See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Precipitation with $AgNO_3$ The question asks which brominated compound will form a precipitate with aqueous $AgNO_3$. This depends on whether the C–Br bond is **ionic** (forms $Br^-$ ions) or **covalent** (remains as C–Br). ## Key Principle $Ag^+$ precipitates as $AgBr$ only when **ionic bromide ions** ($Br^-$) are released in solution. This occurs via **SN1 mechanism**, which requires a stable carbocation intermediate. ## Analysis of Each Option **(A) Bromobenzene (C$_6$H_5$Br$):** - C–Br bond is part of aromatic ring and very strong (sp² hybridization) - Cannot form carbocation; very poor SN1 substrate - No precipitate **(B) Tropylium bromide derivative (7-membered ring):** - The benzylic C–Br is attached to an allylic position on a 7-membered ring - Forms a **highly stable aromatic tropylium cation** ($C_7H_7^+$), which is a resonance-stabilized system - Undergoes rapid SN1, releasing $Br^-$ ions - $Ag^+$ + $Br^- \rightarrow AgBr \downarrow$ (white precipitate) **(C) Cyclopentadienyl bromide:** - Although 5-membered ring, less stabilized carbocation than tropylium - Much slower reaction than option (B) **(D) Cyclopropyl bromide:** - Forms unstable primary carbocation - No significant SN1 reaction; no precipitate ## Answer: **(B)** The 7-membered ring allows formation of the exceptionally stable aromatic **tropylium cation**, making it the only substrate that readily undergoes SN1 hydrolysis to release $Br^-$ and form $AgBr$ precipitate.