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NS4. Stereochemistry
Apart from measuring rates of reaction and deducing the rate law, there are other lines of evidence that can suggest how the reaction is occurring. For aliphatic nucleophilic substitution, stereochemistry of the products provides some additional evidence.
Suppose you carry out a nucleophilic substitution reaction using a chiral starting material. You decide to convert (S)-3-chlorooctane into the corresponding azide. Azides are pretty widely used reagents (but slightly dangerous and potentially explosive). They are employed in a class of reactions called "click chemistry;" you've just heard about these reactions and you want to try one out for yourself.
You have complete control over the mechanism of the reaction (not so easy in reality, but in this thought experiment you can set the dial on your stir plate to the desired mechanism). You choose to make the reaction occur through an SN2 pathway.
You know the product will be chiral, so you plan to check its optical rotation. The trouble is, once you have finished the reaction, the optical rotation is exactly the opposite of what you were expecting, based on the values of other compounds like this one. You did the reaction successfully but got the unexpected enantiomer.
You're not worried. You've been taking this nifty chemistry class and you have an idea of something else to try. This time you select an SN1 pathway.
You finish the reaction and get the right product, but it shows no optical rotation whatsoever. This time you got a racemic mixture.
This is just a thought experiment, but what would it all mean? Why might changing mechanism influence the stereochemistry?
This is just a thought experiment, but the results are generally true: in an SN2 reaction, the chiral center undergoes an inversion. The three-dimensional arrangement of groups around the chiral center is the opposite of how it started. In an an SN1 reaction, the chiral center undergoes racemization. There is a 50:50 mixture of enantiomers.
Problem NS4.1.
1. Propose reasons why the stereochemistry would flip in an SN2 reaction. A drawing will help.
2. Propose reasons why an equal mixture of stereoisomers would result from an SN1 reaction. A drawing will help.
Problem NS4.2.
You may have noticed that two different solvents were used in the two reactions above. Propose a reason why this change in solvent may lead to a change in mechanism.
In an SN2 reaction, the nucleophile donates electrons to the electrophilic carbon, displacing the leaving group from the other side. The nucleophile donates its electrons to the lowest unoccupied molecular orbital, which displays a lage lobe on the side of the carbon opposite the leaving group.
As a result, the nucleophile always approaches from the opposite side of the electrophilic carbon as the location of the leaving group, and ends up on the opposite side from where the leaving group group was.
• SN2 reactions at chiral centers lead to reversal or "inversion" of the chiral center.
On the other hand, in an SN1 reaction, the nucleophile enters only after the leaving group has left. At that point, the electrophilic carbon is a cation, so it is trigonal planar because it only has three groups attached to it. The nucleophile could easily approach from either side.
• SN1 reactions at chiral centers lead to racemic mixtures.
Problem NS4.3.
Show the products of the following reactions, and indicate stereochemical configuration of bothe the starting material and the product in each case.
Problem NS4.4.
The lesson here can be restated as follows: the mechanism affects the stereochemical outcome of the reaction. Explain why that fact is important in the context of making a chiral drug for the pharmaceutical market. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS4._Stereochemistry_in_NS.txt |
NS5. Regiochemistry
Regiochemistry is the term for where changes take place in a reaction. It can be another indication of how the reaction occurred.
In aliphatic nucleophilic substitution, the answer seems pretty obvious: the reaction takes place at the electrophilic carbon, the one attached to the electronegative halogen. That's where the leaving group is. When the leaving group is replaced, that's where the nucleophile will be. But this isn't always true.
• In an SN2 reaction, the nucleophile is always found on the carbon where the leaving group used to be.
• In an SN1 reaction, the nucleophile is usually found on the carbon where the leaving group used to be. Sometimes it moves.
Under some circumstances, unexpected changes occur. The following two reactions are examples of such surprises. These reactions happen to take place via an SN1 mechanism.
So, the regiochemistry of this reaction may be more complicated than we thought.
What is happening in these two reactions? In one of them, the bromine is just hopping from one place to another along the molecule. Some of the original compound remains, too, so there is a mixture. If you look carefully, though, the bromine has switched places with a hydrogen atom. It doesn't seem like that hydrogen atom could come off very easily.
In the other reaction, something very similar is happening. Bromide and chloride both have lone pairs, so they can both be nucleophiles as well as leaving groups, and one can replace the other. There is lots of chloride around, so it beats any bromide to the electrophilic carbon. Once again, though, some of the chloride seems to end up in the wrong place.
This sort of behaviour is characteristic of carbocations. It is called a rearrangement, in which part of the molecule unexpectedly switches places.
Again, one of the products forms in a simple enough way.
The formation of the other product involves a "1,2-hydride shift." In this event, a hydrogen anion hops from one carbon to the next, leaving a cation where it used to be.
Problem NS5.1.
Draw the reaction above with all the hydrogens drawn in the structures, to confirm the formal charges and the positions of the hydrogens.
Problem NS5.2.
Explain the observed product ratios in the above reaction.
Problem NS5.3.
Predict the products of the following reactions.
The hydride can hop one carbon away because of the proximity to the empty p orbital with which it can overlap and form a new bond.
The barrier for a hydride shift is not very high, provided a carbocation is available on the very next carbon. As a result, an equilibrium between cations is established pretty quickly. Below, there is an equilibrium between two secondary cations on the 2-methylpentyl skeleton.
However, that particular structure has another possible cation that is more stable. Once a tertiary cation forms, the hydride isn't likely to hop back.
On the other hand, there is also a primary position. A hydride shift could give a primary cation, but that isn't likely to happen, because it would be too far uphill.
Altogether, there is an energy surface linking several different possible cations. In this case, however, one tertiary cation would quickly dominate. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS5._Regiochemistry_in_NS.txt |
NS6. Structural Factors Influencing the Mechanism
If two possible mechanisms can occur, there may be some factors that have an influence on the course of the reaction, tipping it one way or the other. One of the most important factors determining the mechanism of aliphatic nucleophilic substitution is the structure of the alkyl halide.
The SN1 mechanism involves formation of a carbocation. Ion stability is often a very important factor influencing how easily a reaction occurs. It stands to reason that, the more stable the cation that forms, the more easily an SN1 mechanism can occur.
There is a simple trend that more substituted carbocations are more stable than less-substituted ones. By more-substituted, we mean carbocations in which the carbon bearing the positive charge is attached to more carbons and fewer hydrogens. A tertiary carbocation is more stable than a secondary, a secondary is more stable than a primary carbocation, and a primary is more stable than a methyl cation.
This trend is usually explained by hyperconjugation in the more substituted cation. In hyperconjugation, neighbouring σ-bonding orbitals overlap with the empty π-orbital that is the center of the carbocation.
That electronic donation from the occupied σ-bonding orbitals helps delocalize the positive charge, lowering the positive charge on the central carbon and placing a little of it on the surrounding ones. Energetically, the interaction is favorable because the electrons in the σ-bonding orbitals are lowered stabilized by delocalization.
Overall, SN1 reactions occur much more easily when the halide is attached to a more substituted carbon. The resulting carbocation forms more readily in what is otherwise the hardest step in the reaction. SN1 reactions occur most easily at tertiary carbons, moderately well at secondary ones, and very sluggishly at primary ones, if at all.
Problem NS6.1.
Although they may be considered primary alkyl halides, compounds like benzyl chloride and allyl bromide are capable of undegoing SN1 reactions. Show why.
SN2 reactions do not undergo highly charged intermediates, so ion stability is less important in that pathway. On the other hand, there is a stark contrast between SN1 reactions and SN2 reactions in terms of steric effects. SN1 reactions occur via a trigonal planar intermediate, which is less sterically crowded than the starting material. SN2 reactions do not occur via an intermediate, but the transition state through which they proceed is actually five-coordinate; the electrophilic carbon gets more crowded as the reaction proceeds. Steric effects are much more important in this concerted pathway.
Comparing the approach to the transition state in a series of alkyl halides of different substitution, we can see some steric differences. The pictures shows a snapshot very early in the reaction, when the nucleophile is just approaching the electrophile but the reaction is not really committed yet. Comparing these pictures, it seems most likely that the nucleophile will keep going forward in the top case, with the methyl. In the bottom case, with the tertiary halide, chances are that the nucleophile will just bounce off the methyl groups before it can connect with the electrophilic carbon.
As a result, SN2 reactions are more likelyto occur with less substituted alkyl halides. They occur very easily with methyl halides and primary alkyl halides. They occur moderately well at secondary alkyl halides, but only with difficulty at tertiary alkyl halides.
Problem NS6.2.
In each of the following cases, is nucleophilic substitution likely to proceed via an SN1 or an SN2 reaction? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS6._Structural_Effects_in_NS.txt |
NS7. Solvent Effects
There are often several factors that can influence the course of a reaction. Probably the most important is the structure of the alkyl halide, but the solvent can also play a role.
The crucial difference between SN1 and SN2 reacions is the ionization step in the SN1 pathway. Factors that stabilize ions, and assist in ionization, promote this pathway.
In general, more polar solvents are often helpful in nucleophilic substitutions; the nucleophile may be an ionic compound itself, and a more polar solvent will help it to dissolve. However, especially polar solvents may provide additional stability to ions.
The most polar solvents tend to be those that are capable of hydrogen bonding, such as water and alcohols. These are sometimes called "polar, protic solvents." The "protic" part refers to hydrogen bonding; in hydrogen bonding, the hydrogen atom attached to a very electronegative oxygen or nitrogen develops a significant positive charge, like a proton.
Polar, protic solvents can stabilize ions through very strong intermolecular attractions. The protic hydrogen can strongly interact with anions, whereas the lone pair on the oxygen atom can stabilize cations.
These stabilizing interactions can strongly stabilize the intermediates in SN1 reactions. In the same way, the transition state leading into the intermediate is also significantly stabilized. The barrier for this reaction is lowered and the reaction can occur more quickly.
In addition, polar, protic solvents may play an additional role in stabilizing the nucleophile. If the nucleophile is stabilized, it is less likely to react until a sufficiently strong electrophile becomes available. As a result, polar, protic solvents may also depress the rate of SN2 reactions. Once the alkyl halide ionizes and a more attractive electrophile becomes available, the nucleophile can spring into action.
Problem NS7.1.
The following are a baker's dozen of potential solvents.
1. Identify the protic solvents in this group.
2. Identify the completely non-polar solvents in this group.
3. Identify the polar, aprotic solvents.
4. Rank the polar solvents from most polar to least polar.
5. Identify two basic compounds, frequently used to facilitate proton transfers in reactions.
Problem NS7.2.
Describe the SN1 reaction as slow, medium or fast in the following cases.
1. NaCN and PhCH2Cl in acetonitrile, CH3CN.
2. NaSH and 2-bromo-2-methylheptane in methanol.
3. KI and 1-chloropentane in 2-propanol.
Problem NS7.3.
Describe the SN2 reaction as slow, medium or fast in the following cases.
1. LiCCCH3 and 2-bromopentane in THF.
2. NaN3 and 3-chloro-3-methyloctane in DMF.
3. LiOPh and 1-bromohexane in methanol.
NS8. Nucleophilicity in NS
NS8. Nucleophilicity
The nucleophile can sometimes play a pronounced role in nucleophilic substitutions. The following relative rates have been observed when these nucleophiles reacted with methyl bromide in methanol:
note: Ph = phenyl, C6H5; Ac = acetyl, CH3C=O; Et = ethyl, CH3CH2.
Presumably, some of the species react much more quickly with methyl bromide because they are better nucleophiles than others.
Problem NS8.1.
Sometimes we can draw general conclusions about kinetic factors by looking at sub-groups among the data. Determine how the following factors influence nucleophilicity (the ability of a species to act as a nucleophile). Support your ideas with groups of examples from the data (preferably more than just a pair of entries).
1. charge on the nuclophile
2. size of the atom bearing the charge
3. electronegativity of the atom bearing the charge
4. delocalization of charge
Problem NS8.2.
Nucleophilicity plays a strong role in the rate of one type of substitution mechanism, but not the other.
1. In which mechanism is it important? Support your idea.
2. Is the reaction of methyl bromide likely to proceed via this mechanism? Why or why not?
Problem NS8.3.
A trend very similar to the data above is found in substitution reactions of py2PtCl2 (py = pyridine) in methanol. Draw a mechanism for this substitution and explain why nucleophilicity plays an important role.
Problem NS8.4.
Very fast nucleophiles are sometimes more likely to undergo SN2 reactions than SN1 reactions. Explain why. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS7._Solvent_Effects_in_NS.txt |
NS9. Enolate Nucleophiles
Enolates and related nucleophiles deserve a closer look because they are and because they have their own issues of regiochemistry.
Remember, an enolate is just the conjugate base of an enol. An enolate can also be thought of as the conjugate base of a related carbonyl. Because the enolate is a delocalized anion, it can be protonated in two different places to get two different conjugates.
Enols typically are not seen because of a rapid equilibrium with that related carbonyl compound. As soon as an enol forms, if there is any way for it to transfer a proton to get to the carbonyl, it will do so. This kind of equilibrium is called "tautomerism," involving the transfer of one proton from one place to another within the molecule. The enol and its related carbonyl are referred to as "tautomers." "Tautomers" describes the relationship between these two molecules.
Enamines are very similar to enolates, but with a nitrogen atom in place of the oxygen. Hence, they are amines instead of alcohols.
Enamine, enolates and enols are all turbo-charged nucleophiles. The nucleophilic atom is the alpha carbon. Although that carbon can be thought of as a double bonded carbon, with no lone pair, that position is motivated to donate electrons because of pi donation from the oxygen (or nitrogen).
One of the issues with these nucleophiles has to do with asymmetry about the carbonyl (or the would-be carbonyl). If one alpha position next to the carbonyl isn't the same as the other one, two possible enolates could result from removal of a proton.
That means we potentially have two different nucleophiles from the same starting compound. Sometimes, mixtures of products result from enolate reactions.
Nevertheless, enolates and enamines are very broadly used in the synthesis of important things like pharmaceuticals, precisely because they can be controlled so well. If you want the enolate on one side of the carbonyl -- we'll call it the more-substituted side -- then you can have it. If you want the enolate on the other side of the carbonyl -- the less-substituted side -- you can have that, instead.
Let's think about what is different about those two sides of the carbonyl. One side is more substituted. It has more stuff on it. It's more crowded.
We will focus on the formation of enolate ions. To get the proton off and turn a carbonyl compound into an enolate requires a base. Some control over which proton is removed might come from the choice of base. Maybe to get the proton off the more crowded position, you need a smaller base.
Conversely, to get the proton exclusively from the least crowded position, and have very little chance of getting it from the more crowded spot, you could use a really big base.
But there's something else about enolates that is apparent only when you look at the ions in one resonance form. Enolate ions can be thought of as alkenes, of course. Depending on which proton we remove, we get two different alkenes. There may be factors that make one of these two alkenes more stable. If so, there may be ways to form that one instead of the other.
In general, more-substituted alkenes are more stable than less-substituted ones. The more substituted alkene is formed via loss of the proton at the more crowded position.
Forming a product based on its relative stability means relying on thermodynamics. One way to do that is to allow the deprotonation happen reversibly. Given multiple chances, the more stable enolate will form eventually.
On the other hand, if you intend to take the proton off the least substituted position, you don't want any reversibility. Given the chance, the wrong enolate will eventually form.
This is a case in which we need kinetic control to get one product: we want the least-substituted enolate, and we depend on it forming more quickly than the ther enolate. After all, if the other enolate forms , it is more stable; it isn't likely to come back.
For the other product, we need themodynamic control. We depend on the eventual product stability of the more substituted enolate to pull the reaction through. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Nucleophilic_Substitution_at_Tetrahedral_Carbon/NS9._Enolate_Nucleophiles.txt |
OA5. Coupling Reactions in Organic Synthesis
Oxidative addition and reductive elimination are key steps in industrial catalysis. For example, both steps are featured in palladium-catalyzed cross-coupling reactions, the subject of the 2010 Nobel Prize in Chemistry. The prize was awarded to Richard Heck of the University of Delaware, Ei-Ichi Negishi of Purdue University and Akira Suzuki of Hokkaido University. With these reactions, workers in a variety of fields can make molecules that otherwise would be quite difficult to make. These molecules in turn may be important pharmaceuticals or useful compounds for electronic displays in computers and other devices, to name just a couple.
Figure OA5.2. Catalytic cycle for the Negishi cross-coupling reaction.
Oxidative addition is used to activate substrates. Substrates that normally might not react get ready to react with something. Then they sit on the metal atom and wait for something else to come along and react with them.
Figure OA1.1. A generalized oxidative addition.
Bonds can be broken via oxidative addition that cannot easily be broken via other reactions. For example, although the H-H bond is very strong, it can be cleaved in the presence of a variety of metal atoms and ions. Transition metals possess the right tools to coax the two hydrogen atoms apart from each other.
Figure OA1.2. A generalized oxidative addition of dihydrogen.
Figure OA1.3. A specific case of oxidative addition of dihydrogen. When hydrogen adds to Vaska's complex, the oxidation state of iridium changes from +1 to +3.
Reductive elimination, in turn, is used to couple different groups together to form useful products. Once two groups are sitting beside each other on a transition metal atom or ion, they can bond to each other rather than the metal and go off together as a new molecule.
Figure OA1.5. A general scheme for reductive elimination.
Clearly, a reductive elimination is just an oxidative addition in reverse. The reaction can go in either direction. That means it can, conceivably, occur in equilibrium. At this level, you would not be expected to know which direction would be favoured for a particular reaction. However, you might be able to predict which direction a reaction would proceed based on factors such as le Chatelier's principle.
• le Chatelier's principle says that a change in the reaction conditions will lead to a shift in product:reactant ratio that offsets the change
For example, adding more reactant to the reaction shifts equilibrium to the right. More product is made, and some of the extra reactant is used up, so that the system can come back to its natural equilibrium. If products are somehow removed from the system, the reaction will also shift to the right, using up reactants and replacing the missing product. If the reaction is exothermic (produces heat) and more heat is added to the system, the reaction would shift to the left, using up some products and making more reactants in order to remove excess heat.
The reversibility of oxidative addition / reductive elimination actually serves very well in catalytic processes. For example, one of the most important catalytic processes in the world is catalytic hydrogenation, in which two hydrogen atoms are added across a double bond (usually a C=C bond, but sometimes a C=O or C=N bond). The process requires oxidative addition of hydrogen to a metal, but it also requires reductive elimination of an alkyl and a hydride to form the final product, forming a hydrocarbon.
Figure OA1.5. A reductive elimination of a methyl and a hydride to form methane.
The addition of dihydrogen to Vaska's complex and other transition metals is a reversible reaction. The hydrogen can be released again if the reaction moves to the left in a reductive elimination. That reversibility makes transition metal compounds useful for hydrogen storage. Hydrogen gas is voluminous, flammable and generally dangerous. By cleaving H2 and binding hydrogen to metal atoms, hydrogen can be more safely stored and released again under the right conditions.
In reductive elimination, bonds can be made that cannot be formed via other reactions. That makes it a useful part of strategies to make commodity chemicals and complex organic molecules such as pharmaceuticals.
Problem OA1.1.
Based on le Chatelier's principle, propose conditions under which:
1. Vaska's complex could bind hydrogen
2. the resulting dihydride adduct of Vaska's complex could release dihydrogen again.
Problem OA1.2.
Draw products of oxidative addition of the following compounds to (PPh3)2Pd.
a) HBr b) H2 c) I2 d) CH3-Br
OA2. Overview of Oxidative Addition
OA2. Reaction Overview
Oxidative addition is a general term for the insertion of a metal between two atoms that were previously bonded together.
Figure OA2.1. Our general scheme for an oxidative addition of hydrogen to a metal.
Oxidative addition is a reaction type rather than mechanism. Several different mechanisms are possible, including polar reactions, non-polar / concerted reactions and radical reactions. For example the formation of Grignard nucleophiles by treatment of alkyl halides with magnesium could be described as an oxidative addition.
Figure OA2.2. Formation of a Grignard reagent could be described as an oxidative addition, although the mechanism does not resemble the ones that we will look at here.
However, Grignard formation is believed to occur via a radical reaction, and has very little relationship to the addition of hydrogen to Vaska's complex, for instance. In this section, we will look at polar oxidative additions and concerted oxidative additions. Radical oxidative additions will be left for a later chapter on radical chemistry.
Where do the terms "oxidative addition" and "reductive elimination" come from? Think back to how we learned to count valence electrons in transition metal complexes. One of the first things we did was remove the ligands from the complex to see whether there would be a charge on the metal without the ligands.
Figure OA2.3. Deconstruction of a metal complex to discern the charge on the metal.
If each H ligand in Figure OA2.2 is a hydride (reasonable because H is more electronegative than the metal), then removing them would leave the metal with a 2+ charge. That means the metal is in the (II) oxidation state, even if it doesn't formally have a charge on it in the complex. Addition of H2 to a metal atom, M, is accompanied by increase in formal oxidation state at the metal (by +2).
Reductive Elimination is microscopic reverse of oxidative addition.
Figure OA2.3. Deconstruction of a metal complex to discern the charge on the metal.
Two atoms that were bonded to one metal atom become bonded to each other, instead. Keep in mind the formal oxidation state of the metal. This elimination is accompanied by decrease in formal oxidation state at the metal (by -2).
Remember, oxidation state can often be determined by giving each ligand the pair of electrons it shares with the metal. Sort out the formal charge on the donor atom in the ligand, and you will know the charge or oxidation state of the metal.
For example, in the example of MH2 in Figure OA2.3., assume each bond between M and H is a pair of electrons that belongs with the hydrogen. A hydrogen with two electrons is an anion. As a result, the metal must have an oxidation state of +2 (usually written with Roman numeral II).
Problem OA2.1.
Determine the oxidation state on the metal before and after each of the following reactions.
Problem OA2.2.
Propose a reason why the addition of a ligand such as a phosphine can sometimes result in reductive elimination from a coordination complex. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Oxidative_Addition_and_Reductive_Elimination/OA1%3A_Intro_to_Oxidative_Addition.txt |
OA3. Polar Oxidative Addition
The polar oxidative addition mechanism is very similar to an aliphatic nucleophilic substitution (SN1 or SN2) reaction.
Figure OA3.1. An example of a polar oxidative addition.
In an oxidative addition, the metal can act as a nucleophile in the first step in an SN2 process. In the second step, the liberated halide binds to the metal. That doesn't happen in a normal nucleophilic substitution. In this case, the metal has donated its electrons and is able to accept another pair from the halide.
Figure OA3.2. Mechanistic steps in a polar oxidative addition.
Polar oxidative addition has some requirements similar to a regular SN1 or SN2 reaction:
• Requires good leaving group
• Requires tetrahedral carbon (or a proton) as electrophile
Problem OA3.1.
1. What do you think is the most difficult step (i.e. the rate-determining step) for the reaction in Figure OA3.2? Why?
2. Suggest the probable rate law for this reaction.
Problem OA3.2.
The platinum compound shown below is capable of reductively eliminating a molecule of iodobenzene.
a) Show the products of this reaction.
The starting platinum compound is completely stable in benzene; no reaction occurs in that solvent. However, reductive elimination occurs quickly when the compound is dissolved in methanol instead.
b) Explain why the solvents may play a role in how easily this compound reacts.
The reaction in methanol is inhibited by added iodide salts, such as sodium iodide.
c) Provide a mechanism for the reductive elimination of iodobenzene from the platinum complex, taking into account the solvent dependence and the inhibition by iodide ion.
Problem OA3.3.
For the following reaction,
1. Identify the oxidation state at platinum in the reactant and the products.
2. Assign stereochemical configuration in the product and the reactant.
3. Explain the steresochemistry of the reaction.
Problem OA3.4.
Reaction of the following deuterium-labeled alkyl chloride with tetrakis(triphenylphosphine) palladium produces an enantiomerically pure product (equation a). Draw the expected product.
However, reaction of a very similar alkyl halide produces a compound that is only 90% enantiomerically pure. Draw the major product and explain the reason that there is some racemization.
Problem OA3.5.
Frequently, oxidative additions and reductive eliminations are preceded or followed other reactions. Draw a mechanism for the following transformation.
OA4. Concerted Oxidative Addition
OA4. Concerted Oxidative Addition
Concerted oxidative addition is a more general reaction than polar addition, in the sense that it is not restricted to compounds that can undergo aliphatic nucleophilic substitution. It could also be thought of as non-polar oxidative addition, because it does not involve charged intermediates as seen in the polar mechanism.
Aryl halides, for example, do not undergo nucleophilic substitution, but they do undergo concerted oxidative addition.
Instead of proceeding step by step, the addition of both fragments is synchronized. They add to the metal at the same time.
Figure OA4.1.
At first, it's difficult to understand this mechanism in terms of nucleophiles and electrophiles. The reaction is generally explained in terms of molecular orbital interactions, however, that can be thought of as nucleophile-electrophile interactions.
There are interactions involved in a concerted or non-polar oxidative addition.
• There is sigma bond donation from a bonding orbital in the substrate into a metal p orbital. This interaction is shown on the left of figure OA4.2.
• There is donation from a metal d orbital into an antibonding orbital on the substrate. This interaction is shown in the middle of figure OA4.2.
• Overall, a pair of electrons are donated from the substrate to the metal, and a pair of electrons are donated from the metal to the substrate.
Figure OA4.2. Molecular orbital interactions in a non-polar oxidative addition.
Figure OA4.3. A curved arrow representation of non-polar oxidative addition.
Problem OA4.1.
Provide a mechanism, with curved arrows, for the following reaction.
Problem OA4.2.
Propose a reason for the fact that one of the following dimethyl palladium compounds undergoes reductive elimination, but the other one does not.
Problem OA4.3.
Frequently, oxidative addition and reductive elimination are combined with other reactions into catalytic cycles. These cycles form the basis of important processes used to make valuable materials. Propose catalytic cycles for the following reactions. You don't need to draw curved arrows; just provide the intermediate formed after each reaction step.
OA5. Coupling Reactions
Oxidative addition and reductive elimination are key steps in industrial catalysis. For example, both steps are featured in palladium-catalyzed cross-coupling reactions, the subject of the 2010 Nobel Prize in Chemistry. The prize was awarded to Richard Heck of the University of Delaware, Ei-Ichi Negishi of Purdue University and Akira Suzuki of Hokkaido University. With these reactions, workers in a variety of fields can make molecules that otherwise would be quite difficult to make. These molecules in turn may be important pharmaceuticals or useful compounds for electronic displays in computers and other devices, to name just a couple.
The Negishi reaction involves catalytic addition of alkylzinc nucleophiles to vinyl halides.
Figure OA5.1. The Negishi cross-coupling reaction.
The catalytic cycle believed to operate for this reaction involves the crucial oxidative addition of the vinyl halide to the metal. The alkylzinc probably delivers the alkyl nucleophile to the metal via more conventional nucleophilic substitution. Once both pieces are both on the metal, they can reductively eliminate together.
Figure OA5.2. Catalytic cycle for the Negishi cross-coupling reaction.
Problem OA5.1.
Explain why the vinyl halide pictured in the example above would not react directly with the alkylzinc reagent.
Problem OA5.2.
Label each step of the catalytic cycle for the Negishi reaction with the appropriate term (oxidative addition, etc).
Problem OA5.3
Draw the mechanism for the Negishi reaction using curved arrow notation. There are many other examples of coupling reactions in organic synthesis. The Suzuki reaction is somewhat similar to the Negishi reaction.
Figure OA5.4. The Suzuki reaction.
The Heck reaction involves activation of a vinylic or aryl C-H bond.
Figure OA5.5. The Heck reaction.
Problem OA5.4.
By analogy with the Negishi reaction, propose a catalytic cycle for the Suzuki reaction.
Problem OA5.5.
Propose a catalytic cycle for the Heck reaction.
Read more about these Nobel Prize-winning reactions at the Nobel Prize website. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Oxidative_Addition_and_Reductive_Elimination/OA3._Polar_Oxidative_Addition.txt |
Let's think first about the interaction of light with matter. We have all seen light shine on different objects. Some objects are shiny and some are matte or dull. Some objects are different colors. Light interacts with these objects in different ways. Sometimes, light goes straight through an object, such as a window or a piece of glass.
Because chemical reactions are frequently conducted in solution, we will think about light entering a solution.
Imagine sunlight shining through a glass of soda. Maybe it is orange or grape soda; it is definitely coloured. We can see that as sunlight shines through the glass, colored light comes out the other side. Also, less light comes out than goes in.
Maybe some of the light just bounces off the glass, but some of it is definitely absorbed by the soda.
So, what is the soda made of? Molecules. Some of these molecules are principally responsible for the colour of the soda. There are others, such as the ones responsible for the flavor or the fizziness of the drink, as well as plain old water molecules. The soda is a solution; it has lots of molecules (the solute) dissolved in a solvent (the water).
Light is composed of photons. As photons shine through the solution, some of the molecules catch the photons. They absorb the light. Generally, something in the molecule changes as a result. The molecule absorbs energy from the photon and is left in an excited state.
The more of these molecules there are in the solution, the more photons will be absorbed. If there are twice as many molecules in the path of the light, twice as many photons will be absorbed. If we double the concentration, we double the absorbance.
• The amount of light absorbed depends on the concentration of the solution.
Alternatively, if we kept the concentration of molecules the same, but doubled the length of the vessel through which the light traveled, it would have the same effect as doubling the concentration. Twice as much light would be absorbed.
• The absorbance depends on the length of cell holding the solution.
These two factors together make up part of a mathematical relationship, called Beer's Law, describing the absorption of light by a material:
$A = \epsilon c l$
in which A = Absorbance, the percent of light absorbed; c = the concentration; l = the length of the light's path through the solution; ε = the "absorptivity" or "extinction coeficient" of the material, which is a measure of how easily it absorbs a photon that it encounters.
That last factor, ε, suggests that not all photons are absorbed easily, or that not all materials are able to absorb photons equally well. There are a couple of reasons for these differences.
Remember, often a particular soda will absorb light of a particular colour. That means, only certain photons corresponding to a particular colour of light are absorbed by that particular soda.
How does that affect what we see? If the red light is being absorbed by the material, it isn't coming back out again. The blue and yellow light still are, though. That means the light coming out is less red, and more yellowy-blue. We see green light emerging from the glass.
A "colour wheel" or "colour star" can help us keep track of the idea of complementary colours. When a colour is absorbed on one side of the star, we see mostly the colour on the opposite side of the star.
Problem PC1.1.
What colour of photon is probably most strongly absorbed by a glass of orange soda?
Why do certain materials absorb only certain colours of light? That has to do with the properties of photons. Photons have particle-wave duality, just like electrons. They have wave properties, including a wavelength.
That wavelength corresponds to the energy of a photon, according to the Planck-Einstein equation:
$E = {h c \over \lambda }$
in which E = energy of the photon, h = Planck's constant (6.625 x 10-34 J s-1), c = speed of light (3.0 x 108 m s-1), λ = wavelength of light in m.
Alternatively, the Planck-Einstien equation can be thought of in terms of frequency of thr photon: as a photon passes through an object, how frequently does one of its "crests" or "troughs" encounter the object? How frequently does one full wavelength of the photon pass an object? That parameter is inversely proportional to the wavelength. The equation becomes:
$E = h v$
in which ν = the frequency of the photon, in s-1.
The visible spectrum, shown below, contains a very limited range of photon wavelengths, between about 400 and 700 nm.
The higher the frequency, the higher the energy of the photon. The longer the wavelength, the lower the energy of the photon.
As a result of this relationship, different photons have different amounts of energy, because different photons have different wavelengths.
Problem PC1.2.
The visible spectrum ranges from photons having wavelengths from about 400 nm to 700 nm. The former is the wavelength of violet light and the latter is the wavlength of red light. Which one has higher energy: a photon of blue light or a photon of red light?
Problem PC1.3.
Ultraviolet light -- invisible to humans and with wavelengths beyond that of violet light -- is associated with damage to skin; these are the cancer-causing rays from the sun. Explain their danger in terms of their relative energy.
Different materials absorb photons of different wavelengths because absorption of a photon is an absorption of energy. Something must be done with that energy. In the case of ultraviolet and visible light, the energy is of the right general magnitude to excite an electron to a higher energy level.
However, we know that energy is quantized. That means photons will be absorbed only if they have exactly the right amount of energy to promote an electron from its starting energy level to a higher one (producing an "excited state"). Just like Goldilocks, a photon with too much energy won't do the trick. Neither will a photon with too little. It has to be just right.
If the absorption of a UV-visible photon is coupled to the excitation of an electron, what happens when the electron falls back down to the ground state? You might expect a photon to be released.
This phenomenon was observed during the late nineteenth century, when scientists studied the "emission spectra" of metal ions. In these studies, the metal ions would be heated in a flame, producing characteristic colours. In that event, the electron would be thermally promoted to a higher energy level, and when it relaxed, a photon would be emitted corresponding to the energy of relaxation.
By passing this light through a prism or grating, scientists could separate the observed colour into separate lines of different wavelengths. This evidence led directly to the idea of Niels Bohr and others that atoms had electrons in different energy levels, whci is part of our view of electronic structure today. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/PC._Photochemistry/PC1._Absorbance.txt |
So far, we have come across one big rule of photon absorbance. In order to be absorbed, a photon's energy has to match an energy difference within the compound that is absorbing it.
In the case of visible or ultraviolet light, the energy of a photon is roughly in the region that would be appropriate to promote an electron to a higher energy level. Different wavelengths would be able to promote different electrons, depending on the energy difference between an occupied electronic energy level and an unoccupied one. Other types of electromagnetic radiation would not be able to promote an electron, but they would be coupled to other events. For example, absorption of infrared light is tied to vibrational energy levels. Microwave radiation is tied to rotational energy levels in molecules. Thus, one reason a photon may or may not be absorbed has to do with whether its energy corresponds to the available energy differences within the molecule or ion that it encounters.
Franck-Condon: Electronic and Vibrational Coupling
Photons face other limitations. One of these is a moderate variation on our main rule. It is called the Frank Condon Principle. According to this idea, when an electron is excited from its normal position, the ground state, to a higher energy level, the optimal positions of atoms in the molecule may need to shift. Because electronic motion is much faster than nuclear motion, however, any shifting of atoms needed to optimize positions as they should be in the excited state will have to wait until after the electron gets excited. In that case, when the electron lands and the atoms aren't yet in their lowest energy positions for the excited state, the molecule will find itself in an excited vibrational state as well as an excited electronic state.
That means the required energy for excitation doesn't just correspond to the difference in electronic energy levels; it is fine-tuned to reach a vibrational energy level, which is quantized as well.
• The Franck Condon Principle states that electronic transitions are vertical.
• A vertical transition is one in which non of the nuclei move while the electron journeys from one state to another.
• A vertical transition may begin in a vibrational ground state of an electronic ground state and end in a vibrational excited state of an electronic excited state.
LaPorte: Orbital Symmetry
There are other restrictions on electronic excitation. Symmetry selection rules, for instance, state that the donor orbital (from which the electron comes) and the acceptor orbital (to which the electron is promoted) must have different symmetry. The reasons for this rule are based in the mathematics of quantum mechanics. What constitutes the same symmetry vs. different symmetry is a little more complicated than we will get into here. Briefly, let's just look at one "symmetry element" and compare how two orbitals might differ with respect to that element.
If an orbital is centrosymmetric, one can imagine each point on the orbital reflecting through the very centre of the orbital to a point on the other side. At the end of the operation, the orbital appears unchanged. That means the orbital is symmetric with respect to a centre of inversion..
If we do the same thing with a sigma antibonding orbital, things turn out differently.
In the drawing, the locations of the atoms are labelled A and B, but the symmetry of the orbital itself doesn't depend on that. If we imagine sending each point on this orbital through the very centre to the other side, we arrive at a picture that looks exactly the opposite of what we started with. These two orbitals have different symmetry. A transition from one to the other is allowed by symmetry.
Problem RO2.1.
Decide whether each of the following orbitals is centrosymmetric.
a) an s orbital b) a p orbital c) a d orbital d) a π orbital e) a π* orbital
Problem RO2.2.
Decide whether each of the following transitions would be allowed by symmetry.
a) π → π* b) p → π* c) p → σ* d) d → d
Symmetry selection rules are in reality more like "strong suggestions." They depend on the symmetry of the molecule remaining strictly static, but all kinds of distortions occur through molecular vibrations. Nevertheless, these rules influence the likelihood of a given transition. The likelihood of a transition, similarly, has an influence upon the extinction coefficient, ε.
transition ε, extinction coefficient
π → π* 3,000 - 25,000 M-1 cm-1
p → π* 20 - 150 M-1 cm-1
p → σ* 100 - 7,000
d → d 5 - 400 M-1 cm-1
Spin State
Let's take a quick look at one last rule about electronic emissions. This rule concerns the spin of the excited electron, or more correctly, the "spin state" of the excited species. The spin state describes the number of unpaired electrons in the molecule or ion.
number of unpaired electrons spin state
0 singlet
1 doublet
2 triplet
3 quartet
The rule says that in an electronic transition, the spin state of the molecule must be preserved. That means if there are no unpaired electrons before the transition, then the excited species must also have no unpaired electrons. If there are two unpaired electrons before the transition, the excited state must also have two unpaired electrons. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/PC._Photochemistry/PC2._Rules_of_Electronic_Excitation.txt |
Sometimes, when an excited state species relaxes, giving off a photon, the wavelength of the photon is different from the one that initially led to excitation. When this happens, the photon is invariably red-shifted; its wavelength is longer than the initial one. This situation is called "fluorescence."
How can that be? Isn't energy quantized? How is the molecule suddenly taking a commission out of the energy the original photon brought with it?
Relaxation and Fluorescence
This discrepancy is related to the Franck-Condon principle from the previous page. When an electron is promoted to an electronic excited state, it often ends up in an excited vibrational state as well. Thus, some of the energy put into electronic excitation is immediately passed into vibrational energy. Vibrational energy, however, doesn't just travel in photons. It can be gained or lost through molecular collisions and heat transfer.
The electron might simply drop down again immediately; a photon would be emitted of exactly the same wavelength as the one that was previously absorbed. On the other hand, if the molecule relaxes into a lower vibrational state, some of that initial energy will have been lost as heat. When the electron relaxes, the distance back to the ground state is a little shorter. The photon that is emitted will have lower energy and longer wavelength than the initial one.
Just how does a molecule undergo vibrational relaxation? Vibrational energy is the energy used to lengthen or shorten bonds, or to widen or squeeze bond angles. Given a big enough molecule, some of this vibrational energy could be transferred into bond lengths and angles further away from the electronic transition. Otherwise, if the molecule is small, it may transfer some of its energy in collisions with other molecules.
There are lots of examples of energy being transferred this way in everyday life. In a game of pool, one billiard ball can transfer its energy to another, sending it toward the pocket. Barry Bonds can transfer a considerable amount of energy through his bat into a baseball, sending it out of the park, just as Serena Williams can send a whole lot of energy whizzing back at her sister. In curling, one stone can transfer its energy to another, sending it out of the house and giving Canada the gold over Sweden.
In molecules, as one molecule drops to a lower vibrational state, the other will hop up to a higher vibrational state with the energy it gains. In the drawing below, the red molecule is in an electronic excited and vibrational state. In a collision, it transfers some of its vibrational energy to the blue molecule.
Radiationless Transitions: Internal Conversion
If electrons can get to a lower energy state, and give off a little energy at a time, by hopping down to lower and lower vibrational levels, do they need to give off a giant photon at all? Maybe they can relax all the way down to the ground state via vibrational relaxation. That is certainly the case. Given lots of vibrational energy levels, and an excited state that is low enough in energy so that some of its lower vibrational levels overlap with some of the higher vibrational levels of the ground state, the electron can hop over from one state to the other, without releasing a photon.
This event is called a "radiationless transition", because it occurs without release of a photon. The electron simply slides over from a low vibrational state of the excited electronic state to a high vibrational state of the electronic ground state. We will see a couple of iinds of radiationless transitions. Specifically, if the electron simply keeps dropping a vibrational level at a time back to the ground state, the process is called "internal conversion".
Internal conversion has an important consequence. Because the absorption of UV and visible light can result in energy transfer into vibrational states, much of the energy that is absorbed from these sources is converted into heat. That can be a good thing if you happen to be a marine iguana trying to warm up in the sun after a plunge in the icy Pacific. It can also be a tricky thing if you are a process chemist trying to scale up a photochemical reaction for commercial production of a pharmaceutical, because you have to make sure the system has adequate cooling available.
Radiationless Transitions: Intersystem Crossing
There is a very similar event, called "intersystem crossing", that leads to the electron getting caught between the excited state and the ground state. Just as, little by little, vibrational relaxation can lead the electron back onto the ground state energy surface, it can also lead the electron into states that are intermediate in energy.
For example, suppose an organic molecule undergoes electronic excitation. Generally, organic molecules have no unpaired electrons. Their ground states are singlet states. According to one of our selection rules for electronic excitation, the excited state must also have no unpaired electrons. In other words, the spin on the electron that gets excited is the same after excitation as it was before excitation.
However, that's not the lowest possible energy state for that electron. When we think about atomic orbital filling, there is a rule that governs the spin on the electrons in degenerate orbitals: in the lowest energy state, spin is maximized. In other words, when we draw a picture of the valence electron configuration of nitrogen, we show nitrogen's three p electrons each in its own orbital, with their spins parallel.
The picture with three unpaired electrons, all with parallel spins, shows a nitrogen in the quartet spin state. Having one of those spins point the other way would result in a different spin state. One pair of electrons in the p level would be spin-paired, one up and one down, even though they are in different p orbitals. That would leave one electron without an opposite partner. The nitrogen would be in a doublet spin state.
That isn't what happens. The quartet spin state is lower in energy than the doublet state. That's just one of the rules of quantum mechanics: maximize spin when orbitals are singly occupied.
It's the same in a molecule. The triplet state is lower in energy than the singlet state. Why didn't the electron get excited to the triplet state in the first place? That's against the rules. But sliding down vibrationally onto the triplet state from the singlet excited state isn't, because it doesn't involve absorption of a photon.
Intersystem crossing can have important consequences in reaction chemistry because it allows access to triplet states that are not normally avaiable in many molecules. Because triplet states feature unpaired electrons, their reactivity is often typified by radical processes. That means an added suite of reactions can be accessed via this process.
Phosphorescence: A Radiationless Transition Followed by Emission
Intersystem crossing is one way a system can end up in a triplet excited state. Even though this state is lower in energy than a singlet excited state, it can't be accessed directly via electronic excitation because that would violate the spin selection rule.
That's where the electron gets stuck, though. The quick way back down to the bottom is by emitting a photon, but because that would involve a change in spin state, it isn't allowed. Realistically speaking, that means it takes a long time. By "a long time", we might mean a few seconds, several minutes, or possibly even hours. Eventually, the electron can drop back down, accompanied by the emission of a photon. This situation is called "phosphorescence".
Molecules that display phosphorescence are often incorporated into toys and shirts so that they will glow in the dark.
Photosensitization
We have already seen that an excited state molecule can transfer some vibrational energy to another molecule via a collision. What about the energy of the electroic excited state? Can a molecule transfer a large quantum of energy to another -- essentially a photon's worth, but without the photon? The answer is yes.
In a collision, one molecule in an electronic excited state can transfer its energy to another. In the process, the first molecule returns to the ground state and the second is excited. This process is called "photosensitization".
Photosensitization can occur in a couple of different ways. Because photosensitization does not involve absorption or emission of a photon, it can also lead to formation of a triplet excited state.
The significance of photosensitization is that compounds that do not have strong chromophores can still access electronic excited states if they come into contact with other molecules that do have strong chromophores. There are a number of compounds that are routinely used to induce excitation in other molecules; these photochemical enablers are referred to as photosensitizers.
PC4. Photolysis
We have seen that the absorption of photons (especially in the ultraviolet-visible spectrum) is connected to the excitation of electrons. After excitation, a number of different relaxation pathways lead back to the ground state. Sometimes, absorption of a photon leads to a vastly different outcome. Instead of just relaxing again, the molecules may undergo bond-breaking reactions, instead.
An example of this phenomenon is observed in the complex ion [Co(NH3)]63+. Addition of UV light to this complex results in loss of ammonia. In the absence of UV light, however, the complex ion is quite stable.
In many cases, loss of a ligand is followed by replacement by a new one. For example, if an aqueous solution of [Co(NH3)]63+ is photolysed, an ammonia ligand is easily replaced by water.
Problem PC4.1.
Draw a d orbital splitting diagram for [Co(NH3)]63+. Explain why this complex is normally inert toward substitution.
Problem PC4.2.
Use the d orbital splitting diagram for [Co(NH3)]63+ to explain why this complex is undergoes substitution upon irradiation with UV light.
Photolysis is the term used to describe the use of light to initiate bon-breaking events. Photolysis frequently involves the use of high-intensity ultraviolet lamps. The high intensity light is needed in order to provide enough photons to get higher conversion of reactant into a desired product. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/PC._Photochemistry/PC3._Fluorescence_and_Phosphorescence.txt |
Pericyclic reactions differ from the ones we have looked at so far because they are not easily understood in Lewis acid- Lewis base terms. There is not always a clear nucleophile and electrophile in these reactions. In fact, they may appear to involve completely non-polar reactants. The classic example of a pericyclic reaction is a Diels Alder reaction. A Diels Alder reaction is a reaction between two alkenes.
Figure PR1.1. A Diels Alder reaction between ethene and 1,3-butadiene.
Normally, we think of both of these compounds as nucleophiles. It isn't easy to see why one would react with the other. It isn't easy to see how electrons would be attracted from one molecule to the other.
Instead, pericyclic reactions rely on weak attractions between (or within) molecules that can lead to electronic interactions that result in new bond formation. Normally, pericyclic reactions are studied using molecular orbital calculations to map out these electronic interactions. They are also explained qualitatively using molecular orbital tools.
PR2. Cope and Claisen Rearrangements
A rearrangement is a reaction in which one molecule undergoes bonding changes, with the transfer of one atom or group from one position in the molecule to another.
Proton tautomerism is a kind of rearrangement. A proton is removed from one site in the molecule and put back in a different site nearby. Tautomerism generally requires a couple of proton transfer steps in a row. A proton is removed from one site and then a proton is placed on the new site. (In another variation, a proton is added at one site and then a proton is removed from the old site.)
However, rearrangements often involve the concerted transfer of a group from one site to another within the molecule. The group loses its bond to one site and gains its bond to the other site at the same time.
A Cope rearrangement happens that way.
Figure PR2.1. A Cope rearrangement
At first it may not seem like much has happened in this reaction. The two pi bonds have changed position, however, and so has one of the sigma bonds. That means a total of six electrons have moved (two electrons per bond).
Figure PR2.2. A diagram of electron movement in a Cope rearrangement.
It does not matter which directions you draw the arrows moving in figure PR2.2. They could be shown going clockwise or counterclockwise. There is no electrophile or nucleophile. However, the arrows help with "electron book-keeping." The number of electrons is significant, however.
• Six electrons move in a circle of six atoms.
That patterns is reminiscent of benzene. This reaction may be related to the unusual stability of benzene. The transition state for this reaction is considered to be somewhat like benzene. Halfway between one structure and the other, the electrons are delocalized around the ring of atoms.
Figure PR2.3. The transition state in a Cope rearrangement.
A Cope rearrangement can be considered to occur via a rearrangement of overlap between a group of orbitals around this ring. Two orbitals forming a sigma bond tilt away from each other while two orbitals that are pi bonding tilt toward each other.
Figure PR2.4. Orbital rearrangement in a Cope rearrangement
Now there are p orbitals parallel to each other on the left, able to form new pi bonds.
Many concerted rearrangements can be thought of in terms of these orbital reorganizations.
Problem PR2.1.
A Claisen rearrangement is very similar to a Cope rearrangement, but oxygen is involved.
1. Draw curved arrows to keep track of electrons in this Claisen rearrangement.
2. Draw the aromatic transition state of the Claisen rearrangement.
3. Draw the orbital reorganization in the Claisen rearrangement. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/Pericyclic_Reactions/PR1._Introduction_to_Pericyclic_Reactions.txt |
As noted previously, the Diels Alder reaction is a classic example of a pericyclic reaction.
Figure PR3.1. The Diels Alder reaction.
Unlike the Cope and Claisen rearrangements, this reaction often occur intermolecularly (between two molecules). It always occurs between an alkene on one molecule and a conjugated diene on the other molecule. The alkene is referred to as a "dienophile"; it reacts with the conjugated pair of double bonds.
Figure PR3.2. The diene and dienophile in the Diels Alder reaction.
Problem PR3.1.
Draw curved arrows to keep track of electrons in the Diels Alder reaction.
Problem PR3.2.
Draw the aromatic transition state of the Diels Alder reaction.
Problem PR3.3.
Many pericyclic reactions are reversible. The reversible process is usually named the same way as the forward reaction, but with the prefix "retro". For example, a retro-Diels Alder reaction is shown below.
1. Draw curved arrows for the retro-Diels Alder reaction.
2. The forward reaction is favoured at low temperature, whereas the retro reaction is favoured at high temperature. Explain why using the expression for free energy of a reaction, DG = DH - T DS.
Once again, the reaction can be thought of in terms of a reorganization of electrons between these two molecules. In the Diels Alder reaction, we can think of an interaction between the LUMO on one molecule and the HOMO on the other. As it happens, the LUMO on one molecule has the correct symmetry such that it can overlap and form a bonding interaction with the HOMO on the other molecule.
Figure PR3.3. Qualitative molecular orbital picture of the Diels Alder reaction.
Pay attention to the p orbital drawings on the carbons that will bond to each other to form the six-membered ring. It is important that those orbitals are able to overlap with each other to form an in-phase interaction. In that way, these carbon atoms at the ends of the diene and dienophile are able to bond with each other.
A Diels Alder reaction is sometimes called a [2+4] addition reaction. A 2-carbon unit on one molecule interacts with a 4-carbon unit on another molecule.
In contrast, the addition of one regular alkene to another regular alkene would be called a [2+2] addition reaction. If this reaction occurred, two alkenes would come together to form a four-membered ring.
Figure PR3.4. A [2+2] addition reaction.
However, [2+2] addition reactions don't occur without special circumstances. There are a couple of reasons why, and you may be able to suggest some at this point.
You might say that the four-membered ring would be much more strained than the six-membered ring formed by the Diels Alder reaction. That is true, but it may not be reason enough to prevent the reaction from happening. Four-membered rings do occur in nature despite their strain energy.
You might also say that the benzene-like transition state that stabilizes the pathway through a Cope or Diels Alder reaction isn't possible in a [2+2] addition. In fact, the transition state would be more like antiaromatic cyclobutadiene. The transition state would be very high in energy.
Another problem shows up if we look at the orbital interactions in a [2+2] addition reaction. The HOMO on one alkene and LUMO on the other alkene do not overlap so that bonds can form between the two ends. If the p orbitals on one end are in phase, the p orbitals on the other end must be out of phase. The concerted reorganization of bonding possible for the Diels Alder reaction can't happen here.
Figure PR3.5. Qualitative molecular orbital picture of [2+2] addition reaction.
In fact, there is a way around that problem. Irradiating an alkene with UV light leads to promotion of an electron from the LUMO to the HOMO. The alkene is now in an "excited state".
Figure PR3.6. Excitation of an electron in an alkene.
This does not happen with 100% efficiency, so only some of the alkenes will become excited. In the excited state alkene, the HOMO now resembles the LUMO of the ground state alkene. Because of the matching symmetry between these orbitals, the addition reaction can proceed.
Figure PR3.7. HOMO-LUMO interaction between a ground-state alkene and an excited-state alkene.
A [4+2] reaction is sometimes referred to as "thermally-allowed", whereas a [2+2] addition is sometimes referred to as "photolytically-allowed." This distinction refers to the need for electronic excitation to accomplish the latter type of reaction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/Pericyclic_Reactions/PR3._The_Diels_Alder_Reaction.txt |
Olefin metathesis, or alkene metathesis, is an important process in petroleum refining and in the synthesis of important compounds such as pharmaceuticals. The mechanism of olefin metathesis is related to pericyclic reactions like Diels Alder and [2+2] reactions. In other words, it occurs through the concerted interaction of one molecule with another.
In petroleum refining, heating alkenes over metal oxide surfaces results in the formation of longer-chain alkenes. In particular, terminal olefins (with the double bond at the end of the chain) are converted into internal olefins (with the double bond somewhere in the middle of the chain).
Figure PR4.1. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins".
What does that reaction have to do with addition reactions involving double bonds?
Clearly, the alkenes have double bonds. In addition, so do the metal oxides. Metal atoms inside the metal oxides are bridged together by oxygen atoms. The surface of the metal oxides may be covered with a mixture of hydroxyl groups as well as terminal oxides (M=O groups). The terminal oxides on the surface are the important part of the catalyst.
Figure PR4.2. A simplified structure of a chunk of metal oxide surface.
When metal alkylidene complexes were developed in the 1970's, it was found that they, too, could catalyze this reaction.
Figure PR4.3. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins"
In fact, scientists working in petroleum chemistry soon came to believe that metal oxides on the catalyst surface were converted to alkylidenes, which then carried out the work of olefin metathesis. The reaction, it turns out, involves a [2+2] cycloaddition of an alkene to a metal alkylidene (to the metal-carbon double bond). This reaction results in a four-membered ring, called a metallacyclobutane. The [2+2] cycloaddition is quickly followed by the reverse reaction, a retro-[2+2]. The metallacyclobutane pops open to form two new double bonds.
The Chauvin Mechanism
This mechanism is called the Chauvin mechanism, after its first proponent, Yves Chauvin of the French Petroleum Institute. Chauvin's proposal of this mechanism shortly after the discovery of metal alkylidenes by Dick Schrock at DuPont earned him a Nobel Prize in 2005. Chauvin and Schrock shared the prize with Bob Grubbs, who made it possible for the reaction to be adapted easily to the synthesis of complex molecules such as pharmaceuticals.
Figure PR4.4. The Chauvin mechanism for olefin metathesis.
Why does olefin metathesis lead to the formation of internal alkenes? The [2+2] addition and retro-[2+2] reactions occur in equilibrium with each other. Each time the metallacyclobutane forms, it can form two different pairs of double bonds through the retro reaction. In the presence of terminal alkenes, one of those pairs of alkenes will eventually include ethene. Since ethene is a gas, it is easily lost from the system, and equilibrium shifts to the right in the equation below.
Figure PR4.5. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins".
That leaves a longer-chain alkylidene on the metal, ready to be attached to another long chain through subsequent [2+2] addition and reversion reactions.
Figure PR4.6. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins".
In most cases, a [2+2] addition will not work unless photochemistry is involved, but it does work with metal alkylidenes. The reason for this exception is thought to involve the nature of the metal-carbon double bond. In contrast to an orbital picture for an alkene, an orbital picture for an alkylidene pi bond suggests orbital symmetry that can easily interact with the LUMO on an alkene. That's because a metal-carbon \(pi\) bond likely involves a d orbital on the metal, and the d orbital has lobes alternating in phase like a \(pi\) antibonding orbital.
Figure PR4.7. Olefin metathesis produces longer-chain "internal olefins" from shorter chain "terminal olefins". | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/Pericyclic_Reactions/PR6._Olefin_Metathesis.txt |
Electron transfer is one of the most basic processes that can happen in chemistry. It simply involves the movement of an electron from one atom to another. Many important biological processes rely on electron transfer, as do key industrial transformations used to make valuable products. In biology, for example, electron transfer plays a central role in respiration and the harvesting of energy from glucose, as well as the storage of energy during photosynthesis. In society, electron transfer has been used to obtain metals from ores since the dawn of civilization.
Oxidation state is a useful tool for keeping track of electron transfers. It is most commonly used in dealing with metals and especially with transition metals. Unlike metals from the first two columns of the periodic table, such as sodium or magnesium, transition metals can often transfer different numbers of electrons, leading to different metal ions. Sodium is generally found as Na+ and magnesium is almost always Mg2+, but manganese could be Mn2+, Mn3+, and so on, as far as Mn7+.
At first glance, "oxidation state" is often synonymous with "charge on the metal". However, there is a subtle difference between the two terms. For example, in a coordination complex, a metal atom that is ostensibly an ion with a charge of +2 may have very little charge on it at all. Instead, the positive charge may be delocalized onto the ligands that are donating their own electrons to the metal. Oxidation state is used instead to describe what the charge on the metal ion would be if the coordinated ligands were removed and the metal ion left by itself.
Oxidation state is commonly denoted by Roman numerals after the symbol for the metal atom. This designation can be shown either as a superscript, as in MnII, or in parentheses, as in Mn(II); both of these descriptions refer to a Mn2+ ion, or what might have been a Mn2+ ion before it got into a bonding situation.
Problem RO1.1.
Translate the following oxidation state descriptions into charges on the metal.
a) AgI b) Ni(II) c) MnVII d) Cr(VI) e) Cu(III) f) FeIV g) OsVIII h) Re(V)
Problem RO1.2.
1. Provide the valence shell electron configuration for each metal species in the previous question (e.g. oxygen's is 2s22px22py12pz1).
2. Draw an energy level diagram showing the occupation of valence s, p and d levels for each metal species in the previous question.
The oxidation state of a metal within a compound can be determined only after the other components of the compound have been conceptually removed. For example, metals are frequently found in nature as oxides. An oxide anion is O2-, so every oxygen in a compound will correspond to an additional 2- charge. In order to balance charge, the metal must have a corresponding plus charge.
For example, sodium oxide has the formula Na2O. If the oxygen ion is considered to have a 2- charge, and there is no charge overall, there must be a corresponding charge of +2. That means each sodium ion has a charge of +1.
Problem RO1.3.
Determine the charge on the metal in each of the following commercially valuable ores. Note that sulfur, in the same column of the periodic table as oxygen, also has a 2- charge as an anion.
a) galena, PbS b) cassiterite, SnO2 c) cinnabar, HgS d) pyrite, FeS2 e) haematite, Fe2O3 f) magnetite, Fe3O4
Problem RO1.4.
Sphalerite is a common zinc ore, ZnS. However, sphalerite always has small, variable fractions of iron in place of some of the zinc in its structure. What is the likely oxidation state of the iron?
Problem RO1.5.
Sometimes non-metals such as carbon are thought of in different oxidations states, too. For example, the coke used in smelting metal ores is roughly C, in oxidation state 0. Determine the oxidation state of carbon in each case, assuming oxygen is always 2- and hydrogen is always 1+.
a) carbon monoxide, CO b) carbon dioxide, CO2 c) methane, CH4 d) formaldehyde, H2CO e) oxalate, C2O42-
Problem RO1.6.
Sometimes it is useful to know the charges and structures of some of the earth's most common anions. Draw Lewis structures for the following anions:
a) hydroxide, HO- b) carbonate, CO32- c) sulfate, SO42- d) nitrate, NO3-
e) phosphate, PO43- f) silicate, SiO44- g) inosilicate, SiO32-
Problem RO1.7.
Use your knowledge of common anions to determine the oxidation states on the metals in the following ores.
a) dolomite, MgCO3 b) malachite, Cu2CO3(OH)2 c) manganite, MnO(OH)
d) gypsum, CaSO4 e) rhodochrosite, MnCO3 f) rhodonite, MnSiO3
Problem RO1.8.
In mixed-metal species, the presence of two different metals may make it difficult to assign oxidation states to each. For the following ores, propose one solution for the oxidation states of the metals.
a) chalcopyrite, CuFeS2 b) franklinite, ZnFe2O4 c) beryl, Be3Al2(SiO3)6 d) bornite or peacock ore, Cu5FeS4
e) turquoise, CuAl6(PO4)4(OH)8
Problem RO1.9.
Feldspars are believed to make up about 60% of the earth's crust. The alkali, alkaline earth and aluminum metals in these tectosilicates are typically found in their highest possible oxidation states. What are the charges on the silicates in the following examples?
a) orthoclase, KAlSi3O8 b) anorthite, CaAl2Si2O8 c) celsian, BaAl2Si2O8 d) albite, NaAlSi3O8
Problem RO1.10.
Frequently, minerals are solid solutions in which repeating units of different compositions are mixed together homogeneously. For example, labradorite is a variation of anorthite in which about 50% of the aluminum ions are replaced by silicon ions and about 50% of the calcium ions are replaced by sodium ions. Show that this composition would still be charge neutral overall.
Problem RO2.1.
a) Cu → Cu(I) + e-
b) Fe(III) + 3 e- → Fe
c) Mn → Mn(III) + 3 e-
d) Zn(II) + 2 e- → Zn
e) 2 F- → F2 + 2 e-
f) H2 → 2 H+ + 2 e- | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO1._Oxidation_State.txt |
Cyclic voltammetry is a commonly used method of measuring the reduction potential of a species in solution. The species may be a coordination complex or a redox-active organic compound, for example. Cyclic voltammetry provides additional data that can be interpreted to make conclusions about the reduction / oxidation reaction and the stability of the species resulting from the electron transfer.
In cyclic voltammetry, rather than measuring the voltage produced by a reaction as we discussed before, a voltage is instead applied to the solution. The voltage is changed over time and current through a circuit is monitored. When the voltage reaches a point at which a reduction/oxidation is induced, current begins to flow. A cyclic voltammogram is a plot of current versus applied voltage.
In the experiment, the species of interest is dissolved along with some electrolyte, which promotes conductivity in the solution. Three electrodes are inserted into the solution. The working electrode, where the reduction / oxidation reaction takes place, is mostly covered with an insulator, but has a small disc of electrode exposed so the reaction can take place in a carefully controlled area. The counter electrode completes the circuit. A reference electrode with a known potential is also used in order to measure the potential applied to the cell. Unusually, the solution must not be stirred.
To begin the experiment, a potential is applied that is much more positive than the potential of the reference electrode. This step ensures that the species of interest is completely oxidized to begin with. The voltage is then swept in the negative direction at a constant rate. That is, the potential at the working electrode gets lower and lower, possibly until it becomes negative compared to the reference electrode. At some point, the voltage sweep is reversed, and it becomes more and more positive until it returns to the initial setting.
The (simulated) results of such an experiment are shown below.
At point A, the potential is very positive but is then swept to lower and lower values. At point B, current begins to flow as the voltage reaches a point that allows reduction to occur. Current keeps increasing until, at point C, all of the species in the vicinity of the working electrode has already been reduced. This point is called "cathodic peak potential". Current then begins to decrease, although some still keeps flowing as more of the species slowly diffuses over to the working electrode (point D).
The reasons for two of the features of the experimental design are now apparent. The reason for the very small exposed surface of the working electrode (usuallyan exposed disc about 1 mm wide) is to limit the area in which reaction takes place, so that we can observe when a controlled population of species has been reduced. The reason for not stirring the solution is similar; if we stirred the solution, more species would be continually and quickly fed to the working electrode and we would never observe a point at which the reaction was "finished".
At some point, the potential is increased again (point E). Current keeps decreasing; that trend is reversed as the previously reduced species is again oxidized. This time, current flows in the opposite direction, and a negative peak is observed. At point F, all of the species in the vicinity of the working electrrode has been oxidized, and current begins to "drop" again. This point is called "anodic peak potential". The "formal potential" is the mean of anodic and cathodic peak potential.
Problem RO10.1.
What is different about the following cyclic voltammogram compared to the previous one? Explain what is happening in this sample.
Problem RO10.2.
What is different about the following cyclic voltammogram? Explain what is happening in this sample. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO10._Cyclic_Voltammetry.txt |
The idea of oxidation states is not normally applied to organic compounds, but it can be useful to do so. When we do, we can gain some insight into certain reactions of organic molecules.
For example,carbon dioxide, CO2, can be thought of as having carbon in an oxidized state. If we apply the usual oxidation state rule, carbon dioxide is overall neutral and contains two oxygens, each with 2- charge. To counter that charge, carbon must be in oxidation state 4+.
On the other hand, methane, CH4, can be thought of as having carbon in a reduced state. If we apply the usual oxidation state rule here, methane is overall neutral but contains four protons. That means the carbon must be in a 4- oxidation state. Of course, the carbon does not behave as if it has a minus four charge. But we will see that this sort of exercise can be useful for book-keeping purposes.
Problem RO11.1.
Assign the formal oxidation state to carbon in the following molecules.
a) methanol, CH3OH b) formaldehyde or methanal, CH2O c) carbonate, CO32- d) hydrogen cyanide, HCN
e) ethane, CH3CH3 f) ethene, CH2CH2 g) ethyne, CHCH
The general trend here is that the more bonds there are to oxygen, the more oxidized is carbon. The more bonds there are to hydrogen, the more reduced is carbon.
Adding a hydrogen nucleophile to a carbonyl electrophile is routinely referred to as a reduction. For example, adding sodium borohydride to methanal would result in reduction to form methanol. Of course, a hydride is really a proton plus two electrons. We could write an equation for the reduction of methanal that looks a lot like the redox reactions we see in a table of standard reduction potentials.
$CH_2O + H^- + H^+ \rightarrow CH_3OH$
or
$CH_2O + 2e^- + 2H^+ \rightarrow CH_3OH$
It stands to reason that the opposite reaction, the conversion of methanol to methanal, is a two electron oxidation.
$CH_3OH \rightarrow CH_2O + 2e^- + 2H^+$
We know how to accomplish the reduction of methanal, at least on paper. We just add a complex metal hydride, such as lithium aluminum hydride or sodium borohydride, to the carbonyl compound. After an acidic aqueous workup to remove all the lithium and aluminum compounds, we get methanol. For practical reasons, methanol may be difficult to isolate this way, but that's the general idea of the reaction.
How do we accomplish the reverse reaction?
One way would be to provide a hydride acceptor in the reaction, so that we could catch hydride as it comes off the methanol. The most well-known such entity is NAD+, of course. There are biological oxidations that employ NAD+ for this reason.
More generally, the reaction can be accomplished in a number of ways, on paper, by separating out the two tasks involved. We need something to accept the two protons: that's a base. We need something to accept the two electrons: that's an oxidizing agent.
For the latter task, there are a number of high oxidation state transition metal compounds that are quite willing to accept two electrons. One of the most widely employed is Cr(VI), which accepts two electrons to become Cr(IV). A number of other methods are available, having been developed partly to avoid the toxicity of chromium salts, but let's look at the chromium case as an example.
A simple chromium(VI) compound is chromium trioxide. A simple base is pyridine. If we took these two reagents together with benzyl alcohol in a solvent such as dichloromethane, what would happen? OK, you might not want to try this, because chromium trioxide has an alarming capacity to cause spontaneous combustion in organic compounds, but we can do it on paper.
Is chromium trioxide a nucleophile or an electrophile? That Cr(VI) is pretty electrophilic, surely. So what part of the benzyl alcohol is nucleophilic? The oxygen atom. When we mix these things, the oxygen atom would likely coordinate to the chromium.
When the oxygen atom coordinates to the chromium, the oxygen gets a positive formal charge. It is now motivated to lose a proton. That's what the pyridine is for.
Now we have accomplished one of the goals of the reaction. We have removed a proton from benzyl alcohol. We have one more proton and two electrons left. The second proton will have to come from the carbon attached to the oxygen; that's the place where we need to form a carbonyl.
But wait a minute. You can't take two protons off the same molecule, can you? And certainly not from two atoms that are right next to each other. Doesn't that generate an unstable dianion?
Not this time. The chromium is there to accept two electrons. We won't generate an anion at all, as far as the benzyl alcohol is concerned. It is oxidized to benzaldehyde.
Problem RO11.2.
A completely different outcome to this reaction would be obtained in aqueous solution because of the equilibrium that exists between a carbonyl and the geminal diol (or hydrate) in water. Instead of obtaining an aldehyde, a carboxylic acid would be obtained via a second reduction. Provide a mechanism for this reaction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO11._Redox_in_Organic_Carbonyls.txt |
Problem RO1.1.
a) Ag+ b) Ni2+ c) Mn7+ d) Cr6+ e) Cu3+ f) Fe3+ g)Os8+ h) Re5+
Problem RO1.3.
a) Pb2+ & S2- b) Sn4+ & 2 O2- c) Hg2+ & S2- d) Fe4+ & 2 S2- e) 2 Fe3+ & 3 O2- f) 2 Fe3+ & 1 Fe2+ & 4 O2-
Problem RO1.4.
Probably Fe2+, to replace Zn2+ ions.
Problem RO1.5.
a) C 2+ b) C4+ c) 4- d) C0 e) C3+
Problem RO1.7.
a) Mg2+ b) Cu2+ c) Mn3+ d) Ca2+ e) Mn2+ f) Mn2+
Problem RO1.8.
a) Cu(II), Fe(II) b) Zn(II), Fe(III) c) Be(II), Al(III) d) Cu(I), Fe(III) e) Cu(II), Al(III)
Problem RO2.1.
1. Cu → Cu(+ + e-
2. Fe3+ + 3 e-→ Fe
3. Mn → Mn3+ + 3 e-
4. Zn2+ + 2 e- → Zn
5. 2 F- → F2 + 2 e-
6. H2 → 2 H+ + 2 e-
Problem RO2.2.
1. Cu(I) + Fe(III) → Cu(II) + Fe(II)
2. Cu(I) + Ag(0) → Cu(0) + Ag(I)
3. 3 F2 + 2 Fe → 6 F- + 2 Fe(III)
4. 2 Mo3+ + 3 Mn → 2 Mo + 3 Mn2+
Problem RO2.3.
1. MnO2 + 2 H+ + 2 e- → Mn(OH)2
2. 2 NO + 2 e- + 2 H+ → N2O + H2O
3. HPO32- + 2 e- + 3 H+ → H2PO2- + H2O
4. Sn(OH)62- + 2 e- + 3 H+ → HSnO2- + 4 H2O
Problem RO3.1.
MnO4- : 4 x O2- (= 8-) + Mn7+ = 1- overall
MnO2 : 2 x O2- (= 4-) + Mn4+ = neutral overall
difference = 3 e-
Problem RO3.2.
a) SO42- : 4 x O2- (= 8-) + S6+ = 2- overall
S2O82- : 8 x O2- (= 16-) + 2 x S7+ (= 14+) = 2- overall
difference = 1 e- per S, or 2 e- overall
S2O82- + 2 e- → 2 SO42-
b) HPO32- : 3 x O2- (= 6-) + H+ + P3+ = 2- overall
P : P(0)
difference = 3 e-
HPO32- + 3 e- + 5 H+ → P + 3 H2O
c) Ti2O3 : 3 x O2- (= 6-) + 2 x Ti3+ (= 6+) = neutral overall
TiO : O2- + Ti2+ = neutral overall
difference = 1 e- per Ti, or 2 e- overall
Ti2O3 + 2 e- + 2 H+ → 2 TiO + H2O
d) N2 : N(0)
NH2OH : O2- + 3 x H+ (= 3+) + N- = neutral overall
difference = 1 e- per N, or 2 e- overall
N2 + 2 e- + 2 H+ + 2 H2O → 2 NH2OH
Problem RO3.3.
Lithium is an alkali metal, in the first column of the periodic table. It has a relatively low ionization energy because it has a noble gas configuration as a cation. That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons. In lithium metal, the outermost electron is relatively far from the nucleus and so it is at a relatively high energy, and easily lost.
Problem RO3.4.
Fluorine is a halogen, with a relatively high electron affinity. It easily gains an electron to get to a noble gas configuration as a fluoride anion. That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons.
Problem RO3.5.
From most easily oxidized to least easily oxidized: Li > Al > Fe > Cu > Au
Problem RO3.6.
1. E0 = + 0.796 (Ag+/Ag) - 1.83 (Au/Au+) = -1.034 V (no forward reaction)
2. E0 = - 0.44 (Fe2+/Fe) + 0.762 (Zn/Zn2+) = + 0.0322 V (forward reaction)
3. E0 = + 0.52 (Cu+/Cu) + 3.04 (Li/Li+) = + 3.56 V (forward reaction)
4. E0 = + 0.77 (Fe3+/Fe2+) - 0.796 (Ag/Ag+) = - 0.026 V (no forward reaction)
Problem RO3.7.
When the table of standard reduction potentials is displayed with the most negative value at the top and the most positive value at the bottom, any given half-reaction will go forward if it is coupled with the reverse of a half-reaction that lies above it in the table. The opposite is not the case; no half reaction will go forward if it is coupled with the reverse of a half-reaction below it in the table.
a)
a
b)
c)
Problem RO4.2.
a) 2 Li + F2 → 2 Li+ + 2 F-
b) Eo = +5.91 V
c) Things look pretty grim.
d)
e) This scheme would result in the release of a small amount of energy at each stage. Each step could be harnessed to perform a task more efficiently, with less heat loss.
Problem RO4.5.
There are really two significant departures from expectation here. Lithium is much more active than expected based on electronegativity. The larger alkali metals, cesium, rubidium and francium, are all less active than expected on that basis.
We will see that another factor the influences activity in redox is the stability of ions in aqueous solution. Lithium cation is a small ion; water molecules bind very strongly to the ion because the electrons get relatively close to lithium's nucleus. That strong binding stabilizes this ion especially, tipping the malance of the reaction more strongly towards oxidation of lithium. The larger alkali metal ions are not nearly as stabilized by water ligands in aqueous solution, so the balance of their reactions does not tilt as strongly towards aqueous ions.
Problem RO5.1.
Li+/Li: E0 = - 3.04 V; ΔHvap = 147 kJ/mol; IE = 520 kJ/mol; ΔHh = -520 kJ/mol
Na+/Na: E0 = - 2.71 V; ΔHvap = 97 kJ/mol; IE = 495 kJ/mol; ΔHh = -406 kJ/mol
K+/K: E0 = - 2.931 V; ΔHvap = 77 kJ/mol; IE = 419 kJ/mol; ΔHh = -320 kJ/mol
Potassium should be the easiest of the three to oxidize. It is easier to oxidize than sodium. However, lithium's high heat of hydration reverses the trend and tips the balance of reaction in favour of ion formation.
Problem RO5.2.
Cu2+/Cu: E0 = + 0.340 V; ΔHvap = 300 kJ/mol; IE = 745 kJ/mol & 1958 kJ/mol; ΔHh = - 2099 kJ/mol
Ni2+/Ni: E0 = - 0.25 V; ΔHvap = 377 kJ/mol; IE = 737 kJ/mol & 1753 kJ/mol; ΔHh = - 2096 kJ/mol
Zn2+/Zn: E0 = - 0.7618 V; ΔHvap = 123 kJ/mol; IE = 906 kJ/mol & 1733 kJ/mol; ΔHh = - 2047 kJ/mol
In this case, zinc may be considered the outlier. Copper should be easier to reduce than nickel based solely on electronegativity. However, zinc's very low heat of vaporization suggests that formation of the solid metal is less favoured in that case, helping to tilt the balance toward zinc ion instead.
Problem RO8.1.
The bonds to iron would contract because the increased charge on the iron would attract the ligand donor electrons more strongly. The bonds to copper would lengthen because of the lower charge on the copper.
Problem RO8.2.
1. Most likely there are repulsive forces between ligands if the bonds get too short.
2. Insufficient overlap between metal and ligand orbitals would weaken the bond and raise the energy.
3. The range of possible bond lengths gets broader as energy is increased. The bond has more latitude, with both longer and shorter bonds allowed at higher energy. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO13._Solutions_for_Selected_Problems.txt |
"Oxidation state" implies a description that can change: a metal can go from one oxidation state to another. For example, a Cu(I) can become a Cu(0). It does so by an electron transfer from one place to another. In the case of Cu(I)/Cu(0), an electron would have to be donated by some other species.
A loss of electrons is called an "oxidation", whereas a gain of electrons is called a "reduction" (as immortalized in the mnemonic: LEO the lion goes GER). Since an electron always goes from somewhere to somewhere else, one thing is always oxidized when something else is reduced. (Note that this is a little like proton transfer reactions: a proton is always transferred from one basic site to another, and is never really by itself.) These paired processes are called "reduction-oxidation" reactions, or "redox" for short.
So the reduction of Cu(I) to Cu(0) is just a "half-reaction"; it needs a corresponding oxidation to make it happen. Li could donate an electron, for example, to become Li+. In a biological setting, a Fe(III) in an important protein may need an extra electron to become Fe(II) in order to do its job (your life may be at stake here). It could get the electron from a nearby Cu(I), which becomes Cu(II).
Problem RO2.1.
Balance the following half reactions by adding the right number of electrons to one side or the other.
a) Cu → Cu(I) b) Fe(III) → Fe
c) Mn → Mn3+ d) Zn2+ → Zn
e) F- → F2 f) H2 → H+
Sometimes, redox reactions work out very neatly: one participant needs an electron or two, and the other participant has one or two electrons to give. For example, a Cu2+ ion in need of two electrons to become a Cu atom might get them from a Zn atom, which would become a Zn2+ ion.
In other words (or other symbols):
$Cu^{2+} + 2e^- \rightarrow Cu$
$Zn^{2+} \rightarrow Zn + 2 e^-$
Adding those together:
$Cu^{2+} + Zn \rightarrow Cu + Zn^{2+}$
Note that the electrons on each side just cancel each other, much like adding the same thing to both sides of an equals sign.
Other times, things may be slightly more complicated. There may be an issue of conserving matter, for instance. For example, hydrogen gas, H2, can be oxidised to give proton, H+. We can't have more hydrogen atoms before the reaction than afterwards; matter can't just be created or destroyed. To solve that problem, the oxidation of hydrogen gas, H2, produces two protons, not just one, and so two electrons are involved as well.
Alternatively, two half-reactions may actually involve different numbers of electrons, and so proportions of each half reaction need to be adjusted in order to match the number of electrons properly.
Problem RO2.2.
Put the following pairs of half reactions together to make a full reaction in each case. Make sure you have balanced the half reactions first.
a) Cu(I) → Cu(II) and Fe(III) → Fe(II) b) Cu(I) → Cu(0) and Ag(0) → Ag(I)
c) F2 → F- and Fe → Fe3+ d) Mo3+ → Mo and Mn → Mn2+
In many cases, redox reactions do not just involve simple metal ions or atoms. Often, the metal atom is found within a compound or a complex ion. For example, one of the most common oxidizing agents in common use is permanganate ion, MnO4-, which is usually converted to manganese dioxide, MnO2 during a reaction.
That means the half reaction here is:
$MnO_4^- \rightarrow MnO_2$
In order to sort out how many electrons are being traded, we need to know the oxidation state of manganese before and after the reaction. That turns out to be Mn(VII) before and Mn(IV) afterwards. That means 3 electrons are added to permanganate to produce manganese dioxide.
Now we have:
$MnO_4^- + 3 e^- \rightarrow MnO_2$
But now we just have a mass balance problem again. There are oxygen atoms before the reaction that have just disappeared after the reaction. Where could those oxygen atoms have gotten to? On this planet, the simplest answer to that question is always water. So maybe 2 waters were produced as part of the reaction.
That gives us:
$MnO_4^- + 3 e^- \rightarrow MnO_2 + 2 H_2O$
Only now we have more problems. First of all, now we have some hydrogen atoms on the right that we didn't have on the left. Where did these things come from? Also, there is this niggling problem of negative charges that we had before the reaction that we don't have after the reaction. Charge, like matter, doesn't just appear or disappear. It has to go someplace, and we have to explain where.
We'll kill two birds with one stone. Maybe some protons were added to the reaction at the beginning, giving us those hydrogen atoms for the water and balancing out the charge.
We are left with:
$MnO_4^- + 3 e^- + 4 H^+ \rightarrow MnO_2 + 2H_2O$
It all works out. There are four negative charges on the left, and four plus charges, so no charge overall. There are no charges on the right. There is one manganese on the left and on the right. There are four oxygens on the left and on the right. There are four hydrogens on the left and on the right.
An Alternative Situation is Possible
Now let's take a little detour. We're going to go back in time, to the point where we realized we had a problem with our oxygen atoms.
Now we have:
$MnO_4^- + 3 e^- \rightarrow MnO_2$
But now we just have a mass balance problem again. There are oxygen atoms before the reaction that have just disappeared after the reaction. Where could those oxygen atoms have gotten to? And while we're at it, there are four negative charges on the left and none on the right. Charge doesn't just appear or disappear during a reaction. It has to be balanced.
One solution for this problem involves the production of hydroxide ions, HO-, during the reaction. Hydroxide ions are pretty common; there are a few in every glass of water. The charge in the reaction would be balanced if four hydroxide ions were produced, and it would explain where those oxygen atoms went.
That gives us:
$MnO_4^- + 3 e^- \rightarrow MnO_2 + 4 OH^-$
Now the charge is balanced! But the oxygen atoms aren't. And where did those hydrogen atoms come from?
Well, on this planet, a good source of hydrogen and oxygen atoms is water. Maybe the reaction needs water.
That means:
$MnO_4^- + 3 e^- + 2 H_2O \rightarrow MnO_2 + 4 OH^-$
One manganese on each side. Four hydrogens on each side. Six oxygens on each side. Four negative charges on each side. Nothing has appeared or disappeared during the reaction. We know where everything went.
There is actually a shortcut to get to this solution. If we already know how to balance the reaction in acididc media, we just add enough hydroxides to neutralize the acid (H+ + -OH = H2O). But we have to add those hydroxides to both sides. Some of the waters will then cancel out to leave the balanced reaction.
Start with acid
$MnO_4^- + 3 e^- + 4 H^+ \rightarrow MnO_2 + 2 H_2 O$
Add base to both sides
$MnO_4^- + 3 e^- + 4 H^+ + 4OH^- \rightarrow MnO_2 + 2 H_2O + 4 ^- OH$
Neutralize
$MnO_4^- + 3 e^- + 4 H_2O \rightarrow MnO_2 + 2 H_2O + 4 OH^-$
Cancel the extra waters
$MnO_4^- + 3 e^- + 2 H_2O \rightarrow MnO_2 + 4 OH^-$
Now we're back. We have seen two different outcomes to this problem. The moral of the story is that sometimes there is more than one right answer. In the case of redox reactions, sometimes things happen a little differently depending on whether things are occurring under acidic conditions (meaning, in this context, that there are lots of protons around) or in basic conditions (meaning there aren't many protons around but there is hydroxide ion).
Apart from helping us to keep track of things, the presence of acids (protons) and bases (hydroxide ions) in redox reactions are common in reality. For example, batteries rely on redox reactions to produce electricity; the electricity is just a current of electrons trying to get from one site to another to carry out a redox reaction. Many batteries, such as car batteries, contain acid. Other batteries, like "alkaline" batteries, for example, contain hydroxide ion.
Problem RO2.3.
Balance the following half reactions. Assume the reactions are in acidic conditions.
a)MnO2 → Mn(OH)2 b) NO → N2O
c)HPO32- → H2PO2- d) Sn(OH)62- → HSnO2- | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO2._Redox_Reactions.txt |
Iron and copper are two common metals in biology, and they are both involved in electron relays in which electrons are passed from one metal to another to carry out transformations on substrates in cells. But which one passes the electron to which? Or can it go either way?
How easily one metal can pass an electron to another, or how easily one metal can reduce another, is a pretty well-studied question. There are some preferences, some of which can be easily understood.
In order to look at this question, electrochemists typically measure the voltage produced when a circuit is set up that includes an electron made of the metal in question and an electrode made of a "standard hydrogen electrode". In a standard hydrogen electrode, hydrogen gas (H2) reacts on a platinum surface to produce two electrons and two protons. The electrons travel along a wire to the other electrode. The other electrode sits in an aqueous solution containing a salt of the metal in question. How easily is the metal ion reduced by accepting the electrons from the standard hydrogen electrode? The more downhill energetically this process is, the more positive is the voltage measured in the circuit.
For example, we might put some copper(II) salts, such as CuSO4, into the solution together with a copper electrode. Then we would see whether the copper in solution is spontaneously reduced to copper metal. That would happen, essentially, if the copper ion is more easily reduced than a proton. Thus, an electron released by hydrogen flows from the platinum electrode to the copper electrode and is picked up by copper ions waiting in solution.
Otherwise, what would happen if the reaction were not spontaneous? If no reaction occurred at all, maybe there would be no voltage. However, if the opposite reaction were to occur -- if copper were able to provide an electron to convert protons into molecular hydrogen -- then current would flow through the circuit in the opposite direction. A voltage would register, but it would be negative.
As it happens, the reaction is spontaneous, the hydrogen does send electrons through the wire, turning into protons in the process, and at the other end of the wire, the copper ion is converted into copper metal, putting another layer on the surface of the electrode. The voltage is about 0.34V.
This is a comparative, rather than absolute, measurement. We are measuring the intrinsic potential of an electron to be transferred from one species, a hydrogen molecule, to another, a Cu(II) ion.
It is also an intensive, rather than extensive, property. It does not matter how much copper we have, or how much hydrogen; the electron still has the same natural tendency to flow from the hydrogen to the copper.
The results of many such studies, carefully measured under specific conditions for maximum reproducibility, are gathered in a table of reduction potentials. The reactions are referred to as "half-reactions" because they each provide only half the picture of what is going on. The electron in each reaction doesn't come from nowhere; every reaction in the table would involve transfer of an electron from elemental hydrogen to form a proton.
Half Reaction (species are aqueous unless noted otherwise) Potential, Volts
Li+ + e- → Li (s) - 3.04
Al(OH)3(s) + 3e- → Al(s) + 3OH- - 2.31
Al3+ + 3e- → Al (s) - 1.662
Zn2+ + 2e- → Zn (s) - 0.762
Fe2+ + 2e- → Fe (s) - 0.44
CO2(g) + 2H+ + 2 e- → CO(g) + H2O - 0.11
SnO (s) + 2H+ + 2 e- → Sn (s) + H2O - 0.10
2 H+ + 2 e- → H2(g) 0.00
Fe3O4(s) + 8 H+ + 8 e- → 3Fe(s) + 4 H2O + 0.085
Cu2+ + 2e- → Cu0 + 0.34
CO(g) + 2H+ + 2 e- → C(s) + H2O + 0.52
Cu+ + e- → Cu (s) + 0.52
MnO4- + 2 H2O + 3e- → MnO2(s) + 4 OH- + 0.59
Fe3+ + e- → Fe2+ (aq) + 0.77
Ag+ + e- → Ag (s) + 0.796
MnO2 +4 H+ + 2e- → Mn2+ + 2 H2O + 1.23
MnO4- + 4 H+ + 3e- → MnO2(s) + 2 H2O + 1.70
Au+ + e- → Au (s) + 1.83
F2 (g) + e- → 2F- + 2.87
Much more extensive tables of reduction potentials can be found; for example, see the following Wikipedia data page.
