Total number of optical active stereoisomers of the following compound is :
`CH_(3)-CH=CH-CHOH-CH=CH-CH_(3)`

Options

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

Answer

To determine the total number of optical active stereoisomers for the compound `CH3-CH=CH-CHOH-CH=CH-CH3`, we will follow these steps:

### Step 1: Identify the Structure
The given compound is `CH3-CH=CH-CHOH-CH=CH-CH3`. This compound has two double bonds and a hydroxyl group (-OH).

### Step 2: Count the Chiral Centers
A chiral center (or chiral carbon) is a carbon atom that is attached to four different groups. We need to analyze the structure to identify any chiral centers.

- The first carbon (C1) is attached to two hydrogens and a methyl group (CH3), so it is not chiral.
- The second carbon (C2) is part of a double bond and is attached to two hydrogens, so it is not chiral.
- The third carbon (C3) is part of a double bond and is attached to a hydrogen and a hydroxyl group, but it also has another carbon attached, making it not chiral.
- The fourth carbon (C4) has a hydroxyl group and is attached to two other carbons, but it is not chiral as it has two identical groups (the two adjacent carbons).
- The fifth carbon (C5) is part of a double bond and is attached to two hydrogens, so it is not chiral.
- The sixth carbon (C6) is attached to two hydrogens and a methyl group, so it is not chiral.

After analyzing all the carbons, we find that there are **no chiral centers** in this compound.

### Step 3: Count the Double Bonds
The compound has **two double bonds**.

### Step 4: Apply the Formula for Stereoisomers
The formula for calculating the number of stereoisomers is given by:
\[ \text{Number of stereoisomers} = 2^{(n + c)} \]
where:
- \( n \) = number of double bonds
- \( c \) = number of chiral centers

In our case:
- \( n = 2 \) (from the two double bonds)
- \( c = 0 \) (since there are no chiral centers)

Thus, substituting the values:
\[ \text{Number of stereoisomers} = 2^{(2 + 0)} = 2^2 = 4 \]

### Conclusion
The total number of optical active stereoisomers for the compound `CH3-CH=CH-CHOH-CH=CH-CH3` is **4**.

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