Question

Why do tertiary alkyl halides react in an `"S"_N1` mechanism more easily than `"S"_N2`?

Answer 1

Because they are bulky (kinetically stable), and hence block against `"S"_N2` backside-attack, giving the alternative mechanism of `"S"_N1` a greater percentage of success than `"S"_N2`.

They also form the most thermodynamically stable carbocation.

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`"S"_N1` is first-order nucleophilic substitution, so nucleophile strength is an important factor to consider.

Nucleophilicity is a kinetic phenomenon, so a good nucleophile is fast.

• For `"S"_N1` reactions of alkyl halides, the `stackrel("central carbon")overbrace("C")-stackrel("leaving group")overbrace"LG"` bond is weak, blocked, and the nucleophile is slow. So the `"LG"` leaves first in an `"S"_N1` fashion, giving a first-order process.

Also, because tertiary carbocations are among the most thermodynamically stable, this departure is favorable (not nonspontaneous).

• For `"S"_N2` reactions of alkyl halides, the `stackrel("central carbon")overbrace("C")-stackrel("leaving group")overbrace"LG"` bond is weak, open, and the nucleophile is fast. So the nucleophile influences the `"LG"`'s leaving.

As less substituted carbocations are less thermodynamically stable, the leaving group cannot easily depart on its own, so the joint process makes it a second-order process.

A bulky alkyl halide like a `3^@` alkyl halide blocks the nucleophile from attacking the central carbon, decreasing the percentage of successful `"S"_N2` and increasing the percentage of success in the alternative choice, `"S"_N1`.

Basically, the site to be attacked is heavily cluttered, which is hard to get past, and the significant thermodynamic stability of the carbocation promotes its formation during the `"S"_N1` process.