Strength of alkenes

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I'm looking at a question with a passage that contains the heats of hydrogenation of various alkenes, ranging from least to most substituted. The numbers become progressively less negative.
One question asks that "The greatest amount of energy is released by the oxidative cleavage of an alkene that is:"
The answer was unsubstituted

Another question asks, "The greatest amount of energy is required to break which of the following carbon carbon bonds?""
And the answer was the fully substituted alkene.


First of all, what do the values in the table mean? Why are they negative? Is that the energy that must be absorbed by the bond to break it?

How can the unsubstituted alkene release the most energy but the fully substituted require the most energy?
 
No, when a bond is broken, energy is given off. When bonds are formed, it requires energy, so the energy level decreases. It all depends on the given situation though.
 
No, when a bond is broken, energy is given off. When bonds are formed, it requires energy, so the energy level decreases. It all depends on the given situation though.
If you're going to answer questions, please try to do so a little more thoroughly; your responses are pretty incomplete, and that can be pretty confusing.
 
No, when a bond is broken, energy is given off. When bonds are formed, it requires energy, so the energy level decreases. It all depends on the given situation though.

This is not true.

To break a bond, you must add energy into the system (which explains the increase in energy during the activation part of most energy diagrams). Energy is then released as new bonds are formed. This is why in general chemistry ∆H = Energy of bonds broken - Energy of bonds formed. From the system's perspective, energy is coming in to break the bonds, so it is a positive value (from the system's perspective). From the system's perspective, energy is going out when new bonds form, so it is a negative value (from the system's perspective). It's a sign convention.
 
To break a bond, you must add energy into the system (which explains the increase in energy during the activation part of most energy diagrams). Energy is then released as new bonds are formed. This is why in general chemistry ∆H = Energy of bonds broken - Energy of bonds formed. From the system's perspective, energy is coming in to break the bonds, so it is a positive value (from the system's perspective). From the system's perspective, energy is going out when new bonds form, so it is a negative value (from the system's perspective). It's a sign convention.

Great response but you should omit the bolded part. You're confusing a thermodynamic effect with a purely kinetic one. In other words, adding energy to a system is a pre-requisite for any chemical reaction, whether endothermic or exothermic. You can't get away from an activation energy and the activation energy does not explain why breaking bonds is endothermic. The endothermic nature of bond-breaking is a purely thermodynamic effect and exists because of a change in the internal energy of the system.
 
The activation energy is a compilation of several factors, including overcoming the steric hindrance needed to align the reactants and the breaking of the bond to the leaving group. For nearly every bimolecular reaction, you have this process occurring, which is why all energy diagrams show an initial climb to the local maxima (referred to as the activation energy). Transition state formation is an energy-absorbing process. I have chosen "energy-absorbing" as opposed to "endothermic" to make it more palatable, but they are the same thing in this instance. There should be no confusion in terms of defining the climb of the energy diagram to the transition state between kinetics and thermodynamics, although in general undergraduate courses emphasize kinetics and thermodynamics as separate but related entities. If this concept is causing you an issue, then think about a simple SN2 reaction where backside attack requires the nucleophile align and the electrophile rehybridize. As this occurs, the leaving group starts to distance itself from the electrophilic carbon, which is the start of the bond-breaking process. The activation energy of the reaction is a sum of the alignment energy, the rehybridization energy, the bond partial dissociation energy minus the bond partial formation energy of the nucleophile to the electrophilic carbon. As the reaction proceeds, the hybridization returns to sp3, so that energy is returned as you move left to right along the reaction coordinate. You can see this in more detail by looking up ab initio calculations for substitution reactions. It is easiest to start with gas phase, although solvent effects are not really that complicated. Hope this helps.
 
The activation energy is a compilation of several factors, including overcoming the steric hindrance needed to align the reactants and the breaking of the bond to the leaving group. For nearly every bimolecular reaction, you have this process occurring, which is why all energy diagrams show an initial climb to the local maxima (referred to as the activation energy). Transition state formation is an energy-absorbing process. I have chosen "energy-absorbing" as opposed to "endothermic" to make it more palatable, but they are the same thing in this instance. There should be no confusion in terms of defining the climb of the energy diagram to the transition state between kinetics and thermodynamics, although in general undergraduate courses emphasize kinetics and thermodynamics as separate but related entities. If this concept is causing you an issue, then think about a simple SN2 reaction where backside attack requires the nucleophile align and the electrophile rehybridize. As this occurs, the leaving group starts to distance itself from the electrophilic carbon, which is the start of the bond-breaking process. The activation energy of the reaction is a sum of the alignment energy, the rehybridization energy, the bond partial dissociation energy minus the bond partial formation energy of the nucleophile to the electrophilic carbon. As the reaction proceeds, the hybridization returns to sp3, so that energy is returned as you move left to right along the reaction coordinate. You can see this in more detail by looking up ab initio calculations for substitution reactions. It is easiest to start with gas phase, although solvent effects are not really that complicated. Hope this helps.

