Reaction pathways are the HSC exam skill that separates Band 4 from Band 6 — not because the individual steps are hard, but because you need every reaction from L06 to L18 working together simultaneously, and one missing condition or wrong arrow invalidates the whole chain.
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Today's hook: In 2022, the global aspirin shortage raised questions about supply chains — aspirin (acetylsalicylic acid) is synthesised from salicylic acid via esterification, which itself comes from phenol. Every step in that factory pathway is an organic reaction you now know. By the end of this lesson you will be able to map a multi-step synthesis from a starting material to a target molecule — the same skill industrial chemists use every day.
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Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A food scientist needs to make banana flavouring (ethyl ethanoate) in three steps from ethene and ethanol. A pharmaceutical chemist needs to synthesise an ester from a haloalkane starting material in four steps.
Before reading: draw the functional group changes needed to get from ethene (CH₂=CH₂) to ethyl ethanoate (CH₃COOC₂H₅). How many steps? What changes at each step?
KEY: Distillation = stop at aldehyde | Reflux + excess oxidant = go to carboxylic acid
Learning Intentions
goals
Know
All 15+ reaction types and their conditions from L06–L18
Which functional group conversions are possible vs impossible in Module 7
The Markovnikov limitation in hydration of asymmetrical alkenes
Understand
Why distillation vs reflux determines aldehyde vs carboxylic acid
Why secondary alcohol oxidation is a dead end (ketone)
How to choose the shortest valid path between two functional groups
Can Do
Apply the four-step algorithm to any pathway problem
Write balanced equations with full conditions for each step
Construct a four-step synthesis from haloalkane to ester
Scan these before reading
vocab
Reaction pathwayA sequence of organic reactions connecting starting material to a target product via specified intermediate compounds.
Functional group interconversionSystematic approach to organic synthesis: identify what functional group the product has and which reaction produces it.
Retrosynthetic analysisWorking backwards from the target molecule to identify suitable precursors and reagents at each step.
SelectivityChoosing conditions that convert one functional group without affecting others (e.g., mild oxidant for aldehyde only).
Yield in multi-step synthesisOverall yield = product of individual step yields; more steps = lower overall yield, increasing cost and waste.
Core Concepts
01
1 — The Complete Reaction Map: Every Arrow You Have Built
By L19 you have learned 15+ individual reaction types. This card assembles them into a single connected map so you can see which functional groups connect to which and what the shortest path between any two points is.
HSC Must-Do: Print or draw this reaction map and keep it visible during all Module 7 revision. When you encounter a pathway question, trace the route on the map FIRST before writing any equations. The map shows all possible routes and lets you choose the shortest one.
Common Error: Students draw arrows that do not exist — for example, alkane → alcohol directly (no such route in Module 7 — must go alkane → haloalkane → alcohol). Or ketone → carboxylic acid (impossible — ketones cannot be oxidised further). Every arrow must have known conditions behind it.
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.
+5 XPQuick Check
A student wants to go from an alkane to an alcohol. How many steps are required in Module 7?
02
2 — The Systematic Approach: Four-Step Algorithm for Pathway Problems
A multi-step pathway problem is not solved by intuition — it is solved by a systematic algorithm: identify functional groups of start and end, find the shortest connected path on the map, then write each step with full conditions.
1
Identify functional groups. Look at the starting material and target product. Write down the functional group class of each, and note if the carbon skeleton changes (if it does — the pathway is invalid for Module 7 reactions that don't alter chain length).
2
Locate on the reaction map. Find the start and target functional groups on the map. Count the minimum number of arrows between them. Each arrow = one step = one set of conditions.
3
Check each step. Can this transformation be done in one reaction? Are there Markovnikov/anti-Markovnikov issues? Does distillation vs reflux matter (aldehyde vs carboxylic acid)? Is there a dead end (secondary alcohol → ketone)?
4
Write equations with full conditions. For each step: (1) balanced structural equation; (2) conditions box — REAGENT / CATALYST / EQUIPMENT; (3) name of the intermediate compound produced.
