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Here's my thinking:

  1. Atomic oxygen in its ground state is a free radical because there are two unpaired electrons in p-orbitals.

  2. When species react new orbitals are formed, and perhaps a situation can occur where the amount of degenerate orbitals is changed by a chemical reaction. This would implicate the filling of these orbitals, following the aufbau principle and hunds rule of filling.

  3. Therefore it is thinkable that two reactant species (i.e. not homolytic bond fission) which are not radicals, can react in such a way that the new electron orbitals are filled in such a way that all electrons are no longer paired => radical species formed from non-radical species.
  4. Likewise it is conceivable that a radical species can react with a non-radical species (i.e. not an even number of radical species terminating eachother) to form only non-radical species.

Is my argument and results 3 & 4 correct? Can someone give an example of 3 and/or 4?

Edit: My main takeaway - when considering the feasibility of a radical creation/destruction, look at conservation of spin

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    $\begingroup$ NOTE TO ALL: As of the time of this comment, the answers from Klaus, myself, and porphyrin were posted before the OP edited his question to include the two constraints excluding homolytic bond fission for #3 and radical termination reactions for #4. Klaus updated his answer in response to a comment from OP. Jan's answer was posted after the above edits to the question were made. $\endgroup$
    – hBy2Py
    Dec 7, 2016 at 1:49
  • $\begingroup$ Allow me to re-reiterate that while your’s and Klaus’ were, Porphyrin’s was not. (Albeit by 3 minutes) $\endgroup$
    – Jan
    Dec 8, 2016 at 21:06

4 Answers 4

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A free radical cannot be created in a simple, one-step process from non-radical reactants other than by bond dissociation (which you explicitly excluded).

A fundamental principle of quantum chemistry is the conservation of spin. Over the course of a simple reaction, the total spin quantum number $S$ must remain constant — excluding more complex effects like spin-orbit coupling. Non-radical compounds have a singlet state, i.e. their total spin quantum number $S = 0$ or $2S+1=1$; all electrons are paired. Typical organic radicals such as the tert-butyl radical have one unpaired spin and are therefore in a doublet state $S = 1/2$, $2S+1 = 2$. Diradicals such as $\ce{O2}$ are in a triplet state with $S = 1$ and $2S+1 = 3$. To arrive at a product state with nonzero $S$, you would either need to generate a pair of opposing spin compounds (simple bond dissociation) or start of with a nonzero $S$.

This is why reactions that start with peroxides often generate singlet oxygen rather than triplet oxygen. By complex mechanisms and emitting a faint red glow, this then slowly isomerises to triplet oxygen.

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  • $\begingroup$ So because of the same conservation law, 4. is also impossible? $\endgroup$
    – Adroit
    Dec 7, 2016 at 5:25
  • $\begingroup$ to continue with spin ideas, it is spin allowed for two singlet states to combine to produce two triplets, which at a push might be considered as radicals, for example $\ce{^1M + ^1O_2 \rightarrow ^3M + ^3O_2}$ $\endgroup$
    – porphyrin
    Dec 7, 2016 at 12:07
  • $\begingroup$ @porphyrin Doesn't that completely contradict the conservation law Jan stated? You start with S=0 (singlet states) and end with S=2 (two triplet states) $\endgroup$
    – Adroit
    Dec 7, 2016 at 16:47
  • $\begingroup$ @Adroit It doesn’t contradict much like the creation of two dublet radicals from a singlet molecule (i.e. homolytic bond-cleavage) doesn’t contradict. $\endgroup$
    – Jan
    Dec 7, 2016 at 16:51
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    $\begingroup$ @Adroit, it easier to start with two triplets. Total spin has series $|S_1+S_2|..|S_1-S_2|$ which is $2,1,0$ spin or multiplicity $5, 3, 1$. Assume pair collide then complex formed has spin multiplicity one of $5, 3, 1$. Thus for singlet spin complex $^3A+^3B <=> ^1(^3A ^3B) <=> ^1A + ^1B $ is spin allowed. For triplet complex products are $^3A+^1B$ and probably no products from quintet complex. Triplet-triplet annihilation is well known in photochemistry as is singlet -singlet annihilation but a intense laser is needed as singlets have a short lifetime wrt triplets. $\endgroup$
    – porphyrin
    Dec 8, 2016 at 9:07
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As for #3, photoinduced electron transfer (PET) reactions do what you want.

Imagine that 2,4,6-triphenylpyrylium tetrafluoroborate ($\ce{P+BF4-}$) is irradiated at $\lambda = 350~\mathrm{nm}$ in the presence of an alkene and a nucleophile ($\ce{ROH}$).

PET of alkene and a pyrylium salt

In the excited singlet state, $\ce{^1P+^*}$ is a strong electron acceptor, that oxidizes the alkene to the corresponding radical cation an is itself turned to a pyranyl radical, $\ce{P^{\cdot}}$. (Here is the first radical.)

Then the alcohol adds to the alkene radical cation, the adduct deprotonates spontaneously. (Here is the second radical.)

