P orbitals in pi bonds

First- and Second-Row Anomalies 858 The Use of p Orbitals in Pi Bonding 861 The Use (or Not) of d Orbitals by Nonmetals 86ft Reactivity and d Orbital Participation 875... 

Rotation of the right carbon by 90° produces the middle structure. The plane defined by the left CH3—C—H is now perpendicular to the plane defined by the CH3—C—H on the right. The red p orbital on the left C is in the plane of the page, and the blue p orbital on the right C is pointed directly at you, so the two p orbitals are also perpendicular to one another. Therefore, there is no stabilizing overlap of these p orbitals—the pi bond has been broken. 

An atom involved in resonance generally will not be sp3 hybridized because it needs at least one unhybridized p orbital for pi-bonding overlap. 

A bond formed by sideways overlap of two p orbitals. A pi bond has its electron density in two lobes, one above and one below the line joining the nuclei, (p. 47)... 

An idealized single bond is a sigma bond—one that has cylindrical symmetry. In contrast, a p-orbital or pi-bond orbital has pi symmetry—one that is antisymmetric with respect to reflection in a plane passing through the atomic centers with which it is associated. In ethene, the pi-bonding orbital is symmetric with respect to reflection in a plane perpendicular to and bisecting the C-C bond, whereas the pi-star-anti bonding orbital is antisymmetric with respect to this operation. 

A covalent bond is formed by the overlap of two atomic orbitals each with one electron. There are two types sigma and pi. A sigma bond involves the overlap of two atomic orbitals head-to-head in one position (such as two s-orbitals, an s and a p-orbital, or two p-orbitals). A pi-bond involves the overlap of parallel p-orbitals at both lobes. 

To predict the hybridization and geometry of an atom in a resonance hybrid, consider the resonance form with the most pi bonds to that atom. An atom involved in resonance generally will not be sp hybridized because it needs at least one unhybridized p orbital for pi-bonding overlap. 

This solvent is called tetrahydrofuran, or THF for short. Even though it somewhat stabilizes the empty p orbital on the boron atom in BH3, nevertheless the boron atom is very eager to look for any other sources of electron density that it can find. It is an electrophile—it is scavenging for sites of high electron density to fill its empty orbital. A pi bond is a site of high electron density, and therefore, a pi bond can attack borane. In fact, this is the hrst step of our mechanism. A pi bond attacks the empty p orbital of boron, which triggers a simultaneous hydride shift ... 

When two p orbitals overlap in a side-by-side configuration, they form a pi bond, shown in Figure 7.7. This bond is named after the Greek letter 7t. The electron clouds in pi bonds overlap less than those in sigma bonds, and they are correspondingly weaker. Pi bonds are often found in molecules with double or triple bonds. One example is ethene, commonly known as ethylene, a simple double-bonded molecule (Figure 7.8). The two vertical p orbitals form a pi bond. The two horizontal orbitals form a sigma bond. 

Pi (ti) bonds are formed by the side by side overlap of two parallel p orbitals. In the n bond, the electron cloud lies above and below the plane formed by o bonds, n bonds are weaker than o bonds. 

The Greek letter pi, ir, is the equivalent of our letter p. When we imagine looking along the intemuclear axis, a ir-bond resembles a pair of electrons in a p-orbital. All n-bonds will be colored yellow in this text. 

E2 elimination requires partial formation of a new pi bond, with its parallel p orbitals, in the transition state. The electrons that once formed a C—H bond must begin to overlap with the orbital that the leaving group is vacating. Formation of this new pi bond implies that these two sp3 orbitals must be parallel so that pi overlap is possible as the hydrogen and halogen leave and the orbitals rehybridize to the p orbitals of the new pi bond. 

Parallel p orbitals in ethylene. The pi bond in ethylene is formed by overlap of the unhybridized p orbitals on the sp2 hybrid carbon atoms. This overlap requires the two ends of the molecule to be coplanar. 

Now we are ready to construct the molecular orbitals of buta-1,3-diene. The p orbitals on Cl through C4 overlap, giving an extended system of four p orbitals that form four pi molecular orbitals. Two MOs are bonding, and two are antibonding. To represent the four p orbitals, we draw four p orbitals in a line. Although buta-1,3-diene is not linear, this simple straight-line representation makes it easier to draw and visualize the molecular orbitals. 

Double bonds that alternate with single bonds, with interaction by overlap of the p orbitals in the pi bonds, (p. 667)... 

Finally, there is the case for sp hybridization and the formation of a second pi bond. The second pi bond is the result of the overlap of the p orbitals in the z axis. Because a sigma bond is formed in the x axis and two pi bonds are formed in the y and z axes, a triple bond is formed. The triple bond will be shorter than the double bond and the triple bond will be stronger than the double bond as well. (See Figure 5.16.)... 

The remaining p orbitals in the product are parallel to the newly formed bond. It is less clear in the reverse reaction, which follows the same path but in the opposite direction, that the lone pair orbitals of O and L must be aligned parallel to each other and to the breaking bond at the transition state. The orbitals are then lined up so that they can easily become the allylic pi system of the ester. If one of the lone pairs were not lined up, the allylic system could not be established at the transition state, and that transition state would be much higher in energy. Allylic stabilization is about 14 to 25 kcal/mol. 

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