Rauk Orbital Interaction Theory of Organic Chemistry

EXERCISES 279

6.Singlet carbenes react with carbonyl compounds to produce zwitterionic intermediates which can cyclize to epoxides or react further with another carbonyl to form 1,3-dioxacyclopentanes. Show which orbitals are involved and predict the geometry of approach of the carbene and carbonyl compound in the initial interaction that produces the zwitterion.

7.Explain the bisected geometry of cyclopropyl aldehyde.

8.Alkali metal reduction of ketones may involve the following steps:

(a)Discuss the electronic structure of each intermediate in terms of the orbitals involved.

(b)Show a plausible geometry for the transition state for the disproportionation step, according to orbital interaction theory.

(c)What additional features emerge from the study by Rautenstrauch et al. (Rautenstrauch, V.; MeÂgard, P.; Bourdin, B.; Furrer, A., J. Am. Chem. Soc., 1992, 114, 1418) on the reduction of camphors with potassium in liquid ammonia?

(d)Present a brief summary of stereochemical considerations in metal reductions of carbonyls as deduced in the theoretical study of Wu and Houk (Wu, Y.-D.; Houk, K. N., J. Am. Chem. Soc., 1992, 114, 1656).

9.Hahn and le Noble have discovered strongly enhanced stereoselectivity in the reduction of 5-substituted adamantanones where the substitution at C5 is by positive nitrogen (Hahn, J. M.; le Noble, W. J., J. Am. Chem. Soc., 1992, 114, 1916). Thus E=Z ratios for borohydride reduction of compounds 1±3 were in the range 20±25. Suggest a possible reason for these observations on the basis of orbital interaction theory.

Answer. A possible explanation for the observed diastereoselectivity of nucleophilic addition to the carbonyl involves a static distortion of the carbonyl group so as to improve p donation of the b s CÐC bonds into the p orbital of the carbonyl

280 EXERCISES

…c†

Figure B8.1. (a, b) Static distortion of carbonyl group to favor overlap with the highlighted bonds. The bonds in (a) are poorer p donors because of electrostatic e¨ects of the quaternary N center and interaction with sNR. Thus interaction (b) is favored. (c) Distortion in the transition state to favor donation into sCNu bond.

group, as shown in Figure B8.1. The consequent polarization of the orbital at carbon results in more favorable overlap with the HOMO of the nucleophile (Hÿ or HÐBHÿ3 , in the present case) and attack from the direction indicated by the arrow. Just such a distortion was observed theoretically in 2-adamantyl cation (Dutler, R.; Rauk, A.; Sorensen, T. S.; Whitworth, S. M., J. Am. Chem. Soc., 1989, 111, 9024± 9029). It may be argued that the static distortion in the ground state of the carbonyl is insu½cient to account for the high selectivity observed. According to the theory of Cieplak et al. (Cieplak, A. S.; Tait, B. D.; Johnson, C. R., J. Am. Chem. Soc., 1989, 111, 8447±8462), the distortion occurs naturally as one approaches the transition state, one p face being preferred by donation into the sCNu orbital (Figure B8.1c).

10. Unlike carbonyl compounds, thiocarbonyl compounds such as thioketones, R1R2CÐS, are very reactive and often impossible to isolate in a pure state. Apply principles of orbital interaction theory to explain:

(a)The high reacticity of thioketones toward nucleophiles

(b)The tendency of thioketones to dimerize. Predict the structure of the dimer.

EXERCISES 281

Chapter 9

1.Cyclopropanes may be obtained from the Michael condensation of bromoand chloromalonate carbanions (Le Menn, J.-C.; Tallec, A.; Sarrazin, J., Can. J. Chem., 1991, 69, 761±767).

The mechanism involves nucleophilic addition to a Z-substituted ole®n followed by an intramolecular bimolecular nucleophilic substitution. Several side reactions also occur. Discuss the chemistry involved in this reaction, pointing out substituent effects at each stage.

