Covers Lessons 5–8: reactions of alkenes, alkynes and alkanes, including addition, substitution, combustion and the bromine-water test.
Mercury-catalysed hydration and Lindlar partial hydrogenation are specialist conditions beyond the required reactions. The core here is addition, substitution, combustion and the bromine-water test.
1. Propyne (CH₃C≡CH) reacts with excess water in the presence of dilute H₂SO₄ and Hg²⁺ at ~60 °C. What is the organic product? Extension
2. A student reacts but-2-ene with HBr. What is the major product?
3. Which set of conditions correctly produces ethene from ethyne, NOT ethane? Extension
4. Bromine water is added to two unknown compounds. Compound X decolourises the bromine water; Compound Y does not. Which conclusion is MOST accurate?
5. Which equation correctly represents the complete combustion of butane (C₄H₁₀)?
MC Answers: 1-C | 2-B | 3-B | 4-B | 5-B
1-C: Alkyne hydration (H₂SO₄ + Hg²⁺) gives a ketone via Markovnikov's rule. H adds to the terminal C1, –OH to C2 → enol → tautomerises instantly to propanone. Propanal requires a terminal alkyne hydration with special conditions.
2-B: But-2-ene is symmetrical (CH₃CH=CHCH₃), C2 and C3 are equivalent. H can add to either carbon; Br goes to the other. Both directions give 2-bromobutane (the Markovnikov product by symmetry).
3-B: Partial hydrogenation with the Lindlar catalyst (poisoned Pd) stops at the alkene stage. Option A (excess H₂, Ni) → ethane (full reduction). Option D gives ethanal, not ethene.
4-B: Alkynes also decolourise Br₂ water, "X is an alkene" (option A) is too specific. Bromine water tests for unsaturation OR easily oxidisable groups (e.g. aldehydes). Y has none of these features.
5-B: Balance: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. Check: C: 8=8 ✓, H: 20=20 ✓, O: 26=26 ✓. Option A has insufficient O₂.
SA1 (5 marks):
(a) CH₂=CHCH₂CH₂CH₃ + H₂ → CH₃CH₂CH₂CH₂CH₃. Reagent: H₂; catalyst: Ni (or Pd/Pt); ~150–200 °C, high pressure. [1+1 marks]
(b) CH₂=CHCH₂CH₂CH₃ + H₂O → CH₃CH(OH)CH₂CH₂CH₃ (pentan-2-ol, Markovnikov). Reagent: steam; catalyst: H₃PO₄ or H₂SO₄; ~300 °C, ~65 atm. [1 mark]
(c) CH₂=CHCH₂CH₂CH₃ + HBr → CH₃CHBrCH₂CH₂CH₃ (2-bromopentane, Markovnikov). Reagent: HBr; no catalyst; room temperature; fume cupboard. [1 mark]
SA2 (4 marks): Once initiated by UV (Cl₂ → 2Cl•), Cl• abstracts H from CH₄ → CH₃• → reacts with Cl₂ → CH₃Cl [1]. CH₃Cl remains in the mixture and Cl• can also abstract H from CH₃Cl → CH₂Cl₂, then CHCl₃, then CCl₄ [1]. All four products form simultaneously; relative amounts depend on CH₄:Cl₂ ratio but selectivity is always poor [1]. Industrial implication: expensive separation steps needed → more selective routes (e.g. methanol + HCl) are preferred for pure chloromethane production [1].
Tick when you have finished all questions.