# How can you tell the rate determining step?

Aug 19, 2016

When you first learn about it in high school, it is often just told to you which step is the slow (rate-limiting/rate-determining) step.

If you want to guess at which step it is, we assume you know the mechanism already. Often, the slow step is the step that is least favorable, so it takes a while for the fairly unstable intermediate to form before it gets propelled to form the product(s).

An example is the ${\text{S}}_{N} 1$ mechanism on tert-butyl bromide, where bromide (${\text{Br}}^{-}$) departs as a leaving group, and then a slow/poor nucleophile attacks the planar carbocation.

Attacking the carbocation is faster than forming it, because:

• ${\text{Br}}^{-}$ leaves on its own, when the nucleophile is slow. The slow nucleophile makes it so that ${\text{Br}}^{-}$ has to participate by itself, i.e. in a first-order process, for this step, and thus ${\text{Br}}^{-}$ has little to no assistance to leave faster.

Therefore, this is the slow step.

• The steric hindrance of tert-butyl bromide is lessened when the carbocation forms, since a planar molecule is inherently less sterically-hindered than its tetrahedral conformation.

Therefore, this is a faster step.