Problem 61 - Chemistry An Atoms First Approach

Question

A proposed mechanism for a reaction is $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{Br}^{-} \quad \text { Slow }$$ $$\mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+} \quad \text { Fast }$$ $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}_{3} \mathrm{O}^{+}\quad \text { Fast }$$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction? What are the intermediates in the proposed mechanism?

Step-by-Step Solution

Answer

The rate law for the proposed mechanism is Rate = \(k[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br}]\), the overall balanced equation is: $$\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br} + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH} + \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{Br}^{-}$$ And the intermediates are: 1. $$\mathrm{C}_{4}\mathrm{H}_{9}^{+}$$ 2. $$\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^{+}$$

Step 1: The rate-determining step, or slow step, is the slowest step in the mechanism, which dictates the rate law. In this case, the slow step is provided: $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{Br}^{-} \quad \text { Slow }$$ #Step 2: Write the rate law for the rate-determining step#

Since the rate-determining step is the slowest step, its rate law will be the overall rate law for the mechanism. The rate of the reaction can be written as k times the concentration of the reactant in the slow step: Rate = \(k[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br}]\) #Step 3: Identify the overall balanced equation for the reaction#

Step 2: To find the overall balanced equation for the reaction, we need to add all the elementary steps of the mechanism: 1. $$\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}^{+} + \mathrm{Br}^{-}$$ 2. $$\mathrm{C}_{4}\mathrm{H}_{9}^{+}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^{+}$$ 3. $$\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^{+}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH} + \mathrm{H}_{3}\mathrm{O}^{+}$$ Add them together and cancel out any intermediate species that appear on both sides of the equation: $$\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br} + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH} + \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{Br}^{-}$$ #Step 4: Identify the intermediates in the mechanism#

Intermediates are species that are produced in one step and consumed in another. In this mechanism, the intermediates are: 1. $$\mathrm{C}_{4}\mathrm{H}_{9}^{+}$$: Produced in the first step and consumed in the second step. 2. $$\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^{+}$$: Produced in the second step and consumed in the third step. In conclusion, the rate law for the proposed mechanism is Rate = \(k[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br}]\), the overall balanced equation is: $$\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br} + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH} + \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{Br}^{-}$$ And the intermediates are: 1. $$\mathrm{C}_{4}\mathrm{H}_{9}^{+}$$ 2. $$\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^{+}$$

Key Concepts

Rate Law

The rate law is a crucial component in understanding chemical reaction kinetics. It provides an equation that relates the rate of a reaction to the concentration of the reactants. In this context, the rate law is determined by the rate-determining step, which is often the slowest step in a reaction mechanism.

For a given reaction mechanism, the rate-determining step dictates the rate law expression. This is because, in a multi-step reaction, the slowest step acts as a bottleneck, controlling the overall reaction speed. In this exercise, the slow step is:

• \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{Br}^{-}\)

The rate law derived from this step can be written as follows: Rate = \(k[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br}]\). Here, \(k\) is the rate constant, and its value is determined experimentally. The concentration of \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br}\) affects how quickly the reaction can proceed. It’s important that this concentration appears in the rate law because the formation of the products depends directly on it.

Understanding rate law not only helps in predicting how a change in concentration could affect the reaction rate but also aids in verifying whether a proposed mechanism can explain the observed kinetics.

Intermediates

In chemical reactions, intermediates are transient species that form during the conversion of reactants to products. They are created in one step and consumed in the subsequent step, making them crucial players in the reaction mechanism.

For this particular mechanism, two intermediates have been identified:

• \(\mathrm{C}_{4}\mathrm{H}_{9}^{+}\)
• \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^{+}\)

These intermediates play a vital role by providing insight into the step-by-step progression of the reaction. The \(\mathrm{C}_{4}\mathrm{H}_{9}^{+}\) is formed in the first slow step and then reacts quickly with water to form \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^{+}\) in the second step. This, in turn, reacts with another water molecule to form the final product \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}\).

Although intermediates are essential to the reaction mechanism, they do not appear in the overall balanced equation, as they are used up as quickly as they are created. Understanding intermediates helps chemists to propose and confirm mechanisms, illustrating how complex transformations occur on a molecular level.

Balanced Equation

The overall balanced equation of a chemical reaction provides a simplified representation of the transformation from reactants to products. It combines all the steps of the reaction mechanism into a single equation that reflects the conservation of mass and atoms throughout the process.

For the given mechanism, by carefully adding all the elementary steps and canceling out the intermediates, we derive the balanced equation:

• \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br} + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH} + \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{Br}^{-}\)

This equation captures the net chemical change without detailing the steps involved. It shows how one bromobutane molecule \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br}\) reacts with one water molecule \(\mathrm{H}_{2}\mathrm{O}\) to produce butanol \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}\), a hydronium ion \(\mathrm{H}_{3}\mathrm{O}^{+}\), and a bromide ion \(\mathrm{Br}^{-}\).

Balancing equations is fundamental in chemical literacy, ensuring chemical descriptions are accurate and feasible. It also aids in the stoichiometric calculations needed to determine quantities for reactions in lab and industry settings.