Year 12 Chemistry Module 7, Organic Chemistry ★ Consolidation ⏱ ~50 min Lesson 8 of 23 Optional consolidation

Hydrocarbon Reactions Mastery, Conditions, Products & Spot the Error

Every mark lost in Module 7 hydrocarbon reaction questions comes down to one of three errors, wrong conditions, wrong product, or wrong reaction type. This lesson fixes all three.

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Today's hook: In 2021, a chemistry student's exam paper included an error in a reaction scheme, a product labelled as a primary alcohol was actually a secondary alcohol based on Markovnikov's rule. That single classification error cost 3 marks. By the end of this lesson you will be able to catch and correct errors like these in any hydrocarbon reaction question, because you understand not just the products but why those products form.
0/5TASKS
Optional consolidation

This lesson does not introduce new compulsory content. It consolidates the hydrocarbon reactions from Lessons 5 to 7. Work through it if those reactions are not yet automatic; the core route does not require it, and questions here that turn on Lindlar catalysis, tautomerism or specialist product terminology are extension.

1

Connect: diagnose your reaction-condition gaps

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
No New Syllabus Dot Points

Consolidation Lesson, No new syllabus dot points. This lesson deepens and stress-tests the content from L06 (alkene addition reactions) and L07 (alkyne addition, alkane halogenation, combustion). Work through the conditions table, the Spot the Error cards, and the recall challenge. Any condition set you cannot reproduce from memory is a potential lost mark in an exam.

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Fix These Before You Go Further
Wrong: Alkene hydration only requires water and heat, no catalyst is needed.
Right: Alkene hydration requires dilute H₂SO₄ as a catalyst. The acid protonates the double bond first, creating a carbocation intermediate that water then attacks. Without acid catalysis, the reaction is impractically slow. For alkynes, both H₂SO₄ AND Hg²⁺ catalyst are required, and the reaction produces a ketone (not an alcohol) via enol tautomerisation.
Before You Begin

A student is given five reaction equations on a test. They get the reaction type right every time but lose marks on every single question. The marker has written the same comment five times: "conditions incomplete." The student used the right reagents but forgot the catalyst on two questions, wrote "UV light (catalyst)" on one, forgot to mention high pressure on the hydration, and wrote "heat" instead of "H2SO4 + Hg²⁺" on the alkyne hydration.

Before working through this lesson, write down from memory how many hydrocarbon reactions from L06 and L07 you can fully describe, reagent, catalyst, conditions, equipment, without looking at your notes. Count them. That number will be higher at the end.

Learning Intentions
goals

Know

  • Reaction conditions for addition, substitution, combustion, and elimination reactions
  • Vicinal dihalides form on adjacent carbons; geminal dihalides form on the same carbon
  • Multi-step synthesis requires planning each transformation sequentially

Understand

  • How to work backwards from a product to identify the starting material and conditions
  • Why high pressure and catalyst conditions are required for specific transformations
  • Common structural errors in reaction equations and how to fix them

Can Do

  • Identify reaction type from conditions and predict the product
  • Work backwards from product to determine starting material and reagents
  • Spot errors in reaction equations and provide corrected versions
Scan these before reading
vocab
Vicinal dihalideA compound with two halogen atoms on adjacent carbon atoms, formed by addition of halogen (X₂) to an alkene.
Geminal dihalideA compound with two halogen atoms on the same carbon atom, formed by two-step hydrohalogenation of an alkyne.
Markovnikov's ruleIn addition of HX to an unsymmetrical alkene, H attaches to the carbon with more H atoms already attached (more stable carbocation forms on more substituted carbon).
HydrationAddition of water across a double or triple bond. Alkene: H₂SO₄ catalyst, steam, ~300°C, high pressure. Alkyne: H₂SO₄ + Hg²⁺, ~60°C.
HydrogenationAddition of H₂ across a double or triple bond. Full: Ni/Pd/Pt, 150–200°C. Partial (to alkene): Lindlar catalyst.
Multi-step synthesisA sequence of chemical reactions designed to convert a starting material into a target product through planned intermediate steps.
CONDITIONS REFERENCE TABLE
2

Explain: match each reaction to its complete conditions

01
Conditions Reference, All Eight Reactions

Every exam question tests whether you know the complete conditions set. Reproduce this table from memory as part of your consolidation. These reaction patterns build directly on Lessons 5–7, any reaction involving alkanes, alkenes, or alkynes covered there applies here, and this lesson connects them into a single systematic framework.

