See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Stability of Carbonium Ion from C–Cl Ionization **Key Principle:** Carbonium ion stability follows: tertiary > secondary > primary > methyl ## Analysis of Each Compound: **(1) $(CH_3)_3C$–Cl (tert-butyl chloride)** - Ionization gives: $(CH_3)_3C^+$ — **tertiary carbocation** - Three alkyl groups provide maximum hyperconjugation and inductive stabilization **(2) Benzyl chloride: C$_6$H$_5$CH$_2$–Cl** - Ionization gives: C$_6$H$_5$CH$_2^+$ — **primary benzylic carbocation** - Benzene ring provides resonance stabilization through multiple resonance structures - Positive charge delocalized onto aromatic ring ($\pi$ electrons) **(3) $O_2N$-C$_6$H$_4$–CH–Cl (nitrobenzyl chloride)** - Ionization gives: $O_2N$-C$_6$H$_4$–CH$_2^+$ — **primary carbocation** - Electron-withdrawing $NO_2$ group destabilizes the carbocation - Resonance stabilization is minimal **(4) $(CH_3)_2CH$–Cl (isopropyl chloride)** - Ionization gives: $(CH_3)_2CH^+$ — **secondary carbocation** - Less stabilization than tertiary ## Conclusion: **Option (2) is correct** because the benzylic carbocation achieves exceptional stability through **resonance delocalization** of the positive charge onto the aromatic ring. Although it's technically primary, resonance stabilization makes it more stable than the secondary carbocation in (4) and comparable to or exceeding some tertiary carbocations.