Coursework

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 3 hour session allocated to you, and return to finish the rest if you wish at your convenience. At the end of the 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-like conformations of Cyclohexane

Construct chair and boat-like conformations of cyclohexane. Compare the energies of both forms. Check carefully if your boat really is a boat, or whether it has any apparent distorsion. Try changing one or more of the CH2 groups into an oxygen and see if that affects things. For the record, the point group symmetries of the various species which may be involved are D3d for the chair conformation, C2v for a boat form (if it exists?), and D2 for any twisted boat form.

References

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 . Its pretty clear from this article that a boat form of cyclohexane does not actually exist as a stable species, it instead being a transition state connecting two twist-boat structures.
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 .

Enantiomers vs Diastereomers Part 1: Butanes and Helicenes.

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

1. 

   2-bromo-3-chlorobutane
   

   Pentahelicene
   The compound 2-bromo-3-chlorobutane has two chiral centres, and four isomers (2²) are therefore possible. Calculate all four isomers. For each isomer, you will have to think about whether you have obtained the lowest energy conformer.
2. Can your four energies be grouped? The expected result is you get two pairs of energies. Each pair should correspond to enantiomers, and the two enantiomers should have identical energies. Any two compounds which have different energies should instead be diastereomers (or different conformers of enantiomers, which is why you should strive to find the lowest energy conformer). Can you reproduce this pattern?
3. Armed with the rotatable 3D models, does it now become easier to assign the (R) and (S) configuration to each of your four isomers?

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. (Note: the smallest helicene which can be resolved experimentally into enantiomers is in fact [5]helicene]). 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). [7]circulene is a known molecule, with a unique saddle-shaped structure, shown on the left (there is no real need for you to build this model, but do please do so if you are curious). Graphene is a related polymeric molecule, of much topical interest in the semi-conducting and other industries.

References

1. Wikipedia article on Diastereomers
2. Wikipedia article on Helicenes and related molecules
3. 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, Steroids and Podcasts!

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 two of 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). 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 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. Optional: 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). Optional: 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. Wikipedia article
3. 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 .
4. For a video-Podcast of Barton and Woodward (and other Nobel prize winners), subscribe here
5. 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 .
6. 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

Axial/Equatorial preferences in cyclohexane and cyclohexanone and Hydrogen Bonding

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.

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

Menthone Beckmann (of rearrangement fame) in 1889 dissolved optically active levorotatory (-)-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) (+)-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, only one of which allows the [S] asymmetric carbon to become planar and then protonate to the [R] epimer. This is the so called thermodynamic enol. The other, which leaves the [S]-centre untouched is the kinetic enol. Find out if simple molecular modelling correctly predicts that the thermodynamic enol is indeed the more stable of the two. 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).

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 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!.
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.

Additional 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.

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 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 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 2: 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
   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/jp013160
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.

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! (OK, the last example is NOT a molecule).

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.

1. The last word!

Follow ups to this Course

The molecular mechanics procedure is quick and simple, but not always accurate. Different molecular mechanics force fields also vary in their accuracy. The most accurate tend to be part of complex programs, or commercial. The one you are using in Ghemical is a relatively basic one, and may exhibit more artefacts than e.g a commercial one such as found in the Chem3D program.

A proper molecular model must also take into account electrons, as noted above. But solving the necessary equations takes much more computer time. In later courses in 2nd year, you will be shown how to do this, using programs such as Gaussview, Gaussian, GAMESS, and the like. Third and fourth year courses deal with the theory and practice in much more detail.

Further Documentation, Reading and Viewing

1. Ghemical Manual gives more advanced options, but be aware it relates to an earlier version of Ghemical.
2. Second year modelling experiment on the thermal expansion of MgO.
3. Third year modelling experiment undertaken in the third year organic chemistry laboratory.
4. Third year modelling lab on Inorganic Chemistry, including three advanced individual projects on Mo(CO)₄L₂, boron based acids and Gold interactions with Water.
5. A local third year course on organic molecular modelling with a number of more elaborate case studies illustrating the application of molecular modelling.
6. Some further local examples of molecular models deriving from first and second year problem classes and tutorials.
7. The Wikipedia page on molecular modelling, a short summary which gives some good further leads.
8. The Wikipedia page on molecular graphics, a technique that goes hand in hand with molecular modelling.
9. A Wikibook on organic chemistry
10. The grand Daddy of all molecular models, invented at Imperial College around 1860, and now in the archives of the Royal Institution. These models are the source of the familar colour scheme now used, i.e. Hydrogen=White, Oxygen=Red, Nitrogen=blue, etc.
11. Another father of molecular modelling, but only on paper!, also achieved in 1861. Loschmidt constructed these models in the same sense that Watson and Crick did for DNA, as proposals, and not representing structural proof in any way.
12. For an interesting way of presenting scientific genealogies of scientists, see J. Andraos, Scientific genealogies of physical and mechanistic organic chemists, Can. J. Chem./Rev. Can. Chim., 2005, 83, 1400-1414. DOI:
13. The preception of the 3D character of many molecules can be enhanced by viewing using stereoscopic systems. One such system is available for student use, and lecture theatre C is equipped with stereoscopic projection.

