The vapure of a carboxylic acid `HA` when passed over `MnO` at 573 K`yields Propanone. The acid `HA` is

Options

A. Methanoic acid
B. Ethanoic acid
C. Propanoic acid
D. Butanoic acid

Answer

To determine the carboxylic acid \( HA \) that yields propanone when passed over manganese oxide (\( MnO \)) at 573 K, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- We know that the reaction involves a carboxylic acid \( HA \) being converted into propanone (\( CH_3COCH_3 \)) when passed over \( MnO \) at a specific temperature (573 K).

2. **Identifying the Product**:
- The product of the reaction is propanone, which is a ketone with the structure \( CH_3COCH_3 \).

3. **Analyzing the Options**:
- We have four options for the carboxylic acid:
1. Mythenoic acid
2. Ethenoic acid
3. Propenoic acid
4. Biotenoic acid

4. **Determining the Structure of Ethenoic Acid**:
- Ethenoic acid (also known as acetic acid) has the formula \( CH_2=COOH \) and can be represented as \( CH_3COOH \).

5. **Reaction of Ethenoic Acid with \( MnO \)**:
- When ethenoic acid is heated with \( MnO \), it undergoes oxidation. The reaction can be represented as:
\[ CH_3COOH \xrightarrow{MnO, 573 K} CH_3COCH_3 + CO_2 + H_2O \]
- This reaction shows that ethenoic acid yields propanone, carbon dioxide, and water.

6. **Conclusion**:
- Since the reaction of ethenoic acid produces propanone, we conclude that the carboxylic acid \( HA \) is **ethenoic acid**.

### Final Answer:
The acid \( HA \) is **ethenoic acid**.