The Critical Distillation vs Reflux Decision (primary alcohol oxidation):
Primary Alcohol R–CH₂OH
→ K₂Cr₂O₇ distillation
Aldehyde R–CHO
→ K₂Cr₂O₇ excess reflux
Carboxylic Acid R–COOH
Secondary Alcohol R–CHOH–R'
→ K₂Cr₂O₇ reflux
Ketone ⚠ R–CO–R'
Cannot oxidise further — dead end in Module 7
Trap 1 — Oxidation level Aldehyde and carboxylic acid both come from primary alcohol — distillation gives aldehyde; reflux + excess gives acid. Specify which every time.
Trap 2 — Alkane to alcohol Alkane → alcohol is NOT a direct Module 7 reaction. Must go: alkane → haloalkane → alcohol (two steps).
Trap 3 — Markovnikov Hydration or hydrohalogenation of asymmetrical alkene puts functional group at more substituted carbon. Check if the target has OH at C1 — if so, Markovnikov is the wrong approach.
Trap 4 — Esterification arrow Esterification is REVERSIBLE (⇌). Always write ⇌, state conc. H₂SO₄ catalyst and reflux, and note yield <100%.
HSC Must-Do: For every pathway step, write the name AND structural formula of each intermediate — not just the final product. The marker needs to see that you know what is produced at each stage.
Common Error: Students write steps in the wrong order — they work backwards in their heads but write forward in the wrong sequence. Always write forward: Step 1 → Intermediate 1; Step 2 → Intermediate 2; ... → Target. Number each step clearly.
+5 XPTrue / False
True or False: To obtain an aldehyde from a primary alcohol, you use K₂Cr₂O₇/H₂SO₄ with REFLUX equipment.
03
3 — Pathway Drills: Building From Simple to Complex
The best way to internalise the algorithm is to apply it to progressively more complex examples — starting with two-step paths and building toward four-step syntheses.
Drill 1 — Ethanol → Ethyl Ethanoate (2 steps, same starting material in both roles):
Conditions:Conc. H₂SO₄ (catalyst), heat under reflux. Reversible — yield ~65%.
Intermediate:Ethyl ethanoate
Note: Ethanol is used in two roles — oxidise part to ethanoic acid, keep the rest for esterification. This "split the starting material" strategy is common in HSC pathway questions.
Drill 2 — Propan-1-ol → Propyl Propanoate (3 steps, same starting material in two roles):
Step 1
Equation:CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O
Conditions:K₂Cr₂O₇/H₂SO₄, distillation (collect propanal before over-oxidation).
HSC Must-Do: When a pathway says "starting from compound X only," you may need to use it in two different roles — oxidise part to a carboxylic acid, then esterify another portion of the original alcohol with that acid.
Common Error: Reversing oxidation steps — writing "aldehyde + K₂Cr₂O₇ → alcohol" (this is reduction, not in Module 7). You can only go uphill (oxidise) in Module 7.
+5 XPComplete the Sentence
In Drill 1, ethanol is used in two roles: part is ________ to ethanoic acid (using K₂Cr₂O₇/H₂SO₄/reflux), and the rest is used as the ________ for esterification with that acid.
The hardest pathway questions give you an unfamiliar starting material and expect you to use functional group identification + the reaction map to construct a valid route — this card builds that skill explicitly.
Approach for unknown compounds: ALWAYS identify the functional group class of the unknown first. From the functional group, locate it on the map, then trace the path.
Four-Step Worked Pathway — 1-Bromobutane → Butyl Butanoate:
H₂SO₄ activates acid and dehydrates water produced (Le Chatelier → shift right → higher yield). Butan-1-ol from Step 1 is used here.
HSC Must-Do: In every multi-step pathway answer, present steps as a numbered sequence with: (1) the reaction equation; (2) conditions box; (3) the name of the intermediate or final product. A paragraph without numbered steps is very difficult for the marker to follow.
