This is #77 from Achiever Test 1.

Which of the following sets of hydrocarbons correctly displays free radicals in descending order of stability?



A. CH3H2C. > CH2=CHĊH2 > (CH3)3C. > (CH3)2HC.
B. CH2=CHĊH2 > (CH3)3C. > (CH3)2HC. > CH3H2C.
C. CH3H2C. > CH2=CHĊH2 > (CH3)3C. > (CH3)2HC.
D. (CH3)2HC. > CH3H2C. > CH2=CHĊH2 > (CH3)3C.
E. (CH3)3C. > CH2=CHĊH2 > (CH3)2HC. > CH3H2C.

The correct answer is B, but I thought that the most stable radical was a tertiary radical, making the answer E. Could someone please explain to me why I'm wrong? Thanks.

The radical (or a carbocation) is most stable when in a conjugated system.

The tertiary readical is very stable, yes. But the conjugated radical is even more stable. This is because the lone electron can diffuse through a conjugated allylic system. Studies have shown that a conjugated radical is more stable than the tertiary radical.

when you hear radical and stability, think resonance structures first. the one with the most resonance will probably be the most stable

Ok, I got it. Thanks.