Alkanes

Alkanes are the simplest family of organic compounds. They contain only carbon and hydrogen atoms joined by single covalent bonds, making them saturated hydrocarbons. Despite their apparent simplicity, alkanes are enormously important — they are the primary components of crude oil and natural gas, and their combustion provides most of the world's energy.

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Structure and Bonding

Alkanes have the general formula CₙH₂ₙ₊₂. Each carbon atom forms four single (sigma, σ) bonds arranged in a tetrahedral geometry with bond angles of approximately 109.5°. The bonds are formed by the overlap of sp³ hybridised orbitals on carbon with either another sp³ orbital (C-C bond) or the 1s orbital of hydrogen (C-H bond).

Because all bonds are sigma bonds with free rotation, alkanes adopt a variety of conformations. The carbon backbone is often drawn as a zigzag to represent the tetrahedral angles.

Sigma Bonds

A sigma bond is formed by the head-on overlap of atomic orbitals along the bond axis. This creates a cylindrically symmetrical region of electron density between the two nuclei. The free rotation around sigma bonds is why alkanes are flexible molecules.

Bond Enthalpies

Understanding the strength of bonds in alkanes helps explain their reactivity:

Bond	Bond Enthalpy (kJ mol⁻¹)	Comment
C-C	348	Moderately strong
C-H	412	Strong — difficult to break
C-C in ethane	346	Very similar across all alkanes

The high bond enthalpies of both C-C and C-H bonds contribute to the low reactivity of alkanes.

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Physical Properties

Boiling Points

Alkane molecules are non-polar, so the only intermolecular forces between them are van der Waals (London dispersion) forces. These temporary dipole–induced dipole forces arise from the random movement of electrons.

The strength of van der Waals forces depends on:

1. Number of electrons (and therefore molecular size) — more electrons mean stronger temporary dipoles
2. Surface area of contact — longer, straighter chains can pack more closely, increasing the area over which van der Waals forces act

As a result:

• Boiling points increase with increasing chain length (methane −162 °C, ethane −89 °C, propane −42 °C, butane −1 °C, pentane 36 °C)
• Branched isomers have lower boiling points than straight-chain isomers of the same molecular formula (reduced surface area)

Boiling Point Data Table

Alkane	Formula	Boiling Point (°C)	State at 25 °C
Methane	CH₄	−162	Gas
Ethane	C₂H₆	−89	Gas
Propane	C₃H₈	−42	Gas
Butane	C₄H₁₀	−1	Gas
Pentane	C₅H₁₂	36	Liquid
Hexane	C₆H₁₄	69	Liquid
Octane	C₈H₁₈	126	Liquid
Eicosane	C₂₀H₄₂	343	Solid

Solubility

Alkanes are insoluble in water because they are non-polar and cannot form hydrogen bonds with water molecules. They are soluble in other non-polar solvents. This principle — "like dissolves like" — is a recurring theme in chemistry.

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Chemical Reactivity

Alkanes are relatively unreactive. The C-C and C-H bonds are strong (bond enthalpies approximately 348 and 412 kJ mol⁻¹ respectively) and non-polar, so they are not readily attacked by common reagents such as acids, bases, or oxidising agents under normal conditions.

The reason alkanes are unreactive towards ionic and polar reagents is that they lack any electron-rich or electron-poor sites. There are no lone pairs, no pi bonds, and no significant dipoles. Electrophiles have nothing to be attracted to, and nucleophiles have no positive centre to attack.

However, alkanes do undergo two important types of reaction:

1. Combustion

Complete combustion occurs with a plentiful supply of oxygen:

CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O

For example, methane: CH₄ + 2O₂ → CO₂ + 2H₂O

Complete combustion is highly exothermic and is the basis for using alkanes as fuels.

Incomplete combustion occurs when the oxygen supply is limited, producing carbon monoxide (CO) and/or carbon (soot, C) instead of carbon dioxide:

2CH₄ + 3O₂ → 2CO + 4H₂O (producing carbon monoxide) CH₄ + O₂ → C + 2H₂O (producing carbon/soot)

2. Free Radical Substitution

Alkanes react with halogens (Cl₂, Br₂) in the presence of ultraviolet light via a free radical substitution mechanism. This is covered in detail in the next lesson.

