Every mark lost in Module 7 hydrocarbon reaction questions comes down to one of three errors — wrong conditions, wrong product, or wrong reaction type. This lesson fixes all three.
Wrong: Alkene hydration only requires water and heat — no catalyst is needed.
Right: Alkene hydration requires dilute H₂SO₄ as a catalyst. The acid protonates the double bond first, creating a carbocation intermediate that water then attacks. Without acid catalysis, the reaction is impractically slow. For alkynes, both H₂SO₄ AND Hg²⁺ catalyst are required, and the reaction produces a ketone (not an alcohol) via enol tautomerisation.
Consolidation Lesson — No new syllabus dot points. This lesson deepens and stress-tests the content from L06 (alkene addition reactions) and L07 (alkyne addition, alkane halogenation, combustion). Work through the conditions table, the Spot the Error cards, and the recall challenge. Any condition set you cannot reproduce from memory is a potential lost mark in an exam.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A student is given five reaction equations on a test. They get the reaction type right every time but lose marks on every single question. The marker has written the same comment five times: "conditions incomplete." The student used the right reagents but forgot the catalyst on two questions, wrote "UV light (catalyst)" on one, forgot to mention high pressure on the hydration, and wrote "heat" instead of "H2SO4 + Hg²⁺" on the alkyne hydration.
Before working through this lesson, write down from memory how many hydrocarbon reactions from L06 and L07 you can fully describe — reagent, catalyst, conditions, equipment — without looking at your notes. Count them. That number will be higher at the end.
(a) Propane: Full hydrogenation. CH3C≡CH + 2H2 → CH3CH2CH3. Ni (or Pd/Pt) catalyst, ~150–200°C, high pressure, pressure vessel.
(b) Propene: Partial hydrogenation. CH3C≡CH + H2 → CH3CH=CH2. Lindlar catalyst (poisoned Pd — stops at alkene stage), mild conditions. Gives specifically cis-propene.
(c) Propanone: Alkyne hydration. CH3C≡CH + H2O → CH3COCH3. BOTH dilute H2SO4 AND Hg²⁺ catalyst, ~60°C, heated glassware, fume cupboard (Hg toxic). Enol intermediate tautomerises to ketone.
(d) 2-bromopropene: Step 1 hydrohalogenation only. CH3C≡CH + HBr (1 eq) → CH3CBr=CH2. No catalyst, room temperature. Must specify ONE equivalent of HBr — with excess HBr, step 2 occurs and 2,2-dibromopropane forms instead.
(e) 2,2-dibromopropane: Two-step hydrohalogenation (full). Step 1: CH3C≡CH + HBr → CH3CBr=CH2 (2-bromopropene). Step 2: CH3CBr=CH2 + HBr → CH3CBr2CH3 (2,2-dibromopropane). Both steps: no catalyst, room temperature, Markovnikov each time. Two equivalents of HBr total.
(a) Error: 1-chloropropane is the anti-Markovnikov product. Propene: C1 (=CH2) has 2H; C2 (=CH-) has 1H. Markovnikov: H → C1, Cl → C2. Correct Markovnikov product: 2-chloropropane (CH3CHClCH3). Conditions are otherwise correct.
(b) Error: The statement "confirms propene is an alkene" is incomplete. Decolourisation of bromine water confirms unsaturation (C=C or C≡C present). Alkynes also decolourise bromine water. This test cannot distinguish between an alkene and an alkyne. The equation and conditions (Br₂, no catalyst, r.t.) are correct. Corrected conclusion: "Decolourisation confirms unsaturation — the presence of C=C or C≡C. Additional evidence (e.g. molecular formula, or testing with two equivalents of Br₂) is needed to specifically confirm an alkene."
(c) Two errors: (1) Product is wrong — Markovnikov: H → C1 (=CH2, 2H), OH → C2 (=CH-). Correct product: propan-2-ol (CH3CH(OH)CH3), not propan-1-ol. (2) Conditions are incomplete — high pressure (~65 atm) is missing. Full conditions: H2O (steam), H3PO4 catalyst, ~300°C, HIGH PRESSURE.
1. C — Pent-2-ene: C2 and C3 are both bonded to one H each. Markovnikov is determined by substitution: C3 is bonded to an ethyl group (more substituted); C2 is bonded to a methyl group. H adds to C2 (gives more stable secondary carbocation at C3); Cl adds to C3 → 3-chloropentane. Option B (2-chloropentane) is the anti-Markovnikov product. Option D is halogenation product (Cl2 not HCl).
