The number of stereoisomers of the given compound
`CH_(3)-overset(Cl)overset(|)CH-overset(Cl)overset(|)CH-overset(Cl)overset(|)CH-CH_(3)`
will be

Answer

To determine the number of stereoisomers for the compound \( CH_3-CHCl-CHCl-CHCl-CH_3 \), we will follow these steps:

### Step 1: Identify Chiral Centers
First, we need to identify the chiral centers in the compound. A chiral center is typically a carbon atom that is bonded to four different substituents.

In the given compound, we can see that there are three carbon atoms that are bonded to chlorine (Cl) and hydrogen (H) atoms, which makes them chiral centers.

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

### Step 3: Apply the Stereoisomer Formula
The number of stereoisomers can be determined using the formula:
\[ \text{Number of stereoisomers} = 2^N \]
where \( N \) is the number of chiral centers. However, we need to consider the symmetry of the molecule.

Since the compound has a plane of symmetry due to the presence of identical groups on either side of the molecule, we will adjust our calculation. The formula for the number of stereoisomers in the presence of symmetry is:
\[ \text{Number of stereoisomers} = 2^{N-1} \]
Thus, substituting \( N = 3 \):
\[ \text{Number of stereoisomers} = 2^{3-1} = 2^2 = 4 \]

### Step 4: Conclusion
Therefore, the total number of stereoisomers for the compound \( CH_3-CHCl-CHCl-CHCl-CH_3 \) is **4**.

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