Once again, exam results were uniformly good. I have written comments on your exams and posted answers/comments on selected problems here. I don’t assign numerical scores to exams in this class, so if you would like additional feedback, please come see me before you leave, sometime this summer, or next fall.

The instructions for this assignment can be found here.

To summarize, you were asked to predict the regiochemistry of the following cycloaddition using FMO (note: the left and right-hand products will be referred to as meta and ortho, respectively):

Then you were asked to compare the FMO predictions to those based on calculated energy barriers. Finally, you were asked to look at other properties of the transition states (geometry, electron distribution) that might support the overall picture. My results follow …

Frontier MO Analysis

The key to this analysis is to locate the relevant orbitals, identify the dominant FMO interaction, and predict regiochemistry.

The molecular formulas tell us that each molecule contains two orthogonal pi systems. The two pi systems play different roles - one participates in the cycloaddition, the other is a spectator - but we can be overlook this distinction because each pi system contains the same information. In other words, it doesn’t matter if I select HOMO or HOMO-1 as the donor orbital because

they are degenerate. Likewise, it doesn’t matter if I select LUMO or LUMO+1 as the acceptor orbital. Furthermore, it doesn’t matter if my donor and acceptor orbitals are orthogonal because this won’t change the outcome of my analysis.

Some caution is still required. I should actually look at my orbital surfaces in order to characterize each orbital’s nodal pattern. I should never rely solely on orbital energies to assign donor and acceptor roles.

The relevant nitrile oxide (1,3-dipole component) orbitals are shown below.

• Donor (HOMO, -10.16 eV, orbital coefficients C -.32, N -.09, O +.54). Things to notice:
  • The MO contains one “pi” node (Ignore the node between the two C). This is what we expect for the second MO in any linear conjugated system.
  • The isosurface suggests that C makes a larger contribution than O. The orbital coefficients, however, say otherwise. Remember that orbital decay increases with increasing electronegativity: C < N < O. So even though O makes a larger contribution to the MO than C does, the MO decays faster (looks smaller) near O. When in doubt, always check the orbital coefficients.

• Acceptor (LUMO, +5.11 eV, orbital coefficients C +.57, N -.59, O +.25). Things to notice:
  • The MO contains two “pi” nodes. This is what we expect for the third MO in any linear conjugated system.
  • N makes the largest contribution to this orbital (see orbital coefficient). However, N is not one of the reacting atoms. Pay attention only to the reacting atoms.

The relevant alkyne (dipolaraphile) orbitals are:

• Donor (HOMO, -11.67 eV, orbital coefficients CH -.29, CCN -.30). Things to notice:
  • The MO contains one “pi” node overall (second orbital in linear conjugated system), but no nodes within the alkyne (CC π).
  • The orbital coefficients on the two carbons are essentially identical. This MO cannot be used to predict regioselectivity.

• Acceptor (LUMO, +2.98 eV, orbital coefficients CH +.62, CCN -.48). Things to notice:
  • The MO contains two “pi” nodes (third orbital in linear conjugated system), and one node inside the alkyne (CC π*).
  • The terminal C makes a slightly larger contribution to the MO (this is not readily apparent from the isosurface).

Putting it all together, we find that the dominant interaction involves the nitrile oxide HOMO (donor) and alkyne LUMO (acceptor) and we expect O to add to the terminal C of the alkyne. That is, we expect the methyl and cyano groups to be adjacent in the final product.

Calculated Energy Barriers

HF/3-21G energies for the reactants and transition states, and calculated energy barriers, are listed below. Reactant energies were obtained from equilibrium geometry models. Transition state energies were obtained from models generated by transition state searches. IR frequency calculations were used to verify that the transition state models were stationary points (one imaginary vibration frequency).

• nitrile oxide -205.51566 au
• alkyne -167.61366 au
• meta transition state -373.08533 au
  • meta energy barrier = 27.6 kcal/mol
• ortho transition state -373.09821 au
  • ortho energy barrier = 19.5 kcal/mol

Naturally these calculated barrier should be taken with several grains of salt:

• the HF method fails to include electron correlation
• the 3-21G basis set does not place enough basis functions on each atom
• the barriers reflect only calculated changes in electronic energy and do not incorporate changes in nuclear kinetic energy
• likewise, entropy and solvent effects have been ignored.

