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When acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is treated with a trace of acid in water labeled with \({ }^{18} \mathrm{O}\), the label gradually appears in both oxygen atoms of the carboxylic acid. Draw a mechanism that explains this phenomenon.

Short Answer

Expert verified
Labeling occurs via a nucleophilic substitution mechanism involving a tetrahedral intermediate with water.

Step by step solution

01

Understanding the Reaction

Acetic acid CH鈧僀OOH in the presence of a trace of an acid catalyst in water labeled with 鹿鈦窸 exchanges its oxygen atoms through a process involving the formation of an intermediate that allows the 鹿鈦窸 to incorporate into the molecule.
02

Protonation of Acetic Acid

The reaction begins with the protonation of the carbonyl oxygen of acetic acid by the acid catalyst, making the carbonyl carbon more electrophilic.
03

Nucleophilic Attack by Water

The labeled water molecule (H鈧偮光伕O) attacks the electrophilic carbon atom of the carbonyl group, leading to the formation of a tetrahedral intermediate.
04

Tetrahedral Intermediate Formation

The nucleophilic attack leads to the formation of a tetrahedral intermediate. In this intermediate, one oxygen is now from the labeled water, holding an 鹿鈦窸 label.
05

Rearrangement and Reprotonation

The tetrahedral intermediate can collapse back, leading to the elimination of the initially protonated water, which now carries one of the original oxygen atoms that can undergo an exchange reaction with the solvent water.
06

Equilibrium and Exchange

The mechanism allows the labeled oxygen from the labeled water to be incorporated into both oxygen positions within the acetic acid through a reversible equilibrium process.

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Key Concepts

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

Protonation of Carbonyl Group
Protonation is the initial step in many reaction mechanisms, including the oxygen labeling process. For acetic acid, this involves the carbonyl group: a structure featuring a carbon atom double-bonded to an oxygen atom. The protonation step occurs when an acid catalyst donates a proton (H鈦) to the carbonyl oxygen. This addition makes the carbon more electrophilic, meaning it's ready to attract electrons and thus react with nucleophiles.
Protonation serves as a preparatory step, making the subsequent nucleophilic attack easier. This transformation is crucial because it alters the electronic configuration of the molecule, facilitating the other steps of the mechanism.
It is important to remember the role of the acid catalyst in this process: it is not consumed in the reaction but rather helps drive the mechanism forward by repeatedly getting involved and reformed.
Nucleophilic Attack
Once the carbonyl group is protonated, the stage is set for the nucleophilic attack. Here, a nucleophile, often a molecule or ion with a pair of electrons eager to form a bond, targets the electrophilic carbon atom of the carbonyl.
In our mechanism, this nucleophile is the labeled water molecule (H鈧偮光伕O). As it approaches, it uses its electron pair to attack the electrophilic carbon. This breaks the double bond between the carbon and the oxygen in the carbonyl group, creating an intermediate known as the tetrahedral intermediate.
The nucleophilic attack is a critical step because it introduces the labeled oxygen atom into the acetic acid, setting the stage for the label to end up in both oxygen positions of acetic acid during the mechanism.
Tetrahedral Intermediate
The formation of a tetrahedral intermediate marks a central point in the mechanism. This intermediate is a result of the nucleophilic attack, where the incoming nucleophile temporarily creates a four-bonded carbon atom, resembling a tetrahedron shape.
In the context of acetic acid labeling, the tetrahedral intermediate features one of its bonds with the newly introduced oxygen from the labeled water, which holds the 鹿鈦窸 label.
This intermediate is significant because it represents a turning point where the original carbonyl structure is disrupted and prepares to facilitate the removal of one oxygen atom while incorporating the labeled one into the molecule.
It becomes more susceptible to further rearrangement, eventually leading to the formation and exchange of labeled and unlabeled oxygen atoms in future steps.
Oxygen Exchange Mechanism
The final concept, oxygen exchange mechanism, explains how the labeled oxygen atom from the water ends up in both oxygen positions of the acetic acid. After the tetrahedral intermediate forms, it can revert and release a water molecule, eventually allowing the labeled oxygen to replace one of the acid's original oxygen atoms.
This exchange occurs through a dynamic equilibrium; the mechanism cycles through several possible structural formats of acetic acid, enabling continuous exchange. Such reversible reactions help introduce the labeled oxygen across different positions.
In practical terms, it means that both oxygen atoms in acetic acid can eventually become labeled with 鹿鈦窸 as the mechanism proceeds through many cycles. This oxygen exchange mechanism is crucial for understanding how isotopic labeling studies work and their role in revealing pathways of chemical processes.

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Most popular questions from this chapter

Convert 1-bromohexane \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\right)\) into each compound. More than one step may be required. You may use any other organic or inorganic reagents. a. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CN}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COCl}\) d. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) e. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COCH}_{3}\) f. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) g. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) h. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NHCOCH}_{3}\)

Draw the product formed when pentanoyl chloride \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COCl}\right)\) is treated with each reagent. a. \(\mathrm{H}_{2} \mathrm{O}\), pyridine c. \(\mathrm{CH}_{3} \mathrm{COO}\) e. \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{NH}\) (excess) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), pyridine d. \(\mathrm{NH}_{3}\) (excess) f. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (excess)

Devise a synthesis of each labeled compound using \(\mathrm{H}_{2}{ }^{18} \mathrm{O}\) and \(\mathrm{CH}_{3}{ }^{13} \mathrm{CH}_{2} \mathrm{OH}\) as the only sources of labeled starting materials. You may use any other unlabeled organic compounds and inorganic reagents.

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