Exam #2 went quite well. You can pick up your exams from the box outside my office and download selected answers.

I apologize for the long delay in returning this exam. You know the main reason why everything got held up, but I regret the delay just the same.

A couple of comments beyond those in the answer sheet:

• The first step in problem #1 is quite difficult to think about. The other steps are more straightforward and everyone dealt with them more reliably. You are still having some problems with labels. Cycloadditions get [m+n], sigmatropic rearrangements get [m,n], and electrocyclizations get DIS/CON (and maybe number of electrons).
• The first step in problem #2 is nearly impossible to think about. Well, I exaggerate. You can think about it, but it is too difficult for an exam problem. Sorry. Interesting note: most of the remaining steps did not need to be analyzed because they duplicated steps in problem #1, but most of you analyzed these steps just the same. So it goes.
• Only one person built all four transition state models correctly. Two people built two models correctly. Two people were stumped or misled by the models that they had (which weren’t transition state models). I was hoping that the last problem on HW #8 would prepare everyone adequately for this exam problem, but I guess not. Is there something I could have done differently that would have helped?

Exam #2 (3 page PDF). Due Monday, April 27, at 4 pm in my mailbox or my office (slip your exam under my door). Notice that you must also email models to me for one problem so that I can examine them.

Ground rules are the same as those for exam #1.

Since we didn’t have a full class today, and since several important items were discussed, I’m sending this out to everyone so that you have a clear idea of where we are going in the next two weeks.

1. Final project.

According to the instructions for the final project, I was supposed to meet with each of you this week to discuss the research paper you had chosen and your plan of attack. The fact is, aside from one very brief conversation, I haven’t met with anyone yet. I would like everyone to use this weekend to find a paper, formulate a hypothesis, and outline a series of calculations that would test this. It’s okay to be ambiguous about some of these steps. It’s also okay to have two or three papers. I can help you choose. But I would like you to do some planning before you come meet with me.

I would like to meet with each of you between Monday and Wednesday of next week.

2. Exam #2.

This is a take-home exam. It was supposed to be released today and returned next Friday. I would rather have you focus on your final project this weekend so I am going to delay the deadline for returning the exam. You can expect the exam to come out sometime between now and Monday, and it will be due the following Monday, April 27.

3. Teaching evaluations.

Your evaluations of this course are incredibly important to me. I’m proud to say that all of you have nearly perfect attendance records and nearly perfect participation in assignments. (This is the only course at Reed that I have been connected with where I can make either of these statements.) Therefore, I think all of you are in an excellent position to consider carefully what has been presented this semester and furnish a thoughtful evaluation. Unfortunately, we have only one more scheduled class meeting, a modeling session on Monday.

I don’t want to make you come to class just to fill out teaching evaluations. So here is what I will do instead: I will hand out the teaching evaluation forms on Monday. I will rely on you to fill these out on your own time and return them to Cameron Kellett’s thesis office, no later than noon on Wednesday. He will take all of the forms that he has received and return them to the registrar’s office (thank you, Cameron!). If you miss that deadline, please turn your form in to the registrar’s office directly.

4. Spending some Vollum funds.

If you’ve looked at our weekly schedule online, you will have noticed that I designated one day as “breakfast with Alan”. I’d love to do that, but many of you have 9 AM classes and can’t get free for breakfast. So let’s do lunch. We can either try to find a day on which everyone can make it or we can go in two waves. Either way, lunch is on me courtesy of the generous Vollum family.

As far as I know, I am free for lunch Monday, Wednesday, and Friday next week and the week after. So talk among yourselves and find a day (or days) that you would like to go (maybe pick a destination too) and let me know. I will coordinate my schedule with yours.

A previous post described how the ability of a molecule to absorb light can be expressed using a transition dipole moment integral (TDMI) between two quantum mechanical states, the ground state and the excited state. One part of the TDMI involved integration over electron spin coordinates and I want to explore the “spin integral” more deeply.

Practically all ground state organic molecules are closed shell species. This means that every spin-up electron is paired with a spin-down electron and these electron pairs move in the same way (occupy the same orbital). When we say spin we are using a shortcut for the electron’s intrinsic angular momentum. This means I could have also started out by saying: practically all ground state organic molecules have a total (electronic) angular momentum of zero.

It turns out that the angular momentum determines the molecule’s magnetic properties.  Since the total angular momentum is zero, the molecule is diamagnetic and the ground state is non-degenerate. This means the multiplicity of any zero angular momentum state is one. For convenience, we refer to all of these states as singlets.

The ground state, which is the lowest energy singlet, is represented by the symbol S₀. Excited singlet states with zero angular momentum are also singlets and they are represented by the symbol S_i, where i tells us the state’s position relative to the ground state, e.g., i = 1 is the first singlet above the ground state, i = 2 is the second singlet above the ground state, and so on.

