What is the decreasing order of the stabilities of the following free radicals?
(iii) > (i) > (ii)
Let us draw the open forms of there free radicals.
For the carbon that contains the odd electron, there are 4∝ hydrogens. So there will be 4 hyperconjugation. More on that later.
CH3−∙H C|CH3=CH−CH3↔ CH3−H|C|CH3−CH=∙H C|H−H↔
CH3−H|C|CH3−CH=H|C|H ∙H ↔ CH3−H|C|CH3−CH=H C|H∙−H↔
As a general rule, more the number of ∝ hydrogen atoms, more the number of hyperconjugative structures.
⇒ More hyperconjugation (greater degree of hyper conjugation)
Greater desire of hyper conjugation equates to more stability. When we say ∝- hydrogen, here we refer to the σ C-H bond of the carbon atoms directly attached to the carbon atom with unshared p-oribital. Similarly structure (ii) has only 2 ∝ hydrogen atom and (iii) has the maxmium (from the given) of 8 ∝-hydrogen.
Thus (iii) has maximum contributing structures and has the highest stability of the three.
Ans: b (iii) > (i) > (ii)