Question

Identify the major product when anisole is treated with $\text{Br}_2$ in presence of acetic acid.

• A. m-Bromo anisole
• B. p-Bromo anisole
• C. o-Bromo anisole
• D. 2,4,6-Tribromo anisole

Hint:

For almost all activated mono-substituted benzene rings (like anisole, phenol, or toluene) undergoing standard electrophilic substitutions, the para-isomer will always be your major product due to the elimination of steric crowding present at the ortho position.

Answer 1

The Correct Option is B

Step 1: Identify the reaction type.
Anisole ($\text{C}_6\text{H}_5\text{OCH}_3$) with $\text{Br}_2$ in acetic acid is an electrophilic aromatic substitution (bromination).
Step 2: Classify the substituent.
The $-\text{OCH}_3$ group donates lone-pair density into the ring by the $+R$ effect, so it is a strong ring activator and an ortho/para director.
Step 3: Locate the electron-rich sites.
Resonance pushes electron density specifically onto the ortho and para carbons, so the electrophile $\text{Br}^+$ attacks there.
Step 4: Compare ortho versus para.
The ortho positions sit right beside the bulky $-\text{OCH}_3$ group and suffer steric crowding, which discourages attack there.
Step 5: Favour the para position.
The para position lies opposite the methoxy group, free of steric strain, so substitution happens mainly there.
Step 6: Note the mild conditions.
Acetic acid is a mild medium, so only one bromine is introduced; full tribromination needs harsher conditions and is not the answer.
Step 7: Conclude.
The major product is p-bromoanisole.
\[ \boxed{\text{p-Bromo anisole, option (2)}} \]