Organic Chemistry: Concepts and Applications - Headley Allan D. 2020
Bonding and Structure of Organic Compounds
1.5 The Covalent Bond
The VSEPR theory has worked very well to assist in determining the geometry of molecules, but there is another model that can be used to not only assist in determining the geometry of molecules but can also be used to explain important properties of molecules, such as why a hydrogen that is bonded to a triple bonded carbon is more acidic than one that is bonded to a double-bonded carbon or even a single-bonded carbon. In this model, the molecular orbital (MO) theory, atomic orbitals of different atoms form molecular orbitals, which are generated by mixing atomic orbitals.
1.5.1 The Single Bond to Hydrogen
The hydrogen molecule (H2) is the simplest molecule; it contains only a single covalent bond (two electrons, one from each hydrogen atom) and no nonbonding electrons. As a result, it is an ideal molecule to be used to illustrate how atomic orbitals are mixed to form molecular orbitals. To form a hydrogen molecule, the 1s orbital of each atom interacts to form two new orbitals, a molecular orbital, referred to as a sigma (σ) orbital and another molecular orbital, which is higher in energy than the σ orbital and is called a sigma star orbital (σ*). Since there can only be a maximum of two electrons in any of these two orbitals, the two electrons available (one from each hydrogen atom) will occupy the lowest molecular orbital, creating a covalent bond, a sigma (σ) bond. This bond is known as a sigma (σ) bond since the bonding electrons occupy the sigma (σ) molecular orbital. An illustration of the molecular orbitals and the bonding formed in the H2 molecule is shown in Figure 1.17.
Figure 1.17 Molecular orbital diagram of hydrogen molecule.
In this illustration, short horizontal lines are used to represent atomic and molecular orbitals. The short horizontal lines in the middle of the diagram represent molecular orbitals, and the short horizontal lines at the outer ends of the diagram represent atomic orbitals. Note also that the molecular and atomic orbitals have different energy levels. Molecular orbitals that contain the bonding electrons are lower in energy (more stable) than either of the atomic orbitals that contributed to the molecular orbital. As pointed out earlier, there is another type of molecular orbital that is generated from the mixing of the two atomic orbitals. This orbital is higher in energy than either of the atomic orbitals or the sigma molecular orbital; this orbital is called an anti-bonding molecular orbital. If electrons are placed in this orbital, the molecule is not stable.
1.5.2 The Single Bond to Carbon
This theory (molecular orbital theory) can be used also to explain the bonding and geometry of methane (CH4) and other organic molecules. It is known from the VSEPR theory, molecular modeling and other experimental data that the H─C─H bond angles of methane are 109.5° and that the geometry of methane is tetrahedral, and each C─H bond length is 1.10Å. To account for this observation, four equivalent molecular orbitals from the carbon atom to each of the hydrogen atoms of methane must be used; and in each of these molecular orbitals, there must be two electrons — one from carbon and one from each hydrogen to form the four equivalent covalent bonds. In order to create four equivalent orbitals from the carbon atom, there has to be mixing (hybridization) of four atomic orbitals. Atomic orbitals that contain the valence electrons are the most important orbitals to be considered for covalent bonds. As a result, the 2s and all three of the 2p orbitals of carbon are the most important orbitals to be considered for the four new molecular orbitals. In Figure 1.18, the hybridization (mixing) of the 2s and 2p orbitals of carbon to create four new atomic orbitals is shown.
The new hybridized orbitals that are created are called sp3 orbitals. The name is derived from the orbitals that are used to generate these orbitals: one 2s orbital and the three 2p orbitals, hence the name s1p3, or just simply sp3. These orbitals are higher in energy than either the 1s or the 2s orbitals, but lower in energy than the 2p orbitals, as illustrated in Figure 1.18. Also, the mixing of the orbitals results in hybridized orbitals that have 75% p character and 25% scharacter since these are essentially the contributions from these orbitals. The significance of this percentage mixing of atomic orbitals will be important when the bond lengths and electronegativities of atoms in molecules that have different hybridized orbitals are discussed.
Figure 1.18 Hybridization of the 2s and 2p orbitals of carbon to form four equivalent 2sp3 orbitals.
Each C─H bond of methane can be represented by a similar molecular orbital diagram as that used for hydrogen (Figure 1.17), but in this case, four molecular orbitals are needed since there are four covalent bonds in methane. Figure 1.19 gives the illustration of just one of those molecular orbitals of methane in which only one of the sp3 hybridized orbitals of the four shown in Figure 1.18 is used to bond with one hydrogen atom.
