3:30, Tu – Lauren

11:00, W – Harry

1:30, W – Stephen

3:30, W – Alex

The exam will be held in my office unless another room proves more convenient.

Scheme 1. The drawing of transition state 9 is inconsistent with the drawings of 10 and 7. As the old man said to the tourists, “you can’t get there from here.”

Minor point: the structure of 10 is ambiguous because the configuration of the quarternary C isn’t spelled out in Scheme 1, but the reaction of 10 with HF shouldn’t affect this C so it is reasonable to assume that same configurations for 10 and 7. Once you make this assumption, you should find that if you push forward from 9, you will end up with the mirror image of 7. Or, if you choose to work backwards from 7, you have to find that it requires the mirror image of 9.

Scheme 1 (partial)

Switching enantiomers on the reader is very annoying, but not evil incarnate. Although 9 is a chiral transition state, it is derived from an achiral reactant so 7 must be obtained as a racemic mixture. Still, if the intent of the drawing is to help the reader understand the reaction, this is a bad drawing. The authors should have made 9, 10, and 7 internally consistent.

Scheme 2. Since 13 is drawn in the same orientation as 6 (the precursor to 9/10/7) and Scheme 1 is the only “explanation” in the paper, one is sorely tempted to invoke Scheme 1 again. Obviously, as we saw above, this will produce the wrong configuration at both chiral carbons in 14. Relying on Scheme 1 is less forgivable here because the authors report >97% ee for the reaction.  They owe us a scheme that fits the stereochemistry from start to finish.

But the problem turns out to be even worse because Scheme 1 predicts the wrong diastereomer. Notice that the methyl group in 13 becomes an alkene substituent in 14? If we push 13 through the sequence in Scheme 1, the methyl group occupies a pseudo-axial position in the transition state and a cis alkene is the product. 14 is trans. There is simply no way to account for the conversion of 13 to 14 using Scheme 1.

How can we explain the conversion of 13 to 14? It turns out that the “ring flip conformer” of 9 (not the mirror image of 9) can account for the observed reaction. If we flip the ring, we not only change the configuration at the chiral atoms that are being generated in 14, we also move the methyl group into a pseudo-equatorial position which yields a trans alkene. So, when it comes down to it, it is possible to account for the chemistry that is observed, but Scheme 1 is a useless tool for doing this. One wonders if the authors even used this drawing to think about their chemistry or if its contents were just a slip of the mouse.

The two chair transition states discussed above for 13 are shown below. The one on the left looks like the chair in Scheme 1. The one on the right is the “ring flip conformer”. (Note: the caption says ‘paper 6′ because this was paper 6 in 2009.)

transition state models, compd 13, paper 6

First, any compound’s structure is fair game as an exam question, i.e., if a paper refers to a compound by name or by formula or by abbreviation, a careful reader should care enough to figure out what the structure is. I’m not recommending that you draw every compound that a paper refers to (see third point below), but you should know how to draw every compound.

Second, reactions are often described in a paper because the reactions are what the paper is about. However, if a reaction is not described, it’s because the authors assume a careful reader will know this information or will get the necessary information from an appropriate source. Two reactions in this paper deserve a deeper look than what the authors provided:

• Reaction #1 – the transfer of silylene (tBu2Si) from a silacyclopropane to an α-keto ester. The reaction is catalyzed by AgOTs. The mechanism is not described in this paper (because this is not what this paper is about). A careful reader (like you) is either expected to know this reaction (which seems rather unlikely in this case) or get the information from an appropriate source. In this case, that source would be ref. 4. You are expected to look at every reference that appears to contain vital information.
• Reaction #2 – the Ireland-Claisen that lies at the heart of Monday’s assignment. I think all of you copied the drawings from the paper to your sheet, but no one mentioned that it showed an impossible transformation or that the configurations of certain atoms weren’t provided. You shouldn’t expect to find errors like this in an exam paper, but you should be testing each of the ideas in a paper out in your own mind, i.e., you should be enough of a critical reader, that you can catch errors if they are present.

