Carbylamine forms from aliphatic or aromatic primary amine via which of the following intermediates?

The carbylamine reaction, also known as the isocyanide test, is used to detect primary amines (both aliphatic and aromatic). The reaction involves heating a primary amine with chloroform (CHCl₃) and a strong base like potassium hydroxide (KOH). The overall reaction is:

$$ R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O $$

Here, R can be an alkyl or aryl group, and R-NC is the carbylamine (isocyanide). To determine the intermediate, we need to understand the mechanism step by step.

First, the base (OH⁻ from KOH) reacts with chloroform. Chloroform has three chlorine atoms, making the C-H bond acidic due to the electron-withdrawing effect of the chlorines. The base deprotonates chloroform:

$$ CHCl_3 + OH^- \rightarrow CCl_3^- + H_2O $$

The trichloromethanide ion (CCl₃⁻) formed is unstable and undergoes alpha-elimination. This means it loses a chloride ion (Cl⁻) to form dichlorocarbene (:CCl₂), a neutral species with a divalent carbon atom:

$$ CCl_3^- \rightarrow :CCl_2 + Cl^- $$

So, the net reaction for carbene formation is:

$$ CHCl_3 + OH^- \rightarrow :CCl_2 + Cl^- + H_2O $$

Note that this step requires one equivalent of base. In the overall reaction, three equivalents are used because additional steps consume more base.

Next, the primary amine (RNH₂) is deprotonated by another equivalent of base to form the amide ion (RNH⁻):

$$ RNH_2 + OH^- \rightarrow RNH^- + H_2O $$

The amide ion (RNH⁻) acts as a nucleophile and attacks the electrophilic carbene intermediate (:CCl₂). Carbenes are electron-deficient and highly reactive, so they readily accept electrons. This forms a new carbon-nitrogen bond:

$$ RNH^- + :CCl_2 \rightarrow [RNH-CCl_2]^- $$

The intermediate [RNH-CCl₂]⁻ is a carbanion because the carbon atom (originally from the carbene) has three bonds (one to N and two to Cl) and a negative charge. However, this carbanion is unstable and quickly loses a chloride ion to form a neutral dichloroamine:

$$ [RNH-CCl_2]^- \rightarrow RN=CCl_2 + Cl^- $$

The dichloroamine (RN=CCl₂) is still unstable under the basic conditions. It undergoes dehydrohalogenation (loss of HCl) facilitated by another equivalent of base. The base deprotonates the nitrogen, forming an anion that eliminates chloride:

$$ RN=CCl_2 + OH^- \rightarrow [RN-CCl_2]^- + H_2O \quad \text{(deprotonation)} $$ $$ [RN-CCl_2]^- \rightarrow R-NC + Cl^- \quad \text{(elimination)} $$

Alternatively, RN=CCl₂ can directly lose HCl to form the isocyanide, but the base assists in the removal of the proton.

The key point is that the reaction proceeds through the dichlorocarbene (:CCl₂) intermediate, which is a type of carbene. Carbenes are neutral, divalent carbon species with two unshared electrons. They are highly reactive intermediates in organic reactions.

Now, evaluating the options:

• A. Carbanion: While a carbanion ([RNH-CCl₂]⁻) appears later in the mechanism, it is not the primary intermediate responsible for initiating the reaction with the amine. The carbene is formed first and is essential for the reaction.
• B. Carbene: This is the intermediate formed from chloroform and base, and it directly reacts with the amine. It is the characteristic intermediate of the carbylamine reaction.
• C. Carbocation: A carbocation has a positive charge and is not involved in this reaction. The mechanism does not generate any carbocation.
• D. Carbon radical: Radicals have unpaired electrons and are not part of this mechanism, which proceeds via ionic and concerted steps.

Therefore, the intermediate via which carbylamine forms is carbene.

Hence, the correct answer is Option B.