How many monochlorinated isomers would result form the reaction of chlorine with n-butane in the presence of UV light?

Options

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

Explanation

To determine how many monochlorinated isomers can be formed from the reaction of chlorine with n-butane in the presence of UV light, we can follow these steps:

### Step 1: Understand the Structure of n-Butane
n-Butane (C4H10) has the following structure:
- CH3-CH2-CH2-CH3

This means it has four carbon atoms arranged in a straight chain.

### Step 2: Identify the Types of Hydrogen Atoms
In n-butane, there are different types of hydrogen atoms based on their position:
- The terminal carbon atoms (the two CH3 groups) each have three hydrogen atoms.
- The two middle carbon atoms (the two CH2 groups) each have two hydrogen atoms.

### Step 3: Determine Possible Monochlorination Sites
When chlorine reacts with n-butane in the presence of UV light, it can replace one hydrogen atom with a chlorine atom (Cl) at different positions:
- Chlorination at the terminal CH3 groups (2 possibilities):
1. CH3-CH2-CH2-CH2Cl (Chlorobutane)
2. CH3Cl-CH2-CH2-CH3 (Chloro-Butane)

- Chlorination at the internal CH2 groups (2 possibilities):
1. CH3-CHCl-CH2-CH3 (2-chlorobutane)
2. CH3-CH2-CHCl-CH3 (3-chlorobutane)

### Step 4: Count the Unique Isomers
Now we can summarize the possible monochlorinated products:
1. 1-chlorobutane (from chlorination of a terminal carbon)
2. 2-chlorobutane (from chlorination of the second carbon)
3. 3-chlorobutane (from chlorination of the third carbon)

Since the terminal chlorination produces identical products, we only count it once. Therefore, we have:
- 1-chlorobutane
- 2-chlorobutane
- 3-chlorobutane

### Conclusion
Thus, the total number of unique monochlorinated isomers that can be formed from the reaction of chlorine with n-butane is **3**.

---