The benzene ring's H can be replaced by a Br via Free Radical Substitution. Explain.

If we were to assume that benzene reacts with Br₂ without the presence of FeBr₃ or Fe (halogen carriers), we will have to assume that benzene reacts with Br₂ via a free radical substitution mechanism.

If that is the case, we will expect the following to be the mechanism.

However, the lack of the dibenzene compound, which two benzene rings are joined together via a single C-C bond, indicates that the reaction between benzene and Br₂ cannot occur via Free Radical substitution. Hence, we must now account/explain why this is so.

Why is it not possible for the benzene ring's H to be replaced by a Br via Free Radical Substitution? If we examine the proposed propagation step. In order to create the benzene radical, the bromine radical reacts with the C-H bond of benzene causing the latter to break homolytically. However, this C-H bond is stronger than your alkane's C-H bond.

This is because in benzene, its C-H bond is the overlapping between a sp² orbital (from C) and a s orbital (from H). While in the alkanes, the C-H bond is formed by overlapping the sp³ orbital (from C) and a s orbital (from H). The sp² orbital is smaller than the sp³ orbital because the former has a smaller percentage of p in its hybrid orbital, hence overlapping between the sp² and the s is more effective hence a stronger C-H bond.

This stronger C-H bond, makes the breaking of the C-H bond in benzene to be less favourable than the breaking of C-H bond in alkane. Hence, from the angle of differing C-H bond strength, we can account for why when benzene and Br₂ is added together, the reaction will not be a free radical substitution.

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^{Article written by Kwok YL 2010.}

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