Lecture Notes on Aromaticity - Mechanisms | CHEM 6311, Study notes of Mechanics

Download Lecture Notes on Aromaticity - Mechanisms | CHEM 6311 and more Study notes Mechanics in PDF only on Docsity! 175 IX. Aromaticity Before we actually discuss what aromaticity is and the ways that people have tried to measure it, we need to know how p orbitals interact with each other. A. Linear (acyclic) π systems 1. Use the LCAO-MO approach a. We can proceed exactly as before, except this time using only those p orbitals on adjacent atoms which overlap in a π fashion. b. In essence, we are once again ignoring all of the filled level interactions, and taking the σ bond frame as given. c. Example: constructing the three π orbital (allyl) system; combining π & p. i) These orbitals are drawn to indicate relative signs only. Their sizes do not reflect their relative coeffi- cients. ii) They look just like those for H2 and H3 . (a) Look at the p orbitals from the top of the chain (b) For the middle level of the allyl system: 9.1a q 176 d. The π orbitals of butadiene can be constructed from two ethylenes. This is shown in detail in Lowry and Richardson -- please read it carefully. e. There is a set of rules associated with the shapes of the orbitals (see Lowry and Richardson, p. 78). i. The orbitals alternate in symmetry with respect to a mirror plane passing through the middle of the π system. The lowest energy orbital is always S. ii. The lowest energy orbital has no nodes perpendicu- lar to the chain (parallel to the mirror plane). iii. The number of nodes increases by one going from one orbital to the next highest. iv. The highest MO has nodes between each adjacent atom. v. To preserve the symmetry of the system, the nodes must be symmetrically placed with respect to the central mirror plane. vi. In chains with an odd-number of carbons, the A levels must always have a nodal plane at the central carbon atom (which lies in the mirror plane). vii. There are the same number of MO’s as there are atoms in the chain (provided that each atom has one and only one p orbital that it contributes to the p system). viii.The number of electrons are equal to the number of p orbitals minus the charge (taken algebra- ically) of the system (if all atoms in the chain are carbons). f. Example - the pentadienyl system. 179 a. α = energy of a p orbital on carbon before any interaction occurs. b. β = interaction energy - defined as a negative num- ber, typically = -18 kcal/mol. 3. For any acyclic polyene (in units of β) Ej = 2 cos ( j π / n + 1) crj = (2 / n + 1)1/2 (sin rjπ / n + 1) ETOT= Σ nj Ej where, Ej = orbital energy of molecular orbital j n = # of p orbitals j = orbital index = 1, 2 ... n r = atom index = 1, 2 ... n nj = number of e¯s in orbital j crj = coefficient on atom r in molecular orbital j 4. Two more examples: C. Cyclic π systems (These are pictured below. Please learn all of them.) 1. Again, there are a number of patterns that emerge: 180 a. The number of nodes increases with each increase in orbital energy. b. The lowest orbital contains no nodes. c. The lowest orbital is nondegenerate. The next orbitals on up come in degenerate pairs, except in systems with an even number of atoms, where the highest orbital is nondegenerate. i.e. d. A linear combination of two degenerate orbitals gives two orbitals which are equivalent in every re- spect, e.g. + = = 9.10 181 d. The number of MO’s is equal to the number of p orbitals in the cyclic polyene. The number of electrons is equal to the number of atoms minus the (algebraic) charge (for cases with all carbon atoms). e. The MO’s of benzene, below, are a good example: 2. There are two simple ways to get the orbital energies within the Hückel approximation: a. Frost’s circle - construct a circle of diameter 4β and inscribe the cyclic polyene so that one of the verticies of the polygon is exactly at the bottom of the circle. The points at which the verticies of the polygon touch the circle are the orbital energies. e.g. 184 2n+1 orbitals 4n π electrons 9.17 d. The key feature of an aromatic compound, then, is that it is “closed shell.” i) All degenerate orbitals are doubly occupied. ii) There is a (probably large) HOMO - LUMO gap. It should be very stable. e. An antiaromatic structure has two electrons in a degenerate set. i) Therefore, the HOMO-LUMO gap is zero. ii) This means that there must be a very high lying HOMO and a low-lying LUMO. iii) In other words, antiaromatic compounds are ex- pected to be extraodinarily reactive. 