Now, looking at the excited state with electron configuration (HO)¹(LU)¹, we find four different wave functions. One wave function is non-degenerate (singlet!) and has zero total angular momentum. This wave function represents the S₁ state. The other three wave functions have the same electron distribution and energy in the absence of an external magnetic field. They are components of a triply degenerate state or triplet, also called T₁ (i = 1 because this is the first triplet state above the ground state). The three wave functions in the triplet state differ from each other in only one way: the z-component of the angular momentum is different for each wave function.

If you have taken quantum mechanics, you have learned that spin angular momentum can be described by two operators, one for the total angular momentum and another for the z-component of the angular momentum. You has also learned that each operator has a unique quantum number associated with it. The three triplet wave functions have the same quantum number for the total angular momentum, but different quantum numbers for the z-component of the angular momentum. (The situation is analogous to the one that describes the angular momentum of an electron in a p orbital. The l quantum number refers to the total orbital angular momentum and the m_l quantum number refers to the z-component.)

Finally, to return to the TDMI, we find that transitions are allowed only between states of the same total angular momentum (or same total angular momentum quantum number). So a molecule might be able to jump from S_i to S_k, but the leap from S_i to T_k and its reverse, are forbidden.

Today’s lecture contained three slides dealing with different aspects of transition probabilities: molecular geometry, electron spin, and orbital shape. I don’t think I made it very clear, however, how these aspects relate to one another. Let’s see if I can fix that.

First, let’s return to our basic postulate: photochemistry requires a molecule to absorb a photon. If we describe light absorption using the Beer-Lambert law, the molecule must have a fairly large molar extinction coefficient.

Quantum mechanics describes light absorption using different terminology. From this perspective we ask, what is the probability that a photon can cause a system to undergo a transition between state #1 and state #2? The standard answer (see McQuarrie, “Quantum Chemistry 2e”, p. 404-9) is that the probability is related to something called a transition dipole moment integral (TDMI).

Like all quantum mechanical integrals, the TDMI involves two wave functions, one for state #1 and one for state #2, and a quantum mechanical operator. Actually, it’s even simpler than that. The TDMI is an integral over the product of these entities:

TDMI = integral over [wave function (state #1) x operator x wave function (state #2)]

If you have the necessary background, I encourage you to take a look at the actual mathematics (it’s not that bad), but I’m going to try and simplify things even more.

Since we are dealing with an integral it makes sense to look at the variables that we must integrate over. In this case, we must integrate over three separate kinds of coordinates: 1) nuclear spatial coordinates, 2) electron spatial coordinates, and 3) electron spin coordinates. A common approximation that we make is to assume a molecular wave function can be factored into a product of three functions, each of which depends on a different kind of coordinate:

wave function (state #i) = f(nuclear spatial coordinates) x g(electron spatial coordinates) x h(electron spin coordinates)

If this approximation is valid, the integral involving a wave function can also be factored:

TDMI = integral #1 x integral #2 x integral #3

where the three integrals are over nuclear spatial coordinates (#1), electron spatial coordinates (#2), and electron spin coordinates (#3), respectively.

This brings us to the heart of the matter. If any one of these integrals is small or vanishes, the TDMI will also be small or vanish, and the molecule won’t absorb a photon. The probability for absorbing the photon will be high only if all three integrals are large for the transition in question.

The conditions needed to make each integral large were described in class, but I didn’t use the word “integral” so let me continue.

My picture of vibration states inside a ground state surface and an excited state surface was supposed to convey information about integral #1. For this integral to be large, the two vibration states must have large values at the same geometry.

My picture of HOMO and LUMO locations was supposed to convey information about integral #2. For this integral to be large, the two orbitals must have large values in the same region of space (a necessary, but not sufficient condition).

And, my slide on S and T states refers to integral #3. These states have different total spin angular momenta. If you prefer, they have different S² quantum numbers. The integrals between states of different total spin angular momenta vanish.

Bottom-line: the TDMI is a product of three integrals, each of which refers to a different part of the wave function. If any of the functions associated with the ground and excited states makes one of these integrals vanish, light absorption won’t occur.

I started my presentation of photochemistry in the last 10 minutes of today’s lecture – a questionable decision. Apparently, I had just enough time to say some really confusing things before it was time to stop talking completely. After chatting with Cameron after lecture, I thought it might be a good idea to try and set the record straight for everyone.

The two points that I want to review are 1) how thermal energy excites a molecule for chemical reaction, and 2) how photochemical energy excites a molecule for chemical reaction.

Thermal excitation. A big problem here is that we have some deeply engrained mental images of thermally excited molecules and they are not necessarily compatible with each other. They are not necessarily incompatible either, but we rarely stop to examine them so let’s take a look.

