Question

The major product formed in the following reaction is 

• A. (A)
• B. (B)
• C. (C)
• D. (D)

Hint:

In nucleophilic substitution reactions, a nucleophile (like ethoxide) replaces a leaving group (like chloride) to form a new bond.

Answer 1

The Correct Option is C

Step 1: Generation of Dichlorocarbene

The base ($\text{EtONa}$) reacts with chloroform ($\text{CHCl}_3$) to form the unstable trichloromethyl carbanion, which then quickly eliminates a chloride ion ($\text{Cl}^-$) to form the neutral, highly reactive intermediate, dichlorocarbene ($\ddot{\text{C}}\text{Cl}_2$):

$$\text{EtONa} + \text{CHCl}_3 \rightleftharpoons \text{EtOH} + \text{Na}^+\text{C}^-\text{Cl}_3$$

$$\text{Na}^+\text{C}^-\text{Cl}_3 \xrightarrow{-\text{Cl}^-,\ -\text{Na}^+} \ddot{\text{C}}\text{Cl}_2$$

Step 2: Dichlorocarbene Reaction with Pyrrole Anion

The reaction of a carbene with an electron-rich aromatic system, especially an aromatic anion like the potassium salt of pyrrole, is a ring expansion reaction, similar to the Reimer-Tiemann reaction (where the carbene reacts with phenol) or the reaction between carbene and pyrrole to form pyridine derivatives.

The highly nucleophilic pyrrole anion reacts with the electrophilic dichlorocarbene.

The carbene typically adds across the double bond of the electron-rich ring to form an intermediate bicyclic cyclopropane-like species.

This highly strained intermediate then rearranges, followed by the elimination of $\text{HCl}$ (or $\text{Cl}^-$) and aromatization.

The overall process leads to the expansion of the five-membered pyrrole ring into a six-membered ring, resulting in a pyridine derivative.

The product of the reaction between pyrrole and dichlorocarbene is typically $\text{3-chloropyridine}$ (or $\text{2-chloropyridine}$ as a minor product), or a substituted pyridine depending on the mechanism. Given the options, the product is a pyridine with a substituent.

Step 3: Analyzing the Options

The four options are all substituted pyridines:

(A) $\text{2-Chloropyridine}$

(B) $\text{2-Ethoxypyridine}$

(C) $\text{3-Chloropyridine}$

(D) $\text{3-Ethoxypyridine}$

The initial reaction of pyrrole with $\ddot{\text{C}}\text{Cl}_2$ forms $\text{3-chloropyridine}$ ($\mathbf{C}$) as the major product. The intermediate is usually a 1-aza-bicyclo[4.1.0]heptadienyl cation, which eliminates a proton and aromatizes to $\text{3-chloropyridine}$.

The reaction mixture, however, contains a strong nucleophile, ethoxide ($\text{EtO}^-$). The product, $\text{3-chloropyridine}$, is an aryl halide.

$\text{3-Chloropyridine}$ is not highly reactive toward Nucleophilic Aromatic Substitution ($\text{S}_{\text{N}}\text{Ar}$) because the chlorine is not adjacent (ortho/para) to the ring nitrogen (the strongest electron-withdrawing group).

$\text{2-Chloropyridine}$ and $\text{4-chloropyridine}$ undergo $\text{S}_{\text{N}}\text{Ar}$ much more easily.

Since the ethoxide is present during and after the ring expansion, the major product is simply the initially formed stable compound, $\text{3-chloropyridine}$.

Therefore, the major product is $\text{3-chloropyridine}$.

Answer: (C)