What are addition reactions of alkenes?

Addition reactions break the weaker π bond and add two new groups across the former double bond, leaving a saturated product.

What are reactions of alkynes?

Alkynes undergo combustion and addition like alkenes, but each π bond can be added across in turn. So ethyne plus excess bromine gives 1,1,2,2-tetrabromoethane:

What is markovnikov direction wrong?

The H adds to the carbon with more Hs already. Think "the rich get richer" for H atoms.

What are soot from incomplete combustion of alkanes?

Alkanes generally burn cleanly; alkenes and alkynes are sootier. If asked to compare flames, mention the C:H ratio.

§-Syllabus dot point

NSWChemistrySyllabus dot point

Inquiry Question 2: How are hydrocarbons classified and what do their reactions reveal about their structure?

Investigate the structural formulae, properties and reactions of alkanes, alkenes and alkynes, including combustion and addition reactions of alkenes

A focused answer to the HSC Chemistry Module 7 dot point on hydrocarbons. Comparing alkanes, alkenes and alkynes by structure and reactivity, combustion equations, addition reactions of alkenes with halogens, hydrogen halides and water, and worked HSC past exam questions.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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In plain English

Think of the three hydrocarbon families as three types of handshake between carbon atoms. Alkanes only ever do a single, firm handshake ( $C-C$ ), so they are "fully booked" with hydrogens and chemically lazy, only reacting if you force the issue with UV light. Alkenes do a double handshake ( $C=C$ ), which is really one firm grip plus one loose, exposed grip (the pi bond) that anything electron-hungry, like bromine, can grab onto instantly, no UV needed. Alkynes go further with a triple handshake, two loose grips stacked on the one firm one, so they can react twice over, adding bromine or hydrogen not once but twice before becoming fully saturated. The bromine water test is just a colour-changing "unsaturation detector": if the orange-brown bromine disappears the instant you add it, there was a loose grip (a pi bond) ready to be grabbed; if the colour stays orange-brown in the dark, every handshake in that molecule was already firm and single.

What this dot point is asking

NESA wants you to distinguish alkanes, alkenes and alkynes by structure, predict and write equations for their characteristic reactions (combustion, addition, substitution), and explain why a $C=C$ bond makes alkenes much more reactive than alkanes. The dot point also covers the trends in physical properties down each series and the test that distinguishes saturated from unsaturated hydrocarbons.

The answer

Structural comparison

Series	Bond type	General formula	Saturated?	Example
Alkane	$C-C$ single	$C_nH_{2n+2}$	Yes	$CH_3CH_3$ ethane
Alkene	$C=C$ double	$C_nH_{2n}$	No	$CH_2=CH_2$ ethene
Alkyne	$C \equiv C$ triple	$C_nH_{2n-2}$	No	$HC \equiv CH$ ethyne

A double bond is one $\sigma$ plus one $\pi$ bond; a triple bond is one $\sigma$ plus two $\pi$ bonds. The $\pi$ electrons are loosely held and are the site of attack in addition reactions.

Physical properties

Across all three series, increasing chain length raises both melting point and boiling point because the dispersion forces between molecules grow with molecular size. C1 to C4 hydrocarbons are gases at room temperature, C5 to about C16 are liquids, and longer chains are waxy solids. All hydrocarbons are non-polar, immiscible with water, and float on water because their density is below 1 g/mL.

Comparing series at the same carbon number: an alkane and the corresponding alkene or alkyne have very similar boiling points, because the molecules differ only in two or four hydrogens. The double bond does not introduce significant polarity.

Combustion (all hydrocarbons)

Complete combustion (excess $O_2$ ) gives carbon dioxide and water. The general equation for any $C_xH_y$ :

C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O

Combustion is highly exothermic and is the basis for using hydrocarbons as fuels. For methane:

CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H = -890 \text{ kJ/mol}

Incomplete combustion (limited $O_2$ ) gives carbon monoxide $CO$ or soot $C$ plus water. Alkenes and alkynes burn with a sootier flame than alkanes because they have a higher carbon to hydrogen ratio, so less oxygen reaches the inner part of the flame.

