Chemistry • Year 12 • Module 7 • Lesson 20

Organic Reactions Mastery

Critique Band 3–4 student responses to complex synthesis questions, identify every missing or incorrect element, and rewrite each response to Band 6 standard. Then tackle two high-level source critiques requiring evaluation of scientific claims against HSC evidence.

Master • Band 5–6

1. Critique and rewrite — Band 3–4 response A (7 marks)

Band 5–6

Original question (5 marks): Starting from but-1-ene and methanol, outline a synthesis of methyl butanoate. For each step write a balanced equation, full conditions, and explain why those conditions are chosen.

Student A’s response (Band 3–4 — received 2/5)

Step 1: Add water to but-1-ene using sulfuric acid at high temp. Product: butan-1-ol.
Step 2: Oxidise butan-1-ol using K₂Cr₂O₇. Product: butanoic acid.
Step 3: React butanoic acid with methanol to make methyl butanoate. Use heat.

1.1 List every error or omission in Student A’s response. Use the Band 6 four-component format from Lesson 20 Card 3 as your checklist. 3 marks

1.2 Rewrite the response to Band 6 standard. Include: step heading, balanced equation, full conditions box (reagent / catalyst / conditions / equipment), product name, observable colour changes where applicable, one explanation sentence per step for why those conditions produce the desired intermediate. 4 marks

Checklist: (i) Is the hydration arrow reversible? (ii) Are both catalyst and pressure stated for hydration? (iii) Is it distillation (aldehyde) or reflux (acid)? (iv) Is the esterification arrow reversible? (v) Is conc. H₂SO₄ named as catalyst? (vi) Are balanced equations present for every step?

2. Critique and rewrite — Band 3–4 response B (8 marks)

Band 5–6

Original question (6 marks): A chemist wishes to convert propene (CH₃CH=CH₂) into sodium propanoate (CH₃CH₂COO⁻Na⁺) in the fewest possible steps. Outline the complete pathway, write equations, and state all conditions. Explain the purpose of each step.

Student B’s response (Band 3–4 — received 3/6)

Step 1: Hydrate propene with water and an acid catalyst at high temperature to get propan-1-ol or propan-2-ol (Markovnikov’s rule — would give propan-2-ol).
Step 2: Oxidise propan-2-ol with K₂Cr₂O₇ to give propanoic acid.
Step 3: React propanoic acid with NaOH to give sodium propanoate.

2.1 Student B makes a significant chemical error in Step 2. Identify this error and explain fully why it makes the pathway fail. State what actually happens in Step 2 as written by Student B. 3 marks

2.2 Explain whether Student B’s Step 1 correctly applies Markovnikov’s rule, and whether propan-2-ol or propan-1-ol is the correct hydration product of propene. 2 marks

2.3 Rewrite the complete pathway to Band 6 standard. Use a different starting point for Step 2 that avoids the dead end, and include all four components per step (equation, conditions, product, explanation). 3 marks

Key insight: Propene hydration under Markovnikov = propan-2-ol (secondary alcohol). Secondary alcohol + K₂Cr₂O₇ → ketone (propanone), NOT propanoic acid. Ketones are dead ends. To reach propanoic acid you need a primary alcohol (propan-1-ol) — requires anti-Markovnikov addition, or a different starting material. Alternative: use propene + HBr → 1-bromopropane (anti-Markovnikov would give 1-BrPr); or add to step 1 a different route via propan-1-ol.

3. Source critique A — student study guide claim (6 marks)

Band 5–6

Source: Fictional student study guide — “Year 12 Chemistry Quick Notes” (for critique purposes only)

“The oxidation of any alcohol with acidified potassium dichromate will always produce a carboxylic acid, because the dichromate is a powerful oxidising agent. The orange-to-green colour change confirms that carboxylic acid has formed. Tertiary alcohols react more slowly, but eventually give carboxylic acids under strong conditions. Esters can also be easily oxidised to carboxylic acids using the same reagent.”

3.1 Identify all three scientific errors in this passage. For each error: name the specific false claim, explain the correct chemistry, and describe a simple experimental observation that would show the claim is false. 6 marks (2 per error)

Errors to find: (1) Claims ALL alcohols → carboxylic acid. (2) Claims tertiary alcohols eventually react under “strong” conditions. (3) Claims esters can be oxidised to carboxylic acids by K₂Cr₂O₇.

4. Source critique B — science news article claim (6 marks)

Band 5–6

Source: Fictional online science news article, 2024 (for critique purposes only)

“Australian researchers have developed a new single-step process to convert ethane directly into ethanol by adding water to the molecule. The reaction uses a phosphoric acid catalyst at standard atmospheric pressure and room temperature. Because it goes to completion, it produces a 100% yield of ethanol — a breakthrough over traditional multi-step fermentation, which typically yields less than 70% ethanol. The new process means industrial alcohol production can now skip fermentation entirely.”