I'm not confused with kinetics vs. thermodynamics but you seemed confused in your post above that I quoted. You say that to break a bond, energy must be added to the system. This much is correct. But then you qualify that with "which explains the increase in energy during the activation part of most energy diagrams." The energy required to break a bond does not explain the requirement for activation energy. The energy required to break a bond has to do with the energy required to bring the nuclei involved in the bond to infinite separation - this is a Coulombic argument and is a thermodynamic argument. The additional energy required for the reaction to go is what "explains" the activation energy. If my response is still not clear, imagine a system where the breaking of a bond is endothermic by 20 kcal/mol. The activation energy, however, which has enthalpic as well as entropic components (e.g. coming together of two molecules in the TS or breaking apart) is 50 kcal/mol. This reaction will not go at room temperature because the activation energy is well outside the room temperature limit for spontaneous reaction (which is 20-25 kcals) even though the energy that is required to bring the nuclei to infinite separation is within the room temperature limit - that is, 20 kcal/mol. So the increase in energy during the "activation" phase of a reaction coordinate diagram does not have to do with the energy required to break a bond per se because even in exothermic, bond-forming reactions, energy must initially be added to the system.

Solvent effects are actually much harder than you think. Most methods at the DFT level consider solvent as a continuum and do not model solvent effects explicitly. This is why among chemists, computations are not considered too reliable. In other words, errors of 10 kcal or more would not be too surprising because of the imperfections in modeling solvent interactions. For instance, if you're trying to do a catalytic reaction and your catalyst binds to the solvent (iron-aquas, for example) then your DFT results may not be accurate if you use the solvent continuum that is standard.
 
I'm assuming that your lower division class did not go into much detail in terms of the reaction versus reaction coordinate (few do). That appears to be the root of your reluctance to consider that the breaking of a bond (partial breaking as it were) is part of the activation energy in most chemical reactions. Set aside your kinetics is different from thermodynamics perspective for a moment and consider a simple reaction where the activation energy is merely a descriptive term for the sum of all energies involved in the formation of the transition state. What I am suggesting, and what you'll find in advanced organic chemistry or first year graduate school classes, is an analysis of the energy over the course of the pathway. First ask yourself what exactly you think activation energy is. Then consider why the reaction needs this input energy. You have introduced a theoretical description of bond dissociation as two atoms starting at a given distance and then going to infinite separation, but that is not at all what is going on within the confines of a flask. All atoms are interacting with other atoms if they are restricted to the volume of a solution. In a simple substitution reaction, such as an SN1 or SN2, the transition state is formed as the leaving group begins to leave (where it is facilitated by nucleophilic attack in an SN2 and solvent displacement in SN1). The leaving group stretching away from the electrophilic carbon represents a bond breaking process, which requires the input of energy into the system. It is not like the case you describe of infinite separation defining complete bond dissociation requiring a threshold amount of energy. It is the bond breaking process, where the interaction between the carbon and the leaving group goes from stronger to weaker, requiring energy. At its most fundamental level, this is energy absorbing. It is one of the multiple features occurring during the formation of the transition state.

Your bimolecular bond-formation example (something like ammonia with BF3 for instance) does not have a bond breaking aspect, but does require rehybridization of boron during the formation of the transition state. Because no bond is being broken, the activation energy is quite low. But again, that is the point above, where activation energy takes into account multiple factors. Given that most reactions involve the breaking and forming of bonds, you should consider more complex systems.