Third, notes should serve a purpose. They should lighten the mental load that you carry. They should never make your load heavier.

• Too many notes are a burden. Notes that duplicate information provided elsewhere are a burden. Notes should be short.
• Notes that are incomplete, or hard to read, or hard to fathom in the heat of an exam are a burden. Notes should be easy to read and easy to use.
• Anything that you figure out, but will have a hard time bringing to mind, is a burden. Good notes that describe these kinds of items can lighten your mental load during the exam.

• First, examine formulas 9, 10 and 7 in Scheme 1. If the transition state geometry is really the one shown in 9, will 10 and 7 be obtained? If the Scheme contains geometrical inconsistencies, accept the stereochemistry of 7 as correct and work backwards. What should 10 look like? What should 9 look like?
• Second, examine the formulas of 13 and 14 in Scheme 2. This transformation is more complicated because 13 contains a chiral center. Draw a chair transition state that is consistent with this transformation.  Can 9 account for this transformation?

Due Monday. Suggestion: if you are having trouble visualizing these structures, use models. You should be able to figure out how to build a basic Claisen transition state using SPARTAN and you can decorate it from there.

First, a couple of reminders about FMO analysis:

• identify both FMO interactions and then identify the dominant one
• regioselectivity depends on transition state geometry, which depends on orbital overlap, so you need to give me orbital coefficients for the important FMO (note: I didn’t list the coefficients in the answer key, but I looked at them when positioning the * symbols for problem #1)
• the HOMO and LUMO are not always the “FMO”

Wait, aren’t the HOMO and LUMO always the frontier MOs by definition? Yes, they are always the FMO, but adhering to that definition is unwise if either the HOMO or LUMO has the wrong shape. Always use the “frontier” MO that will guide the reaction. If the reaction is pericyclic, you must pay attention to the kinds of bonds that are being broken and find the orbitals that correspond to these bonds. If you’re breaking a π bond, you need to find a π-type MO. All of you fell into this “trap” on problems #1 and #3.

Second, one of you pointed out that, if the reactants in problem #2 (3a, 3b, 3c) are always at equilibrium, the Curtin-Hammett principle will apply. Yes! I totally forgot about that when I wrote my answer key. I am going to go home and eat crow (while thinking admiring thoughts about the intelligence of Reedies). Well done!

This principle has been recognized by scientists for generations. Just think of Pasteur looking at chiral crystals of tartaric acid in his microscope, or of Galileo looking at Jupiter through his telescope, or of Pauling looking at molecular models of alpha-helices and beta-sheets. The list of important discoveries that were inspired by looking goes on and on, yet in our haste to complete assignments and produce publishable results, the idea of patient looking, of looking again, and of looking even more, strikes us as a luxury that no productive member of society can afford, and no modern multitasking individual can sit still for.

Here’s a story about two scientists that I came across in a book in the Reed College library called, amazingly enough, “One Dharma” (Ch. 7) by Joseph Goldstein:

Louis Agassiz was a Swiss-born American naturalist who helped train his students in careful observation. The following is a story told about Samuel Scudder, one of his students:

[Agassiz] intended, he said, to teach the student to see — to observe and compare — and he intended to put the burden of study on them. Probably he never said what he is best known for: “Study nature, not books,” or not in those exact words. But such certainly was the essence of his creed, and for his students the idea was firmly implanted by what they would refer to as “the incident of the fish.”

His initial interview at an end, Agassiz would ask the student when he would like to begin. If the answer was now, the student was immediately presented with a dead fish — usually a very long-dead, pickled, evil-smelling specimen, personally selected by the “master” from one of the wide-mouthed jars that lined his shelves. The fish was placed before the student in a tin pan. He was to look at the fish, the student was told, whereupon Agassiz would leave, not to return until later in the day, if at all.

Samuel Scudder, one of the many from the school who would go on to do important work of their own (his in entomology), described the experience as one of life’s memorable turning points.

In ten minutes, I had seen all that could be seen in that fish…. Half an hour passed — an hour — another hour. The fish began to look loathesome. I turned it over and around: looked it in the face — ghastly; from behind, beneath, above, sideways, at three-quarters view — just as ghastly. I was in despair.

