28 CHAPTER 2 Bonding and Molecular Structure Problem 2.31 Nitrogen trifluoride (NF)and ammonia (NH,)have an electron pair at the fourth corner of a tetrahedron and have similar electronegativity differences between the elements(1.0 for N and F and 0.9 for N and H).Explain the larger dipole moment of ammonia(1.46 D)as compared with that of NF(0.24 D). see Fig.2.13(b),and add to the effect of the electron pair. Net dipole momen is not known et mome a) (6 Figure 2.13 Problem 2.32 NHsalts are much more soluble in water than are the corresponding Na'salts.Explain. Na'is solvated merely by an ion-dipole interaction.NHis solvated by H-bonding HgN+-H--:6-H which is a stronger attractive force. Problem 2.33 The F-of dissolved NaF is more reactive in dimethyl sulfoxide: CHSCH and in acetonitrile,CH C=N.than in CH,OH.Explain H-bonding revails in CH,OH(a protic solvent).CH,OH---F-,thereby decreasing the reactivity of F- CH SOCH,and CH,CN are aprotic solvents:their C-HH's do not H-bond. Find the oxidation of the Cin(a)CH,Cl.(b)CH,Cl.(c)HCO.(d)HCOOH.and (e)CO. From Section 2.5: (a(ON)+(3×1)+(-1)=0: (ON)e=-2 (d(ON)e+2+(-4)=0:(ONc=2 (b)(ONe+(2×1)+[2(-1J=0:(ONe=0 (e)(ONe+(-4④=0 (ON)c=4 ((ON+2×1)+1(-2=0,(ONe=0
28 CHAPTER 2 Bonding and Molecular Structure Problem 2.31 Nitrogen trifluoride (NF3) and ammonia (NH3 ) have an electron pair at the fourth corner of a tetrahedron and have similar electronegativity differences between the elements (1.0 for N and F and 0.9 for N and H). Explain the larger dipole moment of ammonia (1.46 D) as compared with that of NF3 (0.24 D). The dipoles in the three N—F bonds are toward F, see Fig. 2.13(a), and oppose and tend to cancel the effect of the unshared electron pair on N. In NH3, the moments for the three N—H bonds are toward N, see Fig. 2.13(b), and add to the effect of the electron pair. Problem 2.32 NH+ 4 salts are much more soluble in water than are the corresponding Na+ salts. Explain. Na+ is solvated merely by an ion-dipole interaction. NH+ 4 is solvated by H-bonding Figure 2.13 which is a stronger attractive force. Problem 2.33 The F of dissolved NaF is more reactive in dimethyl sulfoxide: and in acetonitrile, CH3C ⎯N, than in CH3OH. Explain. H-bonding prevails in CH3OH (a protic solvent), CH3OH---F, thereby decreasing the reactivity of F. CH3SOCH3 and CH3CN are aprotic solvents; their C—H H’s do not H-bond. Problem 2.34 Find the oxidation of the C in (a) CH3Cl, (b) CH2Cl2, (c) H2CO, (d) HCOOH, and (e) CO2, if (ON)Cl 1. From Section 2.5: (a) (ON)C (3 1) + (1) 0; (ON)C 2 (d) (ON)C 2 (4) 0; (ON)C 2 (b) (ON)C (2 1) + [2(1)] 0; (ON)C 0 (e) (ON)C (4) 0; (ON)C 4 (c) (ON)C (2 1) + [1(2)] 0; (ON)C 0
CHAPTER 2 Bonding and Molecular Structure 29 Problem 2.35 Give a true or False a (a)Since i anions (such as SOand BF7).the central atom is usually the ipheral aton Y.it tends to acquire a positive oxidation number.(b)Oxidation numbers tend to be smaller values than formal charges.(c)A bond between dissimilar atoms always leads to nonzero oxidation numbers.(d)Fluorine never has a positive oxidation number. (a)True.The bonding electrons will be allotted to the more electronegative peripheral atoms,leaving the cen tral atoms with a positive oxidation number. (b)False.In determining formal charges,an electron of each shared pair is assigned to each bonded atom.In determining oxidation numbers ectrons are involved,and more electrons are moved to or away oman atom.Hence.lar pairs of el on nul Il be s resul ctronegativity,as in PH Problem 2.36 Which of the following transformations of organic compounds are oxidations,which are reductions,and which are neither? (@)H,C=CH, →CH,CH,OH()CH,CHO→CH,COOH (e)HC=CH →H,C=CH (b)CHCH,OH CH,CH=O (d)H,C=CH, →CH,CH,C To a h。 on dete ine the a nbers(ON)of the Cat o)eo 2.(b)and (c)are oxidations,the respective changes Problem 2.37 Irradiation with ultraviolet (uv)light permits rotation about a bond.Explain in terms of bonding and antibonding MO's. rons in the te). 共nd ae)立法 e two electrons cancel now only a sigma Problem 2.38 Write the contributing resonance structures and the delocalized hybrid for(a)BCl(b)H.CN, (diazomethane). 一民一一“良 H,C=N=:←→H,民-N=e=&N=0
