Lock in every Module 7 reaction, conditions, equipment, and colour changes, and learn to write Band 6 multi-step synthesis responses under exam pressure.
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
A Band 6 response to a 7-mark pathway question typically contains: seven balanced equations, fourteen conditions boxes, seven named intermediates, and correct arrows, all written under exam conditions in under 12 minutes.
Before reading: open your reaction map from L19 and time yourself. How long does it take you to write, from memory, the reagents and conditions for every reaction in Module 7? That time is your current exam readiness.
This is the authoritative reference table. Study it, then cover it and reproduce it from memory. ★ Priority rows (highlighted) are the most frequently penalised.
| Reaction | Starting material | Reagent | Catalyst | Conditions / temp | Equipment | Arrow | Product(s) |
|---|---|---|---|---|---|---|---|
| Hydrogenation (full) | Alkene or alkyne | H₂ (gas; 2 eq for alkyne) | Ni, Pd, or Pt | ~150–200°C, high pressure | Pressure vessel | → | Alkane |
| Partial hydrogenation | Alkyne | H₂ (gas, 1 eq) | Lindlar (poisoned Pd) | Mild, room temp | Standard glassware | → | Alkene (cis) |
| Halogenation | Alkene or alkyne | Br₂ or Cl₂ | None | Room temperature | Fume cupboard | → | Dihalo (alkene) or tetrahalo (alkyne) |
| Hydrohalogenation | Alkene or alkyne | HCl, HBr, or HI | None | Room temperature | Fume cupboard | → | Monohalo (Markovnikov); geminal dihalo from alkyne |
| ★ Hydration of alkene | Alkene | H₂O (steam) | H₃PO₄ or dil. H₂SO₄ | ~300°C, HIGH PRESSURE (~65 atm) | High-pressure reactor | ⇌ | Alcohol (Markovnikov) |
| ★ Hydration of alkyne | Alkyne | H₂O | dil. H₂SO₄ AND Hg²⁺ (BOTH required) | ~60°C | Heated glassware | → | Ketone (or ethanal from HC≡CH) |
| ★ Halogen subst. (alkane) | Alkane | Cl₂ or Br₂ | UV light (energy source, NOT catalyst) | Room temp, UV light | Transparent glassware | → | Haloalkane + HX (mixture) |
| Dehydration | Alcohol | conc. H₂SO₄ or H₃PO₄ | As above (acid cat.) | ~170–230°C, atmospheric P | Distillation apparatus | → | Alkene + H₂O |
| Haloalkane → alcohol | Haloalkane | NaOH(aq), AQUEOUS | None | Reflux | Reflux condenser | → | Alcohol + NaX |
| Alcohol → haloalkane | Alcohol | HCl, HBr, or HI | None (ZnCl₂ sometimes) | Reflux | Reflux condenser | → | Haloalkane + H₂O |
| ★ 1° alcohol → aldehyde | Primary alcohol | K₂Cr₂O₇/H₂SO₄ | H₂SO₄ (acidified) | Gentle heat; DISTILLATION | Distillation apparatus | → | Aldehyde + H₂O (orange → green) |
| ★ 1° alcohol → carb. acid | Primary alcohol | K₂Cr₂O₇/H₂SO₄ (excess) | H₂SO₄ | Heat; REFLUX | Reflux condenser | → | Carboxylic acid + H₂O (orange → green) |
| 2° alcohol → ketone | Secondary alcohol | K₂Cr₂O₇/H₂SO₄ | H₂SO₄ | Reflux | Reflux condenser | → | Ketone + H₂O (orange → green) |
| 3° alcohol | Tertiary alcohol | K₂Cr₂O₇/H₂SO₄ | Any | NO REACTION (stays orange) | |||
| Aldehyde → carb. acid | Aldehyde | K₂Cr₂O₇/H₂SO₄ (excess) | H₂SO₄ | Reflux | Reflux condenser | → | Carboxylic acid (orange → green) |
| ★ Esterification | Carboxylic acid + alcohol | Carboxylic acid + alcohol | conc. H₂SO₄ (catalyst) | Heat, reflux | Reflux condenser | ⇌ | Ester + H₂O (yield <100%) |
| Ester hydrolysis (acid) | Ester + H₂O | H₂O + dil. H₂SO₄ | H₂SO₄ | Reflux | Reflux condenser | ⇌ | Carboxylic acid + alcohol |
| Saponification | Ester (fat/oil) | conc. NaOH(aq) or KOH(aq) | None (NaOH = reagent) | Reflux | Reflux condenser | → | Carboxylate salt + alcohol (glycerol for fat) |
| Amide formation | Carboxylic acid + amine | Amine (R-NH₂) | None (or acid activation) | Heat | Heated flask | → | Amide + H₂O |
| Fermentation | Glucose solution | Glucose (C₆H₁₂O₆) | Yeast (zymase enzyme) | ~35°C, ANAEROBIC | Sealed vessel | → | Ethanol + CO₂ |
Which reagents are BOTH required for alkyne hydration? +5 XP
By Lesson 20 you have built every individual reaction in Module 7. This card assembles them into a single connected system, every functional group as a node, every transformation as a labelled arrow, so you can see the shortest path between any two points at a glance.
