1559T_ch01_01-1710/22/051:48Pa9e】 ⊕ EQA 1 Structure and Bonding in Organic Molecules s associated with the bondins ms to mak have some familiarity from freshman chemistry.In other words it describes just those topics from freshm lems read the comments below.and,if necessary.look to other supplementary sources for additional problem Outline of the Chapter 1-1 Overview 1-2 Coulomb Forces The simple physical basis of bonding between atoms.Conceptually important. 1-4 Lewis Structures h most critical scction of the chapter.You must leam how todrw comct Lewis structures 1-5 Resonance Forms Applies to species for which no single Lewis picture adequately describes the true structure 1-6 Atomic Orbitals Review material. 17oechrotias 1-8 Hybrid Orbitals rstand the geometry of molecules in the contet of the atomic orbitals involved 19
1 Structure and Bonding in Organic Molecules The first chapter of the text covers the basic features associated with the bonding together of atoms to make molecules. Much of the material (at least through Section 1-8) is really a review of topics with which you may have some familiarity from freshman chemistry. In other words, it describes just those topics from freshman chemistry that are the most important to know in order to get off to a good start in organic chemistry: bonds, Lewis structures, resonance, atomic and molecular orbitals, and hybrid orbitals. Read the chapter, try the problems, read the comments below, and, if necessary, look to other supplementary sources for additional problems and examples. For instance, Pushing Electrons: A Guide for Students of Organic Chemistry by D. Weeks (Saunders, 1997) contains extensive coverage of Lewis structures and resonance. Outline of the Chapter 1-1 Overview 1-2 Coulomb Forces The simple physical basis of bonding between atoms. Conceptually important. 1-3 Ionic and Covalent Bonds Review of principles involved. 1-4 Lewis Structures Perhaps the most critical section of the chapter. You must learn how to draw correct Lewis structures of molecules. 1-5 Resonance Forms Applies to species for which no single Lewis picture adequately describes the true structure. 1-6 Atomic Orbitals Review material. 1-7 Molecular Orbitals Review material. 1-8 Hybrid Orbitals Simplest way to understand the geometry of molecules in the context of the atomic orbitals involved in bonding. 1-9 Structures and Formulas of Organic Molecules General information section. Conventions for drawing formulas. 1 1559T_ch01_01-17 10/22/05 1:48 Page 1
1559Tch0101-1710/22/051:48Page2 EQA 2.chapter 1 STRUCTURE AND BONDNG IN ORGANIC MOLECULES Keys to the Chapter 1-2 and 1-3.Coulomb Forces;Bonds d to each othe r (as de d in this chapter).but difference betw en the atoms.the bondins electrons tend to be closer to the more electr onegative atom,thereby creating a partial charge separation.In general,for A less electronegative than B.we have A:B See the in the vicinity of more partial negative charge it themore in its vicinity.the more partial posi arge it be to compare colors fro t maps.how e co anic chem oms with opposite charges or polarities are attra h other.Then. lectrons mo up of electrons.it's necessary to keep track of how many electrons are involved.and where they are located. Lewis structres are of paramount importance in this bookkeeping process. 1-4.Lewis Structures Whether you've ever done Le wis structures before or not.follow the rules in Section 1-4 very closely.Becom aroun mon a or the that you will quickly and confidently be able to picture a Lewis structure for any y of the tyres of species vo u gain confidence through practice.you will be able to use shorthand notations nic compounds and other anic or ing anic species Thes kkeeping system to help us keep track of electrons in reactions 1-5. Resonance Forms arrows are in kind of n is the se of oe-eade play in organic chemistry.As shown in this section,many species have structures that cannot be represented hich can only b descr inte▣ nnature betweent wo or more co a molecule by drawing the resonance forms separated by double-beaded arrows and enclosed in brackets.The eis called the resonance hybrid.The only difference between the reson ce forms is a differen one Lewis structure.In cases like this you need to be only and if t The second convention in Section 1-5 is the use of show the movement of electron pairs In this section the only gplicationisinsho g ho w the elec se arrows you leam and understand or on movement The guidelines in the textbook for determining e degree to which each resonance form contributes to the actual structure are the ones you will use the most when dealing with the most common atoms in organi