Figure 1.19 Molecular orbital diagram of one of the bonds of methane, note that one electron is from hydrogen and the other is from one of the sp3 hybridized orbitals of carbon.
Based on this molecular orbital model, there are four equivalent sigma bonds from carbon to four hydrogen atoms, and the result is a molecule with a tetrahedral geometrical arrangement, as shown in Figure 1.20. Note that the geometry is the same as that predicted by the VSEPR theory.
1.5.3 The Single Bond to Heteroatoms
A similar approach using the molecular orbital theory can be used to explain the experimental observations of ammonia (NH3). It is known that the bond angles of ammonia are slightly less than that of methane, actually 107°. The molecular orbital picture shown in Figure 1.21 best explains this observation in which sp3 hybridized orbitals from the nitrogen atom are used to bond to the hydrogen atoms from nitrogen and the unshared electrons are in a sp3 hybridized molecular orbital. Since these electrons are nonbonding and hence occupy more space than bonding electrons, they will slightly compress the N─H bond angles, compared to that of methane.
Figure 1.20 Methane showing the tetrahedral arrangement of the four equivalent bonds from carbon to four hydrogen atoms.
Figure 1.21 Hybridization of the 2s and 2p orbitals of nitrogen to form four equivalent 2sp3 orbitals. Note that the number of electrons is different from that of carbon; there are a total of seven electrons in nitrogen and a total of six electrons in carbon.
The orbitals of nitrogen are sp3 hybridized, with bond angles close to that of methane, but slightly less as explained above. The model for ammonia is shown in Figure 1.22.
Figure 1.22 Ammonia showing the trigonal pyramidal arrangement of the three equivalent bonds from nitrogen to three hydrogen atoms and the fourth orbital, which contains the unshared pair of electrons (shown in blue).
The molecular orbital theory can be used to explain the bonding and other properties of water (H2O). From the VSEPR theory, there are two bonds from oxygen to two hydrogen atoms in water and there are also two pairs of nonbonding electron. Similar to ammonia, there must be molecular orbitals available for not only the bonding electrons but also nonbonding electrons. As a result, the orbitals of water are sp3, in which two are used to bond to hydrogens and the other two contain two pairs of nonbonding electrons.
Problem 1.16
1. The oxygen atom of water (H2O) also has four sp3 hybridized orbitals, two to the hydrogen atoms and two that contain two lone pairs of nonbonding electrons. Use a figure similar to that in Figure 1.21 to represent the orbitals and electrons in water.
2. Explain why the observed bond angle of water is 104.5° and not 109.5°, similar to methane (CH4) or ammonia (NH3), which is 107°.
1.5.4 The Carbon—Carbon Double Bond
The expanded structure of ethylene, C2H4, is shown below.
Based on the VSEPR theory, ethylene is a trigonal planar (flat) molecule, with bond angles about each carbon atom of 120°. Since the bond angles about the carbon for CH4 are 109.5°, it is obvious that sp3 hybridized orbitals cannot be used to explain the bonding and geometry of ethylene. For ethylene, there are three atoms bonded to each carbon — two bonds to two hydrogen atoms and the other to another carbon atom; note that for ethylene, there are no nonbonding electrons. Thus, for each carbon atom, three equivalent orbitals are needed to bond to three atoms. The creation of three equivalent orbitals from each carbon atom requires the hybridization (mixing) of the 2s and 2p orbitals that must be considered; but in this case, only the hybridization of the one 2s and only two of the 2p orbitals are needed to create three new orbitals. The three new orbitals that are created are called s1p2 or simply, sp2 orbitals. Figure 1.23 gives an illustration of the hybridization needed to form three equivalent sp2 orbitals from one of the carbon atoms of ethylene.
Note that not all three 2p orbitals are used as in the case to create four sp3 hybridized orbitals for CH4. For ethylene, there is one 2p orbital that is left unhybridized. Also note that the three hybridized orbitals of ethylene have 66.6% p character and 33.3% s character; hence, there is more s character for the carbon atom of molecules that utilize sp2 orbitals, compared to molecules, such as methane, that use sp3 hybridized orbitals. Also, note that when the electrons are placed in the orbitals, they are distributed across all three hybridized orbitals and the 2p orbital before pairing occurs. Based on Hunds rule, the expectation is that since the 2p orbital is a bit higher in energy than the 2sp2 orbitals that the electrons would pair before occupying the vacant 2p orbital. In this case, however, the energy difference between the 2sp2 and 2p orbitals is extremely small and the energy needed to pair the electrons is a slightly more, and as a result, the electrons occupy separate orbitals before pairing as shown in Figure 1.23.
