Question

Why does the reaction with `("CH"_3)_3"Br"` proceed with an `"S"_N1` and/or `"E"1` mechanism? What does it have to do with steric hindrance?

Answer 1

It can be considered as basically spatial crowding.

---

Steric hindrance is a kinetic factor that limits the ease to which a nucleophile (electron pair donor) can approach an electrophile (electron pair acceptor).

As an example, consider the reaction seen below, which one might hope is `"S"_N2`:

Here, the nucleophile is cyanide (`""^(-):"CN"`) and the electrophile is the central carbon on the alkyl halide (tert-butyl bromide).

We should predict that reaction does not work via an `"S"_N2` mechanism, based on the idea of steric hindrance---the three `"CH"_3` groups are blocking the cyanide from performing its backside-attack.

(LEFT: steric hindrance; RIGHT: reduced steric hindrance)

That reduces the ease to which a successful collision can occur between two reactants, and slows down the first step in a given substitution mechanism (making it the rate-limiting step).

Hence, this reaction proceeds more easily as a first-order mechanism (e.g. `"S"_N1` or `E1`), since the `"Br"^(-)` has the time (during the slow step) to come off on its own.

The rate law for this could then be approximated as first-order based on the slow step, since it dominates the extent of the reaction time:

`r(t) ~~ k_1["Br"^(-)]`

(Depending on the choice of solvent, either `"S"_N1` or `E1` could occur.)