Total number of optically stereoisomers of `CH_(3)-underset(overset(|)(CI))(C)H-underset(overset(|)(CI))(C)H-underset(overset(|)(CI))(C)H-CH_(3)`

Options

A. 2
B. 4
C. 6
D. 8

Answer

To determine the total number of optically active stereoisomers for the compound `CH₃-CH(CI)-CH(CI)-CH₃`, we will follow these steps:

### Step 1: Identify the Structure
The given compound can be represented as:
```
CH₃
|
CH - CI
|
CH - CI
|
CH₃
```
This structure indicates that there are multiple carbon atoms, some of which are bonded to chlorine (Cl) atoms.

### Step 2: Identify Chiral Centers
Chiral centers (or stereogenic centers) are carbon atoms that are bonded to four different groups. In this compound, we need to analyze each carbon atom:

1. The two terminal carbons (attached to CH₃ groups) are not chiral because they are bonded to three hydrogen atoms.
2. The two central carbons (each bonded to a Cl, a hydrogen, and a CH group) are chiral centers because they are bonded to four different groups:
- For the first chiral center: Cl, H, and the two CH groups.
- For the second chiral center: Cl, H, and the two CH groups.

### Step 3: Count the Chiral Centers
From the analysis, we find that there are **2 chiral centers** in the compound.

### Step 4: Use the Formula for Optical Stereoisomers
The number of optically active stereoisomers can be calculated using the formula:
\[ \text{Number of stereoisomers} = 2^n \]
where \( n \) is the number of chiral centers.

Substituting the value of \( n \):
\[ \text{Number of stereoisomers} = 2^2 = 4 \]

### Conclusion
Thus, the total number of optically active stereoisomers for the compound is **4**.

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