Choose the correct set of reagents for the following conversion.

Step 1: Analyze the target structure

Target: Ethyl benzene with Br on the benzene ring

This means we need:

1. Benzene ring
2. Ethyl group ($\ce{-CH2CH3}$) attached to the ring
3. Bromine on the ring (not on the side chain)

Step 2: Plan the synthesis

Starting from $\ce{CH≡CH2}$ (which seems to be ethyne/acetylene, $\ce{HC≡CH}$):

Route:

1. Convert acetylene to benzene (cyclotrimerization)
2. Add ethyl group to benzene (Friedel-Crafts alkylation)
3. Brominate the ring (electrophilic aromatic substitution)

Wait, the starting material notation is unclear. Let me assume it's benzene or we start from benzene.

Actually, looking at the answer options, they all start with halogenation reagents, suggesting we're modifying benzene.

Step 3: Correct sequence

To get ethyl benzene with Br on ring:

1. First bromination of benzene: $\ce{Br2/Fe}$ → bromobenzene
2. Friedel-Crafts alkylation: Add ethyl group using $\ce{CH3CH2Cl}$ or $\ce{CH2=CH2}$ with acid
3. Dehydrohalogenation: If needed, use alc. KOH

Wait, let me reconsider by looking at option (a):

• $\ce{Br2/Fe}$ - bromination of benzene ring
• $\ce{Cl2, \Delta}$ - chlorination (possibly of side chain after alkylation)
• alc. KOH - elimination to form double bond or remove HCl

Actually, the sequence might be:

1. Brominate benzene ring: $\ce{Br2/Fe}$
2. Add ethyl group via Friedel-Crafts
3. Use alc. KOH for any elimination if needed

Answer: (a)

Key Reactions:

• $\ce{Br2/Fe}$ or $\ce{Br2/FeBr3}$ → electrophilic aromatic bromination
• $\ce{RCl/AlCl3}$ → Friedel-Crafts alkylation
• Alc. KOH → elimination (E2) to form alkene
• Aq. KOH → substitution (SN2/SN1) to form alcohol