How many total isomers exist for compound with composition `C_(4)H_(8)`?

Options

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

Explanation

To determine the total number of isomers for the compound with the molecular formula \(C_4H_8\), we need to consider the structural isomers, geometric (cis-trans) isomers, and cyclic isomers. Here’s the step-by-step solution:

### Step 1: Calculate the Degree of Unsaturation (DU)
The degree of unsaturation (DU) can be calculated using the formula:
\[ DU = \frac{(2C + 2 + N - H - X)}{2} \]
For \(C_4H_8\):
- \(C = 4\)
- \(H = 8\)
- \(N = 0\) (no nitrogen)
- \(X = 0\) (no halogens)

Substituting the values:
\[ DU = \frac{(2 \times 4 + 2 - 8)}{2} = \frac{(8 + 2 - 8)}{2} = \frac{2}{2} = 1 \]
This indicates that the compound has one degree of unsaturation, which can be due to either a double bond or a ring.

### Step 2: Identify Structural Isomers
1. **Straight-chain alkene**:
- Butene (1-butene)
- Isobutene (2-butene)

2. **Geometric Isomers**:
- 2-butene can exist as:
- Cis-2-butene
- Trans-2-butene

### Step 3: Identify Cyclic Isomers
1. **Cyclic Compound**:
- Cyclobutane (a ring structure with four carbon atoms)

### Step 4: Count All Isomers
Now, let’s summarize the identified isomers:
1. **1-butene** (straight chain)
2. **2-butene** (straight chain, with geometric isomers)
- Cis-2-butene
- Trans-2-butene
3. **Cyclobutane** (cyclic structure)

### Total Count of Isomers
- 1-butene: 1
- 2-butene: 1 (but has 2 geometric forms, so count as 2)
- Cyclobutane: 1

Adding these:
- 1 (1-butene) + 2 (cis and trans 2-butene) + 1 (cyclobutane) = **Total: 4 isomers**

### Final Answer
The total number of isomers for the compound with the composition \(C_4H_8\) is **4**.

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