Question

What is the decreasing order of the stabilities of the following free radicals?

• A. ((i) > (ii) > (iii)
• B. (iii) > (i) > (ii)
• C. (ii) > (iii) > (i)
• D. All thru are equally stable

Answer 1

The correct option is B

(iii) > (i) > (ii)

Let us draw the open forms of there free radicals.

For the carbon that contains the odd electron, there are $4\propto$ hydrogens. So there will be $4$ hyperconjugation. More on that later.

$C{H}_3-\stackrel{\bullet }{\stackrel{H}{\stackrel{}{\underset{C{H}_3}{\underset{|}{C}}}}}=CH-C{H}_3\leftrightarrow$ $C{H}_3-\stackrel{H}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-CH=\stackrel{\bullet }{\stackrel{H}{\stackrel{}{\underset{H}{\underset{|}{C}}}}}-H\leftrightarrow$

$C{H}_3-\stackrel{H}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-CH=\stackrel{H}{\stackrel{|}{\underset{H}{\underset{|}{C}}}}\stackrel{\bullet }{H}\leftrightarrow C{H}_3-\stackrel{H}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-CH=\stackrel{H}{\stackrel{}{\underset{\bullet }{\underset{H}{\underset{|}{C}}}}}-H\leftrightarrow$

As a general rule, more the number of $\propto$ hydrogen atoms, more the number of hyperconjugative structures.
$\Rightarrow$ More hyperconjugation (greater degree of hyper conjugation)
Greater desire of hyper conjugation equates to more stability. When we say $\propto$- hydrogen, here we refer to the $\sigma$ C-H bond of the carbon atoms directly attached to the carbon atom with unshared p-oribital. Similarly structure (ii) has only 2 $\propto$ hydrogen atom and (iii) has the maxmium (from the given) of 8 $\propto$-hydrogen.
Thus (iii) has maximum contributing structures and has the highest stability of the three.

Ans: b (iii) > (i) > (ii)