Number of meso forms and recemic forms are possible in the following structure is `CH_(3)COCH_(2)CO_(2)H`

Options

A. 2,0
B. 0,4
C. 0,2
D. 0,0

Explanation

To determine the number of meso forms and racemic forms possible for the compound `CH3COCH2COOH`, we will analyze its structure step by step.

### Step 1: Identify the structure
The given structure is `CH3COCH2COOH`. Let's break it down:
- The compound has four carbon atoms.
- The functional groups present are a carbonyl (C=O) and a carboxylic acid (-COOH).

### Step 2: Check for chiral centers
A chiral center is a carbon atom that is bonded to four different groups. We need to examine each carbon in the structure:
1. **C1 (CH3)**: This carbon is bonded to three hydrogen atoms and one carbon (C2). Not chiral.
2. **C2 (C=O)**: This carbon is bonded to one carbon (C1), one carbon (C3), and one oxygen (O). Not chiral.
3. **C3 (CH2)**: This carbon is bonded to one carbon (C2), one carbon (C4), and two hydrogen atoms. Not chiral.
4. **C4 (COOH)**: This carbon is bonded to one carbon (C3), one oxygen (O), and one hydroxyl group (OH). Not chiral.

Since none of the carbons in the molecule are chiral, we conclude that there are **no chiral centers**.

### Step 3: Determine the presence of meso forms
A meso compound has a plane of symmetry, which means that it can be superimposed on its mirror image. Since we have established that there are no chiral centers, there can be no meso forms.

### Step 4: Determine the presence of racemic forms
Racemic forms arise from the presence of enantiomers, which are non-superimposable mirror images of each other. Since there are no chiral centers in the molecule, it cannot form enantiomers. Therefore, there are no racemic forms.

### Conclusion
Based on the analysis:
- **Number of meso forms**: 0
- **Number of racemic forms**: 0

Thus, the final answer is:
- **Zero meso forms and zero racemic forms.**