See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: To convert a benzylic methyl group (Ar-CH3) into a side chain Ar-CH2-C≡C-CH3, we need two steps: (1) convert CH3 to CH2X (benzylic halide), then (2) perform an SN2 reaction with an acetylide nucleophile. Step-by-step reasoning: Step 1 - Convert Ar-CH3 to Ar-CH2Br (benzylic bromination): NBS (N-bromosuccinimide) in CCl4 under heat/light conditions performs free-radical benzylic bromination selectively. This converts the -CH3 group on the ring to -CH2Br, giving 4-bromobenzyl bromide (note: the aryl Br on the ring is unaffected because NBS under radical conditions selectively brominates the benzylic position, not the aromatic ring). Step 2 - SN2 with sodium acetylide: The sodium salt of propyne, CH3C≡C(-)Na(+) (sodium propynylide), is a strong nucleophile. It attacks the benzylic carbon of Ar-CH2Br in an SN2 reaction, displacing bromide and forming Ar-CH2-C≡C-CH3. The aryl bromide is not reactive under these mild SN2 conditions. Why other options fail: (a) KOH and heat would cause elimination of HBr from an alkyl halide, but there is no alkyl halide present in the starting material (Ar-CH3). Also, reacting with CH3C≡C-Br (propynyl bromide) as an electrophile doesn't make sense for forming the target bond in the right order. (b) KMnO4 and heat would oxidize the methyl group to a carboxylic acid (Ar-COOH), destroying the CH3 group entirely and giving the wrong functional group. This cannot lead to the desired product. (d) Mg in ether would attempt to form a Grignard reagent from the aryl bromide, but the aryl Br would react with Mg preferentially. Reacting the resulting aryl Grignard with CH3C≡CBr and then H3PO4 does not give the target product in a logical synthetic sequence. Only option (c) correctly uses NBS/CCl4/heat for selective benzylic bromination followed by SN2 with sodium propynylide to install the -C≡C-CH3 unit via the newly formed benzylic bromide. Therefore, the correct answer is C.