4-Oxobutanoic acid is reduced with Na-borohydride and the product is treated with aqueous acid. The final product is :

Options

A.
B.
C.
D.

Answer

To solve the problem step by step, we will analyze the reduction of 4-oxobutanoic acid with sodium borohydride (NaBH4) and the subsequent treatment with aqueous acid.

### Step 1: Understand the structure of 4-Oxobutanoic Acid
4-Oxobutanoic acid has the following structure:
- It is a four-carbon chain (butanoic acid) with a ketone group (C=O) at the 4th carbon.
- The structure can be represented as:
```
HOOC-CH2-CH2-C(=O)-H
```
where the carbonyl group (C=O) is at the 4th position.

### Step 2: Reduction with Sodium Borohydride (NaBH4)
When 4-oxobutanoic acid is treated with sodium borohydride, the ketone group (C=O) is reduced to an alcohol (C-OH). The reaction can be summarized as follows:
- The ketone group is reduced to an alcohol, resulting in the formation of 4-hydroxybutanoic acid:
```
HOOC-CH2-CH2-CH(OH)-H
```

### Step 3: Treatment with Aqueous Acid
After the reduction, the product (4-hydroxybutanoic acid) is treated with aqueous acid. In this step:
- The hydroxyl group (OH) can be protonated due to the acidic environment, making it a better leaving group.
- The protonation leads to the formation of water (H2O) and the carbon atom that was attached to the hydroxyl group can form a double bond with the adjacent carbon atom.

### Step 4: Final Product Formation
The final product after the treatment with aqueous acid will be:
- The compound will lose the water molecule, resulting in the formation of a double bond between the second and third carbon atoms:
```
HOOC-CH=CH-COH
```
- The final product is 4-hydroxybut-2-enoic acid.

### Conclusion
The final product after the reduction of 4-oxobutanoic acid with NaBH4 and subsequent treatment with aqueous acid is 4-hydroxybut-2-enoic acid.

### Final Answer
The final product is 4-hydroxybut-2-enoic acid.

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