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Bonding in Methane and Ethane: Single Bonds

We will begin the discussion of bonding in organic compounds by looking at the bonding in methane, a compound with only one carbon atom. Then we will examine the bonding in ethane (a compound with two carbons and a carbon–carbon single bond), in ethene (a compound with two carbons and a carbon–carbon double bond), and in ethyne (a compound with two carbons and a carbon–carbon triple bond). Next, we will look at bonds formed by atoms other than carbon that are commonly found in organic compounds—bonds formed by oxygen, nitrogen, and the halogens. Because the orbitals used in bond formation determine the bond angles in a molecule, you will see that if we know the bond angles in a molecule, we can figure out which orbitals are involved in bond formation.

Bonding in Methane
Methane (CH4) has four covalent C−H bonds. Because all four bonds have the same length and all the bond angles are the same (109.5°), we can conclude that the four C−H bonds in methane are identical.

Four different ways to represent a methane molecule are shown here.

Methane Molecule representation

In a perspective formula, bonds in the plane of the paper are drawn as solid lines, bonds protruding out of the plane of the paper toward the viewer are drawn as solid wedges, and those protruding back from the plane of the paper away from the viewer are drawn as hatched wedges.

The potential map of methane shows that neither carbon nor hydrogen carries much of a charge: There are neither red areas, representing partially negatively charged atoms, nor blue areas, representing partially positively charged atoms. (Compare this map with the potential map for water on p. 14). The absence of partially charged atoms can be explained by the similar electronegativities of carbon and hydrogen, which cause carbon and hydrogen to share their bonding electrons relatively equally. Methane is a nonpolar molecule.
You may be surprised to learn that carbon forms four covalent bonds since you know that carbon has only two unpaired electrons in its ground-state electronic configuration (Table 1.2). But if carbon were to form only two covalent bonds, it would not complete its octet. Now we need to come up with an explanation that accounts for carbon’s forming four covalent bonds. If one of the electrons in the 2s orbital were promoted into the empty 2p atomic orbital, the new electronic configuration would have four unpaired electrons; thus, four covalent bonds could be formed. Let’s now see whether this is feasible energetically.
Bonding in Methane and Ethane
Because a p orbital is higher in energy than an s orbital, promotion of an electron from an s orbital to a p orbital requires energy. The amount of energy required is 96 kcal/mol. The formation of four C−H bonds releases 420 kcal/mol of energy because the bond dissociation energy of a single C−H bond 105 kcal/mol. If the electron were not promoted, carbon could form only two covalent bonds, which would release only 210 kcal/mol. So, by spending 96 kcal/mol (or 402 kJ/mol) to promote an electron, an extra 210 kcal/mol (or 879 kJ/mol) is released. In other words, promotion is energetically advantageous (Figure 1.9).
Electron Promotion

We have managed to account for the observation that carbon forms four covalent bonds, but what accounts for the fact that the four C−H bonds in methane are identical? Each has a bond length of 1.10 Å, and breaking any one of the bonds requires the same amount of energy(105 kcal/mol or 439 kJ/mol). If carbon used an s orbital and three p orbitals to form these four bonds, the bond formed with the s orbital would be different from the three bonds formed with p orbitals. How can carbon form four identical bonds, using one s and three p orbitals? The answer is that carbon uses hybrid orbitals.

Linus Carl Pauling

Linus Carl Pauling (1901–1994) was born in Portland, Oregon. A friend’s home chemistry laboratory sparked Pauling’s early interest in science. He received a Ph.D. from the California Institute of Technology and remained there for most of his academic career. He received the Nobel Prize in chemistry in 1954 for his work on molecular structure. Like Einstein, Pauling was a pacifist, winning the 1964 Nobel Peace Prize for his work on behalf of nuclear disarmament.

