Alkanes - Free Radical Substitution Mechanism

When we learn how to draw the Free-Radical Substitution mechanism. It is useful to think about the classification of the reactions first. This helps us see the pattern and think abit more about this pattern. In this manner, we can understand the reaction better.

From the classification of reactions, we can recognise that when the C-H bond of the alkane molecule undergoes a substitution, this reaction goes via the free-radical substitution mechanism.

The C-H is a non-polar and a very strong covalent bond (Bond Energy of 410 kJ mol^-1, hence it is a chemical bond that is reluctant to be broken. Therefore, the C-H bond is unreactive.

The difficulty to break a C-H bond, implies that in order for a chemical reaction to break it, a highly reactive reactant is needed. This how the relevance of free radicals is understood.

Free radicals are highly reactive substance. They contain an unpaired electron, generally one short of the complete octet. This deficiency in electrons make the free radical extremely reactive. In addition, because free radicals are very reactive, they are indiscriminate with whom they attack.

The above illustration show how to describe the free radical substitution mechanism, which generally have three steps. The video below provides a step-by-step guide to drawing the mechanism.

In conclusion, there are a few more points to take note about the free radical substitution reaction.

(1) The propagation step involving the alkane molecule reacting with the halogen radical can be seen as the Rate-Determinating Step, because it is the step that breaks the stronger bond hence it should have the highest activation energy.

(2) Despite radicals being indiscriminate with what they attack, the stability of the product will determine whether the reaction is feasible.

For example, as the Hydrogen radical is less stable than an alkyl radical (i.e. a radical which has the Carbon with the unpaired electron), therefore the former is not formed in the reaction between the halogen radical and alkane molecule. This implies that the radical attacks the Hydrogen instead of the Carbon.

(3) When we are using an alkane molecule that contains the C-C bond. Since a radical is extremely reactive, it will still attack the H of the C-H bond, because it is the easier to reach hydrogen since it is more exposed than the carbon atom. In addition, the H-X bond formed will always be more stable than the C-X bond formed.

(4) There are other radical reactions, such free radical addition reaction, but it is not in our syllabus.

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^{Article written by Kwok YL 2009.}

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