4.1 This article contains multiple scientific errors and misrepresentations about organic chemistry. Identify three specific errors. For each: state the false claim, correct the chemistry with reference to HSC Module 7 content, and explain how you would verify the correct chemistry experimentally or by reference to reaction conditions. 6 marks (2 per error)

Errors to find: (1) Ethane (alkane) cannot undergo direct hydration — only alkenes can. (2) Hydration of alkene is NOT 100% yield — it is a reversible equilibrium ⇌ and requires high pressure (not standard atmospheric). (3) Fermentation comparison is misleading — the “<70%” claim conflates equilibrium yield with actual industrial recovery; fermentation converts glucose, not ethane.

Marking criteria — Do not peek before attempting

Q1.1 — Errors in Student A (marking criteria)

Award 1 mark per valid error or omission identified, up to 3 marks. Acceptable errors include:

Step 1: “high temp” is incomplete — must specify HIGH PRESSURE (~65 atm); arrow should be ⇌ (reversible equilibrium); catalyst not specified (H₃PO₄ or dil. H₂SO₄); no balanced equation.
Step 2: “Oxidise… to butanoic acid” — full conditions missing (H₂SO₄ as acidifying agent, REFLUX, excess oxidant); K₂Cr₂O₇ formula incomplete (subscript error common); no balanced equation; no colour-change observation.
Step 3: “Use heat” — conc. H₂SO₄ not named as catalyst; arrow must be ⇌; no equation; no yield note; no explanation of why catalyst used.

Q1.2 — Band 6 rewrite criteria for Student A

Award 1 mark per step for equation + conditions + explanation, up to 4 marks. Model answer:

Step 1: Hydration — but-1-ene → butan-1-ol (or butan-2-ol, Markovnikov).
CH₃CH₂CH=CH₂ + H₂O ⇌ CH₃CH₂CH(OH)CH₃ (Markovnikov — butan-2-ol).
Conditions: H₂O (steam), H₃PO₄ catalyst, ~300°C, HIGH PRESSURE (~65 atm), high-pressure reactor.
Explanation: High pressure is required because the reaction is gas-phase and the equilibrium favours the alcohol product at elevated pressure; reversible arrow required as equilibrium yield is <100%.

Step 2: Full oxidation — butan-2-ol → butanone (NOTE: this is a DEAD END — secondary alcohol gives ketone).
Teacher note: A fully Band 6 response recognises that butan-2-ol (Markovnikov product) gives butanone (a dead end), not butanoic acid. To reach butanoic acid the student must plan to use an anti-Markovnikov route (via butan-1-ol) or note the limitation. Award 4/4 only if the student addresses this.
If using butan-1-ol (anti-Markovnikov or noting this requires a different route): CH₃CH₂CH₂CH₂OH + 2[O] → CH₃CH₂CH₂COOH + H₂O. Conditions: K₂Cr₂O₇/H₂SO₄ (excess, acidified), heat under reflux, reflux condenser. Observable: orange → green. Explanation: Excess oxidant + reflux prevents removal of intermediate aldehyde, ensuring full oxidation to carboxylic acid.

Step 3: Esterification — butanoic acid + methanol → methyl butanoate.
CH₃CH₂CH₂COOH + CH₃OH ⇌ CH₃CH₂CH₂COOCH₃ + H₂O.
Conditions: conc. H₂SO₄ (catalyst, a few drops), heat under reflux, reflux condenser. Arrow: ⇌ (reversible). Explanation: H₂SO₄ acts as acid catalyst (activates carboxylic acid for nucleophilic attack) AND dehydrating agent (absorbs water product, shifting equilibrium right — Le Chatelier). Yield <100% because reaction is reversible.

Q2.1 — Error in Student B, Step 2 (marking criteria)

Award 3 marks for: (1) Identifying the specific error (1 mark): propan-2-ol is a secondary alcohol — K₂Cr₂O₇ oxidises secondary alcohols to ketones, not carboxylic acids. (2) Explaining why it fails (1 mark): the product is propanone (a ketone = dead end), not propanoic acid; ketones cannot be further oxidised under Module 7 conditions. (3) What actually happens (1 mark): K₂Cr₂O₇ turns orange→green, and propanone (CH₃COCH₃) forms — the synthesis terminates here without reaching the carboxylic acid target.

Q2.2 — Markovnikov’s rule check (marking criteria)

Award 2 marks: (1) Student B correctly applies Markovnikov’s rule — the OH attaches to the carbon with more H atoms (C-2 of propene), giving propan-2-ol (secondary). (2) Propan-2-ol IS the correct hydration product of propene; Student B is correct on this point. The error is downstream (in Step 2, treating a secondary alcohol as if it were primary).

