1.8 sp2 HYBRID ORBITALS AND THE STRUCTURE OF ETHYLENE 15 sp3hybrids.Im rem 1.The hThee e sp bitals lie in the the remaining p orbital wn in Figure 1.14 FIGURE 114 An so2-hybridized arhon The three equivalent s hybrid orbitals(green)lie in a plane at angles of1 pendicular to the spplane. Side view Top view When two sp2-hybridized carbons approach each other,they form a a bond by sp2-sp2 overlap.At the same time,the unhybridized p orbitals tomatlon f bond Th ommo o bond and a 2p-2p z bond results in the sharing of four electrons and the formation of a carbon-carbon double bond(Figure 1.15).Note that the elec- trons in the o bond occupy the region centered between nuclei,while the electrons in the m bond occupy regions above and below a line drawn between nuclei. To complete the structure of ethylene.four hydrogen atoms form o bonds with the remaining four sp2 orbitals.Ethylene thus has a planar structure,with d 121 BCband angl H bond angle and the H orbitals bond ridized car bons forms a carbon head6bondone bon dou and the other part ults from(sid ays) bond p carbor p carbon Carbon-carbon double bond itap of bond has regions ofelec tron density above and ealinedawmbetren
sp3 hybrids. Imagine instead that the 2s orbital combines with only two of the three available 2p orbitals. Three sp2 hybrid orbitals result, and one 2p orbital remains unchanged. The three sp2 orbitals lie in a plane at angles of 120° to one another, with the remaining p orbital perpendicular to the sp2 plane, as shown in Figure 1.14. sp2 sp2 sp2 sp2 sp2 sp2 p p 90 Side view Top view 120 When two sp2-hybridized carbons approach each other, they form a bond by sp2–sp2 overlap. At the same time, the unhybridized p orbitals approach with the correct geometry for sideways overlap, leading to the formation of what is called a pi () bond. The combination of an sp2–sp2 bond and a 2p–2p bond results in the sharing of four electrons and the formation of a carbon–carbon double bond (Figure 1.15). Note that the electrons in the bond occupy the region centered between nuclei, while the electrons in the bond occupy regions above and below a line drawn between nuclei. To complete the structure of ethylene, four hydrogen atoms form bonds with the remaining four sp2 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). 121.3 C C 117.4 H H H H 134 pm 108.7 pm Carbon–carbon double bond C C sp2 sp carbon 2 carbon sp2 orbitals p orbitals bond bond bond FIGURE 1.14 An sp2-hybridized carbon. The three equivalent sp2 hybrid orbitals (green) lie in a plane at angles of 120° to one another, and a single unhybridized p orbital (red/blue) is perpendicular to the sp2 plane. FIGURE 1.14 An sp2-hybridized carbon. The three equivalent sp2 hybrid orbitals (green) lie in a plane at angles of 120° to one another, and a single unhybridized p orbital (red/blue) is perpendicular to the sp2 plane. FIGURE 1.15 The structure of ethylene. Orbital overlap of two sp2-hybridized carbons forms a carbon– carbon double bond. One part of the double bond results from (head-on) overlap of sp2 orbitals (green), and the other part results from (sideways) overlap of unhybridized p orbitals (red/blue). The bond has regions of electron density above and below a line drawn between nuclei. FIGURE 1.15 The structure of ethylene. Orbital overlap of two sp2-hybridized carbons forms a carbon– carbon double bond. One part of the double bond results from (head-on) overlap of sp2 orbitals (green), and the other part results from (sideways) overlap of unhybridized p orbitals (red/blue). The bond has regions of electron density above and below a line drawn between nuclei. 1.8 sp2 hybrid orbitals and the structure of ethylene 15 39144_01_0001-0032.indd 15 7/27/09 1:28:41 PM
CHAPTER 1 STRUCTURE AND BONDING 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 154 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 m part of the double bond is not as great as the head-on overlap in the o part. saCa aw the line-bond structure of form isnedooSe onway th wo hydrogens,one carbon,and on oxygnc combine: 0 Formaldehyde H Like the carbon atoms in ethylene,the carbon atom in formaldehyde is in a double bond and is therefore sp2-hybridized. Problem 1.10 Draw a line-bond structure for propene.CH3CH=CH2:indicate the hybridiza- tion of each carbon:and predict the value of each bond angle. Problem 1.11 H2C- =CH-CH CH2:indicate bond struture for buta-1difondn Problem1.12 molecular Follo entify the Aspirin (acetylsalicylic acid)
