See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: 2-(Chloromethyl)furan undergoes SN1 solvolysis because the carbocation intermediate (furfuryl cation) formed after loss of Cl- is stabilized by resonance with the furan ring oxygen. This delocalization spreads positive charge to multiple positions of the furan ring. Step 1 - Ionization: 2-(Chloromethyl)furan loses Cl- under SN1 conditions to form a resonance-stabilized furfuryl carbocation. The positive charge is delocalized not only on the benzylic-type CH2+ carbon but also onto C3 and C5 of the furan ring via the oxygen lone pairs. Step 2 - Resonance structures of the carbocation: The primary carbocation at the exocyclic CH2+ position resonates to place positive charge at C5 (and C3) of the furan ring through the oxygen. This gives three major resonance contributors with electrophilic character at the exocyclic carbon, at C3, and at C5 of the ring. Step 3 - Nucleophilic attack by CH3O-: Methoxide (from CH3ONa in CH3OH) attacks each electrophilic site: - Attack at exocyclic CH2+ gives 2-(methoxymethyl)furan → product (a) - Attack at C5 of the ring carbocation gives a product where OCH3 is at C5 and CH3 (or CH2) rearrangement gives the structure shown in (b) - Attack at C3 of the ring carbocation gives a product where OCH3 is at C3 and the CH3 group ends up at C2, giving structure (c) Step 4 - Why all three are formed: Because the SN1 mechanism produces a delocalized carbocation with electrophilic sites at multiple positions, methoxide can attack any of these positions, yielding all three products (a), (b), and (c). Step 5 - Why other individual options fail: Options (a), (b), and (c) alone are each only partially correct because they account for only one of the possible attack sites on the delocalized cation. Since all three products are simultaneously possible, option (d) 'All of these' is the correct and complete answer. Therefore, the correct answer is D.