# Regioselectivity in opening of cationic three-membered rings

The mechanism of electrophilic addition on alkenes often involves an electrophile attacking the double bond and causing an electromeric shift of electrons. The electrophile then forms a bridged cation if possible.

The carbon on the left side has an alkyl group which shows inductive effect (and hyperconjugation effect) which increases the electron density on the left carbon. The right carbon is not bonded by any such alkyl group. Therefore, the electron density on the left carbon must be higher than that on the right carbon.

Why does the nucleophile $\ce{Nu}$ attack the carbon which has higher electron density?

There are three factors that you should consider in nucleophilic attack.

1. Steric effects
2. Electronic effects
3. Stereoelectronic effects

Two of these are related to lowering the overall energy of the transition state. The third relates to how productive a collision can be.

The one that matters less here (or is already accounted for) is stereoelectronic effects. Notice that you proposed a backside attack for a $$\mathrm{S_N2}$$ reaction. Therefore, you have considered the proper alignment of orbitals, which is what is meant by stereoelectronic effects.

Steric effects government repulsion of large groups. In your example, attacking the more substituted side would result in greater steric repulsion since the $$\ce{R}$$ group is larger than $$\ce{H}$$. Steric arguments therefore argue for attack at the other carbon.

Frequently, the three factors do not say the same thing, and you need to use some intuition and/or experience to figure out which one dominates (in reality, the relative energies will determine the "winner").

In this case, it turns out with a carbocation and a negative nucleophile, perhaps not surprisingly, electronic effects dominate. There are actually two related electronic effects. One is that the partial positive is greater on the more substituted carbon (cf. Markovnikov rule). This carbon is hyperconjugatively stabilized (relative to the other carbon) by greater substitution. In addition, the greater positive charge also implies a weaker $$\ce{C-E}$$ bond, which further implies a lower energy $$\ce{C-E}$$ $$\sigma^{*}$$ orbital. We can then argue that the transition state energy will be lower given that the two interacting orbitals are closer in energy. And indeed, the electronic effects dominate this reaction giving the reactivity shown in your diagram.