Pi bond in ethylene

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. 

It takes approximately 264 kj/mol (63 kcal/mol) to break the pi bond in ethylene— that is, to rotate one carbon by 90° with respect to the other so that no overlap occurs between 2p orbitals on adjacent carbons (Figure 4.1). This energy is considerably greater than... 

According to the orbital overlap model, a carbon-carbon double bond consists of one sigma bond formed by the overlap of sp hybrid orbitals and one pi bond formed by the overlap of parallel 2p atomic orbitals. It takes approximately 264 kJ/mol (63 kcal/mol) to break the pi bond in ethylene. 

Figure 4.14 Bonding in ethylene, C2H4. (a) Top view of the sigma bonds between carbon atoms and between carbon and hydrogen atoms. All of the atoms lie in the same plane, making C2H4 a linear molecule, (b) Side view showing how the two 2p orbitals on the two carbon atoms overlap, leading to the formation of a pi bond. Blue and red colors denote + and -regions of the orbitals, (c) The interactions in (a) and (b) lead to the formation of the sigma bonds and pi bond in ethylene. Note that the pi bond lies above and below the plane of the molecule.

In contrast to the severe difficulty of cracking a sigma bond, insertion of a transition metal into a pi bond can proceed in facile fashion. This can be illustrated by the attack of Ti on the pi bond of ethylene, which leads to metallacycle formation in the reaction... 

In ethylene, there are two types of bonds. Sigma (tr) bonds have the overlap of the orbitals on a line between the two atoms involved in the covalent bond. In ethylene, the C-H bonds and one of the C-C bonds are sigma bonds. Pi (ir) bonds have the overlap of orbitals above and below a line through the two nuclei of the atoms involved in the bond. A double bond is always composed of one sigma and one pi bond. A carbon-to-carbon triple bond results from the... 

The double bond between carbon and oxygen looks just like the double bond in ethylene. There is a sigma bond formed by overlap of sp2 hybrid orbitals and a pi bond formed by overlap of the unhybridizedp orbitals on carbon and oxygen (Figure 2-19). 

Two more electrons must go into the carbon-carbon bonding region to form the double bond in ethylene. Each carbon atom still has an unhybridized p orbital, and these overlap to form a pi-bonding molecular orbital. The two electrons in this orbital form... 

Atomic orbitals that do not participate in the hybridization are then used for bonding with other atomic orbitals on adjacent centers as long as there is nonzero overlap of the atontic orbitals. For example, a hybridization description for the bonding in ethylene accounts for a so-called sigma (a) framework of bonding utihzing sp hybrids and a second interaction called a pi (tt) interaction between pure p atontic orbitals on the carbon atoms (8). 

Figure 2.3 The bonding in ethylene is explained by combining two sp hybridized carbon atoms. The C—H sigma bonds all lie in the same plane. The unhybridized p orbitals of the carbon atoms overlap above and below the molecular plane to form the pi bond.

Two more electrons must go into the carbon-carbon bonding region to form the double bond in ethylene. Each carbon atom still has an unhybridized p orbital, and these overlap to form a pi-bonding molecular orbital. The two electrons in this orbital form the second bond between the double-bonded carbon atoms. For pi overlap to occur, these p orbitals must be ptarallel, which requires that the two carbon atoms be oriented with all their C—H bonds in a single plane (Figure 7-1). Half of the pi-bonding... 

Restricted rotation about the carbon-carbon double bond in ethylene, (a) Orbital overlap model showing the pi bond. (b)The pi bond is broken by rotating the plane of one H — C—H group by 90° with respect to the plane of the other H — C — H group. 

The key difference between poly( styrene) and styrene is the presence of a double bond in the latter. In poly(styrene), there are two sigma bonds cormecting neighboring styrene units as opposed to the one pi bond in monomeric styrene. The polymerization of alkenes like styrene represents one of the most common polymer synthesis techniques and leads to ubiquitous materials like poly(ethylene), poly(propylene), and poly(vinyl chloride) in addition to poly(styrene). There are many ways to convert alkenes into polymers by chain polymerization and while they are all different in detail, there are some common features. As a starting point, let us first consider the anionic polymerization of styrene to prepare the poly(styr-ene) molecule depicted in Figure 1(a). 

This leaves two unhybiidized p orbitals (each containing an electron) on each C atom. Figure 9.9 shows the sigma and pi bonds in the acetylene molecule (also known as ethyne). Just as one sigma bond and one pi bond make up a double bond, one sigma bond and two pi bonds make up a triple bond. Figure 9.10 (pp. 340-341) summarizes the formation of bonds in ethane, ethylene, and acetylene. 

A reaction takes place when the positive end of the HCl molecule approache.s the double bond in ethylene. The pi bond breaks, and the electrons it contained move as indicated by the curved arrows shown in the following equation, forming a sigma bond between the H atom of the HCl molecule and one of the C atoms. As this new bond forms, two things happen ... 

Reaction (12) is known to be more exothermic than (10). The reason is that addition of the first H (in 10) necessitates promotion of one electron from 7t to tt in CHj=CH2. We call this spatial excitation and we recognize that this has a mono- as well as a bi-electronic component. Addition of a second H (in 12) [after H has already been inserted (in 11) and has caused spatial excitation of CH2 = CHj] is more exothermic than simply addition of H to CH2 = CH2 (in 10) because spatial excitation of a reactant is no longer required. The difference between AEj and AEj is nothing else but the thermodynamically defined strength of the pi bond of ethylene and all this basic VB argumentation is probably already familiar to the reader. 

What is the MM3 enthalpy of formation at 298.15 K of styrene Use the option Mark all pi atoms to take into account the conjugated double bonds in styrene. Is the minimum-energy structure planar, or does the ethylene group move out of the plane of the benzene ring ... 

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. 

Bonding orbitals in ethylene (CH2=CH2) and acetylene (CH=CH). The sigma bond backbones are shown in blue. The pi bonds (one in ethylene and two in acetylene) are shown in red. Note that a pi bonding orbital consists of two lobes. 

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. 

To illustrate the importance of vicinal connectivity of conjugating units, we consider two dienes in non-vicinal relationships 1,5-hexadiene, 9, and allene, 10. As shown in Table 3.19, the direct diene conjugations are negligible in both species, on account of spatial separation in 9 and symmetry-imposed orthogonality of the two pi planes in 10. Consistently with the essential absence of conjugation, the unsaturated C—C bonds of 9 and 10 have calculated bond orders characteristic of ethylene or other unconjugated systems and the ficc NLMOs have essentially localized character ... 

Even cursory inspection will show that conjugation alters the properties of the participating pi bonds. For example, the vinyl 7tcc NBO of 6 is not homopolar as in ethylene, but instead becomes rather strongly polarized toward C2,... 

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