A compound is formed by substitution of two chlorine for two hydrogens in propane. The number of possible isomeric compounds is

Options

A. 4
B. 3
C. 5
D. 2

Answer

To solve the problem of determining the number of possible isomeric compounds formed by the substitution of two chlorine atoms for two hydrogen atoms in propane, we can follow these steps:

### Step 1: Understand the Structure of Propane
Propane is a three-carbon alkane with the molecular formula \( C_3H_8 \). Its structure can be represented as:
```
H H H
| | |
H--C--C--C--H
| | |
H H H
```

### Step 2: Identify the Substitution Reaction
In this reaction, we are replacing two hydrogen atoms with two chlorine atoms. The substitution can occur at different positions on the propane molecule.

### Step 3: Determine Possible Substitution Positions
1. **Substitution on the same carbon atom**:
- Both chlorine atoms can be substituted on the first carbon atom:
```
Cl Cl
| |
H--C--C--C--H
| |
H H
```
- Both chlorine atoms can be substituted on the second carbon atom:
```
H Cl
| |
H--C--C--C--H
| |
H Cl
```
- Both chlorine atoms can be substituted on the third carbon atom:
```
H H
| |
H--C--C--C--H
| |
Cl Cl
```

2. **Substitution on different carbon atoms**:
- One chlorine on the first carbon and one on the second carbon:
```
Cl H
| |
H--C--C--C--H
| |
H Cl
```
- One chlorine on the first carbon and one on the third carbon:
```
Cl H
| |
H--C--C--C--H
| |
H Cl
```
- One chlorine on the second carbon and one on the third carbon:
```
H Cl
| |
H--C--C--C--H
| |
Cl H
```

### Step 4: Count the Isomers
From the above substitutions, we can summarize the possible isomers:
1. **1,1-Dichloropropane**: Both Cl on the first carbon.
2. **1,2-Dichloropropane**: One Cl on the first carbon, one on the second carbon.
3. **1,3-Dichloropropane**: One Cl on the first carbon, one on the third carbon.
4. **2,2-Dichloropropane**: Both Cl on the second carbon.
5. **2,3-Dichloropropane**: One Cl on the second carbon, one on the third carbon.
6. **3,3-Dichloropropane**: Both Cl on the third carbon.

### Step 5: Identify Chiral Centers
Among these isomers, we need to check for chirality:
- **1,2-Dichloropropane** has a chiral center, leading to two enantiomers (R and S configurations).
- **2,2-Dichloropropane** and **3,3-Dichloropropane** do not have chiral centers.

### Conclusion
The total number of isomeric compounds is:
- 1 (1,1-Dichloropropane)
- 2 (1,2-Dichloropropane - R and S)
- 1 (2,2-Dichloropropane)
- 1 (3,3-Dichloropropane)
- 1 (1,3-Dichloropropane)
- 1 (2,3-Dichloropropane)

Thus, the total number of isomeric compounds formed is **5**.

### Final Answer
The number of possible isomeric compounds is **5**.

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