Reaction Reagent(s) Catalyst Conditions Product type
Alkene hydrogenation H₂ Ni (or Pd/Pt) 150–200°C, high pressure Alkane
Alkene halogenation X₂ (Cl₂ or Br₂) None Room temp, no solvent or CCl₄ Vicinal dihalide
Alkene hydrohalogenation HX (HCl or HBr) None Room temp; Markovnikov applies Haloalkane
Alkene hydration H₂O (steam) H₃PO₄ or dil. H₂SO₄ ~300°C, high pressure (~65 atm) Alcohol (Markovnikov)
Alkyne hydrogenation (full) 2H₂ Ni/Pd/Pt 150–200°C, high pressure Alkane
Alkyne hydrogenation (partial) H₂ (1 eq) Lindlar catalyst Mild; stops at cis-alkene cis-Alkene
Alkyne hydration H₂O H₂SO₄ AND Hg²⁺ ~60°C, fume cupboard (Hg toxic) Ketone (via enol tautomerism)
Alkane halogen substitution X₂ (Cl₂ or Br₂) UV light (energy source) UV irradiation (not heat), gas phase Haloalkane + HX
UV light is an energy source for free radical initiation, not a catalyst. Writing "UV light (catalyst)" is a common error that loses marks.

Quick Check: True or False, The UV light in alkane halogenation is acting as a catalyst because it speeds up the reaction without being consumed.

3

Represent: distinguish vicinal and geminal products

02
Vicinal vs Geminal Dihalides, Telling Them Apart

Both dihalide types arise from hydrocarbon reactions, but they form by different mechanisms from different starting materials. Being able to identify which type is present, and work backwards to the starting material, is a key HSC skill.

Vicinal Dihalide

Location: Halogen atoms on adjacent (neighbouring) carbons, C1 and C2

Formed by: Halogenation of an alkene (X₂ addition across C=C)

Example: Propene + Cl₂ → 1,2-dichloropropane (Cl on C1 and C2)

Conditions: Cl₂ or Br₂, room temp, no catalyst

Geminal Dihalide

Location: Both halogen atoms on the same carbon

Formed by: Two-step hydrohalogenation of an alkyne (2 × HX, Markovnikov each step)

Example: Propyne + 2HBr → 2,2-dibromopropane (both Br on C2)

Conditions: Excess HX, room temp, no catalyst, two equivalents

To identify which alkyne gives a specific geminal dihalide, remove both halogens from the same carbon and reform C≡C between that carbon and the adjacent one. E.g., 2,2-dibromopropane → C2 has both Br → remove both → reform C≡C at C2–C1 → propyne (CH₃C≡CH).

Microtask: A compound has the structure CH₃CBr₂CH₃ (2,2-dibromopropane). What type of dihalide is this, and what is its alkyne precursor?

4

Respond: plan transformations in sequence

03
Multi-Step Synthesis, Planning Each Transformation

Multi-step synthesis questions give you a starting material and a target product and ask you to plan the sequence of reactions. The key skill is choosing the right reaction type at each step.

