20

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, ...


18

I liked this question because I had never thought much about it. However, it's not such a mystery because the answer is on Wikipedia's “radical” page: Historically, the term radical was also used for bound parts of the molecule, especially when they remain unchanged in reactions. These are now called functional groups. For example, methyl alcohol was ...


18

I believe that you are asking if there are any reactions that are essentially "non-polar". There are two classes that immediately come to mind: (1) radical reactions and (2) pericyclic reactions. A common radical reaction studied in introductory organic chemistry is the radical halogenation of alkanes. Instead of cationic and anionic reactants/intermediates/...


18

The electron didn't go anywhere. It's in an unhybridized p orbital on the central bromine, and yes, $\ce{Br3O8}$ is a free radical. That is why it decomposes above -80ºC.$^{[1]}$ $^{[1]}$ Cotton, F. A. Progress in Inorganic Chemistry - Volume 2; Interscience Publishers: New York, NY, 1960.


15

This is a tough question. I think it might even be unfair to ask such a question on a test in non-advanced classes. In advanced classes it could make an interesting topic of discussion, but I'm still not sure that the "real" answer is known. What can be said is that due to resonance, both the allylic and benzylic radicals are more stable than the t-butyl ...


13

Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like. According to R. ...


13

First off, it was very astute of you to recognize that if the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible. In a typical carbocation, the carbon bearing ...


13

This reaction is taken from Tetrahedron Lett. 1995, 36 (17), 3015. (DOI: 10.1016/0040-4039(95)00502-4) The heat (provided by refluxing) is required to initiate the reaction by the decomposition pathway you have drawn. The cyanopropyl radical thus formed abstracts a hydrogen atom from the tin hydride, a consequence of the weak Sn–H bond: The ...


12

First off, your observation is correct, rearrangements involving free radicals and carbanions are noticeably less common than rearrangements involving carbocations. Many common carbocation rearrangements involve 1,2 shifts. If we look at the MO diagram for a 1,2 shift and compare shifting a substituent to an empty p orbital (carbocation, case A in ...


12

In its ground state, naked carbon is triplet $^3P$, with two metastable singlet states $^1D$ and $^1S$ ($^1D$ being the one that participates in most reactions) while the tetraradical is the least stable one: It can be synthesized by several different methods. Graphite vaporization by an electric arc or laser can be employed. Photolysis of suitable ...


11

$\beta$-Carotene is a highly delocalised $\pi$ system. According to George Britton there are numerous mechanisms to achieve this. Oxidation Oxidising radicals can remove an electron from the carotene molecule. $$\ce{CAR + .R+ -> .CAR+ + :\!R}$$ Reduction Analogous it can accept an additional electron. $$\ce{CAR + .R- -> .CAR- + R}$$ Hydrogen ...


11

Usually the dot is put there to emphasize that the nitric oxide is a free radical that includes an unpaired electron. This is especially notable by comparison with $\ce{NO^+}$, which does not have the unpaired electron. Note that the nomenclature $\ce{·NO}$ should not be rendered as showing the unpaired electron on nitrogen. The unpaired electron is ...


10

We can draw the 3 Lewis structures (or the corresponding resonance structures) pictured below for $\ce{O_2}$ Since an oxygen atom has 6 electrons, A would correspond to a structure with a single bond between the oxygen atoms, 2 lone pairs on each oxygen and an unpaired electron on each oxygen; however A does not have an octet around each oxygen, in fact, ...


9

For free radical reactions, the most important parameter in assessing bond strength is bond dissociation enthalpy (BDE). Typical values for $\ce {C-F}$ bonds are around $\mathrm {100\ kcal/mol}$, while for $\ce {C-Cl}$ bonds are around $\mathrm {80\ kcal/mol}$; $\ce {C-Cl}$ bonds are therefore weaker. This can be rationalized by considering the poorer ...


9

Imagine your favorite activities are playing Video games (Most Favourite), and other ones are reading chemistry, spending time in a chemical lab. (these two are activities are those activities in between you can't choose one that you want to do, or say you like them equally.) You dislike equally: reading history, reading civics, reading economics. Let's ...


9

As Martin has mentioned, carbenes are a good starting point if you are looking for organic compounds with a triplet ground state. In these carbenes, the HOMO is not twofold degenerate, as your question suggests. Instead, there are two singly occupied orbitals with different energy. The energy gap, though, is typically small, and since there is an energy ...


