Chemistry • Year 12 • Module 7 • Lesson 7
Reactions of Alkynes & Alkanes
Apply two-step alkyne addition, Markovnikov's rule, and combustion balancing to data tables, sequencing tasks, and real Australian contexts.
1. Reaction conditions comparison table
The table below lists five reaction types for alkynes and alkanes. Fill in the missing conditions, reagents, and products for each reaction. 10 marks
| Reaction type | Typical equation (using ethyne or methane) |
Reagent(s) | Conditions / catalyst | Primary organic product(s) & class |
|---|---|---|---|---|
| Alkyne full halogenation (two steps) | HC≡CH + 2Br2 → ? | Br2 (2 eq) | ||
| Alkyne partial hydrogenation to cis-alkene | HC≡CH + H2 → ? | Ethene (cis-alkene) | ||
| Alkyne hydration | CH3C≡CH + H2O → ? | H2O | ||
| Alkyne hydrohalogenation (full, 2 eq HBr) | HC≡CH + 2HBr → ? | HBr (2 eq) | No catalyst, room temp | |
| Alkane halogen substitution | CH4 + Cl2 → ? |
2. Sequence the steps — two-step halogenation of ethyne
The events below describe the two-step halogenation of HC≡CH with excess Br2. They are shuffled. Write the correct order (1–6) in the “Order” column. 6 marks
| Order | Event |
|---|---|
| CHBr=CHBr reacts with the second equivalent of Br2; the remaining pi bond opens. | |
| The product CHBr2CHBr2 (1,1,2,2-tetrabromoethane) forms — a fully saturated tetrahaloalkane. | |
| Ethyne (HC≡CH) with its two pi bonds is exposed to the first equivalent of Br2. | |
| Bromine water is decolourised for a second time — confirming two pi bonds were present. | |
| The first pi bond opens; one Br adds to each carbon of the triple bond. | |
| CHBr=CHBr (1,2-dibromoethene) forms — a dihaloalkene still containing one pi bond. |
3. Data interpretation — acetylene use in Australian welding
The bar chart below shows estimated acetylene (ethyne) consumption by sector in Australia in 2023, based on industrial gas distribution data from BOC Australia. Use the chart and your knowledge of combustion chemistry to answer the questions. 9 marks
Adapted from BOC Australia industrial gas usage estimates (2023). Values are approximate.
3.1 Identify the largest and smallest sectors of acetylene use in Australia and state the percentage share of each. 2 marks
3.2 Write the balanced equation for complete combustion of ethyne (HC≡CH) in an oxy-acetylene torch. Show your working using the C → H → O balancing method. 3 marks
3.3 A welder using an oxy-acetylene torch in a poorly ventilated workshop restricts oxygen flow. Explain what change would be observed in the flame and describe the hazard this creates, using specific combustion products. 2 marks
3.4 The “chemical synthesis” sector uses ethyne as a starting material for PVC production (via Qenos, Australia’s sole PVC manufacturer). Outline the two-step reaction sequence from ethyne to chloroethene (vinyl chloride), including equations and conditions for each step. 2 marks
4. Cause-and-effect chain — chlorination of drinking water
NSW Health uses chlorine gas (Cl2) to disinfect drinking water. In the absence of UV light, Cl2 dissolves in water to form hypochlorous acid (HOCl) which kills pathogens. However, chlorine can also react with organic matter in water to produce chlorinated by-products such as chloromethane (CH3Cl) when trace methane is present under UV exposure. Trace the cause-and-effect chain below. Fill in the empty boxes. 5 marks
5. Predict and justify — Markovnikov’s rule applied to alkynes
Predict the major organic product when propyne (CH3C≡CH) reacts with one equivalent of HBr under Markovnikov addition. Justify your prediction by explaining which carbon receives the Br and why. 4 marks
Equation: CH3C≡CH + HBr →
Now predict the product when a second equivalent of HBr is added (Markovnikov, step 2 of full hydrohalogenation). Name the product and state whether it is a geminal or vicinal dihalide. 2 marks
Q1 — Reaction conditions comparison table
Row 1 (Full halogenation): Conditions: no catalyst, room temperature; Product: CHBr2CHBr2 (1,1,2,2-tetrabromoethane, tetrahaloalkane).
Row 2 (Partial hydrogenation): Reagent: H2 (1 eq); Conditions: Lindlar catalyst (poisoned Pd), room temperature.
Row 3 (Hydration): Conditions: dilute H2SO4 AND Hg2+, ~60°C; Product: propanone (ketone) from propyne.
Row 4 (Full hydrohalogenation): Product: CH3CHBr2 (1,1-dibromoethane, geminal dihalide).
Row 5 (Substitution): Reagent: Cl2; Conditions: UV light (energy source — NOT a catalyst); Products: CH3Cl (chloromethane) + HCl.
Q2 — Sequence (correct order)
3 → 5 → 6 → 1 → 2 → 4
That is: (3) Ethyne exposed to first Br2; (5) first pi bond opens; (6) CHBr=CHBr forms; (1) reacts with second Br2; (2) CHBr2CHBr2 forms; (4) bromine water decolourised second time.
Q3 — Data graph interpretation
3.1 Largest: oxy-acetylene welding (48%); smallest: other industrial (10%).
3.2 C: 2C → 2CO2. H: 2H → 1H2O. O right: 4 + 1 = 5 → O2 = 5/2. Multiply by 2: 2HC≡CH + 5O2 → 4CO2 + 2H2O.
3.3 The flame would change from a clean blue flame to a yellow/sooty flame. Incomplete combustion produces CO (toxic — binds haemoglobin and blocks O2 transport) and C (soot) instead of CO2, creating an acute health hazard for the welder.
3.4 Step 1: HC≡CH + HCl → CH2=CHCl (chloroethene/vinyl chloride) via hydrohalogenation (Markovnikov, room temp). Alternatively, industrial route: HC≡CH + Cl2 → CHCl=CHCl, then pyrolysis at 500°C gives CH2=CHCl + HCl. Either route acceptable.
Q4 — Cause-and-effect chain
Effect 1: homolysis (homolytic fission) of the Cl—Cl bond; generating Cl• (chlorine radicals).
Effect 2: H; HCl.
Effect 3: CH3Cl (chloromethane); Cl• (which propagates the chain).
Overall outcome: Free radical substitution converts CH4 + Cl2 (under UV light) to CH3Cl + HCl; this reaction is an energy-initiated chain mechanism, not a catalytic one.
Q5 — Markovnikov prediction
One equivalent HBr (Markovnikov): CH3C≡CH + HBr → CH3CBr=CH2 (2-bromopropene). Markovnikov: H adds to C3 (terminal, has 1 H), Br adds to C2 (more substituted, bonded to CH3 — which stabilises the intermediate vinyl cation). 4 marks: equation (1), product name (1), which C gets Br and why (2).
Second equivalent HBr (Markovnikov, step 2): CH3CBr=CH2 + HBr → CH3CBr2CH3 (2,2-dibromopropane). H adds to C3 (=CH2, more H), Br adds to C2 (more substituted, 0 H). The product is a geminal dihalide (both Br on the same carbon, C2). 2 marks: equation/name (1), geminal identification (1).