Can a non-rate determining step control what products form in a reaction?

Question

In Organic Chemistry by Clayden, the stereoselectivity and regioselectivity of the E1 mechanism is explained in terms of the differing transition state energies (and therefore rates) of the possible deprotonation steps.

However, the rate-determining step of the E1 mechanism is the formation of the carbocation, not the deprotonation. How is it that the deprotonation is able to control the stereoselectivity and regioselectivity of the reaction despite not controlling the overall rate of reaction?

I have a potential explanation for this, though I am not sure if it is correct. I think that the rate-determining step only limits the rate of the overall elimination, and has no impact on the stereoselectivity. Although carbocation deprotonation has to occur much faster than the carbocation formation, when comparing the two deprotonation routes, one route is still much faster than the other (even though both are quite fast compared to the carbocation formation), which means that the deprotonation is stereo and regioselective, because the majority of the carbocations will react via the faster route. Is this thinking correct?

Answer 1

The rate-determining step determines the rate of the overall reaction. However, that is not to say that there is only one transition state that determines the regio- and stereo-selectivity of an entire reaction.

With these mechanisms, many transition states can form, and in chemistry, reactions tend to follow pathways that flow with lower energy demands.

For example, take a waterslide with large bumps, which represent the different transition states you have to pass. Climbing to the top of the waterfall is your first step, and that requires the most energy, just like the formation of the carbocation which has the highest activation barrier. On the top deck, there are several options for slides, and again, on each slide, there are many bumps. On some rides, they are much larger, corresponding to less favoured pathways. Although you will get past them since you are at the top, your ride would be much slower, which would not as fun. The ride with the smallest bumps will get you down the fastest, and so you'd choose that one.

For all the pathways, forming E/Z-product or more the more/less substituted product, you will encounter the formation of a carbocation. From there, it is simply that the pathways that result in E and more substituted products go through lower-in-energy activation barriers. Thus, those pathways are favoured (kinetically and thermodynamically) and the observed regio and stereo-selectivity become present.