16 chapter 1 structure and bonding 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 154 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 part of the double bond is not as great as the head-on overlap in the part. WORKED EXAMPLE 1.2 Predicting the Structures of Simple Molecules from Their Formulas Commonly used in biology as a tissue preservative, formaldehyde, CH2O, contains a carbon–oxygen double bond. Draw the line-bond structure of formaldehyde, and indicate the hybridization of the carbon atom. 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: Formaldehyde C HH O Like the carbon atoms in ethylene, the carbon atom in formaldehyde is in a double bond and is therefore sp2-hybridized. Problem 1.10 Draw a line-bond structure for propene, CH3CHUCH2; indicate the hybridization of each carbon; and predict the value of each bond angle. Problem 1.11 Draw a line-bond structure for buta-1,3-diene, H2CUCHXCHUCH2; indicate the hybridization of each carbon; and predict the value of each bond angle. Problem 1.12 Following is a molecular model of aspirin (acetylsalicylic acid). Identify the hybridization of each carbon atom in aspirin, and tell which atoms have lone pairs of electrons (gray C, red O, ivory H). Aspirin (acetylsalicylic acid) 39144_01_0001-0032.indd 16 7/27/09 1:28:42 PM
1.9 sp HYBRID ORBITALS AND THE STRUCTURE OF ACETYLENE 17 19 sp Hybrid Orbitals and the Structure of Acetylene In addition to forming single and double bonds by sharing two and four elec ocan form a triple bond by sharing six electrons as acetylene, rhital an sp hyb hybrid orbita 180 t.and two porbitals while obitals a the y-a axis and the axis.as shown in Figure 116 FIGURE 1.16 An sp-hybridized om.Ihe t o sp hybr dicular to the two remainingp orbitals (red/blue). One sp hybrid Another sp hybri When two sp-hybridized carbon atoms approach each other.sp hybrid orbitals on each carbon overlap head-on to form a strong sp-sp o bond.In addition,the pz orbitals from each carbon form a pz-pz bond by sideways overlap,and the py orbitals overlap similarly to form a py-py a bond.The net effect is the sharing of six electrons and formation of a carbon- carbon triple bond.The two remaining sp hybrid orbitals each form aobond with hydrogen to complete the acetylene molecule (Figure 1.17). p orbital FIGURE 1.17 The structure of acetylene.The two sp-hybrid. ized carbon bond ond and two p-p bond a rbon triple bond 120 80 onds h nave a length of 106 pm an
1.9 sp Hybrid Orbitals and the Structure of Acetylene In addition to forming single and double bonds by sharing two and four electrons, respectively, carbon also can form a triple bond by sharing six electrons. To account for the triple bond in a molecule such as acetylene, HXCmCXH, we need a third kind of hybrid orbital, an sp hybrid. Imagine that, instead of combining with two or three p orbitals, a carbon 2s 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 x-axis, while the remaining two p orbitals are perpendicular on the y-axis and the z-axis, as shown in Figure 1.16. 180 One sp hybrid Another sp hybrid sp sp p p When two sp-hybridized carbon atoms approach each other, sp hybrid orbitals on each carbon overlap head-on to form a strong sp–sp bond. In addition, the pz orbitals from each carbon form a pz–pz bond by sideways overlap, and the py orbitals overlap similarly to form a py–py bond. The net effect is the sharing of six electrons and formation of a carbon–carbon triple bond. The two remaining sp hybrid orbitals each form a bond with hydrogen to complete the acetylene molecule (Figure 1.17). H C CH 120 pm 106 pm 180° Carbon–carbon triple bond sp orbital sp orbital sp orbitals bond bond bond p orbitals p orbitals 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 FIGURE 1.16 An sp-hybridized carbon atom. The two sp hybrid orbitals (green) are oriented 180° away from each other, perpendicular to the two remaining p orbitals (red/blue). FIGURE 1.16 An sp-hybridized carbon atom. The two sp hybrid orbitals (green) are oriented 180° away from each other, perpendicular to the two remaining p orbitals (red/blue). FIGURE 1.17 The structure of acetylene. The two sp-hybridized carbon atoms are joined by one sp–sp bond and two p–p bonds. FIGURE 1.17 The structure of acetylene. The two sp-hybridized carbon atoms are joined by one sp–sp bond and two p–p bonds. 1.9 sp hybrid orbitals and the structure of acetylene 17 39144_01_0001-0032.indd 17 7/27/09 1:28:43 PM
