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0 | CHAPTER 1 <br> Structure and Bonding -
FIGURE 1.1 The enzyme HMG-CoA reductase, shown here as a so-called ribbon model, catalyzes a crucial step in the body's synthesis of cholesterol. Understanding how this enzyme functions has led to the development of drugs credited with saving millions of lives. (credit: image from the RCSB PDB (rcsb.org) of PBD ID 1HW9 (E.S. Istvan, J. Deisenhofer) (2001) Structural mechanism for statin inhibition of HMGCoA reductase Science 292: 1160-1164/RCSB PDB, CC BY 1.0) |
1 | CHAPTER CONTENTS -
1.1 Atomic Structure: The Nucleus
1.2 Atomic Structure: Orbitals
1.3 Atomic Structure: Electron Configurations
1.4 Development of Chemical Bonding Theory
1.5 Describing Chemical Bonds: Valence Bond Theory
$1.6 s p^{3}$ Hybrid Orbitals and the Structure of Methane
$1.7 s p^{3}$ Hybrid Orbitals and the Structure of Ethane
$1.8 s p^{2}$ Hybrid Orbitals and the Structure of Ethylene
1.9 sp Hybrid Orbitals and the Structure of Acetylene
1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur
1.11 Describing Chemical Bonds: Molecular Orbital Theory
1.12 Drawing Chemical Structures
WHY THIS CHAPTER? We'll ease into the study of organic chemistry by first reviewing some ideas about atoms, bonds, and molecular geometry that you may recall from your general chemistry course. Much of the material in this chapter and the next is likely to be familiar to you, but it's nevertheless a good idea to make sure you understand it before moving on.
What is organic chemistry, and why should you study it? The answers to these questions are all around you.
Every living organism is made of organic chemicals. The proteins that make up your hair, skin, and muscles; the DNA that controls your genetic heritage; the foods that nourish you; and the medicines that heal you are all organic chemicals. Anyone with a curiosity about life and living things, and anyone who wants to be a part of the remarkable advances taking place in medicine and the biological sciences, must first understand organic chemistry. Look at the following drawings for instance, which show the chemical structures of some molecules whose names might be familiar to you. Although the drawings may appear unintelligible at this point, don't worry. They'll make perfectly good sense before long, and you'll soon be drawing similar structures for any substance you're interested in.
Oxycodone (OxyContin)
Cholesterol
Benzylpenicillin |
2 | CHAPTER CONTENTS -
Oxycodone (OxyContin)
Cholesterol
Benzylpenicillin
Historically, the term organic chemistry dates to the mid-1700s, when it was used to mean the chemistry of substances found in living organisms. Little was known about chemistry at that time, and the behavior of the "organic" substances isolated from plants and animals seemed different from that of the "inorganic" substances found in minerals. Organic compounds were generally low-melting solids and were usually more difficult to isolate, purify, and work with than high-melting inorganic compounds.
By the mid-1800s, however, it was clear that there was no fundamental difference between organic and inorganic compounds. The only distinguishing characteristic of organic compounds is that all contain the element carbon.
Organic chemistry, then, is the study of carbon compounds. But why is carbon special? Why, of the more than 197 million presently known chemical compounds, do almost all of them contain carbon? The answers to these questions come from carbon's electronic structure and its consequent position in the periodic table (FIGURE 1.2). As a group 4A element, carbon can share four valence electrons and form four strong covalent bonds. Furthermore, carbon atoms can bond to one another, forming long chains and rings. Carbon, alone of all elements, is able to form an immense diversity of compounds, from the simple methane, with one carbon atom, to the staggeringly complex DNA, which can have more than 100 million carbons. |
3 | CHAPTER CONTENTS -
| Grou 1A | | | | | | | | | | | | | | | | | 8A |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| H | 2A | | | | | | | | | | | 3A | 4A | 5A | 6A | 7A | He |
| Li | Be | | | | | | | | | | | B | C | N | 0 | F | Ne |
| Na | Mg | | | | | | | | | | | Al | Si | P | S | Cl | Ar |
| K | Ca | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | Ga | Ge | As | Se | Br | Kr |
| Rb | Sr | Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd | In | Sn | Sb | Te | I | Xe |
| Cs | Ba | La | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg | TI | Pb | Bi | Po | At | Rn |
| Fr | Ra | Ac | | | | | | | | | | | | | | | |
FIGURE 1.2 Carbon, hydrogen, and other elements commonly found in organic compounds are shown in the colors typically used to represent them.
Not all carbon compounds are derived from living organisms, however. Modern chemists have developed a remarkably sophisticated ability to design and synthesize new organic compounds in the laboratory-medicines, dyes, polymers, and a host of other substances. Organic chemistry touches the lives of
everyone. Its study can be a fascinating undertaking. |
4 | CHAPTER CONTENTS - 1.1 Atomic Structure: The Nucleus
As you might remember from your general chemistry course, an atom consists of a dense, positively charged nucleus surrounded at a relatively large distance by negatively charged electrons (FIGURE 1.3). The nucleus consists of subatomic particles called neutrons, which are electrically neutral, and protons, which are positively charged. Because an atom is neutral overall, the number of positive protons in the nucleus and the number of negative electrons surrounding the nucleus are the same.
FIGURE 1.3 A schematic view of an atom. The dense, positively charged nucleus contains most of the atom's mass and is surrounded by negatively charged electrons. The three-dimensional view on the right shows calculated electron-density surfaces. Electron density increases steadily toward the nucleus and is 40 times greater at the blue solid surface than at the gray mesh surface.
Although extremely small-about $10^{-14}$ to $10^{-15}$ meter ( m ) in diameter-the nucleus nevertheless contains essentially all the mass of the atom. Electrons have negligible mass and circulate around the nucleus at a distance of approximately $10^{-10} \mathrm{~m}$. Thus, the diameter of a typical atom is about $2 \times 10^{-10} \mathrm{~m}$, or 200 picometers (pm), where $1 \mathrm{pm}=10^{-12} \mathrm{~m}$. To give you an idea of how small this is, a thin pencil line is about 3 million carbon atoms wide. Although most chemists throughout the world use the International System (SI) of units and describe small distances in picometers, many organic chemists and biochemists in the United States still use the unit angstrom ( $\AA$ ) to express atomic distances, where $1 \AA=100 \mathrm{pm}=10^{-10} \mathrm{~m}$. As you probably did in your general chemistry course, however, we'll stay with SI units in this book. |
5 | CHAPTER CONTENTS - 1.1 Atomic Structure: The Nucleus
A specific atom is described by its atomic number ( $Z$ ), which gives the number of protons (or electrons) it contains, and its mass number ( $\boldsymbol{A}$, which gives the total number of protons and neutrons in its nucleus. All the atoms of a given element have the same atomic number: 1 for hydrogen, 6 for carbon, 15 for phosphorus, and so on; but they can have different mass numbers depending on how many neutrons they contain. Atoms with the same atomic number but different mass numbers are called isotopes. The element carbon, for instance, has three isotopes that occur naturally, with mass numbers of 12,13 , and 14. Carbon- 12 has a natural abundance of $98.89 \%$, carbon-13 has a natural abundance of $1.11 \%$, and carbon-14 has only a negligible natural abundance.
The weighted-average of an element's naturally occurring isotopes is called atomic weight and is given in unified atomic mass units (u) or daltons (Da) where 1 u or 1 Da is defined as one twelfth the mass of one atom of carbon-12. Thus, the atomic weight is 1.008 u for hydrogen, 12.011 u for carbon, 30.974 u for phosphorus, and so on. Atomic weights of all elements are given in the periodic table in Appendix $D$. |
6 | CHAPTER CONTENTS - 1.2 Atomic Structure: Orbitals
How are the electrons distributed in an atom? You might recall from your general chemistry course that, according to the quantum mechanical model, the behavior of a specific electron in an atom can be described by a mathematical expression called a wave equation-the same type of expression used to describe the motion of waves in a fluid. The solution to a wave equation is called a wave function, or orbital, and is denoted by the lowercase Greek letter psi ( $\psi$ ).
When the square of the wave function, $\psi^{2}$, is plotted in three-dimensional space, an orbital describes the volume of space around a nucleus that an electron is most likely to occupy. You might therefore think of an orbital as looking like a photograph of the electron taken at a slow shutter speed. In such a photo, the orbital would appear as a blurry cloud, indicating the region of space where the electron has been. This electron cloud doesn't have a sharp boundary, but for practical purposes we can set the limits by saying that an orbital represents the space where an electron spends $90 \%$ to $95 \%$ of its time.
What do orbitals look like? There are four different kinds of orbitals, denoted $s, p, d$, and $f$, each with a different shape. Of the four, we'll be concerned primarily with $s$ and $p$ orbitals because these are the most common in
organic and biological chemistry. An $s$ orbital has a spherical shape, with the nucleus at its center; a $p$ orbital has a dumbbell shape with two parts, or lobes; and four of the five $d$ orbitals have a cloverleaf shape with four lobes, as shown in FIGURE 1.4. The fifth $d$ orbital is shaped like an elongated dumbbell with a doughnut around its middle.
An $s$ orbital
A p orbital
Ad orbital
FIGURE 1.4 Representations of $\boldsymbol{s}, \boldsymbol{p}$, and $\boldsymbol{d}$ orbitals. An $s$ orbital is spherical, a $p$ orbital is dumbbell-shaped, and four of the five $d$ orbitals are cloverleaf-shaped. Different lobes of $p$ orbitals are often drawn for convenience as teardrops, but their actual shape is more like that of a doorknob, as indicated. |
7 | CHAPTER CONTENTS - 1.2 Atomic Structure: Orbitals
The orbitals in an atom are organized into different layers around the nucleus called electron shells, which are centered around the nucleus and have successively larger size and energy. Different shells contain different numbers and kinds of orbitals, and each orbital within a shell can be occupied by two electrons. The first shell contains only a single $s$ orbital, denoted $1 s$, and thus holds only 2 electrons. The second shell contains one $2 s$ orbital and three $2 p$ orbitals and thus holds a total of 8 electrons. The third shell contains a $3 s$ orbital, three $3 p$ orbitals, and five $3 d$ orbitals, for a total capacity of 18 electrons. These orbital groupings and their energy levels are shown in FIGURE 1.5.
FIGURE 1.5 Energy levels of electrons in an atom. The first shell holds a maximum of 2 electrons in one $1 s$ orbital; the second shell holds a maximum of 8 electrons in one $2 s$ and three $2 p$ orbitals; the third shell holds a maximum of 18 electrons in one $3 s$, three $3 p$, and $3 d$ orbitals; and so on. The two electrons in each orbital are represented by five up and down arrows, $\uparrow \downarrow$. Although not shown, the energy level of the $4 s$ orbital falls between $3 p$ and $3 d$.
The three different $p$ orbitals within a given shell are oriented in space along mutually perpendicular directions, denoted $p_{\mathrm{x}}, p_{\mathrm{y}}$, and $p_{\mathrm{z}}$. As shown in FIGURE 1.6, the two lobes of each $p$ orbital are separated by a region of zero electron density called a node. Furthermore, the two orbital regions separated by the node have different algebraic signs, + and -, in the wave function, as represented by the different colors in FIGURE 1.4 and FIGURE 1.6. As we'll see in Section 1.11, these algebraic signs for different orbital lobes have important consequences with respect to chemical bonding and chemical reactivity.
A $2 p_{\mathrm{x}}$ orbital
A $2 p_{\mathrm{y}}$ orbital |
8 | CHAPTER CONTENTS - 1.2 Atomic Structure: Orbitals
A $2 p_{\mathrm{x}}$ orbital
A $2 p_{\mathrm{y}}$ orbital
A $\mathbf{2} p_{z}$ orbital
FIGURE 1.6 Shapes of the $2 p$ orbitals. Each of the three mutually perpendicular, dumbbell-shaped orbitals has two lobes separated by a node. The two lobes have different algebraic signs in the corresponding wave function, as indicated by the different colors. |
9 | CHAPTER CONTENTS - 1.3 Atomic Structure: Electron Configurations
The lowest-energy arrangement, or ground-state electron configuration, of an atom is a list of the orbitals occupied by its electrons. We can predict this arrangement by following three rules. |
10 | RULE 1 -
The lowest-energy orbitals fill up first, $1 s \rightarrow 2 s \rightarrow 2 p \rightarrow 3 s \rightarrow 3 p \rightarrow 4 s \rightarrow 3 d$, according to the following graphic, a statement called the Aufbau principle. Note that the $4 s$ orbital lies between the $3 p$ and $3 d$ orbitals in energy.
RULE 2
Electrons act in some ways as if they were spinning around an axis, somewhat as the earth spins. This spin can have two orientations, denoted as up ( $\uparrow$ ) and down ( $\downarrow$ ). Only two electrons can occupy an orbital, and they must have opposite spins, a statement called the Pauli exclusion principle.
RULE 3
If two or more empty orbitals of equal energy are available, one electron occupies each with spins parallel until all orbitals are half-full, a statement called Hund's rule.
Some examples of how these rules apply are shown in TABLE 1.1. Hydrogen, for instance, has only one electron, which must occupy the lowest-energy orbital. Thus, hydrogen has a $1 s$ ground-state configuration. Carbon has six electrons and the ground-state configuration $1 s^{2} 2 s^{2} 2 p_{x}{ }^{1} 2 p_{y}{ }^{1}$, and so forth. Note that a superscript is used to represent the number of electrons in a particular orbital.
| TABLE 1.1 <br> Elements | Ground-State Electron Configurations of Some | | | |
| :--- | :--- | :--- | :--- | :--- |
| Element | Atomic number | Configuration | | |
| Hydrogen | 1 | $1 s$ | $\uparrow$ | |
| Carbon | 6 | $2 p$ | $\uparrow \uparrow$ | $\uparrow$ |
PROBLEM What is the ground-state electron configuration of each of the following elements:
1-1 (a) Oxygen (b) Nitrogen (c) Sulfur
PROBLEM How many electrons does each of the following biological trace elements have in its outermost $\mathbf{1 - 2}$ electron shell?