Chapter 10

1.Predict the structure of the complex formed between oxirane and HCl. Use orbital interaction arguments and draw an orbital interaction diagram. How does your prediction compare with the experimental structure (Legon, A. C.; Rego, C. A.; Wallwork, A. L., J. Chem. Phys., 1992, 97, 3050±3059)?

2.In principle, an alkyl halide, R1R2CHÐCClR3R4, may react with a nucleophile,

Nuÿ, by SN1, SN2, E1, E1cb, or E2 mechanisms. Choose R1, R2, R3, R4, and Nuÿ so that the course of the reaction may be expected to follow each of the mechanisms. The groups and nucleophile should be su½ciently di¨erent so that the stereochemical consequences of the reaction are obvious in your answer. Justify your choices using perturbative MO arguments but do not draw orbital interaction diagrams.

3.(a) Suggest a reason that ¯uoboric acid is a very strong acid.

(b)Why does Ph3CÐOH dehydrate in the presence of HBF4? Would CH3OH do the same thing?

(c)What is the function of acetic anhydride in the Ph3C‡ synthesis?

(d)Why does cycloheptatriene lose hydride readily? Would cyclopentadiene do the same?

(e)Why can one not prepare tropylium tetra¯uoborate by the more straightforward route shown below?

4.Suggest an explanation based on orbital interactions for the observed stereochemistry for E2 elimination reactions, that is, the strong stereoelectronic preference that the CÐH and CÐX bonds be anti-coplanar.

5.Using simple orbital interaction considerations, explain the following observations. Use orbital interaction diagrams and show the orbitals clearly.

(a)The acidity of the CÐH bond in cyclopentadiene …pKa ˆ 16†.

282 EXERCISES

…a†

…b†

Figure B10.1. The frontier orbitals of an ÐNH2 group and an ÐOH group. The stronger hydro- gen-bonding interaction (b) determines the most stable conformation of ethanolamine.

(b)Ethanolamine exists in a number of hydrogen-bonded conformations, two of which are shown below.

One of the two is substantially more stable than the other. Which is more stable? Explain. (RaÈsaÈnen, M.; Aspiala, A.; Homanen, L.; Murto, J., J. Mol. Struct., 1982, 96, 81.)

Answer to 5(b). The orbitals and orbital energies of generic ÐNH2 and ÐOH groups are shown side by side in Figure B10.1. Conformations a and b exhibit hydrogen-bonding interactions a and b, respectively. Interaction b is favored by the smaller energy gap and additionally by increased polarization of both nN and sOH orbitals.

6.The bond dissociation energy (BDE) and acidity constant …pKa† of cubane (CubÐH) have been determined experimentally by Eaton and co-workers (Hare, M.; Emrick, T.; Eaton, P. E.; Kass, S. R., J. Am. Chem. Soc., 1997, 119, 237±238). The BDE is unusually high for a tertiary CÐH bond, 427 kJ/mol, about 25 kJ/mol higher than in isobutane. The CÐH bond is also quite acidic, comparable to the acidity of the NÐH bond in ammonia, pKa ˆ 36, indicating an unusual stability for the anion

EXERCISES 283

Figure B10.2. Frontier orbitals of a CÐH bond in cubane. The high s character of the C orbital leads to a lower energy sCH.

Cub:ÿ. The normal pKa for a hydrocarbon is about 50. Use orbital interaction diagrams to explain these features.

Answer. The orbitals and orbital energies of a CÐH bond of cubane are shown in Figure B10.2. The key here is the nearly 90 angles of the CÐC sigma bonds. This suggests a high p character in the CÐC bonding orbitals and high s character for formally spn orbital of the CÐH bond. As a consequence of the greater energy separation of the C and H orbitals, the sCH orbital is lower in energy and more polarized than for a normal alkyl CÐH bond, and therefore more susceptible to attack by a base (more kinetic acidity). The resulting carbanion is more stable because of the lower energy of the spn orbital. The lower energy of this orbital, and consequently the lower energy of the sCH orbital, implies the greater CÐH bond dissociation energy of the cubane CÐH bond. The arguments respecting the acidity and BDE of the cubane CÐH bond are derived from the discussion of heterolytic and

284 EXERCISES

homolytic bond dissociations in Chapter 4. Entirely analogous arguments apply to the increasing acidity and increasing BDEs of CÐH bonds in the series, alkyl CH, alkenyl CH, and alkynyl CH in Chapter 10. In this series, increased s character is due to the hybridization state of the C atom.

