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★ IQ3–IQ5 · Consolidation · Lesson 20 of 23 · 60 min

Organic Reactions Mastery — Conditions, Pathways & Band 6 Responses

Reaction pathways are the highest-mark skill in HSC Chemistry — and the gap between Band 4 and Band 6 is almost entirely about conditions recall and structured annotation. This consolidation lesson gives you the complete reference table, the spot-the-error drill, the Band 6 template, and a 7-mark model answer to build genuine exam fluency.

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★ Think First — Before You Read

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.

Key Terms — scan these before reading
HydrocarbonAn organic compound containing only carbon and hydrogen atoms.
Functional groupA specific atom arrangement determining characteristic chemical reactions.
Homologous seriesA family of compounds with the same functional group, differing by CH₂.
Addition polymerA polymer formed by monomers adding together without loss of atoms.
Condensation polymerA polymer formed with elimination of a small molecule such as water.
EsterificationA condensation reaction between a carboxylic acid and an alcohol forming an ester.

Misconceptions to Fix

Wrong: Chemical equations can be balanced by changing subscripts in formulas.

Right: Chemical equations must be balanced by changing coefficients only. Subscripts in chemical formulas define the identity of the compound — changing them creates a different substance. If you cannot balance an equation with whole-number coefficients, check that your formulas are correct.

01

Complete Module 7 Reaction Conditions Reference

This is the authoritative reference table. Study it, then cover it and reproduce it from memory using the blank table in Card 4. ★ Priority rows (highlighted) are the most frequently penalised.

ReactionStarting materialReagentCatalystConditions / tempEquipmentArrowProduct(s)
Hydrogenation (full)Alkene or alkyneH₂ (gas; 2 eq for alkyne)Ni, Pd, or Pt~150–200°C, high pressurePressure vesselAlkane
Partial hydrogenationAlkyneH₂ (gas, 1 eq)Lindlar (poisoned Pd)Mild, room tempStandard glasswareAlkene (cis)
HalogenationAlkene or alkyneBr₂ or Cl₂NoneRoom temperatureFume cupboardDihalo (alkene) or tetrahalo (alkyne)
HydrohalogenationAlkene or alkyneHCl, HBr, or HINoneRoom temperatureFume cupboardMonohalo (Markovnikov); geminal dihalo from alkyne
★ Hydration of alkeneAlkeneH₂O (steam)H₃PO₄ or dil. H₂SO₄~300°C, HIGH PRESSURE (~65 atm)High-pressure reactorAlcohol (Markovnikov)
★ Hydration of alkyneAlkyneH₂Odil. H₂SO₄ AND Hg²⁺ (BOTH required)~60°CHeated glasswareKetone (or ethanal from HC≡CH)
★ Halogen subst. (alkane)AlkaneCl₂ or Br₂UV light (energy source — NOT catalyst)Room temp, UV lightTransparent glasswareHaloalkane + HX (mixture)
DehydrationAlcoholconc. H₂SO₄ or H₃PO₄As above (acid cat.)~170–230°C, atmospheric PDistillation apparatusAlkene + H₂O
Haloalkane → alcoholHaloalkaneNaOH(aq) — AQUEOUSNoneRefluxReflux condenserAlcohol + NaX
Alcohol → haloalkaneAlcoholHCl, HBr, or HINone (ZnCl₂ sometimes)RefluxReflux condenserHaloalkane + H₂O
★ 1° alcohol → aldehydePrimary alcoholK₂Cr₂O₇/H₂SO₄H₂SO₄ (acidified)Gentle heat; DISTILLATIONDistillation apparatusAldehyde + H₂O (orange → green)
★ 1° alcohol → carb. acidPrimary alcoholK₂Cr₂O₇/H₂SO₄ (excess)H₂SO₄Heat; REFLUXReflux condenserCarboxylic acid + H₂O (orange → green)
2° alcohol → ketoneSecondary alcoholK₂Cr₂O₇/H₂SO₄H₂SO₄RefluxReflux condenserKetone + H₂O (orange → green)
3° alcoholTertiary alcoholK₂Cr₂O₇/H₂SO₄AnyNO REACTION (stays orange)
Aldehyde → carb. acidAldehydeK₂Cr₂O₇/H₂SO₄ (excess)H₂SO₄RefluxReflux condenserCarboxylic acid (orange → green)
★ EsterificationCarboxylic acid + alcoholCarboxylic acid + alcoholconc. H₂SO₄ (catalyst)Heat, refluxReflux condenserEster + H₂O (yield <100%)
Ester hydrolysis (acid)Ester + H₂OH₂O + dil. H₂SO₄H₂SO₄RefluxReflux condenserCarboxylic acid + alcohol
SaponificationEster (fat/oil)conc. NaOH(aq) or KOH(aq)None (NaOH = reagent)RefluxReflux condenserCarboxylate salt + alcohol (glycerol for fat)
Amide formationCarboxylic acid + amineAmine (R-NH₂)None (or acid activation)HeatHeated flaskAmide + H₂O
FermentationGlucose solutionGlucose (C₆H₁₂O₆)Yeast (zymase enzyme)~35°C, ANAEROBICSealed vesselEthanol + CO₂
Four most-penalised omissions in HSC:
1. Alkyne hydration — missing Hg²⁺. Both dil. H₂SO₄ AND Hg²⁺ required, every time.
2. Alkene hydration — missing high pressure. Catalyst + temperature alone = partial marks only.
3. Alkane halogenation — writing "UV light (catalyst)". UV is an energy source, not a catalyst.
4. Distillation (→ aldehyde) vs reflux (→ carboxylic acid) — this distinction is tested almost every year.
Exam TipFor organic chemistry questions, draw full structural formulas showing all atoms and bonds — condensed or skeletal formulas alone may lose marks in HSC extended-response questions.

