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 labeled with oxygen- 18 , the oxygen- 18 ends up in the methyl acetate: 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? Explain.

Short answer

During the reaction of acetic acid with labeled methyl alcohol, the O-H bond of the acid and the C-OH bond of the alcohol break, resulting in the formation of methyl acetate with the oxygen-18 atom in it.

Step-by-step solution

Write the reaction equation with the labeled oxygen-18.

The reaction between acetic acid and methyl alcohol is as follows: \[\mathrm{CH_3COOH} + \mathrm{CH_3OH} \rightarrow \mathrm{CH_3COOCH_3} + \mathrm{H_2O}\] However, we are given that the methyl alcohol is labeled with oxygen-18. Representing a labeled oxygen atom as O*, the reaction proceeds as: \[\mathrm{CH_3COOH} + \mathrm{CH_3O^*H} \rightarrow \mathrm{CH_3COO^*CH_3} + \mathrm{H_2O}\] By seeing that the oxygen-18 labeled atom ends up in the methyl acetate product, we can determine which bonds break during the reaction.

Analyze the possibility of breaking the C-OH bond of the acid and the O-H bond of the alcohol

If the C-OH bond of the acid and the O-H bond of the alcohol break, the reaction would be represented as: \[\mathrm{CH_3CO}^-\ +\ \mathrm{H^+} + \mathrm{CH_3O^*}^-\ +\ \mathrm{H^+} \rightarrow \mathrm{CH_3COO^*CH_3} + \mathrm{H_2O}\] This possibility suggests that the oxygen-18 should end up in the water, not in the methyl acetate product. Therefore, this possibility can be discarded.

Analyze the possibility of breaking the O-H bond of the acid and the C-OH bond of the alcohol

If the O-H bond of the acid and the C-OH bond of the alcohol break, the reaction would be represented as: \[\mathrm{CH_3COO}^-\ +\ \mathrm{H^+} + \mathrm{CH_3}^+ +\ \mathrm{O^*H}\ \rightarrow \mathrm{CH_3COO^*CH_3} + \mathrm{H_2O}\] In this case, the oxygen-18 surrounds itself in the methyl acetate product, matching the given information.

Conclusion

Based on the given information and the analysis of the possible bond breakage, we conclude that during the reaction of acetic acid with labeled methyl alcohol, the O-H bond of the acid and the C-OH bond of the alcohol break to form methyl acetate with the oxygen-18 atom in it.

Key concepts

Reaction Mechanism

In organic chemistry, understanding a reaction mechanism is like unraveling a story. It tells us step by step how reactants turn into products. When dealing with the reaction of acetic acid and methyl alcohol to form methyl acetate, the mechanism involves various bond breaking and forming steps that lead to a new molecule. In our case, we start by looking at the reactants, acetic acid \((\mathrm{CH_3COOH})\) and methyl alcohol \((\mathrm{CH_3OH})\), and how they interact. We explore which bonds break and what new bonds form to pave the way for methyl acetate.
This insight helps chemists predict the outcome of reactions and tailor conditions to synthesize desired compounds efficiently.

Isotope Labeling

Isotope labeling is a nifty technique used to trace the path of atoms through a chemical reaction. By including an isotope, like oxygen-18, in the methyl alcohol, we can track where this atom appears in the product. In this reaction, the labeled oxygen-18 started in methyl alcohol and ended up in methyl acetate. This label acts as a molecular GPS, helping chemists determine which bonds were broken and rearranged.
Analyzing its final position gives clues about the elusive molecular dance steps that occurred, enhancing our understanding of the reaction mechanism. This process is crucial in research and academia for pinning down intricate pathways in chemistry.

Esterification

Esterification is the process where acids and alcohols combine to form esters, along with water. This is a common reaction in organic synthesis, widely used to create flavors, fragrances, and polymers. In our case, acetic acid \((\mathrm{CH_3COOH})\) reacts with methyl alcohol \((\mathrm{CH_3OH})\) in a condensation reaction to form methyl acetate \((\mathrm{CH_3COOCH_3})\).
This process involves breaking and forming specific bonds, which we explored earlier, to convert reactants into the desired ester. The presence of the labeled oxygen helps to identify which specific bonds undergo transformation to achieve the final product. Understanding esterification is fundamental in both industrial applications and academic research.

Bond Breaking

Bond breaking is a critical aspect of chemical reactions, dictating which atoms will interact to form new compounds. In this problem, different bonds in acetic acid and methyl alcohol break to ultimately produce methyl acetate. We specifically analyzed the breaking of the \(\mathrm{O-H}\) bond in acetic acid and the \(\mathrm{C-OH}\) bond in methyl alcohol.
This revelation tells us that during the esterification, the oxygen-18 label transfers from the alcohol to form part of the ester linkage. Understanding which bonds break helps in predicting product formation and reaction pathways. This aspect of chemistry provides valuable information for controlling and optimizing chemical reactions efficiently.