Radicals are unstable intermediates. So we’re going to have to discuss some ways to stabilize them.
Concept: The radical stability trend.4m
Radicals are very high energy and very short lived, so anything that we can do to stabilize them whatsoever, will have a really big effect in their likelihood to be formed. What that means is that we have to figure out what is the trend of stability for radicals.
I just want to show you guys right now, basically, this is the trend. What you're going to notice is that I'm going to compare this trend to the trend for carbocations. Now if you don't know the trend for carbocations yet, that's okay. I'm just going to point out the major difference here.
First of all, radicals are electron deficient. Now what I mean by that is that there's an orbital. And usually, each orbital has space for how many electrons? Two. That's the Pauli Exclusion principle. But, in this case, we have a radical with just one electron so that would be what we call a partially-filled orbital. That's not very stable. So anyway that we can push electrons into that orbital, that will make it more stable. There is an effect that does that and that effect is called hyperconjugation.
What hyperconjugation says is that the more R groups you have around an empty orbital or a partially-filled orbital, the more stable it will be. So in this case, what I want to do is I want to basically say the more R groups are on my radical, the more stable it's going to be. Easy.
Notice that that trend does hold true. This actually holds true for both carbocations, which are empty orbitals completely, there's nothing in there, and radicals. They're really the same thing, so notice that for my increasing stability, I have here that I have a tertiary here, a tertiary carbocation is very stable, and I also have a tertiary radical kind of at the top here.
But notice there's a slight difference here. It turns out that tertiary is the best type of carbocation that I can form, but it's not the best type of radical. I actually have a different type of radical here that's more stable. That's because it turns out that unlike carbocations, allylic and benzylic radicals are actually going to be the most stable. Now allylic and benzylic are just words to mean that you're next to a double bond or you're next to a benzene ring. What that's saying is that if you can resonate, that's going to make the radical more stable than anything.
Here I have drawn the allylic and the benzylic. This would be allylic. This would be benzylic. Notice that both of these are directly next to a double bond, so a double bond and the radical could switch places through resonance structure. What that would do is that would delocalize that electron's efficiency over several atoms, stabilizing it.
What I want you guys to be mindful of is that this is actually going to be important for reactions. These sites here are very crucial for reactions that we're going to learn later because they're very stable.
What I want you guys to do here is determine which of the following radicals would be the most stable by looking at this trend, just basically looking – forget the carbocation one because we're not talking about those. We're just talking about radicals. Figure out which of these would be the most stable and why. Don't forget to look at resonance structures to make sure that you're looking at both of the ways that the radical could be represented because remember that in a resonance structure, they're constantly in hybrid of each other. So you can't determine stability just based on one of the resonance structures.
Unlike carbocations, allylic and benzylic radicals are ALWAYS most stable.
Example: Determine which of the following radicals is the most stable.4m
List the stability of different kinds of radicals: 1°, 2°, 3° and methyl radicals. Why? (Use only one word/terminology to explain)
Identify the most stable and least stable radical of the following.
For each of the following, choose the best answer (give the letter).
(i) Which one has the smaller heat of hydrogenation: (a) 1-methylcyclohexene or (b) 3-methylcyclohexene?
(ii) Which one has the larger heat of combustion: (a) (E)-cyclononene or (b) (Z)-cyclononene?
(iii) Which one is a concerted reaction: (a) acid-catalyzed dehydration of an alcohol; (b) reaction of a tertiary alkyl chloride
with potassium tert-butoxide; or (c) addition of Cl 2 to an alkene?
(iv) Which one is thermodynamically more stable: (a) isobutyl radical or (b) tert-butyl radical?
(v) Which one is the rate law for an E2 reaction of an alkyl halide: (a) rate = k[RX] or (b) rate = k[RX][base]?
(vi) The major product of the reaction of methylcyclopentane with Br 2 in the presence of heat and light is which of
the following type of alkyl bromide: (a) primary; (b) secondary; or (c) tertiary?
(vii) The major product of the reaction of 1-pentene with HBr in the presence of peroxides and heat is formed via which of
the following: (a) a secondary carbocation; (b) a secondary radical; or (c) a secondary carbanion?
(viii) Stereoisomers that are not mirror images of each other and are not superimposable are:
(a) diastereomers; (b) enantiomers; (c) conformers; (d) polymers.
Which is the most stable radical?
Rank the following carbon radicals according to their increasing stability. (least stable most stable)
A. 1, 4, 3, 2
B. 1, 4, 2, 3
C. 2, 3, 1, 4
D. 3, 4, 2, 1
E. 4, 1, 2, 3
F. 3, 2, 1, 4
Which of the following is the most stable radical?
Rank the following radicals in order of stability (1 = most stable, 4 = least stable).
Circle the most easily abstracted hydrogen (H a, H b, Hc, or H d) in free radical bromination and explain why.
In the molecule shown below, determine which of the labeled bonds is expected to have the lowest bond dissociation energy.