Common Error: Carbon chain length changing unexpectedly. Every Module 7 functional group interconversion preserves the carbon skeleton. The only step where two molecules combine is esterification (and amide formation).
+5 XPQuick Check
In the 4-step synthesis of butyl butanoate from 1-bromobutane, why is DISTILLATION used in Step 2 but REFLUX used in Step 3?
Common Misconceptions — Pathway Problems
Misconception 1: "I can go directly from alkane to alcohol." Reality: No Module 7 reaction converts alkane → alcohol directly. Must go alkane → haloalkane (X₂/UV) → alcohol (NaOH(aq)/reflux) — two steps.
Misconception 2: "Reflux always gives an aldehyde." Reality: Reflux with excess oxidant gives a carboxylic acid. Distillation gives an aldehyde. Equipment determines the product.
Misconception 3: "A secondary alcohol can be oxidised to a carboxylic acid." Reality: Secondary alcohol → ketone only. Ketones CANNOT be oxidised further to carboxylic acids under Module 7 conditions — dead end.
Misconception 4: "Esterification is irreversible." Reality: Esterification is REVERSIBLE (⇌). Must write ⇌ in the equation. Saponification with NaOH is irreversible (→).
Worked Examples
05
Example 1 — Two-Step Pathway Identification (Straightforward)
Problem: Identify the reagents and conditions for each step. Name all intermediates. Ethanol → [Step 1] → Intermediate A → [Step 2] → Ethyl ethanoate
G
Given: Starting material: ethanol. Target: ethyl ethanoate (CH₃COOC₂H₅). Two steps.
F
Find: Identify Intermediate A and conditions for both steps.
(a) NaOH(aq), reflux → propan-1-ol. (b) then K₂Cr₂O₇/distillation → propanal. (c) then K₂Cr₂O₇ excess/reflux → propanoic acid. Key distinction: distillation stops at aldehyde; reflux + excess goes to acid.
07
Example 3 — Extended Four-Step Synthesis (Hard, 8 marks)
Problem: Starting from 1-bromobutane, describe a four-step synthesis of butyl butanoate. For each step write the balanced equation, all conditions, name the intermediate, and explain why those conditions give the desired intermediate rather than an alternative product.
1
Step 1 — 1-bromobutane → butan-1-ol: CH₃CH₂CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaBr Conditions: NaOH(aq), reflux. Intermediate: butan-1-ol. Why aqueous NaOH: Aqueous NaOH provides OH⁻ for nucleophilic substitution at the C-Br bond → alcohol. Alcoholic (ethanolic) NaOH would instead promote elimination (E2) → alkene — wrong product.
2
Step 2 — butan-1-ol → butanal: CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O Conditions: K₂Cr₂O₇/H₂SO₄, distillation. Intermediate: butanal. Why distillation: Butanal (BP 75°C) has a lower boiling point than butan-1-ol (BP 118°C). Distilling removes butanal as it forms, preventing further oxidation to butanoic acid by the excess dichromate.
3
Step 3 — butanal → butanoic acid: CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOH Conditions: K₂Cr₂O₇/H₂SO₄ (excess), reflux. Intermediate: butanoic acid. Why reflux: Reflux keeps the aldehyde in contact with excess oxidant until complete oxidation to the carboxylic acid occurs. We want full oxidation, so the reflux condenser returns vapour to the flask.
4
Step 4 — butanoic acid + butan-1-ol → butyl butanoate: CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂O Conditions: conc. H₂SO₄ (catalyst), heat under reflux. Reversible (⇌). Why conc. H₂SO₄: H₂SO₄ provides H⁺ to activate the carboxylic acid for nucleophilic attack by the alcohol. It is also a dehydrating agent — absorbing water produced, shifting equilibrium right by Le Chatelier's principle → higher ester yield. Note: Butan-1-ol from Step 1 serves both as starting material for Steps 2–3 AND as the alcohol for Step 4.
