Year 12 Chemistry Module 7, Organic Chemistry ⏱ ~50 min Lesson 19 of 23 Core

Organic Reaction Pathways, Synthesis & Multi-Step Problems

Reaction pathways are the HSC exam skill that separates a developing answer from a top-band, 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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Connect: analyse the target molecule

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
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Think First, Before You Read

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?

Complete Reaction Conditions Reference

ALKENE → ALCOHOL: + H₂O (steam), H₃PO₄ cat., ~300°C, high pressure
ALKENE → HALOALKANE: + HX, no cat., r.t. (Markovnikov, X to more substituted C)
ALKANE → HALOALKANE: + X₂, UV light, r.t.
HALOALKANE → ALCOHOL: + NaOH(aq), reflux
PRIMARY ALCOHOL → ALDEHYDE: + K₂Cr₂O₇/H₂SO₄, distillation (remove aldehyde before over-oxidation)
PRIMARY ALCOHOL → CARBOXYLIC ACID: + K₂Cr₂O₇/H₂SO₄ (excess), reflux
SECONDARY ALCOHOL → KETONE: + K₂Cr₂O₇/H₂SO₄, reflux (DEAD END, ketone cannot be further oxidised)
CARBOXYLIC ACID + ALCOHOL ⇌ ESTER + H₂O: conc. H₂SO₄ cat., reflux
ESTER + NaOH → CARBOXYLATE + ALCOHOL: conc. NaOH, reflux, → (saponification, irreversible)
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.
Oxidation state changesPrimary alcohol → aldehyde → carboxylic acid (oxidation); carboxylic acid → alcohol (reduction).
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
2

Explain: navigate the complete reaction map

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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 XP Quick Check

A student wants to go from an alkane to an alcohol. How many steps are required in Module 7?

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Represent: use a pathway algorithm

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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 a valid connected path on the map, then write each step with full conditions.

Route quality: Fewer steps often improve overall yield and simplicity, but the shortest route is not automatically best. Also evaluate safety, selectivity, purification, available reagents and waste.
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 XP True / False

True or False: To obtain an aldehyde from a primary alcohol, you use K₂Cr₂O₇/H₂SO₄ with REFLUX equipment.

4

Respond: build pathways from simple to complex

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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):

Step 1

Equation: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
Conditions: K₂Cr₂O₇/H₂SO₄ (excess), heat, reflux. Orange → green.
Intermediate: Ethanoic acid

Step 2

Equation: CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O
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).
Intermediate: Propanal

Step 2

Equation: CH₃CH₂CHO + [O] → CH₃CH₂COOH
Conditions: K₂Cr₂O₇/H₂SO₄ (excess), reflux.
Intermediate: Propanoic acid

Step 3

Equation: CH₃CH₂COOH + CH₃CH₂CH₂OH ⇌ CH₃CH₂COOC₃H₇ + H₂O
Conditions: Conc. H₂SO₄ catalyst, reflux. Reversible (⇌).
Intermediate: Propyl propanoate
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 XP Complete 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.

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Feedback: avoid dead ends and condition traps

04

4, Complex Pathway Problems: Four-Step Syntheses and Unknown Starting Materials

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:

StepEquationConditionsWhy these conditions?
Step 1CH₃CH₂CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaBrNaOH(aq), refluxAqueous NaOH → substitution → alcohol. Alcoholic NaOH → elimination → alkene instead.
Step 2CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂OK₂Cr₂O₇/H₂SO₄, distillationButanal (BP 75°C) is more volatile than butan-1-ol (BP 118°C), distilling removes aldehyde before excess oxidant converts it to carboxylic acid.
Step 3CH₃CH₂CH₂CHO + [O] → CH₃CH₂CH₂COOHK₂Cr₂O₇/H₂SO₄ (excess), refluxReflux keeps everything in the flask, excess oxidant completes the conversion to carboxylic acid.
Step 4CH₃CH₂CH₂COOH + CH₃CH₂CH₂CH₂OH ⇌ CH₃CH₂CH₂COOC₄H₉ + H₂OConc. H₂SO₄ (catalyst), reflux, reversible (⇌)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 XP Quick 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 (→).
6

Apply: reveal a complete synthesis step by step

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.
1
Identify Intermediate A: Ethyl ethanoate = ethyl (from ethanol) + ethanoate (from ethanoic acid). Esterification needs a carboxylic acid. Therefore Intermediate A = ethanoic acid (CH₃COOH).
2
Step 1, ethanol → ethanoic acid:
CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
Conditions: K₂Cr₂O₇/H₂SO₄ (excess), heat under reflux. Colour change orange → green.
3
Step 2, esterification:
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOC₂H₅ + H₂O
Conditions: conc. H₂SO₄ (catalyst), heat under reflux. Reversible (⇌), yield ~65%.
A
Intermediate A = ethanoic acid. Step 1: excess K₂Cr₂O₇/H₂SO₄, reflux. Step 2: conc. H₂SO₄ cat., reflux, ⇌.
06

Example 2, Three Parallel Pathways from 1-Bromopropane (Intermediate)

Problem: Starting from 1-bromopropane (CH₃CH₂CH₂Br), outline: (a) One-step synthesis of propan-1-ol. (b) Two-step synthesis of propanal. (c) Two-step synthesis of propanoic acid.

1
(a) 1-bromopropane → propan-1-ol (1 step):
CH₃CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂OH + NaBr
Conditions: NaOH(aq), reflux. Intermediate: propan-1-ol.
2
(b) → propanal (2 steps):
Step 1: same as (a) → propan-1-ol.
Step 2: CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O
Conditions: K₂Cr₂O₇/H₂SO₄, distillation. Intermediate: propanal (aldehyde).
3
(c) → propanoic acid (2 steps):
Step 1: same as (a) → propan-1-ol.
Step 2: CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O
Conditions: K₂Cr₂O₇/H₂SO₄ (excess), reflux. Product: propanoic acid.
A
(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.
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: select and justify a valid pathway

🔬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
Check Your Understanding, Multiple Choice
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.

1

Review: retrieve the whole reaction map

Check what actually stuck

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)?

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