Lecture 20. Energetics of Reactions
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Nowick, James S.
Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. 
 Good morning. Good morning. So today we're finishing up or hopefully finishing up our discussion of chapter six, understanding organic reactions. And what I'd like to talk about today are some of the specifics of the energetics of reactions. We talked last time in very general terms with energy diagrams talking about whether reactions went in a forward direction or didn't go, whether the products were higher in energy than the reactants or lower. We talked about the relative energies of intermediates and the relative energies of transition states. Today we're going to talk a little bit quantitatively about some of the energies of reactions. In most cases you won't be able to just look at a reaction and tell exactly how many kilocalories per mole of enthalpy it releases. But in certain simple reactions where you're just breaking and making a couple of bonds where there aren't too many, where there aren't ionic species present that have big energies of salvation, we can make some very accurate calculations about the energies of reactions. I'd also, if we have time at the end of the class, like to give you one number that's extremely useful to know, it's 1.36 and I'll put that into some context. But let's start by looking at a very simple reaction, the chlorination of methane and see if we can make some sense of the energy or more specifically the enthalpy of the reaction. So if you take methane, CH4, and you mix it with chlorine, they're both gases and usually this is done in the presence of some heater light, I'll talk more about that in a moment. You can get chloromethane in HCl. And I'd like to take a moment to talk about the energy of this reaction. Now we talked about the broad classes of reactions before and this falls into the class of a substitution reaction. Remember in a substitution reaction you break a bond and you replace it with something else. And so in this particular case we're breaking a bond between a carbon and a hydrogen or more specifically a methyl carbon and a hydrogen. And we're forming a bond between a carbon and a chlorine or more specifically a methyl carbon and a chlorine. So these are the bonds that are being made and broken. And if we can tally up all of the bonds, the energies of all of the bonds that we're making and we're breaking in this reaction, we can figure out the enthalpy of the reaction. Well these are the bonds that are perhaps obvious from an organocentric point of view meaning I'm looking just at the carbon atom which is what I do as an organic chemist. But of course we're also breaking the chlorine-chlorine bond and we're also forming an HCl bond. And so the question that I want to ask is does this reaction take or make energy? And I want to be a little more specific because the one thing that we really can address very well here specifically is enthalpy. We've been a little fast and loose in our energy diagrams between generic energy, enthalpy and free energy. And today I'd like to focus a little bit. So specifically we can focus this question and say is the reaction endothermic? Remember an endothermic reaction is one in which a heat of reaction is positive so I'll put parenthetically delta H0 is positive. Or is the reaction exothermic? And of course that's delta H0 is negative. So really if you can keep in mind that negative means heat given off products lower in energy than the reactants and positive means heat absorbed, then everything I'll talk about can make sense without us having to really look at any equations. In fact there are just two equations I think organic chemists need to know and I'll show you those later on. You've already seen them in general chemistry. So let's imagine for a moment that we're going to go through a series of breaking and making the appropriate bonds. So imagine if you will that we're going to break the methyl hydrogen bond. Now the great thing about this, this process actually occurs but in a slight different order, the great thing about it since we're only interested in the end result that we've broken chlorine-chlorine bond, we've broken a methyl hydrogen bond, we formed a methyl chlorine bond, we formed a hydrogen chlorine bond. It doesn't matter what really occurs in the mechanism of the reaction, all that we need to do is tally up the energy of the bonds and make sure that we've gotten to the right molecules. So imagine for a moment that we break this methyl hydrogen bond and we get CH3 dot. CH3 dot remember is a methyl radical. It's a carbon with an electron in a vacant p orbital, it's trigonal planar, roughly trigonal planar in geometry and you have an electron in a vacant p orbital but it doesn't matter what's happening here. Just imagine because we're going to have a