Transformation
Reagent/Catalyst
Equipment
Key point
Alkene → Alcohol
H₂O, H₃PO₄ cat.
High T (300°C), high P
Markovnikov — OH to more substituted C
Alcohol → Alkene
Conc. H₂SO₄ or H₃PO₄
Heat (~170°C)
Reverse of hydration
Alkane → Haloalkane
X₂, UV light
Room temp.
Radical substitution; mixture of products
Haloalkane → Alcohol
NaOH(aq)
Reflux
Aqueous = substitution (not elimination)
Primary alcohol → Aldehyde
K₂Cr₂O₇/H₂SO₄
Distillation
Remove product to prevent over-oxidation
Primary alcohol → Carboxylic acid
K₂Cr₂O₇/H₂SO₄ (excess)
Reflux
Keep in flask for full oxidation
Secondary alcohol → Ketone
K₂Cr₂O₇/H₂SO₄
Reflux
DEAD END — cannot oxidise further
Acid + Alcohol ⇌ Ester + H₂O
Conc. H₂SO₄ (cat.)
Reflux
Reversible; H₂SO₄ also dehydrates
Ester + NaOH → Soap + Alcohol
NaOH(aq)
Reflux
Saponification — IRREVERSIBLE (→)
RCOOH + RNH₂ → Amide + H₂O
Heat
Condensation
Amide bond = peptide bond
Complete the microtasks above to unlock Practice phase (? XP needed)
The Organic Synthesis tool shows that to convert a primary alcohol to an ALDEHYDE (not acid), you would use…
🔬Predict — Then Reveal+8 XP
Target molecule: butan-2-ol. Starting material: but-1-ene. Predict the complete two-step synthesis pathway, including reagents, conditions, and the intermediate compound formed at each step.
Your predictionExpert answerCompare
Step 1: Add HBr to but-1-ene under room temperature conditions. By Markovnikov's rule, Br adds to C2 (more substituted), giving 2-bromobutane (CH₃CHBrCH₂CH₃). Step 2: React 2-bromobutane with NaOH(aq) under reflux — nucleophilic substitution (SN2) replaces Br⁻ with OH⁻, giving butan-2-ol (CH₃CHOHCH₂CH₃). Overall: but-1-ene → 2-bromobutane → butan-2-ol.
Check Your Understanding — Multiple Choice
Q1. A student wants to synthesise ethyl propanoate. Which combination of starting materials and pathway is correct?
Q2. In a multi-step synthesis, a student needs to convert a primary alcohol to an aldehyde (not a carboxylic acid). Which combination of oxidising agent and equipment is correct?
Q3. Which of the following is NOT a valid two-step pathway within Module 7 scope?
Q4. A student wants to convert 1-chloropropane to propanoic acid in three steps. What is the correct sequence?
Q5. Which statement best explains why conc. H₂SO₄ is used as a catalyst in esterification rather than being a reactant?
Short Answer Questions
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-bromobutane, outline a four-step synthesis of butyl butanoate. For each step, write the balanced equation, conditions, and name the intermediate. Explain why distillation is used in one step and reflux in another during the oxidation sequence.
Q8. (6 marks) A student proposes the following 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 alternative pathway starting from butan-1-ol that achieves butyl butanoate. Write equations and conditions for each step.
Q1 — Answer: C
Ethyl propanoate = ethyl group (C₂H₅–, from ethanol) + propanoate (from propanoic acid). Option C correctly identifies: ethanol from ethene (hydration: H₂O, H₃PO₄, 300°C) + propanoic acid from propan-1-ol (excess K₂Cr₂O₇/H₂SO₄, reflux); then esterification (conc. H₂SO₄, reflux, ⇌). Option B is wrong — esterification requires a CARBOXYLIC ACID, not an aldehyde. Option D names the wrong ester (ethanoic acid + propan-1-ol → propyl ethanoate, not ethyl propanoate).