2. B — Alkyne hydration requires BOTH dil. H2SO4 AND Hg²⁺ at ~60°C; product is ketone (or ethanal from ethyne). Option A has wrong conditions (those describe alkene hydration) and wrong product. Option C has correct conditions but wrong product (alcohol not ketone). Option D has wrong conditions and omits Hg²⁺.
3. C — Ethane + Br2 + UV light → halogen substitution. One H replaced by Br → bromoethane + HBr. UV light is the energy source (not catalyst). Option B is wrong: ethane has no C=C (saturated alkane), so addition is impossible. Option D correctly identifies the reaction but mislabels UV light as a catalyst.
4. B — A geminal dihaloalkane (both Br on same C — C2) from 2 eq HBr → two-step hydrohalogenation of an alkyne. Working backwards from CH3CBr2CH3: remove both Br from C2, reform C≡C → propyne (CH3C≡CH). Markovnikov step 1: H to C1 (terminal carbon), Br to C2 → 2-bromopropene. Step 2: H to C1 again, Br to C2 → 2,2-dibromopropane ✓. Propene (options A/C) would give 1,2-dibromopropane with HBr, not a geminal dihalide.
5. B — To stop at the alkene stage from an alkyne, use partial hydrogenation with the Lindlar catalyst (1 eq H2, mild conditions). Option A (full hydrogenation with Ni/excess H2) would continue to ethane, not stop at ethene.
Q1 (3 marks): The observation that compound X decolourises bromine water confirms that X contains unsaturation — a C=C or C≡C bond [1]. However, the conclusion that X is specifically an alkene is not fully justified. Alkynes also contain C=C equivalent sites (two pi bonds) and also decolourise bromine water by addition [1]. Additionally, cycloalkanes (which have the same CnH2n formula as alkenes) do NOT decolourise bromine water, so if the formula is CnH2n, bromine water decolourising actually rules out a cycloalkane and supports an alkene — but alkynes cannot be ruled out from bromine water alone. Additional evidence is needed to distinguish an alkene from an alkyne [1].
Q2 (4 marks): Reaction type: alkene hydration [1]. The product 2-methylbutan-2-ol has the -OH on C2 (a tertiary carbon — bonded to three carbon groups). Two starting alkenes are possible: 2-methylbut-1-ene (CH2=C(CH3)CH2CH3) and 2-methylbut-2-ene (CH3C(CH3)=CHCH3) [1]. For 2-methylbut-1-ene: Markovnikov — H to C1 (=CH2, 2H), OH to C2 (=C(CH3)-, 0H, more substituted) → OH at C2 → 2-methylbutan-2-ol ✓ [1]. For 2-methylbut-2-ene: C2 (=C(CH3), 0H) and C3 (=CH-, 1H). Markovnikov — H to C3, OH to C2 (more substituted) → OH at C2 → same product, 2-methylbutan-2-ol ✓. Both alkenes give the same Markovnikov product because both have C2 as the more substituted C=C carbon [1].
Q3 (5 marks): (a) The first attempt failed because the reaction requires BOTH dilute H2SO4 AND Hg²⁺ catalyst — omitting Hg²⁺ means the reaction cannot proceed [1]. Hg²⁺ acts as a Lewis acid catalyst that activates the C≡C toward attack by water — without it, the activation barrier for alkyne hydration cannot be overcome under these mild conditions (60°C is insufficient without Hg²⁺) [1]. (b) The second attempt used H3PO4 at 300°C and high pressure — these are the conditions for alkene hydration, not alkyne hydration [1]. Propyne is an alkyne, but under these conditions (if the reaction occurred at all with a solid acid at high temperature), the conditions would not give propanone. More likely, these conditions describe a different reaction pathway entirely. If propyne did react with steam and H3PO4 at 300°C, it would likely partially hydrolyse or not react cleanly — but the conditions describe alkene hydration. The student has confused the conditions for alkene hydration (H3PO4, 300°C, high pressure → alcohol) with alkyne hydration (H2SO4 + Hg²⁺, 60°C → ketone). The product obtained would be an alcohol (propan-2-ol, if propyne reacted as if it were propene) rather than propanone [1]. The two reactions use completely different catalysts, temperatures, and pressures, and give products of different functional group classes [1].
Return to your opening count of how many reaction condition sets you could recall from memory. Now test yourself again — without looking at the reference table — and count how many complete condition sets you can reproduce:
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Defend your ship by blasting the correct answers for Hydrocarbon Reactions Mastery — Conditions, Products & Spot the Error. Scores count toward the Asteroid Blaster leaderboard.
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Tick when you can reproduce the full conditions table from memory and have corrected all six Spot the Error equations.