`C_(2)H_(5)Cl+Naunderset(-NaCI)overset("dry ether")(rarr)A`. How many isomers are formed by monohalogenation of A including stereo isomers

Options

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

Explanation

To solve the problem of how many isomers are formed by the monohalogenation of butane (C₂H₅Cl + Na → A, where A = butane), we will follow these steps:

### Step 1: Identify the structure of A
The reaction given is a Wurtz reaction where ethyl chloride (C₂H₅Cl) reacts with sodium (Na) in dry ether to form butane (C₄H₁₀). The structure of butane can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \]

### Step 2: Understand monohalogenation
Monohalogenation refers to the substitution of one hydrogen atom in butane with a halogen atom (in this case, chlorine). This can occur at different positions in the butane molecule.

### Step 3: Identify possible positions for halogenation
1. **1-chlorobutane**: Chlorine replaces one of the terminal hydrogen atoms.
- Structure: CH₃-CH₂-CH₂-CH₂Cl
2. **2-chlorobutane**: Chlorine replaces one of the hydrogen atoms on the second carbon.
- Structure: CH₃-CH(Cl)-CH₂-CH₃

### Step 4: Determine stereoisomers
For 2-chlorobutane, there is a possibility of forming stereoisomers because it has a chiral center (the second carbon). The two enantiomers of 2-chlorobutane are:
- (R)-2-chlorobutane
- (S)-2-chlorobutane

### Step 5: Count all isomers
Now, we can summarize the isomers formed:
1. **1-chlorobutane** (achiral)
2. **2-chlorobutane** (chiral, with 2 enantiomers)

Thus, the total isomers including stereoisomers are:
- 1-chlorobutane: 1 isomer
- 2-chlorobutane: 2 enantiomers

### Final Count
Total isomers = 1 (from 1-chlorobutane) + 2 (from 2-chlorobutane) = **3 isomers**.

### Conclusion
The total number of isomers formed by the monohalogenation of A (butane) including stereoisomers is **3**.

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