How can you tell the rate determining step?

1 Answer
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 #"S"_N1# mechanism on tert-butyl bromide, where bromide (#"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:

  • #"Br"^(-)# leaves on its own, when the nucleophile is slow. The slow nucleophile makes it so that #"Br"^(-)# has to participate by itself, i.e. in a first-order process, for this step, and thus #"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.