Keys to the Chapter 1-2 and 1-3. Coulomb Forces; Bonds “Unlike charges attract” and “like charges repel.” These consequences of elementary physics dealing with electrostatics and Coulomb’s law are central to a basic understanding of chemistry. Not only do they determine whether, and how strongly, atoms will bond to each other (as described in this chapter), but they also influence an even more complicated process: whether two molecules are likely to react with each other. Time and time again we will return to simple electrostatics, in the context of the properties of the individual elements, to explain the reactions of organic chemistry. Most organic molecules contain polarized covalent bonds. In bonds of this type, one or more pairs of electrons are shared between two atoms, but because of an electronegativity difference between the atoms, the bonding electrons tend to be closer to the more electronegative atom, thereby creating a partial charge separation. In general, for A less electronegative than B, we have A : B. See the specific examples in Section 1-3. The computer-generated electrostatic potential maps in Section 1-3 are visual representations of relative charge distribution and, therefore, bond polarity: In any such map, the more red in the vicinity of an atom, the more partial negative charge it contains, and the more blue in its vicinity, the more partial positive charge it bears. (Be careful not to compare colors from different maps, however. The colors have been scaled for each map individually to bring out even small polarity differences as clearly as possible.) As you will see later on, most of the reactions in organic chemistry follow a general pattern. First, two nonbonded atoms with opposite charges or polarities are attracted to each other. Then, electrons move from the “electron-rich” to the “electron-poor” atom to form a new covalent bond between them. Because bonds are made up of electrons, it’s necessary to keep track of how many electrons are involved, and where they are located. Lewis structures are of paramount importance in this bookkeeping process. 1-4. Lewis Structures Whether you’ve ever done Lewis structures before or not, follow the rules in Section 1-4 very closely. Become familiar with the number of electrons around common atoms and the common arrangements of these electrons in the bonds of molecules. This familiarity, brought about by doing lots of examples, is the best way to ensure that you will quickly and confidently be able to picture a Lewis structure for any of the types of species you will encounter later on. As you gain confidence through practice, you will be able to use shorthand notations, such as lines instead of dots for bonding electron pairs. Organic chemistry involves reactions between organic compounds and other organic or inorganic species. These reactions can involve both bond-breaking and bond-forming processes, and the key to both is the movement of electrons. Lewis structures provide the bookkeeping system to help us keep track of electrons in reactions. 1-5. Resonance Forms Two important conventions involving arrows are introduced in Section 1-5. The first is the use of double-headed arrows between resonance forms. This is a special kind of notation because of the special role resonance forms play in organic chemistry. As shown in this section, many species have structures that cannot be represented by a single Lewis structure. They can only be described as intermediate in nature between two or more contributing forms, each of which by itself is an incomplete picture of the molecule’s structure. We represent such a molecule by drawing the resonance forms separated by double-headed arrows and enclosed in brackets. The