(a) Magnesium
(b) Cobalt
(c) Selenium |
11 | RULE 1 - 1.4 Development of Chemical Bonding Theory
By the mid-1800s, the new science of chemistry was developing rapidly, especially in Europe, and chemists had begun to probe the forces holding compounds together. In 1858, the German chemist August Kekulé and the Scottish chemist Archibald Couper independently proposed that, in all organic compounds, carbon is tetravalent-it always forms four bonds when it joins other elements to form stable compounds. Furthermore, said Kekulé, carbon atoms can bond to one another to form extended chains of linked atoms. In 1865, Kekulé provided another major advance when he suggested that carbon chains can double back on themselves to form rings of atoms.
Although Kekule and Couper were correct in describing the tetravalent nature of carbon, chemistry was still viewed in a two-dimensional way until 1874. In that year, the Dutch chemist Jacobus van't Hoff and French chemist Joseph Le Bel added a third dimension to our ideas about organic compounds when they proposed that the four bonds of carbon are not oriented randomly but have specific spatial directions. Van't Hoff went even further and suggested that the four atoms to which carbon is bonded sit at the corners of a regular tetrahedron, with carbon in the center.
A representation of a tetrahedral carbon atom is shown in FIGURE 1.7. Note the conventions used to show threedimensionality: solid lines represent bonds in the plane of the page, the heavy wedged line represents a bond coming out of the page toward the viewer, and the dashed line represents a bond receding back behind the page, away from the viewer. Get used to them; these representations will be used throughout the text.
FIGURE 1.7 A representation of van't Hoff's tetrahedral carbon atom. The solid lines represent bonds in the plane of the paper, the heavy wedged line represents a bond coming out of the plane of the page toward the viewer, and the dashed line represents a bond going back behind the plane of the page away from the viewer. |
12 | RULE 1 - 1.4 Development of Chemical Bonding Theory
Why, though, do atoms bond together, and how can chemical bonds be described electronically? The why question is relatively easy to answer: atoms bond together because the compound that results is more stable and lower in energy than the separate atoms. Energy-usually as heat-is always released and flows out of the chemical system when a bond forms. Conversely, energy is added to the chemical system when a bond breaks. Making bonds always releases energy, and breaking bonds always absorbs energy. The how question is more difficult. To answer it, we need to know more about the electronic properties of atoms.
We know through observation that eight electrons (an electron octet) in an atom's outermost shell, or valence shell, impart special stability to the noble-gas elements in group 8A of the periodic table: $\mathrm{Ne}(2+8)$; $\mathrm{Ar}(2+8+$ 8 ); $\mathrm{Kr}(2+8+18+8)$. We also know that the chemistry of the main-group elements on the left and right sides of the periodic table is governed by their tendency to take on the electron configuration of the nearest noble gas. The alkali metals such as sodium in group 1A, for example, achieve a noble-gas configuration by losing the single $s$ electron from their valence shell to form a cation, while the halogens such as chlorine in group 7A achieve a noble-gas configuration by gaining a $p$ electron to fill their valence shell and form an anion. The resultant ions are held together in compounds like $\mathrm{Na}^{+} \mathrm{Cl}^{-}$by the electrical attraction of unlike charges that we call an ionic bond. |
13 | RULE 1 - 1.4 Development of Chemical Bonding Theory
But how do elements closer to the middle of the periodic table form bonds? Look at methane, $\mathrm{CH}_{4}$, the main constituent of natural gas, for example. The bonding in methane is not ionic because it would take too much energy for carbon ( $1 s^{2} 2 s^{2} 2 p^{2}$ ) either to gain or lose four electrons to achieve a noble-gas configuration. Instead, carbon bonds to other atoms, not by gaining or losing electrons, but by sharing them. Such a sharedelectron bond, first proposed in 1916 by the American chemist G. N. Lewis, is called a covalent bond. The neutral collection of atoms held together by covalent bonds is called a molecule. Ionic compounds such as sodium chloride, however, are not called molecules.
A simple way of indicating the covalent bonds in molecules is to use what are called Lewis structures, or electron-dot structures, in which the valence-shell electrons of an atom are represented as dots. Thus, hydrogen has one dot representing its $1 s$ electron, carbon has four dots ( $2 s^{2} 2 p^{2}$ ), oxygen has six dots ( $2 s^{2} 2 p^{4}$ ), and so on. A stable molecule results whenever a noble-gas configuration of eight dots (an octet) is achieved for all main-group atoms or two dots for hydrogen. Even simpler than Lewis structures is the use of Kekulé structures, or line-bond structures, in which the two-electron covalent bonds are indicated as lines drawn between atoms. |
14 | RULE 1 - 1.4 Development of Chemical Bonding Theory
| Electron-dot structures (Lewis structures) | <smiles></smiles> | $\begin{gathered} H: \ddot{\mathrm{N}}: \mathrm{H} \\ \ddot{\mathrm{H}} \end{gathered}$ | $H: \ddot{O}: H$ | <smiles></smiles> |
| :---: | :---: | :---: | :---: | :---: |
| Line-bond structures (Kekulé structures) | <smiles>C</smiles> | <smiles>[NH]</smiles> | $\mathrm{H}-\ddot{\mathrm{O}}-\mathrm{H}$ | <smiles>CO</smiles> |
| | Methane $\left(\mathrm{CH}_{4}\right)$ | Ammonia $\left(\mathrm{NH}_{3}\right)$ | Water $\left(\mathrm{H}_{2} \mathrm{O}\right)$ | Methanol $\left(\mathrm{CH}_{3} \mathrm{OH}\right)$ |
The number of covalent bonds an atom forms depends on how many additional valence electrons it needs to reach a noble-gas configuration. Hydrogen has one valence electron (1s) and needs only one more to reach the helium configuration ( $1 s^{2}$ ), so it forms one bond. Carbon has four valence electrons ( $2 s^{2} 2 p^{2}$ ) and needs four more to reach the neon configuration $\left(2 s^{2} 2 p^{6}\right)$, so it forms four bonds. Nitrogen has five valence electrons ( $2 s^{2}$ $2 p^{3}$ ), needs three more, and forms three bonds; oxygen has six valence electrons ( $2 s^{2} 2 p^{4}$ ), needs two more, and forms two bonds; and the halogens have seven valence electrons, need one more, and form one bond. |
15 | RULE 1 - 1.4 Development of Chemical Bonding Theory
Valence electrons that are not used for bonding remain as dots in structures and are called lone-pair electrons, or nonbonding electrons. The nitrogen atom in ammonia, $\mathrm{NH}_{3}$, for instance, shares six valence electrons in three covalent bonds and has its remaining two valence electrons as two dots in a nonbonding lone pair. As a time-saving shorthand, nonbonding electrons are often omitted when drawing line-bond structures, but you still have to keep them in mind since they're often crucial in chemical reactions.
Ammonia |
16 | Predicting the Number of Bonds Formed by Atoms in Molecules -
How many hydrogen atoms does phosphorus bond to in forming phosphine, $\mathrm{PH}_{\text {? }}$ ? |
17 | Strategy -
Identify the periodic group of phosphorus, and find from that how many electrons (bonds) are needed to make an octet. |
18 | Solution -
Phosphorus is in group 5A of the periodic table and has five valence electrons. It thus needs to share three more electrons to make an octet and therefore bonds to three hydrogen atoms, giving $\mathrm{PH}_{3}$. |
19 | Drawing Electron-Dot and Line-Bond Structures -
Draw both electron-dot and line-bond structures for chloromethane, $\mathrm{CH}_{3} \mathrm{Cl}$. |
20 | Strategy -
Remember that a covalent bond-that is, a pair of shared electrons-is represented as a line between atoms. |
21 | Solution -
Hydrogen has one valence electron, carbon has four valence electrons, and chlorine has seven valence electrons. Thus, chloromethane is represented as
Chloromethane
PROBLEM Draw a molecule of chloroform, $\mathrm{CHCl}_{3}$, using solid, wedged, and dashed lines to show its tetrahedral 1-3 geometry.
PROBLEM Convert the following representation of ethane, $\mathrm{C}_{2} \mathrm{H}_{6}$, into a conventional drawing that uses solid, 1-4 wedged, and dashed lines to indicate tetrahedral geometry around each carbon (black = C, gray = H). |
22 | Ethane -
PROBLEM What are likely formulas for the following substances?
1-5 (a) $\mathrm{CCl}_{\text {? }}$
(b) AlH ?
(c) $\mathrm{CH}_{?} \mathrm{Cl}_{2}$
(d) SiF
(e) $\mathrm{CH}_{3} \mathrm{NH}_{?}$
PROBLEM Write line-bond structures for the following substances, showing all nonbonding electrons:
1-6 (a) $\mathrm{CHCl}_{3}$, chloroform (b) $\mathrm{H}_{2} \mathrm{~S}$, hydrogen sulfide (c) $\mathrm{CH}_{3} \mathrm{NH}_{2}$, methylamine
(d) $\mathrm{CH}_{3} \mathrm{Li}$, methyllithium
PROBLEM Why can't an organic molecule have the formula $\mathrm{C}_{2} \mathrm{H}_{7}$ ? |
23 | 1-7 - 1.5 Describing Chemical Bonds: Valence Bond Theory
How does electron sharing lead to bonding between atoms? Two models have been developed to describe covalent bonding: valence bond theory and molecular orbital theory. Each model has its strengths and weaknesses, and chemists tend to use them interchangeably depending on the circumstances. Valence bond theory is the more easily visualized of the two, so most of the descriptions we'll use in this book derive from that approach.
According to valence bond (VB) theory, a covalent bond forms when two atoms approach each other closely and a singly occupied orbital on one atom overlaps a singly occupied orbital on the other atom. The electrons are now paired in the overlapping orbitals and are attracted to the nuclei of both atoms, thus bonding the atoms together. In the $\mathrm{H}_{2}$ molecule, for instance, the $\mathrm{H}-\mathrm{H}$ bond results from the overlap of two singly occupied hydrogen $1 s$ orbitals.
The overlapping orbitals in the $\mathrm{H}_{2}$ molecule have the elongated egg shape we might get by pressing two spheres together. If a plane were to pass through the middle of the bond, the intersection of the plane and the overlapping orbitals would be a circle. In other words, the $\mathrm{H}-\mathrm{H}$ bond is cylindrically symmetrical, as shown in FIGURE 1.8. Such bonds, which are formed by the head-on overlap of two atomic orbitals along a line drawn between the nuclei, are called sigma ( $\boldsymbol{\sigma}$ ) bonds.
FIGURE 1.8 The cylindrical symmetry of the $\mathbf{H - H} \boldsymbol{\sigma}$ bond in an $\mathbf{H}_{\mathbf{2}}$ molecule. The intersection of a plane cutting through the $\boldsymbol{\sigma}$ bond is a circle. |
24 | 1-7 - 1.5 Describing Chemical Bonds: Valence Bond Theory
During the bond-forming reaction $2 \mathrm{H} \cdot \longrightarrow \mathrm{H} 2,436 \mathrm{~kJ} / \mathrm{mol}(104 \mathrm{kcal} / \mathrm{mol})$ of energy is released. Because the product $\mathrm{H}_{2}$ molecule has $436 \mathrm{~kJ} / \mathrm{mol}$ less energy than the starting $2 \mathrm{H} \cdot$ atoms, the product is more stable than the reactant and we say that the $\mathrm{H}-\mathrm{H}$ bond has a bond strength of $436 \mathrm{~kJ} / \mathrm{mol}$. In other words, we would have to put $436 \mathrm{~kJ} / \mathrm{mol}$ of energy into the $\mathrm{H}-\mathrm{H}$ bond to break the $\mathrm{H}_{2}$ molecule apart into two H atoms (FIGURE 1.9). For convenience, we'll generally give energies in both kilocalories (kcal) and the SI unit kilojoules (kJ): $1 \mathrm{~kJ}=0.2390$ kcal; $1 \mathrm{kcal}=4.184 \mathrm{~kJ}$.
FIGURE 1.9 Relative energy levels of two $\mathbf{H}$ atoms and the $\mathbf{H}_{\mathbf{2}}$ molecule. The $\mathrm{H}_{2}$ molecule has $436 \mathrm{~kJ} / \mathrm{mol}(104 \mathrm{kcal} / \mathrm{mol})$ less energy than the two separate H atoms, so $436 \mathrm{~kJ} / \mathrm{mol}$ of energy is released when the $\mathrm{H}-\mathrm{H}$ bond forms. Conversely, $436 \mathrm{~kJ} / \mathrm{mol}$ is absorbed when the $\mathrm{H}-\mathrm{H}$ bond breaks. |
25 | 1-7 - 1.5 Describing Chemical Bonds: Valence Bond Theory
How close are the two nuclei in the $\mathrm{H}_{2}$ molecule? If they are too close, they will repel each other because both are positively charged. Yet if they're too far apart, they won't be able to share the bonding electrons. Thus, there is an optimum distance between nuclei that leads to maximum stability (FIGURE 1.10). Called the bond length, this distance is 74 pm in the $\mathrm{H}-\mathrm{H}$ molecule. Every covalent bond has both a characteristic bond strength and bond length.