Strategy: Propyne as a Synthesis Hub
Propyne (CH₃C≡CH) can be converted to five different products depending on which reaction you apply:
Target productReaction typeKey condition
Propane (CH₃CH₂CH₃)Full hydrogenation2H₂, Ni, 150–200°C, high P
cis-Propene (CH₃CH=CH₂)Partial hydrogenation1H₂, Lindlar catalyst, mild
Propanone (CH₃COCH₃)Alkyne hydrationH₂O, H₂SO₄ + Hg²⁺, ~60°C
2-bromopropene (CH₃CBr=CH₂)Hydrohalogenation (1 eq)1 eq HBr, r.t., Markovnikov step 1
2,2-dibromopropane (CH₃CBr₂CH₃)Hydrohalogenation (2 eq)2 eq HBr, r.t., Markovnikov both steps
Beyond the syllabus (extension): Selective partial hydrogenation of an alkyne to an alkene using a Lindlar catalyst (poisoned palladium) is not part of NSW HSC Chemistry Module 7 and is not assessed; it is shown here as enrichment. NSW requires only full hydrogenation of an unsaturated hydrocarbon with H₂ and a Ni/Pd/Pt catalyst.
Beyond the syllabus (extension): Catalytic hydration of an alkyne to a ketone (H₂SO₄ with an Hg²⁺ catalyst, via enol–keto tautomerism) is beyond NSW HSC Chemistry Module 7 and is not assessed. The NSW-required hydration reaction is addition of H₂O across a C=C double bond (alkene → alcohol).

Microtask: A student wants to convert propyne to propanone. Which set of conditions would achieve this correctly?

5

Feedback: find and repair reaction errors

04
Spot the Error, Six Common Exam Mistakes

Each of the errors below matches a real exam-style mistake. Identify the error before reading the explanation, the act of spotting it first reinforces the correct version.

Error 1: Propene + HCl → 1-chloropropane (HCl addition, Markovnikov)

Fix: Markovnikov says H adds to C1 (has 2H), Cl to C2 (has 1H, more substituted). Product is 2-chloropropane, not 1-chloropropane. The product name given is the anti-Markovnikov product.

Error 2: "Propene decolourises bromine water, therefore propene is confirmed to be an alkene (not an alkyne)."

Fix: Bromine water decolourisation confirms unsaturation (C=C or C≡C), but it cannot distinguish an alkene from an alkyne, alkynes also decolourise Br₂(aq). The conclusion is too strong. Additional evidence is needed.

Error 3: Propene + H₂O → propan-1-ol. Conditions: H₂SO₄, 300°C.

Fix: Two errors. (1) Markovnikov: OH goes to C2 (more substituted), so the product is propan-2-ol, not propan-1-ol. (2) Conditions are incomplete, high pressure (~65 atm) is missing. Full: H₂O/steam, H₃PO₄, ~300°C, HIGH PRESSURE.

Error 4: Alkyne hydration: propyne + H₂O → propan-2-ol. Conditions: H₂SO₄, ~60°C.

Fix: Two errors. (1) The product of alkyne hydration is a ketone (propanone), not an alcohol. The enol intermediate tautomerises to a ketone. (2) Hg²⁺ catalyst is missing, both H₂SO₄ AND Hg²⁺ are required.

Error 5: Ethane + Br₂ → bromoethane. Conditions: UV light (catalyst), room temp.

Fix: UV light is not a catalyst, it is an energy source that initiates the free radical chain by providing energy to break Br–Br homolytically. A catalyst is regenerated and lowers activation energy; UV light is absorbed to produce radicals. Write: "UV light (energy source for radical initiation)", not "catalyst".

Error 6: But-2-yne + excess Br₂ → 2,3-dibromobutene (stopping at the alkene stage).

Fix: Excess Br₂ will not stop at the alkene stage, the second pi bond also undergoes addition. With excess Br₂, but-2-yne gives 2,2,3,3-tetrabromobutane. You must specify 1 equivalent (1 eq) of Br₂ to obtain the dibromoalkene product.

Odd One Out: Three of the following statements about hydrocarbon reactions are correct. Which one contains an error?

6

Apply: reverse a reaction pathway

05
Working Backwards, Identifying Starting Materials from Products

Many HSC questions give you the product and ask for the starting material and conditions. The strategy is to identify the functional group on the product, determine what bond it replaced, and reverse the reaction.

