See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: SbCl5 is a strong Lewis acid and oxidizing agent. When an allylic/vinyl chloride substrate reacts with SbCl5, it can act as a hydride/chloride abstractor (oxidant), removing electrons from the substrate. Step 1 - Identify the starting material: The starting material is a cyclooctadienyl chloride (an 8-membered carbocyclic ring containing two double bonds and a C-Cl bond). This is consistent with a structure like chlorocycloocta-2,5-diene or similar, which upon ionization and rearrangement can give cyclooctatetraene (COT)-derived cation. Step 2 - Role of SbCl5: SbCl5 abstracts the chloride ion from the allylic chloride to form a carbocation. The reaction: R-Cl + SbCl5 → R(+) + SbCl6(-). With 2 equivalents of SbCl5, two chloride abstractions can occur, or one chloride abstraction followed by further oxidation. Step 3 - Formation of the dication: The cyclooctadienyl chloride, upon treatment with 2 equivalents of SbCl5, undergoes ionization. The chloride is abstracted by SbCl5 to give SbCl6(-) and a carbocation. The resulting 8-membered ring carbocation can lose another hydride or undergo further oxidation/elimination to generate the cyclooctatetraene dication (COT2+). COT2+ is a 6 pi-electron aromatic system (Hückel: 4n+2 with n=1), making it stable. Step 4 - Why COT2+ is favorable: Cyclooctatetraene dication (COT2+) has 6 pi electrons and is planar aromatic, satisfying Hückel's rule (4n+2, n=1). The two SbCl5 molecules each abstract one Cl or facilitate two-electron oxidation, giving [COT]2+ with 2 SbCl6(-) as counterions. Step 5 - Why other options fail: - Option (a): COT2- (dianion) would require reduction, not oxidation by SbCl5. SbCl5 is an oxidant, not a reductant. - Option (c): Simple neutral COT would require only elimination of HCl, not involvement of 2 SbCl5 as oxidizing agents. - Option (d): A mixture of dianion and dication is chemically unreasonable under these oxidizing conditions. Therefore, the correct answer is B.