Give the decreasing order of the stabilities of the following:

i i i > i i > i

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i i > i i i > i

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i i i > i > i i

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i > i i i > i i

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Solution

The correct option is C $i i i > i > i i$
The correct answer is option(c). (i), (ii) and (iii) are secondary, primary and tertiary free radical respectively. The electron donating groups increase the stability of free radical. So alkyl group being the electron donating group increase the stability of free radical.
In tertiary free radical there are 3 alkyl groups present, so it is the most stable among the three. In secondary free radical there are two alkyl groups are present, so it is more stable than primary free radical.

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