Degree of Unsaturation of Colored Product from Reaction of Compound P

To determine the degree of unsaturation (also known as the Index of Hydrogen Deficiency, IHD) of compound $\mathbf{Q}$, we analyze the structure of the starting compound $\mathbf{P}$ and the chemical transformation taking place in the presence of concentrated $\text{H}_2\text{SO}_4$.

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Step 1: Degree of Unsaturation of Starting Compound $\mathbf{P}$

Compound $\mathbf{P}$ is $1$-hydroxy-$2,4,6$-triphenylcyclohexa-$2,5$-dienecarboxylic acid methyl ester. Its structure consists of:

1. Three phenyl rings ($-\text{C}_6\text{H}_5$): Each phenyl ring contains $1$ ring and $3$ double bonds. $\text{DU}_{\text{phenyl}} = 3 \times (1 + 3) = 12$

2. Central 6-membered ring: Contains $1$ ring and $2$ endocyclic double bonds. $\text{DU}_{\text{central ring}} = 1 + 2 = 3$

3. Ester group ($-\text{CO}_2\text{CH}_3$): Contains $1$ carbonyl double bond ($\text{C}=\text{O}$). $\text{DU}_{\text{ester}} = 1$

Summing these contributions gives the total degree of unsaturation for compound $\mathbf{P}$: $\text{DU}(\mathbf{P}) = 12 + 3 + 1 = 16$

Alternatively, using the molecular formula of $\mathbf{P}$, $\text{C}_{26}\text{H}_{22}\text{O}_3$: $\text{DU}(\mathbf{P}) = C + 1 - \frac{H}{2} = 26 + 1 - \frac{22}{2} = 16$

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Step 2: Reaction Mechanism with Concentrated $\text{H}_2\text{SO}_4$

When compound $\mathbf{P}$ is treated with concentrated $\text{H}_2\text{SO}_4$, it undergoes a series of acid-catalyzed ionic transformations:

1. Protonation and Dehydration: The hydroxyl group ($-\text{OH}$) at $\text{C}_1$ is protonated to form a good leaving group ($-\text{OH}_2^+$). Loss of a water molecule ($\text{H}_2\text{O}$) generates a highly resonance-stabilized carbocation at $\text{C}_1$.

   • Elimination of $\text{H}_2\text{O}$ increases the degree of unsaturation by $+1$.

2. First Intramolecular Cyclization: The carbocation at $\text{C}_1$ undergoes an intramolecular electrophilic aromatic substitution (EAS) attack onto the ortho-position of an adjacent phenyl group (at $\text{C}_2$). Subsequent loss of $\text{H}^+$ restores aromaticity, yielding a fused 5-membered ring (fluorene derivative).

   • Formation of this new ring increases the degree of unsaturation by $+1$.

3. Acylium Ion Formation and Second Cyclization: The methyl ester group ($-\text{CO}_2\text{CH}_3$) at $\text{C}_1$ gets protonated under strongly acidic conditions and loses methanol ($\text{CH}_3\text{OH}$) to form an acylium ion ($-\text{C}^+=\text{O}$). This acylium ion undergoes a second intramolecular electrophilic aromatic substitution onto the ortho-position of the phenyl group at $\text{C}_6$, forming a fused polycyclic aromatic ketone.

   • Formation of this second fused ring increases the degree of unsaturation by $+1$.

4. Aromatization to Form Colored Product $\mathbf{Q}$: The extended polycyclic system undergoes final aromatization/conjugation to form the deeply colored product $\mathbf{Q}$ with formula $\text{C}_{25}\text{H}_{14}\text{O}$.

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Step 3: Calculation of Degree of Unsaturation of Compound $\mathbf{Q}$

Using the molecular formula of the final product $\mathbf{Q}$ ($\text{C}_{25}\text{H}_{14}\text{O}$): $\text{DU}(\mathbf{Q}) = C + 1 - \frac{H}{2}$ $\text{DU}(\mathbf{Q}) = 25 + 1 - \frac{14}{2} = 26 - 7 = 19$

Or directly from the change in unsaturation during the reaction: $\text{DU}(\mathbf{Q}) = \text{DU}(\mathbf{P}) + 3 = 16 + 3 = 19$

The degree of unsaturation present in compound $\mathbf{Q}$ is 19.