Q2.3 — Band 6 rewrite for Student B (marking criteria)

Award 1 mark per step, 3 steps maximum. Model pathway to reach sodium propanoate from propene via primary alcohol route:

Option A — via 1-bromopropane intermediate (3 steps):
Step 1: CH₃CH=CH₂ + HBr → CH₃CH(Br)CH₃ (Markovnikov: 2-bromopropane). [Or use anti-Markovnikov route for 1-bromopropane, but this is not covered in Module 7 — best to note the limitation.]
Alternative approach: Step 1: Hydrate propene under Markovnikov → propan-2-ol. Step 2: Accept that you get propanone (dead end). For Band 6: explicitly state “propene hydration under Markovnikov gives a secondary alcohol dead end; to reach propanoic acid a different route is required.” Award full marks for recognising and explaining the dead end, then proposing an alternative valid route (e.g. use propan-1-ol as starting material, or note that CH₃CH₂COOH can be made from propan-1-ol via oxidation).
Step 3 (all routes): CH₃CH₂COOH + NaOH(aq) → CH₃CH₂COO⁻Na⁺ + H₂O. Conditions: NaOH(aq), no catalyst, neutralisation (no heat required). Explanation: NaOH is a strong base that fully deprotonates the carboxylic acid (acid-base reaction, single arrow →, irreversible).

Q3 — Source critique A (marking criteria)

Award 2 marks per error (1 for identifying false claim + 1 for correct chemistry + experimental evidence). Three errors:

Error 1: “Any alcohol always gives carboxylic acid.” False: secondary alcohols give ketones (dead end); tertiary alcohols give no reaction. Correction: only primary alcohols (with excess oxidant + reflux) give carboxylic acids. Experimental evidence: treat butan-2-ol (secondary) with K₂Cr₂O₇/H₂SO₄ → orange turns green but NO carboxylic acid forms (product is butanone, a ketone, which gives no Tollens’ silver mirror).

Error 2: “Tertiary alcohols react more slowly but eventually give carboxylic acids.” False: tertiary alcohols are completely resistant to K₂Cr₂O₇/H₂SO₄ under Module 7 conditions; the solution stays orange regardless of concentration or reaction time. Correction: no oxidation is possible because the tertiary carbon has no C–H bond to break. Experimental evidence: treat 2-methylpropan-2-ol with excess K₂Cr₂O₇/H₂SO₄ under extended reflux — solution remains orange; no acidic product detected (no fizzing with Na₂CO₃).

Error 3: “Esters can also be easily oxidised to carboxylic acids using the same reagent.” False: esters do not contain an oxidisable C–H group adjacent to a heteroatom; K₂Cr₂O₇/H₂SO₄ does not oxidise esters. Correction: esters are hydrolysed (not oxidised) — ester + H₂O → carboxylic acid + alcohol (using dil. H₂SO₄, reflux) or saponified (NaOH, reflux). Experimental evidence: treat ethyl ethanoate with K₂Cr₂O₇/H₂SO₄ under reflux — solution remains orange; no colour change, confirming no oxidation.

Q4 — Source critique B (marking criteria)

Award 2 marks per error (1 claim + 1 correction + verification). Three errors:

Error 1: “Convert ethane to ethanol by adding water.” False: ethane is an alkane — it has no C=C double bond and does not undergo hydration. Only alkenes (with C=C) react with water in a hydration reaction. Correction: to produce ethanol from ethane, you would first need to dehydrogenate or crack ethane to ethene (industrial process), then hydrate ethene. Verification: attempt to react ethane gas with steam/H₃PO₄ at 300°C — no colour change in bromine water, no alcohol product detected by IR spectroscopy.

Error 2: “100% yield; room temperature and atmospheric pressure.” False: alkene hydration is a reversible reaction (⇌) and never achieves 100% yield; it requires HIGH PRESSURE (~65 atm) and ~300°C. At room temperature and atmospheric pressure the reaction would not proceed at any useful rate. Verification: industrial hydration of ethene (Qenos, Altona) operates at ~300°C and ~65 atm — these conditions are publicly documented and are required to achieve practical (but still sub-100%) yields. At atmospheric pressure and room temperature, almost no ethanol is produced.

Error 3: “Fermentation typically yields less than 70% ethanol.” Misleading comparison: fermentation converts glucose to ethanol — a different feedstock (grain/sugar vs ethane). The <70% figure conflates equilibrium yield with industrial recovery rate; modern distillation recovers >99% purity ethanol. The comparison implies fermentation is inefficient vs. this “new process” but the feedstocks are entirely different. Correction: fermentation is the dominant industrial route for renewable ethanol (Manildra Group, NSW, uses grain starch); the comparison with a non-existent single-step ethane-to-ethanol process is scientifically unfounded and misleading. Verification: Manildra Group publicly reports ~65 ML/year ethanol from grain fermentation at commercial scale, demonstrating viability.