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. RULE 1 The lowest-energy orbitals fill up first, 1 s → 2 s → 2 p → 3 s → 3 p → 4 s → 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 ( ↑ ) and down ( ↓ ). 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 . 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. Ground-State Electron Configurations of Some Elements Element Atomic number Configuration Hydrogen 1 Carbon 6 Phosphorus 15 Problems

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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 Kekulé 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 . Note the conventions used to show three-dimensionality: 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. 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. 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: Ne (2 + 8 ); Ar (2 + 8 + 8 ); 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 Na + Cl – by the electrical attraction of unlike charges that we call an ionic bond . But how do elements closer to the middle of the periodic table form bonds? Look at methane, 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 shared-electron 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. 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 (1 s ) 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 (2 s 2 2 p 6 ), 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. 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, 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. Predicting the Number of Bonds Formed by Atoms in Molecules How many hydrogen atoms does phosphorus bond to in forming phosphine, PH ? ? Strategy Identify the periodic group of phosphorus, and find from that how many electrons (bonds) are needed to make an octet. 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 PH 3 . Drawing Electron-Dot and Line-Bond Structures Draw both electron-dot and line-bond structures for chloromethane, CH 3 Cl. Strategy Remember that a covalent bond—that is, a pair of shared electrons—is represented as a line between atoms. Solution Hydrogen has one valence electron, carbon has four valence electrons, and chlorine has seven valence electrons. Thus, chloromethane is represented as Problems

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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 H 2 molecule, for instance, the H–H bond results from the overlap of two singly occupied hydrogen 1 s orbitals. The overlapping orbitals in the 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 H–H bond is cylindrically symmetrical, as shown in . Such bonds, which are formed by the head-on overlap of two atomic orbitals along a line drawn between the nuclei, are called sigma ( σ ) bonds . The cylindrical symmetry of the H–H σ bond in an H 2 molecule. The intersection of a plane cutting through the σ bond is a circle. During the bond-forming reaction 2H · ⟶ H2, 436 kJ/mol (104 kcal/mol) of energy is released. Because the product H 2 molecule has 436 kJ/mol less energy than the starting 2 H · atoms, the product is more stable than the reactant and we say that the H–H bond has a bond strength of 436 kJ/mol. In other words, we would have to put 436 kJ/mol of energy into the H–H bond to break the H 2 molecule apart into two H atoms ( ). For convenience, we’ll generally give energies in both kilocalories (kcal) and the SI unit kilojoules (kJ): 1 kJ = 0.2390 kcal; 1 kcal = 4.184 kJ. Relative energy levels of two H atoms and the H 2 molecule. The H 2 molecule has 436 kJ/mol (104 kcal/mol) less energy than the two separate H atoms, so 436 kJ/mol of energy is released when the H–H bond forms . Conversely, 436 kJ/mol is absorbed when the H–H bond breaks . How close are the two nuclei in the 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 ( ). Called the bond length , this distance is 74 pm in the H–H molecule. Every covalent bond has both a characteristic bond strength and bond length. A plot of energy versus internuclear distance for two hydrogen atoms. The distance between nuclei at the minimum energy point is the bond length .

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sp 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, CH 4 , for instance. As we’ve seen, carbon has four valence electrons (2 s 2 2 p 2 ) 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 C–H bonds. In fact, though, all four C–H bonds in methane are identical and are spatially oriented toward the corners of a regular tetrahedron, as shown previously in . 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 , these tetrahedrally oriented orbitals are called sp 3 hybrid orbitals . Note that the superscript 3 in the name sp 3 tells how many of each type of atomic orbital combine to form the hybrid, not how many electrons occupy it. Four sp 3 hybrid orbitals , oriented toward the corners of a regular tetrahedron, are formed by the combination of an s orbital and three p orbitals (red / blue) . The sp 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. 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 sp 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, sp 3 hybrid orbitals form stronger bonds than do unhybridized s or p orbitals. The asymmetry of sp 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 sp 3 hybrid orbitals of a carbon atom overlaps with the 1 s orbital of a hydrogen atom, four identical C–H bonds are formed and methane results. Each C–H bond in methane has a strength of 439 kJ/mol (105 kcal/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 H–C–H is 109.5°, the so-called tetrahedral angle. Methane thus has the structure shown in . The structure of methane, showing its 109.5° bond angles.

sp3Hybrid Orbitals and the Structure of Methane

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sp 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, C 2 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 ( σ ) overlap of an sp 3 hybrid orbital from each ( ). The remaining three sp 3 hybrid orbitals on each carbon overlap with the 1 s orbitals of three hydrogens to form the six C–H bonds. The C–H bonds in ethane are similar to those in methane, although a bit weaker: 421 kJ/mol (101 kcal/mol) for ethane versus 439 kJ/mol for methane. The C–C bond is 153 pm in length and has a strength of 377 kJ/mol (90 kcal/mol). All the bond angles of ethane are near, although not exactly at, the tetrahedral value of 109.5°. The structure of ethane. The carbon–carbon bond is formed by σ overlap of two sp 3 hybrid orbitals . For clarity, the smaller lobes of the sp 3 hybrid orbitals are not shown. Problems

