HSC Exam Practice

Sample response. Markovnikov’s rule states that in the addition of HX (or H₂O) to an unsymmetrical alkene or alkyne, the hydrogen atom attaches to the carbon atom that already has the greater number of hydrogen atoms bonded to it. It applies to addition reactions of HX and H₂O (hydrohalogenation and hydration).

Marking notes. 1 mark for correctly stating that H goes to the carbon with more H atoms; 1 mark for stating the reaction type (addition / hydrohalogenation / hydration).

Sample response. A vicinal dihalide has two halogen atoms on adjacent (neighbouring) carbon atoms; it forms when Br₂ (or Cl₂) adds across the C=C double bond of an alkene (halogenation). A geminal dihalide has both halogen atoms on the same carbon atom; it forms by two sequential additions of HX to an alkyne (two-step hydrohalogenation), with Markovnikov’s rule applied at each step so both additions direct X to the same carbon.

Marking notes. 1 mark for correct definition of vicinal (adjacent C atoms); 1 mark for correct starting material and reaction (alkene + X₂); 1 mark for correct definition of geminal (same C atom); 1 mark for correct starting material and reaction (alkyne + 2 HX, Markovnikov each step).

Sample response. Reagents: water (H₂O). Catalysts: dilute H₂SO₄ AND Hg²⁺ (mercury(II) sulfate — both are required). Temperature: ~60 °C. Pressure: ambient. Both H₂SO₄ and Hg²⁺ must be specified; neither alone is sufficient. The product belongs to the ketone functional group class (the initial enol intermediate tautomerises to the ketone).

Marking notes. 1 mark for dil. H₂SO₄ AND Hg²⁺ (both required for this mark); 1 mark for ~60 °C and ambient pressure; 1 mark for identifying the product as a ketone.

Sample response. A catalyst increases the rate of a reaction by providing an alternative pathway of lower activation energy and is not consumed in the reaction — it is regenerated. UV light is consumed (absorbed) by Cl₂ (or Br₂) molecules, splitting them into two chlorine (or bromine) radicals; UV light does not appear in the product and is not regenerated. It provides the initiation energy to start the chain reaction rather than lowering the activation energy throughout.

Marking notes. 1 mark for explaining that a catalyst is not consumed / is regenerated; 1 mark for explaining that UV is consumed (absorbed) to produce radicals and initiate the chain, not regenerated.

Sample response. Ethene (CH₂=CH₂) is suitable because: (1) it contains a C=C double bond (pi bond) that can open to allow new C–C bonds to form between monomers; and (2) it is a small, relatively simple molecule that can link repetitively without producing a by-product (addition polymerisation, unlike condensation).

Marking notes. 1 mark for C=C double bond (pi bond) that opens during polymerisation; 1 mark for any valid second feature (small size / no by-product / ability to repeat).

Sample response. Bromine water reacts with unsaturated compounds by electrophilic addition across a pi bond (C=C or C≡C); the Br₂ is consumed and the orange colour disappears. Alkenes (one C=C pi bond) and alkynes (two pi bonds in C≡C) both contain pi bonds and therefore decolourise bromine water. Cycloalkanes have the same molecular formula (C_nH_2n) as alkenes but contain only C–C single bonds (sigma bonds) in a ring — there is no pi bond available for addition. Without a pi bond, no electrophilic addition occurs and bromine water remains orange.

Marking notes. 1 mark for identifying bromine water reacts by addition across a pi bond; 1 mark for stating both alkenes and alkynes possess pi bonds (hence decolourise); 1 mark for explaining cycloalkanes lack pi bonds (sigma bonds only in the ring) and therefore do not react.

Sample response. As temperature increases from 0 °C to 80 °C, the percentage yield of 2-bromobutane (Markovnikov product) decreases from 91% to 64%, while the yield of the anti-Markovnikov product (3-bromobutane) increases from 9% to 36%.

Marking notes. 1 mark for describing the decrease in Markovnikov product yield (with values); 1 mark for describing the corresponding increase in anti-Markovnikov product yield (with values).

Sample response. But-2-ene is CH₃CH=CHCH₃. The C=C carbons are C2 (bonded to CH₃ and CH₃CH—, i.e. one H, more substituted) and C3 (also one H, bonded to CH₃). Because the molecule is symmetric about the double bond, both C2 and C3 have the same substitution. By Markovnikov’s rule, H adds to the carbon with more H atoms already attached; both carbons here have one H each, so either C can receive the Br. However, by convention and stability reasoning, Br adds to the more substituted internal carbon (C2), giving 2-bromobutane [1]. The C2 position is the Markovnikov addition site [1]. 2-bromobutane is the expected major product because this is the only addition product when both carbons of the C=C are equivalent or when C2 is deemed more substituted [1].

Marking notes. 1 mark for applying Markovnikov’s rule correctly (H to the carbon with more H, Br to less H); 1 mark for identifying C2 as the Br-receiving carbon; 1 mark for naming 2-bromobutane as the major product.

Sample response. But-2-ene: C₄H₈. C: 4C → 4CO₂ [½ mark]. H: 8H → 4H₂O [½ mark]. O right: 8 + 4 = 12 → O₂ = 6 [½ mark]. Balanced equation: C₄H₈ + 6O₂ → 4CO₂ + 4H₂O [½ mark for correct final equation].

Marking notes. 1 mark for correct working using C→H→O method; 1 mark for the correctly balanced final equation.

Error: The product is wrong. The conditions given (H₃PO₄, steam, 300 °C, 65 atm) are the correct conditions for alkene hydration, and the reaction is addition of H₂O across propene. By Markovnikov’s rule: C3 (=CH₂) has 2 H; C2 (=CH–) has 1 H. H → C3, OH → C2. The correct product is propan-2-ol (CH₃CH(OH)CH₃), not propan-1-ol.

