See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The sulfite dianion SO3^2- has the structure with sulfur bearing a lone pair, one S=O (the exocyclic oxygen giving the dianion its charge distribution), and two S-O^- bonds. The key question is which site (oxygen or sulfur) acts as the nucleophile/site of protonation or alkylation. Step 1 - Structure of sulfite dianion: The sulfite ion SO3^2- (shown as O^- - S(=O) - O^-) has sulfur as the central atom with a lone pair. It is ambident: it can react at oxygen (O-attack) or at sulfur (S-attack). Step 2 - Reaction with H+ (Product A): Protonation is thermodynamic/kinetic and occurs at the more electronegative oxygen atom (O-protonation). This gives O^- - S(=O) - O - H, which is the bisulfite/hydrogen sulfite ion (HSO3^-). This matches option (b)'s first product: O^- - S(=O) - O - H. Step 3 - Reaction with CH3I (Product B): Alkylation with soft electrophiles like alkyl halides (CH3I) follows HSAB theory. CH3I is a soft electrophile, and sulfur is a soft nucleophile. Therefore, S-alkylation occurs, giving a product where CH3 attaches to sulfur. The sulfur, now bonded to CH3, O^-, =O, and still one O^- results in a zwitterionic/charged sulfur species: O^- - S^+(=O)(O^-) - CH3. This is methylation at sulfur, producing methyl sulfonate-type species with a positively charged sulfur bearing O^-, =O, CH3, and maintaining the other O^-. This matches option (b)'s second product: O^- - S^+(=O)(O^-) - CH3. Step 4 - Why other options fail: - Option (a): Shows O-methylation for product B (O^- - S(=O) - O - CH3), but CH3I being a soft electrophile prefers S-alkylation over O-alkylation. Incorrect. - Option (c): Shows O-methylation for A and S-methylation+H on S for B, which is inconsistent with protonation giving O-H product. - Option (d): Shows protonation at sulfur for A and O-methylation for B, both incorrect by HSAB and standard sulfite chemistry. Therefore, the correct answer is B.