/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 92 Isotopes are often used as "trac... [FREE SOLUTION] | 91Ó°ÊÓ

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Isotopes are often used as "tracers" to follow an atom through a chemical reaction, and the following is an example. Acetic acid reacts with methanol by eliminating a molecule of water and forming methyl acetate (See Chapter 11 ). $$\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{CH}_{3} \mathrm{OH} \quad \longrightarrow \quad \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O}$$ Explain how you could use the isotope \(^{18} \mathrm{O}\) to show whether the oxygen atom in the water comes from the - OH of the acid or the - OH of the alcohol.

Short Answer

Expert verified
Use \(^{18}O\) labeling on the \(-OH\) group of each reactant sequentially and analyze the product to trace the source of oxygen in water.

Step by step solution

01

Identify the isotopic tracer

To determine the source of the oxygen atom in the water molecule, we can use an isotope, such as \(^{18}O\), as a tracer. This isotope is a stable, non-radioactive variant of oxygen that can substitute for the regular \(^{16}O\).
02

Assign the isotope to a reactant

The first method involves labeling the \(-OH\) group of the acetic acid (\(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)) with \(^{18}O\). This results in the acetic acid formula being \(\mathrm{CH}_{3} \mathrm{C}^{18}\mathrm{O}_{2} \mathrm{H}\). Then, proceed with the reaction.
03

Analyze the product

After running the labeled reaction, analyze the products. If the water produced (\(\mathrm{H}_2\mathrm{O}\)) contains \(^{18}O\), then the oxygen atom from the \(-OH\) of the acid contributed to the water molecule. This would indicate the oxygen originated from the acetic acid.
04

Repeat with alternate labeling

To confirm the source of the oxygen atom in water, repeat the reaction but label the \(-OH\) group of the methanol (\(\mathrm{CH}_{3} \mathrm{OH}\)) with \(^{18}O\), resulting in \(\mathrm{CH}_{3} \mathrm{^{18}OH}\). Run the reaction again.
05

Confirm conclusions

Analyze the products again. If this time, the labeled \(^{18}O\) ends up in the water produced, it confirms that the oxygen in water originated from \(-OH\) of the methanol. Consequently, it can be confirmed which reactant's oxygen is involved in the water formation.

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Key Concepts

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

Chemical Reaction Mechanisms
Understanding chemical reaction mechanisms is crucial in studying how chemical reactions proceed, step-by-step, from reactants to products. These mechanisms offer a detailed picture of the molecular changes occurring during a reaction. By knowing the specific steps and interactions, chemists can better predict reaction outcomes and manipulate conditions to achieve desired results.

An essential part of these mechanisms is distinguishing between reactants, intermediates, and the final products. This differentiation helps identify which bonds break and which form during the reaction. In our exercise involving acetic acid and methanol, the formation of methyl acetate and water is a classic example of an esterification mechanism.

Here, speculation exists about which oxygen—either from acetic acid ( −OH ) or methanol ( −OH )—ends up in the water. By incorporating isotopes as tracers, scientists can trace these molecular transformations, providing a seamless insight into the reaction's pathway. This method highlights the invaluable tools that isotopes offer in uncovering reaction mechanisms and improving our understanding of chemistry.
Oxygen Isotopes
Isotopes are variants of a particular chemical element that contain the same number of protons but differ in the number of neutrons. Oxygen isotopes, such as ^{16}O , ^{17}O , and ^{18}O , are naturally occurring and vary based on their atomic mass.

Among these, ^{18}O is particularly useful for scientific studies as it is stable and non-radioactive. Scientists often employ ^{18}O as a tracer in chemical reactions to track the movement and origins of oxygen atoms through a chemical process. This isotope's distinct mass allows it to be separated and identified from the more common ^{16}O using specialized analysis techniques like mass spectrometry.

