Chemistry • Year 12 • Module 7 • Lesson 8
Hydrocarbon Reactions Mastery
Apply Markovnikov’s rule, interpret reaction data, distinguish test reactions (bromine water, Baeyer’s) and correct condition errors in novel contexts.
1. Interpret laboratory data — bromine water distinguishing tests
A NATA-accredited laboratory tested four unlabelled hydrocarbon samples (A–D) with bromine water and with acidified KMnO4 solution. Results are shown below. 8 marks
| Sample | Bromine water (Baeyer’s) | Acidified KMnO4 | Molecular formula |
|---|---|---|---|
| A | No decolourisation (stays orange) | No decolourisation (stays purple) | C4H10 |
| B | Rapid decolourisation | Rapid decolourisation (purple → brown) | C4H8 |
| C | Rapid decolourisation | Rapid decolourisation | C4H6 |
| D | No decolourisation | No decolourisation | C4H8 |
1.1 Identify the homologous series most likely represented by each sample. Justify each identification using both the test result and the molecular formula. 4 marks
1.2 A student concludes that Sample B is “definitely an alkene because it decolourises bromine water.” Assess the validity of this conclusion. 2 marks
1.3 Explain why Sample D does not decolourise bromine water despite having the same molecular formula as Sample B. 2 marks
2. Interpret graph — rate of bromine decolourisation for different hydrocarbons
The graph below shows how the absorbance of bromine (at 420 nm) changes over time when equal concentrations of bromine water are mixed with four hydrocarbons under identical conditions. Lower absorbance = more bromine consumed = faster reaction. 7 marks
Figure 2.1. Bromine absorbance (420 nm) vs time for four hydrocarbons under identical conditions. Illustrative data; cf. standard spectrophotometric unsaturation tests used in Australian NATA-accredited laboratories.
2.1 Describe the trend for propane and cyclohexane. Explain these results in terms of molecular structure. 2 marks
2.2 Compare the rate of bromine consumption for propene and propyne during the first 30 seconds. Account for any difference using knowledge of the degree of unsaturation. 3 marks
2.3 A student claims this experiment proves that propene and propyne are both alkenes. Evaluate this claim with reference to the data and reaction chemistry. 2 marks
3. Markovnikov’s rule — predict the major product
For each reaction below, predict the major organic product using Markovnikov’s rule. State which carbon receives the H and which receives the X (or OH). 6 marks (2 each)
3.1 But-1-ene + HBr → [major product = ?]
Structure of but-1-ene: CH2=CH–CH2–CH3
3.2 2-methylpropene + HCl → [major product = ?]
Structure of 2-methylpropene: (CH3)2C=CH2
3.3 Propyne + HBr (1 eq) → [major product = ?]
Structure of propyne: CH3–C≡CH
4. Case study — petroleum refinery reactions at Ampol Lytton (Brisbane)
Ampol’s Lytton refinery (one of Australia’s last remaining refineries) processes crude petroleum fractions that are rich in alkanes, with smaller quantities of alkenes and aromatic compounds. In the fluid catalytic cracking (FCC) unit, large alkane molecules are broken into smaller ones, and significant quantities of propene and butene are produced as by-products. These alkenes can undergo further industrial reactions. 5 marks
4.1 Identify the most appropriate reaction and full conditions to convert propene to propan-2-ol on an industrial scale. Justify why propan-2-ol rather than propan-1-ol is the major product. 3 marks
4.2 Butene produced at Lytton can be polymerised to polybutene, a synthetic oil additive. State the reaction type and two essential conditions (one physical, one chemical) required. 2 marks
Q1.1 — Identify homologous series
A — alkane (C4H10 = CnH2n+2; no unsaturation, so no reaction with either reagent).
B — most likely alkene (C4H8 = CnH2n; decolourises both bromine water and KMnO4, consistent with C=C). Note: a cycloalkane also has CnH2n but does not decolourise; the positive tests confirm C=C.
C — alkyne (C4H6 = CnH2n−2; rapid decolourisation of both reagents due to C≡C).
D — cycloalkane (C4H8 = CnH2n, same as alkene; but no reaction with either reagent indicates no C=C, so it is the saturated cyclic isomer, cyclobutane).
