Question

Explain the mechanism of alkaline hydrolysis of t-butyl bromide with energy profile diagram.

Answer 1

The alkaline hydrolysis to tert-butyl bromide with aqueous alkali such as NaOH or KOH is as follows.
$\underset{\underset{bromide}{tert-butyl}}{(C{H}_3{)}_{3}C}-Br+NaOH\left(aq\right)\to \underset{\underset{alcohol}{tert-butyl}}{(C{H}_3{)}_{3}C}-OH+NaBr$.
The rate of this reaction depends only on the concentration of the tert-butyl bromide and is independent of the concentration of alkali added.
Rate $\alpha \left[\right(C{H}_3{)}_{3}C-Br]$
Rate $=K\left[\right(C{H}_3{)}_{3}C-Br]$
This is a first order reaction because rate of hydrolysis of $(C{H}_3{)}_3-Br$ is independent of the concentration of alkali or $O{H}^{-}$ ions. This can be explained by two-step mechanism shown below. Each step is an elementary reaction with its own rate constant, step $1$ proceeds much more slowly than step $2$.
Step $1$. $(C{H}_3{)}_{3}C-Br\stackrel{slow}{\underset{k_1}{\to }}(C{H}_3)C^{+}+B{r}^{-}$
Rate of reaction $=k_1\left[\right(C{H}_3{)}_{3}C-Br]$
The first step consists of breaking of C-Br bond and it determines the rate of overall reaction. So, step $1$ is called the rate-determining step. The rate determining step in this reaction involves only a single molecule, therefore, it is said to be unimolecular. Also, this type of mechanism is known as $S{N}^1$ mechanism(substitution, nucleophilic, unimolecular).
Step $2$. $(C{H}_3{)}_3{C}^{+}+O{H}^{-}\stackrel{fast}{\underset{k_2}{\to }}(C{H}_3)C-OH$
Rate of reaction$=k_2\left[\right(C{H}_3{)}_3{C}^{+}\left]\right[O{H}^{-}]$
The second step involves the attack of $O{H}^{-}$ ion. This is the fast step, since it is the bond formation step.
Energy profile diagram of $S{N}^1$ mechanism shows that rate of a reaction is independent of the concentration of nucleophile. The first step requires larger activation energy $\left(\Delta E_1\right)$ than the second step $\left(\Delta E_2\right)$. The first step to form carbocation determines the rate of overall reaction. The second step, which is the attack of nucleophile on carbocation is exothermic i.e., it is a lower energy transition state. The intermediate carbocation appears at a low point in the diagram. The conditions and reagents which favour the formation of carbocation will accelerate the $S{N}^1$ reaction. The energy difference between products and reactants is $\Delta$H, i.e., Heat of reaction.