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Environmental Issues from Burning Alkanes

The combustion of fossil fuels (primarily alkanes) creates several environmental pollutants:

Pollutant	Source	Environmental Effect
CO₂	Complete combustion	Greenhouse gas — contributes to global warming and climate change
CO	Incomplete combustion	Toxic — binds to haemoglobin, preventing oxygen transport
Particulates (soot)	Incomplete combustion	Respiratory problems, contributes to smog
NOₓ (NO and NO₂)	N₂ + O₂ reacting at high engine temperatures	Acid rain, photochemical smog, respiratory irritant
SO₂	Sulfur impurities in fuel	Acid rain
Unburnt hydrocarbons	Incomplete combustion	Contribute to photochemical smog, potential carcinogens

Carbon monoxide is particularly dangerous because it is colourless and odourless. It binds irreversibly to haemoglobin with approximately 200 times the affinity of oxygen, forming carboxyhaemoglobin and reducing the blood's oxygen-carrying capacity.

Nitrogen oxides form when nitrogen and oxygen from the air react at the very high temperatures inside internal combustion engines. Catalytic converters in vehicle exhausts reduce NOₓ emissions by converting them back to N₂ and O₂:

2CO + 2NO → 2CO₂ + N₂

The catalytic converter simultaneously removes CO and NOₓ.

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Crude Oil and Fractional Distillation

Crude oil is a complex mixture of hundreds of different hydrocarbons, mainly alkanes. Fractional distillation separates crude oil into useful fractions based on boiling point.

The process works because:

1. Crude oil is heated to approximately 350 °C so that most hydrocarbons vaporise.
2. The vapour enters a fractionating column that is hot at the bottom and cool at the top.
3. Vapours rise up the column and condense when they reach the level where the temperature is equal to their boiling point.
4. Shorter-chain hydrocarbons (lower boiling points) are collected near the top; longer-chain hydrocarbons (higher boiling points) condense lower down.

Typical fractions include: refinery gases (C₁–C₄), petrol/gasoline (C₅–C₁₀), naphtha, kerosene, diesel, fuel oil, and bitumen.

graph TD
    A["Crude Oil (heated to ~350 °C)"] --> B["Fractionating Column"]
    B --> C["Refinery Gases C₁-C₄ (top, coolest)"]
    B --> D["Petrol / Gasoline C₅-C₁₀"]
    B --> E["Naphtha C₈-C₁₂"]
    B --> F["Kerosene C₁₁-C₁₅"]
    B --> G["Diesel C₁₅-C₂₀"]
    B --> H["Fuel Oil C₂₀-C₃₀"]
    B --> I["Bitumen C₃₀+ (bottom, hottest)"]

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Cracking

Fractional distillation produces more long-chain fractions than there is market demand for, and too few short-chain fractions. Cracking breaks long-chain alkanes into shorter, more useful molecules — including alkenes, which are important feedstocks for the chemical industry.

Thermal Cracking

• High temperature (400–900 °C) and high pressure (up to 70 atm)
• Produces a high proportion of alkenes (useful for making polymers)
• Homolytic fission of C-C bonds (bonds break evenly, forming radicals)

Catalytic Cracking

• Lower temperature (approximately 450 °C) and slight pressure
• Uses a zeolite catalyst (aluminosilicate)
• Produces branched alkanes and cyclic hydrocarbons (useful for petrol)
• Heterolytic fission assisted by the catalyst

For example, decane can be cracked:

C₁₀H₂₂ → C₈H₁₈ + C₂H₄

This produces octane (a petrol component) and ethene (used to make polyethene and other chemicals).

graph LR
    A["Long-chain alkane (e.g. C₁₀H₂₂)"] --> B{"Cracking"}
    B -->|"Thermal: 400-900 °C, high pressure"| C["Shorter alkane + Alkenes"]
    B -->|"Catalytic: ~450 °C, zeolite"| D["Branched alkanes + Cyclic hydrocarbons"]
    C --> E["Alkenes → Polymers, chemical feedstocks"]
    D --> F["Petrol (high octane rating)"]

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Supply and Demand

The demand for different fractions does not match what crude oil naturally provides. There is high demand for petrol and chemical feedstocks (short chains) but the crude oil contains a large proportion of long-chain fractions. Cracking addresses this mismatch by converting less useful long chains into more valuable short chains and alkenes. This is an example of how industrial chemistry responds to economic pressures.