Q2 — Answer: B
To stop oxidation at the aldehyde, the aldehyde must be removed from contact with oxidant before further oxidation occurs. Distillation achieves this — aldehyde has a lower boiling point than the parent alcohol, so it can be collected as it forms. Option A (reflux + excess) keeps aldehyde in contact with oxidant → over-oxidation to carboxylic acid. Option C (KMnO₄, reflux) is a stronger oxidant under reflux → carboxylic acid product. Option D (H₂SO₄, heat) gives dehydration → alkene, not oxidation.
Q3 — Answer: C
Reduction of a carboxylic acid to an alcohol is NOT in the Module 7 reaction set. Module 7 covers oxidation only — going up the oxidation ladder. There is no Module 7 reagent for the reverse direction. Options A, B, D are all valid two-step pathways using known Module 7 conditions.
Q4 — Answer: A
1-chloropropane → propanoic acid in three steps: Step 1: NaOH(aq)/reflux → propan-1-ol (substitution). Step 2: K₂Cr₂O₇/H₂SO₄/distillation → propanal (mild oxidation to aldehyde). Step 3: K₂Cr₂O₇ excess/H₂SO₄/reflux → propanoic acid (further oxidation). Option C includes conc. H₂SO₄/reflux in step 3 which is esterification conditions, not oxidation. Option B starts with an oxidation step on a haloalkane which is not a Module 7 reaction.
Q5 — Answer: B
H₂SO₄ is a catalyst — it is consumed at the start and regenerated at the end. H⁺ from H₂SO₄ protonates the carbonyl oxygen of the carboxylic acid, activating it for nucleophilic attack by the alcohol's lone pair. Additionally, concentrated H₂SO₄ acts as a dehydrating agent — absorbing water produced, reducing water concentration, and shifting equilibrium right (Le Chatelier) to increase ester yield.
Step 1: CH₃CH₂CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaBr. Conditions: NaOH(aq), reflux. Product: butan-1-ol. [1 mark] Step 2: K₂Cr₂O₇/H₂SO₄, distillation → butanal. Distillation removes butanal (lower BP than butan-1-ol) as it forms, preventing excess oxidant from further oxidising it to butanoic acid. [1.5 marks] Step 3: K₂Cr₂O₇/H₂SO₄ (excess), reflux → butanoic acid. Reflux keeps butanal in contact with excess oxidant until complete conversion to carboxylic acid. [1.5 marks] Step 4: conc. H₂SO₄ (catalyst), reflux, ⇌ → butyl butanoate. [1 mark]
Q8 — Sample Answer (6 marks)
(a) Error: "butanone → butanoic acid" is not achievable in Module 7. Butanone (butan-2-one) is a ketone — the product of oxidising a secondary alcohol (butan-2-ol). Ketones cannot be further oxidised to carboxylic acids because the carbonyl carbon has no C–H bond available for oxidation. Dead end at butanone. [2 marks] (b) From butan-1-ol: Step 1: K₂Cr₂O₇/H₂SO₄, distillation → butanal. Step 2: K₂Cr₂O₇ excess/H₂SO₄, reflux → butanoic acid. Step 3: conc. H₂SO₄ cat., reflux, ⇌ → butyl butanoate (using butan-1-ol from stock). [4 marks]
Revisit Your Think First Response
Back at the start, you were asked how an industrial chemist plans a synthesis — working backwards from a target. Now you can answer that precisely: identify the functional group in the target, find a reaction that forms it, identify what starting material that reaction requires, and repeat. For aspirin: ester group → esterification of salicylic acid with ethanoic anhydride → salicylic acid available from phenol oxidation. Each arrow in that pathway is a reaction you know.
For the Think First example: ethene → ethyl ethanoate requires three steps: (1) hydration → ethanol; (2) full oxidation → ethanoic acid; (3) esterification with more ethanol → ethyl ethanoate. Did you identify all three steps? Did you notice that ethanol plays two separate roles (starting material for oxidation AND the alcohol for esterification)?