Problem 138 - Introductory Chemistry Atoms First

Question

Consider the basic hydrolysis (reaction with aqueous base) of \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\). \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(a q)+\mathrm{OH}^{-}(a q) \rightarrow\) $$\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{COH}(a q)+\mathrm{Br}^{-}(a q)$$ The rate law is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^{-}\). What does this imply about the mechanism of this reaction?

Step-by-Step Solution

Answer

The rate law indicates that the hydrolysis of \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\) occurs in a two-step process. The first step, likely the rate-determining step, involves the dissociation of \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\) into a carbocation and a bromide ion. The second step, assumed to be faster than the first, involves the reaction between the carbocation and hydroxide ion to form the alcohol product. This interpretation of the reaction mechanism is consistent with the given rate law and the overall reaction.

Step 1: Understanding the rate law

The rate law given for this reaction is first order with respect to \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^{-}\). Mathematically, it can be expressed as: $$\text{Rate} = k\left[\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\right]$$ where k is the rate constant. This means that the rate of the reaction is directly proportional to the concentration of \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\) and does not depend on the concentration of \(\mathrm{OH}^{-}\).

Step 2: Understanding the mechanism of the reaction

A reaction mechanism is the detailed step-by-step description of a chemical reaction that explains how the reactants are converted into products. In this case, we are given the overall reaction and the rate law, so we need to make an interpretation about the underlying mechanism.

Step 3: Interpreting the rate law with respect to the reaction mechanism

Due to the fact that the rate law does not depend on the concentration of \(\mathrm{OH}^{-}\), we can suggest that the hydrolysis reaction occurs in a two-step process: 1. In the first step, \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\) undergoes a dissociation into a carbocation and a bromide ion: $$\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr} \rightarrow \left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{C}^{+} + \mathrm{Br}^{-}$$ This step is likely the rate-determining step, as the overall rate depends only on the concentration of \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\). 2. In the second step, the carbocation reacts with the hydroxide ion to form the alcohol product: $$\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{C}^{+} + \mathrm{OH}^{-} \rightarrow \left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{COH}$$ Since this step does not affect the rate law, it is assumed to be faster than the first step.

Step 4: Conclusions about the reaction mechanism

Based on the information given and the analysis of the rate law, we can make the following conclusions about the mechanism of the basic hydrolysis of \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\): 1. The reaction likely proceeds in a two-step process. 2. The first step is the rate-determining step, involving the dissociation of \(\left(\mathrm{CH}_{3}\right){ }_{3} \mathrm{CBr}\) into carbocation and bromide ion. 3. The second step is faster than the first step and involves the reaction between the carbocation and hydroxide ion to form the alcohol product. This interpretation of the reaction mechanism is consistent with the given rate law and the overall reaction.

Key Concepts

Rate Law

The rate law for a chemical reaction helps in understanding how the rate of reaction is affected by the concentration of the reactants. In the case of the hydrolysis of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\), the given rate law is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\). This can be summarized mathematically as:

• \( \text{Rate} = k\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\right] \)

Here, \(k\) is the rate constant. The rate law shows that the reaction rate depends solely on the concentration of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and not on \(\mathrm{OH}^{-}\).
This suggests a scenario where the initial step involving \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) dictates the speed of the reaction.

First Order Reaction

In a first order reaction, the rate of reaction is directly proportional to the concentration of one reactant. This means that any change in that reactant's concentration will have a direct and proportional effect on the rate of reaction.

• For example, if the concentration of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) is doubled, the reaction rate will also double.

This reaction exhibits first order kinetics concerning \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\), highlighting its significant influence in the reaction mechanism, especially in determining the rate.

Rate-Determining Step

The rate-determining step is a crucial concept in understanding reaction mechanisms, particularly for multi-step processes. In this element of the reaction, a bottleneck forms because this step is slower than the rest, effectively controlling the overall rate of the reaction.
For the hydrolysis of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\), the first step is believed to be the rate-determining step, which involves the dissociation of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) to form a carbocation and a bromide ion:
\[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr} \rightarrow \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+} + \mathrm{Br}^{-}\]
This step largely influences the reaction rate, as indicated by the rate law being independent of \(\mathrm{OH}^{-}\) concentration.

Carbocation Formation

A carbocation is a positively charged ion featuring a carbon atom that lacks a full octet of electrons. The formation of a carbocation is a critical part of many organic reaction mechanisms, frequently acting as the intermediate.

• In this specific reaction, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) transforms into \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}\), creating a carbocation.
• This carbocation is highly reactive and quickly reacts with \(\mathrm{OH}^{-}\) to form the final product, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\).

Carbocation formation is vital in determining both the rate and the pathway of the reaction, especially since it's part of the rate-determining step.