Arrange the following radicals in decreasing order of stability: (I) (C6H5)3C., (II) (C6H5)2CH., (III) C6H5-CH2., (IV) CH3..

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JEE Advanced levelSingle correct optionDifficulty 3 of 3g1094

Arrange the following radicals in decreasing order of stability: (I) $(C X_{6} H X_{5}) X_{3} C \cdot$ , (II) $(C X_{6} H X_{5}) X_{2} CH \cdot$ , (III) $C X_{6} H X_{5} - CH X_{2} \cdot$ , (IV) $CH X_{3} \cdot$ .

A $I > II > III > IV$
B $IV > III > II > I$
C $I > III > II > IV$
D $II > I > III > IV$

Answer: A

The unpaired electron in each aryl-substituted methyl radical is delocalised into the aromatic rings. Triphenylmethyl delocalises over three rings, diphenylmethyl over two, benzyl over one, and the methyl radical has no delocalisation. The stability follows the number of rings, giving I > II > III > IV.

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Arrange the following radicals in decreasing order of stability: (I) (CX6​HX5​)X3​C⋅, (II) (CX6​HX5​)X2​CH⋅, (III) CX6​HX5​−CHX2​⋅, (IV) CHX3​⋅.

Answer: A

More questions on Stability of resonating structures and conjugated species

Arrange the following radicals in decreasing order of stability: (I) $(C X_{6} H X_{5}) X_{3} C \cdot$ , (II) $(C X_{6} H X_{5}) X_{2} CH \cdot$ , (III) $C X_{6} H X_{5} - CH X_{2} \cdot$ , (IV) $CH X_{3} \cdot$ .