7t*-orbital

T orbital for benzene obtained from spin-coupled valence bond theory. (Figure redrawn from Gerratt ], D L oer, P B Karadakov and M Raimondi 1997. Modem valence bond theory. Chemical Society Reviews 87 100.) figure also shows the two Kekule and three Dewar benzene forms which contribute to the overall wavefunction Kekuleform contributes approximately 40.5% and each Dewar form approximately 6.4%. 

Figure 6-2 t he 7t orbital of Ethylene. Both ititeractioiis---------between... 

As orbitals spread into bands, orbitals oriented for a or a bonds spread into the widest bands. 7t orbitals form narrower bands and 5 bonding orbitals form the narrowest bands. 

The bonding in aminoboranes has been represented as 2 BR2 X2N i m 2 show the contribution of some 7t-orbital interaction... 

The 7T orbital extends over the entire enone system. As a consequence the C —CO bond is strengthened in the excited states and a-cleavage, the important primary photoprocess in nonconjugated ketones, is not observed. The most important unimolecular reaction types are processes which involve... 

Orbitals 7 and 9 (the latter is the LUMO) of formaldehyde exhibit this same character. Orbital 7 is a bonding 7t orbital, and orbital 9 is a Tt . However, the n orbital formed of the pj orbitals from the carbon and the oxygen (which also lie in the YZ plane) is not the HOMO. Instead, an orbital formed from Pj, orbitals from the carbon and the oxygen and from the s orbitals on the hydrogens is the highest occupied orbital. The contributions from the carbon and oxygen are situated along the double bond while the HOMO in ethylene was perpendicular to this bond. 

First, try to draw resonance contributors for both ground state and triplet anthrone. Then display a spin density surface for the triplet state of anthrone. (Note that the spin density surface shows the location of both unpaired electrons, one of which may be in a 7t orbital and one of which may be in a o orbital.) Where are the two unpaired electrons Are they localized or delocalized Given that spin delocalization generally leads to stabilization, would you expect the triplet state of anthrone to be stable ... 

An explanation for the stereoselectivity of the reaction involves optimal overlap of the 7t-orbital of the carbonyl with the developing electron rich p-orbital on C2 during the Sj,j2 displacement of the chloride by the alkoxide (24). Thus, orbital overlap imposes conformational constraints in the transition state that leads to nonbonding interactions disfavoring transition state 15P... 

In the 1,3-dipolar cycloaddition reactions of especially allyl anion type 1,3-dipoles with alkenes the formation of diastereomers has to be considered. In reactions of nitrones with a terminal alkene the nitrone can approach the alkene in an endo or an exo fashion giving rise to two different diastereomers. The nomenclature endo and exo is well known from the Diels-Alder reaction [3]. The endo isomer arises from the reaction in which the nitrogen atom of the dipole points in the same direction as the substituent of the alkene as outlined in Scheme 6.7. However, compared with the Diels-Alder reaction in which the endo transition state is stabilized by secondary 7t-orbital interactions, the actual interaction of the N-nitrone p -orbital with a vicinal p -orbital on the alkene, and thus the stabilization, is small [25]. The endojexo selectivity in the 1,3-dipolar cycloaddition reaction is therefore primarily controlled by the structure of the substrates or by a catalyst. 

In molecular orbital terms, the stability of the allyl radical is due to the fact that the unpaired electron is delocalized, or spread out, over an extended 7T orbital network rather than localized at only one site, as shown by the computer-generated MO in Fig 10.3. This delocalization is particularly apparent in the so-called spin density surface in Figure 10.4, which shows the calculated location, of the unpaired electron. The two terminal carbons share the unpaired electron equally. 

Pyridine is a flat, hexagonal molecule with bond angles of 120°. It undergoes substitution rather than addition and generally behaves like benzene. Draw a picture of the 7T orbitals of pyridine to explain its properties. Check your answer by looking ahead to Section 15.7. 

The chemisorptive bond is a chemical bond. The nature of this bond can be covalent or can have a strong ionic character. The formation of the chemisorptive bond in general involves either donation of electrons from the adsorbate to the metal (donation) or donation of electrons from the metal to the adsorbate (backdonation).2 In the former case the adsorbate is termed electron donor, in the latter case it is termed electron acceptor.3 In many cases both donation and backdonation of electrons is involved in chemisorptive bond formation and the adsorbate behaves both as an electron acceptor and as an electron donor. A typical example is the chemisorption of CO on transition metals where, according to the model first described by Blyholder,4 the chemisorptive bond formation involves both donation of electrons from the 7t orbitals of CO to the metal and backdonation of electrons from the metal to the antibonding n orbitals of CO. 

The complex might be best described by saying that each oxygen atom could form a double bond with the chromium atom with the. use of either one of its two 7t orbitals and the accompanying pair of electrons, extra stability being achieved through the resonance between the two kinds of double bonds. 

According to this theory, an acid is defined as a proton donor and a base as a proton acceptor (a base must have a pair of electrons available to share with the proton this is usually present as an unshared pair, but sometimes is in a 7t orbital). An acid-base reaction is simply the transfer of a proton from an acid to a base. (Protons do not exist free in solution but must be attached to an electron pair). When the acid gives up a proton, the species remaining still retains the electron pair to which the proton was formerly attached. Thus the new species, in theory at least, can reacquire a proton and is therefore a base. It is referred to as the conjugate base of the acid. All acids have a conjugate base, and all bases have a conjugate acid. All acid-base reactions fit the equation... 

Enolate anions (4e) that have been heated by infiared multiple photon absorption for which torsional motion about the H2C-C bond, which destabilizes the 7t orbital containing the extra electron, is the mode contributing most to vibration-to-electronic energy transfer and thus to ejection. 

The orbital mixing rules are applied to the polarization of 7t of ethene by a C=0 group on the assumption that is lowered below The 7t orbital has mix in phase and the low lying tc orbital mix out of phase with (Scheme 16). As a result, the phase relation between t( and n is fixed. The amplitude is larger on C than on C and the carbonyl carbon. 

The two t7-bonding orbitals have identical energies, because the corresponding atomic p orbitals have identical energies. Likewise, the two 7T orbitals have identical energies. 

Overall, it can be seen from Figs. 4 and 5 that in predicting the p parameters almost identical results are obtained in five out of the eight cases. Significantly perhaps, these include the more highly localized O- lone pair and C H orbitals. The worst agreement in both molecules is seen for the more delocalized ring 7t orbital. 

Clearly, several aspects of the orbital optimisation remain to be clarified. Firstly a numerical test using a system more complex than Ilj should be made. What happens to 7T orbitals or strongly hybridized orbitals should be also examined. It would be also interesting to explain how the optimisation - as described here - is related to an energy lowering, as well as the practical use of the present description in actual calculations, etc. .. These different aspects will be examined in forthcoming publications. 

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