Substitution reactions of alkanes

Alkanes are unreactive towards most reagents at room temperature. With halogens (chlorine or bromine) in UV light, they undergo free radical substitution:

CH_{4(g)} + Cl_{2(g)} \xrightarrow{UV} CH_3Cl + HCl

The mechanism has three steps: initiation ( $Cl_2 \rightarrow 2Cl \cdot$ under UV), propagation ( $Cl \cdot + CH_4 \rightarrow HCl + CH_3 \cdot$ , then $CH_3 \cdot + Cl_2 \rightarrow CH_3Cl + Cl \cdot$ ), and termination (radicals combine).

Addition reactions of alkenes

Addition reactions break the weaker $\pi$ bond and add two new groups across the former double bond, leaving a saturated product.

1. Hydrogenation (H₂, Ni catalyst, heat). Alkene plus hydrogen gives the corresponding alkane:

CH_2=CH_{2(g)} + H_{2(g)} \xrightarrow{Ni, 150 \text{ degrees C}} CH_3CH_{3(g)}

2. Halogenation ( $Br_2$ or $Cl_2$ , room temperature, no catalyst). Alkene decolourises bromine water (orange-brown to clear) instantly. This is the standard test for unsaturation:

CH_2=CH_{2(g)} + Br_{2(aq)} \rightarrow CH_2BrCH_2Br

3. Hydrohalogenation (HX, e.g. HCl or HBr). Alkene plus a hydrogen halide gives a haloalkane. Asymmetric alkenes follow Markovnikov's rule: H adds to the carbon already carrying more hydrogens, X adds to the more substituted carbon.

CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3 \text{ (major)}

4. Hydration ( $H_2O$ , dilute $H_2SO_4$ catalyst, heat). Alkene plus water gives an alcohol. Markovnikov also applies.

CH_2=CH_{2(g)} + H_2O_{(g)} \xrightarrow{H_2SO_4, 300 \text{ degrees C}} CH_3CH_2OH

This is industrially how ethanol is made from ethene.

Reactions of alkynes

Alkynes undergo combustion and addition like alkenes, but each $\pi$ bond can be added across in turn. So ethyne plus excess bromine gives 1,1,2,2-tetrabromoethane:

HC \equiv CH + 2Br_2 \rightarrow CHBr_2CHBr_2

Ethyne $C_2H_2$ burns at very high temperatures with oxygen, which is why oxyacetylene torches are used for welding and metal cutting.

The bromine water test

To distinguish a saturated hydrocarbon (alkane) from an unsaturated one (alkene or alkyne), add a few drops of bromine water and shake.

Alkene or alkyne: orange-brown colour rapidly disappears (clear/colourless) due to addition.
Alkane: colour persists, unless exposed to UV light in which case it fades slowly with HBr fumes (substitution).

A second confirmatory test is acidified $KMnO_4$ : alkenes and alkynes decolourise purple permanganate at room temperature; alkanes do not react.

Reaction and test checkpoints

Bromine water test: instant decolourisation (orange-brown to colourless) = unsaturated ( $C=C$ or $C \equiv C$ present); colour persists in the dark = alkane (saturated).
General formulae: alkane $C_nH_{2n+2}$ , alkene $C_nH_{2n}$ , alkyne $C_nH_{2n-2}$ . Each series differs from the last by loss of 2 hydrogens (one more degree of unsaturation).
Substitution needs UV; addition does not. Alkanes only react with halogens under UV (free radical substitution); alkenes and alkynes react with halogens at room temperature in the dark (electrophilic addition).
Markovnikov's rule: for $HX$ or $H_2O$ addition to an asymmetric alkene, H adds to the carbon already carrying more hydrogens ("the rich get richer").
Complete combustion, general form: $C_xH_y + \left(x + \dfrac{y}{4}\right)O_2 \rightarrow xCO_2 + \dfrac{y}{2}H_2O$ .

An owned energy profile compares the activation energy pathway for alkane substitution (needs UV to overcome a high barrier) against alkene addition (a much lower barrier, reacting readily at room temperature):

Worked classification and reaction examples

Worked example 1: classify and predict.

