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

Reactions of Alkynes & Alkanes, Hydration, Substitution & Combustion

Master two-step alkyne addition, the surprising ketone from alkyne hydration, why UV light is an energy source not a catalyst, and systematic combustion balancing.

Today's hook: Ethyne burns at ~3500°C in an oxy-acetylene torch, yet the same molecule is a starting material for PVC and paints. What happens when you add bromine to that triple bond, one product or two?
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Connect: map alkyne and alkane reactions

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

Ethyne (acetylene) is the fuel in an oxy-acetylene welding torch, it burns hotter than almost any other common fuel (~3500°C) because it releases enormous energy as both its C≡C triple bond and C-H bonds combust. The same compound is also the industrial starting material for making PVC plastic, vinyl acetate (used in paints and adhesives), and ethanol.

Before you read on, write down what you predict would happen if you added bromine (Br₂) to ethyne. Would you expect one product or two? Why? You will return to this at the end of the lesson.

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Reactions of Alkynes and Alkanes, Key Equations
Alkyne + 2H2 → Alkane
Full hydrogenation Ni/Pd/Pt, 150–200°C
Alkyne + H2 → Alkene (cis)
Partial hydrogenation Lindlar catalyst
Alkyne + H2O → Ketone (or ethanal)
Hydration, NOT an alcohol Dil. H2SO4 + Hg²⁺, ~60°C
Alkane + X2 → Haloalkane + HX
Halogen substitution UV light (energy source, NOT catalyst)
CxHy + O2(excess) → CO2 + H2O
Complete combustion Balance: C first, H second, O last
A triple bond has TWO pi bonds, so alkyne addition always goes in two steps. Alkane reactions produce two products (haloalkane + HX for substitution; CO₂ + H₂O for combustion).

In Lesson 8 you will apply these reaction pathways to unfamiliar molecules, predicting products for both addition and substitution reactions with full conditions.

Learning Intentions
Goals
Know
  • Alkyne addition is a two-step process, each pi bond of the triple bond reacts separately
  • (Enrichment, beyond core NSW Module 7) Alkyne hydration requires dilute H2SO4 AND Hg²⁺; the product is a ketone (not an alcohol)
  • UV light is an energy source for alkane halogenation, NOT a catalyst
Understand
  • Why alkynes need 2 equivalents of reagent for full addition but only 1 to stop at the alkene
  • Why alkyne hydration gives a ketone (enol tautomerism) rather than an alcohol
  • Why UV light initiates radical halogenation while providing energy, not lowering activation energy like a catalyst
Can Do
  • Write step 1 and step 2 equations for alkyne addition reactions
  • Identify whether a compound is an alkyne, alkene, or alkane from its molecular formula and reactivity with Br₂
  • Balance complete and incomplete combustion equations using the systematic C→H→O method
Scan these before reading
Vocab
two-step additionThe sequential addition of two equivalents of a reagent to a triple bond, converting alkyne → alkene → alkane (or alkane equivalent).
Lindlar catalystPalladium supported on calcium carbonate and deliberately deactivated with lead compounds and quinoline; it stops hydrogenation at the cis-alkene stage.
enol tautomerismThe spontaneous rearrangement of a vinyl alcohol (enol, -C(OH)=C-) to the more stable carbonyl form (keto, -CO-CH-). This is why alkyne hydration gives a ketone, not an alcohol.
geminal dihalideA compound with two halogen atoms on the same carbon (e.g. CH₃CHBr₂). Formed from full hydrohalogenation of an alkyne.
free radical substitutionA chain reaction where UV light breaks a halogen molecule (X₂) into radicals (X•) that then replace hydrogen atoms on an alkane. UV light is the energy source, not a catalyst.
incomplete combustionCombustion with limited oxygen supply, producing CO and/or soot (C) rather than CO₂. Incomplete combustion products are toxic and/or harmful to health and environment.
📚 Core Content
2

Explain: two pi bonds allow two additions

Beyond the syllabus. Alkyne addition reactions, alkane substitution and combustion are core. Lindlar catalyst, controlled partial hydrogenation, enol–keto tautomerism, mercury-catalysed hydration mechanisms and vicinal/geminal terminology are extension — exam questions stay within the required reaction families.
1
Alkyne Addition, Two Steps, Two Pi Bonds

A triple bond reacts twice, controlling which step you stop at controls the product

A triple bond has two pi bonds, so where an alkene reacts once with one equivalent of a reagent, an alkyne reacts twice, and controlling which step you stop at is the key to controlling what product you make.

