The answer lies in the fact that many (probably most) reactions do not occur in one step. In many cases there is not a smooth transition from reactants to products with a single transition state and activation energy, as we have been (simplistically) portraying it, but rather a series of steps, each with their own transition state and activation energy. Here, we will only consider one step and two step reactions, but in reality there could be many distinct steps from reactant to product. Each step represents a kind of subreaction, with its own activation energy and equilibrium state. The kinetics of a reaction is generally determined by the slowest of these subreactions - if forms a kind of “bottleneck” or “rate-limiting step”.

The reaction between methyl bromide (CH3Br) and hydroxide (–OH) is a one step reaction, that involves a nucleophilic attack by the hydroxide on the carbon. A nucleophile being a molecule that can donate electrons (here the hydroxide group) to an electrophile, a molecule that can accept them (methyl bromide). How do we know? Well the rate equation: rate = k [CH3Br] [–OH] tells us that the rate depends on both reactants, and we also know the rate depends on k - which is related to the height of the activation energy barrier. What this equation tells us is that in the transition state (the species at the highest energy on the energy profile) both reactants are present. We can imagine (and spectroscopically detect, because the structure of a molecule will determine how it interacts with light) what the structure of the transition state might look like.

The nucleophile (–OH) is attracted to the δ+ on the methyl carbon. At the same time the bromide ion starts to leave, so that at the “top” of the transition state (the most unstable point, requiring the most energy to form) we have a carbon that is coordinated to five other atoms by partial or full bonds. Given that carbon normally makes four bonds, it is no wonder that this “pentavalent” species sits at the reaction’s highest energy point)

In contrast, the reaction of t-butyl bromide and hydroxide is a two step reaction.
(CH₃)₃CBr + –OH ↔ (CH₃)₃COH + Br

How do we know? Because the rate equation for this reaction is first order: rate = k[(CH3)3CBr], and this tells us is that only (CH3)3CBr is involved in the step that determines the rate.

That is the transition state at the highest point on the energy diagram involves only the t-butyl bromide molecule: there is no nucleophile (the hydroxide) present during the step that determines how fast the reaction goes. The mechanism for this reaction involves two discrete steps, the first is the ionization of the t-butyl bromide, which involves breaking the C–Br bond. This results in a positively charged carbon (the bromine takes all the electrons and becomes bromide ion) - a very unstable (but distinct) species known as a carbocation.

The resulting carbocation is intermediate, it sits in an energy well between two less stable states. This distinguishes it from the transition state, which precariously “sits” at the highest local energy state (surrounded by lower energy states). Intermediates are in energy “valleys”, transitions states are at the summit of an energy “hill”.  The carbocation can react with the hydroxide, to form the t-butyl alcohol, or it can react with the bromide to reform the original product (or a variety of other “side“ reactions can occur). The important point here is that ideas about how the reaction proceeds can be deduced from the kinetic rate data on each reaction.

The rate equation gives us information about what reactants are present in the rate determining step of the reaction. The reaction can only go as fast as the slowest step - the step with the highest activation energy barrier. By analogy, imagine you are traveling at 70 mph on a five lane highway, if the lanes suddenly narrow to allow only one lane of traffic, all the cars slow down. Even though they are capable of traveling faster, no-one can get past the slowest cars.

Questions to answer:

Draw a reaction energy diagram for a two step reaction in which the second (or first) step is rate determining.
What is the rate equation for a reaction which occurs in the following steps:
       A + B → C (fast)
       A + C → D (slow)
Explain why it is not possible to write a rate equation from the reaction equation.