Question

Which is the most stable free radical?
A.${H_2}C$ = $CH\dot C{H_2}$
B.$C{H_3}\dot C{H_2}$
C.${(C{H_3})_2}\dot CH$
D.${(C{H_3})_3}\dot C$

Answer 1

If the molecules containing at least one unpaired electrons, that type of molecule is known as free radicals. Generally most of the molecules have an even number of molecules and it makes covalent bonds with other compounds. But some of the molecules will contain one or more unpaired electrons and it is known as free radicals. And there is a relation between resonance and the stability of free radicals.

Complete answer:
This free radical is an allylic free radical. And there is a delocalization of electrons that will take place to attain the stability. And this is the most stable free radical in these compounds. And the order of stability is,
$C{H_2}$ = $CH\dot C{H_2} > {(C{H_3})_3}\dot C > {(C{H_3})_2}\dot CH > C{H_3}\dot C{H_2}$
And, here the stability is attained by resonance effect. Therefore, the unpaired electron will be delocalized through the conjugated double bond. Hence, option (A) is correct.
This radical is a primary free radical and it is the least stable free radical. Hence, the option (B) is incorrect.
The secondary free radical is less stable than the tertiary and allylic free radical. Hence, option (C) is incorrect.
This is a tertiary free radical. And these radicals are attained stability by hyperconjugation. But it is less stable than allylic free radicals. Hence, the option (D) is incorrect.

Hence, option (A) is correct.

Note:
Among these given free radicals, allylic free radicals are the most stable free radical. Because, there is a delocalization of electrons that takes place in that allyl free radical and the resonance will affect the stability of this free radical. But, primary, secondary and tertiary free radicals are less stable than allyl free radicals. Because, there is no delocalization of electrons present in these free radicals.