Compound: $CH_3CH_2CH=CH_2$ (but-1-ene).
Classification: contains one $C=C$ double bond, so it is an alkene, general formula $C_nH_{2n}$ (here $C_4H_8$ ).
Bromine water test prediction: instant decolourisation, orange-brown to colourless, because the pi bond of the double bond is attacked by $Br_2$ in an electrophilic addition.
Balanced equation: $CH_3CH_2CH=CH_{2(g)} + Br_{2(aq)} \rightarrow CH_3CH_2CHBrCH_2Br_{(l)}$ .

Marker's note: full marks require stating BOTH the classification reasoning (bond type, general formula) and the specific observation (colour change), not just the equation.

Worked example 2: Markovnikov addition of $HBr$ .

Compound: propene, $CH_3CH=CH_2$ , reacting with $HBr$ .
Step 1: identify the more substituted carbon of the double bond. $C2$ (the middle carbon) already bears one H and is attached to two other carbons' worth of substitution character in the transition state; by Markovnikov's rule the H of $HBr$ adds to the carbon with more hydrogens already, which is the terminal $CH_2$ carbon ( $C1$ ).
Step 2: $Br$ adds to the other (more substituted) carbon, $C2$ .
Major product: $CH_3CHBrCH_3$ (2-bromopropane).
Balanced equation: $CH_3CH=CH_{2(g)} + HBr_{(g)} \rightarrow CH_3CHBrCH_{3(l)}$ .

Marker's note: markers reward explicitly naming Markovnikov's rule and identifying which carbon has more hydrogens BEFORE stating the product; giving only the product without the reasoning risks losing a mark even if the structure drawn is correct.

Examples in context

Example 1. Cracking at the Kurnell refinery (legacy operations). Until its 2014 conversion to an import terminal, Caltex's Kurnell refinery cracked long-chain alkanes from Bass Strait crude into shorter alkenes and alkanes for the petrol pool. Steam cracking ethane gave ethene plus hydrogen: $C_2H_6 \rightarrow C_2H_4 + H_2$ . The ethene was the feedstock for the adjacent ethanol and polyethene units. Combustion of the residual fuel gas in the cracker furnace followed the standard alkane equation $C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O$ in excess oxygen. The HSC framework for combustion and addition reactions is the same chemistry the plant ran at industrial scale.

Example 2. Bromine water test in NSW HSC depth study. A school lab issued unlabelled samples of hexane and hex-1-ene asks students to identify them. Adding bromine water (orange-brown) to hex-1-ene gives instant decolourisation as the bromine adds across the double bond: $CH_2=CH(CH_2)_3CH_3 + Br_2 \rightarrow CH_2BrCHBr(CH_2)_3CH_3$ . Hexane does not decolourise bromine water in the absence of UV. The contrast is the canonical HSC test for unsaturation. The same bromine number method is used at Hunter Valley wineries to measure unsaturated lipids in grape-seed oil and at Manildra to check ethanol purity for fuel-grade specifications.

Try this

Q1. State the general molecular formula for an alkane, an alkene and an alkyne, and give one observation that distinguishes an alkene from an alkane chemically. [3 marks]

Cue. Alkane $C_nH_{2n+2}$ , alkene $C_nH_{2n}$ , alkyne $C_nH_{2n-2}$ ; alkene decolourises bromine water instantly without UV.

Q2. Write a balanced equation for the complete combustion of octane ( $C_8H_{18}$ ) and calculate the volume of $CO_2$ at 25 degrees C and 100 kPa produced from 5.00 g of octane (molar volume 24.79 L mol $^{-1}$ ). [3 marks]

Cue. $C_8H_{18} + 12.5 O_2 \rightarrow 8 CO_2 + 9 H_2O$ ; $n(\text{octane}) = 5.00 / 114.23 = 0.0438$ ; $n(CO_2) = 0.350$ ; $V = 0.350 \times 24.79 = 8.69$ L.

