In the reaction sequence
`CH_(2)=CH_(2)underset(C Cl_(4))overset(HBr)(rarr)Boverset(C_(2)H_(5)ONa) (rarr)`Ethoxyethane, B is:-

Options

A. `C_(2)H_(6)Br`
B. `CH_(3)Br`
C. `C_(2)H_(5)Br`
D. `CH_(2)=CHBr`

Explanation

To solve the question, we will analyze the reaction sequence step by step.

### Step 1: Identify the initial reactant
The initial reactant is ethylene (CH₂=CH₂), which is an alkene.

### Step 2: Reaction with HBr
When ethylene reacts with HBr in the presence of CCl₄ (carbon tetrachloride), the reaction proceeds via electrophilic addition. The H⁺ ion from HBr will add to one of the carbon atoms of the double bond, forming a carbocation.

- **Carbocation formation**:
\[ \text{CH}_2=CH_2 + HBr \rightarrow \text{CH}_3\text{C}^+\text{H}_2 + Br^- \]
Here, we have formed a carbocation (CH₃-CH₂⁺).

### Step 3: Nucleophilic attack by Br⁻
The bromide ion (Br⁻) will then attack the carbocation, resulting in the formation of bromoethane.

- **Formation of B**:
\[ \text{CH}_3\text{C}^+\text{H}_2 + Br^- \rightarrow \text{C}_2\text{H}_5\text{Br} \]
Thus, B is bromoethane (C₂H₅Br).

### Step 4: Reaction with C₂H₅ONa
Next, bromoethane (C₂H₅Br) reacts with sodium ethoxide (C₂H₅ONa). The ethoxide ion (C₂H₅O⁻) acts as a nucleophile and attacks the carbon atom bonded to bromine.

- **Nucleophilic substitution**:
\[ \text{C}_2\text{H}_5\text{Br} + \text{C}_2\text{H}_5\text{O}^- \rightarrow \text{C}_2\text{H}_5\text{O}\text{C}_2\text{H}_5 + Br^- \]
This leads to the formation of ethoxyethane (C₂H₅O-C₂H₅).

### Conclusion
Thus, the final answer is that B is bromoethane (C₂H₅Br).

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