chart and if you've read your chapter 6 we've already seen on page 207, there's a table that gives us lots of bond dissociation energies for specific molecules but there are also archetypes for general bonds where you have say a primary carbon bound to a hydrogen or a secondary carbon bound to a hydrogen or a tertiary carbon bound to a hydrogen. So imagine that we break the methyl hydrogen bond to get a methyl radical in a hydrogen atom. The delta H naught, we're just going to tally up the delta H naughts for all the bonds involved. So the delta H naught for this particular process is the bond dissociation energy. We can refer to that as BDE, it'll make things simpler because we'll be using it a couple of times. I'm just going to abbreviate it as BDE just like I sometimes abbreviate reaction RXN. And we can think of this as the bond strength. This is the energy to literally rip a bond apart. Now I don't know if this is going to work. I wanted to connect things to your textbook and let's see if I can wake up the document camera and give us a table here. If not, I'll just read it off of your textbook for you. Let's see if this works. Well, the good news about technology, the good news about technology foiling me is that I do have a copy of the page in your textbook. So on page 207, it gives a whole bunch of bond dissociation energies. It gives bond dissociation energies for hydrogen halide bonds, HF, HCl, et cetera. It gives all the halogens, the energies to rip their bonds apart. It gives various types of alcohol hydrogen bonds. It gives a methyl hydrogen bond. It gives an ethyl hydrogen bond. It gives an isopropyl hydrogen bond, a secondary bond to hydrogen from the isopropyl group to the hydrogen, various other bonds. It gives carbon carbon bond strengths, carbon halogen bond strengths. You'll have a chance to use this table in your discussion this week and I urge you to bring your textbook to the discussion section. We'll have a problem involving the hydrogenation of ethylene which is going to be very similar to what we're seeing here today. So as I said, table 6.2 page 207 and I already looked it up. The bond dissociation enthalpy for methane is 104 kilocalories per mole. And we're going to use kilocalories per mole in this class. It's a nice way of keeping track of things. All right, I want to do this for some others and then we can tally things up. Let me put this ELMO or whatever they call it away. All right, so four reactions that we have to worry about in this particular thought experiment. We already worried about breaking the methyl hydrogen bond. We said we paid a price of 104 kilocalories per mole. Other bonds that we break, we break the chlorine bond. If you look in your table on page 207, chlorine-chlorine bonds are a lot less strong than carbon-carbon bonds or carbon-hydrogen bonds. This is one of the reasons why chlorine and such a reactive gas bromine and iodine are as well. The bond dissociation enthalpy for the chlorine-chlorine bond is 58 kilocalories per mole. Now, again, thinking about this in a thought process, if we're forming a carbon-chlorine bond, at least formally what we're doing is we're taking a methyl group and a chlorine and bringing them together. The point is now we're looking at the negative of the bond dissociation enthalpy over here. In other words, the bond dissociation enthalpy of a methyl-chlorine bond, again, taken from your table, is 84 kilocalories per mole. But, of course, it's going to be in the opposite sense when we tally everything up. In other words, we're paying the price of 104. We're paying the price of 58. But we're getting a reward of 84 kilocalories per mole because formally we're forming that bond. And then lastly, we're going to be dealing with a hydrogen atom. And again, this doesn't matter whether the reaction occurs with ever seeing free hydrogen atoms. In fact, it does not occur with free hydrogen atoms. But we're forming a hydrogen-chlorine bond. And so we have to think about the energy, the price that we pay for breaking this bond. And then just realize that we reap the reward of that energy when we tally everything up. So the bond dissociation enthalpy here is 103 kilocalories per mole. And now all we have to realize is that the energy of the reaction is the price we pay in bonds broken minus the rewards that we get in bonds made. That makes sense, right? If you form a bond, it makes heat. Negative is good. Enthalpy, negative enthalpy means heat is given off. If you break a bond, you have a positive value. You have to put energy in to break a bond. So let's tally things up for this particular case. We pay a price of 104. We pay a price of 58. And then conversely, we reap a reward of 84. And we reap a reward of 103, all in kilocalories per mole. So the tally on this is negative 25 kilocalories per mole. That's an exothermic reaction. So let's give you a calibration scale. We saw that a kilocalorie per mole wasn't a whole heck of a lot of energy. That was the energy of two methyl groups banging into each other. 