Problem 120
Question
Which of the following statements is/are correct? 1\. \(\mathrm{CH}_{3} \mathrm{CHCl}_{2} \mathrm{CH}_{3}\) is a secondary alkyl halide. 2\. Acetylene is formed when \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl}\) is heated with water. 3\. Iodoform gives a precipitate with AgNO, solution on heating, whereas chloroform does not. 4\. Freon \(\left(\mathrm{CCl}_{2} \mathrm{~F}_{2}\right)\) is prepared by the action of \(\mathrm{CCl}_{4}\) and \(\mathrm{SbF}_{3}\) in the presence of \(\mathrm{SbCl}_{5}\) as a catalyst. (a) 1,2 and 3 (b) 2,3 and 4 (c) 1,3 and 4 (d) 3 and 4
Step-by-Step Solution
Verified Answer
The correct statements are 1, 3, and 4, so the answer is (c).
1Step 1: Analyze Statement 1
Statement 1 claims that \( \mathrm{CH}_{3} \mathrm{CHCl}_{2} \mathrm{CH}_{3} \) is a secondary alkyl halide. To check this, identify the type of halide by looking at the carbon atom bonded to the halogen. In \( \mathrm{CHCl}_{2} \), the carbon atom with the chloride (\( \mathrm{Cl} \)) is connected to two other carbon atoms (making it a secondary carbon), confirming that it is indeed a secondary alkyl halide. Therefore, statement 1 is correct.
2Step 2: Analyze Statement 2
Statement 2 claims that acetylene is formed when \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl} \) (chloroethylene) is heated with water. The actual reaction involves the elimination of hydrogen chloride by hydrolysis, which does not directly produce acetylene. Instead, water would not typically transform \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Cl} \) into acetylene. Thus, statement 2 is incorrect.
3Step 3: Analyze Statement 3
Statement 3 claims that iodoform gives a precipitate with AgNO3 solution on heating, whereas chloroform does not. Iodoform, \( \mathrm{CHI}_{3} \), reacts with silver nitrate to form silver iodide \( \mathrm{AgI} \) which is a yellow precipitate. However, chloroform \( \mathrm{CHCl}_{3} \) does not give such a reaction with AgNO3, as \( \mathrm{AgCl} \) formation does not happen under these conditions. Therefore, statement 3 is correct.
4Step 4: Analyze Statement 4
Statement 4 describes the preparation of Freon \( \left(\mathrm{CCl}_{2} \mathrm{~F}_{2}\right) \) from \( \mathrm{CCl}_{4} \) using \( \mathrm{SbF}_{3} \) in the presence of a catalyst \( \mathrm{SbCl}_{5} \). This is a correct description of a method used to produce Freon, as antimony trifluoride and antimony pentachloride are used in the halogen exchange process for this preparation. So, statement 4 is correct.
Key Concepts
Secondary Alkyl HalideHalogenation ReactionsIodoform ReactionFreon Preparation
Secondary Alkyl Halide
A secondary alkyl halide is an organic compound in which a halogen atom, such as chlorine, bromine, iodine, or more rarely fluorine, is bonded to a secondary carbon. A secondary carbon is a carbon atom attached to two other carbon atoms and at least one hydrogen. This type of halide is generally denoted as R-CHX-R', where X represents a halogen.
Secondary alkyl halides are pivotal in organic synthesis due to their reactivity. One common example is the compound \(\mathrm{CH}_{3} \mathrm{CHCl}_{2} \mathrm{CH}_{3} \), also known as 2,3-dichlorobutane. In this compound, the middle carbon atom is bonded to the chlorine atom, making it a secondary carbon, as it is connected to two adjoining carbon atoms, qualifying it as a secondary alkyl halide.
These compounds play a crucial role in substitution and elimination reactions, where they interact with nucleophiles or bases to further form a plethora of other organic molecules.