FIGURE 1.10 A plot of energy versus internuclear distance for two hydrogen atoms. The distance between nuclei at the minimum energy point is the bond length. |
26 | $1.6 s p^{3}$ Hybrid Orbitals and the Structure of Methane -
The bonding in the hydrogen molecule is fairly straightforward, but the situation is more complicated in organic molecules with tetravalent carbon atoms. Take methane, $\mathrm{CH}_{4}$, for instance. As we've seen, carbon has four valence electrons $\left(2 s^{2} 2 p^{2}\right)$ and forms four bonds. Because carbon uses two kinds of orbitals for bonding, $2 s$ and $2 p$, we might expect methane to have two kinds of $\mathrm{C}-\mathrm{H}$ bonds. In fact, though, all four $\mathrm{C}-\mathrm{H}$ bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron, as shown previously in FIGURE 1.7. How can we explain this?
An answer was provided in 1931 by Linus Pauling, who showed mathematically how an $s$ orbital and three $p$ orbitals on an atom can combine, or hybridize, to form four equivalent atomic orbitals with tetrahedral orientation. Shown in FIGURE 1.11, these tetrahedrally oriented orbitals are called $\boldsymbol{s p}^{\mathbf{3}}$ hybrid orbitals. Note that the superscript 3 in the name $s p^{3}$ tells how many of each type of atomic orbital combine to form the hybrid, not how many electrons occupy it.
FIGURE 1.11 Four $s p^{3}$ hybrid orbitals, oriented toward the corners of a regular tetrahedron, are formed by the combination of an $s$ orbital and three $\boldsymbol{p}$ orbitals (red/blue). The $s p^{3}$ hybrids have two lobes and are unsymmetrical about the nucleus, giving them a directionality and allowing them to form strong bonds to other atoms. |
27 | $1.6 s p^{3}$ Hybrid Orbitals and the Structure of Methane -
The concept of hybridization explains how carbon forms four equivalent tetrahedral bonds but not why it does so. The shape of the hybrid orbital suggests the answer to why. When an $s$ orbital hybridizes with three $p$ orbitals, the resultant $s p^{3}$ hybrid orbitals are unsymmetrical about the nucleus. One of the two lobes is larger than the other and can therefore overlap more effectively with an orbital from another atom to form a bond. As a result,
$s p^{3}$ hybrid orbitals form stronger bonds than do unhybridized $s$ or $p$ orbitals.
The asymmetry of $s p^{3}$ orbitals arises because, as noted previously, the two lobes of a $p$ orbital have different algebraic signs, + and - in the wave function. Thus, when a $p$ orbital hybridizes with an $s$ orbital, the positive $p$ lobe adds to the $s$ orbital but the negative $p$ lobe subtracts from the $s$ orbital. The resultant hybrid orbital is therefore unsymmetrical about the nucleus and is strongly oriented in one direction.
When each of the four identical $s p^{3}$ hybrid orbitals of a carbon atom overlaps with the $1 s$ orbital of a hydrogen atom, four identical $\mathrm{C}-\mathrm{H}$ bonds are formed and methane results. Each $\mathrm{C}-\mathrm{H}$ bond in methane has a strength of $439 \mathrm{~kJ} / \mathrm{mol}(105 \mathrm{kcal} / \mathrm{mol})$ and a length of 109 pm . Because the four bonds have a specific geometry, we also can define a property called the bond angle. The angle formed by each $\mathrm{H}-\mathrm{C}-\mathrm{H}$ is $109.5^{\circ}$, the so-called tetrahedral angle. Methane thus has the structure shown in FIGURE 1.12.
FIGURE 1.12 The structure of methane, showing its $109 . \mathbf{5}^{\circ}$ bond angles. |
28 | $1.7 s p^{3}$ Hybrid Orbitals and the Structure of Ethane -
The same kind of orbital hybridization that accounts for the methane structure also accounts for the bonding together of carbon atoms into chains and rings to make possible many millions of organic compounds. Ethane, $\mathrm{C}_{2} \mathrm{H}_{6}$, is the simplest molecule containing a carbon-carbon bond.
We can picture the ethane molecule by imagining that the two carbon atoms bond to each other by head-on sigma ( $\sigma$ ) overlap of an $s p^{3}$ hybrid orbital from each (FIGURE 1.13). The remaining three $s p^{3}$ hybrid orbitals on each carbon overlap with the 1 s orbitals of three hydrogens to form the six $\mathrm{C}-\mathrm{H}$ bonds. The $\mathrm{C}-\mathrm{H}$ bonds in ethane are similar to those in methane, although a bit weaker: $421 \mathrm{~kJ} / \mathrm{mol}$ ( $101 \mathrm{kcal} / \mathrm{mol}$ ) for ethane versus $439 \mathrm{~kJ} / \mathrm{mol}$ for methane. The C-C bond is 153 pm in length and has a strength of $377 \mathrm{~kJ} / \mathrm{mol}(90 \mathrm{kcal} / \mathrm{mol})$. All the bond angles of ethane are near, although not exactly at, the tetrahedral value of $109.5^{\circ}$.
FIGURE 1.13 The structure of ethane. The carbon-carbon bond is formed by $\sigma$ overlap of two $s p^{3}$ hybrid orbitals. For clarity, the smaller
lobes of the $s p^{3}$ hybrid orbitals are not shown.
PROBLEM Draw a line-bond structure for propane, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$. Predict the value of each bond angle, and 1-8 indicate the overall shape of the molecule.
PROBLEM Convert the following molecular model of hexane, a component of gasoline, into a line-bond 1-9 structure (black $=\mathrm{C}$, gray $=\mathrm{H}$ ).
Hexane |
29 | $1.8 s p^{2}$ Hybrid Orbitals and the Structure of Ethylene -
The bonds we've seen in methane and ethane are called single bonds because they result from the sharing of one electron pair between bonded atoms. It was recognized nearly 150 years ago, however, that carbon atoms can also form double bonds by sharing two electron pairs between atoms or triple bonds by sharing three electron pairs. Ethylene, for instance, has the structure $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$ and contains a carbon-carbon double bond, while acetylene has the structure $\mathrm{HC} \equiv \mathrm{CH}$ and contains a carbon-carbon triple bond.
How are multiple bonds described by valence bond theory? When we discussed $s p^{3}$ hybrid orbitals in Section 1.6, we said that the four valence-shell atomic orbitals of carbon combine to form four equivalent $s p^{3}$ hybrids. Imagine instead that the $2 s$ orbital combines with only $t w o$ of the three available $2 p$ orbitals. Three $\boldsymbol{s p}^{\mathbf{2}}$ hybrid orbitals result, and one $2 p$ orbital remains unchanged. Like $s p^{3}$ hybrids, $s p^{2}$ hybrid orbitals are unsymmetrical about the nucleus and are strongly oriented in a specific direction so they can form strong bonds. The three $s p^{2}$ orbitals lie in a plane at angles of $120^{\circ}$ to one another, with the remaining $p$ orbital perpendicular to the $s p^{2}$ plane, as shown in FIGURE 1.14.
Side view
Top view
FIGURE $1.14 \boldsymbol{s p} \boldsymbol{p}^{2}$ Hybridization. The three equivalent $\boldsymbol{s p ^ { 2 }}$ hybrid orbitals lie in a plane at angles of $120^{\circ}$ to one another, and a single unhybridized $\boldsymbol{p}$ orbital (red/blue) is perpendicular to the $s p^{2}$ plane. |
30 | $1.8 s p^{2}$ Hybrid Orbitals and the Structure of Ethylene -
When two carbons with $s p^{2}$ hybridization approach each other, they form a strong $\sigma$ bond by $s p^{2}-s p^{2}$ head-on overlap. At the same time, the unhybridized $p$ orbitals interact by sideways overlap to form what is called a pi ( $\boldsymbol{\pi}$ ) bond. The combination of an $s p^{2}-s p^{2} \sigma$ bond and a $2 p-2 p \pi$ bond results in the sharing of four electrons and the formation of a carbon-carbon double bond (FIGURE 1.15). Note that the electrons in the $\sigma$ bond occupy the region centered between nuclei, while the electrons in the $\pi$ bond occupy regions above and below a line drawn between nuclei.
To complete the structure of ethylene, four hydrogen atoms form $\sigma$ bonds with the remaining four $s p^{2}$ orbitals. Ethylene thus has a planar structure, with $\mathrm{H}-\mathrm{C}-\mathrm{H}$ and $\mathrm{H}-\mathrm{C}-\mathrm{C}$ bond angles of approximately $120^{\circ}$. (The actual values are $117.4^{\circ}$ for the $\mathrm{H}-\mathrm{C}-\mathrm{H}$ bond angle and $121.3^{\circ}$ for the $\mathrm{H}-\mathrm{C}-\mathrm{C}$ bond angle.) Each $\mathrm{C}-\mathrm{H}$ bond has a length of 108.7 pm and a strength of $464 \mathrm{~kJ} / \mathrm{mol}(111 \mathrm{kcal} / \mathrm{mol})$.
FIGURE 1.15 The structure of ethylene. One part of the double bond in ethylene results from $\sigma$ (head-on) overlap of $s p^{2}$ hybrid orbitals, and the other part results from $\pi$ (sideways) overlap of unhybridized $\boldsymbol{p}$ orbitals (red/blue). The $\pi$ bond has regions of electron density above and below a line drawn between nuclei. |
31 | $1.8 s p^{2}$ Hybrid Orbitals and the Structure of Ethylene -
As you might expect, the carbon-carbon double bond in ethylene is both shorter and stronger than the single bond in ethane because it has four electrons bonding the nuclei together rather than two. Ethylene has a $\mathrm{C}=\mathrm{C}$ bond length of 134 pm and a strength of $728 \mathrm{~kJ} / \mathrm{mol}$ ( $174 \mathrm{kcal} / \mathrm{mol}$ ) versus a C-C length of 153 pm and a strength of $377 \mathrm{~kJ} / \mathrm{mol}$ for ethane. The carbon-carbon double bond is less than twice as strong as a single bond because the sideways overlap in the $\pi$ part of the double bond is not as great as the head-on overlap in the $\sigma$ part. |
32 | Drawing Electron-Dot and Line-Bond Structures -
Commonly used in biology as a tissue preservative, formaldehyde, $\mathrm{CH}_{2} \mathrm{O}$, contains a carbon-oxygen double bond. Draw electron-dot and line-bond structures of formaldehyde, and indicate the hybridization of the carbon orbitals. |
33 | Strategy -
We know that hydrogen forms one covalent bond, carbon forms four, and oxygen forms two. Trial and error, combined with intuition, is needed to fit the atoms together. |
34 | Solution -
There is only one way that two hydrogens, one carbon, and one oxygen can combine:
Like the carbon atoms in ethylene, the carbon atom in formaldehyde is in a double bond and its orbitals are therefore $s p^{2}$-hybridized.
PROBLEM Draw a line-bond structure for propene, $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}$. Indicate the hybridization of the orbitals on 1-10 each carbon, and predict the value of each bond angle.
PROBLEM Draw a line-bond structure for 1,3-butadiene, $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}$. Indicate the hybridization of the 1-11 orbitals on each carbon, and predict the value of each bond angle.
PROBLEM A molecular model of aspirin (acetylsalicylic acid) is shown. Identify the hybridization of the
1-12 orbitals on each carbon atom in aspirin, and tell which atoms have lone pairs of electrons (black = C, red $=0$, gray $=H$ ). |
35 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene -
In addition to forming single and double bonds by sharing two and four electrons, respectively, carbon can also form a triple bond by sharing six electrons. To account for the triple bond in a molecule such as acetylene, $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}$, we need a third kind of hybrid orbital, an $\boldsymbol{s p}$ hybrid. Imagine that, instead of combining with two or three $p$ orbitals, a carbon $2 s$ orbital hybridizes with only a single $p$ orbital. Two sphybrid orbitals result, and two $p$ orbitals remain unchanged. The two $s p$ orbitals are oriented $180^{\circ}$ apart on the right-left ( $x$ ) axis, while the p orbitals are perpendicular on the up-down $(y)$ axis and the in-out $(z)$ axis, as shown in FIGURE 1.16.
One $s p$ hybrid Another $s p$ hybrid
FIGURE 1.16 sp Hybridization. The two $s p$ hybrid orbitals are oriented $180^{\circ}$ away from each other, perpendicular to the two remaining $\boldsymbol{p}$ orbitals (red/blue).
When two $s p$-hybridized carbon atoms approach each other, $s p$ hybrid orbitals on each carbon overlap headon to form a strong $s p-s p \sigma$ bond. At the same time, the $p_{\mathrm{Z}}$ orbitals from each carbon form a $p_{\mathrm{Z}}-p_{\mathrm{Z}} \pi$ bond by sideways overlap, and the $p_{\mathrm{y}}$ orbitals overlap similarly to form a $p_{\mathrm{y}}-p_{\mathrm{y}} \pi$ bond. The net effect is the sharing of six electrons and formation of a carbon-carbon triple bond. Each of the two remaining $s p$ hybrid orbitals forms a $\sigma$ bond with hydrogen to complete the acetylene molecule (FIGURE 1.17). |
36 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene -
FIGURE 1.17 The structure of acetylene. The two carbon atoms are joined by one $s p-s p \sigma$ bond and two $p-p \pi$ bonds.