That said, the calculated Δ(ΔE*) is large and consistent with the FMO analysis. The ortho product should form more rapidly.

Supporting analysis - transition state geometry

Geometry & orbital overlap. According to the FMO analysis, the ortho transition state should create better CO orbital overlap. This should manifest itself as a shorter CO distance in the ortho transition state. The distances for the forming (partial) single bonds are:

• ortho transition state: 1.870 (CO) and 2.455 Å (CC)
• meta transition state: 2.032 (CO) and 2.286 Å (CC)

consistent with our prediction.

Geometry & asynchronicity. According to the FMO analysis, the ortho transition state is characterized by a “large-large” (CO) and a “small-small” (CC) pair of orbital overlaps. The meta at transition state, on the other hand, is characterized by two “large-small” orbital overlaps. Therefore, the ortho transition state should be more asynchronous. This is borne out by the calculated transition state geometries (see above).

Geometry & Hammond postulate. The HF/3-21G energies of the products (equilibrium geometry models) and the corresponding reaction energies are:

• ortho product -373.25670 au (ΔE = -79.9 kcal/mol)
• meta product -373.24611 au (ΔE = -73.2 kcal/mol)

According to the Hammond postulate, the transition states for these exothermic reactions should be reactant-like. Furthermore, we can expect that the transition state for the more exothermic ortho reaction will be more reactant-like.

A large number of geometrical changes occur during these reactions. Some of these are relevant, but others are not. For example, a cycloaddition necessarily brings the two reactants close together so the interatomic distances used in the previous sections are not reliable guides to whether a transition state is reactant-like or product-like (and, we noted that these distances can be explained in other ways). It is better to look at geometry changes that occur within each reactant. There are many distances and angles that we could look at, but here are some that appear in the nitrile oxide:

• CN distance. Reactant = 1.136 Å. Products = 1.452 (ortho) and 1.438 Å (meta). The transition state distances are clearly reactant-like, 1.152 (ortho) and 1.161(meta) Å, and the shorter ortho distance is clearly more reactant-like.
• NO distance. Reactant = 1.317 Å. Products = 1.289 (ortho) and 1.294 Å (meta). The transition state distances are clearly reactant-like, 1.333 (ortho) and 1.311 Å (meta), and the longer ortho distance is clearly more reactant-like.
• CNO bond angle. Reactant = 180°. Products = 104.9° (ortho) and 105.5° (meta). The transition states appear to be slightly closer to products than reactant by this measure, 140.9° (ortho), and 139.6° (meta).
• CCN bond angle. Reactant = 180°. Products = 122.1° (ortho) and 121.3° (meta). This angle makes the transition states look reactant-like, 165.3° (ortho), and 158.7° (meta), and the ortho transition state looks more reactant-like.

Although there is not unanimous agreement on the position of the transition state (and I have not reported any parameters for the alkyne), it would seem that, on balance, the data support the Hammond postulate.

Supporting analysis - charge transfer

The dominant FMO interaction involves a nitrile oxide donor MO and an alkyne acceptor MO. Therefore, in the transition state, we expect to see positive charge accumulate on the nitrile oxide and negative charge accumulate on the alkyne.

Since the reactants are both neutral, the simplest (but most tedious) way to assess this prediction is to look at the overall charge on each “reactant” in the transition state. In other words, I will sum the atomic charges of all of the alkyne atoms, including the hydrogen and cyano group. Fortunately, it is not necessary to sum the atomic charges for the nitrile oxide (which contains more atoms) because the transition state has no net charge. (Note: atomic charges are obtained by selecting Display: Properties and then selecting the atom of interest. The following values are based on Natural Charges.)

