Total number of stereoisomer possible for the compound
`H_(3)C-CH=CH-CH=CH-CH=CH-Ph`

Answer

To determine the total number of stereoisomers for the compound `H3C-CH=CH-CH=CH-CH=CH-Ph`, we will follow these steps:

### Step 1: Identify the Structure
The compound has a long carbon chain with alternating double bonds and ends with a phenyl group (Ph). The structure can be represented as:
```
H3C-CH=CH-CH=CH-CH=CH-Ph
```

### Step 2: Count the Double Bonds
In the given compound, we can see that there are 3 double bonds (C=C) present in the carbon chain:
1. Between C1 and C2
2. Between C3 and C4
3. Between C5 and C6

### Step 3: Determine the Number of Chiral Centers
Next, we need to check for chiral centers in the compound. A chiral center is typically a carbon atom that is bonded to four different groups. In this case, the compound does not have any chiral centers since the carbon atoms involved in the double bonds do not meet the criteria for chirality.

### Step 4: Calculate the Total Number of Stereoisomers
The total number of stereoisomers can be calculated using the formula:
\[ \text{Total Stereoisomers} = 2^n \]
where \( n \) is the number of double bonds (since each double bond can exist in a cis or trans configuration).

In our case, we have:
- \( n = 3 \) (the number of double bonds)

So, we calculate:
\[ \text{Total Stereoisomers} = 2^3 = 8 \]

### Conclusion
The total number of stereoisomers possible for the compound `H3C-CH=CH-CH=CH-CH=CH-Ph` is **8**.

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