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A-Level Deep Dive: Alkanes

Spec mapping

Edexcel 9CH0 specification Topic 6 — Organic Chemistry I, sub-topic 6.3 covers the alkane homologous series: structure (saturated, sp3-hybridised, σ-bonded), low reactivity arising from non-polar C-H and C-C bonds, complete and incomplete combustion, and free-radical substitution with halogens (refer to the official specification document for exact wording). Sub-topic 6.3 is examined directly on Paper 2 (organic and physical chemistry) and synoptically on Paper 1 (energetics, where bond enthalpies are used to calculate combustion energies) and on Paper 3 (general and practical principles, where alkane fractional distillation is the basis of CP6). The data booklet provides mean bond enthalpies for C-H (412), C-C (348), C=O (805), O=O (496) and O-H (463) kJ mol$^{-1}$−1 — these values must be applied confidently.

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Worked example with full mark scheme

Question (8 marks):

(a) Calculate the standard enthalpy of combustion of butane, $C_4H_{10}$C4​H10​, using mean bond enthalpies. Give your answer to three significant figures. (5)

(b) A poorly ventilated gas heater is observed to emit a yellow flame. Identify the gas responsible for the colour, name the harmful product of incomplete combustion, and explain why incomplete combustion is dangerous in a domestic setting. (3)

Solution with mark scheme:

(a) Step 1 — write the balanced equation.

$C_4H_{10}(g) + \dfrac{13}{2}O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)$C4​H10​(g)+213​O2​(g)→4CO2​(g)+5H2​O(l)

M1 — balanced equation with correct stoichiometric coefficients (commonly written with $\tfrac{13}{2}$213​ or doubled to $2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$2C4​H10​+13O2​→8CO2​+10H2​O).

Step 2 — identify bonds broken and made. Bonds broken (per mole of butane): 3 × C-C (in the propane skeleton: actually butane has 3 C-C bonds), 10 × C-H, $\tfrac{13}{2}$213​ × O=O. Bonds made: 8 × C=O (4 CO2 × 2 each), 10 × O-H (5 H2O × 2 each).

M1 — bond inventory correct: 3 C-C, 10 C-H, 6.5 O=O broken; 8 C=O, 10 O-H made.

Step 3 — sum bond enthalpies.

Bonds broken: $3(348) + 10(412) + 6.5(496) = 1044 + 4120 + 3224 = 8388$3(348)+10(412)+6.5(496)=1044+4120+3224=8388 kJ.

Bonds made: $8(805) + 10(463) = 6440 + 4630 = 11070$8(805)+10(463)=6440+4630=11070 kJ.

M1 — correct arithmetic for bonds broken (8388 kJ) or bonds made (11070 kJ); allow ±50 for rounding.

Step 4 — calculate $\Delta H$ΔH.

$\Delta H = \text{(bonds broken)} - \text{(bonds made)} = 8388 - 11070 = -2682 \text{ kJ mol}^{-1}$ΔH=(bonds broken)−(bonds made)=8388−11070=−2682 kJ mol−1

A1 — $\Delta H = -2680$ΔH=−2680 kJ mol$^{-1}$−1 (3 s.f.).

A1 — explicit comment that the value is negative because combustion is exothermic, and the bond-enthalpy method gives an approximate value (mean bond enthalpies are averaged across compounds; the true $\Delta H_c$ΔHc​ of butane is approximately $-2877$−2877 kJ mol$^{-1}$−1).

(b) Step 1 — yellow flame.

A1 — yellow flame indicates incomplete combustion; the colour comes from incandescent soot (carbon particulates).

Step 2 — harmful product.

A1 — carbon monoxide (CO) is produced; it is colourless and odourless, binds to haemoglobin with ~210× the affinity of oxygen, and prevents oxygen transport.

Step 3 — explain the danger.

A1 — incomplete combustion in a domestic setting is dangerous because CO is undetectable by smell and causes asphyxiation; complete combustion requires a plentiful supply of O2 and gives only CO2 and H2O.

Total: 8 marks (M3 A5).

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Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): Methane is the principal component of natural gas. When burned in a domestic boiler, the flue gas is monitored for CO content as a safety check.

(a) Write the balanced equations for (i) complete combustion of methane and (ii) incomplete combustion producing CO and water. (2)

(b) Explain, with reference to the C-H bond, why alkanes have similar combustion enthalpies per CH2 unit. (2)

(c) State two environmental concerns associated with combustion of alkane fuels, beyond CO production. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1) — $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$CH4​(g)+2O2​(g)→CO2​(g)+2H2​O(l).
• A1 (AO1.1) — $2CH_4(g) + 3O_2(g) \rightarrow 2CO(g) + 4H_2O(l)$2CH4​(g)+3O2​(g)→2CO(g)+4H2​O(l) (or any correctly balanced incomplete-combustion equation producing CO).