Rauk Orbital Interaction Theory of Organic Chemistry

EXERCISES 259

p-type interaction is maximized if the two orbitals are coplanar, that is, if the OF bond is perpendicular to the OOF plane. The occupied orbital that results from the interaction is bonding in the OÐO region and antibonding in the OÐF region, thus accounting for the shortened OÐO separation and the increased OF separation. The interaction is mirrored by the other OÐOF pair, and the two pairs of interactions are synergistic, that is, they feed on each other. Transfer of electrons from one end to the other raises the nO orbital energy of the acceptor side and lowers the sOF on the donor side, thereby promoting the back donation of charge.

6.The propertiesof DABCO (1,4-diazabicyclo[2.2.2]octane) are of considerable interest. The inand out-of-phase interaction of the two occupied sp3 hybridized lone pairs on the two nitrogen atoms may be described by an interaction diagram such as shown in Figure 3.5b.

As expected, DABCO is a much stronger base than triethylamine, has two ionization potentials separated by several electron-volts, and is more easily oxidized than trimethylamine. What is unusual about DABCO is that the in-phase combination of the nonbonded pairs is the HOMO, the lowest IP is lower than that of trimethylamine, and the next higher IP is higher than that of trimethylamine. Explain.

Answer. This question asks us to explain some unexpected features of a fairly simple molecule. Based on the simplest application of orbital interaction theory, as shown in Figure B3.7a, one would expect the HOMO and HOMO-1 of DABCO to be the

out-of-phase …a200† and in-phase …a10 † combinations of the interacting nitrogen lone pairs, respectively. The reference point may be taken to be the energy of one of the N lone pairs, nN, which can also be taken as the energy of the HOMO of trimethylamine, that is, the IP by Koopmans' theorem. Each nitrogen atom is a pyramidal tricoordinated atom, and the single remaining valence orbital is an spn hybrid directed away from the point of the pyramid. The two interacting spn hybrid orbitals are pointed away from each other and so will interact only weakly through the smaller back lobes. Nevertheless, one would expect to observe that DABCO has one IP which is lower than that of trimethylamine (ionization from the HOMO) and one which is higher (ionization from the HOMO-1). Everything is as expected until one realizes that the actual symmetries of HOMO and HOMO-1 of DABCO are reversed! The explanation must be that there is a secondary interaction with the CÐC sigma bonds which cannot be neglected. This is shown in Figure B3.7b. Be-

cause of the threefold symmetry, the three CÐC bonding orbitals interact to give

three group orbitals, one of which has a10 symmetry and a degenerate pair of e0 symmetry. Likewise, the sCC orbitals combine to form an orbital of symmetry a200 and a degenerate pair of symmetry e00. These are given on the right-hand side of Figure

B3.7b, although only the a10 and a200 combinations are drawn out explicitly. Interaction of the occupied a10 combination of nN with the occupied a10 combination of sCC e¨ectively raises the a10 nN combination, mixed out of phase with the a10 sCC combination. At the same time, the occupied a200 nN combination is lowered by in-phase interaction of the a200 sCC combination, evidently enough to end up below the raised a10 orbital. This interaction with the s bonding orbitals is strong enough to ``undo'' the weaker primary nN ±nN interaction.

260 EXERCISES

Figure B3.7. (a) Four-electron, two-orbital interaction diagram for the N lone pairs, nN; (b) secondary interactions with the in-phase combinations of the CÐC s and s orbitals which reverse the order of the HOMO and HOMO-1. The orbitals symmetries are speci®ed in D3h.

7.(a) Draw a two-orbital interaction diagram for the BÐC bond, as in H2BÐCH3. Label the diagram clearly. Sketch all of the orbitals. Identify the orbitals (e.g., n; p; p ; s; s , etc., as appropriate).

(b)Draw a separate two-orbital interaction diagram for H2BÐCH3, and explain on the basis of the diagram how the presence of the BH2 group may serve to acidify a CH bond toward proton donation.

8.Use a two-orbital interaction diagram to explain the observed feature of any one of the following systems. Note: A clear orbital interaction diagram includes pictures of the orbitals before and after the interaction and shows the disposition of the electrons. A brief verbal explanation of the diagram, in addition to the interpretation, is also desirable. If more than two orbitals seem to be involved, use your judgment to choose the two most important orbitals.

(a)The barrier to rotation about the CÐN bond of formamide, NH2CHO, is 80 kJ/mol, substantially larger than the value usually found for a CÐN single bond, 8 kJ/mol.

(b)Reaction of alkenes with BrCl proceeds via the intermediate [alkene-Br]‡ rather than the intermediate [alkene-Cl]‡.

(c)Ammonia and ClF form a 1 : 1 complex in the gas phase. Predict the structure of the donor±acceptor complex.

