1-38 0g=3-g:→0-g=9←→g=8=过 o)过=0-：→过-9=可 element,must adhere strictly to the octet rule. 1-39 #3 (a) #1 #2 CH-N -C-NH2 H+to #1 H+to #3 、H+to#2 H NH N CH-N- H H no other resonance forms H no other resonance forms NH NH CH-N=C-H,→cH,-N -a,←→cH,-N-C= H (b)Prot interpret as being more basic. 1-40 aCH,-E-C=N:←→CH,-C=C=N H minor l) :0: 0 (b)CH3-C= t-CH→CH,- C=C-CH3→CH,-C-C=C-CH HH HH HH minor minor gpn 19
1-38 . . . . .. (a) O==S-O: .. • :O-S==O .. • O==S==O + + (b) 0=0-0: .. • :0-0=0 + + (c) The last resonance form of S02 has no equivalent form in 03. Sulfur, a third row element, can have more than eight electrons around it because of d orbitals, whereas oxygen, a second row element, must adhere strictly to the octet rule. 1-39 (a) H NH #3 � .. #1 NH #2 � •• II •• / CH3 -N-C-NH2 I H I H+to#2 H+to#3 � NH + II + I II CH3-N- C-NH2 + NH2 II CH3-N - C - NH3 I H I H CH3 - � -C-NH2 no other resonance forms H t no other resonance forms NH2 NH2 NH2 + I I I + CH3 - N = C - NH2 .... ... t---'l... CH3 - N - C - NH2 .... ... t---'l.... CH3 - N - C = NH2 I I + I H H H (b) Protonation at nitrogen #3 gives four resonance forms that delocalize the positive charge over all three nitrogens and a carbon-a very stable condition. Nitrogen #3 will be protonated preferentially, which we interpret as being more basic. 1-40 (a) CH3 - C -C N: .... .. t----I .... CH3-C == C = N: I I •• H H minor major (negative charge on electronegative atom) . . - :0 : :0 : :0 : I + I II (b) CH3 - C==C-C-CH3 ...... t----I ... CH -C-C==C-CH ...... t----I ... CH3-C-C==C-CH3 I I 3 + I I 3 I I H H H H H H minor mmor 19 major-full octets, no charge separation
1-40 continued 0 :0: CH3-C- -C-CH3←→CH-C=C-C-CH→CH,-C C=C-CH3 H H minor major major negative charge on electronegative atoms-equal energy (d) H,-Cc=c-N=c-c=c-N=9cH-=c-c=N- HH: HHH:O: HHH: minor minor g限oegmon d that its resonance forms until they become second nature.The importance of understanding resonance forms cannot be overemphasized. cH,--c=c-N-:→cH-c=c--0的 HHH:O: HHH:O: NH2 NH2 e)CH,CH--iH·→CH,CH-C-H2→CH,CH-C=NH minor major-full octets major-full octets 1-4 equal energy H-c-HcH---a4→cH-=8-aH 1 H H no resonance stabilization more stable-resonance stabilized (b CH2=C-C-CH3CH2-C=C-CH3 CH2=C- -CH, HH HH more stable-resonance stabilized no resonance stabilization 20
1-40 continued (c) :0 : :0 : .. - :0: :0: .. - :0: :0: II - . . II CH3-C- ? -C-CH3 I II II I ..... ... 1-----1 ... CH3 - C == C - C - CH3 ..... .. 1-----1.. CH3 - C - C == C - CH3 I I H minor H H \ major major } �--------- ----�--_/ Y negative charge on electronegative atoms-equal energy (d) + CH3-C-C I I I I ==C-N==0 - •• • • - + + CH3-C==C-C-N=0 -CH3-C== C-C==N-0 : I I I I • • I I I I •• H H H :0: minor H H H :0 : H H H :0: minor major-negative charge on electrone gati ve atoms NOTE: The two structures below are resonance forms, varying from the first two structures in part (d) by the diff erent positions of the double bonds in the N02. Usually, chemists omit drawing the second form of the N02 group although we all understand that its presence is implied. It is good idea to draw all the resonance forms until they become second nature. The importance o/understanding resonance forms cannot be overemphasized. + •• - + CH3 - C - C == C - N -0: ..... .. f-- .