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) during halogenation, we can follow these steps:

### Step 1: Understand the Halogenation Process
Halogenation involves the substitution of hydrogen atoms in ethane with halogen atoms (like Cl or Br). Since ethane has two carbon atoms, we will consider the possible substitutions on each carbon atom.

### Step 2: Identify Possible Substitution Sites
Ethane (C2H6) has the following structure:
```
H H
\ /
C
/ \
H C
/ \
H H
```
Each carbon atom can have hydrogen atoms replaced by halogens.

### Step 3: Count the Monosubstituted Products
1. **Monosubstitution**:
- Substituting one hydrogen atom with a halogen on either carbon:
- For the first carbon: CH3-CH2-Cl (1-chloroethane)
- For the second carbon: CH2Cl-CH3 (1-chloroethane, same as above)
- Total from monosubstitution: **1 unique product**.

### Step 4: Count the Disubstituted Products
2. **Disubstitution**:
- Substituting two hydrogen atoms with halogens:
- Two halogens on the same carbon:
- CH2Cl-CH2Cl (1,2-dichloroethane)
- CHCl2-CH3 (1,1-dichloroethane)
- CH3-CHCl2 (1,1-dichloroethane, same as above)
- One halogen on each carbon:
- CHCl-CH2Cl (1,1-dichloroethane)
- Total from disubstitution: **3 unique products**.

### Step 5: Count the Trisubstituted Products
3. **Trisubstitution**:
- Substituting three hydrogen atoms with halogens:
- CHCl2-CHCl2 (1,1,2-trichloroethane)
- CH3-CHCl3 (1,1,1-trichloroethane)
- Total from trisubstitution: **2 unique products**.

### Step 6: Count the Tetrasubstituted Products
4. **Tetrasubstitution**:
- Substituting all hydrogen atoms with halogens:
- CCl3-CCl3 (1,1,1,1-tetrachloroethane)
- Total from tetrasubstitution: **1 unique product**.

### Step 7: Total the Unique Products
Now we can sum the unique products from each substitution type:
- Monosubstituted: 1
- Disubstituted: 3
- Trisubstituted: 2
- Tetrasubstituted: 1

**Total = 1 + 3 + 2 + 1 = 7 unique halogenated products.**

### Final Answer
The total number of halogenated products likely to be formed by ethane is **7**.

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