true structure is called the resonance hybrid. The only difference between the resonance forms is a different location for the electrons from one to the next. The same geometrical arrangement of the atoms is maintained in all the resonance forms. Molecules that actually exist as resonance hybrids are often represented by only one Lewis structure. In cases like this you need to be aware of the fact that this is a shortcut used for convenience purposes only and that the real structure is still the resonance hybrid—the other resonance forms are implied even if they aren’t written down. The second convention in Section 1-5 is the use of curved arrows to show the movement of electron pairs. In this section the only application is in showing how the electron pairs shift in going from one Lewis structure of a resonance hybrid to another. Pictorial descriptions of electron movement using these arrows will be very useful tools to help you learn and understand organic chemistry. The guidelines in the textbook for determining the relative degree to which each resonance form contributes to the actual structure are the ones you will use the most when dealing with the most common atoms in organic 2 • Chapter 1 STRUCTURE AND BONDING IN ORGANIC MOLECULES 1559T_ch01_01-17 10/22/05 1:48 Page 2
1559T_ch01_01-1710/22/051:48Page3 ⊕ EQA Keys to the Chapter·3 chemis ALWAYS TRUE (these are your reminders): 1.Individual contributing resonance formsdoexist.Only the resonance ybrid.which isa"weighted 2.Altrage"or nd of s have the same total number of valence electron and the same total charge their valence shell in any .e geometry and set of atomic positions for the actual chemical substance being represented. USUALLY TRUE(these are mostly implied by the guidelines in the textbook-we're just spelling them out): differ ony in the positions ofand/or nonbonding elctos Thebond stay put. ane form into another h tron de rs from places where there is an 3. ms witl are u ally more im t contributors than resonance usually more important contributors. )and below(r. 1)are nor limited to octets 1-6,1-7,and 1-8.Orbitals Atomic orbitals are a convenient way to represent the distribution of electrons in atoms.Note that the and ced with the the char ital an alterative to the Lewis electron-dot method for umber of m nbing a bond is a onding atomic o ls and will give rise to st Hybrid orbitals are d ved by mixing atomic wave functions.They are used to explain the geo several ac is long e ny by contributing to bonding.The participation of different numbers ofs and porbitals in the hybridization al. ange of bor angles.thereby permitting electron pairs to get as far away from each other as pos- Keep in mind some points for bookkeeping with hybridization.If an atom starts with ones orbital and three p orbitals,it will always end up with a total of four orbitals,no matter how they have hybridized for bonding contains 50%s and 50 character,whereas an sp2hybrid iss andp in na ature fore hybric 1.sp hybridized atom (linear geometry):contains two sp orbitals (each one iss andp in character)and two ordinary p orbitals 2(s+D)+2D=1s+3
chemistry. It is somewhat abbreviated. Some additional considerations regarding resonance forms (and some reminders of the basic rules) follow. ALWAYS TRUE (these are your reminders): 1. Individual contributing resonance forms do not exist. Only the resonance hybrid, which is a “weighted average” or blend of the contributing forms, is real. 2. All resonance forms of a single chemical species must have the same total number of valence electrons and the same total charge. 3. Second-row atoms (i.e., up through neon) can never exceed an octet in their valence shell in any resonance form. In other words, the rules for drawing Lewis structures apply to drawing resonance forms. 4. Atom positions and geometries do not change from one resonance form to another—there is only one geometry and set of atomic positions for the actual chemical substance being represented. USUALLY TRUE (these are mostly implied by the guidelines in the textbook—we’re just spelling them out): 1. Resonance forms differ only in the positions of and/or nonbonding electrons. The bond electrons normally stay put. 2. We convert one resonance form into another by moving electron pairs from places where there is an excess of electron density to places where there is an electron deficiency. 