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.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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.
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.
Every exam question tests whether you know the complete conditions set. Reproduce this table from memory as part of your consolidation. These reaction patterns build directly on Lessons 5–7 — any reaction involving alkanes, alkenes, or alkynes covered there applies here, and this lesson connects them into a single systematic framework.
| Reaction | Reagent(s) | Catalyst | Conditions | Product type |
|---|---|---|---|---|
| Alkene hydrogenation | H₂ | Ni (or Pd/Pt) | 150–200°C, high pressure | Alkane |
| Alkene halogenation | X₂ (Cl₂ or Br₂) | None | Room temp, no solvent or CCl₄ | Vicinal dihalide |
| Alkene hydrohalogenation | HX (HCl or HBr) | None | Room temp; Markovnikov applies | Haloalkane |
| Alkene hydration | H₂O (steam) | H₃PO₄ or dil. H₂SO₄ | ~300°C, high pressure (~65 atm) | Alcohol (Markovnikov) |
| Alkyne hydrogenation (full) | 2H₂ | Ni/Pd/Pt | 150–200°C, high pressure | Alkane |
| Alkyne hydrogenation (partial) | H₂ (1 eq) | Lindlar catalyst | Mild; stops at cis-alkene | cis-Alkene |
| Alkyne hydration | H₂O | H₂SO₄ AND Hg²⁺ | ~60°C, fume cupboard (Hg toxic) | Ketone (via enol tautomerism) |
| Alkane halogen substitution | X₂ (Cl₂ or Br₂) | UV light (energy source) | UV irradiation (not heat), gas phase | Haloalkane + HX |
Quick Check: True or False — The UV light in alkane halogenation is acting as a catalyst because it speeds up the reaction without being consumed.
Both dihalide types arise from hydrocarbon reactions, but they form by different mechanisms from different starting materials. Being able to identify which type is present — and work backwards to the starting material — is a key HSC skill.
Location: Halogen atoms on adjacent (neighbouring) carbons — C1 and C2
Formed by: Halogenation of an alkene (X₂ addition across C=C)
Example: Propene + Cl₂ → 1,2-dichloropropane (Cl on C1 and C2)
Conditions: Cl₂ or Br₂, room temp, no catalyst
Location: Both halogen atoms on the same carbon
Formed by: Two-step hydrohalogenation of an alkyne (2 × HX, Markovnikov each step)
Example: Propyne + 2HBr → 2,2-dibromopropane (both Br on C2)
Conditions: Excess HX, room temp, no catalyst, two equivalents
Microtask: A compound has the structure CH₃CBr₂CH₃ (2,2-dibromopropane). What type of dihalide is this, and what is its alkyne precursor?
Multi-step synthesis questions give you a starting material and a target product and ask you to plan the sequence of reactions. The key skill is choosing the right reaction type at each step.
| Target product | Reaction type | Key condition |
|---|---|---|
| Propane (CH₃CH₂CH₃) | Full hydrogenation | 2H₂, Ni, 150–200°C, high P |
| cis-Propene (CH₃CH=CH₂) | Partial hydrogenation | 1H₂, Lindlar catalyst, mild |
| Propanone (CH₃COCH₃) | Alkyne hydration | H₂O, H₂SO₄ + Hg²⁺, ~60°C |
| 2-bromopropene (CH₃CBr=CH₂) | Hydrohalogenation (1 eq) | 1 eq HBr, r.t., Markovnikov step 1 |
| 2,2-dibromopropane (CH₃CBr₂CH₃) | Hydrohalogenation (2 eq) | 2 eq HBr, r.t., Markovnikov both steps |
Microtask: A student wants to convert propyne to propanone. Which set of conditions would achieve this correctly?
Each of the errors below matches a real exam-style mistake. Identify the error before reading the explanation — the act of spotting it first reinforces the correct version.
Error 1: Propene + HCl → 1-chloropropane (HCl addition, Markovnikov)
Fix: Markovnikov says H adds to C1 (has 2H), Cl to C2 (has 1H, more substituted). Product is 2-chloropropane, not 1-chloropropane. The product name given is the anti-Markovnikov product.
Error 2: "Propene decolourises bromine water, therefore propene is confirmed to be an alkene (not an alkyne)."
Fix: Bromine water decolourisation confirms unsaturation (C=C or C≡C), but it cannot distinguish an alkene from an alkyne — alkynes also decolourise Br₂(aq). The conclusion is too strong. Additional evidence is needed.
Error 3: Propene + H₂O → propan-1-ol. Conditions: H₂SO₄, 300°C.
