Total number of optically active stereoisomers of `CH_(3)-underset(overset(|)(CI))(C)H-CH_(2)-CH_(2)-underset(overset(|)(CI))(C)H-CH_(3)` are :

Answer

To determine the total number of optically active stereoisomers of the compound `CH3-CH(Cl)-CH2-CH2-CH(Cl)-CH3`, we will follow these steps:

### Step 1: Identify the chiral centers
A chiral center is a carbon atom that is bonded to four different groups. In the given compound, we will analyze each carbon atom to see if it meets this criterion.

- The first carbon (C1) is bonded to three hydrogens and one carbon (C2), so it is not chiral.
- The second carbon (C2) is bonded to a chlorine (Cl), a hydrogen (H), a carbon (C3), and another carbon (C1), making it a chiral center.
- The third carbon (C3) is bonded to two hydrogens and two carbons (C2 and C4), so it is not chiral.
- The fourth carbon (C4) is bonded to a chlorine (Cl), a hydrogen (H), a carbon (C5), and another carbon (C3), making it another chiral center.
- The fifth carbon (C5) is bonded to three hydrogens and one carbon (C4), so it is not chiral.

Thus, the compound has **two chiral centers** (C2 and C4).

### Step 2: Determine the number of stereoisomers
The number of stereoisomers can be calculated using the formula:
\[ \text{Number of stereoisomers} = 2^n \]
where \( n \) is the number of chiral centers.

In this case, since we have 2 chiral centers:
\[ \text{Number of stereoisomers} = 2^2 = 4 \]

### Step 3: Identify optically active stereoisomers
Not all stereoisomers are optically active. To determine the number of optically active stereoisomers, we need to consider the presence of any meso compounds. A meso compound has multiple stereocenters but is superimposable on its mirror image, meaning it is not optically active.

In our case, since the two chiral centers (C2 and C4) are not symmetrical with respect to each other, all stereoisomers are optically active.

### Conclusion
Thus, the total number of optically active stereoisomers of the compound `CH3-CH(Cl)-CH2-CH2-CH(Cl)-CH3` is **4**.

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