Question

"P" is a hydrocarbon of molecular formula:- \(C_8H_{14}\). On ozonolysis, "P" forms "Q". "Q" on treatment with alkali under reflux condition produces "R", which on treatment with \(I_2 / NaOH\) gives a yellow precipitate. Acidification of the solution gives "S". The structure of "S" is given below:-}

• A. A
• B. B
• C. C
• D. D

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This is a roadmap problem. We need to identify the starting hydrocarbon \(P\) (\(C_8H_{14}\)) that eventually leads to a specific cyclopentene carboxylic acid \(S\) through ozonolysis, intramolecular aldol condensation, and an iodoform reaction.
Step 2: Key Formula or Approach:
1. Ozonolysis of a cyclic alkene typically opens the ring.
2. Alkali/reflux suggests intramolecular aldol condensation if a dialdehyde or diketone is formed.
3. Iodoform test (\(I_2/NaOH\)) confirms the presence of a methyl ketone (\(-COCH_3\)).
Step 3: Detailed Explanation:
Assume \(P\) is 1,2-dimethylcyclohexene (Option D).
1. Ozonolysis of P: Cleaving the double bond in 1,2-dimethylcyclohexene gives octane-2,7-dione (\(Q\)):
\[ CH_3-CO-(CH_2)_4-CO-CH_3 \]
2. Intramolecular Aldol of Q: In the presence of alkali, one of the methyl groups of the diketone forms a carbanion and attacks the other carbonyl. Closing a 5-membered ring is kinetically favorable:
\[ \text{octane-2,7-dione} \xrightarrow{OH^-/\Delta} \text{1-acetyl-2-methylcyclopentene (R)} \]
3. Iodoform reaction on R: The acetyl group (\(-COCH_3\)) in \(R\) reacts with \(I_2/NaOH\) to form a yellow precipitate (\(CHI_3\)) and a carboxylate salt.
4. Acidification: The salt is converted to the free acid, 2-methylcyclopent-1-ene-1-carboxylic acid (S). This matches the structure provided in the prompt.
Step 4: Final Answer:
The correct structure of "P" is 1,2-dimethylcyclohexene.