`CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)-CH_(2)-CH_(3)underset(hv)overset(Cl_(2))rarr`
Mono-chloro product (inculding steroisomers) are :

Options

A. 6
B. 7
C. 8
D. 9

Answer

To determine the number of monochloro products (including stereoisomers) formed from the reaction of the given hydrocarbon with chlorine (Cl₂) under UV light (hv), we will follow these steps:

### Step 1: Identify the structure of the hydrocarbon
The given hydrocarbon is:
\[ CH_3 - CH(CH_3) - CH_2 - CH_2 - CH_3 \]
This structure indicates that we have a branched alkane with a total of 6 carbon atoms.

### Step 2: Identify the types of hydrogen atoms present
In the hydrocarbon, we need to identify the different types of hydrogen atoms based on their environment:
1. **Terminal CH₃ groups**: There are two equivalent terminal methyl groups (CH₃).
2. **Branched CH (attached to CH₃)**: This carbon has one hydrogen and is attached to two different groups (CH₃ and CH₂).
3. **CH₂ groups**: There are two CH₂ groups in the chain, and they are equivalent.

### Step 3: Determine the possible monochloro products
Now, we will replace each type of hydrogen with chlorine (Cl) to find the different monochloro products:
1. **Replacing one of the terminal CH₃ hydrogens**: This gives us one product, which is the same regardless of which terminal CH₃ is replaced.
- Product: \( CH_2Cl - CH(CH_3) - CH_2 - CH_2 - CH_3 \)

2. **Replacing the hydrogen on the branched CH**: This gives us another unique product.
- Product: \( CH_3 - CClH - CH_2 - CH_2 - CH_3 \)
- This carbon is a chiral center, leading to two stereoisomers (R and S configurations).

3. **Replacing one of the CH₂ hydrogens**: This gives us another product.
- Product: \( CH_3 - CH(CH_3) - CClH - CH_2 - CH_3 \)
- This carbon is also a chiral center, leading to two stereoisomers (R and S configurations).

### Step 4: Count the total products including stereoisomers
Now, let's summarize the products:
1. From the terminal CH₃: 1 product
2. From the branched CH: 2 products (1 chiral center gives 2 stereoisomers)
3. From the CH₂: 2 products (1 chiral center gives 2 stereoisomers)

### Total Products:
- Total = 1 (from CH₃) + 2 (from branched CH) + 2 (from CH₂) = 5 products.

### Conclusion
The total number of monochloro products (including stereoisomers) formed is **5**.

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