The C≡C triple bond consists of one sigma bond and two pi bonds. Each pi bond can undergo addition independently. This means alkyne addition reactions proceed in two distinct steps:

  • Step 1: First pi bond opens, alkyne → alkenyl product (still has one pi bond)
  • Step 2: Second pi bond opens, alkenyl product → fully saturated product (no pi bonds)

Hydrogenation, Full and Partial

Full hydrogenation (2 eq H₂, Ni/Pd/Pt, ~150–200°C)

R-C≡C-R' + 2H2 → R-CH2-CH2-R' (alkane)

Partial hydrogenation (1 eq H₂, Lindlar catalyst)

R-C≡C-R' + H2 → R-CH=CH-R' (cis-alkene only, Lindlar stops here)

The Lindlar catalyst is palladium supported on calcium carbonate and deliberately deactivated with lead compounds and quinoline. This reduces its activity so hydrogenation stops after the first addition step rather than continuing to the alkane. Syn addition selectively produces the cis (Z) alkene.

Example: HC≡CH + H2 → CH2=CH2 (Lindlar)  |  HC≡CH + 2H2 → CH3CH3 (Ni, full)

Halogenation, Two Steps

Step 1

Equation (ethyne): HC≡CH + Br2 → CHBr=CHBr
Product name: 1,2-dibromoethene
Conditions: No catalyst, r.t.

Step 2

Equation (ethyne): CHBr=CHBr + Br2 → CHBr2CHBr2
Product name: 1,1,2,2-tetrabromoethane
Conditions: No catalyst, r.t.

Both steps decolourise bromine water an alkyne will decolourise bromine water in two sequential additions, consuming two equivalents of Br2.

Hydrohalogenation, Two Steps (Markovnikov applies each time)

Step 1

Equation (ethyne + HBr): HC≡CH + HBr → CH2=CHBr
Product name: Bromoethene (vinyl bromide)
Notes: Markovnikov

Step 2

Equation (ethyne + HBr): CH2=CHBr + HBr → CH3CHBr2
Product name: 1,1-dibromoethane (geminal dihalide)
Notes: Both Br on C2

Note the product is a geminal dihalide (both halogens on the same carbon), distinct from the vicinal dihalide produced by halogenation of an alkene (one halogen on each carbon).

Two-step halogenation of ethyne migrated to a sequence layout
Two-step halogenation of ethyne: each Br₂ molecule opens one pi bond. Full addition requires 2 equivalents of Br₂.
Exam Rule: For every alkyne addition question, state clearly whether you are showing step 1 only, step 2 only, or the full two-step result. "Excess reagent" or "full addition" means go to step 2. "One equivalent" or "partial hydrogenation" means stop at step 1. Writing only one step when the question asks for full addition costs marks.
Common Error: Students apply two-step addition to alkenes. Alkenes have only ONE pi bond, they react once and stop. Only alkynes, with TWO pi bonds, can be fully halogenated to a tetrahalo product. If you are writing a second addition step for an alkene reaction, you have confused the two compound classes.
How many equivalents of Br₂ are consumed in the FULL halogenation of ethyne (HC≡CH) to 1,1,2,2-tetrabromoethane?
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Represent: hydration produces a carbonyl compound

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Alkyne Hydration, The Special Case That Produces a Ketone

The only addition in this module where the expected product does not form

Every other alkene or alkyne addition in this module produces what you might expect, hydration of an alkyne is the one reaction that surprises students every time, because the product is not the alcohol you would predict but a ketone.

Reactants: alkyne + H2O. Conditions: dilute H2SO4 AND Hg²⁺ catalyst (mercury(II) sulfate, both required); ~60°C. Product: ketone (NOT an alcohol).

Conditions, Alkyne Hydration

REAGENT
H2O (water or dilute aqueous solution)
CATALYST
Dilute H2SO4 AND Hg²⁺ (mercuric sulfate, HgSO4), BOTH required
CONDITIONS
~60°C
EQUIPMENT
Heated glassware; fume cupboard (Hg compounds are toxic)
Why a Ketone, Not an Alcohol?