Know

  • All 20 reaction types and their exact conditions
  • Which functional group conversions are impossible in Module 7
  • The four most-penalised omissions in HSC marking

Understand

  • Why distillation vs reflux determines aldehyde vs carboxylic acid
  • Why ketone and tertiary alcohol are dead ends
  • How to structure a Band 6 pathway response

Can Do

  • Reproduce the complete conditions table from memory
  • Identify and correct errors in a flawed synthesis pathway
  • Write a 7-mark pathway response in Band 6 format
02

1 — The Complete Reaction Pathway Map: Every Arrow in One View

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.

Alkyne Alkene Alkane Haloalkane +H₂, Lindlar +Br₂/HX +H₂, Ni X₂/UV light +HX, r.t. (Markovnikov) Alcohol (hub) Aldehyde Carboxylic Acid H₂O, H₃PO₄ 300°C, HIGH P ⇌ H₂SO₄, Δ (dehydration) NaOH(aq), reflux +HX, reflux K₂Cr₂O₇/H₂SO₄ DISTILLATION ★ K₂Cr₂O₇ excess REFLUX ★ K₂Cr₂O₇ excess/H₂SO₄, reflux (direct) Ketone ⛔ DEAD END 2° Alcohol (secondary) hydration K₂Cr₂O₇/H₂SO₄ reflux 3° Alcohol ⛔ no oxidation Ester conc. H₂SO₄ reflux ⇌ ★ (alcohol component in esterification) Soap (RCOO⁻Na⁺) NaOH(aq), reflux → (saponification) H₂O/H₂SO₄ reflux ⇌ Amide + amine, heat → Glucose → Ethanol yeast, 35°C, anaerobic H₂O, H₂SO₄ + Hg²⁺ ★ 60°C → ketone Legend: Forward reaction Reverse/alternative Ester reactions (⇌) ⛔ = dead end (cannot oxidise further) ★ = most tested condition: DISTILLATION (→ aldehyde) vs REFLUX (→ carboxylic acid); HIGH PRESSURE (alkene hydration); Hg²⁺ (alkyne hydration); ⇌ (esterification) Shortest paths: 2-step: haloalkane → alcohol → alkene ∥ 3-step: alkene → alcohol → carboxylic acid → ester ∥ 4-step: haloalkane → alcohol → aldehyde → carboxylic acid → ester
Must-do: Draw this map by hand, from memory, on a blank sheet of paper. Include every arrow with abbreviated conditions (e.g. "K₂Cr₂O₇/H⁺, distil" — not just "oxidation"). Time yourself. When you can reproduce the complete map in under 5 minutes with all conditions, you have internalised the core skill of Module 7.
Common error — arrows that do not exist: (1) Direct arrow from alkane to alcohol — 2 steps required. (2) Arrow from carboxylic acid back to aldehyde/alcohol — reduction is not in Module 7. (3) Arrow from ketone to carboxylic acid — ketones are dead ends. Check every arrow on your hand-drawn map against the table before the exam.
03

