Coursework

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Molecular modelling Coursework to be attempted during Scheduled Sessions

These projects are arranged in increasing order of difficulty, and time taken to complete. You should do as many as you can in the 2 hour session allocated to you, and return to finish the rest if you wish at your convenience. At the end of each session, we will conduct a number auction. For each project, the bidding will start with the first volunteer offering an energy for the system (or one of the isomers). If anyone has a lower energy for that molecule, they will then bid that energy. The winner will be the one with the lowest energy.

Conformational analysis I: Chair and Boat conformations of Cyclohexane

Cyclohexane

Construct chair-like and boat-like conformations of cyclohexane. Compare the energies of both forms. Check carefully the geometry of the boat; does it have any unusual feature?

Chiralane

Optional: Try changing one or more of the CH2 groups into an oxygen and see if that affects things. Optional: The molecule on the left is called chiralane. Are its rings boats or chairs?

References and further exploration

1. The first suggestion of two forms for cyclohexane goes as far back as H. Sachse, Chem. Ber, 1890, 23, 1363 and Z. Physik. Chem., 1892, 10, 203. This is nicely explained here. E. Mohr, J. Prakt. Chem., 1918, 98, 315 and Chem. Ber., 1922, 55, 230, translated Sachse's argument into a pictorial one.
2. The article that put conformational analysis on the map: D. H. R. Barton and R. C. Cookson, The principles of conformational analysis, Q. Rev. Chem. Soc., 1956, 10, 44. DOI:10.1039/QR9561000044
3. Wikipedia article
4. D. A. Dixon and A. Komornicki, Ab initio conformational analysis of cyclohexane, J. Phys. Chem., 1990, 94, 5630 - 5636; DOI:10.1021/j100377a041 .
5. For a more modern application of this technique, see I. Columbus, R. E. Hoffman, and S. E. Biali, Stereochemistry and Conformational Anomalies of 1,2,3- and 1,2,3,4-Polycyclohexylcyclohexanes. J. Am. Chem. Soc., 1996, 118, 6890 - 6896; DOI:10.1021/ja960380h .
6. Animations of the process can be seen at this blog: http://www.ch.imperial.ac.uk/rzepa/blog/?p=7926
7. For the record, the point group symmetries of the various species which may be involved are D_3d for the chair conformation, C_2v for a boat geometry, and D₂ for any twisted boat form. Is any of these forms chiral?
8. The second molecule shown in this section is called [6.6]chiralane. It is peculiar for having many six-membered saturated rings, all of them as twist-boats rather than chairs! (a chair has a plane of symmetry, a twist boat only axes, which of course allows it to be chiral). See here for more details.
9. More detail on the conformation of rings (and acyclic systems) will be found in the lecture course on the topic to be given in the spring term.

Enantiomers vs Diastereomers Part 1: Butanes

This problem illustrates, using models, the difference between an enantiomer and a diastereomer.

2-chloro-3-bromobutane

2-bromo-3-chlorobutane The compound 2-bromo-3-chlorobutane has two chiral centres, and four isomers (22) are therefore possible. Calculate all four isomers, and for each be careful to label each of the two stereo centres R or S as you go. For each of the four isomers R,R, S,S, R,S, S,R you will have to think about whether you have obtained the lowest energy conformer. Can your four energies be grouped in any sensible way? Another example of relevance to the problem sheets associated with the lecture course is shown on the right. Can you use modelling to predict the relative energies of the four possible diastereoisomers resulting from formation of the aldol product? You will have to consider the conformations of your stereoisomers, and whether there are any other factors which may intervene in stabilising one stereoisomer over the other?

A reality check: A search of the Cambridge database

search query

To find out what the optimal conformation of a 2,3-dihalobutane actually is (in the solid state), we can search for this in the Cambridge database.

References and further exploration

1. Wikipedia article on Diastereomers
2. The Cahn-Ingold-Prelog Priority rules

Enantiomers vs Diastereomers Part 2: Helicenes

Pentahelicene

Circulene Circulene Circulene

Construct some helicenes (pentahelicene or [5]helicene is shown on the right), using conjugated bonds for all the ring bonds. Benzene, naphthalene, phenanthrene and benzophenanthrene are in fact the first four members of this series. At what point in this series can you detect helicity cropping up? This is manifested by a non-planar helical wind of the molecule. If you do detect it, note how the wind is either left or right handed, ie the two forms are enantiomers of each other. Try displaying the molecule in spacefill mode (see above) to see if you can identify the source of the helicity. Optional: [7]circulene is a known molecule shown on the left. Is it flat? (see DOI:10.1021/ja00219a036 ). An [8]circulene is also known (DOI:10.1021/jp806134u ) [5]circulene has instead a five-membered central ring. It too is puckered, but not nearly so much as [7]circulene. See DOI:10.1021/ja306992k These are all related to graphene (Nobel Prize 2010).