True or false: A direct single-step conversion from alkane to alcohol exists in Module 7. +5 XP
A marker reading 30 HSC responses to the same pathway question sees the same errors repeatedly. This card presents all five, embedded in a realistic flawed pathway, so you recognise and fix them before the exam finds them.
Complete the gap: To obtain an aldehyde from a primary alcohol using K₂Cr₂O₇/H₂SO₄, use _____ (not reflux). +5 XP
A Band 6 pathway response is not just chemically correct, it is structured, annotated, and gives the marker every component needed to award full marks on the first read.
CH₃CH₂CH₂Cl + NaOH(aq) → CH₃CH₂CH₂OH + NaClWhich colour change confirms that oxidation has occurred with K₂Cr₂O₇/H₂SO₄? +5 XP
The difference between 65% and 85% in Module 7 is almost entirely about conditions recall, the chemistry is correct, but missing "high pressure" or "aqueous NaOH" or "distillation" costs marks systematically across multiple questions.
Which one is the ODD ONE OUT, the only reaction that uses REFLUX but does NOT produce a carboxylic acid? +5 XP
The odd ones out are haloalkane → alcohol (produces an alcohol, not carboxylic acid) and esterification (produces an ester).
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
Product: ethanoic acid. Observable: K₂Cr₂O₇ orange → green. Excess oxidant + reflux ensures complete oxidation past ethanal to ethanoic acid.
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O
Product: ethyl ethanoate. Arrow: reversible (⇌). Yield ~65–67% at equilibrium. Note: ethanol plays two roles, partially oxidised to ethanoic acid (Step 1), and used directly as the alcohol component for esterification (Step 2). Split the batch.
CH₃CH₂CH₂Cl + NaOH(aq) → CH₃CH₂CH₂OH + NaCl
Product: propan-1-ol (primary alcohol). Explanation: NaOH must be aqueous, aqueous OH⁻ acts as a nucleophile, displacing Cl⁻ and producing the alcohol. Alcoholic NaOH would cause elimination → propene. Reflux is required because the reaction is slow at room temperature; the condenser keeps volatile 1-chloropropane in the flask.
CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O
Product: propanoic acid (carboxylic acid). Observable: K₂Cr₂O₇ orange → green. Explanation: Excess dichromate + reflux ensures the propanal intermediate is fully oxidised to propanoic acid. If distillation were used, propanal would be removed before further oxidation, giving propanal, not the target propanoic acid.
CH₃CH₂COOH + CH₃OH ⇌ CH₃CH₂COOCH₃ + H₂O
Product: methyl propanoate (ester). Arrow: reversible (⇌), equilibrium yield ~67% for equimolar reactants. Explanation: Conc. H₂SO₄ performs two roles: (1) acid catalyst, H⁺ activates the carboxylic acid for nucleophilic attack by methanol; (2) dehydrating agent, absorbs water produced, shifting equilibrium right (Le Chatelier's Principle) to increase ester yield. Reflux retains volatile methanol in the flask.
