In the energy profile diagrams below, it can be clearly seen that the free radical chlorination of an alkane

$$\ce{RH + Cl2 -> RCl + HCl}$$

is exothermic. However, the corresponding bromination

$$\ce{RH + Br2 -> RBr + HBr}$$

is endothermic. Why is there a difference?

enter image description here

  • 5
    $\begingroup$ That's because both reactions products, hydrohalogenide and halogenoalkane are more stable with chlorine. The orbital overlap is not as good in the case of the larger bromine. $\endgroup$
    – Karl
    Dec 3, 2017 at 11:13
  • $\begingroup$ This earlier answer may be helpful. $\endgroup$
    – ron
    Dec 3, 2017 at 14:56
  • $\begingroup$ It looks like your energy diagrams are for the first propagation steps not the full reaction [RH + X(dot) ---> R(dot) + HX]. You also need R(dot) + X2 ---> RX + X(dot) for the second step. Chlorination is more exothermic than bromination but both are exothermic. $\endgroup$
    – user55119
    Dec 3, 2017 at 17:51

1 Answer 1


First off, we can see that both reactions are exothermic If we do the thermodynamic analysis.

Given that your starting materials have the same heat of formation:

$$\Delta H^\circ_f(Products) = \Delta H^\circ_f (\ce{R-H}) + 0, \quad{\Delta H^\circ_f(\ce{Cl}) = \Delta H^\circ_f(\ce{Br})= 0}$$

$$\ce{R = Et}\\ \Delta H^\circ_f (\ce{EtH}) = \pu{-84 kJ mol^{-1}}$$ $$ \begin{array}{cccccc} \hline \text{Product 1} & \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] & \text{Product 2} & \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] & \Sigma\ \Delta \text{H}^\circ_f\ [\pu{kj mol^{-1}}] &\text{Result}\\ \hline \ce{HBr} & -\text{36.45} & \ce{Et-Br} & -\text{97.6} &-\text{134.05}<-\text{84} &\text{Exothermic} \\ \ce{HCl} & -\text{92.31} & \ce{Et-Cl} & -\text{137} & -\text{229.31}<-\text{84} &\text{Exothermic}\\ \hline \end{array} $$

Back to your question:

Why is there a difference?

The graphs you are looking at only consider the reaction:

$$\ce{R-H + X^. -> R^. + HX}$$

For which we get: $$ \begin{array}{cccccc} \hline \text{X} & \Delta \text{H}^\circ_f(\ce{X^.}) & \Delta \text{H}^\circ_f(\ce{RH}) & \Delta \text{H}^\circ_f(\ce{R^.}) & \Delta \text{H}^\circ_f(\ce{HX}) & \Delta \text{H}^\circ_{rxn} & \text{Result}\\ \hline \ce{Br} & -\text{96.94}^{**} & -\text{84} & -\text{119}^\dagger &-\text{36.45} &\text{25.49}&\text{Endothermic} \\ \ce{Cl} & -\text{121.29}^{**} & -\text{84} & -\text{119}^\dagger & -\text{92.31} &-\text{6.02}&\text{Exothermic}\\ \hline \end{array} $$ *all units in $[\pu{kj mol^{-1}}]$ all values from Wikipedia unless otherwise stated
** Table of Bond Dissociation Energies
$^\dagger$ Values from NIST WebBook

Here the enthalpy of radicalization of the alkane is different for chlorine or bromine, but the halogenation in total is exothermic


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