For nucleophilic aromatic substitution ($\text{S}_\text{N}\text{Ar}$):
1. The rate of substitution increases with the number of electron-withdrawing groups (especially $-\text{NO}_2$) located ortho and para to the leaving group (halide).
2. Under heating with excess nucleophile, all sufficiently activated leaving groups will be replaced.
Step 1: Understanding the Question:
We need to determine the major product when 1,3-dichloro-2,5-dinitrobenzene is treated with methoxide ion ($\text{MeO}^-$) in methanol ($\text{MeOH}$) under heating.
Step 2: Key Formula or Approach:
This reaction is a nucleophilic aromatic substitution ($\text{S}_\text{N}\text{Ar}$) reaction. In $\text{S}_\text{N}\text{Ar}$, chlorine atoms act as excellent leaving groups if they are activated by strong electron-withdrawing groups (like $-\text{NO}_2$) located at ortho or para positions relative to them.
Step 3: Detailed Explanation:
Let us analyze the structure of the starting material:
- The compound has two chlorine atoms at positions 1 and 3.
- It has two nitro groups ($-\text{NO}_2$) at positions 2 and 5.
- Looking at each chlorine atom, they are both ortho to the nitro group at position 2 and para to the nitro group at position 5.
- The strongly electron-withdrawing $-\text{NO}_2$ groups significantly decrease the electron density of the benzene ring, especially at the carbons bearing the chlorine atoms.
- This activation allows the methoxide nucleophile ($\text{MeO}^-$) to attack the chlorine-bearing carbons, forming highly stable anionic Meisenheimer intermediates where the negative charge is delocalized onto the electronegative oxygen atoms of the nitro groups.
- Since both chlorine atoms are equally and highly activated, and the reaction is carried out under heating with excess methoxide, both chlorine atoms undergo substitution.
- This results in the replacement of both chlorine atoms by methoxy groups ($-\text{OMe}$), yielding the diether product shown in structure C.
