Absolutely Small: How Quantum Theory Explains Our Everyday World - Michael D. Fayer (2010)

Chapter 14. Bigger Molecules: The Shapes of Polyatomic Molecules

THE WORLD AROUND US is composed of polyatomic molecules. Polyatomic molecules are molecules with more than two atoms. These range in size from triatomic molecules, such as carbon dioxide (CO2), which is a major greenhouse gas, to molecules with thousands of atoms, such as large proteins that are responsible for most biological functions. As discussed in Chapter 13, diatomic molecules can only have one shape, linear. Larger molecules, however, can have very complicated shapes and structures. For example, saturated fats, unsaturated fats, or polyunsaturated fats differ by their shapes and structures, which are determined by the nature of chemical bonds. A given large molecule can have more than one shape. Trans fats, which are currently being at least partially removed from food (see Chapter 16), only differ in their shape from molecules made up of the same sequence of atoms that are not trans fats. A central question in molecular matter is, how do molecules achieve their shapes and how are different shapes possible for molecules made of the same atoms connected together in the same way?

Before proceeding, it is worth stating that the covalent bond, which is responsible for holding atoms together to form molecules, is an intrinsically quantum mechanical phenomenon. It was not possible to explain the nature of chemical bonds or the structure of molecules before the advent of quantum theory. Linus Pauling (1901-1994) won the Nobel Prize in Chemistry in 1954 “for his research into the nature of the chemical bond and its application to the elucidation of the structure of complex substances.” It is noteworthy that Linus Pauling is one of the few people to win two Nobel Prizes. In 1962 he won the Nobel Peace Prize.


To examine the shapes of molecules and how chemical bonding can give rise to different shapes, we need to introduce some new ideas about atomic orbitals. We will use methane as an example of a relatively simple polyatomic molecule to bring out the important issues.

Methane (natural gas) is CH4. Now, look again at carbon’s position in the Periodic Table (Figure 11.4). Note that carbon needed to make four electron pair sharing covalent bonds to achieve the neon closed shell configuration. In methane, carbon makes four bonds with four hydrogen atoms. The connectivity of the atoms can be shown in a simple diagram, which is the left portion of Figure 14.1. Each line represents an electron pair covalent bond. However, this diagram does not tell much about methane’s shape. Methane is not flat. The right side of the figure shows a ball-and-stick model of the three-dimensional shape of methane. (See the end of Chapter 13 for a discussion of ball-and-stick and space-filling models.) Methane is a perfect tetrahedron. Imagine having a model like the right side of Figure 14.1 and gluing a triangular piece of paper with sides of equal length that just covered three of the hydrogen atoms. You can glue on four such triangles, three sides and the bottom. These four triangles form a perfect triangular pyramid with the hydrogens at the four apices and the carbon in the middle. The angle formed by the line from a hydrogen to the carbon and then a line from the carbon to another hydrogen is exactly 109.5°. This is true of all four angles. They are identical. The angle for a perfect tetrahedral molecule is 109.5°.


FIGURE 14.1. Left: diagram showing the bond connectivity of methane, but not its three-dimensional shape. Right: a three-dimensional ball-and-stick model of methane that shows the tetrahedral shape of the molecule.

Minimizing Repulsion Between Bonds Determines the Shape

Why does methane have a tetrahedral shape? In Chapter 13, we saw that bonding molecular orbitals concentrate electron density between the atomic nuclei. The concentration of electron density between the nuclei is shown in Figures 13.2 and 13.3 for σ and π bonds. Chapter 13 discussed diatomic molecules in which only two atoms are bonded. We did not have to worry about how multiple atoms with some set of bonds connecting them would be arranged. While quantum theory can calculate the details of the shapes of molecules, the basic reason why a molecule will have a particular shape, such as tetrahedral, is quite simple. In polyatomic molecules, sharing electrons between two atomic nuclei to form a bond concentrates electron density between the nuclei, just as it does with diatomic molecules. However, in polyatomics, we have a number of bonds, each of which has a high density of negatively charged electrons. The negatively charged regions, the bonds, repel each other. Basically, the bonds want to stay as far away from each other as possible. These regions of high negatively charged electron density want to overlap as little as possible. Forming bonds lowers the energy of the system relative to the separated atoms. If the energy is not lowered, bonds will not be formed. But to make the energy as low as possible, the system of atoms assumes a configuration that minimizes the electron repulsions by keeping the bonds away from each other.

