COMEDK 2025 Evening Solved Paper

Concept:Free radical stability increases with the delocalization of the unpaired electron.Explanation:The stability of a free radical depends on how effectively the unpaired electron can be spread out.Resonance delocalization over aromatic rings provides much stronger stabilization than hyperconjugation.Compare each radical:• Methyl radical ($\dot{\mathrm{C}}\mathrm{H}_3$): Only three hyperconjugative C–H bonds, no resonance.• Vinyl radical ($\mathrm{CH}_2=\dot{\mathrm{C}}\mathrm{H}$): Unpaired electron on an sp² carbon; very little stabilization.• Tertiary butyl radical ($(\mathrm{CH}_3)_3\dot{\mathrm{C}}$): Nine hyperconjugative C–H bonds, moderate stabilization.• Diphenylmethyl radical ($(\mathrm{C}_6\mathrm{H}_5)_2\dot{\mathrm{C}}\mathrm{H}$): The unpaired electron is delocalized over two entire phenyl rings via resonance. This offers the greatest possible stabilization among the given options.Because resonance over two aromatic rings outweighs any hyperconjugation effect, the diphenylmethyl radical is the most stable.Answer:Option A: $\left(\mathrm{C}_6\mathrm{H}_5\right)_2\dot{\mathrm{C}}\mathrm{H}$