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Methyl acetate \(\left(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right)\) is formed by the reaction of acetic acid with methyl alcohol. If the methyl alcohol is labeled with oxygen-18, the oxygen-18 ends up in the methyl acetate: \(\begin{array}{l}{\text { (a) Do the } \mathrm{C}-\mathrm{OH} \text { bond of the acid and the } \mathrm{O}-\mathrm{H} \text { bond }} \\ {\text { of the alcohol break in the reaction, or do the } \mathrm{O}-\mathrm{H} \text { bond }} \\ {\text { of the acid and the } \mathrm{C}-\mathrm{OH} \text { bond of the alcohol break? }}\end{array}\) \(\begin{array}{l}{\text { (b) Imagine a similar experiment using the radioisotope }^{3} \mathrm{H} \text { , }} \\ {\text { which is called tritium and is usually denoted T. Would the }} \\ {\text { reaction between } \mathrm{CH}_{3} \mathrm{COH} \text { and } \mathrm{TOCH}_{3} \text { provide the same }} \\ {\text { information about which bond is broken as does the above }} \\ {\text { experiment with } \mathrm{H}^{18} \mathrm{OCH}_{3} ?}\end{array}\)

Short Answer

Expert verified
In the reaction between acetic acid (CH3COOH) and methyl alcohol (CH3OH), the O-H bond of the alcohol breaks, as evidenced by the presence of the oxygen-18 label in the methyl acetate product. Using tritium (T) as a radioisotope would provide the same information as it also reveals that the O-H bond of the alcohol is broken.

Step by step solution

01

(a) Identifying Broken Bonds

To understand which bonds are breaking, let's write the balanced chemical equation for this reaction. The reaction between acetic acid (CH3COOH) and methyl alcohol (CH3OH) can be written as: \(CH_3COOH + CH_3OH \rightarrow CH_3COOCH_3 + H_2O\) Now, in the given exercise, it is given that the oxygen-18 label ends up in the methyl acetate. Let's denote oxygen-18 with O* for simplicity. So, when the methyl alcohol is labeled with oxygen-18, it will be written as CH3O*H. The reaction becomes: \(CH_3COOH + CH_3O*H \rightarrow CH_3COO*CH_3 + H_2O\) Since the oxygen-18 labeled atom is present in the methyl acetate product, this suggests that the O-H bond of alcohol is being broken.
02

(b) Using Tritium as a Radioisotope

Now, let's consider the reaction between CH3COH and TOCH3, where T is the tritium isotope: \(CH_3COH + TOCH_3 \rightarrow CH_3COTCH_3 + H_2O\) In this case, the tritium isotope label ends up in methyl acetate in the same manner. Comparing this with the previous labeled oxygen-18 experiment, we can infer that the information about which bond is broken is the same for these two reactions, revealing that the O-H bond of the alcohol is broken in both cases.

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Key Concepts

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

Chemical Reaction
The synthesis of methyl acetate involves a chemical reaction known as esterification, where an alcohol and an acid react to form an ester and water. In the provided exercise, acetic acid (\textsc{CH}\(_3\)\textsc{COOH}) reacts with methyl alcohol (\textsc{CH}\(_3\)\textsc{OH}) to yield methyl acetate (\textsc{CH}\(_3\)\textsc{COOCH}\(_3\)) and water (\textsc{H}\(_2\)\textsc{O}).

To ensure the concept of a chemical reaction is clear, it's important to note that it involves breaking and forming chemical bonds. This transformation leads to one or more new substances with different properties from the reactants. Additionally, a balanced chemical equation represents the stoichiometry of the reaction, which means the number of atoms of each element is conserved on both sides of the reaction.
Isotope Labeling
Isotope labeling is a technique used to track the movement of atoms through a chemical reaction. In this exercise, methyl alcohol is labeled with oxygen-18, a stable isotope of oxygen. This method allows us to observe that, after the reaction, the oxygen-18 appears in the methyl acetate molecule, not in the by-product water. This information is crucial as it highlights which specific bond breaks during the reaction—namely, the bond between oxygen and hydrogen in the alcohol.