Figure 1.23 Hybridization of the 2s and two 2p orbitals of carbon to form three equivalent 2sp2 orbitals and one 2p orbital. Note that since the 2p orbitals are degenerate, there are several options to get three equivalent 2sp2 orbitals.
Thus, for each carbon atom of ethylene, there are three sigma bonds, one is bonded to each of the two hydrogen atoms and the third to another sp2 hybridized carbon as shown in Figure 1.24. The carbon—carbon bond length is shorter than a sp3 carbon—carbon length; the C─C bond length of CH3CH3 is 1.54 Å and that of ethylene is 1.33 Å. This observation can be explained due to the different types of hybridized orbitals that are used to make the molecular orbitals. The sp2 orbital has more s character and electrons that are in those orbitals are closer to the nucleus, compared to the sp3 orbital, which has less s character resulting in the electrons in the sp3 orbital being further from the nucleus. Figure 1.24 gives a graphical representation of ethylene.
As shown in Figure 1.24, each carbon atom in ethylene has an unhybridized 2p orbital, which contains one electron. There is a second type of bond, which is created between the p atomic orbitals of the adjacent carbon atoms as illustrated in Figure 1.25. This new molecular orbital that is created is called the pi (π) molecular orbital and the bond that is formed is called a pi (?) bond and is different from a sigma bond in that it is created using electrons from the p orbitals.
Another representation of this type of bonding is shown in Figure 1.26, in which the atomic orbitals of each p orbital form a molecular orbital, which is low in energy, and the other molecular orbital, which is higher in energy, is known as an antibonding or pi* (π*) molecular orbital.
Since the p orbitals have an hourglass shape as described earlier in the chapter, a single pi (π) bond has electron distribution above and below the plane of the sigma bond. Figure 1.27 gives another illustration of ethylene showing the pi (π) bond, where the electron density is above and below the plane of the molecule.
Figure 1.24 Graphical representation of sp2 sigma bonds and unhybridized p orbital of ethylene.
Figure 1.25 Graphical illustration of the pi (π) bond of ethylene.
Figure 1.26 Molecular orbital diagram of the pi (π) bond of ethylene
This newly created bond, the pi (π) bond, in combination with the sigma (σ) bond, result in a carbon—carbon double bond. This carbon—carbon double bond has four electrons, but in different molecular orbitals, two are in the σ molecular orbital and two in the π molecular orbital. Figure 1.28 gives another representation of the relative energies of these two orbitals, a sigma (σ) and pi (π). Note that the antibonding orbitals, σ* and π*, have no electrons.
Figure 1.27 Graphical illustration of the pi (π) orbital of ethylene showing the pi (π) bond, where the electron density is above and below the plane of the molecule.
Figure 1.28 Relative energies of the molecular orbitals of ethylene
Problem 1.17
1. What type of orbitals would you expect for the bonding from the central atom in BF3?
2. Give a diagram similar to that in Figure 1.17 for BF3.
3. Based on the location of Al in the periodic table, predict the geometry that would exist for AlCl3.
1.5.5 The Carbon—Heteroatom Double Bond
It is possible for carbon to form a double bond to heteroatoms, such as nitrogen or oxygen. It is known that the geometry around the carbon of methyleneimine and formaldehyde is 120 °C and that these molecules are flat and described as trigonal planar; the Lewis dot structures of these molecules are shown below.
For methyleneimine, the carbon atom is bonded to two hydrogen atoms and a nitrogen atom; as a result, there should be three sp2 orbitals from carbon to these atoms. For the nitrogen atom, which is bonded to a sp2 carbon, a hydrogen has one pair of nonbonded electrons; thus, three equivalent orbitals must be created: one to form a sigma bond to carbon, one to hydrogen, and another orbital for the unshared pair of electrons. As a result, the orbitals of the nitrogen atom are sp2 hybridized orbitals. Similarly, the carbon atom of formaldehyde is bonded to three atoms (two hydrogen atoms and an oxygen atom); thus, the carbon atom uses sp2 hybridized orbitals to form sigma bonds to these atoms. However, for the oxygen atom, which is bonded to an sp2 carbon and has two pairs of nonbonded electrons, three equivalent orbitals must be created: one to form a sigma bond to carbon and the other two orbitals for the two unshared pairs of electrons. As a result, the orbitals of the oxygen atom are sp2. For these two molecules, there are two electrons in each of the unhybridized p orbitals. These electrons form a pi (π) bond similar to that described for ethylene.
