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 ethem and see what you 9-H 9-H on oxygen ⊕ don't the rul 't mean any of them are in to b y of the resonance foms arat os rep (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) 6 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