In methane, the tetrahedral shape minimizes the repulsions between the bonds. Look at the left-hand drawing in Figure 14.1. The four hydrogens are in a plane. If I keep them in the plane, they are as well separated as possible. The angle formed by two adjacent bonds is 90°. With the atoms in the plane, if I make one of the angles larger than 90° to further separate two of the hydrogen-carbon bonds, these bonds get closer to the other two bonds. So if the four hydrogen atoms are kept in the same plane as the carbon atom, the best that can be done is to have 90° bond angles.

However, there is no reason that all of the atoms have to lie in a plane. In the structure on the left side of Figure 14.1, the angle between the top and bottom H atoms is 180°, and these two C-H bonds are far apart; the same is true for the left and right H atoms. Imagine pulling the top and bottom H atoms above the plane of the page, while keeping the bond lengths the same and pushing the right and left H atoms below the plane of the page. The top and bottom C-H bonds are getting closer together and the angle is reduced to less than 180°, but they are getting further away from the left and right C-H bonds. Bringing the top and bottom C-H bonds above the plane and pushing the left and right C-H bonds below the plane reduces the overall bond-bond interaction. The interaction between the top and bottom bonds is increased, but they were very far apart to begin with. This is also true of the left and right bonds. However, if the top-bottom and left-right angles are reduced too much, the repulsion will go up again. There is a best angle, and that is 109.5°, the tetrahedral angle. This is the angle that keeps the electrons in the four C-H bonds as far apart as possible.

Lone Pairs Also Matter

In Chapter 11, we found that to obtain a closed shell configuration, C needs to form four bonds, N needs to form three bonds, and O needs to form two bonds. If the bonds are to hydrogens, then we have methane, ammonia, and water, CH4, NH3, and H2O. In discussing HF at the end of Chapter 13, we noted that some of the F electrons were not involved in bonding at all. The electrons were paired in what are essentially atomic orbitals called lone pairs. Lone pairs are pairs of nonbonding electrons that give rise to high electron density in the region of space they occupy. Electrons in bonds do not want to be near electrons in lone pairs. So, although they are not bonds, lone pairs also influence the shapes of molecules. In a bond, the electrons are shared and more or less concentrated between two atoms. Lone pairs do not have a second atom to hang on to. As a result, the lone pair electron distribution is bunched closer to the atom they belong to, and it is somewhat “fatter” than a bond pair distribution.

Figure 14.2 shows models of methane, ammonia, and water. Ammonia has one lone pair, and water has two lone pairs. If you include the lone pairs, all three molecules are basically tetrahedral in shape. However, ammonia and water are not perfect tetrahedrons. The lone pair in ammonia is more spread out than the bond pairs. To minimize the overall electron-electron repulsion to produce the lowest energy, the bonds move away from the lone pair, thereby bringing the bonds closer together. The HNH angle in ammonia is 107.3°, slightly less than the perfect tetrahedral angle. Water has two lone pairs, causing the angle between the hydrogen-oxygen bonds to be reduced further to 104.5°.


FIGURE 14.2. Methane (left), ammonia (center), and water (right). Lone pair electrons repel the bond pair electrons, pushing the bonds closer together, which reduces the angle between bonds to H atoms from the central atom.

Trigonal-Shaped Molecules

If a central atom is bound to only three other atoms, it will have a trigonal shape with the four atoms lying in a plane. Figure 14.3 shows two trigonal molecules, BH3 and H2CO (formaldehyde). BH3 exists, but it is very reactive because it is two electrons short of the neon closed shell configuration. In BH3, each H has a single bond to the B. The HBH angle is exactly 120°. The hydrogens form a perfect equilateral triangle. This is the shape that keeps the bonds as far apart as possible, which lowers the energy by reducing the repulsive interactions between the bonding electrons in each bond.

In Chapter 13, the MO diagram for O2 (Figure 13.8) showed that the oxygen molecule has a double bond. In formaldehyde (the smelly liquid in the jars containing dead things in biology class), the O is double bonded to the C. The double bond is shown in the ball-and-stick figure model as two cylinders joining the atoms rather than one. The double bond gives the O the neon closed shell configuration as in the O2 molecule. The C needs to share two additional electrons to obtain the neon closed shell configuration, which it does by single bonding to two H atoms. We will discuss double bonds in detail to see how they can be formed from atomic orbitals. Here, we only need to recognize that a double bond concentrates two pairs of electrons between the C and the O. Because of the extra electron density, a double bond is fatter than a single bond. The fatter C=O double bond pushes the C—H single bonds away from it and toward each other. The angles are shown in Figure 14.3. The result is that formaldehyde is still a planar trigonal molecule, but it is not a perfect equilateral triangle.