Enol Tautomerism Explained

The immediate product of adding H2O across C≡C is a vinyl alcohol (enol) a compound with an -OH group attached directly to a C=C carbon. Vinyl alcohols are thermodynamically unstable and spontaneously rearrange (tautomerise) to the more stable carbonyl form.

For all alkynes except ethyne, the stable keto form is a ketone (C=O flanked by two carbon groups).

For ethyne specifically: HC≡CH + H2O → [CH2=CHOH] → CH3CHO (ethanal, an aldehyde, not a ketone, because the carbonyl ends up at the terminal carbon).

Alkyne hydration sequence
Alkyne hydration: propyne → enol (vinyl alcohol, unstable) → propanone (ketone). The OH never survives, enol tautomerism gives the keto form instantly.
AlkyneEquationProductClass
Ethyne (HC≡CH)HC≡CH + H2O → CH3CHOEthanalAldehyde ⚠️
Propyne (CH3C≡CH)CH3C≡CH + H2O → CH3COCH3PropanoneKetone
But-1-yne (HC≡CCH2CH3)HC≡CCH2CH3 + H2O → CH3COCH2CH3Butan-2-oneKetone
But-2-yne (CH3C≡CCH3)CH3C≡CCH3 + H2O → CH3COCH2CH3Butan-2-oneKetone
Beyond the syllabus (extension): Catalytic hydration of alkynes (dilute H2SO4 with an Hg²⁺ catalyst) is not part of NSW HSC Chemistry Module 7 and is not assessed (it is included as enrichment). The syllabus requires only hydration of alkenes (addition of H2O across a C=C double bond to form an alcohol).
Extension note: If you do explore alkyne hydration, note the product is a KETONE (or ethanal from ethyne), not an alcohol. Copying the alkene hydration rule straight across to alkynes gives the wrong functional-group class.
True or False: When propyne reacts with water (H₂SO₄ + Hg²⁺, 60°C), the product is propan-2-ol (an alcohol).
4

Respond: substitution and combustion of alkanes

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Alkane Reactions, Halogen Substitution

UV light is the energy source, not a catalyst, and this distinction is heavily tested

Alkanes are the least reactive organic class, but they do react with halogens under one specific condition, and that condition is a UV light source, not a catalyst, a fact the HSC tests repeatedly.

Alkanes have only sigma bonds, no pi bonds, no polar functional groups. They resist most reagents at room temperature. Halogen substitution under UV light is one of only two reaction types available to alkanes.

Conditions, Alkane Halogenation

REAGENT
Cl2 or Br2 (gas or vapour phase with the alkane)
CATALYST
NONE, UV light is NOT a catalyst
CONDITIONS
UV light (sunlight or UV lamp) as energy source
EQUIPMENT
Transparent glassware (to allow UV penetration); fume cupboard

In this reaction, one hydrogen atom on the alkane is replaced by a halogen atom. HX is produced as a second product, this is what makes it a substitution reaction, not addition.

  • CH4 + Cl2 → CH3Cl + HCl  (UV light)
  • CH3CH3 + Br2 → CH3CH2Br + HBr  (UV light)

️ Critical: UV Light Is an Energy Source, NOT a Catalyst

A catalyst is a substance that lowers the activation energy of a reaction and is regenerated unchanged, it is matter. UV light is electromagnetic radiation, it is energy, not matter. UV light provides the energy to break the Cl-Cl or Br-Br bond by homolysis, generating halogen radicals (X•) that initiate the chain reaction. It is absorbed in the initiation step and is not regenerated. Calling UV light a "catalyst" is factually wrong and is one of the most penalised errors in Module 7.

The reaction proceeds by free radical chain mechanism UV light initiates the chain by breaking X-X bonds homolytically. Alkyl radicals then propagate the chain by reacting with further X2. The mechanism is not required for HSC, but knowing UV light = energy source (initiation) is required.

