Question:

Choose the correct set of reagents for the following conversion:

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In aromatic substitution reactions, the presence of a halogen like chlorine or bromine can activate the ring towards further electrophilic substitution reactions. Use specific reagents and conditions (e.g., heat and alcoholic KOH) to control the outcome.
Updated On: Apr 14, 2025
  • \( \text{Cl}_2/\text{Fe}; \text{Br}_2/\text{anhy.} \text{AlCl}_3; \text{aq. KOH} \)
  • \( \text{Br}_2/\text{Fe}; \text{Cl}_2, \Delta; \text{alc. KOH} \)
  • \( \text{Cl}_2/\text{anhy.} \text{AlCl}_3; \text{Br}_2/\text{Fe}; \text{alc. KOH} \)
  • \( \text{Br}_2/\text{anhy.} \text{AlCl}_3; \text{Cl}_2, \Delta; \text{aq. KOH} \)
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The Correct Option is B

Solution and Explanation

The given transformation involves the conversion of ethyl benzene (C\(_6\)H\(_5\)C\(_2\)H\(_5\)) to the desired product where a Br atom is added to the benzenoid ring. 
- First, bromination of ethylbenzene occurs using \( \text{Br}_2 \) in the presence of iron (Fe), which acts as a catalyst.
The electrophilic aromatic substitution reaction results in the formation of a brominated product. 
- After bromination, the next step involves chlorination using \( \text{Cl}_2 \) under heat (denoted by \( \Delta \)), leading to the introduction of the chlorine atom at the desired position. 
- Finally, the product undergoes dehydrohalogenation in the presence of alcoholic potassium hydroxide (alc.
KOH), leading to the formation of the double bond as required. 

Thus, the correct set of reagents to form the desired product is option (2).

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