Question

In the following reaction, the major product T is 
\[ \text{(i) NaOMe} \quad \text{(ii) H}_3\text{O}^+, \text{reflux} \quad \text{(iii) Polyphosphoric acid} \]

• A. A
• B. B
• C. C
• D. D

Hint:

Polyphosphoric acid commonly promotes intramolecular Friedel–Crafts acylation to form fused-ring ketones.

Answer 1

The Correct Option is C

Step 1: Reaction with NaOMe.
The bromoketone undergoes intramolecular nucleophilic substitution:
O$^-$ (from NaOMe) attacks the benzylic carbon containing Br, forming a cyclic intermediate (O-alkylation).
This produces a substituted $\beta$-hydroxy ketone after protonation.
Step 2: Acidic heating (H$_3$O$^+$, reflux).
This converts the $\beta$-hydroxy ketone into the corresponding $\alpha,\beta$-unsaturated ketone through dehydration (E1cb mechanism).
A conjugated enone is formed.
Step 3: Cyclization with polyphosphoric acid.
Polyphosphoric acid promotes intramolecular Friedel–Crafts acylation.
The aromatic ring attacks the carbonyl carbon of the enone, forming a fused bicyclic ketone (indone-type structure).
This yields a six-membered ring fused with a five-membered ring, giving the product shown in option (D).
Step 4: Identify correct product.
Option (D) corresponds exactly to the indanone derivative expected after intramolecular acylation.
Other structures (A–C) represent incorrect orientations of methyl substituents or incorrect ring fusion.
Step 5: Conclusion.
Thus, the major product T is structure (D).