Further substitution: With excess Cl2, multiple substitutions occur:

CH4 → CH3Cl → CH2Cl2 → CHCl3 → CCl4

Exam Rule: In every response about alkane halogen substitution, write "UV light (energy source)" never "UV light (catalyst)." A useful memory rule: a catalyst is MATTER (a substance); UV light is ENERGY. Energy cannot be a catalyst.
Common Error: Students write "alkanes react with bromine water under UV light." Bromine WATER does not react with alkanes the same way pure Br₂ does, the aqueous environment interferes. The reaction requires Br2 in vapour or organic solvent phase. Bromine water is used for the unsaturation test (alkenes/alkynes), not for alkane halogenation.
Methane reacts with Cl₂ under UV light. Which statement about UV light in this reaction is correct?
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Combustion of Hydrocarbons, Complete and Incomplete

Balance using C → H → O order; know the products and impacts of each type

All hydrocarbons undergo combustion with oxygen. The products depend entirely on whether oxygen supply is sufficient.

ConditionProductsObservationImpact
Complete combustion (excess O2)CO2 + H2O onlyClean blue flameCO2 → greenhouse effect
Incomplete combustion (limited O2)CO + H2O (or C + H2O)Yellow/sooty flameCO = toxic; C (soot) = respiratory harm
Systematic Method

Balancing Combustion Equations: C → H → O

Step 1: Balance C, coefficient on CO2 equals number of C atoms in fuel.

Step 2: Balance H, coefficient on H2O equals half the number of H atoms in fuel.

Step 3: Balance O, count all O atoms on the right, divide by 2 to get the O2 coefficient. If this is a fraction, multiply the whole equation by 2.

Example, propane combustion:

  • C3H8: 3C → 3CO2  |  8H → 4H2O  |  O right: 3×2 + 4×1 = 10 → O2 = 5
  • C3H8 + 5O2 → 3CO2 + 4H2O ✓

Example, butane combustion:

  • C4H10: 4C → 4CO2  |  10H → 5H2O  |  O right: 8 + 5 = 13 → O2 = 13/2
  • Multiply by 2: 2C4H10 + 13O2 → 8CO2 + 10H2O ✓

Environmental impacts of combustion products:

  • CO: binds haemoglobin 250× more strongly than O2 → blocks oxygen transport → toxic
  • C (soot/PM2.5): respiratory and cardiovascular disease; probable carcinogen
  • CO2: greenhouse gas → enhanced greenhouse effect → climate change
  • NOx: formed when atmospheric N2 reacts with O2 at high combustion temperatures (~1500°C), the nitrogen comes from the air, NOT the fuel
Exam Rule: When balancing combustion equations, always use the systematic C → H → O method. Balancing O first leads to a cycle of adjustments. Check whether the question specifies "complete" or "incomplete", complete means products are ONLY CO2 + H2O; incomplete means state whether CO or C (or both) forms depending on the degree of oxygen restriction.
Common Error: "NOx is produced because the fuel contains nitrogen." This is wrong. Almost all common hydrocarbon fuels contain no nitrogen. NOx forms because high combustion temperatures (~1500°C) cause atmospheric N2 (from the air) to react with atmospheric O2 (from the air). The nitrogen source is the air, not the fuel.
Complete the systematic method: to balance C₃H₈ + O₂ → CO₂ + H₂O, first balance ______ (gives 3CO₂), then balance ______ (gives 4H₂O), then balance ______ to get O₂ = ______.
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Feedback: compare alkene and alkyne evidence

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Alkene vs Alkyne Addition Reactions, Key Differences

Use this comparison to avoid mixing up alkene and alkyne reactions in exam questions.

Alkene (C=C, one pi bond)

  • Reacts with 1 eq of H2, X2, or HX
  • Hydrogenation: needs Ni/Pd/Pt, 150–200°C
  • Halogenation: gives vicinal dihaloalkane
  • Hydrohalogenation: gives monohalogenated alkane
  • Hydration: gives alcohol (H3PO4, 300°C, high P)
  • Decolourises Br₂ in one addition

Alkyne (C≡C, two pi bonds)

  • Reacts with 2 eq for full addition; 1 eq to stop at alkene
  • Partial hydrogenation to cis-alkene: needs Lindlar catalyst
  • Full halogenation: gives tetrahalo alkane (2 X on each C)
  • Full hydrohalogenation: gives geminal dihaloalkane
  • Hydration: gives ketone (dil. H2SO4 + Hg²⁺, 60°C)
  • Decolourises Br₂ in two sequential additions
Three of the following are true about alkynes. Which one is false?
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High-Frequency Misconceptions
  • ✕

    "Alkyne + water → alcohol." Wrong. Alkyne hydration produces a ketone (or ethanal from ethyne) because the enol intermediate tautomerises. Alcohol is only produced from alkene hydration.

