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.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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.
In Lesson 8 you will apply these reaction pathways to unfamiliar molecules — predicting products for both addition and substitution reactions with full conditions.
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:
R-C≡C-R' + 2H2 → R-CH2-CH2-R' (alkane)
R-C≡C-R' + H2 → R-CH=CH-R' (cis-alkene only — Lindlar stops here)
The Lindlar catalyst is palladium "poisoned" with lead acetate and quinoline — this deactivates it just enough so the reaction stops after the first addition step. It cannot continue reducing the alkene to an alkane. It selectively produces the cis (Z) alkene.
Example: HC≡CH + H2 → CH2=CH2 (Lindlar) | HC≡CH + 2H2 → CH3CH3 (Ni, full)
Both steps decolourise bromine water — an alkyne will decolourise bromine water in two sequential additions, consuming two equivalents of Br2.
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).
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).
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 | Equation | Product | Class |
|---|---|---|---|
| Ethyne (HC≡CH) | HC≡CH + H2O → CH3CHO | Ethanal | Aldehyde ⚠️ |
| Propyne (CH3C≡CH) | CH3C≡CH + H2O → CH3COCH3 | Propanone | Ketone |
| But-1-yne (HC≡CCH2CH3) | HC≡CCH2CH3 + H2O → CH3COCH2CH3 | Butan-2-one | Ketone |
| But-2-yne (CH3C≡CCH3) | CH3C≡CCH3 + H2O → CH3COCH2CH3 | Butan-2-one | Ketone |
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.
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.
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
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.
| Condition | Products | Observation | Impact |
|---|---|---|---|
| Complete combustion (excess O2) | CO2 + H2O only | Clean blue flame | CO2 → greenhouse effect |
| Incomplete combustion (limited O2) | CO + H2O (or C + H2O) | Yellow/sooty flame | CO = toxic; C (soot) = respiratory harm |
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:
Example — butane combustion:
Environmental impacts of combustion products:
Use this comparison to avoid mixing up alkene and alkyne reactions in exam questions.
"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.
Problem: Write equations for both steps of the reaction between but-2-yne and bromine (Br2). Name all products.
Identify but-2-yne: CH3C≡CCH3. Triple bond between C2 and C3. The compound is symmetrical — both C≡C carbons have one CH3 each.
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).
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).
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.
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.
(a) Identify propyne: CH3C≡CH. C2 (bonded to CH3, more substituted) and C1 (terminal, bonded to H, less substituted). Markovnikov: H → C1; OH → C2.
Immediate product: enol CH3C(OH)=CH2 (prop-1-en-2-ol — vinyl alcohol, unstable). This tautomerises instantly to the keto form.
Keto form: CH3COCH3 = propanone (ketone). Equation: CH3C≡CH + H2O → CH3COCH3. Conditions: dilute H2SO4 AND Hg²⁺ catalyst, ~60°C, heated glassware/fume cupboard.
(b) Evaluate the prediction: The student predicts propan-2-ol (CH3CHOHCH3 — a 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.
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.
(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.
(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).
(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 ✓
(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.
(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.
After 1 mol H₂: propyne → propene (CH₂=CHCH₃) — the triple bond becomes a double bond. After 2 mol H₂: propene → propane (CH₃CH₂CH₃) — the double bond is fully reduced. Lindlar's catalyst (Pd poisoned with lead acetate/quinoline) is selectively deactivated so it can only catalyse the first hydrogenation (alkyne → cis-alkene). It cannot facilitate the second addition (alkene → alkane), allowing isolation of the cis-alkene product.
Complete the Learn phase to unlock Practice.
Write the equation(s), state all conditions, and name the product(s) for each reaction.
For each scenario, identify the compound class, write any relevant equations, and evaluate any student claims.
1. What is the product of the full hydrohalogenation of propyne (CH3C≡CH) with excess HBr?
2. Which statement about UV light in the reaction CH4 + Cl2 → CH3Cl + HCl is correct?
3. But-1-yne (HC≡CCH2CH3) reacts with water under dilute H2SO4 and Hg²⁺ at 60°C. What is the organic product?
4. Which piece of evidence alone can distinguish an alkyne from an alkene?
5. Which is the correct balanced equation for complete combustion of propane (C3H8)?
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)
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 Lindlar catalyst is poisoned Pd (with lead acetate and quinoline) — it is deactivated enough to stop after one H2 addition, preventing further reduction of the alkene to alkane.
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.
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.
1. C — 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. B — 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. C — 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. D — 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. A — 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.
Q1 (3 marks): The Lindlar catalyst is palladium that has been poisoned with lead acetate and quinoline [1]. This deactivates the catalyst sufficiently so that it can catalyse the addition of the first equivalent of H2 across a triple bond (alkyne → alkene), but lacks enough activity to continue reducing the resulting C=C bond to a C-C single bond (alkene → alkane) [1]. A standard Ni, Pd, or Pt catalyst has full activity and continues the hydrogenation to the fully saturated alkane product [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].
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:
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?
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