See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Free Radical Stability Analysis **Radical Stability Order:** Tertiary > Secondary > Primary > Methyl **Evaluating each option:** **(A) Benzyl radical** ($C_6H_5-\dot{CH}_2$) - Primary radical with resonance stabilization through the benzene ring - Moderate stability due to resonance delocalization **(B) Nitro-substituted benzyl radical** ($NO_2$-$C_6H_4$-$\dot{CH}_2$) - Primary radical with electron-withdrawing $NO_2$ group - $NO_2$ destabilizes the radical by pulling electron density away - Less stable than (A) **(C) Tertiary radical** ($(CH_3)_3\dot{C}$) - Three alkyl groups attached to the radical carbon - **Maximum hyperconjugation and inductive stabilization** from three methyl groups - Highest stability due to: - Three $\sigma$-bonds donating electron density to the unpaired electron - Hyperconjugation from all three methyl C-H bonds **(D) Secondary radical** ($(CH_3)_2\dot{CH}$) - Two alkyl groups provide stabilization - Less stable than tertiary (only 2 hyperconjugation sources vs. 3) **Answer: (C)** — The tertiary radical $(CH_3)_3\dot{C}$ is most stable because the three methyl groups provide maximum hyperconjugation and inductive electron donation to stabilize the unpaired electron.