A positive reduction potential indicates a spontaneous reaction. That makes sense, for instance, in the reaction of fluorine to give fluoride ion. For that reaction, E0 = 2.87 V. Of course, fluorine is a very electronegative element, and it will spontaneously accept an electron to obtain a noble gas configuration.
A negative reduction potential, on the other hand, indicates a reaction that would not occur spontaneously. For example, we would not expect lithium cation to accept an electron. We are used to thinking about alkali metals easily giving up their electrons to become cations. The reduction of lithium ion has a reduction potential E0 = -3.04 V. This reaction would only occur if it were driven by an expenditure of energy.
The opposite reaction, on the other hand, would be the oxidation of lithium metal to give a lithium cation. That reaction would occur spontaneously, and would have a spontaneous "oxidation potential". In fact, that value is + 3.04 V. The oxidation potential is always the same magnitude of the reduction potential for the reverse reaction, but with the opposite sign.
These signs may seem counter-intuitive if you are used to thinking of free energy changes. A negative free energy change means energy is lost in a reaction. A positive free energy change means energy must be put into a reaction to drive it forward. In fact, reduction potential and free energy are closely linked by the following expression:
$\Delta G = - nFE^0$
in which n = number of electrons transferred in the reaction; F = Faraday's constant, 96 500 Coulombs/mol.
So, a positive reduction potential translates into a negative free energy change.
Note that reduction potentials are pretty sensitive to changes in the environment. Factors that may stabilize one particular metal ion may not have the exact same effect on another, and so the preference for one state versus another will be altered slightly under different conditions.
For example, permanganate ion (MnO4-) has a more positive reduction potential under "acidic conditions" (with excess protons in solution) compared to "basic conditions" (with a paucity of protons in solution and instead an excess of hydroxide ion). The reduction potential under acidic conditions is +1.23V, compared to +0.59 V under basic conditions.
Tables of reduction potentials are also useful in assessing the opposite reaction. For example, lithium metal spontaneously reduces protons to produce hydrogen gas, becoming lithium ion in turn. The potential for that reaction is simply the opposite of the reduction potential of lithium ion; this is called the oxidation potential of lithium metal. The more positive a metal's oxidation potential, the more easily it is oxidized. However, we don't need a separate table of those values; they are just the opposite of the reduction potentials. Reactions with negative reduction potentials easily go backwards, reducing the proton to hydrogen gas by taking an electron from the reducing agent.
However, the most important use of standard reduction potentials is combining them to find out the potential of new reactions.
For example, when it says in the table that
$Cu^+ + e^- \rightarrow Cu (s) E^0 = 0.53 V$
It really means that is the potential produced for a specific reaction involving electron transfer between hydrogen and copper ion:
$H_2 (g) + 2 Cu^+ \rightarrow 2 Cu (s) + 2 H^+ E^0 = 0.53 V$
And when it says that
$Fe ^{2+} + 2 e^- \rightarrow Fe (s) E^0 = - 0.44 V$
It really means that
$H_2 (g) + Fe^{2+} \rightarrow Fe (s) + 2 H^+ E^0 = -0.44 V$
But now we know that reaction is endergonic, with a negative reduction potential and a positive free energy change. However, the reverse reaction
$Fe (s) + 2 H^+ \rightarrow H_2 (g) + Fe^{2+} E^0 = 0.44 V$
has a positive reduction potential and would proceed easily.
Now, if we combine those previous reactions, simply by adding them together
$H_2 (g) + 2 Cu^+ + Fe (s) + 2 H^+ \rightarrow H_2 (g) + Fe^{2+} + 2 Cu (s) + 2 H^+ E^0 = 0.44 + 0.53 V$
and simplifying
$2 Cu^+ + Fe (s) \rightarrow Fe^{2+} + 2 Cu (s) E^0 = 0.97 V$
That means the hydrogen reaction doesn't need to be involved at all. It's just a common reference point for all the other reactions. If we know how far uphill (or downhill) any two reactions are compared to that one, then we know how they compare to each other, too.
We should note that electrochemistry is a business that demands great care. There are a number of factors that can cause variations in the potential that is measured, and so we need to be very careful to control for those factors. For example, if there is a buildup of charge in one solution or another (because we are taking cations out of solution in one case and putting them into solution in the other), the ability to remove more electrons at one electrode and deliver them at another may be hindered. For that reason, a "salt bridge" is incorporated into the design of the system; this bridge allows ions to diffuse from one cell to the other in order to keep charge balanced. Also, the solutions are maintained at a standard concentration to make sure measurements are always made in comparable circumstances. Finally, non-reactive electrolytes (salts) are added to solution to aid in conductivity and maintain a constant ionic strength.
Problem RO3.1.
Calculate oxidation states to confirm that the manganese ion is being reduced in the following reaction:
MnO4- + 2 H2O + 3e- → MnO2(s) + 4 OH-
Problem RO3.2.
Balance the following half reactions by adding the right number of electrons to one side or the other, based on oxidation state. Then add water molecules and protons to help balance oxygens and overall charge.
1. S2O82- → 2 SO42-
2. HPO32- → P
3. Ti2O3 → 2 TiO
4. N2 → 2 NH2OH
Problem RO3.3.
Why is the reduction potential of Li+ so negative?
Problem RO3.4.
Why is the reduction potential of F2 so positive?
Problem RO3.5.
Rank the following metals from most easily oxidized to least easily oxidized: gold, iron, aluminum, copper, lithium.
Problem RO3.6.
Calculate reduction potentials for the following reactions:
1. Au + Ag+ → Au+ + Ag
2. Zn + Fe2+ → Zn2+ + Fe
3. Li + Cu+ → Li+ + Cu
4. Ag + Fe3+ → Ag+ + Fe2+
Problem RO3.7.
In general, if one reaction is combined with the reverse of a reaction above it in the table, will the overall reaction be spontaneous? What about if a reaction is combined with the reverse of a reaction below it? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO3._Reduction_Potentials.txt |
Metal ores are typically salts, such as oxides, carbonates or sulfides. Conversion of these ores into metals requires oxidation/reduction reactions.
That's not always the case. Some early forays into metallurgy involved native gold (native meaning the metal is found in its elemental state in nature). Gold is relatively soft. It could be easily worked and shaped by heating it. Occasionally, native silver and copper can also be found.
Problem RO6.1.
Explain, with the help of a table of standard reduction potentials, why silver and gold can sometimes be found as elements rather than salts.
However, a major leap forward came when people learned to make alloys, mixing in small amounts of other metals to make harder, sturdier materials. For example, the addition of tin to copper ushered in "the bronze age". Tin itself had to be made from its ore via smelting; the earliest evidence for this process comes from what is now Turkey, where it was performed over eight thousand years ago. However, alloys were apparently not discovered until several thousand years later.
In smelting, ore is heated to a high temperature in the presence of carbon sources, such as charcoal or coke. The partial combustion of the carbon source produces carbon monoxide which acts as a reducing agent.
Problem RO6.2.
Show the half-reactions involved in the reduction of tin oxide with carbon monoxide. Use them to come up with a balanced reaction for the process, and calculate the standard potential for the reaction.
Another major advance in metallurgy involved the conversion of iron ores into iron and steel. There is evidence that iron smelting in sub-Saharan Africa and Sri Lanka about three thousand years ago. Archaeological evidence in Sri Lanka shows that smelters were located on mountainsides facing the ocean, where constant winds provided ample oxygen to produce fires hot enough for smelting.
In the United States, the discovery of iron ores in the states along the Great Lakes, the use of the Great Lakes as a transportation network, and the availability of anthracite coal in Pennsylvania fueled the development of an American steel industry and the rise of a major industrial power. The fact that the great lakes states are still referred to as the "rust belt" is a testament to the manufacturing prowess of the region throughout the twentieth century, which proceeded from having all the necessary features for an iron-based economy in close geographic proximity.
Problem RO6.3.
Show the half-reactions involved in the reduction of iron oxide with carbon monoxide. Use them to come up with a balanced reaction for the process, and calculate the standard potential for the reaction.
Problem RO6.4.
Show the half-reactions involved in the reduction of aluminum oxide with carbon monoxide. Use them to come up with a balanced reaction for the process, and calculate the standard potential for the reaction.
Aluminum is a very important material in our economy. It is lightweight, strong, and forms a very hard oxide coating when exposed to the elements, rather than the rust that results from weathering steel. In contrast to the steel industry, the aluminum industry is a far-flung operation in which ore mined on one continent may be shipped to another for processing. However, aluminum metal isn't accessible via smelting. So how is it done?
Just as a thermodynamically favoured redox reaction can produce a voltage in a circuit, if we already have a voltage produced by another source, we can drive an unfavourable redox reaction to completion. We can drive the reaction backwards.
Quebec is a major producer of aluminum, despite being endowed with virtually no aluminum ore. Bauxite, the major aluminum-containing ore, is a mixture of minerals of formulae Al(OH)3 or AlO(OH) found amalgamated with other clays and minerals. It is found near the earth's surface in tropical and sub-tropical areas, left behind after of millenia of erosion and drainage of more soluble materials from underlying bedrock. The major producers of bauxite are Vietnam, Australia and Guinea, as well as a number of countries in South America.
Why ship bauxite all the way to the taiga to make aluminum? Aluminum production requires a lot of electrons, and those electrons can't be provided by coal or coke. Instead, they usually come from massive hydroelectric generating stations, such as the 16,000 megawatt James Bay Project in northern Quebec. To make aluminum, you go where electricity is cheap and plentiful.
The bauxite is first processed to help remove all those other materials that come mixed with the aluminum ore. It is dissolved in base, filtered and re-precipitated with acid. The residue is heated to drive off water, leaving pure alumina (Al2O3).
Instead of performing this redox reaction in aqueous solution, it is done in the molten state. Alumina has a melting point around 2,000 oC, but that temperature drops to a much more manageable 1,000 oC if a "flux" is added. To avoid contaminating the aluminum ions, cryolite has often been used as the flux, because it is also an aluminum salt.
The alumina is melted in an iron vat, which conveniently functions as one of the electrodes in the redox reaction. It is the cathode, supplying electrons. Graphite anodes draw electrons out of the bath to complete a circuit. Two reactions occur: aluminum ions are reduced to aluminum at the cathode, which drops to the bottom of the vat and is drained away periodically. Oxide ions are oxidized to molecular oxygen at the anodes. However, at these temperatures, the oxygen quickly reacts with the carbon anodes to produce carbon dioxide -- that is, the anodes actually disappear as the reaction proceeds.
Problem RO6.5.
Take a look at the redox reaction happening in the vat.
1. Provide a half reaction for reduction of aluminum ion.
2. Provide a half reaction for oxidation of the oxide anion.
3. Provide an overall, balanced reaction.
4. Calculate the standard potential for this reaction. (Don't worry about the lack of an aqueous solution; we'll just get an estimate of the real potential.)
Problem RO6.6.
Cryolite is added to get alumina to melt at a lower temperature. Unlike bauxite, it's a somewhat rare mineral found in Greenland and Quebec. Presumably, the aluminum ions in the cryolite also get reduced. Wouldn't the rare cryolite quickly get used up? Explain why this isn't a problem. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO6._Reduction_of_Metal_Ores.txt |
How does an electron get from one metal to another? This might be a more difficult task than it seems. In biochemistry, an electron may need to be transfered a considerable distance. Often, when the transfer occurs between two metals, the metal ions may be constrained in particular binding sites within a protein, or even in two different proteins.
That means the electron must travel through space to reach its destination. Its ability to do so is generally limited to just a few Angstroms (remember, an Angstrom is roughly the distance of a bond). Still, it can react with something a few bond lengths away. Most things need to actually bump into a partner before they can react with it.
This long distance hop is called an outer sphere electron transfer. The two metals react without ever contacting each other, without getting into each others' coordination spheres. Of course, there are limitations to the distance involved, and the further away the metals, the less likely the reaction. But an outer sphere electron transfer seems a little magical.
Barrier to Reaction: A Qualitative Picture of Marcus Theory
So, what holds the electron back? What is the barrier to the reaction?
Rudy Marcus at Caltech has developed a mathematical approach to understanding the kinetics of electron transfer, in work he did beginning in the late 1950's. We will take a very qualitative look at some of the ideas in what is referred to as "Marcus Theory".
An electron is small and very fast. All those big, heavy atoms involved in the picture are lumbering and slow. The barrier to the reaction has little to do with the electron's ability to whiz around, although even that is limited by distance. Instead, it has everything to do with all of those things that are barely moving compared to the electron.
Imagine an iron(II) ion is passing an electron to an iron(III) ion. After the electron transfer, they have switched identities; the first has become an iron(III) and the second has become an iron(II) ion.
Nothing could be simpler. The trouble is, there are big differences between an iron(II) ion and an iron(III) ion. For example, in a coordination complex, they have very different bond distances. Why is that a problem? Because when the electron hops, the two iron atoms find themselves in sub-optimal coordination environments.
Problem RO8.1.
Suppose an electron is transferred from an Fe(II) to a Cu(II) ion. Describe how the bond lengths might change in each case, and why. Don't worry about what the specific ligands are.
Problem RO8.2.
In reality, a bond length is not static. If there is a little energy around, the bond can lengthen and shorten a little bit, or vibrate. A typical graph of molecular energy vs. bond length is shown below.
1. Why do you think energy increases when the bond gets shorter than optimal?
2. Why do you think energy increases when the bond gets longer than optimal?
3. In the following drawings, energy is being added as we go from left to right. Describe what is happening to the bond length as available energy increases.
Problem RO8.3.
The optimum C-O bond length in a carbon dioxide molecule is 1.116 Å. Draw a graph of what happens to internal energy when this bond length varies between 1.10 Å and 1.20 Å. Don't worry about quantitative labels on the energy axis.
Problem RO8.4.
The optimum O-C-O bond angle in a carbon dioxide molecule is 180 °. Draw a graph of what happens to internal energy when this bond angle varies between 170 ° and 190 °. Don't worry about quantitative labels on the energy axis.
The barrier to electron transfer has to do with reorganizations of all those big atoms before the electron makes the jump. In terms of the coordination sphere, those reorganizations involve bond vibrations, and bond vibrations cost energy. Outside the coordination sphere, solvent molecules have to reorganize, too. Remember, ion stability is highly influenced by the surrounding medium.
Problem RO8.5.
Draw a Fe(II) ion and a Cu(II) ion with three water molecules located somewhere in between them. Don't worry about the ligands on the iron or copper. Show how the water molecules might change position or orientation if an electron is transferred from iron to copper.
Thus, the energetic changes needed before electron transfer can occur involve a variety of changes, including bond lengths of several ligands, bond angles, solvent molecules, and so on. The whole system, involving both metals, has some optimum set of positions of minimum energy. Any deviations from those positions requires added energy. In the following energy diagram, the x axis no longer defines one particular parameter. Now it lumps all changes in the system onto one axis. This picture is a little more abstract than when we are just looking at one bond length or one bond angle, but the concept is similar: there is an optimum set of positions for the atoms in this system, and it would require an input of energy in order to move any of them move away from their optimum position.
It is thought that these kinds of reorganizations -- involving solvent molecules, bond lengths, coordination geometry and so on -- actually occur prior to electron transfer. They happen via random motions of the molecules involved. However, once they have happened, there is nothing to hold the electron back. Its motion is so rapid that it can immediately find itself on the other atom before anything has a chance to move again.
Consequently, the barrier to electron transfer is just the amount of energy needed for all of those heavy atoms to get to some set of coordinates that would be accessible in the first state, before the electron is transferred, but that would also be accessible in the second state, after the electron is transfered.
Problem RO8.6.
Describe some of the changes that contribute to the barrier to electron transfer in the following case.
In the drawing below, an electron is transferred from one metal to another metal of the same kind, so the two are just switching oxidation states. For example, it could be an iron(II) and an iron(III), as pictured in the problem above. In the blue state, one iron has the extra electron, and in the red state it is the other iron that has the extra electron. The energy of the two states are the same, and the reduction potential involved in this trasfer is zero. However, there would be some atomic reorganizations needed to get the coordination and solvation environments adjusted to the electron transfer. The ligand atoms and solvent molecules have shifted in the change from one state to another, and so our energy surfaces have shifted along the x axis to reflect that reorganization.
That example isn't very interesting, because we don't form anything new on the product side. Instead, let's picture an electron transfer from one metal to a very different one. For example, maybe the electron is transferred from cytochrome c to the "copper A" center in cytochrome c oxidase, an important protein involved in respiratory electron transfer.
Problem RO8.7.
In the drawing above, some water molecules are included between the two metal centres.
1. Explain what happens to the water molecules in order to allow electron transfer to occur, and why.
2. Suppose there were a different solvent, other than water, between the complexes. How might that affect the barrier to the reaction?
The energy diagram for the case involving two different metals is very similar, except that now there is a difference in energy between the two states. The reduction potential is no longer zero. We'll assume the reduction potential is positive, and so the free energy change is negative. Energy goes down upon electron transfer.
Compare this picture to the one for the degenerate case, when the electron is just transferred to a new metal of the same type. A positive reduction potential (or a negative free energy change) has the effect of sliding the energy surface for the red state downwards. As a result, the intersection point between the two surfaces also slides downwards. Since that is the point at which the electron can slide from one state to the other, the barrier to the reaction decreases.
What would happen if the reduction potential were even more positive? Let's see in the picture below.
The trend continues. According to this interpretation of the kinetics of electron transfer, the more exothermic the reaction, the lower its barrier will be. It isn't always the case that kinetics tracks along with thermodynamics, but this might be one of them.
But is all of this really true? We should take a look at some experimental data and see whether it truly works this way.
Oxidant k (M-1s-1) (margin of error shown in parentheses)
Co(diene)(NH3)23+ 0.12 3.0(4)
Co(diene)H2O)NCS2+ 0.38 11(1)
Co(diene)(H2O)23+ 0.53 800(100)
Co(EDTA) 0.60 6000(1000)
As the reduction potential becomes more positive, free energy gets more negative, and the rate of the reaction dramatically increases. So far, Marcus theory seems to get things right.
Problem RO8.8.
1. Plot the data in the above table.
2. How would you describe the relationship? Is it linear? Is it exponential? Is it direct? Is it inverse?
3. Plot rate constant versus free energy change. How does this graph compare to the first one?
Marcus Inverted Region
When you look a little closer at Marcus theory, though, things get a little strange. Suppose we make one more change and see what happens when the reduction potential becomes even more positive.
So, if Marcus is correct, at some point as the reduction potential continues to get more positive, reactions start to slow down again. They don't just reach a maximum rate and hold steady at that plateau; the barrier gets higher and higher and the reactions get slower and slower. If you feel a little skeptical about that, you're in good company.
Marcus always maintained that this phenomenon was a valid aspect of the theory, and not just some aberration that should be ignored. The fact that nobody had ever actually observed such a trend didn't bother him. The reason we didn't see this kind of thing, he said, was that we just hadn't developed technology that was good enough to measure these kind of rates accurately.
But technology did catch up. Just take a look at the following data (from Miller, J. Am.Chem. Soc. 1984, 3047).
Don't worry that there are no metals involved anymore. An electron transfer is an electron transfer. Here, an electron is sent from the aromatic substructure on the right to the substructure on the left. By varying the part on the left, we can adjust the reduction potential (or the free energy change, as reported here.
Problem RO8.9.
1. Plot the data in the above table.
2. How would you describe the relationship?
As the reaction becomes more exergonic, the rate increases, but then it hits a maximum and decreases again. Data like this means that the "Marcus Inverted Region" is a real phenomenon. Are you convinced? So were other people. In 1992, Marcus was awarded the Nobel Prize in Chemistry for this work.
Problem RO8.10.
Take a look at the donor/acceptor molecule used in Williams' study, above. a) Why do you suppose the free energy change is pretty small for the first three compounds in the table? b) Why does the free energy change continue to get bigger over the last three compounds in the table?
Problem RO8.11.
The rates of electron transfer between cobalt complexes of the bidentate bipyridyl ligand, Co(bipy)3n+, are strongly dependent upon oxidation state in the redox pair. Electron transfer between Co(I)/Co(II) occurs with a rate constant of about 109 M-1s-1, whereas the reaction between Co(II)/Co(III) species proceeds with k = 18 M-1s-1.
1. What geometry is adopted by these complexes?
2. Are these species high spin or low spin?
3. Draw d orbital splitting diagrams for each complex.
4. Explain why electron transfer is so much more facile for the Co(I)/Co(II) pair than for the Co(II)/Co(III) pair. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO8._Electron_Transfer_Mechanisms%3A_Oute.txt |
In some cases, electron transfers occur much more quickly in the presence of certain ligands. For example, compare the rate constants for the following two electron transfer reactions, involving almost exactly the same complexes:
$Co (NH_3)_6^{3+} + Cr^{2+} \rightarrow Co^{2+} + Cr^{3+} + 6 NH_3 K = 10^{-4} M^{-1} s^{-1}$
$Co (NH_3)_5 Cl^{2+} + Cr^{2+} \rightarrow Co^{2+} + CrCl^{2+} + 6 NH_3 K = 6 \times 10^5 M^{-1} s^{-1}$
(Note: aqua ligands are omitted for simplicity. Ions, unless noted otherwise, are aqua complexes.)
Notice two things: first, when there is a chloride ligand involved, the reaction is much faster. Second, after the reaction, the chloride ligand has been transferred to the chromium ion. Possibly, those two events are part of the same phenomenon.
Similar rate enhancements have been reported for reactions in which other halide ligands are involved in the coordination sphere of one of the metals.
In the 1960’s, Henry Taube of Stanford University proposed that halides (and other ligands) may promote electron transfer via bridging effects. What he meant was that the chloride ion could use one of its additional lone pairs to bind to the chromium ion. It would then be bound to both metals at the same time, forming a bridge between them. Perhaps the chloride could act as a conduit for electron transfer. The chloride might then remain attached to the chromium, to which it had already formed a bond, leaving the cobalt behind.
Electron transfers that occur via ligands shared by the two metals undergoing oxidation and reduction are termed "inner sphere" electron transfers. Taube was awarded the Nobel Prize in chemistry in 1983; the award was based on his work on the mechanism of electron transfer reactions.
Problem RO9.1.
Take another look at the two electron transfer reactions involving the cobalt and chromium ion, above.
1. What geometry is adopted by these complexes?
2. Are these species high spin or low spin?
3. Draw d orbital splitting diagrams for each complex.
4. Explain why electron transfer is accompanied by loss of the ammonia ligands from the cobalt complex.
5. The chloride is lost from the cobalt comples after electron transfer. Why does it remain on the chromium?
Other ligands can be involved in inner sphere electron transfers. These ligands include carboxylates, oxalate, azide, thiocyanate, and pyrazine ligands. All of these ligands have additional lone pairs with which to bind a second metal ion.
Problem RO9.2.
Draw an example of each of the ligands listed above bridging between a cobalt(III) and chromium(II) aqua complex.
Once the bridge is in place, the electron transfer may take place via either of two mechanisms. Suppose the bridging ligand is a chloride. The first step might actually involve an electron transfer from chlorine to the metal; that is, the chloride could donate one electron from one of its idle lone pairs. This electron could subsequently be replaced by an electron transfer from metal to chlorine.
Alternatively, an electron might first be transferred from metal to chlorine, which subsequently passes an electron along to the other metal. In the case of chlorine, this idea may be unsatisfactory, becuase chlorine already has a full octet. Nevertheless, some of the other bridging ligands may have low-lying unoccupied molecular orbitals that could be populated by this extra electron, temporarily. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Part_V%3A__Reactivity_in_Organic%2C_Biological_and_Inorganic_Chemistry_3/RO._Reduction_and_Oxidation/RO9._Electron_Transfer%3A_Inner_Sphere.txt |
Radicals are species that have unpaired electrons. They can be atoms or molecules and they can be neutral species or ions. Frequently, radicals are very reactive. However, their mode of reactivity does not fall neatly within the normal patterns of Lewis acids and bases, or nucleophiles and electrophiles.
Radicals play common roles in atmospheric chemistry, including equilibration of the ozone layer. They are also found in a variety of biochemical pathways. In addition, radicals are employed in a number of useful processes, such as the polymerization of methyl methacrylate or vinyl chloride, commonly used to make shatter-resistant "glass" and pipes for plumbing, respectively.
Compounds of p-block elements form radicals if one of the atoms has seven electrons in its valence shell rather than the usual eight.
Problem RR1.1.
Draw structures for the following neutral radicals, making sure to fill in the correct number of electrons.
a) Br b) OH c) CH3 d) CH3CH2S e) CH3CHCH2 f) NO g) (CH3)2NO h) NO2
A molecule could become a radical in a number of different ways. A bond may break in half via the addition of energy, in the form of either heat or light. Otherwise, it may simply transfer one of its electrons elsewhere. Again, this event may be precipitated by the addition of heat or light energy. Of course, a molecule that receives an additional electron from elsewhere may also become a radical.
Problem RR1.2.
Draw structures for the following cationic radicals, making sure to fill in the correct number of electrons and the formal charge.
a) H2C=O b) CH3NH2 c) CH3OCH3 d) CH3CH2CH2Br e) CH3CH2CHCH2
Problem RR1.3.
Draw structures for the following anionic radicals, making sure to fill in the correct number of electrons.
a) O2 b) H2CO c) CH3CCCH3 d) cyclo-C6H6
The compounds above are all simple radicals, containing one unpaired electron. Compounds may also have more than one unpaired electron. For example, elemental oxygen, O2, is a diradical. Although its Lewis structure does not suggest anything unusual, its molecular orbital diagram reveals that oxygen actually has two unpaired electrons.
Problem RR1.4.
Show a molecular orbital interaction diagram illustrating the origins of the molecular orbitals on O2 from the atomic orbitals of oxygen.
A diradical could take two different forms. For example, molecular oxygen has two singly-occupied molecular orbitals. The single electron in each of those orbitals could adopt one of two different spin states. Both could adopt the same spin state (designated with arrow "up", for example), or they could be "spin-paired" (one "up", one "down"). The former situation is called a "triplet state", whereas the latter case is termed a "singlet state". These two situations result in some physical differences, such as different interactions with a magnetic field.
Subsequent pages will focus on the reactivity of radicals, with an emphasis on the stages of radical chain reactions.
RR2. Initiation: Bond Homolysis
Sometimes, radicals form because a covalent bond simply splits in half. Two atoms that used to be bonded to each other go their separate ways. Each atom takes with it one electron from the former bond. This process is called homolysis, meaning the bond is breaking evenly. In contrast, heterolysis is the term for a bond that breaks via ionization, with one atom getting both electrons from the bond.
• Homolysis describes breaking a bond in half, with one electron going to each side of the former bond.
In pictures, we show this process using curved arrows, but the arrows we use are slightly different from the ones you may be used to seeing in polar reaction chemistry. Instead of a regular arrowhead, we use a half arrowhead. This kind of arrow looks a little more like a fish hook. It is easy to remember the roles of the two kinds of arrows, because a full arrowhead describes the movement of an electron pair, whereas a half arrowhead describes the movement of only one electron.
Why would a covalent bond simply break apart? There are really a number of factors and a number of events that may result in this situation. The simple part of the story is that the bond must have been weak in the first place. There was enough energy available in the form of heat transferred from the surroundings (or sometimes in the form of light) to overcome the stabilization energy of the bond.
What makes a bond weak or strong? That is a complicated question. Many factors influence bond strength. However, two of the main factors responsible for covalent bond strength are the degree of electron sharing because of "overlap" and the degree of bond polarity resulting from "exchange". Most strong covalent bonds rely on a mixture of these two factors.
One fairly common feature in homolysis is a bond between two atoms of the same kind. For example, elemental halogens often undergo homolysis pretty easily. The ease with which these bonds can be split in half is illustrated by their low bond dissociation energies. Not much energy needs to be added in order to overcome the bonds between these atoms.
Bond Bond
Dissociation
Energy
(kcal/mol)
C-C 85
F-F 37
Cl-Cl 57
Br-Br 45
I-I 35
This propensity for radical formation can be understood in terms of the lack of a polar component in these bonds. These atoms rely solely on atomic overlap to share electrons with each other.
There is a notable exception to the rule that homoatomic bonds are inherently weak, and that is a carbon-carbon bond. Its bond dissociation energy is listed in the table for comparison with the halogens. The relative strength of carbon-carbon bonds gives rise to a multitude of carbon-based "organic" compounds in nature. The formation of bonds between like atoms is called "catenation"; carbon is the world champion.
Problem RR2.1.
Draw structures for the following reagents and show curved arrows to illustrate the initiation of radicals in each case.
a) Br2 b) H2O2 c) (CH3)3CO2H d) (CH3CH2)2S2
Silicon (BDESi-Si = 53 kcal/mol) and sulfur (BDES-S = 54 kcal/mol) are also capable of catenation, but the bonds that these atoms form between themselves are much weaker than C-C bonds. It seems to be generally true that larger atoms form weaker bonds, at least in the main group of the periodic table. After all, I-I bonds are weaker than Br-Br bonds, which are weaker than Cl-Cl bonds. It is sometimes argued that this trend is a result of poor spatial overlap between the more diffuse p orbitals nearer the bottom of the periodic table.
Problem RR2.2.
Draw structures of the following reagents and show curved arrows to illustrate the initiation of radicals in each case.
a) (CH3CH2)4Pb b) (CH3)2SbSb(CH3)2 c) (CH3CH2CH2CH2)3SnH
More information on bond strengths is available at Wired Chemist. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/RR._Radical_Reactions/RR1._Introduction_to_Radicals.txt |
Bond strength isn't just about the interaction of the two fragments bonded together. It is also influenced by the stability of those two species on their own. When the bond is broken, what pieces are left over?
The formation of radicals may be driven by the weakness of a particular bond. In terms of radical formation via bond homolysis, the reaction is more product-favoured if the bond being broken is weak. In other words, the bond is not very low in energy, so the overall reaction may become more downhill (or at least less uphill). In that case, forward reaction is favoured because of reactant destabilization.
However, a downhill reaction could also occur through product stabilization.
We have seen that the stability of anions and cations is strongly influenced by delocalization. Factors that spread the excess charge onto multiple atoms, rather than allowing charge to concentrate on one atom, make charged species much more stable.
For example, carbon-based anions are relatively unstable, but a delocalized carbanion is within the realm of possibility. Enolate ions are particularly easy to obtain because negative charge is partially delocalized onto a more electronegative oxygen atom.
Delocalization also strongly stabilizes radicals. It is one of the most important factors in the stability of carbon-based radicals.
Problem R3.1.
Illustrate the resonance stabilization in the following radicals
a) allyl, CH2CHCH2 b) benzyl, CH2C6H5 c) cyclopentadienyl, C5H5
Roadmap Problems in Natural Product synthesis
Topics required for successful completion are listed under each link.
Abbysomicin, Sorensen, 2006
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution, Alkene addition, Pericyclics
Annamoxic Acid, Corey, 2004
Carbonyl addition, Elimination, Alkene addition, Pericyclics
Anominine, Nicolaou, 2012
Carbonyl Addition, Aldol, Carboxyl Substitution, Oxidation, Nucleophilic Substitution, Elimination, Radicals
Aplykurodinone, Danishefsky, 2010
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination
Aza-epothilone B, Danishefsky, 2000
Carbonyl addition, Carboxylic substitution, Nucleophilic Substitution, Bioassay study
Callipeltoside, MacMillan, 2008
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution, Alkene addition
Dendrobine, Carreira, 2012
Carbonyl addition, Carboxylic substitution
Dynemicin, Danishefsky, 1996
Carbonyl addition, Carboxylic substitution, Nucleophilic substitution & elimination, Oxidation, Alkene oxidation, Pericyclics
Echinopine, Nicolaou, 2010
Carbonyl addition, Carboxylic substitution, Nucleophilic Substitution
Fischerindole, Baran, 2008
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution, Alkene addition, Radicals
Ginkgolide, Corey, 1988
Carbonyl addition, Conjugate addition, Enolates, Carboxylic substitution, Nucleophilic substitution, Alkene addition, Rearrangements, Radicals
Glucosylceramide, Overkleeft, 2007
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination
Contains a bioassay exercise.
Histrionicotoxin, Corey, 1975
Carbonyl addition, Carboxylic substitution, Aldol & Claisen, A-lpha-alkylation, Conjugate addition, Oxidation, Nucleophilic Substitution & Elimination
Juvenile Hormone 1, Corey, 1968
Carbonyl addition, Phosphorus ylides, Carboxylic substitution, Oxidation, Nucleophilic substitution, Cuprate addition, Alkene addition
Longifolene, Corey, 1961
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination, Alkene oxidation
Nakadomarin, Kerr, 2007
Carbonyl addition, Conjugate addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination
Okilactomycin, Smith, 2007
Carbonyl addition, Aldols, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination, Pericyclics
Perhydrohistrionicotoxin, Corey, 1975
Carbonyl addition, Enolates, Conjugate addition, Carboxylic substitution, Oxidation, Nucleophilic substitution & elimination, Radicals
Periplanone, Still, 1979
Carbonyl addition, Aldols, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination, Pericyclics
Prostaglandin A2, Corey, 1972
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination
Quinine, Stork, 2001
Carbonyl addition, Carboxylic substitution, Nucleophilic Substitution
Quinocarcin, Stoltz, 2008
Carbonyl addition, Carboxylic substitution
Salvileucalin, Reisman, 2011
Carbonyl addition, Carboxylic substitution
Saxitoxin, Du Bois, 2006
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution, Alkene addition
Contains a bioassay exercise.
Serotobenine, Fukuyama & Kan, 2008
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination, Alkene oxidation, Pericyclics
Solamine, Stark, 2006
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination, Alkene oxidation
Strychnine, Overman, 1993
Carbonyl addition, Enolates, Conjugate addition, Carboxylic substitution, Nucleophilic substitution & elimination, Alkene oxidation
Exercise: catalytic cycles in transition metal organometallics.
Taxol, Nicolaou, 1994
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution & elimination, Alkene oxidation, Pericyclics, Radicals
Tubingensin, Li, 2012
Carbonyl addition, Carboxylic substitution, Aldol
Tetrodotoxin, Kishi, 1972
Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution, Alkene addition, Alkene oxidation, Pericyclics, Rearrangements
Zaragozic acid, Nicolaou, 1994
Carbonyl addition, Carboxylic Substitution, Sulfur ylides, Oxidation, Nucleophilic substitution, Alkene oxidation, Radicals, Organo-transition metal reactions
Zincophorin, Hsung, 2007
Nucleophilic Substitution, Alkene addition | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/RR._Radical_Reactions/RR3._Initiation%3A_Radical_Stability.txt |
The alpha carbon is the first carbon atom attached to a functional group, such as a carbonyl. The carbonyl group makes the alpha hydrogen (the hydrogen on the alpha carbon) slightly acidic via a resonance stabilization mechanism. This results in several different reactions of note.
• Acetoacetic Ester Synthesis
The acetoacetic ester synthesis allows for the conversion of ethyl acetoacetate into a methyl ketone with one or two alkyl groups on the alpha carbon.
• Acidity of Alpha Hydrogens & Keto-enol Tautomerism
• Aldol Reaction
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule.
• Alpha Alkylation
The alpha alkylation reaction involves an α hydrogen being replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as other Sn2 reactions.
• Alpha Halogenation
A carbonyl containing compound with α hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate.
• Claisen Reactions
Because esters can contain αα hydrogens they can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed; the product is a β-keto ester.
• Deuterium Exchange
Due to the acidic nature of α hydrogens they can be exchanged with deuterium by reaction with D2O (heavy water). The process is accelerated by the addition of an acid or base; an excess of D2O is required. The end result is the complete exchange of all α hydrogens with deuterium.
• Enamine Reactions
As previously seen, aldehydes and ketones react with 2 o amines to reversibly form enamines. Example Reversible Enamines act as nucleophiles in a fashion similar to enolates. Because of this enamines can be used as synthetic equivalents as enolates in many reactions. This process requires a three steps: 1) Formation of the enamine, 2) Reaction with an eletrophile to form an iminium salt, 3) Hydrolysis of the iminium salt to reform the aldehyde or ketone. Some of the advantages of using an enamine over and enolate are enamines are neutral, easier to prepare, and usually prevent the overreaction problems plagued by enolates. These reactions are generally known as the Stork enamine reaction after Gilbert Stork of Columbia University who originated the work. Typically we use the following 2 o amines for enamine reactions
• Malonic Ester Synthesis
• Michael Additions & Robinson Annulation
Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds is a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942).
• Synthesis of Enols and Enolates
Reactivity of Alpha Hydrogens
The acetoacetic ester synthesis allows for the conversion of ethyl acetoacetate into a methyl ketone with one or two alkyl groups on the alpha carbon.
Steps
Step 1: Deprotonation with ethoxide
Step 2: Alkylation via and SN2 Reaction
3) Hydrolysis and decarboxylation
Addition of a second alky group
After the first step and additional alkyl group can be added prior to the decarboxylation step. Overall this allows for the addition of two different alkyl groups. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Acetoacetic_Ester_Synthesis.txt |
Alkyl hydrogen atoms bonded to a carbon atom in a a (alpha) position relative to a carbonyl group display unusual acidity. While the pKa values for alkyl C-H bonds is typically on the order of 40-50, pKa values for these alpha hydrogens is more on the order of 19-20. This can most easily be explained by resonance stabilization of the product carbanion, as illustrated in the diagram below.
In the presence of a proton source, the product can either revert back into the starting ketone or aldehyde or can form a new product, the enol. The equilibrium reaction between the ketone or aldehyde and the enol form is commonly referred to as "keto-enol tautomerism". The ketone or aldehyde is generally strongly favored in this reaction.
Because carbonyl groups are sp2 hybridized the carbon and oxygen both have unhybridized p orbitals which can overlap to form the C=O $\pi$ bond.
The presence of these overlapping p orbitals gives $\alpha$ hydrogens (Hydrogens on carbons adjacent to carbonyls) special properties. In particular, $\alpha$ hydrogens are weakly acidic because the conjugate base, called an enolate, is stabilized though conjugation with the $\pi$ orbitals of the carbonyl. The effect of the carbonyl is seen when comparing the pKa for the $\alpha$ hydrogens of aldehydes (~16-18), ketones (~19-21), and esters (~23-25) to the pKa of an alkane (~50).
Of the two resonance structures of the enolate ion the one which places the negative charge on the oxygen is the most stable. This is because the negative change will be better stabilized by the greater electronegativity of the oxygen.
Keto-enol Tautomerism
Because of the acidity of α hydrogens carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
Contributors
Prof. Steven Farmer (Sonoma State University)
• Clarke Earley (Department of Chemistry, Kent State University Stark Campus)
Aldol Reaction
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction is yet another example of electrophilic substitution at the alpha carbon in enolate anions. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. Due to the carbanion like nature of enolates they can add to carbonyls in a similar manner as Grignard reagents. For this reaction to occur at least one of the reactants must have α hydrogens.
Aldol Reaction Mechanism
A three step mechanism:
Step 1: Enolate formation
Step 2: Nucleophilic attack by the enolate
Step 3: Protonation
Aldol Condensation: the dehydration of Aldol products to synthesize α, β unsaturated carbonyl (enones)
The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this $\beta$-elimination reaction is an α,β-unsaturated aldehyde or ketone. Base-catalyzed elimination occurs with heating. The additional stability provided by the conjugated carbonyl system of the product makes some aldol reactions thermodynamically and mixtures of stereoisomers (E & Z) are obtained from some reactions. Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water, are termed Condensations. Hence, the following examples are properly referred to as aldol condensations. Overall the general reaction involves a dehydration of an aldol product to form an alkene:
Figure: General reaction for an aldol condensation
Going from reactants to products simply
Figure: The aldol condensatio example
Aldol Condensation Mechanism
1) Form enolate
2) Form enone
When performing both reactions together always consider the aldol product first then convert to the enone. Note! The double bond always forms in conjugation with the carbonyl.
Intramolecular aldol reaction
Molecules which contain two carbonyl functionalities have the possibility of forming a ring through an intramolecular aldol reaction. In most cases two sets of $\alpha$ hydrogens need to be considered. As with most ring forming reaction five and six membered rings are preferred.
As with other aldol reaction the addition of heat causes an aldol condensation to occur.
Mixed Aldol Reaction and Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective.
Example 4: Mixed Aldol Reactions
The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens.
The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
AACH2CHO + BCH2CHO + NaOH → AA + BB + AB + BA
The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Acidity_of_Alpha_Hydrogens_and_Keto-enol_Tautomerism.txt |
Enolates can act as a nucleophile in Sn2 type reactions. The alpha alkylation reaction involves an α hydrogen being replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as Sn2 reactions previously discussed. Good leaving groups like chloride, bromide, iodide, tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for this reaction. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylation’s occurring.
Mechanism
Stpe 1: Enolate formation
Step 2: Sn2 attack
Example \(2\)
Please write the structure of the product for the following reactions.
Alpha Halogenation
A carbonyl containing compound with $\alpha$ hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. For this reaction Cl2, Br2 or I2 can be used as the halogens.
General reaction
Acid Catalyzed Mechanism
Under acidic conditions the reaction occurs thought the formation of an enol which then reacts with the halogen.
Step 1: Protonation of the carbonyl
Step 2: Enol formation
Step 3: SN2 attack
Step 4: Deprotonation
Base Catalyzed Mechanism
Under basic conditions the enolate forms and then reacts with the halogen. Note! This is base promoted and not base catalyzed because an entire equivalent of base is required.
Step 1: Enolate formation
Step 2: SN2 attack
Overreaction during base promoted α halogenation
The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used.
The Haloform Reaction
Methyl ketones typically undergo halogenation three times to give a trihalo ketone due to the increased reactivity of the halogenated product as discussed above. This trihalomethyl group is an effective leaving group due to the three electron withdrawing halogens and can be cleaved by a hydroxide anion to effect the haloform reaction. The product of this reaction is a carboxylate and a haloform molecule (CHCl3, CHBr3, CHI3). Overall the haloform reaction represents an effective method for the conversion of methyl ketones to carboxylic acids. Typically, this reaction is performed using iodine because the subsequent iodoform (CHI3) is a bright yellow precipitate which is easily filtered off.
General reaction
Mechanism
1) Formation of the trihalo species
2) Nulceophilic attack on the carbonyl carbon
3) Removal of the leaving group
4) Deprotonation
Exercise $1$
Please draw the products of the following reactions
Answer
Claisen Reactions
Because esters can contain $\alpha$ hydrogens they can undergo a condensation reaction similar to the aldol reaction called a Claisen Condensation. In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed; the product is a β-keto ester. A major difference with the aldol reaction is the fact that hydroxide cannot be used as a base because it could possibly react with the ester. Instead, an alkoxide version of the alcohol used to synthesize the ester is used to prevent transesterification side products.
Claisen Condensation
Basic reaction
Going from reactants to products simply
Claisen Condensation Mechanism
1) Enolate formation
2) Nucleophilic attack
3) Removal of leaving group
Dieckmann Condensation
A diester can undergo an intramolecular reaction called a Dieckmann condensation.
Crossed Claisen Condensation
Claisen condensations between different ester reactants are called Crossed Claisen reactions. Crossed Claisen reactions in which both reactants can serve as donors and acceptors generally give complex mixtures. Because of this most Crossed Claisen reactions are usually not performed unless one reactant has no alpha hydrogens.
Deuterium Exchange
Due to the acidic nature of α hydrogens they can be exchanged with deuterium by reaction with D2O (heavy water). The process is accelerated by the addition of an acid or base; an excess of D2O is required. The end result is the complete exchange of all α hydrogens with deuterium.
Mechanism in basic conditions
Step 1: Enolate Formation
Step 2: Deuteration
Exercise \(1\)
Please draw the product for the following reactions.
Answer | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Alpha_Alkylation.txt |
As previously seen, aldehydes and ketones react with 2o amines to reversibly form enamines.
Example
Reversible
Enamines act as nucleophiles in a fashion similar to enolates. Because of this enamines can be used as synthetic equivalents as enolates in many reactions. This process requires a three steps:
1. Formation of the enamine,
2. Reaction with an eletrophile to form an iminium salt,
3. Hydrolysis of the iminium salt to reform the aldehyde or ketone.
Some of the advantages of using an enamine over and enolate are enamines are neutral, easier to prepare, and usually prevent the overreaction problems plagued by enolates. These reactions are generally known as the Stork enamine reaction after Gilbert Stork of Columbia University who originated the work.
Typically we use the following 2o amines for enamine reactions
Alkylation of an Enamine
Enamined undergo an SN2 reaction with reactive alkyl halides to give the iminium salt. The iminium salt can be hydrolyzed back into the carbonyl.
Step 1: Formation of an enamine
Step 2: SN2 Alkylation
Step 3: Reform the carbonyl by hydrolysis
All three steps together:
Acylation of Enamines
Enamine can react with acid halides to form β-dicarbonyls
1) Formation of the enamine
2) Nucleophilic attack
3) Leaving group removal
4) Reform the carbonyl by hydrolysis
All three steps together:
Michael Addition using Enamines
Enamines, like other weak bases, add 1,4 to enones. The end product is a 1,5 dicarbonyl compound.
Malonic Ester Synthesis
Malonic ester is a reagent specifically used in a reaction which converts alkyl halides to carboxylic acids called the Malonic Ester Synthesis. Malonic ester synthesis is a synthetic procedure used to convert a compound that has the general structural formula 1 into a carboxylic acid that has the general structural formula 2.
• reaction 1:
reaction 2:
reaction 3:
reaction 4:
A more direct method to convert 3 into 4 is the reaction of 3 with the enolate ion (5) of ethyl acetate followed by hydrolysis of the resultant ester.
However, the generation of 5 from ethyl acetate quantitatively in high yield is not an easy task because the reaction requires a very strong base, such as LDA, and must be carried out at very low temperature under strictly anhydrous conditions.
Malonic ester synthesis provides a more convenient alternative to convert 3 to 4.
Malonic ester synthesis can be adapted to synthesize compounds that have the general structural formula 6.
R3, R4 = identical or different alkyl groups
eg:
reaction 1:
reaction 2:
reaction 1 (repeat):
reaction 2 (repeat):
reaction 3:
reaction 4:
Malonic Ester Synthesis
Due to the fact that Malonic ester’s α hydrogens are adjacent to two carbonyls, they can be deprotonated by sodium ethoxide (NaOEt) to form Sodio Malonic Ester.
Because Sodio Malonic Ester is an enolate, it can then be alkylated with alkyl halides.
After alkylation the product can be converted to a dicarboxylic acid through saponification and subsequently one of the carboxylic acids can be removed through a decarboxylation step.
Mechanism
1) Saponification
2) Decarboxylation
3) Tautomerization
All of the steps together form the Malonic ester synthesis.
$RX \rightarrow RCH_2CO_2H$
Example
Michael Additions and Robinson Annulation
Enolates undergo 1,4 addition to α, β-unsaturated carbonyl compounds is a process called a Michael addition. The reaction is named after American chemist Arthur Michael (1853-1942).
Robinson Annulation
Many times the product of a Michael addition produces a dicarbonyl which can then undergo an intramolecular aldol reaction. These two processes together in one reaction creates two new carbon-carbon bonds and also creates a ring. Ring-forming reactions are called annulations after the Latin work for ring annulus. The reaction is named after English chemist Sir Robert Robinson (1886-1975) who developed it. He received the Nobel prize in chemistry in 1947. Remember that during annulations five and six membered rings are preferred.
Synthesis of Enols and Enolates
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing SN2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pKa > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pKa > 45), NaNH2 (sodium amide, pKa = 34), and LiN[CH(CH3)2]2 (lithium diisopropylamide, LDA, pKa 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane.
Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced.
Examples
If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide.
NaOCH2CH3 = Na+ -OCH2CH3 = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
However under acidic and basic conditions the equilibrium can be shifted to the right
Mechanism for Enol Formation
Acid conditions
1) Protonation of the Carbonyl
2) Enol formation
Basic conditions
1) Enolate formation
2) Protonation | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Enamine_Reactions.txt |
A rearrangement reaction is a broad class of organic reactions where the carbon skeleton of a molecule is rearranged to give a structural isomer of the original molecule. Often a substituent moves from one atom to another atom in the same molecule.
Rearrangement Reactions
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing SN2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pKa > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pKa > 45), NaNH2 (sodium amide, pKa = 34), and LiN[CH(CH3)2]2 (lithium diisopropylamide, LDA, pKa 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane.
Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced.
Examples
If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide.
NaOCH2CH3 = Na+ -OCH2CH3 = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples.
However under acidic and basic conditions the equilibrium can be shifted to the right
Mechanism for Enol Formation
Acid conditions
1) Protonation of the Carbonyl
2) Enol formation
Basic conditions
1) Enolate formation
2) Protonation | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Rearrangement_Reactions/Keto-Enol_Tautomerism.txt |
After proteins are synthesized via transcription of DNA, their structures may be modified in a number of ways. This process, in general, is referred to as "post-translational modification." It might involve the attachment of small organic groups to the protein, attachment of other biomolecules, or even the complete alteration of an amino acid residue into another form, so that the protein actually contains a residue that does not exist in the basic beginner's amino acid alphabet.
Minor modifications of proteins can be very important in regulating enzyme activity. Modest changes in structure can have a big impact on the conformation of the protein, which in turn might open or close access into the enzyme's interior or alter the shape of the active site. Alternatively, these changes might affect the location of a protein by moving it from the cell membrane into the cell interior or vice versa. They might also result in the formation or dissociation of supramolecular assemblies, by increasing or decreasing the attraction between two separate biomolecules.
The physical reasons for these structure-property changes are based in simple intermolecular attractions or, in the case of a large, complicated protein, intramolecular attractions between different parts of the same protein. A hydrogen bonding interaction or ion-dipole interaction that holds the protein in one conformation may be disrupted if a key player in that interaction is suddenly masked. For example, a cationic ammonium side chain may react to become an amide group; the absence of a full positive charge on this group significantly alters its intermolecular attractions.
There are probably hundreds of ways in which proteins are modified. We will look at just a few different modifications that occur in proteins, some of which are closly tied to carboxylate substitution. Note that any of these modifications might act to turn an enzyme "off" or "on"; the details depend on the individual case.
Acetylation
Acetylation is the attachment of an acetyl group, CH3C=O, to another compound such as an amino acid residue. Serine and threonine groups might be acetylated to make esters, cysteine side chains might be acetylated to make thioesters, or amino groups. Frequently, acetylation refers specifically to reaction at an amino group, such as the N-terminus of a protein or in a lysine side chain, forming an amide.
The source of the acetyl group is acetyl coenzyme A (AcSCoA, below), the thioester workhorse of the cell. This structure may seem a little bit complicated, but at the most basic level it is just a carboxyl electrophile (CH3C=O) attached to a very large thiolate leaving group.
Of course, lysine side chains and the N-termini of proteins are usually in a protonated state under biological conditions. That means that there must be a deprotonation step along the reaction pathway.
One example of the role of acetylation is seen in histones. Histones are proteins found in chromatin, the mixture of DNA and proteins in the cell nucleus. The already-coiled DNA helix is further wrapped around histones, bundling it up into a smaller package. DNA has many, many negative charges all along it because the phosphate units in its phosphate-sugar copolymer backbone are negatively charged at typical biological pH. The DNA interacts easily with the histones because they are rich in positively charged lysine residues.
Storing DNA in smaller bundles lets you keep more junk in your nucleus, but you don't want to just let the DNA sit there. Once in a while you want to take it out and do something with it, but how can you do that when it's stuck to those darned histones? The answer is, you just have to turn off the histones' force field. Get rid of that positive charge, and the DNA won't be stuck anymore. It's easy to do that by acetylating the histones.
This change in charge results because, although amines are easily protonated, amides are not. Thus, amines are likely to carry positive charge under biological conditions, whereas amides are likely to be neutral.
Problem CX10.1.
Explain why an amine is easily protonated but an amide is not.
Problem CX10.2.
Provide a mechanism for the acetylation of a lysine side chain. Assume the presence of histidine and its conjugate acid, histidinium ion; these species are common proton shuttles in biological reactions.
Phosphorylation
Serine, tyrosine and threonine residues are frequently modified by phosphorylation, which is the formation of a phosphate ester. The phosphorylation of these groups results in a change from a neutral side chain to an anionic side chain. The phosphate donor in these cases is another ubiquitous cellular performer, ATP.
The complete structure of ATP is shown here. It is a little less complicated than AcSCoA. Once again, at the most basic level of reactivity, it provides a phosphate electrophile, attached to a phosphate leaving group.
The anionic nature of the phosphate may not be apparent in the above drawing of ATP, but in reality, at biological pH, a phosphate would be deprotonated. Phosphates actually undergo multiple equilibria involving the loss of two protons.
Problem CX10.3.
Provide a mechanism for phosphorylation of serine in the presence of an appropriate biological proton shuttle.
Alkylation (Farnesylation)
There are other modifications that occur via different mechanisms. In a farnesylation, reaction does not even take place at a carbonyl. It does not take place at a phosphonate or a sulfonate.
Problem CX10.4.
Sulfur is the nucleophile in the farnesylation shown above. Identify the electrophile and show a mechanism with curved arrows.
Problem CX10.5.
Palmitoylation is closely related to acetylation, but the electrophile in this case is a palmitoyl group instead of an acetyl.
1. Provide a mechanism for palmitoylation of cysteine in the presence of an appropriate biological proton shuttle.
2. Explain how palmitoylation might influence the interaction of a protein with a cell membrane, composed of phospholipids like this one: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX11._Protein_Modifications.txt |
Problem CX1.1.
These heteroatoms are all found in the upper right-hand corner of the periodic table. They are all pretty electronegative and they all have lone pairs.
Problem CX2.2.
1. The electronegativity of the heteroatom attached to the carbonyl group in a carboxyloid is one factor that allows it to leave and form its own stable anion.
2. Although carbon and hydrogen are more electronegative than many of the elements in the periodic table, they are not stable enough as anions to form easily on their own.
Problem CX3.1.
We might expect carboxyloids with the most electronegative elements attached to the carbonyl to be the most reactive and least stable towards substitution (in other words, carboxyloids with the most electronegative heteroatoms would become substituted the most easily).
In that case, we would predict that the carboxyloids with the most electronegative substituent (oxygen) would be the most reactive. There are a number of different kinds and we will think about how they relate to each other shortly.
After the oxygen derivatives we would predict either the nitrogen derivatives or the chloride, depending on what electronegativity scale we happen to use (remember, electronegativity is not an experimentally pure property, but the result of a calculation that can be performed in different ways). The sulfur derivative would be least reactive.
There are still several different oxygen derivatives to compare: carboxylic acids (OH), carboxylates (O-), esters (OR, in which R is an alkyl or carbon chain) and acid anhydrides (OC=O). The easiest to differentiate is the carboxylate, because of its negative charge. It must be less attractive to a nucleophile than the other oxygen derivatives, because it would offer more repulsion to an incoming lone pair.
However, we can't really predict whether it would be any less reactive than the nitrogen, chlorine or sulfur analogues, because who knows whether the charge or the nature of the atom matters more?
As it happens, the charge probably matters more. We learn that simply by looking at the experimental trend and seeing that the carboxylate is the least reactive of all the carboxyloids.
Turning to the other three oxygen derivatives, it would be difficult to differentiate between the effect of a remote hydrogen atom versus an alkyl chain in the ester versus the carboxylic acid, so we'll say those two are about the same. On the other hand, the additional electron-withdrawing carbonyl group in the acid anhydride probably has a profound effect, so we would expect that compound to attract nucleophiles more strongly.
Of course, the series we have produced above is not the "right answer". It does not match the experimentally observed series of carboxyloid reactivities. Nevertheless, it is very useful in terms of building an understanding of carboxyloids. It tells us that electronegativity may play a role here, but that it can't be the only factor.
Some other factor is putting some of the derivatives out of order. In particular, the acid chloride (C=OCl) and the thioester (C=OSR) do not fit.
Problem CX3.2.
Electronegativity is an abvious factor that could influence an atom's ability to π-donate, but we just looked at that factor in the previous section, so let's look at another atomic property instead. Of course, different atoms have different sizes. In particular, if we look at the atoms involved in carboxyloid substituents, we can divide them into 2nd row atoms and 3rd row atoms.
It's actually well-documented that the degree of overlap between two orbitals influences how well they bond together. Since carbon is in the second row, it is about the same size as, and overlaps pretty well with, other second row atoms. Third row atoms are a little too big, on the other hand.
That factor breaks the carboxyloids into two different groups. Assuming π-donation is a major factor, sulfur and chlorine may be placed above the others in tems of reactivity. They cannot donate as well as oxygen or nitrogen can.
From there, differences among the atoms from the same row may be sorted out based on electronegativity differences.
Problem CX3.3.
Amide bonds are among the most stable carboxyloids possible. That stability makes them well-suited to form useful structures that will not decompose easily. Remember, any change that occurs in matter occurs through chemical reactions, including the formation and decomposition of biomaterials. Shutting down a potential chemical reaction means a material will be more durable.
a)
b)
Problem CX6.1.
Because acid chlorides are at the top of the carboxyloid reactivity diagram (the ski hill), and other halides are likely to be similar in reactivity to the chloride, this reaction would be uphill from the other carboxyloids.
Problem CX6.5.
Amides and carboxylates are the least reactive carboxyloids, so it might not be too surprising that they do not react with these nucleophiles.
Problem CX6.6.
Acid chlorides typically react with these cuprate reagents.
Problem CX6.7.
Borohydrides could presumably react with acid chlorides, anhydrides and thioesters, which are the most reactive carboxyloids. They probably can't react with amids or carboxylate ions, which are even farther downhill than esters.
Problem CX10.1.
This change in charge results because, although amines are easily protonated, amides are not. Protonation of an amide would result in a cation adjacent to the very positive carbonyl carbon, leading to a buildup of localized positive charge. That wouldn't be easy. Furthermore, the amide nitrogen is not very likely to donate its electrons to a proton in the first place. Its protons are too busy. They are tied up in conjugation with the carbonyl, so they really aren't available to act as the lone pair of a base. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX12._Solutions_For_Selected_Problems.txt |
Just as a simple carbonyl is an electrophile, so is a carboxyloid. Carboxyloids react with many of the same nucleophiles that react with aldehydes and ketones. The overall result of the reaction is very different. However, the mechanism of the reaction is really quite similar.
Nucleophiles have the overall effect of adding across the carbonyl group of an aldehyde or ketone. Addition of a proton produces an alcohol. The C=O bond is parlty broken to a C-O bond. In some cases, the C=O bond is completely broken and after several steps the carbonyl oxygen is replaced by some other heteroatom. On the other hand, nucleophiles add to carboxyloids and end up replacing the heteroatom next to the carbonyl. The carbonyl itself remains intact after the reaction is complete.
The general pattern in carboxyloid chemistry is for nucleophiles to substitute for the heteroatomic group next to the carbonyl. In these reactions, the group next to the carbonyl is sometimes referred to as a "leaving group". That term simply means that this group has been replaced by the end of the reaction.
Note that there is no reason to believe that the initial elementary reaction between a carboxyloid and a nucleophie is any different than that of a simple carbonyl with a nucleophile. An examination of the frontier orbitals in a carboxyloid suggests the pi antibonding level would be the site of population by a nucleophilic lone pair. The carbonyl pi bond would break as the nucleophile approaches.
However, an important feature of carbonyl chemistry is that two heteroatoms on one tetrahedral carbon cannot last. One always pushes the other off. If the former carbonyl oxygen pushes off the nucleophile, the system returns to the starting materials. However, if the former carbonyl oxygen displaces the heteroatomic group next to it, there is an overall change in bonding and a different product is formed. The net result is replacement of the group next to the carbonyl.
Problem CX2.1.
The overall result of reaction with a carboxyloid is to displace the heteroatom group next to the carbonyl. This group is typically liberated as an anion.
1. What feature of the heteroatoms typically found attached to the carbonyl in carboxyloids allows them to be displaced from the molecule as anions?
2. Why doesn't this same reaction happen with aldehydes and ketones?
Problem CX2.2.
Suggest an order of reactivity for the carboxyloids: rank them from most reactive to least reactive. Provide a reason for your trend.
CX5. Getting Towed Uphill
The uphill carboxyloids are useful materials in terms of being able to make other compounds. For example, a thioester such as acetyl coenzyme A may be able to make a variety of acyl esters. In the laboratory, acid chlorides are very common starting materials to make other carboxyloids. However, if they are so far uphill, how are they formed in the first place?
The two most common methods of making acid chlorides are treatment with thionyl chloride or with oxalyl chloride.
Figure CX5.1. Possible syntheses of an acid chloride.
The key part of making the uphill acid chloride out of the downhill carboxylic acid is the reagent used. Structurally, the reagents can be compared to acid chlorides themselves. They can be though of as being a little bit like uphill carboxyloids themselves. Thus, as one compound gives its chloride and gets oxygenated on its way downhill, it provides the energy needed to drive the carboxylic acid uphill.
Figure CX5.2. Conversion of thionyl chloride to sulfur dioxide and hydrochloric acid.
Figure CX5.3. Conversion of oxalyl chloride to carbon dioxide, carbon monoxide and hydrochloric acid.
Problem CX5.1.
Reaction of a carboxylic acid with oxalyl chloride starts with a carboxyloid substitution using the oxalyl chloride as electrophile and carboxylic acid as nucleophile. The chloride ion that is liberated then acts as a nucleophile in a cascade reaction that releases CO2 and CO as an acid chloride forms. Draw the mechanism.
Problem CX5.2.
Reaction of a carboxylic acid with thionyl chloride is very similar to reaction with oxalyl chloride, described above. Draw the mechanism of the reaction.
Problem CX5.3.
Provide additional factors (such as energetics, equilibrium concepts) that explain why oxalyl chloride and thionyl chloride can drive the conversion of a carboxylic acid to an acid chloride.
Acid anhydrides are usually made from the corresponding carboxylic acids. One molecule of carboxylic acid acts as a nucleophile and a second acts as an electrophile. Because the same kind of molecule acts as nucleophile and electrophile, acid anhydrides are typically symmetric: they havae a (C=O)O(C=O) unit in the middle, with the same alkyl groups on either side of it.
To aid formation of acid anhydrides, carboxylic acids are often heated strongly (well above 100 oC). Otherwise, they are sometimes heated in the presence of a strong drying agent, such as phosphorus pentoxide (empirically, P2O5). In the presence of water, phosphorus pentoxide is converted to phosphoric acid, H3PO4.
Problem CX5.4.
Draw a mechanism for the conversion of ethanoic acid to ethanoic anhydride.
Problem CX5.5.
Show the reverse reaction to the conversion of ethanoic acid to ethanoic anhydride (the same reaction, in the other direction). Which of the two directions do you think is favourable, based on the ski hill?
Problem CX5.6.
Draw a mechanism for the conversion of phosphorus pentoxide to phosphoric acid.
Problem CX5.7.
Explain why the conditions outlined above lead to acid anhydride formation. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX2._General_Reactivity_Patterns.txt |
Carboxyloids can be interconverted through addition of typical heteroatomic nucleophiles: amines, alcohols, thiols and water. In addition, other nucleophiles can displace the "leaving group" on a carboxyloid, provided the nucleophile is reactive enough.
Problem CX6.1.
Could a halide, such as bromide or chloride, replace a carboxyloid leaving group easily? Explain.
Carbon and hydrogen anions (or "semianions") are very good nucleophiles. Earlier, we saw how they can react with simple carbonyls. Although the lone pair on a carbon or a hydrogen is often masked in a covalent bond with a moderately electropositive metal such as aluminum or magnesium, that bonding pair of electrons is still nucleophilic enough to donate to a good electrophile. These nucleophiles can often react with carboxylic acid derivatives.
Problem CX6.2.
Draw a mechanism for the replacement of the chloride in propanoyl chloride with a hydride from lithium aluminum hydride.
Problem CX6.3.
Draw a mechanism for the replacement of the chloride in propanoyl chloride with a methyl from methylmagnesium chloride.
Addition of a Grignard reagent (alkylmagnesium halide) to a carboxyloid results in the formation of a ketone. Addition of a complex hydride reagent (such as lithium aluminum hydride) to a carboxyloid results in the formation of an aldehyde. We have already seen that these reagents can add to aldehydes and ketones to afford alcohols.
So, what happens if a Grignard reagent is added to an acid chloride or an acid anhydride? The carboxyloid would be converted to a ketone. If there are still more Grignard molecules around, they would probably convert the ketone into an alkoxide ion (and ultimately an alcohol via protonation). The thing is, it is very likely that there will be more Grignard molecules around. A reaction tends to involve millions of reactant molecules, so by the time the first thousand or so molecules of carboxyloid have been converted to ketone, hundreds of those ketone molecules have already been converted to alkoxide.
In most cases, alkyl reagents and hydride reagents will add twice to carboxyloids. They will convert the carboxyloid into an aldehyde or ketone. Because aldehydes and ketones also react with hydride and alkyl nucleophiles, they will react a second time.
Figure CX6.1. A modified potential energy surface that includes aldehydes and ketones.
Problem CX6.4.
Grignard reagents will not effect leaving group replacement in carboxylic acids. Show why that particular reaction does not occur, with the help of a mechanism.
Problem CX6.5.
Grignard reagents generally do not react with either amides or carboxylate ions. Explain why.
When a series of compounds varies from more reactive to less reactive in a particular reaction type, the possibility for selectivity arises. Some reagents may react with a few carboxyloids, but not with others. For example, organocuprates such as (CH3)2CuLi, which do not generally react well with aldehydes and ketones, are very selective in terms of which carboxyloids they will react with.
Problem CX6.6.
Which carboxyloids do you think will react with (CH3)2CuLi? Show the reaction in each case (the reactant, reagent, and a reaction arrow going to the product). Explain your reasoning.
Problem CX6.7.
Complex hydride reagents can be very selective towards carboxyloids. For example, sodium borohydride is not powerful enough to react with esters.
1. Which of the carboxyloids can sodium borohydride react with? Explain.
2. What other carboxyloids can sodium borohydride NOT react with? Explain.
Problem CX6.8.
Lithium aluminum hydride can induce carboxylic substitution with carboxylate salts such as sodium octanoate.
1. What would be the ultimate product of this reaction? Explain.
2. What other carboxyloids can lithium aluminum hydride react with? Explain.
Problem CX6.9.
1. Which is more reactive: lithium aluminum hydride or sodium borohydride?
2. Which is more selective: lithium aluminum hydride or sodium borohydride?
3. How can you explain the difference in reactivity between lithium aluminum hydride and sodium borohydride?
Problem CX6.10.
The reaction of lithium aluminum hydride with amides is unusual in that the final product of the reaction is generally an amine.
1. Why does this reaction seem to be different from other carboxyloid reactions?
2. Draw a mechanism for this reaction. (Hint: at some point, an oxygen atom donates a pair of electrons to aluminum.)
3. Propose a reason why this hydride reaction follows a different path than other reactions of hydrides with carboxyloids.
CX7. Enolates: Substitution and Decarboxylation
Alkylmagnesium reagents, alkylcuprates and complex hydrides can all react with carboxyloids. When they do, a carbon or hydrogen nucleophile bonds to the carbonyl carbon, usually replacing the leaving group at that position. Another common carbon nucleophile is an enolate ion. Enolate ions can also react with carboxyloids, although not typically with amides.
Probably the most common enolate reaction involving carboxyloids is the reaction of esters. If a strong base is added to solution of ester, some of the esters will become deprotonated, forming enolate anions. These ions will be nucleophiles. Some esters will remain protonated. These esters will be electrophiles. Donation of the enolate to the ester, with subsequent loss of the leaving group, leads to a beta-ketoester.
In principle, this reaction could conceivably go backwards. The enolate ion could potentially be displaced by an alkoxide to get back to an ester and an enolate ion. That's because the enolate ion is a relatively stable ion, and a moderately good leaving group. However, that generally doesn't happen.
Under basic conditions, the beta-ketoester is usually deprotonated, forming a particularly stable ion. This ion formation acts as a "thermodynamic sink" for the reaction, pulling it forward until all of the ester has been consumed.
Problem CX7.1.
Show why the ion that results from deprotonation of the beta-ketoester is particularly stable.
The formation of a beta-ketoester from two esters is called a "Claisen condensation". It is often followed by another important reaction: decarboxylation. If a beta-ketoester is treated with aqueous acid and heated, a couple of reactions take place. First, the ester portion of the molecule is converted into a carboxylic acid. Second, the carboxylic acid is decarboxylated. Carbon dioxide is formed, and the organic molecule becomes a ketone.
Decarboxylation is related to the retro-aldol reaction; formally, it can be thought of as leading to an enolate leaving group. Decarboxylation most commonly occurs in beta-ketoacids, rather than in other carboxylic acids. Otherwise, that leaving group could not occur. The ease of decarboxylation in beta-ketoacids is related to the stability of the enolate anion. Under acidic conditions, of course, an enolate anion does not occur; instead, an enol is formed.
Problem CX7.2.
Draw a mechanism for:
1. Conversion of ethyl-3-oxyhexanoate into 3-oxyhexanoic acid. (Oxy is a prefix meaning a ketone or aldehyde is foundalong the chain).
2. Decarboxylation of the resulting 3-oxyhexanoic acid. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX6._Semi-Anionic_Nucleophiles.txt |
Carboxyloids, such as esters, can interconvert with each other in the presence of the appropriate nucleophile. In the case of esters, an equilibrium will result. If an alcohol is added to an ester, under the right conditions we might get a new ester. A new alcohol would also appear, originating from the old OR group of the original ester. Two different esters and two different alcohols would be in equilibrium. This equilibrium might be perturbed in one direction or another, for example, by the addition of an excess of one nucleophile.
In the jargon of synthetic organic chemistry, an ester is a functional group. It is a site on the molecule at which reactions take place. It is also a site on the molecule that is easily subject to synthetic transformations. In other words, one functional group might easily be converted into another. In the case of a trans-esterification reaction, one ester simply gets converted into another.
Some compounds have more than one functional group. A compound could be both an ester and an alcohol, for instance. We might call this compound a hydroxyester. But wait -- if an alcohol can react with an ester, to make a new ester, doesn't this compound have both components of a reaction built in? Could it react with itself?
There are a couple of ways that could happen. If the chain between the carbonyl and the hydroxyl group is long enough (remember, a six-atom interaction between the hydroxyl oxygen and the carbonyl carbon may be optimal), the hydroxyl could wrap around and form a cyclic ester. That's an intramolecular reaction -- a reaction within one molecule.
Alternatively, if there is another one of these molecules around someplace, an intermolecular reaction might occur. That's a reaction between two different molecules. The hydroxyl group on one molecule can reach out and react with the carbonyl on another molecule.
Now there are two molecules of the same kind bonded to each other. This double molecule is called a dimer. The individual molecules that have been linked togethr to make the dimer are called monomers.
That dimer has two esters in it, not just one. Of course, it still has a hydroxyl group on one end. That hydroxyl group can still react with another carbonyl on another molecule.
Now there are three molecules bonded together. This molecule is called a trimer.
This process could keep going on indefinitely, of course. We might end up with a very large molecule, composed of many individual (former) molecules that have bonded together. This very large molecule made up of repeating units is called a polymer. A polymer is built up from many monomers linked together. This particular kind of polymer is called a polyester.
This polymer is frequently drawn in a way that emphasizes the repeating pattern of monomers that have been incorporated into the chain.
Polyesters can also be made by co-polymerizing two different monomers together. One monomer could be a diester, for example. The other monomer could be a diol. These two molecules are ready to react together, with one molecule acting as an electrophile and the other molecule acting as the nucleophile.
Together, the diester and the diol could be polymerized. The result would be an alternating copolymer, in which diester and diol monomer units alternate all along the polymer chain.
Polymers make up an important class of materials with many uses. Many polymers are lightweight, strong materials used to make parts for automobiles and other products. Polymers can also be very flexible or elastic. The physical properties of polymers are very different from the properties of other molecular compounds. These differences are a direct result of the very large size of polymer molecules. A polymer molecule might be thousands of monomers long, with a molecular weight in the millions.
Problem CX8.1.
Express the following polymer structures in abbreviated structures showing n repeating units in parentheses.
Problem CX8.2.
Show the structures of the polymers that would result from the following monomers. In each case, show a drawing with several enchained monomers.
Problem CX8.3.
Ring-opening polymerization involves a multi-step reaction in which a cyclic compound, such as a lactone (below) is opened into a chain through the addition of a nucleophile (called the "initiator"). The resulting chain is able to act as a nucleophile and open the next lactone, and so on, until a polymer has formed. Show a mechanism for formation of the oligomer in which n = 3.
Problem CX8.4.
Ring-opening polymerizations are frequently accelerated through the addition of small amounts of metal compounds, such as diethylzinc (Et2Zn) or tin octoate (Sn(O2CCH(CH2CH3)CH2CH2CH2CH3)2. Explain the role of these compounds in the reaction mechanism.
Problem CX8.5.
Karen Wooley at Texas A&M recently reported the following synthesis of a polyphosphoramidate for use as a pharmaceutical delivery agent. The goal is to use a benign delivery agent that is easily broken down and excreted by the body, resulting in low toxicity and minimal side effects.
1. Provide a mechanism for the synthesis.
2. Explain why the polyphosphoramidate is expected to be broken down and excreted easily by the body. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX8._Condensation_Polymers.txt |
Peptides and proteins are very important in biology. As a result, synthesis of these molecules has become very important, allowing for the laboratory study of model compounds that can give us insight into how proteins work, as well as pharmaceutically important compounds.
Structurally, amide or peptide bonds are very stable and resistant to carboxyl substitution. That stability makes optimal structures from which to construct proteins. Despite being composed of very long chains of linked amino acids, proteins actually have some limits on their conformational flexibility (their "floppiness"). That allows proteins to more reliably hold a particular shape.
Both the stability and the structural rigidity of peptides arises from the nature of the peptide bond. The pi donation that hinders nucleophiles from substituting at the carbonyl is pronounced enough that it can be considered to form an additional bond. Thus, peptides behave as though they contain C=N bonds rather than C-N bonds. X-ray structure determinations show that the peptide nitrogens in proteins are trigonal planar, not pyramidal. In addition, many peptides exhibit cis-trans isomerism. For every peptide bond, two different isomers can occur, depending on whether a substituent attached to nitrogen is on the same side of the C=N bond as the carbonyl oxygen or the opposite side.
The great stability of these structures does not mean they are easy to make. Part of the difficulty stems from the fact that amino acids are difunctional. In order to form long chain structures, amino acids must be able to react twice: once with an amine, to grow in one direction, and once with a carboxylic acid to grow in the other direction. In other words, an amino acid contains both a nucleophile and an electrophile.
Suppose we were to try to make the dipeptide, ala-phe. This peptide contains an alanine connected to a phenylalanine through a peptide bond. The peptide bond is formed between the carboxylic acid of alanine and the amine of phenylalanine. Assuming the amino acids do react together to form the peptide, combining these two reactants would likely produce a mixture of four dipeptides:
ala-phe ala-ala phe-phe phe-ala
In other words, peptide formation from amino acids is non-selective.
Problem CX9.1.
Draw structures for the four peptides formed by combining phe and ala.
Problem CX9.2.
What tripeptides would be produced by mixing ala, gly and val?
Problem CX9.3.
Simply combining these peptides might not result in any peptide formation at all. Why not?
In laboratory syntheses, a number of techniques have been used to make peptide synthesis selective. Most frequently, protecting groups are used. A protecting group "masks" one of the two functional groups on an amino acid, but leaves the other one open. If one amino acid has its amine protected, it can only react via its carboxylic acid. If the other amino acid has its carboxylic acid protected, it can only react via its amino group. Only one combination will result.
The key to protecting groups is that the reaction used to mask one of the functional groups must be reversible. You must be able to take the protecting group back off when it is no longer needed.
Carboxylic acids are normally protected as esters. Esters can be removed via acid- or base-catalyzed hydrolysis (as can amides, but esters are more reactive, being farther up the ski hill). Amines are normally protected as amides. However, we need to be able to remove specific amides her: the ones that mask the amines, not the ones that we have formed to link two amino acids together. As a result, in peptide synthesis, amines are usually protected as carbamates. Carbamates can be cleaved more easily than amides.
Problem CX9.4.
Propose a reason for the relative reactivity of carbamates and amides.
An additional complication in peptide synthesis is that amines and carboxylic acids do not really exist together. Instead, a proton is transferred from the carboxylic acid to the amine, forming a salt. The carboxylate is no longer very electrophilic, due to its negative charge. Because of its positive charge, the ammonium ion is no longer very nucleophilic.
To get around this problem, a number of coupling agents have been developed. A coupling agent can temporarily convert the carboxylate anion into a more reactive electrophile. To do so, it exploits the nucleophilicity of the carboxylate anion. After donating to the coupling agent, the carbonyl compound becomes more electrophilic. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/CX9._Peptides_and_Proteins%3A_Laboratory_Synthesis.txt |
In most cases, the reactivity of carboxyloids involves converting one carboxyloid into another. This is done by replacing one heteroatom substituent with another. For example, the chloride in an acid chloride may be replaced by an alcohol or alkoxide ion to make an ester.
The fact that one carboxyloid can be converted into another suggests that there would be an equilibrium between them. The ratio of two carboxyloids at equilibrium would be determined by their relative stability, as well as the stability of other associated species in solution.
We can map out the stability of carboxyloids on a potential energy surface, as shown below. The higher energy, less stable, more reactive carboxyloids are shown at the top of the potential energy curve. The lower energy, more stable, less reactive carboxyloids are found lower down on the potential energy curve.
Figure CX3.1. A potential energy curve showing relative reactivity of carboxloids.
The heteroatom attached to the carbonyl in a carboxyloid is always an electronegative atom with a lone pair. Either of those two features might be useful in understanding the reactivity trend illustrated above. For example, an electronegative atom would make the carbonyl carbon more positive. That carbon is already very positive because of the double bond to oxygen. Adding an additional electronegative atom should make it even more so. The amount of positive charge on the carbonyl carbon would be even greater if the atom attached to it were exceptionally electronegative.
On the other hand, a nearby lone pair might counteract the electron-attracting power of the carbonyl carbon. In a sense, we might think about that lone pair as competing with donation from a potential nucleophile.
The ability of an atom to π-donate, then, might have an influence on how strongly the carbonyl will attract nucleophiles. Of course, there is some trade-off involved in π-donation. Usually the atom that donates must take on a positive charge, since it is lending a pair of its own electrons to another atom. Factors that influence how easily this may happen could be important in determining carboxyloid reactivity.
Problem CX3.1.
Based on electronegativity of the atom attached to the carbonyl carbon, we might expect a specific trend in carboxyloid reactivity. Explain how this factor would affect electrophilicity at the carbonyl carbon and predict the corresponding trend in reactivity. Compare this trend with the information in Figure CX3.1.
Problem CX3.2.
Lone pair donation from the atom attached to the carbonyl carbon could also influence carboxyloid reactivity. Explain how this factor would affect electrophilicity at the carbonyl carbon and predict the corresponding trend in reactivity. Compare this trend with the information in Figure CX3.1.
Problem CX3.3.
Using the information in Figure CX3.1, explain why peptides (containing a number of amide bonds, R(C=O)N) are such a common structural feature in biology.
The potential energy curve in Figure CX3.1 is a useful index for the interconversion of carboxyloids. In general, it is easy to go downhill on the curve, but more difficult to go uphill. That means that compounds lower down on the ski hill can be made easily from compounds farther up the ski hill.
In general, pi donation from the heteroatom attached to the carbonyl is a primary factor that determines carboxyloid reactivity. The more able the heteroatom is to donate its pi electrons, the less electrophilic is the carbonyl. Nitrogen is very good at donating its lone pair. It is about the same size as the carbon atom it needs to donate to, and it only a little more electronegative than the carbon.
Oxygen (in esters and carboxylic acids) is next in line, since oxygen is more electronegative than nitrogen.
Chlorine and sulfur are a little too large to donate very well to a carbon atom. The size and energy mismatch between these atoms leads to poor pi bonding, and poor pi donation. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/Comparative_Energies%3A_The_Ski_Hill.txt |
The potential energy curve linking the carboxyloids can be used as a guide to how these compounds can be interconverted. In general, it is possible to take a compound that is higher on the ski hill and convert it to a compound that is lower on the ski hill.
Figure CX4.1. The potential energy surface linking carboxloids.
For example, acid chlorides are widely used to make other carboxyloids. By choosing the correct nucleophile, an acid chloride could be converted to any of the other derivatives. This is really the whole point of an acid chloride; it has no other function other than to provide an easy way to make other derivatives.
Acid anhydrides, also high on the potential energy curve, are also used in the same way. They could be used to make any of the derivatives lower than they are on the ski hill. In turn, acid anhydrides could conceivably be made from acid chlorides. However, because an acid anhydride plays the same role as an acid chloride -- providing a source with which to make the other derivatives -- we wouldn't normally make an acid anhydride from an acid chloride. We would either make one of the other derivatives directly from the acid chloride, or else make it from an acid anhydride that was obtained in another way.
Exercise CX4.1.
Show all of the ways that you could make the following compounds by going down the ski hill. (Show the starting material, the reagent added and a reaction arrow going to the product.) For help with names, see the appendix.
1. ethyl propanoate
2. butanoic acid
3. N-methylhexanamide
4. thioethyl pentanoate
5. potassium octanoate
Exercise CX4.2.
Converting a carboxylic acid into an amide is complicated by a side reaction, as a result of which amide formation becomes an uphill process.
1. Show the mechanism for the conversion of butanoic acid into N-ethylbutanamide via the addition of ethylamine.
2. Show the side reaction that would easily occur between these two reactants.
3. Explain why amide formation becomes an uphill process as a result of this reaction.
Sometimes, two carboxyloids are close enough on the ski hill that it may be possible to convert in either direction between them. In other words, there is an equilibrium between these two compounds, and the equilibrium constant is close enough to unity (K = 1) that the equilibrium can be pushed in either direction.
For selected cases, we can choose which direction the equilibrium will go by changing the reaction conditions. To do so, we can use an important concept of equilibrium: le Chatelier's principle (luh sha-TELL-yay). According to le Chatelier, if conditions in the reaction cause the reaction to move away from equilibrium, the reaction will shift direction until it is back at equilibrium again. In other words, if the actual ratio of reactants to products strays from what it ought to be, the correct reaction will occur so that the ratio returns to normal.
le Chatelier's principle should be partly intuitive. If a reaction is occurring in both reactions, there are really two reactions, one going in each direction. The ability of these reactions to occur depends partly on the amount of reactants available for the forward reaction or the reverse reaction. If extra starting materials are added (on the left side of the reaction), there is too much reactant as defined by the equilibrium constant. The denominator gets bigger and the ratio of products to reactants goes down. However, because there is extra starting material for the forward reaction, more product is quickly made, until the ratio returns to normal.
The opposite situation applies if too much product is made. From the point of view of the reverse reaction, those products of the forward reaction are really the reactants needed to go in the opposite direction. The reverse reaction has more material to work with, and this material can quickly be converted into the stuff on the left hand side of the reaction.
Exercise CX4.3.
Draw the mechanism for the conversion of:
1. hexyl propanoate to propanoic acid.
2. propanoic acid to hexyl propanoate.
In the interconversion of carboxyloids, equilibrium can be influenced in different ways. The conversion between esters and carboxylic acids can be influenced by the solvent used for the reaction. For example, an ester might be converted to a carboxylic acid under aqueous conditions, but a carboxylic acid might be converted to an ester using the appropriate alcohol as the solvent. In other words, because water and alcohol can be viewed as reactants in these cases, adding more can shift the reaction to one side. Solvent is usually present in concentrations many tens or hundreds or thousands of times higher than the reactants and products. Changing the solvent thus has a big impact on the direction of equilibrium.
Another strategy used to influence the equilibrium involves removal of product. For example, if water is a product of the reaction, a drying agent can be added to absorb the water. Drying agents include compounds such as MgSO4 or zeolites (mixed aluminosilicates containing Al, Si, O and other metal ions such as Mg2+, Ca2+ or Ti4+).
If a carboxylic acid is a product of a reaction, its concentration can be lowered by converting it to a carboxylate salt; this would happen easily in the presence of base. In either case, the disappearance of a product of the reaction (or a side product) would draw the reaction to the right in order to replace those products an re-establish the correct equilibrium ratio.
Exercise CX4.4.
Show, with curved arrows, how a magnesium ion could remove water from solution.
Exercise CX4.5.
Show, with curved arrows, how sodium hydroxide would remove butanoic acid from solution.
Introduction to Carboxyloids (Acid Derivatives)
Aldehydes and ketones are carbonyl compounds in which the carbonyl carbon is connected only to carbons atoms (in the case of a ketone) or to one carbon and one hydrogen atom (in the case of an aldehyde). Carboxyloids, or carboxylic acid derivatives, are carbonyl compounds in which the carbonyl carbon is attached to one carbon and one heteroatom. Most commonly, the heteroatom is an oxygen, nitrogen, sulfur or chlorine
Carboxylic acid derivatives were historically thought of as being made from carboxylic acids; hence the name. That name is a mouthful; we will use the term "carboxyloids", a term coined by the 20th century physical organic chemist, Christopher Ingold. The reactivity of carboxyloids is typically different from aldehydes and ketones Because of this difference it is useful to study these compounds separately from the simple carbonyls.
