Total number of isomeric alkene possible with compound having molecular formula `C_(4)H_(8)` is

Answer

To determine the total number of isomeric alkenes possible with the molecular formula C₄H₈, we can follow these steps:

### Step 1: Understand the General Formula for Alkenes
Alkenes follow the general formula CₙH₂ₙ, where n is the number of carbon atoms. For C₄H₈, n = 4.

### Step 2: Determine the Degree of Unsaturation
The degree of unsaturation (or the number of double bonds) can be calculated using the formula:
\[ \text{Degree of Unsaturation} = \frac{(2n + 2 - m)}{2} \]
where n is the number of carbons and m is the number of hydrogens. For C₄H₈:
\[ \text{Degree of Unsaturation} = \frac{(2(4) + 2 - 8)}{2} = \frac{(8 + 2 - 8)}{2} = \frac{2}{2} = 1 \]
This indicates that there is one double bond in the compound.

### Step 3: Identify Possible Structures
Now, we will identify the possible structural isomers for C₄H₈ with one double bond. The possible structures include:
1. **But-1-ene**: CH₂=CH-CH₂-CH₃
2. **But-2-ene**: CH₃-CH=CH-CH₃ (which can exist in two geometric forms: cis and trans)
3. **Isobutylene (2-methylpropene)**: CH₂=C(CH₃)-CH₃

### Step 4: Count the Isomers
Now, let's count the isomers:
1. **But-1-ene**: 1 isomer
2. **But-2-ene**: 2 isomers (cis and trans)
3. **Isobutylene (2-methylpropene)**: 1 isomer

Adding these together gives:
- But-1-ene: 1
- But-2-ene (cis and trans): 2
- Isobutylene: 1

Total = 1 + 2 + 1 = 4 isomers.

### Conclusion
The total number of isomeric alkenes possible with the molecular formula C₄H₈ is **4**.

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