5.2 Exercise
Consider a nucleophilic substitution reaction:(CH3 )3 CCl + OH− → (CH3 )3 COH + Cl−
Let us first assume the following mechanism
k1 ,k−1(CH3 )3 CCl *) (CH3 )3 C+ + Cl−
k2(CH3 )3 COH(CH3 )3 C+ + OH− −→Q1. Apply the steady-state approximation for the intermediate (CH3 )3 C+ and derive
the rate law.
Q2. Show that the rate law is 1st-order when [OH− ] is in excess or when k−1  k2Let us next consider an alternative mechanism
kf ,kbkr(CH3 )3 COH + Cl−(CH3 )3 CCl + OH− *) [Cl · · · (CH3 )3 C · · · OH]− −→Q3. Apply the steady-state approximation for the intermediate complex and show that
this mechanism gives a 2nd-order rate law.
Answers:
1.
Rate =

k1 k2 [(CH3 )3 CCl][OH− ]k−1 [Cl− ] + k2 [OH− ]2.
Rate ' k1 [(CH3 )3 CCl] when k−1 [Cl− ]  k2 [OH− ]3.
Rate =

kf kr[(CH3 )3 CCl][OH− ]kb + kr

5.3 Michaelis-Menten equation
Let us consider an enzyme catalysis reaction in which the enzyme E and a substrate Sform an intermediate complex ES and then the reaction occurs to release the product P:
k1 ,k−1

k2E+PE+S *) ES −→27(51)This looks similar to the scheme in Eq (49), and hence the steady-state approximation
will givek1 k2
Rate =[E][S]
k2 + k−1(It will be a good exercise to derive and confirm this.) However, this looks very different
from the famous Michaelis-Menten equation:
Rate =

Vmax [S][S] + KM

We notice that one of the major difference is that the enzyme concentration [E] is not
involved in the latter. We therefore try to eliminate [E] in the course of the application of
the steady-state approximation. To this end, it is essential to recognize that the enzyme
is a catalyst. Therefore, it is not destroyed by the reaction and the total number ofenzyme molecules is kept constant. This is described by[E] + [ES] = constant = [E]0(52)in which [E]0 is the initially prepared concentration of the enzyme. Now we apply therecipe of the steady-state approximation. First we identify that the ES complex is theintermediate. Its change in the scheme (51) is written asd[ES]= k1 [E][S] − k−1 [ES] − k2 [ES]dt
The steady-state approximation sets the right hand side to be equal to zero. But before
doing it, we will eliminate [E] by using Eq (52):k1 ([E]0 − [ES])[S] − k−1 [ES] − k2 [ES] ' 0
k1 [E]0 [S](53)
k1 [S] + k−1 + k2
The overall rate is that of the product formation determined by the final step of thescheme (51):d[P]= k2 [ES]Rate =dt
We replace [ES] by Eq (53):⇒ [ES]ss =Rate =

k2 k1 [E]0 [S]
k2 [E]0 [S]Vmax [S]==k+k−1
2
k1 [S] + k−1 + k2
[S] + KM
[S] + k1in which Vmax = k2 [E]0 is the maximum rate and KM = (k−1 + k2 )/k1 is the Michaelis
constant.
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Further Readings
Now there would be two ways for you to proceed. One is to review the basics again byreading the recommended textbooks. In particular, it would be a good idea to attempt theworked examples and problems in order to see how the theories work in more practicalsituations and to confirm your understanding of the theories. It is also useful to read
explanations by different authors. This is because no one can explain things perfectly
and you will have to build your own way of understanding by repeating reading andthinking.
Another way forward would be to read advanced books and research papers. One ofrecommendable books is:• A Fersht, Structure and Mechanism in Protein Science, Freeman.
(There are 8 long loan and 2 week loan copies in Barnes Library, Medical School,Shelfmark QP 551/F.)
I trust that this lecture course has provided sufficient background to get ready for thisand other books and papers.

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