References and further exploration

1. Wikipedia article on Helicenes and related molecules
2. The higher helicenes are well known (up to about [14]helicene) and amongst the most chiral molecules known (in terms of how much they rotate the plane of polarised light).
3. The smallest helicene which can be resolved experimentally into enantiomers is in fact [5]helicene].
4. R. H. Janke, G. Haufe, E.-U. Würthwein, and J. H. Borkent, Racemization Barriers of Helicenes: A Computational Study, J. Am. Chem. Soc., 1996, 118 6031 - 6035 DOI:10.1021/ja950774t

Conformational analysis II: cis and trans-decalins

Cis decalin Elimination Woodward

cis Decalin This is the famous molecule that started the whole molecular mechanics modelling ball rolling. Barton in 1948 sought to find out which conformation of cis-decalin was the most stable (see here for video). You should be able to find at least three conformations of this molecule. Try locating these, and conclude which is the most stable. Identify any chair rings and any boat. Measure some dihedral angles to see if the staggered relationships hold (i.e. for such a relationship, the dihedral angle should be close to 60 degrees). Optional: A key step in Woodward's famous synthesis of cortisone is a quinone+butadiene Diels-Alder reaction to give a cis-decalin (left), with an assumption that epimerisation to a trans-decalin is thermodynamically favourable. cis Cortisone Can you verify whether the trans-isomer is indeed more stable? Its not so obvious, since this compound has two extra double bonds in the rings and six sp2 centres which might perturb things. trans Decalin Optional: The two diastereomeric trans-decalin tosylates react quite differently with NaBH4. Construct models for both isomers (use methoxy as a model for the Tosyl group) and from the antiperiplanar alignments of bonds that you can find in each isomer, can you make a connection to the reactivity of each form? Consider very carefully where you would put a lone pair located on the nitrogen (i.e. include the N-Lp "bond" in your antiperiplanar alignments) asuming the this atom is tetrahedral rather than planar. Does this lone pair play any part in either reaction in this position?. Note that the relative energy of the axial/equatorial N-Methyl group will not be an accurate reflection of any antiperiplanar alignments, since these are predominantly electronic in origin, and this mechanics method does not take these into account. The second (elimination) reaction is very slow compared to the first. Discuss with tutors why this might be so (for Hints, see here or here). These reactions do not appear to occur for the corresponding cis-decalins6. Why not?

References and Footnotes

1. D. H. R. Barton, Interactions between non-bonded atoms, and the structure of cis-decalin, J. Chem. Soc., 1948, 340-342. DOI:10.1039/JR9480000340
2. It took Barton months using a hand-cranked calculator to evaluate the relative energies of the various conformations. You will do it in seconds.
3. Wikipedia article
4. For a modern application of mechanics to this molecule, see J. M. A. Baas, B. Van de Graaf, D. Tavernier, and P. Vanhee, Empirical force field calculations. 10. Conformational analysis of cis-decalin, J. Am. Chem. Soc., 1981, 103, 5014 - 5021; DOI:10.1021/ja00407a007 .
5. For a video-Podcast of Barton and Woodward (and other Nobel prize winners), subscribe here
6. R. B. Woodward, F. Sondheimer, and D. Taub, The total Synthesis of Cortisone, J. Am. Chem. Soc., 1951, 73, 4057 - 4057. DOI:10.1021/ja01152a551 .
7. P.-W. Phuan and M. C. Kozlowski, Control of the Conformational Equilibria in Aza-cis-Decalins: Structural Modification, Solvation, and Metal Chelation, J. Org. Chem., 2002, 67, 6339 - 6346; DOI:10.1021/jo025544t

Enantiomers vs Diastereomers Part 3: Stereochemistry of conjugate addition

The reaction on the right is taken from a lecture problem sheet. Part of the problem involves assigning the stereochemistry of the methyl group.

