot the type quickly:

• CₙH₂ₙ₊₂ = alkane (saturated, no reaction with bromine water)
• CₙH₂ₙ = alkene (unsaturated, decolourises bromine water and KMnO₄)
• CₙH₂ₙ₋₂ = alkyne (unsaturated, decolourises bromine water and KMnO₄)

If the formula has two fewer hydrogens than the alkane, it is unsaturated.