3. Resonance forms with the most covalent bonds are usually more important contributors than resonance forms with fewer covalent bonds (but don’t forget that octets take priority). 4. Resonance forms with fewer charged atoms are usually more important contributors. 5. Atoms in the third row (P, S, Cl) and below (Br, I) are not limited to octets of electrons in their outer shells. In fact, Lewis structures with 10 or 12 valence electrons are frequently written for these elements. 1-6, 1-7, and 1-8. Orbitals Atomic orbitals are a convenient way to represent the distribution of electrons in atoms. Note that the and signs associated with parts of these orbitals do not refer to electrical charges. They refer to mathematical signs of functions (wave functions) associated with the distribution of the electrons. Molecular orbitals are similar but are spread out over more than one atom. They provide an alternative to the Lewis electron-dot method for picturing bonds. The number of molecular orbitals involved in describing a bond is always exactly equal to the number of atomic orbitals contributed by the individual atoms. Overlap of atomic orbitals results in bonding, antibonding, and sometimes also nonbonding molecular orbitals. Bonding orbitals are always lower in energy (more stable) than the original constituent atomic levels, and antibonding orbitals are always higher in energy. Thus bonding electrons will be more stable than electrons in nonbonding atomic orbitals and will give rise to strong bonds. Electrons in antibonding orbitals will reduce bonding. Hybrid orbitals are derived by mixing atomic wave functions. They are used to explain the geometrical shapes of molecules. Hybridization provides several advantages for bonding. With the larger lobe of the hybrid orbital located in between the bonded atoms, more electron density is located where it can “do some good” by contributing to bonding. The participation of different numbers of s and p orbitals in the hybridization allows a wide range of bond angles, thereby permitting electron pairs to get as far away from each other as possible and minimizing unfavorable electrostatic repulsion. Keep in mind some points for bookkeeping with hybridization. If an atom starts with one s orbital and three p orbitals, it will always end up with a total of four orbitals, no matter how they have hybridized for bonding purposes. Depending on the ratio of atomic orbitals used in the hybridization, we can describe the resulting hybrid orbitals as consisting of certain percentages of “s character” and “p character.” For example, an sp orbital contains 50% s and 50% p character, whereas an sp2 hybrid is 1 3 s and 2 3 p in nature. The total s and p character around an atom after hybridization always equals that which was present in the orbitals before hybridization: 1. sp hybridized atom (linear geometry): contains two sp orbitals (each one is 1 2 s and 1 2 p in character) and two ordinary p orbitals 2( 1 2 s 1 2 p) 2p 1s 3p Keys to the Chapter • 3 1559T_ch01_01-17 10/22/05 1:48 Page 3
1559r_ch0101-1710/22/051:48Page4 EQA 4.chapter 1 STRUCTURE AND BONDNG IN ORGANIC MOLECULES 2.sphybridized atom (trigonal planar geometry):contains three sporbitals (each one iss andp in character)and one ordinary p orbita 3附s+p)+p=1s+3p 3.sphybridized atom(tetrahedral geometry):contains four sporbitals (each one iss andp in character) 4哈s+p)=1s+3p So in all case exactly four orbitals are pre sent and add up to the equivalent of one s and three p.even though each type of hybridization leads toa form of bonding and molecular shape very different from any of the o show,the mathema al nature of hybridiza on is very flexible.to maximize favorabl yourl o e the mea hen Thee r the biand ything e illh Solutions to Problems 21.(and 22 and 25 -see below a): (b):C::N Triple bond is needed to give octets for C and N atoms. eththe5 (d)H 9 o有 侧H是&0 HN:N::一H:N @::6:一N:6 atom 8.based on the electronegativities in the periodic table
