Animated Solution for Chemistry - Organic Chemistry: In the following reaction, compound Q is obtained from compound P via an ionic intermediate What is the degree of unsaturation of Q ?

Unraveling the Fluorenone Core

Let's embark on a fascinating journey through an elegant organic transformation. We begin by carefully examining our starting material, Compound P, which is methyl 2,4,6-triphenylbenzoate. Notice the central benzene ring. It is heavily substituted, bearing an ester group at the top, and three bulky phenyl rings occupying the ortho, ortho, and para positions. This intense steric crowding is not just a structural detail; it is going to play a crucial role in dictating the pathway of the reaction.

Now, we introduce concentrated sulfuric acid (${\text{H}}_2{\text{SO}}_4$). What is its primary function here? It acts as a powerful source of protons and a strong dehydrating agent. In an organic molecule with multiple oxygen atoms, the acid will hunt for the most basic site to protonate. Here, the ester group is the prime target.

Once the ester oxygen is protonated, it transforms into an excellent leaving group. Under these strongly acidic and dehydrating conditions, a molecule of methanol (${\text{CH}}_3\text{OH}$) departs. This departure leaves behind a positively charged carbon, creating an acylium ion ($\text{-C}\equiv {\text{O}}^{+}$). Look closely at that carbon—it is highly electron-deficient and desperately wants electrons. It has become a super electrophile, ready to react with any available electron source.

So, where will this acylium ion get its electrons? It doesn't have to look far. Right next door, at the ortho position, sits an electron-rich phenyl ring. This perfect spatial proximity sets the stage for an intramolecular Friedel-Crafts acylation.

The $\pi$ electrons of the adjacent phenyl ring swing out and attack the acylium carbon. This nucleophilic attack closes a new ring. Because the attacking carbon was ortho to the original attachment point, we form a new five-membered ring containing a ketone group. This beautiful, fused tricyclic system is known as a fluorenone core. Notice that the other two phenyl rings (at the other ortho and para positions) remain completely untouched. This newly formed, highly conjugated system is our colored Compound Q.

The Degree of Unsaturation

Now for the final, critical question: What is the degree of unsaturation, or Double Bond Equivalent (DBE), of Compound Q? Let's break it down systematically to avoid any double-counting errors.

First, let's analyze the fluorenone core itself: - It contains 2 Benzene rings, each contributing a DBE of 4 (1 ring + 3 $\pi$ bonds). That gives us $2\times 4=8$. - There is 1 Cyclopentane ring in the center, adding 1 to the DBE. - There is 1 Ketone double bond ($\text{C=O}$), adding another 1. - Therefore, the core alone has a DBE of $8+1+1=10$.

But we are not done yet! We must not forget the two isolated phenyl groups hanging off the side of the original central ring. Each of these phenyl rings has a DBE of 4. Since there are two of them, that's an additional $2\times 4=8$.

Adding it all together, $10\text{ (core)}+8\text{ (substituents)}$ gives us a total degree of unsaturation of 18. A perfect, elegant integer answer derived from a beautiful molecular rearrangement.