I might not use a magnifying glass; instruments of all kinds were interdicted. My two hands, my two eyes, and the fish; it seemed a most limited field. I pushed my finger down its throat to feel how sharp the teeth were. I began to count the scales in different rows until I was convinced that that was nonsense. At last, a happy thought struck me — I would draw the fish, and now with surprise, I began to discover new features in the creature.

When Agassiz returned later and listened to Scudder recount what he had observed, his only comment was that the young man must look again.

I was piqued; I was mortified. Still more of that wretched fish! But now I set myself to my task with a will, and discovered one new thing after another…. The afternoon passed quickly and toward its close, the professor inquired: “Do you see it yet?”

“No,” I replied. “I am certain I do not. But I see how little I saw before.”

The day following, having thought of the fish most of the night, Scudder had a brainstorm. The fish, he announced to Agassiz, had symmetrical sides with paired organs.

“Of course, of course,” Agassiz said, obviously pleased. Scudder asked what he might do next, and Agassiz replied, “Oh, look at your fish!”

In Scudder’s case, the lesson lasted a full three days. “Look, look, look,” was the repeated injunction, and the best lesson he ever had, Scudder recalled, “a legacy that professor has left me, as he has left it to many others, of inestimable value, which we could not buy, with which we cannot part.”

Converting published transition state models into SPARTAN’08 models

1. Go to the journal article’s download page: dx.doi.org/10.1021/ol061085x. Follow the link to the Supporting Information and save the PDF file, ol061085xsi20060607_101706.pdf (97 KB), to your computer.
2. Open the PDF file. Notice that the model numbers of the transition states are given in the paper and also in the supporting information. The coordinates for the model follow the model number. Find a model of interest and select just the lines of text that contain atom labels + atomic coordinates. Copy these lines of text (and only these lines) into an empty file and save them as text. (Windows users: Notepad is a good program for creating text files, but it will set the file extension to “.txt” which will need to be fixed in the next step.)
3. Rename the text file so that it has the extension “.xyz”. For example, you might name the file that contains the coordinates of transition state 7 as “ts7.xyz”. (If your computer is set up to hide file extensions, tell it to show you all file extensions before trying to set file extensions.)
4. Move the text file to the computer where you plan to use SPARTAN.
5. Open SPARTAN’08 and click File: Open. SPARTAN looks for .spartan and .spinput files by default so change the entry in the Files of type menu (located at the bottom of the Open window) to read All Files. Find your .xyz file, select it, click Open, and voilá.
6. You should see a complete transition state model, but only if you had copied all of the atomic coordinates into your text file. SPARTAN cannot tell you if your model is complete or not (GIGO) so carefully examine the model (and/or the text file) to make sure everything is there.
7. Repeat steps #2-#6 for a second transition state.

Examine the transition state’s geometry closely and identify the 2 or 3 most important structural motifs.

What is a structural motif? Let me give you a few examples: a geometry that indicates a hydrogen bond, a staggered conformation, a chair conformation, an α-helix, a planar conjugate system, an octahedral coordination environment, etc.. In other words, a structural motif refers to a geometrical pattern that shows that a set of atoms (not just one or two) have organized their positions in some recognizable way and/or according to some rule. It does not have to be a particular stable ensemble, e.g., a boat conformation and an eclipsed conformation are both relatively unstable, but they are still structural motifs.

I want you to look carefully at your transition state models for structural motifs. There will be many in each model so I want you to impress me, not with the number that you find, but with their novelty, subtlety, and significance. To give you a chance, I’ll accept 2 or 3 ideas for each transition state. Your job is to really LOOK and think, “what is trivial and what is really gold“? (Or, more to the point, what will Alan think is trivial/gold? How can I impress him? FYI – I am very hard to impress so really LOOK.) FYI2 – I have been  impressed by comparisons of structural motifs between models.

Orient your model so that the structural motifs can be seen easily, print your image, and attach some text that clearly describes the motifs.