EXERCISES 261

9.The structure of the radical cation of ethylene, [CH2CH2]‡., is not D2h, like ethylene, but rather D2, with a twist of 25 (Merer, A. J.; Schoonveld, L., Can. J. Phys., 1969, 47, 1731. See also (a) KoÈppel, H.; Domcke, W.; Cederbaum, L. S.; von Niessen, W.,

J. Chem. Phys., 1978, 69, 4252±4263. (b) Eriksson, L. A.; Lunell, S.; Boyd, R. J., J. Am. Chem. Soc., 1993, 115, 6896±6900). The equilibrium geometry of the p±p state of ethylene has D2d symmetry. Explain, in terms of orbital interaction theory.

10.Photoexcitation, in orbital terms, can be described as the promotion of a single electron from HOMO to LUMO by the interaction with light. When either transor cis-2-butene is subjected to UV light, the same mixture of cis and trans isomers results. Explain.

Answer to 9 and 10. The orbital interaction diagram for the bonding is shown in Figure B3.8. The geometric changes described for ethylene, its cation radical, and its excited state reinforce our conclusions regarding the e¨ect of overlap, namely that

Figure B3.8. Interaction diagrams: (a) between 2p orbital with the antisymmetric group orbital of the adjacent CH2 group; (b) of p orbitals ethylene in its ground state; (c) of twisted ethylene cation; (d ) of the p±p state of ethylene.

262 EXERCISES

DeU > DeL. It would be su½cient to explain the change in geometry of the p±p state on this basis. The p±p state has one electron in each of the p and p orbitals. If the electrons must be separated, it would be energetically better not to have the interaction between the p orbitals so the geometry changes to D2d where the p orbitals are at right angles and do not overlap by symmetry (Figure B3.8d ). However, if this were the whole story, then the geometry of the ethylene radical cation should also be planar, like ethylene itself. The surprising twisted geometry reveals that a second interaction, forbidden by symmetry in the planar D2h structure, is turned on as twisting about the CÐC bond takes place. This is the interaction between a p orbital of one C atom with the p-like group orbital of the adjacent s bonds which reaches a maximum in the 90 D2d geometry (Figure B3.8a). This interaction occurs at each end (only one is shown in Figure B3.8a). It raises the energy of the p orbital and is attractive with two or three electrons, although in the latter case, less so than a full-¯edged p interaction. Evidently, in the case of the cation radical, the loss of a single electron's worth of p bonding is partially compensated by the interaction with the s bonds, hence the twisting. The situation in the case of the photoexcited cisor trans-2-butene is explained by Figures B3.8b and B3.8d. Both reach the same p±p state geometry, which deviates slightly from local D2d symmetry toward the trans-2- butene structure because of the steric interaction between the methyl groups. When it reverts to the ground state (both electrons go into the same orbital), more of it becomes trans than cis. Actually, in pure alkene, isomerism takes second place to another much faster photoreaction, dimerization. This reaction is also readily understood in orbital terms, but discussion of it is reserved until Chapter 14.

Chapter 4

1.This question deals with the description of a typical dative bond.

(a)Draw a two-orbital interaction diagram for the BÐN bond, as in H3BÿÐN‡H3. Label the diagram clearly.

(b)Sketch all of the orbitals. Identify the orbitals (e.g., n; p; p ; s; s , etc., as appropriate).

(c)Comment on the polarity of the BÐN bond. Why is the boron atom shown with a formal charge of ÿ1?

2.Reaction of alkenes with BrCl proceeds via the intermediate [alkene-Br]‡ rather than the intermediate [alkene-Cl]‡. Explain using a two-orbital interaction diagram. (The structure of the complex between ethylene and BrCl in the gas phase has been determined by microwave spectroscopy. It is T-shaped, with the BrCl molecule lying

perpendicular to the ethylene plane and pointing bromine-end ®rst toward the midpoint of the CÐC bond: Legon, A. C.; Bloemink, H. I.; Hinds, K.; Thorn, J. C.,

Chem. Eng. News, 1994 Nov. 7, 26±29.)

3.Suggest a reason for the rapid expulsion of atomic bromine upon one-electron reduction of N-bromosuccinimide (Lind, J.; Shen, X.; Eriksen, T. E.; MereÂnyi, G.; Eberson, L., J. Am. Chem. Soc., 1991, 113, 4629).

4.Predict the structures and bonding in donor±acceptor complexes of ammonia with F2, Cl2, and ClF. What are the expected relative bond strengths? Are your predictions in accord with the results of high-level theoretical calculations of Rùggen and Dahl (Rùggen, I.; Dahl, T., J. Am. Chem. Soc., 1992, 114, 511)?