FIGURE 14.3. Left: BH3. The atoms lie in a plane. The HB bonds are single, and the hydrogens form a perfect equilateral triangle. Each HBH bond angle is 120. Right: H2CO (formaldehyde). The atoms lie in a plane. The CO bond is a double bond. The angles are unequal.


Returning to methane, the question is how does methane make four tetrahedrally configured bonds? In Chapter 11, we discussed the electronic configuration of the atoms (see Figure 11.1). Carbon has six electrons, two in the 1s, two in the 2s, and two in 2p orbitals. The valence electrons, the electrons used in bonding, are the 2s and 2p electrons. The top portion of Figure 14.4 shows the atomic orbital energy levels with the four valence electrons filled in. The 1s electrons are not shown. As discussed in Chapter 11 and earlier in this chapter, carbon will form four bonds. In methane, it forms four electron pair sharing bonds to four hydrogen atoms. Each H atom contributes one electron. So the carbon must have four unpaired electrons to form bonds. Each carbon unpaired electron can join with one electron from an H atom to yield an electron pair bond. To have four unpaired electrons, carbon “promotes” a 2s electron to a 2p orbital, as shown in the bottom part of Figure 14.4. For an isolated carbon atom, the configuration shown in the bottom of the figure would not occur unless a lot of energy was pumped into the atom. For a carbon atom, putting a 2s electron into a 2p orbital is a high energy configuration. However, when atoms form molecules, the electrons and nuclei of the different atoms affect each other. Imagine bringing four H atoms toward a C atom. Now, the system wants to assume the lowest energy configuration for all five atoms. Forming the four bonds lowers the energy more than putting a 2s electron into a 2p orbital raises the energy.


FIGURE 14.4. Top: Atomic carbon valence orbitals with the four valence electrons. Bottom: When bonding, a carbon atom “promotes” a 2s to a 2p electron to give four unpaired electrons used to form four bonds to other atoms.


So we see how carbon can make four bonds to yield methane, but why is the shape tetrahedral? The three 2p orbitals are px, py, and pz. These three orbitals are perpendicular to each other, with a 90° angle between any pair. If three H atoms bonded to the three 2p orbitals, the bond angles would be 90°. Furthermore, the 2s orbital is spherical. The fourth H atom’s 1s orbital would have to combine with the carbon 2s orbital. Without something else happening, it is pretty clear that using the carbon 2s and the three 2p orbitals is not going to give methane four identical C-H bonds in a tetrahedral configuration. In addition, how does carbon make the trigonal molecule formaldehyde, or the linear molecule carbon dioxide (O = C = O). In each case, tetrahedral, trigonal, or linear, carbon bonding involves the same 2s and 2p orbitals.

Formaldehyde and carbon dioxide involve double bonds, which we will get to shortly. To bring out the important features of the nature of atomic orbitals that can give linear, trigonal, and tetrahedral shapes, we will examine bonding in BeH2, BH3, and CH4, that is, beryllium dehydrate, boron trihydride (borane), and methane. Be and B in BeH2 and BH3 do not have closed shell neon rare gas configurations. Therefore, they are very reactive. These molecules can be made, but they will react with virtually anything they come in contact with to make new molecules in which the Be and B have closed shell configurations. Here we will use them as convenient examples.

