Animated Solution for Chemistry - Organic Chemistry: Match the major products obtained in the reactions given in List-I with the corresponding structures in List-II and choose the correct option.

At first glance, the four reactions presented in this problem look deceptively similar. We are dealing with various oxime derivatives of 2-bromo-5-nitrobenzaldehyde and 2-bromo-5-nitroacetophenone, all being treated with basic reagents. However, this is a classic JEE Advanced trap. The entire reaction pathway hinges on a single, critical geometric detail: stereochemistry.

Specifically, we must determine whether the oxime $-OH$ group (or its acetate derivative) is positioned syn (pointing towards) or anti (pointing away from) the leaving group, which in this case is the bromine atom. Let's decode these reactions one by one.

The Intramolecular $S_{N}Ar$

In reaction P, we start with an aldoxime where the $-OH$ group is syn to the bromine atom. This spatial proximity is the key that unlocks the mechanism. When we introduce aqueous $NaOH$, the strong base deprotonates the oxime, converting it into a powerful oximate nucleophile right next door to the bromine.

The negatively charged oxygen attacks the aromatic ring at the carbon holding the bromine. Because the electron-withdrawing nitro group ($-N{O}_2$) is positioned para to this site, it stabilizes the resulting negative charge in the Meisenheimer intermediate. This allows a smooth intramolecular Nucleophilic Aromatic Substitution ($S_{N}Ar$). The bromide ion is expelled, and a fused five-membered ring—a benzisoxazole intermediate—is formed.

But the reaction doesn't stop there! The carbon atom between the oxygen and nitrogen in this newly formed ring possesses a highly acidic proton. The base abstracts this proton, causing a cascade of electron movement: a carbon-nitrogen triple bond forms, and the oxygen-carbon bond breaks to restore the stable aromaticity of the benzene ring. The final product is 2-hydroxy-5-nitrobenzonitrile (Structure 1).

The Simple Elimination

Reaction Q uses the exact same starting material as P, but the reagents are different. First, acetic anhydride ($(C{H}_{3}CO{)}_{2}O$) is used to acetylate the oxime, transforming the $-OH$ into an acetate group ($-OAc$), which is a much better leaving group.

Next, we introduce aqueous $N{a}_{2}C{O}_3$. This is a mild base, significantly weaker than $NaOH$. It is not strong enough to initiate the complex $S_{N}Ar$ cyclization we saw in reaction P. Instead, it takes the path of least resistance. It abstracts the aldehydic proton, kicking out the acetate group in a straightforward E2 elimination. The bromine atom remains completely untouched, yielding 2-bromo-5-nitrobenzonitrile (Structure 2).

The Blocked Mechanism

Moving to reaction R, we are dealing with a ketoxime. Notice the methyl group replacing the aldehydic proton. Just like in reaction P, the $-OH$ group is syn to the bromine.

When we add $NaOH$, the exact same intramolecular $S_{N}Ar$ cyclization occurs. The oxygen attacks, bromine leaves, and we form the fused benzisoxazole ring. However, here is the catch: why doesn't this ring open up like it did in reaction P?

Look closely at the $C_3$ position of the benzisoxazole ring. In reaction P, it held an acidic proton. Here, it holds a methyl group! There is no acidic proton for the base to abstract. The ring-opening mechanism is physically blocked. The reaction halts at the stable 5-nitro-3-methyl-1,2-benzisoxazole (Structure 4).

The Anti-Geometry

Finally, in reaction S, we have an O-acetyl ketoxime. But observe the stereochemistry carefully: the acetate group is pointing away from the bromine. It is anti.

When treated with aqueous $N{a}_{2}C{O}_3$, the mild base simply hydrolyzes the ester back into an oxime. Because the original geometry was anti, the resulting oxime $-OH$ is also anti to the bromine. It is physically too far away to attack the ring! Cyclization is geometrically impossible. The reaction simply yields the uncyclized anti-oxime (Structure 5).

Summarizing our mechanistic journey: - P yields the ring-opened nitrile (1). - Q yields the simple elimination product (2). - R yields the stable benzisoxazole (4). - S yields the uncyclized anti-oxime (5).

Matching these up, we get $P\to 1$, $Q\to 2$, $R\to 4$, and $S\to 5$. This perfectly aligns with Option (B). A brilliant problem that beautifully intertwines stereochemistry with reaction mechanisms!