20.24 : アルケンへのラジカル反マルコフニコフ付加: 熱力学

Recall that anti-Markovnikov addition is only observed with HBr. The radical addition with HCl and HI is thermodynamically unfavorable.

But why?

Let's examine the thermodynamics for each propagation step involved in the radical mechanism.

For a reaction to be spontaneous, the change in Gibbs free energy, ∆G, must be negative.

For the first propagation step, the sign of ∆G is dictated by the competition between the enthalpy and the entropy terms.

In the case of HBr and HCl, the enthalpy term dominates at low temperatures, and ∆G becomes negative, implying a thermodynamically favorable reaction.

However, in the case of HI, both the enthalpy and entropy terms are positive, implying that ∆G is positive, regardless of temperature, indicating a thermodynamically unfavorable reaction.

For the second propagation step, the entropy term is close to zero because the number of reactants and product molecules are equal. So, the enthalpy term dictates the sign of ∆G.

As evident, ∆G is positive for HCl, rendering the reaction thermodynamically unfavorable.

Overall, the two propagation steps are thermodynamically favorable only in the case of HBr.