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#Carbon reactivity how to#

Yet another form, called amorphous carbon, has no crystalline structure. A fourth form, called Q-carbon, is crystalline and magnetic. Spheroidal, closed-cage fullerenes are called buckerminsterfullerenes, or “buckyballs,” and cylindrical fullerenes are called nanotubes. A third form, called fullerene, consists of a variety of molecules composed entirely of carbon. Two of its well-defined forms, diamond and graphite, are crystalline in structure, but they differ in physical properties because the arrangements of the atoms in their structures are dissimilar.

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#Carbon reactivity how to#

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It had just came to my mind that the radical is very stabilized in the reactant and I just wanted to know if my reasoning was correct.

I have the energies of the reactant, ts and product and they are ok, nothing wrong. Below I show the optimized structures of the reactant and transition state calculated at the B3LYP/6-31++G(d,p) level of theory.ĮDIT: Well, I think I wasn't very clear about what's bothering me. So my questions are:ġ) Am I correct to say that the more stable a radical is, the less reactive it is?Ģ) Am I ignoring some other effects which can affect reactivity? In addition to that, amide groups are also EWG, which again would lower the energy of SOMO orbital. This alone would lower the energy of SOMO orbital, wouldn't it? Also, chlorine atoms can be considered EWG due to their electronegativity. As far as I know, chlorine atoms attached to a carbon-centered radical stabilize it due to inductive effect. Here I show one of the reactions i'm studying:Īs you can see, the radical is centered on a carbon atom which is bonded to two chlorine atoms and also to an amide group.

So, i'm currently working on a computational chemistry project on radical cyclization reactions and I'm having trouble trying to understand the reactivity of the radical. I've read this site for a long time but I had never asked a question, so this is my first time :).