Q1.2 — Assess the student’s conclusion
The conclusion is partially valid but not fully justified [1]. Sample B decolourises bromine water, which confirms the presence of unsaturation (C=C or C≡C), but the test alone cannot distinguish an alkene from an alkyne. In this case the molecular formula (C4H8 = CnH2n) and the failure of C to show as an alkene (C has C4H6, the alkyne formula) together support the alkene identification, but bromine water alone is insufficient evidence [1].
Q1.3 — Why Sample D does not react
Sample D is a cycloalkane (cyclobutane). Cycloalkanes contain only C–C single bonds and C–H bonds — no pi bonds. Bromine water reacts by electrophilic addition across a pi bond; without a pi bond, no addition reaction occurs and the bromine solution remains orange [1]. The molecular formula C4H8 is the same as an alkene’s general formula CnH2n, but the structure is cyclic (all single bonds), not unsaturated [1].
Q2.1 — Propane and cyclohexane trend
Both show a flat line near absorbance 1.0, indicating no change over time [1]. Neither propane (alkane) nor cyclohexane (cycloalkane) undergoes electrophilic addition because they contain no pi bonds. Bromine water does not react with saturated hydrocarbons under these mild conditions (no UV light, no heat) [1].
Q2.2 — Compare propene and propyne rate
Both propene and propyne show rapid decolourisation, but propyne’s curve initially declines slightly faster (steeper at t = 0–15 s) [1]. Propyne has a triple bond (C≡C), meaning it has two pi bonds available for addition; 2 moles of Br2 can be consumed per mole of propyne (first step: Br2 adds across one pi bond giving a vinyl dihalide; second step: Br2 adds again giving a saturated dihalide). Propene has only one pi bond and consumes only 1 mole of Br2 per mole [1]. Both reach near-zero absorbance because both are present in equal concentration and sufficient bromine is present to react fully [1].
Q2.3 — Evaluate the student’s claim
The claim is incorrect [1]. The data show that both propene and propyne decolourise bromine water, but this confirms only that both contain unsaturation (pi bonds). Propyne is an alkyne, not an alkene. The bromine water test cannot differentiate between an alkene and an alkyne [1].
Q3 — Markovnikov products
3.1 But-1-ene + HBr: C1 (=CH2) has 2 H atoms; C2 (=CH–) has 1 H atom. Markovnikov: H to C1, Br to C2. Major product: 2-bromobutane (CH3CHBrCH2CH3).
3.2 2-methylpropene + HCl: The C=C is (CH3)2C= (0 H on this C) and =CH2 (2 H on this C). Markovnikov: H to =CH2 (2 H, more H), Cl to the tertiary carbon. Major product: 2-chloro-2-methylpropane ((CH3)3CCl). The tertiary carbocation intermediate is most stable.
3.3 Propyne + HBr (1 eq): The C≡C is at C1–C2. C1 is the terminal carbon (=CH, 1 H); C2 is internal (=C–CH3, 0 H). Markovnikov: H to C1 (more H), Br to C2. Major product: 2-bromopropene (CH3CBr=CH2). Note: this is a vinyl halide (haloalkene), not a haloalkane.
Q4.1 — Propene to propan-2-ol
Reaction type: alkene hydration (addition of water). Conditions: H2O (steam), H3PO4 solid acid catalyst (on silica support, or dil. H2SO4), ~300 °C, ~65 atm [1]. Markovnikov’s rule: propene is CH3–CH=CH2; C2 (=CH–) has 1 H atom and is more substituted; C3 (=CH2) has 2 H. H (from H2O) goes to C3 (more H), OH goes to C2 [1]. Product is propan-2-ol (CH3CH(OH)CH3), not propan-1-ol, because OH attaches to the more substituted (secondary) carbon per Markovnikov [1].
Q4.2 — Butene polymerisation conditions
Reaction type: addition polymerisation [1]. Physical condition: high temperature and high pressure (to overcome the activation barrier and maintain gaseous/liquid phase). Chemical condition: a peroxide initiator (or Ziegler–Natta catalyst in industrial settings) that generates free radicals to initiate the chain reaction [1].