sp3Hybrid Orbitals and the Structure of Ethane

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sp 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 H 2 C=C H 2 and contains a carbon–carbon double bond, while acetylene has the structure HC≡CH and contains a carbon–carbon triple bond. How are multiple bonds described by valence bond theory? When we discussed sp 3 hybrid orbitals in Section 1.6 , we said that the four valence-shell atomic orbitals of carbon combine to form four equivalent sp 3 hybrids. Imagine instead that the 2 s orbital combines with only two of the three available 2 p orbitals. Three sp 2 hybrid orbitals result, and one 2 p orbital remains unchanged. Like sp 3 hybrids, sp 2 hybrid orbitals are unsymmetrical about the nucleus and are strongly oriented in a specific direction so they can form strong bonds. The three sp 2 orbitals lie in a plane at angles of 120° to one another, with the remaining p orbital perpendicular to the sp 2 plane, as shown in . sp 2 Hybridization. The three equivalent sp 2 hybrid orbitals lie in a plane at angles of 120° to one another, and a single unhybridized p orbital (red / blue) is perpendicular to the sp 2 plane. When two carbons with sp 2 hybridization approach each other, they form a strong σ bond by sp 2 – sp 2 head-on overlap. At the same time, the unhybridized p orbitals interact by sideways overlap to form what is called a pi ( π ) bond . The combination of an sp 2 – sp 2 σ bond and a 2 p –2 p π bond results in the sharing of four electrons and the formation of a carbon–carbon double bond ( ). Note that the electrons in the σ bond occupy the region centered between nuclei, while the electrons in the π bond occupy regions above and below a line drawn between nuclei. To complete the structure of ethylene, four hydrogen atoms form σ bonds with the remaining four sp 2 orbitals. Ethylene thus has a planar structure, with H–C–H and H–C–C bond angles of approximately 120°. (The actual values are 117.4° for the H–C–H bond angle and 121.3° for the H–C–C bond angle.) Each C–H bond has a length of 108.7 pm and a strength of 464 kJ/mol (111 kcal/mol). The structure of ethylene. One part of the double bond in ethylene results from σ (head-on) overlap of sp 2 hybrid orbitals , and the other part results from π (sideways) overlap of unhybridized p orbitals (red / blue) . The π bond has regions of electron density above and below a line drawn between nuclei. 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 C=C bond length of 134 pm and a strength of 728 kJ/mol (174 kcal/mol) versus a C–C length of 153 pm and a strength of 377 kJ/mol for ethane. The carbon–carbon double bond is less than twice as strong as a single bond because the sideways overlap in the π part of the double bond is not as great as the head-on overlap in the σ part. Drawing Electron-Dot and Line-Bond Structures Commonly used in biology as a tissue preservative, formaldehyde, CH 2 O, contains a carbon– oxygen double bond. Draw electron-dot and line-bond structures of formaldehyde, and indicate the hybridization of the carbon orbitals. 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. 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 sp 2 -hybridized. Problems

sp2Hybrid Orbitals and the Structure of Ethylene

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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, H−C≡C−H , we need a third kind of hybrid orbital, an sp 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 sp hybrid orbitals result, and two p orbitals remain unchanged. The two sp orbitals are oriented 180° 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 . sp Hybridization. The two sp hybrid orbitals are oriented 180° away from each other, perpendicular to the two remaining p orbitals (red / blue) . When two sp -hybridized carbon atoms approach each other, sp hybrid orbitals on each carbon overlap head-on to form a strong sp – sp σ bond. At the same time, the p z orbitals from each carbon form a p z – p z π bond by sideways overlap, and the p y orbitals overlap similarly to form a p y – p y π bond. The net effect is the sharing of six electrons and formation of a carbon–carbon triple bond. Each of the two remaining sp hybrid orbitals forms a σ bond with hydrogen to complete the acetylene molecule ( ). The structure of acetylene. The two carbon atoms are joined by one sp – sp σ bond and two p – p π bonds. As suggested by sp hybridization, acetylene is a linear molecule with H–C–C bond angles of 180°. The C–H bonds have a length of 106 pm and a strength of 558 kJ/mol (133 kcal/mol). The C–C bond length in acetylene is 120 pm, and its strength is about 965 kJ/mol (231 kcal/mol), making it the shortest and strongest of any carbon–carbon bond. A comparison of sp, sp 2 , and sp 3 hybridization is given in . Comparison of C−C and C−H Bonds in Methane, Ethane, Ethylene, and Acetylene Molecule Bond Bond strength Bond length (pm) (kJ/mol) (kcal/mol) Methane, CH 4 ( sp 3 ) C−H 439 105 109 Ethane, CH 3 CH 3 ( sp 3 ) C−C ( sp 3 ) 377 90 153 ( sp 3 ) C−H 421 101 109 Ethylene, H 2 C=CH 2 ( sp 2 ) C=C ( sp 2 ) 728 174 134 ( sp 2 ) C−H 464 111 109 Acetylene, HC≡CH ( sp ) C≡C ( sp ) 965 231 120 ( sp ) C−H 558 133 106 Problems

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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 (CH 3 NH 2 ), an organic derivative of ammonia (NH 3 ) and the substance responsible for the odor of rotting fish. The experimentally measured H–N–H bond angle in methylamine is 107.1°, and the C–N–H bond angle is 110.3°, both of which are close to the 109.5° tetrahedral angle found in methane. We therefore assume that nitrogen forms four sp 3 -hybridized orbitals, just as carbon does. One of the four sp 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 sp 3 hybrid orbital of nitrogen occupies as much space as an N–H bond does and is very important to the chemistry of methylamine and other nitrogen-containing organic molecules. 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 sp 3 -hybridized. The C–O–H bond angle in methanol is 108.5°, very close to the 109.5° tetrahedral angle. Two of the four sp 3 hybrid orbitals on oxygen are occupied by nonbonding electron lone pairs, and two are used to form bonds. 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, CH 3 OPO 3 2− , is the simplest example. The O–P–O bond angle in such compounds is typically in the range 110° to 112°, implying sp 3 hybridization for phosphorus orbitals. 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, C–S–H or in sulfides, which have a sulfur atom bonded to two carbons, C–S–C. Produced by some bacteria, methanethiol (CH 3 SH) is the simplest example of a thiol, and dimethyl sulfide, H 3 C–S–CH 3 , is the simplest example of a sulfide. Both can be described by approximate sp 3 hybridization around sulfur, although both have significant deviation from the 109.5° tetrahedral angle. Problems

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Describing Chemical Bonds: Molecular Orbital Theory We said in 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 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 ( ). 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. Molecular orbitals of H 2 . Combination of two hydrogen 1 s atomic orbitals leads to two H 2 molecular orbitals. The lower-energy, bonding MO is filled, and the higher-energy, antibonding MO is unfilled. 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 σ molecular orbitals result from the head-on combination of two s atomic orbitals in H 2 , so bonding and antibonding π molecular orbitals result from the sideways combination of two p atomic orbitals in ethylene. As shown in , the lower-energy, π bonding MO has no node between nuclei and results from the combination of p orbital lobes with the same algebraic sign. The higher-energy, π 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 π bonds in compounds that have more than one double bond. A molecular orbital description of the C–C π bond in ethylene. The lower-energy, π bonding MO results from an additive combination of p orbital lobes with the same algebraic sign and is filled. The higher-energy, π antibonding MO results from a subtractive combination of p orbital lobes with opposite algebraic signs and is unfilled.

