Step 1: Understanding the Question:
The question asks for the major organic product formed when anisole (methoxybenzene) undergoes electrophilic aromatic bromination using bromine dissolved in acetic acid.
Step 2: Key Formula or Approach:
The methoxy group ($-\text{OCH}_3$) attached to the benzene ring contains lone pairs on the oxygen atom that are delocalized into the aromatic ring via resonance ($+R$ effect). This highly activates the ring toward electrophilic aromatic substitution and directs incoming electrophiles to the ortho- and para- positions.
Step 3: Detailed Explanation:
1. Anisole ($\text{C}_6\text{H}_5\text{OCH}_3$) reacts with bromine ($\text{Br}_2$) in ethanoic acid (acetic acid) solvent.
2. The oxygen atom's $+R$ effect increases electron density specifically at the ortho and para positions of the benzene ring.
3. The electrophile ($\text{Br}^+$) attacks these electron-rich centers, yielding a mixture of two structural isomers: o-bromoanisole (minor) and p-bromoanisole (major).
4.
Major vs. Minor Selectivity: The ortho position is structurally crowded because it sits right next to the bulky methoxy group ($-\text{OCH}_3$). This causes significant steric hindrance, which opposes the attack. In contrast, the para position is located on the opposite side of the ring and is completely free from steric crowding. Therefore, p-bromoanisole is formed preferentially as the major product ($\approx 90\%$ yield).
Step 4: Final Answer:
The major product of the reaction is p-bromo anisole, matching option (B).