Bring your images + text to class on Friday (Feb 19). If you prefer, send me an electronic document anytime ahead of class. Some hints for making this part work:

• SPARTAN gives you a number of tools for showing atom labels and changing model types in order to create the clearest image. Play with entries under the Model menu.
• SPARTAN also gives you some tools for changing colors used by the program. This includes atom colors, background color (white is a good choice for images that will be printed on paper), and so on. Play with Options: Colors.
• When you have an image that you like, click once in the SPARTAN window (somehow this updates SPARTAN‘s awareness of the image), then use copy-and-paste to put the image in your favorite word processor. You will probably have to crop and resize the image.
• Add some text describing your motif and selling me on its interesting character.
• Print the image + text using CutePDF Writer.
• Email it to me.

Yes, that’s exactly how I felt around 9 PM tonight after I had tried multiple times without success to generate a list of H2O2 conformations. Using a host of variations on the procedure that we had tried in class, I tried over and over again to get a list of 13 models with optimized geometries and evenly spaced dihedral angles, but it never worked. Just like you saw in class, I wound up with several models that failed to reach their equilibrium geometry and/or several models with the wrong dihedral angles. I was on the verge of quitting when it occurred to me: all of my attempts had started from one model. Maybe I needed to try something new right from the beginning?

In answering this question, I ultimately found an “automatic” method that worked, but not before I had tediously constructed each model by hand. Arrrgh.

The following paragraphs take you through the following: 1) an apparently reliable (but not completely straightforward) automatic method for calculating an energy profile for internal rotation in H2O2 (demonstrated using HF/3-21G), 2) a completely reliable (but tedious) manual method for generating an energy profile for the same rotation (demonstrated using B3LYP/6-31G**), and 3) some thoughts about the structure of H2O2. Enjoy.

Succeeding at automatic ‘energy profile’ calculations. Here is the procedure I followed:

1. Built H2O2
2. Defined dihedral constraint by selecting four atoms
3. Selected constraint (magenta changes to gold) and, using the Properties window, converted the static constraint into a 13 step dynamic constraint ranging from 180 to 0 degrees
4. Calculated an energy profile using Hartree-Fock/3-21G without symmetry (this gave optimized geometries for 12 of 13 models)
5. Calculated the equilibrium geometry of all 13 models subject to constraints (this gave optimized geometries for 13 of 13 models)

This procedure is nearly identical to the one that I demonstrated in class with two key differences. First, I removed the checkmark from the Symmetry box (step #4). This led to a significant improvement. Instead of ending with 4 or 5 partially optimized models with bad dihedral angles, the calculation ended with only one partially optimized model with an appropriate dihedral angle. Second, I re-opened Setup: Calculations and submitted an equilibrium geometry calculation (HF/3-21G) subject to constraints. This had no effect on the 12 models that had already been optimized, but it replaced the partially optimized model for the 13th model with a fully optimized one. Finally, I had obtained a complete list of fully optimized conformations!

Please pay close attention to the models used in, and the computational consequences of, steps #4 and #5. Step #4 is initiated using a single model that contains a dynamic constraint. The energy profile calculation produces a new list model containing 13 separate conformations and each model in the list is given a dihedral constraint with the appropriate angle.

Step #5, on the other hand, is initiated using the list model. The calculation can be set up with any member of the list. Just click Setup: Calculations to open the usual window and you will see the following: calculate equilibrium geometry using HF/3-21G subject to constraints. The Global Calculations box is also checked. There are three important differences between steps #4 and #5. First, energy profile (#4) vs. equilibrium geometry (#5). Second, optimization subject to constraints (#5). Third, a global calculation (#5). The last item is especially important because it makes sure that the same instructions are applied to every model in the list, not just the one that is currently selected. Therefore, the end result will be to optimize each model at its uniquely determined dihedral angle.

