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Free Radical Reactions

By James Ashenhurst

What Factors Destabilize Free Radicals?

Last updated: April 12th, 2019

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?

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 four major factors. I’ll list them in order of importance for the purposes of a typical student encountering free radicals in a typical class.

Factor #1: Hybridization

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? Going from sp3 to sp2 to sp, we’d be holding the partially filled (electron deficient) orbital progressively closer to the nucleus. This is destabilizing!


For that reason alkyl (sp3) radicals are the most stable, followed by alkenyl (sp2) and then alkynyl (sp).

Factor # 2: Electronegativity

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.

Factor #3 : Polarizability

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.



Factor #4 : Electron Withdrawing Groups

From the previous discussion, this should come as no surprise: an electron withdrawing group adjacent to a free radical will draw electron density away from it, resulting in destabilization.

Watch out, however – this final factor can be tricky! In the last post we talked about how it’s easy to think that an oxygen atom adjacent to a free radical would be destabilizing. It’s electron withdrawing, right? Fact is, there’s a second factor in play when the “electron withdrawing” group has a lone pair or two; that lone pair can actually donate into the partially filled orbital, resulting in stabilization.

It’s difficult to predict with certainty what the effect of each group will be in this case.  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

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Comment section

9 thoughts on “What Factors Destabilize Free Radicals?

  1. Dear sir

    I don’t understand why it is destabilizing for a radical to be close to the nucleus? The radical is an unpaired electron and the nucleus is positively charged. I always thought that a postive and negative charged stabilized each other…

    With kind regards

    1. First I would like to remind you that delocalization(spreading out) of free radicals or unpaired electrons due to hyperconjugation makes the atom (bearing negative charge) stable. When the radicals comes closer to the nucleus due to greater electronegativity of the atom it gets localized or concentrated and its can’t spread itself out, so becomes unstable.

  2. I have a question. Why is a methyl radical sp2 instead of sp3? If it was sp3, wouldn’t the radical be further from the nucleus, and therefore more stable?

    (I am also on the same boat as Tim. It doesn’t make sense to me that an unpaired electron would be more stable if it was further from the nucleus.)

  3. Just wanted to say a big thanks!

    This page (and the few before it) have been an unbelievable help :)

    Everything is well laid out and sufficiently explained!


  4. When you combine the LUMO of an EWG with the SOMO of a radical, you get a lowering of the SOMO energy and therefore stabilisation. The fluoromethyl radical you cite is indeed less stable than the corresponding methyl radical, but this is because the shape of the radical goes from planar(ish) to more pyramidal, meaning that the radical is more sigma in character than pi. See Organic Chemistry Clayden et al Oxford University Press (2nd Ed) P. 1025-1027

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