Explanation for the trends in nucleophilicity—Orbital interactions or electrostatic attraction?

Question

The trend of halide nucleophilicity in polar protic solvents is $$\\ce{I- > Br- > Cl- > F-}$$ The reasons given by Solomons and Fryhle^\\[1\\], and by Wade^\\[2\\] are basically as follows. 1. Smaller anions are more solvated than bigger ones because of their 'charge to size ratio'. Thus, smaller ions are strongly held by the hydrogen bonds of the solvent. 2. Larger atoms/ions are more polarizable. Thus, larger nucleophilic atoms/ions can donate a greater degree of electron density to the substrate than a smaller nucleophile whose electrons are more tightly held. This helps to lower the energy of the transition state. Clayden et al.^\\[3\\] give the reason that in S_N2 reactions, the HOMO–LUMO interaction is more important than the electrostatic attraction. The lone pairs of a bigger anion are higher in energy compared to a smaller anion. So, the lone pairs of a bigger anion interact more effectively with the LUMO σ*. Thus, soft nucleophiles react well with saturated carbon. What I don't understand is the trend of halide nucleophilicity in polar aprotic solvents. $$\\ce{F- > Cl- > Br- > I-}$$ In aprotic solvents, only the solvation factor is absent. Bigger anions are still more polarizable and can interact in a better way with the LUMO. Then, what is the reason for the trend reversal in aprotic solvents? Solomons and Fryhle^\\[1\\] say >In these[aprotic] solvents anions are unencumbered by a layer of solvent molecules and they are therefore poorly stabilized by solvation. These “naked” anions are highly reactive both as bases and nucleophiles. In DMSO, for example, the relative order of reactivity of halide ions is opposite to that in protic solvents, and it follows the same trend as their relative basicity. **Shouldn't the orbital interactions matter more than the electrostatic attractions even in polar aprotic solvents? Why does nucleophilicity follow the basicity trend in polar aprotic solvents?** References: 1. Solomons, T. W. Graham; Fryhle, C. B. *Organic Chemistry*, 10th ed.; Wiley: Hoboken, NJ, 2011; pp. 258–260. 2. Wade, L. G. *Organic Chemistry*, 8th ed.; Pearson Education: Glenview, IL, 2013; pp. 237–240. 3. Clayden, J.; Greeves, N.; Warren, S. *Organic Chemistry*, 2nd ed.; Oxford UP: Oxford, U.K., 2012; pp. 355–357.

Answer 1

The way I like to look at that trend is about how "aggressive" the nucleophile is. In the absence of solvation (stabilization of the anion), the nucleophilicity parallels basicity because stronger bases are just looking for something positive (or partially positive) to stick to!

The reason the trend is reversed in protic solvents is that the anions/bases are satisfied with the hydrogen bonding to stabilize the anion, so nucleophilicity then hinges on which anion is more capable of donating its electrons to something else - the larger, more polarizable (and higher HOMO) anions more readily act as nucleophiles under those conditions.

Look at it this way - the simplistic concept of nucleophilicity is about how readily the anion wants to act as a Lewis Base, right? What makes a good Lewis Base? "Strong bases" are less stable anions which really want to grab protons, but having Polarizable anions also help with nucleophilicity. So you have BOTH aspects to consider for nucleophilicity - base strength AND polarizability.

These trends are opposite for the halides, so the key is figuring out which trend "wins". If the solvent is aprotic, the base strength is more relevant and therefore the smaller halides are better Nucleophiles. Thus the order is: F- > Cl- > Br- > I-

However, if the solvent is protic, the base strength is less relevant because of the stabilization of the anion weakening it as a base, therefore the more polarizable/larger halides are better Nucleophiles and the order reverses to: I- > Br- > Cl- > F-



Hope that helps...