Problem CX1.1.
What are some things that the heteroatoms involved in carboxyloids have in common?
Like simple carbonyls, carboxyloids react with nucleophiles Just like simple carbonyls, the LUMO of a carboxyloid is usually the C=O pi antibonding orbital (the π*) Populating this orbital with a pair of electrons from the nucleophile results in breaking the C=O pi bond, leaving only a C-O sigma bond.
Remember that this addition to the C=O bond is often reversible In the case of carboxyloids, however, re-forming the pi bond can go through two pathways In one pathway, the nucleophile can be displaced, returning to starting materials In another pathway, the group attached to the carbonyl can be displaced instead In that pathway, a new product results. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/Interconversion_of_Carboxyloids%3A_Going_Downhill.txt |
Substitution reactions of benzene and other simple arenes.
Electrophilic Substitution Reactions
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and ethanoyl chloride in the presence of an aluminium chloride catalyst.
The electrophilic substitution reaction between benzene and ethanoyl chloride
What is acylation?
An acyl group is an alkyl group attached to a carbon-oxygen double bond. If "R" represents any alkyl group, then an acyl group has the formula RCO-. Acylation means substituting an acyl group into something - in this case, into a benzene ring.
The most commonly used acyl group is CH3CO-. This is called the ethanoyl group. In the example which follows we are substituting a CH3CO- group into the ring, but you could equally well use any other alkyl group instead of the CH3.
The facts
The most reactive substance containing an acyl group is an acyl chloride (also known as an acid chloride). These have the general formula RCOCl.
Benzene is treated with a mixture of ethanoyl chloride, CH3COCl, and aluminium chloride as the catalyst. A ketone called phenylethanone is formed.
or better:
The aluminium chloride isn't written into these equations because it is acting as a catalyst. If you wanted to include it, you could write AlCl3 over the top of the arrow.
The formation of the electrophile
The electrophile is CH3CO+. It is formed by reaction between the ethanoyl chloride and the aluminium chloride catalyst.
The electrophilic substitution mechanism
Stage one
Stage two
The hydrogen is removed by the AlCl4- ion which was formed at the same time as the CH3CO+ electrophile. The aluminium chloride catalyst is re-generated in this second stage.
E. An Industrial Alkylation of Benzene
This page gives you the facts and simple, uncluttered mechanisms for the electrophilic substitution reaction between benzene and alkenes in the presence of a mixture of aluminium chloride and hydrogen chloride as the catalyst.
The electrophilic substitution reaction between benzene and ethene
The facts
Industrially, alkyl groups can be substituted into a benzene ring using a variant on Friedel-Crafts alkylation. One possibility is that instead of using a chloroalkane with an aluminium chloride catalyst, they use an alkene and a mixture of aluminium chloride and hydrogen chloride as the catalyst.
This is a cheaper method because it saves having to make the chloroalkane first.
To put an ethyl group on the ring (to make ethylbenzene), benzene is treated with a mixture of ethene, HCl and aluminium chloride.
or better:
The aluminium chloride and HCl aren't written into these equations because they are acting as catalysts. If you wanted to include them, you could write AlCl3 and HCl over the top of the arrow.
The formation of the electrophile
The electrophile is CH3CH2+. It is formed by reaction between the ethene and the HCl - exactly as if you were beginning to add the HCl to the ethene.
The chloride ion is immediately picked up by the aluminium chloride to form an AlCl4- ion. That prevents the chloride ion from reacting with the CH3CH2+ ion to form chloroethane.
The electrophilic substitution mechanism
Stage one
Stage two
The hydrogen is removed by the AlCl4- ion which was formed at the same time as the CH3CH2+ electrophile. The aluminium chloride and hydrogen chloride catalysts are re-generated in this second stage.
The electrophilic substitution reaction between benzene and propene
The facts
The problem with more complicated alkenes like propene is that you have to be careful about the structure of the product. In each case, you can only really be sure of that structure if you work through the mechanism first.
For example, the propyl group becomes attached to the ring via its middle carbon atom - and not its end one.
You still need a mixture of aluminium chloride and hydrogen chloride as catalysts.
The formation of the electrophile
When the propene reacts with the HCl, the hydrogen becomes attached to the end carbon atom. A secondary carbocation (carbonium ion) is formed because it is more stable than the primary one which would have been formed if the addition was the other way round.
Because the positive charge is on the centre carbon atom, that is the one which will become attached to the ring.
The electrophilic substitution mechanism
Stage one
Stage two
Again, the hydrogen is removed by the AlCl4- ion. The aluminium chloride and hydrogen chloride catalysts are re-generated in this second stage. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/C._The_Friedel-Crafts_Acylation_of_Benzene.txt |
Electrophilic substitution happens in many of the reactions of compounds containing benzene rings - the arenes. For simplicity, we'll only look for now at benzene itself. This is what you need to understand for the purposes of the electrophilic substitution mechanisms:
• Benzene, C6H6, is a planar molecule containing a ring of six carbon atoms each with a hydrogen atom attached.
• There are delocalized electrons above and below the plane of the ring.
• The presence of the delocalized electrons makes benzene particularly stable.
• Benzene resists addition reactions because that would involve breaking the delocalization and losing that stability.
• Benzene is represented by this symbol, where the circle represents the delocalized electrons, and each corner of the hexagon has a carbon atom with a hydrogen attached.
Electrophilic substitution reactions involving positive ions
Benzene and electrophiles
Because of the delocalized electrons exposed above and below the plane of the rest of the molecule, benzene is obviously going to be highly attractive to electrophiles - species which seek after electron rich areas in other molecules. The electrophile will either be a positive ion, or the slightly positive end of a polar molecule. The delocalized electrons above and below the plane of the benzene molecule are open to attack in the same way as those above and below the plane of an ethene molecule. However, the end result will be different.
If benzene underwent addition reactions in the same way as ethene, it would need to use some of the delocalized electrons to form bonds with the new atoms or groups. This would break the delocalization - and this costs energy. Instead, it can maintain the delocalization if it replaces a hydrogen atom by something else - a substitution reaction. The hydrogen atoms aren't involved in any way with the delocalized electrons. In most of benzene's reactions, the electrophile is a positive ion, and these reactions all follow a general pattern.
The general mechanism
The first stage
Suppose the electrophile is a positive ion \(X^+\). Two of the electrons in the delocalized system are attracted towards the \(X^+\) and form a bond with it. This has the effect of breaking the delocalization, although not completely.
The ion formed in this step isn't the final product. It immediately goes on to react with something else. It is just an intermediate.
There is still delocalization in the intermediate formed, but it only covers part of the ion. When you write one of these mechanisms, draw the partial delocalization to take in all the carbon atoms apart from the one that the \(X\) has become attached to. The intermediate ion carries a positive charge because you are joining together a neutral molecule and a positive ion. This positive charge is spread over the delocalized part of the ring. Simply draw the "+" in the middle of the ring. The hydrogen at the top isn't new - it's the hydrogen that was already attached to that carbon. We need to show that it is there for the next stage.
The second stage
Here we've introduced a new ion, \(Y^-\). W here did this come from? You have to remember that it is impossible to get a positive ion on its own in a chemical system - so \(Y^-\) is simply the negative ion that was originally associated with \(X^+\). Don't w orry about this at the moment - it's much easier to see when you've got a real example in front of you.
A lone pair of electrons on Y- forms a bond with the hydrogen atom at the top of the ring. That means that the pair of electrons joining the hydrogen onto the ring aren't needed any more. These then move down to plug the gap in the delocalized electrons, so restoring the delocalized ring of electrons which originally gave the benzene its special stability.
The energetics of the reaction
The complete delocalization is temporarily broken as \(X\) replaces \(H\) on the ring, and this costs energy. However, that energy is recovered when the delocalization is re-established. This initial input of energy is simply the activation energy for the reaction. In this case, it is going to be high (something around 150 kJ mol-1), and this means that benzene's reactions tend to be slow.
Electrophilic substitution reactions not involving positive ions
Halogenation and sulfonation
In these reactions, the electrophiles are polar molecules rather than fully positive ions. Because these mechanisms are different from what's gone before (and from each other), there isn't any point in dealing with them in a general way.
Contributor
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/Electrophilic_Substitution.txt |
This page discusses the problems which arise if you try to write the mechanism for an electrophilic substitution reaction into a benzene ring which already has something else attached to it. There are two problems you might come across:
• Whereabouts in the ring does the substitution happen? Does this make a difference to how you draw the mechanisms?
• Can the group already attached to the ring get involved in any way?
Electrophilic substitution in methylbenzene
The nitration of methylbenzene
If you substitute a nitro group, -NO2, into the benzene ring in methylbenzene, you could possibly get any of the following products:
The carbon with the methyl group attached is thought of as the number 1 carbon, and the ring is then numbered around from 1 to 6. You number in a direction (in this case, clockwise) which produces the smaller number in the name - hence 2-nitromethylbenzene rather than 6-nitromethylbenzene.
In the case of methylbenzene, whatever you attach to the ring, you always get a mixture consisting mainly of the 2- and 4- isomers. The methyl group is said to be 2,4-directing, in the sense that it seems to "push" incoming groups into those positions. Some other groups which might already be on the ring (for example, the -NO2 group in nitrobenzene) "push" incoming groups into the 3- position.
How to write the mechanism for the nitration of methylbenzene
Reacting methylbenzene with a mixture of concentrated nitric and sulphuric acids gives both 2-nitromethylbenzene and 4-nitromethylbenzene. The mechanism is exactly the same as the nitration of benzene. You just have to be careful about the way that you draw the structure of the intermediate ion.
Making 2-nitromethylbenzene (the first step)
This just shows the first step of the electrophilic substitution reaction. Notice that the partial delocalisation in the intermediate ion covers all the carbon atoms in the ring except for the one that the -NO2 group gets attached to. That is the only point of interest in this example - everything else is just the same as with the nitration of benzene. The hydrogen atom is then removed by an HSO4- ion - exactly as in the benzene case.
Making 4-nitromethylbenzene (the first step)
Once again, the only point of interest is in the way the partial delocalisation in the intermediate ion is drawn - again, it covers all the carbon atoms in the ring apart from the one with the -NO2 group attached.
The electrophilic substitution reaction between methylbenzene and chlorine
This is a good example of a case where what is already attached to the ring can also get involved in the reaction. It is possible to get two quite different substitution reactions between methylbenzene and chlorine depending on the conditions used. The chlorine can substitute into the ring or into the methyl group. Here we are only interested in substitution into the ring. This happens in the presence of aluminium chloride or iron, and in the absence of UV light. Substituting into the ring gives a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene.
The mechanisms are exactly the same as the substitution of chlorine into benzene - although you would have to be careful about the way you draw the intermediate ion.
For example, the complete mechanism for substitution into the 4- position is:
Stage one
Stage two
Electrophilic substitution in nitrobenzene
Substitution into the 3- position (the first step)
Methyl groups direct new groups into the 2- and 4- positions, but a nitro group, -NO2, already on the ring directs incoming groups into the 3- position. For example, if the temperature is raised above 50°C, the nitation of benzene doesn't just produce nitrobenzene - it also produces some 1,3-dinitrobenzene. A second nitro group is substituted into the ring in the 3- position.
The mechanism is exactly the same as the nitration of benzene or of methylbenzene - you just have to be careful in drawing the intermediate ion. Draw the partial delocalisation to include all the carbons except for the one the new -NO2 group gets attached to.
In the second stage, the hydrogen atom is then removed by an HSO4- ion - exactly as in the benzene case. This isn't shown because there's nothing new. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/H._Some_Substitution_Reactions_of_Methylbenzene.txt |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and chloromethane in the presence of an aluminium chloride catalyst. Any other chloroalkane would work similarly.
The electrophilic substitution reaction between benzene and chloromethane
Alkylation means substituting an alkyl group into something - in this case into a benzene ring. A hydrogen on the ring is replaced by a group like methyl or ethyl and so on.
Benzene is treated with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminum chloride as a catalyst. On this page, we will look at substituting a methyl group, but any other alkyl group could be used in the same way. Substituting a methyl group gives methylbenzene - once known as toluene.
$C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl$
or better:
The aluminium chloride isn't written into these equations because it is acting as a catalyst. If you wanted to include it, you could write AlCl3 over the top of the arrow.
The formation of the electrophile
The electrophile is CH3+. It is formed by reaction between the chloromethane and the aluminum chloride catalyst.
$CH_3Cl + AlCl_3 \rightarrow ^+CH_3 + AlCl_4^-$
The electrophilic substitution mechanism
Stage one
Stage two
The hydrogen is removed by the $AlCl_4^-$ ion wh ich was formed at the same time as the $CH_3^+$ ele ctrophile . The aluminum chloride catalyst is re-generated in this second stage.
Contributors
Jim Clark (Chemguide.co.uk)
The Halogenation of Benzene
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and chlorine or bromine in the presence of a catalyst such as aluminum chloride or iron.
The electrophilic substitution reaction between benzene and chlorine or bromine
Benzene reacts with chlorine or bromine in an electrophilic substitution reaction, but only in the presence of a catalyst. The catalyst is either aluminum chloride (or aluminum bromide if you are reacting benzene with bromine) or iron. Strictly speaking iron is not a catalyst, because it gets permanently changed during the reaction. It reacts with some of the chlorine or bromine to form iron(III) chloride, $FeCl_3$, or iron(III) bromide, $FeBr_3$.
$2Fe + 3Cl_2 \rightarrow 2FeCl_3$
$2Fe + 3Br_2 \rightarrow 2FeBr_3$
These compounds act as the catalyst and behave exactly like aluminum chloride in these reactions.
The reaction with chlorine
The reaction between benzene and chlorine in the presence of either aluminum chloride or iron gives chlorobenzene.
$C_6H_6 + Cl_2 \rightarrow C_6H_5Cl + HCl$
or:
The reaction with bromine
The reaction between benzene and bromine in the presence of either aluminum bromide or iron gives bromobenzene. Iron is usually used because it is cheaper and more readily available.
$C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr$
or:
The formation of the electrophile
We are going to explore the reaction using chlorine and aluminum chloride. If you want one of the other combinations, all you have to do is to replace each $Cl$ by $Br$, or each $Al$ by $Fe$. As a chlorine molecule approaches the benzene ring, the delocalized electrons in the ring repel electrons in the chlorine-chlorine bond.
It is the slightly positive end of the chlorine molecule which acts as the electrophile. The presence of the aluminum chloride helps this polarization.
The electrophilic substitution mechanism
Stage one
Stage two
The hydrogen is removed by the $AlCl_4^-$ ion which was formed in the first stage. The aluminum chloride catalyst is re-generated in this second stage.
Contributors
Jim Clark (Chemguide.co.uk)
The Nitration of Benzene
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and a mixture of concentrated nitric acid and concentrated sulfuric acid.
The electrophilic substitution reaction between benzene and nitric acid
Benzene is treated with a mixture of concentrated nitric acid and concentrated sulfuric acid at a temperature not exceeding 50°C. As temperature increases there is a greater chance of getting more than one nitro group, -NO2, substituted onto the ring. Nitrobenzene is formed:
$C_6H_6 + HNO_3 \rightarrow C_6H_5NO_2 + H_2O$
or:
The concentrated sulfuric acid is acting as a catalyst.
The formation of the electrophile
The electrophile is the "nitronium ion" or the "nitryl cation", $NO_2^+$. This is formed by reaction between the nitric acid and the sulphuric acid.
$HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ 2HSO_4^- + H_3O^+$
Stage one
Stage two
The Sulfonation of Benzene
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and sulfuric acid (or sulfur trioxide).
The electrophilic substitution reaction between benzene and sulfuric acid
There are two equivalent ways of sulfonating benzene:
• Heat benzene under reflux with concentrated sulfuric acid for several hours.
• Warm benzene under reflux at 40°C with fuming sulfuric acid for 20 to 30 minutes.
$C_6H_6 + H_2SO_4 \rightarrow C_6H_5SO_3H + H_2O$
Or:
The product is benzenesulfonic acid. The electrophile is actually sulfur trioxide, SO3, and you may find the equation for the sulfonation reaction written:
The formation of the electrophile
The sulfur trioxide electrophile arises in one of two ways depending on which sort of acid you are using. Concentrated sulfuric acid contains traces of SO3 due to slight dissociation of the acid.
$H_2SO_4 \rightleftharpoons H_2O + SO_3$
Fuming sulfuric acid, $H_2S_2O_7$, can be thought of as a solution of $SO_3$ i n sulfuric acid - and so is a much richer source of the SO3. Sulfur trioxide is an electrophile because it is a highly polar molecule with a fair amount of positive charge on the sulfur atom. It is this which is attracted to the ring electrons.
The electrophilic substitution mechanism
Stage one
Stage two
The second stage of the reaction involves a transfer of the hydrogen from the ring to the negative oxygen. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/The_Friedel-Crafts_Alkylation_of_Benzene.txt |
Substitution reactions of halogenoalkanes such as bromoethane.
IV. Nucleophilic Substitution Reactions
Halogenoalkanes are also called haloalkanes or alkyl halides. All halogenoalkanes contain a halogen atom - fluorine, chlorine, bromine or iodine - attached to an alkyl group. There are three different kinds of halogenoalkanes: Primary, secondary and tertiary.
Primary halogenoalkanes
In a primary (1°) halogenoalkane, the carbon which carries the halogen atom is only attached to one other alkyl group. Some examples of primary halogenoalkanes include:
Notice that it doesn't matter how complicated the attached alkyl group is. In each case there is only one linkage from the CH2 group holding the halogen to an alkyl group. There is an exception to this. CH3Br and the other methyl halides are often counted as primary halogenoalkanes even though there are no alkyl groups attached to the carbon with the halogen on it.
Secondary halogenoalkanes
In a secondary (2°) halogenoalkane, the carbon with the halogen attached is joined directly to two other alkyl groups, which may be the same or different. Examples:
Tertiary halogenoalkanes
In a tertiary (3°) halogenoalkane, the carbon atom holding the halogen is attached directly to three alkyl groups, which may be any combination of same or different. Examples:
Contributors
Jim Clark (Chemguide.co.uk)
B. What is Nucleophilic Substitution
Halogenoalkanes (also known as haloalkanes or alkyl halides) are compounds containing a halogen atom (fluorine, chlorine, bromine or iodine) joined to one or more carbon atoms in a chain. The interesting thing about these compounds is the carbon-halogen bond, and all the nucleophilic substitution reactions of the halogenoalkanes involve breaking that bond. With the exception of iodine, all of the halogens are more electronegative than carbon.
Table 1: Electronegativity values (Pauling scale)
C 2.5
F 4.0
Cl 3.0
Br 2.8
I 2.5
That means that the electron pair in the carbon-halogen bond will be dragged towards the halogen end, leaving the halogen slightly negative (-) and the carbon slightly positive (+) - except in the carbon-iodine case. Although the carbon-iodine bond doesn't have a permanent dipole, the bond is very easily polarized by anything approaching it. Imagine a negative ion approaching the bond from the far side of the carbon atom:
The fairly small polarity of the carbon-bromine bond will be increased by the same effect.
The strengths of the carbon-halogen bonds
In all of these nucleophilic substitution reactions, the carbon-halogen bond has to be broken at some point during the reaction. In general, the harder it is to break, the slower the reaction will be. Look at the strengths of various bonds (all values in kJ mol-1).
C-H 413
C-F 467
C-Cl 346
C-Br 290
C-I 228
The carbon-fluorine bond is very strong (stronger than C-H) and isn't easily broken. It doesn't matter that the carbon-fluorine bond has the greatest polarity - the strength of the bond is much more important in determining its reactivity. You might therefore expect fluoroalkanes to be very unreactive - and they are! We shall simply ignore them from now on. In the other halogenoalkanes, the bonds get weaker as you go from chlorine to bromine to iodine. That means that chloroalkanes react most slowly, bromoalkanes react faster, and iodoalkanes react faster still.
Rates of reaction: RCl < RBr < RI
Where "<" is read as "is less than" - or, in this instance, "is slower than", and R represents any alkyl group.
Nucleophilic substitution in primary halogenoalkanes
You will need to know about this if your syllabus talks about "primary halogenoalkanes" or about SN2 reactions. If the syllabus is vague, check recent exam papers and mark schemes, and compare them against what follows.
A nucleophile is a species (an ion or a molecule) which is strongly attracted to a region of positive charge in something else. Nucleophiles are either fully negative ions, or else have a strongly - charge somewhere on a molecule. Common nucleophiles are hydroxide ions, cyanide ions, water and ammonia.
Notice that each of these contains at least one lone pair of electrons, either on an atom carrying a full negative charge, or on a very electronegative atom carrying a substantial - charge.
The nucleophilic substitution reaction - an SN2 reaction
We'll talk this mechanism through using an ion as a nucleophile, because it's slightly easier. The water and ammonia mechanisms involve an extra step which you can read about on the pages describing those particular mechanisms. We'll take bromoethane as a typical primary halogenoalkane. The bromoethane has a polar bond between the carbon and the bromine.
We'll look at its reaction with a general purpose nucleophilic ion which we'll call Nu-. This will have at least one lone pair of electrons. Nu- could, for example, be OH- or CN-.
The lone pair on the Nu- ion will be strongly attracted to the + carbon, and will move towards it, beginning to make a co-ordinate (dative covalent) bond. In the process the electrons in the C-Br bond will be pushed even closer towards the bromine, making it increasingly negative.
The movement goes on until the -Nu is firmly attached to the carbon, and the bromine has been expelled as a Br- ion.
Note
The Nu- ion approaches the $\delta^+$ carbon from the side away from the bromine atom. The large bromine atom hinders attack from its side and, being $\delta^-$, would repel the incoming Nu- anyway. This attack from the back is important if you need to understand why tertiary halogenoalkanes have a different mechanism. We'll discuss this later on this page.
There is obviously a point in which the Nu- is half attached to the carbon, and the C-Br bond is half way to being broken. This is called a transition state. It isn't an intermediate. You can't isolate it - even for a very short time. It's just the mid-point of a smooth attack by one group and the departure of another.
Nucleophile effects
A major factor affecting the SN2 reaction is strength of the nucleophile. Recall that a nucleophile is a molecule which is attracted to positive charge. We can describe several general trends which determine the strength of a nucleophile:
1. A molecule which has lost a proton and is negatively charged (base) is stronger than the neutral, protonated version of the same molecule (conjugate acid).
2. Nucleophilicity of the attacking atom decreases from left to right in the periodic table because atoms because atoms are more electronegative (hold electrons more closely) going from left to right.
3. Nucleophilicity of the attacking atom increases down the periodic table as electronegativity decreases and polarizibity increases. Polarizibility refers to the ability of larger atoms lower in the periodic table to share electrons more easily because their electrons are more loosely held.
In addition to the strength of the nucleophile, the structure of the nucleophile also influences the capacity of an SN2 reaction to proceed. As stated many times previously, steric effects are the most important factor in determining whether an SN2 reaction can proceed. We have discussed these effects with regard to the substrate, but they also apply to the nucleophile. A small, unbranched nucleophile will be more effective in an SN2 reaction than a large, branched nucleophile.
Substrate effects
Based on the factors described above, we can outline several trends that affect the likelihood of an SN2 reaction occurring. As described above, the most important factor in deciding whether an SN2 reaction occurs is steric effects. When we describe the nature of the molecule being attacked, called the substrate, we can look at how many alkyl substituents are present on the molecule. For example, the molecule may contain a carbon atom attached to a leaving group and three hydrogen atoms. This would constitute a methyl carbon atom, in analogy to methane. Given that such a molecule has almost no steric hindrance (substituents blocking access to the carbon atom), an SN2 reaction will be highly favored in a molecule with a methyl carbon.
We call a carbon atom with one alkyl substituent a primary carbon atom. SN2 reactions will proceed easily with a primary carbon atom, though not as fast as with a methyl carbon. If there are two alkyl substituents, we call the carbon atom secondary. It is possible for an SN2 reaction to proceed at a secondary carbon atom, but it is not as favorable as in the methyl or primary cases. Finally, if there are three alkyl substituents attached to the carbon atom, we have the case of a tertiary carbon atom. In this case, there is far too much steric hindrance, and the SN2 reaction cannot proceed. The principle that increasing substitution leads to decreasing reactivity is outlined in the following table:
Relative rate of SN2 reaction
Methyl > Primary > Secondary > Tertiary
The degree of substitution of the molecule undergoing attack is not the only factor that influences the rate of an SN2 reaction. Recall that steric effects (the likelihood that attack by the nucleophile will be blocked) are of paramount importance in determining whether an SN2 reaction will take place. Degree of substitution is one factor in determining steric effects, but it is not the only one. We also need to consider the bulkiness of the substituents.
How to write the mechanism
The simplest way is:
Technically, this is known as an SN2 reaction. S stands for substitution, N for nucleophilic, and the 2 is because the initial stage of the reaction involves two species - the bromoethane and the Nu- ion. If your syllabus doesn't refer to SN2 reactions by name, you can just call it nucleophilic substitution.
Some examiners like you to show the transition state in the mechanism, in which case you need to write it in a bit more detail - showing how everything is arranged in space.
Be very careful when you draw the transition state to make a clear difference between the dotted lines showing the half-made and half-broken bonds, and those showing the bonds going back into the paper. Notice that the molecule has been inverted during the reaction - rather like an umbrella being blown inside-out.
Nucleophilic substitution in tertiary halogenoalkanes
Remember that a tertiary halogenoalkane has three alkyl groups attached to the carbon with the halogen on it. These alkyl groups can be the same or different, but in this section, we shall just consider a simple one, (CH3)3CBr - 2-bromo-2-methylpropane.
The nucleophilic substitution reaction - an SN1 reaction
Once again, we'll talk this mechanism through using an ion as a nucleophile, because it's slightly easier, and again we'll look at the reaction of a general purpose nucleophilic ion which we'll call Nu-. This will have at least one lone pair of electrons.
Why is a different mechanism necessary?
You will remember that when a nucleophile attacks a primary halogenoalkane, it approaches the + carbon atom from the side away from the halogen atom. With a tertiary halogenoalkane, this is impossible. The back of the molecule is completely cluttered with CH3 groups.
Since any other approach is prevented by the bromine atom, the reaction has to go by an alternative mechanism.
The alternative mechanism
The reaction happens in two stages. In the first, a small proportion of the halogenoalkane ionises to give a carbocation and a bromide ion.
This reaction is possible because tertiary carbocations are relatively stable compared with secondary or primary ones. Even so, the reaction is slow. Once the carbocation is formed, however, it would react immediately it came into contact with a nucleophile like Nu-. The lone pair on the nucleophile is strongly attracted towards the positive carbon, and moves towards it to create a new bond.
How fast the reaction happens is going to be governed by how fast the halogenoalkane ionises. Because this initial slow step only involves one species, the mechanism is described as SN1 - substitution, nucleophilic, one species taking part in the initial slow step.
Why don't primary halogenoalkanes use the SN1 mechanism?
If a primary halogenoalkane did use this mechanism, the first step would be, for example:
A primary carbocation would be formed, and this is much more energetically unstable than the tertiary one formed from tertiary halogenoalkanes - and therefore much more difficult to produce. This instability means that there will be a very high activation energy for the reaction involving a primary halogenoalkane. The activation energy is much less if it undergoes an SN2 reaction - and so that's what it does instead.
Nucleophilic substitution in secondary halogenoalkanes
There isn't anything new in this. Secondary halogenoalkanes will use both mechanisms - some molecules will react using the SN2 mechanism and others the SN1.
The SN2 mechanism is possible because the back of the molecule isn't completely cluttered by alkyl groups and so the approaching nucleophile can still get at the + carbon atom. The SN1 mechanism is possible because the secondary carbocation formed in the slow step is more stable than a primary one. It isn't as stable as a tertiary one though, and so the SN1 route isn't as effective as it is with tertiary halogenoalkanes.
Contributors
Jim Clark (Chemguide.co.uk)
Jonathan Mooney (McGill University) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/IV._Nucleophilic_Substitution_Reactions/A._Types_of_Halogenoalkanes.txt |
This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and hydroxide ions (from, for example, sodium hydroxide).
The reaction of primary halogenoalkanes with hydroxide ions
The facts
If a halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide, the halogen is replaced by -OH and an alcohol is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture.
The solvent is usually a 50/50 mixture of ethanol and water, because everything will dissolve in that. The halogenoalkane is insoluble in water. If you used water alone as the solvent, the halogenoalkane and the sodium hydroxide solution wouldn't mix and the reaction could only happen where the two layers met.
For example, using 1-bromopropane as a typical primary halogenoalkane:
You could write the full equation rather than the ionic one, but it slightly obscures what's going on:
The bromine (or other halogen) in the halogenoalkane is simply replaced by an -OH group - hence a substitution reaction. In this example, propan-1-ol is formed.
The mechanism
Here is the mechanism for the reaction involving bromoethane:
This is an example of nucleophilic substitution.
Because the mechanism involves collision between two species in the slow step (in this case, the only step) of the reaction, it is known as an SN2 reaction.
If your examiners want you to show the transition state, draw the mechanism like this:
The reaction of tertiary halogenoalkanes with hydroxide ions
The facts
The facts of the reaction are exactly the same as with primary halogenoalkanes. If the halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide in a mixture of ethanol and water, the halogen is replaced by -OH, and an alcohol is produced.
For example:
Or if you want the full equation rather than the ionic one:
The mechanism
This mechanism involves an initial ionisation of the halogenoalkane:
followed by a very rapid attack by the hydroxide ion on the carbocation (carbonium ion) formed:
This is again an example of nucleophilic substitution.
This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an SN1 reaction.
The reaction of secondary halogenoalkanes with hydroxide ions
The facts
The facts of the reaction are exactly the same as with primary or tertiary halogenoalkanes. The halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide in a mixture of ethanol and water.
For example:
The mechanism
Secondary halogenoalkanes use both SN2 and SN1 mechanisms. For example, the SN2 mechanism is:
Should you need it, the two stages of the SN1 mechanism are:
D. Substitution Reactions Involving Water
This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and water.
The reaction of primary halogenoalkanes with water
The facts
There is only a slow reaction between a primary halogenoalkane and water even if they are heated. The halogen atom is replaced by -OH.
For example, using 1-bromoethane as a typical primary halogenoalkane:
An alcohol is produced together with hydrobromic acid. Be careful not to call this hydrogen bromide. Hydrogen bromide is a gas. When it is dissolved it in water (as it will be here), it's called hydrobromic acid.
The mechanism
The mechanism involves two steps. The first is a simple nucleophilic substitution reaction:
Because the mechanism involves collision between two species in this slow step of the reaction, it is known as an SN2 reaction.
The nucleophilic substitution is very slow because water isn't a very good nucleophile. It lacks the full negative charge of, say, a hydroxide ion.
The second step of the reaction simply tidies up the product. A water molecule removes one of the hydrogens attached to the oxygen to give an alcohol and a hydroxonium ion (also known as a hydronium ion or an oxonium ion).
The hydroxonium ion and the bromide ion (from the nucleophilic substitution stage of the reaction) make up the hydrobromic acid which is formed as well as the alcohol.
The reaction of tertiary halogenoalkanes with water
The facts
If the halogenoalkane is heated under reflux with water, the halogen is replaced by -OH to give an alcohol. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture. The reaction happens much faster than the corresponding one involving a primary halogenoalkane.
For example:
The mechanism
This mechanism involves an initial ionisation of the halogenoalkane:
followed by a very rapid attack by the water on the carbocation (carbonium ion) formed:
This is again an example of nucleophilic substitution.
This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an SN1 reaction.
Now there is a final stage in which the product is tidied up. A water molecule removes one of the hydrogens attached to the oxygen to give an alcohol and a hydroxonium ion - exactly as happens with primary halogenoalkanes.
The rate of the overall reaction is governed entirely by how fast the halogenoalkane ionises. The fact that water isn't as good a nucleophile as, say, OH- doesn't make any difference. The water isn't involved in the slow step of the reaction.
The reaction of secondary halogenoalkanes with water
Secondary halogenoalkanes use both an SN2 mechanism and an SN1.
Make sure you understand what happens with primary and tertiary halogenoalkanes, and then adapt it for secondary ones should ever need to.
E. Substitution Reactions Involving Cyanide Ions
Reacting Primary Halogenoalkanes with Cyanide Ions
If a halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol, the halogen is replaced by a -CN group and a nitrile is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture. The solvent is important. If water is present you tend to get substitution by -OH instead of -CN.
For example, using 1-bromopropane as a typical primary halogenoalkane:
$CH_3CH_2CH_2Br + CN^- \rightarrow CH_3CH_2CH_2CN + Br^-$
The full equation could be written rather than the ionic one, but it obscures what's going on:
$CH_3CH_2CH_2Br + KCN \rightarrow CH_3CH_2CH_2CN + KBr$
The bromine (or other halogen) in the halogenoalkane is simply replaced by a -CN group - hence a substitution reaction. In this example, butanenitrile is formed. Here is the mechanism for the reaction involving bromoethane:
This is an example ofnucleophilic substitution. Because the mechanism involves collision between two species in the slow step (in this case, the only step) of the reaction, it is known as an SN2 reaction. If your examiners want you to show the transition state, draw the mechanism like this:
Reacting Tertiary Halogenoalkanes with Cyanide Ions
The facts of the reaction are exactly the same as with primary halogenoalkanes. If the halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol, the halogen is replaced by -CN, and a nitrile is produced.
For example:
$(CH_3)_3CBr + CN^- \rightarrow (CH_3)_3CCN + Br^-$
Or if you want the full equation rather than the ionic one:
$(CH_3)_3CBr + KCN \rightarrow (CH_3)_3CCN + KBr$
This mechanism involves an initial ionization of the halogenoalkane:
followed by a very rapid attack by the cyanide ion on the carbocation (carbonium ion) formed:
This is again an example of nucleophilic substitution. This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an SN1 reaction.
The reaction of secondary halogenoalkanes with cyanide ions
The facts of the reaction are exactly the same as with primary or tertiary halogenoalkanes. The halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol.
For example:
Secondary halogenoalkanes use both SN2 and SN1 mechanisms. For example, the SN2 mechanism is:
The two stages of the SN1 mechanism are: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/IV._Nucleophilic_Substitution_Reactions/C._Substitution_Reactions_Involving_Hydroxide_Ions.txt |
This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and ammonia to produce primary amines.
Reaction of Primary halogenoalkanes with ammonia
The halogenoalkane is heated with a concentrated solution of ammonia in ethanol. The reaction is carried out in a sealed tube. You couldn't heat this mixture under reflux, because the ammonia would simply escape up the condenser as a gas. We'll talk about the reaction using 1-bromoethane as a typical primary halogenoalkane. The reaction happens in two stages. In the first stage, a salt is formed - in this case, ethylammonium bromide. This is just like ammonium bromide, except that one of the hydrogens in the ammonium ion is replaced by an ethyl group.
$CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_3^+Br^-$
There is then the possibility of a reversible reaction between this salt and excess ammonia in the mixture.
$CH_3CH_2NH_3^+Br^- + NH_3 \rightleftharpoons CH_3CH_2NH_2 + NH_4Br^-$
The ammonia removes a hydrogen ion from the ethylammonium ion to leave a primary amine - ethylamine. The more ammonia there is in the mixture, the more the forward reaction is favored as predicted by Le Chatelier's principle. The mechanism involves two steps. The first is a simple nucleophilic substitution reaction:
Because the mechanism involves collision between two species in this slow step of the reaction, it is known as an SN2 reaction. In the second step of the reaction an ammonia molecule may remove one of the hydrogens on the -NH3+. An ammonium ion is formed, together with a primary amine - in this case, ethylamine.
This reaction is, however, reversible. Your product will therefore contain a mixture of ethylammonium ions, ammonia, ethylamine and ammonium ions. Your major product will only be ethylamine if the ammonia is present in very large excess. Unfortunately the reaction doesn't stop here. Ethylamine is a good nucleophile, and goes on to attack unused bromoethane. This gets so complicated that it is dealt with on a separate page. You will find a link at the bottom of this page.
Reaction of tertiary halogenoalkanes with ammonia
The facts of the reactions are exactly the same as with primary halogenoalkanes. The halogenoalkane is heated in a sealed tube with a solution of ammonia in ethanol. For example:
Followed by:
This mechanism involves an initial ionisation of the halogenoalkane:
followed by a very rapid attack by the ammonia on the carbocation (carbonium ion) formed:
This is again an example of nucleophilic substitution. This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an SN1 reaction. There is a second stage exactly as with primary halogenoalkanes. An ammonia molecule removes a hydrogen ion from the -NH3+ group in a reversible reaction. An ammonium ion is formed, together with an amine.
Reaction of secondary halogenoalkanes with ammonia
It is very unlikely that any of the current UK-based syllabuses for 16 - 18 year olds will ask you about this. In the extremely unlikely event that you will ever need it, secondary halogenoalkanes use both an SN2 mechanism and an SN1. Make sure you understand what happens with primary and tertiary halogenoalkanes, and then adapt it for secondary ones should ever need to.
Kinetics of Nucleophilic Substitution Reactions
We will be contrasting about two types of nucleophilic substitution reactions. One type is referred to as unimolecular nucleophilic substitution (SN1), whereby the rate determining step is unimolecular and bimolecular nucleophilic substitution (SN2), whereby the rate determining step is bimolecular. We will begin our discussion with SN2 reactions, and discuss SN1 reactions elsewhere.
Biomolecular Nucleophilic Substitution Reactions and Kinetics
In the term SN2, the S stands for substitution, the N stands for nucleophilic, and the number two stands for bimolecular, meaning there are two molecules involved in the rate determining step. The rate of bimolecular nucleophilic substitution reactions depends on the concentration of both the haloalkane and the nucleophile. To understand how the rate depends on the concentrations of both the haloalkane and the nucleophile, let us look at the following example. The hydroxide ion is the nucleophile and methyl iodide is the haloalkane.
If we were to double the concentration of either the haloalkane or the nucleophile, we can see that the rate of the reaction would proceed twice as fast as the initial rate.
If we were to double the concentration of both the haloalkane and the nucleophile, we can see that the rate of the reaction would proceed four times as fast as the initial rate.
The bimolecular nucleophilic substitution reaction follows second-order kinetics; that is, the rate of the reaction depends on the concentration of two first-order reactants. In the case of bimolecular nucleophilic substitution, these two reactants are the haloalkane and the nucleophile. For further clarification on reaction kinetics, the following links may facilitate your understanding of rate laws, rate constants, and second-order kinetics:
Bimolecular Nucleophilic Substitution Reactions Are Concerted
Bimolecular nucleophilic substitution (SN2) reactions are concerted, meaning they are a one step process. This means that the process whereby the nucleophile attacks and the leaving group leaves is simultaneous. Hence, the bond-making between the nucleophile and the electrophilic carbon occurs at the same time as the bond-breaking between the electophilic carbon and the halogen.
The potential energy diagram for an SN2 reaction is shown below. Upon nucleophilic attack, a single transition state is formed. A transition state, unlike a reaction intermediate, is a very short-lived species that cannot be isolated or directly observed. Again, this is a single-step, concerted process with the occurrence of a single transition state.
Contributors
• Racheal Curtis | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/IV._Nucleophilic_Substitution_Reactions/F._Substitution_Reactions_Involving_Ammonia.txt |
Just as with SN2 reactions, the nucleophile, solvent and leaving group also affect SN1 (Unimolecular Nucleophilic Substitution) reactions. Polar protic solvents have a hydrogen atom attached to an electronegative atom so the hydrogen is highly polarized. Polar aprotic solvents have a dipole moment, but their hydrogen is not highly polarized. Polar aprotic solvents are not used in SN1 reactions because some of them can react with the carbocation intermediate and give you an unwanted product. Rather, polar protic solvents are preferred.
Introduction
Since the hydrogen atom in a polar protic solvent is highly positively charged, it can interact with the anionic nucleophile which would negatively affect an SN2, but it does not affect an SN1 reaction because the nucleophile is not a part of the rate-determining step (See SN2 Nucleophile). Polar protic solvents actually speed up the rate of the unimolecular substitution reaction because the large dipole moment of the solvent helps to stabilize the transition state. The highly positive and highly negative parts interact with the substrate to lower the energy of the transition state. Since the carbocation is unstable, anything that can stabilize this even a little will speed up the reaction. Sometimes in an SN1 reaction the solvent acts as the nucleophile. This is called a solvolysis reaction (see example below). The polarity and the ability of the solvent to stabilize the intermediate carbocation is very important as shown by the relative rate data for the solvolysis (see table below). The dielectric constant of a solvent roughly provides a measure of the solvent's polarity. A dielectric constant below 15 is usually considered non-polar. Basically, the dielectric constant can be thought of as the solvent's ability to reduce the internal charge of the solvent. So for our purposes, the higher the dielectric constant the more polar the substance and in the case of SN1 reactions, the faster the rate.
Below is the same reaction conducted in two different solvents and the relative rate that corresponds with it.
The figure below shows the mechanism of an SN1 reaction of an alkyl halide with water. Since water is also the solvent, this is an example of a solvolysis reaction.
Examples of polar protic solvents are: acetic acid, isopropanol, ethanol, methanol, formic acid, water, etc.
Effects of Nucleophile
The strength of the nucleophile does not affect the reaction rate of SN1 because, as stated above, the nucleophile is not involved in the rate-determining step. However, if you have more than one nucleophile competing to bond to the carbocation, the strengths and concentrations of those nucleophiles affects the distribution of products that you will get. For example, if you have (CH3)3CCl reacting in water and formic acid where the water and formic acid are competing nucleophiles, you will get two different products: (CH3)3COH and (CH3)33COCOH. The relative yields of these products depend on the concentrations and relative reactivities of the nucleophiles.
Effects of Leaving Group
An SN1 reaction speeds up with a good leaving group. This is because the leaving group is involved in the rate-determining step. A good leaving group wants to leave so it breaks the C-Leaving Group bond faster. Once the bond breaks, the carbocation is formed and the faster the carbocation is formed, the faster the nucleophile can come in and the faster the reaction will be completed.
A good leaving group is a weak base because weak bases can hold the charge. They're happy to leave with both electrons and in order for the leaving group to leave, it needs to be able to accept electrons. Strong bases, on the other hand, donate electrons which is why they can't be good leaving groups. As you go from left to right on the periodic table, electron donating ability decreases and thus ability to be a good leaving group increases. Halides are an example of a good leaving group whos leaving-group ability increases as you go down the column.
The two reactions below is the same reaction done with two different leaving groups. One is significantly faster than the other. This is because the better leaving group leaves faster and thus the reaction can proceed faster.
Other examples of good leaving groups are sulfur derivatives such as methyl sulfate ion and other sulfonate ions (See Figure below)
Methyl Sulfate Ion Mesylate Ion Triflate Ion Tosylate Ion
References
• Uggerud E. "Reactivity trends and stereospecificity in nucleophilic substitution reactions." J. Phys. Org. Chem. 2006; 19; 461-466.
• Vollhardt, K. Peter C., and Neil E. Schore. Organic Chemistry Structure and Function. New York: W. H. Freeman, 2007.
• Petrucci, Ralph H. General Chemistry: Principles and Modern Applications. Upper Saddle River, NJ: Pearson Education, 2007.
• Suggs, William J. Organic Chemistry. Canada: Barron's Educational Series Inc., 2002.
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Problems
1. Put the following leaving groups in order of decreasing leaving group ability
2. Which solvent would an SN1 reaction occur faster in? H2O or CH3CN
3. What kind of conditions disfavor SN1 reactions?
4. What are the products of the following reaction and does it proceed via SN1 or SN2?
5. How could you change the reactants in the problem 4 to favor the other substitution reaction?
6. Indicate the expected product and list why it occurs through SN1 instead of SN2?
Answers
1.
2. An SN1 reaction would occur faster in H2O because it's polar protic and would stailize the carbocation and CH3CN is polar aprotic.
3. Polar aprotic solvents, a weak leaving group and primary substrates disfavor SN1 reactions.
4.
Reaction proceeds via SN1 because a tertiary carbocation was formed, the solvent is polar protic and Br- is a good leaving group.
5. You could change the solvent to something polar aprotic like CH3CN or DMSO and you could use a better base for a nucleophile such as NH2- or OH-.
6.
This reaction occurs via SN1 because Cl- is a good leaving group and the solvent is polar protic. This is an example of a solvolysis reaction because the nucleophile is also the solvent.
• Ashiv Sharma | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/SN1/Effects_of_Solvent_Leaving_Group_and_Nucleophile_on_Unimolecular_Substitution.txt |
In contrast to an SN2 reaction, in which the bond-making addition of the nucleophile and the bond-breaking departure of the leaving group occur in a single step, the SN1 reaction involves two separate steps: first the departure of the leaving group and then the addition of the nucleophile.
In the SN1 reaction, the bond between the substrate and the leaving group is broken when the leaving group departs with the pair of electrons that formerly composed the bond. As a result, the carbon atom to which the bond was formerly made is left with a positive charge. This positive charge on a carbon atom is called a carbocation, from "carbon" and "cation", the word for a positively charged atom. The formation of a carbocation is not energetically favored, so this step in the reaction is the slowest step and determines the overall rate of the reaction. The step which controls the overall rate of a reaction is called the rate-determining step.
Only after the leaving group has departed and a carbocation has formed, a nucleophile forms a bond to the carbocation, completing the substitution. This step is more energetically favorable and proceeds more quickly.
There are several important consequences to the unimolecular nature of the rate-determining step in the the SN1 reaction. First, since the rate is controlled by the loss of the leaving group and does not involve any participation of the nucleophile, the rate of the reaction is dependent only on the concentration of the substrate, not on the concentration of the nucleophile. Second, since the nucleophile attacks the carbocation only after the leaving group has departed, there is no need for back-side attack. The carbocation and its substituents are all in the same plane (Figure \(1\)), meaning that the nucleophile can attack from either side. As a result, both enantiomers are formed in an the SN1 reaction, leading to a racemic mixture of both enantiomers. Finally, since the nucleophile does not participate in the rate-determining step, the strength of the nucleophile does not affect the rate of the SN1 reaction.
What factors govern the rate the probability of an SN1 reaction? The single most important factor is the stability of the carbocation. Alkyl substituents increase the stability of a carbocation, so increasing alkyl substitution of the carbon atom increases the probability of an SN1 reaction occurring.
Relative rate of SN1 reaction
Tertiary>Secondary>Primary>Methyl
Recall that, as in the case of an SN1 reaction, the above trend regarding degree of substitution is just a trend and the real factor that determines whether an SN1 reaction can occur is the stability of the carbocation. From above, we would expect an SN1 reaction not to occur at the site of a primary carbon atom. Indeed, such reactions do not occur in ordinary alkanes. However, in molecules in which the carbon next to the site of substitution contains a double bond, the SN1 reaction is possible. The reason is that the positive charge on the carbocation can be delocalized among multiple possible resonance structures (see Resonance and delocalization), making the carbocation dramatically more stable. This effect can occur when the carbon atom of interest is next to one double bond (allylic) or a benzene ring (benzylic). Note that in the allylic case, because of the delocalization of the positive charge, the nucleophile can attack at multiple sites (Figure \(2\)); this effect is absent in the benzylic case due to the need to preserve aromaticity. In summary, the key to the SN1 reaction is the stability of the carbocation.
SN1 Reaction
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Nucleophilic Substitution
Previously (Physical Properties of Haloalkanes), we learned that haloalkanes contain a polarized C-X bond, leaving a carbon that is partially positive and a halogen that is partially negative.
• An electrophile is an electron poor species that can accept a pair of electrons. A carbon that is connected to a halogen in a haloalkane, for example, is an electrophilic carbon.
• A nucleophile is an electron rich species that can donate a pair of electrons. You will be exposed to many different kinds of nucleophiles throughout your course of study. Some nucleophiles will be negatively charged species; others will be neutral.
• Nucleophiles can react with electrophiles. One way in which this occurs is through a process called nucleophilic substitution. In nucleophilic substitution reactions, an electron rich nucleophile bonds with or attacks an electron poor electrophile, resulting in the displacement of a group or atom called the leaving group.
• Nucleophilic substitution of haloalkanes can be described by two reactions. These two types of reactions are shown in the diagram below. In the first reaction, a negatively charged nucleophile attacks the electrophilic carbon of a haloalkane. Upon attack, the leaving group, which is the halogen of the haloalkane, leaves. The end result is a neutral R-Nu species and an anion. In the second reaction, a neutral nucleophile attacks the electrophilic carbon of a haloalkane. The end result of this attack however, is positively charged product and an anion.
Examples of Negatively Charged and Neutral Nucleophiles
In the SN2 reaction, the addition of the nucleophile and the departure of the leaving group occur in a concerted(taking place in a single step) manner, hence the name SN2: substitution, nucleophilic, bimolecular. In the SN2 reaction, the nucleophile approaches the carbon atom to which the leaving group is attached. As the nucleophile forms a bond with this carbon atom, the bond between the carbon atom and the leaving group breaks. The bond making and bond breaking actions occur simultaneously. Eventually, the nucleophile has formed a complete bond to the carbon atom and the bond between the carbon atom and the leaving group is completely broken.
In the image below, we introduce the concepts of arrow pushing and of a reaction mechanism. Recall that electrons compose the bonds in molecules. Hence for any reaction to occur, electrons must move. In arrow pushing, the movement of electrons is indicated by arrows. Arrows may show electrons forming or breaking bonds or traveling as lone pairs or negative charges on atoms. A complete schematic showing all steps in a reaction, including arrow pushing to indicate the movement of electrons, constitutes a reaction mechanism. In the reaction mechanism below, observe that the electrons from a negative charge on the nucleophile "attack" or form a bond with the carbon atom to which the leaving group is attached. In the center image, the partially formed bond is visible, as well as the partially broken bond to the leaving group. In the final image, the bond to the leaving group is broken when its electrons become a negative charge on the leaving group.
Figure 1: SN2 reaction showing concerted, bimolecular participation of nucleophile and leaving group
A consequence of the concerted, bimolecular nature of the SN2 reaction is that the nucleophile must attack from the side of the molecule opposite to the leaving group. This geometry of reaction is called back side attack. In a back side attack, as the nucleophile approaches the molecule from the side opposite to the leaving group, the other three bonds move away from the nucleophile and its attacking electrons. Eventually, these three bonds are all in the same plane as the carbon atom (center image). As the bond to the leaving group breaks, these bonds retreat farther away from the nucleophile and its newly formed bond to carbon atom. As a result of these geometric changes, the stereochemical configuration of the molecule is inverted during an SN2 reaction to the opposite enantiomer. This stereochemical change is called inversion of configuration.
The concerted mechanism and nature of the nucleophilic attack in an SN2 reaction give rise to several important results:
1. The rate of the reaction depends on the concentration of both the nucleophile and the molecule undergoing attack. The reaction requires a collision between the nucleophile and the molecule, so increasing the concentration of either will increase the rate of the reaction.
2. Since the unique geometry of back side attack is required, the most important factor in determining whether an SN2 reaction will occur is steric effects. Steric effects refer to the unfavorable interaction created when atoms are brought too close together. In effect, if the nucleophile or the molecule undergoing attack have too many substituents or substituents which are too bulky, the reaction cannot occur since the nucleophile will be unable to get close enough to the molecule to do a backside attack.
Now let's look at two actual examples of these two general equations. In the first reaction shown below, the negative nucleophile, hydroxide, reacts with methyl iodide. Hydroxide takes the place of the leaving group, iodide, forming neutral methanol and an iodide ion. This reaction is the same as the first type of nucleophilic substitution shown above. In the second reaction shown below, the nuetral nucleophile, ammonia, reacts with iodoethane. Ammonia takes the place of the leaving group, iodide, forming the positively charged product, ethylammonium iodide, and an iodide ion. This reaction is the same as the second type of nucleophilic substitution shown above.
Figure 2. SN2 reaction of methyl chloride and hydroxide ion (left) and ammonia reaction with HCl (right).
Four Factors to Consider in Determining the Relative Ease at Which SN2 Displacement Occurs
Next section: Using Electron-Pushing Arrows
SN2
The Nature of the Leaving Group
In order to understand the nature of the leaving group, it is important to first discuss factors that help determine whether a species will be a strong base or weak base. If you remember from general chemistry, a Lewis base is defined as a species that donates a pair of electrons to form a covalent bond. The factors that will determine whether a species wants to share its electrons or not include electronegativity, size, and resonance.
As Electronegativity Increases, Basicity Decreases: In general, if we move from the left of the periodic table to the right of the periodic table as shown in the diagram below, electronegativity increases. As electronegativity increases, basicity will decrease, meaning a species will be less likely to act as base; that is, the species will be less likely to share its electrons.
As Size Increases, Basicity Decreases: In general, if we move from the top of the periodic table to the bottom of the periodic table as shown in the diagram below, the size of an atom will increase. As size increases, basicity will decrease, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons.
Resonance Decreases Basicity: The third factor to consider in determining whether or not a species will be a strong or weak base is resonance. As you may remember from general chemistry, the formation of a resonance stabilized structure results in a species that is less willing to share its electrons. Since strong bases, by definition, want to share their electrons, resonance stabilized structures are weak bases.
Weak Bases are the Best Leaving Groups
Now that we understand how electronegativity, size, and resonance affect basicity, we can combine these concepts with the fact that weak bases make the best leaving groups. Think about why this might be true. In order for a leaving group to leave, it must be able to accept electrons. A strong bases wants to donate electrons; therefore, the leaving group must be a weak base. We will now revisit electronegativity, size, and resonance, moving our focus to the leaving group, as well providing actual examples.
Note
As Electronegativity Increases, The Ability of the Leaving Group to Leave Increases.
As mentioned previously, if we move from left to right on the periodic table, electronegativity increases. With an increase in electronegativity, basisity decreases, and the ability of the leaving group to leave increases. This is because an increase in electronegativity results in a species that wants to hold onto its electrons rather than donate them. The following diagram illustrates this concept, showing -CH3 to be the worst leaving group and F- to be the best leaving group. This particular example should only be used to facilitate your understanding of this concept. In real reaction mechanisms, these groups are not good leaving groups at all. For example, fluoride is such a poor leaving group that SN2 reactions of fluoroalkanes are rarely observed.
As Size Increases, The Ability of the Leaving Group to Leave Increases: Here we revisit the effect size has on basicity. If we move down the periodic table, size increases. With an increase in size, basicity decreases, and the ability of the leaving group to leave increases. The relationship among the following halogens, unlike the previous example, is true to what we will see in upcoming reaction mechanisms.
Resonance Increases the Ability of the Leaving Group to Leave: As we learned previously, resonance stabilized structures are weak bases. Therefore, leaving groups that form resonance structures upon leaving are considered to be excellent leaving groups. The following diagram shows sulfur derivatives of the type ROSO3- and RSO3-. Alkyl sulfates and sulfonates like the ones shown make excellent leaving groups. This is due to the formation of a resonance stabilized structure upon leaving. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/SN2/Leaving_Groups.txt |
Four Factors in Affecting SN2 Reactions
1. The nature of the leaving group (SN2 Reactions-The Leaving Group)
2. The reactivity of the nucleophile (SN2 Reactions-The Nucleophile)
3. The solvent (SN2 Reactions-The Nucleophile)
4. The structure of the alkyl portion of the substrate (SN2 Reactions-The Substrate)
The Reactivity of the Nucleophile
Now that we have determined what will make a good leaving group, we will now consider nucleophilicity. That is, the relative strength of the nucleophile. Nucleophilicity depends on many factors, including charge, basicity, solvent, polarizability, and the nature of the substituents.
Increasing the Negative Charge Increases Nucleophilicity
Nucleophiles can be neutral or negatively charged. In either case, it is important that the nucleophile be a good Lewis base, meaning it has electrons it wants to share. The following diagram is just a reminder of some of the nucleophiles that were presented in the section covering nucleophilic substitution. In looking at these two types of nucleophiles, you should notice that a reactive atom, such as oxygen, in a neutral species can also be a reactive atom in a negatively charged species. For example, the O in OH- is negatively charged, but the O in H2O is neutral.
It has been experimentally shown that a nucleophile containing a negatively charged reactive atom is better than a nucleophile containing a reactive atom that is neutral. The next diagram illustrates this concept. Notice that when oxygen is part of the hydroxide ion, it bears a negative charge, and when it is part of a water molecule, it is neutral. The O of -OH is a better nucleophile than the O of H2O, and results in a faster reaction rate. Similarly, when nitrogen is part of NH2, it bears a negative charge, and when it is part of NH3, it is neutral. The N of NH2 is a better nucleophile than the N of NH3, and results in a faster reaction rate.
When Moving Across a Row, Nucleophilicity Follows basicity
To say that nucleophilicity follows basicity across a row means that, as basicity increases from right to left on the periodic table, nucleophilicity also increases. As basicity decreases from left to right on the periodic table, nucleophilicity also decreases. When it comes to nucleophilicity, do not assign this same rule when making comparisons between the halogens located in a column. In this case of moving up and down a column, nucleophilicity does not always follow basicity. It depends on the type of solvent you are using.
In the section Nucleophilic Substitution, we assigned a relationship to leaving groups containing C, N, O, and F, showing that the strength of the leaving group follows electronegativity. This is based on the fact that the best leaving groups are those that are weak bases that do not want to share their electrons. The best nucleophiles however, are good bases that want to share their electrons with the electrophilic carbon. The relationship shown below, therefore, is the exact opposite of that shown for the strength of a leaving group.
Solvents and Nucleophilicity
In general chemistry, we classified solvents as being either polar or nonpolar. Polar solvents can be further subdivided into protic and and aprotic solvents.
Protic Solvents
A protic solvent is a solvent that has a hydrogen atom bound to an oxygen or nitrogen. A few examples of protic solvents include H2O, ROH, RNH2, and R2NH, where water is an example of an inorganic protic solvent, and alcohols and amides are examples of organic solvents. The diagram below shows a few examples of protic solvents we will see.
Since oxygen and nitrogen are highly electronegative atoms, the O-H and N-H bonds that are present in protic solvents result in a hydrogen that is positively polarized. When protic solvents are used in nucleophilic substitution reactions, the positively polarized hydrogen of the solvent molecule can interact with the negatively charged nucleophile. In solution, molecules or ions that are surrounded by these solvent molecules are said to be solvated. Solvation is the process of attraction and association of solvent molecules with ions of a solute. The solute, in this case, is a negatively charged nucleophile.
The following diagram depicts the interaction that can occur between a protic solvent and a negatively charged nucleophile. The interactions are called hydrogen bonds. A hydrogen bond results from a from a dipole-dipole force between between an electronegative atom, such as a halogen, and a hydrogen atom bonded to nitrogen, oxygen or fluorine. In the case below, we are using an alcohol (ROH) as an example of a protic solvent, but be aware that this interaction can occur with other solvents containing a positively polarized hydrogen atom, such as a molecule of water, or amides of the form RNH2 and R2NH.
Why is this important? Solvation weakens the nucleophile; that is, solvation decreases nucleophilicity. This is because the solvent forms a "shell" around the nucleophile, impeding the nucleophile's ability to attack an electrophilic carbon. Furthermore, because the charge on smaller anions is more concentrated, small anions are more tightly solvated than large anions.
The picture below illustrates this concept. Notice how the smaller fluoride anion is represented as being more heavily solvated than the larger iodide anion. This means that the fluoride anion will be a weaker nucleophile than the iodide anion. In fact, it is important to note that fluoride will not function as a nucleophile at all in protic solvents. It is so small that solvation creates a situation whereby fluoride's lone pair of electrons are no longer accessible, meaning it is unable to participate in a nucleophilic substitution reaction.
Previously we learned how nucleophilicity follows basicity when moving across a row. In our discussion on the effect of protic solvents on nucleophilicity, we learned that solvation weakens the nucleophile, having the greatest effect on smaller anions. In effect, when using protic solvents, nucleophilicity does not follow basicity when moving up and down a column. In fact, it's the exact opposite: when basicity decreases, nucleophilicity increases and when basicity increases, nucleophilicity decreases.
Aprotic Solvents
An aprotic solvent is a solvent that lacks a positively polarized hydrogen. The next diagram illustrates several polar aprotic solvents that you should become familiar with.
Aprotic solvents, like protic solvents, are polar but, because they lack a positively polarized hydrogen, they do not form hydrogen bonds with the anionic nucleophile. The result, with respect to solvation, is a relatively weak interaction between the aprotic solvent and the nucleophile.
The consequence of this weakened interaction is two-fold. First, by using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol.
A second consequence that results from the weak interaction that occurs between aprotic solvents and nucleophiles is that, under some conditions, there can be an inversion of the reactivity order. An inversion would result in nucleophilicity following basicity up and down a column, as shown in the following diagram. When we considered the effects of protic solvents, remember that the iodide anion was the strongest nucleophile. Now, in considering aprotic solvents under some conditions, the fluoride anion is the strongest nucelophile.
Increasing Atomic Size Increases Nucleophilicity
Thus far, our discussion on nucleophilicity and solvent choice has been limited to negatively charged nucleophiles, such as F-, Cl-, Br-, and I-. With respect to these anions we learned that, when using protic solvents, nucleophilicity does not follow basicity, and when using aprotic solvents, the same relationship can occur, or there could be an inversion in the order of reactivity.
What happens as we move up and down a column when considering uncharged nucleophiles? It turns out that, in the case of uncharged nucleophiles, size dictates nucleophilicity. This is because larger elements have bigger, more diffuse, and more polarizable electron clouds. This cloud facilitates the formation of a more effective orbital overlap in the transition state of bimolecular nucleophilic substitution (SN2) reactions, resulting in a transition state that is lower in energy and a nucleophilic substitution that occurs at a faster rate.
Sterically Hindered Nucleophiles React More Slowly
In the section Kinetics of Nucleophilic Substitution Reactions, we learned that the SN2 transition state is very crowded. Recall that there are a total of 5 groups around the electrophilic center.
For this reason, sterically hindered nucleophiles react more slowly than those lacking steric bulk.
Next section: SN2 Reactions-The Substrate | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/SN2/Nucleophile.txt |
Frontside vs. Backside Attacks
A biomolecular nucleophilic substitution (SN2) reaction is a type of nucleophilic substitution whereby a lone pair of electrons on a nucleophile attacks an electron deficient electrophilic center and bonds to it, resulting in the expulsion of a leaving group. It is possible for the nucleophile to attack the electrophilic center in two ways.
• Frontside Attack: In a frontside attack, the nucleophile attacks the electrophilic center on the same side as the leaving group. When a frontside attack occurs, the stereochemistry of the product remains the same; that is, we have retention of configuration.
• Backside Attack: In a backside attack, the nucleophile attacks the electrophilic center on the side that is opposite to the leaving group. When a backside attack occurs, the stereochemistry of the product does not stay the same. There is inversion of configuration.
The following diagram illustrates these two types of nucleophilic attacks, where the frontside attack results in retention of configuration; that is, the product has the same configuration as the substrate. The backside attack results in inversion of configuration, where the product's configuration is opposite that of the substrate.
Experimental Observation: All SN2 Reactions Proceed With Nucleophilic Backside Attacks
Experimental observation shows that all SN2 reactions proceed with inversion of configuration; that is, the nucleophile will always attack from the backside in all SN2 reactions. To think about why this might be true, remember that the nucleophile has a lone pair of electrons to be shared with the electrophilic center, and the leaving group is going to take a lone pair of electrons with it upon leaving. Because like charges repel each other, the nucleophile will always proceed by a backside displacement mechanism.
SN2 Reactions Are Stereospecific
The SN2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. For example, if the substrate is an R enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the R enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the S enantiomer.
Conversely, if the substrate is an S enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the S enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the R enantiomer.
In conclusion, SN2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are cis and substrates that are trans. If the cis configuration is the substrate, the resulting product will be trans. Conversely, if the trans configuration is the substrate, the resulting product will be cis.
Next section: SN2 Reactions-The Leaving Group
Substrate
Four Factors to Consider in Determining the Relative Ease at Which SN2 Displacement Occurs
• The nature of the leaving group (SN2 Reactions-The Leaving Group)
• The reactivity of the nucleophile (SN2 Reactions-The Nucleophile)
• The solvent (SN2 Reactions-The Nucleophile)
• The structure of the alkyl portion of the substrate (SN2 Reactions-The Substrate)
Sterically Hindered Substrates Will Reduce the SN2 Reaction Rate
Now that we have discussed the effects that the leaving group, nucleophile, and solvent have on biomolecular nucleophilic substitution (SN2) reactions, it's time to turn our attention to how the substrate affects the reaction. Although the substrate, in the case of nucleophilic substitution of haloalkanes, is considered to be the entire molecule circled below, we will be paying particular attention to the alkyl portion of the substrate. In other words, we are most interested in the electrophilic center that bears the leaving group.
In the section Kinetics of Nucleophilic Substitution Reactions, we learned that the SN2 transition state is very crowded. Recall that there are a total of five groups around the electrophilic center, the nucleophile, the leaving group, and three substituents.
If each of the three substituents in this transition state were small hydrogen atoms, as illustrated in the first example below, there would be little steric repulsion between the incoming nucleophile and the electrophilic center, thereby increasing the ease at which the nucleophilic substitution reaction can occur. Remember, for the SN2 reaction to occur, the nucleophile must be able to attack the electrophilic center, resulting in the expulsion of the leaving group. If one of the hydrogens, however, were replaced with an R group, such as a methyl or ethyl group, there would be an increase in steric repulsion with the incoming nucleophile. If two of the hydrogens were replaced by R groups, there would be an even greater increase in steric repulsion with the incoming nucleophile.
How does steric hindrance affect the rate at which an SN2 reaction will occur? As each hydrogen is replaced by an R group, the rate of reaction is significantly diminished. This is because the addition of one or two R groups shields the backside of the electrophilic carbon, impeding nucleophilic attack.
The diagram below illustrates this concept, showing that electrophilic carbons attached to three hydrogen atoms results in faster nucleophilic substitution reactions, in comparison to primary and secondary haloalkanes, which result in nucleophilic substitution reactions that occur at slower or much slower rates, respectively. Notice that a tertiary haloalkane, that which has three R groups attached, does not undergo nucleophilic substitution reactions at all. The addition of a third R group to this molecule creates a carbon that is entirely blocked.
Substitutes on Neighboring Carbons Slow Nucleophilic Substitution Reactions
Previously we learned that adding R groups to the electrophilic carbon results in nucleophilic substitution reactions that occur at a slower rate. What if R groups are added to neighboring carbons? It turns out that the addition of substitutes on neighboring carbons will slow nucleophilic substitution reactions as well.
In the example below, 2-methyl-1-bromopropane differs from 1-bromopropane in that it has a methyl group attached to the carbon that neighbors the electrophilic carbon. The addition of this methyl group results in a significant decrease in the rate of a nucleophilic substitution reaction.
If R groups were added to carbons farther away from the electrophilic carbon, we would still see a decrease in the reaction rate. However, branching at carbons farther away from the electrophilic carbon would have a much smaller effect. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/SN2/Sterochemistry.txt |
The light our eyes see is but a small part of a broad spectrum of electromagnetic radiation. On the immediate high energy side of the visible spectrum lies the ultraviolet, and on the low energy side is the infrared. The portion of the infrared region most useful for analysis of organic compounds is not immediately adjacent to the visible spectrum, but is that having a wavelength range from 2,500 to 16,000 nm, with a corresponding frequency range from 1.9*1013 to 1.2*1014 Hz.
Introduction
Photon energies associated with this part of the infrared (from 1 to 15 kcal/mole) are not large enough to excite electrons, but may induce vibrational excitation of covalently bonded atoms and groups.
The covalent bonds in molecules are not rigid sticks or rods, such as found in molecular model kits, but are more like stiff springs that can be stretched and bent. The mobile nature of organic molecules was noted in the chapter concerning conformational isomers. We must now recognize that, in addition to the facile rotation of groups about single bonds, molecules experience a wide variety of vibrational motions, characteristic of their component atoms. Consequently, virtually all organic compounds will absorb infrared radiation that corresponds in energy to these vibrations. Infrared spectrometers, similar in principle to the UV-Visible spectrometer described elsewhere, permit chemists to obtain absorption spectra of compounds that are a unique reflection of their molecular structure. An example of such a spectrum is that of the flavoring agent vanillin, shown below.
The complexity of this spectrum is typical of most infrared spectra, and illustrates their use in identifying substances. The gap in the spectrum between 700 & 800 cm-1 is due to solvent (CCl4) absorption. Further analysis (below) will show that this spectrum also indicates the presence of an aldehyde function, a phenolic hydroxyl and a substituted benzene ring. The inverted display of absorption, compared with UV-Visible spectra, is characteristic. Thus a sample that did not absorb at all would record a horizontal line at 100% transmittance (top of the chart).
The frequency scale at the bottom of the chart is given in units of reciprocal centimeters (cm-1) rather than Hz, because the numbers are more manageable. The reciprocal centimeter is the number of wave cycles in one centimeter; whereas, frequency in cycles per second or Hz is equal to the number of wave cycles in 3*1010 cm (the distance covered by light in one second). Wavelength units are in micrometers, microns (μ), instead of nanometers for the same reason. Most infrared spectra are displayed on a linear frequency scale, as shown here, but in some older texts a linear wavelength scale is used. A calculator for interconverting these frequency and wavelength values is provided on the right. Simply enter the value to be converted in the appropriate box, press "Calculate" and the equivalent number will appear in the empty box.
Infrared spectra may be obtained from samples in all phases (liquid, solid and gaseous). Liquids are usually examined as a thin film sandwiched between two polished salt plates (note that glass absorbs infrared radiation, whereas NaCl is transparent). If solvents are used to dissolve solids, care must be taken to avoid obscuring important spectral regions by solvent absorption. Perchlorinated solvents such as carbon tetrachloride, chloroform and tetrachloroethene are commonly used. Alternatively, solids may either be incorporated in a thin KBr disk, prepared under high pressure, or mixed with a little non-volatile liquid and ground to a paste (or mull) that is smeared between salt plates.
Frequency - Wavelength Converter
Frequency in cm-1
Wavelength in μ
Vibrational Spectroscopy
A molecule composed of n-atoms has 3n degrees of freedom, six of which are translations and rotations of the molecule itself. This leaves 3n-6 degrees of vibrational freedom (3n-5 if the molecule is linear). Vibrational modes are often given descriptive names, such as stretching, bending, scissoring, rocking and twisting. The four-atom molecule of formaldehyde, the gas phase spectrum of which is shown below, provides an example of these terms. If a ball & stick model of formaldehyde is not displayed to the right of the spectrum, press the view ball&stick model button on the right. We expect six fundamental vibrations (12 minus 6), and these have been assigned to the spectrum absorptions. To see the formaldehyde molecule display a vibration, click one of the buttons under the spectrum, or click on one of the absorption peaks in the spectrum.
Gas Phase Infrared Spectrum of Formaldehyde, H2C=O
Click Here. In practice, infrared spectra do not normally display separate absorption signals for each of the 3n-6 fundamental vibrational modes of a molecule. The number of observed absorptions may be increased by additive and subtractive interactions leading to combination tones and overtones of the fundamental vibrations, in much the same way that sound vibrations from a musical instrument interact. Furthermore, the number of observed absorptions may be decreased by molecular symmetry, spectrometer limitations, and spectroscopic selection rules. One selection rule that influences the intensity of infrared absorptions, is that a change in dipole moment should occur for a vibration to absorb infrared energy. Absorption bands associated with C=O bond stretching are usually very strong because a large change in the dipole takes place in that mode.
Some General Trends:
1. Stretching frequencies are higher than corresponding bending frequencies. (It is easier to bend a bond than to stretch or compress it.)
2. Bonds to hydrogen have higher stretching frequencies than those to heavier atoms.
3. Triple bonds have higher stretching frequencies than corresponding double bonds, which in turn have higher frequencies than single bonds.(Except for bonds to hydrogen).
The general regions of the infrared spectrum in which various kinds of vibrational bands are observed are outlined in the following chart. Note that the blue colored sections above the dashed line refer to stretching vibrations, and the green colored band below the line encompasses bending vibrations. The complexity of infrared spectra in the 1450 to 600 cm-1 region makes it difficult to assign all the absorption bands, and because of the unique patterns found there, it is often called the fingerprint region. Absorption bands in the 4000 to 1450 cm-1 region are usually due to stretching vibrations of diatomic units, and this is sometimes called the group frequency region.
Group Frequencies
Detailed information about the infrared absorptions observed for various bonded atoms and groups is usually presented in tabular form. The following table provides a collection of such data for the most common functional groups. Following the color scheme of the chart, stretching absorptions are listed in the blue-shaded section and bending absorptions in the green shaded part. More detailed descriptions for certain groups (e.g. alkenes, arenes, alcohols, amines & carbonyl compounds) may be viewed by clicking on the functional class name. Since most organic compounds have C-H bonds, a useful rule is that absorption in the 2850 to 3000 cm-1 is due to sp3 C-H stretching; whereas, absorption above 3000 cm-1 is from sp2 C-H stretching or sp C-H stretching if it is near 3300 cm-1.
Typical Infrared Absorption Frequencies
Stretching Vibrations
Bending Vibrations
Functional Class
Range (cm-1)
Intensity
Assignment
Range (cm-1)
Intensity
Assignment
Alkanes
2850-3000 str CH3, CH2 & CH
2 or 3 bands
1350-1470
1370-1390
720-725
med
med
wk
CH2 & CH3 deformation
CH3 deformation
CH2 rocking
Alkenes
3020-3100
1630-1680
1900-2000
med
var
str
=C-H & =CH2 (usually sharp)
C=C (symmetry reduces intensity)
C=C asymmetric stretch
880-995
780-850
675-730
str
med
med
=C-H & =CH2
(out-of-plane bending)
cis-RCH=CHR
Alkynes
3300
2100-2250
str
var
C-H (usually sharp)
C≡C (symmetry reduces intensity)
600-700 str C-H deformation
Arenes
3030
1600 & 1500
var
med-wk
C-H (may be several bands)
C=C (in ring) (2 bands)
(3 if conjugated)
690-900 str-med C-H bending &
ring puckering
Alcohols & Phenols
3580-3650
3200-3550
970-1250
var
str
str
O-H (free), usually sharp
O-H (H-bonded), usually broad
C-O
1330-1430
650-770
med
var-wk
O-H bending (in-plane)
O-H bend (out-of-plane)
Amines
3400-3500 (dil. soln.)
3300-3400 (dil. soln.)
1000-1250
wk
wk
med
N-H (1°-amines), 2 bands
N-H (2°-amines)
C-N
1550-1650
660-900
med-str
var
NH2 scissoring (1°-amines)
NH2 & N-H wagging
(shifts on H-bonding)
Aldehydes & Ketones
2690-2840(2 bands)
1720-1740
1710-1720
med
str
str
str
str
str
str
C-H (aldehyde C-H)
C=O (saturated aldehyde)
C=O (saturated ketone)
aryl ketone
α, β-unsaturation
cyclopentanone
cyclobutanone
1350-1360
1400-1450
1100
str
str
med
α-CH3 bending
α-CH2 bending
C-C-C bending
Carboxylic Acids & Derivatives
2500-3300 (acids) overlap C-H
1705-1720 (acids)
1210-1320 (acids)
str
str
med-str
str
str
str
str
str
str
O-H (very broad)
C=O (H-bonded)
O-C (sometimes 2-peaks)
C=O
C=O (2-bands)
O-C
C=O
O-C (2-bands)
C=O (amide I band)
1395-1440
1590-1650
1500-1560
med
med
med
C-O-H bending
N-H (1°-amide) II band
N-H (2°-amide) II band
Nitriles
Isocyanates,Isothiocyanates,
Diimides, Azides & Ketenes
2240-2260
2100-2270
med
med
C≡N (sharp)
-N=C=O, -N=C=S
-N=C=N-, -N3, C=C=O
To illustrate the usefulness of infrared absorption spectra, examples for five C4H8O isomers are presented below their corresponding structural formulas. Try to associate each spectrum (A - E) with one of the isomers in the row above it.Answers
Other Functional Groups
Infrared absorption data for some functional groups not listed in the preceding table are given below. Most of the absorptions cited are associated with stretching vibrations. Standard abbreviations (str = strong, wk = weak, brd = broad & shp = sharp) are used to describe the absorption bands.
Functional Class
Characteristic Absorptions
Sulfur Functions
S-H thiols 2550-2600 cm-1 (wk & shp)
S-OR esters 700-900 (str)
S-S disulfide 500-540 (wk)
C=S thiocarbonyl 1050-1200 (str)
S=O sulfoxide
1030-1060 (str)
1325± 25 (as) & 1140± 20 (s) (both str)
1345 (str)
1365± 5 (as) & 1180± 10 (s) (both str)
1350-1450 (str)
Phosphorous Functions
P-H phosphine 2280-2440 cm-1 (med & shp)
950-1250 (wk) P-H bending
(O=)PO-H phosphonic acid 2550-2700 (med)
P-OR esters 900-1050 (str)
P=O phosphine oxide
1100-1200 (str)
1230-1260 (str)
1100-1200 (str)
1200-1275 (str)
Silicon Functions
Si-H silane 2100-2360 cm-1 (str)
Si-OR 1000-1110 (str & brd)
Si-CH3 1250± 10 (str & shp)
Oxidized Nitrogen Functions
=NOH oxime
3550-3600 cm-1 (str)
1665± 15
945± 15
N-O amine oxide
960± 20
1250± 50
N=O nitroso
1550± 50 (str)
1530± 20 (as) & 1350± 30 (s)
Internal Links
• Organic Chemistry With a Biological Emphasis
Infrared Spectroscopy
The covalent bonds in molecules are not rigid sticks or rods, such as found in molecular model kits, but are more like stiff springs that can be stretched and bent. In addition to the facile rotation of groups about single bonds, molecules experience a wide variety of vibrational motions, characteristic of their component atoms. Consequently, virtually all organic compounds will absorb infrared radiation that corresponds in energy to these vibrations. Infrared spectrometers, similar in principle to the UV-Visible spectrometer, permit chemists to obtain absorption spectra of compounds that are a unique reflection of their molecular structure.
Below are some characteristic vibrational modes of organic molecules:
The energy of molecular vibration is quantized rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the amplitude of the vibration will increase, but the vibrational frequency will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the infrared region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds.
Let's take 2-hexanone as an example. Picture the carbonyl bond of the ketone group as a spring. This spring is constantly bouncing back and forth, stretching and compressing, pushing the carbon and oxygen atoms further apart and then pulling them together. This is the stretching mode of the carbonyl bond. In the space of one second, the spring 'bounces' back and forth 5.15 x 1013 times - in other words, the ground-state frequency of carbonyl stretching for a the ketone group is about 5.15 x 1013 Hz.
If our ketone sample is irradiated with infrared light, the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10-6 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state.
The value of ΔE - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state.
With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 1013 to 1014 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side.
Our 5.15 x 1013 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity.
The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light.
The power of infrared spectroscopy arises from the observation that different functional groups have different characteristic absorption frequencies. The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 1013 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 1013 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne.
Some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. In order for a vibrational mode to absorb infrared light, it must result in a periodic change in the dipole moment of the molecule. Such vibrations are said to be infrared active. In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls.
Some kinds of vibrations are infrared inactive. The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules do absorb IR light).
Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone.
There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called wavenumber (cm-1), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘% transmittance’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you!
Template:ExampleStart
Express the wavenumber value of 3000 cm-1 in terms of wavelength (in meter units). Template:ExampleEnd
The key absorption peak in this spectrum is that from the carbonyl double bond, at 1716 cm-1 (corresponding to a wavelength of 5.86 mm, a frequency of 5.15 x 1013 Hz, and a ΔE value of 4.91 kcal/mol). Notice how strong this peak is, relative to the others on the spectrum: a strong peak in the 1650-1750 cm-1 region is a dead giveaway for the presence of a carbonyl group. Within that range, carboxylic acids, esters, ketones, and aldehydes tend to absorb in the shorter wavelength end (1700-1750 cm-1), while conjugated unsaturated ketones and amides tend to absorb on the longer wavelength end (1650-1700 cm-1).
The jagged peak at approximately 2900-3000 cm-1 is characteristic of tetrahedral carbon-hydrogen bonds. This peak is not terribly useful, as just about every organic molecule that you will have occasion to analyze has these bonds. Nevertheless, it can serve as a familiar reference point to orient yourself in a spectrum.
You will notice that there are many additional peaks in this spectrum in the longer-wavelength 400 -1400 cm-1 region. This part of the spectrum is called the fingerprint region. While it is usually very difficult to pick out any specific functional group identifications from this region, it does, nevertheless, contain valuable information. The reason for this is suggested by the name: just like a human fingerprint, the pattern of absorbance peaks in the fingerprint region is unique to every molecule, meaning that the data from an unknown sample can be compared to the IR spectra of known standards in order to make a positive identification. In the mid-1990's, for example, several paintings were identified as forgeries because scientists were able to identify the IR footprint region of red and yellow pigment compounds that would not have been available to the artist who supposedly created the painting (for more details see Chemical and Engineering News, Sept 10, 2007, p. 28).
Now, let’s take a look at the IR spectrum for 1-hexanol.
As you can see, the carbonyl peak is gone, and in its place is a very broad ‘mountain’ centered at about 3400 cm-1. This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules.
In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm-1.
We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance.
The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm-1 is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens.
Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm-1 due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen.
It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in this table.
As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol.
More examples of IR spectra
To illustrate the usefulness of infrared absorption spectra, examples for five C4H8O isomers are presented below their corresponding structural formulas. Try to associate each spectrum with one of the isomers in the row above it.
Contributors
• Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris)
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Infrared_Spectroscopy/Infrared_spectroscopy_2.txt |
The Mass Spectrometer
In order to measure the characteristics of individual molecules, a mass spectrometer converts them to ions so that they can be moved about and manipulated by external electric and magnetic fields. The three essential functions of a mass spectrometer, and the associated components, are:
1. A small sample is ionized, usually to cations by loss of an electron. The Ion Source
2. The ions are sorted and separated according to their mass and charge. The Mass Analyzer
3. The separated ions are then measured, and the results displayed on a chart. The Detector
Because ions are very reactive and short-lived, their formation and manipulation must be conducted in a vacuum. Atmospheric pressure is around 760 torr (mm of mercury). The pressure under which ions may be handled is roughly 10-5 to 10-8 torr (less than a billionth of an atmosphere). Each of the three tasks listed above may be accomplished in different ways. In one common procedure, ionization is effected by a high energy beam of electrons, and ion separation is achieved by accelerating and focusing the ions in a beam, which is then bent by an external magnetic field. The ions are then detected electronically and the resulting information is stored and analyzed in a computer. A mass spectrometer operating in this fashion is outlined in the following diagram. The heart of the spectrometer is the ion source. Here molecules of the sample (black dots) are bombarded by electrons (light blue lines) issuing from a heated filament. This is called an EI (electron-impact) source. Gases and volatile liquid samples are allowed to leak into the ion source from a reservoir (as shown). Non-volatile solids and liquids may be introduced directly. Cations formed by the electron bombardment (red dots) are pushed away by a charged repellor plate (anions are attracted to it), and accelerated toward other electrodes, having slits through which the ions pass as a beam. Some of these ions fragment into smaller cations and neutral fragments. A perpendicular magnetic field deflects the ion beam in an arc whose radius is inversely proportional to the mass of each ion. Lighter ions are deflected more than heavier ions. By varying the strength of the magnetic field, ions of different mass can be focused progressively on a detector fixed at the end of a curved tube (also under a high vacuum).
When a high energy electron collides with a molecule it often ionizes it by knocking away one of the molecular electrons (either bonding or non-bonding). This leaves behind a molecular ion (colored red in the following diagram). Residual energy from the collision may cause the molecular ion to fragment into neutral pieces (colored green) and smaller fragment ions (colored pink and orange). The molecular ion is a radical cation, but the fragment ions may either be radical cations (pink) or carbocations (orange), depending on the nature of the neutral fragment. An animated display of this ionization process will appear if you click on the ion source of the mass spectrometer diagram.
The Nature of Mass Spectra
A mass spectrum will usually be presented as a vertical bar graph, in which each bar represents an ion having a specific mass-to-charge ratio (m/z) and the length of the bar indicates the relative abundance of the ion. The most intense ion is assigned an abundance of 100, and it is referred to as the base peak. Most of the ions formed in a mass spectrometer have a single charge, so the m/z value is equivalent to mass itself. Modern mass spectrometers easily distinguish (resolve) ions differing by only a single atomic mass unit, and thus provide completely accurate values for the molecular mass of a compound. The highest-mass ion in a spectrum is normally considered to be the molecular ion, and lower-mass ions are fragments from the molecular ion, assuming the sample is a single pure compound.
Atomic mass is given in terms of the unified atomic mass unit (symbol: μ) or dalton (symbol: Da). In recent years there has been a gradual change towards using the dalton in preference to the unified atomic mass unit. The dalton is classified as a "non-SI unit whose values in SI units must be obtained experimentally". It is defined as one twelfth of the rest mass of an unbound atom of carbon-12 in its nuclear and electronic ground state, and has a value of 1.660538782(83)x10-27 kg.
The following diagram displays the mass spectra of three simple gaseous compounds, carbon dioxide, propane and cyclopropane. The molecules of these compounds are similar in size, CO2 and C3H8 both have a nominal mass of 44 Da, and C3H6 has a mass of 42 Da. The molecular ion is the strongest ion in the spectra of CO2 and C3H6, and it is moderately strong in propane. The unit mass resolution is readily apparent in these spectra (note the separation of ions having m/z=39, 40, 41 and 42 in the cyclopropane spectrum). Even though these compounds are very similar in size, it is a simple matter to identify them from their individual mass spectra. By clicking on each spectrum in turn, a partial fragmentation analysis and peak assignment will be displayed. Even with simple compounds like these, it should be noted that it is rarely possible to explain the origin of all the fragment ions in a spectrum. Also, the structure of most fragment ions is seldom known with certainty.