Marking notes. 1 mark for identifying the product error (propan-1-ol is anti-Markovnikov); 1 mark for writing the correct product (propan-2-ol) with Markovnikov justification.

Error: The conditions are wrong. The product (propanone) and the equation are correct for alkyne hydration, but the conditions stated (H₃PO₄, steam, 300 °C, 65 atm) are the conditions for alkene hydration, not alkyne hydration. Correct conditions for alkyne hydration: water (H₂O), dilute H₂SO₄ AND Hg²⁺, ~60 °C, ambient pressure. H₃PO₄ at high temperature would not catalyse the alkyne’s pi bond activation; Hg²⁺ is essential.

Marking notes. 1 mark for identifying that the conditions are those for alkene (not alkyne) hydration; 1 mark for providing the correct conditions (dil. H₂SO₄ + Hg²⁺, ~60 °C).

Error: The product is wrong. Br₂ adds across the C=C double bond of propene by halogenation; both Br atoms add, one to each carbon of the C=C. The product is a vicinal dihalide: 1,2-dibromopropane (CH₃CHBrCH₂Br). The student has written CH₃CH(Br)CH₃ (2-bromopropane), which would be the product of HBr addition (hydrohalogenation), not Br₂ addition. The conditions (no catalyst, r.t.) are correct for halogenation.

Marking notes. 1 mark for identifying the product is wrong (CH₃CH(Br)CH₃ is the HBr product, not the Br₂ product); 1 mark for writing the correct product, 1,2-dibromopropane (both Br atoms added, vicinal dihalide).

Sample response. Step 1 (full hydrogenation of propyne to propane) is valid: Ni catalyst at 150–200 °C and high pressure with excess H₂ gives propane. [1] Step 2 (hydration of propane to propan-2-ol) is flawed: propane is a saturated alkane with no C=C or C≡C bond; it cannot undergo addition of water. Alkene hydration requires a C=C bond to be present. The conditions given (H₃PO₄, steam, high pressure) are specifically for alkene hydration and are inapplicable to an alkane. [2] Step 3 (propan-2-ol + HBr → 2-bromopropane) would be valid if propan-2-ol were actually produced, but since Step 2 fails, Step 3 cannot occur either. The reaction of an alcohol with HBr to give a haloalkane (and water) is an elimination/substitution and the conditions are broadly reasonable, though the reaction typically requires concentrated HBr at elevated temperature. [1] More efficient alternative pathway from propyne to 2-bromopropane: direct hydrohalogenation of propene. First partially hydrogenate propyne to propene using the Lindlar catalyst (1 eq H₂, mild conditions, Lindlar catalyst — poisoned Pd). Then add HBr to propene (no catalyst, r.t.): by Markovnikov’s rule, H → C3, Br → C2, giving 2-bromopropane directly in two steps rather than three, and avoiding the need for an alkane intermediate that cannot be hydrated. [2] Alternatively, two-step HBr addition to propyne gives 2,2-dibromopropane (not the target), so that pathway is not appropriate. [1 overall evaluation mark]

Marking notes. 1 mark — identifies Step 1 as valid with correct reasoning. 2 marks — identifies Step 2 as flawed (propane is saturated, cannot undergo addition of water; conditions are only for alkene) and explains correctly. 1 mark — assesses Step 3 (valid in principle if Step 2 were possible). 2 marks — proposes a valid more-efficient pathway from propyne to 2-bromopropane (partial hydrogenation to propene using Lindlar, then HBr addition via Markovnikov). 1 mark — evaluative overall judgement (the student’s pathway fails at Step 2 and the alternative avoids the unworkable alkane intermediate).

Sample response. The claim contains three errors. Error 1: the monomer is ethene (CH₂=CH₂), not ethane (CH₃–CH₃). Ethane is a saturated alkane with no C=C double bond and cannot undergo addition polymerisation. [1] Error 2: it is the C=C pi bond that opens during addition polymerisation, not a C–C single bond. Single sigma bonds are very strong and stable; they do not spontaneously break at the temperatures and pressures of industrial polyethylene production. [1] Error 3: a catalyst is required (and the reaction is not spontaneous at room temperature). Industrial polyethylene production uses either Ziegler–Natta catalysts (TiCl₄/AlEt₃) at moderate temperature and pressure (HDPE process) or peroxide radical initiators at high temperature and high pressure (LDPE process). Without an initiator or catalyst to generate radicals (or coordinate the monomer), the polymerisation does not proceed. [1] Correct chemistry: ethene monomers are linked by opening of the C=C pi bond, forming a repeating −CH₂–CH₂− backbone. Each monomer contributes both its carbons to the growing chain, and no by-product is released (addition polymerisation). High temperature, high pressure, and a peroxide initiator or Ziegler–Natta catalyst are required for industrial production. [1] Evaluative statement: the claim confuses ethane (alkane) with ethene (alkene), misidentifies which bond breaks, and incorrectly states no catalyst is needed — three fundamental errors that reflect a misunderstanding of both hydrocarbon structure and polymerisation mechanism. [1]

Marking notes. 1 mark — identifies monomer error (ethane vs ethene). 1 mark — identifies bond-type error (pi bond opens, not C–C sigma). 1 mark — identifies catalyst/initiator error and states what is actually required. 1 mark — describes correct addition polymerisation chemistry (pi bond opening, backbone formation, no by-product). 1 mark — evaluative summary or explicit statement that the claim contains three significant errors reflecting fundamental misunderstanding.