Running Ghemical on your own Computer

1. Go get the software from here. It installs on either Windows XP or MacOS X. For installation notes see here
2. Although there are many available molecular modelling programs, many are commercial, and a fair proportion handle only the visualization part of the modelling, and not the geometry optimization part. Two general purpose programs that are licensed for use in the department are Chem3D and CAChe. The combination Gaussview/Gaussian 03 is available for high level calculations.

Submitting more accurate calculations to the Departmental SCAN Cluster

Export from Ghemical

The Chemistry department runs a SCAN (Supercomputer at Night) system, whereby teaching computers which would otherwise only idle in the middle of the night, can be used to run more time consuming calculations than is possible interactively on a single computer whilst sitting in front of it.

One far more reliable and quantitative way of modelling a molecule is to subject it to quantum mechanical modelling using Density Functional theory. In practice, this is implemented here using a program called Gaussian 03. The procedure to submit such a job is as follows:

Creating an Input file

• After you have optimised your sketched molecule using Ghemical, as described above, right click in the black display window. This will produce the floating menu, from which you select file and then Export. Select Gaussian 98/03 Cartesian Input for the type and type a name for the file (make sure that the name of the file ends with .gjf). It will be saved in your H: drive by default.
• 

  Typical Gaussian input
  The file will have to be edited before it can be submitted. You can do this either with Gaussview as the program, but a much simpler method is to open the file (pentahelicene.gjf in this example) using eg the Windows Wordpad editor. This is invoked simply by double clicking on the file. Remove any existing lines starting with % or # and replace them with one of the following single lines (the second example also results in the vibrational frequences and from these the entropy being computed, and hence the zero-point and free-energy corrected value, ΔG). This latter option will take significantly longer however.
  

# B3LYP/6-31G(d) opt
or
# B3LYP/6-31G(d) opt freq
to produce a file that looks like the one shown on the right.

• For a molecule the size of e.g. pentahelicene, the calculation will take about 4-5 hours overnite. If for some reason, your molecule is taking longer, you can always reduce the size of the basis set to e.g. B3LYP/3-21G*, or submit the job on a Friday, when it will have the entire weekend available to it. If you want greater accuracy (but for longer computing time), try e.g. # B3LYP/cc-pVTZ opt freq.

Submitting the Input file

• 

  Create a new job
  You will have to login as yourself. You can submit as many jobs as you wish through this mechanism, but you must prepare the input (.gjf) file for each first. The SCAN operates during the period 23.00-07.30 overnight. If a job is not completed during this period, it will be scheduled to run again (from the beginning) the next night. For this reason, you should only schedule jobs that can complete in an 8 hour window. In practice this means submitting molecules only a little bit larger than pentahelicene.
• 

  Create a project
  

  Select a pool
  After you are logged in you should organise your jobs by project. Create a suitable new project, then select New job, the Application (currently only Gaussian) the Project, and press continue.
  
  
• 

  Upload your input file
  You now have to find the Gaussian input file, as prepared above. You should Browse to drive H: to find this file. Add a description which will help you identify the job.
• 

  The Chemistry Condor Pool
  The job will be added to your list of jobs, andyou can view its status (but this depends on there being a vacant machine in the Condor pool).
  
  
• 

  Viewing the outputs
  When the job has completed, click on the Job List link. This will show all available outputs. Download the program Log file (this will help you chart whether the calculation was successfull) or the Gaussian Formatted Checkpoint file onto the desktop of the computer you are using, and the file should open up Gaussview, where the molecule can be viewed and checked. You can use the latter file to e.g. plot molecular orbitals for the molecule, view vibrational modes, etc. Full details of these procedures are described in the Gaussview manuals.

Archiving the output into a digital repository

Depositing an entry in DSpace

A very recent innovation is the Institutional digital repository, a resource for permanently archiving calculations, spectra and crystal structures. You can get a flavour of this by archiving your own calculation in the SPECTRa digital repository. To the right of the Portal display is a link termed Publish. If you click on this, and the calculation is actually in a state to be published (it may for example have failed for some reason) then appropriate metadata for the calculation is collected, and the collection deposited into the repository. From here, it can be retrieved in future.

About this wiki: Opencourseware

This course is presented as a wiki. This differs from conventional hand-outs or web pages in several aspects.

1. Anyone (who has a valid Imperial College login and password) can edit it, for the purpose of correcting errors, clarifying ambiguities, and even adding more examples, or references to existing examples. However, this activity is not anonymous; you can see who has done what by inspecting the history of the article. If you are considering making changes, go read these rules first.
2. You may notice that some terms appear in red. This is because the original author has enclosed the term thus: [[red]], acting as a suggestion or hint that someone may wish to pick up this term, and expand it into something informative. If you think you can add something helpful to others, please go ahead: click on the red section and starting editing! If the result contains inaccuracies, someone may come along and correct them. If you are dubious that this scheme works, just go visit Wikipedia. The idea behind this is that we produce joined up courses and not just isolated islands of information and knowledge.
3. You can also hit the edit button if you want to find out how any particular effect is achieved. You do not have to actually change anything.
4. This is an experiment! If you have any comments on the experiment, or suggestions for improvements, go instead to the discussion page and say something there. Do however remember that anyone in the world (!) can see this (it is opencourseware, go read this stimulating and provocative view of how knowledge may be owned and disseminated in the future), so remember not to write anything inappropriate. You cannot do so anonymously!