Since a molecule of carbon dioxide is composed of only three atoms, its mass spectrum is very simple. The molecular ion is also the base peak, and the only fragment ions are CO (m/z=28) and O (m/z=16). The molecular ion of propane also has m/z=44, but it is not the most abundant ion in the spectrum. Cleavage of a carbon-carbon bond gives methyl and ethyl fragments, one of which is a carbocation and the other a radical. Both distributions are observed, but the larger ethyl cation (m/z=29) is the most abundant, possibly because its size affords greater charge dispersal. A similar bond cleavage in cyclopropane does not give two fragments, so the molecular ion is stronger than in propane, and is in fact responsible for the the base peak. Loss of a hydrogen atom, either before or after ring opening, produces the stable allyl cation (m/z=41). The third strongest ion in the spectrum has m/z=39 (C3H3). Its structure is uncertain, but two possibilities are shown in the diagram. The small m/z=39 ion in propane and the absence of a m/z=29 ion in cyclopropane are particularly significant in distinguishing these hydrocarbons.
Most stable organic compounds have an even number of total electrons, reflecting the fact that electrons occupy atomic and molecular orbitals in pairs. When a single electron is removed from a molecule to give an ion, the total electron count becomes an odd number, and we refer to such ions as radical cations. The molecular ion in a mass spectrum is always a radical cation, but the fragment ions may either be even-electron cations or odd-electron radical cations, depending on the neutral fragment lost. The simplest and most common fragmentations are bond cleavages producing a neutral radical (odd number of electrons) and a cation having an even number of electrons. A less common fragmentation, in which an even-electron neutral fragment is lost, produces an odd-electron radical cation fragment ion. Fragment ions themselves may fragment further. As a rule, odd-electron ions may fragment either to odd or even-electron ions, but even-electron ions fragment only to other even-electron ions. The masses of molecular and fragment ions also reflect the electron count, depending on the number of nitrogen atoms in the species.
Ions with no nitrogen
or an even # N atoms
odd-electron ions
even-number mass
even-electron ions
odd-number mass
Ions having an
odd # N atoms
odd-electron ions
odd-number mass
even-electron ions
even-number mass
This distinction is illustrated nicely by the follwing two examples. The unsaturated ketone, 4-methyl-3-pentene-2-one, on the left has no nitrogen so the mass of the molecular ion (m/z = 98) is an even number. Most of the fragment ions have odd-numbered masses, and therefore are even-electron cations. Diethylmethylamine, on the other hand, has one nitrogen and its molecular mass (m/z = 87) is an odd number. A majority of the fragment ions have even-numbered masses (ions at m/z = 30, 42, 56 & 58 are not labeled), and are even-electron nitrogen cations. The weak even -electron ions at m/z=15 and 29 are due to methyl and ethyl cations (no nitrogen atoms). The fragmentations leading to the chief fragment ions will be displayed by clicking on the appropriate spectrum. Repeated clicks will cycle the display.
4-methyl-3-pentene-2-one
N,N-diethylmethylamine
When non-bonded electron pairs are present in a molecule (e.g. on N or O), fragmentation pathways may sometimes be explained by assuming the missing electron is partially localized on that atom. A few such mechanisms are shown above. Bond cleavage generates a radical and a cation, and both fragments often share these roles, albeit unequally.
Isotopes
Since a mass spectrometer separates and detects ions of slightly different masses, it easily distinguishes different isotopes of a given element. This is manifested most dramatically for compounds containing bromine and chlorine, as illustrated by the following examples. Since molecules of bromine have only two atoms, the spectrum on the left will come as a surprise if a single atomic mass of 80 Da is assumed for Br. The five peaks in this spectrum demonstrate clearly that natural bromine consists of a nearly 50:50 mixture of isotopes having atomic masses of 79 and 81 Da respectively. Thus, the bromine molecule may be composed of two 79Br atoms (mass 158 Da), two 81Br atoms (mass 162 Da) or the more probable combination of 79Br-81Br (mass 160 Da). Fragmentation of Br2 to a bromine cation then gives rise to equal sized ion peaks at 79 and 81 Da.
bromine
vinyl chloride
methylene chloride
The center and right hand spectra show that chlorine is also composed of two isotopes, the more abundant having a mass of 35 Da, and the minor isotope a mass 37 Da. The precise isotopic composition of chlorine and bromine is:
• Chlorine: 75.77% 35Cl and 24.23% 37Cl
• Bromine: 50.50% 79Br and 49.50% 81Br
The presence of chlorine or bromine in a molecule or ion is easily detected by noticing the intensity ratios of ions differing by 2 Da. In the case of methylene chloride, the molecular ion consists of three peaks at m/z=84, 86 & 88 Da, and their diminishing intensities may be calculated from the natural abundances given above. Loss of a chlorine atom gives two isotopic fragment ions at m/z=49 & 51 Da, clearly incorporating a single chlorine atom. Fluorine and iodine, by contrast, are monoisotopic, having masses of 19 Da and 127 Da respectively. It should be noted that the presence of halogen atoms in a molecule or fragment ion does not change the odd-even mass rules given above.
To make use of a calculator that predicts the isotope clusters for different combinations of chlorine, bromine and other elements Click Here. This application was developed at Colby College.
Isotopic Abundance Calculator
C H N O SSi
Molecular Ion
100%
M + 1
M + 2
Two other common elements having useful isotope signatures are carbon, 13C is 1.1% natural abundance, and sulfur, 33S and 34S are 0.76% and 4.22% natural abundance respectively. For example, the small m/z=99 Da peak in the spectrum of 4-methyl-3-pentene-2-one (above) is due to the presence of a single 13C atom in the molecular ion. Although less important in this respect, 15N and 18O also make small contributions to higher mass satellites of molecular ions incorporating these elements.
The calculator on the right may be used to calculate the isotope contributions to ion abundances 1 and 2 Da greater than the molecular ion (M). Simply enter an appropriate subscript number to the right of each symbol, leaving those elements not present blank, and press the "Calculate" button. The numbers displayed in the M+1 and M+2 boxes are relative to M being set at 100%.
Fragmentation Patterns
The fragmentation of molecular ions into an assortment of fragment ions is a mixed blessing. The nature of the fragments often provides a clue to the molecular structure, but if the molecular ion has a lifetime of less than a few microseconds it will not survive long enough to be observed. Without a molecular ion peak as a reference, the difficulty of interpreting a mass spectrum increases markedly. Fortunately, most organic compounds give mass spectra that include a molecular ion, and those that do not often respond successfully to the use of milder ionization conditions. Among simple organic compounds, the most stable molecular ions are those from aromatic rings, other conjugated pi-electron systems and cycloalkanes. Alcohols, ethers and highly branched alkanes generally show the greatest tendency toward fragmentation.
The mass spectrum of dodecane on the right illustrates the behavior of an unbranched alkane. Since there are no heteroatoms in this molecule, there are no non-bonding valence shell electrons. Consequently, the radical cation character of the molecular ion (m/z = 170) is delocalized over all the covalent bonds. Fragmentation of C-C bonds occurs because they are usually weaker than C-H bonds, and this produces a mixture of alkyl radicals and alkyl carbocations. The positive charge commonly resides on the smaller fragment, so we see a homologous series of hexyl (m/z = 85), pentyl (m/z = 71), butyl (m/z = 57), propyl (m/z = 43), ethyl (m/z = 29) and methyl (m/z = 15) cations. These are accompanied by a set of corresponding alkenyl carbocations (e.g. m/z = 55, 41 &27) formed by loss of 2 H. All of the significant fragment ions in this spectrum are even-electron ions. In most alkane spectra the propyl and butyl ions are the most abundant.
The presence of a functional group, particularly one having a heteroatom Y with non-bonding valence electrons (Y = N, O, S, X etc.), can dramatically alter the fragmentation pattern of a compound. This influence is thought to occur because of a "localization" of the radical cation component of the molecular ion on the heteroatom. After all, it is easier to remove (ionize) a non-bonding electron than one that is part of a covalent bond. By localizing the reactive moiety, certain fragmentation processes will be favored. These are summarized in the following diagram, where the green shaded box at the top displays examples of such "localized" molecular ions. The first two fragmentation paths lead to even-electron ions, and the elimination (path #3) gives an odd-electron ion. Note the use of different curved arrows to show single electron shifts compared with electron pair shifts.
The charge distributions shown above are common, but for each cleavage process the charge may sometimes be carried by the other (neutral) species, and both fragment ions are observed. Of the three cleavage reactions described here, the alpha-cleavage is generally favored for nitrogen, oxygen and sulfur compounds. Indeed, in the previously displayed spectra of 4-methyl-3-pentene-2-one and N,N-diethylmethylamine the major fragment ions come from alpha-cleavages. Further examples of functional group influence on fragmentation are provided by a selection of compounds that may be examined by clicking the left button below. Useful tables of common fragment ions and neutral species may be viewed by clicking the right button.
Assorted Mass Spectra
View Fragment Tables
The complexity of fragmentation patterns has led to mass spectra being used as "fingerprints" for identifying compounds. Environmental pollutants, pesticide residues on food, and controlled substance identification are but a few examples of this application. Extremely small samples of an unknown substance (a microgram or less) are sufficient for such analysis. The following mass spectrum of cocaine demonstrates how a forensic laboratory might determine the nature of an unknown street drug. Even though extensive fragmentation has occurred, many of the more abundant ions (identified by magenta numbers) can be rationalized by the three mechanisms shown above. Plausible assignments may be seen by clicking on the spectrum, and it should be noted that all are even-electron ions. The m/z = 42 ion might be any or all of the following: C3H6, C2H2O or C2H4N. A precise assignment could be made from a high-resolution m/z value (next section).
Odd-electron fragment ions are often formed by characteristic rearrangements in which stable neutral fragments are lost. Mechanisms for some of these rearrangements have been identified by following the course of isotopically labeled molecular ions. A few examples of these rearrangement mechanisms may be seen by clicking the following button.
Assorted Rearrangement Fragmentations
High Resolution Mass Spectrometry
In assigning mass values to atoms and molecules, we have assumed integral values for isotopic masses. However, accurate measurements show that this is not strictly true. Because the strong nuclear forces that bind the components of an atomic nucleus together vary, the actual mass of a given isotope deviates from its nominal integer by a small but characteristic amount (remember E = mc2). Thus, relative to 12C at 12.0000, the isotopic mass of 16O is 15.9949 Da (not 16) and 14N is 14.0031 Da (not 14).
Formula
C6H12 C5H8O C4H8N2
Mass
84.0939 84.0575 84.0688
By designing mass spectrometers that can determine m/z values accurately to four decimal places, it is possible to distinguish different formulas having the same nominal mass. The table on the right illustrates this important feature, and a double-focusing high-resolution mass spectrometer easily distinguishes ions having these compositions. Mass spectrometry therefore not only provides a specific molecular mass value, but it may also establish the molecular formula of an unknown compound.
Tables of precise mass values for any molecule or ion are available in libraries; however, the mass calculator provided below serves the same purpose. Since a given nominal mass may correspond to several molecular formulas, lists of such possibilities are especially useful when evaluating the spectrum of an unknown compound. Composition tables are available for this purpose, and a particularly useful program for calculating all possible combinations of H, C, N & O that give a specific nominal mass has been written by Jef Rozenski.
Mass Spectrometry
In mass spectrometry (MS), we are interested in the mass - and therefore the molecular weight - of our compound of interest, and often the mass of fragments that are produced when the molecule is caused to break apart. There are many different types of MS instruments, but they all have the same three essential components. First, there is an ionization source, where the molecule is given a positive electrical charge, either by removing an electron or by adding a proton. Depending on the ionization method used, the ionized molecule may or may not break apart into a population of smaller fragments. In the figure below, some of the sample molecules remain whole, while others fragment into smaller pieces.
Next in line there is a mass analyzer, where the cationic fragments are separated according to their mass. Finally, there is a detector, which detects and quantifies the separated ions.
One of the more common types of MS techniques used in the organic laboratory is electron ionization. In the ionization source, the sample molecule is bombarded by a high-energy electron beam, which has the effect of knocking a valence electron off of the molecule to form a radical cation. Because a great deal of energy is transferred by this bombardment process, the radical cation quickly begins to break up into smaller fragments, some of which are positively charged and some of which are neutral. The neutral fragments are either adsorbed onto the walls of the chamber or are removed by a vacuum source. In the mass analyzer component, the positively charged fragments and any remaining unfragmented molecular ions are accelerated down a tube by an electric field. This tube is curved, and the ions are deflected by a strong magnetic field. Ions of different mass to charge (m/z) ratios are deflected to a different extent, resulting in a ‘sorting’ of ions by mass (virtually all ions have charges of z = +1, so sorting by the mass to charge ratio is the same thing as sorting by mass). A detector at the end of the curved flight tube records and quantifies the sorted ions.
Below is typical output for an electron-ionization MS experiment (MS data below is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan.
The sample is acetone. On the horizontal axis is the value for m/z (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the base peak, is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak corresponds to a fragment with m/z = 43 - . The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the molecular ion peak, or parent peak. Notice that there is a small peak at m/z = 59: this is referred to as the M+1 peak. How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the 13C rather than the 12C isotope. The 13C isotope is, of course, heavier than 12C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the 2H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a 13C or 2H.
The fragmentation of molecular ions into an assortment of fragment ions is a mixed blessing. The nature of the fragments often provides a clue to the molecular structure, but if the molecular ion has a lifetime of less than a few microseconds it will not survive long enough to be observed. Without a molecular ion peak as a reference, the difficulty of interpreting a mass spectrum increases markedly. Fortunately, most organic compounds give mass spectra that include a molecular ion, and those that do not often respond successfully to the use of milder ionization conditions. Among simple organic compounds, the most stable molecular ions are those from aromatic rings, other conjugated pi-electron systems and . Alcohols, ethers and highly branched generally show the greatest tendency toward fragmentation.
The mass spectrum of dodecane illustrates the behavior of an unbranched alkane. Since there are no heteroatoms in this molecule, there are no non-bonding valence shell electrons. Consequently, the radical cation character of the molecular ion (m/z = 170) is delocalized over all the covalent bonds. Fragmentation of C-C bonds occurs because they are usually weaker than C-H bonds, and this produces a mixture of alkyl radicals and alkyl carbocations. The positive charge commonly resides on the smaller fragment, so we see a homologous series of hexyl (m/z = 85), pentyl (m/z = 71), butyl (m/z = 57), propyl (m/z = 43), ethyl (m/z = 29) and methyl (m/z = 15) cations. These are accompanied by a set of corresponding alkenyl carbocations (e.g. m/z = 55, 41 &27) formed by loss of 2 H. All of the significant fragment ions in this spectrum are even-electron ions. In most alkane spectra the propyl and butyl ions are the most abundant.
The presence of a functional group, particularly one having a heteroatom Y with non-bonding valence electrons (Y = N, O, S, X etc.), can dramatically alter the fragmentation pattern of a compound. This influence is thought to occur because of a "localization" of the radical cation component of the molecular ion on the heteroatom. After all, it is easier to remove (ionize) a non-bonding electron than one that is part of a covalent bond. By localizing the reactive moiety, certain fragmentation processes will be favored. These are summarized in the following diagram, where the green shaded box at the top displays examples of such "localized" molecular ions. The first two fragmentation paths lead to even-electron ions, and the elimination (path #3) gives an odd-electron ion. Note the use of different curved arrows to show single electron shifts compared with electron pair shifts.
The charge distributions shown above are common, but for each cleavage process the charge may sometimes be carried by the other (neutral) species, and both fragment ions are observed. Of the three cleavage reactions described here, the alpha-cleavage is generally favored for nitrogen, oxygen and sulfur compounds. Indeed, in the spectra of 4-methyl-3-pentene-2-one and N,N-diethylmethylamine the major fragment ions come from alpha-cleavage.
The complexity of fragmentation patterns has led to mass spectra being used as "fingerprints" for identifying compounds. Environmental pollutants, pesticide residues on food, and controlled substance identification are but a few examples of this application. Extremely small samples of an unknown substance (a microgram or less) are sufficient for such analysis. The following mass spectrum of cocaine demonstrates how a forensic laboratory might determine the nature of an unknown street drug. Even though extensive fragmentation has occurred, many of the more abundant ions (identified by magenta numbers) can be rationalized by the three mechanisms shown above. Plausible assignments may be seen by clicking on the spectrum, and it should be noted that all are even-electron ions. The m/z = 42 ion might be any or all of the following: C3H6, C2H2O or C2H4N. A precise assignment could be made from a high-resolution m/z value (next section).
Molecules with lots of oxygen atoms sometimes show a small M+2 peak (2 m/z units greater than the parent peak) in their mass spectra, due to the presence of a small amount of 18O (the most abundant isotope of oxygen is 16O). Because there are two abundant isotopes of both chlorine (about 75% 35Cl and 25% 37Cl) and bromine (about 50% 79Br and 50% 81Br), chlorinated and brominated compounds have very large and recognizable M+2 peaks. Fragments containing both isotopes of Br can be seen in the mass spectrum of ethyl bromide:
A common fragmentation pattern for larger carbonyl compounds is called the McLafferty rearrangement:
The mass spectrum of 2-hexanone shows a 'McLafferty fragment' at m/z = 58, while the propene fragment is not observed because it is a neutral species (remember, only cationic fragments are observed in MS). The base peak in this spectrum is again an acylium ion.
When alcohols are subjected to electron ionization MS, the molecular ion is highly unstable and thus a parent peak is often not detected. Often the base peak is from an ‘oxonium’ ion.
Other functional groups have predictable fragmentation patterns as well. By carefully analyzing the fragmentation information that a mass spectrum provides, a knowledgeable spectroscopist can often ‘put the puzzle together’ and make some very confident predictions about the structure of the starting sample.
Exercise \(1\)
The mass spectrum of an aldehyde gives prominent peaks at m/z = 59 (12%, highest value of m/z in the spectrum), 58 (85%), and 29 (100%), as well as others. Propose a structure, and identify the three species whose m/z values were listed
Gas Chromatography - Mass Spectrometry
Quite often, mass spectrometry is used in conjunction with a separation technique called gas chromatography (GC). The combined GC-MS procedure is very useful when dealing with a sample that is a mixture of two or more different compounds, because the various compounds are separated from one another before being subjected individually to MS analysis. We will not go into the details of gas chromatography here, although if you are taking an organic laboratory course you might well get a chance to try your hand at GC, and you will almost certainly be exposed to the conceptually analogous techniques of thin layer and column chromatography. Suffice it to say that in GC, a very small amount of a liquid sample is vaporized, injected into a long, coiled metal column, and pushed though the column by helium gas. Along the way, different compounds in the sample stick to the walls of the column to different extents, and thus travel at different speeds and emerge separately from the end of the column. In GC-MS, each purified compound is sent directly from the end of GC column into the MS instrument, so in the end we get a separate mass spectrum for each of the compounds in the original mixed sample. Because a compound's MS spectrum is a very reliable and reproducible 'fingerprint', we can instruct the instrument to search an MS database and identify each compound in the sample.
The extremely high sensitivity of modern GC-MS instrumentation makes it possible to detect and identify very small trace amounts of organic compounds. GC-MS is being used increasingly by environmental chemists to detect the presence of harmful organic contaminants in food and water samples. Airport security screeners also use high-speed GC-MS instruments to look for residue from bomb-making chemicals on checked luggage. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Mass_Spectrometry/Mass_spectrometry_1.txt |
Over the past fifty years nuclear magnetic resonance spectroscopy, commonly referred to as NMR, has become the preeminent technique for determining the structure of organic compounds. Of all the spectroscopic methods, it is the only one for which a complete analysis and interpretation of the entire spectrum is normally expected. Although larger amounts of sample are needed than for mass spectroscopy, NMR is non-destructive, and with modern instruments good data may be obtained from samples weighing less than a milligram. To be successful in using NMR as an analytical tool, it is necessary to understand the physical principles on which the methods are based.
Introduction
The nuclei of many elemental isotopes have a characteristic spin (I). Some nuclei have integral spins (e.g. I = 1, 2, 3 ....), some have fractional spins (e.g. I = 1/2, 3/2, 5/2 ....), and a few have no spin, I = 0 (e.g. 12C, 16O, 32S, ....). Isotopes of particular interest and use to organic chemists are 1H, 13C, 19F and 31P, all of which have I = 1/2. Since the analysis of this spin state is fairly straight forward, our discussion of NMR will be limited to these and other I = 1/2 nuclei. For a table of nuclear spin characteristics
The following features lead to the NMR phenomenon
1. A spinning charge generates a magnetic field, as shown by the animation on the right. The resulting spin-magnet has a magnetic moment (μ) proportional to the spin.
2. In the presence of an external magnetic field (B0), two spin states exist, +1/2 and -1/2. The magnetic moment of the lower energy +1/2 state is aligned with the external field, but that of the higher energy -1/2 spin state is opposed to the external field. Note that the arrow representing the external field points North.
3. The difference in energy between the two spin states is dependent on the external magnetic field strength, and is always very small. The following diagram illustrates that the two spin states have the same energy when the external field is zero, but diverge as the field increases. At a field equal to Bx a formula for the energy difference is given (remember I = 1/2 and μ is the magnetic moment of the nucleus in the field).
Strong magnetic fields are necessary for NMR spectroscopy. The international unit for magnetic flux is the tesla (T). The earth's magnetic field is not constant, but is approximately 10-4 T at ground level. Modern NMR spectrometers use powerful magnets having fields of 1 to 20 T. Even with these high fields, the energy difference between the two spin states is less than 0.1 cal/mole. To put this in perspective, recall that infrared transitions involve 1 to 10 kcal/mole and electronic transitions are nearly 100 time greater.
For NMR purposes, this small energy difference (ΔE) is usually given as a frequency in units of MHz (106 Hz), ranging from 20 to 900 Mz, depending on the magnetic field strength and the specific nucleus being studied. Irradiation of a sample with radio frequency (rf) energy corresponding exactly to the spin state separation of a specific set of nuclei will cause excitation of those nuclei in the +1/2 state to the higher -1/2 spin state. Note that this electromagnetic radiation falls in the radio and television broadcast spectrum. Nmr spectroscopy is therefore the energetically mildest probe used to examine the structure of molecules.
The nucleus of a hydrogen atom (the proton) has a magnetic moment μ = 2.7927, and has been studied more than any other nucleus. The previous diagram may be changed to display energy differences for the proton spin states (as frequencies) by mouse clicking anywhere within it.
4. For spin 1/2 nuclei the energy difference between the two spin states at a given magnetic field strength will be proportional to their magnetic moments. For the four common nuclei noted above, the magnetic moments are: 1H μ = 2.7927, 19F μ = 2.6273, 31P μ = 1.1305 & 13C μ = 0.7022. These moments are in nuclear magnetons, which are 5.05078•10-27 JT-1. The following diagram gives the approximate frequencies that correspond to the spin state energy separations for each of these nuclei in an external magnetic field of 2.35 T. The formula in the colored box shows the direct correlation of frequency (energy difference) with magnetic moment (h = Planck's constant = 6.626069•10-34 Js).
2. Proton NMR Spectroscopy
This important and well-established application of nuclear magnetic resonance will serve to illustrate some of the novel aspects of this method. To begin with, the NMR spectrometer must be tuned to a specific nucleus, in this case the proton. The actual procedure for obtaining the spectrum varies, but the simplest is referred to as the continuous wave (CW) method. A typical CW-spectrometer is shown in the following diagram. A solution of the sample in a uniform 5 mm glass tube is oriented between the poles of a powerful magnet, and is spun to average any magnetic field variations, as well as tube imperfections. Radio frequency radiation of appropriate energy is broadcast into the sample from an antenna coil (colored red). A receiver coil surrounds the sample tube, and emission of absorbed rf energy is monitored by dedicated electronic devices and a computer. An NMR spectrum is acquired by varying or sweeping the magnetic field over a small range while observing the rf signal from the sample. An equally effective technique is to vary the frequency of the rf radiation while holding the external field constant.
As an example, consider a sample of water in a 2.3487 T external magnetic field, irradiated by 100 MHz radiation. If the magnetic field is smoothly increased to 2.3488 T, the hydrogen nuclei of the water molecules will at some point absorb rf energy and a resonance signal will appear. An animation showing this may be activated by clicking the Show Field Sweep button. The field sweep will be repeated three times, and the resulting resonance trace is colored red. For visibility, the water proton signal displayed in the animation is much broader than it would be in an actual experiment.
Since protons all have the same magnetic moment, we might expect all hydrogen atoms to give resonance signals at the same field / frequency values. Fortunately for chemistry applications, this is not true. By clicking the Show Different Protons button under the diagram, a number of representative proton signals will be displayed over the same magnetic field range. It is not possible, of course, to examine isolated protons in the spectrometer described above; but from independent measurement and calculation it has been determined that a naked proton would resonate at a lower field strength than the nuclei of covalently bonded hydrogens. With the exception of water, chloroform and sulfuric acid, which are examined as liquids, all the other compounds are measured as gases.
Why should the proton nuclei in different compounds behave differently in the NMR experiment ?
The answer to this question lies with the electron(s) surrounding the proton in covalent compounds and ions. Since electrons are charged particles, they move in response to the external magnetic field (Bo) so as to generate a secondary field that opposes the much stronger applied field. This secondary field shields the nucleus from the applied field, so Bo must be increased in order to achieve resonance (absorption of rf energy). As illustrated in the drawing on the right, Bo must be increased to compensate for the induced shielding field. In the upper diagram, those compounds that give resonance signals at the higher field side of the diagram (CH4, HCl, HBr and HI) have proton nuclei that are more shielded than those on the lower field (left) side of the diagram.
The magnetic field range displayed in the above diagram is very small compared with the actual field strength (only about 0.0042%). It is customary to refer to small increments such as this in units of parts per million (ppm). The difference between 2.3487 T and 2.3488 T is therefore about 42 ppm. Instead of designating a range of NMR signals in terms of magnetic field differences (as above), it is more common to use a frequency scale, even though the spectrometer may operate by sweeping the magnetic field. Using this terminology, we would find that at 2.34 T the proton signals shown above extend over a 4,200 Hz range (for a 100 MHz rf frequency, 42 ppm is 4,200 Hz). Most organic compounds exhibit proton resonances that fall within a 12 ppm range (the shaded area), and it is therefore necessary to use very sensitive and precise spectrometers to resolve structurally distinct sets of hydrogen atoms within this narrow range. In this respect it might be noted that the detection of a part-per-million difference is equivalent to detecting a 1 millimeter difference in distances of 1 kilometer.
Chemical Shift
Unlike infrared and UV-visible spectroscopy, where absorption peaks are uniquely located by a frequency or wavelength, the location of different NMR resonance signals is dependent on both the external magnetic field strength and the rf frequency. Since no two magnets will have exactly the same field, resonance frequencies will vary accordingly and an alternative method for characterizing and specifying the location of NMR signals is needed. This problem is illustrated by the eleven different compounds shown in the following diagram. Although the eleven resonance signals are distinct and well separated, an unambiguous numerical locator cannot be directly assigned to each.
One method of solving this problem is to report the location of an NMR signal in a spectrum relative to a reference signal from a standard compound added to the sample. Such a reference standard should be chemically unreactive, and easily removed from the sample after the measurement. Also, it should give a single sharp NMR signal that does not interfere with the resonances normally observed for organic compounds. Tetramethylsilane, (CH3)4Si, usually referred to as TMS, meets all these characteristics, and has become the reference compound of choice for proton and carbon NMR.
Since the separation (or dispersion) of NMR signals is magnetic field dependent, one additional step must be taken in order to provide an unambiguous location unit. This is illustrated for the acetone, methylene chloride and benzene signals by clicking on the previous diagram. To correct these frequency differences for their field dependence, we divide them by the spectrometer frequency (100 or 500 MHz in the example), as shown in a new display by again clicking on the diagram. The resulting number would be very small, since we are dividing Hz by MHz, so it is multiplied by a million, as shown by the formula in the blue shaded box. Note that νref is the resonant frequency of the reference signal and νsamp is the frequency of the sample signal. This operation gives a locator number called the Chemical Shift, having units of parts-per-million (ppm), and designated by the symbol δ. Chemical shifts for all the compounds in the original display will be presented by a third click on the diagram.
The compounds referred to above share two common characteristics:
The hydrogen atoms in a given molecule are all structurally equivalent, averaged for fast conformational equilibria.
The compounds are all liquids, save for neopentane which boils at 9 °C and is a liquid in an ice bath.
The first feature assures that each compound gives a single sharp resonance signal. The second allows the pure (neat) substance to be poured into a sample tube and examined in a NMR spectrometer. In order to take the NMR spectra of a solid, it is usually necessary to dissolve it in a suitable solvent. Early studies used carbon tetrachloride for this purpose, since it has no hydrogen that could introduce an interfering signal. Unfortunately, CCl4 is a poor solvent for many polar compounds and is also toxic. Deuterium labeled compounds, such as deuterium oxide (D2O), chloroform-d (DCCl3), benzene-d6 (C6D6), acetone-d6 (CD3COCD3) and DMSO-d6 (CD3SOCD3) are now widely used as NMR solvents. Since the deuterium isotope of hydrogen has a different magnetic moment and spin, it is invisible in a spectrometer tuned to protons.
For the properties of some common NMR solvents Click Here.
From the previous discussion and examples we may deduce that one factor contributing to chemical shift differences in proton resonance is the inductive effect. If the electron density about a proton nucleus is relatively high, the induced field due to electron motions will be stronger than if the electron density is relatively low. The shielding effect in such high electron density cases will therefore be larger, and a higher external field (Bo) will be needed for the rf energy to excite the nuclear spin. Since silicon is less electronegative than carbon, the electron density about the methyl hydrogens in tetramethylsilane is expected to be greater than the electron density about the methyl hydrogens in neopentane (2,2-dimethylpropane), and the characteristic resonance signal from the silane derivative does indeed lie at a higher magnetic field. Such nuclei are said to be shielded. Elements that are more electronegative than carbon should exert an opposite effect (reduce the electron density); and, as the data in the following tables show, methyl groups bonded to such elements display lower field signals (they are deshielded). The deshielding effect of electron withdrawing groups is roughly proportional to their electronegativity, as shown by the left table. Furthermore, if more than one such group is present, the deshielding is additive (table on the right), and proton resonance is shifted even further downfield.
Compound
(CH3)4C
(CH3)3N
(CH3)2O
CH3F
δ
0.9 2.1 3.2 4.1
Compound
(CH3)4Si
(CH3)3P
(CH3)2S
CH3Cl
δ
0.0 0.9 2.1 3.0
Proton Chemical Shifts (ppm)
Cpd. / Sub.
X=Cl
X=Br
X=I
X=OR
X=SR
CH3X
3.0 2.7 2.1 3.1 2.1
CH2X2
5.3 5.0 3.9 4.4 3.7
CHX3
7.3 6.8 4.9 5.0
The general distribution of proton chemical shifts associated with different functional groups is summarized in the following chart. Bear in mind that these ranges are approximate, and may not encompass all compounds of a given class. Note also that the ranges specified for OH and NH protons (colored orange) are wider than those for most CH protons. This is due to hydrogen bonding variations at different sample concentrations.
Proton Chemical Shift Ranges*
Low Field
Region
High Field
Region
* For samples in CDCl3 solution. The δ scale is relative to TMS at δ = 0.
To make use of a calculator that predicts aliphatic proton chemical shifts Click Here. This application was developed at Colby College.
Signal Strength
The magnitude or intensity of NMR resonance signals is displayed along the vertical axis of a spectrum, and is proportional to the molar concentration of the sample. Thus, a small or dilute sample will give a weak signal, and doubling or tripling the sample concentration increases the signal strength proportionally. If we take the NMR spectrum of equal molar amounts of benzene and cyclohexane in carbon tetrachloride solution, the resonance signal from cyclohexane will be twice as intense as that from benzene because cyclohexane has twice as many hydrogens per molecule. This is an important relationship when samples incorporating two or more different sets of hydrogen atoms are examined, since it allows the ratio of hydrogen atoms in each distinct set to be determined. To this end it is necessary to measure the relative strength as well as the chemical shift of the resonance signals that comprise an NMR spectrum. Two common methods of displaying the integrated intensities associated with a spectrum are illustrated by the following examples. In the three spectra in the top row, a horizontal integrator trace (light green) rises as it crosses each signal by a distance proportional to the signal strength. Alternatively, an arbitrary number, selected by the instrument's computer to reflect the signal strength, is printed below each resonance peak, as shown in the three spectra in the lower row. From the relative intensities shown here, together with the previously noted chemical shift correlations, the reader should be able to assign the signals in these spectra to the set of hydrogens that generates each. If you click on one of the spectrum signals (colored red) or on hydrogen atom(s) in the structural formulas the spectrum will be enlarged and the relationship will be colored blue. Hint: When evaluating relative signal strengths, it is useful to set the smallest integration to unity and convert the other values proportionally.
rapid OH exchange with the deuterium of heavy water to assign hydroxyl proton resonance signals . As shown in the following equation, this removes the hydroxyl proton from the sample and its resonance signal in the NMR spectrum disappears. Experimentally, one simply adds a drop of heavy water to a chloroform-d solution of the compound and runs the spectrum again. The result of this exchange is displayed below.
Nuclear Magnetic Resonance Spectroscopy
We would expect the NMR spectrum of two spin-coupled protons, A & B, to display a pair of doublets. However, if the ratio of Δν to J (both in Hz) decreases to less than 10 a significant distortion of this expected pattern will take place, as shown in the following diagram. By repeated clicking on the diagram four examples of this behavior will be presented, starting with a compound in which Δν/J = 7.9 and followed by cases having smaller ratios. The fourth example, 2-methyl-2-phenyl-1,3-propanediol is especially interesting. Here, the two identical methylene groups have a pair of diastereotopic protons (Ha & Hb). Because these hydrogens have similar but different chemical shifts, the expected doublet pair experiences a second order distortion.
Second order splitting distortions are not limited to single proton pairs. The following spectrum of 2-dimethylaminoethyl acetate shows what appears to be a well behaved pair of triplets coming from adjacent methylene groups. Such cases are often deceptively simple, and become very complex when Δν/J is less than 10. Clicking on the diagram will display a simple example.
For additional examples of Second Order splitting patterns Click Here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Nuclear_Magnetic_Resonance_Spectroscopy/Lost.txt |
Nuclear spin may be related to the nucleon composition of a nucleus in the following manner:
• Odd mass nuclei (i.e. those having an odd number of nucleons) have fractional spins.
• Examples are I = 1/2 ( 1H, 13C, 19F ), I = 3/2 ( 11B ) & I = 5/2 ( 17O ).
• Even mass nuclei composed of odd numbers of protons and neutrons have integral spins. Examples are I = 1 ( 2H, 14N ).
• Even mass nuclei composed of even numbers of protons and neutrons have zero spin ( I = 0 ). Examples are 12C, and 16O.
Spin Properties of Nuclei
Spin 1/2 nuclei have a spherical charge distribution, and their NMR behavior is the easiest to understand. Other spin nuclei have nonspherical charge distributions and may be analyzed as prolate or oblate spinning bodies. All nuclei with non-zero spins have magnetic moments (μ), but the nonspherical nuclei also have an electric quadrupole moment (eQ). Some characteristic properties of selected nuclei are given in the following table.
Isotope
Natural %
Abundance
Spin (I)
Magnetic
Moment (μ)*
Magnetogyric
Ratio (γ)
1H
99.9844 1/2 2.7927 26.753
2H
0.0156 1 0.8574 4,107
11B
81.17 3/2 2.6880 --
13C
1.108 1/2 0.7022 6,728
17O
0.037 5/2 -1.8930 -3,628
19F
100.0 1/2 2.6273 25,179
29Si
4.700 1/2 -0.5555 -5,319
31P
100.0 1/2 1.1305 10,840
* μ in units of nuclear magnetons = 5.05078•10-27 JT-1
γ in units of 107rad T-1 sec-1
A Model for NMR Spectroscopy
The model of a spinning nuclear magnet aligned with or against an external magnetic field (for I = 1/2 nuclei) must be refined for effective interpretation of NMR phenomena. Just as a spinning mass will precess in a gravitational field (a gyroscope), the magnetic moment μ associated with a spinning spherical charge will precess in an external magnetic field. In the following illustration, the spinning nucleus has been placed at the origin of a cartesian coordinate system, and the external field is oriented along the z-axis. The frequency of precession is proportional to the strength of the magnetic field, as noted by the equation: ωo = γBo. The frequency ωo is called the Larmor frequency and has units of radians per second. The proportionality constant γ is known as the gyromagnetic ratio and is proportional to the magnetic moment (γ = 2pm/hI). Some characteristic γ's were listed in the preceding table of nuclear properties.
A Spinning Gyroscope in a Gravity Field
magnetic moment
μ
A Spinning Charge in a Magnetic Field
If rf energy having a frequency matching the Larmor frequency is introduced at a right angle to the external field (e.g. along the x-axis), the precessing nucleus will absorb energy and the magnetic moment will flip to its I = _1/2 state. This excitation is shown in the following diagram. Note that frequencies in radians per second may be converted to Hz (cps) by dividing by 2π.
The energy difference between nuclear spin states is small compared with the average kinetic energy of room temperature samples, and the +1/2 and _1/2 states are nearly equally populated. Indeed, in a field of 2.34 T the excess population of the lower energy state is only six nuclei per million. Although this is a very small difference , when we consider the number of atoms in a practical sample (remember the size of Avogadro's number), the numerical excess in the lower energy state is sufficient for selective and sensitive spectroscopic measurements. The diagram on the left below illustrates the macroscopic magnetization of a sample containing large numbers of spin 1/2 nuclei at equilibrium in a strong external magnetic field (Bo). A slight excess of +1/2 spin states precess randomly in alignment with the external field and a smaller population of _1/2 spin states precess randomly in an opposite alignment. An overall net magnetization therefore lies along the z-axis.
Net Macroscopic Magnetization of a Sample in an External Magnetic Field B0
Excitation by RF Energy and Subsequent Relaxation
The diagram and animation on the right show the changes in net macroscopic magnetization that occur as energy is introduced by rf irradiation at right angles to the external field. It is convenient to show the rf transmitter on the x-axis and the receiver-detector coil on the y-axis. On clicking the "Introduce RF Energy" button the animation will begin, and will repeat five times.
• First, the net magnetization shifts away from the z-axis and toward the y-axis. This occurs because some of the +1/2 nuclei are excited to the _1/2 state, and the precession about the z-axis becomes coherent (non-random), generating a significant y component to the net magnetization (M). The animation pauses at this stage.
• After irradiation the nuclear spins return to equilibrium in a process called relaxation. As the xy coherence disappears and the population of the +1/2 state increases, energy is released and detected by the receiver. The net magnetization spirals back, and eventually the equilibrium state is reestablished.
An inherent problem of the NMR experiment must be pointed out here. We have noted that the population difference between the spin states is proportionally very small. A fundamental requirement for absorption spectroscopy is a population imbalance between a lower energy ground state and a higher energy excited state. This can be expressed by the following equation, where A is a proportionality constant. If the mole fractions of the spin states are equal (η+ = η- ) then the population difference is zero and no absorption will occur. If the rf energy used in an NMR experiment is too high this saturation of the higher spin state will result and useful signals will disappear.
Relaxation Mechanisms
For NMR spectroscopy to be practical, an efficient mechanism for nuclei in the higher energy _1/2 spin state to return to the lower energy +1/2 state must exist. In other words, the spin population imbalance existing at equilibrium must be restored if spectroscopic observations are to continue. Now an isolated spinning nucleus will not spontaneouly change its spin state in the absence of external perturbation. Indeed, hydrogen gas (H2) exists as two stable spin isomers: ortho (parallel proton spins) and para (antiparallel spins). NMR spectroscopy is normally carried out in a liquid phase (solution or neat) so that there is close contact of sample molecules with a rapidly shifting crowd of other molecules (Brownian motion). This thermal motion of atoms and molecules generates local fluctuating electromagnetic fields, having components that match the Larmor frequency of the nucleus being studied. These local fields stimulate emission/absorption events that establish spin equilibrium, the excess spin energy being detected as it is released. This relaxation mechanism is called Spin-Lattice Relaxation (or Longitudinal Relaxation). The efficiency of spin-lattice relaxation depends on factors that influence molecular movement in the lattice, such as viscosity and temperature. The relaxation process is kinetically first order, and the reciprocal of the rate constant is a characteristic variable designated T1, the spin-lattice relaxation time. In non-viscous liquids at room temperature T1 ranges from 0.1 to 20 sec. A larger T1 indicates a slower or more inefficient spin relaxation. Another relaxation mechanism called spin-spin relaxation (or transverse relaxation) is characterized by a relaxation time T2. This process, which is actually a spin exchange, will not be discussed here.
Pulsed Fourier Transform Spectroscopy
In a given strong external magnetic field, each structurally distinct set of hydrogens in a molecule has a characteristic resonance frequency, just as each tubular chime in percussion instrument has a characteristic frequency. To discover the frequency of a chime we can strike it with a mallet and measure the sound emitted. This procedure can be repeated for each chime in the group so that all the characteristic frequencies are identified.
An alternative means of aquiring the same information is to strike all the chimes simultaneously, and to subject the complex collection of frequencies produced to mathematical analysis. In the following diagram the four frequencies assigned to our set of chimes are added together to give a complex summation wave. This is a straightforward conversion; and the reverse transformation, while not as simple, is readily accomplished, provided the combination signal is adequately examined and characterized.
A CW NMR spectrometer functions by irradiating each set of distinct nuclei in turn, a process analagous to striking each chime independently. For a high resolution spectrum this must be done slowly, and a 12 ppm sweep of the proton region takes from 5 to 10 minutes. It has proven much more efficient to excite all the proton nuclei in a molecule at the same time, followed by mathematical analysis of the complex rf resonance frequencies emitted as they relax back to the equilibrium state. This is the principle on which a pulse Fourier transform spectrometer operates. By exposing the sample to a very short (10 to 100 μsec), relatively strong (about 10,000 times that used for a CW spectrometer) burst of rf energy along the x-axis, as described above, all of the protons in the sample are excited simultaneously. The macroscopic magnetization model remains useful if we recognize it is a combination of megnetization vectors for all the nuclei that have been excited.
The overlapping resonance signals generated as the excited protons relax are collected by a computer and subjected to a Fourier transform mathematical analysis. As shown in the diagram on the left, the Fourier transform analysis, abbreviated FT, converts the complex time domain signal emitted by the sample into the frequency (or field) domain spectrum we are accustomed to seeing. In this fashion a complete spectrum can be acquired in a few seconds. Because the relaxation mechanism is a first order process, the rf signal emitted by the sample decays exponentially. This is called a free induction decay signal, abbreviated FID.
Free Induction Decay Signal
Since, the FID signal collected after one pulse, may be stored and averaged with the FID's from many other identical pulses prior to the Fourier transform, the NMR signal strength from a small sample may be enhanced to provide a useable spectrum. This has been essential to acquiring spectra from low abundance isotopes, such as 13C. In practice, the pulse FT experiment has proven so versatile that many variations of the technique, suited to special purposes, have been devised and used effectively.
Examples of Anisotropy Influences on Chemical Shift
The compound on the left has a chain of ten methylene groups linking para carbons of a benzene ring. Such bridged benzenes are called paracyclophanes. The meta analogs are also known. The structural constraints of the bridging chain require the middle two methylene groups to lie over the face of the benzene ring, which is a NMR shielding region. The four hydrogen atoms that are part of these groups display resonance signals that are more than two ppm higher field than the two methylene groups bonded to the edge of the ring (a deshielding region).
The 14 π-electron bridged annulene on the right is an aromatic (4n + 2) system, and has the same anisotropy as benzene. Nuclei located over the face of the ring are shielded, and those on the periphery are deshielded. The ring hydrogens give resonance signals in the range 8.0 to 8.7 δ, as expected from their deshielded location (note that there are three structurally different hydrogens on the ring). The two propyl groups are structurally equivalent (homotopic), and are free to rotate over the faces of the ring system (one above and one below). On average all the propyl hydrogens are shielded, with the innermost methylene being the most affected. The negative chemical shifts noted here indicate that the resonances occurs at a higher field than the TMS reference signal.
A remarkable characteristic of annulenes is that antiaromatic 4n π-electron systems are anisotropic in the opposite sense as their aromatic counterparts. A dramatic illustration of this fact is provided by the dianion derivative of the above bridged annulene. This dianion, formed by the addition of two electrons, is a 16 π-electron (4n) system. In the NMR spectrum of the dianion, the ring hydrogens resonate at high field (they are shielded), and the hydrogens of the propyl group are all shifted downfield (deshielded). The innermost methylene protons (magenta) give an NMR signal at +22.2 ppm, and the signals from the adjacent methylene and methyl hydrogens also have unexpectedly large chemical shifts.
Compounds in which two or more benzene rings are fused together include examples such as naphthalene, anthracene and phenanthrene, shown in the following diagram, present interesting insights into aromaticity and reactivity. The resonance stabilization of these compounds, calculated from heats of hydrogenation or combustion, is given beneath each structure.
Unlike benzene, the structures of these compounds show measurable double bond localization, which is reflected in their increased reactivity both in substitution and addition reactions. However, the 1HNMR spectra of these aromatic hydrocarbons do not provide much insight into the distribution of their pi-electrons. As expected, naphthalene displays two equally intense signals at δ 7.46 & 7.83 ppm. Likewise, anthracene shows three signals, two equal intensity multiplets at δ 7.44 & 7.98 ppm and a signal half as intense at δ 8.4 ppm. Thus, the influence of double bond localization or competition between benzene and higher annulene stabilization cannot be discerned.
The much larger C48H24 fused benzene ring cycle, named "kekulene" by Heinz Staab and sometimes called "superbenzene" by others, serves to probe the relative importance of benzenoid versus annulenoid aromaticity. A generic structure of this remarkable compound is drawn on the left below, together with two representative Kekule contributing structures on its right. There are some 200 Kekule structures that can be drawn for kekulene, but these two canonical forms represent extremes in aromaticity. The central formula has two [4n+2] annulenes, an inner [18]annulene and an outer [30]annulene (colored pink and blue respectively). The formula on the right has six benzene rings (colored green) joined in a ring by meta bonds, and held in a planar configuration by six cis-double bond bridges.
The coupled annulene contributor in the center has an energetically equivalent canonical form in which the single and double bonds making up the annulenes are exchanged. If these contributors dominate the aromatic character of kekulene, the 6 inside hydrogens should be shielded by the ring currents, and the 18 hydrogens on the periphery should be deshielded. Furthermore, the C:C bonds composing each annulene ring should have roughly equal lengths.
If the benzene contributor on the right (and its equivalent Kekule form) dominate the aromaticity of kekulene, all the benzene hydrogens will be deshielded, and the six double bond links on the periphery will have bond lengths characteristic of fixed single and double bonds
The extreme insolubility of kekulene made it difficult to grow suitable crystals for X-ray analysis or obtain solution NMR spectra. These problems were eventually solved by using high boiling solvents, the 1HNMR spectrum being taken at 150 to 200° C in deuterated tetrachlorobenzene solution. The experimental evidence demonstrates clearly that the hexa-benzene ring structure on the right most accurately represents kekulene. This evidence is shown below. The extremely low field resonance of the inside hydrogens is assigned from similar downfield shifts in model compounds.
It is important to understand that the shielding and deshielding terms used throughout our discussion of relative chemical shifts are themselves relative. Indeed, compared to a hypothetical isolated proton, all the protons in a covalent compound are shielded by the electrons in nearby sigma and pi-bonds. Consequently, it would be more accurate to describe chemical shift differences in terms of the absolute shielding experienced by different groups of hydrogens. There is, in fact, good evidence that the anisotropy of neighboring C-H and C-C sigma bonds, together with that of the bond to the observed hydrogen, are the dominate shielding factors influencing chemical shifts. The anisotropy of pi-electron systems augments this sigma skeletal shielding. Nevertheless, the deceptive focus on anisotropic pi-electron influences is so widely and commonly used that this view has been retained and employed in these pages.
Hydrogen Bonding Influences
Hydrogen bonding of hydroxyl and amino groups not only causes large variations in the chemical shift of the proton of the hydrogen bond, but also influences its coupling with adjacent C-H groups. As shown on the right, the 60 MHz proton NMR spectrum of pure (neat) methanol exhibits two signals, as expected. At 30° C these signals are sharp singlets located at δ 3.35 and 4.80 ppm, the higher-field methyl signal (magenta) being three times as strong as the OH signal (orange) at lower field. When cooled to -45 ° C, the larger higher-field signal changes to a doublet (J = 5.2 Hz) having the same chemical shift. The smaller signal moves downfield to δ 5.5 ppm and splits into a quartet (J = 5.2 Hz). The relative intensities of the two groups of signals remains unchanged. This interesting change in the NMR spectrum, which is shown in the two spectra below, is due to increased stability of hydrogen bonded species at lower temperature. Since hydrogen bonding not only causes a resonance shift to lower field, but also decreases the rate of intermolecular proton exchange, the hydroxyl proton remains bonded to the alkoxy group for a sufficient time to exert its spin coupling influence.
Under routine conditions, rapid intermolecular exchange of the OH protons of alcohols often prevents their coupling with adjacent hydrogens from being observed. Intermediate rates of proton exchange lead to a broadening of the OH and coupled hydrogen signals, a characteristic that is useful in identifying these functions. Since traces of acid or base catalyze this hydrogen exchange, pure compounds and clean sample tubes must be used for experiments of the kind described here.
Methanol 1H NMR at approximately Room Temperature
Methanol 1H NMR at -45 oC
Another way of increasing the concentration of hydrogen bonded methanol species is to change the solvent from chloroform-d to a solvent that is a stronger hydrogen bond acceptor. Examples of such solvents are given in the following table. In contrast to the neat methanol experiment described above, very dilute solutions are used for this study. Since chloroform is a poor hydrogen bond acceptor and the dilute solution reduces the concentration of methanol clusters, the hydroxyl proton of methanol generates a resonance signal at a much higher field than that observed for the pure alcohol. Indeed, the OH resonance signal from simple alcohols in dilute chloroform solution is normally found near δ 1.0 ppm.
The exceptionally strong hydrogen bond acceptor quality of DMSO is demonstrated here by the large downfield shift of the methanol hydroxyl proton, compared with a slight upfield shift of the methyl signal. The expected spin coupling patterns shown above are also observed in this solvent. Although acetone and acetonitrile are better hydrogen-bond acceptors than chloroform, they are not as effective as DMSO.
1H Chemical Shifts of Methanol in Selected Solvents
Solvent CDCl3 CD3COCD3 CD3SOCD3 CD3C≡N
CH3–O–H
CH3
O–H
3.40
1.10
3.31
3.12
3.16
4.01
3.28
2.16
The solvent effect shown above suggests a useful diagnostic procedure for characterizing the OH resonance signals from alcohol samples. For example, a solution of ethanol in chloroform-d displays the spectrum shown on the left below, especially if traces of HCl are present (otherwise broadening of the OH and CH2 signals occurs). Note that the chemical shift of the OH signal (red) is less than that of the methylene group (blue), and no coupling of the OH proton is apparent. The vicinal coupling (J = 7 Hz) of the methyl and methylene hydrogens is typical of ethyl groups. In DMSO-d6 solution small changes of chemical shift are seen for the methyl and methylene group hydrogens, but a dramatic downfield shift of the hydroxyl signal takes place because of hydrogen bonding. Coupling of the OH proton to the adjacent methylene group is evident, and both the coupling constants can be measured. Because the coupling constants are different, the methylene signal pattern is an overlapping doublet of quartets (eight distinct lines) rather than a quintet. Note that residual hydrogens in the solvent give a small broad signal near δ 2.5 ppm.
For many alcohols in dilute chloroform-d solution, the hydroxyl resonance signal is often broad and obscured by other signals in the δ 1.5 to 3.0 region. The simple technique of using DMSO-d6 as a solvent, not only shifts this signal to a lower field, but permits 1°-, 2 °- & 3 °-alcohols to be distinguished. Thus, the hydroxyl proton of 2-propanol generates a doublet at δ 4.35 ppm, and the corresponding signal from 2-methyl-2-propanol is a singlet at δ 4.2 ppm. The more acidic OH protons of phenols are similarly shifted – from δ 4 to 7 in chloroform-d to δ 8.5 to 9.5 in DMSO-d6.
Spin-Spin Coupling
vicinal (joined by three sigma bonds). In this case a neighboring proton having a +1/2 spin shifts the resonance frequency of the proton being observed to a slightly higher value (up to 7 Hz), and a _1/2 neighboring spin shifts it to a lower frequency. Remember that the total population of these two spin states is roughly equal, differing by only a few parts per million in a strong magnetic field. If several neighboring spins are present, their effect is additive.
In the spectrum of 1,1-dichloroethane shown on the right, it is clear that the three methyl hydrogens (red) are coupled with the single methyne hydrogen (orange) in a manner that causes the former to appear as a doublet and the latter as a quartet. The light gray arrow points to the unperturbed chemical shift location for each proton set. By clicking on one of these signals, the spin relationship that leads to the coupling pattern will be displayed. Clicking elsewhere in the picture will return the original spectrum.
Full Spectrum of 1,1-dichloroethane
2.06 ppm Signal Explained
5.89 ppm Signal Explained
The statistical distribution of spins within each set explains both the n+1 rule and the relative intensities of the lines within a splitting pattern. The action of a single neighboring proton is easily deduced from the fact that it must have one of two possible spins. Interaction of these two spin states with the nuclei under observation leads to a doublet located at the expected chemical shift. The corresponding action of the three protons of the methyl group requires a more detailed analysis. In the display of this interaction four possible arrays of their spins are shown. The mixed spin states are three times as possible as the all +1/2 or all _1/2 collection. Consequently, we expect four signals, two above the chemical shift and two below it. This spin analysis also suggests that the intensity ratio of these signals will be 1:3:3:1. The line separations in splitting patterns are measured in Hz, and are characteristic of the efficiency of the spin interaction; they are referred to as coupling constants (symbol J). In the above example, the common coupling constant is 6.0 Hz.
Multipliticy Relative Line Intensity (Starting at the top: Singlet, Doublet, Triplet, Quartet, and Quintet)
A simple way of estimating the relative intensities of the lines in a first-order coupling pattern is shown on the right. This array of numbers is known as Pascal's triangle, and is easily extended to predict higher multiplicities. The number appearing at any given site is the sum of the numbers linked to it from above by the light blue lines. Thus, the central number of the five quintet values is 3 + 3 = 6. Of course, a complete analysis of the spin distributions, as shown for the case of 1,1-dichloroethane above, leads to the same relative intensities.
Coupling constants are independent of the external magnetic field, and reflect the unique spin interaction characteristics of coupled sets of nuclei in a specific structure. As noted earlier, coupling constants may vary from a fraction of a Hz to nearly 20 Hz, important factors being the nature and spatial orientation of the bonds joining the coupled nuclei. In simple, freely rotating alkane units such as CH3CH2X or YCH2CH2X the coupling constant reflects an average of all significant conformers, and usually lies in a range of 6 to 8 Hz. This conformational mobility may be restricted by incorporating the carbon atoms in a rigid ring, and in this way the influence of the dihedral orientation of the coupled hydrogens may be studied.
The structures of cis and trans-4-tert-butyl-1-chlorocyclohexane, shown above, illustrate how the coupling constant changes with the dihedral angle (φ) between coupled hydrogens. The inductive effect of chlorine shifts the resonance frequency of the red colored hydrogen to a lower field (δ ca. 4.0), allowing it to be studied apart from the other hydrogens in the molecule. The preferred equatorial orientation of the large tert-butyl group holds the six-membered ring in the chair conformation depicted in the drawing. In the trans isomer this fixes the red hydrogen in an axial orientation; whereas for the cis isomer it is equatorial. The listed values for the dihedral angles and the corresponding coupling constants suggest a relationship, which has been confirmed and clarified by numerous experiments. This relationship is expressed by the Karplus equation shown below.
Geminal couplings are most commonly observed in cyclic structures, but are also evident when methylene groups have diastereomeric hydrogens.
Spin Decoupling
We have noted that rapidly exchanging hydroxyl hydrogens are not spin-coupled to adjacent C-H groups. The reason for this should be clear. As each exchange occurs, there will be an equal chance of the new proton having a +1/2 or a _1/2 spin (remember that the overall populations of the two spin states are nearly identical). Over time, therefore, the hydroxyl hydrogen behaves as though it is rapidly changing its spin, and the adjacent nuclei see only a zero spin average from it. If we could cause other protons in a molecule to undergo a similar spin averaging, their spin-coupling influence on adjacent nuclei would cease. Such NMR experiments are possible, and are called spin decoupling.
When a given set of nuclei is irradiated with strong rf energy at its characteristic Larmor frequency, spin saturation and rapid interconversion of the spin states occurs. Neighboring nuclei with different Larmor frequencies are no longer influenced by specific long-lived spins, so spin-spin signal splitting of the neighbors vanishes. The following spectrum of 1-nitropropane may be used to illustrate this technique. The three distinct sets of hydrogens in this molecule generate three resonance signals (two triplets and a broad sextet). A carefully tuned decoupling signal may be broadcast into the sample while the remaining spectrum is scanned. The region of the decoupling signal is obscured, but resonance signals more than 60 Hz away may still be seen. By clicking on one of the three signals in the spectrum, the results of decoupling at that frequency will be displayed.
The Influence of Magnetic Field Strength
The presence of symmetrical, easily recognized first-order splitting patterns in a NMR spectrum depends on the relative chemical shifts of the spin-coupled nuclei and the magnitude of the coupling constant. If the chemical shift difference (i.e. Δδ in Hz) is large compared to J the splitting patterns will be nearly first order. If, on the other hand, the difference is relatively small (less than 10 J) second order distortion of the signal splitting will be observed. One important advantage in using very high field magnets for NMR is that the separation (or dipersion) of different sets of protons is proportional to field strength, whereas coupling constants do not change.
It is important to remember that structurally different sets of nuclei do not always produce distinctly different signals in an NMR spectrum. For example, the hydrocarbon octane has four different sets of protons, as shown in the following formula:
CH3CH2CH2CH2CH2CH2CH2CH3
Now methyl hydrogens have a smaller chemical shift than methylene hydrogens, so methyl groups (colored black here) can usually be distinguished. However, the chemical shifts of the different methylene groups (blue, red & green) are so similar that many NMR spectrometers will not resolve them. Consequently, a 90 MHz proton spectrum of octane shows a distorted triplet at δ 0.9 ppm, produced by the six methyl protons, and a strong broad singlet at δ 1.2 ppm coming from all twelve methylene protons. A similar failure to resolve structurally different hydrogen atoms occurs in the case of alkyl substituted benzene rings. The chemical shift difference between ortho, meta and para hydrogens in such compounds is often so small that they are seen as a single resonance signal in an NMR spectrum. The 90 MHz spectrum of benzyl alcohol in chloroform-d solution provides an instructive example, shown below. A broad strong signal at δ 7.24 ppm is characteristic of the aromatic protons on alkylbenzenes. Since the chemical shifts of these hydrogens are nearly identical, no spin coupling is observed. If the magnetic field strength is increased to 400 Mz (lower spectrum) the aromatic protons are more dispersed (orange, magenta and green signals), and the spin coupling of adjacent hydrogens (J = 7.6 Hz) causes overlap of the signals (gray shaded enlargement).
1H NMR of Benzyl alcohol
1H NMR of Anisole
Anisole, an isomer of benzyl alcohol, has a more dispersed set of aromatic signals, thanks to the electron donating influence of the methoxy substituent. The 90 MHz spectrum of anisole shows this greater dispersion, but the spin coupling of adjacent hydrogens still results in signal overlap. The 400 Mz spectrum at the bottom illustrates the greater dispersion of the chemical shifts, and since the coupling constants remain unchanged, the splitting patterns no longer overlap. In all these examples a very small meta-hydrogen coupling has been ignored.
Example \(1\)
Not all simple compounds have simple proton NMR spectra. The following example not only illustrates this point, but also demonstrates how a careful structural analysis can rationalize an initially complex spectrum.
The 100 MHz 1H NMR spectrum of a C3H5ClO compound is initially displayed. This spectrum is obviously complex and not easily interpreted, except for concluding that no olefinic C-H protons are present.
With a higher field spectrum it is clear that each of the five hydrogen atoms in the molecule is structurally unique, and is producing a separate signal. Also, it is clear there is considerable spin coupling of all the hydrogens. To see the coupling patterns more clearly it is necessary to expand and enhance the spectrum in these regions. For purposes of our demonstration, this can be done by clicking on any one of the signal multiplets. Clicking in an open area should return the original 500 MHz display. In some of the expanded displays two adjacent groups of signals are shown. Once an enlarged pattern is displayed, the line separations in Hz can be measured (remember that for a 500 MHz spectrum 1 ppm is 500 Hz). The middle signal at 3.2 ppm is the most complex, and overlap of some multiplet lines has occurred.
Solution
This spectrum has several interesting features. First, hydrogens A, B & C are clearly different, and are spin-coupled to each other. Hydrogens B & C are geminally related, whereas A is oriented to B & C in a vicinal manner. Since JAB and JBC are similar, the HB signal is a broad triplet. Although hydrogens D & E might seem identical at first glance, they are diastereotopic, and should therefore have different chemical shifts. The DE geminal coupling constant is 11.7 Hz, so each of these hydrogens appears as a doublet of doublets.
The splitting of the HA signal is complex and not immediately obvious. The diagram shows the consequences of the four operating coupling constants.
Additional Information from 13C NMR Spectroscopy
Broad band decoupling of the hydrogen atoms in a molecule was an essential operation for obtaining simple (single line) carbon NMR spectra. The chemical shifts of the carbon signals provide useful information, but it would also be very helpful to know how many hydrogen atoms are bonded to each carbon. Aside from the fact that carbons having no bonded hydrogens generally give weak resonance signals, this information is not present in a completely decoupled spectrum.
Clever methods of retaining the hydrogen information while still enjoying the benefits of proton decoupling have been devised. The techniques involved are beyond the scope of this discussion, but the overall results can still be appreciated. The 13C NMR spectrum of camphor shown below will serve as an illustration. It will be helpful to view an expanded section of this spectrum from δ 0.0 to 50.0 ppm, and this will be presented in the High Field Expansion spectrum. The two lowest field signals are missing in the expanded display.
High Field Expansion of Camphor
Even though the expanded display now shows the distinct carbon signals clearly, the origin of each is ambiguous. An early method of regaining coupling information was by off-resonance decoupling. In this approach a weaker and more focused proton decoupling frequency is applied as the carbon spectrum is acquired. Vestiges of the C-H coupling remain in the carbon signals, but the apparent coupling constants are greatly reduced. View the Off-Resonance Decouple spectrum. The results of such an experiment will be displayed. Notice that all the methyl groups are quartets (three coupled hydrogens), the methylene groups are triplets and methine carbons are doublets. Overlap of two quartets near δ 19 ppm and the doublet and triplet near δ 43 ppm are complicating factors.
Off-Resonance Decouple Spectrum of Camphor
A better way for classifying the carbon signals is by a technique called INEPT (insensitive nuclear enhancement by polarization transfer). This method takes advantage of the influence of hydrogen on 13C relaxation times, and can be applied in several modes. One of the most common applications of INEPT separates the signals of methyl and methine carbons from those of methylene carbons by their sign. Carbons having no hydrogen substituents have a zero signal.
INEPT Spectrum of Camphor
Properties of Some Deuterated NMR Solvents
Solvent B.P. °C Residual
1H signal (δ)
Residual
13C signal (δ)
acetone-d6 55.5 2.05 ppm 206 & 29.8 ppm
acetonitrile-d3 80.7 1.95 ppm 118 & 1.3 ppm
benzene-d6 79.1 7.16 ppm 128 ppm
chloroform-d 60.9 7.27 ppm 77.2 ppm
cyclohexane-d12 78.0 1.38 ppm 26.4 ppm
dichloromethane-d2 40.0 5.32 ppm 53.8 ppm
dimethylsulfoxide-d6 190 2.50 ppm 39.5 ppm
nitromethane-d3 100 4.33 ppm 62.8 ppm
pyridine-d5 114 7.19, 7.55 & 8.71 ppm 150, 135.5 & 123.5 ppm
tetrahydrofuran-d8 65.0 1.73 & 3.58 ppm 67.4 & 25.2 ppm
Contributors
• Layne Morsch (University of Illinois Springfield) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Nuclear_Magnetic_Resonance_Spectroscopy/Supplemental_NMR_Topics.txt |
An obvious difference between certain compounds is their color. Thus, quinone is yellow; chlorophyll is green; the 2,4-dinitrophenylhydrazone derivatives of aldehydes and ketones range in color from bright yellow to deep red, depending on double bond conjugation; and aspirin is colorless. In this respect the human eye is functioning as a spectrometer analyzing the light reflected from the surface of a solid or passing through a liquid. Although we see sunlight (or white light) as uniform or homogeneous in color, it is actually composed of a broad range of radiation wavelengths in the ultraviolet (UV), visible and infrared (IR) portions of the spectrum. As shown on the right, the component colors of the visible portion can be separated by passing sunlight through a prism, which acts to bend the light in differing degrees according to wavelength.
Electromagnetic radiation such as visible light is commonly treated as a wave phenomenon, characterized by a wavelength or frequency. Wavelength is defined on the left below, as the distance between adjacent peaks (or troughs), and may be designated in meters, centimeters or nanometers (10-9 meters). Frequency is the number of wave cycles that travel past a fixed point per unit of time, and is usually given in cycles per second, or hertz (Hz). Visible wavelengths cover a range from approximately 400 to 800 nm. The longest visible wavelength is red and the shortest is violet. Other common colors of the spectrum, in order of decreasing wavelength, may be remembered by the mnemonic: ROY G BIV. The wavelengths of what we perceive as particular colors in the visible portion of the spectrum are displayed and listed below. In horizontal diagrams, such as the one on the bottom left, wavelength will increase on moving from left to right.
• Violet: 400 - 420 nm
• Indigo: 420 - 440 nm
• Blue: 440 - 490 nm
• Green: 490 - 570 nm
• Yellow: 570 - 585 nm
• Orange: 585 - 620 nm
• Red: 620 - 780 nm
When white light passes through or is reflected by a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed. This relationship is demonstrated by the color wheel shown below. Here, complementary colors are diametrically opposite each other. Thus, absorption of 420-430 nm light renders a substance yellow, and absorption of 500-520 nm light makes it red. Green is unique in that it can be created by absoption close to 400 nm as well as absorption near 800 nm.
Early humans valued colored pigments, and used them for decorative purposes. Many of these were inorganic minerals, but several important organic dyes were also known. These included the crimson pigment, kermesic acid, the blue dye, indigo, and the yellow saffron pigment, crocetin. A rare dibromo-indigo derivative, punicin, was used to color the robes of the royal and wealthy. The deep orange hydrocarbon carotene is widely distributed in plants, but is not sufficiently stable to be used as permanent pigment, other than for food coloring. A common feature of all these colored compounds, displayed below, is a system of extensively conjugated $\pi$-electrons.
The Electromagnetic Spectrum
The visible spectrum constitutes but a small part of the total radiation spectrum. Most of the radiation that surrounds us cannot be seen, but can be detected by dedicated sensing instruments. This electromagnetic spectrum ranges from very short wavelengths (including gamma and x-rays) to very long wavelengths (including microwaves and broadcast radio waves). The following chart displays many of the important regions of this spectrum, and demonstrates the inverse relationship between wavelength and frequency (shown in the top equation below the chart).
The energy associated with a given segment of the spectrum is proportional to its frequency. The bottom equation describes this relationship, which provides the energy carried by a photon of a given wavelength of radiation.
To obtain specific frequency, wavelength and energy values use this calculator.
UV-Visible Absorption Spectra
To understand why some compounds are colored and others are not, and to determine the relationship of conjugation to color, we must make accurate measurements of light absorption at different wavelengths in and near the visible part of the spectrum. Commercial optical spectrometers enable such experiments to be conducted with ease, and usually survey both the near ultraviolet and visible portions of the spectrum. For a description of a UV-Visible spectrometer Click Here.
The visible region of the spectrum comprises photon energies of 36 to 72 kcal/mole, and the near ultraviolet region, out to 200 nm, extends this energy range to 143 kcal/mole. Ultraviolet radiation having wavelengths less than 200 nm is difficult to handle, and is seldom used as a routine tool for structural analysis.
The energies noted above are sufficient to promote or excite a molecular electron to a higher energy orbital. Consequently, absorption spectroscopy carried out in this region is sometimes called "electronic spectroscopy". A diagram showing the various kinds of electronic excitation that may occur in organic molecules is shown on the left. Of the six transitions outlined, only the two lowest energy ones (left-most, colored blue) are achieved by the energies available in the 200 to 800 nm spectrum. As a rule, energetically favored electron promotion will be from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO), and the resulting species is called an excited state. For a review of molecular orbitals click here.
When sample molecules are exposed to light having an energy that matches a possible electronic transition within the molecule, some of the light energy will be absorbed as the electron is promoted to a higher energy orbital. An optical spectrometer records the wavelengths at which absorption occurs, together with the degree of absorption at each wavelength. The resulting spectrum is presented as a graph of absorbance (A) versus wavelength, as in the isoprene spectrum shown below. Since isoprene is colorless, it does not absorb in the visible part of the spectrum and this region is not displayed on the graph. Absorbance usually ranges from 0 (no absorption) to 2 (99% absorption), and is precisely defined in context with spectrometer operation.
Because the absorbance of a sample will be proportional to the number of absorbing molecules in the spectrometer light beam (e.g. their molar concentration in the sample tube), it is necessary to correct the absorbance value for this and other operational factors if the spectra of different compounds are to be compared in a meaningful way. The corrected absorption value is called "molar absorptivity", and is particularly useful when comparing the spectra of different compounds and determining the relative strength of light absorbing functions (chromophores). Molar absorptivity ($\epsilon$) is defined as:
$\epsilon = \dfrac{A}{c\; l}$
wwhere
• $A$ is the sample absorbance
• $c$ is the sample concentration in moles/liter
• $l$ is the length of light path through the sample in cm
If the isoprene spectrum on the right was obtained from a dilute hexane solution (c = 4 * 10-5 moles per liter) in a 1 cm sample cuvette, a simple calculation using the above formula indicates a molar absorptivity of 20,000 at the maximum absorption wavelength, symbolized as λmax. Indeed the entire vertical absorbance scale may be changed to a molar absorptivity scale once this information about the sample is in hand. Clicking on the spectrum will display this change in units.
Chromophore
Example
Excitation
λmax, nm
ε @
λmax
Solvent
C=C Ethene
π → π*
171 15,000 hexane
C≡C 1-Hexyne
π → π*
180 10,000 hexane
C=O Ethanal
n → π*
π → π*
290
180
15
10,000
hexane
hexane
N=O Nitromethane
n → π*
π → π*
275
200
17
5,000
ethanol
ethanol
C-X X=Br
X=I
Methyl bromide
Methyl Iodide
n → σ*
n → σ*
205
255
200
360
hexane
hexane
From the chart above it should be clear that the only molecular moieties likely to absorb light in the 200 to 800 nm region are pi-electron functions and hetero atoms having non-bonding valence-shell electron pairs. Such light absorbing groups are referred to as chromophores. A list of some simple chromophores and their light absorption characteristics is provided on the left above. The oxygen non-bonding electrons in alcohols and ethers do not give rise to absorption above 160 nm. Consequently, pure alcohol and ether solvents may be used for spectroscopic studies.
The presence of chromophores in a molecule is best documented by UV-Visible spectroscopy, but the failure of most instruments to provide absorption data for wavelengths below 200 nm makes the detection of isolated chromophores problematic. Fortunately, conjugation generally moves the absorption maxima to longer wavelengths, as in the case of isoprene, so conjugation becomes the major structural feature identified by this technique.
Molar absorptivities may be very large for strongly absorbing chromophores (>10,000) and very small if absorption is weak (10 to 100). The magnitude ofε reflects both the size of the chromophore and the probability that light of a given wavelength will be absorbed when it strikes the chromophore. For further discussion of this topic Click Here.
The Importance of Conjugation
A comparison of the absorption spectrum of 1-pentene, λmax = 178 nm, with that of isoprene (above) clearly demonstrates the importance of chromophore conjugation. Further evidence of this effect is shown below. The spectrum on the left illustrates that conjugation of double and triple bonds also shifts the absorption maximum to longer wavelengths. From the polyene spectra displayed in the center diagram, it is clear that each additional double bond in the conjugated pi-electron system shifts the absorption maximum about 30 nm in the same direction. Also, the molar absorptivity (ε) roughly doubles with each new conjugated double bond. Spectroscopists use the terms defined in the table on the right when describing shifts in absorption. Thus, extending conjugation generally results in bathochromic and hyperchromic shifts in absorption.
The appearance of several absorption peaks or shoulders for a given chromophore is common for highly conjugated systems, and is often solvent dependent. This fine structure reflects not only the different conformations such systems may assume, but also electronic transitions between the different vibrational energy levels possible for each electronic state. Vibrational fine structure of this kind is most pronounced in vapor phase spectra, and is increasingly broadened and obscured in solution as the solvent is changed from hexane to methanol.
Terminology for Absorption Shifts
Nature of Shift
Descriptive Term
To Longer Wavelength Bathochromic
To Shorter Wavelength Hypsochromic
To Greater Absorbance Hyperchromic
To Lower Absorbance Hypochromic
To understand why conjugation should cause bathochromic shifts in the absorption maxima of chromophores, we need to look at the relative energy levels of the pi-orbitals. When two double bonds are conjugated, the four p-atomic orbitals combine to generate four pi-molecular orbitals (two are bonding and two are antibonding). This was described earlier in the section concerning diene chemistry. In a similar manner, the three double bonds of a conjugated triene create six pi-molecular orbitals, half bonding and half antibonding. The energetically most favorable $\pi \rightarrow \pi^*$ excitation occurs from the highest energy bonding pi-orbital (HOMO) to the lowest energy antibonding pi-orbital (LUMO).
The following diagram illustrates this excitation for an isolated double bond (only two pi-orbitals) and, on clicking the diagram, for a conjugated diene and triene. In each case the HOMO is colored blue and the LUMO is colored magenta. Increased conjugation brings the HOMO and LUMO orbitals closer together. The energy (ΔE) required to effect the electron promotion is therefore less, and the wavelength that provides this energy is increased correspondingly (remember λ = h c/ΔE ).
Examples of $\pi \rightarrow \pi^*$ transitions. Click on the Diagram to Advance
Many other kinds of conjugated pi-electron systems act as chromophores and absorb light in the 200 to 800 nm region. These include unsaturated aldehydes and ketones and aromatic ring compounds. A few examples are displayed below. The spectrum of the unsaturated ketone (on the left) illustrates the advantage of a logarithmic display of molar absorptivity. The $\pi \rightarrow \pi^*$ absorption located at 242 nm is very strong, with an ε = 18,000. The weak $n \rightarrow \pi^*$ absorption near 300 nm has an ε = 100.
Benzene exhibits very strong light absorption near 180 nm (ε > 65,000) , weaker absorption at 200 nm (ε = 8,000) and a group of much weaker bands at 254 nm (ε = 240). Only the last group of absorptions are completely displayed because of the 200 nm cut-off characteristic of most spectrophotometers. The added conjugation in naphthalene, anthracene and tetracene causes bathochromic shifts of these absorption bands, as displayed in the chart on the left below. All the absorptions do not shift by the same amount, so for anthracene (green shaded box) and tetracene (blue shaded box) the weak absorption is obscured by stronger bands that have experienced a greater red shift. As might be expected from their spectra, naphthalene and anthracene are colorless, but tetracene is orange.
The spectrum of the bicyclic diene (above right) shows some vibrational fine structure, but in general is similar in appearance to that of isoprene, shown above. Closer inspection discloses that the absorption maximum of the more highly substituted diene has moved to a longer wavelength by about 15 nm. This "substituent effect" is general for dienes and trienes, and is even more pronounced for enone chromophores. A set of empirical rules for predicting the λmax of such chromophores has been developed. These rules may be viewed by Clicking Here.
Visible and Ultraviolet Spectroscopy
Molar absoptivities may be very large for strongly absorbing chromophores (>10,000) and very small if absorption is weak (10 to 100). The magnitude of $\epsilon$ reflects both the size of the chromophore and the probability that light of a given wavelength will be absorbed when it strikes the chromophore. A general equation stating this relationship may be written as follows:
$\epsilon = 0.87 \times 10^{20} \times Ρ \times a$
with
• $Ρ$ is the transition probability (between 0 to 1) and
• $a$ is the chromophore area in $cm^2$
The factors that influence transition probabilities are complex, and are treated by what spectroscopists refer to as "Selection Rules". A rigorous discussion of selection rules is beyond the scope of this text, but one obvious factor is the overlap of the orbitals involved in the electronic excitation. This is nicely illustrated by the two common transitions of an isolated carbonyl group. The $n \rightarrow \pi^*$ transition is lower in energy ($\lambda_{max}=290\; nm$) than the $\pi \rightarrow \pi^*$ transition ($\lambda_{max}=180\; nm$), but the ε of the former is a thousand times smaller than the latter. The spatial distribution of these orbitals suggests why this is so. As illustrated in the following diagram, the n-orbitals do not overlap at all well with the $\pi*$ orbital, so the probability of this excitation is small. The $\pi \rightarrow \pi^*$ transition, on the other hand, involves orbitals that have significant overlap, and the probability is near 1.0. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Visible_and_Ultraviolet_Spectroscopy/Absorption_Intensity.txt |
Woodward-Fieser Rules for Calculating the λmax of Conjugated Dienes and Polyenes
Core Chromophore
Substituent and Influence
Transoid Diene
215 nm
R- (Alkyl Group) .... +5 nm
RO- (Alkoxy Group) .. +6
X- (Cl- or Br-) ......... +10
RCO2- (Acyl Group) .... 0
RS- (Sulfide Group) .. +30
R2N- (Amino Group) .. +60 Further π -Conjugation C=C (Double Bond) ... +30
C6H5 (Phenyl Group) ... +60
Cyclohexadiene*
260 nm
(i) Each exocyclic double bond adds 5 nm. In the example on the right, there are two exo-double bond components: one to ring A and the other to ring B.
(ii) Solvent effects are minor.
* When a homoannular (same ring) cyclohexadiene chromophore is present, a base value of 260 nm should be choosen. This includes the ring substituents. Rings of other size have a lesser influence.
λmax (calculated) = Base (215 or 260) + Substituent Contributions
Woodward-Fieser Rules for Calculating the π → π* λmax of Conjugated Carbonyl Compounds
Core Chromophore
Substituent and Influence
R = Alkyl 215 nm
R = H 210 nm
R = OR' 195 nm
α- Substituent
R- (Alkyl Group) +10 nm
Cl- (Chloro Group) +15
Br- (Chloro Group) +25
HO- (Hydroxyl Group) +35
RO- (Alkoxyl Group) +35
RCO2- (Acyl Group) +6
β- Substituent
R- (Alkyl Group) +12 nm
Cl- (Chloro Group) +12
Br- (Chloro Group) +30
HO- (Hydroxyl Group) +30
RO- (Alkoxyl Group) +30
RCO2- (Acyl Group) +6
RS- (Sulfide Group) +85
R2N- (Amino Group) +95
γ & δ- Substituents
R- (Alkyl Group) +18 nm (both γ & δ)
HO- (Hydroxyl Group) +50 nm (γ)
RO- (Alkoxyl Group) +30 nm (γ)
Further π -Conjugation C=C (Double Bond) ... +30
C6H5 (Phenyl Group) ... +60
Cyclopentenone
202 nm
(i) Each exocyclic double bond adds 5 nm. In the example on the right, there are two exo-double bond components: one to ring A and the other to ring B.
(ii) Homoannular cyclohexadiene component adds +35 nm (ring atoms must be counted separately as substituents)
(iii) Solvent Correction: water = –8; methanol/ethanol = 0; ether = +7; hexane/cyclohexane = +11
λmax (calculated) = Base + Substituent Contributions and Corrections
Examples
UV-Visible Spectroscopy
1. Contributors
A diagram of the components of a typical spectrometer are shown in the following diagram. The functioning of this instrument is relatively straightforward. A beam of light from a visible and/or UV light source (colored red) is separated into its component wavelengths by a prism or diffraction grating. Each monochromatic (single wavelength) beam in turn is split into two equal intensity beams by a half-mirrored device. One beam, the sample beam (colored magenta), passes through a small transparent container (cuvette) containing a solution of the compound being studied in a transparent solvent. The other beam, the reference (colored blue), passes through an identical cuvette containing only the solvent. The intensities of these light beams are then measured by electronic detectors and compared. The intensity of the reference beam, which should have suffered little or no light absorption, is defined as I0. The intensity of the sample beam is defined as I. Over a short period of time, the spectrometer automatically scans all the component wavelengths in the manner described. The ultraviolet (UV) region scanned is normally from 200 to 400 nm, and the visible portion is from 400 to 800 nm.
If the sample compound does not absorb light of of a given wavelength, I = I0. However, if the sample compound absorbs light then I is less than I0, and this difference may be plotted on a graph versus wavelength, as shown on the right. Absorption may be presented as transmittance (T = I/I0) or absorbance (A= log I0/I). If no absorption has occurred, T = 1.0 and A= 0. Most spectrometers display absorbance on the vertical axis, and the commonly observed range is from 0 (100% transmittance) to 2 (1% transmittance). The wavelength of maximum absorbance is a characteristic value, designated as λmax.
Different compounds may have very different absorption maxima and absorbances. Intensely absorbing compounds must be examined in dilute solution, so that significant light energy is received by the detector, and this requires the use of completely transparent (non-absorbing) solvents. The most commonly used solvents are water, ethanol, hexane and cyclohexane. Solvents having double or triple bonds, or heavy atoms (e.g. S, Br & I) are generally avoided. Because the absorbance of a sample will be proportional to its molar concentration in the sample cuvette, a corrected absorption value known as the molar absorptivity is used when comparing the spectra of different compounds. This is defined as:
Molar Absorptivity
ε = A/ c l
( where A= absorbance, c = sample concentration in moles/liter
& l = length of light path through the cuvette in cm.)
For the spectrum on the right, a solution of 0.249 mg of the unsaturated aldehyde in 95% ethanol (1.42 10-5 M) was placed in a 1 cm cuvette for measuement. Using the above formula, ε = 36,600 for the 395 nm peak, and 14,000 for the 255 nm peak. Note that the absorption extends into the visible region of the spectrum, so it is not surprising that this compound is orange colored.
Molar absoptivities may be very large for strongly absorbing compounds (ε >10,000) and very small if absorption is weak (ε = 10 to 100). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Visible_and_Ultraviolet_Spectroscopy/Empirical_Rules_for_Absorption_Wavelengths_of_Conjugated_Systems.txt |
While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo electronic transitions. What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital.
Electronic transitions
Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO).
If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as a σ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm.
When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen.
The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated pi systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores.
Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system . Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2pz atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding.
Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm.
As conjugated pi systems become larger, the energy gap for a π - π* transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π* transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a ΔE of 111 kcal/mol.
In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange.
The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π* transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π* antibonding MO:
This is referred to as an n - π* transition. The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an n - π* transition is smaller that that of a π - π* transition – and thus the n - π* peak is at a longer wavelength. In general, n - π* transitions are weaker (less light absorbed) than those due to π - π* transitions.
Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well).
Template:ExampleEnd
Here is the absorbance spectrum of the common food coloring Red #3:
Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λmax of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes.
Now, take a look at the spectrum of another food coloring, Blue #1:
Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue.
Applications of UV spectroscopy in organic and biological chemistry
UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the Beer - Lambert Law to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance. If we divide the observed value of A at λmax by the concentration of the sample (c, in mol/L), we obtain the molar absorptivity, or extinction coefficient (ε), which is a characteristic value for a given compound.
ε = A/c
The absorbance will also depend, of course, on the path length - in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are mol * L-1cm-1. If we look up the value of e for our compound at λmax, and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD+ the literature value of ε at 260 nm is 18,000 mol * L-1cm-1. In our NAD+ spectrum we observed A260 = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10-5 M.
Template:ExampleStart
The literature value of ε for 1,3-pentadiene in hexane is 26,000 mol * L-1cm-1 at its maximum absorbance at 224 nm. You prepare a sample and take a UV spectrum, finding that A224 = 0.850. What is the concentration of your sample?
Template:ExampleEnd
The bases of DNA and RNA are good chromophores:
Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of ε = 0.020 ng-1×mL for double-stranded DNA at its λmax of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology).
Template:ExampleStart
50 mL of an aqueous sample of double stranded DNA is dissolved in 950 mL of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in mg/mL?
Template:ExampleEnd
Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’).
As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other.
Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores.
Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Visible_and_Ultraviolet_Spectroscopy/Ultraviolet_and_visible_spectroscopy.txt |
Nomenclature of Mercaptans
Mercaptans can be named by naming the parent compound immediately followed by the word thiol. The -SH group can also be named as a substituent using the group name, sulfhydryl. Mercaptans can also be named by naming the carbon group as a separate word followed by the word mercaptan.
• CH3-SH
• methanethiol
• sulfhydrylmethane
• methyl mercaptan
Try to name the following compounds using these conventions-
Try to draw structures for the following compounds-
• 2-sulfhydrylbutane J
• 3-pentanethiol J
b. A common name that you should know is...
thiophenol
- Try to name this compound in two other ways-J
Nomenclature of Sulfides
Sulfides can be named most readily by naming each of the two carbon groups as a separate word followed by a space and the word sulfide.
