HSCScienceExam practice
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Chemistry  •  Year 12  •  Module 7  •  Lesson 7

HSC Exam Practice

Reactions of Alkynes & Alkanes — Hydration, Substitution & Combustion

10 questions / 3 sections / 35 marks total
Section 1

Short answer

1.Short answer — Band 3–4

1.1

Define free radical substitution and identify the role of UV light in this reaction.

3marks Band 3
1.2

Describe how the number of pi bonds in an alkyne explains why its halogenation reaction proceeds in two distinct steps.

2marks Band 3
1.3

Identify the organic product and its functional group class when but-1-yne (HC≡CCH2CH3) reacts with water in the presence of dilute H2SO4 and Hg2+. Write the equation for the overall reaction.

3marks Band 4
1.4

Outline the difference between a geminal dihalide and a vicinal dihalide, giving one example of a reaction that produces each type from an alkyne starting material.

3marks Band 4
1.5

Distinguish between complete and incomplete combustion of a hydrocarbon, with reference to oxygen supply, products formed, and one environmental or health impact of each.

4marks Band 4
1.6

Write the balanced equation for complete combustion of pentane (C5H12). Show your working using the C → H → O method.

3marks Band 4
Section 2

Data response

2.Data response — Band 4–5

2.1

The table below lists four hydrocarbon fuels. For each, write the balanced equation for complete combustion in excess oxygen using the systematic C → H → O method, then identify the products. 6 marks (1.5 per row)

Fuel (formula) Balanced combustion equation Products
Methane (CH4)    
Ethyne (HC≡CH)    
Propane (C3H8)    
Butane (C4H10)    
6marks Band 4
2.2

A gas torch uses propyne (CH3C≡CH) as its fuel in an oxy-acetylene-type system.

(a) Write the balanced equation for complete combustion of propyne using the C → H → O method. Show your working for each step. 3 marks

(b) The torch is operated in a poorly ventilated workshop. State the products formed under limited oxygen supply and identify one health hazard of each product. 2 marks

(c) Propyne also reacts with bromine. Write the equation for Step 1 of the bromination of propyne (one equivalent of Br2) and name the product. State whether the product decolourises bromine water. 1 mark

6marks Band 4–5
Section 3

Extended response

3.Extended response — Band 5–6

3.1

“Adding HBr to propyne produces 1,2-dibromopropane because bromine atoms always add to adjacent (vicinal) carbons, regardless of whether you start from an alkyne or an alkene.”

— Chemistry revision website, accessed 2024.

Identify the scientific error in the claim above. Explain the correct chemistry, including the full two-step mechanism of HBr addition to propyne, and use Markovnikov’s rule to identify the actual product of each step. 5 marks

5marks Band 5
3.2

Evaluate the statement: “The two reactions available to alkanes — halogenation and combustion — both have significant industrial and environmental relevance in Australia.” In your response, discuss the conditions, products, and at least one named Australian industrial or environmental context for each reaction type. 6 marks

6marks Band 5–6
Marking criteria — Do not reveal until after submission

Q1.1 — Free radical substitution (3 marks)

Free radical substitution is a chain reaction in which a halogen molecule (X2) reacts with an alkane, with one H atom replaced by a halogen atom and HX produced as a co-product [1]. UV light provides the energy (it is an energy source, not a catalyst) to break the X—X bond by homolysis, producing two halogen radicals (X•) in the initiation step [1]. UV light is electromagnetic radiation, not matter, so it cannot be a catalyst; it is absorbed in initiation and is not regenerated [1].

Q1.2 — Two pi bonds (2 marks)

A C≡C triple bond has two pi bonds [1]. Each addition step opens one pi bond; the first equivalent of X2 opens the first pi bond to give a dihaloalkene (which still has one pi bond), and the second equivalent opens the second pi bond to give the fully saturated tetrahaloalkane [1].

Q1.3 — But-1-yne hydration (3 marks)

HC≡CCH2CH3 + H2O → CH3COCH2CH3 [1]. Product: butan-2-one [1]. Functional group class: ketone (not alcohol — the enol intermediate tautomerises to the keto form) [1].

Q1.4 — Geminal vs vicinal dihalide (3 marks)

A geminal dihalide has both halogen atoms on the same carbon (e.g. CH3CHBr2); a vicinal dihalide has halogen atoms on adjacent carbons (e.g. CHBrCHBr) [1]. Geminal: full hydrohalogenation of an alkyne (e.g. HC≡CH + 2HBr → CH3CHBr2) [1]. Vicinal: halogenation of an alkyne (step 1, 1 eq Br2; e.g. HC≡CH + Br2 → CHBr=CHBr) [1].

