Introductory organic chemistry invariably features the mechanism of haloalkane solvolysis, and introduces both the Sn1 two-step mechanism, and the Sn2 one step mechanism to students. They are taught to balance electronic effects (the stabilization of carbocations) against steric effects in order to predict which mechanism prevails. It was whilst preparing a tutorial on this topic that I came across what was described as the special case of neopentyl bromide, the bimolecular solvolysis of which has been identified (DOI: 10.1021/ja01182a117) as being as much as 3 million times slower than methyl bromide. This is attributed to a very strong steric effect on the reaction, greater even than that which might be experienced by t-butyl bromide! Time I thought, to take a look at what might make neopentyl bromide so special, and what those supposed electronic and steric effects were really up to.
How does one construct a quantitative model? Well, a method which incorporates both van der Waals effects (dispersion attractions) and solvation in computing a potential energy surface seems appropriate. I used ωB97XD/6-311G(d,p) with SCRF correction for methanol as solvent. This predicts the following transition state structure. The calculated (free energy) barrier from the reactant is 30.2 kcal/mol.
Compare this with that for methyl bromide itself, for which a free energy barrier of 20.8 kcal/mol is calculated, a reasonably facile reaction at room temperature. What do these models tell us?So in this simple example, we can see illustrated many interesting effects; the balance between Sn1 and Sn2, the close approaches between atom pairs that characterize steric effects, the difference between fully ionic and less ionic structures, how the reaction normal mode (coordinate) changes with this changing mechanism. The text books have not gotten it wrong, but its nice to have some numbers associated with these concepts.
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