2. Hoffmann - Goldstein Approach a. This uses the interaction of two acyclic polyene chains, “ribbons”, to form a cyclic system. i.e. is this interaction stabilizing or destabilizing in terms of HOMO-LUMO interactions ? + 9.18 b. Symmetry properties i) Any polyene ribbon will have the following symmetry properties: m S A HOMO LUMO ψp ψd S A HOMO LUMO ψp ψd m m Hoffmann-Goldstein nomenclature 9.19 185 ii) For any polyene ribbon with an even or odd number of n electrons there will be four possibilities: HOMO LUMO ψp ψp ψp ψpψd ψd ψd ψd mode 1 2 3 0 (=4) 2+ + 3+ 2+ + 2- 3- + - 2- (1e-) (1e-) (1e-) (5e-) (5e-) (2e-) (2e-) (2e-) (6e-) (6e-) (7e-) (7e-) (3e-) (3e-) (4e-) (4e-) (8e-) (8e-) - - + 9.20 The pattern of these orbital combinations is important 2+ 9.21 butadiene2+ has the same HOMO-LUMO symmetry properties as hexatriene iii ) Modes (a) We can symbolize each ribbon by a shorthand notation - nz i.e. where: n = # of carbons = # p orbitals z = molecular charge 2+ = 42 = 60 2- = 5 2- 9.22 186 (b) The mode of each ribbon is (n - z) modulo 4. (i) Modulo 4 is the operation in which the quantity n - z is divided by 4, and the remainder is the value used. (ii) Example: For the ribbons shown on the previ- ous page: 42: mode = (4 - 2) modulo 4 = (2) modulo 4 = 2 60: mode = (6 - 0) modulo 4 = 2 5-2: mode = (5 - (-2)) modulo 4 = 3 c. Interacting ribbons i) There are three ways that mode 0 and mode 2 ribbons can interact: S A HOMO LUMO ψp ψd m S A HOMO LUMO ψp ψd m m net destabilization mode (2+2) S A HOMO LUMO ψp ψd m S A HOMO LUMO ψp ψd m net destabilization mode (0+0) 9.23 m S A HOMO LUMO ψp ψd m S A HOMO LUMO ψp ψd m stabilization mode (2+0) 9.24 Therefore, with these two types, only mode (2+0) [or (0+2)] interactions are stabilizing. 189 X 9.27 3. Diamagnetic anisotropy. This is nmr information -- usually 1H nmr. Energetic issues are avoided. a. Aromatic protons resonate at lower fields than nonaromatic protons. 10 9 8 7 6 5 4 aromatic protons olefinic protons δ ppm 9.28 b. This deshielding is attributed to diamagnetic anisot- ropy. i) The general model that is used is shown below. Orienting an aromatic molecule perpendicular to the applied magnetic field, Ho, causes the π electrons to circulate like the free electrons in a circular copper wire. ii) The circulating electrons create their own magnetic field, H’ (which is much, much smaller than Ho). (a) Within the polyene ring (where the e¯s are circulating), the direction of the field is opposite to that at Ho, and it extends outward as shown. (b) Ha, a proton outside of the polyene ring, experi- ences an increased field (Ho + H’), and is expected to resonate at a higher applied frequency. (c) This model also predicts that a proton inside of the polyene ring (such as Hb) will feel a diminished field, Ho - H’. Thus it should resonate at a lower frequency. This has been experimentally tested and found to be true. Ha Hb HoH' 9.29 190 c. Examples: H H H H H H H H H H H H H HH H H H 9.30 [18] annulene, prepared by Frans Sondheimer. 18 π electrons = 4n+2 when n=4. Accordingly the 12 outside protons resonate at 10.2δ, wheras, the six inside protons are found at -3.0δ! Compare this with: H H δ = 5.6 ppm δ = 5.75 ppm 9.31 However, this is still not the whole story, for example : H H H H H H H H H H H H H H H H 9.32 For [16] annulene, 16π electrons = 4n with n=4. Fur- thermore, the 12 outside protons resonate at 5.1 δ The 4 inside protons resonate at 10.3 δ! d. Aromatic vs. antiaromatic i) Do the “aromatic” electron circulate in the presence of an applied magnetic field opposite to “antiaromatc” electrons? This is clearly nonsense. Electrons are electrons. ii ) While this classical model of a current in the wire works beautifully for polyene rings with 4n + 2 π 191 electrons, it breaks down for other electron counts. iii) We will not cover this in detail, but diamagnetic anisotropy really depends on the mixing of excited states into the ground state under the influence of a magnetic field (a) For a 4n + 