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Question

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


A

((i) > (ii) > (iii)

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B

(iii) > (i) > (ii)

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C

(ii) > (iii) > (i)

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D

All thru are equally stable

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Solution

The correct option is C

(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.

CH3H C|CH3=CHCH3 CH3H|C|CH3CH=H C|HH

CH3H|C|CH3CH=H|C|H H CH3H|C|CH3CH=H C|HH

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)


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