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Answers to Selected Problems Answers to Selected Problems Chapter 1 1.2. 0.55 A 1.3. +-y-.z- ++ y+, z+ t-t-y+,~+ -!-!y+.z+ *x+ lx+ t-x+ *s -ls i- s- 1- s+ -l s+ t- s+ BeO triplet CO singlet NO doublet 1.5. "Ionic bonds" using 6p electrons are stronger than those using 6s electrons (Pb2+2Br- p Pb4 +4Br-). "Covalent bonds" with 6s electrons are stronger than those with 6p electrons (PbMe4 > PbMe2)' The energy gap between 6p and 6s orbitals is larger than that for 2p and 2s orbitals. 1.6. SOl = Cda + b + c + d + e + f) S02 = C2(a + b - c - d + e +f) S03 = C3(a - b + c - d + e + f) 541 542 Answers to Selected Problems S04 = C4 (a + b + c + d - e +f) SOs = Cs(a + b - c - d - e +f) SOb = C6 (a - b - c + d - e + f) 1.9. B. c. D d. 0 1.10. There is no vibrational distortion of a homonuclear diatomic that can lower its symmetry. Chapter 2 2.2. Charge density qi = 1; n-bond order Pii = I. 2.3. Il.E = 1l.E,. + IlE,; 1l.E, = 2(1 - i 12) {J. 2.4. Second-order perturbations are not additive. In double union of even AHs, the second­ order terms are large and cannot be ignored. Chapter 3 3.1. 4 J -jill jm Answers to Selected Problems 543 3.2. a. Ii) I1E n = 2-=( 2 + --3) {J = 0.92{J " ,,/17.7 j17.7 I1Enp = 2 ( --~- + 1 ) {J = 1.28{J j17.7 )17.7 (ii) I1En, = 2 ( -~= + ~-- ) {J = 0.84{J )20.7 fti0 (iii) I1Enp = 2( -- 6- + -~){J = 1.47{J ,,/17.7 v17.7 (iv) I1En, = 2 ( -/ + 2 ) {J = 1.35{J ..;20.7 )20.7 Thus (iii) > (ii) > Ii): (iv) > (ii). b. Ii) (ii) (iii) (iv) (i) > (ii) > (iv):> (iii). ;- 3.4. I1E n values: (i) :2 (I - 3 v 17) fJ = 0.54{J. (ii) 2 (I :1, 20) {J = 0.66{J. (iii) 4.20fi + 2(1 - 3/" 20){J = 0.86{J. 544 Answers to Selected Problems (iv) 4/17/3 + 2(1 - 3/)17)/3 = 0.78/3. (v) 4;20/3 + 2(1 - 3;,/20)/3 = 0.86/3. (vi) 1 17/3 + 2(1 - 3/v 17) /3 = 0.60/3. 3.5. a. > b. < c. > d. < 3.6. x = 0.471 /3 = 0.471 /3 x = 0.638 TsX -t;x -x ' fs-x f '-LI"X x = 0.484 NorrD-- rrx "hx x = 0.433 3.7. For example. ratio of numbers of classical structures: NBMO coefficient, cc6 2:(2/)42) Answers to Selected Problems 545 l l/j4I) ~ ~ . cCO~ ~ ~ I o:x) 1:(1/)42) However, 3.8. -p ~ ~ > I > ~ ~ ct:o~ # # cSM. = (8/7) P 0 Chapter 4 4.2. They are stabilized by mutual conjugation. 4.3. CHr-OH CHz E01 6 + 2H zSO. ~ 6 + H30+ + 2HSO.- H 0 EO, 0+ 02S0• ~ G+DSO,- EE 20 ~-- D=o Br Li+ I 6 + BuLi ~ o +B~Br 546 Answers to Selected Problems 4.4. a. tlErr{ionization) (i) 2(1 - 3/Jl7) Ii = 0.541i CH20H (ii) ~I 2{1 - 5/}46) Ii = 0.531i cB/" ~ I § (iii) 2(1-5/fo)Ii=0.60Ii (iv) 2(1 - 4/}34) Ii = 0.63tJ (V) 2(1 - 3/fo) Ii = 0.691i (vi) 2(1 - 3/,,118) Ii = 0.591i Thus the order of equilibrium constants is (vi> (iv) > (iii) '> (vi) > (i) '> (ii) b. The anionic equilibrium constants should follow the same order because in the ab­ sence of heteroatoms, the change in 1t energy for anions, cations, and radicals should be the same. 'R 4.5. a. S' -GI Ill, / (i) All bond lengths are in units of the standard bond length (1.40 AI. Iii) The field effect Fs is obtained on the assumption that the substituent acts as a dipole with charge 4 at atom i and charge - 0.94 one standard bond length from i along the bond to the substituent. The magnitude of the charge is accounted for by the Fs parameter. Thus Answers to Selected Problems 547 for p-F, 0- ¢ F FF = 4.85, F F = 4.85 ( ! _ Q2) = 0.525 rim 3 4 (iii) q'.m is the formal negative charge at the reaction site when the substituent is replaced by CH2 -; for the p-F this is 1/7. Thus, (iv) The q's in the term for the mesomeric field effect are obtained in the same way. :r\\ 2.65 .tmd.ro bood 1m,"" 1. ~ ~ ,.x 0~ 7X F MFF L ;i' = -0.577(~ +~) = -0.062 .. m km 2.65 2.65 a F 1.4 = 0.525 - 0.30 - 0.062 = 0.16 a l ,4: Br = 0.23, CH3 = -0.11 a l .3 : CI = 0.50, CH30 = 0.12 b.1og(KF IKH ) = pa (- pK F ) - (- pK H ) = 1.74(0.16) = 0.27 pK r = 12.80 - 0.27 = 12.53 (calc) = 12.82 (obs). 4.6. These will follow the reactivity numbers, with appropriate statistical corrections: N, = 2(ao, + (/0,)' 4.7. This is a long exercise. The structure of the dimer is not known. 