As suggested by $s p$ hybridization, acetylene is a linear molecule with $\mathrm{H}-\mathrm{C}-\mathrm{C}$ bond angles of $180^{\circ}$. The $\mathrm{C}-\mathrm{H}$ bonds have a length of 106 pm and a strength of $558 \mathrm{~kJ} / \mathrm{mol}(133 \mathrm{kcal} / \mathrm{mol})$. The $\mathrm{C}-\mathrm{C}$ bond length in acetylene is 120 pm , and its strength is about $965 \mathrm{~kJ} / \mathrm{mol}(231 \mathrm{kcal} / \mathrm{mol})$, making it the shortest and strongest of any carbon-carbon bond. A comparison of $s p, s p^{2}$, and $s p^{3}$ hybridization is given in TABLE 1.2. |
37 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene -
| Molecule | Bond | Bond strength | | Bond length (pm) |
| :---: | :---: | :---: | :---: | :---: |
| | | (kJ/mol) | ( $\mathrm{kcal} / \mathrm{mol}$ ) | |
| Methane, $\mathrm{CH}_{4}$ | $\left(s p^{3}\right) \mathrm{C}-\mathrm{H}$ | 439 | 105 | 109 |
| Ethane, $\mathrm{CH}_{3} \mathrm{CH}_{3}$ | $\left(s p^{3}\right) \mathrm{C}-\mathrm{C}\left(s p^{3}\right)$ | 377 | 90 | 153 |
| | $\left(s p^{3}\right) \mathrm{C}-\mathrm{H}$ | 421 | 101 | 109 |
| Ethylene, $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$ | $\left(s p^{2}\right) \mathrm{C}=\mathrm{C}\left(s p^{2}\right)$ | 728 | 174 | 134 |
| | $\left(s p^{2}\right) \mathrm{C}-\mathrm{H}$ | 464 | 111 | 109 |
| Acetylene, $\mathrm{HC} \equiv \mathrm{CH}$ | $(s p) \mathrm{C} \equiv \mathrm{C}(s p)$ | 965 | 231 | 120 |
| | (sp) $\mathrm{C}-\mathrm{H}$ | 558 | 133 | 106 |
PROBLEM Draw a line-bond structure for propyne, $\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}$. Indicate the hybridization of the orbitals on
1-13 each carbon, and predict a value for each bond angle. |
38 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur
The valence-bond concept of orbital hybridization described in the previous four sections is not limited to carbon. Covalent bonds formed by other elements can also be described using hybrid orbitals. Look, for instance, at the nitrogen atom in methylamine $\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)$, an organic derivative of ammonia ( $\mathrm{NH}_{3}$ ) and the substance responsible for the odor of rotting fish.
The experimentally measured $\mathrm{H}-\mathrm{N}-\mathrm{H}$ bond angle in methylamine is $107.1^{\circ}$, and the $\mathrm{C}-\mathrm{N}-\mathrm{H}$ bond angle is
$110.3^{\circ}$, both of which are close to the $109.5^{\circ}$ tetrahedral angle found in methane. We therefore assume that nitrogen forms four $s p^{3}$-hybridized orbitals, just as carbon does. One of the four $s p^{3}$ orbitals is occupied by two nonbonding electrons (a lone pair), and the other three hybrid orbitals have one electron each. Overlap of these three half-filled nitrogen orbitals with half-filled orbitals from other atoms ( C or H ) gives methylamine. Note that the unshared lone pair of electrons in the fourth $s p^{3}$ hybrid orbital of nitrogen occupies as much space as an $\mathrm{N}-\mathrm{H}$ bond does and is very important to the chemistry of methylamine and other nitrogen-containing organic molecules.
Methylamine
Like the carbon atom in methane and the nitrogen atom in methylamine, the oxygen atom in methanol (methyl alcohol) and many other organic molecules can be described as $s p^{3}$-hybridized. The $\mathrm{C}-\mathrm{O}-\mathrm{H}$ bond angle in methanol is $108.5^{\circ}$, very close to the $109.5^{\circ}$ tetrahedral angle. Two of the four $s p^{3}$ hybrid orbitals on oxygen are occupied by nonbonding electron lone pairs, and two are used to form bonds. |
39 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur
In the periodic table, phosphorus and sulfur are the third-row analogs of nitrogen and oxygen, and the bonding in both can be described using hybrid orbitals. Because of their positions in the third row, however, both phosphorus and sulfur can expand their outer-shell octets and form more than the typical number of covalent bonds. Phosphorus, for instance, often forms five covalent bonds, and sulfur often forms four.
Phosphorus is most commonly encountered in biological molecules in compounds called organophosphates, which contain a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to carbon. Methyl phosphate, $\mathrm{CH}_{3} \mathrm{OPO}_{3}{ }^{2-}$, is the simplest example. The $\mathrm{O}-\mathrm{P}-\mathrm{O}$ bond angle in such compounds is typically in the range $110^{\circ}$ to $112^{\circ}$, implying $s p^{3}$ hybridization for phosphorus orbitals.
Methyl phosphate (an organophosphate)
Sulfur is most commonly encountered in biological molecules either in compounds called thiols, which have a sulfur atom bonded to one hydrogen and one carbon, $\mathrm{C}-\mathrm{S}-\mathrm{H}$ or in sulfides, which have a sulfur atom bonded to two carbons, $\mathrm{C}-\mathrm{S}-\mathrm{C}$. Produced by some bacteria, methanethiol $\left(\mathrm{CH}_{3} \mathrm{SH}\right)$ is the simplest example of a thiol, and dimethyl sulfide, $\mathrm{H}_{3} \mathrm{C}-\mathrm{S}-\mathrm{CH}_{3}$, is the simplest example of a sulfide. Both can be described by approximate $s p^{3}$ hybridization around sulfur, although both have significant deviation from the $109.5^{\circ}$ tetrahedral angle.
Methanethiol
Dimethyl sulfide |
40 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur
Methanethiol
Dimethyl sulfide
PROBLEM Identify all nonbonding lone pairs of electrons in the following molecules, and tell what geometry
1-14 you expect for each of the indicated atoms.
(a) The oxygen atom in dimethyl ether, $\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}$ (b) The nitrogen atom in trimethylamine,
(c) The phosphorus atom in phosphine, $\mathrm{PH}_{3}$
(d) The sulfur atom in the amino acid methionine, |
41 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.11 Describing Chemical Bonds: Molecular Orbital Theory
We said in Section 1.5 that chemists use two models for describing covalent bonds: valence bond theory and molecular orbital theory. Having now seen the valence bond approach, which uses hybrid atomic orbitals to account for geometry and assumes the overlap of atomic orbitals to account for electron sharing, let's look briefly at the molecular orbital approach to bonding. We'll return to this topic in Chapters 14, 15, and 30 for a more in-depth discussion.
Molecular orbital (MO) theory describes covalent bond formation as arising from a mathematical combination of atomic orbitals (wave functions) on different atoms to form molecular orbitals, so called because they belong to the entire molecule rather than to an individual atom. Just as an atomic orbital, whether unhybridized or hybridized, describes a region of space around an atom where an electron is likely to be found, so a molecular orbital describes a region of space in a molecule where electrons are most likely to be found.
Like an atomic orbital, a molecular orbital has a specific size, shape, and energy. In the $\mathrm{H}_{2}$ molecule, for example, two singly occupied $1 s$ atomic orbitals combine to form two molecular orbitals. There are two ways for the orbital combination to occur-an additive way and a subtractive way. The additive combination leads to formation of a molecular orbital that is lower in energy and roughly egg-shaped, while the subtractive combination leads to a molecular orbital that is higher in energy and has a node between nuclei (FIGURE 1.18). Note that the additive combination is a single, egg-shaped, molecular orbital; it is not the same as the two overlapping 1 s atomic orbitals of the valence bond description. Similarly, the subtractive combination is a single molecular orbital with the shape of an elongated dumbbell.
FIGURE 1.18 Molecular orbitals of $\mathbf{H}_{\mathbf{2}}$. Combination of two hydrogen $1 s$ atomic orbitals leads to two $\mathrm{H}_{2}$ molecular orbitals. The lowerenergy, bonding MO is filled, and the higher-energy, antibonding MO is unfilled. |
42 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.11 Describing Chemical Bonds: Molecular Orbital Theory
The additive combination is lower in energy than the two hydrogen $1 s$ atomic orbitals and is called a bonding MO because electrons in this MO spend most of their time in the region between the two nuclei, thereby bonding the atoms together. The subtractive combination is higher in energy than the two hydrogen $1 s$ orbitals and is called an antibonding MO because any electrons it contains can't occupy the central region between the nuclei, where there is a node, and thus can't contribute to bonding. The two nuclei therefore repel each other.
Just as bonding and antibonding $\sigma$ molecular orbitals result from the head-on combination of two $s$ atomic orbitals in $\mathrm{H}_{2}$, so bonding and antibonding $\pi$ molecular orbitals result from the sideways combination of two $p$ atomic orbitals in ethylene. As shown in FIGURE 1.19, the lower-energy, $\pi$ bonding MO has no node between nuclei and results from the combination of $p$ orbital lobes with the same algebraic sign. The higher-energy, $\pi$ antibonding MO has a node between nuclei and results from the combination of lobes with opposite algebraic signs. Only the bonding MO is occupied; the higher-energy, antibonding MO is vacant. We'll see in Chapters 14, 15 , and 30 that molecular orbital theory is particularly useful for describing $\pi$ bonds in compounds that have more than one double bond.
FIGURE 1.19 A molecular orbital description of the C-C $\boldsymbol{\pi}$ bond in ethylene. The lower-energy, $\boldsymbol{\pi}$ bonding MO results from an additive combination of $p$ orbital lobes with the same algebraic sign and is filled. The higher-energy, $\boldsymbol{\pi}$ antibonding MO results from a subtractive combination of $p$ orbital lobes with opposite algebraic signs and is unfilled. |
43 | 1.9 sp Hybrid Orbitals and the Structure of Acetylene - 1.12 Drawing Chemical Structures
Let's cover just one more point before ending this introductory chapter. In the structures we've been drawing until now, a line between atoms has represented the two electrons in a covalent bond. Drawing every bond and every atom is tedious, however, so chemists have devised several shorthand ways for writing structures. In condensed structures, carbon-hydrogen and carbon-carbon single bonds aren't shown; instead, they're understood. If a carbon has three hydrogens bonded to it, we write $\mathrm{CH}_{3}$; if a carbon has two hydrogens bonded to it, we write $\mathrm{CH}_{2}$; and so on. The compound called 2-methylbutane, for example, is written as follows:
2-Methylbutane
Note that the horizontal bonds between carbons aren't shown in condensed structures-the $\mathrm{CH}_{3}, \mathrm{CH}_{2}$, and CH units are simply placed next to each other-but vertical carbon-carbon bonds like that of the first of the condensed structures drawn above is shown for clarity. Notice also in the second of the condensed structures that the two $\mathrm{CH}_{3}$ units attached to the CH carbon are grouped together as $\left(\mathrm{CH}_{3}\right)_{2}$.
Even simpler than condensed structures are skeletal structures such as those shown in TABLE 1.3. The rules for drawing skeletal structures are straightforward. |
44 | RULE 1 -
Carbon atoms aren't usually shown. Instead, a carbon atom is assumed to be at each intersection of two lines (bonds) and at the end of each line. Occasionally, a carbon atom might be indicated for emphasis or clarity. |
45 | RULE 2 -
Hydrogen atoms bonded to carbon aren't shown. Because carbon always has a valence of 4, we mentally supply the correct number of hydrogen atoms for each carbon. |
46 | RULE 3 -
Atoms other than carbon and hydrogen are shown.
One further comment: Although such groupings as $-\mathrm{CH}_{3},-\mathrm{OH}$, and $-\mathrm{NH}_{2}$ are usually written with the C , O , or N atom first and the H atom second, the order of writing is sometimes inverted to $\mathrm{H}_{3} \mathrm{C}-, \mathrm{HO}-$, and $\mathrm{H}_{2} \mathrm{~N}$ - if needed to make the bonding connections clearer. Larger units such as $-\mathrm{CH}_{2} \mathrm{CH}_{3}$ are not inverted, though; we don't write $\mathrm{H}_{3} \mathrm{CH}_{2} \mathrm{C}$ - because it would be confusing. There are, however, no well-defined rules that cover all cases; it's largely a matter of preference.
| TABLE 1.3 Line-bond and Skeletal Structures for Some Compounds |
| :--- |
| Compound |
| Isoprene, $\mathrm{C}_{5} \mathrm{H}_{8}$ |
| Line-bond structure |
| Shethylcyclohexane, $\mathrm{C}_{7} \mathrm{H}_{14}$ | |
47 | Interpreting a Line-Bond Structure -
Carvone, a substance responsible for the odor of spearmint, has the following structure. Tell how many hydrogens are bonded to each carbon, and give the molecular formula of carvone.
Carvone |
48 | Strategy -
The end of a line represents a carbon atom with 3 hydrogens, $\mathrm{CH}_{3}$; a two-way intersection is a carbon atom with 2 hydrogens, $\mathrm{CH}_{2}$; a three-way intersection is a carbon atom with 1 hydrogen, CH ; and a four-way intersection is a carbon atom with no attached hydrogens.
Solution
Carvone ( $\mathrm{C}_{\mathbf{1 0}} \mathrm{H}_{\mathbf{1 4}} \mathrm{O}$ )
PROBLEM How many hydrogens are bonded to each carbon in the following compounds, and what is the 1-15 molecular formula of each substance?
(a)
Adrenaline
Estrone (a hormone)
PROBLEM Propose skeletal structures for compounds that satisfy the following molecular formulas: There is
1-16 more than one possibility in each case.