Problem 1.18
Formic acid has two oxygen atoms that use different types of orbitals for bonding. (i) For each oxygen atom, determine the type of orbitals used for bonding? (ii) Determine the type of orbitals used for bonding around the carbon atom?
1.5.6 The Carbon—Carbon Triple Bond
The Lewis dot structure of acetylene, C2H2, is: H:C:::C:H and is also shown below.
Based on the VSEPR theory, acetylene is a linear molecule. It is known that the bond angle around each carbon is 180° and that the C─H bond length is 1.061 Å, which is shorter than either the C─H bond length of ethylene or ethane. These observations imply that sp2 or sp3 molecular orbitals cannot be used to describe the geometry and bond lengths of acetylene. Since there are two atoms bonded to each carbon atom (one to hydrogen and the other to another carbon atom), two equivalent orbitals are needed. In order to create two equivalent orbitals from one of the carbon atoms to a hydrogen atom and to another carbon atom, there needs to be hybridization (mixing) of one 2s orbital and only one of the 2p orbitals to create s1p1 or simply sp orbitals. The hybridization to create an sp orbital is illustrated in Figure 1.29.
In this case, the orbitals are 50% s in character and 50% p in character. Thus, these orbitals have the most s character, compared to the sp3 or the sp2 orbitals. Since acetylene has two similar carbon atoms, there must be two sp hybridized carbon atoms. For each carbon, one is bonded to a hydrogen atom and to the other sp hybridized carbon resulting in two sigma bonds, one to another carbon and one to a hydrogen atom, as illustrated in Figure 1.30.
The two unhybridized atomic 2p orbitals on each carbon atom of acetylene are perpendicular to each other, i.e. 90°; another term that is often used to describe the geometric relationship between these two orbitals is that the orbitals are orthogonal to each other. The electrons in these orbitals create two new pi (π) bonds that are orthogonal to each other, as shown in Figure 1.31.
Figure 1.29 Hybridization of the 2s and one 2p orbital of carbon to form two equivalent 2sp orbitals. Note that since the 2p orbitals are degenerate, there are several options to get two equivalent 2sp orbitals.
Figure 1.30 Representation of the sigma bonds and two unhybridized 2p orbitals (each with one electron) of acetylene.
Figure 1.31 Representation of the sigma bonds and two perpendicular pi (π) bonds of acetylene.
The combination of two pi bonds and one sigma bond that is created between the carbon atoms is called a carbon—carbon triple bond — two pi bonds and one sigma bond.
Since the geometry of acetylene is similar to that of carbon dioxide, with bond angles of 180° about the central carbon atom of carbon dioxide, the orbitals that are used for the carbon atom of carbon dioxide are also sp.
Problem 1.19
i. What type of orbitals would you expect for the bonding in BeCl2?
ii. Give a diagram similar to that in Figure 1.29 for this molecule.
1.5.7 The Carbon—Heteroatom Triple Bond
It is possible to form a similar triple bond from carbon to nitrogen or oxygen. The structures of examples of molecules, in which a triple bond is formed from carbon to nitrogen atom and from carbon to oxygen atom, are shown below.
Note that the case of carbon monoxide is unique in that there are formal charges on the atoms as shown above. For carbon monoxide there is one carbon atom that is bonded to an oxygen atom by a triple bond, which is a very strong bond. This triple bond consists of two covalent bonds as well as one dative covalent bond, also known as a coordinate covalent bond. A dative covalent bond is a covalent bond in which one of the atoms provides a lone pair of electrons to a make a bond. For carbon monoxide, oxygen is the atom that provides a pair of electrons to form this bond. We will not encounter many of these types of bonds throughout this course of organic chemistry. For hydrogen cyanide, the carbon atom is bonded to hydrogen and nitrogen; as a result, sp orbitals must be used to form bonds from carbon to these atoms. For the nitrogen atom, which is bonded to only the sp carbon and has one pair of nonbonded electrons, two equivalent orbitals must be created: one to form a sigma bond to carbon and another orbital for the unshared pair of electrons. As a result, the orbitals of the nitrogen atom are sp.