Match the major products obtained in the reactions given in List-I...

For the reaction in (P), 2-bromo-5-nitrobenzaldehyde oxime reacts with aqueous NaOH. The strong base deprotonates the oxime group. The resulting oximate anion undergoes intramolecular nucleophilic aromatic substitution ($S_NAr$), attacking the ortho-carbon and displacing the bromide ion to form a 1,2-benzisoxazole intermediate. Since it is derived from an aldoxime, this intermediate has a hydrogen atom at the C3 position. In the presence of NaOH, it undergoes base-catalyzed Kemp elimination (ring opening) to form 2-hydroxy-5-nitrobenzonitrile, which corresponds to structure (1). Thus, P $\rightarrow$ 1.

In reaction (Q), the aldoxime is treated with acetic anhydride, which acetylates the oxime group to form an O-acetate. Subsequent treatment with the mild base $Na_2CO_3$ causes the elimination of acetic acid, converting the aldoxime acetate into a nitrile. The bromide is not displaced under these mild conditions. The product is 2-bromo-5-nitrobenzonitrile, which corresponds to structure (2). Thus, Q $\rightarrow$ 2.

For the reaction in (R), 2-bromo-5-nitroacetophenone oxime reacts with aqueous NaOH. Similar to (P), the oximate anion is formed and displaces the ortho-bromide via $S_NAr$ to form a 1,2-benzisoxazole ring. However, because it is derived from a ketoxime, there is a methyl group at the C3 position instead of a hydrogen. Therefore, Kemp elimination cannot occur, and the reaction stops at the stable 3-methyl-5-nitro-1,2-benzisoxazole, which corresponds to structure (4). Thus, R $\rightarrow$ 4.

In reaction (S), the O-acetate of the ketoxime is treated with aqueous $Na_2CO_3$. Since it is a ketoxime derivative, it cannot undergo elimination to form a nitrile. The mild base simply hydrolyzes the acetate group back to the oxime. The base is not strong enough to promote the subsequent $S_NAr$ ring closure. The product is 2-bromo-5-nitroacetophenone oxime, which corresponds to structure (5). Thus, S $\rightarrow$ 5.

Combining these results gives the sequence P $\rightarrow$ 1, Q $\rightarrow$ 2, R $\rightarrow$ 4, S $\rightarrow$ 5.

Answer: P $\rightarrow$ 1; Q $\longrightarrow$ 2; R $\rightarrow$ 4; S $\longrightarrow$ 5