Q3. But-1-ene reacts with HBr to give a major and minor product. (a) State Markovnikov's rule. (b) Identify the major product. (c) Explain why no second product forms with the alkane equivalent unless UV is supplied. [1+2+2 marks]

Cue. (a) H adds to the carbon with more H atoms. (b) 2-bromobutane (major). (c) Alkane requires free-radical substitution initiated by UV homolysis of $Br_2$ .

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC4 marksCompare the reactivity of ethane and ethene by writing balanced equations for one reaction of each with bromine and explaining the difference in mechanism and observation.

Show worked answer →

A 4 mark answer needs two balanced equations, the mechanism labels, and the observational difference.

Ethane (alkane) with bromine. A substitution reaction that needs UV light to initiate.

C_2H_{6(g)} + Br_{2(g)} \xrightarrow{UV} C_2H_5Br_{(l)} + HBr_{(g)}

Mechanism: free radical substitution. Slow without UV. Observation: brown bromine vapour fades slowly only when illuminated, and HBr fumes form.

Ethene (alkene) with bromine. An addition reaction across the $C=C$ double bond, occurring at room temperature in the dark.

C_2H_{4(g)} + Br_{2(aq)} \rightarrow C_2H_4Br_{2(l)}

Mechanism: electrophilic addition. Fast at room temperature. Observation: orange-brown bromine water is rapidly decolourised (clear).

Comparison. Ethene reacts much faster because the $\pi$ bond is electron-rich and attacks the electrophile $Br_2$ . Ethane has only C-H and C-C $\sigma$ bonds, which are unreactive without high-energy UV initiation.

Markers reward (1) both balanced equations, (2) naming substitution vs addition, (3) the visual observation, (4) explaining why the $\pi$ bond drives the difference.

2018 HSC3 marksWrite a balanced equation for the complete combustion of propene (C₃H₆) and calculate the volume of CO₂ produced at 25°C and 100 kPa when 5.6 g of propene is fully combusted.

Show worked answer →

Step 1: Balanced equation.

2C_3H_{6(g)} + 9O_{2(g)} \rightarrow 6CO_{2(g)} + 6H_2O_{(l)}

Or per mole of propene: $C_3H_6 + 4.5 O_2 \rightarrow 3 CO_2 + 3 H_2O$ .

Step 2: Moles of propene. $M(C_3H_6) = 42.08$ g/mol.

n(C_3H_6) = \frac{5.6}{42.08} = 0.133 \text{ mol}

Step 3: Moles of $CO_2$ . Ratio 3:1.

n(CO_2) = 3 \times 0.133 = 0.399 \text{ mol}

Step 4: Volume at 25 degrees C, 100 kPa. Molar volume = 24.79 L/mol.

V = 0.399 \times 24.79 = 9.89 \text{ L}

Markers reward (1) the balanced equation, (2) correct mole calculation, (3) correct molar volume at HSC standard conditions.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksClassify each hydrocarbon as an alkane, alkene or alkyne and state the general formula that confirms your classification: (a)

CH_3CH_2CH_2CH_3

; (b)

CH_2=CHCH_3

; (c)

HC \equiv CCH_3

Show worked solution →

A 3-mark classify question needs the correct family AND the matching general formula for each.

(a) $CH_3CH_2CH_2CH_3$ (butane): All bonds are single $C-C$ bonds, so this is an alkane, matching $C_nH_{2n+2}$ (here $C_4H_{10}$ ).
(b) $CH_2=CHCH_3$ (propene): Contains one $C=C$ double bond, so this is an alkene, matching $C_nH_{2n}$ (here $C_3H_6$ ).
(c) $HC \\equiv CCH_3$ (propyne): Contains one $C \\equiv C$ triple bond, so this is an alkyne, matching $C_nH_{2n-2}$ (here $C_3H_4$ ).

Marking criteria: 1 mark per compound for correctly naming the family AND citing the matching general formula; family name alone without the formula reasoning earns partial credit only.

foundation3 marksState the reagent, condition and observation for a test that distinguishes hex-1-ene from hexane, and explain the chemical reason for the difference.