1. Assuming that the reaction is thermodynamically controlled, which of the two possible diastereoisomers is lower in energy?
2. In fact, the reaction proceeds via a Cu-alkene π-complex (DOI:10.1021/ja067533d ) and a more accurate model would involve computing the energy of that model rather than the product of the reaction as above. In general, the molecular mechanics method does not handle transition elements, and one would have to use quantum mechanics instead.
3. It may indeed be that it is the transition state for the reaction of the Cu complex that needs to be modelled, not the complex itself. This too is beyond the scope of the current lab.
4. A second example from the problem sheet is shown below the first. There are two regiochemical outcomes, each with an unknown stereochemistry for the methyl group. Does molecular mechanics modelling cast any light on the possible answers?
   • The difference in behaviour of Mg and Cu is again not something that mechanics can easily address. Here again, this aspect of the chemistry has to be tackled using an electronic theory of chemistry, not a mechanical one.

Menthone/isomenthone and Bridgehead enols: Thermodynamic vs Kinetic Control Part 1.

Menthone Menthone

Menthone Beckmann (of rearrangement fame) in 1889 dissolved optically active levorotatory (-) (S,R)-menthone ([α]D -28°) in conc. sulfuric acid, followed by quenching on ice to give what Beckmann assumed was pure (and what we would nowadays call diastereomeric) (+) (R,R)-isomenthone, [α]D +28°. He suggested for the first time that such an isomerisation, involving epimerisation at the asymmetric centre next to the keto group, proceeded via an intermediate enol in which the tetrahedral asymmetric carbon becomes planar. But this famous (perhaps even notorious2) early example of a reaction mechanism makes an interesting assumption, which can be tested by molecular modelling. Two possible enols can be formed, one of which is the so called thermodynamic enol, the other being the kinetic enol. Only one of these enols can resulting in the menthone isomerising to isomenthone. Find out if simple molecular modelling correctly predicts that the thermodynamic enol is indeed the more stable of the two. Hint: Model the enol and not the ketone. Consider carefully any conformational isomers possible. Given that the optical rotation3 of pure (+)-isomenthone is now known to be [α]D +101° rather than +28°, we can infer that Beckmann's product contains only 43% isomenthone and hence still contains 57% of original menthone, corresponding to an equilibrium constant of K= 0.75. This can be related to a (free energy) difference using the equation ΔG = -RT ln K, or ΔG = 0.7 kJ/mol (menthone being lower in energy by this amount compared to isomenthone). Can this energy difference be verified using molecular mechanics modelling? Can you explain why menthone is the more stable? (For another hint, or possibly a fright, visit this page). To study the kinetic process, we cannot use the Molecular mechanics approach, since we will need to find the transition state for proton removal. Mechanics by its nature cannot break bonds, and so the only vialble method is quantum mechanics, which you will encounter next year. Enols Optional: The molecule on the right is taken from Don Craig's problem sheet 1, associated with the lecture course on enols in synthesis. Four different hydrogens could potentially be removed by treating the compound with base. Construct the model, and from inspection, can you decide which of the four possible enols might form? Calculate the energies of all four. Which is lowest? (Hint: the carbanion orbital resulting from proton removal has to be able to conjugate with the π-system of the carbonyl group).

References and footnotes

1. E. Beckmann, Annalen, 1889, 250, 322. DOI:10.1002/jlac.18892500306 .
2. Many of Beckmann's misconceptions were corrected by O. Wallach, Annalen, 1893, 276, 296. DOI:10.1002/jlac.18932760306 . The notoriety is because the coincidence of equal but opposite optical rotations obtained in this experiment led Beckmann to believe that he had obtained the enantiomer of menthone, and not as we now know, the impure (R,R) diastereomer. It should be borne in mind that the concept of tetrahedral and asymmetric carbon was only 15 years old at this time (see Jacobus Henricus van't Hoff and Joseph Achille Le Bel). Nevertheless confusion over this aspect persisted for some time after, and was often evident in the writings of even very famous chemists of the time (and Beckmann was very famous)!.
3. Wikipedia article
4. From about 1890-1935, mechanistic organic chemistry was born. In the absence of UV, IR, NMR, MS and X-Ray techniques, the polarimeter occupied a pivotal role. Many of the great discoveries in reaction mechanisms (keto-enol tautomerism as seen here, carbocations, the Walden inversion, etc) relied on polarimetric measurements.
5. A notorious modern example of (unwanted) epimerisation of a ketone is Thalidomide, where one epimer inhibits morning sickness in pregnant women, and the other epimer is teratogenic, causing fetal abnormalities. The equilibrium in this case does not require conc. sulfuric acid, but can occur at physiological pH.