100 kilocalories per mole, that's a lot of energy. That's the energy of a bond roughly. 10 kilocalories per mole, that's a reasonably sizable amount of energy and here we're at 25. So this reaction releases a good deal of heat. Now a general rule is exothermic reactions tend to proceed. In most reactions you need some energy to get over energy barriers. In the actual mechanism of this reaction, which you'll learn about later, it's called a free radical halogenation reaction, a free radical coronation reaction. We need enough activation energy to break the weakest bond in the whole gamish. The weakest bond is the chlorine-chlorine bond and that's what gets the reaction started. Most reactions have some type of activation energy and that activation energy can be overcome usually by heat, sometimes by light. Now when I say heat, we have heat in this room and I don't mean just because it's a little warm here today. I mean that some reactions proceed at room temperature because molecules are in motion at room temperature. Some like this reaction need a little bit more heat, a couple of hundred degrees to get it going. Some reactions even will go at negative 78 degrees at the temperature of dry ice. There's still enough heat. Virtually no reactions would proceed at liquid helium temperatures 4 degrees Kelvin, colder than any place on Earth, very cold, but most reactions can get enough energy to overcome activation barriers with heat or with light if they are exothermic. And so let me finish the equation that I had written at the very start just so you have a complete equation for the chlorination of methane. CH3 plus Cl2 goes to, and I'll use the abbreviation delta, heat or H nu, a photon, light right at the cusp of the UV spectrum right at the edge of the visible. Sunlight's a good way of doing this. Goes to chloromethane plus HCl. Some reactions, as you'll see this week, also go with catalysis. And so the hydrogenation of ethylene often is carried out in the presence of a catalyst like palladium or platinum. You'll learn this week about a reaction, actually you won't learn about a reaction that involves potassium cyanide as a catalyst. We ended up putting that one off for a future time. All right. So what good is all of this? Let's take a look at another example, and I'm going to write a perfectly reasonable reaction on the blackboard. As a matter of fact, it's one that you should have been able to predict by analogy. After all, the halogens are pretty similar. Iodine is in many ways related to bromine or chlorine. So let's write the reaction iodine plus methane goes to methyl iodide plus HI. Let's see if we could expect this reaction to occur. And the question we want to ask is simply, is this reaction exothermic? So we're going to do the same thing but a little bit faster. The bond dissociation enthalpy we already saw was 104 kilocalories per mole for the methyl hydrogen bond. If you look in your table on page 207, the iodine bond is really weak. Its bond dissociation enthalpy is 36 kilocalories per mole. If you're watching your wallet, you know that having to pay less is good, right? We'd rather get a textbook for $36 than for $58. So you say you're well on your way to having an exothermic reaction here. But now we look at what we get back when we sell our textbooks. And the methyl iodine bond is worth a lot less than the methyl chlorine bond. The bond dissociation enthalpy is only 56 kilocalories per mole. And the hydrogen iodine bond bond dissociation enthalpy, again from table 6.2 on page 207, is only 71 kilocalories per mole. And so we can go ahead and figure out the enthalpy of this reaction, delta H naught. Now same calculation, 104 plus 36, the amount we've spent, minus 56 minus 71, the amount we've gotten back. And now we're at positive 13 kilocalories per mole. The reaction is endothermic. And endothermic reactions tend not to go. Highly endothermic reactions tend not to proceed. And as I said, 10 kilocalories per mole, 13 kilocalories per mole, that's a good bit. So I can go back to my original equation here and annotate it to ask whether I expected to proceed and the answer is I would draw an X through this arrow. This reaction would tend not to go. All right, the big, the wild card in all of this is that I've been talking about enthalpy rather than free energy. And really, really what we want to be talking about is free energy in the position of equilibria. Because really what I'm asking, every reaction you can at least nominally think of is an equilibrium. The real question is does an equilibrium lie to the left or the right? And in these reactions we're basically here saying, well let's ignore enthalpy. And that's actually a reasonable thing. The