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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 CH 3 ; if a carbon has two hydrogens bonded to it, we write CH 2 ; and so on. The compound called 2-methylbutane, for example, is written as follows: Note that the horizontal bonds between carbons aren’t shown in condensed structures—the CH 3 , 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 CH 3 units attached to the CH carbon are grouped together as (CH 3 ) 2 . Even simpler than condensed structures are skeletal structures such as those shown in . The rules for drawing skeletal structures are straightforward. 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. 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. RULE 3 Atoms other than carbon and hydrogen are shown. One further comment: Although such groupings as –CH 3 , –OH, and –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 H 3 C–, HO–, and H 2 N– if needed to make the bonding connections clearer. Larger units such as –CH 2 CH 3 are not inverted, though; we don’t write H 3 CH 2 C– because it would be confusing. There are, however, no well-defined rules that cover all cases; it’s largely a matter of preference. Line-bond and Skeletal Structures for Some Compounds Compound Line-bond structure Skeletal structure Isoprene, C 5 H 8 Methylcyclohexane, C 7 H 14 Phenol, C 6 H 6 O 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. Strategy The end of a line represents a carbon atom with 3 hydrogens, CH 3 ; a two-way intersection is a carbon atom with 2 hydrogens, 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 Problems 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? 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. 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 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 LD 50 value is between 1 and 4 g/kg depending on the animal species. Aspirin, for comparison, has an LD 50 of 1.1 g/kg, and ethanol (ethyl alcohol) has an LD 50 of 10.6 g/kg. lists the LD 50 for some other familiar substances. The lower the value, the more toxic the substance. Note, though, that 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. Some LD 50 Values Substance LD 50 (g/kg) Substance LD 50 (g/kg) Strychnine 0.005 Chloroform 1.2 Arsenic trioxide 0.015 Iron(II) sulfate 1.5 DDT 0.115 Ethyl alcohol 10.6 Aspirin 1.1 Sodium cyclamate 17 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 banned in the European Union. 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 ( π ) bond sigma ( σ ) bond skeletal structure sp hybrid orbital sp 2 hybrid orbital sp 3 hybrid orbital valence bond (VB) theory valence shell 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 ground-state 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 ( σ ) bonds ; bonds formed by sideways interaction of p orbitals are called pi ( π ) 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 sp 3 hybrid orbitals . When forming a double bond with planar geometry, carbon uses three equivalent sp 2 hybrid orbitals and one unhybridized p orbital. When forming a triple bond with linear geometry, carbon uses two equivalent sp 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. 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. 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. Additional Problems Visualizing Chemistry Electron Configurations Electron-Dot and Line-Bond Structures Hybridization Skeletal Structures General Problems

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Why This Chapter? 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) 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.

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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 Na + and Cl – ions, which are held together in the solid by electrostatic attractions between unlike charges. The C–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 ( ). The continuum in bonding from covalent to ionic is a result of an unequal distribution of bonding electrons between atoms. The symbol δ (lowercase Greek letter delta) means partial charge, either partial positive ( δ +) for the electron-poor atom or partial negative ( δ –) 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 , electronegativities are based on an arbitrary scale, with fluorine the most electronegative (EN = 4.0) and cesium the least (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. 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 F = 4.0 and 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 (EN = 2.5) and hydrogen (EN = 2.1) have similar electronegativities. Bonds between carbon and more electronegative elements such as oxygen (EN = 3.5) and nitrogen (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, δ +, and the electronegative atom with a partial negative charge, δ –. An example is the C–O bond in methanol, CH 3 OH ( a). 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 C–Li bond in methyllithium, CH 3 Li ( b). Polar covalent bonds. (a) Methanol, CH 3 OH, has a polar covalent C–O bond, and (b) methyllithium, CH 3 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; δ −) to blue (electron-poor; δ +) . Note in the representations of methanol and methyllithium in that a crossed arrow 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 ( δ +), and the head of the arrow is electron-rich ( δ –). Note also in 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; δ –) and electron-poor (blue; δ +) 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 σ 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. Problems

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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 , μ (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, μ = Q × r . Dipole moments are expressed in debyes (D), where 1 D = 3.336 × 10 –30 coulomb meters (C · m) in SI units. For example, the unit charge on an electron is 1.60 × 10 –19 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 × 10 –29 C · m, or 4.80 D. μ = Q × r μ = ( 1.60 × 10 − 19 C ) ( 100 × 10 − 12 m ) 1 D 3.336 × 10 − 30 C ⋅ m = 4.80 D Dipole moments for some common substances are given in . Of the compounds shown in the table, sodium chloride has the largest dipole moment (9.00 D) because it is ionic. Even small molecules like water ( μ = 1.85 D), methanol (CH 3 OH; μ = 1.70 D), and ammonia ( μ = 1.47 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. Dipole Moments of Some Compounds Compound Dipole moment (D) Compound Dipole moment (D) NaCl 9.00 NH 3 1.47 CH 2 O 2.33 CH 3 NH 2 1.31 CH 3 Cl 1.87 CO 2 0 H 2 O 1.85 CH 4 0 CH 3 OH 1.70 CH 3 CH 3 0 CH 3 CO 2 H 1.70 0 CH 3 SH 1.52 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. Predicting the Direction of a Dipole Moment Make a three-dimensional drawing of methylamine, CH 3 NH 2 , and show the direction of its dipole moment ( μ = 1.31). 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. Solution Methylamine contains an electronegative nitrogen atom with a lone pair of electrons. The dipole moment thus points generally from –CH 3 toward the lone pair. Problems

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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 (CH 3 SOCH 3 ), 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. 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 C–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 N–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 N–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 S–C single bonds, one in the S–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. A summary of commonly encountered formal charges and the bonding situations in which they occur is given in . 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. A Summary of Common Formal Charges Atom C N O S P Structure 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 Problems

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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 C–O single bond (135 pm) and a typical C=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 π 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 three-dimensional 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. 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.

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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: 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. RULE 2 Resonance forms differ only in the placement of their π 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 sp 2 -hybridized and the oxygen atoms remain in exactly the same place in both resonance forms. Only the positions of the π electrons in the C=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 π 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 C=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: 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.