CHAPTER 1 STRUCTURE AND BONDING TABLE12 Comparison of C-C and C-H Bonds in Methane,Ethane,Ethylene,and Acetylene Bond strength Molecule Bond (kJ/mol)(kcal/mol)Bond length (pm) Methane,CHa (sp)C-H 439 105 109 Ethane,CHgCH3 (sp)C-C(sp 377 90 154 (sp)C-H 420 100 109 Ethylene,H2C-CH2 (sp2)C-C(sp2) 728 174 134 (sp2)C-H 464 111 109 Acetylene,HC=CH (sp)C=C(sp) 965 231 120 (sp)C-H 558 133 106 -carbon bond.A comparison of sp.sp2 Problem 1.13 moiecmdnapme2encddneehetitia 110 Hybridization of Nitrogen,Oxygen, Phosphorus,and Sulfur The valence-bond cone of oital hybridization descibed in the previ id orb ent bo d by ihe at the nitr nsibl。for th odo of rotting fish The e experimentally measured H-N-H bond angle in methvlamine 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 hybridizes to form four sp3 orbitals,just as carbon does.One of the four orbitals is occupied by two nonbonding electrons,and the other three hybrid orbitals have one electron each.Overlap of these half-filled orbitals with half- filled orbitals from other atoms(C or H)gives methylamine.Note that the unshared lone pair of electrons in the fourth sp3 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. one pai CH3 110.3 Methylamine
18 chapter 1 structure and bonding shortest and strongest of any carbon–carbon bond. A comparison of sp, sp2, and sp3 hybridization is given in Table 1.2. Problem 1.13 Draw a line-bond structure for propyne, CH3CCH; indicate the hybridization of each carbon; and predict a value for each bond angle. 1.10 Hybridization of Nitrogen, Oxygen, Phosphorus, and Sulfur The valence-bond concept of orbital hybridization described in the previous four sections is not limited to carbon compounds. Covalent bonds formed by other elements can also be described using hybrid orbitals. Look, for instance, at the nitrogen atom in methylamine, CH3NH2, an organic derivative of ammonia (NH3) 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 hybridizes to form four sp3 orbitals, just as carbon does. One of the four sp3 orbitals is occupied by two nonbonding electrons, and the other three hybrid orbitals have one electron each. Overlap of these half-filled orbitals with halffilled orbitals from other atoms (C or H) gives methylamine. Note that the unshared lone pair of electrons in the fourth sp3 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. Methylamine H H CH3 Lone pair 107.1° 110.3° N Bond strength Molecule Bond (kJ/mol) (kcal/mol) Bond length (pm) Methane, CH4 (sp3) CXH 439 105 109 Ethane, CH3CH3 (sp3) CXC (sp3) 377 90 154 (sp3) CXH 420 100 109 Ethylene, H2CUCH2 (sp2) CUC (sp2) 728 174 134 (sp2) CXH 464 111 109 Acetylene, HCmCH (sp) CmC (sp) 965 231 120 (sp) CXH 558 133 106 TABLE 1.2 Comparison of C–C and C–H Bonds in Methane, Ethane, Ethylene, and Acetylene 39144_01_0001-0032.indd 18 7/27/09 1:28:44 PM
1.10 HYBRIDIZATION OF NITROGEN,OXYGEN,PHOSPHORUS,AND SULFUR Like the carbon atom in meth and the nitrogen atom in methylami ih。ethan thy ay other organic 08.6 s very clos eto the 109.5 tetrahedra ang Two of the occupied by nonbonding electron I Lone pairs_ 108.5° nmhnanonoal 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 car expand their outer-shell octets and form more than the typical number o COV bonds sphorus,for instance,often forms five covalent bonds. organophosph e oxygens alsc yI pho he c P o bon for the phosph ange110P011 implying sp ￥1109 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,or in sulfides,which have a sulfur atom bonded to two carbon ate sp Lone pairs 96. 99.1 Methanethio Dimethyl sulfide
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 sp3-hybridized. The C–O–H bond angle in methanol is 108.5°, very close to the 109.5° tetrahedral angle. Two of the four sp3 hybrid orbitals on oxygen are occupied by nonbonding electron lone pairs, and two are used to form bonds. Methanol (methyl alcohol) Lone pairs 108.5° O H CH3 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 occasionally forms four. Phosphorus is most commonly encountered in biological molecules in organophosphates, compounds that contain a phosphorus atom bonded to four oxygens, with one of the oxygens also bonded to carbon. Methyl phosphate, CH3OPO3 2, is the simplest example. The O–P–O bond angle in such compounds is typically in the range 110° to 112°, implying sp3 hybridization for the phosphorus. –O –O P O O CH3 Methyl phosphate (an organophosphate) 110° 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, or in sulfides, which have a sulfur atom bonded to two carbons. Produced by some bacteria, methanethiol (CH3SH) is the simplest example of a thiol, and dimethyl sulfide [(CH3)2S] is the simplest example of a sulfide. Both can be described by approximate sp3 hybridization around sulfur, although both have significant deviation from the 109.5° tetrahedral angle. H3C Methanethiol 96.5° S Lone pairs H CH3 Dimethyl sulfide 99.1° S CH3 Lone pairs 1.10 hybridization of nitrogen, oxygen, phosphorus, and sulfur 19 39144_01_0001-0032.indd 19 7/27/09 1:28:44 PM