115 (a)HCOOH +CN -→HCo0+HCN acionger FRODUCTS acid pK23.76 pKa 9.22 (b)CH:COO- + FAVORS REACTANTS pKa 15.5 pK4.74 (c)CH3OH + NaNH, CH3O-Na* +NH FAVORS ke PRODUCTS base ker aci pKa15.5 pKa33 (d)Nat -OCH+ HCN HOCH, NaCN FAvORS stronger ker PRODUCTS acid base pKa9.22 pK215.5 (e)HCI H2O -→H0*+C stronger weaker weaker base acid base The first reaction in Table 1-5 shows the Keg for this reaction is 160,favoring products. (f)HO+CHO-H20 CH:OH FAVORS stronger stronger weaker weaker PRODUCTS p-17 base base acid The seventh reaction in Table 1-5 shows the K for the reverse of this reaction is 3.2 x 10-16. Therefore. must be the inverse.or3.,strongly favoring Products. 1-16 CH-C-0-H +H' CH,-C- 6-H H Protonation -5(c)): on of the single ance forms (as s cso the proton would bond i the oxygen that gives a more more oxygen 9
1-15 (a) HCOOH + -CN .- -- HCOO- + HCN FAVORS (b) (c) (d) (e) (f) 1-16 stronger stronger weaker weaker PRODUCTS acid base base acid pKa 3.76 pKa 9.22 CH3COO- + CH30H .. -- CH3COOH + CH3O- FAVORS weaker weaker stronger stronger REACTANTS base acid acid base pKa 15.5 pKa 4.74 .- CH30H + NaNH2 -- CH30- Na+ + NH3 FAVORS stronger stronger weaker weaker PRODUCTS acid base base acid pKa 1 5.5 pKa 33 .- Na+ -OCH3 + HCN -- HOCH3 + NaCN FAVORS stronger stronger weaker weaker PRODUCTS base acid acid base pKa 9.22 pKa 15.5 .- HCl + H2O -- H3O+ + CI- FAVORS stronger stronger weaker weaker PRODUCTS acid base acid base The first reaction in Table 1-5 shows the Keq for this reaction is 160, favoring products. H30+ .- + CH3O- -- H2O + CH30H FAVORS stronger stronger weaker weaker PRODUCTS acid base base acid pKa-1 .7 pKa 1 5.5 The seventh reaction in Table 1-5 shows the Keq for the reverse of this reaction is 3.2 x 10- 16 . Therefore, Keq for this reaction as written must be the inverse, or 3. 1 x 1015 , strongly favoring products. :0: II CH 3 -C-O- H •• :0: II CH 3 -C- O-H +\ H Protonation of the double-bonded oxygen gives three resonance forms (as shown in Solved Problem 1-5(c)); protonation of the single-bonded oxygen gives only one. In general , the more resonance forms a species has, the more stable it is, so the proton would bond to the oxygen that gives a more stable species, that is, the double-bonded oxygen. 9
metanol moe We mu nhae po rem (a)This problem can be viewed in two ways.1)Quantitatively,the pK.values determine the order of acidity.2)Qualitatively,the stabilities of the conjugate bases determine the order of acidity(see Solved Problem acid is structures)the conjugate base of d,acetate ion,is resonance-stabilized,so gen at has er a mildly clectroncgative nitrogen atom and is therefore the least stabilizd so methylamine is the least acidic acetic acid ethanol methylamine pKa4.74 pKa=15.5 pKa=33 strongest acid weakest acid hoyaue h The basicity of methylamine is between the basicity of ethoxide ion and ethanol cthoxide ion methylamine cthanol 1-18 Curved arrows show electron movement.as described in text section 1-14 CHCH,-82H4CH一对H一→cHCH,-s+cH6-N-H stronger acid stronger base H conjugate acid equilibrium favors PRODUCTS weaker acid 0: (b)CH;CH2-C- 2HcH-N-cH←→cH,-N-cH+CH,CH-C-9 stronger acid stronger base coniugate acid equilibrium favors PRODUCTS weaker acid :o: CH;CH2-C=0 - H s--H-→CH-QH+ stronger base 0: stronger acid PRODUCTSavors 0: 0: -9 8之88"&起8-即 conjugate base.weaker base 10
