SolutionsStability of radicals
1. Order the radicals
Arrange from most to least stable: (CH3)2CH·, CH3·, (CH3)3C· and CH3CH2·. Classify them as methyl, primary, secondary or tertiary.
(CH3)3C· (tertiary) > (CH3)2CH· (secondary) > CH3CH2· (primary) > CH3· (methyl).
2. Dissociation energies
The energies of the primary, secondary and tertiary C–H bonds of 2-methylbutane are 98, 94.5 and 93 kcal/mol. Which hydrogen is the easiest to remove? How does this relate to the stability of the radical?
The tertiary one, because its bond is the weakest. The more stable the radical formed, the less energy it costs to break the C–H bond.
3. Geometry of the methyl radical
What is the hybridization of the carbon in the methyl radical? What is its geometry and where is the unpaired electron?
sp2 hybridization, trigonal planar geometry, and the unpaired electron occupies the unhybridized p orbital, perpendicular to the plane.
4. Count the stabilizing bonds
How many neighbouring C–H bonds can overlap with the radical p orbital in the methyl, ethyl, isopropyl and tert-butyl radicals?
Methyl: 0; ethyl: 3; isopropyl: 6; tert-butyl: 9.
The more neighbouring σ bonds, the more the electron deficiency is delocalized and the less unstable the radical.