See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Friedel-Crafts Acylation (Step 1: butanoyl chloride / AlCl3): Benzene undergoes Friedel-Crafts acylation with butanoyl chloride (CH3CH2CH2COCl) in the presence of AlCl3. This introduces a butanoyl group (-C(=O)CH2CH2CH3) onto the benzene ring, giving butyrophenone (phenyl propyl ketone, specifically 1-phenyl-1-butanone). Step 2 - Clemmensen Reduction (Zn(Hg), HCl): The ketone carbonyl is reduced to a methylene group using Clemmensen reduction conditions (Zn amalgam in concentrated HCl). This converts the -C(=O)CH2CH2CH3 group into -CH2CH2CH2CH3 (a butyl group). The product is n-butylbenzene. Step 3 - Electrophilic Aromatic Bromination (Br2, FeBr3): n-Butylbenzene undergoes electrophilic aromatic substitution with Br2/FeBr3. The butyl group is an alkyl substituent, which is an ortho/para director. Since the para position is less sterically hindered than the ortho positions, bromination occurs predominantly at the para position, giving 1-bromo-4-butylbenzene. This sequence - acylation to avoid carbocation rearrangements, reduction of ketone, then bromination directed para to the alkyl group - is the classic synthetic strategy to attach an unbranched alkyl group and then brominate para to it. Why other options fail: (a) Shows a ketone still present (no reduction occurred) with meta bromination - incorrect because ketones direct meta, but the ketone was reduced in step 2, and meta bromination would only apply before reduction. (b) Shows a ketone still present with para bromination and Ac2O - incorrect sequence and reagents. (d) Shows ortho bromination relative to a ketone - incorrect regiochemistry and the ketone was not reduced. Therefore, the correct answer is C.