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Q.

S $\to_{(2) Pyridine, ∆}^{(1) I_{2} {NaHCO}_{3}}$
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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