1 Identify the functional group on the product, alcohol, haloalkane, dihalide, ketone, etc.
2 Determine what was added H₂ (hydrogenation), X₂ (halogenation), HX (hydrohalogenation), H₂O (hydration)
3 Count the atoms added one molecule of X₂ → 2 halogens (vicinal); two molecules of HX → geminal dihalide from alkyne
4 Remove the added atoms from the product and reform the pi bond(s) at the correct position
5 Identify conditions from the reaction table, state reagent, catalyst, temperature, pressure
For an alcohol product: if the OH is on a secondary or tertiary carbon, it came from Markovnikov hydration of an alkene. Identify which C=C position gives that OH location.

Fill the Gap: The product 2-methylbutan-2-ol has an OH group on a [tertiary / secondary / primary] carbon. This is consistent with Markovnikov addition of water to a(n) [alkene / alkyne / alkane]. The catalyst required is [Ni / H₂SO₄ or H₃PO₄ / H₂SO₄ + Hg²⁺].

Interactive Tool, Hydrocarbon Reactions Predictor Open fullscreen ↗
True or false? Combustion is common to all hydrocarbons, but addition reactions occur only with unsaturated hydrocarbons (alkenes and alkynes).
7

Practice: build, diagnose and explain reaction pathways

🔀 Sort the Steps +7 XP
To solve a reverse-synthesis problem (identify starting material given the product), place these steps in the correct order:
Identify the required reagent/condition for that reaction type
Work backwards to determine what functional group the starting material must have
Check that the identified starting material and conditions are chemically realistic
Identify the functional group in the target product
Determine which reaction type produces that functional group
ACTIVITIES
A1
Multi-Step Synthesis from Propyne

Propyne (CH₃C≡CH) is the starting material for each of the following target products. For each one, write: (i) the balanced equation, (ii) the full conditions (reagent, catalyst, temperature, pressure, equipment), and (iii) the mechanism type (addition/substitution/other).

  1. Target: Propane (CH₃CH₂CH₃)
  2. Target: cis-Propene (CH₃CH=CH₂)
  3. Target: Propanone (CH₃COCH₃)
  4. Target: 2-bromopropene (CH₃CBr=CH₂)
  5. Target: 2,2-dibromopropane (CH₃CBr₂CH₃)
A2
Find and Fix All Errors

Each equation or statement below contains one or more errors. Identify every error and write the fully corrected version. Do not assume there is only one error per item.

  1. (a) Propene + HCl → CH₃CH₂CH₂Cl (1-chloropropane). Conditions: HCl(g), room temp.
  2. (b) Propene + Br₂(aq) → decolourises. Conclusion: this confirms propene is an alkene, not an alkyne.
  3. (c) Propene + H₂O → propan-1-ol. Conditions: H₂SO₄ catalyst, 300°C.
CHECK YOUR UNDERSTANDING
MC
Test Your Knowledge

The bank draws a fresh set of five cumulative hydrocarbon-reaction questions for each attempt.

SA
Extended Response Practice

Q1 (3 marks): A student concludes that compound X is an alkene because it decolourises bromine water. Evaluate this conclusion.

Q2 (4 marks): 2-methylbutan-2-ol is produced by hydration. (a) Identify the reaction type and two possible alkene starting materials. (b) For each starting material, apply Markovnikov's rule to show both give the same product.

Q3 (5 marks): A student attempts to make propanone from propyne. In their first attempt, they use H₂O with H₂SO₄ at 60°C, no reaction occurs. In their second attempt, they use H₂O with H₃PO₄ at 300°C and high pressure, a reaction occurs but the product is not propanone. (a) Explain why the first attempt failed (2 marks). (b) Explain what product the second conditions would likely give and why the student has confused two different reactions (3 marks).

ANSWERS
Worked solutions
Reveal comprehensive answers

Activity 1, Multi-Step Synthesis from Propyne

(a) Propane: Full hydrogenation. CH₃C≡CH + 2H₂ → CH₃CH₂CH₃. Ni (or Pd/Pt) catalyst, ~150–200°C, high pressure, pressure vessel. Addition reaction (two-step hydrogenation, alkyne → alkene → alkane).