7. The Lewis acid±base complex BH3NH3 crystallizes in a form which has a short

ˆ Ê

H H distance (dHH 2:02 A), which has been described as a ``dihydrogen bond'' (Klooster, W. T.; Koetzle, T. F.; Siegbahn, P. E. M.; Richardson, T. B.; Crabtree, R. H., J. Am. Chem. Soc., 1999, 121, 6337±6343). Provide a description of the BÐH HÐN ``bond'' using orbital interaction theory. Does your description also explain why the geometry is bent at BÐH H and nearly linear at H HÐN?

Chapter 11

1.Use orbital interaction diagrams to explain why methoxybenzene (anisole) prefers a conformation in which the methyl group lies in the plane of the aromatic ring.

2.Use orbital interaction diagrams to explain why benzyne is an excellent dienophile in Diels±Alder reactions.

Indicate the probable orbitals involved in the 343 and 500 nm absorption of this species (see Kim, E. K.; Kochi, J. K., J. Am. Chem. Soc., 1991, 113, 4962).

Answer. We postulate that the structure of the complex will be that which provides the most favorable HOMO±LUMO interaction(s), while minimizing interactions between occupied orbitals. The NO‡ is isoelectronic with CO. Its LUMO will be very low lying, a degenerate pair of p orbitals polarized toward the nitrogen atom. The benzene HOMOs also are a degenerate pair with a single nodal surface bisecting the ring. Two structures may be postulated for the complex. An attractive possibility involves both HOMOs and both LUMOs, as shown in Figure B11.1a, analogous to transition metal carbonyl complexes. This complex, which has symmetry C6v, has two serious strikes against it. First, there is no obvious bonding contribution from the sp hybrid orbital on N, as there is in transition metal carbonyl complexes, since the benzene LUMO has the wrong symmetry to interact with it. The sp hybrid orbital interactions are destabilizing. Second, the p orbitals of NO‡ are not as highly polarized as they are in CO, leaving more of the orbital not involved in bonding. An alternative structure which does not have these disadvantages but uses only one of the HOMO±LUMO pairs for bonding is shown in Figure B11.1b. The corresponding interaction diagram (Figure B11.1c) suggests a possible secondary interaction of the component LUMOs, which lowers the LUMO for the system. Proposals for the 343and 500-nm transitions are shown as dashed lines.

5.The cyclopentadienyl cation 1 has been studied extensively both theoretically and experimentally and was recently generated under solvolytic conditions (Allen, A. D.;

EXERCISES 285

…a†

Figure B11.1. (a) Bonding interactions and structure of benzene±NO‡ complex with C6v symmetry. (b) Bonding interactions and structure of benzene±NO‡ complex with Cs symmetry. (c) Interaction diagram for the Cs complex. The dashed lines correspond to possible assignments for the two lowest transitions.

Sumonja, M.; Tidwell, T. T., J. Am. Chem. Soc., 1997, 119, 2371±2375, and references therein). It has the properties expected of an antiaromatic compound. It is very di½cult to generate 1 under solvolytic (SN1) conditions from the precursor iodide 2. Nevertheless, 2 undergoes SN2 reaction unusually rapidly (2 reacts with bromide ion about 10 times faster than 3).

286 EXERCISES

Use principles of orbital interaction theory to explain:

(a)Why the ground state of 1 is not singlet but triplet

(b)Why the C5H5 ring is usually found as an anion

(c)Why nucleophilic substitution of 2 by bromide ion may be faster than substitution of the saturated compound cyclopentyl iodide 3 by bromide

6.A study by Abu-Raqabah and Symons (Abu-Raqabah, A.; Symons, M. C. R., J. Am.