Q1.5 — Complete vs incomplete combustion (4 marks)

Complete combustion occurs in excess O2; products are CO2 + H2O only; clean blue flame [1]. Environmental impact: CO2 is a greenhouse gas contributing to the enhanced greenhouse effect and climate change [1]. Incomplete combustion occurs in limited O2; products include CO and/or soot (C) + H2O; yellow/sooty flame [1]. Health/environmental impact: CO binds haemoglobin ≈250× more strongly than O2, blocking O2 transport; soot (PM2.5) causes respiratory and cardiovascular disease [1].

Q1.6 — Pentane combustion (3 marks)

C: 5C → 5CO2 [1]. H: 12H → 6H2O [1]. O right: 10 + 6 = 16 → O2 = 8. Balanced equation: C5H12 + 8O2 → 5CO2 + 6H2O [1].

Q2.1 — Combustion equations table (6 marks)

Methane: C: 1C → 1CO2. H: 4H → 2H2O. O: 2+2=4 → O2=2. CH4 + 2O2 → CO2 + 2H2O. Products: CO2 + H2O (complete combustion only).

Ethyne: C: 2C → 2CO2. H: 2H → 1H2O. O: 4+1=5 → O2=5/2. Multiply by 2: 2HC≡CH + 5O2 → 4CO2 + 2H2O. Products: CO2 + H2O.

Propane: C: 3C → 3CO2. H: 8H → 4H2O. O: 6+4=10 → O2=5. C3H8 + 5O2 → 3CO2 + 4H2O. Products: CO2 + H2O.

Butane: C: 4C → 4CO2. H: 10H → 5H2O. O: 8+5=13 → O2=13/2. Multiply by 2: 2C4H10 + 13O2 → 8CO2 + 10H2O. Products: CO2 + H2O.

Marking notes: 1 mark per row for correct balanced equation; 0.5 mark per row for correct products if equation is unbalanced but products are correct. Accept fractional coefficients for ethyne row.

Q2.2 — Propyne combustion and reactions (6 marks)

(a) Balanced equation for complete combustion of propyne (3 marks):
C: 3C → 3CO2 [1]. H: 4H → 2H2O [1]. O right: 6 + 2 = 8 → O2 = 4 [1]. CH3C≡CH + 4O2 → 3CO2 + 2H2O.

(b) Limited oxygen (2 marks): Incomplete combustion produces CO (carbon monoxide) and/or C (soot/carbon) [1]. CO binds haemoglobin ~250× more strongly than O2, blocking oxygen transport — toxic. Soot (C, PM2.5) causes respiratory and cardiovascular disease [1].

(c) Step 1 bromination (1 mark): CH3C≡CH + Br2 → CH3CBr=CHBr (1,2-dibromopropene, a dihaloalkene). The product still has one C=C pi bond, so it does decolourise bromine water in a further addition (step 2) [1].

Q3.1 — Source critique: HBr addition to propyne (5 marks)

Error identified (1 mark): The claim that HBr addition always gives a vicinal (1,2-) dihalide is incorrect for hydrohalogenation. Unlike halogenation (Br2 addition, which gives a vicinal dihalide), HBr addition follows Markovnikov’s rule, which places both Br atoms on the same (more substituted) carbon in full addition, giving a geminal dihalide.

Step 1 with Markovnikov (2 marks): CH3C≡CH + HBr → CH3CBr=CH2 (2-bromopropene). H adds to C3 (terminal, more H), Br adds to C2 (more substituted, fewer H) per Markovnikov [1 for equation, 1 for Markovnikov reasoning].

Step 2 with Markovnikov (2 marks): CH3CBr=CH2 + HBr → CH3CBr2CH3 (2,2-dibromopropane — a geminal dihalide). H adds to C3 (=CH2, more H), Br adds to C2 (already has Br) [1 for equation, 1 for identifying geminal product & distinction from vicinal].

Q3.2 — Alkane reactions: marking criteria (6 marks)

Halogenation — conditions and products (2 marks): Alkane + X2 + UV light (energy source, not catalyst) → haloalkane + HX. UV light initiates radical chain; HX is a co-product. Example: CH4 + Cl2 → CH3Cl + HCl. [1 for conditions including UV as energy source; 1 for products including HX].

Halogenation — Australian industrial context (1 mark): NSW Health uses chlorination (Cl2 in water) for drinking water disinfection; chlorination of organic matter can produce chlorinated by-products. PVC synthesis via Qenos (Altona, VIC) involves chloroethene (vinyl chloride) production, which requires halogenation chemistry.

Combustion — conditions and products (2 marks): Complete combustion (excess O2): alkane + O2 → CO2 + H2O; incomplete combustion (limited O2): produces CO and/or soot + H2O [1 for each combustion type with products].

Combustion — Australian environmental context (1 mark): Natural gas (predominantly methane) combustion in Australian energy generation and vehicles contributes CO2 to the enhanced greenhouse effect; bushfire smoke (incomplete combustion of biomass) produces CO and PM2.5 which are health hazards for Australian communities (e.g. 2019–20 Black Summer fires).