(a) $\mathrm{C}_{5} \mathrm{H}_{12}$
(b) $\mathrm{C}_{2} \mathrm{H}_{7} \mathrm{~N}$
(c) $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$
(d) $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}$
PROBLEM The following molecular model is a representation of para-aminobenzoic acid (PABA), the active
1-17 ingredient in many sunscreens. Indicate the positions of the multiple bonds, and draw a skeletal structure (black $=\mathrm{C}$, red $=\mathrm{O}$, blue $=\mathrm{N}$, gray $=\mathrm{H}$ ).
para-Aminobenzoic acid
(PABA) |
49 | Organic Foods: Risk versus Benefit -
Contrary to what you may hear in supermarkets or on television, all foods are organic-that is, complex mixtures of organic molecules. Even so, when applied to food, the word organic has come to mean an absence of synthetic chemicals, typically pesticides, antibiotics, and preservatives. How concerned should we be about traces of pesticides in the food we eat? Or toxins in the water we drink? Or pollutants in the air we breathe?
Life is not risk-free-we all take many risks each day without even thinking about it. We decide to ride a bike rather than drive, even though there is a ten times greater likelihood per mile of dying in a bicycling accident than in a car. We decide to walk down stairs rather than take an elevator, even though 32,000 people die from falls each year in the United States. Some of us decide to smoke cigarettes, even though it increases our chance of getting cancer by $50 \%$. But what about risks from chemicals like pesticides?
FIGURE 1.20 How dangerous is the pesticide being sprayed on this crop? (credit: "NRCSAR83001(265)" by USDA Natural Resources Conservation Service/Wikimedia Commons, Public Domain)
One thing is certain: without pesticides, whether they target weeds (herbicides), insects (insecticides), or molds and fungi (fungicides), crop production would drop significantly, food prices would increase, and famines would occur in less developed parts of the world. Take the herbicide atrazine, for instance. In the United States alone, approximately 100 million pounds of atrazine are used each year to kill weeds in corn, sorghum, and sugarcane fields, greatly improving the yields of these crops. Nevertheless, the use of atrazine continues to be a concern because traces persist in the environment. Indeed, heavy atrazine exposure can pose health risks to humans and some animals. Because of these risks, the United States Environmental Protection Agency (EPA) has decided not to ban its use because doing so would result in lower crop yields and increased food costs, and because there is no suitable alternative herbicide available. |
50 | Organic Foods: Risk versus Benefit -
Atrazine
How can the potential hazards from a chemical like atrazine be determined? Risk evaluation of chemicals is carried out by exposing test animals, usually mice or rats, to the chemical and then monitoring the animals for signs of harm. To limit the expense and time needed, the amounts administered are typically hundreds or thousands of times greater than those a person might normally encounter. The results obtained in animal tests are then distilled into a single number called an $\mathrm{LD}_{50}$, the amount of substance per kilogram of body weight that
is a lethal dose for $50 \%$ of the test animals. For atrazine, the $\mathrm{LD}_{50}$ value is between 1 and $4 \mathrm{~g} / \mathrm{kg}$ depending on the animal species. Aspirin, for comparison, has an $\mathrm{LD}_{50}$ of $1.1 \mathrm{~g} / \mathrm{kg}$, and ethanol (ethyl alcohol) has an $\mathrm{LD}_{50}$ of $10.6 \mathrm{~g} / \mathrm{kg}$.
TABLE 1.4 lists the $\mathrm{LD}_{50}$ for some other familiar substances. The lower the value, the more toxic the substance. Note, though, that $\mathrm{LD}_{50}$ values only pertain to the effects of heavy exposure for a relatively short time. They say nothing about the risks of long-term exposure, such as whether the substance can cause cancer or interfere with development in the unborn.
| TABLE 1.4 Some LD 50 | | | |
| :--- | :--- | :--- | :---: |
| Substance | $\mathrm{LD}_{50}(\mathrm{~g} / \mathrm{kg})$ | Substance | $\mathrm{LD}_{50}(\mathrm{~g} / \mathrm{kg})$ |
| Strychnine 0.005 Chloroform 1.2 <br> Arsenic trioxide 0.015 Iron(II) sulfate 1.5 <br> DDT 0.115 Ethyl alcohol 10.6 <br> Aspirin 1.1 Sodium cyclamate 17 | | | | |
51 | Organic Foods: Risk versus Benefit -
So, should we still use atrazine? All decisions involve tradeoffs, and the answer is rarely obvious. Does the benefit of increased food production outweigh possible health risks of a pesticide? Do the beneficial effects of a new drug outweigh a potentially dangerous side effect in a small number of users? Different people will have different opinions, but an honest evaluation of facts is surely the best way to start. As of June 2022, atrazine was still approved for continued use in the United States because the EPA believes that the benefits of increased food production outweigh possible health risks. At the same time, atrazine is little used, though not banned, in the European Union. |
52 | Key Terms -
- antibonding MO
- atomic number ( $Z$ )
- Aufbau principle
- bond angle
- bond length
- bond strength
- bonding MO
- condensed structure
- covalent bond
- electron shell
- electron-dot structure
- ground-state electron configuration
- Hund's rule
- ionic bond
- isotope
- Kekulé structure
- Lewis structure
- line-bond structure
- lone-pair electrons
- mass number ( $A$ )
- molecular orbital (MO) theory
- molecule
- node
- nonbonding electron
- orbital
- organic chemistry
- Pauli exclusion principle
- pi ( $\pi$ ) bond
- sigma ( $\sigma$ ) bond
- skeletal structure
- sp hybrid orbital
- $s p^{2}$ hybrid orbital
- $s p^{3}$ hybrid orbital
- valence bond (VB) theory
- valence shell |
53 | Summary -
The purpose of this chapter has been to get you up to speed-to review some ideas about atoms, bonds, and molecular geometry. As we've seen, organic chemistry is the study of carbon compounds. Although a division into organic and inorganic chemistry occurred historically, there is no scientific reason for the division.
An atom consists of a positively charged nucleus surrounded by one or more negatively charged electrons. The electronic structure of an atom can be described by a quantum mechanical wave equation, in which electrons are considered to occupy orbitals around the nucleus. Different orbitals have different energy levels and different shapes. For example, $s$ orbitals are spherical and $p$ orbitals are dumbbell-shaped. The groundstate electron configuration of an atom can be found by assigning electrons to the proper orbitals, beginning with the lowest-energy ones.
A covalent bond is formed when an electron pair is shared between atoms. According to valence bond (VB) theory, electron sharing occurs by the overlap of two atomic orbitals. According to molecular orbital (MO) theory, bonds result from the mathematical combination of atomic orbitals to give molecular orbitals, which belong to the entire molecule. Bonds that have a circular cross-section and are formed by head-on interaction are called sigma ( $\boldsymbol{\sigma}$ ) bonds; bonds formed by sideways interaction of $p$ orbitals are called $\mathbf{p i}$ ( $\boldsymbol{\pi}$ ) bonds.
In the valence bond description, carbon uses hybrid orbitals to form bonds in organic molecules. When forming only single bonds with tetrahedral geometry, carbon uses four equivalent $\boldsymbol{s p}^{\mathbf{3}}$ hybrid orbitals. When forming a double bond with planar geometry, carbon uses three equivalent $\boldsymbol{s p}^{\mathbf{2}}$ hybrid orbitals and one unhybridized $p$ orbital. When forming a triple bond with linear geometry, carbon uses two equivalent $\boldsymbol{s p h}$ hybrid orbitals and two unhybridized $p$ orbitals. Other atoms such as nitrogen, phosphorus, oxygen, and sulfur also use hybrid orbitals to form strong, oriented bonds. |
54 | Summary -
Organic molecules are usually drawn using either condensed structures or skeletal structures. In condensed structures, carbon-carbon and carbon-hydrogen bonds aren't shown. In skeletal structures, only the bonds and not the atoms are shown. A carbon atom is assumed to be at the ends and at the junctions of lines (bonds), and the correct number of hydrogens is supplied mentally. |
55 | WHY YOU SHOULD WORK PROBLEMS -
There's no surer way to learn organic chemistry than by working problems. Although careful reading and rereading of this text are important, reading alone isn't enough. You must also be able to use the information you've read and be able to apply your knowledge in new situations. Working problems gives you practice at doing this.
Each chapter in this book provides many problems of different sorts. The in-chapter problems are placed for immediate reinforcement of ideas just learned, while end-of-chapter problems provide additional practice and come in several forms. They often begin with a short section called "Visualizing Chemistry," which helps you see the microscopic world of molecules and provides practice for working in three dimensions. After the visualizations are many further problems, which are organized by topic. Early problems are primarily of the drill type, providing an opportunity for you to practice your command of the fundamentals. Later problems tend to be more thought-provoking, and some are real challenges.
As you study organic chemistry, take the time to work the problems. Do the ones you can, and ask for help on the ones you can't. If you're stumped by a particular problem, check the accompanying Study Guide and Student Solutions Manual for an explanation that should help clarify the difficulty. Working problems takes effort, but the payoff in knowledge and understanding is immense. |
56 | Visualizing Chemistry -
PROBLEM Convert each of the following molecular models into a skeletal structure, and give the formula of
1-18 each. Only the connections between atoms are shown; multiple bonds are not indicated (black = C, red $=0$, blue $=\mathrm{N}$, gray $=\mathrm{H}$ ) .
(a)
Coniine (the toxic substance in poison hemlock)
(b)
Alanine (an amino acid)
PROBLEM The following model is a representation of citric acid, the key substance in the so-called citric acid
1-19 cycle, by which food molecules are metabolized in the body. Only the connections between atoms are shown; multiple bonds are not indicated. Complete the structure by indicating the positions of multiple bonds and lone-pair electrons (black = C, red = O, gray $=\mathrm{H}$ ).
PROBLEM The following model is a representation of acetaminophen, a pain reliever sold in drugstores
1-20 under a variety of names, including Tylenol. Identify the hybridization of each carbon atom in acetaminophen, and tell which atoms have lone pairs of electrons (black = C, red = O, blue = N, gray $=\mathrm{H})$.
PROBLEM The following model is a representation of aspartame, $\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{5}$, known commercially under
1-21 many names, including NutraSweet. Only the connections between atoms are shown; multiple bonds are not indicated. Complete the structure for aspartame, and indicate the positions of multiple bonds (black $=\mathrm{C}$, red $=\mathrm{O}$, blue $=\mathrm{N}$, gray $=\mathrm{H}$ ). |
57 | Electron Configurations -
PROBLEM How many valence electrons does each of the following dietary trace elements have?
1-22
(a) Zinc (b) Iodine (c) Silicon (d) Iron
PROBLEM Give the ground-state electron configuration for each of the following elements:
1-23
(a) Potassium (b) Arsenic
(c) Aluminum
(d) Germanium |
58 | Electron-Dot and Line-Bond Structures -
PROBLEM What are likely formulas for the following molecules?
1-24 (a) $\mathrm{NH}_{?} \mathrm{OH}$
(b) $\mathrm{AlCl}_{\text {? }}$
(c) $\mathrm{CF}_{2} \mathrm{Cl}_{?}$
(d) $\mathrm{CH}_{?} \mathrm{O}$
PROBLEM Why can't molecules with the following formulas exist?
$\mathbf{1 - 2 5}$ (a) $\mathrm{CH}_{5}$ (b) $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{~N}$ (c) $\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Br}_{2}$
PROBLEM Draw an electron-dot structure for acetonitrile, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{~N}$, which contains a carbon-nitrogen triple
1-26 bond. How many electrons does the nitrogen atom have in its outer shell? How many are bonding, and how many are nonbonding?
PROBLEM Draw a line-bond structure for vinyl chloride, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}$, the starting material from which PVC
1-27 poly(vinyl chloride) plastic is made.
PROBLEM Fill in any nonbonding valence electrons that are missing from the following structures:
1-28
(a)
Dimethyl disulfide
(b)
Acetamide
(c)
Acetate ion
PROBLEM Convert the following line-bond structures into molecular formulas:
1-29
(a)
(b)
(c)
Vitamin C (ascorbic acid)
Nicotine
(d)
Glucose |
59 | Electron-Dot and Line-Bond Structures -
Acetate ion
PROBLEM Convert the following line-bond structures into molecular formulas:
1-29
(a)
(b)
(c)
Vitamin C (ascorbic acid)
Nicotine
(d)
Glucose
PROBLEM Convert the following molecular formulas into line-bond structures that are consistent with valence
1-30 rules:
(a) $\mathrm{C}_{3} \mathrm{H}_{8}$
(b) $\mathrm{CH}_{5} \mathrm{~N}$
(c) $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$ (2 possibilities)
(d) $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{Br}$ (2 possibilities)
(e) $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}$ (3 possibilities) (f) $\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{~N}$ (4 possibilities)
PROBLEM Draw a three-dimensional representation of the oxygen-bearing carbon atom in ethanol,
1-31 $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$, using the standard convention of solid, wedged, and dashed lines. |
60 | Electron-Dot and Line-Bond Structures -
PROBLEM Oxaloacetic acid, an important intermediate in food metabolism, has the formula $\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{5}$ and
1-32 contains three $\mathrm{C}=\mathrm{O}$ bonds and two $\mathrm{O}-\mathrm{H}$ bonds. Propose two possible structures.
PROBLEM Draw structures for the following molecules, showing lone pairs:
1-33 (a) Acrylonitrile, $\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}$, which contains a carbon-carbon double bond and a carbon-nitrogen triple bond
(b) Ethyl methyl ether, $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}$, which contains an oxygen atom bonded to two carbons
(c) Butane, $\mathrm{C}_{4} \mathrm{H}_{10}$, which contains a chain of four carbon atoms
(d) Cyclohexene, $\mathrm{C}_{6} \mathrm{H}_{10}$, which contains a ring of six carbon atoms and one carbon-carbon double bond
Hybridization
PROBLEM What is the hybridization of each carbon atom in acetonitrile (Problem 1-26)?
1-34
PROBLEM What kind of hybridization do you expect for each carbon atom in the following molecules?