  • ✕

    "UV light is the catalyst for alkane halogenation." Wrong. UV light is an energy source that initiates radical formation, it is not a substance, not regenerated, and not a catalyst.

  • ✕

    "The Hg²⁺ catalyst alone is enough for alkyne hydration." Wrong. Both dilute H2SO4 AND Hg²⁺ must be stated. Omitting either earns partial marks at best.

  • ✕

    "An alkyne decolourises bromine water to a colourless product in a single step." Wrong. An alkyne reacts with Br₂ in two steps, it can decolourise two portions of bromine water, one after the other.

🧠 Worked Examples

Worked Example 1

Two-Step Alkyne Addition

Problem: Write equations for both steps of the reaction between but-2-yne and bromine (Br2). Name all products.

1

Identify but-2-yne: CH3C≡CCH3. Triple bond between C2 and C3. The compound is symmetrical, both C≡C carbons have one CH3 each.

2

Step 1 addition (first Br2 equivalent): First pi bond opens. One Br adds to each carbon of the C≡C.
CH3C≡CCH3 + Br2 → CH3CBr=CBrCH3
Product: 2,3-dibromobut-2-ene (a dihaloalkene, still has one C=C).

3

Step 2 addition (second Br2 equivalent): Second pi bond opens. One Br adds to each carbon of the C=C.
CH3CBr=CBrCH3 + Br2 → CH3CBr2CBr2CH3
Product: 2,2,3,3-tetrabromobutane (tetrahaloalkane, fully saturated).

4

Both steps decolourise bromine water. The alkyne consumed 2 equivalents of Br2 in full halogenation.

Answer: Step 1: CH3C≡CCH3 + Br2 → CH3CBr=CBrCH3 (2,3-dibromobut-2-ene). Step 2: CH3CBr=CBrCH3 + Br2 → CH3CBr2CBr2CH3 (2,2,3,3-tetrabromobutane). Both steps decolourise bromine water.

Worked Example 2

Alkyne Hydration, Evaluating a Prediction

Problem: (a) Write the equation for the reaction of propyne (CH3C≡CH) with water and identify the organic product. State all conditions. (b) A student predicts the product is propan-2-ol. Evaluate this prediction.

1

(a) Identify propyne: CH3C≡CH. C2 (bonded to CH3, more substituted) and C1 (terminal, bonded to H, less substituted). Markovnikov: H → C1; OH → C2.

2

Immediate product: enol CH3C(OH)=CH2 (prop-1-en-2-ol, vinyl alcohol, unstable). This tautomerises instantly to the keto form.

3

Keto form: CH3COCH3 = propanone (ketone). Equation: CH3C≡CH + H2O → CH3COCH3. Conditions: dilute H2SO4 AND Hg²⁺ catalyst, ~60°C, heated glassware/fume cupboard.

4

(b) Evaluate the prediction: The student predicts propan-2-ol (CH3CHOHCH3a secondary alcohol). This is incorrect. Propan-2-ol would form only if the enol intermediate remained as an enol (vinyl alcohol). Vinyl alcohols are thermodynamically unstable, they spontaneously tautomerise to the more stable carbonyl form (propanone). The student has confused alkyne hydration with alkene hydration. Alkene + water → alcohol. Alkyne + water → ketone. The functional group class of the product is completely different.

Answer: (a) CH3C≡CH + H2O → CH3COCH3 (propanone). Conditions: dilute H2SO4 + Hg²⁺ catalyst, ~60°C. (b) Incorrect, propan-2-ol is not formed because the enol intermediate tautomerises to propanone (a ketone). Alkyne hydration always gives a carbonyl compound, not an alcohol.

Worked Example 3

Extended Response, Compound Identification & Integration

Problem (7 marks): Four unknown compounds W, X, Y, Z are given: W (C4H6) decolourises Br₂ in two sequential additions; X (C4H8) decolourises Br₂ in one addition; Y (C4H10) does not decolourise Br₂ but burns cleanly in excess O2; Z is produced when Y reacts with Cl₂ under UV light. (a) Identify the functional group class of W, X, Y. (b) Give the molecular formula and name of Z. (c) Write the balanced equation for complete combustion of Y (butane). (d) W reacts with H2O + H2SO4 + Hg²⁺ at 60°C, identify the product and explain why it is not an alcohol. (e) X reacts with H2 over Ni at 150°C, identify the product and reaction type.

