The number of stereoisomers of compound `CH_(3)-CH=CH-CH(Br)CH_(3)` is :

Options

A. 2
B. 3
C. 4
D. 6

Answer

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

### Step 1: Identify the structure of the compound
The compound can be represented as follows:
- The first carbon (C1) is attached to three hydrogens and one carbon (C2).
- The second carbon (C2) is attached to one hydrogen, one carbon (C3), and one carbon (C1) with a double bond.
- The third carbon (C3) is attached to one hydrogen, one carbon (C4), and one carbon (C2) with a double bond.
- The fourth carbon (C4) is attached to one bromine atom, one carbon (C5), and one hydrogen.
- The fifth carbon (C5) is attached to three hydrogens and one carbon (C4).

### Step 2: Count the number of chiral centers
A chiral carbon atom is defined as a carbon atom that is bonded to four different groups.
- C1: Has three hydrogens and one carbon (not chiral).
- C2: Has one hydrogen, one carbon (C1), and one carbon (C3) (not chiral due to the double bond).
- C3: Has one hydrogen, one carbon (C2), and one carbon (C4) (not chiral due to the double bond).
- C4: Has one bromine, one hydrogen, and one carbon (C5) (this is a chiral center).
- C5: Has three hydrogens and one carbon (not chiral).

Thus, there is **one chiral carbon atom** (C4).

### Step 3: Count the number of double bonds
In the compound, there is **one double bond** between C2 and C3.

### Step 4: Calculate the total number of stereoisomers
The formula for calculating the number of stereoisomers is given by:
\[ \text{Number of stereoisomers} = 2^{(n)} \]
where \( n \) is the sum of the number of chiral centers and the number of double bonds.

Here, we have:
- Number of chiral centers = 1 (from C4)
- Number of double bonds = 1 (between C2 and C3)

Thus, \( n = 1 + 1 = 2 \).

Now, substituting into the formula:
\[ \text{Number of stereoisomers} = 2^2 = 4 \]

### Conclusion
The number of stereoisomers for the compound \( CH_3-CH=CH-CH(Br)CH_3 \) is **4**.

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