Identify the bonds broken and formed in the following addition...

Question

Identify the bonds broken and formed in the following addition reaction.

(a) Would you expect this reaction to be favored based on entropy?

(b) Based on enthalpy [qualitatively]?

(c) Overall?

Chemical reaction showing the formation of an imine from an aldehyde and an amine, with bonds broken and formed indicated.

Accepted Answer

Hello everyone. And welcome back. Our next problem says, identify the bonds broken and formed in the reaction shown below, identify whether the reactions are favored or not. Based on qualitative analysis of the following factors. Number one entropy, number two, enthalpy, number three overall. And the reaction we have shows one molecule of cyclone reacting with one molecule of isopropyl amine. And our product is a cyclops ring. And on the same carbon, we now have as substituents an alcohol group and then the isopropyl amine with a bond from the ring to the nitrogen. So let's think about which bonds have been broken and which formed. So looking at the cyclops, no, we see that we had a carbon bond or carbonel group and that now was changed into an alcohol group. So the bonds broken in that reactant are a carbon oxygen pi bond. So it didn't break both bonds, just one of them. Then in our Amin, we had NH two. So two hydrogens on the nitrogen. Now that that nitrogen has a new bond to the carbon, one of the hydrogens is gone. So in terms of that reactant, the bond broken is a nitrogen hydrogen bond. So is that it for our reactants? Yes, when we look at the molecules, we can see that we have still ac o bond from the carbon hill group. And the isopropyl amine is pretty much all intact except for the removal of one hydrogen and the new bond to the carbon. So now let's look at the bonds formed. So when our carbonel group became an alcohol group, there's a new oxygen hydrogen bond. And then when the amine added to the psychopath ring, there was a new carbon nitrogen bomb formed. So that's that part of the question pretty straightforward. The bonds broken are ac O pi bond and an NH bond, the bonds formed are an oh bond and AC N bond. So now let's move on to part two of the question. It says, determine whether the reactions are favored or not based on a qualitative ana analysis. So we're not looking at actual numbers but thinking through the characteristics of this reaction of the following factors, entropy entropy and overall. So entropy, of course, we recall is a measure of the disorder of the system. And in organic chemistry terms, when we think about the disorder of a reaction, a nice way to estimate is how many molecules would we have because the more molecules that are present, the more the this potential disorder. So in the case of this reaction, we have two molecules of reactant going to one molecule of product So is this reaction favored based on qualitative analysis of entropy. And the answer is it is not favored based on entropy because we're going from more disorder to less disorder. So highlight that. Now, let's think about enthalpy. Well, we've talked about the bonds broken and formed. And if we were doing a quantitative analysis, we could look up the bond association energies of these and we could add up the bonds broken because that would take energy that needed to be put in. And then we could add up the bonds form because that would release energy and look at whether our reaction overall is exothermic or endothermic. But this tells us to do a qualitative analysis. So in terms of entropy, we're just going to look in general at the driving forces for this reaction to take place and to do that, we shouldn't just look at the reaction. Let's look at the mechanism to think about what is driving it to either go forward or not. So, as a tentative mechanism, we know that we have lone pairs on this carbon, oxygen, oxygen is more electron negative than carbon. So we've got a partial negative charge on the oxygen and a partial positive charge on the carbon in that carbonel group. Then r amine, there's also a low pair on that nitrogen and a partial negative charge on the more electron nitrogen. So now we've got this electron nitrogen with these lone pairs and that can actually, we know that a new nitrogen carbon bond is going to be formed. So we can draw those electrons from the lone pair on the nitrogen coming over and attacking the carbon carbon. And then that bumps the electrons from the C pi bond up onto the oxygen. Then we'll get an intermediate where we've got both groups on here. So we have c double bonded to an O actually, that's very, not a lot of space there. So I'm going to just draw the intermediate below. We'll draw a little arrow pointing down and write intermediate. So we'll have our cyclops ring and we now have a negatively charged oxygen on there from where we broke that co pi bond. And then we have our new carbon nitrogen bond that nitrogen still has two hydrogens. So our nitrogen has a positive charge and then it's got a second hy hydrogen and then the isopropyl group. So the negative facility charged oxygen, it's one of those lone pairs can come on over and deproteinate that nitrogen which will give us that final product. So let's look at this mechanism and think about in general, what is the entropy of this situation? And as we said, that comes from thinking about the driving forces or toy. Well, neither of our product, pro neither of our reactants, excuse me, nor our product are charge or radicals or Carbocaine, anything like that that would be charged as well. So neither product nor reactant are particularly unstable. So we're not being driven by needing to stabilize an inherently unstable molecule. And we know that our reaction has to proceed by the mechanism we posited through a very unstable intermediate, our intermediate has two opposite charges on it relatively close to each other. So we'll put unstable intermediate. So the unstable intermediate, when you don't have a driving force of stabilizing, an inherently unstable initial reactant is going to make this reaction not favorable. So you'd need to have a really, you know, strong driving force in terms of your product being much more stable than your reactant to get past this unstable intermediate. So based on just what we've analyzed about the possible entropy factors, we would say that this is not favorable based on the fact that we don't need to stabilize a product or reactant and we have an unstable intermediate. So we'll say that this reaction is not favored due to entropy. So now our last step overall, well, it's not favored due to entropy. We're guessing it's not favored due to entropy. So it definitely would not be favored overall. So with our reaction not favored on the basis of entropy and most likely not favored on the basis of entropy, we can say that overall, this reaction would likely be not favorable to take place. See you in the next video

Identify the bonds broken and formed in the following addition reaction.
(a) Would you expect this reaction to be favored based on entropy?
(b) Based on enthalpy [qualitatively]?
(c) Overall?

Key Concepts

Addition Reactions

Entropy

Enthalpy

Watch next