Show worked solution →

Reagent and condition: Add a few drops of bromine water to each sample at room temperature, in the dark (no UV).
Observation: Hex-1-ene: the orange-brown bromine water is instantly decolourised (becomes colourless). Hexane: the orange-brown colour persists (no reaction).
Chemical reason: Hex-1-ene has a $C=C$ double bond, so it contains a reactive, electron-rich pi bond that undergoes electrophilic addition with $Br_2$ at room temperature. Hexane has only $C-C$ and $C-H$ sigma bonds, which are unreactive towards $Br_2$ without UV light to initiate free radical substitution.

Marking criteria: 1 mark for the correct reagent/condition, 1 mark for both observations stated correctly, 1 mark for the pi bond/electrophilic addition reasoning explaining the difference.

core5 marksA student reacts 8.40 g of hex-1-ene (

C_6H_{12}

) with excess bromine water. Calculate the mass of bromine (

Br_2

) theoretically consumed, to 3 significant figures. (

M(C_6H_{12}) = 84.16\ \text{g mol}^{-1}

M(Br_2) = 159.80\ \text{g mol}^{-1}

Show worked solution →

Step 1: write the balanced equation (1:1 stoichiometry).

CH_2=CH(CH_2)_3CH_{3(l)} + Br_{2(aq)} \rightarrow CH_2BrCHBr(CH_2)_3CH_{3(l)}

One mole of hex-1-ene reacts with one mole of $Br_2$ .

Step 2: moles of hex-1-ene.

n(C_6H_{12}) = \frac{m}{M} = \frac{8.40\ \text{g}}{84.16\ \text{g mol}^{-1}} = 0.09981\ \text{mol}

Step 3: moles of $Br_2$ (1:1 ratio).

n(Br_2) = 0.09981\ \text{mol}

Step 4: mass of $Br_2$ .

m = n \times M = 0.09981\ \text{mol} \times 159.80\ \text{g mol}^{-1} = 15.95\ \text{g}

Step 5: round to 3 significant figures (matching 8.40 g, which has 3 s.f.).

m(Br_2) = 16.0\ \text{g}

Marking criteria: 1 mark for the correct balanced 1:1 equation, 1 mark for correct moles of hex-1-ene, 1 mark for correctly carrying the 1:1 mole ratio, 1 mark for the mass calculation, 1 mark for the correct final answer to 3 significant figures with units. This is a THEORETICAL value; a real bromine-water addition would need at least this much $Br_2$ present for complete reaction.

core5 marksThe energy profile diagram above compares alkane free radical substitution with alkene electrophilic addition. (a) Identify which pathway has the higher activation energy and state one piece of evidence for this from the diagram. (b) Explain, in terms of bonding, why the two pathways have different activation energies. (c) State one experimental condition that must be supplied for the higher-Ea pathway to proceed at a useful rate.

Show worked solution →

(a) Identification and evidence: The alkane substitution pathway (alkane + $Cl_2$ ) has the higher activation energy. Evidence: its peak sits much closer to the "high" energy line on the diagram than the peak of the alkene addition (alkene + $Br_2$ ) pathway, which peaks much lower, closer to the reactant energy level.
(b) Bonding explanation: Alkane substitution requires breaking a strong $C-H$ sigma bond and a $Cl-Cl$ sigma bond homolytically to generate radicals, which needs a large energy input, giving a high activation energy. Alkene addition only requires breaking the weaker, more exposed pi bond of the $C=C$ double bond, which has electron density above and below the bond axis and is far more accessible to an electrophile like $Br_2$ , so much less energy is needed to start the reaction.
(c) Condition required: UV light (to homolytically cleave $Cl_2$ into radicals and initiate the substitution mechanism).

Marking criteria: (a) 1 mark for correct pathway identified, 1 mark for evidence quoted from the diagram. (b) 2 marks for the bonding-based explanation (sigma vs pi bond strength/accessibility). (c) 1 mark for UV light (or equivalent high-energy initiation).

core6 marksBut-2-yne (

CH_3C \equiv CCH_3

) is fully hydrogenated to butane using excess hydrogen gas and a nickel catalyst. (a) Write a balanced equation for the full hydrogenation. (b) Calculate the volume of hydrogen gas, at 25 degrees C and 100 kPa, required to fully hydrogenate 2.16 g of but-2-yne, to 3 significant figures. (

M(C_4H_6) = 54.09\ \text{g mol}^{-1}

, molar volume

= 24.79\ \text{L mol}^{-1}

Show worked solution →

(a) Balanced equation. Each $\pi$ bond of the triple bond needs its own $H_2$ molecule, so full hydrogenation of a triple bond to a single bond needs 2 mol of $H_2$ per mole of alkyne.

