![]() ![]() More of a contribution, more of an electron donating effect, than in our previous example. So when we think about overlapping our orbitals for oxygen and carbon, this is a better situation than before, because carbon and oxygen are the same period on the periodic table. When we think about resonance, I could move this lone pair of electrons from oxygen into here and This is your Y substituent, you have this other oxygen So once again we thinkĪbout induction first, so this oxygen is withdrawing some electron density from this carbon. So let's look at our nextĬarboxylic acid derivative, which is an acid anhydrite. They will react with water, sometimes violently, at room temperature. And if induction dominates, then we would expectĪcyl or acid chlorides to be extremely reactive. So this is not a major contributor in the overall resonance hybrid. Overlap means that chlorine is not donating a lot of electron density to our carb needle carbon here. You mismatch these sizes they can't overlap as well. This resonance structure, we have a pi bond betweenĬarbon and chlorine, and if we draw the P orbital- carbon's in the second period, so we draw a P orbitalįor the second period, and the thing about chlorine,Ĭhlorine's in the third period so it has a bigger P orbital. Possible resonance structure, once again a negative oneįormal charge on the oxygen, and a plus one formalĬharge on the chlorine. Possibility of resonance, I would move these electrons into here, and push those electrons ![]() Our carb needle carbon, and so does our chlorine. We know the oxygen withdraws some electron density from So we start with an acyl or acid chloride. And whichever one is going to win- we can think about thisīalance for helping us to determine the reactivity of our carboxylic acid derivatives. So we have these two competing effects, induction versus resonance. So resonance will decrease the reactivity of a carboxylic acid derivative. Making it less electrophilic, and therefore making it less ![]() And if you're donating electron density, you're decreasing the To our carb needle carbon, that increases the electron density. Increasing the electron density from this lone pair of electrons And we would have a pi bond between our carbon and our Y substituent. Oxygen is going to have a negative one formal charge. When we draw our resonance structure we can see that our top So our Y substituent withĪ lone pair of electrons can donate some electron density And so induction increases the reactivity of carboxylic acid derivatives. So it's more electrophilic and better able to react That makes our carb needleĬarbon more partially positive. We're withdrawing electron density from our carb needle carbon. So let's go ahead and writeĭown the first effect, the inductive effect. So that's going to withdrawĮven more electron density from our carb needle carbon. And for carboxylic acid derivatives our Y substituent is anĮlectronegative atom too. Some electron density away from our carb needle carbon, making it partially positive. We know that carb needlesĪre reactive because this oxygen is withdrawing 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.Voiceover: Here we have a representative carboxylic acid derivative with this Y substituent hereīonded to the carb needle. 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 :). ![]()
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