IB Chemistry: Alkanes

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10.2.1 Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarity.

Alkanes are very unreactive – this can be attributed to two factors.

 Firstly, alkanes, as we saw in 10.1, contains only C-C and C-H bonds. These bonds are quite strong, therefore quite a lot of energy is required to break these bonds.

Additionally, the C-C and C-H bonds are non-polar. If you look in the data-booklet, the difference in electronegativity values between either the C-C or C-H bonds are very similar (actually, in the C-C bond, electronegativity values are the same, for some odd reason ;).) This similarity in electronegativity, as we learned in Unit 4, allows the electrons in the covalent bond to be shared quite equally. As these electrons are equally shared, it’s unlikely that any other reactants will interfere with the alkane.

This unreactiveness is often why Alkanes are such desirable compounds – they can be moved around and compressed whilst maintaining it’s stability.

10.2.2 Describe using equations the complete and incomplete combustion of alkanes.

 Alkanes react with oxygen in a reaction called combustion, to produce both carbon dioxide and water.

 Let’s take the example of the reaction of methane and oxygen. Please note here, for combustion to work properly, we need an excess supply of oxygen. Reacting alkanes with an insufficient supply of oxygen can lead to fatal results, but we will talk about this in a second.

Methane + oxygen → Carbon Dioxide + Water

CH4 (g) + 2O2 (g)→  CO2 (g) + 2H2O (l)

Tip: Make sure subscripts are balanced in the symbolic equation! You can learn how to do this in Unit 1.

Now, what happens when we burn an alkane with limited amounts of oxygen. Instead of carbon dioxide, carbon monoxide (CO) will be produced.

2CH4 (s) + 2O2 (g) → 2CO (g) +4H2O (l)

Carbon monoxide is highly toxic.

10.2.3 Describe, using equations, the reactions of methane and ethane with chlorine and bromine. In addition, describe the mechanisms in these reactions.

 Alkanes react with halogens in the presence of UV violet. The products formed are a halogenoalkane and a hydrogen halide. However, we must also be familiar with the mechanisms involved in the process.

 There are three steps which drive this process, and they are:

    •  Initiation
    • (Chain) Propagation
    • Termination


 Take an element like chlorine as an example. The bonds between the chlorine atoms are quite weak. As a result, UV light often has sufficient energy to break these bonds, so each of the atoms have an unbonded pair of electrons.

Methane chlorination: initiation

This process is known as homolytic fission, and the chlorine bond is split, producing two chlorine free radicals. Free radicals are species with an unpaired non-bonded electron – this makes free radicals highly reactive species.

 This step involves the generation of free radicals, so we call this step “Initiation”.


 The next step is called “Propagation”. Here, these free radicals react with an alkane (we’ll use methane here) to form a halogenoalkane and a hydrogen halide. The mechanism can be seen below:

Methane chlorination: propagation

CH4 + Cl* →CH3 * + HCl

CH3* + Cl2 → CH3Cl + Cl*

Here, CH3 * is the halogenoalkane (Chloromethane), and Hydrogen Chloride is formed. Now, the CH3* free radical will undergo another reaction with Cl2 (or the initial halogen), and hence causing a chain reaction. Since this reaction involves both using and generating free radicals, we call this step Propagation.


 This is the last step, which removes the free radicals by allowing them to react together, essentially producing a Cl-Cl atom.  The mechanism can be seen below:

Methane chlorination: termination

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