Optional: Additional Molecular modelling Coursework

Please feel free to try these problems in your own time, and to discuss these with your organic tutors and lecturers. Note also that the relevant lectures may occur in the spring as well as autumn terms.

Axial/Equatorial preferences in cyclohexane and cyclohexanone and Hydrogen Bonding

Cyclohexanone

Construct a chair cyclohexane and replace firstly one of the axial hydrogens with the following groups: methyl, t-butyl, OH. Calculate the energy of the axial isomer. Then repeat (either by deleting/redrawing or by moving) for the equatorial forms. Compare the energies of the two isomers. Does any energy difference increase with the size of the group? Does OH fit into this in terms of size? thiomethyl cyclohexanone The dissolving metal reduction of cyclohexanones in a protic solvent (i.e. one capable of hydrogen bonding) is thermodynamically controlled and gives the more stable, equatorial alcohol. In fact, its probably the alkoxide that is the product, not the free alcohol. It is thought the alkoxide is actually a lot larger than the alcohol, accounting for the substantial equatorial preference. Can you think why its larger? [Ghemical cannot in fact model this, since the force field does not include parameters for the alkoxide anion]. Determine the axial/equatorial preference of 2-methylthio-cyclohexanone (Hint: there are many conformations possible, and you should try a few to see if you can get the lowest).

References and Footnotes

1. A. H. Lewin and S. Winstein, NMR. Spectra and Conformational Analysis of 4-Alkylcyclohexanols J. Am. Chem. Soc.; 1962, 84, 2464 - 2465; DOI:10.1021/ja00871a049
2. F. R. Jensen and L. H. Gale, The Conformational Preference of the Bromo and Methyl Groups in Cyclohexane by IR Spectral Analysis, J. Org. Chem., 1960, 25, 2075 - 2078. DOI:10.1021/jo01082a001
3. K. B. Wiberg, J. D. Hammer, H. Castejon, W. F. Bailey, E. L. DeLeon, and R. M. Jarret, Conformational Studies in the Cyclohexane Series. 1. Experimental and Computational Investigation of Methyl, Ethyl, Isopropyl, and tert-Butylcyclohexanes, J. Org. Chem., 1999, 64, 2085 - 2095; DOI:10.1021/jo990056f . The salient point here is that the enthalpy and entropy of this series differ in their trends.
4. Just when you are starting to think that things are quite simple, along comes the observation: S. E. Biali, Axial monoalkyl cyclohexanes, J. Org. Chem., 1992, 57, 2979 - 2980; DOI:10.1021/jo00037a001
5. And this one with knobs on: In all-trans-1,2,3,4,5,6-hexaisopropylcyclohexane, all the alkyl groups are located at axial rather than equatorial positions: O. Golan, Z. Goren, and S. E. Biali, Axial-equatorial stability reversal in all-trans-polyalkylcyclohexanes, J. Am. Chem. Soc., 1990, 112, 9300 - 9307. DOI:10.1021/ja00181a036 .
6. J. A. Anderson, K. Crager, Kelly, L.Fedoroff, G. S. Tschumper, Gregory S. Anchoring the potential energy surface of the cyclic water trimer. J. Chem. Physics, 2004, 121, 11023-11029. DOI:10.1063/1.1799931 .
7. R. R. Fraser, N. C. Faibish, On the purported axial preference in 2-methylthio- and 2-methoxycyclohexanones: steric effects versus orbital interactions, Can. J. Chem., 1995, 73, 88-94.

How to induce room temperature hydrolysis of a peptide

Pentahelicene

Peptide hydrolysis This introduces a further example of how simple conformational analysis can quickly rationalize kinetic behaviour. At neutral pH and 25° the half life for hydrolysis of a peptide bond is around 500 years (and thank goodness, or we would ourselves all rapidly hydrolise to a mush!). Some enzymes however can achieve this in less than 1 second, an acceleration of 1013! Organic chemists are not quite so clever, but they can achieve room temperature hydrolysis of a peptide in 21 minutes by careful conformational design. The two isomers shown on the right differ only in their stereochemistry, one hydrolysing quickly, the other slowly. Build a model of each compound, and calculate two isomers for each, varying in whether the ring N-substituent is oriented axial or equatorial with respect to the decalin ring. On the basis of your two pairs of energies, can you rationalise the observed kinetic behaviour? Do you know why both of these compounds take very much less than 500 years to hydrolise the peptide bond? Hint1: Use the chair-chair conformation for cis-decalin as your template for constructing this system. Hint2: When constructing your models, think if there are any hydrogen bonds that might stabilize the structure! Hint3: Hydrolysis can only occur when the OH group can approach the carbonyl of the peptide bond close enough to react, and at the right angle of approach.

