`CH_3 CH=CHCHO` is oxidised to `CH_3 CH=CHCOOH` using :

Options

A. Alkaline `KMnO_4`
B. Selenium dioxide
C. Ammoniacal `AgNO_3`
D. All

Explanation

To solve the problem of oxidizing the compound `CH₃CH=CHCHO` to `CH₃CH=CHCOOH`, we need to identify the appropriate oxidizing agent that will selectively oxidize the aldehyde group without affecting the double bond.

### Step-by-Step Solution:

1. **Identify the Functional Groups**:
- The starting compound `CH₃CH=CHCHO` is an aldehyde with an alkene (double bond) present. The target compound `CH₃CH=CHCOOH` is an alpha, beta-unsaturated carboxylic acid.

2. **Understand the Reaction Requirements**:
- We need an oxidizing agent that can convert the aldehyde (−CHO) group to a carboxylic acid (−COOH) without affecting the alkene (C=C) double bond.

3. **Evaluate Possible Oxidizing Agents**:
- **Alkaline KMnO₄**: This reagent is a strong oxidizer that will oxidize both the aldehyde and the double bond. Therefore, it is not suitable for our reaction because it would lead to over-oxidation and produce acetic acid instead.
- **Selenium Dioxide (SeO₂)**: Similar to KMnO₄, selenium dioxide can also oxidize both the aldehyde and the double bond, resulting in over-oxidation. Thus, it is also not suitable.
- **Ammonical Silver Nitrate (Tollens' Reagent)**: This reagent selectively oxidizes aldehydes to carboxylic acids without affecting the double bond. Therefore, it is the appropriate choice for our reaction.

4. **Conclusion**:
- The suitable oxidizing agent for converting `CH₃CH=CHCHO` to `CH₃CH=CHCOOH` is ammonical silver nitrate.

### Final Answer:
The compound `CH₃CH=CHCHO` is oxidized to `CH₃CH=CHCOOH` using ammonical silver nitrate.

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