One image I have is of little ball-like molecules shooting around and rapidly bouncing off of each other. Something like an angry swarm of bees or billiard balls (can billiard balls get angry?). A second and completely different image I have is of a single molecule with all of its atoms are jiggling around like mad. Do either of these sound familiar?

My images clearly represent two different kinds of energy. The “shooting-and-bouncing” image reflects translational or kinetic energy. The “jiggle” image reflects vibrational energy. The images may seem different, but they have something crucial in common: both involve moving nuclei. This means that my images must agree on one thing. When we boost a molecule’s thermal energy, we make the nuclei move faster.

Now how do we transfer this idea, the idea of fast-moving nuclei, to a reaction energy diagram? One way is to show increased movement along the horizontal axis, i.e., along the reaction coordinate. This is commonly done by showing a ball rolling over the hills in the diagram. This view is quite logical. We know the reaction coordinate reflects nuclear positions so, if the nuclei are moving, they are changing their positions and it stands to reason that this will move the molecule along the reaction coordinate.

But wait. What about motion perpendicular to the reaction coordinate? We talked about perpendicular axes in connection with kinetic isotope effects. That discussion revealed that the reaction coordinate shows only one kind of change in nuclear position.

And what about vibration? Whenever we talk about harmonic oscillators and IR spectroscopy we draw a parabolic or quadratic energy diagram and we show the vibrational energy states as horizontal lines inside the diagram. The lines get longer as we move to higher energy states which seems to suggest the molecule can explore more geometries, but the center of the line, the average geometry, doesn’t change.

Let me narrow the discussion a bit. Let’s consider a unimolecular reaction, say a sigmatropic rearrangement or an electrocyclic ring closure. Two things are necessary for this kind of reaction to occur: the reactant molecule must have the right geometry and it must have the right amount of energy. Because the reaction is unimolecular, the energy must be entirely vibrational (a bimolecular reaction, on the other hand, requires translational energy too). In other words, the reactant must be in some excited vibrational energy state comparable in energy to the transition state.

How does a reactant molecule get into this excited vibrational energy state? Two ways. First, by colliding with other molecules. Second, if the molecule is already excited, by shuttling energy between vibration modes. This is incredibly efficient, but it is also totally random. For example, if our reactant is ethanol and a collision excites the OH stretch, we will find that the excitation of this mode is temporary. The energy associated with the OH bond will quickly be redistributed to other vibrations.

Returning to our reaction energy diagram, it seems reasonable to think of the “energy well” near the reactant geometry as having a vibration mode associated with it with many energy states. Excitation means activating, concentrating energy in, this mode. This is probably a rather sudden process and I have drawn it below as a vertical transition, i.e., a change in energy with no change in geometry, but this is terribly simplistic. To be honest, I don’t know how energy collects in this mode. (Click image for full-size.)

Once the reaction coordinate “mode” is excited, two processes compete: decay (downward vertical transition, dashed arrow) and reaction (large horizontal transition). The reaction carries the molecule to a new geometry without changing its energy, so one more decay process must occur (downward vertical transition, solid arrow) to “finish” the reaction.

Photochemical excitation. The excitation process here is quite simple. A photon is absorbed and a new excited electronic state is created. We can safely assume that this occurs without any change in molecular geometry and we draw the excitation as a vertical transition. At this point, the picture becomes more complicated, but I will set these complications aside for another time. The two points of importance are these: 1) the excited state’s energy surface looks nothing like the ground state surface, and 2) once again we must consider a competition between decay and reaction. The tendency to react is governed by the shape of the excited state surface, and can be thought of as a thermal excitation that follows the initial photochemical excitation, so “reaction” is not just a single horizontal transition. The decay process is also more complicated (it’s not as simple as colliding with another molecule), so I’ll just stop here and give you a rough diagram to look at (click image for full-size).

Bottom-line: Thermal reactions and photochemical reactions can both be viewed as excitation-geometry change-decay sequences. Only the nuclei are excited in a thermal reaction, while both electrons and nuclei are excited in a photochemical reaction.

What do you think about that?

A great discussion today of paper #6 – Synthesis of Tertiary α-Hydroxy Acids by Silylene Transfer to α-Keto Esters, OL, 2007, 4651. It was fun to have so many participants and so many ideas.

The take-home lessons stressed the importance of:

• asking yourself hard questions about unfamiliar reaction mechanisms (you know which questions to ask)
• acquainting yourself with unfamiliar technical terms (a frequent problem with organometallics because the reagents are not organic)
• not taking stereochemistry on faith

Going a bit farther with the third point …

the authors’ provided a 3-D drawing of the transition state geometry for the Ireland-Claisen rearrangement that could not possibly lead to the products that they obtained. Some of you were able to work this out for yourselves. How did you do it? Did you use drawings only? Don’t forget that our problem-solving toolkit includes the ability to make transition state models and follow stereochemistry from reactants to products (that was HW #1, if memory serves).

