See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When a vicinal dihalide (1,2-dihalide or in this case a 1,3-dibromo substituted cyclobutene) is treated with a silver salt like AgBF4, the silver ions abstract the halide ions as insoluble AgBr, generating carbocation centers. The driving force here is aromaticity. Step 1: Identify the starting material. The starting material is 1,2-dibromo-3,4-diphenyl... more precisely it is a cyclobutene with Br at C1 and C3 (the sp3 carbons) and Ph groups at all four carbons (two Ph on C1, two Ph on C3, or one Ph each on C1, C2, C3, C4 with Br on C1 and C3). Looking at the structure: C1 has Br and Ph, C2 has Ph (part of double bond), C3 has Ph (part of double bond), C4 has Br and Ph. So it is 1,4-dibromo-1,2,3,4-tetraphenylcyclobut-2-ene (trans-1,4-dibromo-1,2,3,4-tetraphenylcyclobut-2-ene). Step 2: Reaction with 2 AgBF4. Each AgBF4 abstracts one Br^- (forming AgBr precipitate) and one BF4^- is released as counterion. Removing 2 Br^- (each as Br^-) from C1 and C4 generates a dication: the cyclobutadiene ring bears a 2+ charge overall. Step 3: Aromaticity consideration. Cyclobutadienyl dication (cyclobutadiene^2+) has 2 pi electrons (4n+2 with n=0), making it a Hückel aromatic system. This is the driving force for the ionization. The resulting tetraphenylcyclobutadienyl dication is stabilized by aromaticity (2 pi electrons) and by the four phenyl groups. Step 4: The product (x) is the tetraphenylcyclobutadienyl dication with charge +2, with BF4^- as counterions. This corresponds to option (a). Why other options fail: - (b) -2 charge: removing Br^- gives cations, not anions. - (c) +1 charge: two equivalents of AgBF4 are used, abstracting two bromides, giving +2 not +1. - (d) -1 charge: same reasoning, ionization gives positive charge. Therefore, the correct answer is A.