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  • $\begingroup$ I don't fully understand your answer. There are some concepts i have yet to learn, like "singlet state" and "adduct". Furthermore I think I might have been misunderstood (see corrections in bold). Is your example mechanism an example of non-homolytic bond fission leading to non-radical species forming radical-species? If so, please try to make the mechanism more clear for me. $\endgroup$
    – Adroit
    Dec 6, 2016 at 20:50
  • $\begingroup$ Apologies :) I'll add a sketch of the process that is not a homolytical fission to make myself more clear. $\endgroup$
    – Klaus-Dieter Warzecha
    Dec 6, 2016 at 21:08
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Can a free radical be created by chemical reaction of non-radical species?

Yes.

One straightforward example of this is the unimolecular dissociation of hydrogen peroxide into two hydroxyl radicals:

$$ \ce{H2O2 -> 2 HO^\bullet} $$

Related, if you consider this homolytic dissociation reaction as occurring in the presence of a free radical scavenger like ABTS, you end up with two relatively stable radicals in the final solution:

$$ \ce{2 ABTS^{2-} + H2O2 -> 2 ABTS^{2-} + 2 HO^\bullet -> 2 ABTS^{\bullet\,-} + 2 OH-} $$

  1. Likewise it is conceivable that a radical species can react with a non-radical species to form only non-radical species.

Agreed, but only if you have an even number of electrons in the total system. If there's an odd number, you're guaranteed to have a radical in there somewhere, at all times.

That said, one example is the dimerization equilibrium of nitrogen dioxide with dinitrogen tetroxide:

$$ \ce{2 NO2^\bullet <=> N2O4} $$

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  • $\begingroup$ Good answer with use of understandable examples. I do, however, think my question was not phrased clearly enough, as I have been misunderstood, For point 3. I meant specifically non homolytic bond fissions. For point 4. I meant a odd number of radical species reacting with non-radical species to create only non-radical species. $\endgroup$
    – Adroit
    Dec 6, 2016 at 20:45
  • $\begingroup$ @Adroit An odd number of radical species (well, monoradical species, I guess) can never react to form all non-radical species. It's mathematically impossible when you have an odd number of electrons. $\endgroup$
    – hBy2Py
    Dec 6, 2016 at 20:52
  • $\begingroup$ Also, @Adroit, it's bad practice on SE sites to make edits to existing questions that change the meaning of the original question, as it leaves any answers posted prior to the edit as being no longer relevant to the question. IMO you're better off leaving the question as it was, then asking a new question with more precise wording, with a link back to the prior question. $\endgroup$
    – hBy2Py
    Dec 6, 2016 at 20:55
  • $\begingroup$ A radical species need not necessarily have an odd number of electrons? Think for example of O(g) or O2(g). They are radicals because of degenerate orbitals being filled by hunds 1. rule $\endgroup$
    – Adroit
    Dec 6, 2016 at 20:58
  • $\begingroup$ @Adroit <nod>, quite right, I follow you now; that was why you specifically mentioned $\ce{O2}$ in #1. I still recommend reverting this question, and asking a new, better-clarified one, as I was not the only person confused by what you posted. $\endgroup$
    – hBy2Py
    Dec 6, 2016 at 21:01
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Yes very,very many reactions:

$\ce{H_2 + O_2}$ produces H, O, OH and $\ce{HO_2}$ radicals;

$\ce{H_2 + X_2}$, where X is a halogen, produces H & X radicals,

$\ce{CH3CHO \rightarrow \cdot CH3 + \cdot CHO}$ and

$\ce{ \cdot CH3 + CH3CHO \rightarrow CH4 + \cdot CH3CO}$

and many others. These and many more are all radical chain reactions, either initiated by high temperature or light, and the rate laws are often misleadingly simple.

In a chain reaction one radical reacts with non-radical to form a new radical and a non-radical as in the example with $\ce{\cdot CH3}$ and acetaldehyde above.

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  • $\begingroup$ First example isn't great because $\ce{O2}$ is basically a diradical. $\endgroup$
    – Zhe
    Dec 6, 2016 at 21:54
  • $\begingroup$ @zhe, yes it is paramagnetic. It produces OH radicals via $\ce{H2 + O2 \rightarrow 2OH\cdot}$. An alternative is $\ce{RH + O2 \rightarrow \cdot R + \cdot HO2}$ where RH is a hydrocarbon or $\ce{Br2 + CH4}$ which produces Br and $\ce{\cdot CH3}$ radicals $\endgroup$
    – porphyrin
    Dec 6, 2016 at 22:13
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    $\begingroup$ All these examples either start from radicals or include homolytic bond cleavages which was specifically excluded by OP. $\endgroup$
    – Jan
    Dec 6, 2016 at 22:49
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    $\begingroup$ @Jan This answer was posted before OP edited his question to add the constraint excluding homolytic bond cleavage. $\endgroup$
    – hBy2Py
    Dec 7, 2016 at 1:47
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    $\begingroup$ @hBy2Py Hover above the 17 hours ago with your mouse; a tooltip will appear with the exact posting time. Works here, on the questions page and in the revision histories. $\endgroup$
    – Jan
    Dec 7, 2016 at 13:50

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