EXERCISES 263

5.In the dihalogenated ethanes, 1,2-dichloroethane and 1,2-dibromoethane preferentially adopt the conformation in which the two halogens are anti to each other as expected on the basis of steric repulsions. On the other hand, 1,2-di¯uoroethane exists predominantly in the gauche conformation (Friesen, D.; Hedberg, K., J. Am. Chem. Soc., 1980, 102, 3987). This unexpected behavior has been explained as a manifestation of the gauche e¨ect (Wolfe, S., Acc. Chem. Res., 1972, 5, 102) from the formation of ``bent'' bonds (Wiberg, K. B.; Murcko, M. A.; Laidig, K. E.; MacDougall, P. J., J. Phys. Chem., 1990, 94, 6956) or as due to hyperconjugative interaction between the CÐF and CÐH bonds (Radom, L., Prog. Theor. Org. Chem., 1982, 3, 1). The last explanation relies on orbital interaction theory. Can you reproduce the arguments?

6.In a microwave investigation, Lopata and Kuczkowski (Lopata, A. D.; Kuczkowski, R. L., J. Am. Chem. Soc., 1981, 103, 3304±3309) determined that the equilibrium geometry of FCH2 ÐOÐCHO was as shown below. Explain the near 90 orientation of the CF bond relative to the plane of the rest of the molecule (the explanation is the same as for the anomeric e¨ect in sugars.)

7.Low-energy electron impact spectroscopy of [1.1.1]propellane reveals an unusually low energy for electron capture to form the radical anion (2.04 eV compared to about 6 eV for most alkanes):

In addition, appreciable lengthening of the interbridgehead CÐC bond is indicated (Schafer, O.; Allan, M.; Szeimies, G.; Sanktjohanser, M., J. Am. Chem. Soc., 1992, 114, 8180±8186). Can you explain these observations on the basis of orbital interaction theory? Compare your explanation with that o¨ered in the reference.

Answer. The orbitals of a normal CÐC bond and of the central CÐC bond of [1.1.1]propellane are shown in Figures B4.1a and B4.1b, respectively. For both systems, the interacting orbitals are spn hybrids at about the same energy. The direction of polarization is dictated by the rest of the s framework. In the case of the propellane, the hybrid orbitals are pointed away from each other and therefore interact much more weakly, resulting in a much lower s as LUMO. This accounts for the ease of capture of an electron. The anion would have an electron in this orbital, thereby reducing the bond order and weakening the bond further.

8.Solve the HMO equations for the s orbitals and orbital energies of the CÐC and CÐO bonds: assume that h…O† ˆ h…C† ÿ jhCOj; hCC ˆ hCO, and SCO ˆ 0. Sketch the results in the form of an interaction diagram. Which bond is stronger? Calculate the homolytic bond dissociation energies in units of jhCCj. What is the net charge on O, assuming that it arises solely from the polarization of the s bond?

EXERCISES 265

2.The results of an SHMO calculation on (Z)-pentadienone, 1, are given below (Note: aO: ˆ a ÿ 0:97jbj, bCO ˆ ÿ1:09jbj):

(a)Sketch the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) on the basis of the calculation.

(b)Assuming that 1 were to react with an electrophile, where is the most likely site of attack? Compare the Lewis basicity of 1 to that of ethylene.

(c)Assuming that 1 were to react with a nucleophile, where is the most likely site of attack? Compare the Lewis acidity of 1 to that of ethylene.

(d)In terms of jbj, estimate the energy of the lowest p±p electronic excitation of 1. Is the p±p state likely to be higher or lower than the n±p state? Explain in less than three lines.

(e)Calculate the p bond order of the C2 ÐC3 bond of 1. What is the net charge on O1?

(f ) Show the most likely product from a Diels±Alder reaction of 1 with 1,3- butadiene. Pay attention to all relevant aspects of stereoand regioselectivity.

Answer. There are a number of questions of this type which are primarily concerned with the interpretation of the results of MO calculations and their relationship with the ideas of orbital interaction theory. The data are derived from a calculation using SHMO.

(a)Pentadienone 1 has six p electrons. A ``sketch'' of the frontier MOs is shown in Figure B5.1. The relative sizes of the atomic orbital are proportional to the sizes of the MO coe½cients and the relatives phases are shown by shading. The convention is adopted that a positive signed coe½cient corresponds to a p orbital with the unshaded lobe up and a negative coe½cient is portrayed by a p orbital with the shaded lobe up. The actual convention you use is unimportant, but you must be consistent.

(b)Attack by an electrophile will be directed to the part of the HOMO which has the largest coe½cient, in this case, atom 3. Atom 6 is a close second. Basicity relative to ethylene is governed by the relative energy of the HOMO and by the relative magnitudes of the largest coe½cients. The HOMO is substantially higher, suggesting greater basicity, but the largest coe½cient (0.55) is smaller

266 EXERCISES

Figure B5.1. The HOMO and LUMO of pentadienone and ethylene (SHMO).

than in ethylene (0.71), suggesting lower basicity. Since the two factors are in opposition, one might expect that the basicity will be similar, perhaps slightly greater than that of ethylene.