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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 X, Y, and Z in the general structure might be C, N, O, P, 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 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 C=O bond on the left. The O=C–C : − grouping is a typical one for which two resonance structures can be drawn. Just as there is a C=O bond to the left of the lone pair, there is a second C=O bond to the right. Thus, we can draw a total of three resonance structures for the 2,4-pentanedione anion. Drawing Resonance Forms for an Anion Draw three resonance structures for the carbonate ion, CO 3 2– . 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 C=O double bond, giving the grouping . Solution Exchanging the position of the double bond and an electron lone pair in each grouping generates three resonance structures. 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 ( · ). Strategy Find the three-atom groupings that contain a multiple bond next to an atom with a p orbital. Solution The unpaired electron is on a carbon atom next to a C=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. Thus, the three resonance forms for the pentadienyl radical are: Problems

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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, H + , and a Brønsted–Lowry base is a substance that accepts a hydrogen ion. (The name proton is often used as a synonym for 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 (Cl – ) and hydronium cation (H 3 O + ). This and other acid–base reactions are reversible, so we’ll write them with double, forward-and-backward arrows. 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 H 2 O gains a proton, is called the conjugate acid of the base. Other common mineral acids such as H 2 SO 4 and HNO 3 behave similarly, as do organic acids such as acetic acid, CH 3 CO 2 H. In a general sense, For example: Notice that water can act either as an acid or as a base, depending on the circumstances. In its reaction with HCl, water is a base that accepts a proton to give the hydronium ion, H 3 O + . In its reaction with ammonia (NH 3 ), however, water is an acid that donates a proton to give ammonium ion (NH 4 + ) and hydroxide ion, HO – . Problems

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Acid and Base Strength Different acids differ in their ability to donate H + . Stronger acids, such as HCl, react almost completely with water, whereas weaker acids, such as acetic acid (CH 3 CO 2 H), react only slightly. The exact strength of a given acid HA in water solution is described using the acidity constant ( K a ) for the acid-dissociation equilibrium. Recall from general chemistry that the concentration of solvent is ignored in the equilibrium expression and that brackets [ ] around a substance refer to the concentration of the enclosed species in moles per liter. HA + H 2 O ⇌ A − + H 3 O + K a = [ H 3 O + ] [ A - ] [ HA ] Stronger acids have their equilibria toward the right and thus have larger acidity constants, whereas weaker acids have their equilibria toward the left and have smaller acidity constants. The range of K a values for different acids is enormous, running from about 10 15 for the strongest acids to about 10 –60 for the weakest. Common inorganic acids such as H 2 SO 4 , HNO 3 , and HCl have K a ’s in the range of 10 2 to 10 9 , while organic acids generally have K a ’s in the range of 10 –5 to 10 –15 . As you gain experience, you’ll develop a rough feeling for which acids are “strong” and which are “weak” (always remembering that the terms are relative, not absolute). Acid strengths are normally expressed using p K a values rather than K a values, where the p K a is the negative common logarithm of the K a : p K a = − log K a A stronger acid (larger K a ) has a smaller p K a , and a weaker acid (smaller K a ) has a larger p K a . lists the p K a ’s of some common acids in order of their strength, and a more comprehensive table is given in . Notice that the p K a value shown in for water is 15.74, which results from the following calculation. Because water is both the acid and the solvent, the equilibrium expression is H 2 O (acid) + H 2 O (solvent) ⇌ OH − + H 3 O + K a = [ H 3 O + ] [ A − ] [ HA ] = [ H 3 O + ] [ OH − ] [ H 2 O ] = [ 1.0 × 10 − 7 ] [ 1.0 × 10 − 7 ] [ 55.4 ] = 1.8 × 10 − 16 p K a = 15.74 The numerator in this expression is the so-called ion-product constant for water, K w = [H 3 O + ][OH – ] = 1.00 × 10 –14 , and the denominator is the molar concentration of pure water, [H 2 O] = 55.4 M at 25 °C. The calculation is artificial in that the concentration of “solvent” water is ignored while the concentration of “acid” water is not, but it is nevertheless useful for making a comparison of water with other weak acids on a similar footing. Notice also in that there is an inverse relationship between the acid strength of an acid and the base strength of its conjugate base. A strong acid has a weak conjugate base, and a weak acid has a strong conjugate base. To understand this inverse relationship, think about what is happening to the acidic hydrogen in an acid–base reaction. A strong acid is one that loses H + easily, meaning that its conjugate base holds the H + weakly and is therefore a weak base. A weak acid is one that loses H + with difficulty, meaning that its conjugate base holds the proton tightly and is therefore a strong base. The fact that HCl is a strong acid, for example, means that Cl – does not hold H + tightly and is thus a weak base. Water, on the other hand, is a weak acid, meaning that OH – holds H + tightly and is a strong base. Relative Strengths of Some Common Acids and Their Conjugate Bases Acid Name p K a Conjugate base Name CH 3 CH 2 O H Ethanol 16.00 CH 3 CH 2 O – Ethoxide ion H 2 O Water 15.74 HO – Hydroxide ion H CN Hydrocyanic acid 9.31 CN – Cyanide ion H 2 PO 4 – Dihydrogen phosphate ion 7.21 HPO 4 2– Hydrogen phosphate ion CH 3 CO 2 H Acetic acid 4.76 CH 3 CO 2 – Acetate ion H 3 PO 4 Phosphoric acid 2.16 H 2 PO 4 – Dihydrogen phosphate ion H NO 3 Nitric acid –1.3 NO 3 – Nitrate ion H Cl Hydrochloric acid –7.0 Cl – Chloride ion Problems

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Predicting Acid–Base Reactions from p K a Values Compilations of p K a values like those in and are useful for predicting whether a given acid–base reaction will take place, because H + will always go from the stronger acid to the stronger base. That is, an acid will donate a proton to the conjugate base of a weaker acid, and the conjugate base of a weaker acid will remove a proton from a stronger acid. Since water (p K a = 15.74) is a weaker acid than acetic acid (p K a = 4.76), for example, hydroxide ion holds a proton more tightly than acetate ion does. Hydroxide ion will therefore react to a large extent with acetic acid, CH 3 CO 2 H, to yield acetate ion and H 2 O. Another way to predict acid–base reactivity is to remember that the product conjugate acid in an acid–base reaction must be weaker and less reactive than the starting acid and the product conjugate base must be weaker and less reactive than the starting base. In the reaction of acetic acid with hydroxide ion, for example, the product conjugate acid (H 2 O) is weaker than the starting acid (CH 3 CO 2 H), and the product conjugate base (CH 3 CO 2 – ) is weaker than the starting base (OH – ). Predicting Acid Strengths from p K a Values Water has p K a = 15.74, and acetylene has p K a = 25. Which is the stronger acid? Does hydroxide ion react to a significant extent with acetylene? Strategy In comparing two acids, the one with the lower p K a is stronger. Thus, water is a stronger acid than acetylene and gives up H + more easily. Solution Because water is a stronger acid and gives up H + more easily than acetylene, the HO – ion must have less affinity for H + than the HC≡C : − ion. In other words, the anion of acetylene is a stronger base than hydroxide ion, and the reaction will not proceed significantly as written. Calculating K a from p K a According to the data in , acetic acid has p K a = 4.76. What is its K a ? Strategy Since p K a is the negative logarithm of K a , it’s necessary to use a calculator with an ANTILOG or INV LOG function. Enter the value of the p K a (4.76), change the sign (–4.76), and then find the antilog (1.74 × 10 –5 ). Solution K a = 1.74 × 10 –5 . Problems

