Apr 22, 2013
im confused on what my professor said here.
radical stability is ranked 3>2>1 and 3 being the most stable. if 3 is the most stable should it be predominantly the major product? he gave us an example where the the radical replaced H in the primary hydrogen. but in the ranking primary is the least stable. im confused thanks
Oct 20, 2013
Yes, stability for carbocations and radicals are the same order 3>2>1 and yes, least reactive means most stable.
Do you have the example?
Hmm was there a double bond in the example? Allylic carbocations and radicals are really stable (even more than 3) due to resonance.


Company Rep & Bad Singer
10+ Year Member
May 25, 2007
There is a second factor at play here, and that is the number of 1˚, 2˚, and 3˚ Hs on the molecule. If you think about 2-methylbutane for instance, you have nine 1˚ Hs, two 2˚ Hs, and only one 3˚ H. If 3˚ is let's say four times as stable as 1˚ at a given temperature, that isn't a big enough favorability to offset the 9 : 1 ratio. So your product distribution of 3˚ to 1˚ will end up (4 x 1) : (1 x 9) = 4 : 9. You also need to consider the reaction rate. If it's a fast reaction, then there isn't time to be selective and the distribution will resemble that of random probability. If the reaction is slow, then there is plenty of time for the stable intermediate to form, and thus lead to the 3˚ product. This is why chlorination is not very selective (very fast reaction) and bromination is highly selective (slow reaction).

Hope this helps.