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The major organic compound formed by the reaction of \(1,1,1-\) trichloroethane with silver powder is: (a) Acetylene (b) Ethene (c) 2 - Butyne (d) 2 - Butene

Short Answer

Expert verified
The major organic compound formed is acetylene.

Step by step solution

01

Identify the reactant

The reactant given in the problem is 1,1,1-trichloroethane, which has the molecular formula \(C_2H_3Cl_3\). This compound consists of an ethane backbone (two carbon atoms) with three chlorine atoms attached to one of the carbon atoms.
02

Recognize the role of silver powder

The keyword in the problem indicates that silver powder acts as a reducing agent. In reactions involving trihaloalkanes, silver can facilitate the elimination of halogen atoms, leading to the formation of triple bonds between carbon atoms, forming alkynes.
03

Predict the elimination reaction

When 1,1,1-trichloroethane is treated with silver powder, two molecules of chlorine are eliminated from the 1,1,1-trichloroethane, leaving the carbon atoms with a triple bond. This process forms an alkyne.
04

Determine the resultant compound

Removing the three chlorines from 1,1,1-trichloroethane results in the formation of acetylene. Acetylene is an alkyne with the formula \(C_2H_2\) and a carbon-carbon triple bond.

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Key Concepts

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

1,1,1-trichloroethane
1,1,1-Trichloroethane is a volatile and colorless compound. It is based on an ethane structure, which means it has two carbon atoms bonded together. Unique to 1,1,1-trichloroethane, it has three chlorine atoms replacing hydrogen atoms on the same carbon atom. This chemical can be represented with the formula \(C_2H_3Cl_3\).
Understanding the structure is crucial as it relates to its reactivity in chemical processes.
  • It exhibits significant reactivity due to the presence of these chlorine atoms.
  • The chlorine atoms are more electronegative, which makes it easier for them to leave during chemical reactions.
This compound is used in a variety of chemical reactions, including the synthesis of alkynes. Knowing what makes this compound reactive will help you understand the transformations it undergoes during various chemical reactions.
Silver powder reduction
Silver powder is often used as a reducing agent in organic chemistry. Reducing agents are substances that donate electrons in reactions, helping transform compounds by removing specific atoms or groups. In the case of 1,1,1-trichloroethane, silver powder plays a crucial role. It enables the transformation of trichloroethane into a more simplified compound by eliminating chlorine atoms. This process involves stages where chlorines are replaced or removed, creating an opportunity for the remaining carbon atoms to form stronger bonds, specifically a triple bond characterizing an alkyne.
  • The reduction by silver powder involves the elimination of halogens, typically in compounds with multiple halogen substituents.
  • In 1,1,1-trichloroethane, silver helps remove two chlorine atoms, which is essential for the eventual formation of acetylene.
Understanding this process is vital as it clarifies how certain transformations lead to alkyne synthesis.
Alkyne synthesis
Alkyne synthesis involves creating compounds that contain at least one carbon-carbon triple bond. These types of chemical bonds are stronger than single or double bonds and add unique properties to the compound.In the exercise, acetylene is synthesized from 1,1,1-trichloroethane using silver powder. This transformation highlights the concept of dehalogenation, where halogen atoms (in this case, chlorines) are removed.
The synthesis steps include:
  • Reacting 1,1,1-trichloroethane with silver powder.
  • Facilitating the removal of chlorine atoms through this reduction process.
  • Creating a carbon-carbon triple bond, resulting in an alkyne like acetylene with the formula \(C_2H_2\).
Mastering this reaction process is crucial in organic chemistry as alkynes are fundamental building blocks for many industrial and commercial applications.

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Most popular questions from this chapter

Read the following Statement-1 (Assertion) and Statement -2 (Reason) and answer as per the options given below: Statement-1 : Bromobenzene upon reaction with \(\mathrm{Br}_{2} / \mathrm{Fe}\) gives \(1,4-\) dibromobenzene as the major product. [2008S] Statement-2 : In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile. (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement- 1 (c) Statement-1 is True, Statement- 2 is False (d) Statement-1 is False, Statement-2 is True

An organic compound \(X\), on analysis gives \(24.24\) per cent carbon and \(4.04\) per cent hydrogen. Further, sodium extract of \(1.0 \mathrm{~g}\) of \(X\) gives \(2.90 \mathrm{~g}\) of silver chloride with acidified silver nitrate solution. The compound \(X\) may be represented by two isomeric structures, \(Y\) and \(Z\). \(Y\) on treatment with aqueous potassium hydroxide solution gives a dihydroxy compound while \(Z\) on similar treatment gives ethanal. Find out the molecular formula of \(X\) and give the structures of \(Y\) and \(Z\).

An \(\mathrm{S}_{\mathrm{N}} 2\) reaction at an asymmetric carbon of a compound always gives (a) an enantiomer of the substrate (b) a product with opposite optical rotation (c) a mixture of diastereomers (d) a single stereoisomer

Write the structure of all the possible isomers of dichloroethene. Which of them will have zero dipole moment?

In \(\mathrm{S}_{\mathrm{N}} 2\) reactions, the correct order of reactivity for the following compounds: \(\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) and \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) is: (a) \(\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\)

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