Question

does -M effect stabilize free radical

Answer 1

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The effect of substituents on the geometries, apicophilicities, radical stabilization energies, and bond dissociation energies of (*)P(CH(3))(3)X (X = CH(3), SCH(3), OCH(3), OH, CN, CF(3), Ph) were studied via high-level ab initio molecular orbital calculations. Two alternative definitions for the radical stabilization energy (RSE) were considered: the standard RSE, in which radical stability is measured relative to H-P(CH(3))(3)X, and a new definition, the alpha-RSE, which measures stability relative to P(CH(3))(2)X. We show that these alternative definitions yield almost diametrically opposed trends; we argue that alpha-RSE provides a reasonable qualitative measure of relative radical stability, while the standard RSE qualitatively reflects the relative strength of the P-H bonds in the corresponding H-P(CH(3))(3)X phosphines. The (*)P(CH(3))(3)X radicals assume a trigonal-bipyramidal structure, with the X-group occupying an axial position, and the unpaired electron distributed between a 3p(sigma)-type orbital (that occupies the position of the "fifth ligand"), and the sigma orbitals of the axial bonds. Consistent with this picture, the radical is stabilized by resonance (along the axial bonds) with configurations such as X(-) P(*+)(CH(3))(3) and X(*) P(CH(3))(3). As a result, substituents that are strong sigma-acceptors (such as F, OH, or OCH(3)) or have weak P-X bonds (such as SCH(3)) stabilize these configurations, resulting in the largest apicophilicities and alpha-RSEs. Unsaturated pi-acceptor substituents (such as phenyl or CN) are weakly stabilizing and interact with the 3p(sigma)-type orbital via a through-space effect. As part of this work, we challenge the notion that phosphorus-centered radicals are more stable than carbon-centered radicals.


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Answer 2

No, free radical is stabilised by +M. Free radical has incomplete octet.