Be is two electrons past the He closed shell configuration. As an atom, these two electrons would be paired in the 2s orbital. These are Be’s valence electrons. In BeH2, each H has one electron in a 1s orbital. For Be to make two electron pair bonds, one to each H atom, it will promote one of the 2s electrons to a 2p orbital, which we will take to be the 2pz, as shown at the top of Figure 14.5. The second line of the figure shows schematics of the 2s and the 2pzorbitals separated. These actually have the same center, the nucleus of the Be. These orbitals are electron probability amplitude waves. Waves can be added and subtracted to give new waves. We start with two atomic orbitals, the 2s and the 2pz, and by addition and subtraction, we get two new atomic orbitals, called hybrid orbitals. When we add the waves together, we get regions of constructive interference and regions of destructive interference because the lobes of the probability amplitude waves have signs. The third line in Figure 14.5 shows the sum of the 2s and 2pz orbitals. This is called an sp hybrid, which we have labeled spz+ because it is an sp hybrid made from an s orbital and the 2pz orbital, and its big positive lobe points long the +z direction. The second to the bottom line of Figure 14.5 shows the 2s minus the 2pz. One way to think of this is just that we have flipped the 2pz orbital shown in the top line of the figure around so the positive lobe points to the left rather than to the right, and then we add. We have labeled this hybrid orbital spz- because its big positive lobe points in the—z direction. This orbital has the same shape as the spz+ orbital but it points in the opposite direction. We start with two atomic orbitals, 2s and 2pz, and we end with two hybrid atomic orbitals with their positive lobes pointing along the z axis in opposite directions. The bottom line of Figure 14.5 shows the schematic of both hybrid atomic orbitals as they would occur in the Be atom. There is one Be nucleus with the two hybrid orbitals pointing along the +z and -z directions.


FIGURE 14.5. Top: Be valence electrons with one promoted to the 2pzorbital. Next: The 2s and 2pzorbitals of Be shown separated. Next: The sum of 2s and 2pzto form the hybrid atomic orbital spz+. Next: 2s—2pzto form the hybrid orbital spz-. Bottom: The two hybrid orbitals of Be point in opposite directions along z.

To make BeH2, Be will use the two hybrid atomic orbitals to form electron pair bonding molecular orbitals with two hydrogens. The bonding is shown schematically in Figure 14.6. The top of the figure shows two hydrogen atoms, Ha and Hb, approaching a Be. Hydrogen atom electrons are in 1s orbitals, 1sa and 1sb. The beryllium has its two hybrid atomic orbitals, spz- and spz+, pointing at the hydrogen 1s orbitals. The middle portion of the figure shows a schematic illustration of the overlapping atomic orbitals. The hydrogen 1sa orbital on the left side of the figure will form a bonding MO with the Be spz- hybrid atomic orbital. This bonding MO will contain two electrons, one from the hydrogen and one of the two Be valence electrons. The hydrogen 1sb orbital on the right side of the figure will form a bonding MO with the spz+ hybrid orbital. The electron from Hb and the other Be valence electron will form another covalent bond. These are σ bonds since there is electron density along the line connecting the nuclei. The result is the linear BeH2 molecule shown at the bottom of the figure.


FIGURE 14.6. Top: Two H atoms approach a Be. Middle: The H 1s orbitals form electron pair bonds with the two Be sp hybrid orbitals to produce the linear BeH2molecular shown as a ball and stick model at the bottom.

In total the BeH2 molecule has six electrons counting the two pairs of bonding electrons and the two electrons in the Be 1s orbital. It is important to note that all of these electrons are probability amplitude waves that span the entire molecule. When we say that the 1sa electron makes an electron pair bond with the spz- electron, this is bookkeeping. All of the electrons are free to roam over the entire molecule. However, the probability distributions for the electrons are such that at any given instant there is electron density associated with the Be and one of the H atoms that corresponds to one bond and electron density associated with the Be and the other H atom that corresponds to the other bond.


As discussed in connection with Figure 14.3, BH3 is a trigonal molecule with 120° between bonds. Atomic boron has three valence electrons, two in the 2s and one in a 2p orbital. To form three electron pair sharing bonds to three hydrogen atoms, the boron atom needs three unpaired electrons. As shown at the top of Figure 14.7, B will promote one electron from a 2s orbital to a 2p orbital to have the three unpaired electrons. If the molecule lies in the xy plane, then the 2p orbitals involved in bonding will be the 2px and 2py. To form the equilateral trigonal BH3 molecule, the three boron atomic orbit als will hybridize to yield the three hybrid atomic orbitals, 135 , 136 , and 137 . The sp2 notation means that the hybrid orbital is composed of an s orbital and two different p orbitals. We start with three different orbitals, 2s, 2px, and 2py. Orbitals are never gained or lost, so we end up with three different hybrid orbitals. These are shown in the middle portion of Figure 14.7. Each orbital has a positive and a negative lobe. The adjacent lobes have an angle of 120° between them. Each of these contains one of the three unpaired B valence electrons.

