Year 12 Chemistry Module 7 · Organic Chemistry ⏱ ~45 min 5 MC · 3 Short Answer Lesson 20 of 23 Optional consolidation

Organic Reactions Mastery (Consolidation)

Lock in every Module 7 reaction, conditions, equipment, and colour changes, and learn to write top-band multi-step synthesis responses under exam pressure.

Today's hook: In 2023, a Year 12 chemistry student in NSW achieved full marks on the multi-step organic synthesis question, but only after realising that controlling the reflux vs distillation condition for alcohol oxidation was the difference between aldehyde and carboxylic acid. That condition appears in every reaction-pathway question. By the end of this lesson you will be able to apply the complete set of organic reaction conditions with precision.
0/5TASKS
Optional consolidation

A capstone pass over the organic reactions already taught. Valuable practice, but not a new compulsory lesson, and your core readiness should not depend on the advanced alkyne, mechanism or quantitative acid/base material it revisits.

1

Connect: diagnose your reaction recall

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions
Before You Read

A top-band 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.

Learning Intentions
By the end of this lesson
Know

Key facts

  • 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

Concepts

  • Why distillation vs reflux determines aldehyde vs carboxylic acid
  • Why ketone and tertiary alcohol are dead ends
  • How to structure a top-band pathway response
Can do

Skills

  • Reproduce the complete conditions table from memory
  • Identify and correct errors in a flawed synthesis pathway
  • Write a 7-mark pathway response in top-band format
Scan these before reading
Key Terms
Substitution reactionOne atom or group replaces another; alkane + halogen (free-radical), or halogenoalkane + OH⁻ (nucleophilic).
Addition reactionTwo reactants combine across a multiple bond to form one product; alkene + H₂, HX, X₂, or H₂O.
Elimination reactionLoss of atoms/groups from adjacent carbons to form a multiple bond; e.g., alcohol → alkene (dehydration).
Condensation reactionTwo molecules combine with loss of a small molecule (often H₂O); esterification, amide formation, condensation polymerisation.
Oxidation of organic compoundsPrimary alcohol → aldehyde → carboxylic acid; secondary alcohol → ketone; using acidified Cr₂O₇²⁻ or KMnO₄.
Reduction of organic compoundsCarboxylic acid or ester → alcohol using LiAlH₄ or NaBH₄; addition of H₂ across double bonds.
📚 Core Content
2

Explain: organise reactions by family

01
Complete Module 7 Reaction Conditions Reference

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
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 Tip: For 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.

Which reagents are BOTH required for alkyne hydration? +5 XP

3

Represent: rebuild the complete conditions map

02
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 and every transformation as a labelled arrow, so you can compare valid paths between any two points.

Route quality: Fewer steps generally improve yield and reduce waste, but the shortest route is not automatically the best. Also check safety, selectivity, purification, available reagents and waste.
How the functional groups interconvert: the reagents and conditions linking alkynes, alkenes, alkanes, haloalkanes, alcohols, aldehydes, acids, ketones, esters, soaps and amides.
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.

True or false: A direct single-step conversion from alkane to alcohol exists in Module 7. +5 XP

4

Respond: diagnose a flawed synthesis

03
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.

Complete the gap: To obtain an aldehyde from a primary alcohol using K₂Cr₂O₇/H₂SO₄, use _____ (not reflux). +5 XP

5

Feedback: annotate a top-band response

04
top-band Pathway Annotation Guide

A top-band 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."
Developing vs top-band response comparison:
Developing response: "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.
Top-band response: 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.

Which colour change confirms that oxidation has occurred with K₂Cr₂O₇/H₂SO₄? +5 XP

6

Apply: complete a timed synthesis

05
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.

Final revision drill: In your last week before the HSC, complete the blank conditions 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.

Key Rules to Know Cold:

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).

All four of these are reflux reactions. Which is the ODD ONE OUT, the only one that reaches an equilibrium rather than going to completion?

Esterification is written with ⇆ because the ester and water react back to the acid and alcohol. The two oxidations and the haloalkane hydrolysis all go to completion, so they take a single →. Note also what each one makes: two produce a carboxylic acid, the haloalkane produces an alcohol, and esterification produces an ester.

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.

Match each reaction to its arrow type, and to the reason for it.

  • Esterification
  • Haloalkane → alcohol
  • Alkene hydration
  • Aldehyde → carboxylic acid
  • → irreversible, oxidation goes to completion
  • ⇆ reversible, dehydration reverses it
  • → irreversible, OH⁻ substitution goes to completion
  • ⇆ reversible, the ester and water react back
Interactive Tool, Organic Synthesis Pathway Builder Open fullscreen ↗
The Organic Synthesis tool shows that to convert a primary alcohol to an ALDEHYDE (not acid), you would use…
1

Practice: apply the complete reaction map

🔀Sort the Steps+7 XP
Arrange the synthesis-planning steps in the correct order.
Select reactions that achieve each transformation
Check each intermediate and the required conditions
Write the full pathway with structures, reagents and conditions
Identify the functional groups in the starting material and target
Determine the required functional-group transformation
✏️ Activities
A
Organic Compound Identification

For each compound or description below, identify the functional group, name the compound using IUPAC nomenclature, and draw its structural formula.

  1. A compound with the molecular formula C₅H₁₂O reacts with acidified dichromate to produce a ketone. Identify the functional group, name the compound, and draw its structure.
  2. An alkene with the formula C₄H₈ undergoes hydrogenation to form butane. Draw the structure of the original alkene and name it.
  3. Describe the difference between distillation and reflux in the context of alcohol oxidation, including one example of each.
📝 Check Your Understanding
SA
Extended Response Practice

Q6. (6 marks) Outline a three-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 four-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 three-step alternative starting from butan-1-ol that achieves the same target product. Write equations and conditions for each step.

Show All Answers

Multiple Choice

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.

Short Answer Model Answers

Q6, Sample Answer (6 marks)
Step 1: CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH. Conditions: H₂O (steam), H₃PO₄ catalyst, ~300°C, high pressure (~65 atm). Product: ethanol. [2 marks]
Step 2: Oxidise part of the ethanol: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O. Conditions: K₂Cr₂O₇/H₂SO₄ excess, reflux. Product: ethanoic acid. [2 marks]
Step 3: Esterify the ethanoic acid with reserved ethanol: CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O. Conditions: 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) Three-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]

How did your thinking change?

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 top-band 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.

1

Review: retrieve conditions without the table

Check what actually stuck
🏆 Review

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 top-band four-component format for a pathway question, what four things must each step contain?