Chemistry • Year 12 • Module 7 • Lesson 8
Hydrocarbon Reactions Mastery
Evaluate extended-response answers at HSC Band 5–6 level — identifying errors in reasoning, conditions and Markovnikov application, and writing corrected responses.
Instructions. This worksheet presents two flawed student extended-response answers. For each, you must: (a) identify every scientific error and explain why it is wrong; (b) explain the correct chemistry; and (c) write an improved model response. This tests the same skills NESA markers use when awarding — or withholding — marks in Section III.
1. Source critique A — Alkene addition reactions and Markovnikov’s rule (7 marks)
Student Response A — to be critiqued
“When propene reacts with HBr, the reaction is called hydrohalogenation. The H attaches to the carbon with the fewer hydrogen atoms (the more substituted one) and the Br attaches to the carbon with more hydrogen atoms. This means the product is 1-bromopropane (CH2BrCH2CH3), which is the Markovnikov product. Markovnikov’s rule works because the carbocation forms on the carbon that already has fewer H atoms attached, making it more stable because it is bonded to more carbon groups.
The reaction also requires a Ni catalyst and moderate heating (about 100 °C) to proceed. Without the catalyst the activation energy is too high.
Bromine water can be used to confirm that this reaction has occurred because propene decolourises bromine water, whereas propane does not. This test shows that the product 1-bromopropane is saturated and will not decolourise bromine water.”
1.1 Identify and explain all scientific errors in Student Response A. For each error, state what is wrong and what the correct chemistry is. 4 marks
1.2 Write an improved model response for the following question: “Describe the reaction of propene with HBr. State the reaction type, the major product and the conditions. Explain why this product forms rather than the alternative.” 3 marks
2. Source critique B — Alkyne hydration and product identification (8 marks)
Student Response B — to be critiqued
“Propyne can be hydrated to form propan-1-ol. The reaction requires dilute sulfuric acid and mercury(II) sulfate catalyst. The conditions used are the same as those used for hydrating propene: steam at about 300 °C and 65 atmospheres pressure. This is because both propyne and propene undergo addition of water across their unsaturated bonds.
The product forms via Markovnikov’s rule: the –OH group attaches to C1 (the terminal carbon) because C1 has more hydrogen atoms, giving propan-1-ol (CH3CH2CH2OH).
Propan-1-ol can be confirmed as the product by testing with acidified KMnO4. If the purple solution turns brown, it confirms that a primary alcohol is present.
This reaction is industrially important; for example, at Qenos Altona (Melbourne) ethyne is hydrated to make ethanol, which is used as a fuel additive.”
2.1 Identify and explain all scientific errors in Student Response B. For each error, state what is wrong and what the correct chemistry is. 5 marks
2.2 Write an improved model response for the following question: “Describe the hydration of propyne. Identify the reaction type, all conditions required, the major product and explain why that product forms.” 3 marks
Q1.1 — Errors in Student Response A
Error 1 (Product name and Markovnikov statement): The student has the rule backwards. Markovnikov’s rule states that H attaches to the carbon with more hydrogen atoms (not fewer). For propene (CH3–CH=CH2): C3 (=CH2) has 2 H atoms; C2 (=CH–) has 1 H atom. H goes to C3, Br goes to C2. The major product is therefore 2-bromopropane (CH3CHBrCH3), not 1-bromopropane. [1]
Error 2 (Conditions — Ni catalyst and heating): Hydrohalogenation of an alkene with HBr requires no catalyst and no elevated temperature. The reaction occurs at room temperature. Ni and heat are the conditions for hydrogenation (addition of H2), not hydrohalogenation. [1]
