Question:

The major product formed in the given reaction sequence is

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Potassium phthalimide first alkylates (Gabriel-type), but under alkoxide reflux the imide \emph{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.
Updated On: Aug 28, 2025
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The Correct Option is B

Solution and Explanation

Step 1: N-alkylation of potassium phthalimide.
The starting anion is the \(\mathrm{K^+}\) salt of phthalimide (\(\mathrm{PhthN^-K^+}\)). With the primary \(\alpha\)-bromoketone (phenacyl bromide derivative), it undergoes an S\(_\mathrm{N}2\) reaction to give an \(\mathbf{N}\)-phenacethyl phthalimide:
\[ \ce{PhthN^- + PhCOCH2CH2Br→ PhthN-CH2CH2COPh}. \] Step 2: Methoxide-promoted opening of the imide.
Under \(\mathrm{NaOMe/MeOH}\) (reflux), methoxide attacks one imide carbonyl to give a \emph{phthalamate} (ring-opened imide) bearing an amide and a methyl imidate/ester. This step renders the carbonyls \(\beta\) to the benzylic methylene. Step 3: Intramolecular condensation (cyclization).
Deprotonation at the \(\alpha\)-position of the phenacyl unit (\(\ce{PhCO\text{–}CH2\text{–}}\)) generates an enolate that \emph{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 \(\ce{C{=}O(Ph)}\) placed as in option (B). Step 4: Regioselectivity.
Cyclization occurs onto the amide carbonyl that positions the benzyloyl group \emph{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). \[ \boxed{\text{Major product is the hydroxybenzamide shown in option (B).}} \]
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