Predicting Acid–Base Reactions from pKaValues

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Organic Acids and Organic Bases Many of the reactions we’ll be seeing in future chapters, including practically all biological reactions, involve organic acids and organic bases. Although it’s too early to go into the details of these processes now, you might keep the following generalities in mind: Organic Acids Organic acids are characterized by the presence of a positively polarized hydrogen atom (blue in electrostatic potential maps) and are of two main kinds: acids such as methanol and acetic acid that contain a hydrogen atom bonded to an electronegative oxygen atom (O–H) and those such as acetone ( ) that contain a hydrogen atom bonded to a carbon atom next to a C=O bond ( O=C–C–H ). Methanol contains an O–H bond and is a weak acid, while acetic acid also contains an O–H bond and is a somewhat stronger acid. In both cases, acidity is due to the fact that the conjugate base resulting from loss of H + is stabilized by having its negative charge on a strongly electronegative oxygen atom. In addition, the conjugate base of acetic acid is stabilized by resonance ( and ). The acidity of acetone and other compounds with C=O bonds is due to the fact that the conjugate base resulting from loss of H + is stabilized by resonance. In addition, one of the resonance forms stabilizes the negative charge by placing it on an electronegative oxygen atom. Electrostatic potential maps of the conjugate bases from methanol, acetic acid, and acetone are shown in . As you might expect, all three show a substantial amount of negative charge (red) on oxygen. Electrostatic potential maps of the conjugate bases of (a) methanol, (b) acetic acid, and (c) acetone. The electronegative oxygen atoms stabilize the negative charge in all three. Compounds called carboxylic acids, which contain the –CO 2 H grouping, occur abundantly in all living organisms and are involved in almost all metabolic pathways. Acetic acid, pyruvic acid, and citric acid are examples. You might note that at the typical pH of 7.3 found within cells, carboxylic acids are usually dissociated and exist as their carboxylate anions, –CO 2 – . Organic Bases Organic bases are characterized by the presence of an atom (reddish in electrostatic potential maps) with a lone pair of electrons that can bond to H + . Nitrogen-containing compounds such as methylamine are the most common organic bases and are involved in almost all metabolic pathways, but oxygen-containing compounds can also act as bases when reacting with a sufficiently strong acid. Note that some oxygen-containing compounds can act both as acids and as bases depending on the circumstances, just as water can. Methanol and acetone, for instance, act as acids when they donate a proton but as bases when their oxygen atom accepts a proton. We’ll see in Chapter 26 that substances called amino acids, so-named because they are both amines (–NH 2 ) and carboxylic acids (–CO 2 H), are the building blocks from which the proteins in all living organisms are made. Twenty different amino acids go into making up proteins—alanine is an example. Interestingly, alanine and other amino acids exist primarily in a doubly charged form called a zwitterion rather than in the uncharged form. The zwitterion form arises because amino acids have both acidic and basic sites within the same molecule and therefore undergo an internal acid–base reaction.

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Acids and Bases: The Lewis Definition The Lewis definition of acids and bases is more encompassing than the Brønsted–Lowry definition because it’s not limited to substances that donate or accept just protons. A Lewis acid is a substance that accepts an electron pair, and a Lewis base is a substance that donates an electron pair. The donated electron pair is shared between the acid and the base in a covalent bond. Lewis Acids and the Curved Arrow Formalism The fact that a Lewis acid is able to accept an electron pair means that it must have either a vacant, low-energy orbital or a polar bond to hydrogen so that it can donate H + (which has an empty 1 s orbital). Thus, the Lewis definition of acidity includes many species in addition to H + . For example, various metal cations, such as Mg 2+ , are Lewis acids because they accept a pair of electrons when they form a bond to a base. We’ll also see in later chapters that certain metabolic reactions begin with an acid–base reaction between Mg 2+ as a Lewis acid and an organic diphosphate or triphosphate ion as the Lewis base. In the same way, compounds of group 3A elements, such as BF 3 and AlCl 3 , are Lewis acids because they have unfilled valence orbitals and can accept electron pairs from Lewis bases, as shown in . Similarly, many transition-metal compounds, such as TiCl 4 , FeCl 3 , ZnCl 2 , and SnCl 4 , are Lewis acids. The reaction of boron trifluoride, a Lewis acid, with dimethyl ether, a Lewis base. The Lewis acid accepts a pair of electrons, and the Lewis base donates a pair of nonbonding electrons. Note how the movement of electrons from the Lewis base to the Lewis acid is indicated by a curved arrow. Note also how, in electrostatic potential maps, the boron becomes more negative (red) after reaction because it has gained electrons and the oxygen atom becomes more positive (blue) because it has donated electrons. Look closely at the acid–base reaction in , and notice how it's shown. Dimethyl ether, the Lewis base, donates an electron pair to a vacant valence orbital of the boron atom in BF 3 , a Lewis acid. The direction of electron-pair flow from base to acid is shown using a curved arrow, just as the direction of electron flow from one resonance structure to another was shown using curved arrows in . We’ll use this curved-arrow notation throughout the remainder of this text to indicate electron flow during reactions, so get used to seeing it. Some further examples of Lewis acids follow: Lewis Bases The Lewis definition of a base—a compound with a pair of nonbonding electrons that it can use to bond to a Lewis acid—is similar to the Brønsted–Lowry definition. Thus, H 2 O, with its two pairs of nonbonding electrons on oxygen, acts as a Lewis base by donating an electron pair to an H + in forming the hydronium ion, H 3 O + . In a more general sense, most oxygen- and nitrogen-containing organic compounds can act as Lewis bases because they have lone pairs of electrons. A divalent oxygen compound has two lone pairs of electrons, and a trivalent nitrogen compound has one lone pair. Note in the following examples that some compounds can act as both acids and bases, just as water can. Alcohols and carboxylic acids, for instance, act as acids when they donate an H + but as bases when their oxygen atom accepts an H + . Note also that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one atom with a lone pair of electrons and can therefore react at more than one site. Acetic acid, for example, can be protonated either on the doubly bonded oxygen atom or on the singly bonded oxygen atom. Reaction normally occurs only once in such instances, and the more stable of the two possible protonation products is formed. For acetic acid, protonation by reaction with sulfuric acid occurs on the doubly bonded oxygen because that product is stabilized by two resonance forms. Using Curved Arrows to Show Electron Flow Using curved arrows, show how acetaldehyde, CH 3 CHO, can act as a Lewis base. Strategy A Lewis base donates an electron pair to a Lewis acid. We therefore need to locate the electron lone pairs on acetaldehyde and use a curved arrow to show the movement of a pair toward the H atom of the acid. Solution Problems