2 — Spot the Error: A Flawed Five-Step Synthesis

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.

The problem: Synthesise methyl butanoate (CH₃CH₂CH₂COOCH₃) starting from 1-bromobutane (CH₃CH₂CH₂CH₂Br) and methanol (CH₃OH). A student writes the following pathway — find all errors.
Step 1: CH₃CH₂CH₂CH₂Br + NaOH(alc) → CH₃CH₂CH₂CH₂OH + NaBr. "NaOH in ethanol, reflux. Product: butan-1-ol."
Step 2: CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO. "K₂Cr₂O₇/H₂SO₄, reflux. Product: butanal."
Step 3: CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH. "K₂Cr₂O₄/H₂SO₄ (excess), reflux. Product: butanoic acid."
Step 4: CH₃CH₂CH₂COOH + CH₃OH CH₃CH₂CH₂COOCH₃ + H₂O. "conc. H₂SO₄, reflux."
Step 5: "Shake with water in separating funnel. Collect upper layer."

The Five Errors — and Corrections:

ERROR 1 — Step 1: Wrong type of NaOH NaOH in ethanol (alcoholic NaOH) causes elimination — producing an alkene (but-1-ene), NOT the alcohol butan-1-ol. This makes the entire synthesis fail from Step 1.
CORRECTION 1: Use NaOH(aq) — aqueous sodium hydroxide. Aqueous OH⁻ acts as a nucleophile → substitution → butan-1-ol ✓. Memorise: NaOH(aq) = substitution = alcohol; NaOH(alc) = elimination = alkene.
ERROR 2 — Step 2: Wrong equipment for aldehyde Reflux keeps the aldehyde (butanal) in contact with excess oxidant → over-oxidised to butanoic acid (carboxylic acid). To stop at the aldehyde, the aldehyde must be removed from the reaction as it forms.
CORRECTION 2: Use K₂Cr₂O₇/H₂SO₄ with DISTILLATION (not reflux). Butanal has a lower BP than butan-1-ol — distillation removes it before further oxidation ✓.
ERROR 3 — Step 3: Wrong oxidant formula K₂Cr₂O₄ does not exist as a standard reagent. Chromate is CrO₄²⁻ (yellow) and is not used for alcohol/aldehyde oxidation at HSC level.
CORRECTION 3: Always write K₂Cr₂O₇ (potassium dichromate, orange → green). K₂Cr₂O₇, not K₂Cr₂O₄ ✓.
ERROR 4 — Step 4: Single arrow for reversible reaction The student writes → (single arrow) for esterification. Esterification is a reversible equilibrium — all four species are present at equilibrium; yield is less than 100%.
CORRECTION 4: CH₃CH₂CH₂COOH + CH₃OH CH₃CH₂CH₂COOCH₃ + H₂O. Must use ⇌ ✓.
ERROR 5 — Step 5: Incomplete isolation procedure Simply collecting the upper layer leaves residual butanoic acid and conc. H₂SO₄ in the ester layer — impure product.
CORRECTION 5 — Full isolation procedure: (a) Separate layers (ester upper, aqueous lower). (b) Wash with Na₂CO₃(aq) to convert residual butanoic acid to water-soluble sodium butanoate — releases CO₂ (release pressure). (c) Second water wash. (d) Dry with anhydrous Na₂SO₄. (e) Distil — collect pure ester fraction ✓.
Five errors to memorise for the HSC: (1) Alcoholic vs aqueous NaOH; (2) distillation vs reflux for aldehyde; (3) K₂Cr₂O₇ not K₂Cr₂O₄; (4) ⇌ not → for esterification; (5) full isolation procedure (Na₂CO₃ wash, dry, distil). These appear across HSC marking reports as the most consistently lost marks in Module 7.
04

