The number of acyclic isomers of `C_(3)H_(5)Cl` are:

Options

A. 1
B. 2
C. 3
D. 4

Answer

To determine the number of acyclic isomers of the compound \( C_3H_5Cl \), we can follow these steps:

### Step 1: Determine the degree of unsaturation
The formula for calculating the degree of unsaturation (DU) is:
\[ DU = \frac{(2C + 2 + N - H - X)}{2} \]
Where:
- \( C \) = number of carbon atoms
- \( H \) = number of hydrogen atoms
- \( N \) = number of nitrogen atoms
- \( X \) = number of halogen atoms

For \( C_3H_5Cl \):
- \( C = 3 \)
- \( H = 5 \)
- \( Cl = 1 \)

Plugging in the values:
\[ DU = \frac{(2 \times 3 + 2 + 0 - 5 - 1)}{2} = \frac{(6 + 2 - 5 - 1)}{2} = \frac{2}{2} = 1 \]
This indicates that there is one degree of unsaturation, which could correspond to a double bond or a ring.

### Step 2: Identify possible structures
Given that we have one degree of unsaturation, we can have:
1. A double bond between carbon atoms.
2. A cyclic structure (but we are looking for acyclic isomers).

### Step 3: Draw the possible acyclic structures
We can have the following acyclic structures for \( C_3H_5Cl \):
1. **Propene with a chlorine substituent**:
- **1-chloropropene**: \( CH_2=CH-CH_2Cl \)
- **2-chloropropene**: \( CH_3-CH=CHCl \)

2. **Chlorinated propane**:
- **1-chloropropane**: \( CH_3-CH_2-CH_2Cl \)
- **2-chloropropane**: \( CH_3-CH(Cl)-CH_3 \)

### Step 4: Count the isomers
From the structures identified:
1. **1-chloropropene**
2. **2-chloropropene**
3. **1-chloropropane**
4. **2-chloropropane**

Thus, the total number of acyclic isomers of \( C_3H_5Cl \) is **4**.

### Final Answer:
The number of acyclic isomers of \( C_3H_5Cl \) is **4**.

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