In the esterification reaction described, substituting ^{18}O for regular oxygen enables chemists to determine the precise source of the oxygen in the resultant water molecule. This kind of isotopic labeling aids in confirming the mechanistic pathways of reactions by revealing which atoms move from the reactants to the products, thus enriching our understanding of complex chemical dynamics.
Reactant Labeling
Reactant labeling is a powerful technique used in chemical experiments to track specific atoms through a reaction. By substituting atoms in the reactants with isotopically labeled versions, scientists can monitor these atoms' paths and determine their eventual location in the products.

In practical application, isotopically labeling reactants can reveal reaction pathways and atomic transformations. It involves replacing an atom in a molecule's functional group with an isotope. This process helps maintain the original structure while providing a way to monitor changes during reactions.

In the example with acetic acid and methanol, the strategic labeling of −OH groups with ^{18}O helps clarify the source of the oxygen in the reaction's water product. By labeling either acetic acid or methanol, chemists can compare the resulting products after the reaction. Tracking the ^{18}O presence in water allows precise identification of oxygen's origin.
  • If ^{18}O appears in the water after labeling the acetic acid, the oxygen originates from the acid.
  • If it appears when methanol's −OH is labeled, then the alcohol supplies the oxygen.
This labeling process underscores the practical utility of isotopes in exploring chemical pathways and verifying theoretical reaction mechanisms.

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Most popular questions from this chapter

Experimental data are listed here for the reaction \(\mathrm{A} \longrightarrow 2 \mathrm{B}.\) $$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\\\(\mathrm{s})\end{array} & \begin{array}{l}{[\mathrm{B}]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data, connect the points with a smooth line, and calculate the rate of change of [B] for each \(10-\) s interval from 0.0 to 40.0 s. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of \([\mathrm{A}]\) related to the rate of change of \([\mathrm{B}]\) in each time interval? Calculate the rate of change of \([\mathrm{A}]\) for the time interval from 10.0 to \(20.0 \mathrm{s}\) (c) What is the instantaneous rate when \([\mathrm{B}]=0.750\) \(\mathrm{mol} / \mathrm{L} ?\)

For the reaction \(2 \mathrm{C}_{2} \mathrm{F}_{4} \longrightarrow \mathrm{C}_{4} \mathrm{F}_{8},\) a graph of \(1 /\left[\mathrm{C}_{2} \mathrm{F}_{4}\right]\) versus time gives a straight line with a slope of \(+0.04 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} .\) What is the rate law for this reaction?

Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is Step 1 Fast, reversible \(\quad \mathrm{HA} \rightleftarrows \mathrm{H}^{+}+\mathrm{A}^{-}\) Step 2 Fast, reversible \(\quad \mathrm{X}+\mathrm{H}^{+} \rightleftarrows \mathrm{XH}^{+}\) Step 3 Slow \(\mathrm{XH}^{+} \longrightarrow\) products What rate law is derived from this mechanism? What is the order of the reaction with respect to HA? How would doubling the concentration of HA affect the reaction?

Data for the reaction $$\begin{aligned}\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{CH}_{3}\mathrm{CN}\right)\right]^{+}+\mathrm{NC}_{5}\mathrm{H}_{5} \longrightarrow & \\\&\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{NC}_{5}\mathrm{H}_{5}\right)\right]^{+}+\mathrm{CH}_{3} \mathrm{CN}\end{aligned}$$ are given in the table. Calculate \(E_{\mathrm{a}}\) from a plot of \(\ln k\) versus \(1 / T\) $$\begin{array}{ll}\hline T(\mathrm{K}) & k\left(\min ^{-1}\right) \\\\\hline 298 & 0.0409 \\\308 & 0.0818 \\\318 & 0.157 \\\\\hline\end{array}$$

The reaction of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) is thought to occur in two steps: Step 1 Slow \(\quad \mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})\) Step 2 Fast \(\quad \mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Show that the elementary steps add up to give the overall, stoichiometric equation. (b) What is the molecularity of each step? (c) For this mechanism to be consistent with kinetic data, what must be the experimental rate equation? (d) Identify any intermediates in this reaction.

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