Problem 1
Question
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
Step-by-Step Solution
Verified Answer
The major organic compound formed is acetylene.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
4Step 4: 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.
Key Concepts
1,1,1-trichloroethaneSilver powder reductionAlkyne synthesis
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.
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.
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.
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:
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 \(C_2H_2\).
Other exercises in this chapter
Problem 2
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