� CH3 - C == C - C - N -0: I I I II • • I I I II •• H H H :0: H H H :0: + NH2 NH2 I II NH2 I + (e) CH3CH2 - C - NH2 .... .. I---J.,.. CH3CH2 - C - NH2 .... ... I---J.� + CH3CH2 - C == NH2 mmor \.. major-full octets major-ful l octet:; 1-41 (a) + y equal energy + + CH3-C- CH3 I CH3 - C - 0 - CH3 .... .. I---l.,� CH3 - C == 0 - CH3 I • • I •• H H H no resonance stabilization more stable-resonance stabilized �) + + CH2 == C - C - CH3 ..... .. f---l.,� CH2 - C == C - CH3 I I I I H H H H more stable-resonance stabilized 20 H I + CH2==C-C-CH2 I I H H no resonance stabilization
1-41 continued (e)H-C-CH3 H-E二cN:→H-C=C=N H H H no resonance stabilization more stable-resonance stabilized (d) CH2 C二CH CH, C-H H H more stable-resonance stabilized no resonance stabilization y H CH3-N-CH3 CH--CH CH-C-CH3 CH-C-CH3 cH-c-ct CH3-C-CH more stable-resonance stabilized no resonance stabilization 1-42 These pKa values from the text,Table 1-5,and Appendix 5 provide the answers.The lower the PKa thesiad most acidic NH3< CHOH CHCOOH H2S04 33 15.5 4.74 -5 and protonated acedic acid in Appendix 5 of the textbook.) least basic most basic CH;COOH HSO<CH;COO-<CH3O<NaOH <NH2 from-6.1 from-5 from 4.74 from 15.5 from 15.7 from 33 pKa =-logio Ka =-logio(5.2x 10-5)=4.3 for phenylacetic acid for propionic acid.pK4.87:Ka17=1.35x105 (b)phenylacetic acid is 3.9 times stronger than propionic acid 5.2x10-5 =3.9 1.35x10-5 (c) -CH2CO0+CHCH2COOH -CH2COOH CHCH2COO- weaker acid stronger acid Equilibrium favors the weaker acid and base.In this reaction,reactants are favored 21
1-41 continued (c) H-C-CH3 \ H H-C-C - N: \ H ... .. H-C=C=N: \ H no resonance stabi li zation more stable-resonance stabi lized (e) more stable-resonance stabilized CH3 - N 1 - CH3 CH3-C-CH3 + ..... ... I----I .. � + CH3 - � - CH3 CH3-C-CH3 + C c ./ CH2 \ ' H C C ...:/ ' H \ H no resonance stabilization H \ CH3-C-CH3 I CH3-C-CH3 + more stable-resonance stabilized no resonance stabilization 1-42 These pKa values from the text, Table 1-5, and Appendix 5 provide the answers. The lower the pKa ' the stronger the acid. least acidic most acidic NH3 33 < < < 1-43 Conjugate bases of the weakest acids will be the strongest bases. The pKa values of the conjugate acids are listed here. (The relative order of the first two was determined from the pKa values of sulfuric acid and protonated acedic acid in Appendix 5 of the textbook.) least basic from �6.1 from �5 from 4.74 from 15.5 from 1 5.7 1 -44 (a) pKa = � loglO Ka = � loglO (5.2 x 10 - 5) = 4.3 for phenylacetic acid for propionic acid, pKa 4.87: Ka = 10 - 4 .87 = 1 .35 x 10 - 5 (b) phenylacetic acid is 3.9 times stronger than propionic acid 5.2 x 10 - 5 = 3.9 1.35 x l0 - S most basic from 33 (c) < > CH2COO- + CH3CH2COOH ... weaker acid Equilibrium favors the weaker acid and base. -- < > CH2COOH + CH3CH2COOstronger acid In this reaction, reactants are favored. 21
