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

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.

#"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.