Question:
Identify the product 'C' formed in the following series of reactions.
Reaction:
Bromoethane $\xrightarrow{\text{Mg / dry ether}}$ A $\xrightarrow{\text{H}_2\text{O / dry ether}}$ B $\xrightarrow{\text{Br}_2 / \text{UV}}$ C
Identify the product 'C' formed in the following series of reactions.
Reaction:
Bromoethane $\xrightarrow{\text{Mg / dry ether}}$ A $\xrightarrow{\text{H}_2\text{O / dry ether}}$ B $\xrightarrow{\text{Br}_2 / \text{UV}}$ C
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Key idea:
Grignard $\rightarrow$ hydrolysis gives alkane → halogenation gives haloalkane
Updated On: May 8, 2026
- Ethane
- Bromoethane
- Ethyl magnesium bromide
- Ethene
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The Correct Option is B
Solution and Explanation
Step 1: Formation of Grignard reagent \[ \text{C}_2\text{H}_5\text{Br} \xrightarrow{\text{Mg / dry ether}} \text{C}_2\text{H}_5\text{MgBr} \quad (A) \]
Step 2: Hydrolysis \[ \text{C}_2\text{H}_5\text{MgBr} + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_6 + \text{Mg(OH)Br} \] So, \(B = \text{ethane}\)
Step 3: Bromination (free radical substitution) \[ \text{C}_2\text{H}_6 \xrightarrow{\text{Br}_2 / \text{UV}} \text{C}_2\text{H}_5\text{Br} \]
Step 4: Conclusion
Final product \(C = \text{Bromoethane}\) Final Answer: Option (B)
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