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2019 O Level Chem P1 Q34 — Organic Chemistry / Cracking Solution

Organic Chemistry / Cracking 2019 O Level Pure Chemistry · Paper 1 · Q34

Question

The equation for the cracking of an alkane is shown. C₁₀H₂₂ → C₄H₁₀ + X + Y Compound X has two carbon atoms. X decolourises aqueous bromine. What are the formulae of compounds X and Y?Question diagram
  1. A
  2. B
  3. C
  4. D
Correct answer: A

Video Solution

Explanation

Answer: A (X = C₂H₄, Y = C₄H₈). Cracking breaks the long alkane \(\text{C}_{10}\text{H}_{22}\) into smaller molecules: \(\text{C}_{10}\text{H}_{22} \to \text{C}_4\text{H}_{10} + \text{X} + \text{Y}\).

X has two carbon atoms and decolourises aqueous bromine, so X is an alkene with a C=C double bond — ethene, C₂H₄. The decolourising of bromine water is the standard test for unsaturation.

Now balance the equation. Carbon: \(10 = 4 + 2 + (\text{carbons in Y})\), so Y has 4 carbon atoms. Hydrogen: \(22 = 10 + 4 + (\text{hydrogens in Y})\), so Y has 8 hydrogen atoms. That makes Y = C₄H₈, butene, which is also an alkene formed during cracking.

So X = C₂H₄ and Y = C₄H₈, which is row A. The key checks are that an alkene is needed to decolourise bromine, and that the carbon and hydrogen atoms must balance on both sides.

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