HSC Exam Practice

Sample response. Dehydration is an elimination reaction in which a molecule of water (H₂O) is removed from an alcohol molecule, forming an alkene with a C=C double bond. The H comes from a carbon adjacent to the C–OH carbon, and the OH comes from the –OH group. Conditions: concentrated H₂SO₄ (or conc. H₃PO₄) as catalyst and dehydrating agent; approximately 170°C; atmospheric pressure; distillation apparatus to remove the volatile alkene product.

Marking notes. 1 mark for defining dehydration as elimination of water to form an alkene. 1 mark for stating concentrated acid (accept either H₂SO₄ or H₃PO₄). 1 mark for stating approximately 170°C (or “heat” with the concentrated acid specified).

Sample response. CH₃CH₂CH₂OH + HBr → CH₃CH₂CH₂Br + H₂O. Reaction type: nucleophilic substitution. Organic product: 1-bromopropane (or “bromopropane”, accepting any correct IUPAC name).

Marking notes. 1 mark for correct balanced equation including H₂O as product. 1 mark for reaction type (substitution; accept “nucleophilic substitution”). 1 mark for correct product name (1-bromopropane).

Sample response. Butan-2-ol is a secondary alcohol; it is oxidised to butanone (a ketone) by K₂Cr₂O₇/H₂SO₄ under reflux. The solution changes from orange to green as Cr₂O₇²⁻ is reduced to Cr³⁺. Word equation: butan-2-ol + acidified potassium dichromate → butanone + water + chromium(III) sulfate.

Marking notes. 1 mark for identifying the product as butanone (or “a ketone”). 1 mark for correctly describing the colour change (orange → green). 1 mark for a correct word equation or explanation that includes Cr₂O₇²⁻ → Cr³⁺.

Sample response. The oxidation mechanism requires removing one H from the C–OH carbon and one H from the –OH group to form the C=O bond. In 2-methylpropan-2-ol, the C–OH carbon is bonded to three methyl groups and carries no hydrogen atoms. Because there is no C–H bond on the C–OH carbon available to break, the oxidation reaction cannot proceed, and K₂Cr₂O₇/H⁺ solution stays orange.

Marking notes. 1 mark for stating that oxidation requires removing H from the C–OH carbon. 1 mark for identifying that the C–OH carbon in a tertiary alcohol has no H atoms (three C groups instead). 1 mark for stating the consequence: no reaction, solution stays orange.

Sample response. Using distillation: the aldehyde product has a lower boiling point than the original alcohol, so as the aldehyde forms it is immediately vaporised and removed from the reaction flask before excess oxidant can react with it further. Aldehyde is the product. Using reflux: the condenser returns all vapours to the flask, so the aldehyde remains in contact with excess K₂Cr₂O₇/H⁺ for an extended time and is oxidised a second time to a carboxylic acid. Carboxylic acid is the product. The key difference is whether the aldehyde intermediate can be removed from the oxidant before further reaction occurs.

Marking notes. 1 mark for stating distillation gives aldehyde. 1 mark for explaining why (aldehyde removed from flask before further oxidation due to lower boiling point). 1 mark for stating reflux gives carboxylic acid. 1 mark for explaining why (aldehyde retained in flask; further oxidised by excess oxidant).

Sample response. Reactivity order: HI > HBr > HCl (fastest to slowest). The C–I bond formed in the product is weaker than C–Br or C–Cl (bond dissociation energy decreases down Group 17), making the iodoalkane thermodynamically easier to form. Iodide ion (I⁻) is also a better nucleophile because its outer electron cloud is more polarisable (larger, more diffuse). HCl is slowest because C–Cl is the strongest of the three bonds and Cl⁻ is the least polarisable nucleophile.

Marking notes. 1 mark for correct order (HI > HBr > HCl). 1 mark for bond-strength rationale (C–I weakest, easiest to form). 1 mark for nucleophilicity rationale or equivalent electronegativity/polarisability argument. Accept either or both chemical justifications for the second and third marks.

Compound identifications.

