Alkanes

H2 Chemistry (9476), JC1

By the end of this topic, you can answer

1

Butane and 2-methylpropane are both C4H10. How many different alkanes can you draw with five carbons?

2

Methane, petrol and candle wax are all alkanes. Why is one a gas, one a liquid and one a solid?

3

Why do oil refineries crack long alkanes into shorter ones?

4

Why does chlorine react with methane in sunlight but not in the dark?

5

How does a catalytic converter remove carbon monoxide and nitrogen oxides from car exhaust?

Recall: what you know already

1What is the general formula of the alkanes?

CnH2n+2[1]

2From Introduction to Organic Chemistry: what is a free radical?

A species with an unpaired electron.[1]

3From Chemical Bonding: what IMF exists between non-polar molecules?

Instantaneous dipole–induced dipole interactions.[1]
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)
nNameStructural formula
1methaneCH₄
2ethaneCH₃CH₃
3propaneCH₃CH₂CH₃
4butaneCH₃(CH₂)₂CH₃
5pentaneCH₃(CH₂)₃CH₃
6hexaneCH₃(CH₂)₄CH₃
1

Structure and Naming

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".

1

Structure and Naming

1.3Structural isomers of C₅H₁₂

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

Your turnName CH₃CH(CH₃)CH(CH₃)CH₃. 2,3-dimethylbutane

2

Physical Properties

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

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

Physical Properties

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

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

Physical Properties

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
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

Reactivity and Combustion

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

4

Reactivity and Combustion

Your turn10 cm³ of a gaseous hydrocarbon needs exactly 50 cm³ of O₂ for complete combustion and gives 30 cm³ of CO₂. Find its formula.

x = 30 ÷ 10 = 3[1]
x + y4 = 50 ÷ 10 = 5, so y = 8[1]
C₃H₈, propane[1]
5

Free Radical Substitution

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

Step 1: Initiation

UV breaks Cl–Cl homolytically to give a low concentration of Cl•. Use half arrows.

5

Free Radical Substitution

Step 2: Propagation

Cl• takes an H atom from CH₄, giving a new radical, •CH₃

•CH₃ takes a Cl atom from Cl₂, giving the product and Cl• again

Cl• is regenerated, so the reaction is a chain reaction (mechanisms with fish-hook arrows on the next slide)

Step 3: Termination

Two 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

Free Radical Substitution

Step 2: Propagation

A half (fish-hook) arrow moves one electron.

Step 3: Termination
5

Free Radical Substitution

5.1Common explanations for FRS

  • Why does Cl–Cl break, not C–H? BE(Cl–Cl) = 244 kJ mol⁻¹ is the weakest bond, and 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

Free Radical Substitution

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)
6

Ratio of Products

6.1Statistical ratio

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
6

Ratio of Products

Your turnFind the statistical ratios: (a) 1-chloropropane : 2-chloropropane from propane, (b) the two monochloro products of 2-methylpropane.

(a)
6 primary : 2 secondary H = 3 : 1[1]
(b)
9 primary : 1 tertiary H = 9 : 1[1]
6

Ratio of Products

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

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)

2CO + 2NO → 2CO₂ + N₂

CO is oxidised to CO₂, NO is reduced to N₂.

hydrocarbons + NOx → CO₂ + H₂O + N₂

7

Pollution from Combustion

Quick checkMethane and excess chlorine are exposed to UV light. Which is not a possible product?

ACH₂Cl₂
BC₂H₆
CH₂
DCCl₄
NoteC: no H• forms (that step is energetically unfavourable), so H₂ cannot form. C₂H₆ forms in termination, and CH₂Cl₂ and CCl₄ by further substitution.

Exit ticket

1Why does pentane boil higher than 2,2-dimethylpropane?

Pentane has a larger surface area of contact, so stronger instantaneous dipole–induced dipole interactions[1]
more energy is needed to overcome them.[1]

2Write the two propagation steps for CH₄ + Br₂.

Br• + CH₄ → •CH₃ + HBr[1]
•CH₃ + Br₂ → CH₃Br + Br•[1]

3What conditions favour monosubstitution?

Limited halogen (excess alkane).[1]

4Why are alkanes insoluble in water?

Energy released forming permanent dipole–induced dipole interactions is insufficient to overcome the instantaneous dipole–induced dipole interactions between alkane molecules[1]
and the hydrogen bonds between water molecules.[1]