position of an equilibrium depends technically on delta G naught. Delta G naught is the free energy of reaction. Now I said there are two equations that one really needs to know for organic chemistry. And the ones you learned in general chemistry, everything else like the addition and breaking of bonds, I mean that's kind of intuitive. Two equations that I think really, really are the heart of any math in organic chemistry. Delta G naught equals delta H naught. Minus T delta S naught. And the T delta S naught, delta S naught is the change in entropy of a reaction. That's that nebulous term of disorder, randomness. We should know for example that if you break a bond, if you go from an iodine molecule to two iodine atoms from one particle to two particles, you've increased disorder. If you brought two things together, two iodine atoms, in general we've increased order. If we decrease order, if we increase entropy, delta S is positive. If we decrease entropy, delta S is negative. All right. So, and typically the value of this term here, the T delta S term is usually for a reaction less than about 10 kilocalories per mole. That's why I was saying by the time we're up at 13 or 25, chances are enthalpy is going to be enough to get away with it. All right. The other equation that one should know in organic chemistry is delta G equals negative RT natural log of K where K is the equilibrium constant. Because as I said, the real question on every reaction is you could at least nominally think of it as an equilibrium. Does the equilibrium lie to the right or to the left? And remember K is equal to the concentration of products with all of their powers. If you have two of a molecule then it would be to the squared over concentration of reactants. This should be old hat from general chemistry. In other words, if I have a reaction of A plus B is an equilibrium with C plus D, K is equal to concentration of C and D, the product of those two concentrations over the concentration of reactants A and D. Pardon? Which are the universal gas constant? Do you mean what value R is equal to 0.001987 kilocalories? Because as I said, I'm a kilocalories kind of person per mole Kelvin. All right. So what are the implications of this? I want to give you one example and then that generality of that number, 1.36 kilocalories per mole that I mentioned before. So let me give you one specific. We talked about the ring flip reaction of cyclohexane and I said that the energy to put a methyl group axial on cyclohexane is 1.8 kilocalories per mole. In other words, cyclohexane, methylcyclohexane exists as an equilibrium mixture of axial and equatorial conformers. Delta G naught for this reaction equals 1.8 kilocalories per mole. And so our equilibrium constant K equals E to the negative 1.8 divided by 0.001987, the gas constant and its product. And we'll say at 298 Kelvin. So I'll say at 298 K that equilibrium constant equals this value that's E to the negative 3.04 or 0.048. In other words, we have a ratio of 1 to 0.048 and that ratio is 21 to 1 just working the math on that at equilibrium. In other words, methylcyclohexane has a dynamic equilibrium between the axial and the equatorial conformer and at equilibrium you have about 5%, just under 5% of the molecules in the axial conformation and about 95 in the equatorial conformation. All right. The one very useful magic number to keep in mind, 1.36 kilocalories per mole. At room temperature at 298 Kelvin, that corresponds to a 10 to 1 ratio. That corresponds to an equilibrium constant of 1 of 10 or 0.1 depending on which value, whether you're positive or negative. In other words, delta G naught equals 1.36 kilocalories per mole leads to K equals 0.1. I just did the math on that but it's a nice number to have in your head because now with nothing more than arithmetic you can use it. If our delta G naught is negative, of course, negative is good, negative means to the right, that's going to lead to K equals 10. And the beauty of exponents and logarithms is now if you double that, your equilibrium constant is going to be 100. In other words, if it's negative 2.72, I've just taken negative 1.36 times 2, that's going to lead to K equals 100. And if our value is negative 4.08 kilocalories per mole, that's just negative 1.36 times 3, that's going to lead to K equals 1,000. So with nothing without a calculator, without anything, you suddenly can come up with a reasonable scale. Oh yeah, 10 kilocalories per mole really does mean an equilibrium lies pretty darn far in one direction or another. A kilocalorie here, a kilocalorie per there means you've got some of both species present. All right, next time we'll pick up talking about chapter 7. Goodnight, bye!