1-l7 In Solved Problem 1-4, the structures of ethanol and methylamine are shown to be similar to methanol and ammonia, respectively. We must infer that their acid-base properties are also similar. (a) This problem can be viewed in two ways. 1) Quantitatively, the pKa values determine the order of acidity. 2) Qualitatively, the stabilities of the conjugate bases determine the order of acidity (see Solved Problem 1-4 for structures): the conjugate base of acetic acid, acetate ion, is resonance-stabilized, so acetic acid is the most acidic; the conjugate base of ethanol has a negative charge on a very electronegative oxygen atom; the conjugate base of methylamine has a negative charge on a mildly electronegative nitrogen atom and is therefore the least stabilized, so methylamine is the least acidic. acetic acid > ethanol > methylamine pKa 4.74 pKa"" 1 5.5 pKa "" 33 strongest acid weakest acid (b) Ethoxide ion is the conjugate base of ethanol, so it must be a stronger base than ethanol; Solved Problem 1-4 indicates ethoxide is analogous to hydroxide in base strength. Methylamine has pKb 3.36. The basicity of methylamine is between the basicity of ethoxide ion and ethanol. ethoxide ion > methylamine > ethanol strongest base weakest base 1-18 Curved arrows show electron movement, as described in text section 1-14. stronger acid stronger base equilibrium favors PRODUCTS '0' __ .. CH3CH2-�: conjugate base weaker base H . . + CH3-�-H H conjugate acid weaker acid '11' . � � •• (b) CH3CH2 -C-0-H + CH3 -N- CH3 • • I -- 1 + CH3- � - CH3 (c) stronger acid H stronger base equilibrium favors PRODUCTS • O' .. � 0. '11' •• CH3 - O-H + H-O -S-O-H • • II stronger base : 0·: equilibrium favors PRODUCTS stronger acid -- :O-S- O-H ..... ... l----I .. � i-·· :�: .. • • II •• : 0: H conjugate acid weaker acid H 1+ .. CH 3 - O-H •• conjugate acid weaker acid :0 : II O==S- O-H I : 0: .. + .. conjugate base, weaker base 10 conjugate base weaker base :0:- r . . , .. O==S-O-H • • II •• :0 :
1-18 continued (d) N泣H◆C-H =→H-g-H+Na-H stronger base stronger acid equilibrium favors PRODUCTS (e) H CH-NH+ CH-:CH-N-H CH-6-H stronger base conjugate acid stronger acid weaker acid equilibrium favors PRODUCTS 0: 0 c-C-2a-日-H+c--的 stronger acid stronger base equilibrium favors PRODUCTS CH-c= (g) 0 :8: cH,--2m 9-cH,←= -CH←→g= -CH3 weaker acid :0: 0: :0: equilibrium favors REACTANTS weaker base 0: ） 0: ,1cH,-c-←→cH-c=g∫+H-9- :0 conjugate acid stronger acid (⊙)H-BH+CH-文-CH±H-B-H acid H base : CH-C-H+-H=CH3-C-H acid base :9-H 11
1-18 continued (d) - .. � 0· Na+ :O -H + H-S-H .. stronger base stronger acid equilibrium favors PRODUCTS --- .. H -O-H + Na+ .. :S-H conjugate acid weaker acid conjugate base weaker base (e) H � I + /' __ " •• - CH3 - �� H + CH3-O: H stronger base --- CH3- � -H + H conjugate base weaker base CH 3 - O-H •• conjugate acid weaker acid stronger acid equilibrium favors PRODUCTS stronger acid stronger base conjugate acid weaker acid II •• - i :0: . . : 0: � I •• . ......... CH -C=O equilibrium favors PRODUCTS + CH3-C- 0: •• - t conjugate base 3 •• weaker base (g) :0: ~ :0: :0: :0:- } II .� -.. II II •• I CH - C - 3 O-H + :O-S-CH ..... .. f---l ... � 0==S-CH3 ..... .. f-- ... _ 0==S-CH3 .. •• II 3 •• I •• II weaker acid : 0 : : 0 : : 0 : weaker base equilibrium favors REACTANTS : 0: :0: II I CH3 -C -�: .... .. f---l ... � CH3 -C = � conjugate base stronger base :0: II H - O-S-CH • • II 3 : 0: conjugate acid stronger acid 1-19 Solutions for (a) and (b) are presented in the Solved Problem in the text. Here, the newly formed bonds are shown in bold. H • • I _ (c) H-B-H + CH -O - CH " H-B-H 1'- 3 /" 3 1+ (d) acid H �base CH 3 -O - CH • • 3 co II : •• CH 3 -C � -H + :O-H •• acid base .. :0: I CH3-r -H :O -H 11