• Reactant: alkyne = 0, nitrile oxide = 0
• Transition state
  • ortho: alkyne = -0.17, nitrile oxide = +0.17
  • meta: alkyne = -0.10, nitrile oxide = +0.10
• Product
  • ortho: alkyne = +0.20, nitrile oxide = -0.20
  • meta: alkyne = +0.22, nitrile oxide = -0.22

The direction of electron transfer in the transition state is nitrile oxide → alkyne and is consistent with the FMO prediction. Interestingly, the direction of electron transfer reverses as the transition state changes into product. This is not unexpected. The dominant FMO interaction may control the approach to the transition state, but it involves only one electron pair and other electron pairs must eventually participate in the reaction.

In conclusion, while we might have concerns about the low level theory used to construct these models, the fact that several properties of the models (energy, geometry, and electron distribution) seem to be internally consistent, i.e., they can all be rationalized using a single theory, increases our confidence in the qualitative validity of these results.

Todd and Melissa have requested Wed, April 30, 10 am as the date and time for our wrap-up breakfast. Does this work for everyone else? If I don’t hear otherwise, I will assume it does.

I will drive my van so that we can load everyone into it.

A number of you have stopped by to talk about the exam’s problems, or perhaps I should say, problems you’re having with the exam.

In order to minimize any inequities that might arise from me giving information to some students and not others, here is a summary of major items that I have communicated:

Problem #1

1. I mis-lettered the parts of problem #1. It starts with part b instead of part a. Oh well. Please follow the lettering in the exam.
2. C-Li bonds are drawn as covalent in the figures. This is convenient for the artist (me), but for purposes of drawing and analyzing pericyclic reactions, you may find it more useful to view this as an ionic bond.
3. Other aspects of the formulas might also be adjusted to make them more convenient for drawing and analyzing pericyclic reactions. (Just one of several possible hints: there is more than one way to draw a benzene ring…)

Problem #2

1. I have given you incomplete information about stereochemistry. I don’t intend to change that. Do the best you can. Perhaps more than one outcome is possible. Perhaps not.

Overall

1. Although this is a take-home exam, it is not meant to be a marathon exam. It was constructed as a take-home so that I wouldn’t have to sacrifice class time for the exam, and you would be able to do some modeling. I really figured that the non-modeling portion of the exam could be completed in an hour or less.

Download the description of the final project.

When it comes time to write, you may also be interested in my excerpts from the ACS Style Guide. (Also posted on the Miscellaneous page under “Journal articles - described“)

Exam #2 (2 page PDF). Due Monday, April 28, at 3 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.

The article on CuAAC (copper-catalyzed azide-alkyne cycloadditions) can be found in issue #1, volume #40 of Aldrichimica Acta.

• Aldrichimica Acta home page
• Sortable index of all Aldrichimica Acta articles

Click chemistry even has its own Wikipedia page. Click here.

We aren’t tracking very well with the online schedule at this point. Here is what I had planned at the start of the semester:

• M 4/14 - Click 3+2 cycloadditions, distribute paper #8
• W 4/16 - Thermal 3+2 cycloadditions, collect HW #8
• F 4/18 - discuss paper #8, distribute exam #2, distribute final project instructions
• M 4/21 - 3+2 modeling session

I would still like to cover these topics, but we’ve lost a day because I never built student evaluations into my schedule, so I am going to catch up by eliminating paper #8. Here is my current plan for the remainder of the semester:

• M 4/14 - Photochemical 2+2 & singlet oxygen
• W 4/16 - Click 3+2 cycloadditions, collect HW #8
• F 4/18 - Thermal 3+2 cycloadditions, distribute exam #2, distribute final project instructions
• M 4/21 - 3+2 modeling session

After M 4/21, I will also change my office hours to 10-11 instead of 11-11:30 because we won’t have any more scheduled classes. However, I would still like to take all of you to breakfast before the end of the semester.

Here are some links for Friday’s lecture. My computer was goofy and slow so you might want to look these materials over more carefully on your own.

Jablonski diagrams

• Lecture animation
• Another diagram. This one correctly represents internal conversion and intersystem crossing as processes that (more or less) conserve energy.
• Wikipedia also has a short and useful description of internal conversion that contains some material on biochemical systems that I was unaware of

(Hanovia) Photochemical reactor

• Lecture image
• Another version that allows magnetic stirring (power supply and stand also shown; Sigma-Aldrich). Click on image for larger image.