1

(a) Identify W, X, Y: W (C4H6) = CnH2n-2 → alkyne. Two sequential Br₂ additions confirm two pi bonds (C≡C). X (C4H8) = CnH2n → alkene. One Br₂ addition confirms one pi bond (C=C). Y (C4H10) = CnH2n+2 → alkane. No Br₂ reaction confirms no pi bonds; combustion confirms hydrocarbon.

2

(b) Compound Z: Y (C4H10) + Cl2 + UV light → halogen substitution. One H replaced by Cl; HCl also produced. Formula: C4H9Cl (chlorobutane). If Y is butane: a mixture of 1-chlorobutane and 2-chlorobutane forms, because substitution can occur at any C-H bond (poor selectivity).

3

(c) Combustion of butane (C4H10): C: 4C → 4CO2. H: 10H → 5H2O. O: 8+5=13 → O2=13/2. Multiply by 2: 2C4H10 + 13O2 → 8CO2 + 10H2O ✓

4

(d) W (alkyne) + H2O/H2SO4/Hg²⁺: Alkyne hydration. For a C4 alkyne (but-1-yne or but-2-yne), Markovnikov hydration produces butan-2-one (CH3COCH2CH3), a ketone. It is not an alcohol because: the initial addition product is a vinyl alcohol (enol), which is thermodynamically unstable and spontaneously tautomerises to the more stable keto form (ketone). The OH group does not persist, it converts to C=O.

5

(e) X (alkene, C4H8) + H2/Ni/150°C: Hydrogenation. H2 adds across the C=C. Product: C4H10 (butane, now fully saturated). Reaction type: hydrogenation (addition).

Answer: (a) W = alkyne; X = alkene; Y = alkane. (b) Z = C4H9Cl (chlorobutane, mixture of 1-chloro and 2-chloro); UV light is the energy source. (c) 2C4H10 + 13O2 → 8CO2 + 10H2O. (d) Butan-2-one (ketone), not alcohol because enol tautomerises to keto form. (e) Butane (C4H10); hydrogenation.

Interactive Tool, Hydrocarbon Reactions Predictor Open fullscreen ↗
True or false? Acid-catalysed hydration of an alkyne (with an Hg²⁺ catalyst) ultimately gives a ketone rather than a stable alcohol.
🔬 Predict, Then Reveal +8 XP
Predict: propyne (CH≡CCH₃) reacts with excess H₂ in the presence of a Pd/C catalyst. Before reading, predict (a) the product after one mole of H₂ and (b) after two moles of H₂, and (c) why Lindlar's catalyst would stop the reaction at the first addition.
Your predictionExpert answerCompare
6

Apply: tests, equations and product pathways

Complete the Learn phase to unlock Practice.

✏️ Activities
A1
Activity 1, Alkyne and Alkane Reaction Practice

Write the equation(s), state all conditions, and name the product(s) for each reaction.

Reactants / Task
1. Ethyne + 2 eq Br2 (full halogenation, write both steps)
2. Propyne + H2O (alkyne hydration, state both catalysts)
3. But-2-yne + H2 (Lindlar catalyst, partial hydrogenation only)
4. Methane + Cl2 (state conditions correctly, is UV light a catalyst?)
Your equation, conditions, and product names
A2
Activity 2, Identify and Classify from Evidence

For each scenario, identify the compound class, write any relevant equations, and evaluate any student claims.

Scenario
A. Compound P (C₅H₈) decolourises two portions of bromine water.
B. A student attempts alkyne hydration using Hg²⁺ alone at 60°C. The reaction fails.
C. A student reports "burning butane in limited air reduces greenhouse gas emissions."
D. Ethyne + H₂O/H₂SO₄/Hg²⁺ → the student expects ethanal. Evaluate.
Your analysis
📝 Check Your Understanding
MC
Randomised Question Bank

Complete five questions drawn from the ten aligned alkyne-and-alkane reaction items.

