Compound (A) `C_(5)H_(8)O_(2)` liberated `CO_(2)` on reaction with sodium bicarbonate. It exists in two forms neither of which is optically active. It yielded compound (B) `C_(5)H_(10)O_(2)` on hydrogenation. Compound (B) can be separated into enantiomorphs. Write structures of (A) and (B).

Answer

To solve the problem, we need to identify the structures of compounds (A) and (B) based on the information provided.

### Step 1: Analyze Compound (A)
The molecular formula of compound (A) is C₅H₈O₂. It liberates CO₂ when reacted with sodium bicarbonate (NaHCO₃), indicating that it likely contains a carboxylic acid group. Carboxylic acids react with sodium bicarbonate to produce carbon dioxide.

### Step 2: Determine the Structure of Compound (A)
Given that compound (A) is C₅H₈O₂ and is a carboxylic acid, we can propose that it has a double bond (alkene) in addition to the carboxylic acid functional group. The presence of two forms that are not optically active suggests that it could exist as cis and trans isomers.

1. **Cis Structure**:
- The structure can be drawn as follows:
```
CH3
|
C=C
| \
COOH CH3
```
This is 2-methylbut-2-enoic acid (cis form).

2. **Trans Structure**:
- The trans structure would be:
```
CH3 COOH
| |
C=C
| \
CH3 H
```
This is 2-methylbut-2-enoic acid (trans form).

### Step 3: Write the IUPAC Name for Compound (A)
The IUPAC name for compound (A) is **2-methylbut-2-enoic acid**.

### Step 4: Analyze Compound (B)
Compound (B) has the molecular formula C₅H₁₀O₂. It is produced from compound (A) through hydrogenation, which means that the double bond in compound (A) will be converted into a single bond, resulting in a saturated compound.

### Step 5: Determine the Structure of Compound (B)
Upon hydrogenation of compound (A), we add hydrogen (H₂) across the double bond:
- The structure of compound (B) after hydrogenation can be drawn as:
```
CH3
|
CH3-C-COOH
|
CH2
```
This structure corresponds to 2-methylbutanoic acid.

### Step 6: Write the IUPAC Name for Compound (B)
The IUPAC name for compound (B) is **2-methylbutanoic acid**. This compound can exist as enantiomers due to the presence of a chiral center.

### Summary of Structures
- **Compound (A)**:
- Structure:
- Cis:
```
CH3
|
C=C
| \
COOH CH3
```
- Trans:
```
CH3 COOH
| |
C=C
| \
CH3 H
```
- IUPAC Name: **2-methylbut-2-enoic acid**

- **Compound (B)**:
- Structure:
```
CH3
|
CH3-C-COOH
|
CH2
```
- IUPAC Name: **2-methylbutanoic acid**