Introduction

In this reaction sequence, anisole reacts with acetic anhydride
((CH₃CO)₂O) in presence of anhydrous AlCl₃.
We are required to determine the structure of intermediate X (major product).

This reaction belongs to Friedel–Crafts acylation, an important electrophilic aromatic substitution reaction frequently asked in NEET, JEE, and university entrance exams.

 Step 1 – Type of Reaction

Reaction:

Anisole + (CH₃CO)₂O → AlCl₃ → X

This is a Friedel–Crafts acylation reaction where an acetyl group (–COCH₃) is introduced onto the benzene ring.

 Step 2 – Directing Effect of Methoxy Group (–OCH₃)

• Strong electron donating (+M effect)
• Activates benzene ring
• Ortho/Para directing group

Therefore, substitution occurs mainly at:

• Ortho position
• Para position (major product)

Para product dominates due to less steric hindrance and greater stability.

 Step 3 – Product Formed

Acylation introduces:

–COCH₃

Hence the product is:

p-Methoxyacetophenone

 Option-wise Explanation

 Option A – Aromatic Ketone (Para acylation)

• –COCH₃ directly attached to benzene
• Para to methoxy group
• Correct Friedel–Crafts product

Correct Answer ✔

 Option B – Ester Formation

• Shows –OCOCH₃
• Represents ester
• Friedel–Crafts forms ketone, not ester

Incorrect ✘

 Option C – Meta Substitution

• Methoxy never directs meta
• Violates directing rule

Incorrect ✘

 Option D – O-Acylation (Phenyl acetate)

• Acylation on oxygen
• Not favored with AlCl₃
• Ring acylation preferred

Incorrect ✘

 Final Answer

Option A – p-Methoxyacetophenone

Important Exam Concepts

• Friedel–Crafts acylation mechanism
• Electrophilic aromatic substitution
• Ortho/Para directing groups
• Major product prediction

 Quick Memory Trick

“OCH₃ → Ortho/Para → Para major → Ketone formed”

 Conclusion

Thus, anisole undergoes Friedel–Crafts acylation with acetic anhydride to form
para-methoxyacetophenone as the major product (X).
Understanding directing effects makes such questions quick and easy to solve.