The bottom portion of Figure 14.7 shows the bonding of B with three H atoms. Each H atom has a single 1s electron. An H 1s orbital combines with a B sp2 orbital to form a bonding molecular orbital. There are two electrons in this bonding MO, one from the H and one from the B. The result is an electron pair bond. Each of the bonds between B and H is a σ bond since there is electron density along the line connecting the nuclei. A model of BH3 is shown in Figure 14.3.


FIGURE 14.7. Top: B valence electrons with one promoted to a 2p orbital. Middle: the 2s, 2px, and 2pyorbitals of B combine in three combinations to form three hybrid atomic orbitals, 139, 140and 141. The angle between the lobes is 120. Bottom: The three B hybrid orbitals forming bonds with the three H atom 1s orbitals.


In methane, carbon makes four bonds to four hydrogen atoms. As discussed above and shown in Figures 14.1 and 14.2, methane is tetrahedral. As illustrated in Figure 14.4, to make four electron pair sharing covalent bonds, carbon promotes one of its 2s electrons to a 2p orbital. It then has four unpaired electrons in the 2s, 2px, 2py, and 2pz orbitals. As discussed in some detail following Figure 14.4, these four carbon atomic orbitals will not yield four identical bonds to four hydrogens and produce a tetrahedral molecule. Therefore, the 2s, 2px, 2py, and 2pz orbitals combine in four different combina tions to form four atomic hybrid orbitals, 142 , 143 , 144 , and 145. The sp3 designation means that each of the four hybrid atomic orbitals is a combination of an s orbital and the three different p orbitals. Figure 14.8 shows the four sp3 orbitals overlapping with the four hydrogen 1s orbitals. Only the positive lobes of the sp3orbitals are shown. Each of them has a small negative lobe pointing in the opposite direction from the positive lobe in a manner analogous to that shown for the sp2 orbitals in the middle portion of Figure 14.7. The orbitals represented by dashed curves are in the plane of the page. The orbital shown as a solid curve is coming out of the plane of the page at an angle, and the orbital shown as the dot-dash curve is at an angle going into the plane of the page. The angle between any pair of the sp3 lobes is 109.5°, giving the perfect tetrahedral shape discussed in connection with Figures 14.1 and 14.2.

Figure 14.2 shows methane, ammonia, and water. We said that each of these is tetrahedral when the lone pairs are included, but ammonia and water are slightly distorted tetrahedrons. Like methane, ammonia and water also use sp3 hybridization to form bonds. The nitrogen in ammonia, NH3, has five valence electrons. Two of them form a lone pair. The lone pair does not participate in bonding. The nitrogen uses three of its four sp3 hybrid orbitals to bond to the three H atoms. The fourth sp3 orbital contains the lone pair. As discussed, the H-N-H bond angle is a little less than the perfect tetrahedral angle, 109.5°, because the spatial distribution of the lone pair electrons is somewhat fatter than the N-H bond pair electrons, and the fatter lone pair pushes the N-H bonds slightly toward each other. The oxygen in water, H2O, has six valence electrons. Four of them form two lone pairs. These two lone pairs of electrons do not participate in bonding. Oxygen uses two of its sp3 hybrid orbitals to form bonds to the two hydrogens. The other two sp3 orbitals are each occupied by a lone pair. These lone pairs cause the H-O-H angle to be less than the perfect tetrahedral angle of 109.5° (see Figure 14.2).


FIGURE 14.8. Four carbon sp3hybrid atomic orbitals and four hydrogen 1s orbitals for carbon bonding to four hydrogens in methane. Dashed orbitals in the plane of the page. Solid orbitals out of the plane. Dot-dash orbitals into the plane. Only the positive lobes of the sp3hybrids are shown. The four sp3hybrids form a perfect tetrahedron.


A hydrocarbon is a molecule that is solely made up of carbon and hydrogen atoms. First we will discuss more complicated hydrocarbons than methane, but initially only molecules with single bonds. The next simplest hydrocarbon after methane is ethane. Ethane has two carbons and six hydrogens with the chemical formula, C2H6. Figure 14.9 displays the ethane structure in three ways. The top shows only the bonds between the atoms. Each carbon has a single bond to three hydrogens and a single bond to the other carbon. The middle portion of the figure shows the hybrid atomic orbitals used in bonding. To make the four electron pair sharing bonds, each carbon uses four sp3 hybrid orbitals, just like in methane. Three of the orbitals on each carbon are used to make bonds to three hydrogens. A sp3 orbital combines with a hydrogen 1s orbital to form a bonding MO to yield a σ bond. The fourth sp3 orbital belonging to one carbon forms an MO with the sp3 orbital on the other carbon to produce a carbon-carbon σ bonding MO.

