HSC Exam Practice

Sample response. Nucleophilic substitution is a reaction in which a nucleophile (here, OH¹¯ from aqueous NaOH) attacks a carbon atom bonded to a halogen in a haloalkane and replaces the halogen, which leaves as a halide ion (X¹¯). The product is an alcohol (R—OH).

Marking notes. 1 mark for identifying that a nucleophile (or OH¹¯) replaces the halogen; 1 mark for stating the product is an alcohol (or R—OH) with the halogen leaving as a halide/NaX.

Sample response. CH&sub2;=CH&sub2; + H&sub2;O ⇌ CH&sub3;CH&sub2;OH. Conditions: H&sub3;PO&sub4; catalyst (absorbed on silica, heterogeneous); temperature ~300°C; pressure ~65 atm.

Marking notes. 1 mark for correct balanced equation with reversible arrow (⇌) — single arrow (→) loses this mark; 1 mark for H&sub3;PO&sub4; catalyst and ~300°C; 1 mark for ~65 atm. A → instead of ⇌ loses 1 mark.

Sample response. (1) Yeast / zymase enzyme: acts as the biological catalyst that converts glucose to ethanol; without it the reaction is too slow to proceed. (2) ~35°C: the optimum temperature for enzyme activity; above ~45°C zymase denatures and loses function. (3) Anaerobic conditions (no oxygen): with oxygen present, yeast undergoes aerobic respiration (glucose + O&sub2; → CO&sub2; + H&sub2;O) instead of fermentation, producing no ethanol.

Marking notes. 1 mark per condition + explanation (3 marks total). Each must name the condition and explain why it is essential. Stating “~35°C” without explaining why scores 0 for that condition.

Sample response. Aqueous NaOH (NaOH dissolved in water) provides OH¹¯ as a nucleophile in a polar aqueous environment, which replaces the halogen in a haloalkane via nucleophilic substitution to give an alcohol and a sodium halide salt: R—X + NaOH(aq) → R—OH + NaX. Alcoholic NaOH (NaOH dissolved in ethanol) acts as a strong base in a less polar environment, favouring elimination: OH¹¯ removes a hydrogen from a carbon adjacent to the halogen-bearing carbon, forming a C=C double bond and producing an alkene and HX: R—CH&sub2;—CHX—R′ → R—CH=CH—R′ + HX.

Marking notes. 1 mark for aqueous NaOH + substitution + alcohol product; 1 mark for alcoholic NaOH + elimination + alkene product; 1 mark for explaining that the solvent (water vs ethanol) determines the reaction pathway / product. Acceptable to state: aqueous = substitution; alcoholic = elimination, with correct products.

Sample response. The hydration equation CH&sub2;=CH&sub2; + H&sub2;O ⇌ CH&sub3;CH&sub2;OH has 2 moles of gas on the left (ethene + steam) and effectively 0 moles of gas on the right (ethanol condenses as a liquid at process conditions). By Le Chatelier’s Principle, increasing pressure shifts the equilibrium to the side with fewer moles of gas — in this case, the right side (products). Therefore, high pressure (~65 atm) shifts the equilibrium to the right, increasing the equilibrium yield of ethanol.

Marking notes. 1 mark for correct mole count (2 moles gas left, 0 right / fewer moles on product side); 1 mark for Le Chatelier’s Principle applied correctly (increasing pressure favours fewer moles of gas); 1 mark for explicitly stating the result (equilibrium shifts right / ethanol yield increases).

Sample response. The forward hydration reaction is exothermic. By Le Chatelier’s Principle, decreasing temperature favours the exothermic (forward) direction, shifting the equilibrium right and increasing ethanol yield. The temperature that maximises equilibrium yield would therefore be as low as possible. However, lower temperature also means a slower reaction rate and lower production throughput, making the process economically unviable. 300°C is chosen as a compromise: the equilibrium yield is below the maximum (shifted left compared to lower temperatures) but the reaction rate is sufficiently fast for continuous industrial production. This is analogous to the Haber process for ammonia synthesis.