3 — Band 6 Pathway Annotation Guide

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.

The Four-Component Format (one per step):

1
Step description line orientation mark
"Step 1: Substitution — 1-chloropropane → propan-1-ol"
States step number, reaction type, and what is converted to what. Orients the marker before they read the equation.
2
Balanced structural equation 1 mark per step
CH₃CH₂CH₂Cl + NaOH(aq) → CH₃CH₂CH₂OH + NaCl Single arrow (→) for irreversible. Reversible arrow (⇌) for hydration of alkene, esterification, ester hydrolysis.
3
Full conditions box — all four elements 1 mark per step
REAGENT: NaOH(aq) — aqueous sodium hydroxide
CATALYST: none
CONDITIONS: heat under reflux
EQUIPMENT: round-bottom flask, reflux condenser
4
Product name + observable colour change 1 mark per step
"Product: propan-1-ol (primary alcohol)."
For oxidation steps, always state colour change: K₂Cr₂O₇ orange → green (any oxidation); KMnO₄ purple → colourless; Tollens': silver mirror (aldehyde only); Fehling's: brick-red precipitate (aldehyde only); Br₂ water: decolourises (C=C or C≡C).
5
Condition explanation sentence (for 7+ mark responses) discrim. marks
One sentence explaining why the specific conditions give the desired product rather than an alternative.
"Aqueous NaOH is specified because alcoholic NaOH would cause elimination → propene, not substitution → propan-1-ol."
Band 4 vs Band 6 comparison:
Band 4: "Step 1: Oxidise butan-1-ol. Use K₂Cr₂O₇. Get butanal." — Missing: balanced equation, full conditions (catalyst, temperature, equipment), product name, condition explanation. Marks earned: 1 out of potential 3.
Band 6: Numbered steps, balanced equation, full conditions box, product name, colour change, one explanation sentence per step. Every component the marker needs.
Structure earns marks: In a 10-mark extended response, the readability and logical flow of your answer can be worth 2–3 marks in its own right — clear structure lets the marker find the correct chemistry and award partial credit. Write in numbered steps. Never write pathway responses as continuous prose.
05

4 — The Complete Conditions Memory Challenge

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.

Activity: Blank Conditions Table — Complete from Memory

For each reaction below, write the reagent, catalyst, conditions/temperature, equipment, and correct arrow type from memory. Do NOT look at the reference table until you have attempted every row. Then check against the table at the top of this lesson.

Reaction ReagentCatalystConditions / TempEquipmentArrow
Hydrogenation (full)
Partial hydrogenation
★ Hydration of alkene
★ Hydration of alkyne
★ Alkane halogenation
Dehydration
Haloalkane → alcohol
★ 1° alcohol → aldehyde
★ 1° alcohol → acid
★ Esterification
Saponification
Fermentation

★ Priority rows most frequently left incomplete in HSC exams.