Importance of Isotope Labeling

Isotope labeling helps in understanding intricate details of reaction mechanisms. In research, it can shed light on reaction pathways and the rate at which individual atoms or groups of atoms are involved in a reaction. Through exercises like this one, students can visualize and comprehend how isotope labeling aids in the study of chemical processes.
Reaction Mechanism
The reaction mechanism provides insight into the step-by-step sequence of elementary reactions by which overall chemical change occurs. In the case of methyl acetate synthesis, the reaction mechanism involves a nucleophilic attack by the oxygen of the methyl alcohol on the carbonyl carbon of acetic acid, followed by the elimination of water and the formation of the ester bond.

Understanding Reaction Intermediates

Reaction mechanisms often involve intermediates—species that appear in the course of the reaction but are not present in the final products. By using isotopically labeled reactants, such as oxygen-18 or tritium, the fate of particular atoms can be monitored, confirming or refuting proposed reaction mechanisms. As such, these techniques provide vital clues to the precise steps that constitute the reaction and offer a window into the complex transformations that compounds undergo during chemical reactions.

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Most popular questions from this chapter

One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a})_{20}^{40} \mathrm{Ca}\) and \(_{20}^{45} \mathrm{Ca}\) , \((\mathbf{b})^{12} \mathrm{C}\) and \(^{14} \mathrm{C},\) \((\mathbf{c})\) lead-206 and thorium-230. Explain your choice in each case.

Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of 8.02 days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of Nal, in which only a small fraction of the iodide is radioactive. (a) Why is Nal a good choice for the source of iodine? (b) If a Geiger counter is placed near the person's thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about 12\(\%\) of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to 0.01\(\%\) of the original amount?

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{kJ} / \mathrm{min.}\) (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (\mathbf{b} )If the energy released in the reaction \begin{equation}^{235} \mathrm{U}+_{0}^{1} \mathrm{n} \longrightarrow_{56}^{141} \mathrm{Ba}+_{36}^{92} \mathrm{Kr}+3_{0}^{1} \mathrm{n} \end{equation}\(\left(^{235} \mathrm{U}\right.\)nuclear mass,234.9935 amu; \(^{235} \mathrm{Ba}.\) nuclear mass, 140.8833 amu; \(^{92} \mathrm{Kr}\) nuclear mass, 91.9021 amu) is taken as typical of that occurring in a nuclear reactor, what mass of uranium-235 is required to equal 0.10\(\%\) of the solar energy that falls on Earth in 1.0 day?

How much energy must be supplied to break a single \(^{21} \mathrm{Ne}\) nucleus into separated protons and neutrons if the nucleus has a mass of 20.98846 amu? What is the nuclear binding energy for 1 mol of \(^{21} \mathrm{Ne} ?\)

Naturally found uranium consists of 99.274\(\%^{238} \mathrm{U}\) \(0.720 \%^{233} \mathrm{U},\) and 0.006\(\%^{233} \mathrm{U}\) As we have seen, \(^{235} \mathrm{U}\) is the isotope that can undergo a nuclear chain reaction. Most of the \(^{255}\) U used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, UF \(_{6}(g) .\) (a) What is the mass of UF \(_{6}\) in a 30.0 -L vessel of UF \(_{6}\) at a pressure of 695 torr at 350 \(\mathrm{K} ?\) (b) What is the mass of \(^{235} \mathrm{U}\) in the sample described in part (a)? (c) Now suppose that the \(\mathrm{UF}_{6}\) is diffused through a porous barrier and that the change in the ratio of of \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U}\) in the diffused gas can be described by Equation 10.23. What is the mass of \(^{235} \mathrm{U}\) in a sample of the diffused gas analogous to that in part (a)? (d) After one more cycle of gaseous diffusion, what is the percentage of \(^{235} \mathrm{UF}_{6}\) in the sample?

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