The bottom portion of Figure 14.9 is a standard method of schematically displaying the spatial arrangement of atoms in a molecule. The atoms that are bonded and lie in the plane of the page are connected with lines. Bonded atoms that stick out of the plane of the page are connected by narrow filled triangles with the sharp end of the triangle pointing at the atom in the plane and the base of the triangle next to the atom out of the plane. Bonded atoms that stick into the plane of the page are connected by open triangles with the base of the triangle next to the atom in the plane and the point of the triangle next to the atom below the plane. As shown in the bottom portion of the figure, each carbon is at the center of a tetrahedron. Any pair of bonds around a carbon has the tetrahedral angle of 109.5° between them. The C-H bond length is 1.07 Å (1.07 × 10-10 m), and the C-C bond length is 1.54 Å. Carbon atoms are bigger than hydrogen atoms, so the separation of the atomic centers is greater for the two carbon atoms than for a C and an H.


FIGURE 14.9. Three diagrams of ethane, C2H6. Top: The bonds between atoms. Middle: Each carbon has four sp3hybrid atomic orbitals, three bond to hydrogens and the fourth to the other carbon. Solid lines: in the plane of the page; dashed lines: out of the plane; dotted lines: into the plane. Bottom: Method for showing spatial arrangement. Lines: in the plane of the page; filled triangles stick out of the page; open triangles stick into the page. Each carbon’s bonds are tetrahedral, with bond angles 109.5°. The C-C bond is longer than the C-H bonds.

Figure 14.10 shows a ball-and-stick model (top) and a space filling model (bottom) of ethane. In Figures 14.9 and 14.10, we have five different representations of the ethane molecule. Only the spacing-filling molecule gives a real feel for the spatial nature of the molecule. The other four representations exaggerate the separations between the atoms for clarity. The atoms in the ball-and-stick and spacing-filling models in Figure 14.10 are the same sizes. In the ball-and-stick model, the bonds are shown as cylinders, and the atoms are separated by the bonds. It is important to recognize that the bonding comes from the formation of molecular orbitals. The electrons are shared by the atoms, which are not separated as in the ball-and-stick model or the other representations. The surface in the space-filling model contains a large fraction of the electron probability distribution. In the space-filling model, the atoms are shaded differently so that they can be distinguished.


FIGURE 14.10. Ethane ball-and-stick model (top), space-filling model (bottom). The atoms are the same size in the two models.

We need to discuss one more relatively simple molecule, propane, before the molecules get large enough that some general features can be brought out. Propane has three carbon atoms and eight hydrogens. Its chemical formula is C3H8. The formula doesn’t tell how the atoms are connected. It can also be written as H3C-H2C-CH 3. In this notation, it is understood that the hydrogens are bonded to a carbon. The carbons are bonded to each other with single bonds. The end carbons are bonded to three hydrogens and another carbon. The center carbon is bonded to two hydrogens and two carbons. Figure 14.11 shows two representations of propane. The top diagram shows the bonding and angles between the bonds. Each carbon uses four sp3 hybrid orbitals to make four σ bonds. The carbons are tetrahedral, with the C-C-C angle 109.5° and the HCH angles 109.5°. The bottom portion of the figure shows a ball-and-stick model of propane.


Methane, ethane, and propane have only one way that the atoms can be bonded together and only on spatial conformation. Butane and all larger hydrocarbons have multiple structural configurations (the way the atoms are bonded together) and more than one spatial configuration for a particular structural configuration. Butane has four carbons. Its chemical formula is C4H10. There are two distinct structural forms of butane. These are called structural isomers. Figure 14.12 displays the two structural isomers of butane. Both molecules have the same number of carbons and hydrogens, but they have very different shapes. Butane can be n-butane, which is normal butane. If we take propane and add one more carbon on the end, we get n-butane. n-butane is referred to as a linear chain because a carbon is at most bonded to two other carbons, one on either side. As can be seen in the ball-and-stick model, the molecule is not actually linear because each carbon has a tetrahedral arrangement of bonds formed using four sp3 hybrid orbitals.


FIGURE 14.11. A diagram and a ball-and-stick model of propane, C3H8. The carbon centers are tetrahedral.


