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When two moles of ethyne are passed through \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\) dissolved in ammonium chloride solution the product is (1) 1 -butyne (2) vinyl acetylenc (3) divinyl acctylene (4) chloroprenc

Short Answer

Expert verified
The product is divinyl acetylene (option 3).

Step by step solution

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01

- Understand the reactants

Identify ethyne, which is \( C_2H_2 \), and react it with \( Cu_2Cl_2 \) in ammonium chloride solution.
02

- Predict the reaction outcome

When ethyne undergoes dimerization in the presence of \( Cu_2Cl_2 \) and ammonium chloride, it forms diacetylene (also known as butadiyne).
03

- Confirm the product

Recognize that diacetylene is also known as divinyl acetylene, which corresponds to option (3).

Key Concepts

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

dimerization of ethyne
Dimerization is a chemical process where two identical molecules combine to form a single compound. In this reaction, ethyne (acetylene), represented by the formula \(C_2H_2\), undergoes dimerization. When ethyne is passed through a solution of \(Cu_2Cl_2\) (cuprous chloride) dissolved in ammonium chloride, the ethyne molecules react.
During this process, each ethyne molecule connects via their triple bonds. This results in a new compound formed by the linking of two ethyne molecules. By understanding the principle of dimerization, it becomes easier to predict the products of similar reactions.
diacetylene formation
In the reaction, ethyne (\(C_2H_2\)) reacts with cuprous chloride (\(Cu_2Cl_2\)) and ammonium chloride, resulting in the formation of diacetylene. Diacetylene is also referred to as butadiyne, and its chemical formula is \(C_4H_2\).
Here’s how the conversion occurs:
  • Each ethyne molecule (\(C_2H_2\)) links at the triple bond.
  • This dimerization forms a linear chain structure, creating diacetylene (\

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Most popular questions from this chapter

Vinyl acetylene reacts with one equivalent of \(\mathrm{HCl}\) to produce (1) chloroprene (2) neoprene (3) lewisite (4) mesitylene

\(\Lambda\) compound \(\left(\mathrm{C}_{5} \mathrm{I}_{8}\right)\) reacts with ammonical \(\Lambda \mathrm{gNO}_{3}\) to give a white precipitate and reacts with an excess of \(\mathrm{KMnO}_{4}\) solution to give \(\left(\mathrm{CII}_{3}\right)_{2} \mathrm{CH}\) COOII. The compound is (1) \(\mathrm{CII}_{3}=\mathrm{CII} \mathrm{CH}=\mathrm{CII} \mathrm{CII}_{3}\) (2) \(\left(\mathrm{CII}_{3}\right)_{2} \mathrm{CII} \mathrm{C}=\mathrm{CII}\) (3) \(\mathrm{CII}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{C} \equiv \mathrm{CII}\) (4) \(\left(\mathrm{CII}_{3}\right)_{2} \mathrm{C}=\mathrm{C}=\mathrm{CII}_{2}\)

Which of the following reactions will yicld \(2.2\) -dibromopropane? (1) \(\mathrm{HC} \equiv \mathrm{CH}+2 \mathrm{HBr} \longrightarrow\) (2) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHBr}+\mathrm{HBr} \longrightarrow\) (3) \(\mathrm{CH}_{3} \mathrm{C}=\mathrm{CH}+2 \mathrm{HBr} \longrightarrow\) (4) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \longrightarrow\)

Which of the following will convert \(\mathrm{HC}=\mathrm{CCH}_{2} \mathrm{CH}_{3}\) to \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3} ?\) (1) \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}\) (2) \(\mathrm{Hg}^{21} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (3) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (conc) \(/ \mathrm{H}_{3} \mathrm{PO}_{4}\) (4) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{KMnO}_{4}\)

Which of the following statements is correct? (1) alkynes are more reactive than alkenes towards halogen addition (2) alkynes are less reactive than alkenes towards halogen addition (3) both alkynes and alkenes are cqually reactive towards halogen addition (4) primary vinylic cation \((\mathrm{RCII}=\mathrm{CII})\) is more rcactive than sccondary vinylic cation \(\left(\mathrm{RC}=\mathrm{CII}_{2}\right)\)

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