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Q.

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S→(2) Pyridine, ∆(1) I2NaHCO3

Find out the major product for the above reaction.

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a

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b

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c

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d

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answer is C.

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Detailed Solution

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  • Step 1: Haloform Reaction with Iodine (I₂) and NaHCO₃
    • The starting compound is a carboxylic acid derivative attached to a hydroxyl group (−OH) and a carbonyl group (C=O) on a cyclic structure.
    • The first reagent, I₂ (iodine) in the presence of NaHCO₃ (sodium bicarbonate), initiates a haloform reaction. In this reaction:
      • The hydroxyl group (−OH) on the carbon adjacent to the carboxyl group is oxidized by iodine (I₂) to form a haloform intermediate, where the hydrogen from the −OH group is replaced by iodine.
      • The reaction creates an iodo intermediate, which is crucial for the next step in the reaction.

 

  • Step 2: Decarboxylation with Pyridine (Py) and Heat
    • The intermediate undergoes decarboxylation when treated with pyridine (Py) and heat (Δ).
    • Pyridine acts as a base, and under heat, it promotes the loss of the carboxyl group (−COOH) as carbon dioxide (CO₂), leading to the formation of a substituted cyclohexanone.
    • This is a common reaction where a carboxyl group is eliminated via a cyclization mechanism, resulting in a cyclohexanone derivative with an iodine atom attached at a specific position.
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