Match each reaction to its correct arrow type. +5 XP
Complete the Learn phase to unlock Practice.
For each compound or description below, identify the functional group, name the compound using IUPAC nomenclature, and draw its structural formula.
Q1. A student reacts butan-2-ol with excess K₂Cr₂O₇/H₂SO₄ under reflux. What is observed, and what is the organic product?
Q2. Which correctly identifies the single error in this equation: "CH₃CH₂COOH + CH₃OH → CH₃CH₂COOCH₃ + H₂O (conditions: H₂SO₄, reflux)"?
Q3. A student wants to synthesise propanal from propan-1-ol. Which conditions exactly achieve this, no more, no less?
Q4. In the Spot the Error pathway (Card 3), which single error would cause the entire synthesis to produce zero yield of target product, not just reduce yield or produce impure product?
Q5. Compound X (formula C₄H₁₀O) gives no colour change with K₂Cr₂O₇/H₂SO₄. Compound Y (formula C₄H₈) decolourises bromine water. Which two-step sequence converts X to Y?
Q6. (4 marks) Outline a two-step synthesis of ethyl ethanoate starting from ethene only. For each step, write the balanced equation, state all conditions, and name the compound produced.
Q7. (5 marks) Starting from 1-bromopropane and ethanol, outline a three-step synthesis of ethyl propanoate. For each step, write the balanced equation, conditions, and name the intermediate. Explain in one sentence why distillation is used in one step and reflux in another during the oxidation sequence.
Q8. (6 marks) A student proposes the pathway: "butan-2-ol → butanone → butanoic acid → butyl butanoate." (a) Identify the error in this pathway and explain why it is not achievable in Module 7. (b) Propose a valid four-step alternative starting from butan-1-ol that achieves the same target product. Write equations and conditions for each step.
Q1, Answer: C
Butan-2-ol is a secondary alcohol (C-OH at C2, bonded to two other carbons: CH₃ and CH₂CH₃). Secondary alcohols oxidise to ketones (butanone, butan-2-one). K₂Cr₂O₇ orange → green (oxidation did occur). Ketones cannot be further oxidised, no H on the carbonyl carbon. Option A (aldehyde) = primary alcohol product only. Option B misclassifies butan-2-ol as tertiary (tertiary = 3 carbon neighbours; secondary = 2). Option D (carboxylic acid) = primary alcohol full oxidation only.
Q2, Answer: B
The only error is the single arrow (→). Esterification is always a reversible equilibrium (⇌), yield <100% at equilibrium. Naming is correct: methyl (from methanol) + propanoate (from propanoic acid) = methyl propanoate ✓. H₂SO₄ IS a catalyst (not reagent, not consumed) ✓. Any alcohol can esterify, Option D is incorrect.
Q3, Answer: C
K₂Cr₂O₇/H₂SO₄ with DISTILLATION: propanal (lower BP than propan-1-ol) is collected as it forms, preventing excess oxidant from converting it to propanoic acid. Option A (reflux) keeps propanal in contact with oxidant → propanoic acid forms. Option B has wrong oxidant formula (K₂Cr₂O₄ does not exist). Option D (KMnO₄) is too powerful, it over-oxidises even under distillation conditions.
Q4, Answer: B
Alcoholic NaOH in Step 1 produces an alkene (but-1-ene) by elimination, NOT butan-1-ol. Without the primary alcohol, Steps 2–4 (oxidation, further oxidation, esterification) all fail, zero yield of methyl butanoate. Option C (reflux in Step 2) gives butanoic acid instead of butanal, the synthesis can still reach ester via a different intermediate path. Option D (wrong arrow) is a notation error with no effect on the actual chemistry yield. Option A (wrong oxidant formula) is a writing error, the step would still work chemically.
