Topic 11.2 · Alkanes

Alkanes

1

Structure and Naming

1.1The alkane homologous series

  • Alkanes: saturated hydrocarbons, CnH2n+2. Every C is sp³, tetrahedral, 109.5°.
  • Cycloalkanes: closed rings, CnH2n (same general formula as alkenes)
C₃H₆C₄H₈C₅H₁₀C₆H₁₂
nNameStructural formulanNameStructural formula
1methaneCH₄4butaneCH₃(CH₂)₂CH₃
2ethaneCH₃CH₃5pentaneCH₃(CH₂)₃CH₃
3propaneCH₃CH₂CH₃6hexaneCH₃(CH₂)₄CH₃

1.2Naming a branched alkane

  1. Find the longest continuous carbon chain (parent name)
  2. Number from the end nearer the first branch
  3. Name substituents with locants, in alphabetical order (ignore di-, tri-)

3-ethyl-2-methylhexane: 6-carbon main chain, numbered from the left so the branches get the lowest locants. "Ethyl" comes before "methyl".

123456methylethyl

1.3Structural isomers of C₅H₁₂

pentaneb.p. 36 °C2-methylbutaneb.p. 28 °C2,2-dimethylpropaneb.p. 10 °Cmore branching

Same molecular formula, different structural formula. Alkanes with 4 or more C show branching isomerism.

2

Physical Properties

C–H bonds are essentially non-polar → molecules held by weak id–id interactions.

no. of C atomsmethaneethanepropanebutanepentanehexaneb.p. / °C0−150123456

1. b.p. increases with the number of C atoms

  1. No. of electrons ↑ → larger, more polarisable electron cloud
  2. → stronger id–id → more energy needed to overcome them

2. Branched isomers have lower b.p.

  1. More branching → more spherical, smaller surface area of contact
  2. → weaker id–id → less energy needed to overcome them

pentane 36 °C > 2-methylbutane 28 °C > 2,2-dimethylpropane 10 °C

58Topic 11.2 · Alkanes
2

Physical Properties (continued)

2.1Solubility

Soluble in non-polar solvents (e.g. CCl₄). Insoluble in water:

  1. Energy released from forming pd–id interactions between alkane and water
  2. is insufficient to overcome the id–id between alkane molecules and the hydrogen bonds between water molecules

2.2Density and viscosity

Less dense than water

  1. Molecules are held by weaker id–id than the hydrogen bonds between water molecules
  2. → molecules are less closely packed
  • Density and viscosity ↑ with Mr: stronger id–id draws molecules closer
  • More branching → poorer packing → lower density
3

Cracking

DefinitionCracking: the breaking down of long-chain alkanes into shorter-chain alkanes and alkenes, using heat and a catalyst.
  • Conditions: high temperature, Al₂O₃ or SiO₂ catalyst (catalytic cracking)
  • Meets demand for petrol and gives alkenes for making polymers
decane, C₁₀H₂₂decane, C₁₀H₂₂heat, Al₂O₃octane, C₈H₁₈ (petrol)+ethene, C₂H₄
4

Reactivity and Combustion

Why are alkanes unreactive?

  1. C–H bonds are non-polar: no electron-rich or electron-poor sites to attract nucleophiles or electrophiles
  2. C–C and C–H bonds are strong

They react only with O₂ (combustion) and halogens (free radical substitution).

  • Complete (excess O₂): CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = −890 kJ mol⁻¹
  • Incomplete (limited O₂): 2CH₄ + 3O₂ → 2CO + 4H₂O, or C (soot)
  • Alkanes burn as gases, higher alkanes are less volatile and burn with sooty flames

4.1Identify a hydrocarbon from combustion data

Worked example15 cm³ of a gaseous hydrocarbon is burnt completely in 100 cm³ of O₂. 60 cm³ of CO₂ forms and 10 cm³ of O₂ is left over (same conditions). Find its formula.

