In the last post we introduced free radicals – neutral, electron-deficient chemical species with a partially filled orbital – and learned that they are highly reactive intermediates in organic chemistry.
In this post we’ll cover two of the most important concepts concerning these species: their geometry, and their stability.
It’s this latter concept that we’ll see is particularly important for understanding many free-radical reactions in organic chemistry. [Spoiler: the factors that affect free radical stability are the same factors that stabilize carbocations [discussed previously here]
Let’s talk a bit about stability first, and then circle back to their structure. Being electron deficient, you might already have a hunch regarding factors that might stabilize free radicals. Waaaay back, we talked about how a considerable portion of organic chemistry can be explained simply by understanding that: 1) opposite charges attract (and like charges repel), and 2) the stability of charges increases if it can be spread out over a greater volume. These still apply here!
Electron poor species are stabilized by neighboring atoms that can donate electron density. [“if you’re poor, it helps to have rich neighbors”]. The most common way to interpret “rich neighbors” here is the observation that increasing the number of alkyl groups on the carbon bearing the free radical increases its stability. Radical stability increases in the order methyl < primary < secondary < tertiary. [For a second, more conceptually complex example, see the bottom of the post]. **
Secondly, we have also learned that any factor which can lead to the electron deficient site being delocalized [spread out] over a larger area will also stabilize electron poor species. Previously, for example, we saw that the positive charge of a carbocation was considerably stabilized when it was adjacent to a π bond. That’s because the carbocation is sp2 hybridized and bears an empty p orbital, allowing for overlap with the adjacent p orbitals and therefore leading the positive charge to be delocalized over multiple carbon atoms, in a manner that is most easily grasped by drawing resonance structures.
Carbocations are flat – so it’s easy to see how the p orbital could be in line with adjacent p orbitals of a double bond. But what about the geometry of free radicals?
If we draw out the electrons in a typical alkyl free radical, we see that there are three bonding pairs and a single unpaired electron, for a total of four occupied orbitals. By analogy to, say, amines, we might expect that the hybridization of the molecule to be sp3 and geometry of a free radical would be trigonal pyramidal. That’s actually a good approximation, except that the “pyramid” is a little shallower than it is for molecules which have a full lone pair. ** [see note below]
When the free radical is adjacent to a π bond, there’s a significant stabilization to be obtained if the p orbitals are all in line so they can overlap [“conjugation”] with each other. Overlap is increased (and the molecule’s energy lowered) if the “shallow pyramid” is flattened out. It’s a good approximation to think of a free radical adjacent to a π bond as being “sp2” hybridized.
So what does this all boil down to? The electron-deficient free radical can be delocalized over multiple carbons. Therefore, free radicals are stabilized by resonance.
If you read the article on the stabilization of carbocations, you might notice something: the same factors which stabilize free radicals are also the same factors which stabilize carbocations!
Quiz time: one of the most stable free radicals known is the triphenylmethyl radical, discovered by Moses Gomberg in 1900. In the absence of oxygen, this radical is indefinitely stable at room temperature. Can you identify the factors which might make this free radical particularly stable?
Next Post: What Factors Destabilize Free Radicals?
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[Optional – for more advanced students]
In addition to alkyl groups, free radicals are also stabilized by adjacent groups with lone pairs, such as oxygen and nitrogen. At first thought, oxygen might not seem like much of an electron donating group, since it’s quite electronegative. However, oxygen does have two lone pairs of electrons. How might these be involved?
The adjacent oxygen atom can donate electron density to the half-empty p orbital, which is a stabilizing interaction. The orbital picture looks like this.
Yes, there’s an electron in the antibonding orbital, but on the whole the interaction is stabilizing since bonding electrons outnumber antibonding electrons here.
** One note for advanced students – the “shallow pyramid” has a low barrier to inversion. This means that if a free radical is formed from an optically active chiral center, rapid racemization generally ensues.
Aromatic Synthesis (1) - "Order Of Operations"
More Reactions on the Aromatic Sidechain: Reduction of Nitro Groups and the Baeyer Vi
The Retro Diels-Alder Reaction
The Cope Elimination
Nucleophilic Aromatic Substitution (2) - Arynes
What Makes A Good Nucleophile?
3 Trends That Affect Boiling Points
Five Key Factors That Influence Acidity
Polar Protic? Polar Aprotic? Nonpolar? All About Solvents
Table of Functional Group Priorities for Nomenclature
{ 10 comments… read them below or add one }
Could you please explain why are there two unpaired electrons in the MO diagram at the end of your article? Shouldn’t it be only one?
Hi – this is an effect called “hyperconjugation”. It’s an interaction between a filled orbital (the lone pair of oxygen) and a half-empty orbital (the free radical). Donation of electron density to the half-filled orbital results in some stabilization. This is what the orbital picture is trying to present. I haven’t written an article on hyperconjgation yet but there is a wikipedia article.
Thanks for clarifying.
MO diagrams with unpaired electrons and molecules with more than two atoms are unfamiliar to me, so I’ll settle for that for now :)
This “hyperconjugation” that James was talking about is also the reason for the stability increase from primary to tertiary alkyl radicals; it’s worth noting that the lone pair on the oxygen and the two electrons in the sigma bond between the adjacent carbon and hydrogen are carrying out the same function, and aren’t separate entities.
hi..can i ask you one question..why is it when the benzylic undergoes monohalogenation..the halogen attach to the side chain insteaad of the ring itself?
please answer???
Look up the difference between aryl radicals and benzylic radicals. One is resonance stabilized, the other is not.
Does steric hindrance have any affect in whether a radical will form? For example, would ethylbenzene be more likely to undergo radical halogenation than tert-butylbenzene?
Well, t-butylbenzene would certainly not form a radical as easily as ethylbenzene, but that’s because there are no hydrogens that can be removed that will result in a resonance-stabilized radical!
Generally speaking steric effects are not as important in free radical reactions.
One of the reasons is that charged nucleophiles and electrophiles carry around a solvent “shell” with them that helps to stabilize the charges. Kind of like an entourage that surrounds them at all times. Free radical reactions are neutral and this “entourage” effect is not as much of an issue.
I want to ask when O oxygen shows it +M group character and have free radical on itself by making double bond with tha carbon ,at that time numbers of electronic Oxygen is 9 which can not be possible with the 2nd period’s member….Please explain
Hi James,
I have several questions regarding the reactivity of the allyl and benzyl radicals towards association with HO2: allyl/benzyl+ho2–>allyl-ho2/benzyl-ho2.
I have read several posts and papers claiming that although the benzyl radical shows more resonant structures than the allyl one, both radicals show similar stability because some of the resonant structures of benzyl are not very stable as they break the aromaticity. This makes me think that the reactivity of both radicals when it comes to associate with an HO2 radical is similar, and both reactions should show similar rate constants.
However, I wonder if this issue could be addressed in a different way. How about looking into the SOMO orbitals of allyl, benzyl, and HO2? I have run quantum calculations and observed that the energy of the SOMO of allyl is 21.9 kcal mol-1 with respect to that of the HO2 (set as 0.0 kcal mol-1 as a reference energy), while that of the benzyl radical is 28.9 kcal mol-1. The energy gap between the SOMO orbitals of the pair HO2-Allyl is 7.0 kcal mol-1 smaller than that of the HO2-Benzyl one. Does this mean that the overlap between the SOMO orbitals of allyl and HO2 is much larger and thus the resulting MO during the association reaction will be much more stable? If so, does this mean that the association reaction allyl+HO2 should show much larger rate constants than benzyl+HO2 reaction?
Thank you