Covers Lessons 9–12: Alkene and alkyne reactions — addition (hydrogenation, halogenation, hydrohalogenation, hydration), bromine water test, Markovnikov's rule, and combustion.
1. Propyne (CH₃C≡CH) reacts with excess water in the presence of dilute H₂SO₄ and Hg²⁺ at ~60 °C. What is the organic product?
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?
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.