Free Radical Reactions
What Factors Destabilize Free Radicals?
Last updated: December 20th, 2019 |
Three Factors Which Influence The Stability of Free Radicals: Hybridization, Electronegativity, and Polarizability.
In the last post we talked about several factors that stabilize free radicals. We saw that since free radicals can be considered electron deficient, any factor which results in donation of electrons to a free radical helps to stabilize it, as do any factor which results in a delocalization [“spreading out”] of that free radical.
[We also saw that the factors which stabilize free radicals are the same ones that stabilize carbocations].
In this post we’re going to switch things around. What factors might destabilize free radicals?
Table of Contents
- Free Radicals Are Destabilized By Removing Electron Density
- Radical Stability Decreases With Increasing s-Character Of The Orbital
- Radical Stability Decreases With Increasing Electronegativity Of The Atom
- Radical Stability Decreases As Polarizability Is Decreased
- Summary: Factors Which Decrease Free-Radical Stability
If we remember that free radicals are stabilized by electron donating groups, we might reason that they are destabilized when electron density is taken away.
So what factors might result in a free radical being less “electron rich” ?
There are three major factors. I’ll list them in order of importance for the purposes of a typical student encountering free radicals in a typical class.
If you recall some of the factors that affect acidity you might recall that a lone pair of electrons becomes more stable as the hybridization of the carbon goes from sp3 to sp2 to sp. That’s because of the greater s-character of the orbital, which results in the lone pair being held more closely to the (positively charged) nucleus.
What might happen if we’re dealing with a free radical instead? As the s-character of the orbital containing the free radical is increased, the half-filled orbital containing the free radical is held more closely to the nucleus. What effect does this have on the stability of the free radical? It’s actually destabilizing, because being closer to the nucleus, the electron affinity of the orbital will increase.
For that reason alkyl radicals (generally considered to be sp2 hybridized) are the most stable, followed by vinyl and phenyl radicals (sp-hybridized) , followed by alkynyl radicals.
Quiz time: Having read the paragraph above, what might you think is the effect of electronegativity on free radical stability? What might happen to the stability of a free radical as you increase the electronegativity of the atom? For example, compare the sequence H3C , H2N, HO, and F . Which free radical should be the most stable?
Electronegativity, as we’ve talked about before, is like “greed” for electrons. Increasing electronegativity is going to draw a free radical closer to the nucleus, and as we saw above, this results in destabilization.
Going down the periodic table, we also notice an increasing stability in free radicals, going from F < Cl < Br < I . While this can likewise be thought of as resulting from a decrease in electronegativity , another way to look at it is that going down the periodic table results in an increase in the size of the atom, and with that, allows for the electron-deficient orbital to be spread out over a greater volume.
The main factor we’ve seen here that destabilizes free radicals is bringing the half-filled orbital closer to the nucleus (greater s-character, higher electronegativity) or by restricting the delocalization of the free radical (decreasing polarizability). Here, I think it’s important not to focus on the effect of electronegativity on the electron (radical) but on the effect of electronegativity on the “hole” – i.e. the empty orbital. Bringing a half-empty orbital closer to the nucleus will greatly increase its potential energy (the electrostatic attraction of the nucleus for an electron) and increase electron affinity, making that free radical much more reactive (and in this case, reactivity = instability).
So how might we “quantify” the stability of a free radical?
Believe it or not, there’s actually a remarkably simple way to learn how stable free radicals are, using a measurement you’re probably already familiar with! We’ll talk about that in the next post.
Next Post: Bond Strengths And Radical Stability
A previous version of this post included adjacent electron-withdrawing groups as a destabilizing influence on free-radical stability. I’ve removed this because it’s just too complicated for our purposes.
For example, take the methyl radical, H3C• . Replacing hydrogen with fluorine (a strong electron withdrawing group) one might expect radical stability to decrease. Well, it doesn’t. It’s actually more stable due to the ability of the fluorine lone pair to donate to the half-filled orbital. A second fluorine has a similar effect. However, the F3C• radical is less stable than the methyl radical. This is hard to predict with simple rules. See: Homolytic Bond Dissociation Enthalpies of the C−H Bonds Adjacent to Radical Centers, Xian-Man Zhang The Journal of Organic Chemistry 1998 63 (6), 1872-1877 . DOI: 10.1021/jo971768d
A note on hybridization. Vinyl radicals have an (E) and (Z) form and the inversion barrier from one to the other increases as the electronegativity of the substituents increases. Since the molecule must pass through an sp-hybridized geometry in order to invert, this supports the notion that taking electron density away from the sp-orbital destabilizes the radical. This is a very interesting paper where the rates of inversion were studied:
Effect of Substituents on the Structure of the Vinyl Radical: Calculations and Experiments. Carlo Galli,*, Alessandra Guarnieri,Heinz Koch,†, Paolo Mencarelli,* and, and Zvi Rappoport‡ The Journal of Organic Chemistry 1997 62 (12), 4072-4077 DOI: 10.1021/jo962373h