Problem 2.35 Give a True or False answer to each question and justify your answer. (a) Since in polyatomic anions XYn m (such as SO2 4 and BF 4), the central atom X is usually less electronegative than the peripheral atom Y, it tends to acquire a positive oxidation number. (b) Oxidation numbers tend to be smaller values than formal charges. (c) A bond between dissimilar atoms always leads to nonzero oxidation numbers. (d) Fluorine never has a positive oxidation number. (a) True. The bonding electrons will be allotted to the more electronegative peripheral atoms, leaving the central atoms with a positive oxidation number. (b) False. In determining formal charges, an electron of each shared pair is assigned to each bonded atom. In determining oxidation numbers, pairs of electrons are involved, and more electrons are moved to or away from an atom. Hence, larger oxidation numbers result. (c) False. The oxidation numbers will be zero if the dissimilar atoms have the same electronegativity, as in PH3 . (d) True. F is the most electronegative element; in F2 , it has a zero oxidation number. Problem 2.36 Which of the following transformations of organic compounds are oxidations, which are reductions, and which are neither? (a) H2 C=CH2 CH3 CH2 OH (c) CH3 CHO CH3 COOH (e) HC ⎯CH H2 C=CH2 (b) CH3 CH2OH CH3CH=O (d) H2 C=CH2 CH3CH2 Cl To answer the question, determine the average oxidation numbers (ON) of the C atoms in reactant and in product. An increase (more positive or less negative) in ON signals an oxidation; a decrease (more negative or less positive) signals a reduction; no change means neither. (a) and (d) are neither, because (ON)C is invariant at 2. (b) and (c) are oxidations, the respective changes being from 2 to 1 and from 1 to 0. (e) is a reduction, the change being from 1 to 2. Problem 2.37 Irradiation with ultraviolet (uv) light permits rotation about a π bond. Explain in terms of bonding and antibonding MO’s. Two p AO’s overlap to form two pi MO’s, π (bonding) and π* (antibonding). The two electrons in the original p AO’s fill only the π MO (ground state). A photon of UV causes excitation of one electron from π to π* (excited state). CHAPTER 2 Bonding and Molecular Structure 29 ↑↓ π π * (ground state) ↑ ↓ π π * (excited state) uv (Initially, the excited electron does not change its spin.) The bonding effects of the two electrons cancel. There is now only a sigma bond between the bonded atoms, and rotation about the bond can occur. Problem 2.38 Write the contributing resonance structures and the delocalized hybrid for (a) BCl3 , (b) H2 CN2 (diazomethane). (a) Boron has six electrons in its outer shell in BCl3 and can accommodate eight electrons by having a B—Cl bond assume some double-bond character. (b)
30 CHAPTER 2 Bonding and Molecular Structure for mos (a) H,C2CHC:。→HC-CH=C:。→h,t-CH=G I1)>1(2)>Ⅲ(3) 0: H-C--H→H-6--H→H-g--H←→H-g-8-H I n V1)>VI(2)>Vm(3)>VIm(4 Problem 2.40 What is the difference between isomers and contributing resonance structures? Isomers and real compouds differ in the ar m ten to give some ndiction of the electronicstructure of ceran species for which a typical Lewisstructurecan- not be written. Problem 2.41 Use the HON method to determine the hybridized state of the underlined elements (a)HC=CH (b)HC=O (c)HC=C (d)AIC (e)PFs (f)CH;CH3 NUMBER OF GBONDS NUMBER OF UNSHARED ELECTRON PAIRS HON HYBRID STATE 3 0 2 32 1652 1002 654
30 CHAPTER 2 Bonding and Molecular Structure NUMBER OF σ BONDS NUMBER OF UNSHARED ELECTRON PAIRS HON HYBRID STATE 2 02 sp 3 03 sp2 1 12 sp 6 06 sp3d2 5 05 sp3d 2 24 sp3 Problem 2.39 Arrange the contributing structures for (a) vinyl chloride, H2 CCHCl, and (b) formic acid, HCOOH, in order of increasing importance (increasing stability) by assigning numbers starting with 1 for most important and stable. I is most stable because it has no formal charge. III is least stable since it has an electron-deficient C. In III, Cl uses an empty 3d orbital to accommodate a fifth pair of electrons. Fluorine could not do this. The order of stability is I(1) > II(2) > III(3) V and VI have the greater number of covalent bonds and are more stable than either VII or VIII. V has no formal charge and is more stable than VI. VIII is less stable than VII, since VIII’s electron deficiency is on O, which is a more electronegative atom than the electron-deficient C of VII. The order of stability is V(1) > VI(2) > VII(3) > VIII(4) Problem 2.40 What is the difference between isomers and contributing resonance structures? Isomers and real compounds differ in the arrangement of their atoms. Contributing structures have the same arrangement of atoms; they differ only in the distribution of their electrons. Their imaginary structures are written to give some indication of the electronic structure of certain species for which a typical Lewis structure cannot be written. Problem 2.41 Use the HON method to determine the hybridized state of the underlined elements: (a) (b) (a) (b) (c) (d) (e) (f )