 Good morning.  Good morning.  So today we're finishing up or hopefully finishing up our discussion  of chapter six, understanding organic reactions.  And what I'd like to talk about today are some of the specifics  of the energetics of reactions.  We talked last time in very general terms  with energy diagrams talking about whether reactions went  in a forward direction or didn't go, whether the products were  higher in energy than the reactants or lower.  We talked about the relative energies of intermediates  and the relative energies of transition states.  Today we're going to talk a little bit quantitatively about some  of the energies of reactions.  In most cases you won't be able to just look at a reaction  and tell exactly how many kilocalories per mole of enthalpy  it releases.  But in certain simple reactions where you're just breaking  and making a couple of bonds where there aren't too many,  where there aren't ionic species present  that have big energies of salvation,  we can make some very accurate calculations  about the energies of reactions.  I'd also, if we have time at the end of the class,  like to give you one number that's extremely useful to know,  it's 1.36 and I'll put that into some context.  But let's start by looking at a very simple reaction,  the chlorination of methane and see if we can make some sense  of the energy or more specifically the enthalpy  of the reaction.  So if you take methane, CH4, and you mix it with chlorine,  they're both gases and usually this is done in the presence  of some heater light, I'll talk more about that in a moment.  You can get chloromethane in HCl.  And I'd like to take a moment to talk  about the energy of this reaction.  Now we talked about the broad classes of reactions before  and this falls into the class of a substitution reaction.  Remember in a substitution reaction you break a bond  and you replace it with something else.  And so in this particular case we're breaking a bond  between a carbon and a hydrogen  or more specifically a methyl carbon and a hydrogen.  And we're forming a bond between a carbon and a chlorine  or more specifically a methyl carbon and a chlorine.  So these are the bonds that are being made and broken.  And if we can tally up all of the bonds, the energies of all  of the bonds that we're making and we're breaking  in this reaction, we can figure out the enthalpy  of the reaction.  Well these are the bonds that are perhaps obvious  from an organocentric point of view meaning I'm looking just  at the carbon atom which is what I do as an organic chemist.  But of course we're also breaking the chlorine-chlorine bond  and we're also forming an HCl bond.  And so the question that I want to ask is does this reaction take  or make energy?  And I want to be a little more specific because the one thing  that we really can address very well here specifically is  enthalpy.  We've been a little fast and loose in our energy diagrams  between generic energy, enthalpy and free energy.  And today I'd like to focus a little bit.  So specifically we can focus this question  and say is the reaction endothermic?  Remember an endothermic reaction is one  in which a heat of reaction is positive  so I'll put parenthetically delta H0 is positive.  Or is the reaction exothermic?  And of course that's delta H0 is negative.  So really if you can keep in mind that negative means heat  given off products lower in energy than the reactants  and positive means heat absorbed,  then everything I'll talk about can make sense without us having  to really look at any equations.  In fact there are just two equations I think organic  chemists need to know and I'll show you those later on.  You've already seen them in general chemistry.  So let's imagine for a moment that we're going to go  through a series of breaking  and making the appropriate bonds.  So imagine if you will that we're going  to break the methyl hydrogen bond.  Now the great thing about this, this process actually occurs  but in a slight different order, the great thing about it  since we're only interested in the end result  that we've broken chlorine-chlorine bond,  we've broken a methyl hydrogen bond,  we formed a methyl chlorine bond,  we formed a hydrogen chlorine bond.  It doesn't matter what really occurs in the mechanism  of the reaction, all that we need to do is tally  up the energy of the bonds and make sure  that we've gotten to the right molecules.  So imagine for a moment that we break this methyl hydrogen bond  and we get CH3 dot.  CH3 dot remember is a methyl radical.  It's a carbon with an electron  in a vacant p orbital, it's trigonal planar,  roughly trigonal planar in geometry  and you have an electron in a vacant p orbital  but it doesn't matter what's happening here.  Just imagine because we're going to have a chart and if you've  read your chapter 6 we've already seen on page 207,  there's a table that gives us lots of bond dissociation energies  for specific molecules but there are also archetypes  for general bonds where you have say a primary carbon bound  to a hydrogen or a secondary carbon bound to a hydrogen  or a tertiary carbon bound to a hydrogen.  