• A = ethene (CH₂=CH₂): dehydration (elimination) of ethanol; Step 1 balanced equation: CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, ~170°C).
• B = bromoethane (CH₃CH₂Br): electrophilic addition (hydrohalogenation) of HBr across ethene; reaction type = addition.
• C = ethanol (CH₃CH₂OH): nucleophilic substitution of bromoethane with NaOH(aq); reaction type = substitution. Note: this regenerates ethanol, confirming the synthesis is illustrative of functional group interconversions.
• D = ethanoic acid (CH₃COOH): oxidation of ethanol under reflux; Step 4 balanced equation: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O (excess K₂Cr₂O₇/H₂SO₄, reflux).

Part (a) marking (6 marks): 1 mark per correct compound name (4). 1 mark for Step 1 balanced equation. 1 mark for Step 4 balanced equation.

Part (b) overall yield (2 marks): Overall yield = 0.88 × 0.76 × 0.91 × 0.83 = 0.503 ≈ 50%. (1 mark for method: multiplying all four yields; 1 mark for correct answer in range 49–51%.)

Part (c) reflux in Step 4 (2 marks): Compound C is a primary alcohol; if distillation were used, the aldehyde intermediate (ethanal) would be collected before further oxidation, giving ethanal instead of ethanoic acid (1 mark). Using reflux keeps all components in the flask, allowing excess K₂Cr₂O₇/H₂SO₄ to fully oxidise the aldehyde intermediate to the carboxylic acid (1 mark).

Sample response. The acidified K₂Cr₂O₇ test is useful for distinguishing alcohol classes, but has important limitations. For primary alcohols, the test gives an orange-to-green colour change in both distillation and reflux setups; distillation collects an aldehyde product (lower boiling point, removed before further reaction), while reflux drives full oxidation to the carboxylic acid. For secondary alcohols, K₂Cr₂O₇/H⁺ also gives an orange-to-green change under reflux, but the product is a ketone. Ketones cannot be further oxidised under standard conditions (no H on the carbonyl carbon), so there is no further colour change. For tertiary alcohols, the solution stays orange: the C–OH carbon bears no H atom, so no C–H bond is available to break, and the Cr₂O₇²⁻ ion is not reduced. The key observations distinguishing the three classes are: whether colour change occurs at all (tertiary: no; 1° and 2°: yes); the type of organic product (1° vs 2°); and whether an additional distillate is collected (1° aldehyde, lower BP). A limitation is that the test alone cannot name the specific compound — it only classifies it. Another limitation is that non-alcohol reducing agents (e.g. aldehydes already present) would also cause the colour change, reducing specificity. In an Australian context, breathalysers used by Australian police use this chemistry: ethanol in exhaled breath reduces orange Cr₂O₇²⁻ to green Cr³⁺. The extent of colour change (measured as absorbance of orange light) is calibrated against known ethanol concentrations to give a quantitative ethanol reading, not a simple binary colour test. This example illustrates both the strength (reliable colour-change indicator for oxidisable alcohols) and the limitation (any oxidisable compound in breath would also cause a colour change, reducing specificity for ethanol alone) of the dichromate test.

Marking criteria:

• 1 mark — correctly describes the oxidation of primary alcohol: distillation gives aldehyde, reflux gives carboxylic acid; colour change orange→green in both cases.
• 1 mark — correctly describes secondary alcohol: orange→green; product is ketone; cannot be further oxidised.
• 1 mark — correctly describes tertiary alcohol: stays orange; structural reason (no H on C–OH carbon).
• 1 mark — identifies that distillation vs reflux gives different products for primary alcohols and explains the reason (aldehyde removed before further oxidation by distillation).
• 1 mark — identifies at least one limitation (e.g. cannot identify specific compound; other reducing agents also give colour change; requires calibration to be quantitative).
• 1 mark — names a specific Australian application (breathalyser / police alcohol testing) and describes the chemistry involved (Cr₂O₇²⁻ orange → Cr³⁺ green as ethanol is oxidised; quantitative reading compared to legal limit).
• 1 mark — reaches an evaluative judgement: the test is useful for classification (1°/2°/3°) and for detecting ethanol presence, but limited in specificity for compound identity; quantitative calibration is needed for legal applications.