In the monochlorination of 3-methylpentan, let `x` be the number of pairs of isomers which exist as enantiomers, `y` be the number of pairs of isomers which exist as diastereomers, `z` be the number of isomers which are achiral. Calculate the value of `x+y+z`.

Explanation

To solve the problem of monochlorination of 3-methylpentane and calculate the values of \( x \), \( y \), and \( z \), we will follow these steps:

### Step 1: Understand the structure of 3-methylpentane
3-methylpentane is a branched alkane with the following structure:

```
CH3
|
CH3-CH2-C-CH2-CH3
```

### Step 2: Identify possible chlorination sites
In 3-methylpentane, chlorination can occur at different carbon atoms. The possible sites for chlorination are:
- At the 1st carbon (C1)
- At the 2nd carbon (C2)
- At the 3rd carbon (C3)
- At the 4th carbon (C4)
- At the 5th carbon (C5)

### Step 3: Determine the products of chlorination
When chlorination occurs, we will get different isomers based on which carbon the chlorine atom replaces a hydrogen atom.

### Step 4: Identify chirality in the products
1. **Chlorination at C2**: This will give a chiral center, leading to enantiomers.
2. **Chlorination at C3**: This will also give a chiral center, leading to another pair of enantiomers.
3. **Chlorination at C1, C4, and C5**: These positions do not create chiral centers, leading to achiral products.

### Step 5: Count the isomers
- **Pairs of isomers that exist as enantiomers (x)**:
- Chlorination at C2 creates one pair of enantiomers.
- Chlorination at C3 creates another pair of enantiomers.
- Thus, \( x = 2 \).

- **Pairs of isomers that exist as diastereomers (y)**:
- Chlorination at C1 and C2 gives diastereomers.
- Chlorination at C1 and C3 gives diastereomers.
- Chlorination at C2 and C3 gives diastereomers.
- Similarly, combinations of other chlorination sites will yield additional diastereomers.
- Total pairs of diastereomers can be counted as \( y = 8 \).

- **Isomers that are achiral (z)**:
- Chlorination at C1, C4, and C5 leads to achiral products.
- Thus, \( z = 2 \).

### Step 6: Calculate \( x + y + z \)
Now we can add the values:
\[ x + y + z = 2 + 8 + 2 = 12 \]

### Final Answer
The value of \( x + y + z \) is **12**.

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