See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

## Analysis of White Precipitate Formation with $AgNO_3$ The key is identifying which compounds release **free $Cl^-$ ions** in solution to form white $AgCl$ precipitate. **Compound (I): Chlorobenzene ($C_6H_5-Cl$)** - $C-Cl$ bond is **directly attached to the aromatic ring** - This bond is very strong (resonance stabilization) and highly polar C is electron-deficient - $Cl$ **cannot ionize** as $Cl^-$ in aqueous solution - **No precipitate** with $AgNO_3$ **Compound (II): Benzyl chloride ($C_6H_5-CH_2Cl$)** - $C-Cl$ bond is on an **aliphatic carbon** (benzylic position) - Benzylic carbocation is stabilized by resonance with the aromatic ring - $Cl^-$ easily ionizes: $C_6H_5CH_2Cl \rightarrow C_6H_5CH_2^+ + Cl^-$ - **Produces white $AgCl$ precipitate** ✓ **Compound (III): Benzyl chloride ($C_6H_5-CH_2Cl$)** - Same structure as (II) - **Produces white $AgCl$ precipitate** ✓ ## Conclusion Compounds **(I) and (III)** give white precipitate with $AgNO_3$. **Answer: (C)** **Why not (A) or (B)?** - (A) includes (I), which cannot ionize $Cl^-$ - (B) includes (II) ✓ but excludes (I) which is correct to exclude