9

Actually, the sequence of reactions leading to a low but constant concentration of $\ce{Br^.}$ radicals is very simple: $$\begin{align}\ce{R2N-Br &<<=> R2N^. + Br^.}\tag{1}\\[0.4em] \ce{Br^. + Br-NR2 &<=> Br2 + R2N^.}\tag{2}\\[0.4em] \ce{Br2 &<<=> 2 Br^.}\tag{3}\end{align}$$ The $\ce{N-Br}$ bond is very weak and can ...


8

It does not have to do with steric effects as you suggested. Rather, the reason behind this trend is the inductive effect. Figuratively speaking, carbon side-chains attached to an central atom "push" electron density towards this atom, while hydrogen atoms don't. If you break a C-C bond and produce a carbon radical, it will be stabilized if electron density ...


8

If azobisisobutyronitrile (AIBN) underwent homolytic cleavage over time, do you expect the resulting cyanoprop-2-yl radicals would survive as free radicals for a long period of time? If not, what would be the decomposed products? Alkali metals reduce benzophenone in THF to form blue diphenylketyl radical anions. I'm not sure azobisisobutyronitrile (AIBN) or ...


8

Note that most radical mechanisms do not explicitly feature homolysis of $\ce{C-H}$ bonds or $\ce{C-C}$ bonds. If one were to compare these two processes, say for ethane $\ce{CH3CH3}$, we would find that homlysis of $\ce{C-C}$ is favored over homolysis of $\ce{C-H}$. Obligatory bond dissociation energy references for the organics and here for the halogens. ...


8

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.


7

Odd electron species do dimerize. $\ce{NO}$ Nitric oxide dimerizes, but only at low temperature (and probably high pressure}. $\ce{NO2}$ Nitrogen dioxide does dimerize. In fact, this is a well known property of $\ce{NO2}$. $\ce{NO2}$ (orange-brown) is in equilibrium with $\ce{N2O4}$ (colorless). $$\ce{2NO2 <=> N2O4}$$ The position of this ...


7

The halogenation of toluene is possible by two mechanisms. In German, they are referred to by the shorthands KKK and SSS, where KKK stands for Katalysator, Kälte, Kern (catalyst, cold and core) while SSS is Sonne, Siedehitze, Seitenkette (sun, sizzling heat and side chain). At low temperatures and in the presence of an activating catalyst such as $\ce{FeCl3}...


7

There is nearly no thermodynamic aspect to this reaction. The activation energy for any alkane hydrogen abstraction by a fluorine radical is essentially 0$^{[1]}$. This means that there is an equally likely chance of fluorinating at any given position. In this case, we need only know the proportions of homotopic hydrogen. Homotopic hydrogen are labelled with ...


7

There are various examples of molecules in triplet ground state. The smallest (organic) is possibly methylene, $\ce{:\!CH2}$. Carbenes in general may adopt triplet state as their ground states. Other possibilities are biradicals (From the IUPAC gold book): An even-electron molecular entity with two (possibly delocalized) radical centres which act nearly ...


6

Free radicals are usually very reactive. If we bring two free radicals or a free radical and a molecule with electrons available for bonding together a bond will usually form. Yes, it does cost energy to bring those electrons close together. But look at the payoff, a new chemical bond is formed and this is usually quite exothermic (stabilizing), usually ...


6

TLDR; It depends who you are talking to (organic chemist, ESR spectroscopist, or physical chemist/chemical physicist), and when. The way I read the quote below the fold, there have been three major senses of the definition, at least to the best of the knowledge of a giant in the field like Herzberg: The sense as described by Terry Bollinger, where an ...


6

As is currently being discussed in this question, I'm not convinced that the methyl radical is planar, but even if not planar, it is certainly very close to planarity. It is also certainly less pyramidal than the $\ce{.SiH3}$ radical as you state. The $\ce{.SiH3}$ radical is more pyramidal than the $\ce{.CH3}$ radical due to differences in ...


6

This is the Hofmann-Löffler-Freytag reaction, which was later examined by E. J. Corey (J. Am. Chem. Soc., 1960, 82, 1657-1668, DOI). When the reaction is performed in sulfuric acid, it is conceivable that the nitrogen atom of the N-Bromoalkane is protonated. Homolytic cleavage of the $\ce{N-Br}$ thus probably leads to a radical cation. The latter then ...


6

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 ($\ce{ROH}$). In the excited singlet state, $\ce{^1P+^*}$ is a strong electron acceptor, that oxidizes the alkene to ...


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