Q5, Answer: D
X = 2-methylpropan-2-ol (tertiary alcohol, no K₂Cr₂O₇ colour change ✓; C₄H₁₀O ✓). Y = 2-methylpropene (C₄H₈, alkene, decolourises Br₂ ✓). Two-step route: Step 1: 2-methylpropan-2-ol + HBr → 2-bromo-2-methylpropane + H₂O (substitution: HBr, reflux). Step 2: 2-bromo-2-methylpropane + NaOH(alc) → 2-methylpropene + NaBr + H₂O (elimination: alcoholic NaOH, reflux). Direct dehydration (conc. H₂SO₄) would be ONE step, but the question specifies two.
Q6, Sample Answer (4 marks)
Step 1: CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH. Conditions: H₂O (steam), H₃PO₄ catalyst, ~300°C, high pressure (~65 atm). Product: ethanol. [2 marks, equation + conditions incl. high pressure]
Step 2: Use half the ethanol → oxidise: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O (K₂Cr₂O₇/H₂SO₄ excess, reflux → ethanoic acid); then esterify with remaining ethanol: CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄ catalyst, reflux, ⇌). Product: ethyl ethanoate. [2 marks]
Q7, Sample Answer (5 marks)
Step 1: CH₃CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂OH + NaBr. NaOH(aq), reflux. Product: propan-1-ol. [1 mark]
Step 2: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O. K₂Cr₂O₇/H₂SO₄, distillation. Product: propanal. [1 mark]
Step 3: CH₃CH₂CHO + [O] → CH₃CH₂COOH. K₂Cr₂O₇/H₂SO₄ (excess), reflux. Product: propanoic acid. [1 mark]
Step 4: CH₃CH₂COOH + CH₃CH₂OH ⇌ CH₃CH₂COOC₂H₅ + H₂O. Conc. H₂SO₄ (catalyst), reflux, ⇌. Product: ethyl propanoate. [1 mark]
Explanation: Distillation in Step 2 removes propanal (lower BP than propan-1-ol) as it forms, preventing excess K₂Cr₂O₇ from over-oxidising it to propanoic acid; reflux in Step 3 keeps propanal in contact with excess oxidant to ensure complete conversion to propanoic acid. [1 mark]
Q8, Sample Answer (6 marks)
(a) Error: The step "butanone → butanoic acid" is not achievable in Module 7. Butanone is a ketone, the product of oxidising a secondary alcohol. Ketones have no H on the carbonyl carbon; they cannot be further oxidised to carboxylic acids under Module 7 conditions. The pathway reaches a dead end at butanone. [2 marks]
(b) Four-step from butan-1-ol:
Step 1: CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O. K₂Cr₂O₇/H₂SO₄, distillation. Product: butanal. [1 mark]
Step 2: CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH. K₂Cr₂O₇/H₂SO₄ (excess), reflux. Product: butanoic acid. [1 mark]
Step 3: CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂O. Conc. H₂SO₄ (cat.), reflux, ⇌. Product: butyl butanoate. [1 mark + 1 mark for reversible arrow and correct naming]
Back at the start you were asked about reflux versus distillation when oxidising an alcohol. Now you know the complete rule: reflux (condenser returns the product to the flask) → carboxylic acid, because the aldehyde intermediate cannot escape and is fully oxidised by the excess reagent. Distillation (product removed as it forms) → aldehyde, because the product leaves the flask before being oxidised further. This single condition controls two completely different products from one reaction. Apply it to every oxidation question.
At the start of this lesson you timed yourself reproducing all Module 7 reaction conditions from memory. After working through the full reference table, the Spot the Error analysis, the Band 6 template, and the blank table drill, time yourself again on the blank conditions table. How much faster are you now? Any rows that still require more than 5 seconds of recall are your priority targets for the days before the exam.
What are the EXACT conditions for alkene hydration, all four elements?
State the NaOH rule: what is the difference between NaOH(aq) and NaOH(alc) in reactions with haloalkanes?
Explain distillation vs reflux in the oxidation of a primary alcohol. Which gives an aldehyde and which gives a carboxylic acid?
Name the two dead ends in Module 7 and explain why they cannot be further oxidised.
State the Band 6 four-component format for a pathway question, what four things must each step contain?
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