2. sp2 hybridized atom (trigonal planar geometry): contains three sp2 orbitals (each one is 1 3 s and 2 3 p in character) and one ordinary p orbital 3( 1 3 s 2 3 p) 1p 1s 3p 3. sp3 hybridized atom (tetrahedral geometry): contains four sp3 orbitals (each one is 1 4 s and 3 4 p in character) 4( 1 4 s 3 4 p) 1s 3p So in all cases exactly four orbitals are present and add up to the equivalent of one s and three p, even though each type of hybridization leads to a form of bonding and molecular shape very different from any of the others. As these examples show, the mathematical nature of hybridization is very flexible, to maximize favorable bonding attractions and minimize unfavorable electron–electron repulsions. Prepare yourself to use the material in these sections: These are the basics, and everything else will build from them. Solutions to Problems 21. (and 22 and 25—see below) (a) (b) Triple bond is needed to give octets for C and N atoms. (c) Note that the availability of d orbitals allows S to be surrounded by a fifth electron pair. (d) (e) (f) Double bond between nitrogens. (g) A molecule with two double bonds. (h) (i) 22. The symbols and are written above or below the appropriate atoms in the answers to Problem 21 above. In each polar bond, the more electropositive atom is designated , and the more electronegative atom , based on the electronegativities in the periodic table. N N O N N O Major (O more electronegative than N) H N N N H N N N H H C C O H H N N H H H H C O C H H H H H H C N H Cl Cl S Cl Cl S O O Major (octets) Br C N Cl F 4 • Chapter 1 STRUCTURE AND BONDING IN ORGANIC MOLECULES 1559T_ch01_01-17 10/22/05 1:48 Page 4
1559T_ch01_01-1710/22/051:48Page5 ⊕ EQA Solutions to Problems.5 23.(a)H: Hydride ion.Contrast H(a proton)an HH) H (b)H:C:- A.C has an octet and a -l charge H A carbocation.C has only a sextet and a +l chargs H (d)H:C A carbon"radical."C is neutral.bonded to only three other atoms.and surrounded by 7 electrons :H The methylammonium cion.The product of CH,NH,HNH H (h)H:C:::C:- Another carbanion od) (H:0:O:H Hydrogen peroxide 24.How to begin?Look at each atom and compare the bonding pattem with simpler structures that are more familiar to you.Count bonds.Count electrons.This exercise will make it easier for you to recognize similar situations later (as in exams).Then do the formal charge determination as described in the text. (a)The oxygen has three bonds and a lone pair.What simpler species do you know that is similar? 6(group for O)-[3 (half of shared e in bonds)+(unshared e)]=+1 rmal"neutral species.Therefore,based on our models,left and rightwe can derive the solution n the center H (b)The double bond betv en C and o cha hing The the and will have a +1 formal charge: c=0-H
23. (a) Hydride ion. Contrast H (a proton) and HT (H atom). (b) A carbanion. C has an octet and a 1 charge. (c) A carbocation. C has only a sextet and a 1 charge. (d) A carbon “radical.” C is neutral, bonded to only three other atoms, and surrounded by 7 electrons. (e) The methylammonium cation. The product of CH3NH2 H. Compare NH3 H NH4 . (f) Methoxide ion. The product of ionization of methanol, CH3OH CH3O H. Compare H2O HO H. (g) A “carbene.” A neutral carbon, bonded to two other atoms, with only a sextet of electrons. (h) Another carbanion. Carbanions [(b) and (h)], carbocations (c), free radicals (d), and carbenes (g) are reactive species of high energy. They can, however, function as reaction “intermediates.” (i) Hydrogen peroxide. 24. How to begin? Look at each atom and compare the bonding pattern with simpler structures that are more familiar to you. Count bonds. Count electrons. This exercise will make it easier for you to recognize similar situations later (as in exams). Then do the formal charge determination as described in the text. (a) The oxygen has three bonds and a lone pair. What simpler species do you know that is similar? Hydronium ion is one, probably the simplest. We’ve already seen the formal charge determination 1 for the oxygen in hydronium ion, based upon the calculation 6 (group # for O) [3 (half of shared e– in bonds) 2 (unshared e)] 1 The oxygen in the species in this problem is analogous. The carbon has four bonds, just like methane, a “normal” neutral species. Therefore, based on our models, left and right, we can derive the solution in the center: (b) The double bond between C and O changes nothing. The count is the same: carbon has four bonds and will be neutral, as in methane, and oxygen has three bonds and a lone pair, as in hydronium, and will have a 1 formal charge: C O H H H H H H O H H C O H H H H H H C H H H O O H C C H H C H H C O H H H H C N H H H H H C H H H C H H H C H H Solutions to Problems • 5 1559T_ch01_01-17 10/22/05 1:48 Page 5