Reference

1. M. Fernandes, F. Fache, M. Rosen, P.-L. Nguyen, and D. E. Hansen, 'Rapid Cleavage of Unactivated, Unstrained Amide Bonds at Neutral pH', J. Org. Chem., 2008, ASAP: DOI:10.1021/jo800706y

Caryophyllene: The phenomenon of Atropisomerism

1. 

   Caryophyllene ketone
   Caryophyllene, a constituent of many essential oils, include clove oil, has a trans alkene contained in a 9-membered ring. One interesting property is that it has 4 diastereoisomers possible, originating from a total of three asymmetric centres present in the molecule. Two of these are conventional chiral centres, one is present in the form of a disymmetric trans double bond. To understand why such a bond can result in two configurations, one must appreciate that (concurrent) rotation about the two C-C single bonds adjacent to the alkene is in fact restricted, because to the hydrogen labelled H_a cannot easily pass by the edge of the 4-membered ring. Construct this molecule (in fact the ketone rather than the alkene) and optimize its geometry. Note in particular that the ring junction is trans and not cis.
2. You will find you may well have obtained one of two forms. In the first, the H_a hydrogen will be opposite the C=O group, in the other it will be adjacent to it. Record the energy of whatever form you got. At the end of the course, we will try to find the winner with the lowest energy (this is not as trivial as it sounds!).
3. Next, take your structure, and try to flip the trans alkene bond around so that eg if the methyl were previously pointing up, now it will point down. You may find a combination of erasing/redrawing or of moving, will accomplish this. You may also find another trick useful, of deleting all hydrogens, and then re-sprouting them back on again. Re-optimise your structure and compare the energy with your first isomer.
4. Another feature of this model is that you can judge which group is in the so-called shielded region of the carbonyl group magnetic anisotropy. Using this information, you can see if there are any anomalous ¹H chemical shifts that might need explaining!

References

1. M. Clericuzio, G. Alagona, C. Ghio, and L. Toma, Ab Initio and Density Functional Evaluations of the Molecular Conformations of -Caryophyllene and 6-Hydroxycaryophyllene, J. Org. Chem. 2000, 65, 6910 - 6916. DOI:10.1021/jo000404+ .
2. Wikipedia article
3. For a recent application of this phenomenon, see P. C. Bulman Page, B. R. Buckley, S. D.R. Christie, M. Edgar, A. M. Poulton, M. R.J. Elsegood and V. McKee, A new paradigm in N-heterocyclic carbenoid ligands, J. Organometallic Chem., 2005, 690, 6210-6216. D DOI:10.1016/j.jorganchem.2005.09.015 .

Germacrene: Conformational analysis of medium sized rings

Germacrene

Germacrene and the thermal reaction product Germacrene is a natural product with a ten-membered ring; it has the triene structure shown. Assuming that it adopts a crown conformation, build a three-dimensional model. On heating, germacrene is converted into one of the stereoisomers of the divinylcyclohexane, via a [3,3] sigmatropic pericyclic reaction. Predict from your model for Germacrene whether the product will have the two vinyl groups cis or trans to one another.

References

1. K. Shimazaki, M. Mori, K. Okada, T. Chuman, H. Goto, K. Sakakibara and M. Hirota, Conformational analyses of periplanone analogs by molecular mechanics calculations, J. Chem. Ecology, 1991, 17, 779-88. DOI:10.1007/BF00994200 .
2. H. Shirahama, E. Sawa and T. Matsumoto, Conformational aspects of germacrene B. Are the germacrenes resolvable ?, Tetrahedron Lett., 1979, 20, 2245-2246. DOI:10.1016/S0040-4039(01)93687-1 . See also DOI:10.1039/P19750002332 for an explanation of the selective epoxidation of germacrene.

Xestoquinone: Regio and Stereoselectivity in the Diels Alder reaction

1. 