As for solvent effects, they are not as difficult as you may believe. If you wish to take into account non-linear effects, you can make then far more cumbersome than needed for simple ab initio calculations. This is essence is where physical chemists diverge from organic chemists. But as you mention, an 'average' solvent state serves well in these energetic calculations (i.e., continuum). The reality is that for many physical organic chemists, these calculations (DFT as you state) are actually quite well received for predicting structural features during a reaction process. They model systems rather well in fact. With electrocyclic reactions for instance, the error is rarely greater than 2 kcals/mole. As a system gets more complex, the error will increase, but I doubt you can find a specific example involving simple MCAT-level organic chemistry reaction where the error would be as large as you mention in your post.

Going much beyond this post is a moot point. Reaction modeling is a field in and of itself. We have both oversimplified chemical reactions, but that is appropriate at the MCAT level. I'm done responding as it does not serve the community well at this point. I think it would be great if there was a passage on theoretical calculations, but I doubt that will be the case.

I have a great deal of respect for your answers and believe you are one of the smartest people at this site. Your time and energy helps this site go and I do not want an argument of semantics to take away from anything you or I post. Thanks for your input and I'll see you in another thread I'm sure.
 
I'm assuming that your lower division class did not go into much detail in terms of the reaction versus reaction coordinate (few do). That appears to be the root of your reluctance to consider that the breaking of a bond (partial breaking as it were) is part of the activation energy in most chemical reactions. Set aside your kinetics is different from thermodynamics perspective for a moment and consider a simple reaction where the activation energy is merely a descriptive term for the sum of all energies involved in the formation of the transition state. What I am suggesting, and what you'll find in advanced organic chemistry or first year graduate school classes, is an analysis of the energy over the course of the pathway. First ask yourself what exactly you think activation energy is. Then consider why the reaction needs this input energy. You have introduced a theoretical description of bond dissociation as two atoms starting at a given distance and then going to infinite separation, but that is not at all what is going on within the confines of a flask. All atoms are interacting with other atoms if they are restricted to the volume of a solution. In a simple substitution reaction, such as an SN1 or SN2, the transition state is formed as the leaving group begins to leave (where it is facilitated by nucleophilic attack in an SN2 and solvent displacement in SN1). The leaving group stretching away from the electrophilic carbon represents a bond breaking process, which requires the input of energy into the system. It is not like the case you describe of infinite separation defining complete bond dissociation requiring a threshold amount of energy. It is the bond breaking process, where the interaction between the carbon and the leaving group goes from stronger to weaker, requiring energy. At its most fundamental level, this is energy absorbing. It is one of the multiple features occurring during the formation of the transition state.

You assume too much. I am several years into my PhD in chemistry (switching fields now, obviously) and am talking with experience in physical organic chemistry as well as graduate-level PChem. I agree that this is more advanced than needed for the MCAT though, but would like to make the chemistry apparent for other people. I get where our confusion is though. You spoke of an "increase in energy during the activation part" of a reaction coordinate whereas I thought you were talking about an activation energy in general and when you said "to break a bond, energy must be added to the system," I thought you were talking about ground state effects in the substrate and product. The reason this caused a problem was because if you were talking about an activation energy in general, then the energy input required to vibrationally excite the bond to the point of bond breaking does not by itself explain the activation energy (although it explains part of it - hence the increase in energy). Now I believe we are on the same page.

As for solvent effects, they are not as difficult as you may believe. If you wish to take into account non-linear effects, you can make then far more cumbersome than needed for simple ab initio calculations. This is essence is where physical chemists diverge from organic chemists. But as you mention, an 'average' solvent state serves well in these energetic calculations (i.e., continuum). The reality is that for many physical organic chemists, these calculations (DFT as you state) are actually quite well received for predicting structural features during a reaction process. They model systems rather well in fact. With electrocyclic reactions for instance, the error is rarely greater than 2 kcals/mole. As a system gets more complex, the error will increase, but I doubt you can find a specific example involving simple MCAT-level organic chemistry reaction where the error would be as large as you mention in your post.

My field is organometallic chemistry, with a focus on organic synthesis using catalysis. I won't pretend to have any knowledge about physical chemists' computations but I do know that in organic systems - especially with a catalyst - DFT is not seen to be reliable and claiming an error of <2 kcals would be laughable. Again, this is how it is viewed in organics. Ab initio methods serve well for the gas phase, as you say, but very few reactions are carried out in the gas phase. Once you introduce solvent terms into the method, that's where the energy goes awry and if you're modeling a catalyst, that's even harder because metals have an exponentially larger toolbox of weird things to do than organics - something that we do not teach in your freshman organic chemistry class.
 
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