Benzene is 45 kJ/mol more energetic than bromobenzene. I understand that due to resonance in bromine to benzene in bromobenzene increase the stability but such an enormous amount of energy can't only be dealt by resonance. There must be another reason. What is it?

  • 2
    $\begingroup$ Could you state the source of information? $\endgroup$ Jun 29, 2018 at 7:30
  • $\begingroup$ @TanYoungBoon stated in L.G Wade"bromination of benzene gives the product(bromobenzene) that is 45kl/mol more stable than benzene " $\endgroup$
    – Hercules
    Jun 29, 2018 at 15:35

1 Answer 1


The corresponding section in Wade, L. G. Organic Chemistry, 8th Edition; Pearson Education: Glenview, IL, 2013; p 758. reads

Bromination of Benzene Bromination follows the general mechanism for electrophilic aromatic substitution. (…)

Formation of the sigma complex is rate-limiting, and the transition state leading to it occupies the highest-energy point on the energy diagram (Figure 17-1). This step is strongly endothermic because it forms a nonaromatic carbocation. The second step is exothermic because aromaticity is regained and a molecule of HBr is evolved. The overall reaction is exothermic by 45 kJ/mol (10.8 kcal mol).

Thus, the given value of 45 kJ/mol refers to the reaction enthalpy of the total reaction $$\ce{Br2 + C6H6 -> C6H5Br + HBr}$$

The reaction is mainly so exothermic because of the formation of HBr. This can be shown using values for the standard molar enthalpy of formation taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

$$\begin{align} &\Delta_\mathrm f H^\circ(\ce{Br2(l)})&&=0\ \mathrm{kJ/mol}\\ &\Delta_\mathrm f H^\circ(\ce{Br2(g)})&&=30.9\ \mathrm{kJ/mol}\\ &\Delta_\mathrm f H^\circ(\ce{C6H6(l)})&&=49.1\ \mathrm{kJ/mol}\\ &\Delta_\mathrm f H^\circ(\ce{C6H5Br(l)})&&=60.9\ \mathrm{kJ/mol}\\ &\Delta_\mathrm f H^\circ(\ce{HBr(g)})&&=-36.3\ \mathrm{kJ/mol}\\ \end{align}$$

although the exact value of 45 kJ/mol cannot be reproduced using these values.


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