2 polyene: HOMO LUMO 9.33 So lEo - E11, 1Eo - E21, etc. is large. (b) However for a 4n polyene: HOMO LUMO 9.34 IE0 - E1I, IEo - E2I, etc. is small, because the HOMO-LUMO gap (if any) is small. There is therefore a large amount of mixing of excited states into the ground state in this case. 4. Structure a. The idea is that a molecule will always distort itself to the most stable situation i) Thus, if conjugation is stabilizing then the structure should reflect this fact. ii) On the other hand, if conjugation is destabilizing then its structure should be distorted in a way that reflects this. iii) In terms of aromaticity, aromatic compounds should show no alternation of C-C single and double bonds whereas nonaromatic compounds will, i.e. H H H H H H H H H H H H H H 1.398Å 1.33Å 1.46Å 9.35a 194 ∴ butadiene has 0.48β more π stabilization than two totally noninteracting ethylenes. Taking, β = -18 kcal/ mol, this would amount to 8.6 kcal/mol. ii) For cyclohexatriene we would expect three such interactions, for a total of ~ 26 kcal/mol. This seems to explain the bulk of the stabilization! d. A different argument claims that there should be additional stabilization in cyclohexatriene, because the C-C single bonds are made up of sp2 - sp2 hybrids. i) These are stronger that sp3 - sp3 hybrids. ii) This suggests that the “aromaticity” -- that special stability due to conjugation -- may in fact be an illusion. iii) Indeed, some people have argued that there is no stability associated with the regular structure in ben- zene. X X X X X X X X X X X X + X = CH N Li H 9.38a For X = N and H, there is very good evidence that the cyclic, “aromatic!” structure on the left is not stable. When X = Li it turns out that it is. Much work has been undertaken in the last few years for the X=CH case. There are a couple of very good pieces of evi- dence which show that in fact if it were not for the C- C σ system, benzene would be a distorted 1,3,5- cyclohexatriene! e. What then about antiaromatic compounds - is there destabilization in them? i) We talked before about why they should have a distorted structure -- the Jahn - Teller theorem. (a) Destroying π conjugation in this case does lead to stabilization. (b) Does this imply a destabilization associated with a planar, flat structure? That is, is there a π destabi- lization? Probably not. ii) Let’s first take a real system. 195 + H2 + H2 ∆H = -98 kcal/mol ∆H = -23 kcal/mol X 4 -92 kcal/mol 9.39 (a) Cyclooctateraene shows 6 kcal/mol of destabilization. Is this from antiaromaticity? (b) The C-C-C angles in tub conformation of cyclooctatetraene are 126.5°. This molecule cer- tainly has ring strain; cyclooctene probably has much less strain. (c) One can argue that cyclooctatetraene is nonaromatic (there is effectively no π overlap between C=C bonds) so it should not show any antiaromaticity. iii) Now let’s take a look at cyclobutadiene (a) At the Hückel level: 2-versus 0.00 β 2.00 β antiaromatic aromatic 9.40 Where is the stability difference? (a) One can argue reasonably, in fact, that the cyclobutadiene dianion should be more destabilized by resonance than is cyclobutadiene. (b) There are two more electrons, hence there should be more e¯ - e¯ repulsion. (c) A recent, very good V.B. calculation has found that square, singlet cyclobutadiene is stabilized by resonance to the tune of 30 kcal/mol! f. The moral of this story is that there probably is not 196 any “special” stabilization associated with aromaticity. i) π conjugation in any form is stabilizing in an ener- getic sense. ii) All cyclic conjugated polyenes are stablized in this way. iii) For the (hypothetical) antiaromatic ones, however, in the latter we can expect distortions to a yet more stable form, or a very high reactivity. F. Homoaromaticity 1. Homoaromaticity involves conjugation interrupted by one or more (~sp3) insulating carbon atom(s). a. It follows Hückel’s rule. b. Often one observes a special reactivity. c. Example one: protonation on cyclooctatetraene; a homotropylium ion. H+ etc etc 9.41 -0.6 ppm 5.2 ppm 6.6 ppm8.5 ppm i) The endo proton on C-8 is strongly shielded (δ -0.6). ii) The ring protons appear at δ 8.5 (H-2,3,4,5,6) or δ 6.6 (H-1,7). iii) The C-8 exo proton is at δ 5.2. iv) There is a barrier to ring flipping of 22.5 kcal/mol. v) All of this is taken as evidence of homoaromaticity. d. Example two: decomposition of anti-7-norbornenyl-p-toluenesulfonate. 