Chem. Soc., 1990, 112, 8614) has characterized the pyridine±chlorine atom threeelectron bonded species Py Cl by ESR and UV spectroscopy. In an earlier paper, Breslow and co-workers (Breslow, R.; Brandl, M.; Hunger, J.; Adams, A. D., J. Am. Chem. Soc., 1987, 109, 3799) considered ring acylated pyridine±chlorine radicals to be p radicals and anticipated special stability for the 4-carboalkoxypyridine±chlorine radical.

(a)Using an orbital interaction diagram, provide a bonding description that explains the red-shifted s ! s UV absorption and the increased length of the NÐCl bond.

(b)Show by means of orbital interaction arguments whether Breslow's expectation is justi®ed or not. In other words, would a Z substituent in the 4-position stabilize a pyridine±chlorine p radical?

Answer to 6(a). We consider the radical to be derived by addition of an electron to N-chloropyridinium cation. The character of the radical is in doubt because it is uncertain whether the LUMO of the N-chloropyridinium cation will be one of the p orbitals, or sNCl. While s orbitals will, as a rule, be higher than p orbitals, this may not be the case here because of two factors: (a) the two orbitals involved in the s-type interaction, nN and 3pCl, are both low in energy to start with and (b) the fact that a 3p rather than a 2p orbital is involved reduces the magnitude of the intrinsic interaction matrix element as explained in Chapter 4. The experimental evidence of Abu-Raqabah and Symons is consistent with a s-type radical as shown in Figure B11.2. Occupancy of the sNCl orbital would lead to a substantial reduction in the NÐCl bond order. The consequent lengthening of the NÐCl bond is accompanied by reduction of the s±s gap, which is seen as a red shift of the s±s UV absorption.

Answer to 6(b). The p orbitals of N-chloropyridinium cation as calculated by SHMO are shown in Figure B11.3. An extra electron has been added to the LUMO p4. Since p4 has its largest coe½cient at the 4-position, it is clear that Breslow's expectation is correct; the SOMO of a p radical would interact favorably with a Z substituent at the 4-position. This is readily veri®ed by an SHMO calculation on the N-chloropyridinyl radical with a Z-substituent at this position.

7.The ``inorganic benzene'' borazine has been shown to undergo electrophilic substitution in a manner very similar to benzene itself. Does the electrophile attack at B or N? Use orbital interaction theory to predict the site of attack. Compare this with the prediction on the basis of an SHMO calculation and experimental ®ndings (Chiavarino, B.; Crestoni, M. E.; Di Marzio, A.; Fornarini, S.; Rosi, M., J. Am. Chem. Soc., 1999, 121, 11204±11210).

Answer. This is a bit of a trick question since the ``Kekule resonance structure'' of borazine tends to imply that the positively charged electrophile should go for the negatively charged boron atoms. Consideration of the orbital interactions shown in Figure B11.4 quickly assures us that the occupied MO electron density is almost

288 EXERCISES

Figure B11.4. The SHMO p orbitals and orbital energies of borazine, shown as the result of orbital interactions between symmetrized group orbitals of an N3 equilateral triangle and a B3 equilateral triangle.

entirely on the N atoms, so this must be the site of electrophilic attack. The energies and orbital sizes shown in the ®gure are readily derived by use of the SHMO program using parameters appropriate for tricoordinated boron and nitrogen atoms.

Chapter 12

1.In a Diels±Alder reaction, when both p systems are polarized, the more favorable overlap, and therefore the stronger interaction, occurs when the ends with the larger coe½cients get together and the smaller coe½cients get together. Predict the major product of each of the following cycloaddition reactions:

(a)(1E )-1-phenyl-1,3-butadiene ‡ acrolein !

(b)(3E )-2-methyl-1,3-pentadiene ‡ formaldehyde !