Chemical Reactions of Alkanes (OCR A-Level Chemistry A): Revision Notes

Chemical Reactions of Alkanes

Reactivity of alkanes

Alkanes are known for their low reactivity compared to other organic compounds. Understanding why they are so unreactive helps explain their uses and the conditions needed for them to react.

Why alkanes are unreactive

Alkanes show limited reactivity with most common laboratory reagents for three main reasons:

• Strong σ-bonds: The carbon-carbon (C—C) and carbon-hydrogen (C—H) sigma bonds in alkanes require significant energy to break, making them stable and resistant to reaction

• Non-polar C—C bonds: Carbon-carbon bonds are completely non-polar because they form between identical atoms with the same electronegativity

• Effectively non-polar C—H bonds: Carbon and hydrogen have very similar electronegativities (2.5 and 2.1 respectively), meaning the C—H bond has such a small dipole that it can be considered non-polar for practical purposes

The non-polar nature of these bonds means alkanes do not attract polar reagents, and the bond strength means they resist breaking under normal laboratory conditions. This combination explains why alkanes are generally unreactive with acids, bases, oxidising agents, and reducing agents.

Combustion of alkanes

Although alkanes resist most reactions, they readily undergo combustion (burning) in the presence of oxygen. This reactivity makes them valuable as fuels.

Complete combustion

When sufficient oxygen is available, alkanes burn completely to produce carbon dioxide and water. This process releases substantial amounts of heat energy, which is why alkanes are used as fuels for heating, cooking, and transport.

The general word equation for complete combustion is: $\text{alkane} + \text{oxygen} \rightarrow \text{carbon dioxide} + \text{water} \;(+ \text{heat energy})$

Examples of complete combustion reactions include:

Methane (natural gas): $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Ethane: $\text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$

Propane: $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)$

Hexane (in petrol): $\text{C}_6\text{H}_{14}(l) + \frac{19}{2}\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 7\text{H}_2\text{O}(l)$

General equation for complete combustion

For any alkane with molecular formula $\text{C}_x\text{H}_y$, you can use this general equation:

$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}$

Incomplete combustion

When oxygen supply is limited, alkanes cannot burn completely. In these conditions, hydrogen atoms still oxidise to water, but carbon oxidation is incomplete, producing toxic carbon monoxide (CO) or even solid carbon (soot).

This explains why incomplete combustion occurs in enclosed spaces like car engines, faulty heating systems, or poorly ventilated rooms where oxygen supply cannot meet demand.

Real-world applications

The large amount of carbon dioxide produced during combustion contributes to environmental concerns. For instance, burning 1 kg of heptane (typical petrol component) in a car produces approximately 1680 dm³ of CO₂ at room temperature and pressure during just a 10-mile journey.

Reactions of alkanes with halogens

Alkanes undergo substitution reactions with halogens when exposed to ultraviolet (UV) radiation. These reactions are practically important for producing halogenoalkanes.

Halogenation conditions

For the reaction to proceed, two conditions must be met:

1. UV light (or sunlight): Provides the high energy needed to break the halogen-halogen bond and initiate the reaction
2. Suitable halogen: Only chlorine (Cl₂) and bromine (Br₂) are used in practice
   • Fluorine is too reactive and would cause an explosive reaction
   • Iodine is insufficiently reactive and barely reacts with alkanes

Example reaction

Mechanism for bromination of alkanes

The reaction between alkanes and halogens proceeds via a radical substitution mechanism. A mechanism shows the step-by-step movement of electrons during a reaction, helping us understand how products form from reactants.

What are radicals?

The three stages of radical substitution

Radical substitution mechanisms proceed through three distinct stages: initiation, propagation, and termination.

Stage 1: Initiation

The initiation stage starts the reaction by creating reactive radicals from stable molecules. UV radiation provides the energy needed to break the covalent bond in a bromine molecule by homolytic fission.

$\text{Br—Br} \xrightarrow{\text{UV light}} \text{Br}• + •\text{Br}$

The UV radiation supplies sufficient energy to break the Br—Br bond, forming two highly reactive bromine radicals. Each bromine atom takes one electron from the bond.

Stage 2: Propagation

The propagation stage is where the main reaction occurs. It consists of two steps that form a chain reaction – the product of one step becomes the reactant for the next, allowing the cycle to repeat.

Propagation step 1: $\text{CH}_4 + \text{Br}• \rightarrow •\text{CH}_3 + \text{HBr}$

A bromine radical attacks a C—H bond in methane, removing a hydrogen atom to form hydrogen bromide (HBr) and creating a methyl radical (•CH₃).

Propagation step 2: $•\text{CH}_3 + \text{Br}_2 \rightarrow \text{CH}_3\text{Br} + \text{Br}•$

The methyl radical reacts with another bromine molecule, forming the organic product bromomethane (CH₃Br) and regenerating a bromine radical.

Crucially, the bromine radical produced in step 2 can react with another methane molecule (step 1), allowing the propagation cycle to continue. In theory, these propagation steps could repeat until all reactants are consumed. Research suggests up to a million propagation cycles may occur before a termination step stops the reaction.

Stage 3: Termination

The termination stage occurs when two radicals collide and combine, forming a stable molecule with all electrons paired. This removes radicals from the reaction mixture and stops the chain reaction.

Several termination reactions are possible, depending on which radicals meet:

$\text{Br}• + •\text{Br} \rightarrow \text{Br}_2$

$•\text{CH}_3 + •\text{CH}_3 \rightarrow \text{C}_2\text{H}_6$

$•\text{CH}_3 + •\text{Br} \rightarrow \text{CH}_3\text{Br}$

When two radicals combine, both radicals are eliminated from the reaction mixture, halting further propagation steps in that reaction pathway.

Limitations of radical substitution in organic synthesis

Although radical substitution provides a method for synthesizing halogenoalkanes, it has significant drawbacks that limit its usefulness for preparing a single pure organic compound.

Further substitution

Once bromomethane (CH₃Br) forms in the second propagation step, bromine radicals can attack it just as readily as they attack methane. This leads to further substitution, progressively replacing more hydrogen atoms:

$\text{CH}_4 \xrightarrow{+\text{Br}_2} \text{CH}_3\text{Br} \xrightarrow{+\text{Br}_2} \text{CH}_2\text{Br}_2 \xrightarrow{+\text{Br}_2} \text{CHBr}_3 \xrightarrow{+\text{Br}_2} \text{CBr}_4$

The result is a mixture of products: CH₃Br (monobromomethane), CH₂Br₂ (dibromomethane), CHBr₃ (tribromomethane), and CBr₄ (tetrabromomethane), along with unreacted methane and bromine. Separating the desired product from this mixture requires additional purification steps.

Substitution at different positions in carbon chains

For methane and ethane, only one monosubstituted product is possible because all hydrogen atoms are equivalent (bonded to the same carbon environment).

However, for alkanes with three or more carbon atoms, hydrogen atoms exist in different positions along the carbon chain. Radicals can substitute at any of these positions, creating a mixture of structural isomers (compounds with the same molecular formula but different structural arrangements).

For example, pentane (C₅H₁₂) can form three different monosubstituted bromopentane isomers, depending on which hydrogen is replaced:

The three isomers show bromine attached to different carbon positions in the five-carbon chain. These are positional isomers of bromopentane, all with molecular formula C₅H₁₁Br but with different structures.

Remember!