Why doesn't TEMPO react with itself?

Question

TEMPO ((2,2,6,6-tetramethylpiperidin-1-yl)oxyl, or (2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl) has the following structure:

Given that it's a free radical, why doesn't it react with itself, despite being capable of reacting with other radicals in the system itself)?

In addition, according to Wikipedia, it's a red-orange, sublimable solid. How can it be a solid despite being a radical?

Everything can form a solid. Being or not being a radical is irrelevant here. — Ivan Neretin, Feb 19 '18 at 20:04
1,1-Diphenyl-2-picrylhydrazide is a stable, purple, solid free radical. — user55119, Feb 19 '18 at 20:46
You don't need a complicated or sterically hindered molecule. Oxygen and the $\ce{NO_x}$ oxides are quite stable free radicals under typical ambient conditions. Credit goes to molecular orbital structure, in both the radical and any supposed dimer. — Oscar Lanzi, Feb 20 '18 at 01:06

score 22 · Answer 1 · answered Feb 19 '18 at 20:56

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I believe the first part of your question is asking why the oxygen with the unpaired electron doesn't find a second one to form a peroxide.

The answer is actually quite simple. It is by design. Next to the nitrogen, you see the 4 methyl groups that give TEMPO the "TEM" part of it's name. These methyl groups give TEMPO a lot of steric bulk around the oxygen, and this bulk makes it very difficult for two TEMPO radicals to form an oxygen-oxygen bond with each other. There is just not enough space. You can build a model to convince yourself of this.

answered Feb 19 '18 at 20:56

Zhe

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https://chemapps.stolaf.edu/jmol/jmol.php?model=CC1(C)CCCC(C)(C)N1OON2C(C)(C)CCCC2(C)C is a link to the 3D structure of the molecule formed from two TEMPO radicals. – Pritt says Reinstate Monica Mar 28 '18 at 16:44
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It is definitely not only steric repulsion but also a lone pair effect. It is mostly that the formation of a NO-ON bond is quite unstable due to the lone pair repulsion which is larger than the energy gain of forming a sigma bond. (The same reason that peroxides are relatively unstable) – JLT Aug 29 '19 at 11:42

score 9 · Answer 2 · answered Feb 19 '18 at 22:37

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Steric hindrance is the way to make them stable. I have never worked with TEMPO, but I have seen this one in our lab:

Here the steric hindrance is very evident. This is also a nice stable solid.

answered Feb 19 '18 at 22:37

Raoul Kessels

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You don't even need the chlorine substitution. – Oscar Lanzi Feb 20 '18 at 01:01
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@OscarLanzi Right, but the one I have seen making is with chlorine. Surprisingly easy: trichlorobenzene, chloroform and aluminium trichloride. – Raoul Kessels Feb 20 '18 at 06:58

score 0 · Answer 3 · edited Mar 29 '18 at 02:16

The stability of TEMPO and hence is formation is thought to result from resonance involving the nitrogen lone pair as well as hyperconjugation from the numerous methyl groups. See also the Wikipedia article on 3 center 2 electron bond.

The resonance involves oxoammonium contributors as those postulates in oxidation catalyse by TEMPO and carbanions, hence invoking sacrificial hyperconjugation.

As stressed in other answers the steric hindrance of the methyl groups also plays a major role "caging" the radical center and leading to longer than usual bonds when TEMPO reacts.

Consider that coupling two TEMPO would lead to a peroxide. Peroxides are, with few exceptions, unstable and reactive.

The above effects makes a hypothetical $\ce{O-O}$ bond in TEMPO even less stable than that one in other peroxides, where it is already characterised by energy of about $\pu{50 Kcal/mol}$, i.e. less than the already unstable $\ce{O-H}$ bond in the same molecule and generally speaking less than a half of $\ce{C-C}$, $\ce{C-H}$ and $\ce{C-O}$.

Comments are not for extended discussion; this conversation has been moved to chat. — jonsca, Mar 28 '18 at 23:06

Why doesn't TEMPO react with itself?

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