1559Tch0101-1710/22/051:48Page6 6.chapter 1 STRUCTURE AND BONDNG IN ORGANIC MOLECULES 4 (group for C)-13 (half of shared c in bonds)+2 (unsharcd c)l=-1 (d)Proceed as for part(a).by considering simpler.analogous species.Suitable examples are ammo ds,an H-NHH-N-HH-g-H 3 (group for B)-[4 (half of shared e in bonds)]=-1 Therefore.we can arrive at the solution (below,center): H- H-0-H B-0 H-B-H H-6 (f)The oxygens are the same as in water,no problem there.The nitrogen is unusual:With two bonds and one lone pair it has no familiar analogs.Let's do the math: 5(group for N)-[2 (half of shared e-in bonds)+2(unsharede]=+1 The answer is H-6---H. 25.(a)(i)and (ii)Don't move any atoms!Resonance forms differ only in the location of the electron 0 gives .gives HO○o: HOO: favorable as a conributor to the hybrid.The first and third forms have only one charged atom and are the major contributors -H 沙1 (b)Construct on able Lewis structure first:HH All the
(c) A new system, but not complex. You do not have a simpler species for comparison, so just do the calculation. 4 (group # for C) [3 (half of shared e in bonds) 2 (unshared e)] 1 It is a carbon anion, or carbanion. It is isoelectronic—it has the same number of valence electrons (5)—with neutral ammonia and positive hydronium ion. (d) Proceed as for part (a), by considering simpler, analogous species. Suitable examples are ammonium ion for nitrogen with four bonds, and water for oxygen with two bonds and two lone pairs. Thus, we can write the answer (below, center): The species is called hydroxylammonium ion. (e) All three oxygen atoms are “normal”: two bonds and two lone pairs, and therefore all three are neutral. What about boron? We have seen two relevant examples: borane, BH3, neutral boron with three bonds, and borohydride, BH4 – , boron with four bonds and a negative charge, based upon the calculation 3 (group # for B) [4 (half of shared e in bonds)] 1 Therefore, we can arrive at the solution (below, center): (f) The oxygens are the same as in water, no problem there. The nitrogen is unusual: With two bonds and one lone pair it has no familiar analogs. Let’s do the math: 5 (group # for N) [2 (half of shared e in bonds) 2 (unshared e)] 1 The answer is . 25. (a) (i) and (ii) Don’t move any atoms! Resonance forms differ only in the location of the electrons. The forms shown place the negative charge on two of the oxygen atoms. Continue with a Lewis structure that places the charge on the third oxygen: (iii) All three Lewis structures have octets on every large atom, but the middle structure has three charged atoms and two instances of plus–minus charge separations, making it relatively less favorable as a contributor to the hybrid. The first and third forms have only one charged atom and are the major contributors. (b) Construct one reasonable Lewis structure first: . All the atoms are neutral, so we can move electron pairs in a couple of ways to see what we get. Let’s begin by moving a pair of N H H O H C O O O O HO O HO O HO gives gives H O N O H H O H O H O H O B H H H B H H H H H N H H H H N O H O H 6 • Chapter 1 STRUCTURE AND BONDING IN ORGANIC MOLECULES 1559T_ch01_01-17 10/22/05 1:48 Page 6
1559T_ch01_01-1710/22/051:48Page7 EQA Solufions to Problems·7 s from the double bond.Which way?Doesn't ma e them and see what you geIfit's something reasonable,fine.If it isn'tit isn't.So,move the pair toward nitrogen: 9-H 9-H 一 on oxygen to ors that on't the rul of them are (e)Now we have a negatively charged atom.Move electrons away from i 26.Resonance structures for Problem 21 (c).(h).(i)have already been shown.Two other species have 而：：一减C H:C::C:0: (Carbon sextet) Resonance forms may be drawn for (b)and (e)of Problem 24-the structures containing double
