Q.G. Find out the most stable species from the following. Justify.
a. Methyl free radical (CH3•), isopropyl free radical [(CH3)2CH•], tert-butyl free radical [(CH3)3C•]
b. Methyl free radical (CH3•), bromomethyl free radical (CH2Br•), tribromomethyl free radical (CBr3•)
c. Methyl carbocation (CH3⊕), chloromethyl carbocation (CH2Cl⊕), trichloromethyl carbocation (CCl3⊕)

Step 1a. Free radicals -- methyl, isopropyl, tert-butyl. By section 14.6.1's stated order (matching section 14.6.8's hyperconjugation argument), stability rises tert-butyl (3°) > isopropyl (2°) > ethyl/methyl, since more directly-attached alkyl groups mean more α-hydrogens available for hyperconjugative (+I/+hyperconjugation) stabilisation of the radical. The tert-butyl radical, (CH3)3C•, carries 9 α-hydrogens (three from each of its three methyl groups) and is therefore the most stable of the three.

Step 1b. Free radicals -- methyl, bromomethyl, tribromomethyl. Bromine is inductively electron-withdrawing (-I, section 14.6.4), which would seem to destabilise an adjacent radical centre; but bromine ALSO carries lone pairs that can be donated by a resonance-type (+R) mechanism directly into the radical's singly-occupied orbital, exactly the kind of donation section 14.6.6 assigns to halogens generally. With three bromine atoms all able to donate this way, CBr3• ends up more stabilised overall than the plain methyl radical, CH3•, which has no such donation available at all. …

Applying the chapter's own stated stability orders and worked reasoning (hyperconjugation for alkyl-substituted radicals; halogen +R lone-pair donation for radicals; halogen -I withd …

• Applying the SAME reasoning (pure inductive withdrawal) to both the radical set (b) and the cation set (c), when a radical and a cation respond differently to an attached halogen's competing -I and +R effects. …