CH_3C \equiv CCH_{3(g)} + 2H_{2(g)} \xrightarrow{Ni} CH_3CH_2CH_2CH_{3(g)}

(b) Step 1: moles of but-2-yne.

n(C_4H_6) = \frac{m}{M} = \frac{2.16\ \text{g}}{54.09\ \text{g mol}^{-1}} = 0.03993\ \text{mol}

Step 2: moles of $H_2$ (2:1 ratio from the equation).

n(H_2) = 2 \times 0.03993\ \text{mol} = 0.07987\ \text{mol}

Step 3: volume of $H_2$ at 25 degrees C, 100 kPa.

V = n \times V_m = 0.07987\ \text{mol} \times 24.79\ \text{L mol}^{-1} = 1.980\ \text{L}

Step 4: round to 3 significant figures (matching 2.16 g).

V(H_2) = 1.98\ \text{L}

Marking criteria: (a) 1 mark for the correctly balanced equation showing 2 mol $H_2$ per mole of alkyne, with the nickel catalyst condition. (b) 1 mark for correct moles of but-2-yne, 1 mark for correctly doubling to get moles of $H_2$ (recognising the triple bond needs two additions), 1 mark for the volume calculation, 1 mark for the final answer to 3 significant figures with units.

exam6 marksCompare the reactivity of but-1-ene and butane towards bromine, addressing mechanism, conditions, rate and the practical test used to distinguish them, and explain your comparison using bonding concepts.

Show worked solution →

A 6-mark COMPARE needs both similarities/differences addressed across mechanism, conditions, rate and a named test, backed by bonding reasoning.

Band 6 plan.

Name both mechanisms: but-1-ene undergoes electrophilic addition; butane undergoes free radical substitution.
State conditions: but-1-ene reacts with bromine water at room temperature in the dark; butane requires UV light to react at all.
Compare rate: but-1-ene reacts instantly; butane reacts slowly even under UV, because the mechanism needs a three-step radical chain (initiation, propagation, termination) rather than a single electrophilic attack.
Name the distinguishing test: bromine water, decolourised instantly by but-1-ene, unaffected (in the dark) by butane.
Bonding reasoning: but-1-ene's $C=C$ double bond has an exposed, electron-rich pi bond that directly attracts the electrophilic $Br_2$ molecule; butane has only strong, unreactive $C-H$ and $C-C$ sigma bonds that require UV photons to homolytically cleave $Cl$ - $Cl$ / $Br$ - $Br$ bonds before any reaction can begin.

Model answer (excerpt). But-1-ene and butane react with bromine by fundamentally different mechanisms. But-1-ene undergoes electrophilic addition: the electron-rich pi bond of its $C=C$ double bond attacks $Br_2$ , forming 1,2-dibromobutane, and this proceeds instantly at room temperature in bromine water, decolourising it from orange-brown to colourless. Butane, lacking any pi bond, reacts only by free radical substitution, and only when UV light is supplied to homolyse $Br_2$ into radicals; even then the reaction is comparatively slow, since it proceeds through a three-stage radical chain mechanism (initiation, propagation, termination) rather than a single-step electrophilic attack. This difference is exploited directly as the standard laboratory test for unsaturation: instant decolourisation of bromine water identifies an alkene (or alkyne), while a lack of colour change in the dark identifies a saturated alkane.

Marker's note: top-band answers (1) correctly name BOTH mechanisms, (2) state the conditions for both including the UV requirement, (3) explicitly compare rate, not just describe each separately, (4) name bromine water as the confirmatory test, and (5) tie the whole comparison back to the presence or absence of an exposed pi bond. Answers that describe only one hydrocarbon in detail without a direct comparison are capped at half marks.