• CH3-CH2-S-CH3
• ethyl methyl sulfide
Try to name the following compounds-
Try to draw structures for the following compounds-
• 2-butyl 1-propyl sulfide J
• ethyl phenyl sulfide J
b. A common name that you should know is...
thiophene- This compound is aromatic.
Try to draw a structure for the following compound-
• 2-ethenylthiophene J
Nomenclature of Disulfides
Disulfides can be named most readily by naming each of the two carbon groups as a separate word followed by a space and the word disulfide.
• CH3CH2-S-S-CH2CH2CH3
• ethyl 1-propyl disulfide
Try to name the following compound-
Nomenclature of Sulfoxides
Sulfoxides can be named most readily by naming each of the two carbon groups as a separate word followed by a space and the word sulfoxide.
• ethyl methyl sulfoxide
Try to name the following compound-
Nomenclature of Sulfonic Acids
Sulfonic acids can be named most readily by naming the carbon group as a separate word followed by the words sulfonic acid.
phenyl sulfonic acid
Try to draw a structure for the following compound-
• 4-decylphenylsulfonic acid J
Nucleophilicity of Sulfur Compounds
Compounds incorporating a C–S–H functional group are named thiols or mercaptans. Despite their similarity, they are stronger acids and more powerful nucleophiles than alcohols. The IUPAC name of (CH3)3C–SH is 2-methyl-2-propanethiol, commonly called tert-butyl mercaptan.
Nucleophilicity of Sulfur Compounds
Sulfur analogs of alcohols are called thiols or mercaptans, and ether analogs are called sulfides. The chemical behavior of thiols and sulfides contrasts with that of alcohols and ethers in some important ways. Since hydrogen sulfide (H2S) is a much stronger acid than water (by more than ten million fold), we expect, and find, thiols to be stronger acids than equivalent alcohols and phenols. Thiolate conjugate bases are easily formed, and have proven to be excellent nucleophiles in SN2 reactions of alkyl halides and tosylates.
R–S(–) Na(+) + (CH3)2CH–Br (CH3)2CH–S–R + Na(+) Br(–)
Although the basicity of ethers is roughly a hundred times greater than that of equivalent sulfides, the nucleophilicity of sulfur is much greater than that of oxygen, leading to a number of interesting and useful electrophilic substitutions of sulfur that are not normally observed for oxygen. Sulfides, for example, react with alkyl halides to give ternary sulfonium salts (equation # 1) in the same manner that 3º-amines are alkylated to quaternary ammonium salts. Although equivalent oxonium salts of ethers are known, they are only prepared under extreme conditions, and are exceptionally reactive. Remarkably, sulfoxides (equation # 2), sulfinate salts (# 3) and sulfite anion (# 4) also alkylate on sulfur, despite the partial negative formal charge on oxygen and partial positive charge on sulfur.
Oxidation States of Sulfur Compounds
Oxygen assumes only two oxidation states in its organic compounds (–1 in peroxides and –2 in other compounds). Sulfur, on the other hand, is found in oxidation states ranging from –2 to +6, as shown in the following table (some simple inorganic compounds are displayed in orange).
Try drawing Lewis-structures for the sulfur atoms in these compounds. If you restrict your formulas to valence shell electron octets, most of the higher oxidation states will have formal charge separation, as in equation 2 above. The formulas written here neutralize this charge separation by double bonding that expands the valence octet of sulfur. Indeed, the S=O double bonds do not consist of the customary σ & π-orbitals found in carbon double bonds. As a third row element, sulfur has five empty 3d-orbitals that may be used for p-d bonding in a fashion similar to p-p (π) bonding. In this way sulfur may expand an argon-like valence shell octet by two (e.g. sulfoxides) or four (e.g. sulfones) electrons. Sulfoxides have a fixed pyramidal shape (the sulfur non-bonding electron pair occupies one corner of a tetrahedron with sulfur at the center). Consequently, sulfoxides having two different alkyl or aryl substituents are chiral. Enantiomeric sulfoxides are stable and may be isolated.
Thiols also differ dramatically from alcohols in their oxidation chemistry. Oxidation of 1º and 2º-alcohols to aldehydes and ketones changes the oxidation state of carbon but not oxygen. Oxidation of thiols and other sulfur compounds changes the oxidation state of sulfur rather than carbon. We see some representative sulfur oxidations in the following examples. In the first case, mild oxidation converts thiols to disufides. An equivalent oxidation of alcohols to peroxides is not normally observed. The reasons for this different behavior are not hard to identify. The S–S single bond is nearly twice as strong as the O–O bond in peroxides, and the O–H bond is more than 25 kcal/mole stronger than an S–H bond. Thus, thermodynamics favors disulfide formation over peroxide.
Mild oxidation of disufides with chlorine gives alkylsulfenyl chlorides, but more vigorous oxidation forms sulfonic acids (2nd example). Finally, oxidation of sulfides with hydrogen peroxide (or peracids) leads first to sulfoxides and then to sulfones.
The nomenclature of sulfur compounds is generally straightforward. The prefix thio denotes replacement of a functional oxygen by sulfur. Thus, -SH is a thiol and C=S a thione. The prefix thia denotes replacement of a carbon atom in a chain or ring by sulfur, although a single ether-like sulfur is usually named as a sulfide. For example, C2H5SC3H7 is ethyl propyl sulfide and C2H5SCH2SC3H7 may be named 3,5-dithiaoctane. Sulfonates are sulfonate acid esters and sultones are the equivalent of lactones. Other names are noted in the table above.
Oxidation of Alcohols by DMSO
The conversion of 1º and 2º-alcohols to aldehydes and ketones is an important reaction which, in its simplest form, can be considered a dehydrogenation (loss of H2). By providing an oxygen source to fix the product hydrogen as water, the endothermic dehydrogenation process may be converted to a more favorable exothermic one. One source of oxygen that has proven effective for the oxidation of alcohols is the simple sulfoxide solvent, DMSO. The reaction is operationally easy: a DMSO solution of the alcohol is treated with one of several electrophilic dehydrating reagents (E). The alcohol is oxidized; DMSO is reduced to dimethyl sulfide; and water is taken up by the electrophile. Due to the exothermic nature of the reaction, it is usually run at -50 ºC or lower. Co-solvents such as methylene chloride or THF are needed, since pure DMSO freezes at 18º. The reaction of oxalyl chloride with DMSO may generate chlorodimethylsulfonium chloride which then oxidizes the alcohol (Swern Oxidation). Alternatively, a plausible general mechanism for this interesting and useful reaction is drawn below.
Because so many different electrophiles have been used to effect this oxidation, it is difficult to present a single general mechanism. Most of the electrophiles are good acylating reagents, so it is reasonable to expect an initial acylation of the sulfoxide oxygen. (The use of DCC as an acylation reagent was described elsewhere.) The electrophilic character of the sulfur atom is enhanced by acylation. Bonding of sulfur to the alcohol oxygen atom then follows. The remaining steps are eliminations, similar in nature to those proposed for other alcohol oxidations. In some cases triethyl amine is added to provide an additional base. Three examples of these DMSO oxidations are given in the following diagram. Note that this oxidation procedure is very mild and tolerates a variety of other functional groups, including those having oxidizable nitrogen and sulfur atoms. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Thiols_and_Sulfides/Nomenclature_of_Sulfur_Compounds.txt |
Objectives
After completing this section, you should be able to
1. Nomenclature and Reactivity
1. write the IUPAC name of a thiol, given its Kekulé, condensed or shorthand structure.
2. draw the structure of a thiol, given its IUPAC name.
3. write an equation to represent the formation of a thiol by the reaction of hydrosulfide anion with an alkyl halide.
4. write an equation to illustrate the preparation of a thiol by the reaction of thiourea with an alkyl halide.
2. write an equation to show the interconversion between thiols and disulfides.
1. write the name of a sulfide, given its structure.
2. draw the structure of a sulfide, given its name.
3. write an equation showing how a sulfide may be prepared by the reaction of a thiolate anion on an alkyl halide.
4. identify the product from the reaction of a given alkyl halide with a given thiolate anion.
5. identify the reagents necessary to prepare a given sulfide.
6. write an equation to illustrate the formation of a trialkylsulfonium salt from a sulfide and an alkyl halide.
Key Terms
• disulfide
• mercapto group
• (organic) sulfide
• sulfone
• sulfoxide
• thiol
• thiolate anion
• trialkylsulfonium ion (trialkylsulfonium salt)
Study Notes
The chemistry of sulfur-containing organic compounds is often omitted from introductory organic chemistry courses. However, we have included a short section on these compounds, not for the sake of increasing the amount of material to be digested, but because much of the chemistry of these substances can be predicted from a knowledge of their oxygen-containing analogues. A thiol is a compound which contains an SH functional group. The -SH group itself is called a mercapto group. A disulfide is a compound containing an -S-S- linkage. (Organic) sulfides have the structure R-S-R′, and are therefore the sulfur analogues of ethers. The nomenclature of sulfides can be easily understood if one understands the nomenclature of the corresponding ethers. Notice that the term “thio” is also used in inorganic chemistry. For example, SO42 is the sulfate ion; while S2O32, in which one of the oxygen atoms of a sulfate ion has been replaced by a sulfur atom, is called thiosulfate. Thiolate anions, RS- , are analogous to alkoxy anions, RO- . Thiolate anions are better nucleophiles than are alkoxy anions (see Section 11.5, pages 389-394 of the textbook). If you have trouble understanding why trialkylsulfonium ions are formed, think of them as being somewhat similar to the hydronium ions that are formed by protonating water:
Later we shall see examples of tetraalkylammonium ions, R4N+, which again may be regarded as being similar to hydronium ions. sulfoxides and sulfones are obtained by oxidizing organic sulfides. You need not memorize the methods used to carry out these oxidations.
Table 18.1, below, provides a quick comparison of oxygen-containing and sulfur-containing organic compounds.
Oxidation States of Sulfur Compounds
Oxygen assumes only two oxidation states in its organic compounds (–1 in peroxides and –2 in other compounds). Sulfur, on the other hand, is found in oxidation states ranging from –2 to +6, as shown in the following table (some simple inorganic compounds are displayed in orange).
Thiols
Thiols, which are also called mercaptans, are analogous to alcohols. They are named in a similar fashion as alcohols except the suffix -thiol is used in place of -ol. By itself the -SH group is called a mercapto group.
Thiols are usually prepared by using the hydrosulfide anion (-SH) as a neucleophile in an SN2 reaction with alkyl halides.
On problem with this reaction is that the thiol product can undergo a second SN2 reaction with an additional alkyl halide to produce a sulfide side product. This problem can be solved by using thiourea, (NH2)2C=S, as the nucleophile. The reaction first produces an alkyl isothiourea salt and an intermediate. This salt is then hydrolyzed by a reaction with aqueous base.
Disulfides
Oxidation of thiols and other sulfur compounds changes the oxidation state of sulfur rather than carbon. We see some representative sulfur oxidations in the following examples. In the first case, mild oxidation converts thiols to disufides. An equivalent oxidation of alcohols to peroxides is not normally observed. The reasons for this different behavior are not hard to identify. The S–S single bond is nearly twice as strong as the O–O bond in peroxides, and the O–H bond is more than 25 kcal/mole stronger than an S–H bond. Thus, thermodynamics favors disulfide formation over peroxide.
Disulfide bridges in proteins
Disulfide (sulfur-sulfur) linkages between two cysteine residues are an integral component of the three-dimensional structure of many proteins. The interconversion between thiols and disulfide groups is a redox reaction: the thiol is the reduced state, and the disulfide is the oxidized state.
Notice that in the oxidized (disulfide) state, each sulfur atom has lost a bond to hydrogen and gained a bond to a sulfur - this is why the disulfide state is considered to be oxidized relative to the thiol state.
The redox agent that mediates the formation and degradation of disulfide bridges in most proteins is glutathione, a versatile coenzyme that we have met before in a different context (section 14.2A). Recall that the important functional group in glutathione is the thiol, highlighted in blue in the figure below. In its reduced (free thiol) form, glutathione is abbreviated 'GSH'.
In its oxidized form, glutathione exists as a dimer of two molecules linked by a disulfide group, and is abbreviated 'GSSG'.
A new disulfide in a protein forms via a 'disulfide exchange' reaction with GSSH, a process that can be described as a combination of two SN2-like attacks. The end result is that a new cysteine-cysteine disulfide forms at the expense of the disulfide in GSSG.
In its reduced (thiol) state, glutathione can reduce disulfides bridges in proteins through the reverse of the above reaction.
Disulfide bridges exist for the most part only in proteins that are located outside the cell. Inside the cell, cysteines are kept in their reduced (free thiol) state by a high intracellular concentration of GSH, which in turn is kept in a reduced state (ie. GSH rather than GSSG) by a flavin-dependent enzyme called glutathione reductase.
Disulfide bridges in proteins can also be directly reduced by another flavin-dependent enzyme called 'thioredoxin'. In both cases, NADPH is the ultimate electron donor, reducing FAD back to FADH2 in each catalytic cycle.
In the biochemistry lab, proteins are often maintained in their reduced (free thiol) state by incubation in buffer containing an excess concentration of b-mercaptoethanol (BME) or dithiothreitol (DTT). These reducing agents function in a manner similar to that of GSH, except that DTT, because it has two thiol groups, forms an intramolecular disulfide in its oxidized form.
Sulfides
Sulfur analogs of ethers are called sulfides. The chemical behavior of sulfides contrasts with that of ethers in some important ways. Since hydrogen sulfide (H2S) is a much stronger acid than water (by more than ten million fold), we expect, and find, thiols to be stronger acids than equivalent alcohols and phenols. Thiolate conjugate bases are easily formed, and have proven to be excellent nucleophiles in SN2 reactions of alkyl halides and tosylates.
R–S(–) Na(+) + (CH3)2CH–Br (CH3)2CH–S–R + Na(+) Br(–)
Although the basicity of ethers is roughly a hundred times greater than that of equivalent sulfides, the nucleophilicity of sulfur is much greater than that of oxygen, leading to a number of interesting and useful electrophilic substitutions of sulfur that are not normally observed for oxygen. Sulfides, for example, react with alkyl halides to give ternary sulfonium salts (equation # 1) in the same manner that 3º-amines are alkylated to quaternary ammonium salts. Although equivalent oxonium salts of ethers are known, they are only prepared under extreme conditions, and are exceptionally reactive.
sulfides are named using the same rules as ethers except sulfide is used in the place of ether. For more complex substance alkylthio is used instead of alkoxy.
SAM methyltransferases
The most common example of sulfonium ions in a living organism is the reaction of S-Adenosylmethionine. Some of the most important examples of SN2 reactions in biochemistry are those catalyzed by S-adenosyl methionine (SAM) – dependent methyltransferase enzymes. We have already seen, in chapter 6 and again in chapter 8, how a methyl group is transferred in an SN2 reaction from SAM to the amine group on the nucleotide base adenosine:
Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-O-methyltransferase. The substrate here is epinephrine, also known as adrenaline.
Notice that in this example, the attacking nucleophile is an alcohol rather than an amine (that’s why the enzyme is called an O-methyltransferase). In both cases, though, a basic amino acid side chain is positioned in the active site in just the right place to deprotonate the nucleophilic group as it attacks, increasing its nucleophilicity. The electrophile in both reactions is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The methyl carbon is electrophilic because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, as the resulting product will be a neutral and very stable sulfide. All in all, in both reactions we have a reasonably good nucleophile, an electron-poor, unhindered electrophile, and an excellent leaving group.
Because the electrophilic carbon in these reactions is a methyl carbon, a stepwise SN1-like mechanism is extremely unlikely: a methyl carbocation is very high in energy and thus is not a reasonable intermediate to propose. We can confidently predict that this reaction is SN2. Does this SN2 reaction occur, as expected, with inversion of stereochemistry? Of course, the electrophilic methyl carbon in these reactions is achiral, so inversion is not apparent. To demonstrate inversion, the following experiment has been carried out with catechol-O-methyltransferase:
Here, the methyl group of SAM was made to be chiral by incorporating hydrogen isotopes tritium (3H, T) and deuterium (2H, D). The researchers determined that the reaction occurred with inversion of configuration, as expected for an SN2 displacement (J. Biol. Chem. 1980, 255, 9124).
Sulfides can be easily oxidized. Reacting a sulfide with hydrogen peroxide, H2O2, as room termpeature produces a sulfoxide (R2SO). The oxidation can be continued by reaction with a peroxyacid to produce the sulfone (R2SO2)
A common example of a sulfoxide is the solvent dimethyl sulfoxide (DMSO). DMSO is polar aprotic solvent.
Figure AB16.3. DMSO is a very polar, aprotic solvent. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Thiols_and_Sulfides/Thiols_and_Sulfides.txt |
• 1.1: The Lagrangian Formulation of Classical Mechanics
In order to begin to make a connection between the microscopic and macroscopic worlds, we need to better understand the microscopic world and the laws that govern it. We will begin placing Newton's laws of motion in a formal framework which will be heavily used in our study of classical statistical mechanics.
• 1.2: The Hamiltonian formulation of classical mechanics
The Lagrangian formulation of mechanics will be useful later when we study the Feynman path integral. For our purposes now, the Lagrangian formulation is an important springboard from which to develop another useful formulation of classical mechanics known as the Hamiltonian formulation. The Hamiltonian of a system is defined to be the sum of the kinetic and potential energies expressed as a function of positions and their conjugate momenta.
• 1.3: The Microscopic Laws of Motion
• 1.4: Phase Space
We construct a Cartesian space in which each of the 6N coordinates and momenta is assigned to one of 6N mutually orthogonal axes. Phase space is, therefore, a 6N dimensional space. A point in this space is specified by giving a particular set of values for the 6N coordinates and momenta.
• 1.5: Classical microscopic states or microstates and ensembles
• 1.6: Phase space distribution functions and Liouville's theorem
Given an ensemble with many members, each member having a different phase space vector x corresponding to a different microstate, we need a way of describing how the phase space vectors of the members in the ensemble will be distributed in the phase space.
01: Classical mechanics
In order to begin to make a connection between the microscopic and macroscopic worlds, we need to better understand the microscopic world and the laws that govern it. We will begin placing Newton's laws of motion in a formal framework which will be heavily used in our study of classical statistical mechanics.
First, we begin by restricting our discussion to systems for which the forces are purely conservative. Such forces are derivable from a potential energy function $U (r_1, \cdots , r_N)$ by differentiation:
$F_i = - \frac {\partial U}{\partial r_i} \nonumber$
It is clear that such forces cannot contain dissipative or friction terms. An important property of systems whose forces are conservative is that they conserve the total energy
$E = K + U = \frac {1}{2} \sum _{i=1}^N m_i \dot {r} ^2_i + U (r_1, \cdots , r_N ) \nonumber$
To see this, simply differentiate the energy with respect to time:
\begin{align*} \frac {dE}{dt} &= \sum _{i=1}^N m_i \dot {r} _i \cdot \ddot {r}_i + \sum _{i=1}^N \frac {\partial U}{\partial r_i} \cdot \dot {r} _i \[4pt] &=\sum _{i=1}^N \dot {r} _i \cdot F_i - \sum _{i=1}^N F_i \cdot \dot {r} _i \[4pt] &= 0 \end{align*}
where, the second line, the facts that $\ddot {r} _i = \frac {F_i}{m_i}$ (Newton's law) and $F_i = - \frac {\partial U}{\partial r_i}$ (conservative force definition) have been used. This is known as the law of conservation of energy.
For conservative systems, there is an elegant formulation of classical mechanics known as the Lagrangian formulation. The Lagrangian function, $L$, for a system is defined to be the difference between the kinetic and potential energies expressed as a function of positions and velocities. In order to make the nomenclature more compact, we shall introduce a shorthand for the complete set of positions in an $N$-particle system: $r \equiv r_1, \cdots , r_N$ and for the velocities: $\dot {r} \equiv \dot {r}_1, \cdots , \dot {r} _N$. Then, the Lagrangian is defined as follows:
$L (r, \dot {r} ) = K - U = \sum _{i=1}^N \frac {1}{2} m_i \dot {r}^2_i - U (r_1, \cdots , r_N ) \nonumber$
In terms of the Lagrangian, the classical equations of motion are given by the so called Euler-Lagrange equation:
$\frac {d}{dt} \left ( \frac {\partial L}{\partial \dot {r} _i} \right ) - \frac {\partial L}{\partial r_i} = 0 \nonumber$
The equations that result from application of the Euler-Lagrange equation to a particular Lagrangian are known as the equations of motion. The solution of the equations of motion for a given initial condition is known as a trajectory of the system. The Euler-Lagrange equation results from what is known as an action principle. We shall defer further discussion of the action principle until we study the Feynman path integral formulation of quantum statistical mechanics in terms of which the action principle emerges very naturally. For now, we accept the Euler-Lagrange equation as a definition.
The Euler-Lagrange formulation is completely equivalent to Newton's second law. In order to see this, note that
\begin{align*} \frac {\partial L}{\partial \dot {r} _i} &= m_i \dot {r} _i \[4pt] \frac {\partial L}{\partial r _i} &= - \frac {\partial U}{\partial r _i} = F_i \end{align*}
Therefore,
$\frac {d}{dt} \left ( \frac {\partial L}{\partial \dot {r} _i} \right ) - \frac {\partial L}{\partial r _i} = m_i \ddot {r} _i - F_i = 0 \nonumber$
which is just Newton's equation of motion.
An important property of the Lagrangian formulation is that it can be used to obtain the equations of motion of a system in any set of coordinates, not just the standard Cartesian coordinates, via the Euler-Lagrange equation (see problem set #1). | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/01%3A_Classical_mechanics/1.01%3A_The_Lagrangian_Formulation_of_Classical_Mechanics.txt |
The Lagrangian formulation of mechanics will be useful later when we study the Feynman path integral. For our purposes now, the Lagrangian formulation is an important springboard from which to develop another useful formulation of classical mechanics known as the Hamiltonian formulation. The Hamiltonian of a system is defined to be the sum of the kinetic and potential energies expressed as a function of positions and their conjugate momenta. What are conjugate momenta?
Recall from elementary physics that momentum of a particle, $P_i$, is defined in terms of its velocity $\dot {r}_i$ by
$p_i = m_i \dot {r} _i \nonumber$
In fact, the more general definition of conjugate momentum, valid for any set of coordinates, is given in terms of the Lagrangian:
$p_i = \frac {\partial L}{\partial \dot {r} _i } \nonumber$
Note that these two definitions are equivalent for Cartesian variables. In terms of Cartesian momenta, the kinetic energy is given by
$K = \sum _{i=1}^N \frac {P^2_i}{2 m_i} \nonumber$
Then, the Hamiltonian, which is defined to be the sum, $K + U$, expressed as a function of positions and momenta, will be given by
$H (p, r) = \sum _{i=1}^N \frac {P^2_i}{2m_i} + U ( r_1, \cdots , r_N) = H (p, r) \nonumber$
where $p \equiv p_1, \cdots , p_N$. In terms of the Hamiltonian, the equations of motion of a system are given by Hamilton's equations:
$\dot {r} _i = \frac {\partial H}{\partial p_i} \dot {p}_i = - \frac {\partial H}{\partial r_i} \nonumber$
The solution of Hamilton's equations of motion will yield a trajectory in terms of positions and momenta as functions of time. Again, Hamilton's equations can be easily shown to be equivalent to Newton's equations, and, like the Lagrangian formulation, Hamilton's equations can be used to determine the equations of motion of a system in any set of coordinates.
The Hamiltonian and Lagrangian formulations possess an interesting connection. The Hamiltonian can be directly obtained from the Lagrangian by a transformation known as a Legendre transform. We will say more about Legendre transforms in a later lecture. For now, note that the connection is given by
$H (p, r) = \sum _{i=1}^N p_i \cdot \dot {r}_i - L (r , \dot {r} ) \nonumber$
which, when the fact that $\dot {r}_i = \frac {p_i}{m_i}$ is used, becomes
$H (p, r)= \sum _{i=1}^N p_i \cdot \frac {p_i}{m_i} - \sum _{i=1}^N \frac {1}{2} m_i \left (\frac {p_i}{m_i} \right )^2 + U (r_1, \cdots , r_N ) \nonumber$
$= \sum _{i=1}^N \frac {P_i^2}{2m_i} + U(r_1, \cdots , r_N ) \nonumber$
Because a system described by conservative forces conserves the total energy, it follows that Hamilton's equations of motion conserve the total Hamiltonian. Hamilton's equations of motion conserve the Hamiltonian
$H (p (t), r (t) ) = H (p(0), r (0) ) = E \nonumber$
Proof
$H = \text {const} \Rightarrow \frac {dH}{dt} = 0$
$\frac {dH}{dt}= \sum _{i=1}^N \left ( \frac {\partial H}{\partial r_i} \cdot \dot {r} _i + \frac {\partial H}{\partial p_i} \cdot \dot {p} _i \right ) \nonumber$
$=\sum _{i=1}^N \left ( \frac {\partial H}{\partial r_i} \cdot \frac {\partial H}{\partial p_i} - \frac {\partial H}{\partial p_i} \cdot \frac {\partial H}{\partial r_i} \right ) = 0 \nonumber$
QED. This, then, provides another expression of the law of conservation of energy. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/01%3A_Classical_mechanics/1.02%3A_The_Hamiltonian_formulation_of_classical_mechanics.txt |
Consider a system of $N$ classical particles. The particles a confined to a particular region of space by a "container'' of volume $V$. The particles have a finite kinetic energy and are therefore in constant motion, driven by the forces they exert on each other (and any external forces which may be present). At a given instant in time $t$, the Cartesian positions of the particles are $r_1(t), \cdots , r_N(t)$) ) . The time evolution of the positions of the particles is then given by Newton's second law of motion:
$m_i \ddot {r} _i = F_i ( r_1, \cdots , r_N ) \nonumber$
where $F_1, \cdots , F_N$ are the forces on each of the $N$ particles due to all the other particles in the system. The notation $\ddot {r} _i = \frac {d^2 r_i}{dt^2}$.
$N$ Newton's equations of motion constitute a set of $3N$ coupled second order differential equations. In order to solve these, it is necessary to specify a set of appropriate initial conditions on the coordinates and their first time derivatives, $\{r_1 (0), \cdots , r_N(0), \dot {r} _1 (0), \cdots , \dot {r} _N (0) \}$. Then, the solution of Newton's equations gives the complete set of coordinates and velocities for all time $t$.
1.04: Phase Space
We construct a Cartesian space in which each of the $6N$ coordinates and momenta is assigned to one of $6N$ mutually orthogonal axes. Phase space is, therefore, a $6N$ dimensional space. A point in this space is specified by giving a particular set of values for the $6N$ coordinates and momenta. Denote such a point by
$x = (p_1, \cdots , p_N, r_1, \cdots , r_N ) \nonumber$
$x$ is a $6N$ dimensional vector. Thus, the time evolution or trajectory of a system as specified by Hamilton's equations of motion, can be expressed by giving the phase space vector, $x$ as a function of time.
The law of conservation of energy, expressed as a condition on the phase space vector:
$H(x(t)) = \text {const} = E \nonumber$
defines a $6N - 1$ dimensional hypersurface in phase space on which the trajectory must remain.
1.05: Classical microscopic states or microstates and ensembles
A microscopic state or microstate of a classical system is a specification of the complete set of positions and momenta of the system at any given time. In the language of phase space vectors, it is a specification of the complete phase space vector of a system at any instant in time. For a conservative system, any valid microstate must lie on the constant energy hypersurface, $H (x) = E$. Hence, specifying a microstate of a classical system is equivalent to specifying a point on the constant energy hypersurface.
The concept of classical microstates now allows us to give a more formal definition of an ensemble. An ensemble is a collection of systems sharing one or more macroscopic characteristics but each being in a unique microstate. The complete ensemble is specified by giving all systems or microstates consistent with the common macroscopic characteristics of the ensemble.
The idea of ensemble averaging can also be expressed in terms of an average over all such microstates (which comprise the ensemble). A given macroscopic property, $A$, and its microscopic function $a = a (x)$, which is a function of the positions and momenta of a system, i.e. the phase space vector, are related by
$A = \langle a \rangle_{ensemble} = \frac {1}{N} \sum _{\lambda = 1}^N a(x_{\lambda}) \nonumber$
where $x_{\lambda}$ is the microstate of the $\lambda$ th member of the ensemble. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/01%3A_Classical_mechanics/1.03%3A_The_Microscopic_Laws_of_Motion.txt |
Given an ensemble with many members, each member having a different phase space vector $x$ corresponding to a different microstate, we need a way of describing how the phase space vectors of the members in the ensemble will be distributed in the phase space. That is, if we choose to observe one particular member in the ensemble, what is the probability that its phase space vector will be in a small volume $dx$ around a point $x$ in the phase space at time $t$. This probability will be denoted
$f (x,t) dx \nonumber$
where $f (x, t)$ is known as the phase space probability density or phase space distribution function. It's properties are as follows:
$f (x, t ) \ge (0 \nonumber$
$\int dx f (x, t)$ = Number of members in the ensemble
Liouville's Theorem
The total number of systems in the ensemble is a constant. What restrictions does this place on $f (x, t )$? For a given volume $\Omega$ in phase space, this condition requires that the rate of decrease of the number of systems from this region is equal to the flux of systems into the volume.
Let $\hat {n}$ be the unit normal vector to the surface of this region.
The flux through the small surface area element, $dS$ is just $\hat {n}\cdot \dot {x}f(x,t)dS$. Then the total flux out of volume is obtained by integrating this over the entire surface that encloses $\Omega$:
$\int dS \hat {n} \cdot (\dot {x} f(x,t)) = \int_{\Omega}\nabla _{x}\cdot(\dot{x} f(x,t)) \nonumber$
which follows from the divergence theorem. $\nabla _x$ is the $6N$ dimensional gradient on the phase space
$\nabla _x = \left(\frac {\partial}{\partial p_1},\cdots,\frac {\partial}{\partial p_N}, \frac {\partial}{\partial r_1},\cdots, \frac {\partial}{\partial r_N}\right) \nonumber$
$=\left(\nabla_{p_1},\cdots,\nabla_{p_N},\nabla_{r_1},\cdots,\nabla_{r_N}\right) \nonumber$
On the other hand, the rate of decrease in the number of systems out of the volume is
$-\frac {d}{dt}\int_{\Omega} d{x}f(x,t) = -\int_{\Omega}d{x} \frac {\partial}{\partial t} f(x,t) \nonumber$
Equating these two quantities gives
$\int_{\Omega} dx \nabla_{x} \cdot (\dot{x}f( x,t)) = -\int_{\Omega} d{x} \frac {\partial}{\partial t}f(x,t) \nonumber$
But this result must hold for any arbitrary choice of the volume $\Omega$, which we may also allow to shrink to 0 so that the result holds locally, and we obtain the local result:
$\frac {\partial}{\partial t}f(x,t) + \nabla_{x} \cdot (\dot {x} f(x,t)) = 0 \nonumber$
But
$\nabla _{x} \cdot (\dot {x} f (x,t)) = \dot {x} \cdot \nabla _x f(x,t) + f(x,t) \nabla _{x} \cdot \dot {x} \nonumber$
This equation resembles an equation for a "hydrodynamic'' flow in the phase space, with $f (x, t )$ playing the role of a density. The quantity $\nabla _x \cdot \dot {x}$, being the divergence of a velocity field, is known as the phase space compressibility, and it does not, for a general dynamical system, vanish. Let us see what the phase space compressibility for a Hamiltonian system is:
$\nabla_{x}\cdot\dot{x} = \sum_{i=1}^{N}\left[\nabla _{p_i} \cdot \dot {p} _i + \nabla_{r_i}\cdot \dot{r}_i \right] \nonumber$
However, by Hamilton's equations:
$\dot {p}_i = -\nabla_{r_i}H\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\dot{r}_i = \nabla_{p_i}H \nonumber$
Thus, the compressibility is given by
$\nabla_{x}\cdot\dot{x} =\sum_{i=1}^N \left[-\nabla _{p_i} \cdot \nabla _{r_i}H +\nabla_{r_i}\cdot\nabla_{p_i}H\right] = 0 \nonumber$
Thus, Hamiltonian systems are incompressible in the phase space, and the equation for $f (x, t )$ becomes
$\frac {\partial}{\partial t}f(x,t) + \dot{x}\cdot \nabla_{x}f(x,t) = \frac {df}{dt} = 0 \nonumber$
which is Liouville's equation, and it implies that $f (x, t )$ is a conserved quantity when $x$ is identified as the phase space vector of a particular Hamiltonian system. That is, $f (x_t, t)$ will be conserved along a particular trajectory of a Hamiltonian system. However, if we view $x$ is a fixed spatial label in the phase space, then the Liouville equation specifies how a phase space distribution function $f (x, t )$ evolves in time from an initial distribution $f (x, 0 )$. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/01%3A_Classical_mechanics/1.06%3A_Phase_space_distribution_functions_and_Liouville%27s_theorem.txt |
• 2.1: The Ensemble Concept (Heuristic Definition)
Since, from the point of view of macroscopic properties, precise microscopic details are largely unimportant, we might imagine employing a construct known as the ensemble concept in which a large number of systems with different microscopic characteristics but similar macroscopic characteristics is used to "wash out'' the microscopic details via an averaging procedure. This is an idea developed by individuals such as Gibbs, Maxwell, and Boltzmann.
• 2.2: Liouville's Theorem for non-Hamiltonian systems
Non-Hamiltonian dynamical systems are often used to describe open systems, i.e., systems in contact with heat reservoirs or mechanical pistons or particle reservoirs. They are also often used to describe driven systems or systems in contact with external fields.
• 2.3: The Liouville Operator and the Poisson Bracket
The Liouville equation is the foundation on which statistical mechanics rests. It will now be cast in a form that will be suggestive of a more general structure that has a definite quantum analog (to be revisited when we treat the quantum Liouville equation).
• 2.4: Equilibrium ensembles
02: Foundations of classical statistical mechanics
For a typical macroscopic system, the total number of particles $N \sim 10^{23}$. Since an essentially infinite amount of precision is needed in order to specify the initial conditions (due to exponentially rapid growth of errors in this specification), the amount of information required to specify a trajectory is essentially infinite. Even if we contented ourselves with quadrupole precision, however, the amount of memory needed to hold just one phase space point would be about 128 bytes = $2^7 \sim 10^2$ bytes for each number or $10^2 \times 6 \times 10^{23} \sim 10^{17}$ Gbytes. The largest computers we have today have perhaps $10^3$ Gbytes of memory, so we are off by 14 orders of magnitude just to specify 1 point in phase space.
Example $1$
Do we need all this detail?
Yes
There are plenty of chemically interesting phenomena for which we really would like to know how individual atoms are moving as a process occurs. Experimental techniques such as ultrafast laser spectroscopy can resolve short time scale phenomena and, thus, obtain important insights into such motions. From a theoretical point of view, although we cannot follow $10^{23}$ particles, there is some hope that we could follow the motion of a system containing $10^4$ or $10^5$ particles, which might capture most of the features of true macroscopic matter. Thus, by solving Newton's equations of motion numerically on a computer, we have a kind of window into the microscopic world. This is the basis of what are known as molecular dynamics calculations.
No
Intuitively, we would expect that if we were to follow the evolution of a large number of systems all described by the same set of forces but having starting from different initial conditions, these systems would have essentially the same macroscopic characteristics, e.g. the same temperature, pressure, etc. even if the microscopic detailed evolution of each system in time would be very different. This idea suggests that the microscopic details are largely unimportant.
Since, from the point of view of macroscopic properties, precise microscopic details are largely unimportant, we might imagine employing a construct known as the ensemble concept in which a large number of systems with different microscopic characteristics but similar macroscopic characteristics is used to "wash out'' the microscopic details via an averaging procedure. This is an idea developed by individuals such as Gibbs, Maxwell, and Boltzmann.
Ensemble
Consider a large number of systems each described by the same set of microscopic forces and sharing some common macroscopic property (e.g. the same total energy). Each system is assumed to evolve under the microscopic laws of motion from a different initial condition so that the time evolution of each system will be different from all the others. Such a collection of systems is called an ensemble. The ensemble concept then states that macroscopic observables can be calculated by performing averages over the systems in the ensemble. For many properties, such as temperature and pressure, which are time-independent, the fact that the systems are evolving in time will not affect their values, and we may perform averages at a particular instant in time. Thus, let $A$ denote a macroscopic property and let $a$ denote a microscopic function that is used to compute $A$. An example of $A$ would be the temperature, and $a$ would be the kinetic energy (a microscopic function of velocities). Then, $A$ is obtained by calculating the value of $a$ in each system of the ensemble and performing an average over all systems in the ensemble:
$A = \frac {1}{N} \sum _{\lambda = 1}^N a_{\lambda} \nonumber$
where $N$ is the total number of members in the ensemble and $a_{\lambda}$ is the value of $a$ in the $\lambda$ th system.
The questions that naturally arise are:
1. How do we construct an ensemble?
2. How do we perform averages over an ensemble?
3. How many systems will an ensemble contain?
4. How do we distinguish time-independent from time-dependent properties in the ensemble picture?
Answering these questions will be our main objective in Statistical Mechanics. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/02%3A_Foundations_of_classical_statistical_mechanics/2.01%3A_The_Ensemble_Concept_%28Heuristic_Definition%29.txt |
The equations of motion of a system can be cast in the generic form
$\dot {x} = \xi (x) \nonumber$
where, for a Hamiltonian system, the vector function $\xi$ would be
$\xi (x) = \left ( - \dfrac {\partial H}{\partial r_1} , \cdots , - \dfrac {\partial H}{\partial r_N} , \dfrac {\partial H}{\partial p_1}, \cdots , \dfrac {\partial H}{\partial p_N} \right ) \nonumber$
and the incompressibility condition would be a condition on $\xi$:
$\Delta _x \cdot \dot {x} = \Delta _x \cdot \xi = 0 \nonumber$
A non-Hamiltonian system, described by a general vector funciton $\xi$, will not, in general, satisfy the incompressibility condition. That is:
$\Delta _x \cdot \dot {x} = \Delta _x \cdot \xi \ne 0 \nonumber$
Non-Hamiltonian dynamical systems are often used to describe open systems, i.e., systems in contact with heat reservoirs or mechanical pistons or particle reservoirs. They are also often used to describe driven systems or systems in contact with external fields.
The fact that the compressibility does not vanish has interesting consequences for the structure of the phase space. The Jacobian, which satisfies
$\dfrac {d J}{ dt} = J \Delta _x \cdot \dot {x} \nonumber$
will no longer be 1 for all time. Defining $k = \Delta _x \cdot \dot {x}$, the general solution for the Jacobian can be written as
$J ( x_t; x_0 ) = J ( x_0 ; x_0) exp \left ( \int \limits _0 ^t dR k (x_A) \right ) \nonumber$
Note that $J (x_0; x_0 ) = 1$ as before. Also, note that $k = d \ln \dfrac {J}{dt}$. Thus, $k$ can be expressed as the total time derivative of some function, which we will denote W, i.e., $k = \dot {W}$. Then, the Jacobian becomes
\begin{align*} J (x_t ; x_0) &= exp \left ( \int \limits _0^t dR W (x_A) \right ) \[4pt] &= exp ( W (x_t) - W (x_0)) \end{align*}
Thus, the volume element in phase space now transforms according to
$dx_t = exp \left ( W (x_t) - W (x_0) \right ) dx_0 \nonumber$
which can be arranged to read as a conservation law:
$e^{-W(x_t)} dx_t = e^{-W(x_0)} dx_0 \nonumber${
Thus, we have a conservation law for a modified volume element, involving a "metric factor'' $exp (-W (x))$. Introducing the suggestive notation $\sqrt {g} = exp (-W(x))$, the conservation law reads $\sqrt {g(x_t} dx_t = \sqrt {g(x_0} dx_0$. This is a generalized version of Liouville's theorem. Furthermore, a generalized Liouville equation for non-Hamiltonian systems can be derived which incorporates this metric factor. The derivation is beyond the scope of this course, however, the result is
$\partial (f \sqrt {g}) + \nabla _x \cdot (\dot {x} f \sqrt {g} ) = 0 \nonumber$
We have called this equation, the generalized Liouville equation Finally, noting that $\sqrt {g}$ satisfies the same equation as J, i.e.,
$\dfrac {d \sqrt {g}}{dt} = k \sqrt {g} \nonumber$
the presence of $\sqrt {g}$ in the generalized Liouville equation can be eliminated, resulting in
$\dfrac {\partial f}{\partial t} + \dot {x} \cdot \nabla _x f = \dfrac {df}{dt} = 0 \nonumber$
which is the ordinary Liouville equation from before. Thus, we have derived a modified version of Liouville's theorem and have shown that it leads to a conservation law for f equivalent to the Hamiltonian case. This, then, supports the generality of the Liouville equation for both Hamiltonian and non-Hamiltonian based ensembles, an important fact considering that this equation is the foundation of statistical mechanics. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/02%3A_Foundations_of_classical_statistical_mechanics/2.02%3A_Liouville%27s_Theorem_for_non-Hamiltonian_systems.txt |
From the last lecture, we saw that Liouville's equation could be cast in the form
$\frac {\partial f}{\partial t} + \nabla _x \cdot \dot {x} f = 0 \nonumber$
The Liouville equation is the foundation on which statistical mechanics rests. It will now be cast in a form that will be suggestive of a more general structure that has a definite quantum analog (to be revisited when we treat the quantum Liouville equation).
Define an operator
$iL = \dot {x} \cdot \nabla _x \nonumber$
known as the Liouville operator ( $i = \sqrt {-1}$ - the i is there as a matter of convention and has the effect of making $L$ a Hermitian operator). Then Liouville's equation can be written
$\frac {\partial f}{\partial t} + iLf = 0 \nonumber$
The Liouville operator also be expressed as
$iL = \sum _{i=1}^N \left [ \frac {\partial H}{\partial p_i} \cdot \frac {\partial}{\partial r_i} - \frac {\partial H}{\partial r_i} \cdot \frac {\partial}{\partial p_i} \right ] \equiv \left \{ \cdots , H \right \} \nonumber$
where $\{ A, B \}$ is known as the Poisson bracket between $A(x)$ and $B (x)$:
$\left \{ A, B \right \} = \sum _{i=1}^N \left [ \frac {\partial A}{\partial r_i} \cdot \frac {\partial B}{\partial p_i} - \frac {\partial A}{\partial p_i} \cdot \frac {\partial B}{\partial r_i} \right ] \nonumber$
Thus, the Liouville equation can be written as
$\frac {\partial f}{\partial t} + \left \{ f, H \right \} = 0 \nonumber$
The Liouville equation is a partial differential equation for the phase space probability distribution function. Thus, it specifies a general class of functions $f (x, t)$ that satisfy it. In order to obtain a specific solution requires more input information, such as an initial condition on f, a boundary condition on f, and other control variables that characterize the ensemble.
2.04: Equilibrium ensembles
An equilibrium ensemble is one for which there is no explicit time-dependence in the phase space distribution function, $\frac {\partial f}{\partial t} = 0$. In this case, Liouville's equation reduces to
$\left \{ f, H \right \} = 0 \nonumber$
which implies that $f (x)$ must be a pure function of the Hamiltonian
$f (x) = F ( H (x)) \nonumber$
The specific form that $F (H(x))$ has depends on the specific details of the ensemble.
The integral over the phase space distribution function plays a special role in statistical mechanics:
$F = \int dx F ( H (x) ) \label{1}$
It is known as the partition function and is equal to the number of members in the ensemble. That is, it is equal to the number of microstates that all give rise to a given set of macroscopic observables. Thus, it is the quantity from which all thermodynamic properties are derived.
If a measurement of a macroscopic observable $A (x)$ is made, then the value obtained will be the ensemble average:
$\left \langle A \right \rangle = \frac {1}{F} \int dx A (x) F ( H (x) ) \label{2}$
Equations \ref{1} and \ref{2} are the central results of ensemble theory, since they determine all thermodynamic and other observable quantities. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/02%3A_Foundations_of_classical_statistical_mechanics/2.03%3A_The_Liouville_Operator_and_the_Poisson_Bracket.txt |
The microcanonical ensemble is built upon the so called postulate of equal a priori probabilities:
Postulate of equal a priori probabilities: For an isolated macroscopic system in equilibrium, all microscopic states corresponding to the same set of macroscopic observables are equally probable.
03: The Microcanonical Ensemble
In the microcanonical ensemble, the entropy $S$ is a natural function of $N$,$V$ and $E$, i.e., $S=S(N,V,E)$. This can be inverted to give the energy as a function of $N$, $V$, and $S$, i.e., $E=E(N,V,S)$. Consider using Legendre transformation to change from $S$ to $T$ using the fact that
$T= \left(\frac {\partial E}{\partial S}\right)_{N,V} \nonumber$
The Legendre transform $\tilde{E}$ of $E(N,V,S)$ is
$\tilde {E} (N, V, T ) = E (N,V,S(T)) - S \frac {\partial E}{\partial S} \nonumber$
$= E(N,V,S(T)) - TS \nonumber$
The quantity $\tilde{E}(N,V,T)$ is called the Hemlholtz free energy and is given the symbol $A(N,V,T)$ and is the fundamental energy in the canonical ensemble. The differential of $A$ is
$dA = \left( \partial A \over \partial T \right)_{N,V}dT + \left( \partial A \over \partial V \right)_{N,T} dV +\left( \partial A \over \partial N \right)_{T,V} dN \nonumber$
However, from $A = E - TS$, we have
$dA = dE - TdS - SdT \nonumber$
From the first law, $dE$ is given by
$dE = TdS - PdV + \mu dN \nonumber$
Thus,
$dA = - PdV - S dT + \mu dN \nonumber$
Comparing the two expressions, we see that the thermodynamic relations are
$S = -\left(\frac {\partial A}{\partial T}\right)_{N,V} \nonumber$
$P = -\left(\frac {\partial A}{\partial V}\right)_{N,T} \nonumber$
$\mu = -\left(\frac {\partial A}{\partial N}\right)_{V,T} \nonumber$
3.02: The Partition Function
Consider two systems (1 and 2) in thermal contact such that
• $N_2 \gg N_1$
• $E_2 \gg E_1$
• $N= N_1 + N_2$
• $E = E_1 + E_2$
• $\text {dim} (x_1) \gg \text {dim} (x_2)$
and the total Hamiltonian is just
$H (x) = H_1 (x_1) + H_2 (x_2) \nonumber$
Since system 2 is infinitely large compared to system 1, it acts as an infinite heat reservoir that keeps system 1 at a constant temperature $T$ without gaining or losing an appreciable amount of heat, itself. Thus, system 1 is maintained at canonical conditions, $N, V, T$.
The full partition function $\Omega (N, V, E )$ for the combined system is the microcanonical partition function
$\Omega(N,V,E) = \int dx \delta(H(x)-E) = \int dx_1 dx_2 \delta (H_1(x_1) + H_2(x_2)-E) \nonumber$
Now, we define the distribution function, $f (x_1)$ of the phase space variables of system 1 as
$f(x_1) = \int dx_2 \delta (H_1(x_1)+ H_2(x_2)-E) \nonumber$
Taking the natural log of both sides, we have
$\ln f(x_1) = \ln \int dx_2 \delta (H_1(x_1) + H_2(x_2) - E) \nonumber$
Since $E_2 \gg E_1$, it follows that $H_2 (x_2) \gg H_1 (x_1)$, and we may expand the above expression about $H_1 = 0$. To linear order, the expression becomes
\begin{align*} \ln f (x_1) &= \ln \int dx_2 \delta (H_2(x_2)-E) + H_1(x_1) \frac {\partial }{ \partial H_1 (x_1)} \ln \int dx_2 \delta (H_1(x_1) + H_2(x_2) - E) \vert _{H_1(x_1)=0} \[4pt] &= \ln \int dx_2 \delta (H_2(x_2)-E) -H_1(x_1) \frac {\partial}{\partial E} \ln \int dx_2 \delta (H_2(x_2)-E) \end{align*}
where, in the last line, the differentiation with respect to $H_1$ is replaced by differentiation with respect to $E$. Note that
$\ln \int dx_2 \delta (H_2( _2)-E) =\frac {S_2 (E)}{k} \nonumber$
$\frac {\partial}{\partial E} \ln \int dx_2 \delta (H_2(x_2)-E = \frac {\partial}{\partial E} \frac {S_2(E)}{k} = \frac {1}{kT} \nonumber$
where $T$ is the common temperature of the two systems. Using these two facts, we obtain
$\ln f (x_1) = \frac {S_2 (E)}{k} - \frac {H_1 (x_1)}{kT} \nonumber$
$f (x_1) = e^{\frac {S_2(E)}{k}}e^{\frac {-H_1(x_1)}{kT}} \nonumber$
Thus, the distribution function of the canonical ensemble is
$f(x) \propto e^{\frac {-H(x)}{kT}} \nonumber$
The prefactor $exp (\frac {S_2 (E) }{k} )$ is an irrelevant constant that can be disregarded as it will not affect any physical properties.
The normalization of the distribution function is the integral:
$\int dxe^{\frac {-H(x)}{kT}} \equiv Q(N,V,T) \nonumber$
where $Q (N, V, T )$ is the canonical partition function. It is convenient to define an inverse temperature $\beta = \frac {1}{kT}$. $Q (N, V, T )$ is the canonical partition function. As in the microcanonical case, we add in the ad hoc quantum corrections to the classical result to give
$Q(N,V,T) = \frac {1}{N!h^{3N}} \int dx e^{-\beta H(x)} \nonumber$
The thermodynamic relations are thus,
3.03: The Classical Virial Theorem (Microcanonical Derivation)
Consider a system with Hamiltonian $H (x)$. Let $x_i$ and $x_j$ be specific components of the phase space vector.
Theorem $1$: Classical Virial Theorem
The classical virial theorem states that
$\langle x_i \dfrac {\partial H}{ \partial x_j}\rangle = kT\delta_{ij} \nonumber$
where the average is taken with respect to a microcanonical ensemble.
To prove the theorem, start with the definition of the average:
$\langle x_i \dfrac {\partial H}{\partial x_j}\rangle = \dfrac {C}{\Omega (E)} \int dx x_i \dfrac {\partial H}{\partial x_j} \delta(E-H(x)) \nonumber$
where the fact that $\delta (x) = \delta (-x)$ has been used. Also, the $N$ and $V$ dependence of the partition function have been suppressed. Note that the above average can be written as
\begin{align*} \langle x_i \frac {\partial H}{ \partial x_j}\rangle &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E}\int dx x_i \frac {\partial H}{\partial x_j} \theta (E-H(x)) \[4pt] &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx x_i \frac {\partial H}{\partial x_j} \[4pt] &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx x_i \frac {\partial (H-E)}{\partial x_j} \end{align*}
However, writing
$x_i \frac {\partial (H-E)}{\partial x_j} = \frac {\partial}{\partial x_j} \left [x_i(H-E)\right ] - \delta_{ij}(H-E) \nonumber$
allows the average to be expressed as
\begin{align*} \langle x_i \frac {\partial H}{ \partial x_j}\rangle &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E} \int_{H(x)<E} dx \left \{\frac {\partial}{\partial x_j} \left [ x_i(H-E)\right ] + \delta_{ij}(E-H(x))\right \} \[4pt] &= \frac {C}{\Omega(E)} \frac {\partial}{\partial E}\left [ \oint_{H=E} x_i (H-E) dS_j + \delta_{ij} \int_{H<E}d{x} (E-H(x)) \right ] \end{align*}
The first integral in the brackets is obtained by integrating the total derivative with respect to $x_j$ over the phase space variable $x_j$. This leaves an integral that must be performed over all other variables at the boundary of phase space where $H = E$, as indicated by the surface element $dS_j$. But the integrand involves the factor $H - E$, so this integral will vanish. This leaves:
\begin{align*} \langle x_i \frac {\partial H}{ \partial x_j}\rangle &= \frac {C\delta_{ij}}{\Omega(E)} \frac {\partial}{\partial E}\int_{H(x)<E} dx (E-H(x)) \[4pt] &= \dfrac {C\delta_{ij}}{\Omega(E)} \int_{H(x)<E} dx \[4pt] &=\dfrac {\delta_{ij}}{\Omega(E)} \Sigma (E) \end{align*}
where $\Sigma (E)$ is the partition function of the uniform ensemble. Recalling that $\Omega(E) = \frac {\partial}{\partial E} \Sigma (E)$ we have
\begin{align*} \langle x_i \dfrac {\partial H}{ \partial x_j}\rangle &= \delta_{ij} \dfrac {\Sigma(E)}{\frac {\partial \Sigma(E)}{\partial E}} \[4pt] &= \delta_{ij} \frac {1}{\dfrac {\partial \ln \Sigma(E)}{\partial E}} \[4pt] &= k\delta_{ij} \frac {1}{\dfrac {\partial \tilde{S}}{\partial E}} \[4pt]&= kT\delta_{ij} \end{align*}
which proves the theorem.
Example $1$
$x_i = p_i$: and $i = j$ The virial theorem says that
$\langle p_i \frac {\partial H}{ \partial p_j}\rangle = kT \nonumber$
$\langle \frac {p_i^2}{m_i} \rangle = kT \nonumber$
$\langle \frac {p_i^2}{2m_i} \rangle = \frac {1}{2} kT \nonumber$
Thus, at equilibrium, the kinetic energy of each particle must be $\frac {kT}{2}$. By summing both sides over all the particles, we obtain a well know result
$\sum_{i=1}^{3N} \langle \frac {p_i^2}{2m_i} \rangle =\sum_{i=1}^{3N} \langle \frac {1}{2}m_i v_i^2 \rangle = \frac {3}{ 2}NkT \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/03%3A_The_Microcanonical_Ensemble/3.01%3A_Basic_Thermodynamics.txt |
Again, let $x_i$ and $x_j$ be specific components of the phase space vector $x = (p_1,\cdots ,p_{3N},q_1,\cdots,q_{3N})$. Consider the canonical average
$\langle x_i \frac {\partial H}{\partial x_j} \rangle \nonumber$
given by
\begin{align*} \langle x_i \frac {\partial H}{\partial x_j} \rangle &= \frac {1}{Q} C_N \int dx x_i \frac {\partial H}{\partial x_j}e^{-\beta H(x)} \[4pt] &= \frac {1}{Q} C_N \int dx x_i \left(- \frac {1}{\beta} \frac {\partial}{\partial x_j} \right ) e^{-\beta H(x)} \end{align*}
But
\begin{align*} x_i \frac {\partial}{\partial x_j}e^{-\beta H(x)} &= \frac {\partial}{\partial x_j} \left ( x_i e^{-\beta H(x)} \right ) - e^{-\beta H(x)} \frac {\partial x_i}{\partial x_j} \[4pt] &= \frac {\partial}{\partial x_j} \left ( x_i e^{-\beta H(x)} \right ) - \delta_{ij}e^{-\beta H(x)} \end{align*}
Thus,
\begin{align*} \langle x_i \frac {\partial H}{\partial x_j} \rangle &= - \frac {1}{\beta Q} C_N \int dx \frac {\partial}{\partial x_j} \left ( x_i e^{- \beta H (x)} \right ) + \frac {1}{\beta Q} \delta_{ij} C_N \int dx e^{-\beta H(x)} \[4pt] &= - \frac {1}{\beta Q} C_N \int dx' \int dx_j \frac {\partial}{\partial x_j} \left ( x_i e^{-\beta H(x)}\right )+ kT\delta_{ij} \[4pt] &= \int dx' \left.x_i e^{-\beta H(x)} \right \vert _{x_j=-\infty}^{\infty}+ kT\delta_{ij} \end{align*}
Several cases exist for the surface term $x_i exp (-\beta H (x))$:
1. $x_i = p_i$a momentum variable. Then, since $H \sim p^2_i, exp (-\beta H)$ evaluated at $p_i = \pm \infty$ clearly vanishes.
2. $x_i = q_i$and $U \rightarrow \infty$ as $q_i \rightarrow \pm \infty$, thus representing a bound system. Then, $exp (- \beta H )$ also vanishes at $q_i = \pm \infty$.
3. $x_i = q_i$and $U \rightarrow 0$ as $q_i \rightarrow \pm \infty$, representing an unbound system. Then the exponential tends to 1 both at $q_i = \pm \infty$, hence the surface term vanishes.
4. $x_i = q_i$and the system is periodic, as in a solid. Then, the system will be represented by some supercell to which periodic boundary conditions can be applied, and the coordinates will take on the same value at the boundaries. Thus, $H$ and $exp (- \beta H)$ will take on the same value at the boundaries and the surface term will vanish.
5. $x_i = q_i$ and the particles experience elastic collisions with the walls of the container. Then there is an infinite potential at the walls so that $U \rightarrow \infty$ at the boundary and $exp (- \beta H ) \rightarrow 0$ at the boundary.
Thus, we have the result
$\langle x_i \frac {\partial H}{\partial x_j} \rangle = kT\delta_{ij} \nonumber$
The above cases cover many but not all situations, in particular, the case of a system confined within a volume $V$ with reflecting boundaries. Then, surface contributions actually give rise to an observable pressure (to be discussed in more detail in the next lecture).
4.02: Legendre Transforms
The microcanonical ensemble involved the thermodynamic variables $N$, $V$ and $E$ as its variables. However, it is often convenient and desirable to work with other thermodynamic variables as the control variables. Legendre transforms provide a means by which one can determine how the energy functions for different sets of thermodynamic variables are related. The general theory is given below for functions of a single variable.
Consider a function $f(x)$ and its derivative
$y=f'(x) = {df \over dx} \equiv g(x) \nonumber$
The equation $y = g (x)$ defines a variable transformation from $x$ to $y$. Is there a unique description of the function $f(x)$ in terms of the variable $y$? That is, does there exist a function $\phi (y)$ that is equivalent to $f(x)$?
Given a point $x_0$, can one determine the value of the function $f (x_0)$ given only $f' (x_0)$ ? No, for the reason that the function $f (x_0) + c$ for any constant $c$ will have the same value of $f' (x_0 )$ as shown in Figure $1$.
However, the value $f (x_0)$ can be determined uniquely if we specify the slope of the line tangent to $f$ at $x_0$, i.e., $f' (x_0)$ and the $y$-intercept, $b (x_0)$ of this line. Then, using the equation for the line, we have
$f (x_0) = x_0 f' (x_0 ) + b ( x_0 ) \nonumber$
This relation must hold for any general $x$:
$f (x) = x f' (x) + b (x) \nonumber$
Note that $f' (x)$ is the variable $y$, and $x = g^{-1} (y)$, where $g_{-1}$ is the functional inverse of $g$, i.e., $g ( g^{-1} (x) ) = x$. Solving for $b (x) = b ( g^{-1} (y) )$ gives
$b (g^{-1} (y)) = f ( g^{-1} (y)) - y g^{-1} (y) \equiv \phi (y) \nonumber$
where $\phi (y)$ is known as the Legendre transform of $f (x)$. In shorthand notation, one writes
$\phi (y) = f (x) - xy \nonumber$
however, it must be kept in mind that $x$ is a function of $y$.
4.03: Relation between Canonical and Microcanonical Ensembles
We saw that the $E (N, V, S)$ and $A (N, V, T)$ could be related by a Legendre transformation. The partition functions $\Omega (N, V, E)$ and $Q (N, V, T)$ can be related by a Laplace transform. Recall that the Laplace transform $\tilde {f} (\lambda)$ of a function $f (x)$ is given by
$\tilde {f} (\lambda) = \int _{0}^{\infty} dx e^{- \lambda x} f (x) \nonumber$
Let us compute the Laplace transform of $\Omega (N, V, E )$ with respect to $E$:
$\tilde {\Omega} (N, V, \lambda ) = C_N \int _{0}^{\infty} dE e^{- \lambda E} \int dx \delta ( H (x) - E ) \nonumber$
Using the $\delta$-function to do the integral over $E$:
$\tilde {\Omega} (N, V, \lambda ) = C_N \int dx e^{- \lambda H (x) } \nonumber$
By identifying $\lambda = \beta$, we see that the Laplace transform of the microcanonical partition function gives the canonical partition function $Q (N, V, T )$. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/04%3A_The_canonical_ensemble/4.01%3A_Classical_Virial_Theorem_%28canonical_ensemble_derivation%29.txt |
Consider a phase space volume element $dx_0$ at t=0, containing a small collection of initial conditions on a set of trajectories. The trajectories evolve in time according to Hamilton's equations of motion, and at a time t later will be located in a new volume element $dx_t$ as shown in the figure below:
How is $dx_0$ related to $dx_t$dxdd ? To answer this, consider a trajectory starting from a phase space vector $x_0$ in $dx_0$ and having a phase space vector $x_t$ at time $t$ in $dx_t$. Since the solution of Hamilton's equations depends on the choice of initial conditions, $x_t$ depends on $x_0$ :
\begin{align*} x_0 &= \left ( p_1 (0), \cdots , p_N(0), r_1(0), \cdots , r_N (0) \right ) \[4pt] x_0 &= \left ( p_1 (t), \cdots , p_N(t), r_1(t), \cdots , r_N (t) \right ) \[4pt] x^i_t &= x^i_t \left ( x^1_0 , \cdots , x^{6N}_0 \right ) \end{align*}
Thus, the phase space vector components can be viewed as a coordinate transformation on the phase space from $t=0$ to time $t$. The phase space volume element then transforms according to
$dx_t = J (x_t ; x_0 ) dx_0 \nonumber$
where $J (x_t ; x_0 )$ is the Jacobian of the transformation:
$J (x_t ; x_0 ) = \frac {\partial (x^1_t \cdots x^n_t )}{\partial (x^1_0 \cdots x^n_0 )} \nonumber$
where $n=6N$. The precise form of the Jacobian can be determined as will be demonstrated below.
The Jacobian is the determinant of a matrix $M$,
$J (x_t ; x_0 ) = \text {det} (M) = e^{TrlnM} \nonumber$
whose matrix elements are
$M_{ij} = \frac {\partial x^i_t}{\partial x^j_0} \nonumber$
Taking the time derivative of the Jacobian, we therefore have
$\frac {dJ}{dt} = Tr \left ( M^{-1} \frac {dM}{dt} \right ) e^{TrlnM} \nonumber$
$= J \sum _{i=1}^n \sum _{j=1}^n M^{-1}_{ij} \frac {dM_{ij}}{dt} \nonumber$
The matrices $M_{-1}$ and $\frac {dM}{dt}$ can be seen to be given by
$M^{-1}_{ij} = \frac {\partial x^i_0}{\partial x^j_t} \nonumber$
$\frac {dM_{ji}}{dt} = \frac {\partial \dot {x}^i_t}{\partial x^i_0} \nonumber$
Substituting into the expression for $dJ/dt$ gives
\begin{align*} \frac {dJ}{dt} &= J \sum _{i,j=1}^n \frac {\partial x^i_0}{\partial x^j_t} \frac {\partial \dot {x}^i_t}{\partial x^i_0} \[4pt] &= J \sum _{i,j,k=1}^n \frac {\partial x^i_0}{\partial x^j_t} \frac {\partial \dot {x}^i_t}{\partial x^k_t} \frac {\partial x^k_t}{\partial x^i_0} \end{align*}
where the chain rule has been introduced for the derivative $\frac {\partial x^j_t}{\partial x^i_0}$. The sum over i can now be performed:
$\sum _{i=1}^n \frac {\partial x^i_0}{\partial x^j_t} \frac {\partial x^k_t}{\partial x^i_0} = \sum ^n_{i=1} M^{-1}_{ij} M_{ki} = \sum ^n_{i=1} M_{ki}M^{-1}_{ij} = \delta _{kj} \nonumber$
Thus,
$\frac {dJ}{dt} = J \sum ^n_{j,k=1} \delta _{jk} \frac {\partial \dot {x}^j_t}{\partial x^k_0} \nonumber$
$J \sum ^n_{j=1} \frac {\partial \dot {x}^j_t}{\partial x^j_t} = J \nabla _x \cdot \dot {x} \nonumber$
or
$\frac {dJ}{dt} = J \nabla _x \cdot \dot {x} \nonumber$
The initial condition on this differential equation is $J (0) \equiv J (x_0; x_0) = 1$. Moreover, for a Hamiltonian system $\nabla _x \cdot \dot {x} = 0$. This says that $dJ/dt=0$ and $J(0)=1$. Thus, $J (x_t ; x_0 ) = 1$. If this is true, then the phase space volume element transforms according to
$dx_o = dx_t \nonumber$
which is another conservation law. This conservation law states that the phase space volume occupied by a collection of systems evolving according to Hamilton's equations of motion will be preserved in time. This is one statement of Liouville's theorem.
Combining this with the fact that $df/dt=0$, we have a conservation law for the phase space probability:
$f(x_o, o) dx_o = f(x_t,t)dx_t \nonumber$
which is an equivalent statement of Liouville's theorem.
4.05: Energy Fluctuations in the Canonical Ensemble
In the canonical ensemble, the total energy is not conserved. ( $H (x) \ne \text {const}$ ). What are the fluctuations in the energy? The energy fluctuations are given by the root mean square deviation of the Hamiltonian from its average $\langle H \rangle$:
$\Delta E = \sqrt{\langle\left(H-\langle H\rangle\right)^2\rangle} =\sqrt{\langle H^2 \rangle - \langle H \rangle^2} \nonumber$
$\langle H \rangle = - \frac {\partial}{\partial \beta} \ln Q (N,V,T)$
$\langle H^2 \rangle = \frac {1}{Q} C_N \int dx H^2 (x) e^{- \beta H (x)}$ }}
$= \frac{1}{Q} C_N \int dx \frac{\partial^2}{\partial \beta^2}e^{-\beta H(x)}$
$= \frac{1}{Q} \frac {\partial^2}{\partial \beta^2}Q$
$= \frac{\partial^2}{\partial \beta^2}\ln Q + \frac {1}{Q^2} \left( \frac {\partial Q}{\partial \beta}\right)^2$
$= \frac{\partial^2}{\partial \beta^2}\ln Q +\left[\frac{1}{Q} \frac{\partial Q}{\partial \beta}\right]^2$
$= \frac{\partial^2}{\partial \beta^2}\ln Q + \left[ \frac {\partial}{\partial \beta}\ln Q\right]^2$
Therefore
$\langle H^2 \rangle - \langle H\rangle^2 =\frac{\partial^2}{\partial \beta^2}\ln Q \nonumber$
But
$\frac{\partial^2}{\partial \beta^2}\ln Q = kT^2 C_V \nonumber$
Thus,
$\Delta E = \sqrt{kT^2 C_V} \nonumber$
Therefore, the relative energy fluctuation $\frac {\Delta E}{E}$ is given by
$\frac{\Delta E}{E} = \frac{\sqrt{kT^2 C_V}}{E} \nonumber$
Now consider what happens when the system is taken to be very large. In fact, we will define a formal limit called the thermodynamic limit, in which $N\longrightarrow\infty$ and $V\longrightarrow\infty$ such that $\frac {N}{V}$ remains constant.
Since $C_V$ and $E$ are both extensive variables, $C_V\sim N$ and $E \sim N$,
$\frac {\Delta E}{E} \sim \frac{1}{\sqrt{N}} \longrightarrow 0\;\;\;{as}\;\;\;N\rightarrow \infty \nonumber$
But $\frac {\Delta E}{E}$ would be exactly 0 in the microcanonical ensemble. Thus, in the thermodynamic limit, the canonical and microcanonical ensembles are equivalent, since the energy fluctuations become vanishingly small. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/04%3A_The_canonical_ensemble/4.04%3A_Preservation_of_Phase_Space_Volume_and_Liouville%27s_Theorem.txt |
The Helmholtz free energy $A (N, V, T )$ is a natural function of $N, V$ and $T$. The isothermal-isobaric ensemble is generated by transforming the volume $V$ in favor of the pressure $P$ so that the natural variables are $N$, $P$, and $T$ (which are conditions under which many experiments are performed, e.g., `standard temperature and pressure'. Performing a Legendre transformation of the Helmholtz free energy
$\tilde{A}(N,P,T) = A(N,V(P),T) - V(P) \frac {\partial A}{\partial V} \nonumber$
But
$\frac {\partial A}{\partial V} = -P \nonumber$
Thus,
$\tilde{A}(N,P,T) = A(N,V(P),T) + PV \equiv G(N,P,T) \nonumber$
where $G (N, P, T )$ is the Gibbs free energy. The differential of $G$ is
$dG = \left(\frac {\partial G}{\partial P}\right)_{N,T} dP+ \left(\frac {\partial G}{\partial T}\right)_{N,P} dT+ \left(\frac {\partial G}{\partial N}\right)_{P,T} dN \nonumber$
But from $G = A + PV$, we have
$dG = dA + PdV + VdP \nonumber$
but $dA = - SdT - PdV + \mu dN$, thus
$dG = - SdT + VdP + \mu dN \nonumber$
Equating the two expressions for $dG$, we see that
$V=\left(\frac {\partial G}{\partial P}\right)_{N,T} \nonumber$
$S=-\left(\frac {\partial G}{\partial T}\right)_{N,P} \nonumber$
$\mu=\left(\frac {\partial G}{\partial N}\right)_{P,T} \nonumber$
5.02: Pressure and Work Virial Theorems
As noted earlier, the quantity $-\partial H/\partial V$ is a measure of the instantaneous value of the internal pressure $P_{\rm int}$. Let us look at the ensemble average of this quantity
\begin{align*} \langle P_{\rm int} \rangle &= -{1 \over \Delta}C_N\int_0^{\infty}dVe^{-\beta PV} \int d{\rm x}{\partial H \over \partial V}e^{-\beta H({\rm x})} \[4pt] &= {1 \over \Delta} C_N\int_0^{\infty}dVe^{-\beta PV} \int d{\rm x}kT {\partial \over \partial V}e^{-\beta H({\rm x})} \[4pt] &= {1 \over \Delta} \int_0^{\infty}dVe^{-\beta PV} kT {\partial \over \partial V} Q(N,V,T) \end{align*}
Doing the volume integration by parts gives
\begin{align*} \langle P_{\rm int} \rangle &= {1 \over \Delta} \left[e^{-\beta PV} kT Q(N,V,T) \right]\vert _0^{\infty} - {1 \over \Delta } \int _0^{\infty}dVkT \left({\partial \over \partial V} e^{-\beta PV} \right) Q(N,V,T) \[4pt] &=P{1 \over \Delta} \int_0^{\infty}dVe^{-\beta PV} Q(N,V,T) \[4pt] &= P \end{align*}
Thus,
$\langle P_{\rm int}\rangle = P \nonumber$
This result is known as the pressure virial theorem. It illustrates that the average of the quantity $-\partial H/\partial V$ gives the fixed pressure $P$ that defines the ensemble. Another important result comes from considering the ensemble average $-\partial H/\partial V$
$\langle P_{\rm int} V\rangle = {1 \over \Delta} \int_0^{\infty}dVe^{-\beta PV} kTV {\partial \over \partial V}Q(N,V,T) \nonumber$
Once again, integrating by parts with respect to the volume yields
\begin{align*} \langle P_{\rm int}V\rangle &= {1 \over \Delta} \left[e^{-\beta PV} kTV Q(N,V,T) \right]\vert _0^{\infty} - {1 \over \Delta} \int _0^{\infty}dVkT \left({\partial \over \partial V}Ve^{-\beta PV} \right)Q(N,V,T) \[4pt] &={1 \over \Delta} \left[-kT \int_0^{\infty}dVe^{-\beta PV} Q(V) + P \int_0^{\infty}dVe^{-\beta PV} VQ(V)\right] \[4pt] &=-kT + P\langle V \rangle \end{align*}
or $\langle P_{\rm int} V\rangle + kT = P\langle V\rangle \nonumber$
This result is known as the work virial theorem. It expresses the fact that equipartitioning of energy also applies to the volume degrees of freedom, since the volume is now a fluctuating quantity.
5.03: The partition function and relation to thermodynamics
In principle, we should derive the isothermal-isobaric partition function by coupling our system to an infinite thermal reservoir as was done for the canonical ensemble and also subject the system to the action of a movable piston under the influence of an external pressure $P$. In this case, both the temperature of the system and its pressure will be controlled, and the energy and volume will fluctuate accordingly.
However, we saw that the transformation from $E$ to $T$ between the microcanonical and canonical ensembles turned into a Laplace transform relation between the partition functions. The same result holds for the transformation from $V$ to $T$. The relevant "energy'' quantity to transform is the work done by the system against the external pressure $P$ in changing its volume from $V = 0$ to $V$, which will be $PV$. Thus, the isothermal-isobaric partition function can be expressed in terms of the canonical partition function by the Laplace transform:
$\Delta(N,P,T) = {1 \over V_0} \int_0^{\infty} dV e^{-\beta PV} Q(N,V,T) \nonumber$
where $V_0$ is a constant that has units of volume. Thus,
$\Delta (N,P,T) = {1 \over V_0 N! h^{3N}} \int_0^{\infty}dV \int d{\rm x}e^{-\beta (H({\rm x}) + PV)} \nonumber$ The Gibbs free energy is related to the partition function by
$G(N,P,T) = -{1 \over \beta} \ln \Delta(N,P,T) \nonumber$
This can be shown in a manner similar to that used to prove the $A=-(1/\beta)\ln Q$. The differential equation to start with is
$G = A + PV = A + P{\partial G \over \partial P} \nonumber$
Other thermodynamic relations follow:
Volume:
$V = -kT\left({\partial \ln \Delta(N,P,T) \over \partial P}\right)_{N,T} \nonumber$
Enthalpy:
$\bar{H} = \langle H({\rm x}) + PV\rangle = -{\partial \over \partial \beta} \ln \Delta(N,P,T) \nonumber$
Heat capacity at constant pressure
$C_P = \left({\partial \bar{H} \over \partial T}\right)_{N,P} =k\beta^2 {\partial^2 \over \partial \beta}\ln \Delta(N,P,T) \nonumber$
Entropy:
$S$ $=$
$-\left({\partial G \over \partial T}\right)_{N,P}$
$=$
$k\ln \Delta(N,P,T) + {\bar{H} \over T}$
The fluctuations in the enthalpy $\Delta \bar{H}$ are given, in analogy with the canonical ensemble, by
$\Delta \bar{H} = \sqrt{kT^2 C_P} \nonumber$
so that
${\Delta \bar{H} \over \bar{H} } = {\sqrt{kT^2 C_P} \over \bar{H}} \nonumber$
so that, since $C_P$ and $\bar {H}$ are both extensive, $\Delta \bar{H} /\bar{H} \sim 1/\sqrt{N}$ which vanish in the thermodynamic limit. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/05%3A_The_Isothermal-Isobaric_Ensemble/5.01%3A_Basic_Thermodynamics.txt |
From the classical virial theorem
$\langle x_i\frac {\partial H}{\partial x_j}\rangle = kT\delta_{ij} \nonumber$
we arrived at the equipartition theorem:
$\left<\sum_{i=1}^{N}\frac {p_i^2}{2m_i}\right > = \frac {3}{2} NkT \nonumber$
where $\textbf {p}_1, \cdots , \textbf {p}_N$ are the $N$ Cartesian momenta of the $N$ particles in a system. This says that the microscopic function of the $N$ momenta that corresponds to the temperature, a macroscopic observable of the system, is given by
$K( \textbf {p}_1,\cdots ,\textbf {p}_N) = \sum_{i=1}^{N} \frac {\textbf {p}_i^2}{2m_i} \nonumber$
The ensemble average of $K$ can be related directly to the temperature
$T = \frac {2}{3Nk} \left< K({\textbf p}_1,\cdots,{\textbf p}_{N})\right > =\frac {2}{3nR} \left < K({\textbf p}_1,\cdots,{\textbf p}_{3})\right> \nonumber$
$K( \textbf {p}_1,\cdots ,\textbf {p}_N)$ \) is known as an estimator (a term taken over from the Monte Carlo literature) for the temperature. An estimator is some function of the phase space coordinates, i.e., a function of microscopic states, whose ensemble average gives rise to a physical observable.
An estimator for the pressure can be derived as well, starting from the basic thermodynamic relation:
$P = -\left(\frac {\partial A}{\partial V} \right)_{N,T} = kT\left(\frac {\partial \ln Q(N,V,T)}{\partial V}\right)_{N,T} \nonumber$
with
$Q(N,V,T) = C_N\int dx\;e^{-\beta H(x)} =C_N \int d^{N}{\textbf p}\int_{V}d^{N}{\textbf r}\;e^{-\beta H({\textbf p},{\textbf r})} \nonumber$
The volume dependence of the partition function is contained in the limits of integration, since the range of integration for the coordinates is determined by the size of the physical container. For example, if the system is confined within a cubic box of volume $V = L^3$, with $L$ the length of a side, then the range of each $q$ integration will be from 0 to $L$. If a change of variables is made to $s_i = \frac {q_i}{L}$, then the range of each $s$ integration will be from 0 to 1. The coordinates $s_i$ $s_i$ are known as scaled coordinates. For containers of a more general shape, a more general transformation is
${\textbf s}_i = V^{-1/3}{\textbf r}_i \nonumber$
To preserve the phase space volume element, however, we need to ensure that the transformation is a canonical one. Thus, the corresponding momentum transformation is
$\pi_i = V^{1/3}{\textbf p}_i \nonumber$
With this coordinate/momentum transformation, the phase space volume element transforms as
$d^{N}{\textbf p}\;d^{N}{\textbf r}= d^{N}\pi d^{N}{\textbf s} \nonumber$
Thus, the volume element remains the same as required. With this transformation, the Hamiltonian becomes
$H = \sum_{i=1}^{N} \frac {{\textbf p}_i^2}{2m_i} + U({\textbf r}_1,\cdots , {\textbf r} _N) = \sum _{i=1}^N \frac {V^{-2/3} \pi _i^2}{2m_i} U( {V}^{\textbf s}_1,...,V^{1/3}{\textbf s}_{N}) \nonumber$
and the canonical partition function becomes
$Q(N,V,T) = C_N \int \;d^{N}\pi\int\;d^{N}{\textbf s}\;\exp\left \{- \beta \left [ \sum _{i=1}^N \frac {V^{-2/3} \pi ^2_i}{2m_i} +U (V^{1/3}{\textbf s} _1,\cdots,V^{1/3}{\textbf s}_N)\right ]\right \} \nonumber$
Thus, the pressure can now be calculated by explicit differentiation with respect to the volume, $V$:
$P$ $=$ $kT \frac {1}{Q} \frac {\partial Q}{\partial V}$
$=$ $\frac{kT}{Q} C_N\int\;d^{N}\pi\int\;d^{N}{\textbf s}\;\left [ \frac {2}{3} \beta V^{-5/3} \sum _{i=1}^N \frac {\pi ^2_i}{2m_i} - \frac {\beta}{3} V^{-2/3} \sum _{i=1}^N {\textbf s}_i\cdot\frac {\partial U}{\partial (V^{1/3}{\textbf s}_i)}\right ]e^{-\beta H}$
$=$ $\frac {kT}{Q}C_N\int\;d^{N}{\textbf p}\int\;d^{N}{\textbf r}\left[\frac {\beta}{3V} \sum _{i=1}^N \frac {{\textbf P}^2_i}{m_i} - \frac {\beta}{3V} \sum _{i=1}^N {\textbf r}_i\cdot \frac {\partial U}{\partial {\textbf r}_i}\right ]e^{-\beta H({\textbf p},{\textbf r})}$
$=$ $\frac{1}{3V} \left<\sum_{i=1}^{N}\left( \frac {{\textbf p}_i^2}{m_i} + {\textbf r}_i\cdot {\textbf F}_i\right )\right >$
$=$ $\langle-\frac {\partial H}{\partial V}\rangle$
Thus, the pressure estimator is
$\Pi({\textbf p}_1,...,{\textbf p}_N,{\textbf r}_1,...,{\textbf r}_N) = \Pi (x) = \frac {1}{3V} \sum _{i=1}^N \left [ \frac {{\textbf p}_i^2}{2m_i} + {\textbf r}_i\cdot{\textbf F}_i({\textbf r})\right ] \nonumber$
and the pressure is given by
$P = \langle \Pi(x)\rangle \nonumber$
For periodic systems, such as solids and currently used models of liquids, an absolute Cartesian coordinate $q_i$ is ill-defined. Thus, the virial part of the pressure estimator $\sum _i q_i F_i$ must be rewritten in a form appropriate for periodic systems. This can be done by recognizing that the force $F_i$ is obtained as a sum of contributions $F_{ij}$, which is the force on particle $i$ due to particle $j$. Then, the classical virial becomes
$\sum_{i=1}^{N}{\textbf r}_i\cdot {\textbf F}_i$ $=$ $\sum_{i=1}^{N}{\textbf r}_i \cdot \sum_{j\neq i} {\textbf F}_{ij}$
$=$ $\frac {1}{2}\left[\sum_i {\textbf r}_i\sum_{j\neq i}\cdot F_{ij} +\sum_j {\textbf r}_j\cdot\sum_{i\neq j}{\textbf F}_{ji}\right]$
$=$ $\frac {1}{2}\left [ \sum_i{\textbf r}_i\cdot\sum_{j\neq i}{\textbf F}_{ij} - \sum_j{\textbf r}_j\cdot\sum_{i\neq j}{\textbf F}_{ij}\right ]$
$=$ $\frac{1}{2}\sum_{i,j,i\neq j}({\textbf r}_i-{\textbf r}_j)\cdot{\textbf F}_{ij} = \frac {1}{2} \sum_{i,j,i\neq j} {\textbf r}_{ij} \cdot {\textbf F}_{ij}$
where $\textbf {r}_{ij}$ is now a relative coordinate. $\textbf {r} _{ij}$ must be computed consistent with periodic boundary conditions, i.e., the relative coordinate is defined with respect to the closest periodic image of particle $j$ with respect to particle $i$. This gives rise to surface contributions, which lead to a nonzero pressure, as expected. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/05%3A_The_Isothermal-Isobaric_Ensemble/5.04%3A_Temperature_and_pressure_estimators.txt |
In the grand canonical ensemble, the control variables are the chemical potential $\mu$, the volume $V$ and the temperature $T$. The total particle number $N$ is therefore allowed to fluctuate. It is therefore related to the canonical ensemble by a Legendre transformation with respect to the particle number $N$. Its utility lies in the fact that it closely represents the conditions under which experiments are often performed and, as we shall see, it gives direct access to the equation of state.
06: The Grand Canonical Ensemble
In the canonical ensemble, the Helmholtz free energy $A (N, V, T)$ is a natural function of $N , V$ and $T$. As usual, we perform a Legendre transformation to eliminate $N$ in favor of $\mu = \frac {\partial A}{\partial N}$:
\begin{align} \tilde{A}(\mu,V,T) &= A(N(\mu),V,T) - N\left(\frac {\partial A}{\partial N}\right)_{V,T} \[4pt] &= A(N(\mu),V,T) - \mu N \end{align}
It turns out that the free energy $\tilde{A}(\mu,V,T)$ is the quantity $- PV$. We shall derive this result below in the context of the partition function. Thus,
$-PV = A(N(\mu),V,T) - \mu N \nonumber$
To motivate the fact that $PV$ is the proper free energy of the grand canonical ensemble from thermodynamic considerations, we need to introduce a mathematical theorem, known as Euler's theorem:
Euler's Theorem
Let $f(x_1,...,x_N)$ be a function such that
$f(\lambda x_1,...,\lambda x_N) = \lambda^n f(x_1,...,x_N) \nonumber$
Then $f$ is said to be a homogeneous function of degree $n$. For example, the function $f(x) = 3x^2$ is a homogeneous function of degree 2, $f(x,y,z) = xy^2 + z^3$ is a homogeneous function of degree 3, however, $f(x,y) = e^{xy}-xy$ is not a homogeneous function. Euler's Theorem states that, for a homogeneous function $f$,
$nf(x_1,...,x_N) = \sum_{i=1}^N x_i \frac {\partial f}{\partial x_i} \nonumber$
Proof
To prove Euler's theorem, simply differentiate the the homogeneity condition with respect to lambda:
\begin{align*} \frac {d}{d\lambda} f(\lambda x_1,...,\lambda x_N) &= \frac {d}{d\lambda} \lambda^nf(x_1,...,x_N) \[4pt] \sum_{i=1}^N x_i \frac {\partial f}{\partial (\lambda x_i)} &= n\lambda^{n-1}f(x_1,...,x_N) \end{align*}
Then, setting $\lambda = 1$, we have
$\sum_{i=1}^N x_i \frac {\partial f}{\partial x_i} = nf(x_1,...,x_N) \nonumber$
which is exactly Euler's theorem.
Now, in thermodynamics, extensive thermodynamic functions are homogeneous functions of degree 1. Thus, to see how Euler's theorem applies in thermodynamics, consider the familiar example of the Gibbs free energy:
$G = G (N, P, T ) \nonumber$
The extensive dependence of $G$ is on $N$, so, being a homogeneous function of degree 1, it should satisfy
$G (\lambda N, P, T) = \lambda G (N, P, T ) \nonumber$
Applying Euler's theorem, we thus have
$G(N,P,T) = N \frac {\partial G}{\partial N} = \mu N \nonumber$
or, for a multicomponent system,
$G = \sum_{j} \mu_j N_j \nonumber$
But, since
$G = E - TS + PV \nonumber$
it can be seen that $G = \mu N$ is consistent with the first law of thermodynamics.
