Question

E and F in the given reaction scheme are determined for the photolysis (\(h\nu\)) of the fused N–O heterocycle bearing an N-phenyl group, followed by thermolysis (\(\Delta\)).}

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Anthranil/benzisoxazole systems often undergo photochemical N–O bond cleavage to amide (o-acyl anilide) intermediates.
Subsequent heating typically promotes intramolecular recyclization to regenerate an N–O heterocycle (benzisoxazole).
Remember: \(h\nu\) opens the ring (amide), \(\Delta\) closes it (isoxazole).

Answer 1

The Correct Option is A

Step 1: Photochemical N–O bond cleavage and acyl migration.
Irradiation (\(h\nu\)) of N–O heterocycles of the anthranil/benzisoxazole family promotes homolytic/diradical cleavage of the N–O bond to an acyl-nitrene/diradical that rapidly undergoes acyl migration (Bucherer–type/ESIPT assisted) to give the \emph{o-acyl anilide} (ring-opened \(\ce{Ar–C(=O)–NPh}\)). This matches structure \(\mathbf{E}\) in option (A).
Step 2: Thermal recyclization to the N–O five-membered ring.
On heating (\(\Delta\)), the o-acyl anilide cyclizes intramolecularly (nucleophilic attack of the carbonyl oxygen on the adjacent imidoyl carbon followed by N–O bond formation) furnishing the 1,2-benzisoxazole skeleton shown as \(\mathbf{F}\) in option (A).
Step 3: Elimination of other options.
Options (C) and (D) propose oxime or hydroxylamine products for the photolysis step, which are not formed in this photorearrangement; the primary process is N–O cleavage with acyl transfer to the amide (not simple \(\ce{N–O}\) reduction/tautomerization). Option (B) shows an incorrect regiochemistry for the amide intermediate.
Therefore, \(\boxed{\text{(A) is correct: } E=\text{o-acyl anilide (amide), } F=\text{benzisoxazole}}\).