Question

We have considered nucleophilic aromatic substitution of pyridine at the 2-position and 3-position but not at the 4-position. Complete the three possible cases by showing the mechanism for the reaction of methoxide ion with 4-chloropyridine. Show how the intermediate is stabilized by delocalization of the charge onto the nitrogen atom.

Accepted Answer

Identify the type of reaction: This is a nucleophilic aromatic substitution (NAS) reaction, where the methoxide ion (CH₃O⁻) acts as the nucleophile, and 4-chloropyridine is the substrate. The chlorine atom is the leaving group. Determine the site of attack: The methoxide ion will attack the carbon atom at the 4-position of the pyridine ring, where the chlorine atom is attached. This attack occurs because the electron-deficient aromatic ring is activated by the electronegative nitrogen atom in pyridine. Illustrate the addition step: The methoxide ion attacks the 4-position carbon, forming a negatively charged intermediate. This intermediate is a Meisenheimer complex, where the negative charge is delocalized across the ring and onto the nitrogen atom. Use resonance structures to show how the charge is stabilized. For example, one resonance structure will have the negative charge on the nitrogen atom, while others will show the charge on different carbons in the ring. Depict the elimination step: After the intermediate is stabilized, the chlorine atom leaves as a chloride ion (Cl⁻), restoring aromaticity to the pyridine ring. This step completes the substitution reaction. Summarize the overall reaction: The final product is 4-methoxypyridine, where the methoxide group has replaced the chlorine atom at the 4-position. The reaction mechanism demonstrates how the intermediate is stabilized by delocalization of the negative charge onto the nitrogen atom through resonance.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Nucleophilic Aromatic Substitution (NAS)

Mechanism of NAS

Resonance Stabilization