See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound shown is triphenylmethane (Ph3CH), a central methine carbon (CH) bonded to three phenyl rings. Step 2 - Identify the reagents: H+ and AlCl3 are strong Lewis/Bronsted acid conditions capable of generating carbocations and facilitating electrophilic aromatic substitution (Friedel-Crafts type cyclization). Step 3 - Mechanistic concept (intramolecular Friedel-Crafts alkylation): Under acidic conditions (H+/AlCl3), the central C-H bond can be activated, or a hydride shift/protonation gives a triphenylmethyl carbocation (Ph3C+). Alternatively, the reagents promote an intramolecular electrophilic cyclization where one of the ortho positions of one phenyl ring attacks an adjacent phenyl ring's ortho carbon. Step 4 - Cyclization: In triphenylmethane, two of the three phenyl rings are in close proximity. Under acidic conditions, an intramolecular electrophilic attack occurs: the carbocation (or electrophilic carbon) at the methine position facilitates ring closure between two of the ortho-positioned phenyl rings. Specifically, one phenyl ring's ortho carbon forms a new C-C bond with the ortho carbon of an adjacent phenyl ring, creating a five-membered ring fused between two benzene rings — this is the fluorene skeleton. Step 5 - Product formation: The cyclization produces fluorene (dibenzo[a,d][7]annulene's smaller analog — actually dibenzofuran analog without oxygen; fluorene is two benzene rings fused via a cyclopentane ring), with the third phenyl group remaining as a substituent at C-9 (the sp3 carbon of the five-membered ring). This gives 9-phenylfluorene. Step 6 - Why this product: The driving force is the formation of the aromatic-stabilized bicyclic fluorene system. The intramolecular Friedel-Crafts reaction is favored because it creates a stable 5-membered ring fused to two benzene rings, and the reaction is intramolecular (high effective molarity). The remaining phenyl group at C-9 cannot further cyclize under these conditions without significant strain. Therefore, the correct answer is 9-phenylfluorene (fluorene with phenyl substituent at C-9, as drawn p.585).