Problem 6
Question
Acetylene reacts with acetic acid in presence of \(\mathrm{Hg}^{+}\) ions at room temperature to give (a) ethyl acetate (b) acetaldehyde (c) vinyl acetate (d) methyl acetate
Step-by-Step Solution
Verified Answer
Vinyl acetate is formed.
1Step 1: Recognize the Reactants and Their Role
In this reaction, acetylene (\(C_2H_2\) and acetic acid (\(CH_3COOH\) are the primary reactants. Mercury ions (\(Hg^+\)) act as a catalyst at room temperature. Recognizing the reactants and their conditions is crucial to understand the product formation.
2Step 2: Determine the Reaction Mechanism
Under the presence of \(Hg^+\), acetylene reacts with acetic acid through a nucleophilic addition mechanism. Mercury ions facilitate the formation of a carbonyl group by increasing the positive character of the positively charged intermediates, allowing the acetic acid to attack acetylene.
3Step 3: Identify the Product
After the reaction, a vinyl group is formed where acetylene's triple bond gets converted into an acetate group linked with a double bond. This results in the formation of vinyl acetate (\(CH_2=CHOCOCH_3\)), which is the product of this reaction.
Key Concepts
CatalysisNucleophilic AdditionVinyl Acetate Formation
Catalysis
Catalysis significantly enhances the efficiency of chemical reactions by lowering the activation energy needed for the reaction to proceed. In the reaction between acetylene and acetic acid, mercury ions (\( \mathrm{Hg}^{+} \)) are used as a catalyst.
- Catalysts are substances that accelerate a reaction without being consumed in the process. They provide an alternative pathway for the reaction to occur, often leading to increased rates and higher yields.
- Mercury ions work by creating a more reactive environment for acetylene. They do this by stabilizing reaction intermediates, specifically the positively charged ones.
Nucleophilic Addition
Nucleophilic addition is a fundamental chemical reaction mechanism that involves the addition of a nucleophile, which is an electron-rich species, to an electron-deficient site. During the reaction between acetylene and acetic acid, nucleophilic addition plays a crucial role.
- Acetylene, being part of an alkyne group, has a region of high electron density due to its triple bond. This makes it susceptible to reactions involving electrophiles and nucleophiles.
- The mercury catalyst increases the electron-deficient character of acetic acid, making it more receptive to attack by the electron-rich acetylene. This allows acetic acid to effectively add to acetylene in a mechanism driven by nucleophilic addition.
Vinyl Acetate Formation
The formation of vinyl acetate is the end result of the reaction between acetylene and acetic acid under the influence of a mercury catalyst. Understanding this transformation requires comprehension of how acetylene's structure changes during the reaction.
- The process starts with the nucleophilic addition of acetic acid to the electron-rich acetylene, facilitated by the mercury catalyst. This interaction results in the conversion of acetylene's triple bond into a double bond.
- The new double bond is characteristic of the vinyl group, hence the intermediate is referred to as a vinyl compound.
- In this reaction, the vinyl group is combined with the acetate ion from acetic acid to form vinyl acetate, with a full structure of \( \mathrm{CH_2=CHOCOCH_3} \)
Other exercises in this chapter
Problem 4
Addition of HCN to ethyne in the presence of \(\mathrm{Ba}(\mathrm{CN})_{2}\) as a catalyst gives (a) vinyl cyanide (b) ethyl cyanide (c) 1,1 -dicyanoethane (d)
View solution Problem 5
The reagent used for the following conversion, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\)
View solution Problem 6
The addition of \(\mathrm{HCl}\) to \(3,3,3\)-trichloropropene gives (a) \(\mathrm{Cl}_{3} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) (b) \(\mathrm{Cl}_{2}
View solution Problem 7
Which of the following is used for the conversion of 2-hexyne into trans-2-hexane? (a) \(\mathrm{H}_{2} / \mathrm{Pd} / \mathrm{BaSO}_{4}\) (b) \(\mathrm{H}_{2}
View solution