Final revision drill: In your last week before the HSC, complete this blank table every morning before any other chemistry work — without looking at the reference. Any cell that takes more than 5 seconds to recall is a priority gap. The table drill takes 5 minutes and is the highest-return study investment for Module 7.
Exam pressure reveals what was never secure: Conditions recalled effortfully in practice are forgotten under exam conditions. Automatic recall requires approximately 15–20 successful retrievals. If you have read the conditions once or twice, they are not yet automatic. Drill daily.
Distillation vs Reflux rule: K₂Cr₂O₇/H₂SO₄ + DISTILLATION = aldehyde (remove before over-oxidation). K₂Cr₂O₇/H₂SO₄ (excess) + REFLUX = carboxylic acid.
NaOH rule: NaOH(aq) = substitution = alcohol. NaOH(alc) = elimination = alkene.
Alkyne hydration: BOTH dil. H₂SO₄ AND Hg²⁺. Product = ketone (except HC≡CH → ethanal).
Alkene hydration: H₂O (steam), H₃PO₄, ~300°C, HIGH PRESSURE (~65 atm). ⇌ reversible.
Alkane halogenation: UV light = energy source (not catalyst). Produces mixture of haloalkanes.
Esterification: ⇌ reversible. Conc. H₂SO₄ = catalyst (not reagent). Yield <100%.
Dead ends: Ketone (2° alc oxidation) — cannot be further oxidised. Tertiary alcohol — no K₂Cr₂O₇ colour change.
Colour changes: K₂Cr₂O₇ orange → green (any oxidation proceeds). KMnO₄ purple → colourless. Tollens': silver mirror (aldehyde). Br₂ water: decolourises (C=C or C≡C).
06

Worked Example 1 — Two-Step Synthesis from Ethanol

Problem: Starting from ethanol only, synthesise ethyl ethanoate. Write equations and full conditions for each step. Name all intermediates.
1
Ethanol → ethanoic acid (full oxidation) CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
REAGENT: K₂Cr₂O₇/H₂SO₄ (excess, acidified potassium dichromate)
CATALYST: H₂SO₄ (acidified conditions)
CONDITIONS: heat under reflux
EQUIPMENT: round-bottom flask, reflux condenser

Product: ethanoic acid. Observable: K₂Cr₂O₇ orange → green. Excess oxidant + reflux ensures complete oxidation past ethanal to ethanoic acid.

2
Ethanoic acid + ethanol → ethyl ethanoate (esterification) CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O
REAGENT: ethanoic acid + ethanol (use reserved portion of ethanol from starting material)
CATALYST: conc. H₂SO₄ (a few drops — not consumed)
CONDITIONS: heat under reflux
EQUIPMENT: round-bottom flask, reflux condenser

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.


Worked Example 2 — Full 7-Mark Model Response

Problem (7 marks): Starting from 1-chloropropane (CH₃CH₂CH₂Cl) and methanol (CH₃OH), outline the synthesis of methyl propanoate (CH₃CH₂COOCH₃). For each step, write a balanced equation, state all conditions, and explain the role of each condition in producing the desired intermediate.
1
Step 1: Substitution — 1-chloropropane → propan-1-ol CH₃CH₂CH₂Cl + NaOH(aq) → CH₃CH₂CH₂OH + NaCl
REAGENT: NaOH(aq) — aqueous sodium hydroxide
CATALYST: none
CONDITIONS: heat under reflux
EQUIPMENT: round-bottom flask, reflux condenser

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.

2
Step 2: Full oxidation — propan-1-ol → propanoic acid CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O
REAGENT: K₂Cr₂O₇/H₂SO₄ (acidified potassium dichromate, excess)
CATALYST: H₂SO₄ (acidified conditions)
CONDITIONS: heat under reflux
EQUIPMENT: round-bottom flask, reflux condenser

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.

3
Step 3: Esterification — propanoic acid + methanol → methyl propanoate CH₃CH₂COOH + CH₃OH ⇌ CH₃CH₂COOCH₃ + H₂O
REAGENT: methanol (CH₃OH) — provided as starting material
CATALYST: conc. H₂SO₄ (a few drops — catalyst, not consumed)
CONDITIONS: heat under reflux
EQUIPMENT: round-bottom flask, reflux condenser

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.

Interactive

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 2), 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]

Revisit Your Think First Response

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.