Answer: CH3COCl with anhydrous AlCl3.
- A CH3COOH with conc. H2SO4
- B CH3CHO with dilute NaOH
- C CH3COCl with anhydrous AlCl3
- D CH3COONa with soda lime
Correct answer: C. CH3COCl with anhydrous AlCl3
Explanation: Friedel–Crafts acylation needs an acyl halide plus a Lewis acid: anhydrous AlCl3 generates the acylium ion that attacks the ring.
SN2 proceeds in a single step with backside attack and inversion of configuration, while SN1 forms a planar carbocation intermediate first, leading to racemisation.
Concept context
Study carbon-halogen compounds: how they are made, how they react via SN1/SN2 and elimination, their stereochemistry, and their environmental impact.