   Xestoquinone precursor
   This compound is a precursor to a natural product called Xestoquinone. It has four alkene groups, which can individually be considered as the alkene component in a _π2_s + _π4_s Diels Alder cycloaddition. The pair of alkenes a+b or c+d can also act as the diene component in the _π2_s + _π4_s Diels Alder cycloaddition. Construct a model of the product of e.g. forming a bond between alkene a or alkene b and diene c+d, and then reverse the addition by using either c or d adding to the diene a+b. The stereochemistry of addition should always be suprafacial, i.e. preserving the stereochemical relationships of the alkenes. You should very carefully check that this is so in your final model.
2. Whilst you should stop at two models, it is possible to construct many more. For example, one might be able to add to either the top face of alkene b or to its bottom face. Identify the model with the lower energy, and save it for the end of the workshop. We will identify the isomer of lowest energy from everyone's results, this being a communal Monte Carlo experiment to find the global minimum.

References

1. Wikipedia article
2. For the original literature on this synthesis, see R. Carlini, K. Higgs, C. Older, S. Randhawa, and R. Rodrigo, Intramolecular Diels-Alder and Cope Reactions of o-Quinonoid Monoketals and Their Adducts: Efficient Syntheses of (±)-Xestoquinone and Heterocycles Related to Viridin, J. Org. Chem., 1997, 62, 2330 - 2331. DOI:10.1021/jo970394l where you can check to see which isomers actually do form!

Aldol Reaction and anti-Bredt Rings

Aldol

Aldol Reaction When the diketone shown is treated with base, it undergoes an aldol condensation. Two obvious possibililties are elimination of the combination Ha and Oa, or of the alternative combination Hb and Ob. In fact, only a single product is formed. On the basis of energies for both products, can you predict which one is actually formed? Measure a few dihedral angles, ie to find out how planar the alkene present is. Does this suggest a reason why one isomer is less stable than the other? There is a third very remote structural possibility. If you have time, verify that this third product truly is unlikely.

References

1. Bredt's Rule
2. I. Novak, Molecular Modeling of Anti-Bredt Compounds, J. Chem. Inf. Model., 2005, 45, 334 - 338. DOI:10.1021/ci0497354
3. See also this article A. Nickon, D. F. Covey, F.-C. Huang, and Y.-N. Kuo, Unusually facile bridgehead enolization. Locked boat forms in anti-Bredt olefins, J. Am. Chem. Soc., 1975, 97, 904 - 905; DOI:10.1021/ja00837a043 in conjunction with Project 9.

Conformational Preference for asymmetric hydride reduction of a ketone

1. 

   Asymmetric hydride reduction
   The hydride (BH₄, AlH₄, etc) reduction of the ketone shown here is stereospecific, resulting in an alcohol with the stereochemistry shown (known as the Cram or the Felkin-Anh rule). Construct a model of the ketone and establish which of at least two conformations is the lowest in energy.
2. If the hydride anion is delivered from the least hindered position, is the conformation you have consistent with the stereochemistry shown for the product?
3. You can see from Ref 4 that the situation can be far more complex, depending on many other factors.

References

1. Wikipedia article
2. D. J. Cram and D. R. Wilson, Studies in Stereochemistry. XXXII. Models for 1,2-Asymmetric Induction, J. Am. Chem. Soc., 1963, 85, 1245 - 1249. DOI:10.1021/ja00892a008 .
3. Y. Yamamoto, K. Matsuoka, and H. Nemoto, Anti-Cram selective reduction of acyclic ketones via electron transfer initiated processes, J. Am. Chem. Soc., 1988, 110, 4475 - 4476; DOI:10.1021/ja00221a093 .
4. A. Mengel and O. Reiser, Around and beyond Cram's Rule, Chem. Rev., 1999, 99, 1191 - 1224. DOI:10.1021/cr980379w .

Enantiomers vs Diastereomers Part 4: NMR Coupling constants

1. 

   Axial-equatorial interconversion
   In Project 2.2 above, we saw how the energies of diastereomeric compounds could be compared with the corresponding enantiomers. In this extension, we show how molecular modelling can cast light on the conformation adopted by 2-ethyl-4-methyl-1-oxa-cyclopentane-3-carboxylic acid estimated using measured ¹H NMR coupling constants. The (2S,3S,4S) diastereomer has couplings of ³J_H2,H3 8.3 Hz and ³J_H3,H4 9.8 Hz. Two possible conformations of this diastereomer are shown on the right. They differ in that one has Et axial, and Me/COOH equatorial, and the other Et equatorial and Me/COOH axial.
2. 