$1-35$
(a) Propane, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$
(b) 2-Methylpropene,
(c) But-1-en-3-yne, $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{CH}$
(d) Acetic acid,
PROBLEM What is the shape of benzene, and what hybridization do you expect for each carbon?
1-36
PROBLEM What bond angle do you expect for each of the indicated atoms, and what kind of hybridization do
1-37 you expect for the central atom in each molecule?
(a)
(b)
Pyridine
(c)
Lactic acid
(in sour milk) |
61 | Electron-Dot and Line-Bond Structures -
(b)
Pyridine
(c)
Lactic acid
(in sour milk)
PROBLEM Propose structures for molecules that meet the following descriptions:
1-38 (a) Contains two $s p^{2}$-hybridized carbons and two $s p^{3}$-hybridized carbons
(b) Contains only four carbons, all of which are $s p^{2}$-hybridized
(c) Contains two $s p$-hybridized carbons and two $s p^{2}$-hybridized carbons
PROBLEM What kind of hybridization do you expect for each carbon atom in the following molecules:
1-39
(a)
(b)
PROBLEM Pyridoxal phosphate, a close relative of vitamin $\mathrm{B}_{6}$, is involved in a large number of metabolic
1-40 reactions. What is the hybridization and the bond angle for each nonterminal atom?
Pyridoxal phosphate
Skeletal Structures
PROBLEM Convert the following structures into skeletal drawings:
1-41
(a)
Indole
(b)
(c)
(d)
Benzoquinone
PROBLEM How many hydrogens are bonded to each carbon atom in the following substances, and what is the
1-42 molecular formula of each?
(a)
(b)
(c)
PROBLEM Quetiapine, marketed as Seroquel, is a heavily prescribed antipsychotic drug used in the treatment
1-43 of schizophrenia and bipolar disorder. Convert the following representation into a skeletal structure, and give the molecular formula of quetiapine.
PROBLEM How many hydrogens are bonded to each carbon atom in (a) the antiinfluenza agent oseltamivir, 1-44 marketed as Tamiflu, and (b) the platelet aggregation inhibitor clopidogrel, marketed as Plavix? Give the molecular formula of each.
(a)
Oseltamivir
(Tamiflu)
(b)
Clopidogrel
(Plavix) |
62 | General Problems -
PROBLEM Why do you suppose no one has ever been able to make cyclopentyne as a stable molecule?
1-45
Cyclopentyne
PROBLEM Allene, $\mathrm{H}_{2} \mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}$, has two adjacent double bonds. Draw a picture showing the orbitals involved
1-46 in the $\sigma$ and $\pi$ bonds of allene. Is the central carbon atom $s p^{2}$ - or $s p$-hybridized? What about the hybridization of the terminal carbons? What shape do you predict for allene?
PROBLEM Allene (see Problem 1-46) is structurally related to carbon dioxide, $\mathrm{CO}_{2}$. Draw a picture showing the
1-47 orbitals involved in the $\sigma$ and $\pi$ bonds of $\mathrm{CO}_{2}$, and identify the likely hybridization of carbon.
PROBLEM Complete the electron-dot structure of caffeine, showing all lone-pair electrons, and identify the
1-48 hybridization of the indicated atoms.
PROBLEM Most stable organic species have tetravalent carbon atoms, but species with trivalent carbon atoms
1-49 also exist. Carbocations are one such class of compounds. |
63 | A carbocation -
(a) How many valence electrons does the positively charged carbon atom have?
(b) What hybridization do you expect this carbon atom to have?
(c) What geometry is the carbocation likely to have?
PROBLEM A carbanion is a species that contains a negatively charged, trivalent carbon.
1-50
A carbanion
(a) What is the electronic relationship between a carbanion and a trivalent nitrogen compound such as $\mathrm{NH}_{3}$ ?
(b) How many valence electrons does the negatively charged carbon atom have?
(c) What hybridization do you expect this carbon atom to have?
(d) What geometry is the carbanion likely to have?
PROBLEM Divalent carbon species called carbenes are capable of fleeting existence. For example, methylene,
1-51 $: \mathrm{CH}_{2}$, is the simplest carbene. The two unshared electrons in methylene can be either paired in a single orbital or unpaired in different orbitals. Predict the type of hybridization you expect carbon to adopt in singlet (spin-paired) methylene and triplet (spin-unpaired) methylene. Draw a picture of each, and identify the valence orbitals on carbon.
PROBLEM Two different substances have the formula $\mathrm{C}_{4} \mathrm{H}_{10}$. Draw both, and tell how they differ.
1-52
PROBLEM Two different substances have the formula $\mathrm{C}_{3} \mathrm{H}_{6}$. Draw both, and tell how they differ.
1-53
PROBLEM Two different substances have the formula $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$. Draw both, and tell how they differ.
1-54
PROBLEM Three different substances contain a carbon-carbon double bond and have the formula $\mathrm{C}_{4} \mathrm{H}_{8}$. Draw
1-55 them, and tell how they differ.
PROBLEM Among the most common over-the-counter drugs you might find in a medicine cabinet are mild
1-56 pain relievers such ibuprofen (Advil, Motrin), naproxen (Aleve), and acetaminophen (Tylenol).
Ibuprofen
Naproxen |
64 | A carbocation -
Ibuprofen
Naproxen
Acetaminophen
(a) How many $s p^{3}$-hybridized carbons does each molecule have?
(b) How many $s p^{2}$-hybridized carbons does each molecule have?
(c) What similarities can you see in their structures? |
65 | CHAPTER 2 -
Polar Covalent Bonds; Acids and Bases
FIGURE 2.1 The opium poppy is the source of morphine, one of the first "vegetable alkali," or alkaloids, to be isolated. (credit: modification of work "Papaver somniferum" by Liz West/Flickr, CC BY 2.0) |
66 | CHAPTER CONTENTS -
2.1 Polar Covalent Bonds and Electronegativity
2.2 Polar Covalent Bonds and Dipole Moments
2.3 Formal Charges
2.4 Resonance
2.5 Rules for Resonance Forms
2.6 Drawing Resonance Forms
2.7 Acids and Bases: The Brønsted-Lowry Definition
2.8 Acid and Base Strength
2.9 Predicting Acid-Base Reactions from $\mathrm{p} K_{\mathrm{a}}$ Values
2.10 Organic Acids and Organic Bases
2.11 Acids and Bases: The Lewis Definition
2.12 Noncovalent Interactions between Molecules
WHY THIS CHAPTER? Understanding organic chemistry means knowing not just what happens but also why and how it happens at the molecular level. In this chapter, we'll look at some of the ways that chemists describe and account for chemical reactivity, thereby providing a foundation to understand the specific reactions discussed in subsequent chapters. Topics such as bond polarity, the acid-base behavior of molecules, and hydrogen-bonding are a particularly important part of that foundation.
We saw in the previous chapter how covalent bonds between atoms are described, and we looked at the valence
bond model, which uses hybrid orbitals to account for the observed shapes of organic molecules. Before going on to a systematic study of organic chemistry, however, we still need to review a few fundamental topics. In particular, we need to look more closely at how electrons are distributed in covalent bonds and at some of the consequences that arise when the electrons in a bond are not shared equally between atoms. |
67 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
Up to this point, we've treated chemical bonds as either ionic or covalent. The bond in sodium chloride, for instance, is ionic. Sodium transfers an electron to chlorine to produce $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions, which are held together in the solid by electrostatic attractions between unlike charges. The $\mathrm{C}-\mathrm{C}$ bond in ethane, however, is covalent. The two bonding electrons are shared equally by the two equivalent carbon atoms, resulting in a symmetrical electron distribution in the bond. Most bonds, however, are neither fully ionic nor fully covalent but are somewhere between the two extremes. Such bonds are called polar covalent bonds, meaning that the bonding electrons are attracted more strongly by one atom than the other so that the electron distribution between atoms is not symmetrical (FIGURE 2.2).
FIGURE 2.2 The continuum in bonding from covalent to ionic is a result of an unequal distribution of bonding electrons between atoms. The symbol $\delta$ (lowercase Greek letter delta) means partial charge, either partial positive ( $\delta+$ ) for the electron-poor atom or partial negative ( $\delta-$ ) for the electron-rich atom.
Bond polarity is due to differences in electronegativity (EN), the intrinsic ability of an atom to attract the shared electrons in a covalent bond. As shown in FIGURE 2.3, electronegativities are based on an arbitrary scale, with fluorine the most electronegative ( $\mathrm{EN}=4.0$ ) and cesium the least ( $\mathrm{EN}=0.7$ ). Metals on the left side of the periodic table attract electrons weakly and have lower electronegativities, while oxygen, nitrogen, and halogens on the right side of the periodic table attract electrons strongly and have higher electronegativities. Carbon, the most important element in organic compounds, has an intermediate electronegativity value of 2.5 . |
68 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
| $\begin{gathered} \mathrm{H} \\ 2.1 \end{gathered}$ | | | | | | | | | | | | | | | | | He |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\begin{aligned} & \mathrm{Li} \\ & 1.0 \end{aligned}$ | $\begin{aligned} & \hline \mathrm{Be} \\ & 1.6 \end{aligned}$ | | | | | | | | | | | $\begin{gathered} \text { B } \\ 2.0 \end{gathered}$ | $\begin{gathered} C \\ C \\ 2.5 \end{gathered}$ | $\begin{gathered} \mathrm{N} \\ 3.0 \end{gathered}$ | $\begin{gathered} 0 \\ 3.5 \end{gathered}$ | $\begin{gathered} \text { F } \\ 4.0 \end{gathered}$ | Ne | |
69 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
| $\begin{aligned} & \mathrm{Na} \\ & 0.9 \end{aligned}$ | $\begin{aligned} & \hline \mathrm{Mg} \\ & 1.2 \end{aligned}$ | | | | | | | | | | | $\begin{aligned} & \mathrm{Al} \\ & 1.5 \end{aligned}$ | $\begin{gathered} \mathrm{Si} \\ 1.8 \end{gathered}$ | $\begin{gathered} \hline P \\ 2.1 \end{gathered}$ | $\begin{gathered} \mathrm{S} \\ 2.5 \end{gathered}$ | $\begin{gathered} \hline \mathrm{Cl} \\ 3.0 \end{gathered}$ | Ar | |
70 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
| $\begin{gathered} \mathrm{K} \\ 0.8 \end{gathered}$ | $\begin{aligned} & \hline \mathrm{Ca} \\ & 1.0 \end{aligned}$ | $\begin{aligned} & \mathrm{Sc} \\ & 1.3 \end{aligned}$ | Ti 1.5 | $\begin{gathered} \hline \mathrm{V} \\ 1.6 \end{gathered}$ | $\begin{gathered} \mathrm{Cr} \\ 1.6 \end{gathered}$ | $\begin{aligned} & \mathrm{Mn} \\ & 1.5 \end{aligned}$ | $\begin{aligned} & \hline \mathrm{Fe} \\ & 1.8 \end{aligned}$ | Co 1.9 | $\begin{gathered} \hline \mathrm{Ni} \\ 1.9 \end{gathered}$ | Cu 1.9 | Zn 1.6 | $\begin{aligned} & \hline \mathrm{Ga} \\ & 1.6 \end{aligned}$ | $\begin{aligned} & \hline \mathrm{Ge} \\ & 1.8 \end{aligned}$ | As 2.0 | $\begin{aligned} & \hline \mathrm{Se} \\ & 2.4 \end{aligned}$ | $\begin{aligned} & \hline \mathrm{Br} \\ & 2.8 \end{aligned}$ | Kr | |
71 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
| $\begin{aligned} & \hline \mathrm{Rb} \\ & 0.8 \end{aligned}$ | $\begin{gathered} \hline \mathrm{Sr} \\ 1.0 \end{gathered}$ | Y 1.2 | $\begin{aligned} & \mathrm{Zr} \\ & 1.4 \end{aligned}$ | $\begin{aligned} & \hline \mathrm{Nb} \\ & 1.6 \end{aligned}$ | $\begin{aligned} & \hline \text { Mo } \\ & 1.8 \end{aligned}$ | $\begin{aligned} & \hline \text { Tc } \\ & 1.9 \end{aligned}$ | $\begin{aligned} & \mathrm{Ru} \\ & 2.2 \end{aligned}$ | Rh 2.2 | $\begin{aligned} & \hline \mathrm{Pd} \\ & 2.2 \end{aligned}$ | Ag 1.9 | Cd 1.7 | $\begin{aligned} & \hline \text { In } \\ & 1.7 \end{aligned}$ | Sn | Sb | $\begin{aligned} & \hline \mathrm{Te} \\ & 2.1 \end{aligned}$ | $\begin{gathered} \mathrm{I} \\ 2.5 \end{gathered}$ | Xe | |
72 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
| $\begin{gathered} \hline \mathrm{Cs} \\ 0.7 \end{gathered}$ | $\begin{aligned} & \mathrm{Ba} \\ & 0.9 \end{aligned}$ | La 1.0 | Hf 1.3 | Ta 1.5 | W | Re 1.9 | $\begin{aligned} & \hline \text { Os } \\ & 2.2 \end{aligned}$ | Ir 2.2 | Pt 2.2 | Au 2.4 | Hg 1.9 | $\begin{gathered} \hline \mathrm{Tl} \\ 1.8 \end{gathered}$ | $\mathrm{Pb}$ | $\begin{gathered} \mathrm{Bi} \\ 1.9 \end{gathered}$ | $\begin{aligned} & \hline \text { Po } \\ & 2.0 \end{aligned}$ | $\begin{gathered} \hline \text { At } \\ 2.1 \end{gathered}$ | Rn | |
73 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
FIGURE 2.3 Electronegativity values and trends. Electronegativity generally increases from left to right across the periodic table and decreases from top to bottom. The values are on an arbitrary scale, with $\mathrm{F}=4.0$ and $\mathrm{Cs}=0.7$. Elements in red are the most electronegative, those in yellow are medium, and those in green are the least electronegative.