FIGURE 14.12. Two structural isomers of butane, C4H10. In the diagram at the top, CH3represents a carbon bonded to three hydrogens. N-butane is a linear chain in the sense that each carbon is bonded to at most two other carbons. Isobutane is branched. The central carbon is bonded to three other carbons.

As shown in Figure 14.12, butane can have another isomer, called isobutane. Isobutane has a central carbon connected to three other carbons and one hydrogen, with each of the other three carbons only connected to the central carbon and to three hydrogens. All four carbons use sp3 hybrid atomic orbitals for bonding and are tetrahedral. Isobutane is referred to as branched. The fact that butane can have the same number of carbons and hydrogens but two different structures is of great importance. Molecules with more carbon atoms can have many more than two structures.

In addition to the two structural isomers, n-butane has two conformers. Conformers are different shapes, conformations, for the same set of atoms that are connected in the same way. They differ because it is possible for rotation to occur about a C-C single bond. Figure 14.13 shows n-butane in two conformations called trans and gauche. Both conformers shown in the figure are n-butane because the carbon atoms are connected in the same way. If you take the top conformer and rotate the central carbon-carbon bond, as shown by the arrow, 120°, you get the gauche form. The trans conformer has all of the carbons in a plane. The gauche form has three of the carbons in a plane and the fourth carbon is sticking out of the plane of the page. There is actually another gauche form that comes from rotating the trans form around the central C-C bond 120° in the opposite direction from that shown by the arrow. In that case the same three carbons are still in the plane, but the fourth carbon is sticking into the page instead of out of the page. These two gauche forms in some sense have the same shape, but they are not the same. They are like a left hand and a right hand glove. Like the gloves, these two gauche forms cannot be superimposed on each other. They are the mirror images of each other. A carbon center that can have a right hand and left hand form, depending on a rotation, is said to be chiral.


FIGURE 14.13. Two conformers of n-butane. The gauche form is obtained from the trans form by a 120° rotation about the center C-C bond.

The rotations about a C-C single bond that takes a molecule back and forth from trans to gauche can be very fast in a liquid at room temperature. Recent ultrafast infrared laser experiments and theory have determined that the gauche-trans interconversion only takes 50 ps (picoseconds, 10-12 s), or 50 trillionths of a second. So in a liquid at room temperature, the two forms of butane will be switching back and forth so fast that it is not possible to isolate them as distinct molecules.


While it is very easy for rotation to occur about a C-C single bond, this is not true for carbon-carbon double or triple bonds. In Chapter 13, we discussed that O2 has a double bond and N2 has a triple bond. Carbon-carbon bonds can be single, double, or triple. Rotation about a C-C double or triple bond is virtually impossible. Therefore, double bonds can lock molecules that are the same structural isomers into different conformers. As we will see in Chapter 16, this is where the term trans fats comes from. However, before we go on to discuss large molecules, such as trans fats, we first need to talk about C-C double and triple bonds.

In the carbon compounds discussed so far, carbon uses four sp3 hybrid atomic orbitals to make four single σ bonds to other atoms. In such bonding, each carbon has a tetrahedral arrangement of four bonds. In Figure 14.3, formaldehyde is shown. Formaldehyde has a carbon with a double bond. To illustrate the manner in which carbon can make single, double, and triple bonds, we will take a look at the bonding in ethane, ethylene, and acetylene. These three compounds have chemical formulas, H3C—CH3, H2C=CH2, and HC=CH, respectively. Ethane has a single bond, ethylene has a double bond, and acetylene has a triple bond. Figure 14.14 shows the structures of the three molecules. In ethane, each carbon forms four tetrahedral bonds. In ethylene, each carbon forms three trigonal bonds, and in acetylene, each carbon forms two linear bonds.

Although in each of the three molecules the two carbons are bonded to each other, the bond order really makes a difference. Table 14.1 gives the bond orders, C-C bond lengths, and bond energies of the three molecules. As the bond order increases, the bond length decreases substantially and the bond energy almost triples in going from the single to the triple bond.


FIGURE 14.14. Ethane, single bond, carbon tetrahedral bonds. Ethylene, double bond, carbon trigonal bonds. Acetylene, triple bond, carbon linear bonds.


TABLE 14.1. Single, Double, and Triple C-C Bonds.

