Question

The major product formed in the given reaction sequence is 

Hint:

Potassium phthalimide first alkylates (Gabriel-type), but under alkoxide reflux the imide {opens}, enabling intramolecular condensations. With phenacyl bromides, base triggers enolate formation and cyclization to hydroxybenzamide/quinazolinone-type products; regiochemistry follows the most stabilized conjugation path.

Answer 1

Step 1: N-alkylation of potassium phthalimide.
The starting anion is the K⁺ salt of phthalimide (PhthN^-K⁺). With the primary α-bromoketone (phenacyl bromide derivative), it undergoes an S_N2 reaction to give an N-phenacethyl phthalimide:
PhthN^- + PhCOCH₂CH₂Br → PhthN-CH₂CH₂COPh.
Step 2: Methoxide-promoted opening of the imide.
Under NaOMe/MeOH (reflux), methoxide attacks one imide carbonyl to give a phthalamate (ring-opened imide) bearing an amide and a methyl imidate/ester. This step renders the carbonyls β to the benzylic methylene.

Step 3: Intramolecular condensation (cyclization).
Deprotonation at the α-position of the phenacyl unit (PhCO–CH₂–) generates an enolate that intramolecularly attacks the amide carbonyl of the opened phthalamate, closing onto the aromatic ortho position of the phthal framework (acyl transfer to the ring–N). Subsequent proton transfers/tautomerization furnish the 2-hydroxy benzamide skeleton (quinazolinone-type enol), with the C=O(Ph) placed as in option (B).

Step 4: Regioselectivity.
Cyclization occurs onto the amide carbonyl that positions the benzyloyl group adjacent to the ring nitrogen, minimizing strain and maximizing conjugation with the aromatic ring; this gives the regiochemical outcome depicted in (B) rather than (A).

Major product is the hydroxybenzamide shown in option (B).