Answer: CH 3 COCl with anhydrous AlCl 3.
- A CH<sub>3</sub>COOH with conc. H<sub>2</sub>SO<sub>4</sub>
- B CH<sub>3</sub>CHO with dilute NaOH
- C CH<sub>3</sub>COCl with anhydrous AlCl<sub>3</sub>
- D CH<sub>3</sub>COONa with soda lime
Correct answer: C. CH<sub>3</sub>COCl with anhydrous AlCl<sub>3</sub>
Explanation: Friedel–Crafts acylation needs an acyl halide plus a Lewis acid: anhydrous AlCl<sub>3</sub> 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.