Hybrid orbitals are mixed orbitals—they result from combining orbitals. The concept of combining orbitals, called orbital hybridization, was first proposed by Linus Pauling in 1931. If the one s and three p orbitals of the second shell are combined and then apportioned into four equal orbitals, each of the four resulting orbitals will be one part s and three parts p. This type of mixed orbital is called an sp3 (stated “s-p-three” not “s-p-cubed”) orbital. (The superscript 3 means that three p orbitals were mixed with one s orbital to form the hybrid orbitals.) Each sp3 orbital has 25% s character and 75% p character. The four sp3 orbitals are degenerate—they have the same energy.
orbital hybridization
Like a p orbital, an sp3 orbital has two lobes. The lobes differ in size, however, because the s orbital adds to one lobe of the p orbital and subtracts from the other lobe of the p orbital (Figure 1.10). The stability of an sp3 orbital reflects its composition; it is more stable than a p orbital, but not as stable as an s orbital (Figure 1.11). The larger lobe of the sp3orbital is used in covalent bond formation.
sp3 orbital

Figure 1.10 The s orbital adds to one lobe of the p orbital and subtracts from the other lobe of the p orbital.
An s orbital and three p orbitals hybridize to form four sp3 orbitals
Figure 1.11An s orbital and three p orbitals hybridize to form four sp3 orbitals. An sp3 orbital is more stable than a p orbital, but not as stable as an s orbital.

The four sp3 orbitals arrange themselves in space in a way that allows them to get as far away from each other as possible (Figure 1.12a). This occurs because electrons repel each other and getting as far from each other as possible minimizes the repulsion (Section 1.6). When four orbitals spread themselves into space as far from each other as possible, they point toward the corners of a regular tetrahedron (a pyramid with four faces, each an equilateral triangle). Each of the four C−H bonds in methane is formed from overlap of an sp3 orbital of carbon with the s orbital of a hydrogen (Figure 1.12b). This explains why the four C−H bonds are identical.
orbital picture of methane
The angle formed between any two bonds of methane is 109.5°. This bond angle is called the tetrahedral bond angle. A carbon, such as the one in methane, that forms covalent bonds using four equivalent sp3 orbitals is called a tetrahedral carbon.The postulation of hybrid orbitals may appear to be a theory contrived just to make things fit—and that is exactly what it is. Nevertheless, it is a theory that gives us a very good picture of the bonding in organic compounds.

Bonding in Ethane
The two carbon atoms in ethane are tetrahedral. Each carbon uses four sp3 orbitals to form four covalent bonds:
carbon atoms in ethane are tetrahedral
One sp3 orbital of one carbon overlaps an sp3 orbital of the other carbon to form the C−C bond. Each of the remaining three sp3 orbitals of each carbon overlaps the s orbital of a hydrogen to form a C−H bond. Thus, the C−C bond is formed by sp3−sp3 overlap, and each C−H bond is formed by sp3−s overlap (Figure 1.13). Each of the bond angles in ethane is nearly the tetrahedral bond angle of 109.5°, and the length of the C−C bond is 1.54 Å. Ethane, like methane, is a nonpolar molecule.
Ethane Perspective formula ball and stick model
orbital picture of Ethane

Figure 1.13 An orbital picture of ethane. The C−C bond is formed by sp3−sp3 overlap, and each C−H bond is formed by sp3−s overlap. (The smaller lobes of the sp3 orbitals are not shown.)

All the bonds in methane and ethane are sigma (σ) bonds because they are all formed by the end-on overlap of atomic orbitals. All single bonds found in organic compounds are sigma bonds.

All single bonds found in organic compounds are sigma bonds.

Problem:

The MO diagram illustrating the overlap of an sp3 orbital of one carbon with an sp3 orbital of another carbon (Figure 1.14) is similar to the MO diagram for the end-on overlap of two p orbitals, which should not be surprising since sp3 orbitals have 75% p character.
End-on overlap of two sp3 orbital s

Figure 1.14 End-on overlap of two sp3 orbital s to form a σ bonding molecular orbital and a σ* antibonding molecular orbital.

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