Reaction Sequence and Structural Identification of Unsaturated Hydrocarbon P

To determine the correct option(s), we analyze the step-by-step chemical transformations of compound $\mathbf{P}$.

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Step 1: Identification of the Carbon Skeleton and Degree of Unsaturation of Compound $\mathbf{P}$

1. Catalytic Hydrogenation: Compound $\mathbf{P}$ ($\text{C}_9\text{H}_{12}$) undergoes complete hydrogenation over $\text{Pt}/\text{H}_2$ to yield propylcyclohexane ($\text{C}_9\text{H}_{18}$).

   • Molecular formula of propylcyclohexane: $\text{C}_9\text{H}_{18}$ ($\text{Degree of Unsaturation, DU} = 1$, corresponding to one cyclohexane ring).
   • Molecular formula of compound $\mathbf{P}$: $\text{C}_9\text{H}_{12}$ $\text{DU of } \mathbf{P} = 9 + 1 - \frac{12}{2} = 4$
   • Since hydrogenation yields propylcyclohexane, $\mathbf{P}$ contains 1 cyclohexane ring and 3 $\pi$-bonds in total (or 1 double bond + 1 triple bond).

2. Reaction with $\text{NaNH}_2$: $\mathbf{P}$ reacts with $\text{NaNH}_2$ to form an acetylide ion, confirming the presence of a terminal alkyne group ($-\text{C}\equiv\text{CH}$).

   Thus, $\mathbf{P}$ consists of a cyclohexene ring with a propargyl side chain ($-\text{CH}_2-\text{C}\equiv\text{CH}$).

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Step 2: Analysis of Reagent $\mathbf{X}$ and Compound $\mathbf{Q}$

1. Reagent $\mathbf{X}$: The terminal alkyne group in $\mathbf{P}$ is selectively reduced to an alkene group ($-\text{CH}_2-\text{CH}=\text{CH}_2$) without affecting the ring double bond. This selective reduction of an alkyne to a cis-alkene is achieved using Lindlar's catalyst: $\mathbf{X} = \text{Pd-C/quinoline/H}_2$ Hence, Option (B) is correct.

2. Oxidative Cleavage to form $\mathbf{Q}$ ($\text{C}_8\text{H}_{12}\text{O}_6$, Optically Active):

   • The diene intermediate produced after treatment with $\mathbf{X}$ is subjected to strong oxidative cleavage using $\text{KMnO}_4/\text{H}_2\text{SO}_4/\Delta$.
   • The terminal double bond ($-\text{CH}_2-\text{CH}=\text{CH}_2$) is cleaved into a carboxylic acid group ($-\text{CH}_2-\text{COOH}$) with the loss of one carbon as $\text{CO}_2\uparrow$.
   • The double bond in the 6-membered ring is also oxidatively cleaved, opening the ring to generate two additional $-\text{COOH}$ groups.
   • The product $\mathbf{Q}$ is a tricarboxylic acid with 8 carbon atoms ($\text{C}_8\text{H}_{12}\text{O}_6$).

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Step 3: Determining the Position of the Double Bond in $\mathbf{P}$ (Option A vs. Option C)

Let's evaluate the two potential positional isomers for $\mathbf{P}$:

Case 1: 3-(prop-2-yn-1-yl)cyclohex-1-ene (Allylic substitution - Option A)

• Position of side chain: C3 relative to the ring double bond ($\text{C1}=\text{C2}$).
• After reduction with $\mathbf{X}$ and oxidative cleavage with $\text{KMnO}_4/\text{H}_2\text{SO}_4/\Delta$:
  • Ring double bond between C1 and C2 is cleaved, converting C1 and C2 into $-\text{COOH}$ groups.
  • Side chain is converted to $-\text{CH}_2-\text{COOH}$.
• At the C3 carbon, the four attached groups are:
  1. $-\text{H}$
  2. $-\text{COOH}$ (from C2)
  3. $-\text{CH}_2-\text{COOH}$ (from the side chain)
  4. $-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{COOH}$ (from C4–C5–C6–C1)

Since all four attached groups are distinct, C3 is a chiral center. Therefore, compound $\mathbf{Q}$ is optically active, which matches the given experimental condition.

Case 2: 4-(prop-2-yn-1-yl)cyclohex-1-ene (Homoallylic substitution - Option C)

• Position of side chain: C4 relative to the ring double bond ($\text{C1}=\text{C2}$).
• After oxidative cleavage, the groups attached to the C4 carbon are:
  1. $-\text{H}$
  2. $-\text{CH}_2-\text{COOH}$ (from the side chain)
  3. $-\text{CH}_2-\text{COOH}$ (from the C3–C2 path)
  4. $-\text{CH}_2-\text{CH}_2-\text{COOH}$ (from the C5–C6–C1 path)

Here, two of the attached groups are identical ($-\text{CH}_2-\text{COOH}$). Thus, C4 is achiral, making $\mathbf{Q}$ optically inactive, which contradicts the given observation.

Hence, Option (A) is correct and Option (C) is incorrect.

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Step 4: Formation and Identification of Compound $\mathbf{R}$ (Option D)

The sequence of reactions to form $\mathbf{R}$ is as follows:

1. Deprotonation: $\mathbf{P} \xrightarrow{\text{NaNH}_2} \text{Cyclohexenyl}-\text{CH}_2-\text{C}\equiv\text{C}^- \text{Na}^+$
2. Nucleophilic Addition to Acetophenone: $\text{Cyclohexenyl}-\text{CH}_2-\text{C}\equiv\text{C}^- + \text{C}_6\text{H}_5\text{COCH}_3 \rightarrow \text{Cyclohexenyl}-\text{CH}_2-\text{C}\equiv\text{C}-\text{C}(\text{OH})(\text{CH}_3)\text{C}_6\text{H}_5$
3. Acid-Catalyzed Dehydration ($\text{H}_3\text{O}^+/\Delta$): Elimination of water yields an conjugated enyne derivative: $\text{Cyclohexenyl}-\text{CH}_2-\text{C}\equiv\text{C}-\text{C}(=\text{CH}_2)\text{C}_6\text{H}_5$

Since the chiral carbon on the cyclohexene ring remains intact, compound $\mathbf{R}$ is optically active. This structure corresponds precisely to the structure shown in Option (D).

Hence, Option (D) is correct.

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Conclusion

• Option A: Correct ($\mathbf{P}$ is 3-(prop-2-yn-1-yl)cyclohex-1-ene)
• Option B: Correct ($\mathbf{X}$ is $\text{Pd-C/quinoline/H}_2$)
• Option C: Incorrect
• Option D: Correct ($\mathbf{R}$ is the conjugated enyne product)

Correct Options: A, B, D