Marking notes. 1 mark for identifying the forward reaction as exothermic and that lower temperature would increase equilibrium yield (Le Chatelier); 1 mark for explaining that low temperature also decreases reaction rate and throughput; 1 mark for stating that 300°C balances these two competing effects (rate vs yield compromise). Analogous to Haber process: accept as evidence of understanding but not required.

Sample response. Difference 1 — purity: Facility A produces ~95% ethanol directly from the reactor, while Facilities B and C produce only ~10–12%. This reflects the chemical nature of each method: hydration produces ethanol as a near-pure product (ethanol vapour condenses from the reactor stream); fermentation produces a dilute aqueous solution because ethanol is produced in water at only ~15% maximum before yeast toxicity stops the reaction. [2 marks: 1 data identification + 1 chemistry link] Difference 2 — operating temperature: Facility A operates at 300°C, while Facilities B and C operate at 32–35°C. This reflects that alkene hydration requires high temperature to achieve an acceptable reaction rate (rate-yield compromise), whereas fermentation is enzyme-catalysed at near-body temperature, reflecting the optimum temperature for zymase activity. [2 marks: 1 data identification + 1 chemistry link]

Marking notes. 2 marks per difference (1 for identifying from data, 1 for chemistry link). Accept any two valid differences from the table (purity, rate, temperature, feedstock, production volume) provided the chemistry is correctly explained.

Sample response. Additional processing step: fractional distillation is required to concentrate the ~10% ethanol from fermentation to the 96% purity required for pharmaceutical grade. Ethanol (bp 78.4°C) and water (bp 100°C) are separated by repeated distillation fractionation.

Calculation: Facility B produces 100 ML/year at ~10% ethanol by volume. Volume of ethanol in fermented liquid = 100 × 0.10 = 10 ML = 10,000,000 L of pure ethanol equivalent. After 90% distillation efficiency: 10,000,000 × 0.90 = 9,000,000 L = 9 ML of 100% ethanol recovered. At 96% purity, total volume of 96% ethanol product = 9,000,000 / 0.96 ≈ 9.4 ML of 96% ethanol solution per year.

Marking notes. 1 mark for identifying fractional distillation with correct reasoning (different boiling points); 1 mark for correct calculation of ethanol volume from raw ferment (100 × 0.10 = 10 ML); 1 mark for applying 90% efficiency (10 × 0.90 = 9 ML pure ethanol); 1 mark for correct final volume at 96% purity (9/0.96 ≈ 9.4 ML). Accept answers in L or ML provided units are consistent. Allow full marks if working is correct and arithmetic errors are minor.

Sample response. Moles of glucose = 180 g ÷ 180 g mol¹¯ = 1.00 mol. From the equation, 1 mol glucose produces 2 mol ethanol. Moles of ethanol = 2.00 mol. Mass of ethanol = 2.00 × 46 = 92 g.

Marking notes. 1 mark for correct moles of ethanol (2.00 mol); 1 mark for correct mass (92 g). Working must be shown. Award 1/2 if the mole ratio error is the only error (consequential marking).

Sample response. Theoretical ethanol produced = 92 g (from 2.2a). Volume of ethanol = 92 g ÷ 0.789 g mL¹¯ = 116.6 mL. Total volume of solution ≈ 1000 mL (water) + 116.6 mL (ethanol) ≈ 1116.6 mL. Ethanol concentration = 116.6 / 1116.6 × 100% ≈ 10.4% by volume. Since 10.4% < 15%, the 15% toxicity limit would NOT be reached with 1.00 mol glucose in 1 L water — fermentation would theoretically proceed to completion for this batch. [However, if double the glucose were used, ethanol would approach 15% and fermentation would stop before completion.]

Marking notes. 1 mark for correctly converting 92 g to volume of ethanol (116.6 mL); 1 mark for calculating ethanol % by volume correctly (~10.4%); 1 mark for correct conclusion with reasoning (10.4% < 15%, so fermentation would not stop before all glucose is used in this scenario). Consequential marking applies if 2.2(a) answer is carried forward.

