5
$\begingroup$

I've read this site for a long time but I had never asked a question, so this is my first time :).

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. Here I show one of the reactions i'm studying: enter image description here

As you can see, the radical is centered on a carbon atom which is bonded to two chlorine atoms and also to an amide group. As far as I know, chlorine atoms attached to a carbon-centered radical stabilize it due to inductive effect. Also, chlorine atoms can be considered EWG due to their electronegativity. This alone would lower the energy of SOMO orbital, wouldn't it?

In addition to that, amide groups are also EWG, which again would lower the energy of SOMO orbital. So my questions are:

1) Am I correct to say that the more stable a radical is, the less reactive it is?

2) Am I ignoring some other effects which can affect reactivity?

P.S. Below I show the optimized structures of the reactant and transition state calculated at the B3LYP/6-31++G(d,p) level of theory.

Thanks for your attention.

EDIT: Well, I think I wasn't very clear about what's bothering me. I have the energies of the reactant, ts and product and they are ok, nothing wrong. 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. Also, in that same reasoning, could it be that a carbon-centered radical with no chlorines be more reactive? Again, this is a curiosity that came to my mind and I'm want to be sure I'm thinking it in the correct way.

reactant

enter image description here

|improve this question
$\endgroup$
  • $\begingroup$ What part of the radical reactivity is bothering you? I think everything you said is reasonable. $\endgroup$ – Dennis Cao Jun 29 '18 at 17:57
  • $\begingroup$ Reactivity toward cyclization with the alkene part of the molecule. $\endgroup$ – Benjamin233 Jun 29 '18 at 17:59
  • $\begingroup$ Ah so you're wondering why in an equilibrium situation, why/if the cyclized form is preferred? I think your judgement of radical stability is correct, so my first guess is that once the cyclized radical undergoes termination with H* for example, then it's stuck there, whereas a C-H bond in C(=O)CCl2H could still undergo homolysis, eventually driving reaction forward despite the difference in stability between the two radical intermediates. $\endgroup$ – Dennis Cao Jun 29 '18 at 18:05
  • $\begingroup$ I added an edit. $\endgroup$ – Benjamin233 Jun 29 '18 at 19:27
  • $\begingroup$ I question whether it is really an equilibrium. I would argue that the cyclisation occurs simply because it is faster than any other intermolecular process (e.g. H atom abstraction from something else). $\endgroup$ – orthocresol♦ Jun 29 '18 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.