Identifying the Highly Reactive Intermediate in Chloroform Reactions
This explanation focuses on identifying the specific, highly reactive intermediate generated when chloroform ($CHCl_3$) is treated with a strong base, such as alcoholic potassium hydroxide ($KOH$), and its subsequent involvement in the carbylamine reaction.
Detailed Reaction Steps and Mechanism
The reaction proceeds in distinct stages:
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Deprotonation of Chloroform: Chloroform possesses a weakly acidic proton due to the electron-withdrawing effect of the three chlorine atoms. A strong base, like the hydroxide ion ($OH^-$) present in alcoholic $KOH$, abstracts this proton.
$CHCl_3 + OH^- \rightleftharpoons \bar{C}Cl_3 + H_2O$
This step forms the trichloromethyl anion ($\bar{C}Cl_3$), which is a carbanion.
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Alpha-Elimination: The generated trichloromethyl anion ($\bar{C}Cl_3$) is unstable. It rapidly loses a chloride ion ($Cl^-$) from the same carbon atom (an alpha-elimination process).
$\bar{C}Cl_3 \rightarrow :CCl_2 + Cl^-$
This elimination yields a neutral, highly reactive species with a divalent carbon atom, known as dichlorocarbene ($:CCl_2$).
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Carbylamine Reaction Involvement: The dichlorocarbene ($:CCl_2$) generated is the key intermediate that subsequently reacts with primary amines ($RNH_2$) in the characteristic carbylamine reaction (also known as the isocyanide test) to form isocyanides ($RNC$).
$RNH_2 + :CCl_2 \rightarrow \text{Intermediate(s)} \rightarrow RNC + \text{Byproducts}$
Therefore, the highly reactive intermediate formed is dichlorocarbene.
Analysis of Potential Intermediates
Let's examine why dichlorocarbene is the correct intermediate and why the other options are not:
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Dichlorocarbene ($:CCl_2$): This species fits the mechanism described above. The abstraction of a proton followed by the loss of a leaving group ($Cl^-$) from the same carbon leads directly to this divalent intermediate. It is known to be electrophilic and participates in reactions like the carbylamine reaction.
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A free radical ($R_3C \cdot$): Free radicals involve unpaired electrons and are typically formed through homolytic bond cleavage, often initiated by heat or light. The reaction of chloroform with $KOH$ is an ionic process, not a radical one.
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A carbocation ($R_3C^+$): Carbocations have a positive charge on the carbon atom. While the intermediate formed after deprotonation ($\bar{C}Cl_3$) is an anion, the subsequent step involves the loss of $Cl^-$, leading to a neutral divalent carbon species ($:CCl_2$), not a positively charged carbocation.
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A carbanion ($R_3C^-$): A carbanion ($\bar{C}Cl_3$) is formed initially. However, it is not the final, stable reactive intermediate that participates directly in the subsequent steps of the carbylamine reaction. It is a transient species that quickly transforms into dichlorocarbene.
Conclusion on Reactive Intermediate
Based on the reaction mechanism, the treatment of chloroform ($CHCl_3$) with a strong base like alcoholic $KOH$ generates the highly reactive intermediate dichlorocarbene ($:CCl_2$), which is crucial for the carbylamine reaction.