Now, for the Legendre transformed free energy in the grand canonical ensemble, the thermodynamics are
\begin{align*} d\tilde{A} &= dA - \mu dN - Nd\mu \[4pt] &= -PdV - SdT - Nd\mu \end{align*}
But, since
\begin{align*} \tilde{A} &=\tilde{A}(\mu,V,T) \[4pt] d \tilde{A} &= \left(\frac {\partial \tilde{A}}{\partial \mu}\right)_{V,T}d\mu+\left(\frac {\partial \tilde{A}}{\partial V}\right)_{\mu,T}dV+ \left(\frac {\partial \tilde{A}}{\partial T}\right)_{\mu,V}dT \end{align*}
the thermodynamics will be given by
\begin{align*} N &= -\left(\frac {\partial \tilde{A}}{\partial \mu}\right)_{V,T} \[4pt] P &= -\left(\frac {\partial \tilde{A}}{\partial V}\right)_{\mu,T} \[4pt] S &= -\left(\frac {\partial \tilde{A}}{\partial T}\right)_{V, \mu} \end{align*}
Since, $\tilde{A}$ is a homogeneous function of degree 1, and its extensive argument is $V$, it should satisfy
$\tilde{A}(\mu,\lambda V,T) = \lambda \tilde{A}(\mu,V,T) \nonumber$
Thus, applying Euler's theorem,
$\tilde{A}(\mu,V,T) = V \frac {\partial \tilde{A}}{\partial V} = -PV \nonumber$
and since
$\tilde{A} = A-\mu N = E - TS - \mu N \nonumber$
the assignment $\tilde{A} = -PV$ is consistent with the first law of thermodynamics. It is customary to work with $PV$, rather than $- PV$, so $PV$ is the natural free energy in the grand canonical ensemble, and, unlike the other ensembles, it is not given a special name or symbol! | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/06%3A_The_Grand_Canonical_Ensemble/6.01%3A_Thermodynamics.txt |
Consider two canonical systems, 1 and 2, with particle numbers $N_1$ and $N_2$, volumes $V_1$ and $V_2$ and at temperature $T$. The systems are in chemical contact, meaning that they can exchange particles. Furthermore, we assume that $N_2 \gg N_1$ and $V_2 \gg V_1$ so that system 2 is a particle reservoir. The total particle number and volume are
$V = V_1 + V_2 \nonumber$
$N = N_1 + N_2 \nonumber$
The total Hamiltonian $H (x, N )$ is
$H(x,N) = H_1(x_1,N_1) + H_2(x_2,N_2) \nonumber$
If the systems could not exchange particles, then the canonical partition function for the whole system would be
\begin{align*} Q(N,V,T) &= \dfrac {1}{N! h^{3N}}\int dx \, e^{-\beta (H_1(x_1,N_1)+ H_2(x_2,N_2))} \[4pt] &= \dfrac {N_1! N_2!}{N!}Q_1(N_1,V_1,T)Q_2(N_2,V_2,T) \end{align*}
where
$Q_1(N_1,V_1,T) = \dfrac{1}{N_1! h^{3N_1}}\int dx \, e^{-\beta H_1(x_1,N_1)} \nonumber$
$Q_2(N_2,V_2,T) = \dfrac{1}{N_2! h^{3N_2}}\int dx \, e^{-\beta H_2(x_2,N_2)} \nonumber$
However, $N_1$ and $N_2$ are not fixed, therefore, in order to sum over all microstates, we need to sum over all values that $N_1$ can take on subject to the constraint $N = N_1 + N_2$. Thus, we can write the canonical partition function for the whole system as
$Q(N,V,T) = \sum_{N_1=0}^N f(N_1,N) \frac {N_1! N_2!}{N!} Q_1(N_1,V_1,T) Q_2(N_2,V_2,T) \nonumber$
where $f (N_1, N_2 )$ is a function that weights each value of $N_1$ for a given $N$.
Thus,
• $f (0, N )$ is the number of configurations with 0 particles in $V_1$ and $N$ particles in $V_2$.
• $f (1, N)$ is the number of configurations with 1 particles in $V_1$ and $N - 1$ particles in $V_2$.
• etc.
Determining the values of $f (N_1, N )$ amounts to a problem of counting the number of ways we can put $N$ identical objects into 2 baskets. Thus,
\begin{align*} f (0, N ) &= 1 \[4pt] f (1, N) &= N \[4pt] &= \frac {N!}{1! (N - 1)! } \[4pt] f(2,N) &=\frac {N(N-1)}{2} \[4pt] &= \dfrac {N!}{2!(N-2)!} \end{align*}
etc. or generally,
$f(N_1,N) = \frac {N!}{N_1! (N-N_1)!} = \frac {N!}{N_1! N_2!} \nonumber$
which is clearly a classical degeneracy factor. If we were doing a purely classical treatment of the grand canonical ensemble, then this factor would appear in the sum for $Q (N, V, T )$, however, we always include the ad hoc quantum correction $\frac {1}{N !}$ in the expression for the canonical partition function, and we see that these quantum factors will exactly cancel the classical degeneracy factor, leading to the following expression:
$Q(N,V,T) = \sum_{N_1=0}^N Q_1(N_1,V_1,T)Q_2(N_2,V_2,T) \nonumber$
which expresses the fact that, in reality, the various configurations are not distinguishable from each other, and so each one should count with equal weighting. Now, the distribution function $\rho(x)$ is given by
$\rho(x,N) = \frac{\frac{1}{N!h^{3N}}e^{-\beta H(x,N)}}{Q(N,V,T)} \nonumber$
which is chosen so that
$\int dx \rho(x,N) = 1 \nonumber$
However, recognizing that $N_2 \approx N$, we can obtain the distribution for $\rho _1 (x_1, N_1)$ immediately, by integrating over the phase space of system 2:
$\rho_1(x_1,N_1) = \frac{1}{Q(N,V,T)} \frac{1}{N_1! h^{3N_1}}e^{-\beta H_1 (x_1, N_1)} \frac{1}{N_2! h^{3N_2}}\int dx_2 e^{-\beta H_2(x_2,N_2)} \nonumber$
where the $\frac{1}{N_1! h^{3N_1}}$ prefactor has been introduced so that
$\sum_{N_1=0}^N \int dx_1 \rho(x_1,N_1) = 1 \nonumber$
and amounts to the usual ad hoc quantum correction factor that must be multiplied by the distribution function for each ensemble to account for the identical nature of the particles. Thus, we see that the distribution function becomes
$\rho_1(x_1,N_1) = \frac {Q_2(N_2,V_2,T)}{Q(N,V,T)} \frac{1}{N_1! h^{3N_1}}e^{-\beta H_1(x_1,N_1)} \nonumber$
Recall that the Hemlholtz free energy is given by
$A = -\frac {1}{\beta}\ln Q \nonumber$
Thus,
$Q(N,V,T) = e^{-\beta A(N,V,T)}$
$Q_2(N_2,V_2,T) = e^{-\beta A(N_2,V_2,T)} =e^{-\beta A(N-N_1,V-V_1,T)}$
or
$\frac {Q_2(N_2,V_2,T)}{Q(N,V,T)} = e^{-\beta (A(N-N_1,V-V_1,T) -A(N,V,T))} \nonumber$
But since $N \gg N_1$ and $V \gg V_1$, we may expand:
\begin{align*} A(N-N_1,V-V_1,T) &= A(N,V,T) - \frac{\partial A}{\partial N}N_1- \frac{\partial A}{\partial V}V_1 + \cdots \[4pt] &= A(N,V,T) - \mu N_1 + PV_1 + \cdots \end{align*}
Therefore the distribution function becomes
$\rho_1(x_1,N_1) = \frac{1}{N_1! h^{3N_1}}e^{\beta \mu N_1}e^{-\beta PV_1}e^{-\beta H_1(x_1,N_1)} = \frac{1}{N_1! h^{3N_1}} \frac{1}{e^{\beta PV_1}}e^{\beta \mu N_1}e^{-\beta H_1(x_1,N_1)}$
Dropping the "1'' subscript, we have
$\rho(x,N) =\frac{1}{e^{\beta PV}}\left[\frac{1}{N! h^{3N}}e^{\beta \mu N}e^{-\beta H(x,N)}\right] \nonumber$
We require that $\rho (x, N)$ be normalized:
$\sum_{N=0}^{\infty}\int dx\rho(x,N) = 1$
$\frac{1}{e^{\beta PV}}\left[\sum_{N=0}^{\infty} \frac{1}{N! h^{3N}}e^{\beta \mu N}\int dxe^{-\beta H(x,N)}\right] = 1$
Now, we define the grand canonical partition function
${\cal Z}(\mu,V,T) = \sum_{N=0}^{\infty} \frac{1}{N! h^{3N}}e^{\beta \mu N}\int dxe^{-\beta H(x,N)} \nonumber$
Then, the normalization condition clearly requires that
${\cal Z}(\mu,V,T) = e^{\beta PV} \nonumber$
$\ln {\cal Z}(\mu,V,T) = \frac{PV}{kT} \nonumber$
Therefore $PV$ is the free energy of the grand canonical ensemble, and the entropy $S (\mu , V, T )$ is given by
$S(\mu,V,T) = \left(\frac{\partial (PV)}{\partial T}\right)_{\mu, V} = k \ln {\cal Z} (\mu, V, T) - k \beta \left ( \frac {\partial}{\partial \beta } \ln {\cal Z}(\mu,V,T)\right)_{\mu,V} \nonumber$
We now introduce the fugacity $\zeta$ defined to be
$\zeta = e^{\beta \mu} \nonumber$
Then, the grand canonical partition function can be written as
\begin{align*} {\cal Z}(\zeta,V,T) &= \sum_{N=0}^{\infty} \frac{1}{N! h^{3N}} \zeta ^N \int dx e^{-\beta H(x,N)} \[4pt] &=\sum_{N=0}^{\infty} \zeta^N Q(N,V,T) \end{align*}
which allows us to view the grand canonical partition function as a function of the thermodynamic variables $\zeta , V$ and $T$.
Other thermodynamic quantities follow straightforwardly:
$\frac{\partial}{\partial \mu} = \frac{\partial \zeta}{\partial \zeta}= \beta \zeta\frac{\partial}{\partial \zeta} \nonumber$
Thus,
$\langle N \rangle = \zeta \frac{\partial}{\partial \zeta}\ln {\cal Z}(\zeta,V,T) \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/06%3A_The_Grand_Canonical_Ensemble/6.02%3A_Partition_Functions.txt |
Recall the canonical partition function expression for the ideal gas:
$Q(N,V,T) = {1 \over N!} \left[{V \over h^3}\left({2\pi m \over \beta}\right)^{3/2}\right]^{N} \nonumber$
Define the thermal wavelength $\lambda (\beta)$ as $\lambda(\beta) = \left({\beta h^2 \over 2 \pi m}\right)^{1/2} \nonumber$ which has a quantum mechanical meaning as the width of the free particle distribution function. Here it serves as a useful parameter, since the canonical partition can be expressed as
$Q(N,V,T) = {1 \over N!}\left({V \over \lambda^3}\right)^N \nonumber$
The grand canonical partition function follows directly from $Q(N,V,T)$:
${\cal Z}(\zeta,V,T) = \sum_{N=0}^{\infty}{1 \over N!}\left({V\zeta \over \lambda^3}\right)^N = e^{V\zeta/\lambda^3} \nonumber$
Thus, the free energy is
${PV \over kT} = \ln {\cal Z} = {V \zeta \over \lambda^3} \nonumber$
In order to obtain the equation of state, we first compute the average particle number $\langle N \rangle$
$\langle N \rangle = \zeta {\partial \over \partial \zeta}\ln {\cal Z}= {V \zeta \over \lambda^3} \nonumber$
Thus, eliminating $\zeta$ in favor of $\langle N \rangle$ in the equation of state gives
$PV = \langle N \rangle kT \nonumber$
as expected. Similarly, the average energy is given by
$E = -\left({\partial \ln {\cal Z}\over \partial \beta}\right )_{\zeta V} = {3V\zeta \over \lambda^4}{\partial \lambda \over \partial \beta} ={3 \over 2}\langle N \rangle kT \nonumber$
where the fugacity has been eliminated in favor of the average particle number. Finally, the entropy
$S(\mu,V,T) = k\ln {\cal Z}(\mu,V,T) - k\beta\left({\partial\ln {\cal Z} (\mu, V, T) \over \partial \beta} \right)_{\mu, V} = {5 \over 2}\langle N \rangle k + \langle N \rangle k\ln \left[{V\lambda^3 \over \langle N \rangle} \right] \nonumber$
which is the Sackur-Tetrode equation derived in the context of the canonical and microcanonical ensembles.
6.04: Particle Number Fluctuations
In the grand canonical ensemble, the particle number $N$ is not constant. It is, therefore, instructive to calculate the fluctuation in this quantity. As usual, this is defined to be
$\Delta N = \sqrt{\langle N^2 \rangle - \langle N \rangle^2} \nonumber$
Note that
\begin{align*} \zeta{\partial \over \partial \zeta}\zeta {\partial \over \partial \zeta}\ln {\cal Z}(\zeta,V,T) &= {1 \over {\cal Z}}\sum_{N=0}^{\infty}N^2 \zeta^N Q(N,V,T) -{1 \over {\cal Z}^2} \left[\sum_{N=0}^{\infty} N \zeta^N Q(N,V,T)\right]^2 \[4pt] &= \langle N^2 \rangle - \langle N \rangle^2 \end{align*}
Thus,
\begin{align*} \left(\Delta N\right)^2 &=\zeta{\partial \over \partial \zeta} \zeta {\partial \over \partial \zeta} \ln {\cal Z} (\zeta, V, T) \[4pt] &= ({KT}^2){\partial^2 \over \partial \mu^2}\ln {\cal Z}(\mu,V,T) \[4pt] &= kTV{\partial^2 P \over \partial \mu^2} \end{align*}
In order to calculate this derivative, it is useful to introduce the Helmholtz free energy per particle defined as follows:
$a(v,T) = {1 \over N}A(N,V,T) \nonumber$
where $v={V \over N} = {1 \over \rho}$ is the volume per particle. The chemical potential is defined by
\begin{align*} \mu &= {\partial A \over \partial N} \[4pt] &=a(v,T) + N{\partial a \over \partial v}{\partial v \over \partial N} \[4pt] &= a(v,T) - v{\partial a \over \partial v} \end{align*}
Similarly, the pressure is given by
$P = -{\partial A \over \partial V} = -N{\partial a \over \partial v}{\partial v \over \partial V} = -{\partial a \over \partial v} \nonumber$
Also,
${\partial \mu \over \partial v} = -v{\partial^2 a \over \partial v^2} \nonumber$
Therefore,
\begin{align*} \dfrac{\partial P}{\partial \mu} &= {\partial P \over \partial v}{\partial v \over \partial \mu} \[4pt] &= {\partial^2 a \over \partial v^2} \left[v{\partial^2 a \over \partial v^2}\right]^{-1} \[4pt] &= {1 \over v} \end{align*}
and
\begin{align*} \dfrac{\partial^2 P}{\partial \mu^2} &= {\partial \over \partial v}{\partial P \over \partial \mu}{\partial v \over \partial \mu} \[4pt] &= {1 \over v^2} \left[ v {\partial^2 a \over \partial v^2}\right]^{-1} \[4pt] &= -{1 \over v^3 \partial P/\partial v} \end{align*}
But recall the definition of the isothermal compressibility:
$\kappa_T = -{1 \over V}{\partial V \over \partial P}=-{1 \over v \partial p/\partial v} \nonumber$
Thus,
${\partial^2 P \over \partial \mu^2} = {1 \over v^2}\kappa_T \nonumber$
and
$\Delta N = \sqrt{\frac{\langle N \rangle kT \kappa_T}{v}} \nonumber$
and the relative fluctuation is given by
${\Delta N \over N} = {1 \over \langle N \rangle}\sqrt{\frac{\langle N \rangle kT \kappa_T}{v}} \sim {1 \over \sqrt {\langle N \rangle }}\rightarrow 0\;\;{as}\langle N \rangle \rightarrow \infty \nonumber$
Therefore, in the thermodynamic limit, the particle number fluctuations vanish, and the grand canonical ensemble is equivalent to the canonical ensemble. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/06%3A_The_Grand_Canonical_Ensemble/6.03%3A_Ideal_Gas.txt |
Recall the expression for the configurational partition function:
$Z_N = \int d{\textbf r}_1\cdots d{\textbf r}_N e^{-\beta U(r_1,...,r_N)} \nonumber$
Suppose that the potential $U$ can be written as a sum of two contributions
$U({{\textbf r}_1,...,{\textbf r}_N}) = U_0({{\textbf r}_1,...,{\textbf r}_N}) + U_1({{\textbf r}_1,...,{\textbf r}_N}) \nonumber$
where $U_1$ is, in some sense, small compared to $U_0$. An extra bonus can be had if the partition function for $U_0$ can be evaluated analytically.
Let
$Z_N{^{(0)}}= \int {d{\textbf r}_1\cdots d{\textbf r}_N}e^{-\beta U_0({r_1,...,r_N})} \nonumber$
Then, we may express $Z_N$ as
\begin{align*} Z_N &= {Z_N{^{(0)}}\over Z_N{^{(0)}}}\int d{\textbf r}_1\cdots d{\textbf r}_Ne^{-\beta U_0(r_1,...,r_N)}e^{-\beta U_1(r_1,...,r_N)} \[4pt] &= Z_N{^{(0)}}\langle e^{-\beta U_1(r_1,...,r_N)}\rangle_0 \end{align*}
where $\langle \cdots \rangle _0$ means average with respect to $U_0$ only. If $U_1$ is small, then the average can be expanded in powers of $U_1$:
$\langle e^{-\beta U_1}\rangle_0 = { 1 - \beta \langle U_1\rangle_0 +{\beta^2 \over 2!} \langle U_1^2 \rangle_0 - {\beta^3 \over 3!}\langle U_1^3 \rangle_0 +\cdots} \nonumber$
$= { \sum_{k=0}^{\infty} {(-\beta)^k \over k!}\langle U_1^k \rangle_0} \nonumber$
The free energy is given by
$A(N,V,T) = -{1 \over \beta}\ln \left({Z_N \over N!\lambda^{3N}}\right) = -{1 \over \beta}\ln \left({Z_N^{(0)} \over N!\lambda^{3N}}\right)-- {1 \over \beta}\ln \langle e^{-\beta U_1}\rangle_0 \nonumber$
Separating $A$ into two contributions, we have
$A(N,V,T) = A{^{(0)}}(N,V,T) + A{^{(1)}}(N,V,T) \nonumber$
where $A^{(0)}$ is independent of $U_1$ and is given by
$A{^{(0)}}(N,V,T) = -{1 \over \beta}\ln \left({Z_N{^{(0)}}\over N!\lambda^{3N}}\right) \nonumber$
and
\begin{align*} A{^{(1)}}(N,V,T) &= -{1 \over \beta}\ln \langle e^{-\beta U_1}\rangle_0 \[4pt] &=-{1 \over \beta}\ln \langle \sum_{k=0}^{\infty}{(-\beta)^k \over k!}\langle U_1^k \rangle_0 \end{align*}
We wish to develop an expansion for $A^{(1)}$ of the general form
$A{^{(1)}}= \sum_{k=1}^{\infty} {(-\beta)^{k-1} \over k!}\omega_k \nonumber$
where ${\omega _k}$ are a set of expansion coefficients that are determined by the condition that such an expansion be consistent with $\ln\langle \sum_{k=0}^{\infty} (-\beta)^k \langle U_1^k\rangle_0 /k!$.
Using the fact that
$\ln(1+x) = \sum_{k=1}^{\infty} (-1)^{k-1} {x^k \over k} \nonumber$
we have that
\begin{align*} -{1 \over \beta}\ln \left(\sum_{k=0}^{\infty}{(-\beta)^k \over k!} \langle U_1^k \rangle_0\right) &= -{1 \over \beta}\ln \left(1 + \sum_{k=0}^{\infty}{(-\beta)^k \over k!} \langle U_1^k \rangle_0\right) \[4pt] &= { -{1 \over \beta}\sum_{k=1}^{\infty}(-1)^{k-1}{1 \over k}\left(\sum_{l=1}^{\infty}{(-\beta)^l \over l!}\langle U_1^l\rangle_0\right)^k } \end{align*}
Equating this expansion to the proposed expansion for $A^{(1)}$, we obtain
$\sum_{k=1}^{\infty}(-1)^{k-1}{1 \over k}\left(\sum_{l=1}^{\infty} {(-\beta)^l \over l!} \langle U^l_1 \rangle _0\right)^k = \sum_{k=1}^{\infty} (-\beta)^k {\omega_k \over k!} \nonumber$
This must be solved for each of the undetermined parameters ${\omega_k}$, which can be done by equating like powers of $\beta$ on both sides of the equation. Thus, from the $\beta ^1$ term, we find, from the right side:
${\rm Right\ Side}:\;\;\;-{\beta \omega_1 \over 1!} \nonumber$
and from the left side, the $j = 1$ and $k = 1$ term contributes:
${\rm Left\ Side}:\;\;\;-{\beta \langle U_1 \rangle_0 \over 1!} \nonumber$
from which it can be easily seen that
$\omega_1 = \langle U_1 \rangle_0 \nonumber$
Likewise, from the $\beta ^2$ term,
${\rm Right\ Side}:\;\;\; {\beta^2 \over 2!}\omega_2 \nonumber$
and from the left side, we see that the $l = 1, k = 2$ and $l = 2, k = 1$ terms contribute:
${\rm Left\ Side}:\;\;\; {\beta^2 \over 2}\left(\langle U_1^2 \rangle_0- \langle U_1 \rangle_0^2\right) \nonumber$
Thus,
$\omega_2 = \langle U_1^2 \rangle_0 -\langle U_1\rangle_0^2 \nonumber$
For $\beta ^3$, the right sides gives:
${\rm Right\ Side}:\;\;\; -{\beta^3 \over 3!}\omega_3 \nonumber$
the left side contributes the $l = 1, k = 3, k = 2, l = 2$ and $l = 3, k = 1$ terms:
${\rm Left\ Side}: -{\beta^3 \over 6}\langle U_1^3 \rangle + (-1)^2 {1 \over 3}(-\beta \langle U_1\rangle _0 )^3 - {1 \over 2} \left ( -\beta \langle U_1 \rangle _0 + {1 \over 2}\beta^2\langle U_1^2 \rangle\right)^2 \nonumber$
Thus,
$\omega_3 = \langle U_1^3 \rangle_0 + 2\langle U_1 \rangle_0^3- 3\langle U_1 \rangle_0\langle U_1^2 \rangle_0 \nonumber$
Now, the free energy, up to the third order term is given by
\begin{align*} A &= A{^{(0)}}+ \omega_1 - {\beta \over 2}\omega_2 + {\beta^2 \over 6}\omega_3 \cdots \[4pt] &= -{1 \over \beta}\ln \left({Z_N{^{(0)}}\over N! \lambda^{3N}}\right) + \langle U_1 \rangle_0 - {\beta \over 2} \left \langle U_1^2 \rangle_0 - \langle U_1\rangle _0^2 \right ) + {\beta^2 \over 6} \left (\langle U_1^3 \rangle - 3 \langle U_1 \rangle _0\langle U_1^2 \rangle_0 + 2\langle U_1 \rangle_0^3 \right)+ \cdots \end{align*}
In order to evaluate $\langle U_1 \rangle _0$, suppose that $U_1$ is given by a pair potential
$U_1({{\bf r}_1,...,{\bf r}_N}) = {1 \over 2}\sum_{i\neq j}u_1(\vert{\bf r}_i - {\bf r}_j\vert) \nonumber$
Then,
\begin{align*} \langle U_1 \rangle_0 &= {1 \over Z_N{^{(0)}}}\int {d{\textbf r}_1\cdots d{\textbf r}_N}{1 \over 2} \sum_{i \ne j} u_1(\vert{\textbf r}_i-{\textbf r}_j\vert)e^{-\beta U_0( r_1,...,r_N)} \[4pt] &= \dfrac{N(N-1)}{2 Z_N{^{(0)}}} \int d{\textbf r}_1 d{\textbf r}_2 u_1(\vert r_1 - r_2 \vert)\int d{\textbf r}_3\cdots d{\textbf r}_Ne^{-\beta U_0({{\bf r}_1,...,{\bf r}_N})} \[4pt] &= \dfrac{N^2}{2V^2} \int d{\textbf r}_1 d{\textbf r}_2 u_1(\vert {\textbf r}_1-{\textbf r}_2\vert) g_0^{(2)}({\textbf r}_1,{\textbf r}_2) \[4pt] &= \dfrac{\rho^2 V}{2} \int_0^{\infty}4\pi r^2 u_1(r)g_0(r)dr \end{align*}
The free energy is therefore given by
$A(N,V,T) = -{1 \over \beta}\ln\left({Z_N^{(0)} \over N! \lambda ^{3N} } \right ) + {1 \over 2} \rho ^2 V \int _0^{\infty} 4 \pi r^2 u_1 (r) g_0 (r) dr - {\beta \over 2} \left ( \langle U_1^2 \rangle_0 - \langle U_1 \rangle_0^2\right)\cdots \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/07%3A_Distribution_Functions_and_Liquid_Structure/7.01%3A_General_Formulation_of_Distribution_Functions.txt |
We begin by considering a general N-particle system with Hamiltonian
$H = \sum _{i=1}^{3N} {p^2_i \over 2m} + U (r_1, \cdots, r_N) \nonumber$
For simplicity, we consider the case that all the particles are of the same type. Having established the equivalence of the ensembles in the thermodynamic limit, we are free to choose the ensemble that is the most convenient on in which to work. Thus, we choose to work in the canonical ensemble, for which the partition function is
$Q (N, V, T ) = {1 \over N!h^{3N}} \int d^{3N}pd^{3N} re^{-\beta \sum_{i=1}^{3N} \frac {P^2_i}{2m}} e^{-\beta U(r_1, \cdots, r_N)} \nonumber$
The 3N integrations over momentum variables can be done straightforwardly, giving
\begin{align*} Q(N, V, T) &= {1 \over N!\lambda^{3N}} \int dr_1 \cdots dr_N e^{-\beta U(r_1, \cdots, r_N)} \[4pt] &= {Z_N \over N!\lambda^{3N}} \end{align*}
where $\lambda = \sqrt {\frac {\beta h^2}{2\pi m}}$ is the thermal wavelength and the quantity $Z_N$ is known as the configurational partition function
$Z_N = \int dr_1 \cdots dr_N e^{-\beta U(r_1, \cdots, r_N)} \nonumber$
The quantity
${e^{-\beta U(r_1, \cdots, r_N)} \over Z_N} dr_1 \cdots dr_N \equiv P^{(N)} (r_1, \cdots, r_N ) dr_1 \cdots dr_N \nonumber$
represents the probability that particle 1 will be found in a volume element $dr_1$ at the point $r_1$, particle 2 will be found in a volume element $dr_2$ at the point $r_2$,..., particle N will be found in a volume element $dr_N$ at the point $r_N$. To obtain the probability associated with some number n;SPMlt;N of the particles, irrespective of the locations of the remaining n+1,...,N particles, we simply integrate this expression over the particles with indices n+1,...,N:
$P^{(n)} (r_1, \cdots , r_n) dr_1 \cdots dr_n = {1 \over Z_N} \left [ \int dr_{n+1} \cdots dr_N e^{-\beta U(r_1, \cdots, r_N)} \right ] dr_1 \cdots dr_n \nonumber$
The probability that any particle will be found in the volume element $dr_1$ at the point $r_1$ and any particle will be found in the volume element $dr_2$ at the point $r_2$,...,any particle will be found in the volume element $dr_n$ at the point $r_n$ is defined to be
$P^{(n)} (r_1, \cdots , r_n) dr_1 \cdots dr_n = {N! \over (N - n)!} P^{(n)} (r_1, \cdots , r_n) dr_1 \cdots dr_n \nonumber$
which comes about since the first particle can be chosen in N ways, the second chosen in N-1 ways, etc.
Consider the special case of $n=1$. Then, by the above formula,
\begin{align*} P^{(1)} (r_1) &= {1 \over Z_N}{N! \over (N - 1)!} \int dr_2 \cdots dr_N e^{-\beta U(r_1, \cdots, r_N)} \[4pt] &= {N \over Z_N} \int dr_2 \cdots dr_N e^{-\beta U(r_1, \cdots, r_N)} \end{align*}
Thus, if we integrate over all $r_1$, we find that
${1 \over V } \int dr_1 p^{(1)} (r_1) = {N \over V} = p \nonumber$
Thus, $P^{(1)}$ actually counts the number of particles likely to be found, on average, in the volume element $dr_1$ at the point $r_1$. Thus, integrating over the available volume, one finds, not surprisingly, all the particles in the system.
7.03: Structure and distribution functions in classical liquids and gases
So far, we have developed the classical theory of ensembles and applied it to the ideal gas, for which there was no potential of interaction between the particles: $U=0$. We were able to derive all the thermodynamics for this system using the various ensembles available to us, and, in particular, we could compute the equation of state. Now, we wish to consider the general case that the potential is not zero. Of course, all of the interesting physics and chemistry of real systems results from the specific interactions between the particles. Real systems can exhibit spatial structure, undergo phase transitions, undergo chemical changes, exhibit interesting dynamics, basically, a wide variety of rich behavior.
Consider the two snapshots below.
On the left is shown a configuration of an ideal gas, and on the right is shown a configuration of liquid argon. Can you see any inherent structure in the snapshot of liquid argon? While it may not be readily apparent, there is considerable structure in the liquid argon system that is clearly not present in the ideal gas. One way of quantifying spatial structure is through the use of the radial distribution function g(r), which will be discussed in great detail later. For now, it is sufficient to know that g(r) is a measure of the probability that a particle will be located a distance r from a another particle in the system. The figure below shows the function g(r) for the ideal gas and for the liquid argon systems.
It can be seen that the radial distribution function for the ideal gas is completely featureless signifying that it is equally likely to find a particle at any distance $r$ from a given particle. (Since the probability is uniform, $g (r) \sim \frac {1}{r^2}$ for small $r$. This is the particularly normalization condition on $g(r)$ that gives rise to uniform probability. Hence its rapidly rising behavior for small $r$.) For the liquid argon system, $g(r)$ exhibits several peaks, indicating that at certain radial values, it is more likely to find particles than at others. This is a result of the attractive nature of the interaction at such distances. The plot of $g(r)$ also shows that there is essentially zero probability of finding particles at distances less than about 2.5 Å from each other. This is due to the presence of very strong repulsive forces at short distances.
Sometimes, the structure can be readily seen in snapshots of configurations. Consider the following snapshot of a system of water molecules:
The red spheres are oxygen atoms, the grey spheres are hydrogen atoms, the green lines are hydrogen bonds, and the reddish-grey lines are covalent bonds. A good deal of structure can be seen in the form of a complex network of hydrogen bonds. This high degree of structure is characteristic of water and gives rise to the ease which with water can form stable, organized structures around other molecules. In water, one can ask several questions related to structure. For example, what is the probability that an oxygen atom will be found at a distance $r$ away from another oxygen atoms? What is the probability that a hydrogen atom will be located at a distance r from an oxygen atom, etc. The plot below shows the radial distribution functions corresponding to these two scenarios.
The peak in the O-O radial distribution function occurs at roughly 2.8Å which is the well known average hydrogen bond length in water. Of course, one could also ask about structure from the point of view of a hydrogen atom and obtain two other representations of structure in water.
7.04: General Correlation Functions
A general correlation function can be defined in terms of the probability distribution function $p^{(n)} (r_1, \cdots , r_n)$ according to
\begin{align*} g^{(n)}(r_1, \cdots , r_n) &= {1 \over p^n} p^{(n)} (r_1, \cdots , r_n) \[4pt] &= {V^n N! \over Z_N N^n (N - n)!} \int dr_{n+1} \cdots dr_N e^{-\beta U(r_1, \cdots , r_N)} \end{align*}
Another useful way to write the correlation function is
\begin{align*} g^{(n)} (r_1, \cdots , r_n) &= {V^n N! \over Z_N N^n (N - n)!} \int dr'_1 \cdots dr'_N e^{-\beta U(r_1, \cdots , r_N)} \delta (r_1 - r'_1) \cdots \delta (r_n - r'_n) \[4pt] &= {V^n N! \over Z_N N^n (N - n)!} \left < \Pi _{i=1}^n \delta (r_i - r'_i \right >_{r'_1, \cdots , r'_N} \end{align*}
i.e., the general n-particle correlation function can be expressed as an ensemble average of the product of $\delta$ -functions, with the integration being taken over the variables $r'_1, \cdots , r'_N$. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/07%3A_Distribution_Functions_and_Liquid_Structure/7.02%3A_General_Distribution_Functions_and_Correlation_Functions.txt |
In the canonical ensemble, the average energy is given by
$E = - {\partial \over \partial \beta} \ln Q (N, V, \beta ) \nonumber$
$\ln Q (N, V, \beta ) = \ln Z_N - 3N \ln \lambda (\beta ) - \ln N ! \nonumber$
Therefore,
$E = {3N \over \lambda}{\partial \lambda \over \partial \beta } - {1 \over Z_N}{\partial Z_N \over \partial \beta } \nonumber$
Since
$\lambda = \left [ {\beta h^2 \over 2 \pi m } \right ] ^{1/2} \nonumber$
${\partial \lambda \over \partial \beta } = {1 \over 2 \beta } \lambda \nonumber$
Thus,
$E = {3 \over 2} NkT + {1 \over Z_N} \int dr_1 \cdots dr_N U(r_1, \cdots , r_N ) e^{\beta U(r_1, \cdots , r_N)} \nonumber$
$= {3 \over 2} NkT + \langle U \rangle \nonumber$
In order to compute the average energy, therefore, one needs to be able to compute the average of the potential $\langle U \rangle$. In general, this is a nontrivial task, however, let us work out the average for the case of a pairwise-additive potential of the form
$U (r_1, \cdots , r_N ) = {1 \over 2} \sum _{i, j, i \ne j } u(|r_i - r_j|) \equiv U_{pair} (r_1, \cdots , r_N) \nonumber$
i.e., U is a sum of terms that depend only the distance between two particles at a time. This form turns out to be an excellent approximation in many cases. U therefore contains N(N-1) total terms, and $\langle U \rangle$ becomes
$\langle U \rangle = {1 \over 2Z_N} \sum _{i,j, i \ne j} \int dr_1 \cdots dr_N u (|r_i - r_j|)e^{-\beta U_{pair} (r_1, \cdots, r_N)} \nonumber$
$= {N (N - 1) \over 2Z_N} \int dr_1 \cdots dr_N u (|r_1 - r_2|)e^{-\beta U_{pair} (r_1, \cdots, r_N)} \nonumber$
The second line follows from the fact that all terms in the first line are the exact same integral, just with the labels changed. Thus,
$\langle U \rangle = {1 \over 2} \int dr_1 dr_2 u (|r_1 - r_2|) \left [ {N (N - 1) \over Z_N} \int dr_3 \cdots dr_N e^{-\beta U_{pair} (r_1, \cdots, r_N)}\right ] \nonumber$
$= {1 \over 2 } \int dr_1dr_2 u(|r_1 - r_2|)\rho^{(2)} (r_1, r_2 ) \nonumber$
$= {N^2 \over 2V^2} \int \int dr_1dr_2 u(|r_1 - r_2|)p^{(2)} (r_1, r_2 ) \nonumber$
Once again, we change variables to $r = r_1 - r_2$ and $R = {(r_1 + r_2) \over 2}$. Thus, we find that
$\langle U \rangle = {N^2 \over 2V^2} \int drdRu (r)\tilde {g}^{(2)} (r, R) \nonumber$
$= {N^2 \over 2V^2} \int dru (r) \int dR\tilde {g}^{(2)} (r, R) \nonumber$
$= {N^2 \over 2V^2} \int dru (r) \tilde {g} (r) \nonumber$
$= {N^2 \over 2V^2} \int _0^{\infty} dr4 \pi r^2 u (r) g(r) \nonumber$
Therefore, the average energy becomes
$E = {3 \over 2} NkT + {N \over 2} 4\pi \rho \int _0^{\infty} drr^2 u(r) g(r) \nonumber$
Thus, we have an expression for E in terms of a simple integral over the pair potential form and the radial distribution function. It also makes explicit the deviation from "ideal gas'' behavior, where E=3NkT/2.
By a similar procedure, we can develop an equation for the pressure P in terms of g(r). Recall that the pressure is given by
$P = {1 \over \beta } {\partial \ln Q \over \partial V} \nonumber$
$= {1 \over \beta Z_N}{\partial Z_N \over \partial \beta } \nonumber$
The volume dependence can be made explicit by changing variables of integration in $Z_N$ to
$s_i = V^{-1/3} r_i \nonumber$
Using these variables, $Z_N$ becomes
$Z_N = V^N \int ds_1 \cdots ds_N e^{-\beta U(V^{1/R}_1, \cdots, V^{1/R}_N)} \nonumber$
Carrying out the volume derivative gives
\begin{align*} \dfrac{\partial Z_N}{\partial V} &= {N \over V} Z_N - \beta V^N \int ds_1 \cdots ds_N {1 \over 3V} \sum _{i=1}^N r_i \cdot {\partial U \over \partial r_i} e^{-\beta U (V^{1/R}_1, \cdots, V^{1/R}_N)} \[4pt] &= {N \over V} Z_N + \beta \int dr_1 \cdots dr_N {1 \over 3V} \sum _{i=1} r_i \cdot F_i e^{-\beta U (r_1, \cdots, r_N)} \[4pt] &= ds_1 \cdots ds_N {1 \over 3V} \sum _{i=1}^N r_i \cdot ds_1 \cdots ds_N {1 \over 3V} \sum _{i=1}^N r_i \cdot ds_1 \cdots ds_N {1 \over 3V} \sum _{i=1}^N r_i \cdo \end{align*}
Thus,
${1 \over Z_N}{\partial Z_N \over \partial V} = {N\over V} + {\beta \over 3V} \left < \sum _{i=1}^N r_i \cdot F_i \right > \nonumber$
Let us consider, once again, a pair potential. We showed in an earlier lecture that
$\sum _{i=1}^N r_i \cdot F_i = \sum _{i=1} \sum _{j=1, j\ne i}^N r_i \cdot F_{ij} \nonumber$
where $F_{ij}$ is the force on particle i due to particle j. By interchaning the i and j summations in the above expression, we obtain
$\sum _{i=1}^N r_i \cdot F_i = {1 \over 2} \left [ \sum _{i,j,i\ne j} r_i \cdot F_{ij} + \sum _{i,j,i\ne j} r_j \cdot F_{ij} \right ] \nonumber$
However, by Newton's third law, the force on particle i due to particle j is equal and opposite to the force on particle j due to particle i:
$F_{ij} = - F_{ji} \nonumber$
Thus,
$\sum_{i=1}^N r_i \cdot F_i = {1 \over 2} \left [ \sum _{i,j,i\ne j} r_i \cdot F_{ij} - \sum _{i,j,i\ne j} r_j \cdot F_{ij} \right ] = {1 \over 2} \sum _{i,j,i\ne j} (r_i - r_j ) \cdot F_{ij} \equiv {1 \over 2} \sum _{i,j,i\ne j} r_{ij} \cdot F_{ij} \nonumber$
The ensemble average of this quantity is
${\beta \over 3V} \left < \sum _{i=1}^N r_i \cdot F_i \right > = {\beta \over 6V} \left < \sum _{i, j, i \ne j} r_{ij} \cdot F_{ij} \right > = {\beta \over 6VZ_N } \int dr_1 \cdots dr_N \sum _{i, j , i \ne j} r_{ij} \cdot F_{ij} e^{-\beta U_{pair} (r_1, \cdots, r_N)} \nonumber$
As before, all integrals are exactly the same, so that
${\beta \over 3V} \left < \sum _{i=1}^N r_i \cdot F_i \right > = {\beta N (N - 1) \over 6V Z_N} \int d_1 \cdot r_N r_{12} \cdot F_{12} e^{-\beta U_{pair} (r_1, \cdots, r_N)} \nonumber$
$= {\beta \over 6V} \int dr_1 dr_2 r_{12} \cdot F_{12} \left [ {N (N - 1) \over Z_N } \int dr_3 \cdots dr_N e^{-\beta U_{pair} (r_1, \cdots, r_N)} \right ] \nonumber$
$= {\beta \over 6V} \int dr_1 dr_2 r_{12} \cdot F_{12} \rho^{(2)} (r_1, r_2) \nonumber$
$= {\beta N^2 \over 6V^3 } \int dr_1 dr_2 r_{12} \cdot F_{12}g^{(2)} (r_1, r_2) \nonumber$
Then, for a pair potential, we have
$F_{12} = - {\partial U_{pair} \over \partial r_{12} } = - u' (| r_1 - r_2|) {(r_1 - r_2) \over |r_1 - r_2| } = -u' (r_{12}) {r_{12} \over r_{12} } \nonumber$
where u'(r) = du/dr, and $r_{12} = |r_{12}|$. Substituting this into the ensemble average gives
${\beta \over 3V} \left < \sum _{i=1}^N r_i \cdot F_i \right > = - {\beta N^2 \over 6V^3} \int dr_1 dr_2 u' (r_{12} ) r_{12} g^{(2)} (r_1, r_2) \nonumber$
As in the case of the average energy, we change variables at this point to $r = r_1 - r_2$ and $R = {(r_1 + r_2 ) \over 2}$. This gives
${\beta \over 3V } \left < \sum _{i=1}^N r_i \cdot F_i \right > = - {\beta N^2 \over 6V^3} \int dr dR u^t (r) r \tilde {g} ^{(2)} (r, R) \nonumber$
$= - {\beta N^2 \over 6V^2 } \int dr u^t (r) r \tilde {g} (r) \nonumber$
$= - {\beta N^2 \over 6V^2 } \int _0^{\infty} dr 4 \pi r^3 u^t (r) g (r) \nonumber$
Therefore, the pressure becomes
${P \over kT } = \rho - {\rho ^2 \over 6kT } \int _0^{\infty} dr 4 \pi r^3 u^t (r) g(r) \nonumber$
which again gives a simple expression for the pressure in terms only of the derivative of the pair potential form and the radial distribution function. It also shows explicitly how the equation of state differs from the that of the ideal gas ${P \over kT } = \rho$.
From the definition of g(r) it can be seen that it depends on the density $\rho$ and temperature T: $g (r) = g (r; \rho , T )$. Note, however, that the equation of state, derived above, has the general form
${P \over kT } = \rho + B \rho ^2 \nonumber$
which looks like the first few terms in an expansion about ideal gas behavior. This suggests that it may be possible to develop a general expansion in all powers of the density $\rho$ about ideal gas behavior. Consider representing $g (r; \rho, T )$ as such a power series:
$g (r; \rho , T) = \sum _{j=0}^{\infty} \rho ^j g_j (r; T) \nonumber$
Substituting this into the equation of state derived above, we obtain
${P \over kT } = \rho + \sum _{j=0}^{\infty} B_{j+2} (T) \rho ^{j+2} \nonumber$
This is known as the virial equation of state, and the coefficients $B_{j+2} (T)$ are given by
$B_{j+2}(T) = - {1 \over 6kT} \int _0^{\infty} dr 4\pi r^3 u^t (r) g_j (r; T) \nonumber$
are known as the virial coefficients. The coefficient $B_2 (T)$ is of particular interest, as it gives the leading order deviation from ideal gas behavior. It is known as the second virial coefficient. In the low density limit, $g (r; \rho , T) \approx g_0 (r; T)$ and $B_2 (T)$ is directly related to the radial distribution function. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/07%3A_Distribution_Functions_and_Liquid_Structure/7.05%3A_Thermodynamic_quantities_in_terms_of_g%28r%29.txt |
Of particular importance is the case n=2, or the correlation function $g^{(2)} (r_1, r_2)$ known as the pair correlation function. The explicit expression for $g^{(2)} (r_1, r_2)$ is
\begin{align*} g^{(2)} (r_1, r_2) &= {V^2 N! \over N^2 (N - 2)!} \langle \delta (r_1 - r'_1) \delta (r_2 - r'_2)\rangle \[4pt]&= {V^2 (N - 1) \over NZ_N} \int dr_3 \cdots dr_N e^{-\beta U(r_1, \cdots , r_N)} \[4pt] &= {N (N - 1) \over p^2} \langle \delta (r_1 - r'_1) \delta (r_2 - r'_2) \rangle _{r'_1 \cdots r'_N} \end{align*}
In general, for homogeneous systems in equilibrium, there are no special points in space, so that $g^{(2)}$ should depend only on the relative position of the particles or the difference $r_1 - r_2$. In this case, it proves useful to introduce the change of variables
$r = r_1 - r_2 R = {1 \over 2} (r_1 + r_2) \nonumber$
$r_1 = R + {1 \over 2} r, r_2 = R - {1 \over 2} r \nonumber$
Then, we obtain a new function $\tilde {g} ^{(2)}$, a function of $r$ and $R$:
\begin{align*} \tilde {g}^{(2)} (r, R) &= {V^2 (N - 1) \over NZ_N} \int dr_3 \cdots dr_N e^{-\beta U (R + {1 \over 2} r, R - {1 \over 2} r, r_R, \cdots , r_N)} \[4pt] &= {N (N - 1) \over p^2} \left < \delta \left ( R + {1 \over 2} r - r'_1 \right ) \delta \left ( R - {1 \over 2} r - r'_2 \right ) \right >_{r'_1, \cdots r'_N} \end{align*}
In general, we are only interested in the dependence on $r$. Thus, we integrate this expression over $R$ and obtain a new correlation function $\tilde {g} (r)$ defined by
$\tilde {g} (r) = {1 \over V} \int dR \tilde {g} ^{(2)} (r, R) \nonumber$
$= {V (N - 1) \over NZ_N} \int dR dr_3 \cdots dr_N e^{-\beta U (R + \dfrac {1}{2}r, R-\dfrac {1}{2}r, rR, \cdots, rN )} \nonumber$
$= {(N - 1) \over pZ_N } \int dR dr_3 \cdots dr_N e^{-\beta U (R + \dfrac {1}{2}r, R-\dfrac {1}{2}r, rR, \cdots, rN )} \nonumber$
For an isotropic system such as a liquid or gas, where there is no preferred direction in space, only the maginitude or $r, |r| \equiv r$ is of relevance. Thus, we seek a choice of coordinates that involves r explicitly. The spherical-polar coordinates of the vector $r$ is the most natural choice. If $r = (x, y, z )$ then the spherical polar coordinates are
$x = r \sin \theta \cos \phi \nonumber$
$y = r \sin \theta \sin \phi \nonumber$
$x = r \cos \theta \nonumber$
$dr = r^2 \sin \theta \, dr \, d\theta \, d\phi \nonumber$
where $\theta$ and $\phi$ are the polar and azimuthal angles, respectively. Also, note that
$r = r n \nonumber$
where
$n = \left ( \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta \right ) \nonumber$
Thus, the function $g(r)$ that depends only on the distance $r$ between two particles is defined to be
\begin{align*} g(r) &= { 1\over 4 \pi} \int \sin \theta d \theta d \phi \tilde {g} (r) \[4pt] &= {(N - 1) \over 4 \pi p Z_N} \int \sin \theta\, d\theta \, d \phi\, dR dr_3 \cdots dr_N e^{-\beta U (R + \dfrac {1}{2}r, R-\dfrac {1}{2}r, rR, \cdots, rN )} \[4pt] &= {(N - 1) \over 4 \pi p} \left < {\delta (r - r') \over rr'} \right > _{r^t, \theta^t, \phi^t, R^t, r^t_R, \cdots , r^t_N} \end{align*}
Integrating $g(r)$ over the radial dependence, one finds that
$4 \pi p \int _0^{\infty} drr^2 g(r) = N - 1 \approx N \nonumber$
The $g(r)$ function is important for many reasons. It tells us about the structure of complex, isotropic systems, as we will see below, it determines the thermodynamic quantities at the level of the pair potential approximation, and it can be measured in neutron and X-ray diffraction experiments. In such experiments, one observes the scattering of neutrons or X-rays from a particular sample. If a detector is placed at an angle $\theta$ from the wave-vector direction of an incident beam of particles, then the intensity $I (\theta )$ that one observes is proportional to the structure factor
\begin{align*} I (\theta) \sim {1 \over N} \left < \left | \sum _m e^{ik\cdot r_m} \right | ^2 \right > &= {1 \over N} \left < \sum _m e^{ik\cdot (r_m - r_n)} \right >\[4pt] &\equiv S (k) \end{align*}
where $k$ is the vector difference in the wave vector between the incident and scattered neutrons or X-rays (since neutrons and X-rays are quantum mechanical particles, they must be represented by plane waves of the form $exp (ik \cdot r )$). By computing the ensemble average (see problem 4 of problem set #5), one finds that $S (k) = S (k)$ and S(k) is given by
$S(k) = 1 + {4 \pi p \over k} \int _0^{\infty} drr \sin (kr) g (r) \nonumber$
Thus, if one can measure $S(k)$, $g(r)$ can be determined by Fourier transformation. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/07%3A_Distribution_Functions_and_Liquid_Structure/7.06%3A_The_Pair_Correlation_Function.txt |
As a specific example of the application of perturbation theory, we consider the Van der Waals equation of state. Let $U_0$ be given by a pair potential:
$U_0({{\textbf r}_1,...,{\textbf r}_N}) = {1 \over 2}\sum_{i\neq j} u_0(\vert{\textbf r}_i-{\textbf r}_j\vert) \nonumber$
with
$u_0(r) = \left\{\matrix{0 & \;\;\;\;\;\;\;\;r>\sigma \cr\infty & \;\;\;\;\;\;\;\;\;\;r\leq \sigma}\right. \nonumber$
This potential is known as the hard sphere potential. In the low-density limit, the radial distribution function can be shown to be given correctly by $g_0(r) \sim \exp(-\beta u_0(r))$ or
\begin{align*} g_0 (r) &= \theta (r - \sigma ) \[4pt] &= u_1 (r) \end{align*}
$u (r) = u_0 (r) + u_1(r)$ is taken to be some arbitrary attractive potential, whose specific form is not particularly important. Then, the full potential might look like:
$A^{(1)} \approx u(r)=u_0(r)+u_1(r)$
Now, the first term in $A ^{(1)}$ is
\begin{align*} {1 \over 2}\rho^2 V \int _0^{\infty} 4\pi r^2 u_1 (r) g_0 (r) &= {1 \over 2}\rho^2 V\int_0^{\infty}4\pi r^2 u_1(r) \theta (r - \sigma ) \[4pt] &=2 \pi \rho^2V \int_{\sigma}^{\infty} r^2 u_1(r) dr \equiv -aN \rho \[4pt] &=a = -2\pi \int_{\sigma}^{\infty} dr r^2 u_1(r) > 0 \end{align*}
where $\sigma$ is a number that depends on $u_0 (r)$ and the specific form of $u(r) = u_0 (r) + u_1 (r)$.
Since the potential ${Z_N (0) }$ is a hard sphere potential, ${u_0}$ can be determined analytically. If $u_0 (r)$ were 0, then $Z_N^{(0)} = V^N$ would describe an ideal gas and
$Z_N{^{(0)}}= V^N \nonumber$
However, because two particles may not approach each other closer than a distance $u_0 (r)$ between their centers, there is some excluded volume:
If we consider two hard spheres at closest contact and draw the smallest imaginary sphere that contains both particles, then we find this latter sphere has a radius $u_0 (r)$:
$\frac{4}{3} \pi \sigma^3$
Hence the excluded volume for these two particles is
${2 \over 3}\pi \sigma^3 \equiv b \nonumber$
and hence the excluded volume per particle is just half of this:
$Nb \nonumber$
Therefore $Z_N{^{(0)}}= (V-Nb)^N$ is the total excluded volume, and we find that, in the low density limit, the partition function is given approximately by
$A(N,V,T) \approx -{1 \over \beta} \ln \left[{(V-Nb)^N \over N! \lambda^{3N}}\right] - {aN^2 \over V} \nonumber$
Thus, the free energy is
$P = -\left({\partial A \over \partial V}\right)_{N,T} \nonumber$
If we now use this free energy to compute the pressure from
${ {P \over kT}} \nonumber$
we find that
\begin{align*} {N \over V-Nb} - {aN^2 \over kTV^2} &= { {\rho \over 1-\rho b} - {a \rho^2 \over kT} } \[4pt] &= \rho b \gg 1 \end{align*}
This is the well know Van der Waals equation of state. In the very low density limit, we may assume that
${1 \over 1-\rho b} \approx 1 + \rho b, \nonumber$
hence
${P \over kT} \approx \rho + \rho^2\left(b-{a \over kT}\right) \nonumber$
Thus,
$B_2 (T) \approx b - {a \over kT} = {2 \over 3} \pi \sigma^3 + {2 \pi \over kT} \int _{\sigma }^{\infty} dr r^2 u_1 (r) \nonumber$
from which we can approximate the second virial coefficient:
$B_2(T) \approx b-{a \over kT} = {2 \over 3}\pi \sigma^3 +{2\pi \over kT}\int_{\sigma}^{\infty} dr r^2 u_1(r)$
A plot of the isotherms of the Van der Waals equation of state is shown below:
$\frac{\partial P}{\partial V}=0$
$\frac{\partial^2 P}{\partial V}=0$
The red and blue isotherms appear similar to those of an ideal gas, i.e., there is a monotonic decrease of pressure with increasing volume. The black isotherm exhibits an unusual feature not present in any of the ideal-gas isotherms - a small region where the curve is essentially horizontal (flat) with no curvature. At this point, there is no change in pressure as the volume changes. Below this isotherm, the Van der Waals starts to exhibit unphysical behavior. The green isotherm has a region where the pressure decreases with decreasing volume, behavior that is not expected on physical grounds. What is observed experimentally, in fact, is that at a certain pressure, there is a dramatic, discontinuous change in the volume. This dramatic jump in volume signifies that a phase transition has occurred, in this case, a change from a gaseous to a liquid state. The dotted line shows this jump in the volume. Thus, the small flat neighborhood along the black isotherm becomes larger on isotherms below this one. The black isotherm just represents a boundary between those isotherms along which no such phase transition occurs and those that exhibit phase transitions in the form of discontinuous changes in the volume. For this reason, the black isotherm is called the critical isotherm, and the point at which the isotherm is flat and has zero curvature is called a critical point.
A critical point is a point at which
$V_c \nonumber$
Using these two conditions, we can solve for the critical volume ($V_c$) and critical temperature ($(T_c$):
\begin{align*} V_c &= 3Nb \[4pt] kT_c &= {8a \over 27b} \[4pt] P_c &= {a \over 27b^2} \end{align*}
and the critical pressure is therefore
$\kappa_{\rm T} = {1 \over V}{\partial V \over \partial P} \nonumber$
Using these values for the critical pressure, temperature and volume, we can show that the isothermal compressibility, given by
$\left.{\partial P \over \partial V}\right\vert _{V=V_c} =-{NkT \over 2N^2b^2} + {2aN^2 \over 27N^3b^3} = -{kT \over 4Nb^2} + {2a \over 27Nb^3} = {1 \over 4Nb^2}\left({8a \over 27b} - kT\right) \sim (T-T_c) \nonumber$
diverges as the critical point is approached. To see this, note that
$\kappa_{T} \sim (T-T_c)^{-1} \nonumber$
Thus,
${\kappa_T} \nonumber$
It is observed that at a critical point, ${ \vert T-T_c\vert^{-\gamma}}$ diverges, generally, as $Q(N,V,T) = e^{-\beta A(N,V,T)} = {(V-Nb)^N \over N!\lambda^{3N}}e^{\beta aN^2/V}$. To determine the heat capacity, note that
$E \nonumber$
so that
$-{\partial \over \partial \beta}\ln Q(N,V,T) = { -{\partial \over \partial \beta}\left[N\ln (V-Nb) - \ln N! - 3N\ln\lambda + {\beta aN^2 \over V}\right]}$
$= {{3N \over \lambda}{\partial \lambda \over \partial \beta} + {aN^2 \over V} }$
$= {{3N \over \lambda}{\lambda \over \partial \beta} + {aN^2 \over V} }$
$= {{3 NkT \over 2} + {aN^2 \over V} }$
$= C_V = \left({\partial E \over \partial T}\right)_V$
Then, since
$C_V = {3 \over 2}Nk \sim \vert T-T_c\vert^0 \nonumber$
it follows that
${\vert T-T_c\vert^{-\alpha} } \nonumber$
The heat capacity is observed to diverge as $\alpha$. Exponents such as ${\gamma}$ and $\delta$ are known as critial exponents.
Finally, one other exponent we can easily determine is related to how the pressure depends on density near the critical point. The exponent is called ${P \over kT} \sim {\rm const} + C(\rho-\rho_c)^{\delta}$, and it is observed that
${P_c \over kT_c} + {1 \over kT_c}{\partial P \over \partial \rho} \vert _{\rho = \rho _c} (\rho - \rho _c) + {1 \over 2kT_c}{\partial^2 P \over \partial \rho ^2} \vert _{\rho = rho_c} (\rho - \rho _c)^2 + {1 \over 6kT_c}{\partial^3 P \over \partial \rho ^3} \vert _{\rho = \rho_c} (\rho - \rho _c)^3 + \cdots \nonumber$
What does our theory predict for
${P \over kT} \sim const + C ( \rho - \rho _c )^{\delta}$? To determine ${P \over kT} \sim const + C ( \rho - \rho _c )^{\delta}$ we expand the equation of state about the critical density and temperature:
\begin{algin*} {N \over V - Nb} - {aN^2 \over kTV^2} &= {P_c \over kT_c} + {1 \over kT_c}\left.{\partial P \over \partial V}{\partial V \over \partial \rho}_{\rho= \rho_c} (\rho - rho_c) + {1 \over 2kT_c} \left [ {\partial^2 P \over \partial V^2} \left (\partial V \over \partial \rho \right ) + {\partial P \over \partial V}{\partial^2 V \over \partial \rho^2} \right ] \vert_{\rho = \rho_c } (\rho - \rho_c)^2 + {1 \over 6kT_c} {\partial^3 P \over \partial \rho^3}\right\vert _{\rho=\rho_c}(\rho-\rho_c)^3 + \cdots \[4pt] &= {1 \over kT_c}{\partial ^3 P \over \partial \rho ^3} \vert _{\rho = \rho_c} &= {243 \over 8} b^2 \ne 0 \end{align*}
The second and third terms vanish by the conditions of the critical point. The third derivative term can be worked out straightforwardly and does not vanish. Rather
$\delta =0 3 \nonumber$
Thus, we see that, by the above expansion, $\alpha = 0 \gamma = 1 \delta = 3$.
The behavior of these quantities near the critical temperature determine three critical exponents. To summarize the results, the Van der Waals theory predicts that
${\alpha = 0.1, \gamma = 1.45} \nonumber$
The determination of critical exponents such as these is an active area in statistical mechanical research. The reason for this is that such exponents can be grouped into universality classes - groups of systems with exactly the same sets of critical exponents. The existence of universality classes means that very different kinds of systems exhibit essentially the same behavior at a critical point, a fact that makes the characterization of phase transitions via critical exponents quite general. The values obtained above for ${\gamma}$, $\delta$ and ${P \over kT} \sim const + C (\rho - rho_c )^{\delta}$ are known as the mean-field exponents and shows that the Van der Waals theory is really a mean field theory.
These exponents do not agree terribly well with experimental values ( $\delta = 4.2, {\partial P \over \partial V } > 0$ ). However, the simplicity of mean-field theory and its ability to give, at least, qualitative results, makes it, nevertheless, useful. To illustrate universality classes, it can be shown that, within mean field theory, the Van der Waals gas/liquid and a magnetic system composed of spins at particular lattice sites, which composes the so called Ising model, have exactly the same mean field theory exponents, despite the completely different nature of these two systems.
Another problem with the Van der Waals theory that is readily apparent is the fact that it predicts ${\rho }$ for certain values of the density $g_0 (r)$. Such behavior is unstable. Possible ways of improving the approximations used are the following:
1. Improve the approximation to $U_0 (r_1, \cdots , r_N )$.
2. Choose a better zeroth order potential .
3. Go to a higher order in perturbation theory.
Barker and Henderson have shown that going to second order in perturbation theory (see discussion in McQuarrie's book, chapter 14), yields well converged results for a square well fluid, compared to "exact'' results from a molecular dynamics simulation. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/07%3A_Distribution_Functions_and_Liquid_Structure/7.07%3A_Derivation_of_the_Van_der_Waals_equation.txt |
Our treatment of the classical ensembles makes clear that the free energy is a quantity of particular importance in statistical mechanics. Being related to the logarithm of the partition function, the free energy is the generator through which other thermodynamic quantities are obtained, via differentiation. In many cases, the free energy difference between two thermodynamic states is sought. Such differences tell, for example, whether or not a chemical reaction can occur spontaneously or requires input of work and is directly related to the equilibrium constant for the reaction. Thus, for example, from free energy differences, one can compute solvation free energies, acid ionization constants $K_a$ and associated $pK_a$ values, or drug inhibition constants $K_i$, that quantify the ability of a compound to bind to the active site of an enzyme. Another type of free energy often sought is the free energy as a function of one or more generalized coordinates in a system. An example might be the free energy surface as a function of a pair of Ramachandran angles $\phi$ and $\psi$ in an oligopeptide. Such a surface would provide a map of the stable conformations of the molecule, the relative stability of such conformations and the heights of barriers that need to be crossed in a conformational change.
• 8.2: Free-energy Perturbation Theory
We begin our treatment of free energy differences by examining the problem of transforming a system from one thermodynamic state to another.
• 8.3: Adiabatic Switching and Thermodynamic Integration
• 8.4: Reaction Coordinates
It is frequently the case that the progress of some chemical, mechanical, or thermodynamics process can be followed by following the evolution of a small subset of generalized coordinates in a system. When generalized coordinates are used in this manner, they are typically referred to as reaction coordinates, collective variables, or order parameters, often depending on the context and type of system.
• 8.5: Jarzynski's Equality and Nonequilibrium Methods
In this section, the relationship between work and free energy will be explored in greater detail.
• 8.6: The "blue moon'' Ensemble Approach
The blue moon ensemble approach was introduced by Ciccotti and coworkers as a technique for computing the free energy profile along a reaction coordinate direction characterized by one or more barriers high enough that they would not likely be crossed in a normal thermostatted molecular dynamics calculation.
08: Rare-event sampling and free energy calculations
We begin our treatment of free energy differences by examining the problem of transforming a system from one thermodynamic state to another. Let these states be denoted generically as ${\cal A}$ and ${\cal B}$. At the microscopic level, these two states are characterized by potential energy functions $U_{\cal A}({\textbf r}_1,...,{\textbf r}_N)$ and $U_{\cal B}({\textbf r}_1,...,{\textbf r}_N)$. For example, in a drug-binding study, the state ${\cal A}$ might correspond to the unbound ligand and enzyme, while ${\cal B}$ would correspond to the bound complex. In this case, the potential $U_{\cal A}$ would exclude all interactions between the enzyme and the ligand and the enzyme, whereas they would be included in the potential $U_{\cal B }$.
The Helmholtz free energy difference between the states ${\cal A}$ and ${\cal B }$ is simply $A_{\cal A \cal B}= A_{\cal B}-A_{\cal A}$. The two free energies $A_{\cal A}$ and $A_{\cal B}$ are given in terms of their respective canonical partition functions $Q_{\cal A}$ and $Q_{\cal B}$, respectively by $A_{\cal A} = -kT\ln Q_{\cal A}$ and $A_{\cal B} = -kT \ln Q_{\cal B}$, where
\begin{align} Q_{\cal A}(N,V,T) &= C_N\int\;d^N{\textbf p}\;d^N{\textbf r}\;exp\left\{-\beta\left[\sum_{i=1}^N{\textbf p_i^2 \over 2m_i} +U_{\cal A}({\textbf r}_1,...,{\textbf r}_N)\right]\right\} \[4pt] &= Z_{\cal A}(N,V,T) \over N!\lambda^{3N} \end{align}
\begin{align} Q_{\cal B}(N,V,T) &= C_N\int\;d^N{\textbf p}\;d^N{\textbf r}\;exp\left\{-\beta\left[\sum_{i=1}^N{\textbf p_i^2 \over 2m_i} +U_{\cal B}({\textbf r}_1,...,{\textbf r}_N)\right]\right\} \[4pt] &= { {Z_{\cal B}(N,V,T) \over N!\lambda^{3N}} } \label{1} \end{align}
The free energy difference is, therefore,
$A_{\cal A\cal B}= A_{\cal B} - A_{\cal A} = -kT\ln\left ( {Q_{\cal A} \over Q_{\cal A} }\right)= -kT\ln\left({Z_{\cal B} \over Z_{\cal A}}\right) \label{2}$
where $Z_{\cal A}$ and $Z_{\cal B}$ are the configurational partition functions for states $\cal A$ and $\cal B$, respectively,
$Z_{\cal A} = { \int\;d^N{\textbf r}\;e^{-\beta U_{\cal A}({\textbf r}_1,...,{\textbf r}_N)} }$
$Z_{\cal B} = { \int\;d^N{\textbf r}\;e^{-\beta U_{\cal B}({\textbf r}_1,...,{\textbf r}_N)} } \label{3}$
The ratio of full partition functions $Q_{\cal B}/Q_{\cal A}$ reduces to the ratio of configurational partition functions $Z_{\cal B}/Z_{\cal A}$ because the momentum integrations in the former cancel out of the ratio.
Equation \ref{2} is difficult to implement in practice because in any numerical calculation via either molecular dynamics or Monte Carlo, we do not have direct access to the partition function only averages of phase-space functions corresponding to physical observables. However, if we are willing to extend the class of phase-space functions whose averages we seek to functions that do not necessarily correspond to direct observables, then the ratio of configurational partition functions can be manipulated to be in the form of such an average. Consider inserting unity into the expression for $Z_{\cal B}$ as follows:
\begin{align} Z_{\cal B} &= { \int\;d^N{\textbf r}\;e^{-\beta U_{\cal B}({\textbf r}_1,...,{\textbf r}_N)} } \[4pt] &= { \int\;d^N{\textbf r}\;e^{-\beta U_{\cal B}({\textbf r}_1,...,{\textbf r}_N)}e^{-\beta U_{\cal A}({\textbf r}_1,...,{\textbf r}_N)}e^{\beta U_{\cal A}({\textbf r}_1,...,{\textbf r}_N)} } \[4pt] &= { \int\;d^N{\textbf r}\;e^{-\beta U_{\cal A}({\textbf r}_1,...,{\textbf r}_N)}e^{- \beta (U_{\cal B}({\textbf r}_1,...,{\textbf r}_N)-U_{\cal A}({\textbf r}_1,...,{\textbf r}_N))} } \label{4} \end{align}
If we now take the ratio ${Z_{\cal B} \over Z_{\cal A}}$, we find
\begin{align} {Z_{\cal B} \over Z_{\cal A}} &= {1 \over Z_{\cal A}}\int\;d^N{\textbf r}\;e^{-\beta U_{\cal A}({\textbf r}_1,...,{\textbf r}_N)}e^{-\beta(U_{\cal B}({\textbf r}_1,...,{\textbf r}_N)-U_{\cal A}({\textbf r}_1,...,{\textbf r}_N))} \[4pt] &= \left<e^{-\beta (U_{\cal B}({\textbf r}_1,...,{\textbf r}_N)-U_{\cal A}({\textbf r}_1,...,{\textbf r}_N))}\right>_{\cal A} \label{5} \end{align}
where the notation $\left<\cdots\right>_{\cal A}$ indicates an average taken with respect to the canonical configurational distribution of the state $\cal A$. Substituting Equation \ref{5} into Equation \ref{2}, we find
$A_{\cal A \cal B}= -kT\ln\left<e^{-\beta (U_{\cal B}-U_{\cal A})}\right>_{\cal A} \label{6}$
Equation \ref{6} is known as the free-energy perturbation formula; it should be reminiscent of the thermodynamic perturbation formula used to derive the van der Waals equation. Equation \ref{6} can be interpreted as follows: We start with microstates $\{{\textbf r}_1,...,{\textbf r}_N\}$ selected from the canonical ensemble of state ${\cal A}$ and use these to compute $Z_{\cal B}$ by placing them in the state $\cal B$ by simply changing the potential energy from $U_{\cal A}$ to $U_{\cal B}$. In so doing, we need to "unbias'' our choice to sample the configurations from the canonical distribution of state $\cal A$ ${\cal A}$ by removing the weight factor $exp (- \beta U_{\cal A})$ from which the microstates are sample and reweighting the states by the factor $exp (- \beta U_{\cal B})$ corresponding to state $\cal B$. This leads to Equation \ref{5}. The difficulty with this approach is that the microstates corresponding to the canonical distribution of state $\cal A$ may not be states of high probability in the canonical distribution of state $\cal B$. If this is the case, then the potential enegy difference $U_{\cal B} - U_{\cal A}$ will be large, he exponential factor $exp \left [ -\beta (U_{\cal B} - U_{\cal A}) \right ]$ will be negligibly small, and the free energy difference will be very slow to converge in an actual simulation. For this reason, it is clear that the free-energy perturbation formula is only useful for cases in which the two states $\cal A$ and $\cal B$ are not that different from each other.
If $U_{\cal B}$ is not a small perturbation to $U_{\cal A}$, then the free-energy perturbation formula can still be salvaged by introducing a set of $M - 2$ intermediate states with potentials $U_{\alpha} (r_1, \cdots, r_N )$, where $\alpha = 1, \cdots, M$, $\alpha = 1$ corresponds to the state $\cal A$ and $\alpha = M$ corresponds to the state $\cal B$. Let $\Delta U_{\alpha,\alpha+1} = U_{\alpha+1}-U_{\alpha}$. We can now imagine transforming the system from state $\cal A$ to state $\cal B$ by passing through these intermediate states and computing the average of $\Delta U_{\alpha,\alpha+1}$ in the state $\alpha$. Applying the free-energy perturbation formula to this protocol yields the free-energy difference as
$A_{\cal A\cal B}= -kT\sum_{\alpha=1}^{M-1}\ln\left<e^{-\beta\Delta U_{\alpha,\alpha+1}}\right>_{\alpha} \label{7}$
where $\langle\cdots\rangle_{\alpha}$ means an average taken over the distribution $exp (-\beta U_a)$. The key to applying Equation \ref{7} is choosing the intermediate states so as to achieve sufficient overlap between the intermediate states without requiring a large number of them, i.e. choosing the thermodynamic path between states $\cal A$ and $\cal B$ effectively. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/08%3A_Rare-event_sampling_and_free_energy_calculations/8.02%3A_Free-energy_Perturbation_Theory.txt |
The free-energy perturbation approach evokes a physical picture in which configurations sampled from the canonical distribution of state $\cal A$ are immediately "switched'' to the state $\cal B$ by simply changing the potential from $U_{\cal A}$ to $U_{\cal B}$. Such "instantaneous'' switching clearly represents an unphysical path from one state to the other, but we need not concern ourselves with this because the free energy is a state function and, therefore, independent of the path connecting the states. Nevertheless, we showed that the free-energy perturbation theory formula, Equation (6), is only useful if the states $\cal A$ and $\cal B$ do not differ vastly from one another, thus naturally raising the question of what can be done if the states are very different.
The use of a series of intermediate states, by which Equation (7) is derived, exploits the fact that any path between the states can be employed to obtain the free energy difference. In this section, we will discuss an alternative approach in which the system is switched slowly or adiabatically from one state to the other, allowing the system to fully relax at each point along a chosen path from state $\cal A$ to state $\cal B$, rather than instantaneously switching the system between intermediate states, as occurs in Equation (7). In order to effect the switching from one state to the other, we will employ a common trick in the form of an "external'' switching parameter, $\lambda$. This parameter is introduced by defining a new potential energy function
$U({\textbf r}_1,...,{\textbf r}_N,\lambda) \equiv f(\lambda)U_{\cal A} ({\text r}_1,...,{\textbf r}_N)+ g(\lambda)U_{\cal B}({\textbf r}_1,...,{\textbf r}_N) \label{8}$
The functions $f (\lambda)$ and $g (\lambda )$ are referred to as switching functions, and they required to satisfy the conditions $f(0) = 1, f (1) = 0$, corresponding to the state $\cal A$, and $g (0) = 0, g (1) = 1$, corresponding to the state $\cal B$. Apart from these conditions, $f (\lambda)$ and $g (\lambda )$ are completely arbitrary. The mechanism embodied in Equation \ref{8} is one in which some imaginary external controlling influence ("hand of God''), represented by the $\lambda$ parameter, starts the system off in state ${\cal A} (\lambda = 0 )$ and slowly switches off the potential $U_{\cal A}$ while simultaneously switching on the potential $U_{\cal B}$. The process is complete when $\lambda = 1$, when the system is in state $\cal B$. A simple choice for the functions $f (\lambda )$ and $g (\lambda )$ is, for example, $f (\lambda) = 1 - \lambda$ and $g (\lambda ) = \lambda$.
In order to see how Equation \ref{8} can be used to compute the free energy difference $A_{\cal A \cal B}$, consider the canonical partition function of a system described by the potential of Equation \ref{8} for a particular choice of $\lambda$:
$Q(N,V,T,\lambda) = C_N\int\;d^N{\textbf p}\;d^N{\textbf r}\; exp\left \{- \beta \left [\sum_{i=1}^N{{\textbf P}^2_i \over 2m_i}+ U({\textbf r}_1,...,{\textbf r}_N,\lambda)\right]\right\} \label{9}$
This partition function leads to a free energy $A (N, V, T, \lambda)$ via
$A(N,V,T,\lambda) = -kT\ln Q(N,V,T,\lambda) \label{10}$
Recall, however, that the derivatives of the free energy with repsect to $N$, and $V$ and $T$ lead to the chemical potential, pressure and entropy, respectively. What does the derivative of the free energy $A (N, V, T, \lambda )$ with respect to $\lambda$ represent? According to Equation \ref{10}
${\partial A \over \partial \lambda} =-{kT\over Q}{\partial Q \over \partial \lambda} = -{kT \over Z}{\partial Z \over \partial \lambda } \label{11}$
The reader should check that the expressions involving $Q$ and $Z$ are equivalent. Computing the derivative of $Z$ with respect to $\lambda$, we find
\begin{align} {kT \over Z}{\partial Z \over \partial \lambda} &= {kT \over Z}{\partial \over \partial \lambda}\int\;d^N{\textbf r}\;e^{-\beta U({\textbf r}_1,...,{\textbf r}_N,\lambda)} \[4pt] &= {kT \over Z}\int\;d^N{\textbf r}\;\left(-\beta {\partial U \over \partial \lambda}\right)e^{-\beta U({\textbf r}_1,...,{\textbf r}_N,\lambda)} \[4pt] &= -\left<{\partial U \over \partial \lambda}\right> \label{12} \end{align}
Now, the free energy difference $A_{\cal A \cal B}$ can be obtained trivially from the relation
$A_{\cal A\cal B}= \int_0^1 {\partial A \over \partial \lambda}d\lambda \label{13}$
Substituting eqns. \ref{11} and \ref{12} into Equation \ref{13}, we obtain the free energy difference as
$A_{\cal A\cal B}= \int_0^1 \left<{\partial U \over \partial \lambda}\right>_{\lambda}d\lambda \label{14}$
where $\langle\cdots\rangle_{\lambda}$ denotes an average over the canonical ensemble described by the distribution $\exp[-\beta U({\textbf r}_1,...,{\textbf r}_N,\lambda)]$ with $\lambda$ fixed at a particular value. The special choice of $f (\lambda ) = 1 - \lambda$ and $g (\lambda ) = \lambda$ has a simple interpretation. For this choice, Equation \ref{14} becomes
$A_{\cal A\cal B}= \int_0^1\left<U_{\cal B}-U_{\cal A}\right>_{\lambda}d\lambda \label{15}$
The content of Equation (15) can be understood by recalling the relationship between work and free energy from the second law of thermodynamics. If, in transforming the system from state $\cal A$ to state $\cal B$, an amount of work $W$ is performed on the system, then
$W \ge A_{\cal A \cal B} \label{16}$
where equality holds only if the transformation is carried out along a reversible path. Since reversible work is related to a change in potential energy, Equation \ref{15} is actually a statistical version of Equation \ref{16} for the special case of equality. Equation (15) tells us that the free energy difference is the ensemble average of the microscopic reversible work needed to change the potential energy of each configuration from $U_{\cal A}$ to $U_{\cal B}$ along the chosen $\lambda$-path. Note, however, that Equation (14), which is known as the thermodynamic integration formula, is true independent of the choice of $f (\lambda)$ and $g (\lambda )$, which means that Equation (14) always yields the reversible work via the free energy difference. The flexibility in the choice of the $\lambda$-path, however, can be exploited to design adiabatic switching algorithms of greater efficiency that can be achieved with the simple choice $f (\lambda) = 1 - \lambda, g (\lambda ) = \lambda$.
In practice, the thermodynamic integration formula is implemented as follows: A set of $M$ values of $\lambda$ is chosen from the interval $[0, 1]$, and at each chosen value $\lambda _k$, a full molecular dynamics or Monte Carlo calculation is carried out in order to generate the average $\left<{\partial U \over \partial \lambda_k}\right>_{\lambda_k}$. The resulting values of $\left<{\partial U \over \partial \lambda_k } \right>_{\lambda_k}$, $k = 1, \cdots , M$ are then substituted into Equation (14), and the resulted is integrated numerically to produce the free energy difference $A_{\cal A \cal B}$. Thus, we see that the selected values $\left \{ \lambda _k \right \}$ can be evenly spaced, for example, or they could be a set of Gaussian quadrature nodes, depending on how $A (N, V, T, \lambda )$ is expected to vary with $\lambda$ for the chosen $f (\lambda )$ and $g (\lambda )$.
As with free-energy perturbation theory, the thermodynamic integration approach can be implemented very easily. An immediately obvious disadvantage of the method, however, is the same one that applies to Equation (7): In order to perform the numerical integration, it is necessary to perform many simulations of a system at physically uninteresting intermediate values of $\lambda$ where the potential $U({\textbf r}_1,...,{\textbf r}_N,\lambda)$ is, itself, unphysical. Only $\lambda = 0, 1$ correspond to actual physical states and ultimately, we can only attach physical meaning to the free energy difference $A_{\cal A \cal B}= A(N,V,T,1)-A(N,V,T,0)$. Nevertheless, the intermediate averages must be accurately calculated in order for the integration to yield a correct result. The approach to be presented in the next section attempts to reduce the time spent in such unphysical intermediate states and focuses the sampling in the important regions $\lambda = 0, 1$. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/08%3A_Rare-event_sampling_and_free_energy_calculations/8.03%3A_Adiabatic_Switching_and_Thermodynamic_Integration.txt |
It is frequently the case that the progress of some chemical, mechanical, or thermodynamics process can be followed by following the evolution of a small subset of generalized coordinates in a system. When generalized coordinates are used in this manner, they are typically referred to as reaction coordinates, collective variables, or order parameters, often depending on the context and type of system. Whenever referring to these coordinates, we will refer to them as reaction coordinates, although the reader should be aware that the other two designations are also used in the literature.
As an example of a useful reaction coordinate, consider a simple gas-phase diatomic dissociation process
$\ce{AB -> A + B}. \nonumber$
If ${{\textbf r}_A}$ and ${{\textbf r}_B}$ denote the Cartesian coordinates of atom A and B, then a useful generalized coordinate for following the progress of the dissociation is simply the distance $r = \vert{\textbf r}_{B}-{\textbf r}_{A}\vert$. A complete set of generalized coordinates that contains$r$ as one of the coordinates is the set that contains the center of mass
$\textbf R= \dfrac{m_A{\textbf r}_A + m_B{\textbf r}_B}{m_A + m_B}, \nonumber$
the magnitude of the relative coordinate
$r = \vert{\textbf r}_{B}-{\textbf r}_{A}\vert$
and the two angles $\phi = \tan^{-1}(y/x)$ and $\theta = \tan^{-1}(\sqrt{x^2 + y^2}/z)$, where $x$, ${y}$ and $z$ are the components of the relative coordinate
$\textbf r= \textbf r_B-\textbf r_A. \nonumber$
Of course, in the gas-phase, where the potential between $\ce{A}$ and $\ce{B}$ likely only depends on the distance between $\ce{A}$ and $\ce{B}$, $r$ is really the only interesting coordinate. However, if the reaction were to take place in solution, then other coordinate such as $\theta$ and $\phi$ become more relevant as specific orientations might change the mechanism or thermodynamic picture of the process, depending on the complexity of the solvent, and averaging over these degrees of freedom to produce a free energy profile $A (r)$ in $r$ alone will wash out some of this information.
As another example, consider a gas-phase proton transfer reaction
$\ce{A-H ... B -> A ... H-B}. \nonumber$
Here, although the distance $\vert{\textbf r}_H-{\textbf r}_A\vert$ can be used to monitor the progress of the proton away from A and the distance $\vert{\textbf r}_H-{\textbf r}_B\vert$ can be used to monitor the progress of the proton toward B, neither distance alone is sufficient for following the progress of the reaction. However, the difference
$\delta = \vert{\textbf r}_H-{\textbf r}_B\vert - \vert{\textbf r}_H-{\textbf r}_A \vert \nonumber$
can be used to follow the progress of the proton transfer from A to B and, therefore, is a potentially useful reaction coordinate. A complete set of generalized coordinates involving $\delta$ can be constructed as follows. If $\textbf r_A$, $\textbf r_B$ and ${{\textbf r}_H}$ denote the Cartesian coordinates of the three atoms, then first introduce the center-of-mass
$\textbf R = \dfrac{m_A{\textbf r}_A + m_B{\textbf r}_B + m_H{\textbf r}_H}{m_A + m_B + m_H} \nonumber$
the relative coordinate between A and B, ${\textbf r}= {\textbf r}_{\rm B}-{\textbf r}_{\rm A}$, and a third relative coordinate $s$ between H and the center-of-mass of A and B,
${\textbf s}= {\textbf r}- \dfrac{m_{\rm A}{\textbf r}_{\rm A} + m_{\rm B}{\textbf r}_{\rm B}}{m_{\rm A} + m_{\rm B}}. \nonumber$
Finally, ${\textbf r}$ is transformed into spherical polar coordinates, $(r, \theta , \phi )$, and from ${\textbf r}$ and $s$, three more coordinates are formed:
$\sigma = \left\vert{\textbf s}+ {m_{\rm B} \over m_{\rm A} + m_{\rm B} } {\textbf r} \right\vert + \left\vert{\textbf s}- {m_{\rm A} \over m_{\rm A} + m_{\rm B}}{\textbf r} \right\vert \delta = \left\vert{\textbf s}+ {m_{\rm B} \over m_{\rm A} + m_{\rm B} } {\text r} \right\vert - \left \vert{\bf s}- {m_{\rm A} \over m_{\rm A} + m_{\rm B}}{\bf r} \right\vert \nonumber$
and the angle $\alpha$, which measures the "tilt'' of the plane containing the three atoms from the vertical. The coordinates $(\sigma , \delta , \alpha )$ are known as confocal elliptic coordinates. These coordinates could also be used if the reaction takes place in solution. As expected, the generalized coordinates are functions of the original Cartesian coordinates. The alanine-dipeptide example above also employs the Ramachandran angles $\phi$ and $\psi$ as reaction coordinates, and these can also be expressed as part of a set of generalized coordinates that are functions of the original Cartesian coordinates of a system.
While reaction coordinates or collective variables are potentially very useful constructs, they must be used with care, particularly when enhanced sampling methods are applied to them. Enhanced sampling of a poorly chosen reaction coordinate can bias the system in unnatural ways, leading to erroneous predictions of free energy barriers and associated mechanisms. A dramatic example of this is the autodissociation of liquid water following the classic reaction
$\ce{ 2H_2O(l) -> H_3O^{+} (aq) + OH^{-} (aq) } \nonumber$
the ostensibly only requires transferring a proton from one water molecule to another. If this notion of the reaction is pursued, then a seemingly sensible reaction coordinate would simply be the distance between the oxygen and the transferring proton or the number of hydrogens covalently bonded to the oxygen. These reaction coordinates, as it turns out, are inadequate for describing the true nature of the reaction and, therefore, fail to yield reasonable free energies (and hence, values of the autoionization constant $K_w$). Chandler and coworkers showed that the dissociation reaction can only be considered to have occurred when the $\ce{H_3O^+}$ and $\ce{OH^-}$ ions are sufficiently far apart that no contiguous or direct path of hydrogen-bonding in the liquid can allow the proton to transfer back to the water or its origin. In order to describe such a process correctly, a very different type of reaction coordinate would clearly be needed.