SA
Exam-Style Practice

1. Explain why the Lindlar catalyst stops hydrogenation at the alkene stage when a normal Ni catalyst would continue to the alkane. (3 marks)

2. Compare the products formed when propyne reacts with H2O (dilute H2SO4 + Hg²⁺, 60°C) versus when propene reacts with H2O (H3PO4, 300°C, high pressure). Include equations and explain why the products belong to different functional group classes. (4 marks)

3. A student uses only the bromine water test to identify three compounds as an alkane, an alkene, and an alkyne respectively. Evaluate whether this single test is sufficient and describe a complete testing protocol to correctly identify all three compound classes. (5 marks)

Show All Answers

Activity 1, Alkyne and Alkane Practice

1. Ethyne full halogenation: Step 1: HC≡CH + Br2 → CHBr=CHBr (1,2-dibromoethene, dihaloalkene). Step 2: CHBr=CHBr + Br2 → CHBr2CHBr2 (1,1,2,2-tetrabromoethane, tetrahaloalkane). No catalyst; room temperature; fume cupboard. Both steps decolourise bromine water.

2. Propyne hydration: CH3C≡CH + H2O → CH3COCH3 (propanone). Product class: ketone (not alcohol). Conditions: dilute H2SO4 AND Hg²⁺ catalyst, ~60°C, heated glassware, fume cupboard.

3. But-2-yne partial hydrogenation (Lindlar): CH3C≡CCH3 + H2 → CH3CH=CHCH3 (but-2-ene, specifically cis isomer). The supported palladium catalyst is deliberately deactivated with lead compounds and quinoline, so it stops after one H2 addition.

4. Methane + Cl2: CH4 + Cl2 → CH3Cl + HCl. UV light is an energy source, not a catalyst, it provides energy to break the Cl-Cl bond (homolysis), generating Cl radicals that initiate the chain. It is electromagnetic radiation (energy), not matter, and is not regenerated.

Activity 2, Identify and Classify

A. Compound P (C5H8) = alkyne (CnH2n-2 for n=5; two sequential Br₂ additions confirm two pi bonds). Full halogenation product: C5H8Br4 (tetrahaloalkane, 4 Br atoms added across the two pi bonds).

B. The reaction requires BOTH dilute H2SO4 and Hg²⁺, omitting H2SO4 means the reaction does not proceed. Hg²⁺ is needed as a Lewis acid catalyst that activates the C≡C toward nucleophilic attack by water. H2SO4 provides the acidic medium required. Without both, the activation barrier cannot be overcome.

C. The claim is incorrect. Limited oxygen supply gives incomplete combustion, producing toxic carbon monoxide (CO) and/or soot (C) instead of CO2. CO is far more immediately dangerous than CO2. It binds haemoglobin 250× more strongly than O2, blocking oxygen transport and causing toxicity. While less CO2 is produced, the incomplete combustion products are a more acute health hazard.

D. The student is correct. Ethyne (HC≡CH) is the special case, it is symmetrical and terminal. When H2O adds, the enol intermediate is vinyl alcohol (CH2=CHOH), which tautomerises to ethanal (CH3CHO). The carbonyl ends up at C1, making it an aldehyde. Ethyne is the only alkyne that gives an aldehyde; all other alkynes give ketones.

Multiple Choice Answers

1. Correct response: Propyne + excess HBr (two steps, Markovnikov each time). Step 1: H to C1 (terminal, 1H), Br to C2 → CH3CBr=CH2 (2-bromopropene). Step 2: CH3CBr=CH2 + HBr, C1 (=CH2) has 2H, Markovnikov H to C1, Br to C2 (already has Br) → CH3CBr2CH3 = 2,2-dibromopropane (geminal dihalide). Option D would be halogenation (Br₂), not hydrohalogenation (HBr).

2. Correct response: A catalyst is a substance (matter) that lowers activation energy and is regenerated. UV light is energy (electromagnetic radiation), not matter. It is absorbed in the initiation step to break Cl-Cl bonds. Options A and D incorrectly apply the definition of a catalyst to a non-substance.

3. Correct response: But-1-yne hydration: H to C1 (terminal, more H), OH to C2 (Markovnikov). Enol intermediate (but-1-en-2-ol) instantly tautomerises to butan-2-one (CH3COCH2CH3), a ketone. Option B (butan-2-ol) would only form if the enol did not tautomerise. Terminal alkynes beyond ethyne give ketones, not aldehydes; only ethyne gives an aldehyde.