(b) cis-Propene: Partial hydrogenation. CH₃C≡CH + H₂ → CH₃CH=CH₂. Lindlar catalyst (poisoned Pd, stops at alkene stage), mild conditions. Gives specifically cis-propene. Note: using Ni without specifying Lindlar gives the alkane, not the alkene.

(c) Propanone: Alkyne hydration. CH₃C≡CH + H₂O → CH₃COCH₃. BOTH dilute H₂SO₄ AND Hg²⁺ catalyst, ~60°C, heated glassware, fume cupboard (Hg toxic). Enol intermediate (CH₃C(OH)=CH₂) tautomerises to ketone (propanone). Product is a ketone, NOT an alcohol.

(d) 2-bromopropene: Step 1 hydrohalogenation only. CH₃C≡CH + HBr (1 eq) → CH₃CBr=CH₂. No catalyst, room temperature. Must specify ONE equivalent of HBr, with excess HBr, step 2 occurs and 2,2-dibromopropane forms instead.

(e) 2,2-dibromopropane: Two-step hydrohalogenation (full). Step 1: CH₃C≡CH + HBr → CH₃CBr=CH₂ (2-bromopropene). Step 2: CH₃CBr=CH₂ + HBr → CH₃CBr₂CH₃ (2,2-dibromopropane). Both steps: no catalyst, room temperature, Markovnikov each time. Two equivalents of HBr total. Geminal because both Br are on C2 (same carbon).

Activity 2, Find and Fix All Errors

(a) Error: 1-chloropropane is the anti-Markovnikov product. Propene: C1 (=CH₂) has 2H; C2 (=CH-) has 1H. Markovnikov: H → C1, Cl → C2. Correct Markovnikov product: 2-chloropropane (CH₃CHClCH₃). Conditions are otherwise correct.

(b) Error: The conclusion "confirms propene is an alkene" is too strong. Decolourisation of bromine water confirms unsaturation (C=C or C≡C present). Alkynes also decolourise bromine water. This test cannot distinguish between an alkene and an alkyne. Corrected conclusion: "Decolourisation confirms unsaturation, the presence of C=C or C≡C. Additional evidence (e.g., testing with two equivalents of Br₂, or the molecular formula) is needed to confirm specifically an alkene."

(c) Two errors: (1) Product is wrong, Markovnikov: H → C1 (=CH₂, 2H), OH → C2 (=CH-, more substituted). Correct product: propan-2-ol (CH₃CH(OH)CH₃), not propan-1-ol. (2) Conditions are incomplete, high pressure (~65 atm) is missing. Full conditions: H₂O (steam), H₃PO₄ catalyst (or H₂SO₄), ~300°C, HIGH PRESSURE (~65 atm).

Multiple Choice Answers

Markovnikov hydrohalogenation. For an unsymmetrical unsaturated hydrocarbon, track the hydrogen and halogen positions before naming the product.

Alkyne hydration. The extension pathway uses dilute H₂SO₄ and Hg²⁺ at about 60°C and forms a ketone after enol tautomerism.

Alkane halogen substitution. UV light supplies initiation energy; it is not a catalyst.

Geminal dihalide. Two-step hydrohalogenation places both halogens on the same carbon.

Partial hydrogenation. The extension pathway requires a selective catalyst to stop reduction at the alkene.

Short Answer Model Answers

Q1 (3 marks): The observation that compound X decolourises bromine water confirms that X contains unsaturation, a C=C or C≡C bond [1]. However, the conclusion that X is specifically an alkene is not fully justified. Alkynes also contain pi bonds and also decolourise bromine water by addition [1]. Additional evidence is needed to distinguish an alkene from an alkyne, for example, testing with two equivalents of Br₂ (an alkyne would react with both) or determining the molecular formula (CₙH₂ₙ for alkene, CₙH₂ₙ₋₂ for alkyne) [1].

