The number of isomers (including stereoisomers) of `C_5H_(10)` are

Options

A. 10
B. 11
C. 12
D. 13

Answer

To determine the number of isomers (including stereoisomers) for the molecular formula \( C_5H_{10} \), we will follow these steps:

### Step 1: Calculate the Degree of Unsaturation
The degree of unsaturation (DU) can be calculated using the formula:
\[ DU = \frac{(2C + 2 + N - H - X)}{2} \]
Where:
- \( C \) = number of carbon atoms
- \( H \) = number of hydrogen atoms
- \( N \) = number of nitrogen atoms (which is 0 in this case)
- \( X \) = number of halogen atoms (which is also 0 in this case)

Substituting the values for \( C = 5 \) and \( H = 10 \):
\[ DU = \frac{(2 \times 5 + 2 + 0 - 10 - 0)}{2} = \frac{(10 + 2 - 10)}{2} = \frac{2}{2} = 1 \]
This indicates that there is one degree of unsaturation, which means there could be either a double bond or a ring in the structure.

### Step 2: Identify Possible Structures
Given that \( C_5H_{10} \) can have a double bond or a ring, we will explore both options:

#### A. Alkenes (Double Bonds)
1. **Straight-chain alkenes**:
- **1-Pentene**: CH2=CH-CH2-CH2-CH3
- **2-Pentene**: CH3-CH=CH-CH2-CH3 (can have E/Z isomers)

2. **Branched alkenes**:
- **3-Pentene**: CH3-CH2-CH=CH-CH3 (can have E/Z isomers)
- **2-Methyl-1-butene**: CH2=C(CH3)-CH2-CH3
- **2-Methyl-2-butene**: CH3-C(CH3)=CH-CH3 (can have E/Z isomers)

#### B. Cycloalkanes
1. **Cyclopentane**: A 5-membered ring (C5H10).
2. **Methylcyclobutane**: A 4-membered ring with a methyl group.
3. **1,2-Dimethylcyclopropane**: A 3-membered ring with two methyl groups.

### Step 3: Count All Isomers
Now we will count all the unique structures identified:
- **Alkenes**:
- 1-Pentene
- 2-Pentene (E and Z forms)
- 3-Pentene (E and Z forms)
- 2-Methyl-1-butene
- 2-Methyl-2-butene (E and Z forms)

This gives us 7 alkenes.

- **Cycloalkanes**:
- Cyclopentane
- Methylcyclobutane
- 1,2-Dimethylcyclopropane

This gives us 3 cycloalkanes.

### Step 4: Total Isomers
Adding the alkenes and cycloalkanes together:
- Total = 7 (alkenes) + 3 (cycloalkanes) = 10 isomers.

### Step 5: Include Stereoisomers
From the alkenes, we have:
- 2-Pentene has 2 stereoisomers (E and Z).
- 3-Pentene has 2 stereoisomers (E and Z).
- 2-Methyl-2-butene has 2 stereoisomers (E and Z).

Adding these gives us additional stereoisomers:
- Total stereoisomers = 2 (from 2-Pentene) + 2 (from 3-Pentene) + 2 (from 2-Methyl-2-butene) = 6.

### Final Count
Total isomers including stereoisomers:
- 10 (from structural isomers) + 6 (from stereoisomers) = 16 isomers.

### Conclusion
The total number of isomers (including stereoisomers) of \( C_5H_{10} \) is **16**.