4. Correct response: Two sequential Br₂ additions (consuming 2 equivalents) confirms two pi bonds → alkyne. Option A alone cannot distinguish alkyne from alkene (both decolourise Br₂). Options B and C describe reactions shared by both compound classes.

5. Correct response: C: 3 → 3CO2. H: 8 → 4H2O. O right: 6+4=10 → O2 = 5. C3H8 + 5O2 → 3CO2 + 4H2O ✓. Option B has wrong O2 coefficient. Option C gives incomplete combustion product.

Short Answer Model Answers

Q1 (3 marks): The Lindlar catalyst is palladium supported on calcium carbonate and deliberately deactivated with lead compounds and quinoline [1]. It catalyses syn addition of the first equivalent of H2 across a triple bond to produce a cis alkene, but does not continue reducing the resulting C=C bond to an alkane [1]. A standard Ni, Pd, or Pt catalyst continues hydrogenation to the fully saturated alkane [1].

Q2 (4 marks): Propyne: CH3C≡CH + H2O → CH3COCH3 (propanone, a ketone). Conditions: dilute H2SO4 + Hg²⁺, ~60°C [1]. Propene: CH3CH=CH2 + H2O → CH3CH(OH)CH3 (propan-2-ol, an alcohol). Conditions: H3PO4, 300°C, high pressure [1]. The products belong to different functional group classes [1] because propene hydration produces a stable alcohol directly; propyne hydration initially produces a vinyl alcohol (enol) which is thermodynamically unstable and spontaneously tautomerises to propanone (ketone), so the OH group does not persist [1].

Q3 (5 marks): The bromine water test alone is not sufficient [1]. Both alkenes and alkynes decolourise bromine water; an alkane does not. The test can separate the alkane from the other two, but cannot distinguish alkene from alkyne on decolourisation alone [1]. A complete testing protocol: (1) Add Br₂(aq) to each compound. The alkane will not decolourise. The alkene and alkyne will both decolourise [1]. (2) For the two that decolourise: add a second equivalent of Br₂(aq). The alkyne will decolourise again (second pi bond reacts); the alkene will not (it has been fully consumed in step 1) [1]. (3) To confirm the alkane: react with Cl₂ or Br₂ under UV light, a haloalkane + HX product is formed by substitution, confirming the saturated compound is an alkane [1].

How did your thinking change?

Back at the start, you predicted what happens when bromine water is added to ethyne. Now you know: ethyne undergoes two successive electrophilic addition reactions. Step 1: Br2 adds across the triple bond to form 1,2-dibromoethene (CHBr=CHBr), the bromine water decolourises as the first pi bond opens. Step 2: if excess Br2 is present, a second molecule of Br2 adds across the remaining double bond to give 1,1,2,2-tetrabromoethane (CHBr2CHBr2). Each step adds two bromine atoms and converts one degree of unsaturation. The complete decolourisation of two portions of bromine water indicates the triple bond has been fully consumed. An alkene would decolourise only one portion, the two-step behaviour with 2 equivalents of Br2 is the distinguishing test for an alkyne.

Return to your original prediction and compare it to this answer:

  • Ethyne has a C≡C triple bond with two pi bonds so bromine adds in two separate steps
  • Step 1: HC≡CH + Br2 → CHBr=CHBr (1,2-dibromoethene), first pi bond opens
  • Step 2: CHBr=CHBr + Br2 → CHBr2CHBr2 (1,1,2,2-tetrabromoethane), second pi bond opens
  • Both steps produce a colourless product, two portions of bromine water are decolourised
  • This two-step behaviour, consuming 2 equivalents of Br₂, is the key test that distinguishes an alkyne from an alkene
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Review: retrieve the reaction pathways

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🏆 Review

How many pi bonds does an alkyne have? Why does this mean alkyne addition requires two steps?

State all conditions for alkyne hydration and explain why the product is a ketone, not an alcohol.

Why is UV light NOT a catalyst for alkane halogenation? What is it?

Write the balanced equation for complete combustion of butane (C₄H₁₀) using the C→H→O method.

What is the difference between a geminal dihalide and a vicinal dihalide? Which one is produced by full hydrohalogenation of an alkyne?