(c)Attack by a nucleophile will be directed to the part of the LUMO which has the largest coe½cient, in this case, atom 6. Atom 4 is second. Acidity in the Lewis sense relative to ethylene is governed by the relative energy of the LUMO and by the relative magnitudes of the largest coe½cients. The LUMO is much lower, suggesting signi®cantly greater acidity. Again, the largest coe½cient (0.55) is smaller than in ethylene (0.71), suggesting lower acidity. Although the two factors are again in opposition, one might expect that the large energy di¨erence will dominate and that the acidity will be greater than that of ethylene. (In fact, it is extremely di½cult to do a nucleophilic attack on ethylene.)

(d)The energy of the p±p excitation in SHMO theory is just the di¨erence of the LUMO and HOMO orbital energies, 0:961jbj. The ``n'' orbital is a nonbonding orbital on oxygen which lies in the plane of the molecule and is not part of the p system. Its energy is that of a p orbital on a monocoordinated oxygen, a ÿ 0:97jbj. Since the n orbital is lower than the highest occupied p orbital, the excitation should have lower energy than the n±p excitation so the p±p state will be lower in energy than the n±p state.

(e)The bond order of the C2 ÐC3 bond is 2…0:53†…0:45† ‡ 2…ÿ0:25†…0:15† ‡ 2…ÿ0:12†…ÿ0:55† ˆ 0:53. The total electron population on atom 1 is 2…0:57†2 ‡ 2…ÿ0:49†2 ‡ 2…0:45†2 ˆ 1:535 . Since a monocoordinated oxygen atom is neutral with one electron in the p system, the net charge is 1 ÿ 1:535 ˆ ÿ0:535

EXERCISES 267

(f ) The important orbital interaction for the Diels±Alder reaction is the LUMO- (pentadienone)±HOMO(butadiene). The strongest interaction (best overlap) originates if the two molecules are oriented as follows:

The interaction is driven by the overlap of the orbitals with the largest coe½cients of each MO.

3.The results of an SHMO calculation on furan, 2, are given below (Note: aO ˆ aC ÿ 2:09jbCCj, bCO ˆ ÿ0:66jbCCj):

(a)Sketch the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) on the basis of the calculation.

(b)Assuming that 2 were to react with an electrophile, where is the most likely site of attack? Compare the Lewis basicity of 2 to that of ethylene. Show the products of such a reaction (pick a speci®c electrophile).

(c)Assuming that 2 were to react with a nucleophile, where is the most likely site of attack? Compare the Lewis acidity of 2 to that of ethylene. Show the products of such a reaction (pick a speci®c nucleophile).

(d)Consider an equimolar mixture of 2 and 2-cyanofuran, 3. A Diels±Alder product is obtained from the mixture. Predict the structure of the product. The HOMO and LUMO of 3 are shown below.

268 EXERCISES

(e)In terms of a and/or jbj, provide an estimate for the lowest ionization potential of 2 according to the calculation.

(f ) In terms of jbj, estimate the energy of the lowest p±p electronic excitation of 2.

(g)Furan 2 belongs to the point group C2v. Indicate whether the HOMO is symmetric or antisymmetric with respect to each of the three symmetry elements of 2. Do the same for the LUMO. Do the same for the lowest p±p state of 2.

(h)Calculate the net charge on the oxygen atom of 2.

(i)Calculate the p bond order of the CÐO bond of 2.

4.The results of a SHMO calculation on cyclopentadienone, 4, are given below (Note: aO: ˆ a ÿ 0:97jbj, bCO ˆ ÿ1:09jbj):

(a)Sketch the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) on the basis of the calculation.

(b)Assuming that 4 were to react with an electrophile, where is the most likely site of attack? Compare the Lewis basicity of 4 to that of ethylene.

(c)Assuming that 4 were to react with a nucleophile, where is the most likely site of attack? Compare the Lewis acidity of 4 to that of ethylene.

(d)In terms of jbj, estimate the energy of the lowest p±p electronic excitation of 4. Is the p±p state likely to be higher or lower than the n±p state? Explain in less than three lines. Is it plausible that compounds with this structure might be colored (recall tetraphenylcyclopentadienone)?

(e)Calculate the p bond order of the C2 ÐC3 bond of 4.

(f ) Show the most likely product from a Diels±Alder reaction of 4 with 1,3- butadiene. Pay attention to all relevant aspects of stereoand regioselectivity (recall your laboratory chemistry with the tetraphenyl derivative).

(g)Compound 4 is expected to be very reactive and has never been synthesized. Can you suggest a reason? You made a tetraphenyl derivative of 4 in the laboratory. Suggest the manner in which the phenyl groups might stabilize the compound.

5.The results of a SHMO calculation on fulvene, 5 (R ˆ H), are given below: