Question

The most effective reaction for the synthesis of 1-methoxy-4-nitrobenzene is:

• A. \text{1-Bromo-4-nitrobenzene reacting with sodium methoxide}
• B. \text{1-Bromo-4-nitrobenzene reacting with sodium chloride}
• C. \text{1-Bromo-4-nitrobenzene reacting with sodium methoxide and chlorine}
• D. \text{Reaction leading to a different product}

Hint:

When synthesizing substituted aromatic compounds, ensure the nucleophile is appropriate for the halide to be replaced, and consider the positions of substituents on the ring for optimal reaction pathways.

Answer 1

The Correct Option is A

In this reaction, the goal is to synthesize 1-methoxy-4-nitrobenzene, which involves replacing a halogen atom (bromine) with a methoxy group (\(-OCH_3\)) at the para position to the nitro group. This reaction typically proceeds through an S_N2 mechanism, where sodium methoxide (\(NaOCH_3\)) acts as a nucleophile and replaces the bromine atom. The methoxy group is a strong nucleophile and reacts efficiently with the electrophilic carbon bonded to the bromine in 1-bromo-4-nitrobenzene.
Thus, the most effective reaction is the one in option (A), where 1-bromo-4-nitrobenzene reacts with sodium methoxide to give 1-methoxy-4-nitrobenzene. \[ \text{1-Bromo-4-nitrobenzene} + NaOCH_3 \rightarrow \text{1-Methoxy-4-nitrobenzene} \]