Discussion about TEMPO

QUESTION:

TEMPO ((2,2,6,6-tetramethylpiperidin-1-yl)oxyl, or (2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl),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?

Opinion 1:

You don't need a complicated or sterically hindered molecule. Oxygen and the NOx oxides are quite stable free radicals under typical ambient conditions. Credit goes to molecular orbital structure, in both the radical and any supposed dimer.

Opinion 2:

I believe the first part of this 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.

Opinion 3:

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 O?O bond in TEMPO even less stable than that one in other peroxides, where it is already characterised by energy of about 50 Kcal/mol, i.e. less than the already unstable O?H bond in the same molecule and generally speaking less than a half of C?C, C?H and C?O.