Keeping in mind such caveats about the use of reaction coordinates, we now proceed to describe a number of popular methods designed to enhance sampling along pre-selected reaction coordinates. All of these methods are designed to generate, either directly or indirectly, the probability distribution function $P (q_1, \cdots, q_n)$ of a subset of $n$ reaction coordinates of interest in a system. If these reaction coordinates are obtained from a transformation of the Cartesian coordinates $q_{\alpha} = f_{\alpha}({\textbf r}_1,...,{\textbf r}_N)$, ${\alpha = 1, \cdots, n}$, then the probability density that these $n$ coordinates will have values ${q_{\alpha} = s_{\alpha} }$ in the canonical ensemble is
$P(s_1,...,s_n) = {C_N \over Q(N,V,T)}\int\;d^N{\textbf p}\;d^N{\textbf r} e^{-\beta H(p,r)} \Phi_{\alpha=1}^n \delta(f_{\alpha}({\textbf r}_1,...,{\textbf r}_N)-s_{\alpha}) \nonumber$
where the $\delta$-functions are introduced to fix the reaction coordinates at values ${q_1, \cdots, q_n }$ at ${s_1, \cdots, s_n }$. Once $P (s_1, \cdots, s_n )$ is known, the free energy hypersurface in these coordinates is given by
$A(s_1,...,s_n) = -kT\ln P(s_1,...,s_n) \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/08%3A_Rare-event_sampling_and_free_energy_calculations/8.04%3A_Reaction_Coordinates.txt |
In this section, the relationship between work and free energy will be explored in greater detail. We have already introduced the inequality in Equation \ref{16}, which states that if an amount of work $W_{\cal {A}\cal {B}}$ is performed on a system, taking from state ${\cal A}$ to state ${\cal B }$, then $W_{\cal A\cal B}\geq A_{\cal A\cal B}$. Here, equality holds only if the work is performed reversibly. The work referred to here is thermodynamic quantity and, as such, must be regarded as an ensemble average. In statistical mechanics, we can also introduce the mechanical or microscopic work ${\cal W}_{\cal A\cal B}({\rm x})$ performed on one member of the ensemble to drive it from state $\cal A$ to state $\cal B$. Then, $W_{\cal A\cal B}$ is simply an ensemble average of ${\cal W}_{\cal A\cal B}$. However, we need to be somewhat careful about how we define this ensemble average because the work is defined along a particular path or trajectory which takes the system from state $\cal A$ to state $\cal B$, and equilibrium averages do not refer not to paths but to microstates. This distinction is emphasized by the fact that the work could be carried out irreversibly, such that the system is driven out of equilibrium. Thus, the proper definition of the ensemble average follows along the lines already discussed in the context of the free-energy perturbation approach, namely, averaging over the canonical distribution for the state $\cal A$. In this case, since we will be discussing actual paths ${\rm x_t}$, we let the initial condition ${\rm x_0}$ be the phase space vector for the system in the (initial) state $\cal A$. Recall that $x_t = x_t (x_0)$ is a unique function of the initial conditions. Then
\begin{align} W_{\cal A\cal B} &= \langle {\cal W}_{\cal A\cal B}(\rm x_0) \rangle _{\cal A} \[4pt] &= {C_N \over Q_{\cal A} (N, V, T)} \int d \rm x_0 e^{-\beta H_{\cal A}({\rm x}_0)}{\cal W}_{\cal A\cal B}({\rm x}_0) \label{17} \end{align}
and the Clausius inequality can be stated as $\langle {\cal W}_{\cal A\cal B}({\rm x}_0)\rangle_{\cal A}\geq A_{\cal A\cal B}$.
From such an inequality, it would seem that using the work as a method for calculating the free energy is of limited utility, since the work necessarily must be performed reversibly, otherwise one obtains only upper bound on the free energy. It turns out, however, that irreversible work can be used to calculate free energy differences by virtue of a connection between the two quantities first discovered in 1997 by C. Jarzynski that as come to be known as the Jarzynski equality. This equality states that if, instead of averaging ${\cal W}_{\cal A\cal B}({\rm x}_0)$ over the initial canonical distribution (that of state $\cal A$), an average of $\exp[-\beta{\cal W}_{\cal A\cal B}({\rm x}_0)]$ is performed over the same distribution, the result is $\exp[-\beta A_{\cal A\cal B}]$, i.e.
\begin{align} e^{-\beta A_{\cal A\cal B} } &= \langle e^{-\beta {\cal W}_{\cal A \cal B}(\rm x_0)} \rangle_{\cal A} \[4pt] &= \dfrac{C_N}{Q_{\cal A} (N, V, T )} \int d \rm x_0 e^{-\beta H_{\cal A} (\rm x_0)} e^{-\beta {\cal W}_{\cal A\cal B} (\rm x_0)} \label{18} \end{align}
This remarkable result not only provides a foundation for the development of nonequilibrium free-energy methods but also has profound implications for thermodynamics, in general.
The Jarzynski equality be proved using different strategies. Here, however, we will present a proof that is most relevant for the finite-sized systems and techniques employed in molecular dynamics calculations. Consider a time-dependent Hamiltonian of the form
$H({\bf p},{\bf r},t) = \sum_{i=1}^N {{\bf p}_i^2 \over 2m_i} + U({\bf r}_1,...,{\bf r}_N,t) \label{19}$
For time-dependent Hamiltonian's, the usual conservation law $dH/dt=0$ no longer holds, which can be seen by computing
${dH \over dt} = \nabla_{{\rm x}_t}H\dot{{\rm x}_t} +{\partial H \over \partial t} \label{20}$
where the phase space vector ${\rm x}= ({\bf p}_1,...,{\bf p}_N,{\bf r}_1,...,{\bf r}_N)\equiv ({\bf p},{\bf r})$ has been introduced. Integrating both sides over time from $t = 0$ to a final time $t = \tau$, we find
$\int_0^{\tau}\;dt\;{dH \over dt}= \int_0^{\tau}\;dt\;\nabla _{\rm x_t} H{\rm x}_t+ \int_0^{\tau}\;dt\;{\partial H \over \partial t} \label{21}$
Equation \ref{21} can be regarded as a microscopic version of the first law of thermodynamics, in which the first and second terms represent the heat absorbed by the system and the work done on the system over the trajectory, respectively. Note that the work is actually a function of the initial phase-space vector ${{\rm x}_0}$, which can be seen by writing this term explicitly as
$W_{\tau}({\rm x}_0) = \int_0^{\tau}\;dt\;{\partial \over \partial t}H({\rm x}_t({\rm x}_0),t) \label{22}$
where the fact that the work depends explicitly on $\tau$ in Equation \ref{22} is indicated by the subscript. In the present discussion, we will consider that each initial condition, selected from a canonical distribution in ${\rm x _0 }$, evolves according to Hamilton's equations in isolation. In this case, the heat term $\nabla_{{\rm x}_t}H\cdot{\rm x}_t = 0$, and we have the usual addition to Hamilton's equations $dH/dt = \partial H/\partial t$.
With the above condition, we can write the microscopic work as
${\cal W}_{\cal A\cal B}= \int_0^{\tau} {d \over dt}H({\rm x}_t (\rm x_0) , t)dt =H({\rm x}_{\tau}({\rm x}_0),\tau) - H({\rm x}_0,0) \label{23}$
The last term $H({\rm x}_0,0)$ is also $H_{\cal A}({\rm x}_0)$. Thus, the ensemble average of the exponential of the work becomes
$\langle e^{-\beta {\cal W}_{\cal A\cal B}}\rangle_{\cal A} = { {C_N \over Q_{\cal A}(N,V,T)}\int\;d{\rm x}_0\;e^{-\beta H_{\cal A} (\rm x_0)}e^{-\beta [H({\rm x}_{\tau}({\rm x}_0),\tau)-H_{\cal A}({\rm x}_0)]}}$
${{C_N \over Q_{\cal A}(N,V,T)}\int\;d{\rm x}_0\;e^{-\beta H({\rm x}_{\tau}({\rm x}_0),\tau)} }$
The numerator in this expression becomes much more interesting if we perform a change of variables from ${\rm x_0}$ to ${\rm x_{\tau}}$. Since the solution of Hamilton's equations for the time-dependent Hamiltonian uniquely map the initial condition ${\rm x_0}$ onto ${\rm x_t}$, when $t = \tau$, we have a new set of phase-space variables, and by Liouville's theorem, the phase-space volume element is preserved
$d{\rm x}_{\tau} = d{\rm x}_0 \label{24}$
When the Hamiltonian is transformed, we find $H({\rm x}_{\tau},\tau) =H_{\cal B}({\rm x}_{\tau})$. Consequently,
\begin{align} \langle e^{-\beta {\cal W}_{\cal A\cal B}} \rangle_{\cal A} &=\dfrac{C_N}{Q(N,V,T)} \int\;d{\rm x}_{\tau}\;e^{-\beta H_{\cal B} ({\rm x}_{\tau})} \[4pt] &= \dfrac{Q_{\cal B}(N,V,T)}{Q_{\cal A}(N,V,T)} \[4pt] &= e^{-\beta A_{\cal A\cal B}}\end{align}
thus proving the equality. The implication of the Jarzynski equality is that the work can be carried out along a reversible or irreversible path, and the correct free energy will still be obtained.
Note that due to Jensen's inequality:
$\langle e^{-\beta{\cal W}_{\cal A\cal B}} \rangle_{\cal A} \ge e^{-\beta \langle{\cal W}_{\cal A\cal B}\rangle_{\cal A}} \label{25}$
Using Jarzynski's equality, this becomes
$e^{-\beta A_{\cal A\cal B}} \ge e^{-\beta \langle{\cal W}_{\cal A\cal B}\rangle_{\cal A}} \label{26}$
which implies, as expected, that
$A_{\cal A\cal B}\leq \langle{\cal W}_{\cal A\cal B}\rangle_{\cal A} \label{27}$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/08%3A_Rare-event_sampling_and_free_energy_calculations/8.05%3A_Jarzynski%27s_Equality_and_Nonequilibrium_Methods.txt |
The term "blue moon'' in the present context describes rare events, i.e. events that happen once in a blue moon. The blue moon ensemble approach was introduced by Ciccotti and coworkers as a technique for computing the free energy profile along a reaction coordinate direction characterized by one or more barriers high enough that they would not likely be crossed in a normal thermostatted molecular dynamics calculation.
Suppose a process of interest can be monitored by a single reaction coordinate $q_1=f_1({\bf r}_1,...,{\bf r}_N)$ so that eqns. (29) and (30) reduce to
\begin{align} P(s) &= { {C_N \over Q(N,V,T)}\int\;d^N{\bf p}\;d^N{\bf r}e^{-\beta H({\bf p},{\bf r})}\delta(f_1({\bf r}_1,...,{\bf r}_N)-s) } \[4pt] &= { {1 \over N!\lambda^{3N} Q(N,V,T)}\int\;d^N{\bf r}e^{-\beta U({\bf r})}\delta(f_1({\bf r}_1,...,{\bf r}_N)-s) }\[4pt] A (s) &=-kT\ln P(s) \label{31} \end{align}
The "1'' subscript on the value $s$ of ${q_1}$ is superfluous and will be dropped throughout this discussion. In the second line, the integration over the momenta has been performed giving the thermal prefactor factor $\lambda ^{3N}$. In the blue moon ensemble approach, a holonomic constraint $\sigma({\bf r}_1,...,{\bf r}_N) =f_1({\bf r}_1,...,{\bf r}_N)-s$ is introduced in a molecular dynamics calculation as a means of "driving'' the reaction coordinate from an initial value ${s_i}$ to a final value ${s_f}$ via a set of intermediate points ${s_1,...,s_n }$ between ${s_i}$ and ${s_f}$. Unfortunately, the introduction of a holonomic, constraint does not yield the single $\delta$-function condition $\delta(\sigma({\bf r}) =\delta(f_1({\bf r})-s)$, where ${{\bf r}\equiv {\bf r}_1,...,{\bf r}_N }$ required by Equation \ref{31} but rather the product of $\delta$-functions $\delta(\sigma({\bf r}))\delta(\dot{\sigma}({\bf r},{\bf p}))$, since both the constraint and its first time derivative are imposed in a constrained dynamics calculation. We will return to this point a bit later in this section. In addition to this, the blue moon ensemble approach does not yield $A(s)$ directly but rather the derivative
${dA \over ds} = -{kT \over P(s)}{dP \over ds} \label{32}$
from which the free energy profile $A (q)$ along the reaction coordinate and the free energy difference $\Delta A = A(s_f)-A(s_i)$ are given by the integrals
$A(q) = A(s_i) + \int_{s_i}^q ds {dA \over ds}\;\;\;\;\;\;\;\;\;\;\Delta A = \int_{s_i}^{s_f} ds {dA \over ds} \label{33}$
In the free-energy profile expression $A (s_i)$ is just an additive constant that can be left off. The values ${s_1,...,s_n }$ at which the reaction coordinate is constrained can be chosen at equally-spaced intervals between ${s_i}$ and ${s_f}$, in which a standard numerical quadrature can $q=f_1({\bf r})$ be applied for evaluating the integrals in Equation \ref{33}, or they can be chosen according to a more sophisticated quadrature scheme.
We next turn to the evaluation of the derivative in Equation \ref{32}. Noting that $P(s) = \langle \delta(f_1({\bf r})-s)\rangle$, the derivative can be written as
${1 \over P(s)}{dP \over ds} ={C_N \over Q(N,V,T)}{\int\;d^N {\bf p} d^N {\bf r} e^{-\beta H_{(p,r)}} {\partial \over \partial s}\delta(f_1({\bf r})-s)\over \langle \delta(f_1({\bf r})-s)\rangle} \label{34}$
In order to avoid evaluating the derivative of the $\delta$-function, an integration by parts can be used. First, we introduce a complete set of $3N$ generalized coordinates:
$q_{\alpha} = f_{\alpha}({\bf r}_1,...,{\bf r}_N) \label{35}$
and their conjugate momenta ${p_{\alpha} }$. Such a transformation has a unit Jacobian so that $d^N{\bf p}\;d^N{\bf r}= d^{3N}p\;d^{3N}q$. Denoting the transformed Hamiltonian as ${\tilde{H}(p,q) }$, Equation \ref{34} becomes
${1 \over P(s)}{dP \over ds} ={C_N \over Q(N,V,T)}{\int\;d^{3N} P d^{3N} q e^{-\beta \tilde {H} (p, q)}{\partial \over \partial s}\delta(q_1-s)\over \langle \delta(q_1-s)\rangle} \label{36}$
Changing the derivative in front of the $\delta$-function from $\partial/\partial s$ to $\partial/\partial q_1$, which introduces an overall minus sign, and then integrating by parts yields
\begin{align} {1 \over P(s)}{dP \over ds} &= {C_N \over Q(N,V,T)}{\int\;d^{3N}p\;d^{3N}q\;\left[{\partial \over \partial q_1}e^{-\beta \tilde{H}(p,q)}\right]\delta(q_1-s)\over \langle \delta(q_1-s)\rangle} \[4pt] &= -{\beta C_N \over Q(N,V,T)}{\int\;d^{3N}p\;d^{3N}q\;{\partial\tilde {H} \over \partial q_1}e^{-\beta \tilde{H}(p,q)}\delta(q_1-s)\over \langle \delta(q_1-s)\rangle} \[4pt] & = -\beta {\left<\left({\partial \tilde{H} \over \partial q_1}\right)\delta(q_1-s)\right>\over \langle \delta(q_1-s)\rangle} \label{37} \end{align}
The last line defines a new ensemble average, specifically an average subject to the condition (not constraint) that the coordinate ${q_1}$ have the particular value $s$. This average will be denoted $\langle\cdots\rangle^{\rm cond}_{s}$. Thus, the derivative becomes
${1 \over P(s)}{dP \over ds} =-\beta\left<{\partial \tilde{H} \over \partial q_1}\right>^{\rm cond}_s \label{38}$
Substituting Equation \ref{38} yields a free energy profile of the form
$A(q) = A(s_i) + \int_{s_i}^q\;ds\;\left<{\partial \tilde{H} \over \partial q_1}\right>^{\rm cond}_s \label{39}$
from which $\Delta A$ can be computed by letting ${q = s_f}$. Given that - ${\langle \partial \tilde{H}/\partial q_1\rangle^{\rm cond}_s }$ is the expression for the average of the generalized force on ${q_1}$ when ${q_1 = s}$, the integral represents the work done on the system, i.e. the negative of the work done by the system, in moving from ${s_i}$ to an arbitrary final point ${q}$. Since the conditional average implies a full simulation at each fixed value of ${q_1}$, the thermodynamic transformation is certainly carried out reversibly, so that Equation \ref{39} is consistent with the Clausius inequality.
Although Equation \ref{39} provides a very useful insight into the underlying statistical mechanical expression for the free energy, technically, the need for a full canonical transformation of both coordinates and momenta is inconvenient since, from the chain rule
${\partial \tilde{H} \over \partial q_1} =\sum_{i=1}^N\left [ {\partial H \over \partial {\bf p} _i } \cdot {\partial {\bf p}_i \over \partial q_1} + {\partial H \over \partial {\bf r}_i }\cdot{\partial {\bf r}_i \over \partial q_1}\right] \label{40}$
A more useful expression results if the momenta integrations are performed before introducing the transformation to generalized coordinates. Starting again with Equation \ref{34}, we carry out the momentum integrations, yielding
${1 \over P(s)}{dP \over ds} ={1 \over N!\lambda^{3N}Q(N,V,T )} {\int d^N {\bf r} e^{-\beta U (r)} {\partial \over \partial s} \delta(f_1({\bf r})-s)\over \langle \delta(f_1({\bf r})-s)\rangle} \label{41}$
Now, we introduce only the transformation of the coordinates to generalized coordinates $q_{\alpha} = f_{\alpha}({\bf r}_1,...,{\bf r}_N)$. However, because there is no corresponding momentum transformation, the Jacobian of the transformation is not unity. Let $J(q)\equiv J(q_1,...,q_{3N}) =\partial ({\bf r}_1,...,{\bf r}_N)/\partial (q_1,...,q_{3N})$ denote the Jacobian of the transformation. Then, Equation \ref{41} becomes
\begin{align} {1 \over P(s)}{dP \over ds} &= { {1 \over N!\lambda^{3N}Q(N,V,T)}{\int\;d^{3N}q\;J(q)e^{-\beta {\tilde U} (q)}{\partial \over \partial s}\delta(q_1-s)\over \langle \delta(q_1-s)\rangle}} \[4pt] &= {1 \over N!\lambda^{3N}Q(N,V,T)}{\int\;d^{3N}q\;e^{-\beta \left(U(q) - kT J(q) \right )}{\partial \over \partial s}\delta(q_1-s)\over \langle \delta(q_1-s)\rangle} \label{42} \end{align}
where, in the last line, the Jacobian has been exponentiated. Changing the derivative $\partial/\partial s$ to $\partial/\partial q_1$ and performing the integration by parts as was done in Equation \ref{37}, we obtain
\begin{align} {1 \over P(s)}{dP \over ds} &= {1 \over N!\lambda^{3N}Q(N,V,T)}{\int\;d^{3N}q\;{\partial \over \partial q_1}e^{-\beta \left (\tilde{U}(q)-kT\ln J(q)\right)}\delta(q_1-s)\over \langle \delta(q_1-s)\rangle} \[4pt] &= -{\beta \over N!\lambda^{3N}Q(N,V,T)}{\int\;d^{3N}q\;\left[{\partial \tilde {U} \over partial q_1} -KT{\partial \over \partial q_1} \ln J (q) \right]e^{-\beta \left (\tilde{U}(q)-kT\ln J(q)\right)}\delta(q_1-s)\over \langle \delta(q_1-s)\rangle} \[4pt] &= {-\beta\left<\left[{\partial\tilde{U} \over \partial q_1}-kT{\partial \over \partial q_1}\ln J(q)\right]\right>^{\rm cond}_s } \label{43} \end{align}
Therefore, the free energy profile becomes
$A(q) = A(s_i) +\int_{s_i}^q\;ds\;\left<\left[{\partial\tilde {U} \over \partial q_1} - KT {\partial \over \partial q_1}\ln J(q)\right]\right>^{\rm cond}_s \label{44}$
Again, the derivative of ${\tilde U}$, the transformed potential, can be computed form the untransformed potential via the chain rule
${\partial \tilde{U} \over \partial q_1} =\sum_{i=1}^N {\partial U \over \partial {\bf r}_i}\cdot {\partial {\bf r}_i \over \partial q_1} \label{45}$
Equation \ref{44} is useful for simple reaction coordinates in which the full transformation to generalized coordinates is known. We will see shortly how the expression for $A (q)$ can be further simplified in a way that does not require knowledge of the transformation at all. First, however, we must tackle the problem alluded to earlier of computing the conditional ensemble averages from the constrained dynamics employed by the blue moon ensemble method. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/08%3A_Rare-event_sampling_and_free_energy_calculations/8.06%3A_The_blue_moon%27%27_Ensemble_Approach.txt |
• 9.1: Measurement
The result of a measurement of the observable A must yield one of the eigenvalues of A^ . Thus, we see why A is required to be a hermitian operator: Hermitian operators have real eigenvalues.
• 9.2: Physical Observables
• 9.3: The Fundamental Postulates of Quantum Mechanics
• 9.4: The Heisenberg Picture
In all of the above, notice that we have formulated the postulates of quantum mechanics such that the state vector evolves in time, but the operators corresponding to observables are taken to be stationary. This formulation of quantum mechanics is known as the Schrödinger picture. However, there is another, completely equivalent, picture in which the state vector remains stationary and the operators evolve in time. This picture is known as the Heisenberg picture.
• 9.5: The Heisenberg Uncertainty Principle
Because the operators x and p are not compatible, [X,P]≠0, there is no measurement that can precisely determine both x and p simultaneously. Hence, there must be an uncertainty relation between them that specifies how uncertain we are about one quantity given a definite precision in the measurement of the other. Presumably, if one can be determined with infinite precision, then there will be an infinite uncertainty in the other.
• 9.6: The Physical State of a Quantum System
The physical state of a quantum system is represented by a vector denoted |Ψ(t)⟩ which is a column vector, whose components are probability amplitudes for different states in which the system might be found if a measurement were made on it.
• 9.7: Time Evolution of the State Vector
The time evolution of the state vector is prescribed by the Schrödinger equation.
09: Review of the basic postulates of quantum mechanics
The result of a measurement of the observable $A$ must yield one of the eigenvalues of $\hat{A}$. Thus, we see why $A$ is required to be a hermitian operator: Hermitian operators have real eigenvalues. If we denote the set of eigenvalues of $\hat{A}$ by $\{a_i\}$, then each of the eigenvalues ${a_i}$ satisfies an eigenvalue equation
$\hat{A}\vert a_i\rangle = a_i \vert a_i\rangle \nonumber$
where $\vert a_i\rangle$ is the corresponding eigenvector. Since the operator $\hat{A}$ is hermitian and ${a_i}$ is therefore real, we have also the left eigenvalue equation
$\langle a_i\vert \hat{A} = \langle a_i\vert a_i \nonumber$
The probability amplitude that a measurement of $A$ will yield the eigenvalue ${a_i}$ is obtained by taking the inner product of the corresponding eigenvector $\vert a_i \rangle$ with the state vector $\vert\Psi(t)\rangle$, $\langle a_i\vert\Psi(t)\rangle$. Thus, the probability that the value ${a_i}$ is obtained is given by
$P_{a_i} = \vert\langle a_i\vert\Psi(t)\rangle \vert^2 \nonumber$
Another useful and important property of hermitian operators is that their eigenvectors form a complete orthonormal basis of the Hilbert space, when the eigenvalue spectrum is non-degenerate. That is, they are linearly independent, span the space, satisfy the orthonormality condition
$\langle a_i\vert a_j\rangle = \delta_{ij} \nonumber$ and thus any arbitrary vector $\vert\phi\rangle$ can be expanded as a linear combination of these vectors:
$\vert\phi\rangle = \sum_i c_i \vert a_i\rangle \nonumber$ By multiplying both sides of this equation by $\langle a_j\vert$ and using the orthonormality condition, it can be seen that the expansion coefficients are
$c_i = \langle a_i\vert\phi\rangle \nonumber$
The eigenvectors also satisfy a closure relation:
$I = \sum_i \vert a_i\rangle \langle a_i\vert \nonumber$
where $I$ is the identity operator.
Averaging over many individual measurements of $A$ gives rise to an average value or expectation value for the observable $A$, which we denote $\langle A \rangle$ and is given by
$\langle A \rangle = \langle \Psi(t)\vert A\vert\Psi(t)\rangle \nonumber$
That this is true can be seen by expanding the state vector $\vert\Psi(t)\rangle$ in the eigenvectors of $A$:
$\vert\Psi(t)\rangle = \sum_i \alpha_i(t) \vert a_i\rangle \nonumber$
where ${a_i}$ are the amplitudes for obtaining the eigenvalue ${a_i}$ upon measuring $A$, i.e., $\alpha_i = \langle a_i\vert\Psi(t)\rangle$. Introducing this expansion into the expectation value expression gives
\begin{align*} \langle A \rangle (t) &= \sum_{i,j} \alpha_i^*(t) \alpha_j(t) \langle a_i\vert A\vert a_i \rangle \[4pt] &=\sum_{i,j} \alpha_i^*(t) \alpha_j a_i(t) \delta_{ij} \[4pt] &= \sum_i a_i \vert\alpha_i(t)\vert^2 \end{align*}
The interpretation of the above result is that the expectation value of $A$ is the sum over possible outcomes of a measurement of $A$ weighted by the probability that each result is obtained. Since $\vert\alpha_i\vert^2 =\vert\langle a_i\vert\Psi(t)\rangle \vert^2$ is this probability, the equivalence of the expressions can be seen.
Two observables are said to be compatible if $AB = BA$. If this is true, then the observables can be diagonalized simultaneously to yield the same set of eigenvectors. To see this, consider the action of $BA$ on an eigenvector $\vert a_i\rangle$ of $A$. $BA\vert a_i\rangle = a_i B\vert a_i\rangle$. But if this must equal $AB\vert a_i\rangle$, then the only way this can be true is if $B\vert a_i\rangle$ yields a vector proportional to $\vert a_i \rangle$ which means it must also be an eigenvector of $B$. The condition $AB = BA$ can be expressed as
$AB - BA = 0 \nonumber$
that is
$\left[ A, B \right]= 0 \nonumber$
where, in the second line, the quantity $\left [A, B \right ] \equiv AB - BA$ is know as the commutator between $A$ and $B$. If $\left [ A, B \right ] = 0$, then $A$ and $B$ are said to commute with each other. That they can be simultaneously diagonalized implies that one can simultaneously predict the observables $A$ and $B$ with the same measurement.
As we have seen, classical observables are functions of position $x$ and momentum ${P}$ (for a one-particle system). Quantum analogs of classical observables are, therefore, functions of the operators $X$ and $P$ corresponding to position and momentum. Like other observables $X$ and $P$ are linear hermitian operators. The corresponding eigenvalues $x$ and ${P}$ and eigenvectors $\vert x \rangle$ and $\vert P \rangle$ satisfy the equations
$X\vert x\rangle = x\vert x\rangle \nonumber$
$P\vert p\rangle \nonumber =p\vert p\rangle \nonumber$
which, in general, could constitute a continuous spectrum of eigenvalues and eigenvectors. The operators $X$ and $P$ are not compatible. In accordance with the Heisenberg uncertainty principle (to be discussed below), the commutator between $X$ and $P$ is given by
$\left [ X, P \right ] = i \hbar I \nonumber$
and that the inner product between eigenvectors of $X$ and $P$ is
$\langle x\vert p\rangle = {1 \over \sqrt{2\pi\hbar}}e^{ipx/\hbar} \nonumber$
Since, in general, the eigenvalues and eigenvectors of $X$ and $P$ form a continuous spectrum, we write the orthonormality and closure relations for the eigenvectors as:
\begin{align*} \langle x\vert x'\rangle & = \delta(x-x') \[4pt] \langle p\vert p'\rangle &= \delta(p-p') \end{align*}
\begin{align*} \vert \phi \rangle &= \int dx \vert x\rangle \langle x\vert\phi\rangle \[4pt] \vert\phi\rangle &= \int dp \vert p\rangle \langle p\vert\phi \rangle \end{align*}
\begin{align*} I &= \int dx \vert x\rangle \langle x\vert \[4pt] I &= \int dp \vert p\rangle \langle p\vert \end{align*}
The probability that a measurement of the operator $X$ will yield an eigenvalue $x$ in a region $dx$ about some point is
$P(x,t)dx = \vert\langle x\vert\Psi(t)\rangle \vert^2 dx \nonumber$
The object $\langle x \vert \Psi (t) \rangle$ is best represented by a continuous function $\Psi (x, t)$ often referred to as the wave function. It is a representation of the inner product between eigenvectors of $X$ with the state vector. To determine the action of the operator $X$ on the state vector in the basis set of the operator $X$, we compute
$\langle x\vert X\vert\Psi(t)\rangle = x\Psi(x,t) \nonumber$
The action of $P$ on the state vector in the basis of the $X$ operator is consequential of the incompatibility of $x$ and $P$ and is given by
$\langle x\vert P\vert\Psi(t)\rangle = {\hbar \over i}{\partial \over \partial x}\Psi(x,t) \nonumber$
Thus, in general, for any observable $A (X,P)$, its action on the state vector represented in the basis of $X$ is
$\langle x\vert A(X,P)\vert\Psi(t)\rangle = A\left(x,{\hbar\over i}{\partial \over \partial x}\right)\Psi(x,t) \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/09%3A_Review_of_the_basic_postulates_of_quantum_mechanics/9.01%3A_Measurement.txt |
Physical observables are represented by linear, hermitian operators that act on the vectors of the Hilbert space. If $A$ is such an operator, and $\vert\phi\rangle$ is an arbitrary vector in the Hilbert space, then $A$ might act on $\vert\phi\rangle$ to produce a vector $\vert\phi ' \rangle$, which we express as
$A\vert\phi\rangle = \vert\phi'\rangle \nonumber$
Since $\vert\phi\rangle$ is representable as a column vector, $A$ is representable as a matrix with components
$A = \left(\matrix{A_{11} & A_{12} & A_{13} & \cdots \cr A_{21} & A_{22} & A_{23} & \cdots \cr\cdot & \cdot & \cdot & \cdots }\right) \nonumber$
The condition that $A$ must be hermitian means that
$A^{\dagger} = A \nonumber$
or
$A_{ij} = A_{ji}^* \nonumber$
9.03: The Fundamental Postulates of Quantum Mechanics
The fundamental postulates of quantum mechanics concern the following questions:
1. How is the physical state of a system described?
2. How are physical observables represented?
3. What are the results of measurements on quantum mechanical systems?
4. How does the physical state of a system evolve in time?
5. The uncertainty principle.
9.04: The Heisenberg Picture
In all of the above, notice that we have formulated the postulates of quantum mechanics such that the state vector $\vert\Psi(t)\rangle$ evolves in time, but the operators corresponding to observables are taken to be stationary. This formulation of quantum mechanics is known as the Schrödinger picture. However, there is another, completely equivalent, picture in which the state vector remains stationary and the operators evolve in time. This picture is known as the Heisenberg picture. This particular picture will prove particularly useful to us when we consider quantum time correlation functions.
The Heisenberg picture specifies an evolution equation for any operator $A$, known as the Heisenberg equation. It states that the time evolution of $A$ is given by
${dA \over dt} = {1 \over i\hbar}[A,H] \nonumber$
While this evolution equation must be regarded as a postulate, it has a very immediate connection to classical mechanics. Recall that any function of the phase space variables $A (x, p)$ evolves according to
${dA \over dt} = \{A,H\} \nonumber$
where $\{...,...\}$ is the Poisson bracket. The suggestion is that in the classical limit ( $\hbar$ small), the commutator goes over to the Poisson bracket. The Heisenberg equation can be solved in principle giving
\begin{align*} A (t) &= e^{iHt/\hbar}A e^{-iHt/\hbar} \[4pt] &= U^{\dagger}(t)A U(t) \end{align*}
where $A$ is the corresponding operator in the Schrödinger picture. Thus, the expectation value of $A$ at any time $t$ is computed from
$\langle A(t) \rangle = \langle \Psi\vert A(t)\vert\Psi\rangle \nonumber$
where $\vert \Psi \rangle$ is the stationary state vector.
Let's look at the Heisenberg equations for the operators $X$ and $P$. If $H$ is given by
$H = {P^2 \over 2m} + U(X) \nonumber$
then Heisenberg's equations for $X$ and $P$ are
\begin{align*} dX \over dt &= {1 \over i\hbar}[X,H] \[4pt] &= {P \over m} \[4pt] dP \over dt &= {1 \over i\hbar}[P,H] \[4pt] &= -{\partial U \over \partial X} \end{align*}
Thus, Heisenberg's equations for the operators $X$ and $P$ are just Hamilton's equations cast in operator form. Despite their innocent appearance, the solution of such equations, even for a one-particle system, is highly nontrivial and has been the subject of a considerable amount of research in physics and mathematics.
Note that any operator that satisfies $\left [ A(t), H \right ] = 0$ will not evolve in time. Such operators are known as constants of the motion. The Heisenberg picture shows explicitly that such operators do not evolve in time. However, there is an analog with the Schrödinger picture: Operators that commute with the Hamiltonian will have associated probabilities for obtaining different eigenvalues that do not evolve in time. For example, consider the Hamiltonian, itself, which it trivially a constant of the motion. According to the evolution equation of the state vector in the Schrödinger picture,
$\vert\Psi(t)\rangle = \sum_i e^{-iE_it/\hbar}\vert E_i\rangle \langle E_i\vert\Psi(0)\rangle \nonumber$
the amplitude for obtaining an energy eigenvalue $E_j$ at time $t$ upon measuring $H$ will be
\begin{align*} \langle E_j\vert\Psi(t)\rangle &= \sum_i e^{-iE_it/\hbar}\langle E_j\vert E_i\rangle \langle E_i\vert\Psi(0)\rangle \[4pt] &= \sum_i e^{-iE_it/\hbar}\delta_{ij}\langle E_i\vert\Psi(0)\rangle \[4pt] &= e^{-iE_jt/\hbar}\langle E_j\vert\Psi(0)\rangle \end{align*}
Thus, the squared modulus of both sides yields the probability for obtaining $E_j$, which is
$\vert\langle E_j\vert\Psi(t)\rangle \vert^2 = \vert\langle E_j\vert\Psi(0)\rangle \vert^2 \nonumber$
Thus, the probabilities do not evolve in time. Since any operator that commutes with $H$ can be diagonalized simultaneously with $H$ and will have the same set of eigenvectors, the above arguments will hold for any such operator. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/09%3A_Review_of_the_basic_postulates_of_quantum_mechanics/9.02%3A_Physical_Observables.txt |
Because the operators $x$ and $p$ are not compatible, $[\hat{X},\hat{P}]\neq 0$, there is no measurement that can precisely determine both $x$ and $p$ simultaneously. Hence, there must be an uncertainty relation between them that specifies how uncertain we are about one quantity given a definite precision in the measurement of the other. Presumably, if one can be determined with infinite precision, then there will be an infinite uncertainty in the other. Recall that we had defined the uncertainty in a quantity by
$\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle ^2} \tag{1}$
Thus, for $x$ and $p$, we have
$\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle ^2} \tag{2a}$
$\Delta p = \sqrt{\langle p^2 \rangle - \langle p \rangle ^2} \tag{2b}$
These quantities can be expressed explicitly in terms of the wave function $\Psi (x, t)$ using the fact that
$\langle x \rangle = \langle \Psi(t)\vert x\vert\Psi(t)\rangle = \int dx \langle \Psi (t) \vert x \rangle \langle \vert x\vert X\vert\Psi(t)\rangle =\int dx \Psi^*(x,t) x \Psi(x,t) \tag{3}$
and
$\langle x^2 \rangle = \langle \Psi(t)\vert x^2\vert\Psi(t)\rangle = \int \Psi^*(x,t) x^2 \Psi(x,t) \tag{4}$
Similarly,
$\langle p \rangle = \langle \Psi(t)\vert p \vert\Psi(t)\rangle = \int dx \langle \Psi (t) \vert x \rangle \langle \vert p \vert \Psi (t) \rangle = \int dx \Psi^*(x,t){\hbar \over i}{\partial \over \partial x}\Psi(x,t) \tag{5}$
and
$\langle p^2 \rangle = \langle \Psi(t)\vert p^2\vert\Psi(t)\rangle = \int dx \Psi ^* (x, t)\left(-\hbar^2{\partial^2 \over \partial x^2}\right)\Psi(x,t) \tag{6}$
Then, the Heisenberg uncertainty principle states that
$\Delta x \Delta p \stackrel{>}{\sim} \hbar \tag{7}$
which essentially states that the greater certainty with which a measurement of $x$ or $p$ can be made, the greater will be the uncertainty in the other.
9.06: The Physical State of a Quantum System
The physical state of a quantum system is represented by a vector denoted $\vert\Psi(t)\rangle$ which is a column vector, whose components are probability amplitudes for different states in which the system might be found if a measurement were made on it.
A probability amplitude $\alpha$ is a complex number, the square modulus of which gives the corresponding probability $P_{\alpha}$
$P_{\alpha} = \vert\alpha\vert^2 \nonumber$
The number of components of $\vert\Psi(t)\rangle$ is equal to the number of possible states in which the system might observed. The space that contains $\vert\Psi(t)\rangle$ is called a Hilbert space ${\cal H}$. The dimension of ${\cal H}$ is also equal $\vert\Psi(t)\rangle$ to the number of states in which the system might be observed. It could be finite or infinite (countable or not). $\vert\Psi(t)\rangle$ must be a unit vector. This means that the inner product:
$\langle \Psi(t)\vert\Psi(t)\rangle = 1 \nonumber$ In the above, if the vector $\vert\Psi(t)\rangle$, known as a Dirac "ket'' vector, is given by the column
$\vert\Psi(t)\rangle = \left(\matrix{\psi_1 \cr \psi_2 \cr \cdot \cr \cdot \cr \cdot} \right) \nonumber$
then the vector $\langle \Psi(t)\vert$, known as a Dirac "bra'' vector, is given by
$\langle \Psi(t)\vert = (\psi_1^*\;\;\psi_2^*\;\;\cdots ) \nonumber$
so that the inner product becomes
$\langle \Psi(t)\vert\Psi(t)\rangle = \sum_i \vert\psi_i\vert^2 = 1 \nonumber$
We can understand the meaning of this by noting that $\psi_i$, the components of the state vector, are probability amplitudes, and $\vert\psi_i\vert^2$ are the corresponding probabilities. The above condition then implies that the sum of all the probabilities of being in the various possible states is 1, which we know must be true for probabilities.
9.07: Time Evolution of the State Vector
The time evolution of the state vector is prescribed by the Schrödinger equation
$i\hbar {\partial \over \partial t} \vert\Psi(t)\rangle = H\vert\Psi(t)\rangle \nonumber$
where $H$ is the Hamiltonian operator. This equation can be solved, in principle, yielding
$\vert\Psi(t)\rangle = e^{-iHt/\hbar}\vert\Psi(0)\rangle \nonumber$
where $\vert\Psi(0)\rangle$ is the initial state vector. The operator
$U(t) = e^{-iHt\hbar} \nonumber$
is the time evolution operator or quantum propagator. Let us introduce the eigenvalues and eigenvectors of the Hamiltonian $H$ that satisfy
$H\vert E_i\rangle = E_i \vert E_i\rangle \nonumber$
The eigenvectors for an orthonormal basis on the Hilbert space and therefore, the state vector can be expanded in them according to
$\vert\Psi(t)\rangle = \sum_i c_i(t) \vert E_i\rangle \nonumber$
where, of course, $c_i(t) = \langle E_i\vert\Psi(t)\rangle$, which is the amplitude for obtaining the value $E_i$ at time $t$ if a measurement of $H$ is performed. Using this expansion, it is straightforward to show that the time evolution of the state vector can be written as an expansion:
\begin{align*} \vert\Psi(t)\rangle &= \displaystyle e^{-iHt\hbar}\vert\Psi(0)\rangle \[4pt] &= e^{-iHt/\hbar}\sum_i\vert E_i\rangle \langle E_i\vert\Psi(0)\rangle \[4pt] &=\sum_i e^{-iE_i t/\hbar}\vert E_i\rangle \langle E_i\vert\Psi(0)\rangle \end{align*}
Thus, we need to compute all the initial amplitudes for obtaining the different eigenvalues $E_i$ of $H$, apply to each the factor $\exp(-iE_it/\hbar)\vert E_i\rangle$ and then sum over all the eigenstates to obtain the state vector at time $t$.
If the Hamiltonian is obtained from a classical Hamiltonian $H (x, p)$, then, using the formula from the previous section for the action of an arbitrary operator $A (X, P)$ on the state vector in the coordinate basis, we can recast the Schrödiner equation as a partial differential equation. By multiplying both sides of the Schrödinger equation by $\langle x |$, we obtain
\begin{align*} \langle x\vert H(X,P)\vert\Psi(t)\rangle &= i\hbar {\partial \over \partial t}\langle x\vert\Psi(t)\rangle \[4pt] H\left(x,{\hbar \over i}{\partial \over \partial x}\right)\Psi(x,t) &= i\hbar {\partial \over \partial t}\Psi(x,t) \end{align*}
If the classical Hamiltonian takes the form
$H(x,p) = {p^2 \over 2m} + U(x) \nonumber$
then the Schrödinger equation becomes
$\left[-{\hbar^2 \over 2m}{\partial^2 \over \partial x^2} + U(x)\right]\Psi(x,t)= i\hbar {\partial \over \partial t}\Psi(x,t) \nonumber$
which is known as the Schrödinger wave equation or the time-dependent Schrödinger equation. In a similar manner, the eigenvalue equation for $H$ can be expressed as a differential equation by projecting it into the $X$ basis:
\begin{align*} \langle x\vert H\vert E_i\rangle \nonumber &= E_i \langle x\vert E_i\rangle \[4pt] H\left(x,{\hbar \over i}{\partial \over \partial x}\right)\psi_i(x) &= E_i \psi_i(x) \[4pt] \left[-{\hbar^2 \over 2m}{\partial^2 \over \partial x^2} + U(x)\right]\psi_i(x) &= E_i \psi_i(x) \end{align*}
where $\psi_i(x) = \langle x\vert E_i\rangle$ is an eigenfunction of the Hamiltonian. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/09%3A_Review_of_the_basic_postulates_of_quantum_mechanics/9.05%3A_The_Heisenberg_Uncertainty_Principle.txt |
The problem of quantum statistical mechanics is the quantum mechanical treatment of an $N$-particle system. Suppose the corresponding $N$-particle classical system has Cartesian coordinates
$q_1,...,q_{3N} \nonumber$
and momenta
$p_1, \cdots , p_{3N} \nonumber$
and Hamiltonian
$H = \sum_{i=1}^{3N} {p_i^2 \over 2m_i} + U(q_1,...,q_{3N}) \nonumber$
Then, as we have seen, the quantum mechanical problem consists of determining the state vector $\vert \Psi (t) \rangle$ from the Schrödinger equation
$H\vert\Psi(t)\rangle = i\hbar{\partial \over \partial t}\vert\Psi(t)\rangle \nonumber$
Denoting the corresponding operators, $Q_1, \cdots , Q_{3N}$ and $P_1, \cdots , P_{3N}$, we note that these operators satisfy the commutation relations:
$\left[Q_i,Q_j\right] \nonumber$
$=$
$\left[P_i,P_j\right] = 0$
$\left[Q_i,P_j\right] \nonumber$ $=$ $\displaystyle i\hbar I \delta_{ij}$
and the many-particle coordinate eigenstate $\vert q_1...q_{3N}\rangle$ is a tensor product of the individual eigenstate $\vert q_1\rangle ,...,\vert q_{3N}\rangle$:
$\vert q_1...q_{3N}\rangle = \vert q_1\rangle \cdots \vert q_{3N}\rangle \nonumber$
The Schrödinger equation can be cast as a partial differential equation by multiplying both sides by $\langle q_1...q_{3N}\vert$:
$\langle q_1...q_{3N}\vert H\vert\Psi(t)\rangle \nonumber$
$=$
${\partial \over \partial t}\langle q_1...q_{3N}\vert\Psi(t)\rangle$
$\left[-\sum_{i=1}^{3N}{\hbar^2 \over 2m_i}{\partial^2 \over \partial q_i^2} +U(q_1,...,q_{3N})\right]\Psi(q_1,...,q_{3N},t) \nonumber$
$=$
$i\hbar {\partial \over \partial t}\Psi(q_1,...,q_{3N},t)$
where the many-particle wave function is $\Psi(q_1,....,q_{3N},t) =\langle q_1...q_{3N}\vert\Psi(t)\rangle$. Similarly, the expectation value of an operator $A=A(Q_1,...,Q_{3N},P_1,...,P_{3N})$ is given by
$\langle A \rangle = \int dq_1\cdots dq_{3N}\Psi^*(q_1,...,q_{3N}) A \left (q_1, \cdots, q_{3N}, {\hbar \over i}{\partial \over \partial q_1}, \cdots , {\hbar \over i}{\partial \over\partial q_{3N}}\right)\Psi(q_1,...,q_{3N}) \nonumber$
10.02: The Density Matrix and Density Operator
In general, the many-body wave function $\Psi(q_1,...,q_{3N},t)$ is far too large to calculate for a macroscopic system. If we wish to represent it on a grid with just 10 points along each coordinate direction, then for $N = 10^{23}$, we would need $10^{10^{23}}$ total points, which is clearly enormous.
We wish, therefore, to use the concept of ensembles in order to express expectation values of observables $\langle A \rangle$ without requiring direct computation of the wavefunction. Let us, therefore, introduce an ensemble of systems, with a total of $Z$ members, and each having a state vector ${ \vert\Psi^{(\alpha)}\rangle }$, $\alpha = 1, \cdots , Z$. Furthermore, introduce an orthonormal set of vectors $\langle \phi_k \vert\phi_j\rangle = \delta_{ij}$ ) and expand the state vector for each member of the ensemble in this orthonormal set:
$\vert\Psi^{(\alpha)}\rangle = \sum_k C_k^{(\alpha)}\vert\phi_k \rangle \nonumber$
The expectation value of an observable, averaged over the ensemble of systems is given by the average of the expectation value of the observable computed with respect to each member of the ensemble:
$\langle A \rangle = {1 \over Z}\sum_{\alpha=1}^Z \langle \Psi^{(\alpha)}\vert A\vert\Psi^{(\alpha)}\rangle \nonumber$
Substituting in the expansion for $\vert\Psi^{(\alpha)}\rangle$, we obtain
\begin{align*} \langle A \rangle &= {1 \over Z}\sum_{k,l}C_k^{(\alpha)^*}C_l^{(\alpha)}\langle \phi_k \vert A\vert\phi_l \rangle \[4pt] &= \sum_{k,l} \left({1 \over Z}\sum_{\alpha=1}^Z C_l^{(\alpha)}C_k^{(\alpha)^*}\right)\langle \phi_k\vert A\vert\phi_l\rangle \end{align*}
Let us define a matrix
$\rho_{lk} = \sum_{\alpha=1}^Z C_l^{(\alpha)}C_k^{(\alpha)^*} \nonumber$
and a similar matrix
$\tilde{\rho}_{lk} = {1 \over Z}\sum_{\alpha=1}^Z C_l^{(\alpha)}C_k^{(\alpha)^*} \nonumber$
Thus, ${\rho_{lk} }$ is a sum over the ensemble members of a product of expansion coefficients, while ${\rho_{lk} }$ is an average over the ensemble of this product. Also, let $A_{kl} = \langle \phi _k \vert A \vert \phi _l \rangle$. Then, the expectation value can be written as follows:
$\langle A \rangle = {1 \over Z}\sum_{k,l} \rho_{lk}A_{kl} = {1 \over Z} \sum _k (\rho A)_{kk} = {1 \over Z}{\rm Tr}(\rho A)= {\rm Tr}(\tilde{\rho}A) \nonumber$
where ${\rho }$ and $A$ represent the matrices with elements ${\rho _{lk} }$ and $A_{kl}$ in the basis of vectors $\{\vert\phi_k\rangle \}$. The matrix ${\rho _{lk} }$ is known as the density matrix. There is an abstract operator corresponding to this matrix that is basis-independent. It can be seen that the operator
$\rho = \sum_{\alpha=1}^Z \vert\Psi^{(\alpha)}\rangle \langle \Psi^{(\alpha)}\vert \nonumber$
and similarly
$\tilde{\rho} = {1 \over Z}\sum_{\alpha=1}^Z \vert\Psi^{(\alpha)}\rangle \langle \Psi^{(\alpha)}\vert \nonumber$
have matrix elements ${\rho _{lk} }$ when evaluated in the basis set of vectors $\{\vert\phi_k\rangle \}$.
\begin{align*} \langle \phi_l\vert\rho\vert\phi_k\rangle &= \sum_{\alpha=1}^Z \langle \phi _l \vert \Psi ^{(\alpha)} \rangle \langle \Psi ^{(\alpha)} \vert \phi _k \rangle \[4pt] &= \sum_{\alpha=1}^Z C_l^{(\alpha)}C_k^{(\alpha)^*} \[4pt] &= \rho_{lk} \end{align*}
Note that ${\rho}$ is a hermitian operator
$\rho^{\dagger} = \rho \nonumber$
so that its eigenvectors form a complete orthonormal set of vectors that span the Hilbert space. If ${w _k }$ and $\vert w_k \rangle$ represent the eigenvalues and eigenvectors of the operator ${\tilde{\rho} }$, respectively, then several important properties they must satisfy can be deduced.
Firstly, let $A$ be the identity operator $I$. Then, since $\langle I \rangle = 1$, it follows that
$1 = {1 \over Z}{\rm Tr}(\rho) = {\rm Tr}(\tilde{\rho}) = \sum_k w_k \nonumber$
Thus, the eigenvalues of ${\tilde{\rho} }$ must sum to 1. Next, let $A$ be a projector onto an eigenstate of ${\tilde{\rho} }$, $A=\vert w_k\rangle \langle w_k\vert \equiv P_k$. Then
$\langle P_k \rangle = {\rm Tr}(\tilde{\rho} \vert w_k\rangle \langle w_k\vert) \nonumber$
But, since ${\tilde{\rho} }$ can be expressed as
$\tilde{\rho} = \sum_k w_k \vert w_k\rangle \langle w_k\vert \nonumber$
and the trace, being basis set independent, can be therefore be evaluated in the basis of eigenvectors of ${\tilde{\rho} }$, the expectation value becomes
\begin{align*} \langle P_k \rangle &= \sum_j \langle w_j \vert \sum_i w_i \vert w_i \rangle \langle w_i \vert w_k\rangle \langle w_k \vert w_j \rangle \[4pt] &= \sum_{i,j} w_i \delta_{ij}\delta_{ik}\delta_{kj} \[4pt] &={w_k} \end{align*}
However,
\begin{align*} \langle P_k \rangle&= {1 \over Z} \sum_{\alpha=1}^Z \langle \Psi^{(\alpha)}\vert w_k \rangle \langle w_k\vert\Psi^{(\alpha)}\rangle \[4pt] &= {1 \over Z}\sum_{\alpha=1}^Z\vert\langle \Psi^{(\alpha)}\vert w_k\rangle \vert^2 \geq 0 \end{align*}
Thus, ${w_k \ge 0 }$. Combining these two results, we see that, since $\sum _k w_k = 1$ and ${w_k \ge 0 }$, ${0 \le w_k \le 1 }$, so that ${w_k }$ satisfy the properties of probabilities.
With this in mind, we can develop a physical meaning for the density matrix. Let us now consider the expectation value of a projector $\vert a_i\rangle \langle a_i\vert\equiv {\cal P}_{a_i}$ onto one of the eigenstates of the operator $A$. The expectation value of this operator is given by
\begin{align*} \langle {\cal P}_{a_i} \rangle &= {1 \over Z}\sum_{\alpha=1}^Z \langle \Psi^{(\alpha)} \vert P_{a_i} \vert \Psi^{(\alpha)}\rangle \[4pt] &= {1 \over Z} \sum_{\alpha=1}^Z \langle \Psi^{(\alpha)} \vert a_i \rangle \langle a_i \vert \Psi ^{(\alpha)} \rangle \[4pt] &= {1 \over Z}\sum_{\alpha=1}^Z \vert\langle a_i \vert\Psi^{(\alpha)}\rangle \vert^2 \end{align*}
But ${ \vert\langle a_i\vert\Psi^{(\alpha)}\rangle \vert^2 \equiv P_{a_i}^{(\alpha)} }$ is just probability that a measurement of the operator $A$ in the $\alpha$ th member of the ensemble will yield the result ${a_i }$. Thus,
$\langle {\cal P}_{a_i} \rangle = {1 \over Z}\sum_{\alpha=1}^P P_{a_i}^{(\alpha)} \nonumber$
or the expectation value of $P_{a_i}$ is just the ensemble averaged probability of obtaining the value ${a_i}$ in each member of the ensemble. However, note that the expectation value of $P_{a_i}$ can also be written as
\begin{align*} \langle {\cal P}_{a_i} \rangle &= {\rm Tr}(\tilde{\rho}{\cal P}_{a_i}) \[4pt] &= {\rm Tr}(\sum_k w_k\vert w_k\rangle \langle w_k \vert a_i\rangle\langle a_i\vert) \[4pt] &= \sum_{k,l}\langle w_l\vert w_k\vert w_k\vert w_k \rangle \langle w_k\vert a_i \rangle \langle a_i\vert w_l\rangle \[4pt] &= \sum_{k,l} w_k \delta_{kl}\langle w_k a_i \rangle \langle a_i w_l \rangle \[4pt] &= {\sum_k w_k \vert\langle a_i \vert w_k \rangle \vert^2 } \end{align*}
Equating the two expressions gives
${1 \over Z}\sum_{\alpha=1}^Z \langle P^{(\alpha)}_{a_i} \rangle =\sum_k w_k \vert\langle a_i\vert w_k \rangle \vert^2 \nonumber$
The interpretation of this equation is that the ensemble averaged probability of obtaining the value ${a_i}$ if $A$ is measured is equal to the probability of obtaining the value ${a_i}$ in a measurement of $A$ if the state of the system under consideration were the state $\vert w_k \rangle$, weighted by the average probability ${w_k}$ that the system in the ensemble is in that state. Therefore, the density operator ${\rho}$ (or ${\rho}$) plays the same role in quantum systems that the phase space distribution function $f({\bf\Gamma})$ plays in classical systems. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/10%3A_Fundamentals_of_quantum_statistical_mechanics/10.01%3A_Principles_of_quantum_statistical_mechanics.txt |
The time evolution of the operator ${\rho}$ can be predicted directly from the Schrödinger equation. Since $\rho (t)$ is given by
$\rho(t) = \sum_{\alpha=1}^Z \vert\Psi^{(\alpha)}(t)\rangle \langle \Psi^{(\alpha)}(t)\vert \nonumber$
the time derivative is given by
\begin{align*} {\partial \rho \over \partial t} &= \sum_{\alpha=1}^Z\left[\left({\partial \over \partial t}\vert\Psi^{(\alpha)} (t) \rangle \right ) \langle \Psi ^{(\alpha)}(t) \vert + \vert \Psi ^{(\alpha)} (t) \rangle\left({\partial \over \partial t}\langle \Psi^{(\alpha)}(t)\vert\right)\right] \[4pt] &= {1 \over i\hbar}\sum_{\alpha=1}^Z\left[\left(H\vert\Psi^{(\alpha)} (t) \rangle \right ) \langle \Psi^{(\alpha)} (t) \vert - \vert \Psi^{(\alpha)}(t)\langle \left(\langle \Psi^{(\alpha)}(t)\vert H\right)\right] \[4pt] &= {1 \over i\hbar}(H\rho-\rho H) \[4pt] &= {1 \over i\hbar}[H,\rho] \end{align*}
where the second line follows from the fact that the Schrödinger equation for the bra state vector ${\langle \Psi^{(\alpha)}(t)\vert }$ is
$-{i\hbar}{\partial \over \partial t}\langle \Psi^{(\alpha)}(t)\vert = \langle \Psi^{(\alpha)}(t)\vert H \nonumber$
Note that the equation of motion for $\rho (t)$ differs from the usual Heisenberg equation by a minus sign! Since $\rho (t)$ is constructed from state vectors, it is not an observable like other hermitian operators, so there is no reason to expect that its time evolution will be the same. The general solution to its equation of motion is
\begin{align*} \rho (t) &= e^{-iHt/\hbar}\rho(0)e^{iHt/\hbar} \[4pt] &= U(t)\rho(0)U^{\dagger}(t) \end{align*}
The equation of motion for $\rho (t)$ can be cast into a quantum Liouville equation by introducing an operator
$iL = {1 \over i\hbar}[...,H] \nonumber$
In term of $iL$, it can be seen that $\rho (t)$ satisfies
${\partial \rho \over \partial t} = - iL\rho \nonumber$
$\rho (t) = { e^{-iLt}\rho(0) } \nonumber$
What kind of operator is $iL$ ? It acts on an operator and returns another operator. Thus, it is not an operator in the ordinary sense, but is known as a superoperator or tetradic operator (see S. Mukamel, Principles of Nonlinear Optical Spectroscopy, Oxford University Press, New York (1995)).
Defining the evolution equation for ${\rho }$ this way, we have a perfect analogy between the density matrix and the state vector. The two equations of motion are
\begin{align*} {\partial \over \partial t}\vert\Psi(t)\rangle &= {-{i \over \hbar}H\vert\Psi(t)\rangle} \[4pt] {\partial \over \partial t}\rho(t) &= -iL\rho(t) \end{align*}
We also have an analogy with the evolution of the classical phase space distribution $f({\bf\Gamma},t)$, which satisfies
${\partial f \over \partial t} = -iLf \nonumber$
with $iL = \{...,H\}$ being the classical Liouville operator. Again, we see that the limit of a commutator is the classical Poisson bracket.
10.04: A simple example - the quantum harmonic oscillator
As a simple example of the trace procedure, let us consider the quantum harmonic oscillator. The Hamiltonian is given by
$H = {P^2 \over 2m} + {1 \over 2}m\omega^2 X^2 \nonumber$
and the eigenvalues of $H$ are
$E_n = \left(n + {1 \over 2}\right)\hbar\omega,\;\;\;\;\;\;\;\;\;\;n=0,1,2,... \nonumber$
Thus, the canonical partition function is
$Q(\beta) = \sum_{n=0}^{\infty} e^{-\beta (n+1/2)\hbar\omega} = e^{-\beta \hbar \omega/2}\sum_{n=0}^{\infty}\left(e^{-\beta\hbar\omega}\right)^n \nonumber$
This is a geometric series, which can be summed analytically, giving
$Q(\beta) = {e^{-\beta \hbar\omega/2} \over 1-e^{-\beta \hbar \omega}} = {1 \over e^{\beta \hbar \omega / 2} - e^{-\beta \hbar\omega/2}}= {1 \over 2}csch(\beta\hbar\omega/2) \nonumber$
The thermodynamics derived from it as as follows:
1.
Free energy:
The free energy is
$A = -{1 \over \beta}\ln Q(\beta) = {\hbar\omega \over 2} +{1 \over \beta}\ln \left(1-e^{-\beta \hbar \omega}\right) \nonumber$
2.
Average energy:
The average energy $E = \langle H \rangle$ is
$E = -{\partial \over \partial \beta}\ln Q(\beta)= {\hbar\omega \over 2} + {\hbar \omega e^{-\beta \hbar \omega}}= \left({1 \over 2} + \langle n \rangle \right)\hbar\omega \nonumber$
3.
Entropy
The entropy is given by
$S = k\ln Q(\beta) + {E \over T} =-k\ln \left(1-e^{-\beta \hbar \omega} \right ) + {\hbar \omega \over T}{e^{-\beta \hbar \omega}\over 1-e^{-\beta \hbar \omega}} \nonumber$
Now consider the classical expressions. Recall that the partition function is given by
$Q(\beta) = {1 \over h} \int dp dx e^{-\beta \left({p^2 \over 2m} + {1 \over 2} m\omega ^2 x^2 \right )} = {1 \over h} \left ( {2 \pi m \over \beta} \right )^{1/2} = {2\pi \over \beta \omega h} ={1 \over \beta \hbar \omega} \nonumber$
Thus, the classical free energy is
$A_{\rm cl} = {1 \over \beta}\ln(\beta \hbar \omega) \nonumber$
In the classical limit, we may take $\hbar$ to be small. Thus, the quantum expression for $A$ becomes, approximately, in this limit:
$A_{\rm Q} \longrightarrow {\hbar \omega \over 2} + {1 \over \beta}\ln (\beta \hbar \omega) \nonumber$
and we see that
$A_{\rm Q} - A_{\rm cl} \longrightarrow {\hbar \omega \over 2} \nonumber$
The residual ${\hbar \omega \over 2}$ (which truly vanishes when $\hbar \rightarrow 0$) is known as the quantum zero point energy. It is a pure quantum effect and is present because the lowest energy quantum mechanically is not $E = 0$ but the ground state energy $E={\hbar\omega \over 2}$.
10.5.01: The microcanonical ensem
At equilibrium, the density operator does not evolve in time; thus, ${\partial \rho \over \partial t} = 0$. Thus, from the equation of motion, if this holds, then $[H,\rho]=0$, and $\rho (t)$ is a constant of the motion. This means that it can be simultaneously diagonalized with the Hamiltonian and can be expressed as a pure function of the Hamiltonian
$\rho = f(H) \nonumber$
Therefore, the eigenstates of ${\rho}$, the vectors, we called $\vert w_k \rangle$ are the eigenvectors $\vert E_i \rangle$ of the Hamiltonian, and we can write $H$ and ${\rho}$ as
\begin{align*} H &= \sum_i E_i \vert E_i\rangle \langle E_i\vert \[4pt] \rho &= \sum_i f(E_i)\vert E_i\rangle \langle E_i\vert \end{align*}
The choice of the function $f$ determines the ensemble.
10.05: The quantum equilibrium ensembles
Although we will have practically no occasion to use the quantum microcanonical ensemble (we relied on it more heavily in classical statistical mechanics), for completeness, we define it here. The function $f$, for this ensemble, is
$f(E_i)\delta E = \theta(E_i-(E+\delta E))-\theta(E_i-E) \nonumber$
where $\theta (x)$ is the Heaviside step function. This says that $f(E_i)\delta E$ is 1 if $E<E_i<(E+\delta E)$ and 0 otherwise. The partition function for the ensemble is ${\rm Tr}(\rho)$, since the trace of ${\rho }$ is the number of members in the ensemble:
\begin{align*} \Omega(N,V,E) &= {\rm Tr}(\rho) \[4pt] &= \sum_i\left[\theta(E_i-(E+\delta E))-\theta(E_i-E)\right] \end{align*}
The thermodynamics that are derived from this partition function are exactly the same as they are in the classical case:
\begin{align*} S (N, V, E ) &= -k\ln \Omega(N,V,E) \[4pt] 1 \over T &= -k\left({\partial \ln \Omega \over \partial E } \right)_{N,V} \end{align*}
etc.
10.5.02: The canonical ensemble
In analogy to the classical canonical ensemble, the quantum canonical ensemble is defined by
\begin{align*} \rho &= e^{-\beta H} \[4pt] f(E_i) &= e^{-\beta E_i} \end{align*}
Thus, the quantum canonical partition function is given by
\begin{align*} Q(N,V,T) &= {\rm Tr}(e^{-\beta H}) \[4pt] &= \sum_i e^{-\beta E_i} \end{align*}
and the thermodynamics derived from it are the same as in the classical case:
\begin{align*} A (N, V, T ) &= -{1 \over \beta}\ln Q(N,V,T) \[4pt] E (N, V, T ) &=-{\partial \over \partial \beta}\ln Q(N,V,T) \[4pt] P (N, V, T) &= {1 \over \beta}{\partial \over \partial V}\ln Q(N,V,T) \end{align*}
etc. Note that the expectation value of an observable $A$ is
$\langle A \rangle = {1 \over Q}{\rm Tr}(Ae^{-\beta H}) \nonumber$
Evaluating the trace in the basis of eigenvectors of $H$ (and of ${\rho }$ ), we obtain
\begin{align*} \langle A \rangle &= {1 \over Q}\sum_i \langle E_i\vert Ae^{-\beta H} \vert E_i \rangle \[4pt] &= {1 \over Q}\sum_i e^{-\beta E_i} \langle E_i\vert A\vert E_i\rangle \end{align*}
The quantum canonical ensemble will be particularly useful to us in many things to come.
10.5.03: Isothermal-isobaric and
Also useful are the isothermal-isobaric and grand canonical ensembles, which are defined just as they are for the classical cases:
• isothermal-isobaric:
$\Delta(N,P,T) = \int_0^{\infty}dV e^{-\beta PV} Q(N,V,T) =\int_0^{\infty} dV {\rm Tr}(e^{-\beta (H+PV)}) \nonumber$
• grand canonical ensembles
${\cal Z}(\mu,V,T) = \sum_{N=0}^{\infty}e^{\beta \mu N}Q(N,V,T) = \sum_{N=0}^{\infty} {\rm Tr}(e^{-\beta (H-\mu N)}) \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/10%3A_Fundamentals_of_quantum_statistical_mechanics/10.03%3A_Time_evolution_of_the_density_operator.txt |
As in the classical case, we assume a solution of the form
$\rho(t) = \rho_0(H_0) + \Delta \rho(t) \nonumber$
where
$[H_0,\rho_0]=0\;\;\;\;\;\Rightarrow\;\;\;\;\;{\partial \rho_0 \over \partial t}=0 \nonumber$
and we will assume
$\rho_0(H_0) = {e^{-\beta H_0} \over Q(N,V,T)} \nonumber$
Substituting into the Liouville equation and working to first order in small quantities, we find
${\partial \Delta \rho \over \partial t} = {1 \over i\hbar}[H_0,\Delta \rho] -{1 \over i\hbar} [B,\rho_0]F_e(t) \nonumber$
which is a first order inhomogeneous equation that can be solved by using an integrating factor:
$\Delta \rho(t) = -{1 \over i\hbar}\int_{-\infty}^t\;ds\;e^{-iH_0(t-s)/\hbar}[B,\rho_0]e^{iH_0(t-s)/\hbar}F_e(s) \nonumber$
(Note that we have chosen the origin in time to be ${t = - \infty }$, which is an arbitrary choice.)
For an observable $A$, the expectation value is
$\langle A(t)\rangle = {\rm Tr}(\rho A) = \langle A\rangle _0 + {\rm Tr}(\Delta \rho(t) A) \nonumber$
when the solution for $\Delta \rho$ is substituted in, this becomes
$\langle A(t) \rangle$ $=$
$\langle A \rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[Ae^{-iH_0(t-s)/hbar}[B,\rho_0]e^{iH_0(t-s)/\hbar}\right]F_e(s)$
$=$
$\langle A \rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[e^{iH_0(t-s)/\hbar} Ae^{-iH_0(t-s)/hbar}[B,\rho_0]\right]F_e(s)$
$=$
$\langle A\rangle _0 - {1 \over i\hbar} ds\;{\rm Tr}\left[A(t-s)[B,\rho_0]\right]F_e(s)$
where the cyclic property of the trace has been used and the Heisenberg evolution for $A$ has been substituted in. Expanding the commutator gives
$\langle A(t) \rangle$ $=$
$\langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[A(t-s)B\rho_0 - A(t-s)\rho_0B\right]F_e(s)$
$=$
$\langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;{\rm Tr}\left[\rho_0\left(A(t-s)B - BA(t-s)\right)\right]F_e(s)$
$=$
$\langle A\rangle _0 - {1 \over i\hbar}\int_{-\infty}^t\;ds\;F_e(s)\langle [A(t-s),B(0)]_0\rangle$
where the cyclic property of the trace has been used again. Define a function
$\Phi_{AB}(t) = {i \over \hbar}\langle [A(t),B(0)]\rangle _0 \nonumber$
called the after effect function. It is essentially the antisymmetric quantum time correlation function, which involves the commutator between $A (t)$ and $B (0)$. Then the linear response result can be written as
$\langle A(t)\rangle = \langle A \rangle _0 + \int_{-\infty}^t\;ds F_e(s)\Phi_{AB}(t-s) \nonumber$
which is the starting point for the theory of quantum transport coefficients. If we choose to measure the operator $B$, then we find
$\langle B(t)\rangle = \langle B\rangle _0 + \int_{-\infty}^t\;ds\;F_e(s)\Phi_{BB}(t-s) \nonumber$
12.02: Kubo Transform Expression for the Time Correlation Function
We shall derive the following expression for the quantum time correlation function
$\Phi_{AB}(t) = \int_0^{\beta}d\lambda\;\langle \dot{B}(-i\hbar\lambda)A(t)\rangle _0 \nonumber$
known as a Kubo transform relation. Since ${\dot{B} }$ is given by the Heisenberg equation:
$\dot{B} = {1 \over i\hbar}[B,H_0] \nonumber$
it follows that
$\dot{B}(t) = -{1 \over i\hbar}e^{iH_0t/\hbar}[H_0,B(0)]e^{-iH_0t/\hbar} \nonumber$
Evaluating the expression at $t=-i\hbar\lambda$ gives
$\dot{B}(-i\hbar\lambda) = e^{\lambda H_0}{1 \over i\hbar}[B(0),H_0]e^{-\lambda H_0} \nonumber$
Thus,
$\Phi_{AB}(t) = \int_0^{\beta} d\lambda \langle e^{\lambda H_0} \left ({1 \over i\hbar}[B(0),H_0]\right)e^{-\lambda H_0}A(t)\rangle _0 \nonumber$
By performing the trace in the basis of eigenvectors of $H_0$, we obtain
\begin{align*} \Phi_{AB}(t) &= {1 \over Q}\int_0^{\beta}d\lambda\sum_n \langle n\vert e^{\lambda H_0} \left ({1 \over i\hbar}\right)[B(0),H_0]e^{-\lambda H_0}A(t)\vert n\rangle e^{-\beta E_n} \[4pt] &= {1 \over Q}\int_0^{\beta}d\lambda\sum_{m,n} \langle n\vert e^{\lambda H_0}\left ( {1 \over i \hbar } \right ) \left [ B (0), H_0 \right ] e^{-\lambda H_0}\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \[4pt] &= {1 \over Q}\int_0^{\beta}d\lambda\sum_{m,n} e^{\lambda E_n}e^{-\lambda E_m} {1 \over i \hbar } \langle n \vert [B(0),H_0]\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \[4pt] &= {1 \over Q}\sum_{m,n} e^{-\beta E_n}{e^{\beta(E_n-E_m)}-1 \over (E_n - E_m) } {1 \over i\hbar } \langle n\vert[B(0),H_0]\vert m\rangle \langle m\vert A(t)\vert n\rangle e^{-\beta E_n} \end{align*}
But
$\langle n\vert[B(0),H_0]\vert m\rangle = \langle n\vert B(0) H_0 - H_0 B (0)\vert m\rangle = (E_m-E_n)\langle n\vert B(0)\vert m\rangle \nonumber$
Therefore,
\begin{align*} \Phi_{AB}(t) &= -{1 \over i\hbar Q}\sum_{m,n}\left(e^{-\beta E_n}-e^{-\beta E_m}\right)\langle n\vert B(0)\vert m\rangle \langle m\vert A(t)\vert n\rangle \[4pt] &= -{1 \over i\hbar Q}\left[\sum_{m,n}e^{-\beta E_m}\langle m\vert A (t) \vert n \rangle \langle n \vert B (0) \vert m \rangle - \sum _{m,n} e^{-\beta E_n}\langle n\vert B(0)\vert m\rangle \langle m\vert A(t)\vert n\rangle \right] \[4pt] &= { {i \over \hbar}\langle [A(t),B(0)]\rangle _0 } \end{align*} \nonumber
which proves the relation. The classical limit can be deduced easily from the Kubo transform relation:
$\Phi_{AB}(t) \longrightarrow \beta\langle \dot{B}(0)A(t)\rangle _0 \nonumber$
Note further, by using the cylic properties of the trace, that
$\langle \dot{B}(-i\hbar\lambda)B(t)\rangle _0 = -{d \over dt}\langle B(-i\hbar\lambda)B(t)\rangle _0 \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/12%3A_Time-dependent_Processes_-_Classical_case/12.01%3A_Perturbative_solution_of_the_Liouville_equation.txt |
The most general way a system can be driven away from equilibrium by a forcing function $F_e (t)$ is according to the equations of motion:
\begin{align*} \dot {q}_i &= \dfrac{\partial H}{\partial p_i} + C_i({\rm x})F_e(t) \[4pt] \dot {P}_i &=- \dfrac{\partial H}{\partial p_i} + D_i({\rm x})F_e(t) \end{align*}
where the $3N$ functions $C_i$ and $D_i$ are required to satisfy the incompressibility condition
$\sum_{i=1}^{3N}\left[ {\partial C_i \over \partial q_i} + {\partial D_i \over\partial p_i}\right] = 0 \nonumber$
in order to insure that the Liouville equation for $f (x, t)$ is still valid. These equations of motion will give rise to a distribution function $f (x, t)$ satisfying
${\partial f \over \partial t} + iLf = 0 \nonumber$
with $\partial f/\partial t \neq 0$. (We assume that $f$ is normalized so that $\int d{\rm x}f({\rm x},t)=1$.)
What does the Liouville equation say about the nature of $f (x, t)$ in the limit that $C_i$ and $D_i$ are small, so that the displacement away from equilibrium is, itself, small? To examine this question, we propose to solve the Liouville equation perturbatively. Thus, let us assume a solution of the form
$f({\rm x},t) = f_0(H({\rm x})) + \Delta f({\rm x},t) \nonumber$
Note, also, that the equations of motion ${\dot {x} }$ take a perturbative form
$\dot{\rm x}(t) = \dot{\rm x}_0 + \Delta \dot{\rm x}(t) \nonumber$
and as a result, the Liouville operator contains two pieces:
$iL = \dot{\rm x}\cdot\nabla_{\rm x} = \dot{\rm x}_0\cdot \nabla _x +\Delta\dot{\rm x}\cdot\nabla_{\rm x} = iL_0 + i\Delta L \nonumber$
where $iL_0 = \{...,H\}$ and $f_0 (H)$ is assumed to satisfy
$iL_0 f_0(H({\rm x})) = 0 \nonumber$
$\dot {x} _0$ means the Hamiltonian part of the equations of motion
\begin{align*} \dot{q}_i &= \dfrac{\partial H}{\partial p_i} \[4pt] \dot {P}_i &= - \dfrac{\partial H}{\partial q_i} \end{align*}
For an observable $A (x)$, the ensemble average of $A$ is a time-dependent quantity:
$\langle A(t)\rangle = \int d{\rm x}A({\rm x})f({\rm x},t) \nonumber$
which, when the assumed form for $f (x , t)$ is substituted in, gives
$\langle A(t)\rangle = \int d{\rm x}A({\rm x})f_0({\rm x}) + \int dx A (x) \Delta f (x, t ) = \langle A \rangle_0 + \int d{\rm x}A({\rm x})\Delta f({\rm x},t) \nonumber$
where $\langle\cdot\rangle_0$ means average with respect to $f_0 (x)$.
12.04: General Properties of Time Correlation Functions
Define a time correlation function between two quantities $A (x)$ and $B (x)$ by
\begin{align*} C_{AB} (t) &= \langle A(0)B(t)\rangle \[4pt] &= \int d{\rm x}f({\rm x})A({\rm x})e^{iLt}B({\rm x}) \end{align*}
The following properties follow immediately from the above definition:
Property 1
$\langle A(0)B(t)\rangle = \langle A(-t)B(0)\rangle \nonumber$
Property 2
$C_{AB}(0) = \langle A({\rm x})B({\rm x})\rangle \nonumber$ Thus, if $A = B$, then
$C_{AA}(t) = \langle A(0)A(t)\rangle \nonumber$
known as the autocorrelation function of $A$, and
$C_{AA}(0) = \langle A^2\rangle \nonumber$
If we define $\delta A = A - \langle A \rangle$, then
$C_{\delta A\delta A}(0) = \langle (\delta A)^2\rangle =\langle ( A - \langle A \rangle )^2 \rangle = \langle A^2\rangle - \langle A\rangle^2 \nonumber$
which just measures the fluctuations in the quantity $A$.
Property 3
A time correlation function may be evaluated as a time average, assuming the system is ergodic. In this case, the phase space average may be equated to a time average, and we have
$C_{AB}(t) = \lim_{T\rightarrow\infty}{1 \over T-t}\int_0^{T-t}ds A({\rm x}(s))B({\rm x}(t+s)) \nonumber$
which is valid for $t<<T$. In molecular dynamics simulations, where the phase space trajectory is determined at discrete time steps, the integral is expressed as a sum
$C_{AB}(k\Delta t) = {1 \over N-k}\sum_{j=1}^{N-k}A({\rm x}_k)B({\rm x}_{k+j})\;\;\;\;\;\;\;\;\;\;k=0,1,2,...,N_c \nonumber$
where $N$ is the total number of time steps, $\Delta t$ is the time step and $N_c << N$.
Property 4: Onsager regression hypothesis
In the long time limit, $A$ and $B$ eventually become uncorrelated from each other so that the time correlation function becomes
$C_{AB}(t) = \langle A(0)B(t)\rangle \rightarrow \langle A\rangle\langle B\rangle \nonumber$
For the autocorrelation function of $A$, this becomes
$C_{AA}(t)\rightarrow \langle A\rangle^2 \nonumber$
Thus, $C_{AA} (t)$ decays from $\langle A^2 \rangle$ at $t = 0$ to $\langle A^2 \rangle$ as $t \rightarrow \infty$.
An example of a signal and its time correlation function appears in Figure $1$. In this case, the signal is the magnitude of the velocity along the bond of a diatomic molecule interacting with a Lennard-Jones bath. Its time correlation function is shown beneath the signal:
Over time, it can be seen that the property being autocorrelated eventually becomes uncorrelated with itself. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/12%3A_Time-dependent_Processes_-_Classical_case/12.03%3A_Generalized_Equations_of_Motion.txt |
Substituting the perturbative form for $f (x, t )$ into the Liouville equation, one obtains
${\partial \over \partial t}(f_0({\rm x}) + \Delta f({\rm x}, t ) ) + (iL_0 + i\Delta L(t))(f_0({\rm x}) + \Delta f({\rm x},t)) = 0 \nonumber$
Recall $\partial f_0/\partial t=0$. Thus, working to linear order in small quantities, one obtains the following equation for $\Delta f({\rm x},t)$:
$\left({\partial \over \partial t} + iL_0\right)\Delta f({\rm x},t) =-i\Delta L f_0({\rm x}) \nonumber$
which is just a first-order inhomogeneous differential equation. This can easily be solved using an integrating factor, and one obtains the result
$\Delta f({\rm x},t) = -\int_0^t ds e^{-iL_0(t-s)}i\Delta L(s)f_0({\rm x}) \nonumber$
Note that
$i\Delta L f_0({\rm x}) = iL f_0({\rm x}) - iL_0 f_0({\rm x}) = iL f_0({\rm x})=\dot{\rm x}\cdot\nabla_{\rm x}f_0({\rm x}) \nonumber$
But, using the chain rule, we have
\begin{align*} \dot{\rm x}\cdot\nabla_{\rm x}f_0({\rm x}) &= \dot{\rm x}\cdot{\partial f_0 \over \partial H}{\partial H \over \partial {\rm x}} \[4pt] &= {\partial f_0 \over \partial H}\sum_{i=1}^{3N}\left[\dot{p}_i{\partial H \over \partial p_i} +\dot{q}_i {\partial H \over \partial q_i}\right] \[4pt] &={\partial f_0 \over \partial H}\sum_{i=1}^{3N}\left[{\partial H \over \partial p_i} \left (-{\partial H \over \partial q_i} + D_iF_e (t) \right ) +{\partial H \over \partial q_i}\left({\partial H \over \partial p_i} + C_i F_e(t)\right)\right] \[4pt] &={\partial f_0 \over \partial H}\sum_{i=1}^{3N}\left[D_i({\rm x}) {\partial H \over \partial p_i} + C_i({\rm x}){\partial H \over \partial q_i}\right]F_e(t) \end{align*}
Define
$j({\rm x}) = -\sum_{i=1}^{3N}\left[D_i({\rm x}){\partial H \over \partial p_i} + C_i({\rm x}){\partial H \over \partial q_i}\right] \nonumber$
which is known as the dissipative flux. Thus, for a Cartesian Hamiltonian
$H = \sum_{i=1}^N {{\textbf p}_i^2 \over 2m_i} + U({\textbf r}_1,...,{\textbf r}_N) \nonumber$
where ${\textbf F}_i({\textbf r}_1,...,{\textbf r}_N) = -\nabla_iU$ is the force on the ${i}$ th particle, the dissipative flux becomes:
$j({\rm x}) = \sum_{i=1}^N\left[{\bf C}_i({\rm x})\cdot{\textbf F}_i-{\textbf D}_i({\rm x})\cdot {{\textbf p}_i \over m_i}\right] \nonumber$
In general,
$\dot{\rm x}\cdot\nabla_{\rm x}f_0({\rm x}) = -{\partial f_0 \over \partial H}j({\rm x})F_e(t) \nonumber$
Now, suppose $f_0 (x)$ is a canonical distribution function
$f_0(H({\rm x})) = {1 \over Q(N,V,T)}e^{-\beta H({\rm x})} \nonumber$
then
${\partial f_0 \over \partial H} = -\beta f_0(H) \nonumber$
so that
$\dot{\rm x}\cdot\nabla_{\rm x}f_0({\rm x}) = \beta f_0({\rm x})j({\rm x})F_e(t) \nonumber$
Thus, the solution for $\Delta f (x, t)$ is
$\Delta f({\rm x},t) = -\beta\int_0^t ds e^{-iL_0(t-s)}f_0({\rm x})j({\rm x})F_e(s) \nonumber$
The ensemble average of the observable $A (x)$ now becomes
\begin{align*} \langle A (t) \rangle &=\langle A \rangle_0 - \beta\int d{\rm x}A({\rm x})\int_0^t ds e^{-iL_0(t-s)}f_0({\rm x})j({\rm x})F_e(s) \[4pt] &=\langle A \rangle_0 - \beta\int_0^t ds \int d{\rm x}A({\rm x})e^{-iL_0(t-s)}f_0({\rm x})j({\rm x})F_e(s) \[4pt] &=\langle A \rangle_0 - \beta\int_0^t ds \int d{\rm x}f_0({\rm x}) A({\rm x})e^{-iL_0(t-s)}j({\rm x})F_e(s) \end{align*}
Recall that the classical propagator is $exp(iLt)$. Thus the operator appearing in the above expression is a classical propagator of the unperturbed system for propagating backwards in time to $- (t - s)$. An observable $A (x)$ evolves in time according to
\begin{align*} \dfrac{dA}{dt} &= iLA \[4pt] A (t) &= e^{iLt}A(0) \[4pt] A (-t ) &= e^{-iLt}A(0) \end{align*}
Now, if we take the complex conjugate of both sides, we find
$A^*(t) = A^*(0)e^{-iLt} \nonumber$
where now the operator acts to the left on $A^* (0)$. However, since observables are real, we have
$A(t) = A(0) e^{-iLt} \nonumber$
which implies that forward evolution in time can be achieved by acting to the left on an observable with the time reversed classical propagator. Thus, the ensemble average of $A$ becomes
\begin{align*} \langle A(t)\rangle &=\langle A \rangle_0 - \beta\int_0^t ds F_e(s) \int d{\rm x}_0 f_0({\rm x}_0)A({\rm x}_{t-s}({\rm x}_0))j({\rm x}_0) \[4pt] &=\langle A \rangle_0 - \beta\int_0^t ds F_e(s)\langle j(0)A(t-s)\rangle_0 \end{align*}
where the quantity on the last line is an object we have not encountered yet before. It is known as an equilibrium time correlation function. An equilibrium time correlation function is an ensemble average over the unperturbed (canonical) ensemble of the product of the dissipative flux at $t = 0$ with an observable $A$ evolved to a time ${t - s}$. Several things are worth noting:
1. The nonequilibrium average $\langle A(t)\rangle$, in the linear response regime, can be expressed solely in terms of equilibrium averages.
2. The propagator used to evolve $A (x)$ to $A (x, t - s )$ is the operator $exp (iL_0 (t - s ) )$, which is the propagator for the unperturbed, Hamiltonian dynamics with $C_i = D_i = 0$. That is, it is just the dynamics determined by $H$.
3. Since $A({\rm x},t-s) = A({\rm x}(t-s))$ is a function of the phase space variables evolved to a time ${t - s }$, we must now specify over which set of phase space variables the integration $\int dx$ is taken. The choice is actually arbitrary, and for convenience, we choose the initial conditions. Since $x (t)$ is a function of the initial conditions $x (0)$, we can write the time correlation function as $\langle j(0)A(t-s)\rangle_0 = {1 \over Q}\int d{\rm x}_0 e^{-\beta H({\rm x}_0)}j({\rm x}_0)A({\rm x}_{t-s}({\rm x}_0)) \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/12%3A_Time-dependent_Processes_-_Classical_case/12.05%3A_Perturbative_solution_of_the_Liouville_equation.txt |
Suppose that $F_e (t)$ is of the form
$F_e(t) = F_0e^{\epsilon t}\theta(-t) \nonumber$
which adiabatically induces a fluctuation in the system for $t < 0$ and the lets the system evolve in time according to the unperturbed Hamiltonian for $t > 0$. How will the induced fluctuation evolve in time? Combining the Kubo transform relation with the linear response result for $\langle B(t)\rangle$, we find that
\begin{align*} \langle B(t)\rangle &= \int_{-\infty}^0ds\, e^{\epsilon s}\int_0^{\beta} d\lambda\langle \dot{B}(-i\hbar\lambda)B(t-s)\rangle _0 \[4pt] &= -e^{\epsilon t}\int_0^{\beta}d\lambda \int_t^{\infty}du \,e^{-\epsilon u}{d \over du}\langle B(-i\hbar\lambda)B(u)\rangle _0 \end{align*}
where the change of variables ${u=t-s }$ has been made. Taking the limit $\epsilon\rightarrow 0$, and performing the integral over $u$, we find
$\langle B(t)\rangle = -\int_0^{\beta}d\lambda\,\left[\langle B (- i\hbar \lambda) B (\infty )\rangle _0 -\langle B(-i\hbar\lambda)B(t)\rangle _0\right] \nonumber$
Since we assumed that $\langle B\rangle _0 = 0$, we have $\langle B(-i\hbar\lambda)B(\infty)\rangle _0 =\langle B(-i\hbar\lambda)\rangle _0\langle B(\infty)\rangle _0 = 0$. Thus, dividing by $\langle B(0)\rangle$, we find
${\langle B(t)\rangle \over \langle B(0)\rangle } = {\int_0^{\beta} d \lambda B (-i\hbar \lambda) B (t) \rangle _0 \over \int_0^{\beta} d \lambda B (-i\hbar \lambda) B (0) \rangle _0} \rightarrow _{\hbar \rightarrow 0 } {\langle B(0)B(t)\rangle _0 \over \langle B(0)^2 \rangle _0} \nonumber$
Thus at long times in the classical limit, the fluctuations decay to 0, indicting a complete regression or suppression of the induced fluctuation:
${\langle B(t)\rangle \over \langle B(0) \rangle }\rightarrow 0 \nonumber$
12.07: Relation to Spectra
Suppose that $F_e (t)$ is a monochromatic field
$F_e(t) = F_{\omega}e^{i\omega t}e^{\epsilon t} \nonumber$
where the parameter $\epsilon$ insures that field goes to 0 at ${ t = - \infty }$. We will take ${\epsilon\rightarrow 0^+ }$ at the end of the calculation. The expectation value of $B$ then becomes
\begin{align*} \langle B(t)\rangle &= \langle B\rangle _0 + \int_{-\infty}^t\;ds\;\Phi_{BB}(t-s)F_{\omega}e^{i\omega s}e^{\epsilon s} \[4pt] &=\langle B\rangle _0 + F_{\omega}e^{(i\omega + \epsilon)t} \int_0^{\infty}d\tau\Phi_{BB}(\tau)e^{-i(\omega-i\epsilon)\tau} \end{align*}
where the change of integration variables ${\tau=t-s }$ has been made.