   Karplus plot. Click to expand
   By calculating the geometries of both conformations, and measuring the dihedral angle H2-C-C-H3 and H3-C-C-H4, one can assess by using the Karplus equation (left, taken from Ref 2 and relevant for a cyclopentane, but the values for which might be modified by the presence of electronegative substituents), which conformation leads to the best agreement between the calculated angle and the measured coupling constants (Hint: on the basis of the predicted couplings, you should be able to eliminate one of the two conformations shown for this molecule).
3. 

   5-circulene
   In Project 2.2 we also introduced molecules such as helicenes and circulenes. The ¹H NMR of the [5]-circulene shown to the right revealed a complex spectrum at δ 2.98 ppm and again at 3.75 ppm. On the face of it, the four protons labeled H_a and H_b should all be equivalent, and the spectrum should be a single peak, not two complex multiplets. Indeed, if the NMR is recorded at high temperatures, this is exactly what is observed. By constructing a model of the [5]-circulene shown, can you explain why at normal temperatures, the NMR spectrum is so complex?
   
4. 

   Synthesis lab experiment
   A practical application of this technique is to determine the stereochemistry of the product of the reaction between E,E-2,4-hexadien-1-ol and maleic anhydride. You will have the ¹H NMR spectrum of your sample recorded, and evident from that will be peak multiplicities of the various proton resonances. You should endeavour from your analysis to come up with a suggestion for the structure of compound Y, and from this, estimates of the numerical values (but not the signs) of the ²J and ³J couplings visible. Now using the techniques described above, construct a model of your proposed structure for Y. Measure the dihedral angles for all the ³J couplings, and very approximately estimate what the corresponding ³J might be from the diagram above. Does this help you assign the stereochemistry of the product?
5. Advanced topic: Part of the spectroscopic analysis of the compound Y involves interpreting the IR spectrum. Theory can be used in fact to simulate the full IR spectrum. In section 5.3 below, you will find instructions on how to use the model you have calculated here to initiate a so called density functional calculation. This will provide you with the required IR simulation. Follow these instructions, and open the resulting .log file in Gaussview. Go to the Results menu and select vibrations. The IR spectrum will be displayed. Does it match the one you have recorded for yourself?

References

1. M. Karplus, Vicinal Proton Coupling in Nuclear Magnetic Resonance, J. Am. Chem. Soc., 1963, 85, 2870 - 2871; DOI:10.1021/ja00901a059
2. A. Wu, D. Cremer, A. A. Auer, and J. Gauss, Extension of the Karplus Relationship for NMR Spin-Spin Coupling Constants to Nonplanar Ring Systems: Pseudorotation of Cyclopentane, J. Phys. Chem. A,, 2002, 106, 657 -667; DOI:10.1021/jp013160l
3. C. A. Stortz and M. S. Maier, Configurational assignments of diastereomeric γ-lactones using vicinal H–H NMR coupling constants and molecular modelling, J. Chem. Soc., Perkin Trans. 2, 2000, 1832 - 1836. DOI:10.1039/b003862h
4. A. H. Abdourazak, A. Sygula, and P. W. Rabideau Locking the bowl-shaped geometry of corannulene: cyclopentacorannulene, J. Am. Chem. Soc., 1993, 115, 3010 - 3011. DOI:10.1021/ja00060a073

Bridgehead enols: Thermodynamic vs Kinetic Control Part 2.

Bridgehead

Brendanone The ketone Brendan-2-one shown right exhibits unusual behaviour.6 When treated with NaOD/MeOD, deuterium substitution occurs easily and rapidly only in position Hb. Enolisation must of necessity form a bridgehead double bond (anti-Bredt), but clearly one isomer is more stable than the other possible form. Does molecular modelling predict this correctly? The unusually facile enolisation of this ketone (given that it forms an anti-Bredt enol) can also be investigated by molecular modelling. Measure the dihedral angle between the C-Ha or C-Hb vector and the carbonyl group. Assuming that the ideal angle for proton removal is around 90°, which proton is better set up for abstraction? Might this be kinetic rather than thermodynamic control? Cortisone One could also revisit Problem 2.3.3 above. Here, proton abstraction forms an enol which eventually epimerises the bridgehead position to form a trans ring junction. Why should this proton be particularly easy to remove? From what you have learnt above, would this be for kinetic or for thermodynamic reasons (or both?). Are all the relevant effects modelled using the mechanics approach or is consideration of the electrons also necessary?