Carbon-Carbon Double Bond—Ethylene

First let’s look at the bonding in ethylene. As can be seen in Figure 14.15, the carbon centers are trigonal. As discussed, to have trigonal bonding, a carbon atom will use three sp2 hybrid atomic orbitals to form bonding MOs (see Figure 14.7). Carbon has four valence atomic orbitals to use for bonding, 2s, 2px, 2py, and 2pz. In the top part of the figure, the ethylene molecule is in the xy plane. So the carbons and the hydrogens are in the plane of the page, which is xy. To form the trigonal sp2 hybrids to make three bonds, each carbon will use the 2s, 2px, and 2py orbitals. With the three sp2 hybrids, each carbon will make three σ bonds, one to the other carbon and two to hydrogens. The σ bonding is shown in the top portion of Figure 14.15.

When a carbon forms three sp2 hybrids with the 2s, 2px, and 2py orbitals, it has its 2pz orbital left over that does no participate in the σ bonding. In the top part of Figure 14.15, the 2pz orbitals stick out of and into the page. Each carbon has one unpaired electron in its 2pz orbital. The bottom portion of the figure shows the ethylene molecule rotated. The σ bonds are shown as the lines connecting the atoms. The positive lobes of the 2pz orbitals overlap constructively as do the negative lobes. The two 2pz orbitals combine to make a π bonding molecular orbital, as shown in Figure 13.3. This is a π bond because there is no electron density along the line connecting the carbon atomic centers. The net result is that the two carbon atoms have a double bond composed of a σ bond formed from a sp2 orbital on each carbon and a π bond formed from a 2pz orbital on each orbital.


FIGURE 14.15. Ethylene double bond orbitals. Top: Each carbon uses three sp2hybrids to make three σ bonds in a trigonal configuration. The page is the xy plane, with z out of the plane. Bottom: Each carbon has a 2pzorbital not used to form the sp2hybrids. The 2pzorbitals combine to make a π bonding molecular orbital to give a second bond between the carbons.

It is not possible for rotation to occur around the carbon-carbon double bond. Rotation around the bond would require the overlap of the two 2pz orbitals to get worse and worse as the angle got bigger and bigger. For a 90° rotation, the two 2p orbitals would be pointing perpendicular to each other, and have no overlap. Such a rotation would break the π bond, which takes a great deal of energy. As mentioned, measurements and theory have been used to determine that the time for butane in liquid solution to rotate about the C-C single bond is about 50 ps. For ethane the time is about 12 ps. Butane rotates slower than ethane about the single C-C bond because it has two additional methyl groups (CH3), one on each end of the central two carbons. If you put ethylene in the same liquid at room temperature, it can be roughly estimated that it will take about one hundred billion years to rotate about the double bond because of the large amount of energy required to break the π bond. Therefore, for all practical purposes, a double bond (and a triple bond) prevents rotational isomerization between conformers that differ by the configuration about the double bond.

Carbon-Carbon Triple Bond—Acetylene

Acetylene forms a triple bond between the carbons in a manor analogous to the double bond formation in ethylene. Each carbon has four atomic orbitals, 2s, 2px, 2py, and 2pz, to use in bonding. Acetylene is linear (see Figure 14.14). Take the line of the molecule to be the x axis. Then, each carbon will form two sp hybrids from its 2s and 2px orbitals. The two sp hybrids on a carbon will be used to make two σ bonds, one to the other carbon and one to a hydrogen. That leaves two unused 2p orbitals on each carbon, the 2py and the 2pz. The 2pz orbitals on each carbon form one π bonding MO, and the 2py orbitals on each carbon form another π bonding MO. The result is that the two carbons have a triple bond, one σ bond and two π bonds.

In the next several chapters, we will discuss a number of important types of molecules, for example, alcohols, organic acids, large hydrocarbons, and carbon containing fuels, that is, coal, oil, and natural gas. Small alcohols are discussed so that we can see what an alcohol is and how small differences in structure really matter if you decide to drink something other than ethanol (the alcohol in beer). The ideas will show why some molecules dissolve in water while others don’t, and how soap (a type of large organic molecule) makes insoluble grease dissolve in water. We will examine the importance of the inability of double bonds to undergo rotational structural change in connection with fats and trans fats. What happens when carbon-based fuels are burned will be described, and why, for a given amount of energy produced, one fuel produces more of the greenhouse gas carbon dioxide than another. It is well known that carbon dioxide is a greenhouse gas, but why? We will see how the combination of two fundamental quantum mechanical effects makes carbon dioxide a potent greenhouse gas.