Define a frequency-dependent susceptibility by
$\chi_{BB}(\omega-i\epsilon) = \int_0^{\infty}d\tau \Phi_{BB}(\tau) e^{-i(\omega-i\epsilon)\tau} \nonumber$
then
$\langle B(t)\rangle = \langle B\rangle _0 + F_{\omega}e^{i\omega t}e^{\epsilon t}\chi_{BB}(\omega-i\epsilon) \nonumber$
If we let $z=\omega-i\epsilon$, then we see immediately that
$\chi_{BB}(z) = \int_0^{\infty}d\tau\;\Phi_{BB}(\tau) e^{-iz\tau} \nonumber$
i.e., the susceptibility is just the Laplace transform of the after effect function or the time correlation function.
Recall that
\begin{align*} \Phi_{AB}(t) &= {i\over\hbar}\langle [A(t),B(0)]\rangle _0 \[4pt] &= {i \over \hbar}\langle[e^{iH_0t/\hbar} Ae^{-iH_0t\hbar},B]\rangle _0 \end{align*}
Under time reversal, we have
\begin{align*} \Phi_{AB}(-t) &= {i \over \hbar} \langle \left[e^{-iH_0t/\hbar}Ae^{iH_0t/\hbar},B\right]\rangle _0 \[4pt] &= { {i \over \hbar} \langle \left(e^{-iH_0t/\hbar}Ae^{iH_0t/\hbar}B -Be^{-iH_0t/\hbar}Ae^{iH_0t/\hbar}\right)\rangle _0 } \[4pt] &= {i \over \hbar} \langle \left(Ae^{iH_0t/\hbar}Be^{-iH_0t/\hbar} -e^{iH_0t/\hbar}Be^{-iH_0t/\hbar}A\right)\rangle _0 \[4pt] &= {i \over \hbar} \langle \left(AB(t)-B(t)A\right)\rangle _0 \[4pt] &= -{i \over \hbar} \langle \left[B(t),A\right]\rangle \[4pt] &= -\Phi_{BA}(t) \end{align*}
Thus,
$\Phi_{AB}(-t) = -\Phi_{BA}(t) \nonumber$
and if $A = B$, then
$\Phi_{BB}(-t) = -\Phi_{BB}(t) \nonumber$
Therefore
\begin{align*} \chi_{BB}(\omega) &= \lim_{\epsilon\rightarrow 0^+}\int_0^{\infty} dt\;e^{-i(\omega-i\epsilon t)}\Phi_{BB}(t) \[4pt] &=\lim_{\epsilon\rightarrow 0^+}\int_0^{\infty}dt\;e^{-\epsilon t}\left[\Phi_{BB}(t)\cos\omega t - i\Phi_{BB}(t)\sin\omega t\right] \[4pt] &={\rm Re}(\chi_{BB}(\omega)) - i{\rm Im}(\chi_{BB}(\omega)) \end{align*}
From the properties of $\Phi_{BB}(t)$ it follows that
\begin{align*} {\rm Re}(\chi_{BB}(\omega) &= {\rm Re}(\chi_{BB}(-\omega) \[4pt] {\rm Im}(\chi_{BB}(\omega) &= -{\rm Im}(\chi_{BB}(-\omega) \end{align*}
so that ${\rm Im}(\chi_{BB}(\omega))$ is positive for ${ \omega > 0 }$ and negative for ${ \omega < 0 }$. It is a straightforward matter, now, to show that the energy difference $Q (\omega )$ derived in the lecture from the Fermi golden rule is related to the susceptibility by
$Q(\omega) = 2\omega\vert F_{\omega}\vert^2{\rm Im}(\chi_{BB}(\omega)) \nonumber$ | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/12%3A_Time-dependent_Processes_-_Classical_case/12.06%3A_The_Onsager_Fluctuation_Regression_Theorem.txt |
• 12.8.1: Shear Viscosity
The shear viscosity of a system measures is resistance to flow. A simple flow field can be established in a system by placing it between two plates and then pulling the plates apart in opposite directions. Such a force is called a shear force, and the rate at which the plates are pulled apart is the shear rate.
• 12.8.2: The Diffusion Constant
12.08: Time Correlation Functions and Transport Coefficients
The shear viscosity of a system measures is resistance to flow. A simple flow field can be established in a system by placing it between two plates and then pulling the plates apart in opposite directions. Such a force is called a shear force, and the rate at which the plates are pulled apart is the shear rate. A set of microscopic equations of motion for generating shear flow is
$\dot{ \textbf{r}}_i = \dfrac{ \textbf{p}_i}{m_i} + \gamma y_i \hat{\textbf{x}} \nonumber$
$\dot{ \textbf{p}}_i = \textbf{F}_i + \gamma p_{y_i} \hat{\textbf{x}} \nonumber$
where $\gamma$ is a parameter known as the shear rate. These equations have the conserved quantity
$H' = \sum_{i=1}^N \left( \textbf{p}_i + m_i\gamma y_i\hat{\textbf {x}} \right)^2 +U(\textbf{r}_1,..,\textbf{r}_N) \nonumber$
The physical picture of this dynamical system corresponds to the presence of a velocity flow field ${\textbf v}(y) = \gamma y\hat{\textbf x}$ shown in the figure.
The flow field points in the $\hat{\textbf x}$ direction and increases with increasing $y$-value. Thus, layers of a fluid, for example, will slow past each other, creating an anisotropy in the system. From the conserved quantity, one can see that the momentum of a particle is the value of ${{\textbf p}_i}$ plus the contribution from the field evaluated at the position of the particle
${\textbf p}_i \rightarrow {\textbf p}_i + m_i {\textbf v}(y_i) \nonumber$
Such an applied external shearing force will create an asymmetry in the internal pressure. In order to describe this asymmetry, we need an analog of the internal pressure that contains a dependence on specific spatial directions. Such a quantity is known as the pressure tensor and can be defined analogously to the isotropic pressure $P$ that we encountered earlier in the course. Recall that an estimator for the pressure was
$p = {1 \over 3V}\sum_{i=1}^N\left[{{\textbf p}_i^2 \over m_i} + {\textbf r}_i\cdot {\textbf F}_i\right] \nonumber$
and $P = \langle p\rangle$ in equilibrium. Here, $V$ is the volume of the system. By analogy, one can write down an estimator for the pressure tensor ${p_{\alpha\beta} }$:
$p_{\alpha\beta} = {1 \over V}\sum_{i=1}^N\left[{({\textbf p}_i\cdot \hat {e}_{\alpha})({\textbf P}_i \cdot \hat {\textbf e}_{\beta}) \over m_i} + ({\textbf r}_i \cdot \hat {\textbf e}_{\alpha} )({\textbf F}_i\cdot\hat{\textbf e}_{\beta})\right] \nonumber$
and
$P_{\alpha\beta} = \langle p_{\alpha\beta}\rangle \nonumber$
where ${\hat{\textbf e}_{\alpha} }$ is a unit vector in the $\alpha$ direction, ${\alpha=x,y,z }$. This (nine-component) pressure tensor gives information about spatial anisotropies in the system that give rise to off-diagonal pressure tensor components. The isotropic pressure can be recovered from
$P = {1 \over 3}\sum_{\alpha}P_{\alpha\alpha} \nonumber$
which is just 1/3 of the trace of the pressure tensor. While most systems have diagonal pressure tensors due to spatial isotropy, the application of a shear force according to the above scheme gives rise to a nonzero value for the $xy$ component of the pressure tensor $P_{xy}$. In fact, $P_{xy}$ is related to the velocity flow field by a relation of the form
$P_{xy} = -\eta {\partial v_x \over \partial y} = -\eta\gamma \nonumber$
where the coefficient $\eta$ is known as the shear viscosity and is an example of a transport coefficient. Solving for $\eta$ we find
$\eta = -{P_{xy} \over \gamma} = -\lim_{t\rightarrow\infty}{\langle p_{xy}(t) \rangle \over \gamma} \nonumber$
where $\langle p_{xy}(t)\rangle$ is the nonequilibrium average of the pressure tensor estimator using the above dynamical equations of motion.
Let us apply the linear response formula to the calculation of the nonequilibrium average of the $xy$ component of the pressure tensor. We make the following identifications:
$F_e(t) = 1\;\;\;\;\;{\bf C}_i({\rm x}) = \gamma y_i\hat{\textbf x}\;\;\;\;\;{\bf D}_i({\rm x}) = -\gamma p_{y_i}\hat{\textbf x} \nonumber$
Thus, the dissipative flux $j({\rm x})$ becomes
\begin{align*} j({\rm x}) &= \sum_{i=1}^N\left[{\textbf C}_i\cdot {\textbf F}_i - {\textbf D}_i\cdot {{\textbf p}_i \over m_i}\right] \[4pt] &= \sum_{i=1}^N\left[\gamma y_i({\textbf F}_i\cdot \hat{\textbf x}) + \gamma p_{y_i}{{\textbf p}_i\cdot \hat{\textbf x} \over m_i}\right] \[4pt] &= \gamma\sum_{i=1}^N\left[{({\textbf p}_i\cdot \hat{\textbf y})({\textbf p}_i \cdot \hat {\textbf x}) \over m_i} + ({\textbf r}_i\cdot \hat{\textbf y})({\textbf F}_i\cdot \hat{\textbf x})\right] \[4pt] &= \gamma V p_{xy} \end{align*}
According to the linear response formula,
$\langle p_{xy}(t) \rangle = \langle p_{xy}\rangle_0 -\beta\gamma V\int_0^t ds \langle p_{xy}(0) p_{xy}(t-s)\rangle_0 \nonumber$
so that the shear viscosity becomes
$\eta = \lim_{t\rightarrow\infty}\left[-{\langle p_{xy}\rangle _0 \over \gamma } + \beta V \int_0^t ds \langle p_{xy}(0)p_{xy}(t)\rangle_0\right] \nonumber$
Recall that $\langle\cdots\rangle_0$ means average of a canonical distribution with ${\gamma = 0 }$. It is straightforward to show that $\langle p_{xy}\rangle_0=0$ for an equilibrium canonical distribution function. Finally, taking the limit that $t \rightarrow \infty$ in the above expression gives the result
$\eta = {V \over kT}\int_0^{\infty}dt \langle p_{xy}(0) p_{xy}(t)\rangle_0 \nonumber$
which is a relation between a transport coefficient, in this case, the shear viscosity coefficient, and the integral of an equilibrium time correlation function. Relations of this type are known as Green-Kubo relations. Thus, we have expressed a new kind of thermodynamic quantity to an equilibrium time correlation function, which, in this case, is an autocorrelation function of the $xy$ component of the pressure tensor.
12.8.02: The Diff
The diffusive flow of particles can be studied by applying a constant force $f$ to a system using the microscopic equations of motion
\begin{align*} \dot{\textbf r}_i &= { {\textbf p}_i \over m_i} \[4pt] \dot{\textbf p}_i &= {\textbf F}_i({\textbf q}_1,..,{\textbf q}_N) + f\hat{\textbf x} \end{align*}
which have the conserved energy
$H' = \sum_{i=1}^N {{\textbf p}_i^2 \over 2m_i} + U({\textbf q}_1,...,{\textbf q}_N) -f\sum_{i=1}^Nx_i \nonumber$
Since the force is applied in the $\hat{\textbf x}$ direction, there will be a net flow of particles in this direction, i.e., a current $J_x$. Since this current is a thermodynamic quantity, there is an estimator for it:
$u_x = \sum_{i=1}^N \dot{x}_i \nonumber$
and $J_x = \langle u_x \rangle$. The constant force can be considered as arising from a potential field
$\phi(x) = -xf \nonumber$
The potential gradient $\partial \phi/\partial x$ will give rise to a concentration gradient $\partial c / \partial x$ which is opposite to the potential gradient and related to it by
${\partial c \over \partial x} = -{1 \over kT}{\partial \phi \over \partial x} \nonumber$
However, Fick's law tells how to relate the particle current $J_x$ to the concentration gradient
\begin{align*} J_x &= D\dfrac{\partial c}{\partial x} \[4pt] &= -\dfrac{D}{kT} \dfrac{\partial \phi}{\partial x} \[4pt] &= \dfrac{D}{kT}f \end{align*}
where $D$ is the diffusion constant. Solving for $D$ gives
\begin{align*} D &= kT \dfrac{J_x}{f} \[4pt] &= kT\lim_{t\rightarrow\infty}{\langle u_x(t)\rangle \over f} \end{align*}
Let us apply the linear response formula again to the above nonequilibrium average. Again, we make the identification:
$F_e(t) = 1\;\;\;\;\;\;{\textbf D}_i = f\hat{\textbf x}\;\;\;\;\;{\textbf C}_i=0 \nonumber$
Thus,
\begin{align*} \langle u_x(t) \rangle &= \langle u_x\rangle_0 + \beta\int_0^t dsf\langle \left(\sum_{i=1}^N\dot{x}_i(0)\right)\left(\sum_{i=1}^N\dot{x}_i(t-s)\right)\rangle_0 \[4pt] &= \langle u_x \rangle_0 + \beta f\int_0^t ds\sum_{i,j}\langle \dot{x}_i(0)\dot{x}_j(t-s)\rangle_0 \end{align*}
In equilibrium, it can be shown that there are no cross correlations between different particles. Consider the initial value of the correlation function. From the virial theorem, we have
$\langle \dot{x}_i\dot{x}_j\rangle_0 = \delta_{ij}\langle \dot{x}_i^2\rangle_0 \nonumber$
which vanishes for $i \ne j$. In general,
$\langle \dot{x}_i(0)\dot{x}_j(t)\rangle_0 = \delta_{ij}\langle \dot{x}_i(0)\dot{x}_i(t-s)\rangle_0 \nonumber$
Thus,
$\langle u_x(t)\rangle = \langle u_x\rangle_0 + \beta f \int_0^t ds\sum_{i=1}^N\dot{x}_i(0)\dot{x}_i(t-s)\rangle_0 \nonumber$
In equilibrium, $\langle u_x \rangle _0 = 0$ being linear in the velocities (hence momenta). Thus, the diffusion constant is given by, when the limit $t \rightarrow \infty$ is taken,
$D = \int_0^{\infty} \sum_{i=1}^N\langle\dot{x}_i(0)\dot{x}_i(t)\rangle_0 \nonumber$
However, since no spatial direction is preferred, we could also choose to apply the external force in the ${y}$ or $z$ directions and average the result over the these three. This would give a diffusion constant
$D = {1 \over 3}\int_0^{\infty} dt \sum_{i=1}^N\langle \dot{\textbf r}_i(0)\cdot\dot{\textbf r}_i(t)\rangle_0 \nonumber$
The quantity
$\sum_{i=1}^N\langle \dot{\textbf r}_i(0)\cdot\dot{\textbf r}_i(t)\rangle_0 \nonumber$
is known as the velocity autocorrelation function, a quantity we will encounter again in other contexts. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/12%3A_Time-dependent_Processes_-_Classical_case/12.08%3A_Time_Correlation_Functions_and_Transport_Coefficients/12.8.01%3A_Shear_Vi.txt |
Consider a quantum system with a Hamiltonian $H_0$. Suppose this system is subject to an external driving force $F_e (t)$ such that the full Hamiltonian takes the form
$H = H_0 - BF_e(t) = H_0 + H' \nonumber$
where $B$ is an operator through which this coupling occurs. This is the situation, for example, when the infrared spectrum is measured experimentally - the external force $F_e (t)$ is identified with an electric field $E (t)$ and $B$ is identified with the electric dipole moment operator. If the field $F_e (t)$ is inhomogeneous, then $H$ takes the more general form
\begin{align*} H &= H_0 - \int d^3x\; B({\textbf x}) F_e({\textbf x},t) \[4pt] &= H_0 - \sum_{\textbf k} B_{\textbf k}F_{e,{\textbf k}}(t) \end{align*}
where the sum is taken over Fourier modes. Often, $B$ is an operator such that, if $F_e (t) = 0$, then
$\langle B \rangle = {{\rm Tr}\left(Be^{-\beta H}\right) \over {\rm Tr}\left(e^{-\beta H}\right)} \nonumber$
Suppose we take $F_e (t)$ to be a monochromatic field of the form
$F_e(t) = F_{\omega}e^{i\omega t} \nonumber$
Generally, the external field can induce transitions between eigenstates of $H_0$ in the system. Consider such a transition between an initial state $\vert i \rangle$ and a final state $\vert f \rangle$, with energies $E_i$ and $E_f$, respectively:
\begin{align*} H_0\vert i\rangle &= E_i\vert i\rangle \[4pt] H_0\vert f\rangle &= E_f\vert f\rangle \end{align*}
(see figure below).
This transition can only occur if
$E_f = E_i + \hbar\omega \nonumber$
13.1.02: The Transition R
In the next lecture, we will solve the quantum Liouville equation
$i\hbar{\partial \rho \over \partial t} = [H,\rho] \nonumber$
perturbatively and derive quantum linear response theory. However, the transition rate can actually be determined directly within perturbation theory using the Fermi Golden Rule approximation, which states that the probability of a transition's occurring per unit time, $R_{i\rightarrow f}$, is given by
$R_{i\rightarrow f}(\omega) = {2\pi \over \hbar}\vert\langle f\vert B\vert i\rangle \vert^2\delta(E_f-E_i-\hbar\omega ) \nonumber$
The $\delta$-function expresses the fact that energy is conserved. This describes the rate of transitions between specific states $\vert i\rangle$ and $\vert f\rangle$. The transition rate between any initial and final states can be obtained by summing over both $i$ and $f$ and weighting the sum by the probability that the system is found in the initial state $\vert i\rangle$:
$P(\omega) = \sum_{i,f}R_{i\rightarrow f}(\omega) w_i \nonumber$
where ${w_i}$ is an eigenvalue of the density matrix, which we will take to be the canonical density matrix:
$w_i = {e^{-\beta E_i} \over {\rm Tr}\left(e^{-\beta H}\right)} \nonumber$
Using the expression for $R_{i\rightarrow f}(\omega)$, we find
$P(\omega) = {2\pi\over \hbar}\vert F_{\omega}\vert^2\sum_{i, f} w_i \vert \langle i \vert B \vert f \rangle \vert ^2 \delta(E_f-E_i-\hbar\omega ) \nonumber$
Note that
$P(-\omega) = {2\pi\over \hbar}\vert F_{\omega}\vert^2\sum_{i, f } w_i \vert \langle i \vert B \vert f \rangle \vert^2\delta(E_f-E_i+\hbar\omega ) \nonumber$
This quantity corresponds to a time-reversed analog of the absorption process. Thus, it describes an emission event $\vert i\rangle \rightarrow\vert f\rangle$ with $E_f = E_i-\hbar\omega$, i.e., emission of a photon with energy $\hbar \omega$. If can also be expressed as a process $\vert f\rangle \rightarrow \vert i\rangle$ by recognizing that
$w_f = {e^{-\beta E_f} \over {\rm Tr}\left(e^{-\beta H}\right) } = {e^{\beta (E_i-\hbar\omega )} \over {\rm Tr}\left(e^{-\beta H}\right)} \nonumber$
or
$w_f = e^{\beta\hbar\omega }w_i\;\;\;\;\Rightarrow\;\;\;\;w_i = e^{-\beta\hbar\omega }w_f \nonumber$
Therefore
$P(-\omega) = {2\pi\over \hbar}\vert F_{\omega}\vert^2 e^{-\beta \hbar \omega } \sum _{i, f} w_f \vert \langle i\vert B\vert f\rangle \vert^2\delta(E_f-E_i+\hbar\omega ) \nonumber$
If we now interchange the summation indices, we find
\begin{align*} P(-\omega) &= {2\pi\over \hbar}\vert F_{\omega}\vert^2e^{-\beta\hbar\omega}\sum_{i,f}w_i\vert\langle i\vert B\vert f\rangle \vert^2\delta(E_i-E_f+\hbar\omega ) \[4pt] &= {2\pi\over \hbar}\vert F_{\omega}\vert^2\sum_{i,f}w_i\vert\langle i\vert B\vert f\rangle \vert^2e^{-\beta\hbar\omega}\delta(E_i-E_f-\hbar\omega ) \end{align*}
where the fact that $\delta (x) = \delta (- x)$ has been used. Comparing this expression for $P (- \omega )$ to that for $P (\omega )$, we find
$P(-\omega) = e^{-\beta\hbar\omega }P(\omega) \nonumber$
which is the equation of detailed balance. We see from it that the probability of emission is less than that for absorption. The reason for this is that it is less likely to find the system in an excited state $\vert f \rangle$ initially, when it is in contact with a heat bath and hence thermally equilibrated. However, we must remember that the microscopic laws of motion (Newton's equations for classical systems and the Schrödinger equation for quantum systems) are reversible. This means that
$R_{i\rightarrow f}(\omega) = R_{f\rightarrow i}(-\omega) \nonumber$
The conclusion is that, since $P(\omega) > P(-\omega)$, reversibility is lost when the system is placed in contact with a heat bath, i.e., the system is being driven irreversibly in time.
Define
\begin{align*} C_>(\omega) &= \sum_{i,f}w_i \vert\langle i\vert B\vert f\rangle \vert^2\delta(E_f-E_i-\hbar\omega ) \[4pt] C_<(\omega) &= \sum_{i,f}w_i \vert\langle i\vert B\vert f\rangle \vert^2\delta(E_f-E_i+\hbar\omega ) \end{align*}
then
$C_<(\omega) = e^{-\beta\hbar\omega }C_>(\omega) \nonumber$
Now using the fact that the $\delta$-function can be written as
$\delta(E) = {1 \over 2\pi}\int_{-\infty}^{\infty}dt e^{-iEt} \nonumber$
$C_>(\omega)$ becomes
\begin{align*} C_>(\omega) &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\sum_{i,f}w_i \vert\langle i\vert B\vert f\rangle \vert^2 e^{-i(E_f-E_i-\hbar\omega )t/\hbar} \[4pt] &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum_{i,f}w_i \vert\langle i\vert B\vert f\rangle \vert^2 e^{-i(E_f-E_i)t/\hbar} \[4pt] &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum _{i, f} w_i \langle i \vert B\vert f\rangle \langle f\vert B\vert i\rangle e^{-iE_f t/\hbar}e^{iE_it/\hbar} \[4pt] &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum _{i, f} w_i \langle i \vert e^{iH_0t/\hbar} B e^{-iH_0t/\hbar}\vert f\rangle\langle f\vert B\vert i\rangle \end{align*}
Recall that the evolution of an operator in the Heisenberg picture is given by
$B(t) = e^{iH_0t/\hbar}Be^{-iH_0t/\hbar} \nonumber$
if the evolution is determined solely by $H_0$. Thus, the expression for $C > (\omega )$ becomes
\begin{align*} C_>(\omega) &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum_{i,f}w_i \langle i\vert B(t)\vert f\rangle\langle f\vert B\vert i\rangle \[4pt] &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum_{i}w_i \langle i\vert B(t)B(0)\vert i\rangle \[4pt] &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;e^{i\omega t}{\rm Tr}\left[\rho B(t) B(0)\right] \[4pt] &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;e^{i\omega t}\langle B(t) B(0)\rangle \end{align*}
which involves the quantum autocorrelation function $\langle B(t) B(0)\rangle$.
In general, a quantum time correlation function in the canonical ensemble is defined by
$C_{AB}(t) = {{\rm Tr}\left[A(t)B(0)e^{-\beta H}\right] \over{\rm Tr}\left[e^{-\beta H}\right]} \nonumber$
In a similar manner, we can show that
$C_<(\omega) = {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle B(0)B(t)\rangle \neq C_>(\omega) \nonumber$
since
$[B(0),B(t)]\neq 0 \nonumber$
in general. Also, the product $B (0) B (t)$ is not Hermitian. However, a hermitian combination occurs if we consider the energy difference between absorption and emission. The energy absorbed per unit of time by the system is $P(\omega)\hbar\omega$, while the emitted into the bath by the system per unit of time is $P(-\omega)\hbar\omega$. The energy difference $Q ( \omega )$ is just
\begin{align*}Q(\omega) &= [P(\omega)-P(-\omega)]\hbar\omega \[4pt] &= P(\omega)[1-e^{-\beta\hbar\omega }]\hbar\omega \[4pt] &= 2\pi\omega\vert F_{\omega}\vert^2C_>(\omega)[1-e^{-\beta\hbar\omega }] \end{align*}
But since
$C_<(\omega) = e^{-\beta\hbar\omega }C_>(\omega) \nonumber$
it follows that
$C_>(\omega) + C_<(\omega) = \left(1+e^{-\beta\hbar\omega }\right)C_>(\omega) \nonumber$
or
$C_>(\omega) = {C_>(\omega) + C_<(\omega) \over 1+e^{-\beta\hbar\omega}} \nonumber$
Note, however, that
\begin{align*} C_>(\omega) + C_<(\omega) &= {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt e^{i\omega t}\langle B(t)B(0) + B(0) B(t)\rangle \[4pt] &= {1 \over \pi\hbar} dt\;e^{i\omega t}\langle {1 \over 2}[B(0),B(t)]_+\rangle \end{align*}
where $[...,...]_+$ is known as the anticommutator:
$[A,B]_+ = AB+BA .$
The anticommutator between two operators is, itself, hermitian. Therefore, the energy difference is
\begin{align*} Q(\omega) &= { {2\omega \over \hbar}\vert F_{\omega}\vert^2 {1-e^{-\beta\hbar\omega} \over 1 + e^{-\beta \hbar \omega }}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle [B(0),B(t)]_+\rangle } \[4pt] &= {2\omega \over \hbar}\vert F_{\omega}\vert^2 {\rm tanh}(\beta\hbar\omega /2)\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle [B(0),B(t)]_+\rangle \end{align*}
The quantity $\langle [B(0),B(t)]_+\rangle$ is the symmetrized quantum autocorrelation function. The classical limit is now manifest ( ${\rm tanh}(\beta\hbar\omega /2)\rightarrow \beta\hbar\omega /2$ ):
$Q(\omega)\rightarrow \vert F_{\omega}\vert^2 \int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle B(0)B(t)\rangle \nonumber$
The classically, the energy spectrum $Q(\omega)$ is directly related to the Fourier transform of a time correlation function.
13.1.03: Examples
Define
$G(\omega) = {1 \over 2\pi}\int_{-\infty}^{\infty}\;dt\, e^{i\omega t} \langle {1 \over 2}[B(0),B(t)]_+\rangle \nonumber$
which is just the frequency spectrum corresponding to the autocorrelation function of $B$. For different choices of $B$, $G(\omega)$ corresponds to different experimental measurements. Consider the example of a molecule with a transition dipole moment vector $\mu$. If an electric field $\textbf{E}(t)$ is applied, then the Hamiltonian $H'$ becomes
$H' =-\mu \cdot \textbf{E}(t) \nonumber$
If we take $\textbf{E}(t)=E(t) \hat{\text{z}}$, then
$H'=-\mu_z E(t) \nonumber$
Identifying $B=\mu_z$, the spectrum becomes
$G(\omega) = {1 \over 2\pi}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle {1 \over 2}[\mu_z(0),\mu_z(t)]_+\rangle \nonumber$
or for a general electric field, the result becomes
$G(\omega) = {1 \over 2\pi}\int_{-\infty}^{\infty}\;dt\; e^{i \omega t} \langle {1 \over 2} (\mu (0) \cdot \mu (t) + \mu (t) \cdot \mu(0) )\rangle \nonumber$
These spectra are the infrared spectra.
As another example, consider a block of material placed in a magnetic field ${\cal H}(t)$ in the $z$ direction. The spin $S_z$ of each particle will couple to the magnetic field giving a Hamiltonian $H'$
$H' = -\sum_{i=1}^N S_{i,z}{\cal H}(t) \nonumber$
The net magnetization created by the field ${m_z}$ is given by
$m_z = {1 \over N}\sum_{i=1}^N S_{i,z} \nonumber$
so that
$H' = -Nm_z{\cal H}(t) \nonumber$
Identify $B = m_z$ (the extra factor of $N$ just expresses the fact that $H'$ is extensive). Then the spectrum is
$G(\omega) = {1 \over 2\pi}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle {1 \over 2}[m_z(0),m_z(t)]_+\rangle \nonumber$
which is just the NMR spectrum. In general for each correlation function there is a corresponding experiment that measures its frequency spectrum.
To see what some specific lineshapes look like, consider as an ansatz a pure exponential decay for the correlation function $C_{BB} (t)$:
$C_{BB}(t) = \langle B^2\rangle e^{-\Gamma \vert t\vert} \nonumber$
The spectrum corresponding to this time correlation function is
$G(\omega) = {1 \over 2\pi}\int_{-\infty}^{\infty}\;dt e^{i\omega t}C_{BB}(t) \nonumber$
and doing the integral gives
$G(\omega) = {\langle B^2 \rangle \over \pi}{\Gamma \over \omega^2 + \Gamma^2} \nonumber$
which is shown in the figure below:
We see that the lineshape is a Lorentzian with a width $\Gamma$. As a further example, suppose $C_{BB} (t)$ is a decaying oscillatory function:
$C_{BB}(t) = \langle B^2 \rangle e^{-\Gamma\vert t\vert}\cos\omega_0t \nonumber$
which describes well the behavior of a harmonic diatomic coupled to a bath. The spectrum can be shown to be
$G(\omega) = {\langle B^2 \rangle \Gamma \over \pi}\left[{\Gamma ^2 + \omega ^2 + \omega ^2_0 \over \left ( \Gamma ^2 + (\omega - \omega _0 )^2 \right)\left(\Gamma^2 + (\omega+\omega_0)^2\right)}\right] \nonumber$
which contains two peaks at $\omega = \pm \sqrt{\omega_0^2 - \Gamma^2}$ as shown in the figure below: | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/13%3A_Time-dependent_Processes_-_Quantum_Case/13.01%3A_Calculation_of_spectra_from_perturbation_theory/13.1.01%3A_The_Hamiltonian.txt |
• 13.1: Calculation of spectra from perturbation theory
• 13.2: Iterative solution for the interaction-picture state vector
• 13.3: The Interaction Picture
Like the Schrödinger and Heisenberg pictures, the interaction picture is a perfectly valid way of representing a quantum mechanical system. The interaction picture can be considered as "intermediate'' between the Schrödinger picture, where the state evolves in time and the operators are static, and the Heisenberg picture, where the state vector is static and the operators evolve.
• 13.4: Fermi's Golden Rule
Fermi's Golden Rule states that, to first-order in perturbation theory, the transition rate depends only the square of the matrix element of the operator V between initial and final states and includes, via the δ-function, an energy-conservation condition. We will make use of the Fermi Golden Rule expression to analyze the application of an external monochromatic field to an ensemble of systems in order to derive expressions for the observed frequency spectra.
• 13.5: Quantum Linear Response Theory
13: Time-dependent Processes - Quantum Case
The solution to Equation \ref{6} can be expressed in terms of a unitary propagator $U_I(t;t_0)$, the interaction-picture propagator, which evolves the initial state $\vert\Phi(t_0)\rangle$ according to
$\vert\Phi(t)\rangle = U_I(t;t_0)\vert\Phi(t_0)\rangle = U_I(t;t_0)\vert\Psi(t_0)\rangle \label{7}$
Substitution of Equation \ref{7} into Equation \ref{6} yields an evolution equation for the propagator $U_I(t;t_0)$:
$H_I(t)U_I(t;t_0) = i\hbar{\partial \over \partial t}U_I(t;t_0) \label{8}$
The initial condition on Equation \ref{8} is $U_I(t_0;t_0) = I$. In developing a solution to Equation \ref{8}, we assume that $H_I (t)$ is a small perturbation, so that the solution can take the form of a sum of powers of $H_I (t)$.
A solution of this form can be generated by recognizing that Equation \ref{8} can be solved formally in terms of an integral equation:
$U_I(t;t_0) =U_I(t_0;t_0) - {i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t')U_I(t';t_0) =I - {i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t')U_I(t';t_0) \label{9}$
It is straightforward to verify this form solution for $U_I (t; t_0 )$. Computing the time derivative of both sides gives
$i\hbar{\partial \over \partial t}U_I(t;t_0) = -i\hbar{i \over \hbar}{\partial \over \partial t}\int_{t_0}^t\;dt'\;H_I(t')U_I(t';t_0) = H_I(t)U_I(t;t_0) \label{10}$
Thus, Equation \ref{9} is a valid expression of the solution. The implicit nature of the integral equation means that an iterative procedure based on the assumption that $H_I (t)$ is a small perturbation can be easily developed. We start with a zeroth-order solution by setting $H_I (t) = 0$ in Equation \ref{9}, which gives the trivial result
$U_I^{(0)}(t;t_0) = I \label{11}$
This solution is now fed back into the right side of Equation \ref{9} to develop a first-order solution:
\begin{align} U^{(1)}(t;t_0) &= I - {i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t')U_I^{(0)}(t';t_0) \[4pt] &= I - {i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t') \label{12} \end{align}
The first order solution is fed back into the right side of Equation \ref{9} to develop a second-order solution:
$U^{(2)}(t;t_0) = I - {i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t')U_I^{(1)}(t';t_0) = I - {i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t')+ \left({i \over \hbar}\right)^2\int_{t_0}^t\;dt'\;\int_{t_0}^{t'}\;dt''\;H_I(t')H_I(t'') \label{13}$
and so forth, such that the $k$ th-order solution is always generated from the $(k - 1 )$ st-order solution according to the recursion formula:
$U^{(k)}(t;t_0) = I - {i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t')U_I^{(k-1)}(t';t_0) \label{14}$
Thus, the third-order solution is given by
$U^{(2)}(t;t_0) = I - {i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t')+ \left({i \over \hbar}\right)^2\int_{t_0}^t\;dt'\;\int_{t_0}^{t'}\;dt''\;H_I(t')H_I(t'') - \left({i \over \hbar}\right)^3\int_{t_0}^t\;dt'\;\int_{t_0}^{t'}\;dt''\;\int_{t_0}^{t''}\;dt'''\;H_I(t')H_I(t'')H_I(t''') \label{15}$
The exact solution is then just a sum of the solutions obtained at each order:
$U_I(t;t_0) = \sum_{k=0}^{\infty}\left(-1\right)^k \left({i \over \hbar } \right)^{k} \int_{t_0}^t dt' \int_{t_0}^{t'} dt'' \cdots \int_{t_0}^{t^{(k-1)}}\;dt^{(k)}\;H_I(t')H_I(t'')\cdots H_I(t^{(k)}) \label{16}$
Having seen how to generate a solution for the propagator in the interaction picture to arbitrarily high orders in the perturbation, the time evolution of the state vector $\vert\Phi(t)\rangle$ in the interaction picture can be determined from
$\vert\Phi(t)\rangle = U_I(t;t_0)\vert\Phi(t_0)\rangle \label{17}$
and from this expression, the time evolution of the original state vector $\vert\Phi(t)\rangle$ in the Schrödinger picture can be determined
$\vert\Phi(t)\rangle = e^{-iH_0(t-t_0)/\hbar}\vert\Phi(t)\rangle = e^{-iH_0(t-t_0)/\hbar}U_I(t;t_0)\vert\Phi(t_0)\rangle = e^{-iH_0(t-t_0)/\hbar}U_I(t;t_0)\vert\Psi(t_0)\rangle \textstyle \equiv U(t;t_0)\vert\Psi(t_0)\rangle \label{18}$
where we have used the fact that $\vert\Phi(t_0)\rangle = \vert\Psi(t_0)\rangle$ and, in the last line, the full propagator in the Schrödinger picture is identified as
$U(t;t_0) = e^{-iH_0(t-t_0)/\hbar}U_I(t;t_0) \label{19}$
From Equation \ref{19}, the structure of the full propagator for the time-dependent system reveals itself. Let us use Equation \ref{19} to generate the first few lowest order terms in the propagator. Substituting Equation \ref{11} into Equation \ref{19} yields the lowest order contribution to $U (t ; t_0 )$:
$U^{(0)}(t;t_0) = e^{-iH_0(t-t_0)/\hbar}= U_0(t;t_0) \label{20}$
Thus, at zeroth order, Equation \ref{20} implies that the system is to be propagated using the unperturbed propagator $U_0 (t ; t_0 )$ as if the perturbation did not exist. At first order, we obtain
\begin{align} U^{(1)}(t;t_0) &= e^{-iH_0(t-t_0)/\hbar}- {i\over\hbar}e^{-iH_0(t-t_0)/\hbar}\int_{t_0}^t\;dt'\;H_I(t') \[4pt] &= e^{-iH_0(t-t_0)/\hbar}- {i \over \hbar}e^{-iH_0(t-t_0)/\hbar}\int_{t_0}^t\;dt'e^{-iH_0(t'-t_0)/\hbar}{\rm H}_1(t')e^{-iH_0(t'-t_0)/\hbar} \[4pt] &= e^{-iH_0(t-t_0)/\hbar}- {i \over \hbar}\int_{t_0}^t\;dt'\;e^{-iH_0(t-t')/\hbar}{\rm H}_1(t')e^{-iH_0(t'-t_0)/\hbar} \[4pt] &= U_0(t;t_0) - {i \over \hbar}\int_{t_0}^t\;dt'\;U_0(t;t'){\rm H}_1(t')U_0(t';t_0) \label{21} \end{align}
where, in the second line, the definition of $H_I(t)$ in terms of the original perturbation Hamiltonian $H_1 (t)$ has been used. What Equation \ref{21} says is that at first order, the propagator is composed of two terms. The first term is simply the unperturbed propagation from ${t_0}$ to $t$. In the second term, the system undergoes unperturbed propagation from ${t_0}$ to $t'$ and at $t'$, the perturbation $H_1 (t')$ is allowed to act. From $t'$ to $t$, the system undergoes unperturbed propagation. Finally, we need to integrate over all possible intermediate times $t'$.
In a similar manner, it can be shown that up to second order, the full propagator is given by
$U^{(2)} (t; t_0 ) = U_0(t;t_0) - {i \over \hbar}\int_{t_0}^t\;d'\;U_0(t;t'){\rm H}_1(t')U_0(t';t_0) + \left({i \over \hbar}\right)^2\int_{t_0}^t\;dt'\;\int_{t_0}^{t'}\;dt''\;U_0(t;t'){\rm H}_1(t')U_0(t';t''){\rm H}_1(t'')U_0(t'';t_0) \label{22}$
Thus, at second order, the new term involves unperturbed propagation from ${t_0}$ to $t''$, action of $H_1 (t'')$ at $t''$, unperturbed propagation from $t''$ to $t'$, action of $H_1 (t')$ at $t'$ and, finally, unperturbed propagation from $t'$ to $t$. Again, the intermediate times $t'$ and $t''$ must be integrated over. The picture on the left side of the equation indicates that the perturbation causes the system to undergo some undetermined dynamical process between ${t_0}$ and $t$. The terms on the right show how that process is broken down in terms of the action of the perturbation $H_1$ at specific intermediate times. At the $k$ th order, the perturbation Hamiltonian $H_1$ acts on the system at $k$ specific instances in time. Because of the limits of integration, these time instances are ordered chronologically.
The specific ordering of the instances in time when $H_1$ acts on the unperturbed system raises an important point. At each order the expansion for $U_I (t; t_0 )$, the order in which the operators $H_I (t')$, $H_I (t'')$, etc. are multiplied is important. The reason for this is that the operator $H_I (t)$ does not commute with itself at different instances in time
$\left[H_I(t),H_I(t')\right]\neq 0 \label{23}$
Thus, in order to remove any possible ambiguity when specifying the order in which operators are to be applied in a time series, we introduce the time-ordering operator, $T$. The purpose of $T$ is to take a product string of time-dependent operators $A(t_1)B(t_2)C(t_3)\cdots D(t_n)$ which act at different instances in time ${t_1} , {t_2}, \cdots , {t_n}$ and order the operators in the product such that they act chronologically in time from the earliest time to the latest time. For example, the action of $T$ on two operators $A (t_1)$ and $B (t_2)$ is
$T \left ( A (t_1) B (t_2) \right ) = \begin{cases} A(t_1) B(t_2) & t_2 < t_1 \ B (t_2) A (t_1) & t_1 < t_2 \end {cases} \label{24}$
Let us now apply the time-ordering operator to the second-order term. First write the double integral as a sum of two terms generated simply interchanging the names of the dummy variables $t'$ and $t''$:
$\int_{t_0}^t\;dt'\int_{t_0}^{t'}\;dt''\;H_I(t')H_I(t'') ={1 \over 2} \left [ \int_{t_0}^t dt' \int_{t_0}^{t'} dt'' H_I (t') H_I (t'') + \int_{t_0}^t\;dt''\int_{t_0}^{t''}\;dt'\;H_I(t'')H_I(t')\right] \label{25}$
The same region can be covered by choosing $t'\in[t_0,t]$ and $t''\in[t_0,t]$. With this choice, Equation \ref{25} becomes
$\int _{t_0}^t dt' \int _{t_0}^{t'} dt'' H_I (t') H_I (t'' ) = {1 \over 2 } \left [ \int_{t_0}^t dt' \int_{t_0}^{t'} dt'' H_I (t') H_I (t'') + \int_{t_0}^t\;dt'\int_{t'}^{t}\;dt''\;H_I(t'')H_I(t')\right] \label{26}$
In the first term on the right side of Equation \ref{26}, $t'' < t'$ and $H_I (t'' )$ acts first, followed by $H_I (t')$. In the second term, $t' < t''$ and $H_I (t')$ acts first followed by $H_I (t'' )$. The two terms can, thus, be combined with both $t'$ and $t''$ lying in the interval $[ t_0 , t]$ if the time-ordering operator is applied:
$\int_{t_0}^t\;dt'\int_{t_0}^{t'}\;dt''\;H_I(t')H_I(t'') ={1\over 2} \int_{t_0}^t\;dt'\int_{t_0}^{t}\;dt''\;T\left(H_I(t')H_I(t'')\right) \label{27}$
The same analysis can be applied to each order in Equation \ref{16}, recognizing that the number of possible time orderings of a product of $k$ operators is $k!$. Thus, Equation \ref{16} can be rewritten in terms of the time-ordering operator as
$U_I(t;t_0) = \sum_{k=0}^{\infty}\left(-1\right)^k \left({i \over \hbar } \right ) ^k {1 \over k!} \int _{t_0}^t dt_1 \int _{t_0}^t dt_2 \cdots \int_{t_0}^t\;dt_kT\left(H_I(t_1)H_I(t_2)\cdots H_I(t_k)\right) \label{28}$
The sum in Equation \ref{28} resembles the power-series expansion of an exponential, and, indeed, we can write the sum symbolically as
$U_I(t;t_0) = T\left[\exp\left(-{i \over \hbar}\int_{t_0}^t\;dt'\;H_I(t')\right)\right] \label{29}$
which is known as a time-ordered exponential. Equation \ref{29} is really a symbolic representation of Equation \ref{28}, in which it is understood that the time-ordering operator acts to order the operators in each term of the expansion of the exponential.
Given the formalism of time-dependent perturbation theory, we now seek to answer the following question: If the system is initially in an eigenstate of $H_0$ with energy $E_i$, what is the probability as a function of time $t$ that the system will undergo a transition to a new eigenstate of $H_0$ with energy $E_f$? From the statement of the question, it is clear that the initial state vector $\vert \Psi (t_0) \rangle$ is simply the eigenstate of $H_0$ with energy $E_i$
$\vert\Psi(t_0)\rangle = \vert E_i\rangle \label{30}$
The amplitude as a function of time that the system will undergo a transition to the eigenstate $\vert E_f\rangle$ is obtained by propagating this initial state out to time $t$ with the propagator $U (t ; t_0 )$ and then taking the overlap of the resultant state with the eigenstate $\vert E_f\rangle$:
$A_{fi}(t) = \langle E_f\vert U(t;t_0)\vert E_i\rangle \label{31}$
and the probability is just the square magnitude of this complex amplitude:
$P_{fi}(t) = \left\vert\langle E_f\vert U(t;t_0)\vert E_i\rangle\right\vert^2 \label{32}$
Consider, first, the amplitude at zeroth order in perturbation theory. At this order, $U (t ; t_0 ) = U_0 (t; t_0 )$, and the amplitude is simply
\begin{align} A^{(0)}_{fi}(t) &= \langle E_f\vert e^{-iH_0(t-t_0)/\hbar}\vert E_i\rangle \[4pt] &= e^{-iE_i(t-t_0)/\hbar}\langle E_f\vert E_i\rangle\label{33} \end{align}
which clearly vanishes if $E_i \ne E_f$. Thus, at zeroth order, the only possibility is the trivial one in which no transition occurs.
The lowest nontrivial order is first order, where the transition amplitude is given by
\begin{align} {A^{(1)}_{fi}(t)} &=\langle E_f\vert U^{(1)}(t;t_0)\vert E_i\rangle = -{i \over \hbar}\int_{t_0}^t\;dt'\;\langle E_f\vert U_0(t;t'){\rm H}_1(t')U_0(t';t_0)\vert E_i\rangle \[4pt] &=-{i \over \hbar}\int_{t_0}^t\;dt\;\langle E_f\vert e^{-iH_0(t-t')/\hbar}{\rm H}_1(t')e^{-iH_0(t'-t_0)/\hbar}\vert E_i\rangle \[4pt] &= -{i \over \hbar}\int_{t_0}^t\;dt'\;e^{-iE_f(t-t')/\hbar}e^{-iE_i(t'-t_0)/\hbar}\langle E_f\vert{\rm H}_1(t')\vert E_i\rangle \[4pt] &= -{i \over \hbar}e^{-iE_ft/\hbar}e^{iE_it_0/\hbar}\int_{t_0}^t\;dt'\;e^{i(E_f-E_i)t'/\hbar}\langle E_f\vert{\rm H}_1(t')\vert E_i\rangle \label{34} \end{align}
Define a transition frequency ${\omega _{fi} }$ by
$\omega_{fi} = {E_f-E_i \over \hbar} \label{35}$
Then, taking the absolute square of the last line of Equation \ref{34}, we obtain the probability at first-order
$P^{(1)}_{fi}(t) = {1 \over \hbar^2}\left\vert\int_{t_0}^t\; dt' e^{i\omega _{fi} t'}\langle E_f\vert{\rm H}_1(t')\vert E_i\rangle\right\vert^2 \label{36}$
At first order, the probability depends on the matrix element of the perturbation between the initial and final eigenstates. Thus far, the formalism we have derived is valid for any perturbation Hamiltonian $H_1 (t)$. If we consider the use of an external perturbation to probe the eigenvalue spectrum of $H_0$, then the specific type of probe determines the form of $H_1 (t)$, as we saw in the first section and will explore in the next subsection. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/13%3A_Time-dependent_Processes_-_Quantum_Case/13.02%3A_Iterative_solution_for_the_interaction-picture_state_vector.txt |
Consider a quantum system described by a time-dependent Hamiltonian of the form
$H(t) = H_0 + H_1(t) \nonumber$
In the language of perturbation theory, $H_0$ is known as the unperturbed Hamiltonian and describes a system of interest such as a molecule or a condensed-phase sample such as a pure liquid or solid or a solution. $H_1(t)$ is known as the perturbation, and it often describes an external system, such as a laser field, that will be used to probe the energy levels and other properties of $H_0$.
We now seek a solution to the time-dependent Schrödinger equation
${\rm H}(t)\vert\Psi(t)\rangle = \left({\rm H}_0 + {\rm H}_1(t) \right ) \vert \Psi (t) \rangle = i\hbar{\partial \over \partial t}\vert\Psi(t)\rangle \label{1}$
subject to an initial state vector $| \Psi(t_0) \rangle$. In order to solve the equation, we introduce a new state vector $| \Phi(t) \rangle$ related to $| \Psi(t) \rangle$ by
$\vert\Psi(t)\rangle = e^{-i{\rm H}_0(t-t_0)}\vert\Phi(t)\rangle \label{2}$
The new state vector $| \Phi(t) \rangle$ is an equally valid representation of the state of the system. In Chapter 10, we introduced the concept of pictures in quantum mechanics and discussed the difference between the Schrödinger and Heisenberg pictures. Equation \ref{2} represents yet another picture of quantum mechanics, namely the interaction picture. Like the Schrödinger and Heisenberg pictures, the interaction picture is a perfectly valid way of representing a quantum mechanical system. The interaction picture can be considered as "intermediate'' between the Schrödinger picture, where the state evolves in time and the operators are static, and the Heisenberg picture, where the state vector is static and the operators evolve.
However, as we will see shortly, in the interaction picture, both the state vector and the operators evolve in time, however, the time-evolution is determined by the perturbation $H_1(t)$. Equation \ref{2} specifies how to transform between the Schrödinger and interaction picture state vectors. The transformation of operators proceeds in an analogous fashion. If $A$ denotes an operator in the Schrödinger picture, its representation in the interaction picture is given by
${\rm A}_I(t) = e^{i{\rm H}_0(t-t_0)/\hbar}{\rm A}e^{-i{\rm H}_0(t-t_0)/\hbar} \label{3}$
which is equivalent to an equation of motion of the form
${d{\rm A}_I(t) \over dt} = {1 \over i\hbar}[{\rm A}_I(t),{\rm H}_0] \label{4}$
Substitution of Equation \ref{2} into the time-dependent Schrödinger equation yields
\begin{align} \left({\rm H}_0+{\rm H}_1(t)\right)e^{-i{\rm H}_0(t-t_0)/\hbar}\vert\Phi(t)\rangle &= {\rm H}_0e^{-i{\rm H}_0(t-t_0)/\hbar}\vert\Phi(t)\rangle + e^{-i{\rm H}_0(t-t_0)/\hbar}i\hbar{\partial \over \partial t}\vert\Phi(t)\rangle \[4pt] {\rm H}_1(t)e^{-iH_0(t-t_0)/\hbar}\vert\Phi(t)\rangle &= e^{-iH_0(t-t_0)/\hbar}i\hbar{\partial \over \partial t}\vert\Phi(t)\rangle \[4pt] e^{iH_0(t-t_0)/\hbar}{\rm H}_1(t)e^{-iH_0(t-t_0)/\hbar}\vert\Phi(t)\rangle &=i\hbar{\partial \over \partial t}\vert\Phi(t)\rangle \label{5} \end{align}
According to Equation \ref{3}, the $\exp[i{\rm H}_0(t-t_0)/\hbar]{\rm H}_1(t)\exp[-i{\rm H}_0(t-t_0)/\hbar]$ is the interaction-picture representation of the perturbation Hamiltonian, and we will denote this operator as $H_I (t)$. Thus, the time-evolution of the state vector in the interaction picture is given a Schrödinger equation of the form
$H_I(t)\vert\Phi(t)\rangle = i\hbar{\partial \over \partial t}\vert\Phi(t)\rangle \label{6}$
The initial condition to Equation \ref{6}, $\vert\Phi(t_0)\rangle$ is, according to Equation \ref{2}, also $\vert\Phi(t_0)\rangle$. In the next section, we will develop an iterative solution to Equation \ref{6}, which will reveal a rich structure of the propagator for time-dependent systems.
13.04: Fermi's Golden Rule
In the first section, we saw how to formulate the Hamiltonian of a material system coupled to an external electromagnetic field. Moreover, we obtained solutions for the electromagnetic field in the absence of sources or physical boundaries, namely, solutions of the free-field wave equations. In this chapter, we will focus primarily on weak fields. We will also focus on a class of experiments in which the wavelength of electromagnetic radiation is taken to be long compared to the size of the sample under investigation. In this case, the spatial dependence of the electromagnetic field can also be neglected, since
$\cos({\bf k}\cdot{\bf r}-\omega t + \varphi_0) ={\rm Re} \{\exp(i{\bf k}\cdot{\bf r}- i\omega t + \varphi_0)\} \nonumber$
and
$\exp(i{\bf k}\cdot{\bf r})\approx 1 \nonumber$
in the long-wavelength limit. In this case, it is sufficient to consider $H_1 (t)$ to be of the general form
${\rm H}_1(t) = -{\mathcal V}F(\omega)e^{-i\omega t} \label{37}$
where ${\mathcal V}$ is a Hermitian operator.
Although we could use $\sin$ and $\cos$ to express the perturbation, the form in Equation $\ref{37}$ is a particularly convenient one, and since we will be seeking probabilities of transitions, the results we obtain will be real in the end.
Again, the question we seek to answer is given this form for the perturbation, what is the probability that the material system will be excited from an initial eigenstate $\vert E_i \rangle$ with energy $E_i$ to a final state $\vert E_f \rangle$ with energy $E_f$? However, since the perturbation is periodic in time, what we really seek to know is if the perturbation is applied over a long time interval, what is the probability per unit time or rate at which transitions will occur. Thus, in order to make the calculation somewhat easier, let us consider a time interval $T$ and choose $t_0 = - T/2$ and $t = T/2$. At first order, the transition rate ${ R^{(1)}_{fi}(T) }$ is just the total probability ${P^{(1)}_{fi}(T) }$ divided by the interval length $T$:
\begin{align} R^{(1)}_{fi}(T) &= {P^{(1)}_{fi}(T) \over T} \[4pt] &={1 \over T\hbar ^2} \vert F (\omega ) \vert ^2 \left\vert \int _{-T/2}^{T/2} e^{i (\omega _{fi} - \omega )t } dt \right\vert ^2\vert \langle E_f \vert {\mathcal V} \vert E_i\rangle \vert^2 \label{38} \end{align}
For finite $T$, the integral can be carried out explicitly yielding
$\int_{-T/2}^{T/2}e^{i(\omega_{fi}-\omega)t}dt = {\sin(\omega_{fi}-\omega)T/2 \over(\omega_{fi}-\omega)/2} \label{39}$
Thus, the transition rate can be expressed as
$R^{(1)}_{fi}(T) = {1 \over \hbar^2}T\vert F(\omega)\vert^2\vert \langle E_f \vert {\mathcal V} \vert E_i \rangle \vert ^2 { \sin^2(\omega_{fi}-\omega)T/2 \over [(\omega_{fi}-\omega)T/2]^2} \label{40}$
In the limit of $T$ very large, this expression becomes highly peaked only if ${\omega _{fi} = \omega }$. Otherwise, as $T \rightarrow \infty$, the expression vanishes. The condition ${\omega _{fi} = \omega }$ is equivalent to the condition $E_f = E_i + \hbar \omega$, which is a statement of energy conservation. Since $\hbar \omega$ is the energy quantum of the electromagnetic field, the transition can only occur if the energy of the field is exactly "tuned'' for the the transition, and this "tuning'' depends on the frequency of the field. In this way, the frequency of the field can be used as a probe of the allowed transitions, which then serves to probe the eigenvalue structure of $H_0$.
Now, let us consider the $T \rightarrow \infty$ more carefully. We shall denote the rate in this limit simply as $R_{fi}$. In this limit, the integral becomes
\begin{align} \lim_{T\rightarrow\infty}\int_{-T/2}^{T/2}e^{-i(\omega_{fi}-\omega)t}dt &= \int_{-\infty}^{\infty}e^{i(\omega_{fi}-\omega)t}dt \[4pt] &= 2\pi\delta(\omega_{fi}-\omega) \[4pt] &= (2\pi\hbar\delta(E_f-E_i-\hbar\omega) \label{41} \end{align}
Therefore, the expression for the rate in this limit can be written as
\begin{align} R_{fi}(\omega) &= \lim_{T\rightarrow\infty} {P^{(1)}_{fi}(T) \over T}=\lim_{T\rightarrow \infty} {1 \over T\hbar ^2} \left \vert \int _{-T/2}^{T/2} e^{i(\omega _{fi} - \omega )t} dt \right\vert ^2 \left \vert F (\omega ) \right\vert^2 \vert\langle E_f\vert{\mathcal V}\vert E_i\rangle \vert^2 \[4pt] &= \lim_{T\rightarrow\infty}{1 \over T\hbar^2}\left[\int_{-T/2}^{T/2} e^{-i (\omega _{fi} - \omega) t} dt \right ] \left[\int_{-T/2}^{T/2} e^{i (\omega _{fi} - \omega) t} dt \right ]\vert F (\omega )\vert^2 \vert\langle E_f\vert{\mathcal V}\vert E_i\rangle \vert^2 \label{42} \end{align}
where we have dropped the "(1)'' superscript (it is understood that the result is derived from first-order perturbation theory), and indicate explicitly the dependence on the frequency $\omega$. When one the first integral is replaced by the $\delta$-function, the remaining integral becomes simply $T$, which cancels the $T$ in the denominator. Thus, the expression for the rate is finally
$R_{fi}(\omega) = {2\pi \over \hbar}\vert F(\omega)\vert^2\vert \langle E_f \vert {\mathcal V} \vert E_i \rangle \vert ^2 \delta (E_f-E_i-\hbar\omega) \label{43}$
which is known as Fermi's Golden Rule. It states that, to first-order in perturbation theory, the transition rate depends only the square of the matrix element of the operator ${\mathcal V}$ between initial and final states and includes, via the $\delta$-function, an energy-conservation condition. We will make use of the Fermi Golden Rule expression to analyze the application of an external monochromatic field to an ensemble of systems in order to derive expressions for the observed frequency spectra.
13.05: Quantum Linear Response Theory
Consider again the Hamiltonian for a system coupled to a time-dependent field
$H = H_0 - BF_e(t) \nonumber$
We wish to solve the quantum Liouville equation
$i\hbar {\partial \rho \over \partial t} = [H,\rho] \nonumber$
in the linear regime where $F_e(t)$ is small. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/13%3A_Time-dependent_Processes_-_Quantum_Case/13.03%3A_The_Interaction_Picture.txt |
In the theory of chemical reactions, it is often possible to isolate a small number or even a single degree of freedom in the system that can be used to characterize the reaction. This degree of freedom is coupled to other degrees of freedom (for example, reactions often take place in solution). Isomerization or dissociation of a diatomic molecule in solution is an excellent example of this type of system. The degree of freedom of paramount interest is the distance between the two atoms of the molecule - this is the degree of freedom whose detailed dynamics we would like to elucidate. The dynamics of the "bath'' or environment to which is couples is less interesting, but still must be accounted for in some manner. A model that has maintained a certain level of both popularity and success is the so called "harmonic bath'' model, in which the environment to which the special degree(s) of freedom couple is replaced by an effective set of harmonic oscillators. We will examine this model for the case of a single degree of freedom of interest, which we will designate ${q}$. For the case of the isomerizing or dissociating diatomic, ${q}$ could be the coordinate $r - \langle r \rangle$, where $r$ is the distance between the atoms. The particular definition of ${q}$ ensures that $\langle q \rangle = 0$. The degree of freedom ${q}$ is assumed to couple to the bath linearly, giving a Hamiltonian of the form
$H = {p^2 \over 2m} + \phi(q) + \sum_{\alpha}\left[{p_{\alpha}^2 \over 2m_{\alpha} } + {1 \over 2} m_{\alpha} \omega _{\alpha}^2 \left (x_{\alpha} + {g_{\alpha}\over m_{\alpha}\omega_{\alpha}^2}q\right)^2\right] \nonumber$
where the index $\alpha$ runs over all the bath degrees of freedom, ${\omega _{\alpha} }$ are the harmonic bath frequencies, ${m_{\alpha } }$ are the harmonic bath masses, and ${g_{\alpha} }$ are the coupling constants between the bath and the coordinate ${q}$. ${p}$ is a momentum conjugate to ${q}$, and $m$ is the mass associated with this degree of freedom (e.g., the reduced mass ${\mu}$ in the case of a diatomic). The coordinate ${q}$ is assumed to be subject to a potential $\phi (q)$ as well (e.g., an internal bond potential). The form of the coupling between the system ( ${q}$ ) and the bath (${x_{\alpha}}$) is known as bilinear.
Below, using a completely classical treatment of this Hamiltonian, we will derive an equation for the detailed dynamics of ${q}$ alone. This equation is known as the generalized Langevin equation (GLE).
14.02: The Random Force Term
Within the context of a harmonic bath, the term "random force'' is something of a misnomer, since $R (t)$ is completely deterministic and not random at all!!! We will return to this point momentarily, however, let us examine particular features of $R (t)$ from its explicit expression from the harmonic bath dynamics. Note, first of all, that it does not depend on the dynamics of the system coordinate ${q}$ (except for the appearance of $q (0)$). In this sense, it is independent or "orthogonal'' to ${q}$ within a phase space picture. From the explicit form of $R(t)$, it is straightforward to see that the correlation function
$\langle \dot{q}(0)R(t)\rangle = 0 \nonumber$
i.e., the correlation function of the system velocity ${\dot {q} }$ with the random force is 0. This can be seen by substituting in the expression for $R (t)$ and integrating over initial conditions with a canonical distribution weighting. For certain potentials $\phi (q)$ that are even in ${q}$ (such as a harmonic oscillator), one can also show that
$\langle q(0)R(t)\rangle = 0 \nonumber$
Thus, $R (t)$ is completely uncorrelated from both ${q}$ and ${\dot {q}}$, which is a property we might expect from a truly random process. In fact, $R(t)$ is determined by the detailed dynamics of the bath. However, we are not particularly interested or able to follow these detailed dynamics for a large number of bath degrees of freedom. Thus, we could just as well model $R(t)$ by a completely random process (satisfying certain desirable features that are characteristic of a more general bath), and, in fact, this is often done. One could, for example, postulate that $R (t)$ act over a maximum time ${t_{max}}$ at discrete points in time $k \Delta t$, giving $N=t_{\rm max}/\Delta t$ values of $R_k=R(k\Delta t)$, and assume that $R_k$ takes the form of a gaussian random process:
$R_k = \sum_{j=1}^N\left[a_je^{2\pi ijk/N} + b_je^{-2\pi ijk/N}\right] \nonumber$
where the coefficients $\{a_j\}$ and $\{b_j\}$ are chosen at random from a gaussian distribution function. This might be expected to be suitable for a bath of high density, where strong collisions between the system and a bath particle are essentially nonexistent, but where the system only sees feels the relatively "soft'' fluctuations of the less mobile bath. For a low density bath, one might try modeling $R (t)$ as a Poisson process of very strong collisions.
Whatever model is chosen for $R (t)$, if it is a truly random process that can only act at discrete points in time, then the GLE takes the form of a stochastic (based on random numbers) integro-differential equation. There is a whole body of mathematics devoted to the properties of such equations, where heavy use of an calculus is made. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/14%3A_The_Langevin_and_Generalized_Langevin_equations/14.01%3A_The_harmonic_bath_Hamiltonian.txt |
The convolution integral term
$\int_0^t\;d\tau\;\dot{q}(\tau)\zeta(t-\tau) \nonumber$
is called the memory integral because it depends, in general, on the entire history of the evolution of ${q}$. Physically it expresses the fact that the bath requires a finite time to respond to any fluctuation in the motion of the system (${q}$). This, in turn, affects how the bath acts back on the system. Thus, the force that the bath exerts on the system presently depends on what the system coordinate ${q}$ did in the past. However, we have seen previously the regression of fluctuations (their decay to 0) over time. Thus, we expect that what the system did very far in the past will no longer the force it feels presently, i.e., that the lower limit of the memory integral (which is rigorously 0) could be replaced by ${t-t_{\rm mem}}$, where ${t _{mem} }$ is the maximum time over which memory of what the system coordinate did in the past is important. This can be interpreted as a indicating a certain decay time for the friction kernel $\zeta(t)$. In fact, $\zeta(t)$ often does decay to 0 in a relatively short time. Often this decay takes the form of a rapid initial decay followed by a slow final decay, as shown in the figure below:
Consider the extreme case that the bath is capable of responding infinitely quickly to changes in the system coordinate ${q}$. This would be the case, for example, if there were a large mass disparity between the system and the bath ( ${m \gg m_{\alpha} }$). Then, the bath retains no memory of what the system did in the past, and we could take $\zeta(t)$ to be a $\delta$-function in time:
$\zeta(t) = 2\zeta_0\delta(t) \nonumber$
Then
$\int_0^t\;d\tau\;\dot{q}(\tau)\zeta(t-\tau) =\int_0^t\;d\tau \dot {q} (t - \tau ) \zeta (\tau) = 2 \zeta _0 \int _0^t\;d\tau\;\delta(\tau)\dot{q}(t-\tau) = \zeta_0\dot{q}(t) \nonumber$
and the GLE becomes
$m\ddot{q} = -{\partial \phi \over \partial q} - \zeta_0\dot{q} + R(t) \nonumber$
This simpler equation of motion is known as the Langevin equation and it is clearly a special case of the more generalized equation of motion. It is often invoked to describe Brownian motion where clearly such a mass disparity is present. The constant $\zeta _0$ is known as the static friction and is given by
$\zeta_0 = \int_0^{\infty}\;dt\;\zeta(t) \nonumber$
In fact, this is a general relation for determining the static friction constant.
The other extreme is a very sluggish bath that responds slowly to changes in the system coordinate. In this case, we may take $\zeta (t)$ to be a constant $\zeta \equiv \zeta(0)$, at least, for times short compared to the response time of the bath. Then, the memory integral becomes
$\int_0^t\;d\tau\;\dot{q}(\tau)\zeta(t-\tau) \approx \zeta(q(t)-q(0)) \nonumber$
and the GLE becomes
$m\ddot{q} = -{\partial \over \partial q}\left(\phi(q) +{1 \over 2}\zeta (q-q_0)^2\right) + R(t) \nonumber$
where the friction term now manifests itself as an extra harmonic term added to the potential. Such a term has the effect of trapping the system in certain regions of configuration space, an effect known as dynamic caging. An example of this is a dilute mixture of small, light particles in a bath of heavy, large particles. The light particles can get trapped in regions of space where many bath particles are in a sort of spatial 'cage.'' Only the rare fluctuations in the bath that open up larger holes in configuration space allow the light particles to escape the cage, occasionally, after which, they often get trapped again in a new cage for a similar time interval.
14.04: Relation between the Dynamic Friction Kernel and the Random Force
From the definitions of $R(t)$ and $\zeta(t)$, it is straightforward to show that there is a relation between them of the form
$\langle R(0)R(t)\rangle = kT\zeta(t) \nonumber$
This relation is known as the second fluctuation dissipation theorem. The fact that it involves a simple autocorrelation function of the random force is particular to the harmonic bath model. We will see later that a more general form of this relation exists, valid for a general bath. This relation must be kept in mind when introducing models for $R(t)$ and \zeta(t). In effect, it acts as a constraint on the possible ways in which one can model the random force and friction kernel. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/14%3A_The_Langevin_and_Generalized_Langevin_equations/14.03%3A_The_Dynamic_Friction_Kernel.txt |
The GLE can be derived from the harmonic bath Hamiltonian by simply solving Hamilton's equations of motion, which take the form
${\dot{q} } = {P \over m} \nonumber$
${\dot{p}} = {-{\partial \phi \over \partial q} - \sum_{\alpha}g_{\alpha}x_{\alpha}-\sum_{\alpha}{g_{\alpha}^2 \over m_{\alpha}\omega_{\alpha}^2}q } \nonumber$
${\dot{x}_{\alpha} } = {P_{\alpha} \over m_{\alpha }} \nonumber$
${\dot{p}_{\alpha} } = {-m_{\alpha}\omega_{\alpha}^2x_{\alpha}- g_{\alpha}q } \nonumber$
This set of equations can also be written as second order differential equation:
${m\ddot{q} }$ =
${-{\partial \phi \over \partial q} - \sum_{\alpha}g_{\alpha}x_{\alpha}-\sum_{\alpha}{g_{\alpha}^2 \over m_{\alpha}\omega_{\alpha}^2}q}$
${m_{\alpha}\ddot{x}_{\alpha} }$ =
${-m_{\alpha}\omega_{\alpha}^2x_{\alpha}- g_{\alpha}q }$
In order to derive an equation for ${q}$, we solve explicitly for the dynamics of the bath variables and then substitute into the equation for ${q}$. The equation for ${x_{\alpha}}$ is a second order inhomogeneous differential equation, which can be solved by Laplace transforms. We simply take the Laplace transform of both sides. Denote the Laplace transforms of ${q}$ and ${x_{\alpha}}$ as
$\tilde{q}(s)$ = $\int_0^{\infty}\;dt\;e^{-st}q(t)$
${\tilde{x}_{\alpha} }$ = $\int_0^{\infty}\;dt\;e^{-st}x_{\alpha}(t)$
and recognizing that
$\int_0^{\infty}\;dt\;e^{-st}\ddot{x}_{\alpha}(t) = s^2\tilde{x}_{\alpha}(s) - sx_{\alpha}(0) -\dot{x}_{\alpha}(0) \nonumber$
we obtain the following equation for $\tilde{x}_{\alpha}(s)$:
$(s^2 + \omega_{\alpha}^2)\tilde{x}_{\alpha}(s) = sx_{\alpha} (0) + \dot{x}_{\alpha}(0) - {g_{\alpha}\over m_{\alpha}}\tilde{q}(s) \nonumber$
or
$\tilde{x}_{\alpha}(s) = {s \over s^2 + \omega_{\alpha}^2}x_{\alpha}(0) + {1 \over s^2 + \omega _{\alpha}^2} \dot {x} _{\alpha} (0) - {g_{\alpha} \over m_{\alpha}}{\tilde{q}(s) \over s^2 + \omega_{\alpha}^2} \nonumber$
$x_{\alpha} (t)$can be obtained by inverse Laplace transformation, which is equivalent to a contour integral in the complex $s$-plane around a contour that encloses all the poles of the integrand. This contour is known as the Bromwich contour. To see how this works, consider the first term in the above expression. The inverse Laplace transform is
${1 \over 2\pi i}\oint\;ds\;{se^{st} \over s^2 + \omega_{\alpha}^2} = {1 \over 2 \pi i} \oint \;ds\;{se^{st} \over (s+i\omega_{\alpha})(s-i\omega_{\alpha})} \nonumber$
The integrand has two poles on the imaginary $s$-axis at $\pm i \omega _{\alpha}$. Integration over the contour that encloses these poles picks up both residues from these poles. Since the poles are simple poles, then, from the residue theorem:
${1 \over 2\pi i}\oint\;ds\;{se^{st} \over (s+i\omega_{\alpha} )(s - i\omega _{\alpha})} = {1 \over 2 \pi i} \left [ 2 \pi i \left ({i\omega_{\alpha}e^{i\omega _{\alpha}t } \over 2i\omega _{\alpha} } + {-i\omega_{\alpha}e^{i\omega _{\alpha}t } \over -2i\omega_{\alpha}}\right)\right] =\cos\omega_{\alpha}t \nonumber$
By the same method, the second term will give $(\sin\omega_{\alpha}t)/\omega_{\alpha}$. The last term is the inverse Laplace transform of a product of $q (S) \tilde{q}(s)$ and $1/(s^2+\omega_{\alpha}^2)$. From the convolution theorem of Laplace transforms, the Laplace transform of a convolution gives the product of Laplace transforms:
$\int_0^{\infty}\;dt\;e^{-st}\int_0^t\;d\tau\;f(\tau)g(t-\tau) =\tilde{f}(s)\tilde{g}(s) \nonumber$
Thus, the last term will be the convolution of $q (t)$ with $(\sin\omega_{\alpha}t)/\omega_{\alpha}$. Putting these results together, gives, as the solution for $x_{\alpha} (t)$:
$x_{\alpha}(t) = x_{\alpha}(0)\cos\omega_{\alpha}t + {\dot{x} _{\alpha} (0) \over \omega _{\alpha} } \sin \omega _{\alpha} t - {g_{\alpha} \over m_{\alpha}\omega_{\alpha}}\int_0^t d\tau q(\tau)\sin\omega_{\alpha}(t-\tau) \nonumber$
The convolution term can be expressed in terms of ${\dot{q}}$ rather than ${q}$ by integrating it by parts:
${g_{\alpha}\over m_{\alpha}\omega_{\alpha}}\int_0^t\;d\tau\;q (\tau) \sin \omega _{\alpha} (t - \tau ) = {g_{\alpha} \over m_{\alpha} \omega _{\alpha}^2 } \left [ q (t) - q (0) \cos \omega _{\alpha} t \right ] - {g_{\alpha} \over m_{\alpha} \omega _{\alpha}^2}\int_0^t\;d\tau\;\dot{q}(\tau)\cos\omega_{\alpha}(t-\tau) \nonumber$
The reasons for preferring this form will be made clear shortly. The bath variables can now be seen to evolve according to
$x_{\alpha}(t) = x_{\alpha}(0)\cos\omega_{\alpha}t + {\dot{x} (0) \over \omega _{\alpha}} \sin \omega_{\alpha} t + {g_{\alpha} \over m_{\alpha} \omega _{\alpha}^2}\int_0^t\;d\tau\;\dot{q}(\tau)\cos\omega_{\alpha}(t-\tau) - {g_{\alpha} \over m_{alpha}\omega_{\alpha}^2}\left[q(t)-q(0)\cos\omega_{\alpha}t\right] \nonumber$
Substituting this into the equation of motion for ${q}$, we find
$m\ddot{q} = -{\partial \phi \over \partial q} -\sum_{\alpha} g_{\alpha} \left [ x_{\alpha} (0) \cos \omega _{\alpha} t + {P_{\alpha} (0) \over m_{\alpha} \omega _{\alpha} } \sin \omega _{\alpha} t + {g _{\alpha} \over m_{\alpha} \omega _{\alpha}^2 } q(0) \cos \omega _{\alpha} t \right ] - \sum _{\alpha} {g^2_{\alpha} \over m_{\alpha} \omega ^2_{\alpha} } \int _0^t d \tau \dot {q} (\tau) \cos \omega_{\alpha} (t - \tau ) \nonumber$
$+ \sum _{\alpha} {g_{\alpha}^2 \over m_{\alpha}\omega_{\alpha}^2}q(t) - \sum _{\alpha} {g_{\alpha}^2 \over m_{\alpha}\omega_{\alpha}^2}q(t)$
We now introduce the following notation for the sums over bath modes appearing in this equation:
1.
Define a dynamic friction kernel $\zeta(t) = \sum_{\alpha}{g_{\alpha}\over m_{\alpha}\omega_{\alpha}^2}\cos\omega_{\alpha}t \nonumber$
2.
Define a random force $R(t) = -\sum_{\alpha}g_{\alpha}\left[\left(x_{\alpha}(0) + {g_{\alpha} \over m_{\alpha} \omega_{\alpha}^2 } q (0) \right ) \cos \omega_{\alpha} t + {P_{\alpha} (0) \over m_{\alpha}\omega_{\alpha}}\sin\omega_{\alpha}t\right] \nonumber$
Using these definitions, the equation of motion for ${q}$ reads
$m\ddot{q} = -{\partial \phi \over \partial q} - \int_0^t\;d\tau\;\dot{q}(\tau)\zeta(t-\tau)+ R(t) \nonumber$ (1)
Eq. (1) is known as the generalized Langevin equation. Note that it takes the form of a one-dimensional particle subject to a potential $\phi (q)$, driven by a forcing function $R (t)$ and with a nonlocal (in time) damping term ${-\int_0^t\;d\tau\;\dot{q}(\tau)\zeta(t-\tau) }$, which depends, in general, on the entire history of the evolution of ${q}$. The GLE is often taken as a phenomenological equation of motion for a coordinate ${q}$ coupled to a general bath. In this spirit, it is worth taking a moment to discuss the physical meaning of the terms appearing in the equation. | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/14%3A_The_Langevin_and_Generalized_Langevin_equations/14.05%3A_Derivation_of_the_GLE.txt |
A derivation of the GLE valid for a general bath can be worked out. The details of the derivation are given in the book by Berne and Pecora called Dynamic Light Scattering. The system coordinate ${q}$ and its conjugate momentum ${p}$ are introduced as a column vector:
${\textbf A} = \left(\matrix{q \cr p}\right) \nonumber$
and, in addition, one introduces statistical projection operators $P$ and $Q$ that project onto subspaces in phase space parallel and orthogonal to $A$. These operators take the form
$P = {\langle ...{\textbf A}^{\rm T}\rangle \langle {\textbf A}{\textbf A}^{\rm T}\rangle ^{-1} } \nonumber$
$Q = I-P \nonumber$
These operators are Hermitian and satisfy the property of idempotency:
${P^2} = P \nonumber$
${Q^2} = Q \nonumber$
Also, note that
$P {\textbf A} ={\textbf A} \nonumber$
$Q {\textbf A} = {0} \nonumber$
The time evolution of ${\textbf A}$ is given by application of the classical propagator:
${\textbf A}(t) = e^{iLt}{\textbf A}(0) \nonumber$
Note that the evolution of ${\textbf A}$ is unitary, i.e., it preserves the norm of ${\textbf A}$:
$\vert{\textbf A}(t)\vert^2 = \vert{\textbf A}(0)\vert^2 \nonumber$
Differentiating both sides of the time evolution equation for ${\textbf A}$ gives:
${dA \over dt} = e^{iLt} iL{\bf A}(0) \nonumber$
Then, an identity operator is inserted in the above expression in the form $I = P + Q$:
${dA \over dt} = e^{iLt}(P+Q)iL{\textbf A}(0) = e^{iLt}PiL{\textbf A}(0) + e^{iLt}QiL{\textbf A}(0) \nonumber$
The first term in this expression defines a frequency matrix acting on ${\textbf A}$:
${e^{iLt}PiL{\textbf A}(0)}$ $=$
${e^{iLt}\langle iL{\textbf A}{\textbf A}^{\rm T}\rangle\langle {\textbf A}{\textbf A}^{\rm T}\rangle^{-1}{\textbf A} }$
$=$
${\langle iL{\textbf A}{\textbf A}^{\rm T}\rangle\langle {\textbf A}{\textbf A}^{\rm T}\rangle^{-1}e^{iLt}{\textbf A} }$
$=$
${\langle iL{\textbf A}{\textbf A}^{\rm T}\rangle\langle {\textbf A}{\textbf A}^{\rm T}\rangle^{-1}{\textbf A}(t) }$
$\textstyle \equiv$
$i{\bf \Omega}{\textbf A}(t)$
where
${\bf\Omega} = \langle L{\bf A}{\bf A}^{\rm T}\rangle\langle {\bf A}{\bf A}^{\rm T}\rangle^{-1} \nonumber$
In order to evaluate the second term, another identity operator is inserted directly into the propagator:
$e^{iLt} = e^{i(P+Q)Lt} \nonumber$
Consider the difference between the two propagators:
$e^{iLt} - e^{iQLt} \nonumber$
If this difference is Laplace transformed, it becomes
$(s-iL)^{-1} - (s-iQL)^{-1} \nonumber$
which can be simplified via the general operator identity:
${\rm A}^{-1} - {\rm B}^{-1} = {\rm A}^{-1}({\rm B}-{\rm A}){\rm B}^{-1} \nonumber$
Letting
$A = (s - iL ) \nonumber$
$B = (s - iQL ) \nonumber$
we have
${ (s-iL)^{-1}-(s-iQL)^{-1} }$ $=$ ${(s-iL)^{-1}(s-iQL - s + iL)(s-iQL)^{-1} }$
$=$ ${(s-iL)^{-1}iPL(s-iQL)^{-1} }$
or
$(s-iL)^{-1} = (s-iQL)^{-1} + (s-iL)^{-1}(s-iQL - s + iL)(s-iQL)^{-1} \nonumber$
Now, inverse Laplace transforming both sides gives
$e^{iLt} = e^{iQLt} + \int_0^t\;d\tau\;e^{iL(t-\tau)}iPLe^{iQL\tau} \nonumber$
Thus, multiplying fromthe right by $QiL{\textbf A}$ gives
$e^{iLt}QiL{\bf A}= e^{iQLt}QiL{\bf A}+\int_0^t\;d\tau\;e^{iL(t-\tau)}iPLe^{iQL\tau}QiL{\bf A} \nonumber$
Define a vector
${\bf F}(t) = e^{iQLt}QiL{\bf A}(0) \nonumber$
so that
$e^{iLt}QiL{\bf A}= {\bf F}(t) +\int_0^t\;d\tau\;\langle iL {\bf F} (\tau) {\bf A }^T \rangle \langle {\bf A}{\bf A}^{\rm T}\rangle^{-1} {\bf A}(t-\tau) \nonumber$
Because ${\textbf F} (t)$ is completely orthogonal to ${\textbf A} (t)$, it is straightforward to show that
$Q{\bf F}(t) = {\bf F}(t) \nonumber$
Then,
${\langle iL{\bf F}(\tau){\bf A}^{\rm T}\rangle\langle{\bf A}{\bf A}^{\rm T}\rangle^{-1}{\bf A} }$
$=$ ${\langle iLQ{\bf F}(\tau){\bf A}^{\rm T}\rangle\langle{\bf A}{\bf A}^{\rm T}\rangle^{-1}{\bf A} }$
$=$ ${-\langle Q{\bf F}(\tau)(iL{\bf A})^{\rm T}\rangle\langle{\bf A}{\bf A}^{\rm T}\rangle^{-1}{\bf A} }$
$=$ ${-\langle Q^2{\bf F}(\tau)(iL{\bf A})^{\rm T}\rangle\langle{\bf A}{\bf A}^{\rm T}\rangle^{-1}{\bf A} }$
$=$
${-\langle Q{\bf F}(\tau)(QiL{\bf A})^{\rm T}\rangle\langle{\bf A}{\bf A}^{\rm T}\rangle^{-1}{\bf A} }$
$=$
${-\langle {\bf F}(\tau){\bf F}^{\rm T}(0)\rangle\langle{\bf A}{\bf A}^{\rm T}\rangle^{-1}{\bf A} }$
Thus,
$e^{iLt}QiL{\bf A}= {\bf F}(t) - \int_0^t\;d\tau\;\langle {\bf F} (\tau){\bf F}^T (0) \rangle \langle {\bf A}{\bf A}^{\rm T}\rangle^{-1}{\bf A}(t-\tau) \nonumber$
Finally, we define a memory kernel matrix:
${\bf K}(t) = \langle {\bf F}(\tau){\bf F}^{\rm T}(0)\rangle\langle{\bf A}{\bf A}^{\rm T}\rangle^{-1} \nonumber$
Then, combining all results, we find, for ${d {bf A} \over dt }$:
${d{\bf A}\over dt} = i{\bf\Omega}(t){\bf A}-\int_0^t\;d\tau\;{\bf K}(\tau){\bf A}(t-\tau) + {\bf F}(t) \nonumber$
which equivalent to a generalized Langevin equation for a particle subject to a harmonic potential, but coupled to a general bath. For most systems, the quantities appearing in this form of the generalized Langevin equation are
$i{\bf\Omega}$ $=$ ${\bf K} (t)$
${\bf F} (t)$ $=$ ${\bf K} (t)$
$\left(\matrix{0 \cr R(t)}\right)$ $=$ $\phi (q) = {m \omega ^2 q^2 \over 2}$
It is easy to derive these expressions for the case of the harmonic bath Hamiltonian when
$\langle R(0)R(t)\rangle = \langle R(0)e^{iLt}R(0)\rangle = kT\zeta(t) \nonumber$
For the case of a harmonic bath Hamiltonian, we had shown that the friction kernel was related to the random force by the fluctuation dissipation theorem:
$exp (iQLt ) \nonumber$
For a general bath, the relation is not as simple, owing to the fact that ${\textbf F} (t)$ is evolved using a modified propagator $\langle R(0)e^{iQLt}R(0)\rangle = kT\zeta(t)$. Thus, the more general form of the fluctuation dissipation theorem is
$\langle R(0)e^{iQLt}R(0)\rangle \approx \langle R(0)e^{iL_{\rm cons}t}R(0)\rangle \nonumber$
so that the dynamics of $R (t)$ is prescribed by the propagator $\langle R(0)e^{iQLt}R(0)\rangle = kT\zeta(t)$. This more general relation illustrates the difficulty of defining a friction kernel for a general bath. However, for the special case of a stiff harmonic diatomic molecule interacting with a bath for which all the modes are soft compared to the frequency of the diatomic, a very useful approximation results. One can show that
$iL_{cons} \nonumber$
where $C_{vv} (t) = {\langle \dot {q} (0) \dot {q} (t) \rangle \over \langle \dot {q} ^2 \rangle }$ is the Liouville operator for a system in which the diatomic is held rigidly fixed at some particular bond length (i.e., a constrained dynamics). Since the friction kernel is not sensitive to the details of the internal potential of the diatomic, this approximation can also be used for diatomics with stiff, anharmonic potentials. This approximation is referred to as the rigid bond approximation (see Berne, et al, J. Chem. Phys. 93, 5084 (1990)). | textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Advanced_Statistical_Mechanics_(Tuckerman)/14%3A_The_Langevin_and_Generalized_Langevin_equations/14.06%3A_Mori-Zwanzig_Theory-_A_more_general_derivation_of_the_GLE.txt |
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