Assumption: The density of the ethanol–water mixture is assumed to equal the density of pure ethanol (0.789 g mL¹¯), or that the volume of the solution equals water + ethanol volumes additively. (Accept: 100% fermentation conversion is assumed in part (a); yeast remains fully viable throughout; no CO&sub2; occupies solution volume.) Limitation: Fermentation has a maximum ethanol concentration limit (~15%) because ethanol is toxic to yeast above this concentration, meaning glucose cannot be fully converted if the initial glucose concentration is very high — the fermentation stops before completion, so theoretical yield is never fully achieved at high glucose concentrations. [1 mark each]

Sample response. The claim is partly supported but too absolute to be correct as stated. Fermentation has genuine sustainability advantages, but whether it is “always” more sustainable than catalytic hydration depends on context including feedstock, land use, water use, and energy source for purification.

Fermentation: C&sub6;H&sub1;&sub2;O&sub6; → 2C&sub2;H&sub5;OH + 2CO&sub2; (yeast/zymase, ~35°C, anaerobic). Alkene hydration: CH&sub2;=CH&sub2; + H&sub2;O ⇌ CH&sub3;CH&sub2;OH (H&sub3;PO&sub4;, ~300°C, ~65 atm; reversible equilibrium). The hydration reaction uses a ⇌ arrow because it is a reversible equilibrium: both forward (hydration) and reverse (dehydration) reactions occur simultaneously. Le Chatelier’s Principle is applied to explain the choice of high pressure (shifts equilibrium right) and the rate-yield compromise at 300°C (exothermic forward reaction favoured by lower temperature, but lower temperature reduces rate). Fermentation is irreversible (→) and stops due to ethanol toxicity to yeast at ~15%, not because of equilibrium; Le Chatelier does not apply.

Sustainability arguments for fermentation: (1) renewable feedstock — sugarcane in Queensland or wheat starch at the Manildra Group Nowra plant is grown using solar energy and photosynthesis, unlike ethene derived from non-renewable petroleum cracking; (2) near-carbon-neutral — CO&sub2; released during fermentation is offset by CO&sub2; absorbed by the crop during growth; (3) low energy — 35°C vs 300°C + 65 atm for hydration means dramatically lower energy consumption per litre of ethanol produced.

Sustainability arguments against the claim being absolute: (1) low purity — fermentation yields only ~10–15% ethanol; reaching industrial or pharmaceutical grade requires energy-intensive fractional distillation, which partially erodes the energy advantage; (2) land use — large-scale fermentation requires growing feedstock crops, which may involve clearing land, irrigation (high water use), and competition with food crops (food vs fuel debate); (3) scale — industrial hydration (as at Facility A type plants) can produce very large volumes continuously, whereas fermentation is batch-processed and slower, potentially requiring more land and water per unit of product at very large scale. Conclusion: the claim is overstated. In contexts like Brazil or Queensland where renewable feedstock is abundant and cheap, fermentation is genuinely more sustainable. In contexts requiring very high purity, continuous supply, or where distillation energy comes from fossil fuels, the sustainability advantage is reduced or reversed. A context-dependent evaluation is required, not a universal ranking.

Marking criteria: 1 — Balanced fermentation equation with correct coefficients (→) and conditions; 1 — Balanced hydration equation with ⇌ and correct conditions; 1 — Le Chatelier argument correctly applied to hydration (not fermentation) with correct direction of effect for pressure or temperature; 1 — At least two sustainability advantages of fermentation with reasoning (renewable feedstock, low energy, near-carbon-neutral); 1 — At least one valid limitation of fermentation’s sustainability (distillation energy, land use, water use, or purity argument); 1 — Reference to at least one Australian or global industrial context (Manildra Group Nowra / Queensland bioethanol / Brazil sugarcane / German industrial hydration); 1 — Explicit evaluative judgement that rejects the “always” claim and frames sustainability as context-dependent, not a universal ranking.