Question

Why is `3^@ > 2^@ > 1^@ > "methyl"` the trend for reactivity in an `"S"_N1` reaction but `3^@ < 2^@ < 1^@ < "methyl"` for `"S"_N2` reactions?

Answer 1

From an `"S"_N1` perspective instead of an `"S"_N2` perspective, we focus not on the incoming nucleophile but on the `\mathbf("C"-"LG")` bond (where `"LG"` is a leaving group), as the `"C"-"LG"` bond strength is more important than the nucleophile strength in a primarily unimolecular reaction.

In an `"S"_N1` reaction mechanism, the rate-limiting step is the one where the `"C"-"LG"` bond breaks because if that happens, it happens without the coercion of an outside nucleophile. That's the defining characteristic of an `"S"_N1`: unimolecularity in the slow step.

STABILIZATION OF THE INTERMEDIATE

Postulate: We know that if a compound is more stable, it tends to willingly stay in its current state. Similarly, the more stable the intermediate, the more favorably it stays in its current state once it forms.

The greater the number of electron-donating groups (e.g. alkyl) around the carbocation, the more electron density is donated into the electrophilic carbon, stabilizing it via hyperconjugation.

Hyperconjugation is an effect where the `"C"-"H"` bonding electron pairs are delocalized into the empty `p` orbital on the electrophilic carbon.

In general, spreading out electron density is stabilizing, almost like you would see in resonance structures of conjugated `pi` systems like benzene and 1,3,5-hexatriene. Hyperconjugation is ultimately another form of that kind of behavior (though not quite the same thing, so don't equate them).

Thus, the more alkyl groups around the electrophilic carbon, the more stabilized the carbocation intermediate, and the lower in energy it is.

Thus it can stay in that state for long enough that a weak nucleophile can still form a bond with it, making it more favorable for the intermediate to form (remember that `"S"_N2` doesn't have a detectable intermediate!).

Thus, we have the `3^@ > 2^@ > 1^@ > "methyl"` reactivity trend.

This is a thermodynamic effect, as it affects the energy of the intermediate and thus the Gibbs' free energy of the reaction's first step.

AN ADDITIONAL KINETIC EXPLANATION

There can be more than one angle to this.

The strength of the `"C"-"LG"` bond also depends on the basicity of the leaving group and the steric bulk, so we should examine these two main trends.

BASICITY OF LEAVING GROUP

Let us assume comparisons between carbocations of the same kind (`1^@` vs. `1^@`, `2^@` vs. `2^@`, etc).

The stronger the basicity of the leaving group, the stronger the `"C"-"LG"` bond is. That means the reactant becomes more stable (lower in energy) because a stronger bond requires more energy to break, so the larger the activation energy `DeltaE_1^‡` becomes.

Since reactivity is a kinetic factor, and since a more stable reactant is associated with a larger activation energy, it correlates with a slower rate-limiting step, favoring `"S"_N1` over `"S"_N2`.

(It's like trying to drive up a very tall and steep hill, but failing and realizing you need to start further up the hill.)

STERIC BULK

When I say this, I am referring to both the leaving group and everything around the electrophilic carbon.

If the leaving group is bulky (large), it's going to be quite favorable for it to leave because of the steric clash; it's just too "cramped", like you would be amidst a busy New York crowd.

If the stuff around the carbocation is crowding the leaving group, that also facilitates the `"C"-"LG"` bond breaking.

Both work along similar lines, destabilizing the reactant and thus decreasing the activation energy.

Thus, `3^@ > 2^@ > 1^@ > "methyl"` carbocations when it comes to reactivity in `"S"_N1`, and the bulkier alkyl halides react more readily in `"S"_N1` mechanisms than `"S"_N2`.

This is a kinetic effect, since it is associated with the activation energy.