The number of geometrical isomers for
`CH_(3)-CH=CH-CH=CH-CH=CH_(2)` is

Options

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

Explanation

To determine the number of geometrical isomers for the compound \( CH_3-CH=CH-CH=CH-CH=CH_2 \), we will follow these steps:

### Step 1: Identify the double bonds
The compound has three double bonds, which are the sites where geometrical isomerism can occur. The double bonds are located between:
- C1 and C2
- C3 and C4
- C5 and C6

### Step 2: Determine the positions of geometrical isomerism
Geometrical isomerism occurs at double bonds where the substituents attached to the carbon atoms can be arranged differently. For each double bond, we need to check if the substituents on either side of the double bond are different.

In this compound:
- For the first double bond (C1=C2), the groups are \( CH_3 \) (on C1) and \( H \) (on C2), which allows for cis and trans configurations.
- For the second double bond (C3=C4), the groups are \( H \) (on C3) and \( H \) (on C4), which does not allow for geometrical isomerism since both substituents are the same.
- For the third double bond (C5=C6), the groups are \( H \) (on C5) and \( CH_2 \) (on C6), which allows for cis and trans configurations.

Thus, geometrical isomerism is possible at the first and third double bonds.

### Step 3: Count the number of geometrical isomers
Since geometrical isomerism is possible at two double bonds (C1=C2 and C5=C6), we can use the formula for calculating the number of geometrical isomers:

\[ \text{Number of geometrical isomers} = 2^n \]

where \( n \) is the number of double bonds capable of geometrical isomerism.

In our case, \( n = 2 \) (the first and third double bonds).

\[ \text{Number of geometrical isomers} = 2^2 = 4 \]

### Conclusion
Therefore, the total number of geometrical isomers for the compound \( CH_3-CH=CH-CH=CH-CH=CH_2 \) is **4**.

---