Question

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 labcled with oxygen-18, the oxygen-18 ends up in the methyl acetate: CC(=O)CCCCCC(=O)O (a) Do the \(\mathrm{C}-\mathrm{OH}\) bond of the acid and the \(\mathrm{O}-\mathrm{H}\) bond of the alcohol break in the reaction, or do the \(\mathrm{O}-\mathrm{H}\) bond of the acid and the \(\mathrm{C}-\mathrm{OH}\) bond of the alcohol break? (b) Imagine a similar experiment using the radioisotope \({ }^{3} \mathrm{H}\), which is called tritium and is usually denoted \(\mathrm{T}\). Would the reaction between \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{TOCH}_{3}\) provide the same information about which bond is broken as does the above experiment with \(\mathrm{H}^{18} \mathrm{OCH}_{3}\) ?

Short answer

(a) In the formation of Methyl Acetate, the \(\mathrm{O}-\mathrm{H}\) bond of the acid (Acetic Acid) and the \(\mathrm{C}-\mathrm{OH}\) bond of the alcohol (Methyl Alcohol) are broken. (b) No, the reaction between \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{TOCH}_{3}\) with Tritium does not provide the same information about which bond is broken as the experiment with \(\mathrm{H}^{18} \mathrm{OCH}_{3}\).

Step-by-step solution

Analyze the reaction formation of Methyl Acetate

To understand which bond breaks in the reaction, we can start by writing the reaction. We know that Methyl Acetate is formed when Acetic Acid reacts with Methyl Alcohol. Let's write down the molecules: Acetic Acid: \(\mathrm{CH}_{3} \mathrm{COOH}\) Methyl Alcohol: \(\mathrm{CH}_3 \mathrm{OH}\) Methyl Acetate: \(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\) Now, let's write down the labeled Methyl Alcohol with Oxygen-18: Labeled Methyl Alcohol: \(\mathrm{H}^{18} \mathrm{OCH}_{3}\) From this data and knowing that Oxygen-18 ends up in the Methyl Acetate molecule, we can predict which bond breaks when forming the product.

Predict the broken bond

We know that Methyl Acetate is formed with Oxygen-18, as given: \(\mathrm{CH}_{3} \mathrm{COO^{18}CH}_{3}\) Comparing the labeled Methyl Acetate structure with the reactant structures, we can determine that for Oxygen-18 to end up in the Methyl Acetate, the bond breaking in the reaction is: (a) \(\mathrm{O}-\mathrm{H}\) bond of the acid (Acetic Acid), and \(\mathrm{C}-\mathrm{OH}\) bond of the alcohol (Methyl Alcohol). Now, let's analyze the second part of the exercise.

Analyze the experiment with Tritium

In this part, we need to determine whether the reaction between \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{TOCH}_{3}\) using Tritium, would provide the same information about the broken bond as does the experiment with \(\mathrm{H}^{18} \mathrm{OCH}_{3}\). Here, we have labeled Methyl Alcohol with Tritium \(\mathrm{T}\), and the molecular structures are: Acetic Acid: \(\mathrm{CH}_{3} \mathrm{COOH}\) Labeled Methyl Alcohol: \(\mathrm{TOCH}_{3}\)

Compare reactions to determine the same information

Now let's compare the given reaction with the reaction using Tritium: Original Reaction: \(\mathrm{CH}_{3} \mathrm{COOH} + \mathrm{H}^{18} \mathrm{OCH}_{3} \rightarrow \mathrm{CH}_{3} \mathrm{COO^{18}CH}_{3}\) Tritium Reaction: \(\mathrm{CH}_{3} \mathrm{COOH} + \mathrm{TOCH}_{3} \rightarrow ???\) Based on the information given in the exercise, we cannot determine the structure of the product in this tritium-labeled reaction. Thus, we cannot conclude that both experiments would provide the same information about the broken bond in the reaction. (b) No, the reaction between \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{TOCH}_{3}\) with Tritium does not provide the same information about which bond is broken as the experiment with \(\mathrm{H}^{18} \mathrm{OCH}_{3}\).

Key concepts

Bond Breaking in Chemistry

In the formation of methyl acetate through the reaction of acetic acid with methyl alcohol, understanding which bonds break is crucial. Bond breaking is a fundamental concept in chemistry, particularly in organic reactions. When two molecules interact to form a new product, chemical bonds must be broken and/or formed. This process involves energy changes and reconfiguration of atoms.

In our specific reaction, the acetic acid (\( \mathrm{CH}_3 \mathrm{COOH} \)) and methyl alcohol (\( \mathrm{CH}_3 \mathrm{OH} \)) undergo a transformation to create methyl acetate (\( \mathrm{CH}_3 \mathrm{COOCH}_3 \)). Here, for the Oxygen-18 from methyl alcohol to become part of the methyl acetate in its new configuration, the \( \mathrm{O}-\mathrm{H} \) bond of the acetic acid and the \( \mathrm{C}-\mathrm{OH} \) bond of the methyl alcohol must break. It's essential to identify these specific bonds to understand the movement of atoms and isotopes in the reaction mechanism.

This information helps chemists predict reaction outcomes and guides them in designing new substances by carefully selecting reactants and conditions that favor desired bond-breaking pathways. Understanding how bonds break also informs chemists about the energy requirements and potential side reactions that could alter the intended product formation.

Stable Isotopes in Reactions

Stable isotopes, like Oxygen-18, are invaluable tools in tracing chemical pathways. In this reaction between acetic acid and labeled methyl alcohol, the use of Oxygen-18 helps predict molecular changes through bond breakages. Isotopes are atoms of the same element that contain different numbers of neutrons. While they often behave chemically similar, the presence of the stable isotope acts as a marker that can be detected and tracked throughout the reaction.

By labeling methyl alcohol with Oxygen-18, chemists observe that the isotope ultimately appears in methyl acetate. This outcome reveals which bonds in the original reactants broke and how atoms reorganized to form the final product. Using stable isotopes doesn't alter the reaction mechanics, but they offer insight into the molecular dynamics, helping to confirm theoretical predictions about reaction pathways.

• Oxygen-18 labeling provides a precise and non-intrusive way to study complex reactions.
• Detecting this isotope in new compounds confirms the paths molecules traverse during reactions.
• Experiments with stable isotopes assist in validating chemical reaction mechanisms.

Through such isotopic tracing, a clearer understanding of the interaction between reactants in organic chemistry can be achieved.

Organic Chemistry Reactions

Organic chemistry focuses on the study of carbon-containing compounds and their transformations. Reactions in organic chemistry, such as forming methyl acetate, involve intricate changes in molecular structures driven by the breaking and forming of chemical bonds.

In the context of the methyl acetate formation, the process exemplifies an esterification reaction. An esterification reaction occurs when an alcohol and an acid react, producing an ester and water. This particular reaction is catalyzed by the presence of an acid and typically involves the loss of a water molecule as by-products are formed. However, in our case, the crux is not just about creating methyl acetate, but understanding which specific bonds are broken during the transformation.

• In esterification, understanding bond breakages helps predict yields and by-products.
• Knowing the path atoms take in a reaction assists in refining reaction conditions for better efficiency.
• Studying these reactions aids in the synthesis of a wide range of useful compounds, from pharmaceuticals to plastics.

Grasping these basic yet profound concepts in organic reactions is pivotal for aspiring chemists, as they form the bedrock of more advanced studies and applications in chemistry.