One of you wrote “this sucks” next to your NMR assignments. I have to agree. A complete explanation of the 1H and 13C spectra is beyond me.

On other hand, while I wish I had chosen a more accessible set of data, these spectra are typical. Mother Nature is rarely kind. We just have to deal with that.

Here is my quick take on the NMR data along with some subtle relationships that should not be overlooked …

First, two useful ways of looking at compound 9:

1H NMR

First, group protons by chemical shift range:

• aromatics (b-e) - 4H
• alkene (i, j) - 2H
• alkane, O neighbor (p) - 3H
• alkane, unsat’d neighbor (h-l) - 3H
• alkane (n) - 2H

The data contain:

• four 1H signals between 6.86-8.24 ppm. These are b-e. Alkene protons could fall in this region, but only if the alkene is substituted with strong deshielding groups, e.g., C=O. Two signals are doublets with J=8 Hz. These are probably b (8.24) and e (6.86). c and d can’t be assigned reliably to 6.93/7.27 (both multiplets) without spectra to compare with.
• two 1H signals 6.67 and 5.71 ppm. These are i and j. These signals appear in all of the versions of this compound (O bridged, unbridged, Me- and Ph-substituted).
• a 3H singlet at 3.85 ppm (q).
• three 1H signals at 5.32 (d, J=4 Hz), 3.73 (d, J=1.5 Hz), and 3.37 (d, J=2.5 Hz). These are h-l. Two protons, h and l have two unsat’d neighbors and should be extra deshielded, but only one signal (5.32) appears at an extreme shift. The model shows that h and l are nearly coplanar and perpendicular with respect to the aromatic ring, respectively. This geometric difference must cause one proton to be extra deshielded. Other geometric differences: h is near O, while l lies behind C=O. It is hard to assess the shielding and deshielding effects of these geometric factors (note: Spartan can calculate chemical shifts, but I didn’t attempt it - it’s a little time-consuming and I wasn’t expecting you to use this tool).
• two 1H signals at 2.26 (m) and 2.17 (d, J=11 Hz). These are n1 and n2. 11 Hz is a standard geminal coupling constant. The n proton that points towards the aromatic ring is nearly perpendicular to its vicinal neighbors, so it might not couple noticeably with either of them (2.17).

13C NMR

First, group carbons:

• aromatics (a-f) - 6C
• alkenes (gmij) - 4C
• ester carbonyl (o) - 1C
• ester methyl (p) - 1C
• alkyl (hkln) - 4C

Because they give you CH coupling information, some assignments are simple:

• 51.5 CH3 - p
• 41.9 CH2 - n

Fortunately, these assignments are also consistent with the chemical shifts expected for these carbons. Speaking of expectations, Hans Reich’s 13C tables show that benzofuran should have signals at 106 (m), 144 (g), 121-127 (abcd), 111 (e), 155 (f). What you should notice here are the numbers and two patterns, namely, O has a deshielding effect on f and g and a shielding effect on e and m. Another table shows that ester C=O are typically located between 155-165 ppm and attaching an alkyl group to an arene or alkene C has a modest (10 ppm?) deshielding effect (this will apply to g and m).

Now for the observations:

• 166.1 (C) & 163.2 (C) & 145.2 (C) - Based on the above considerations, o must be either 166 or 163, but which? The other two signals must be f and g.
• 131.1 (C) & 124.9 (C) - a and m, but can’t be certain which is which.
• 145.9, 131.7, 128.2, 127.3, 121.2, 110.5 (all CH) - bcdeij. It is tempting to assign 110.5 to e based on the above shift comparisons.
• 83.0, 44.1, 41.9 (all CH) - hkl. First, consider branching - no help, all three C are tertiary. Next consider de/shielding neighbors. Again (see 1H NMR) we have the puzzle of two C attached to two unsaturated (deshielding) neighbors and one C attached to one unsaturated neighbor, but only one unusually deshielded signal.