The total number of halogenated products likely to be formed by ethance is:

Options

A. 4
B. 6
C. 9
D. 8

Explanation

To determine the total number of halogenated products likely to be formed by ethane (C2H6), we will analyze the process of chlorination step by step.

### Step 1: Understand the Structure of Ethane
Ethane has the molecular formula C2H6, which can be represented as:
\[ \text{CH}_3 - \text{CH}_3 \]

### Step 2: Monochlorination
In monochlorination, one hydrogen atom from ethane is replaced by a chlorine atom. This can occur at either of the two carbon atoms, leading to:
1. CH3-CH2Cl (Chloroethane)
2. CH2Cl-CH3 (Chloroethane)

Both products are identical, so we count this as one unique product.

### Step 3: Dichlorination
In dichlorination, two hydrogen atoms are replaced by chlorine atoms. The possible products are:
1. CH3-CHCl2 (1,1-Dichloroethane)
2. CH2Cl-CH2Cl (1,2-Dichloroethane)
3. CHCl2-CH3 (1,2-Dichloroethane)

Here, we have three unique products.

### Step 4: Trichlorination
In trichlorination, three hydrogen atoms are replaced by chlorine atoms. The possible products are:
1. CHCl3-CH3 (1,1,1-Trichloroethane)
2. CH2Cl-CHCl2 (1,1,2-Trichloroethane)
3. CHCl2-CH2Cl (1,2,2-Trichloroethane)

This gives us three unique products.

### Step 5: Tetrachlorination
In tetrachlorination, four hydrogen atoms are replaced by chlorine atoms. The possible products are:
1. CCl4-CH3 (1,1,1,1-Tetrachloroethane)
2. CHCl3-CHCl (1,1,1,2-Tetrachloroethane)

This results in two unique products.

### Step 6: Pentachlorination
In pentachlorination, five hydrogen atoms are replaced by chlorine atoms. The only possible product is:
1. CCl5-CH3 (1,1,1,1,2-Pentachloroethane)

This results in one unique product.

### Step 7: Hexachlorination
In hexachlorination, all hydrogen atoms are replaced by chlorine atoms, leading to:
1. C2Cl6 (Hexachloroethane)

This results in one unique product.

### Step 8: Count All Unique Products
Now, let's summarize the unique products from each step:
- Monochlorination: 1 product
- Dichlorination: 3 products
- Trichlorination: 3 products
- Tetrachlorination: 2 products
- Pentachlorination: 1 product
- Hexachlorination: 1 product

Adding these together gives:
\[ 1 + 3 + 3 + 2 + 1 + 1 = 11 \text{ unique products} \]

### Conclusion
The total number of halogenated products likely to be formed by ethane is **11**.