1-19continued (e)Bronsted-Lowry-proton transfer H:O: O: H:0 H七HH一H。=e+- H- base H acid CH-N-H +CH ：CH,-N-H+ H acid base 1-20 Leaming organic chemistry is similar to leaming a foreign language:new vocabulary,new grammar(the es are pre teonstoevalate your compehension and to guide your review of the 1-21 (a)CARBON!(b)oxygen (c)phosphorus (d)chlorine 1-22 valence e 112345678 H He(2e Li N 0 e Br 1-23 (a)ionic only (b)covalent(H-O-)and ionic(Na*-OH) (c)covalent(H-Cand C-Li).but the C-Li bond is strongly polarized (d)covalent only (e)covalent (H-C and C--)and ionic(Nat-OCH3) (f)covalent (H-C and C=O and C-O)and ionic (HCO2 Na) (g)covalent only 12
1-19 continued (e) Bronsted-Lowry--c-proton transfer •• _ } � :�: i -� :�: H :0: H - C-C-H + :O- H �==� H-C-C - H .... ... H -b == b- H + . ) � /.. .. base (f) 1-20 acid .. � () CH3 - � -H + CH3-S=!: H acid base . . - + :Cl : .. H-O-H Learning organic chemistry is similar to learning a foreign language: new vocabulary, new grammar (the reactions), some new concepts, and even a new alphabet (the symbolism of chemistry). This type of definition question is intended to help you review the vocabulary and concepts in each chapter. All of the definitions and examples are presented in the Glossary and in the chapter, so this Solutions Manual will not repeat them. Use these questions to evaluate your comprehension and to guide your review of the important concepts in the chapter. 1-21 (a) CARBON! (b) oxygen (c) phosphorus (d) chlorine 1-22 valence e- --- 1 2 3 4 5 6 7 8 H He (2e-) Li Be B C N 0 F Ne p S Cl Br I 1-23 (a) ionic only (b) covalent (H-O-) and ionic (Na+ -OH) (c) covalent (H-C and C-Li), but the C-Li bond is strongly polarized (d) covalent only (e) covalent (H-C and C-O-) and ionic (Na+ -OCH3) (f) covalent (H-C and C=O and C-O- ) and ionic (HC02 - Na+) (g) covalent only 12
1-24 y 过、煎 过、 (a) 的萍 (b) 9 9: 油年 CANNOT EXIST orntaisoabcompound H (a)H-N-N-H (b)H-N=N-H (c)H-C-N C-H HH H H H:O: H:O:H C=N: (e)H- C-H 0H-C-- 一H H H 1 0 H 0: (g) H-0- 0 -N=c= (i)H-c-6- S-0-C-H H H 0: H H k) C-N=0 1-26 a)H- C=C-C 13
1-24 (a) : Cl Cl: .. , .. / .. P I :Cl: : Cl Cl: .. " / .. • • p ............ : Cl /' I Cl: :Cl: : Cl Cl: . . , .. / .. (b) N I :Cl: : Cl Cl: . . , / .. • • N ........ .. : c!'" I Cl: . . : CI: .• CANNOT EXIST NCls violates the octet rule; nitrogen can have no more than eight electrons (or four atoms) around it. Phosphorus, a third-row element, can have more than eight electrons because phosphorus can used orbitals in bonding, so PCls is a stable, isolable compound. 1-25 Your Lewis structures may look different from these. As long as the atoms are connected in the same order and by the same type of bond, they are equivalent structures. For now, the exact placement of the atoms on the page is not significant. (a) (d) (g) 1-26 H H H ,1/ H C H , 1+ / . . - H-N-N-H (b) H-N=N -H (c) H - C-N-C-H / I " :CI : I I H H H H :0: I II H C H /1" H H H H :0: H I I II I H-C-C N: (e) H-C-C-H (f) H-C-S-C -H I H : 0: I H H I I I H H H :0: H II I II I H - O-S-O-H (h) H-C-N=C =O (i) H - C-O-S-O-C -H II : 0: H / H :N H I II I H-C- C-C-H I I H H H H :0: I H (k) I I II .• H H H ,,1/ H C " I H-C-C-N=O / H I C /1" H H H I II I H :0: H H :0: H :0: I II I II (a) H-C-C = C-C-C = C-C - O-H (b) :N C-C-C-C-C-H I I I I I I H H H H H H 13 I I H H