So imagine that we break the methyl hydrogen bond  to get a methyl radical in a hydrogen atom.  The delta H naught, we're just going to tally  up the delta H naughts for all the bonds involved.  So the delta H naught for this particular process is the bond  dissociation energy.  We can refer to that as BDE, it'll make things simpler  because we'll be using it a couple of times.  I'm just going to abbreviate it as BDE just  like I sometimes abbreviate reaction RXN.  And we can think of this as the bond strength.  This is the energy to literally rip a bond apart.  Now I don't know if this is going to work.  I wanted to connect things to your textbook and let's see  if I can wake up the document camera and give us a table here.  If not, I'll just read it off of your textbook for you.  Let's see if this works.  Well, the good news about technology,  the good news about technology foiling me is  that I do have a copy of the page in your textbook.  So on page 207, it gives a whole bunch  of bond dissociation energies.  It gives bond dissociation energies  for hydrogen halide bonds, HF, HCl, et cetera.  It gives all the halogens, the energies  to rip their bonds apart.  It gives various types of alcohol hydrogen bonds.  It gives a methyl hydrogen bond.  It gives an ethyl hydrogen bond.  It gives an isopropyl hydrogen bond, a secondary bond  to hydrogen from the isopropyl group to the hydrogen,  various other bonds.  It gives carbon carbon bond strengths,  carbon halogen bond strengths.  You'll have a chance to use this table  in your discussion this week and I urge you  to bring your textbook to the discussion section.  We'll have a problem involving the hydrogenation of ethylene  which is going to be very similar  to what we're seeing here today.  So as I said, table 6.2 page 207 and I already looked it up.  The bond dissociation enthalpy  for methane is 104 kilocalories per mole.  And we're going to use kilocalories per mole in this class.  It's a nice way of keeping track of things.  All right, I want to do this for some others  and then we can tally things up.  Let me put this ELMO or whatever they call it away.  All right, so four reactions that we have to worry  about in this particular thought experiment.  We already worried about breaking the methyl hydrogen bond.  We said we paid a price of 104 kilocalories per mole.  Other bonds that we break, we break the chlorine bond.  If you look in your table on page 207, chlorine-chlorine bonds  are a lot less strong than carbon-carbon bonds  or carbon-hydrogen bonds.  This is one of the reasons why chlorine  and such a reactive gas bromine and iodine are as well.  The bond dissociation enthalpy  for the chlorine-chlorine bond is 58 kilocalories per mole.  Now, again, thinking about this in a thought process,  if we're forming a carbon-chlorine bond,  at least formally what we're doing is we're taking a methyl group  and a chlorine and bringing them together.  The point is now we're looking at the negative  of the bond dissociation enthalpy over here.  In other words, the bond dissociation enthalpy  of a methyl-chlorine bond, again, taken from your table,  is 84 kilocalories per mole.  But, of course, it's going to be in the opposite sense  when we tally everything up.  In other words, we're paying the price of 104.  We're paying the price of 58.  But we're getting a reward of 84 kilocalories per mole  because formally we're forming that bond.  And then lastly, we're going to be dealing  with a hydrogen atom.  And again, this doesn't matter whether the reaction occurs  with ever seeing free hydrogen atoms.  In fact, it does not occur with free hydrogen atoms.  But we're forming a hydrogen-chlorine bond.  And so we have to think about the energy, the price  that we pay for breaking this bond.  And then just realize that we reap the reward  of that energy when we tally everything up.  So the bond dissociation enthalpy here is 103 kilocalories  per mole.  And now all we have to realize is that the energy  of the reaction is the price we pay in bonds broken minus the  rewards that we get in bonds made.  That makes sense, right?  If you form a bond, it makes heat.  Negative is good.  Enthalpy, negative enthalpy means heat is given off.  If you break a bond, you have a positive value.  You have to put energy in to break a bond.  So let's tally things up for this particular case.  We pay a price of 104.  We pay a price of 58.  And then conversely, we reap a reward of 84.  And we reap a reward of 103, all in kilocalories per mole.  So the tally on this is negative 25 kilocalories per mole.  That's an exothermic reaction.  So let's give you a calibration scale.  We saw that a kilocalorie per mole wasn't a whole heck  of a lot of energy.  