Covers Lessons 5–8: Physical properties, IMF trends, hydrocarbons — sources, distillation, cracking, and boiling point comparisons across functional groups.
1. Which explanation correctly accounts for why hexan-1-ol (BP 157 °C) has a much higher boiling point than hexane (BP 69 °C)?
2. In a fractionating column processing crude oil, which fraction would be collected HIGHEST in the column?
3. The cracking of undecane (C₁₁H₂₄) produces ethene (C₂H₄) and one other product. What is the molecular formula of the other product?
4. Why do branched alkanes have lower boiling points than their straight-chain isomers of the same molecular formula?
5. Arrange these C4 compounds in order of INCREASING boiling point: butane, butan-1-ol, butanal, butan-1-amine, butanoic acid
MC Answers: 1-C | 2-D | 3-B | 4-B | 5-A
1-C: The –OH in hexan-1-ol enables H-bond donation and acceptance — far stronger than dispersion-only forces in hexane. Both compounds have 6C chains so dispersion forces are similar; the IMF difference is entirely due to H-bonding.
2-D: Refinery gas (C1–C4) has the lowest boiling point → remains vapour longest → rises highest in column and exits from the top. Bitumen (highest BP) never vaporises and collects at the base.
3-B: Balance C: 11−2=9; balance H: 24−4=20 → C₉H₂₀ (nonane, CₙH₂ₙ₊₂ for n=9 ✓). C₉H₁₈ would be an alkene — cracking gives one alkene + one alkane.
4-B: Same molecular mass (same formula). Branching → compact shape → less surface area → fewer simultaneous dispersion interactions → weaker total IMF → lower BP.
5-A: IMF ranking: dispersion only (butane, −1 °C) < C=O dipole (butanal, 75 °C) < N–H H-bond (butan-1-amine, 78 °C) < O–H H-bond (butan-1-ol, 118 °C) < O–H + dimerisation (butanoic acid, 164 °C).
SA1 (4 marks): Both have –OH that H-bonds with water. For propan-1-ol: the energy released forming H-bonds with water ≈ energy needed to disrupt water network → fully miscible [1]. For hexan-1-ol: the large 6C non-polar chain disrupts water's H-bond network; energy required > energy released from one –OH group → insoluble [1]. General principle: as chain length increases past ~C5, non-polar portion dominates; "like dissolves like" → hexan-1-ol dissolves in non-polar hexane instead [1+1 marks].
SA2 (4 marks): Claim is INCORRECT [1]. A catalyst is a substance that lowers activation energy and is regenerated — UV light is energy (radiation), not matter, and cannot be regenerated [1]. Balanced equation: CH₄ + Cl₂ → CH₃Cl + HCl (conditions: UV light) [1]. Correct role: UV provides energy to break Cl–Cl bond homolytically (Cl₂ → 2Cl•), generating chlorine radicals that initiate the free radical chain reaction [1].
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