Write the overall reaction for the mechanism proposed below and identify any reaction intermediates. Step \(1 \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}+\mathrm{Br}^{-}\) Step \(2 \mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}{ }^{+}\) Step \(3 \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}_{3} \mathrm{O}^{+}\)

Short Answer

Expert verified
\(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br} + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH} + \mathrm{H}_{3}\mathrm{O}^+\), with reaction intermediates \(\mathrm{C}_{4}\mathrm{H}_{9}^+\) and \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^+\).

Step by step solution

01

Write the individual steps of the mechanism

List each step of the reaction mechanism:Step 1: \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}^+ + \mathrm{Br}^-\)Step 2: \(\mathrm{C}_{4}\mathrm{H}_{9}^+ + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^+\)Step 3: \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^+ + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH} + \mathrm{H}_{3}\mathrm{O}^+\)
02

Identify reaction intermediates

Determine the species that appear in the mechanism but not in the overall reaction. Here, the reaction intermediates are \(\mathrm{C}_{4}\mathrm{H}_{9}^+\) and \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^+\) because they are produced and consumed during the reaction mechanism without appearing in the final products or reactants.
03

Cancel intermediates and spectators

Eliminate reaction intermediates and spectator ions to find the overall reaction. \(\mathrm{C}_{4}\mathrm{H}_{9}^+\) and \(\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH}_{2}^+\) do not appear in the overall reaction, so they can be removed. \(\mathrm{Br}^-\) is a spectator ion since it does not participate actively in the reaction, so it is also eliminated.
04

Write the overall reaction

Combine the remaining species to form the overall reaction: \[\mathrm{C}_{4}\mathrm{H}_{9}\mathrm{Br} + \mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{9}\mathrm{OH} + \mathrm{H}_{3}\mathrm{O}^+\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Kinetics
Chemical kinetics is the study of the rates at which chemical reactions occur and the steps through which they proceed. Understanding kinetics is crucial because it doesn't only tell us how fast a reaction occurs, but it also gives insights into the reaction mechanism, which is the step-by-step sequence of elementary reactions by which overall chemical change occurs.

In the given exercise, we follow a multi-step reaction. The time each step takes and their contribution to the overall reaction rate is an essential aspect of kinetics. For instance, a very slow step (also called the rate-determining step) can affect the speed of the overall reaction, even if the other steps are fast. Kinetics also involves studying factors like temperature, concentration, and the presence of catalysts to understand how these can be optimized for faster, more efficient reactions.
Reaction Intermediates
Reaction intermediates are species that are formed during the reaction process but do not appear in the final reaction equation because they are consumed as the reaction proceeds. They are usually highly reactive and have a short lifespan.

As shown in the exercise, the butyl cation (\textbf{C}\(_4\)H\(_9^+\)) and the protonated alcohol (\textbf{C}\(_4\)H\(_9\)OH\(_2^+\)) are intermediates. They are neither reactants nor final products; they temporarily exist during the transition from reactants to products. It's essential to identify intermediates as they can indicate the pathway a reaction takes. In synthesis, knowing reaction intermediates helps chemists design strategies to stabilize them, potentially leading to different outcomes or products.
Nucleophilic Substitution Reactions
Nucleophilic substitution reactions are a fundamental class of reactions in organic chemistry where a nucleophile, a species that is attracted to positive charges, displaces a leaving group in a molecule. These reactions are classified into two main types: S\(_N\)1 and S\(_N\)2.

In this exercise, the mechanism reflects an S\(_N\)1 reaction, characterized by the formation of a carbocation intermediate (the butyl cation). The S\(_N\)1 pathway typically occurs in two steps: first, the leaving group departs, forming a carbocation, and second, the nucleophile attacks the positively charged intermediate. Water acts as the nucleophile attacking the butyl cation, leading to the formation of butanol. This type of reaction is common in organic synthesis, especially in the modification of hydrocarbon chains.

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Most popular questions from this chapter

The mechanism proposed for the oxidation of iodide ion by the hypochlorite ion in aqueous solution is as follows: Step \(1 \mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{HClO}+\mathrm{OH}^{-}\)and its reverse (both fast, equilibrium) Step \(2 \mathrm{I}^{-}+\mathrm{HClO} \longrightarrow \mathrm{HIO}+\mathrm{Cl}^{-}\)(slow) Step \(3 \mathrm{HIO}+\mathrm{OH}^{-} \longrightarrow \mathrm{IO}^{-}+\mathrm{H}_{2} \mathrm{O}\) (fast) Write the rate law for the formation of HIO implied by this mechanism.

The second-order rate constant for the decomposition of \(\mathrm{NO}_{2}\) (to \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) ) at \(573 \mathrm{~K}\) is \(0.54 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}\). Calculate the time for an initial \(\mathrm{NO}_{2}\) concentration of \(0.20 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to decrease to (a) one-half; (b) one-sixteenth; (c) one-ninth of its initial concentration.

The data below were collected for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) at \(780 \mathrm{~K}\). (a) Using a graphing calculator or graphing software, such as that on the Web site for this book, plot the data in an appropriate fashion to determine the order of the reaction. (b) From the graph, determine the rate constant for the rate of consumption of \(\mathrm{I}_{2}\). $$ \begin{array}{lccccc} \text { Time }(\mathrm{s}) & 0 & 1.0 & 2.0 & 3.0 & 4.0 \\ {\left[\mathrm{I}_{2}\right]\left(\mathrm{mmol} \cdot \mathrm{L}^{-1}\right)} & 1.00 & 0.43 & 0.27 & 0.20 & 0.16 \end{array} $$

Complete the following statements for the reaction \(6 \mathrm{Li}(\mathrm{s})+\mathrm{N}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(\mathrm{~s})\). The rate of consumption of \(\mathrm{N}_{2}\) is _______ times the rate of formation of \(\mathrm{Li}_{3} \mathrm{~N}\). (b) The rate of formation of \(\mathrm{Li}_{3} \mathrm{~N}\) is _____times the rate of consumption of Li. (c) The rate of consumption of \(\mathrm{N}_{2}\) is ________ times the rate of consumption of Li.

(a) Using a graphing calculator or graphing software, such as that on the Web site for this book, calculate the activation energy for the acid hydrolysis of sucrose to give glucose and fructose from an Arrhenius plot of the data shown here. (b) Calculate the rate constant at \(37^{\circ} \mathrm{C}\) (body temperature). (c) From data in Appendix 2A, calculate the enthalpy change for this reaction, assuming that the solvation enthalpies of the sugars are negligible. Draw an energy profile for the overall process. $$ \begin{array}{cc} \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k\left(\mathrm{~s}^{-1}\right) \\ \hline 24 & 4.8 \times 10^{-3} \\ 28 & 7.8 \times 10^{-3} \\ 32 & 13 \times 10^{-3} \\ 36 & 20 . \times 10^{-3} \\ 40 . & 32 \times 10^{-3} \\ \hline \end{array} $$

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