Fix: Two errors. (1) Markovnikov: OH goes to C2 (more substituted), so the product is propan-2-ol, not propan-1-ol. (2) Conditions are incomplete — high pressure (~65 atm) is missing. Full: H₂O/steam, H₃PO₄, ~300°C, HIGH PRESSURE.
Error 4: Alkyne hydration: propyne + H₂O → propan-2-ol. Conditions: H₂SO₄, ~60°C.
Fix: Two errors. (1) The product of alkyne hydration is a ketone (propanone), not an alcohol. The enol intermediate tautomerises to a ketone. (2) Hg²⁺ catalyst is missing — both H₂SO₄ AND Hg²⁺ are required.
Error 5: Ethane + Br₂ → bromoethane. Conditions: UV light (catalyst), room temp.
Fix: UV light is not a catalyst — it is an energy source that initiates the free radical chain by providing energy to break Br–Br homolytically. A catalyst is regenerated and lowers activation energy; UV light is absorbed to produce radicals. Write: "UV light (energy source for radical initiation)" — not "catalyst".
Error 6: But-2-yne + excess Br₂ → 2,3-dibromobutene (stopping at the alkene stage).
Fix: Excess Br₂ will not stop at the alkene stage — the second pi bond also undergoes addition. With excess Br₂, but-2-yne gives 2,2,3,3-tetrabromobutane. You must specify 1 equivalent (1 eq) of Br₂ to obtain the dibromoalkene product.
Odd One Out: Three of the following statements about hydrocarbon reactions are correct. Which one contains an error?
Many HSC questions give you the product and ask for the starting material and conditions. The strategy is to identify the functional group on the product, determine what bond it replaced, and reverse the reaction.
Fill the Gap: The product 2-methylbutan-2-ol has an OH group on a [tertiary / secondary / primary] carbon. This is consistent with Markovnikov addition of water to a(n) [alkene / alkyne / alkane]. The catalyst required is [Ni / H₂SO₄ or H₃PO₄ / H₂SO₄ + Hg²⁺].
Complete the microtasks above to unlock Practice. You need ? XP to continue.
Propyne (CH₃C≡CH) is the starting material for each of the following target products. For each one, write: (i) the balanced equation, (ii) the full conditions (reagent, catalyst, temperature, pressure, equipment), and (iii) the mechanism type (addition/substitution/other).
Each equation or statement below contains one or more errors. Identify every error and write the fully corrected version. Do not assume there is only one error per item.
1. Pent-2-ene reacts with HCl. Applying Markovnikov's rule, which product is formed?
2. Which of the following correctly describes alkyne hydration?
3. Ethane reacts with Br₂ in the presence of UV light. What is the correct product and role of UV light?
4. A student isolates a geminal dihaloalkane, 2,2-dibromopropane, from a reaction. What was the starting material and how many equivalents of HBr were used?
5. A chemist wants to convert but-2-yne to cis-but-2-ene (stopping at the alkene stage). Which set of conditions achieves this?
Q1 (3 marks): A student concludes that compound X is an alkene because it decolourises bromine water. Evaluate this conclusion.
Q2 (4 marks): 2-methylbutan-2-ol is produced by hydration. (a) Identify the reaction type and two possible alkene starting materials. (b) For each starting material, apply Markovnikov's rule to show both give the same product.
Q3 (5 marks): A student attempts to make propanone from propyne. In their first attempt, they use H₂O with H₂SO₄ at 60°C — no reaction occurs. In their second attempt, they use H₂O with H₃PO₄ at 300°C and high pressure — a reaction occurs but the product is not propanone. (a) Explain why the first attempt failed (2 marks). (b) Explain what product the second conditions would likely give and why the student has confused two different reactions (3 marks).
(a) Propane: Full hydrogenation. CH₃C≡CH + 2H₂ → CH₃CH₂CH₃. Ni (or Pd/Pt) catalyst, ~150–200°C, high pressure, pressure vessel. Addition reaction (two-step hydrogenation, alkyne → alkene → alkane).
(b) cis-Propene: Partial hydrogenation. CH₃C≡CH + H₂ → CH₃CH=CH₂. Lindlar catalyst (poisoned Pd — stops at alkene stage), mild conditions. Gives specifically cis-propene. Note: using Ni without specifying Lindlar gives the alkane, not the alkene.
(c) Propanone: Alkyne hydration. CH₃C≡CH + H₂O → CH₃COCH₃. BOTH dilute H₂SO₄ AND Hg²⁺ catalyst, ~60°C, heated glassware, fume cupboard (Hg toxic). Enol intermediate (CH₃C(OH)=CH₂) tautomerises to ketone (propanone). Product is a ketone, NOT an alcohol.
(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. Geminal because both Br are on C2 (same carbon).