The two chair transition states one might see starting from compound 13 are shown below. The one on the left looks like the drawing that the authors provided, but it is completely incompatible with the product that they obtained. To get from 13 to the actual product, the atoms must pass through the transition state shown on the right. Notice that this structure puts all of the allylic substituents in pseudo-equatorial positions. (click on image to expand).

transition state models, compd 13, paper 6

(First posted Mar 25, 2008) After class, I built some models (B3LYP/6-31G*) of Claisen rearrangement transition states. The chair transition state is ~4 kcal/mol more stable than the boat when ZPE correction is included (ΔZPE ~0.3 kcal/mol). This is a slightly smaller energy difference than between chair and boat cyclohexane. Therefore, there will be situations where substituent effects, like ring strain or steric hindrance, will make the boat geometry more stable than the chair. You cannot count on the chair geometry always being lower in energy (and remember that there are always two chairs to consider).

Images of the chair (left) and boat (right) models are shown below. Oxygen (red) is coming towards you.

One structural motif that can be seen in both models are quasi-planar three-atom pi systems: OCHCH2 on the left and H2CCHCH2 on the right. The pi systems appear to occupy parallel planes in the chair (left). The pi planes are not quite parallel in the boat (right); the so-called flagpole atoms (the middle atoms in each allylic fragment) tilt away from each other.

Here are the same models, chair (left) and boat (right), from another perspective. Perhaps it is easier to see the arrangement of the pi systems more easily?

Bond distances are given for several bonds (the chair distances don’t match the ones given in class because a different electronic model has been used) and you should spend some time thinking about them. Points of interest:

• The CC partial double bonds in the upper allyl fragment are not quite the same length. Try to persuade yourself that these CC distances are more consistent with a reactant-like transition state.

• The pi systems are closer together in the chair transition state (compare partial single bond distances). Does something draw the systems together in the chair geometry? Or perhaps push them apart in the boat geometry? This will be the subject of another post. Stay tuned.

Homework #6 instructs you to “include zero-point energy (ZPE) corrections” in your calculated energies, both the barrier and the reaction energy. Judging from some questions that have been sent by email and by comment (C. Bailey, 3/22), there are some mysterious parts to this procedure.

First, you need to build three models and obtain their energies. Second, you use the differences between these energies to estimate the reaction energy and barrier. Third, to get a molecule’s “energy”, you need to combine its “total” energy with its “zero-point” energy. We haven’t discussed the procedure for generating or using ZPE so let’s patch this hole right now:

1. Setup and submit a ZPE calculation
2. Find the results of a ZPE calculation
3. Use the results of a ZPE calculation, i.e., combine ZPE with the total energy

Setup and submit. ZPE is defined as the vibration energy of a molecule at absolute zero, 0 K. Therefore, you need to calculate your molecule’s vibration frequencies and then convert these into the molecule’s ZPE. Both steps can be accomplished simply by checking two boxes in the Setup: Calculations window: 1) Compute: IR (this causes frequencies to be calculated), and 2) Print: Thermodynamics (this causes ZPE and several other useful energy corrections to be calculated). After that, Submit.

Find the results. The results of a ZPE calculation can be extracted in two ways. The more traditional way is to examine the output file (Display: Output). If you scroll way down in the Output window, you will see that the Spartan Properties program lists the desired output in tabular form, starting with the line “Standard Thermodynamics quantities at”. The following figure shows output for methane (HF/3-21G model, no scaling of frequencies).

Thermochemistry output

ZPE (126.0070 kJ/mol) can be read directly from the table (red circle). FYI – the Enthalpy column to the right of ZPE (blue circle) contains additional corrections that account for warming the substance from 0 K to 298.15 K. Notice that ZPE is a much larger piece of energy pie for this molecule. ZPE is also more sensitive to the details of molecular structure than other kinds of corrections so ΔZPE can really affect reaction energies and barriers. Other energy corrections have much smaller effects.

Another way to find your molecule’s ZPE is to use Spartan’s spreadsheet tool. Open a spreadsheet and add a ZPE column. That’s it. Choose the energy units that you like best, kJ/mol or kcal/mol (30.1164 kcal/mol for methane).

Use the results. I discovered a really nifty handout that I gave last year’s 324 class and it is still online. It tells you exactly what you need to know about ZPE, as well as other energy corrections. Download it here and check it out (I can’t believe that I forgot to send it around earlier).

If you are still confused, here’s a hint: calculate E(0 K) (= Etotal + ZPE) for each molecule and combine these to calculate the 0 K reaction energy and barrier. And, for you skeptics who think that calculating a reaction energy and barrier for absolute zero is a total crock (nothing reacts at 0 K, brother!), calculate H(298 K) for each molecule and use these values to calculate a reaction enthalpy and activation enthalpy that can actually be compared with experimental measurements.