Question

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

Solution

## The alkaline hydrolysis to tert-butyl bromide with aqueous alkali such as NaOH or KOH is as follows.$$\underset{\underset{\displaystyle bromide}{\displaystyle tert-butyl}}{(CH_3)_3C}-Br+NaOH(aq)\rightarrow \underset{\underset{\displaystyle alcohol}{\displaystyle tert-butyl}}{(CH_3)_3C}-OH+NaBr$$.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 $$\alpha[(CH_3)_3C-Br]$$Rate $$=K[(CH_3)_3C-Br]$$This is a first order reaction because rate of hydrolysis of $$(CH_3)_3-Br$$ is independent of the concentration of alkali or $$OH^-$$ 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 $$1$$. $$(CH_3)_3C-Br\overset{slow}{\underset{k_1}{\rightarrow}}(CH_3)C^++Br^-$$Rate of reaction $$=k_1[(CH_3)_3C-Br]$$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 $$SN^1$$ mechanism(substitution, nucleophilic, unimolecular).Step $$2$$. $$(CH_3)_3C^++OH^-\overset{fast}{\underset{k_2}{\rightarrow}}(CH_3)C-OH$$Rate of reaction$$=k_2[(CH_3)_3C^+][OH^-]$$The second step involves the attack of $$OH^-$$ ion. This is the fast step, since it is the bond formation step.Energy profile diagram of $$SN^1$$ mechanism shows that rate of a reaction is independent of the concentration of nucleophile. The first step requires larger activation energy $$(\Delta E_1)$$ than the second step $$(\Delta E_2)$$. 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 $$SN^1$$ reaction. The energy difference between products and reactants is $$\Delta$$H, i.e., Heat of reaction.Chemistry

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