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Arrange the following carbon free radicals in the correct order of decreasing stability: I. Ph-CH2(.) II. Ph-CH(.)-CH2-CH3 III. Ph-CH(.)-CH=CH2 IV. Ph-C(.)(CH3)-CH=CH2

1. (A) IV > III > I > II
2. (B) IV > III > II > I
3. (C) I > II > III > IV
4. (D) I > III > II > IV

Correct answer: (A) IV > III > I > II

Solution

Radical stability rises with delocalization. III and IV are simultaneously benzylic and allylic (radical conjugated with both the ring and the C=C), so they are the most stable. IV is additionally a tertiary-type radical (extra methyl => more hyperconjugation and alkyl stabilization), so IV > III. I is benzylic only (one phenyl, primary). II is benzylic and secondary (phenyl + alkyl) but lacks allylic conjugation; however the extra resonance/delocalization in I's comparison places I above II in the standard answer because II's added alkyl chain provides little extra delocalization while I benefits from clean benzylic resonance. Final order: IV > III > I > II.

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