1-45 The newly formed bond is shown in bold. (a) cH-:+,CH,- →CH-9-CH+:: nucleophile electrophile Lewis base Lewis acid ()CH0 CH H--H-CH-6-CHs +H-6-H electrophile Lewis Lewis acid (c) :o: H :8: H- H+:N-H H- -H H H-NH Eccegt H (d)CH3-NH2 CH;CH2-CI: CH-N-CH.CH +3. Lew t0: (e)CH3-C-CH3 H- 6一 -H→CH,-C-cH,+: -OH nucleophile Lewis base 0: CH3 0(CH,c∠： Hc-c++:Ci-AI-CI CH3 0: :o: ®CH,-CC+-H→C4-C=cH+H-g-H H Lew pe 22
1-45 The newly formed bond is shown in bold. . . - (a) . . - (' . . CH3- O: + CH3-CI: .. CH3- �- CH3 .. � .. + :CI: (b) (c) nucleophile Lewis base electrophile Lewis acid CH 3 - O-CH + H-O-H +1) 3 •• H3c�ucleoPhile I hO I Lewis base e ectrop Ie Lewis acid c�: � H-C '----/ 1 -H + :N-H H electrophile Lewis acid nucleophile Lewis base nucleophile Lewis base electrophile Lewis acid ·0· � :0: ' , , ' � " (e) CH 3 -C- CH 3 + H-O-S-OH �. " .. nucleophile Lewis base :0: electrophile Lewis acid nucleophile electrophile Lewis base Lewis acid This may also be written in two steps: the C-Cl bond breaks. :R� - . . CH3-?-CH3 + H +� -H CH3 .. - :0 : 1 H-C-H 1+ H-N-H 1 H H + 1 .. - CH3 -N -CH2CH3 1 + :CI: H + :O -H " :0: --t .. � CH3-C-CH3 + " :O-S-OH . . " .. CI 1 - + :Cl - Al - Cl 1 Cl :0: association of the CI with AI, and a second step where . . - : 0: I (g) CH3 - C- CH2 + : O-H '--I � J •• ---t .. � CH3 - C==CH2 + H-O-H H� nucleophile electrophile Lewis base Lewis acid 22
1-45 continued (h)F-B-F CH2=CH2→F-B-CH2-CH2 nucleophile electrophile Lewis base Lewis acid ①BF3-CH2一CH2+CH2=CH2+BF3-CH2一CH2-CH2-CH2 Lewi 1-46 (a)H2SO+CHCOO-HSO+CHCOOH (b)CH,COOH+(CH)3N:±CH,CO0+(CHN一H -C-0-H+OH±〈 C-0+H20 (d(CHhN一H+-OH±(CH3N:+H2O (e)H0-C-0H+20H=0-C-0+2H,0 (0H20+NH±HO+NH (g)HCOOH CH3O-HCOO-+CH3OH 1-47 (a)CH;CH2-O-H CH3-Li- →CH,CH2-OLit+CH (b)The conjugate acid of CH3Li is CH4.Table 1-5 gives the pK of CHa as>40,one of the weakest acids known.The conjugate base of one of the weakest acids known must be one of the strongest bases known. 23
1-45 continu� (h) F-B - F CH2 =CH2 I F electrophile Lewis acid nucleophile Lewis base • F - I + F - f- CH2 - CH2 F +� + (i) BF3 - CH2 -CH2 + CH2 = CH2 .. BF3 - CH2 - CH2 - CH2 - CH2 1-46 electrophile Lewis acid nucleophile Lewis base (a) H2S04 + CH3COO- HS04- + CH3COOH + (b) CH3COOH + (CH3hN: CH3COO- + (CH3hN - H o (e) < ) g - O-H + �OH o < ) g-o- + H20 + (d) (CH3hN - H + -OH o II (e) HO-C-OH + 2 -0H (f) H20 + NH3 (g) HCOOH + CH30- 1-47 (CH3hN: + H20 o II -O-C - O- + 2 H20 HO- + +NH4 HCOO- + CH30H (a) CH3CH2 -O-H + CH3-Li -----l.� CH3CH2 - 0- Li+ + CH4 (b) The conjugate acid of CH3Li is CH4 Table 1-5 gives the pKa of CH4 as > 40, one of the weakest acids known. The conjugate base of one of the weakest acids known must be one of the strongest bases known. 23