See image — Aromatic Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When a C–Cl bond breaks, the chlorine can depart as Cl⁺ (electrofuge), Cl⁻ (nucleofuge/heterolytic with electrons going to carbon), or Cl• (homolytic). The key is determining which mode is favored by the stability of the resulting carbon species or chlorine species. Step 1: Identify the structure. The compound is tricyclopropylcarbinyl chloride — a carbon bearing three cyclopropyl groups and one Cl. This is the central spiro-type carbon. Step 2: Consider heterolytic ionization. If Cl leaves as Cl⁻, the carbon becomes a carbocation: tricyclopropylcarbinyl cation. This cation is extraordinarily stable due to the well-known 'cyclopropylcarbinyl cation' effect. Each cyclopropyl group provides strong homoconjugative (bisected orbital) stabilization to the adjacent carbocation. With three cyclopropyl groups, the resulting carbocation is exceptionally stable — among the most stable carbocations known. Step 3: Because the tricyclopropylcarbinyl cation is so exceptionally stable, the ionization C–Cl → carbocation + Cl⁻ is highly favored. The driving force is the enormous stability of the resulting carbenium ion. Step 4: Evaluate other options. Cl⁺ would require the carbon to become a carbanion (tricyclopropylcarbanion), which is not stabilized and thus unfavorable. Cl• (homolysis) would give a carbon radical, which is less favorable than the highly stabilized carbocation. Cl²⁺ is chemically unreasonable. Step 5: Therefore, Cl is liberated most easily as Cl⁻, with the carbon retaining the bonding electrons to form the highly stabilized tricyclopropylcarbinyl cation. Therefore, the correct answer is B.