That was the energy of two methyl groups banging  into each other.  100 kilocalories per mole, that's a lot of energy.  That's the energy of a bond roughly.  10 kilocalories per mole, that's a reasonably sizable amount  of energy and here we're at 25.  So this reaction releases a good deal of heat.  Now a general rule is exothermic reactions tend to proceed.  In most reactions you need some energy to get  over energy barriers.  In the actual mechanism of this reaction,  which you'll learn about later,  it's called a free radical halogenation reaction,  a free radical coronation reaction.  We need enough activation energy  to break the weakest bond in the whole gamish.  The weakest bond is the chlorine-chlorine bond  and that's what gets the reaction started.  Most reactions have some type of activation energy  and that activation energy can be overcome  usually by heat, sometimes by light.  Now when I say heat, we have heat in this room and I don't mean  just because it's a little warm here today.  I mean that some reactions proceed at room temperature  because molecules are in motion at room temperature.  Some like this reaction need a little bit more heat,  a couple of hundred degrees to get it going.  Some reactions even will go at negative 78 degrees  at the temperature of dry ice.  There's still enough heat.  Virtually no reactions would proceed  at liquid helium temperatures 4 degrees Kelvin,  colder than any place on Earth, very cold,  but most reactions can get enough energy to overcome  activation barriers with heat or with light  if they are exothermic.  And so let me finish the equation that I had written  at the very start just so you have a complete equation  for the chlorination of methane.  CH3 plus Cl2 goes to, and I'll use the abbreviation delta,  heat or H nu, a photon, light right at the cusp  of the UV spectrum right at the edge of the visible.  Sunlight's a good way of doing this.  Goes to chloromethane plus HCl.  Some reactions, as you'll see this week, also go with catalysis.  And so the hydrogenation of ethylene often is carried  out in the presence of a catalyst like palladium or platinum.  You'll learn this week about a reaction,  actually you won't learn about a reaction  that involves potassium cyanide as a catalyst.  We ended up putting that one off for a future time.  All right.  So what good is all of this?  Let's take a look at another example,  and I'm going to write a perfectly reasonable reaction  on the blackboard.  As a matter of fact, it's one  that you should have been able to predict by analogy.  After all, the halogens are pretty similar.  Iodine is in many ways related to bromine or chlorine.  So let's write the reaction iodine plus methane goes  to methyl iodide plus HI.  Let's see if we could expect this reaction to occur.  And the question we want to ask is simply,  is this reaction exothermic?  So we're going to do the same thing but a little bit faster.  The bond dissociation enthalpy we already saw was 104 kilocalories  per mole for the methyl hydrogen bond.  If you look in your table on page 207,  the iodine bond is really weak.  Its bond dissociation enthalpy is 36 kilocalories per mole.  If you're watching your wallet, you know that having  to pay less is good, right?  We'd rather get a textbook for $36 than for $58.  So you say you're well on your way  to having an exothermic reaction here.  But now we look at what we get back when we sell our textbooks.  And the methyl iodine bond is worth a lot less  than the methyl chlorine bond.  The bond dissociation enthalpy is only 56 kilocalories per mole.  And the hydrogen iodine bond bond dissociation enthalpy,  again from table 6.2 on page 207, is only 71 kilocalories per mole.  And so we can go ahead and figure out the enthalpy  of this reaction, delta H naught.  Now same calculation, 104 plus 36, the amount we've spent,  minus 56 minus 71, the amount we've gotten back.  And now we're at positive 13 kilocalories per mole.  The reaction is endothermic.  And endothermic reactions tend not to go.  Highly endothermic reactions tend not to proceed.  And as I said, 10 kilocalories per mole,  13 kilocalories per mole, that's a good bit.  So I can go back to my original equation here and annotate it  to ask whether I expected to proceed and the answer is I would  draw an X through this arrow.  This reaction would tend not to go.  All right, the big, the wild card in all of this is  that I've been talking about enthalpy rather than free energy.  And really, really what we want to be talking  about is free energy in the position of equilibria.  Because really what I'm asking, every reaction you can  at least nominally think of is an equilibrium.  The real question is does an equilibrium lie to the left  or the right?  And in these reactions we're basically here saying,  well let's ignore enthalpy.  And that's actually a reasonable thing.  