electrons from the double bond. Which way? Doesn’t matter—just move them and see what you get! If it’s something reasonable, fine. If it isn’t, it isn’t. So, move the pair toward nitrogen: Well, at least the negative charge is on the more electronegative atom (N). But we’ve separated opposite charges and lost the octet on carbon, so this new resonance form is unlikely to be a major contributor. What if we shift electrons the other way? Now we get something really hideous, having lost nitrogen’s octet and given it a positive charge. However, we can use a lone pair on oxygen to form a double bond with the nitrogen atom, giving the nitrogen back its octet: We’re not getting anything to write home about here, folks. Just because you can draw resonance forms that don’t violate the rules of bonding (like exceeding octets) doesn’t mean any of them are going to be any good. The original Lewis structure, with all neutral atoms, represents this compound best. The rest of the resonance forms are at most only marginal contributors. (c) Now we have a negatively charged atom. Move electrons away from it: Notice that it’s necessary to move an electron pair from the CPN double bond onto the C in order to avoid exceeding an octet on nitrogen. In both forms, all atoms (besides H) have octets. The only difference is the location of the negative charge: it’s better on O (more electronegative than C). So the first Lewis structure is better. 26. Resonance structures for Problem 21 (c), (h), (i) have already been shown. Two other species have additional resonance forms, as shown below. In each case the one below is not nearly as good as that shown in the answer to Problem 21, for the reasons given. (b) (g) Resonance forms may be drawn for (b) and (e) of Problem 24—the structures containing double bonds. (It is always possible to draw a resonance form for a structure with a multiple bond, although the resonance form you get is not necessarily a major contributor.) H H C C O (Carbon sextet) Br C N N Br C (Separation of charge) (Carbon sextet) N H H O C N H H O C N H H O H C N H H O H C N H H H C O N H H O H C N H H O H C Solutions to Problems • 7 1559T_ch01_01-17 10/22/05 1:48 Page 7
1559r_-h0101-1710/22/051:48Page9 8.chapter 1 STRUCTURE AND BONDNG IN ORGANIC MOLECULES H >C6-H一 H (e)H- H-6 B-CO H-6 H-0 2.a6cN一6：C H:H一 Equivalent,major Not as good (oxygen sextet) o是是一是墨 060:一00：0：d:6: Major Major N: H Major
(b) (e) 27. (a) In some of these answers, additional, less favorable forms are shown for comparison purposes. (b) (c) (d) (e) Identical. (f) (g) H H H H C N O (Positive oxygen) (Carbon sextet) Major H H H H C N O H H H H C N O O Cl O O Cl O O Cl O Major Major H H H H H C C C H H H H H C C C Equivalent, major Not as good (oxygen sextet) O O O O O O O O O O O O C O H N H H Major (no separation of charge) C O H N H H H C N H O H H C N H H H C Major (negative charge prefers the more electronegative atom—nitrogen) H C N H H H C O C N N O C Major (negative charge prefers the more electronegative atom—oxygen) H O O H O B H O H O B (Minor contributor: boron lacks an octet) Move electrons away from the negatively charged atom O C H H H C O O H H H (Minor contributor: carbon lacks an octet) Move electrons toward the positively charged atom 8 • Chapter 1 STRUCTURE AND BONDING IN ORGANIC MOLECULES 1559T_ch01_01-17 10/22/05 1:48 Page 8
1559T_ch01_01-1710/22/051:48Page9 ⊕ EQA Solufions to Problems·9 H (Carbon sextet) H 0 H-C-N H oxygens)and we've only shown 12in these 6bonds.We could use the remaining 12 to add three lone pairs to each O.Let's do that.and then figure out the formal charges on the atoms: H H-C- It's a"legal"Lewis structure,we've violated no rules,and we've satisfied the octets for O.but the N is in trouble with only a sextet and a2+charge.Can we make this better?Let's move an elctron pair away from a negative atom and toward a positive one and see what we get. H: H-C-N 日: No double bonds are switched,along with the negative charge: H 一H-C-N H(: : Could we move two electron pairs in toward the N.one from each O?Nope:That would violate the octet rule on N and give us an illegal Lewis structure: H o: H-C H-c H(: ILLEGAL!
