See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 2-iodotoluene (ortho-iodotoluene), a benzene ring bearing an iodo substituent and a methyl group at adjacent (ortho) positions. Step 2 - Reaction with NBS (N-bromosuccinimide): NBS is a benzylic bromination reagent. Under radical (or ionic with light/heat) conditions, NBS selectively brominates the benzylic C-H bond. The methyl group (CH3) adjacent to the ring is at the benzylic position, so NBS converts CH3 into CH2Br. The iodo substituent on the ring is unaffected under these mild radical conditions. Product (A) is therefore 2-iodobenzyl bromide: a benzene ring with I at one position and CH2Br at the ortho position. Step 3 - Reaction with CH3SNa: CH3SNa (sodium methanethiolate, NaSCH3) is a good nucleophile. It undergoes SN2 reaction with the benzylic bromide (CH2Br), displacing bromide and attaching the SCH3 group to the benzylic carbon. This converts CH2Br into CH2-S-CH3. The aryl iodide is not reactive under these mild SN2 conditions (aryl halides do not undergo SN2). Product (B) is therefore a benzene ring with I at one position and CH2-S-CH3 at the ortho position. Step 4 - Match to options: Option (a) shows exactly this structure: benzene ring with I and CH2-S-CH3 at ortho positions. This matches product (B). Why other options fail: - Option (b) shows direct Ar-S-CH3 (no CH2 linker), which would require SNAr or a different mechanism not applicable here; NBS does not remove all three H atoms from CH3. - Option (c) shows CH2Br remaining and SCH3 replacing iodine, which is incorrect because aryl iodides do not undergo SN2 with thiolate under these conditions. - Option (d) is eliminated since option (a) is correct. Therefore, the correct answer is A.