Search any question & find its solution
Question:
Answered & Verified by Expert
The highest peak in energy profile diagram for mechanism of alkaline hydrolysis of tertiary butyl bromide represents
Options:
Solution:
1400 Upvotes
Verified Answer
The correct answer is:
transition state of $1^{\text {st }}$ step
The alkaline hydrolysis to tert-butyl bromide with aqueous alkali such as $\mathrm{NaOH}$ or $\mathrm{KOH}$ is as follows.
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 $\alpha\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\right]$
Rate $=\mathrm{K}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\right]$
This is a first order reaction because rate of hydrolysis of $\left(\mathrm{CH}_{3}\right)_{3}-\mathrm{Br}$ is independent of the concentration of alkali or $\mathrm{OH}^{-}$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 .
Rate of reaction $=\mathrm{k}_{1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\right]$
The first step consists of breaking of $\mathrm{C}-\mathrm{Br}$ 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 $\mathrm{SN}^{1}$ mechanism(substitution, nucleophilic, unimolecular).
Rate of reaction $=\mathrm{k}_{2}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}\right]\left[\mathrm{OH}^{-}\right]$
The second step involves the attack of $\mathrm{OH}^{-}$ion. This is the fast step, since it is the bond formation step.
Energy profile diagram of $\mathrm{SN}^{1}$ mechanism shows that rate of a reaction is independent of the concentration of nucleophile. The first step requires larger activation energy $\left(\Delta \mathrm{E}_{1}\right)$ than the second step $\left(\Delta \mathrm{E}_{2}\right)$. 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 $\mathrm{SN}^{1}$ reaction. The energy difference between products and reactants is $\Delta \mathrm{H}$, i.e., Heat of reaction.
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 $\alpha\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\right]$
Rate $=\mathrm{K}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\right]$
This is a first order reaction because rate of hydrolysis of $\left(\mathrm{CH}_{3}\right)_{3}-\mathrm{Br}$ is independent of the concentration of alkali or $\mathrm{OH}^{-}$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 .
Rate of reaction $=\mathrm{k}_{1}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\right]$
The first step consists of breaking of $\mathrm{C}-\mathrm{Br}$ 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 $\mathrm{SN}^{1}$ mechanism(substitution, nucleophilic, unimolecular).
Rate of reaction $=\mathrm{k}_{2}\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}\right]\left[\mathrm{OH}^{-}\right]$
The second step involves the attack of $\mathrm{OH}^{-}$ion. This is the fast step, since it is the bond formation step.
Energy profile diagram of $\mathrm{SN}^{1}$ mechanism shows that rate of a reaction is independent of the concentration of nucleophile. The first step requires larger activation energy $\left(\Delta \mathrm{E}_{1}\right)$ than the second step $\left(\Delta \mathrm{E}_{2}\right)$. 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 $\mathrm{SN}^{1}$ reaction. The energy difference between products and reactants is $\Delta \mathrm{H}$, i.e., Heat of reaction.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.