`(CH_(3))_(2)C=CHCOCH_(3)` can be oxidised to `(CH_(3))_(2) C=CHCOOH` by

Options

A. Chromic acid
B. `NaOl`
C. `Cu " at" 300^(@)` C
D. `KMnO_(4)`

Answer

To solve the question of how to oxidize `(CH₃)₂C=CHCOCH₃` to `(CH₃)₂C=CHCOOH`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Functional Groups**:
The compound `(CH₃)₂C=CHCOCH₃` contains a double bond (alkene) and a carbonyl group (ketone) in the form of `COCH₃`.

2. **Recognize the Reaction Type**:
The conversion from a ketone to a carboxylic acid indicates an oxidation reaction. Specifically, we are looking to oxidize the ketone group (`COCH₃`) to a carboxylic acid group (`COOH`).

3. **Choose the Oxidizing Agent**:
The iodoform reaction is a specific type of haloform reaction that can be used to oxidize methyl ketones to carboxylic acids. The reagents typically used for this reaction are iodine (I₂) and sodium hydroxide (NaOH), or sodium hypoiodite (NaOI).

4. **Perform the Reaction**:
By treating the compound `(CH₃)₂C=CHCOCH₃` with sodium hypoiodite (NaOI), the methyl ketone group will be oxidized to a carboxylic acid. The overall reaction can be summarized as:
\[ (CH₃)₂C=CHCOCH₃ + NaOI \rightarrow (CH₃)₂C=CHCOOH + Iodoform \]

5. **Conclusion**:
The product of this reaction is `(CH₃)₂C=CHCOOH`, which is the desired carboxylic acid.

### Final Answer:
The compound `(CH₃)₂C=CHCOCH₃` can be oxidized to `(CH₃)₂C=CHCOOH` by using sodium hypoiodite (NaOI) in an iodoform reaction.

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