4.8. Increasing as should increase the half-wave potential in a linear way. The reactive center should be the center of the ring. 4.9. This is an OE, reaction; the change in It energy is given by /'iE. = 2(1 - ao,) p. 548 Answers to Selected Problems Chapter 5 5.1. This reversal illustrates the conflict between the -/ effect of alkyl groups and polarization stabilization of charge. For carbanions. the two effects oppose each other in the gas phase: for cations. they are in the same direction. In solution. the polarizability effects on relative stability are much less important because the medium is highly polarizable, and access of the solvent to the charged center is more important than charge polarization within the molecule. 5.3. Product B would be favored by increasing the solvation power of the medium. 5.3. This question involves the effect of charge delocalization in the transition state on relative solvent effects. The Elcb transition state has much more carbanionic character than that for the E2 reaction. Increasing solvation will favor Elcb reactions. 5.4. Since the BEP principle holds for these reactions. this can be answered in the context of Problems 4.4 and 3.4: H E E ../27 Nt = 4/.... 7 = LSI Nt = 12/ .... 127 = 2.31 N, = 12/.j5i = 1.68 Thus the relative rates are > 4 > 2; for nitration in acetic anhydride. the 1 isomer is formed fastest. Nt = 1.67 Reaction is fastest at the 2 position. 5.6. a. See Problem 4.9; the same principle applies since this is a BEP process. b. The least reactive molecule in the series will show the largest difference between Sv I and SN2 rates. Me0 2C 5.7. a. C02 Me I C ill ~CO'M' III C ~ I HU C02Me Answers to Selected Problems 549 b. d ~, N-N c. +- j/'~ LEi rI ~ /,,,.~ ··-=eN Ph X Ph I CH, CH, , -·-N Ph + N2 d. dJllH D 550 Answers to Selected Problems EE; e. ~< --. ~~ :XY yeO, yD, ""CD, EO! DE! ~~]D'I - ~ <==! + f. C AD, DD Y' D, 'CD, 2 "CH 2 Br Br Br y~/ H! (lJ (pJJr- g. ~ +l Br Br~ ',,-/ = ~ H Br Et:! -, ~-.::::::! m (\ A ~y Br CM" H Br Br o~ o~ EE; EE; ~ <===± h. Ht) \. j ~~~ \, H ~H -/'\ ]I nt '~ n! ~ H H ~ 1 H H~r H ~d a~ H OLL ~~ f50 Answers to Selected Problems 551 CF3 CF3 CF3 CF J I i. EE+ C 1 + ~ + III EE+ ~ I ~ ;6":t / C ~ 0 I I CF3 CF3 CF3 CF3 C02CH3 I ()JJ(CO,CH, C j. M! CH 2 I + III Ii I ~H2 I I (()~ ~ C ~ I C02CH3 C02CH 3 aC02CH3 m + 0> ~ C02CH3 5.S. a. NH z CI Br OH H b. H 20 -- ---> d ---H d Cis 5.12. Cl 5.14. Consider the 1,4 localization energies in the starting material and the intermediate. Chapter 6 6.1. The easiest way to examine the stabilities of open-chain and cyclic excited states using the PMO method is to compare the two by intramolecular union. llE. = 2p"/J,, 552 Answers to Selected Problems The problem is to determine the sign of Prs in the first excited state. The pairing theorem (p. 75) provides a direct basis for doing this. The AO coefficients in open-chain polymers are determined by symmetry and the magnitude of the coefficients on the terminal atoms increases as one approaches a. Thus for butadiene in its first excited state, the orbital occupation and symmetry are A--- S -+--LUMO A-+-1-HOMO S-*-ll- In this state, the 1,4 antibondingness of the singly occupied HOMO is exactly cancelled by the 1,4 bondingness of the singly occupied LUMO since the AO coefficients in these two orbitals are the same by the pairing theorem. The only remaining contribution to PIA is a positive one from the doubly occupied symmetric orbital. This will be the case in all 4n-polyene excited states because the magnitude of the terminal coefficients in­ creases on approaching a, so the symmetric (positive) contributions to P.-w will always be largest. Since P is negative, terminal intramolecular union will always decrease the energies of these systems, i.e., the cyclic systems are aromatic when compared to the excited states of their open-chain analog. 6.3. (See Ref. 22): o m leid+ 6.4.
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