As a rough guide, bonds between atoms whose electronegativities differ by less than 0.5 are nonpolar covalent, bonds between atoms whose electronegativities differ by 0.5 to 2 are polar covalent, and bonds between atoms whose electronegativities differ by more than 2 are largely ionic. Carbon-hydrogen bonds, for example, are relatively nonpolar because carbon ( $\mathrm{EN}=2.5$ ) and hydrogen ( $\mathrm{EN}=2.1$ ) have similar electronegativities. Bonds between carbon and more electronegative elements such as oxygen ( $\mathrm{EN}=3.5$ ) and nitrogen ( $\mathrm{EN}=3.0$ ), by contrast, are polarized so that the bonding electrons are drawn away from carbon toward the electronegative atom. This leaves carbon with a partial positive charge, $\delta$-, and the electronegative atom with a partial negative charge, $\delta$-. An example is the $\mathrm{C}-\mathrm{O}$ bond in methanol, $\mathrm{CH}_{3} \mathrm{OH}$ (FIGURE 2.4a). Bonds between carbon and less
electronegative elements are polarized so that carbon bears a partial negative charge and the other atom bears a partial positive charge. An example is the $\mathrm{C}-\mathrm{Li}$ bond in methyllithium, $\mathrm{CH}_{3} \mathrm{Li}$ (FIGURE 2.4b).
(a)
(b)
Methyllithium |
74 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
(b)
Methyllithium
FIGURE 2.4 Polar covalent bonds. (a) Methanol, $\mathrm{CH}_{3} \mathrm{OH}$, has a polar covalent $\mathrm{C}-\mathrm{O}$ bond, and (b) methyllithium, $\mathrm{CH}_{3} \mathrm{Li}$, has a polar covalent C-Li bond. The computer-generated representations, called electrostatic potential maps, use color to show calculated charge distributions, ranging from red (electron-rich; $\delta$-) to blue (electron-poor; $\delta+$ ).
Note in the representations of methanol and methyllithium in FIGURE 2.4 that a crossed arrow $\longrightarrow$ is used to indicate the direction of bond polarity. By convention, electrons are displaced in the direction of the arrow. The tail of the arrow (which looks like a plus sign) is electron-poor ( $\delta+$ ), and the head of the arrow is electron-rich ( $\delta$-).
Note also in FIGURE 2.4 that calculated charge distributions in molecules can be displayed visually with what are called electrostatic potential maps, which use color to indicate electron-rich (red; $\delta$-) and electron-poor (blue; $\delta+$ ) regions. In methanol, oxygen carries a partial negative charge and is colored red, while the carbon and hydrogen atoms carry partial positive charges and are colored blue-green. In methyllithium, lithium carries a partial positive charge (blue), while carbon and the hydrogen atoms carry partial negative charges (red). Electrostatic potential maps are useful because they show at a glance the electron-rich and electron-poor atoms in molecules. We'll make frequent use of these maps throughout the text and will see many examples of how electronic structure correlates with chemical reactivity.
When speaking of an atom's ability to polarize a bond, we often use the term inductive effect. An inductive effect is simply the shifting of electrons in a $\sigma$ bond in response to the electronegativity of nearby atoms. Metals, such as lithium and magnesium, inductively donate electrons, whereas reactive nonmetals, such as oxygen and nitrogen, inductively withdraw electrons. Inductive effects play a major role in understanding chemical reactivity, and we'll use them many times throughout this text to explain a variety of chemical observations. |
75 | CHAPTER CONTENTS - 2.1 Polar Covalent Bonds and Electronegativity
PROBLEM Which element in each of the following pairs is more electronegative?
2-1 (a) Li or H
(b) B or Br
(c) Cl or I (d) C or H
PROBLEM Use the $\delta+/ \delta$ - convention to indicate the direction of expected polarity for each of the bonds 2-2 indicated.
(a) $\mathrm{H}_{3} \mathrm{C}-\mathrm{Cl}$
(b) $\mathrm{H}_{3} \mathrm{C}-\mathrm{NH}_{2}$
(c) $\mathrm{H}_{2} \mathrm{~N}-\mathrm{H}$
(d) $\mathrm{H}_{3} \mathrm{C}-\mathrm{SH}$
(e) $\mathrm{H}_{3} \mathrm{C}-\mathrm{MgBr}$ (f) $\quad \mathrm{H}_{3} \mathrm{C}-\mathrm{F}$
PROBLEM Use the electronegativity values shown in Figure 2.3 to rank the following bonds from least polar to 2-3 most polar: $\mathrm{H}_{3} \mathrm{C}-\mathrm{Li}, \mathrm{H}_{3} \mathrm{C}-\mathrm{K}, \mathrm{H}_{3} \mathrm{C}-\mathrm{F}, \mathrm{H}_{3} \mathrm{C}-\mathrm{MgBr}, \mathrm{H}_{3} \mathrm{C}-\mathrm{OH}$
PROBLEM Look at the following electrostatic potential map of methylamine, a substance responsible for the $\mathbf{2 - 4}$ odor of rotting fish, and tell the direction of polarization of the $\mathrm{C}-\mathrm{N}$ bond:
Methylamine |
76 | CHAPTER CONTENTS - 2.2 Polar Covalent Bonds and Dipole Moments
Just as individual bonds are often polar, molecules as a whole are often polar as well. Molecular polarity results from the vector summation of all individual bond polarities and lone-pair contributions in the molecule. As a practical matter, strongly polar substances are often soluble in polar solvents like water, whereas less polar substances are insoluble in water.
Net polarity is measured by a quantity called the dipole moment and can be thought of in the following way: assume that there is a center of mass of all positive charges (nuclei) in a molecule and a center of mass of all negative charges (electrons). If these two centers don't coincide, then the molecule has a net polarity.
The dipole moment, $\mu$ (lowercase Greek letter mu), is defined as the magnitude of the charge $Q$ at either end of the molecular dipole times the distance $r$ between the charges, $\mu=Q \times r$. Dipole moments are expressed in debyes $(\mathrm{D})$, where $1 \mathrm{D}=3.336 \times 10^{-30}$ coulomb meters ( $\mathrm{C} \cdot \mathrm{m}$ ) in SI units. For example, the unit charge on an electron is $1.60 \times 10^{-19} \mathrm{C}$. Thus, if one positive charge and one negative charge are separated by 100 pm (a bit less than the length of a typical covalent bond), the dipole moment is $1.60 \times 10^{-29} \mathrm{C} \cdot \mathrm{m}$, or 4.80 D .
$$
\begin{aligned}
& \mu=Q \times r \\
& \mu=\left(1.60 \times 10^{-19} \mathrm{C}\right)\left(100 \times 10^{-12} \mathrm{~m}\right)\left(\frac{1 \mathrm{D}}{3.336 \times 10^{-30} \mathrm{C} \cdot \mathrm{~m}}\right)=4.80 \mathrm{D}
\end{aligned}
$$ |
77 | CHAPTER CONTENTS - 2.2 Polar Covalent Bonds and Dipole Moments
Dipole moments for some common substances are given in TABLE 2.1. Of the compounds shown in the table, sodium chloride has the largest dipole moment $(9.00 \mathrm{D})$ because it is ionic. Even small molecules like water ( $\mu=1.85 \mathrm{D}$ ), methanol ( $\mathrm{CH}_{3} \mathrm{OH} ; \mu=1.70 \mathrm{D}$ ), and ammonia ( $\mu=1.47 \mathrm{D}$ ), have substantial dipole moments, however, both because they contain strongly electronegative atoms (oxygen and nitrogen) and because all three molecules have lone-pair electrons. The lone-pair electrons on oxygen and nitrogen stick out into space away from the positively charged nuclei, giving rise to a considerable charge separation and making a large contribution to the dipole moment.
TABLE 2.1 Dipole Moments of Some Compounds
| Compound | Dipole moment (D) | | Compound |
| :--- | :--- | :--- | :--- |
| NaCl | 9.00 | $\mathrm{NH}_{3}$ | 1.47 |
| $\mathrm{CH}_{2} \mathrm{O}$ | 2.33 | $\mathrm{CH}_{3} \mathrm{NH}_{2}$ | 1.31 |
| $\mathrm{CH}_{3} \mathrm{Cl}$ | 1.87 | $\mathrm{CO}_{2}$ | 0 |
| $\mathrm{H}_{2} \mathrm{O}$ | 1.85 | $\mathrm{CH}_{4}$ | 0 |
| $\mathrm{CH}_{3} \mathrm{OH}$ | 1.70 | $\mathrm{CH}_{3} \mathrm{CH}_{3}$ | 0 |
TABLE 2.1 Dipole Moments of Some Compounds |
78 | CHAPTER CONTENTS - 2.2 Polar Covalent Bonds and Dipole Moments
TABLE 2.1 Dipole Moments of Some Compounds
| Compound | Dipole moment (D) | Compound | Dipole moment (D) |
| :---: | :---: | :---: | :---: |
| $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$ | 1.70 | <smiles>c1ccccc1</smiles> <br> Benzene | 0 |
| $\mathrm{CH}_{3} \mathrm{SH}$ | 1.52 | | |
| <smiles>C1=C[C+]2CC(O2)[Te]1</smiles><smiles>H1([H])C@H[C@H]2C=C[C@H]1C2</smiles><smiles>C[I-]NH+[I-]</smiles>Water Methanol Ammonia <br> $(\mu=1.85 \mathrm{D})$ $(\mu=1.70 \mathrm{D})$ $(\mu=1.47 \mathrm{D})$ | | | |
| | | | |
In contrast with water, methanol, and ammonia, molecules such as carbon dioxide, methane, ethane, and benzene have zero dipole moments. Because of the symmetrical structures of these molecules, the individual bond polarities and lone-pair contributions exactly cancel.
$$
\mathrm{O}=\mathrm{C}=\mathrm{O}
$$
Carbon dioxide ( $\mu=0$ )
Methane
( $\mu=0$ )
Ethane
( $\mu=0$ )
Benzene ( $\mu=0$ ) |
79 | Predicting the Direction of a Dipole Moment -
Make a three-dimensional drawing of methylamine, $\mathrm{CH}_{3} \mathrm{NH}_{2}$, and show the direction of its dipole moment $(\mu=$ 1.31). |
80 | Strategy -
Look for any lone-pair electrons, and identify any atom with an electronegativity substantially different from that of carbon. (Usually, this means O, N, F, Cl, or Br.) Electron density will be displaced in the general direction of the electronegative atoms and the lone pairs. |
81 | Solution -
Methylamine contains an electronegative nitrogen atom with a lone pair of electrons. The dipole moment thus points generally from $-\mathrm{CH}_{3}$ toward the lone pair.
Methylamine
( $\mu=1.31$ )
PROBLEM Ethylene glycol, $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}$, may look nonpolar when drawn, but an internal hydrogen bond 2-5 between the two - OH groups results in a dipole moment. Explain.
PROBLEM Make three-dimensional drawings of the following molecules, and predict whether each has a 2-6 dipole moment. If you expect a dipole moment, show its direction.
(a) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$
(b) $\mathrm{CHCl}_{3}$
(c) $\mathrm{CH}_{2} \mathrm{Cl}_{2}$
(d) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CCl}_{2}$ |
82 | Solution - 2.3 Formal Charges
Closely related to the ideas of bond polarity and dipole moment is the assignment of formal charges to specific atoms within a molecule, particularly atoms that have an apparently "abnormal" number of bonds. Look at dimethyl sulfoxide $\left(\mathrm{CH}_{3} \mathrm{SOCH}_{3}\right)$, for instance, a solvent commonly used for preserving biological cell lines at low temperature. The sulfur atom in dimethyl sulfoxide has three bonds rather than the usual two and has a formal positive charge. The oxygen atom, by contrast, has one bond rather than the usual two and has a formal negative charge. Note that an electrostatic potential map of dimethyl sulfoxide shows the oxygen as negative (red) and the sulfur as relatively positive (blue), in accordance with the formal charges. |
83 | Dimethyl sulfoxide -
Formal charges, as the name suggests, are a formalism and don't imply the presence of actual ionic charges in a molecule. Instead, they're a device for electron "bookkeeping" and can be thought of in the following way: A typical covalent bond is formed when each atom donates one electron. Although the bonding electrons are shared by both atoms, each atom can still be considered to "own" one electron for bookkeeping purposes. In methane, for instance, the carbon atom owns one electron in each of the four $\mathrm{C}-\mathrm{H}$ bonds. Because a neutral, isolated carbon atom has four valence electrons, and because the carbon atom in methane still owns four, the methane carbon atom is neutral and has no formal charge.
The same is true for the nitrogen atom in ammonia, which has three covalent $\mathrm{N}-\mathrm{H}$ bonds and two nonbonding electrons (a lone pair). Atomic nitrogen has five valence electrons, and the ammonia nitrogen also has five-one in each of three shared $\mathrm{N}-\mathrm{H}$ bonds plus two in the lone pair. Thus, the nitrogen atom in ammonia has no formal charge.
