Which is the most suitable reagent for the following conversion ?
` CH_3 -CH=CH -CH_2 - overset(O ) overset(||)C-CH_3 to CH_3 -CH=CH-CH_2 - overset(O ) overset(||)C-OH `

Options

A. Tollen's reagent
B. Benzoyl peroxide
C. `I_2` and NaOH solution
D. Sn and NaOH solution

Answer

To solve the given question, we need to identify the most suitable reagent for converting the compound \( CH_3 - CH = CH - CH_2 - C(=O) - CH_3 \) (a methyl ketone) into \( CH_3 - CH = CH - CH_2 - C(=O) - OH \) (a carboxylic acid).

### Step-by-Step Solution:

1. **Identify the Functional Groups**:
- The starting compound is a methyl ketone due to the presence of the \( C(=O) \) group attached to a methyl group.
- The product is a carboxylic acid, which indicates that the ketone group has been converted to a carboxylic acid.

2. **Recognize the Reaction Type**:
- The conversion from a methyl ketone to a carboxylic acid is characteristic of the haloform reaction. This reaction occurs when a methyl ketone is treated with halogens (like iodine) in the presence of a base (like sodium hydroxide).

3. **Write the Haloform Reaction**:
- In the haloform reaction, the methyl group of the methyl ketone is converted into a haloform (like iodoform, \( CHI_3 \)), and the ketone is oxidized to a carboxylic acid.
- The general reaction can be represented as:
\[ RCOCH_3 + 3I_2 + 4NaOH \rightarrow RCOONa + CHI_3 + 3NaI + 3H_2O \]
- Here, \( R \) represents the rest of the carbon chain.

4. **Select the Reagents**:
- For our specific reaction, we can use iodine (\( I_2 \)) as the halogen and sodium hydroxide (\( NaOH \)) as the base.
- Therefore, the suitable reagents for this conversion are \( I_2 \) and \( NaOH \).

5. **Conclusion**:
- The most suitable reagent for the conversion of the given methyl ketone to the corresponding carboxylic acid is iodine in the presence of sodium hydroxide.

### Final Answer:
The most suitable reagent for the conversion is \( I_2 \) (iodine) in the presence of \( NaOH \) (sodium hydroxide).

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