Q2 (4 marks): Reaction type: alkene hydration [1]. Two possible alkene starting materials: 2-methylbut-1-ene (CH₂=C(CH₃)CH₂CH₃) and 2-methylbut-2-ene (CH₃C(CH₃)=CHCH₃) [1]. For 2-methylbut-1-ene: Markovnikov, H to C1 (=CH₂, 2H), OH to C2 (=C(CH₃)-, 0H, more substituted) → OH at C2 → 2-methylbutan-2-ol ✓ [1]. For 2-methylbut-2-ene: C2 (=C(CH₃), 0H) and C3 (=CH-, 1H). Markovnikov, H to C3, OH to C2 (more substituted, 0H) → same product, 2-methylbutan-2-ol. Both alkenes give the same Markovnikov product because both have C2 as the more substituted alkene carbon [1].

Q3 (5 marks): (a) The first attempt failed because it omitted Hg²⁺ catalyst, alkyne hydration requires BOTH dilute H₂SO₄ AND Hg²⁺ [1]. Hg²⁺ acts as a Lewis acid catalyst that activates the C≡C toward water attack. Without Hg²⁺, the activation energy for alkyne hydration cannot be overcome at 60°C, the reaction does not proceed [1]. (b) H₃PO₄ at 300°C and high pressure are conditions for alkene hydration (of propene → propan-2-ol), not alkyne hydration [1]. The student has confused the two different hydration reactions: alkene hydration (H₃PO₄ or H₂SO₄, 300°C, high pressure → alcohol) vs alkyne hydration (H₂SO₄ + Hg²⁺, 60°C → ketone) [1]. If propyne did react under these conditions at all, it would not give propanone, the product obtained would be an alcohol (propan-2-ol if propyne reacted as a propene-equivalent), not a ketone; the two reactions produce products of entirely different functional group classes [1].

Callback, Spot the Error, Resolved

Back at the start you were introduced to the kind of error that costs marks in an exam: a product labelled as a primary alcohol when it should have been a secondary alcohol, a Markovnikov regiochemistry error.

Now, armed with the full set of hydrocarbon reaction patterns from this lesson, you can diagnose any error of that type:

  • Wrong product class e.g., writing an alcohol instead of a ketone for alkyne hydration. Fix: check whether the starting material is an alkene (→ alcohol) or alkyne (→ ketone via enol tautomerism).
  • Wrong Markovnikov regiochemistry e.g., OH on C1 instead of C2 for propene hydration. Fix: H always adds to the carbon with more H atoms; OH/Br/Cl adds to the more substituted carbon.
  • Wrong reaction type e.g., calling alkyne + Br₂(excess) a substitution. Fix: alkynes undergo addition across the pi bonds; substitution applies to alkanes under UV.

Return to the opening scenario: the student predicted the wrong alcohol by applying Markovnikov regiochemistry. Remember that NSW HSC Chemistry does not require you to predict which isomer forms; writing the hydration reaction itself (alkene + H₂O → alcohol) is what the syllabus assesses. The Markovnikov reasoning here is enrichment for students who want to understand why one product dominates.

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.
← Lesson 7 Lesson 9 →
8

Review: retrieve the conditions from memory

Rapid Recall, Conditions from Memory

Answer each question from memory. Aim for full conditions every time.

  1. State the full conditions for alkyne hydration (reagent, both catalysts, temperature).

  2. Explain why UV light in alkane halogenation is described as an energy source rather than a catalyst.

  3. A compound is CH₃CHBrCH₂Br. Name it, classify it as vicinal or geminal, and state the starting material and reaction that produced it.

  4. Give two reasons why partial hydrogenation of an alkyne requires the Lindlar catalyst specifically (not Ni or Pd/Pt under standard conditions).

  5. A student isolates 2,2-dibromopropane. Working backwards, identify the starting material and write a complete synthesis from it using only reactions from this module.