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Question

Which highly reactive intermediate is generated when chloroform ($CHCl_3$) reacts with a strong base like alcoholic KOH, subsequently reacting with a primary amine in the carbylamine reaction?

The correct answer is

Dichlorocarbene ($:CCl_2$)

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:

  1. 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.

  2. 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 ($:CCl_2$).

  3. 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:

  • 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.
  • 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.
  • 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.
  • 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.

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Important Questions from Organic Chemistry

  1. Which reaction occurs when an alkoxide ion attacks a primary alkyl halide via SN2 mechanism?
  2. Ethanoic acid reacts with alcohols to form what compound?
  3. Life on earth depends on ________ molecules as the main source of energy for all organisms.
  4. Which of the following represents a straight chain hydrocarbon?
  5. Substitution reaction involves:
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