Question

\[ \begin{aligned} &\text { Identify ' } \mathrm{A} \text { ' in the following reaction. }\\ &\underset{\text { (excess) }}{\mathrm{A}}+\underset{\text { (excess) }}{\operatorname{Acetyl} \text { chloride }} \xrightarrow[\mathrm{AlCl}_3]{\text { Anhydrous }} \end{aligned} \]

2-Chloroacetophenone + 4-Chloroacetophenone

• A. Benzene
• B. Chlorobenzene
• C. Toluene
• D. Phenol

Hint:

Friedel-Crafts acylation gives ortho and para products when the ring already carries an ortho/para-directing chlorine.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Acetyl chloride with anhydrous \(\text{AlCl}_3\) acylates an aromatic ring (Friedel-Crafts acylation) and puts a \(\text{COCH}_3\) group on the ring.

Step 2: Read the products
The products are 2-chloroacetophenone and 4-chloroacetophenone.
Both contain a chlorine atom already on the ring, so the starting compound must carry chlorine.

Step 3: Directing effect
Chlorine is deactivating but ortho/para directing, so the acyl group enters at the ortho (2-) and para (4-) positions.

Step 4: Conclude
The starting compound A is chlorobenzene.

Final Answer:
A is chlorobenzene. \[ \boxed{\text{(B) }\text{Chlorobenzene}} \]