CHAPTER 3 Chemical Reactivity and Organic Reactions 3.1 Reaction Mechanism eey (①A+I1+X followed by(②)B+I→Y Substances such as I.formed in intermediate steps and consumed in later steps,are called intermediates. Sometimes the same reactants can give different products via different mechanisms. 3.2 Carbon-Containing Intermediates r example A:B→A*+BoA:+B Homolytic (radical)cleavage.Each separating group takes one electron,for example: AB→A+B rcations are positively charged ions containing a carbon atom having onl six electrons in 31
CHAPTER 3 CHAPTER 3 31 Chemical Reactivity and Organic Reactions 3.1 Reaction Mechanism The way in which a reaction occurs is called a mechanism. A reaction may occur in one step or, more often, by a sequence of several steps. For example, A B X Y may proceed in two steps: (1) A I X followed by (2) B I Y Substances such as I, formed in intermediate steps and consumed in later steps, are called intermediates. Sometimes the same reactants can give different products via different mechanisms. 3.2 Carbon-Containing Intermediates Carbon-containing intermediates often arise from two types of bond cleavage: Heterolytic (polar) cleavage. Both electrons go with one group, for example: A:B A :B or A: B Homolytic (radical) cleavage. Each separating group takes one electron, for example: A:B A· ·B 1. Carbocations are positively charged ions containing a carbon atom having only six electrons in three bonds:
32 CHAPTER3 Chemical Reactivity and Organic Reactions 2.Carbanions are negatively charged ions containing a carbon atom with three bonds and an unshared pair of electrons: -c are just one example. mes are neutral species having a carbon atom with two bonds and two electrons.There are two in which the two electrons have opposit espins and are paired triplet in which the two electrons have the same spin and are in different orbitals 西o的皮 below,and give NUMBER OF + NUMBER OF UNSHARED -HON HYBRID STATE BONDS ELECTRON PAIRS 0 3 sp- 332 0, 2 G hdelctrons of the triplet cabene are not paired and.hence.re no counted they are Problem 3.2 Give three-dimensional representations for the orbitals used by the C's of the five carbon inter- mediates of Problem 3.1.Place all unshared electrons in the appropriate orbitals. (a)Acarboc HO'ofom,and an perpendicular trons.See Fig.3.1(b). (c)A radical has the same orbitals as the carbocation.The difference lies in the presence of the odd electron in the p orbital of the radical.See Fig.3.1(c)
2. Carbanions are negatively charged ions containing a carbon atom with three bonds and an unshared pair of electrons: 32 CHAPTER 3 Chemical Reactivity and Organic Reactions 3. Radicals (or free radicals) are species with at least one unparied electron. This is a broad category in which carbon radicals, are just one example. 4. Carbenes are neutral species having a carbon atom with two bonds and two electrons. There are two kinds: singlet in which the two electrons have opposite spins and are paired in one orbital, and triplet in which the two electrons have the same spin and are in different orbitals. Problem 3.1 Determine the hybrid orbital number (HON) of the five C-containing intermediates tabulated below, and give the hybrid state of the C atom. Unpaired electrons do not require an HO and should not be counted in your determination. NUMBER OF NUMBER OF UNSHARED HON HYBRID STATE σ BONDS ELECTRON PAIRS 3 03 sp2 3 14 sp3 3 03 sp2 2 13 sp2 2 02 sp (a) carbocation (b) carbanion (c) radical (d) singlet carbene (e) triplet carbene Recall that the two unshared electrons of the triplet carbene are not paired and, hence, are not counted; they are in different orbitals. Problem 3.2 Give three-dimensional representations for the orbitals used by the C’s of the five carbon intermediates of Problem 3.1. Place all unshared electrons in the appropriate orbitals. (a) A carbocation has three trigonal planar sp2 HO’s to form three σ bonds, and an empty p AO perpendicular to the plane of the σ bonds. See Fig. 3.1(a). (b) A carbanion has four tetrahedral sp3 HO’s; three form three σ bonds and one has the unshared pair of electrons. See Fig. 3.1(b). (c) A radical has the same orbitals as the carbocation. The difference lies in the presence of the odd electron in the p orbital of the radical. See Fig. 3.1(c).