Pi bonding electrons

For the alternative multi-bonded case, the multivalent ligands may be considered to contribute m additional pi-bonding electrons, where... 

HUckel s 4n -H 2 rule orgchem Aromatic (ring) compounds must have 4n + 2 pi-bonding electrons, where n is a whole number and generally limited to = 0 to 5. When M = 1, for example, there are six pi-electrons, as for benzene. Also known as Huckel s rule. hhk-alz jfor, en plos tu, rul ... 

The pi-bonding electrons give rise to regions of high electron density (red) in the electrostatic potential map of ethylene shown in Figure 7-1. 

This coal of intermediate rank has rather extensive polynuclear formation and polar functional groups. Thus we would expect dispersion, polarizability, and dipolar interactions between the substrate entities and the sorbate molecules. The polar portion of the substrate as well as the highly conjugated pi-bonded electrons most assuredly are involved in the sorption process. Such a concept is quite suggestive and compatible with the polarization theory for sorption processes (9, 13, 14), where the energetics are predicted to follow the relationship... 

After loss of a pi bonding electron throngh electron impact ionization, botii peaks arise from allyhc fragmentations ... 

The carbon—carbon double bond is the distinguishing feature of the butylenes and as such, controls their chemistry. This bond is formed by sp orbitals (a sigma bond and a weaker pi bond). The two carbon atoms plus the four atoms ia the alpha positions therefore He ia a plane. The pi bond which ties over the plane of the atoms acts as a source of electrons ia addition reactions at the double bond. The carbon—carbon bond, acting as a substitute, affects the reactivity of the carbon atoms at the alpha positions through the formation of the aHyUc resonance stmcture. This stmcture can stabilize both positive and... 

Along the bond axis itself, the electron density is zero. The electron pair of a pi (tt) bond occupies a pi bonding orbital. There is one tt bond in the C2H4 molecule, two in QH The geometries of the bonding orbitals in ethylene and acetylene are shown in Figure 7.13. 

The rationale behind this choice of bond integrals is that the radical stabilizing alpha effect of such radicals are explained not by the usual "resonance form" arguments, but by invoking frontier orbital interactions between the singly occupied molecular orbital of the localized carbon radical and the highest occupied molecular orbital (the non-bonding electrons atomic orbital) of the heteroatom (6). For free radicals the result of the SOMO-HOMO interaction Ts a net "one-half" pi bond (a pi bond plus a one-half... 

First we need to locate the part of the molecule where resonance is an issue. Remember that we can push electrons only from lone pairs or bonds. We don t need to worry about all bonds, because we can t push an arrow from a single bond (that would violate the first commandment). So we only care about double or triple bonds. Double and triple bonds are called pi bonds. So we need to look for lone pairs and pi bonds. Usually, only a small region of the molecule will possess either of these features. 

Consider the first arrow in the example below, where we are using the electrons of the pi bond to attack a proton (H ), and expelhng Cl in the process ... 

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 ... 

In the hrst step, we have an alkene reacting with Br2. To understand this step of the mechanism, we must determine which reagent is the nucleophile, and which reagent is the electrophile. The alkene possesses a pi bond, which represents a region in space of electron density. Therefore, the alkene functions as the nucleophile. 

This implies that Br2 is the electrophile. But how does Br2 function as an electrophile The bond between the two bromine atoms is a covalent bond, and we therefore expect the electron density to be equally distributed over both Br atoms. However, an interesting thing happens when a Br2 molecule approaches an alkene. The electron density of the pi bond repels the electron density in the Br2 molecule, creating a temporary dipole moment in Br2. 

As the Br2 molecule gets closer to the alkene, this temporary effect becomes more pronounced. Now we can understand why Br2 functions as an electrophile in this reaction there is a temporary 5+ on the bromine atom that is closer to the pi bond of the alkene. When the electron-rich alkene attacks the electron-poor bromine, we get the following hrst step of our mechanism ... 

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. 

The structures of aromatic hydrocarbons, like benzene (C6H6), involve sp2 hybridized carbon atoms with pi bonds whose electrons are delocalized over the entire benzene ring. 

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