Total number of products form including stereoisomers during monochlorination of 2 - methyl butane?

Options

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

Answer

To determine the total number of products formed, including stereoisomers, during the monochlorination of 2-methylbutane, we can follow these steps:

### Step 1: Identify the structure of 2-methylbutane
2-methylbutane has the following structure:
- It consists of a butane chain (4 carbons) with a methyl group (CH₃) attached to the second carbon.
- The structural formula is:
\[ \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \]

### Step 2: Identify the types of hydrogen atoms available for substitution
In 2-methylbutane, we can identify different types of hydrogen atoms based on their positions:
- **Primary Hydrogens (1°)**: Hydrogens attached to terminal carbons (2 from CH₃ groups).
- **Secondary Hydrogens (2°)**: Hydrogens attached to the second carbon (2 from CH).
- **Tertiary Hydrogens (3°)**: Hydrogens attached to the third carbon (1 from CH₃).

### Step 3: Determine the possible chlorination sites
When monochlorination occurs, chlorine can replace any of the hydrogen atoms. The possible sites for chlorination are:
1. Primary position (1°): 2 equivalent positions (the two CH₃ groups).
2. Secondary position (2°): 2 equivalent positions (the two H on the second carbon).
3. Tertiary position (3°): 1 position (the H on the third carbon).

### Step 4: Count the unique products formed
1. **Chlorination at Primary Carbons (2 products)**:
- CH₃-CH(CH₃)-CH₂-CH₂Cl (from either CH₃)
- CH₂Cl-CH(CH₃)-CH₂-CH₃ (from either CH₃)

2. **Chlorination at Secondary Carbons (2 products)**:
- CH₃-CH(Cl)-CH₂-CH₃ (this carbon is chiral)
- CH₃-CH(CH₃)-CH(Cl)-CH₃ (this carbon is also chiral)

3. **Chlorination at Tertiary Carbon (1 product)**:
- CH₃-CH(CH₃)-C(Cl)-CH₃ (this carbon is chiral)

### Step 5: Identify stereoisomers
- The products from chlorination at the secondary and tertiary positions can lead to stereoisomers:
- For the secondary chlorination, each chiral center can have R/S configurations.
- For the tertiary chlorination, there is also a chiral center.

### Step 6: Total count of products
- From the primary positions: 2 products (no stereoisomers).
- From the secondary positions: 2 products, each with 2 stereoisomers (R and S) = 4 products.
- From the tertiary position: 1 product with 2 stereoisomers (R and S) = 2 products.

### Final Count
Total products = 2 (primary) + 4 (secondary) + 2 (tertiary) = 8 products.

### Conclusion
The total number of products formed, including stereoisomers, during the monochlorination of 2-methylbutane is **8**.

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