(h) 28. Before starting, notice that the last sentence of the problem tells you how the atoms are attached: both compounds have two NOO bonds. So the N is in the middle of nitromethane. We begin with bonds: So far, the valence shells of the carbon and the hydrogens are filled, but the nitrogen and the oxygens are short. But we have 24 electrons to work with (3 from hydrogens 4 from C 5 from N 12 from the oxygens) and we’ve only shown 12 in these 6 bonds. We could use the remaining 12 to add three lone pairs to each O. Let’s do that, and then figure out the formal charges on the atoms: It’s a “legal” Lewis structure, we’ve violated no rules, and we’ve satisfied the octets for O, but the N is in trouble with only a sextet and a 2 charge. Can we make this better? Let’s move an electron pair away from a negative atom and toward a positive one and see what we get. Now we’re doing better: N has an octet, too. We could, of course, have moved an electron pair from the other oxygen instead. The result is identical to what we just got above, but the NOO single and NPO double bonds are switched, along with the negative charge: Could we move two electron pairs in toward the N, one from each O? Nope: That would violate the octet rule on N and give us an illegal Lewis structure: H H H C N O O H H H C N O O ILLEGAL! Octet rule violated— 10 electrons on N H H H C N O O H H H C N O O H H H C N O O H H H C N O O H H H C N O O H H H C N O O H H H C C N O H H H C C N O H H H C C N O (Carbon sextet) Major (oxygen more electronegative than carbon) Solutions to Problems • 9 1559T_ch01_01-17 10/22/05 1:48 Page 9
1559rch0101-1710/22/051:48Page10 10 chapter 1 STRUCTURE AND BONDING IN ORGANIC MOLECULES +8 H-C- H-0 H Since these two forms are identical.they contribute equally to the resonance hybrid.The NO bonds are polar,with a full positive charge on N and a negative charge split,half on eachO. may as hat of al d have hapeN kat the Our starting Lewis structure (below.left)would then have an octet on the Nand one en.but a sextet on H-C- ⊕ cedure,we begin with only single bonds,and then arbitrarily add in the remaining electrons as lone pairs.taking care only to avoid violating the octet rule.The charged O at the end: H-( 6-:一H-C--N=Q: H This i od:All nonby atom with at leas one pair 2y一=Y-艺 Applying this pattem to methyl nitrite.we get H : 一H-C-=N- H The result is the second-best resonance form- -okay for octets.but charges are separated.which makes it less of a contributor than the Lewis structure on the left.The hybrid will more closely resemble the left-hand
So the best two structures are the two we derived above, with octets on all nonhydrogen atoms and a pair of charges. The arrows below show how electron pairs shift to go from one to the other: Since these two forms are identical, they contribute equally to the resonance hybrid. The NO bonds are polar, with a full positive charge on N and a negative charge split, half on each O. You may ask, what would have happened back at the beginning of this exercise if we started by putting one of our extra pairs of electrons on the N, instead of putting all of them on the oxygens? Good question. Our starting Lewis structure (below, left) would then have an octet on the N and one oxygen, but a sextet on the other O. The remedy, shifting the lone pair from the N toward the electron-deficient O, gets us to one of the same final structures that we obtained above: As a rule, as long as all your electrons are in place, and you do not violate the octet rule with the rest, any starting structure will eventually get you to the best answer(s). Now we turn to methyl nitrite. Following the same procedure, we begin with only single bonds, and then arbitrarily add in the remaining electrons as lone pairs, taking care only to avoid violating the octet rule. The structure below left is the result, and it contains a seriously electron-deficient N, just as we found starting out with nitromethane. And, we “fix” it the same way, moving an electron pair “in” from the negatively charged O at the end: This is pretty good: All nonhydrogen atoms have octets, and none are charged. Can we find any other reasonable resonance forms? There is a common pattern described in the text for systems containing an atom with at least one lone pair attached to one of two atoms connected by a multiple bond. You move the lone pair “in” and move a bond “out”: Applying this pattern to methyl nitrite, we get The result is the second-best resonance form—okay for octets, but charges are separated, which makes it less of a contributor than the Lewis structure on the left. The hybrid will more closely resemble the left-hand H H H H C N O O H H C N O O X Y Z X Z Y H H H H C N O O H H C N O O H H H C N O O H H H C N O O H H H C N O O H H H C N O O 10 • Chapter 1 STRUCTURE AND BONDING IN ORGANIC MOLECULES 1559T_ch01_01-17 10/22/05 1:48 Page 10