Succeeding with a (tedious) manual ‘energy profile’ calculation. As I described in class today, another way to assemble an energy profile is to build each model individually and calculate its equilibrium geometry subject to constraints. The following method works and is reasonably quick:

1. Build H2O2
2. Define dihedral constraint (180 degrees)
3. Setup: Calculations to calculate equilibrium geometry (at B3LYP/6-31G**) subject to constraints and submit
4. After calculation is complete, open Display: Properties and follow these steps
5. Edit: Copy
6. File: New Molecule
7. Edit: Paste
8. click “V” (View mode) button
9. select constraint and type new value in Properties window
10. Minimize
11. Setup: Calculations to calculate equilibrium geometry (at B3LYP/6-31G**) subject to constraints and submit
12. Repeat steps #5-#10 as needed with 12 new dihedral angles

There is nothing automatic about this procedure. You must “build” all 13 models and attach dihedral constraints to each one, but copy-and-paste makes this go fairly quickly. The really tedious part is adjusting the value of the dihedral constraint and setting up a new calculation (for some reason, copy-and-paste does not bring the setup parameters along with the rest of the model so all of the menus in the Setup: Calculations window have to be reset to the values you want). But, and this is an important “but” – it works!

Some thoughts about internal rotation and conformational preferences in H2O2. I intentionally used two different methods, the lower quality HF/3-21G method and the higher quality B3LYP/6-31G** method, to construct two energy profiles. It turns out that these methods do not give identical results.

The two energy profiles are shown below (click on thumbnail to get larger image):

HF/3-21G energy profile - H2O2 rotation

B3LYP profile - H2O2 internal rotation

The two profiles look almost the same at first glance: the energy increases as the HOOH angle drops from 180 to 0 degrees (energy maximum). However, the lower quality model (HF/3-21G) predicts a rotation barrier of nearly 12 kcal/mol and, more importantly, it predicts that the optimal geometry occurs when the HOOH dihedral reaches 180 degrees. The higher-quality model (B3LYP/6-31G**), on the other hand, predicts a somewhat smaller barrier (8-9 kcal/mol) and places the optimal geometry at an HOOH dihedral of ~120 degrees. The 180 degree geometry is actually a transition state according to this model.

It turns that there is a very simple way to rationalize the twisted geometry of H2O2. If you remember the MO model of O2 that was presented in Chem 212, you may recall that O2 has a paramagnetic triplet ground state with two unpaired electrons in perpendicular pi systems. If you were going to use this electron configuration to build H2O2, you would need to have each H interact with a different pi system. In other words, the OH bonds would lie in nearly perpendicular planes. Think about it.

I’m sure you all understand what to do: calculate HF/3-21G equilibrium geometries for three conformers of 1, 3-butadiene: planar s-trans, planar s-cis, and skew s-cis.

I suspect, though, that you are not so sure what to do after that. I said you should “rationalize” your results, but what does that mean?

First, I would like you to notice which conformer is most stable, which is least stable, and so on. Second, I would like you to rationalize the energy ordering. Why is the most stable conformer, the most stable? Why is the least stable, the least stable? As you know, I expect you to link energies with structural features and electronic interactions.

Third, all three models are produced by “equilibrium geometry” calculations, but are all three models energy minima? As we saw today, symmetry constraints can make a geometry optimization produce a false minimum. Are any of these models false minima? One way to answer this question is to construct an energy profile for internal rotation and that’s just what I would like you to do. Once you have a profile, and can trace the ups and downs of the energy curve as a function of dihedral angle, you will be able to state unequivocally which models, if any, are minima and which are something else. Which brings me to your last task: explain why certain structures turn out to be false minima, i.e., why the energy profile depicted them as maxima even though the geometry optimization decided they were minima.

• Endo and exo labels used incorrectly.
• Endo and exo compounds drawn incorrectly. Note: the 3-membered ring cannot be wriggled up and down to make different stereoisomers.
• Diastereomeric ratio (dr) does not refer to endo/exo ratio.
• So what are all of the compounds in the product mixture? What are their relative amounts? One or more of these answers will surprise you.
• What are the mechanisms for #5 and #6?
• Did you ever overlook the stereochemistry in a Diels-Alder product? (Yes, all of you did, I think. In #5 and #6.)