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Noncovalent Interactions between Molecules When thinking about chemical reactivity, chemists usually focus their attention on bonds, the covalent interactions between atoms within molecules. Also important, however, particularly in large biomolecules like proteins and nucleic acids, are a variety of interactions between molecules that strongly affect molecular properties. Collectively called either intermolecular forces , van der Waals forces , or noncovalent interactions , they are of several different types: dipole–dipole forces, dispersion forces, and hydrogen bonds. Dipole–dipole forces occur between polar molecules as a result of electrostatic interactions among dipoles. The forces can be either attractive or repulsive depending on the orientation of the molecules—attractive when unlike charges are together and repulsive when like charges are together. The attractive geometry is lower in energy and therefore predominates ( ). Dipole–dipole forces cause polar molecules (a) to attract one another when they orient with unlike charges together, but (b) to repel one another when they orient with like charges together . Dispersion forces occur between all neighboring molecules and arise because the electron distribution within molecules is constantly changing. Although uniform on a time-averaged basis, the electron distribution even in nonpolar molecules is likely to be nonuniform at any given instant. One side of a molecule may, by chance, have a slight excess of electrons relative to the opposite side, giving the molecule a temporary dipole. This temporary dipole in one molecule causes a nearby molecule to adopt a temporarily opposite dipole, resulting in a tiny attraction between the two ( ). Temporary molecular dipoles have only a fleeting existence and are constantly changing, but their cumulative effect is often strong enough to hold molecules close together so that a substance is a liquid or solid rather than a gas. Attractive dispersion forces in nonpolar molecules are caused by temporary dipoles , as shown in these models of pentane, C 5 H 12 . Perhaps the most important noncovalent interaction in biological molecules is the hydrogen bond , an attractive interaction between a hydrogen atom bonded to an electronegative O or N atom and an unshared electron pair on another O or N atom. In essence, a hydrogen bond is a very strong dipole–dipole interaction involving polarized O–H or N–H bonds. Electrostatic potential maps of water and ammonia clearly show the positively polarized hydrogens (blue) and the negatively polarized oxygens and nitrogens (red). Hydrogen bonding has enormous consequences for living organisms. Hydrogen bonds cause water to be a liquid rather than a gas at ordinary temperatures, they hold enzymes in the shapes necessary for catalyzing biological reactions, and they cause strands of deoxyribonucleic acid (DNA) to pair up and coil into the double helix that stores genetic information. One further point before leaving the subject of noncovalent interactions: biochemists frequently use the term hydrophilic , meaning “water-loving,” to describe a substance that is attracted to water and the term hydrophobic , meaning “water-fearing,” to describe a substance that is not strongly attracted to water. Hydrophilic substances, such as table sugar, often have a number of –OH groups in their structure so they can form hydrogen bonds and dissolve in water, whereas hydrophobic substances, such as vegetable oil, do not have groups that form hydrogen bonds and do not dissolve in water. Problems Alkaloids: From Cocaine to Dental Anesthetics Just as ammonia (NH 3 ) is a weak base, there are a large number of nitrogen-containing organic compounds called amines that are also weak bases. In the early days of organic chemistry, basic amines derived from natural sources were known as vegetable alkali, but they are now called alkaloids . More than 20,000 alkaloids are known. Their study provided much of the impetus for the growth of organic chemistry in the nineteenth century and remains today an active and fascinating area of research. The coca bush Erythroxylon coca, native to upland rain forest areas of Colombia, Ecuador, Peru, Bolivia, and western Brazil, is the source of the alkaloid cocaine. (credit: " Erythroxylum coca " by Danna Guevara/Wikimedia Commons, CC BY 4.0) Many alkaloids have pronounced biological properties, and approximately 50% of pharmaceutical agents used today are derived from naturally occurring amines. As just three examples, morphine, an analgesic agent (painkiller), is obtained from the opium poppy Papaver somniferum. Ephedrine, a bronchodilator, decongestant, and appetite suppressant, is obtained from Ephedra sinica , an evergreen shrub native to Mongolia and northeastern China. Cocaine, both an anesthetic and a stimulant, is obtained from the coca bush Erythroxylon coca, endemic to the upland rain forest areas of central South America. (And yes, there really was a small amount of cocaine in the original Coca-Cola recipe, although it was removed in 1906.) Cocaine itself is rarely used medically because it is too addictive, but its anesthetic properties provoked a long search for related but nonaddictive compounds. This search ultimately resulted in the synthesis of the “caine” anesthetics that are commonly used today in dental and surgical anesthesia. Procaine, the first such compound, was synthesized in 1898 and marketed under the name Novocain. It was rapidly adopted and remains in use today as a topical anesthetic. Other related compounds with different activity profiles followed: Lidocaine, marketed as Xylocaine, was introduced in 1943, and mepivacaine (Carbocaine) in the early 1960s. More recently, bupivacaine (Marcaine) and prilocaine (Citanest) have gained popularity. Both are quick-acting, but the effects of bupivacaine last for 3 to 6 hours while those of prilocaine fade after 45 minutes. Notice some structural similarity of all the caines to cocaine itself. An estimate from the U.S. National Academy of Sciences is that less than 1% of all living species have been characterized. Thus, alkaloid chemistry remains an active area of research, and innumerable substances with potentially useful properties have yet to be discovered. Undoubtedly even the caine anesthetics will become obsolete at some point, perhaps supplanted by newly discovered alkaloids. acidity constant ( K a ) alkaloid Brønsted–Lowry acid Brønsted–Lowry base conjugate acid conjugate base dipole moment ( μ ) dispersion force electronegativity (EN) electrostatic potential map formal charge hydrogen bond hydrophilic hydrophobic inductive effect intermolecular force Lewis acid Lewis base noncovalent interaction p K a polar covalent bond resonance form resonance hybrid van der Waals force Summary Understanding organic chemistry means knowing not just what happens but also why and how it happens at the molecular level. In this chapter, we’ve reviewed some of the ways that chemists describe and account for chemical reactivity, thereby providing a foundation for understanding the specific reactions that will be discussed in subsequent chapters. Organic molecules often have polar covalent bonds as a result of unsymmetrical electron sharing caused by differences in the electronegativity of atoms. A carbon–oxygen bond is polar, for example, because oxygen attracts the shared electrons more strongly than carbon does. Carbon–hydrogen bonds are relatively nonpolar. Many molecules as a whole are also polar, owing to the presence of individual polar bonds and electron lone pairs. The polarity of a molecule is measured by its dipole moment , μ . Plus (+) and minus (–) signs are often used to indicate the presence of formal charges on atoms in molecules. Assigning formal charges to specific atoms is a bookkeeping technique that makes it possible to keep track of the valence electrons around an atom and offers some clues about chemical reactivity. Some substances, such as acetate ion and benzene, can’t be represented by a single line-bond structure and must be considered as a resonance hybrid of two or more structures, none of which would be correct by themselves. The only difference between two resonance forms is in the location of their π and nonbonding electrons. The nuclei remain in the same places in both structures, and the hybridization of the atoms remains the same. Acidity and basicity are closely related to the ideas of polarity and electronegativity. A Brønsted–Lowry acid is a compound that can donate a proton (hydrogen ion, H + ), and a Brønsted–Lowry base is a compound that can accept a proton. The strength of a Brønsted–Lowry acid or base is expressed by its acidity constant, K a , or by the negative logarithm of the acidity constant, p K a . The larger the p K a , the weaker the acid. More useful is the Lewis definition of acids and bases. A Lewis acid is a compound that has a low-energy empty orbital that can accept an electron pair; Mg 2 + , BF 3 , AlCl 3 , and H + are examples. A Lewis base is a compound that can donate an unshared electron pair; NH 3 and H 2 O are examples. Most organic molecules that contain oxygen and nitrogen can act as Lewis bases toward sufficiently strong acids. A variety of noncovalent interactions have a significant effect on the properties of large biomolecules. Hydrogen bonding —the attractive interaction between a positively polarized hydrogen atom bonded to an oxygen or nitrogen atom with an unshared electron pair on another O or N atom, is particularly important in giving proteins and nucleic acids their shapes. Additional Problems Visualizing Chemistry Mechanism Problems Electronegativity and Dipole Moments Formal Charges Resonance Acids and Bases General Problems