199 1. Each has its own intrinsic stability patterns - see Goldstein and Hoffmann, J. Am. Chem. Soc. 93, 6193(1971). 2. A couple of examples which show experimental ioniza- tion potentials are shown below: a’a’ e’ a1 b2 a’2 8 9 10 Io n iz at io n P o te n ti al ( eV ) 9.49 8 9 10 a’ a1 a1 b2 b2 1b2 2b2 Io n iz at io n P o te n ti al ( eV ) 9.50 200 3. Work through each interaction with care. I. Heterocycles and Through - Bond Conjugation 1. Using the perturbation theory ideas from the previous chapter, let us work through what happens to the π system on going from benzene to pyridine a. The degenerate p sets of benzene are split, one where the AO coefficient on the carbon atom that is becoming nitrogen is stabilized. The other is left un- changed. b. As mentioned in the previous chapter, electron density is increased on the more electronegative atom for the bonding MO(s) and increased on the more electropositive atom for the antibonding MO(s). This is reflected in the drawings above. The shapes are drawn for one value of Ψ. Therefore, more electron density is associated with contours that are closer to a nucleus. c. The amount of stabilization is proportional to the square of the AO coefficient and how much more elec- tronegative the atom has become relative to carbon. This is demonstrated in the plot below. 1. The ionization potential of X (the atomic first IP - for the 4S3/2 → 3P0 state of X) measures the relative electronegativity of X. 201 2. The slope of the a2 MO is zero. There is a node on the X atom for this MO so it should not be perturbed. The coefficient at X for the 1b1 MO according to Hückel theory is 1/√6 while that for 2b2 is 1/√3. The slopes of the two lines are 0.189 and 0.373, for the 1b1 and 2b1 MOs, respectively. These are very close to the values of the Hückel coefficients squared. 2. The PE spectra of the three di-substituted azines along with pyridine are shown on the next page. a. The assignments of the peaks is a lengthy process and one not without difficulties. Let us accept, how- ever, these results and look at a correlation of the IPs for the two π MOs derived from e1g in benzene 12.0 11.5 11.0 10.5 10.0 9.5 9.0 Io n iz at io n P o te n ti al ( eV ) N N N N N N N 9.54 204 3. Through-bond conjugation. One expects the lone pairs in [2.2.2]-diazabicyclooctane to be that shown below. a. The distance between the lone pairs is enormous so the splitting between the lone pairs is expected to be very small. b. This, however, neglects the C-C σ and σ* orbitlas which have the same symmetry and therefore, can mix with the lone pair combinations. c. Actually the mixing with σ* is quite small since σ* lies at too high of an energy. However, the mixing between σ and n+ is quite strong. The net conse- quence is that n+ is pushed to high energy relative to n- Therefore, the PE spectrum for [2.2.2]- diazabicyclooctane is that shown on the next page. 205 d. The extension to the 1,4-diazine case is straightfor- ward. For 1,3-diazine, there are complications and this analysis needs to take into account not only the C-H bond but also the bond path in the other direction. The actual wavefunctions for the n+ and n- combina- tions are shown below: e. Through-space overlap dies out exponentially, how- ever, through-bond conjugation does not. The trans- mission of electrons in proteins is one example where through-bond conjugation is expected to play a pri- mary role. f. Another example is provided by the stability and singlet-triplet splitting in the three isomeric benzynes: 206 ∆Hf (theo.) ∆Hf (exp.) S-T (kcal/mol)(kcal/mol) 109 106±3 36 123 116±3 17 134 138±1 2 o o 9.60 f. meta and para benzyne are only about 10 and 25 kcal/mol, respectively, less stable than ortho benzyne. Clearly there is a π bond between the two hybrids in ortho benzyne. For the other two compounds it is through bond congugation that makes them so stable (a typical C-C π bond is worth about 60 kcal/mol) and makes the singlet state more stable than the triplet one. J. Buckyballs - what review of π systems would be without them? 1. Definitions and some shapes: