Homolytic fission of the following alkanes forms free radicals CH3−CH3,CH3−CH2−CH3,(CH3)2CH−CH3,CH3−CH2−CH(CH3)2.
The increasing order of stability of the radicals is:
A
(CH3)3.C<(CH3)2.C−CH2CH3<CH3−.CH−CH3<CH3−.CH2
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B
(CH3)2.C−CH2CH3<CH3−.CH−CH3<CH3−.CH2<(CH3)3.C
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C
CH3−.CH2<CH3−.CH−CH3<(CH3)2.C−CH2CH3<(CH3)3.C
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D
CH3−.CH2<CH3−.CH−CH3<(CH3)3.C<(CH3)2.C−CH2CH3
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Solution
The correct option is CCH3−.CH2<CH3−.CH−CH3<(CH3)2.C−CH2CH3<(CH3)3.C Stability of radical follows similar order as carbocation follows.
(methyl<1o<2o<3o).
Moreover, stability of free radical∝Number ofα−H