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Why This Chapter? The bristlecone pine is the oldest living organism on Earth. The waxy coating on its needles contains a mixture of organic compounds called alkanes, the subject of this chapter. (credit: modification of work “Gnarly Bristlecone Pine” by Rick Goldwaser/Flickr, CC BY 2.0) The group of organic compounds called alkanes are simple and relatively unreactive, but they nevertheless provide a useful vehicle for introducing some important general ideas. In this chapter, we’ll use alkanes to introduce the basic approach to naming organic compounds and to take an initial look at some of the three-dimensional aspects of molecules, a topic of particular importance in understanding biological organic chemistry. According to Chemical Abstracts, the publication that abstracts and indexes the chemical literature, there are more than 195 million known organic compounds. Each of these compounds has its own physical properties, such as melting point and boiling point, and each has its own chemical reactivity. Chemists have learned through years of experience that organic compounds can be classified into families according to their structural features and that the members of a given family have similar chemical behavior. Instead of 195 million compounds with random reactivity, there are a few dozen families of organic compounds whose chemistry is reasonably predictable. We’ll study the chemistry of specific families throughout much of this book, beginning in this chapter with a look at the simplest family, the alkanes.

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Functional Groups The structural features that make it possible to classify compounds into families are called functional groups. A functional group is a group of atoms within a molecule that has a characteristic chemical behavior. Chemically, a given functional group behaves in nearly the same way in every molecule it’s a part of. For example, compare ethylene, a plant hormone that causes fruit to ripen, with menthene, a much more complicated molecule found in peppermint oil. Both substances contain a carbon–carbon double-bond functional group, and both therefore react with Br 2 in the same way to give a product in which a Br atom has added to each of the double-bond carbons ( ). This example is typical: the chemistry of every organic molecule, regardless of size and complexity, is determined by the functional groups it contains. The reactions of ethylene and menthene with bromine. In both molecules, the carbon–carbon double-bond functional group has a similar polarity pattern, so both molecules react with Br 2 in the same way. The size and complexity of the molecules are not important. Look at , which lists many of the common functional groups and gives simple examples of their occurrence. Some functional groups have only carbon–carbon double or triple bonds; others have halogen atoms; and still others contain oxygen, nitrogen, or sulfur. Much of the chemistry you’ll be studying is the chemistry of these functional groups. Functional Groups with Carbon–Carbon Multiple Bonds Alkenes, alkynes, and arenes (aromatic compounds) all contain carbon–carbon multiple bonds. Alkenes have a double bond, alkynes have a triple bond, and arenes have alternating double and single bonds in a six-membered ring of carbon atoms. They look different, but because of their structural similarities, they also have chemical similarities. Structures of Some Common Functional Groups Name Structure* Name ending Example Alkene (double bond) - ene H 2 C ═ CH 2 Ethene Alkyne (triple bond) – C ≡ C – - yne HC ≡ CH Ethyne Arene (aromatic ring) None Halide ( X =F, Cl, Br, I) None CH 3 Cl Chloromethane Alcohol - ol CH 3 OH Methanol Ether ether CH 3 O CH 3 Dimethyl ether Monophosphate phosphate CH 3 OPO 3 2− Methyl phosphate Diphosphate diphosphate CH 3 OP 2 O 6 3− Methyl diphosphate Amine - amine CH 3 NH 2 Methylamine Imine (Schiff base) None Nitrile – C ≡ N - nitrile CH 3 C ≡ N Ethanenitrile Thiol - thiol CH 3 SH Methanethiol Sulfide sulfide CH 3 S CH 3 Dimethyl sulfide Disulfide disulfide CH 3 SS CH 3 Dimethyl disulfide Sulfoxide sulfoxide Aldehyde - al Ketone - one Carboxylic acid - oic acid Ester - oate Thioester - thioate Amide -amide Acid chloride -oyl chloride Carboxylic acid anhydride - oic anhydride *The bonds whose connections aren’t specified are assumed to be attached to carbon or hydrogen atoms in the rest of the molecule. Functional Groups with Carbon Singly Bonded to an Electronegative Atom Alkyl halides (haloalkanes), alcohols, ethers, alkyl phosphates, amines, thiols, sulfides, and disulfides all have a carbon atom singly bonded to an electronegative atom—halogen, oxygen, nitrogen, or sulfur. Alkyl halides have a carbon atom bonded to halogen (–X), alcohols have a carbon atom bonded to the oxygen of a hydroxyl group (–OH), ethers have two carbon atoms bonded to the same oxygen, organophosphates have a carbon atom bonded to the oxygen of a phosphate group (–OPO 3 2− ), amines have a carbon atom bonded to a nitrogen, thiols have a carbon atom bonded to the sulfur of an –SH group, sulfides have two carbon atoms bonded to the same sulfur, and disulfides have carbon atoms bonded to two sulfurs that are joined together. In all cases, the bonds are polar, with the carbon atom bearing a partial positive charge ( δ +) and the electronegative atom bearing a partial negative charge ( δ –). Functional Groups with a Carbon–Oxygen Double Bond (Carbonyl Groups) The carbonyl group , C=O (pronounced car-bo- neel ) is common to many of the families listed in . Carbonyl groups are present in a majority of organic compounds and in practically all biological molecules. These compounds therefore behave similarly in many respects but differ depending on the identity of the other atoms bonded to the carbonyl-group carbon. Aldehydes have at least one hydrogen bonded to the C=O , ketones have two carbons bonded to the C=O , carboxylic acids have an –OH group bonded to the C=O , esters have an ether-like oxygen bonded to the C=O , thioesters have a sulfide-like sulfur bonded to the C=O , amides have an amine-like nitrogen bonded to the C=O , acid chlorides have a chlorine bonded to the C=O , and so on. In all these functional groups, the carbonyl carbon atom bears a partial positive charge ( δ +), and the oxygen bears a partial negative charge ( δ –). Problems

