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Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins
Ch. 26 - Transition Metals
Energy Diagram
Gibbs Free Energy
Hammond Postulate
Carbocation Stability
Carbocation Intermediate Rearrangements
Additional Guides
Rank the following carbocations in order of decreasing stability

Bond dissociation energies describe the strength of chemical bonds. They can be determined experimentally. With that said, we’ll need some basic math for this section to determine the Enthalpy (∆H˚) for reactions. 

Concept #1: How to calculate enthalpy using bond dissociation energies. 


All right guys, so let's go ahead and talk about one of the most important topics in thermodynamics which is using bond association energies to calculate the enthalpy of a reaction.
Bond association energies are values that are going to describe the strength of chemical bonds. Now in chemistry courses that are way beyond the level of the course, organic chemistry, we actually can use math to figure out exactly why two atoms have the exact strength of bond that they do. But in this course, you're never going to be asked that. You're never going to be asked to explain why the bond between H and H is 436 kilojoules per mole. That's not going to happen.
So these are going to be basically experimental values that you're going to be given that you're going to use as reference to just figure out one question, is this going to be an exothermic reaction or is this going to be an endothermic reaction? Now keep in mind just because we know the enthalpy, doesn't mean that we actually know the spontaneity. Remember that the enthalpy is just one component of the spontaneity. We're also going to need to know the temperature and the delta S, but for right now all we're worrying about is the delta H. We're just saying is this exothermic or endothermic.
Let's remind ourselves of what exothermic means. It just means that I'm going to get a negative energy at the end and what that means is that I'm making bonds. Remember that it's favorable to make bonds. You release energy as you save energy between those orbitals. Then if I get a positive value, that means I'm breaking bonds because remember it takes energy, I have to put energy into the system to make these atoms fly apart again.
Here I've given you guys my own chart of bond association energies. All these values should be exactly the same to whatever chart you're going to be given. The only thing I did is I just didn't want to take up a full page with every single value, so I just gave you guys some of the most important ones here.
As you can see they're basically broken up into bonds of hydrogen, so I'm just going to put hydrogen bonds. Even though don't get that confused with hydrogen bonding that's just something different. Then I put here methyls, so things bonded to methyl. And then finally, basically things bonded to each other. I don't know how to say that better. But just methyl to methyl, oxygen to oxygen, stuff like that. So you guys can see just the general trends.
And what you can see is that there's a general trend that as my atoms get bigger, what's going to happen is that these bonds are going to get weaker. So you can see how it's not perfect. It's not a perfect trend. And, like I said, you'd have to use very complicated math to defend it. But, as a general rule, what we can say is that as these atoms that are attached to each other get bigger, the bonds are a little bit weaker and that has to do with the distance between them. These bigger atoms are going to have a farther distance than the smaller ones.
So one of the weakest bonds here is going to be I-I. Why? Because those are two humongous atoms that are trying to make a bond to each other. They're going to be much further apart than hydrogen-hydrogen would be. That can partially explain the huge difference in energy between hydrogen-hydrogen and iodine-iodine.
So as I just told you if we make a bond, that's going to be a negative value so that means I would assign a negative value to whatever bonds I'm making. If I make an H-OH, that means I'll put a negative value on that. If I'm breaking a bond, then I'm going to assign a positive value to that.
So what I want you guys to do for the next two problems is I want you guys to analyze the reaction that's taking place. You don't need to know the mechanism. You don't need to know exactly what's going on. But what you do need to know is which bonds are being broken, which bonds are being made. And then assign values to those using the chart that I gave you above. At the end, after you sum all of that together, you're going to know the enthalpy of the reaction and you're going to be able to tell me is this exothermic or endothermic. So go ahead and try it for yourself. 

Enthalpy (ΔH°) is the sum of bond dissociation energies for the reaction.

  • Negative values (-) indicate the making of new bonds = Exothermic
  • Positive values (+) indicate the breaking of new bonds = Endothermic

Example #1: Predict the sign and magnitude of ∆Ho in kj/mol for the following reaction. Identify the reaction as either exothermic or endothermic.


All right guys, so like I said you don't need to know the mechanism of this reaction. What we do need to know is what bonds are being broken and what bonds are being made.
So on my reagents side, reagents, what I have is the bonds that are breaking because what I can see is that I used to have a CH bond here, but now that's turning into C-Br so that means that at some point I must have broken that bond. So that means that this is going to be the breaking side.
If this is the breaking side what sign of enthalpy should this have? Should it be positive or negative? It should be positive because it's taking energy to make these bonds come apart. So all of the values that I put over here are going to be positive. Then all the values I put over here, for the making side, are going to be negative because these are ones that are going to release energy into the system.
So now all I have to do is identify these bonds. One of them I know already is going to be CH3-H. I know that bond is getting broken because right now I see, like I said, it used to be there and now it isn't. Another one that has to get broken is Br2 because Br2 is turning into HBr, so this is going to be Br-Br. That's another one that's going to require energy to break.
Now let's look at my negatives charge, not charge, my negative side. My exothermic side. So basically the bonds that I'm making are equal to CH3-Br because that's a new bond that didn't exist before and another one is H-Br. That's a bond that didn't exist before.
So now all I have to do is pull down the values from my chart and whichever ones are on my reagent side, or my breaking side, are going to be positive, whichever ones are on my product side, or my making side, are going to be negative. So let's go ahead and look up.
So what are the values that I'm bringing down? Well, the CH3-H is going to be 436 so that's going to be positive 436. Br-Br is going to be it looks like, where is that? It's right here, so it's going to be 192, positive 192. Then on the negative side or on the making side, CH3-Br, I'm going to just take myself out of this screen for a second so you guys can see. On the negative side, CH3-Br is going to be right here. That's going to be 923, so I'm going to give that a negative 923. And then H-Br is going to be negative 368. So now I'm pulling it back down. We have all the values we need. Hi, I'm back.
So now all we need to do is plug all these values into a calculator and whatever our end answer is, that's going to be the end answer. And I'm here. Sorry about that. Stop autofocusing. Okay. Let's go ahead. I've got my calculator here. Let's go ahead and enter these values. I have 436 plus 192 minus 293 minus 368 and what that should give you at the end is that your enthalpy, your delta H, is equal to negative 33 kilojoules per mole.
That's going to answer a really big question for us. Is this exothermic or is this endothermic? And the answer is that this has to be exothermic because at the end, the sum of my bond association energies is actually going to be a negative number, which means that I'm getting more stable at the end. I'm making more bonds or I'm gaining more energy than I'm losing by breaking those bonds. So this would be an exothermic reaction.
Do you know if this would be a spontaneous reaction? No, we have no clue yet. We still need the temperature and the delta S to figure that out. But we know with certainty that this is an exothermic reaction. Now I want you guys to do the same thing for the second question. Once again, all on your own. Try to figure it out and then I'll give you guys the answer. 

I hope that made sense! Let's try another similar problem with I, instead of Br and see how that changes things. 

Practice: Predict the sign and magnitude of ∆Ho in kj/mol for the following reaction. Identify the reaction as either exothermic or endothermic.