Secondary alkyl halides are pivotal in organic synthesis due to their reactivity. One common example is the compound \(\mathrm{CH}_{3} \mathrm{CHCl}_{2} \mathrm{CH}_{3} \), also known as 2,3-dichlorobutane. In this compound, the middle carbon atom is bonded to the chlorine atom, making it a secondary carbon, as it is connected to two adjoining carbon atoms, qualifying it as a secondary alkyl halide.
These compounds play a crucial role in substitution and elimination reactions, where they interact with nucleophiles or bases to further form a plethora of other organic molecules.
Halogenation Reactions
Halogenation reactions are fundamental processes in organic chemistry where one or more hydrogen atoms in a hydrocarbon are replaced by halogen atoms (such as chlorine, bromine, etc.). These reactions can be broadly categorized into two main types:
- Free Radical Halogenation: This method involves the homolytic cleavage of a halogen molecule into two halogen radicals. This is followed by the abstraction of a hydrogen atom from the alkane by a halogen radical. The resulting alkyl radical then reacts with another halogen molecule to regenerate the halogen radical and form a halogenated product.
- Electrophilic Halogenation: This typically occurs in aromatic compounds, where the aromatic ring reacts with halogens in the presence of a catalyst under moderate conditions, often forming a halogenated aromatic compound.
Iodoform Reaction
The iodoform reaction is a hallmark test for identifying methyl ketones or alcohols that can be oxidized to methyl ketones. It is a chemical reaction where a methyl ketone is treated with iodine and a base, typically sodium hydroxide, leading to the formation of iodoform, \(\mathrm{CHI}_3\), which appears as a yellow precipitate.
The reaction can be summarized as follows: the methyl ketone structure, \(\mathrm{RCOCH}_3\), when treated with \(\mathrm{I}_2\) and \(\mathrm{NaOH}\), results in the replacement of the hydrogen atoms of the methyl group by iodine, eventually forming a triiodomethyl carbonium ion that rearranges to form iodoform.
This test is unique because iodoform is easily distinguishable due to its distinctive yellow color and characteristic smell, making the iodoform reaction a reliable method in the identification of certain organic compounds in laboratory settings.
The reaction can be summarized as follows: the methyl ketone structure, \(\mathrm{RCOCH}_3\), when treated with \(\mathrm{I}_2\) and \(\mathrm{NaOH}\), results in the replacement of the hydrogen atoms of the methyl group by iodine, eventually forming a triiodomethyl carbonium ion that rearranges to form iodoform.
This test is unique because iodoform is easily distinguishable due to its distinctive yellow color and characteristic smell, making the iodoform reaction a reliable method in the identification of certain organic compounds in laboratory settings.
Freon Preparation
Freons are a series of chlorofluorocarbons (CFCs), common as refrigerants, and are synthesized through specific halogenation reactions. The preparation of Freon-12, designated as \(\mathrm{CCl}_2\mathrm{F}_2\), is a classic example of such a process.
The preparation involves a halogen exchange reaction using carbon tetrachloride (\(\mathrm{CCl}_4\)) and antimony trifluoride (\(\mathrm{SbF}_3\)), with the presence of a catalyst like antimony pentachloride (\(\mathrm{SbCl}_5\)). During the process, fluoride ions replace the chlorine atoms in \(\mathrm{CCl}_4\), producing Freon-12.
The preparation involves a halogen exchange reaction using carbon tetrachloride (\(\mathrm{CCl}_4\)) and antimony trifluoride (\(\mathrm{SbF}_3\)), with the presence of a catalyst like antimony pentachloride (\(\mathrm{SbCl}_5\)). During the process, fluoride ions replace the chlorine atoms in \(\mathrm{CCl}_4\), producing Freon-12.
- This method is favored due to its efficiency in producing a relatively stable compound used in refrigeration.
- However, the environmental impact of such compounds, particularly in ozone layer depletion, has led to a decline in their usage and a search for more sustainable alternatives.
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