The situation is different in dimethyl sulfoxide. Atomic sulfur has six valence electrons, but the dimethyl sulfoxide sulfur owns only five-one in each of the two $\mathrm{S}-\mathrm{C}$ single bonds, one in the $\mathrm{S}-\mathrm{O}$ single bond, and two in a lone pair. Thus, the sulfur atom has formally lost an electron and therefore has a positive formal charge. A similar calculation for the oxygen atom shows that it has formally gained an electron and has a negative charge. Atomic oxygen has six valence electrons, but the oxygen in dimethyl sulfoxide has seven-one in the $O-S$ bond and two in each of three lone pairs. Thus, the oxygen has formally gained an electron and has a negative formal charge.
To express the calculations in a general way, the formal charge on an atom is equal to the number of valence electrons in a neutral, isolated atom minus the number of electrons owned by that bonded atom in a molecule. The number of electrons in the bonded atom, in turn, is equal to half the number of bonding electrons plus the nonbonding, lone-pair electrons. |
84 | Dimethyl sulfoxide -
$$
\begin{aligned}
\text { Formal charge } & =\left(\begin{array}{c}
\text { Number of } \\
\text { valence electrons } \\
\text { in free atom }
\end{array}\right)-\left(\begin{array}{c}
\text { Number of } \\
\text { valence electrons } \\
\text { in bonded atom }
\end{array}\right) \\
& =\left(\begin{array}{c}
\text { Number of } \\
\text { valence electrons } \\
\text { in free atom }
\end{array}\right)-\left(\begin{array}{c}
\text { Number of } \\
\text { bonding electrons } \\
2
\end{array} \begin{array}{c}
\text { Number of } \\
\text { nonbonding } \\
\text { electrons }
\end{array}\right)
\end{aligned}
$$
A summary of commonly encountered formal charges and the bonding situations in which they occur is given in TABLE 2.2. Although only a bookkeeping device, formal charges often give clues about chemical reactivity, so it's helpful to be able to identify and calculate them correctly.
TABLE 2.2 A Summary of Common Formal Charges |
85 | Dimethyl sulfoxide -
TABLE 2.2 A Summary of Common Formal Charges
| Atom | C | | | N | | 0 | | S | | P |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Structure | <smiles>CC(C)(C)C</smiles> | <smiles>CC(C)(C)C</smiles> | <smiles>CC(C)(C)C</smiles> | | <smiles>CN+C</smiles> | <smiles></smiles> | - $\ddot{0}$ | $-\ddot{S}{ }^{+}$ | - | |
| Valence electrons | 4 | 4 | 4 | 5 | 5 | 6 | 6 | 6 | 6 | 5 |
| Number of bonds | 3 | 3 | 3 | 4 | 2 | 3 | 1 | 3 | 1 | 4 |
| Number of nonbonding electrons | 1 | 0 | 2 | 0 | 4 | 2 | 6 | 2 | 6 | 0 |
| Formal charge | 0 | +1 | -1 | +1 | -1 | +1 | -1 | +1 | -1 | +1 |
PROBLEM Calculate formal charges for the nonhydrogen atoms in the following molecules:
2-7 (a)
Diazomethane, $\mathrm{H}_{2} \mathrm{C}=\mathrm{N}=\ddot{\mathrm{N}}$ :
(b) Acetonitrile oxide, $\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{N}-\ddot{\mathrm{O}}$ :
(c) Methyl isocyanide, $\mathrm{H}_{3} \mathrm{C}-\mathrm{N} \equiv \mathrm{C}$ :
PROBLEM Organic phosphate groups occur commonly in biological molecules. Calculate formal charges on
the four O atoms in the methyl phosphate dianion. |
86 | Dimethyl sulfoxide - 2.4 Resonance
Most substances can be represented unambiguously by the Kekulé line-bond structures we've been using up to this point, but an interesting problem sometimes arises. Look at the acetate ion, for instance. When we draw a line-bond structure for acetate, we need to show a double bond to one oxygen and a single bond to the other. But which oxygen is which? Should we draw a double bond to the "top" oxygen and a single bond to the "bottom" oxygen, or vice versa?
Although the two oxygen atoms in the acetate ion appear different in line-bond structures, experiments show that they are equivalent. Both carbon-oxygen bonds, for instance, are 127 pm in length, midway between the length of a typical $\mathrm{C}-\mathrm{O}$ single bond ( 135 pm ) and a typical $\mathrm{C}=\mathrm{O}$ double bond ( 120 pm ). In other words, neither of the two structures for acetate is correct by itself. The true structure is intermediate between the two, and an electrostatic potential map shows that both oxygen atoms share the negative charge and have equal electron densities (red).
The two individual line-bond structures for acetate ion are called resonance forms, and their special resonance relationship is indicated by the double-headed arrow between them. The only difference between the two resonance forms is the placement of the $\pi$ and nonbonding valence electrons. The atoms themselves occupy exactly the same place in both resonance forms, the connections between atoms are the same, and the threedimensional shapes of the resonance forms are the same.
A good way to think about resonance forms is to realize that a substance like the acetate ion is the same as any other. Acetate doesn't jump back and forth between two resonance forms, spending part of the time looking like one and part of the time looking like the other. Rather, acetate has a single unchanging structure that we say is a resonance hybrid of the two individual forms and has characteristics of both. The only "problem" with acetate is that we can't draw it accurately using a familiar line-bond structure-line-bond structures just don't work for resonance hybrids. The difficulty, however, is with the representation of acetate on paper, not with acetate itself. |
87 | Dimethyl sulfoxide - 2.4 Resonance
Resonance is a very useful concept that we'll return to on numerous occasions throughout the rest of this book. We'll see in Chapter 15, for instance, that the six carbon-carbon bonds in aromatic compounds, such as benzene, are equivalent and that benzene is best represented as a hybrid of two resonance forms. Although each individual resonance form seems to imply that benzene has alternating single and double bonds, neither form is correct by itself. The true benzene structure is a hybrid of the two individual forms, and all six carbon-carbon bonds are equivalent. This symmetrical distribution of electrons around the molecule is evident in an
electrostatic potential map.
Benzene (two resonance forms) |
88 | Dimethyl sulfoxide - 2.5 Rules for Resonance Forms
When first dealing with resonance forms, it's useful to have a set of guidelines that describe how to draw and interpret them. The following rules should be helpful: |
89 | RULE 1 -
Individual resonance forms are imaginary, not real. The real structure is a composite, or resonance hybrid, of the different forms. Species such as the acetate ion and benzene are no different from any other. They have single, unchanging structures, and they don't switch back and forth between resonance forms. The only difference between these and other substances is in the way they are represented in drawings. |
90 | RULE 2 -
Resonance forms differ only in the placement of their $\boldsymbol{\pi}$ or nonbonding electrons. Neither the position nor the hybridization of any atom changes from one resonance form to another. In the acetate ion, for instance, the carbon atom is $s p^{2}$-hybridized and the oxygen atoms remain in exactly the same place in both resonance forms. Only the positions of the $\pi$ electrons in the $\mathrm{C}=\mathrm{O}$ bond and the lone-pair electrons on oxygen differ from one form to another. This movement of electrons from one resonance structure to another can be indicated with curved arrows. A curved arrow always indicates the movement of electrons, not the movement of atoms. An arrow shows that a pair of electrons moves from the atom or bond at the tail of the arrow to the atom or bond at the head of the arrow.
The situation with benzene is similar to that with acetate. The $\pi$ electrons in the double bonds move, as shown with curved arrows, but the carbon and hydrogen atoms remain in place.
RULE 3
Different resonance forms of a substance don't have to be equivalent. As an example, we'll see in Chapter 22 that a compound such as acetone, which contains a $\mathrm{C}=\mathrm{O}$ bond, can be converted into its anion by reaction with a strong base. The resultant anion has two resonance forms. One form contains a carbon-oxygen double bond and has a negative charge on carbon; the other contains a carbon-carbon double bond and has a negative charge on oxygen. Even though the two resonance forms aren't equivalent, both contribute to the overall
resonance hybrid.
When two resonance forms are nonequivalent, the actual structure of the resonance hybrid resembles the more stable form. Thus, we might expect the true structure of the acetone anion to be more like that of the form that places the negative charge on the electronegative oxygen atom rather than on the carbon.
RULE 4
Resonance forms obey normal rules of valency. A resonance form is like any other structure: the octet rule still applies to second-row, main-group atoms. For example, one of the following structures for the acetate ion is not a valid resonance form because the carbon atom has five bonds and ten valence electrons: |
91 | RULE 2 -
RULE 5
The resonance hybrid is more stable than any individual resonance form. In other words, resonance leads to stability. Generally speaking, the larger the number of resonance forms a substance has, the more stable the substance is, because its electrons are spread out over a larger part of the molecule and are closer to more nuclei. We'll see in Chapter 15, for instance, that a benzene ring is more stable because of resonance than might otherwise be expected. |
92 | RULE 2 - 2.6 Drawing Resonance Forms
Look back at the resonance forms of the acetate ion and the acetone anion shown in the previous section. The pattern seen there is a common one that leads to a useful technique for drawing resonance forms. In general, any three-atom grouping with a $p$ orbital on each atom has two resonance forms:
The atoms $\mathrm{X}, \mathrm{Y}$, and Z in the general structure might be $\mathrm{C}, \mathrm{N}, \mathrm{O}, \mathrm{P}, \mathrm{S}$, or others, and the asterisk (*) might mean that the $p$ orbital on atom $Z$ is vacant, that it contains a single electron, or that it contains a lone pair of electrons. The two resonance forms differ simply by an exchange in position of the multiple bond and the asterisk from one end of the three-atom grouping to the other.
By learning to recognize such three-atom groupings within larger structures, resonance forms can be systematically generated. Look, for instance, at the anion produced when $\mathrm{H}^{+}$is removed from 2,4-pentanedione by reaction with a base. How many resonance structures does the resultant anion have?
The 2,4-pentanedione anion has a lone pair of electrons and a formal negative charge on the central carbon atom, next to a $\mathrm{C}=\mathrm{O}$ bond on the left. The $\mathrm{O}=\mathrm{C}-\mathrm{C}:^{-}$grouping is a typical one for which two resonance structures can be drawn.
Just as there is a $\mathrm{C}=\mathrm{O}$ bond to the left of the lone pair, there is a second $\mathrm{C}=\mathrm{O}$ bond to the right. Thus, we can draw a total of three resonance structures for the 2,4-pentanedione anion. |
93 | Drawing Resonance Forms for an Anion -
Draw three resonance structures for the carbonate ion, $\mathrm{CO}_{3}{ }^{2-}$. |
94 | Strategy -
Look for three-atom groupings that contain a multiple bond next to an atom with a $p$ orbital. Then exchange the positions of the multiple bond and the electrons in the $p$ orbital. In the carbonate ion, each singly bonded oxygen atom with three lone pairs and a negative charge is adjacent to the $\mathrm{C}=\mathrm{O}$ double bond, giving the grouping Ö: $=C-$ Ö:- |
95 | Solution -
Exchanging the position of the double bond and an electron lone pair in each grouping generates three resonance structures. |
96 | Drawing Resonance Forms for a Radical -
Draw three resonance forms for the pentadienyl radical, where a radical is a substance that contains a single, unpaired electron in one of its orbitals, denoted by a dot $(\cdot)$. |
97 | Strategy -
Find the three-atom groupings that contain a multiple bond next to an atom with a $p$ orbital. |
98 | Solution -
The unpaired electron is on a carbon atom next to a $\mathrm{C}=\mathrm{C}$ bond, giving a typical three-atom grouping that has two resonance forms.
In the second resonance form, the unpaired electron is next to another double bond, giving another three-atom grouping and leading to another resonance form.
Three-atom grouping
Thus, the three resonance forms for the pentadienyl radical are:
PROBLEM Which of the following pairs of structures represent resonance forms, and which do not? Explain. 2-9
(a)
(b)
and
PROBLEM Draw the indicated number of resonance forms for each of the following substances:
2-10 (a) The methyl phosphate anion, $\mathrm{CH}_{3} \mathrm{OPO}_{3}{ }^{2-}$ (3)
(b) The nitrate anion, $\mathrm{NO}_{3}{ }^{-}$(3)
(c) The allyl cation, $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}{ }^{+}$(2)
(d) The benzoate anion (2) |
99 | Solution - 2.7 Acids and Bases: The Brønsted-Lowry Definition
Perhaps the most important of all concepts related to electronegativity and polarity is that of acidity and
basicity. We'll soon see, in fact, that the acid-base behavior of organic molecules explains much of their chemistry. You may recall from a course in general chemistry that two definitions of acidity are frequently used: the Brønsted-Lowry definition and the Lewis definition. We'll look at the Brønsted-Lowry definition in this and the following three sections and then discuss the Lewis definition in Section 2.11.
A Brønsted-Lowry acid is a substance that donates a hydrogen ion, $\mathrm{H}^{+}$, and a Brønsted-Lowry base is a substance that accepts a hydrogen ion. (The name proton is often used as a synonym for $\mathrm{H}^{+}$because loss of the valence electron from a neutral hydrogen atom leaves only the hydrogen nucleus-a proton.) When gaseous hydrogen chloride dissolves in water, for example, a polar HCl molecule acts as an acid and donates a proton, while a water molecule acts as a base and accepts the proton, yielding chloride anion ( $\mathrm{Cl}^{-}$) and hydronium cation $\left(\mathrm{H}_{3} \mathrm{O}^{+}\right)$. This and other acid-base reactions are reversible, so we'll write them with double, forward-andbackward arrows.
Acid
Base
Conjugate base
Conjugate acid
Chloride ion, the product that results when the acid HCl loses a proton, is called the conjugate base of the acid, and hydronium ion, the product that results when the base $\mathrm{H}_{2} \mathrm{O}$ gains a proton, is called the conjugate acid of the base. Other common mineral acids such as $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{HNO}_{3}$ behave similarly, as do organic acids such as acetic acid, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$.
In a general sense, |
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