(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 conclusion "confirms propene is an alkene" is too strong. 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. Corrected conclusion: "Decolourisation confirms unsaturation — the presence of C=C or C≡C. Additional evidence (e.g., testing with two equivalents of Br₂, or the molecular formula) is needed to confirm specifically an alkene."
(c) Two errors: (1) Product is wrong — Markovnikov: H → C1 (=CH₂, 2H), OH → C2 (=CH-, more substituted). 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 (or H₂SO₄), ~300°C, HIGH PRESSURE (~65 atm).
1. C — Pent-2-ene: C2 and C3 are the double bond carbons. C3 is more substituted (bonded to an ethyl group); C2 is less substituted. Markovnikov: 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.
2. B — Alkyne hydration requires BOTH dil. H₂SO₄ AND Hg²⁺ at ~60°C; product is ketone via enol tautomerism. Option A has conditions for alkene hydration (wrong temp/pressure/catalyst). Option C has correct conditions but wrong product (alkyne → ketone, not alcohol). Option D omits Hg²⁺.
3. C — Ethane + Br₂ + UV light → halogen substitution. One H replaced by Br → bromoethane + HBr. UV light is the energy source for radical initiation (not a catalyst — it is consumed in the process of generating radicals). Option B is wrong: ethane has no C=C, 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 C2) comes from two-step hydrohalogenation of an alkyne. Working backwards from CH₃CBr₂CH₃: remove both Br from C2 → reform C≡C at C1–C2 → propyne. Two equivalents of HBr were required. Propene (options A/C) would give 1,2-dibromopropane (vicinal) or 2-bromopropane (with HBr), not a geminal dihalide.
5. B — Partial hydrogenation to stop at the alkene stage requires Lindlar catalyst (1 eq H₂, mild conditions). Option A (excess H₂ + Ni) gives full hydrogenation → butane. Option C (Ni, 1 eq) would not selectively stop at alkene — Ni is not selective enough. Option D (excess H₂ + Lindlar) — excess H₂ with Lindlar could drive full reduction too.
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 pi bonds and also decolourise bromine water by addition [1]. Additional evidence is needed to distinguish an alkene from an alkyne — for example, testing with two equivalents of Br₂ (an alkyne would react with both) or determining the molecular formula (CₙH₂ₙ for alkene, CₙH₂ₙ₋₂ for alkyne) [1].
Q2 (4 marks): Reaction type: alkene hydration [1]. Two possible alkene starting materials: 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, 0H) → same product, 2-methylbutan-2-ol. Both alkenes give the same Markovnikov product because both have C2 as the more substituted alkene carbon [1].
Q3 (5 marks): (a) The first attempt failed because it omitted Hg²⁺ catalyst — alkyne hydration requires BOTH dilute H₂SO₄ AND Hg²⁺ [1]. Hg²⁺ acts as a Lewis acid catalyst that activates the C≡C toward water attack. Without Hg²⁺, the activation energy for alkyne hydration cannot be overcome at 60°C — the reaction does not proceed [1]. (b) H₃PO₄ at 300°C and high pressure are conditions for alkene hydration (of propene → propan-2-ol), not alkyne hydration [1]. The student has confused the two different hydration reactions: alkene hydration (H₃PO₄ or H₂SO₄, 300°C, high pressure → alcohol) vs alkyne hydration (H₂SO₄ + Hg²⁺, 60°C → ketone) [1]. If propyne did react under these conditions at all, it would not give propanone — the product obtained would be an alcohol (propan-2-ol if propyne reacted as a propene-equivalent), not a ketone; the two reactions produce products of entirely different functional group classes [1].
Back at the start you were introduced to the kind of error that costs marks in an exam: a product labelled as a primary alcohol when it should have been a secondary alcohol — a Markovnikov regiochemistry error.
Now, armed with the full set of hydrocarbon reaction patterns from this lesson, you can diagnose any error of that type:
Return to the opening scenario: the student's primary alcohol error was a Markovnikov mistake. Using the hydration rules, the OH group goes to the more substituted carbon — giving a secondary or tertiary alcohol, not a primary one. That is the error that cost 3 marks, and you can now catch it instantly.
Answer each question from memory. Aim for full conditions every time.
State the full conditions for alkyne hydration (reagent, both catalysts, temperature).
Explain why UV light in alkane halogenation is described as an energy source rather than a catalyst.
A compound is CH₃CHBrCH₂Br. Name it, classify it as vicinal or geminal, and state the starting material and reaction that produced it.
Give two reasons why partial hydrogenation of an alkyne requires the Lindlar catalyst specifically (not Ni or Pd/Pt under standard conditions).
A student isolates 2,2-dibromopropane. Working backwards, identify the starting material and write a complete synthesis from it using only reactions from this module.
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