Animated Solution for Chemistry - Organic Chemistry: In the reaction given below, the total number of atoms having $s{p}^2$ hybridization in the major product P is ______.

Let's carefully analyze the structure of our starting material. We are presented with a fascinating tricyclic system. At its core lies a central six-membered ring, specifically a cyclohexane-1,4-dione. Fused to the sides of this central ring are two highly strained cyclobutene rings.

Visualizing this 3D structure is crucial because the position of the double bonds dictates the outcome of the first reaction. The double bonds are located on the far edges of the cyclobutene rings, opposite to the fusion carbons.

Ozonolysis

The first step in our reaction sequence is ozonolysis, followed by a reductive workup using zinc and water ($O_3$ then $Zn/H_{2}O$). Ozonolysis acts like a pair of chemical scissors, specifically targeting and cleaving carbon-carbon double bonds.

When these cyclobutene rings are cleaved, the strain is released, and the rings open up. Each carbon atom that was part of the broken double bond receives an oxygen atom, transforming into an aldehyde (formyl) group. Because we have two cyclobutene rings, this cleavage yields four new $-CHO$ groups attached to the central cyclohexane ring.

Our intermediate now boasts a total of six carbonyl groups: the four newly formed formyl groups and the two original ketone groups from the cyclohexane-1,4-dione core.

Next, we treat this intermediate with an excess of hydroxylamine ($N{H}_{2}OH$). Hydroxylamine is a classic reagent that reacts with both aldehydes and ketones to form oximes. The reaction involves a nucleophilic attack followed by dehydration, effectively replacing the carbon-oxygen double bond ($>C=O$) with a carbon-nitrogen double bond ($>C=N-OH$).

Since we are using an excess of the reagent, we assume complete conversion. All six carbonyl groups are transformed into oxime groups. This gives us our major product, P, which is a hexa-oxime derivative.

Counting the $s{p}^2$ Atoms

Now, we need to count the $s{p}^2$ hybridized atoms in this molecule. Let's examine a single oxime group ($>C=N-OH$). The carbon and the nitrogen atoms are connected by a double bond, meaning they both have a steric number of 3 and are $s{p}^2$ hybridized. The oxygen atom, however, has two single bonds and two lone pairs, making it $s{p}^3$ hybridized.

With six oxime groups in our product P, we simply multiply:

The Catch: Why was 8 also accepted? You might wonder why the official JEE answer key accepted both 8 and 12. The two central ketone groups are flanked by bulky substituents, making them highly sterically hindered. If one assumes these ketones fail to react with hydroxylamine and instead form hydrates ($>C(OH{)}_2$) in the aqueous medium, their $s{p}^2$ atom count drops to zero. In this scenario, only the four unhindered formyl groups form oximes, yielding $4\times 2=8$ $s{p}^2$ atoms. However, 12 remains the most chemically rigorous answer for complete theoretical conversion.