Arrange the following radicals in decreasing order of their relative stability: (I) CH ₃ CH = CH - C H ₂ (II) C H = CHCH ₃ (III) CH ₃ C HCH ₃

Step-by-step solution

Radical (I) is an allylic radical, which is highly stabilized by resonance. Radical (III) is a secondary alkyl radical, which is stabilized by hyperconjugation from six -hydrogen atoms. Radical (II) is a vinylic radical, where the unpaired electron is present in an sp^2 hybridized orbital. It is highly unstable due to the higher electronegativity of the sp^2 carbon. Therefore, the decreasing order of stability is (I) > (III) > (II). Answer: (I) > (III) > (II)