References and Footnotes

1. A. Nickon, D. F. Covey, F.-C. Huang, and Y.-N. Kuo, Unusually facile bridgehead enolization. Locked boat forms in anti-Bredt olefins , J. Am. Chem. Soc., 1975, 97, 904 - 905; DOI:10.1021/ja00837a043 .

Sulfonylation of Naphthalene: Thermodynamic vs Kinetic Control Part 3.

Sulfonylation of naphthalene

The sulfonylation of naphthalene using sulfuric acid is a good example of a mechanism combining both steric and electronic influences. The Molecular mechanics method intrinsic to the Ghemical program can only model the former, and not the latter. It is a worthwhile exercise to establish whether this anticipated deficiency does indeed lead to a model which only partially explains experiment.

It has been known for some time that treating naphthalene with sulfuric acids at low temperatures produces mostly substitution at the 1-position of the naphthalene. Heating the reaction mixture, or conducting the reaction at elevated temperatures produces mostly the 2-isomer. This is indeed a classic example of kinetic vs thermodynamic control, the 1-isomer being the kinetic one and the 2-isomer the thermodynamic one. To model the kinetic reaction, we have to inspect the transition state for the reaction, and here we can approximate this by the Wheland Intermediate. To model the thermodynamic reaction, we have to inspect the product (rather than the transition state) for the reaction.

1. Build models for all four species shown in the diagram on the right. For the two products, define conjugated bond types for all the ring bonds, and define the sulfonyl group with two S=O double bonds and one S-O single bond. Take care to optimise the conformation of the sulfonyl group with respect to the aromatic ring. For the two Wheland intermediates, the limitations of Ghemical will force us to cheat. Ghemical does not have parameters for a carbocation. So define the C2-C3 bond as conjugated (for the 1-Wheland intermediate). When you add hydrogens it will in fact add a second hydrogen to C2. Delete this one hydrogen. Ghemical will calculated the energy regardless of not knowing C2 is actually a carbonium ion! For the 2-Wheland intermediate, ensure that you use exactly the same number of conjugated bond types as you did for the 1-isomer (the two models in a mechanics sense are only comparable if you have the same total number of bond types in each model). You will have to decide whether these (undoubted) approximations have produced reasonable models or not (is the naphthalene framework planar for example, as it should be?).
2. Record the pairs of energies (two for the 1- and 2-products, and two for each preceeding transition (Wheland) state.
3. By turning the spacefilling representation on, which of the two products has the least unfavourable steric interactions between the sulfonic acid group and any adjacent hydrogens? Does this match with their relative energies?
4. Do any unfavourable steric interactions observed in the product(s) also exist in the Wheland intermediates (as models for the transition states)?
5. The relative stability of the Wheland intermediates is always assumed to be an electronic phenomenon. The conventional explanation is that the 1-Wheland isomer is stablized by both one aromatic ring and an allyl cation conjugated to it. The 2-Wheland isomer is stabilised by one aromatic ring conjugated to a secondary carbocation and an alkene. This type of cross conjugation is conventionally assumed to be less favourable. Does a purely mechanical approach to this problem reproduce this expectation? Or is this mechanical approximation to an electronic model too severe? It seems a good point to stop this course, since the next time you will build models, it will indeed be using methods which properly approximate the electronic components.

References

1. R. Lantz, Mechanism of the monosulfonation of naphthalene, Compt. Rend. 1935, 201, 149-52.
2. G. W. Wheland, A Quantum Mechanical Investigation of the Orientation of Substituents in Aromatic Molecules, J. Am. Chem. Soc. 1942, 64, 900 - 908; DOI:10.1021/ja01256a047
3. C. A. Reed, N. L. P. Fackler, K-C. Kim, D. Stasko, D. R. Evans, P. D. W. Boyd, and C. E. F. Rickard, Isolation of Protonated Arenes (Wheland Intermediates) with BArF and Carborane Anions. A Novel Crystalline Superacid, J. Am. Chem. Soc. 1999, 121, 6314 - 6315 DOI:10.1021/ja981861z

Coursework not to be attempted at any time: Antimodelling Molecules

The following represent molecules that should not be modelled under any circumstances! You should instead attempt to NAME them.

If you know of any other antimodelling molecules, please add them here!

Acknowledgements

Some of these cartoons are from here, and six are original. A superb collection of silly names is maintained by Paul May . See DOI:10.1021/jo0349227 for the nanoputians. Music and Chemistry also go together.

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