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Alkanes and Alkane Isomers Before beginning a systematic study of the different functional groups, let’s look first at the simplest family of molecules to develop some general ideas that apply to all families. We saw in that the carbon–carbon single bond in ethane results from σ (head-on) overlap of carbon sp 3 hybrid orbitals. If we imagine joining three, four, five, or even more carbon atoms by C–C single bonds, we can generate the large family of molecules called alkanes . Alkanes are often described as saturated hydrocarbons: hydrocarbons because they contain only carbon and hydrogen; saturated because they have only C–C and C–H single bonds and thus contain the maximum possible number of hydrogens per carbon. They have the general formula C n H 2 n +2 , where n is an integer. Alkanes are also occasionally called aliphatic compounds, a name derived from the Greek aleiphas, meaning “fat.” We’ll see in Section 27.1 that many animal fats contain long carbon chains similar to alkanes. Think about the ways that carbon and hydrogen might combine to make alkanes. With one carbon and four hydrogens, only one structure is possible: methane, CH 4 . Similarly, there is only one combination of two carbons with six hydrogens (ethane, CH 3 CH 3 ) and only one combination of three carbons with eight hydrogens (propane, CH 3 CH 2 CH 3 ). When larger numbers of carbons and hydrogens combine, however, more than one structure is possible. For example, there are two substances with the formula C 4 H 10 : the four carbons can all be in a row (butane), or they can branch (isobutane). Similarly, there are three C 5 H 12 molecules, and so on for larger alkanes. Compounds like butane and pentane, whose carbons are all connected in a row, are called straight-chain alkanes , or normal alkanes. Compounds like 2-methylpropane (isobutane), 2-methylbutane, and 2,2-dimethylpropane, whose carbon chains branch, are called branched-chain alkanes . Compounds like the two C 4 H 10 molecules and the three C 5 H 12 molecules, which have the same formula but different structures, are called Isomers , from the Greek isos + meros, meaning “made of the same parts.” Isomers have the same numbers and kinds of atoms but differ in the way the atoms are arranged. Compounds like butane and isobutane, whose atoms are connected differently, are called constitutional isomers . We’ll see shortly that other kinds of isomers are also possible, even among compounds whose atoms are connected in the same order. As shows, the number of possible alkane isomers increases dramatically with the number of carbon atoms. Number of Alkane Isomers Formula Number of isomers Formula Number of isomers C 6 H 14 5 C 10 H 22 75 C 7 H 16 9 C 15 H 32 4347 C 8 H 18 18 C 20 H 42 366,319 C 9 H 20 35 C 30 H 62 4,111,846,763 Constitutional isomerism is not limited to alkanes—it occurs widely throughout organic chemistry. Constitutional isomers may have different carbon skeletons (as in isobutane and butane), different functional groups (as in ethanol and dimethyl ether), or different locations of a functional group along the chain (as in isopropylamine and propylamine). Regardless of the reason for the isomerism, constitutional isomers are always different compounds with different properties but with the same formula. A given alkane can be drawn in many ways. For example, the straight-chain, four-carbon alkane called butane can be represented by any of the structures shown in . These structures don’t imply any particular three-dimensional geometry for butane; they indicate only the connections among atoms. In practice, as noted in , chemists rarely draw all the bonds in a molecule and usually refer to butane by the condensed structure, CH 3 CH 2 CH 2 CH 3 or CH 3 (CH 2 ) 2 CH 3 . Still more simply, butane can be represented as n -C 4 H 10 , where n denotes normal (straight-chain) butane. Some representations of butane, C 4 H 10 . The molecule is the same regardless of how it’s drawn. These structures imply only that butane has a continuous chain of four carbon atoms; they do not imply any specific geometry. Straight-chain alkanes are named according to the number of carbon atoms they contain, as shown in . With the exception of the first four compounds—methane, ethane, propane, and butane—whose names have historical roots, the alkanes are named based on Greek numbers. The suffix - ane is added to the end of each name to indicate that the molecule identified is an alkane. Thus, pent ane is the five-carbon alkane, hex ane is the six-carbon alkane, and so on. We’ll soon see that these alkane names form the basis for naming all other organic compounds, so at least the first ten should be memorized. Names of Straight-Chain Alkanes Number of carbons ( n ) Name Formula (C n H 2 n +2 ) Number of carbons ( n ) Name Formula (C n H 2 n +2 ) 1 Methane CH 4 9 Nonane C 9 H 20 2 Ethane C 2 H 6 10 Decane C 10 H 22 3 Propane C 3 H 8 11 Undecane C 11 H 24 4 Butane C 4 H 10 12 Dodecane C 12 H 26 5 Pentane C 5 H 12 13 Tridecane C 13 H 28 6 Hexane C 6 H 14 20 Icosane C 20 H 42 7 Heptane C 7 H 16 30 Triacontane C 30 H 62 8 Octane C 8 H 18 Drawing the Structures of Isomers Propose structures for two isomers with the formula C 2 H 7 N. Strategy We know that carbon forms four bonds, nitrogen forms three, and hydrogen forms one. Write down the carbon atoms first, and then use trial and error plus intuition to put the pieces together. Solution There are two isomeric structures. One has the connection C–C–N, and the other has the connection C–N–C. Problems

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