CxHy(g) + (x + y4)O₂(g) → xCO₂(g) + y2H₂O(l)

1. 1 mol CxHy gives x mol CO₂: x = 6015 = 4

2. O₂ used = 100 − 10 = 90 cm³: x + y4 = 9015 = 6

3. 4 + y4 = 6 → y = 8 ∴ C₄H₈

Topic 11.2 · Alkanes59
5

Free Radical Substitution

CH₄ + Cl₂ → CH₃Cl + HCl   Cl₂ or Br₂, UV light or heat (250–400 °C)

Step 1: Initiation
ClClUV light2Cl•UV breaks Cl–Cl homolytically to give a low concentration of Cl•. Use half arrows.
Step 2: Propagation
ClHCH₃HCl+CH₃H₃CClClH₃CCl+ClCl• is regenerated, so the reaction is a chain reaction
Step 3: Termination
ClClCl–ClH₃CCH₃H₃C–CH₃H₃CClH₃C–ClTwo radicals combine, so radicals are used up and not regenerated
Exam tipWrite at least two termination steps. Show the • on the radical carbon (CH₃CH₂•).

5.1Common explanations for FRS

  • Why does Cl–Cl break, not C–H? BE(Cl–Cl) = 244 kJ mol⁻¹ is the weakest bond, UV light does not supply enough energy to break C–H (410 kJ mol⁻¹)
  • Why is no H• formed? Cl• + CH₄ → •CH₃ + HCl has ΔH = −21 kJ mol⁻¹, but Cl• + CH₄ → CH₃Cl + H• has ΔH = +70 kJ mol⁻¹, so it is energetically unfavourable
  • Why does the reaction continue after the UV light is removed? Each propagation step regenerates a radical, so the chain continues until termination uses up the radicals

5.2Different conditions for FRS

  • Limited Cl₂ (excess alkane): mainly monosubstitution, as Cl• is more likely to collide with CH₄
  • Excess Cl₂: polysubstitution: CH₂Cl₂, CHCl₃, CCl₄
  • A mixture of products forms → poor for synthesis
F₂Cl₂, Br₂I₂
explosiveUV light or heatno reaction (ΔH = +22)
60Topic 11.2 · Alkanes
6

Ratio of Products

6.1Statistical ratio

6 primary H4 secondary H Cl1-chlorobutaneCl2-chlorobutane

If every H atom were equally likely to be replaced by Cl, the ratio of products would equal the ratio of the numbers of H atoms that give each product.

  1. Butane has 6 primary H (on the two CH₃) and 4 secondary H (on the two CH₂)
  2. Replacing a primary H gives 1-chlorobutane, replacing a secondary H gives 2-chlorobutane
  3. ∴ statistical ratio 1-chlorobutane : 2-chlorobutane = 6 : 4 = 3 : 2

Propane: 6 primary : 2 secondary H = 3 : 1 for 1- : 2-chloropropane. 2-methylpropane: 9 primary : 1 tertiary H = 9 : 1.

6.2Experimental ratio

Experimentally 30 : 70: more 2-chlorobutane forms than the statistical ratio predicts. Why?

  1. The intermediate that forms 2-chlorobutane is a secondary radical: the electron-deficient radical is stabilised by 2 electron-donating alkyl groups
  2. The intermediate that forms 1-chlorobutane is a primary radical, which is less stable, as it is stabilised by only 1 electron-donating alkyl group
  3. ∴ formation of 2-chlorobutane is favoured due to the more stable intermediate
HH₃CCH₂CH₃C•2° radical: 2 alkyl groups+ Cl₂Cl2-chlorobutane (70%)HHCH₂CH₂CH₃C•1° radical: 1 alkyl group+ Cl₂Cl1-chlorobutane (30%)

Radical stability: methyl < 1° < 2° < 3°. The more electron-donating alkyl groups on the radical carbon, the more stable the radical.

7

Pollution from Combustion

  • CO: binds haemoglobin more strongly than O₂ → less O₂ carried
  • Unburnt hydrocarbons: form photochemical smog
  • NOx: high engine temperature breaks N≡N, N₂ + O₂ → 2NO. Causes smog, low-level O₃, and acid rain via NO₂ + SO₂ → SO₃ + NO
  • CO₂, CH₄: greenhouse gases → enhanced greenhouse effect

Catalytic converter: Pt, Pd, Rh (redox)

Removing CO and NO2CO + 2NO → 2CO₂ + N₂CO is oxidised to CO₂, NO is reduced to N₂.
Removing unburnt hydrocarbonshydrocarbons + NOx → CO₂ + H₂O + N₂
Topic 11.2 · Alkanes61