Error 3 (Carbocation reasoning): The student claims the carbocation forms on the carbon with fewer H atoms and explains stability by “bonded to more carbon groups.” The stability explanation is partially correct (more substitution = more stable carbocation via hyperconjugation/inductive effect), but the student then uses this to justify the wrong product (1-bromopropane). The carbocation does form at C2 (more substituted, 1 H), and Br– attacks C2 — correctly giving 2-bromopropane. The reasoning was correct but was applied backwards to justify the wrong product. [1]
Error 4 (Bromine water conclusion): The bromine water test does not confirm that the reaction occurred or that the product is saturated. The test is a qualitative test for unsaturation (C=C or C≡C). The statement “this test shows that 1-bromopropane is saturated” confuses a structural conclusion (product is saturated because it has no C=C) with the purpose of the test (which is to detect C=C or C≡C in an unknown). The correct use: bromine water confirms that the starting material propene has been consumed because propene decolourises and the haloalkane product does not. [1]
Q1.2 — Model response for propene + HBr
Reaction type: hydrohalogenation (addition of HBr across the C=C double bond) [1]. Conditions: no catalyst required; room temperature; HBr gas or concentrated HBr aqueous solution [1]. Major product: 2-bromopropane (CH3CHBrCH3). Applying Markovnikov’s rule: propene is CH3–CH=CH2; C3 (=CH2) has 2 H atoms; C2 (=CH–) has 1 H atom. H attaches to C3 (more H atoms), Br attaches to C2. This gives the 2-bromo product. Mechanistically, the proton from HBr adds to C3 first, generating the more stable secondary carbocation at C2 (three carbons attached to the positive carbon vs one in the primary alternative); Br– then attacks C2 to give 2-bromopropane [1].
Q2.1 — Errors in Student Response B
Error 1 (Product class — alcohol vs ketone): Alkyne hydration does not give an alcohol as the final product. The initial product is an unstable enol that immediately undergoes keto-enol tautomerisation (enol tautomerisation) to give a ketone. For propyne, the product is propanone (CH3COCH3), not propan-1-ol. [1]
Error 2 (Conditions — incorrect temperature and pressure): The student uses the conditions for alkene hydration (steam, ~300 °C, ~65 atm). Alkyne hydration uses dilute H2SO4 and Hg2+ catalyst at ~60 °C and ambient pressure — completely different. The high temperature and pressure conditions described are for alkene (not alkyne) hydration. [1]
Error 3 (Markovnikov application reversed for product identification): The student states OH attaches to C1 (terminal carbon with more H) to give propan-1-ol. This misapplies Markovnikov’s rule: for propyne (CH3–C≡CH), C1 is the terminal carbon with 1 H, C2 is the internal carbon with 0 H. Markovnikov: OH (from H2O) attaches to C2 (more substituted = fewer H), H attaches to C1. The enol formed has –OH on C2; tautomerisation gives the ketone on C2, yielding propanone — not an alcohol at all. [1]
Error 4 (Qenos Altona — product of ethyne hydration): Hydration of ethyne (CH≡CH) gives ethanal (CH3CHO), not ethanol. Ethanal is an aldehyde (not a fuel additive). Ethanol is produced by fermentation or ethylene hydration, not by ethyne hydration. The claim about Qenos Altona also misrepresents the plant’s operations: Qenos produces polyethylene, not ethanal or ethanol. [1]
Error 5 (Acidified KMnO4 confirmation logic): The student claims KMnO4 decolourisation confirms a “primary alcohol”. Acidified KMnO4 oxidises primary and secondary alcohols, aldehydes, and alkenes/alkynes — it is not specific to primary alcohols. Moreover, since the actual product is a ketone (propanone), not an alcohol, the test’s conclusion about the product identity is doubly incorrect. [1]
Q2.2 — Model response for propyne hydration
Reaction type: alkyne hydration (addition of water across the C≡C bond) [1]. Conditions: water (H2O), dilute H2SO4 as acid catalyst, and Hg2+ (mercury(II) sulfate) as Lewis acid catalyst, ~60 °C, ambient pressure. Both H2SO4 and Hg2+ are required; neither alone is sufficient [1]. Major product: propanone (CH3COCH3). Applying Markovnikov’s rule: for propyne (CH3–C≡CH), the internal C2 is more substituted (0 H), the terminal C1 has 1 H. –OH attaches to C2, giving an unstable enol intermediate (CH3C(OH)=CH2). This enol spontaneously tautomerises to the more stable ketone (propanone, CH3COCH3) via keto-enol tautomerisation [1].