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Arrange the following free radicals in increasing order of stability: (CH3)2CH*, (CH3)3C*, (C6H5)2CH*, (C6H5)3C*.

1. (C6H5)3C* < (C6H5)2CH* < (CH3)3C* < (CH3)2CH*
2. (C6H5)2CH* < (C6H5)3C* < (CH3)3C* < (CH3)2CH*
3. (CH3)2CH* < (CH3)3C* < (C6H5)3C* < (C6H5)2CH*
4. (CH3)2CH* < (CH3)3C* < (C6H5)2CH* < (C6H5)3C*

Correct answer: (CH3)2CH* < (CH3)3C* < (C6H5)2CH* < (C6H5)3C*

Solution

Benzylic-type radicals delocalise the unpaired electron over the aromatic ring(s), so they are much more stable than purely alkyl radicals stabilised only by hyperconjugation. Among the alkyl ones, tertiary (CH3)3C* > secondary (CH3)2CH*. Among the aryl ones, triphenylmethyl (3 rings) > diphenylmethyl (2 rings). The triphenylmethyl radical is famously very stable.

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