Problem 54 - Chemistry and Chemical Reactivity

Question

A three-step mechanism for the reaction of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and \(\mathrm{H}_{2} \mathrm{O}\) is proposed: Step 1 Slow $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}+\mathrm{Br}^{-}$$ Step 2 Fast $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}$$ Step 3 Fast $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{HBr}$$ (a) Write an equation for the overall reaction. (b) Which step is rate-determining? (c) What rate law is expected for this reaction?

Step-by-Step Solution

Answer

(a) The overall reaction is \((\text{CH}_3)_3\text{CBr} + \text{H}_2\text{O} \to (\text{CH}_3)_3\text{COH} + \text{HBr}\). (b) Step 1 is rate-determining. (c) Rate law: \(\text{Rate} = k[(\text{CH}_3)_3\text{CBr}]\).

Step 1: Identify and Write the Overall Reaction

To find the overall reaction, we need to sum up all the individual steps in the mechanism. Initially, identify the intermediates, specifically \( ext{(CH}_3)_3\text{C}^+\), which are species that appear on both sides of the reaction and can be canceled out. Doing this we combine: \[(\text{CH}_3)_3\text{CBr} + \text{H}_2\text{O} \longrightarrow (\text{CH}_3)_3\text{COH} + \text{HBr}\] This is the overall balanced reaction.

Step 2: Determine the Rate-Determining Step (RDS)

The rate-determining step is the slowest step in the mechanism, as it limits the overall rate of the reaction. According to the mechanism provided, Step 1 is designated as 'Slow', indicating that it is the rate-determining step. Therefore, Step 1: \[(\text{CH}_3)_3\text{CBr} \longrightarrow (\text{CH}_3)_3\text{C}^+ + \text{Br}^-\] is the rate-determining step.

Step 3: Derive the Rate Law from the Rate-Determining Step

The rate law is derived from the rate-determining step. Since Step 1 is the slow step, the rate of the entire reaction depends on this step only. The rate law is based on the reactants of the rate-determining step. Therefore, the concentration of \((\text{CH}_3)_3\text{CBr}\) determines the rate: \[ ext{Rate} = k[(\text{CH}_3)_3\text{CBr}]\] Here, \(k\) is the rate constant for the slow step.

Key Concepts

Rate-Determining Step

In a multi-step reaction mechanism, the rate-determining step is crucial. It acts as a bottleneck, governing the speed of the overall reaction. This is because the transition through the slowest step sets the pace for the entire process.

Imagine a multi-step mechanism as a relay race, where a team hands off a baton at each step. If one runner is particularly slow, then the entire team's result is delayed, regardless of the speed of the other runners. Similarly, the slowest step controls the overall reaction rate.

For example, in the given mechanism, Step 1 is labeled as 'Slow,' indicating it is the rate-determining step. Since it involves the breaking of a bond in the reactant \((\text{CH}_3)_3\text{CBr}\), its kinetics significantly impact the rate law derived from this mechanism.

Overall Reaction

The overall reaction in a mechanism represents the net change occurring when all the steps are added together. To determine this, one should sum up all the reactants and products listed across each step and identify any species that appear on both sides, called intermediates, which are then canceled out.

In the provided exercise, the overall reaction is obtained by combining all the steps:

\( (\text{CH}_3)_3\text{CBr} + \text{H}_2\text{O} \rightarrow (\text{CH}_3)_3\text{COH} + \text{HBr} \)

The intermediates, such as \((\text{CH}_3)_3\text{C}^+\), are neither reactants nor final products, and they do not show up in this equation. Hence, the overall reaction provides the full picture of what is consumed and what is formed as the reaction proceeds.

Rate Law

The rate law of a reaction provides an equation that links the rate of the reaction with the concentration of reactants, utilizing a rate constant \(k\). This rate law is derived from the rate-determining step of a mechanism. Therefore, the order of the reaction and involved reactants will reflect those in this step.

For the given mechanism, exclusively Step 1, which is the slowest, dictates the rate law. This step involves only \((\text{CH}_3)_3\text{CBr}\) as the reactant. Thus, as per this slow step, the rate law disclosed is:

• \( \text{Rate} = k[(\text{CH}_3)_3\text{CBr}] \)

This means the reaction is first order with respect to \((\text{CH}_3)_3\text{CBr}\). Through this, we understand how changes in its concentration impact the speed of the reaction.

Intermediates

Intermediates are molecular species that are produced and subsequently consumed during the reaction process. They do not appear in the overall balanced equation of the reaction.

In the context of the given mechanism, intermediates play a hidden yet pivotal role in passing the baton from one reaction step to another. For instance, \((\text{CH}_3)_3\text{C}^+\) is such an intermediate. It is generated in Step 1 and then immediately used up in Step 2 of the mechanism.

Although these intermediates are not present in the overall reaction equation, they are crucial for the progression of the reaction, facilitating the transition from reactants to products. Understanding their role is key to grasping how individual steps interconnect and add up to the full mechanism.