The position of an equilibrium depends technically  on delta G naught.  Delta G naught is the free energy of reaction.  Now I said there are two equations  that one really needs to know for organic chemistry.  And the ones you learned in general chemistry,  everything else like the addition and breaking of bonds,  I mean that's kind of intuitive.  Two equations that I think really, really are the heart  of any math in organic chemistry.  Delta G naught equals delta H naught.  Minus T delta S naught.  And the T delta S naught, delta S naught is the change  in entropy of a reaction.  That's that nebulous term of disorder, randomness.  We should know for example that if you break a bond,  if you go from an iodine molecule to two iodine atoms  from one particle to two particles,  you've increased disorder.  If you brought two things together, two iodine atoms,  in general we've increased order.  If we decrease order, if we increase entropy,  delta S is positive.  If we decrease entropy, delta S is negative.  All right.  So, and typically the value of this term here,  the T delta S term is usually for a reaction less  than about 10 kilocalories per mole.  That's why I was saying by the time we're up at 13 or 25,  chances are enthalpy is going to be enough to get away with it.  All right.  The other equation that one should know  in organic chemistry is delta G equals negative RT natural log  of K where K is the equilibrium constant.  Because as I said, the real question on every reaction is  you could at least nominally think of it as an equilibrium.  Does the equilibrium lie to the right or to the left?  And remember K is equal to the concentration  of products with all of their powers.  If you have two of a molecule then it would be  to the squared over concentration of reactants.  This should be old hat from general chemistry.  In other words, if I have a reaction of A plus B is  an equilibrium with C plus D, K is equal to concentration  of C and D, the product of those two concentrations  over the concentration of reactants A and D. Pardon?  Which are the universal gas constant?  Do you mean what value R is equal to 0.001987 kilocalories?  Because as I said, I'm a kilocalories kind of person  per mole Kelvin.  All right.  So what are the implications of this?  I want to give you one example and then that generality  of that number, 1.36 kilocalories per mole  that I mentioned before.  So let me give you one specific.  We talked about the ring flip reaction of cyclohexane  and I said that the energy to put a methyl group axial  on cyclohexane is 1.8 kilocalories per mole.  In other words, cyclohexane, methylcyclohexane exists  as an equilibrium mixture of axial  and equatorial conformers.  Delta G naught for this reaction equals 1.8 kilocalories per mole.  And so our equilibrium constant K equals E  to the negative 1.8 divided by 0.001987,  the gas constant and its product.  And we'll say at 298 Kelvin.  So I'll say at 298 K that equilibrium constant equals  this value that's E to the negative 3.04 or 0.048.  In other words, we have a ratio of 1 to 0.048  and that ratio is 21 to 1 just working the math  on that at equilibrium.  In other words, methylcyclohexane has a dynamic equilibrium  between the axial and the equatorial conformer  and at equilibrium you have about 5%, just under 5%  of the molecules in the axial conformation  and about 95 in the equatorial conformation.  All right.  The one very useful magic number to keep in mind,  1.36 kilocalories per mole.  At room temperature at 298 Kelvin,  that corresponds to a 10 to 1 ratio.  That corresponds to an equilibrium constant of 1 of 10  or 0.1 depending on which value, whether you're positive  or negative.  In other words, delta G naught equals 1.36 kilocalories  per mole leads to K equals 0.1.  I just did the math on that but it's a nice number to have  in your head because now with nothing more  than arithmetic you can use it.  If our delta G naught is negative, of course,  negative is good, negative means to the right,  that's going to lead to K equals 10.  And the beauty of exponents and logarithms is now  if you double that, your equilibrium constant is going  to be 100.  In other words, if it's negative 2.72,  I've just taken negative 1.36 times 2,  that's going to lead to K equals 100.  And if our value is negative 4.08 kilocalories per mole,  that's just negative 1.36 times 3,  that's going to lead to K equals 1,000.  So with nothing without a calculator, without anything,  you suddenly can come up with a reasonable scale.  Oh yeah, 10 kilocalories per mole really does mean an  equilibrium lies pretty darn far in one direction or another.  A kilocalorie here, a kilocalorie per there means you've got some  of both species present.  All right, next time we'll pick up talking about chapter 7.  Goodnight, bye!
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