The Iodoform test is specific for \(CH_3-CO-\) or \(CH_3-CH(OH)-\) groups. However, in conjugated systems, haloform-like oxidative cleavage can occur, leading to the formation of carboxylic acids.
Step 1: Understanding the Concept:
The sequence involves:
1. Base-catalyzed Aldol condensation between a cyclic ketone and an aldehyde.
2. Iodoform reaction (\(I_2/NaOH\)) followed by acidification. The iodoform reaction typically converts methyl ketones (\(R-CO-CH_3\)) into carboxylic acids (\(R-COOH\)) and \(CHI_3\). Step 2: Key Formula or Approach:
1. Identify the acidic \(\alpha\)-hydrogens in 2,2-dimethylcyclopentanone.
2. Conduct the Aldol condensation with acetaldehyde (\(CH_3CHO\)).
3. Apply oxidative cleavage/iodoform logic to the product 'P'. Step 3: Detailed Explanation:
1. Aldol Condensation: 2,2-dimethylcyclopentanone has \(\alpha\)-protons only at the C-5 position. These react with \(CH_3CHO\) in the presence of \(NaOH\) to form 5-ethylidene-2,2-dimethylcyclopentanone after dehydration.
2. Iodoform Reaction: The ethylidene group (\(=CH-CH_3\)) in \(\alpha,\beta\)-unsaturated ketones can undergo oxidative cleavage under iodoform conditions (\(I_2/NaOH\)) if it can tautomerize or be further oxidized. Specifically, if the reaction produces a \(\beta\)-hydroxy ketone intermediate that can be cleaved, or if the system mimics a methyl ketone structure, a carboxylic acid is formed on the ring.
3. Workup: The resulting sodium salt is acidified by \(HCl\) to the free carboxylic acid. Step 4: Final Answer:
The final product 'X' is the cyclopentanone ring with a carboxylic acid group at the 5th position, as shown in Option (A).
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