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CHAPTER 19: Carboxylic Acids 525 SOLUTIONS TO TEXT PROBLEMS In Chapter Problems 19.1 (b) The four carbon atoms of crotonic acid form a continuous chain. Because there is a double bond between C-2 and C-3, crotonic acid is one of the stereoisomers of 2-butenoic acid. The stereochemistry of the double bond is E. ara 0 (E}-2-Butenoic acid or (E)-But-2-enoic acid (crotonic acid) (c) Oxalic acid is a dicarboxylic acid that contains two carbons. It is ethanedioic acid. ie) OH wry 10) Ethanedioic acid (oxalic acid) (d) The name given to C,H,CO,H is benzoic acid. Because it has a methyl group at the para position, the compound shown is p-methylbenzoic acid, or 4-methylbenzoic acid. Oo OH p-Methylbenzoic acid or 4-methylbenzoic acid (p-toluic acid) 19.2, (b) As in part (a) of this problem we can calculate the conjugate base/acid ratio using the Henderson-Hasselbalch relationship. [conjugate base] _ \(pH-pKa) [acid] 10 [lactate] ——— = (2.5-3.9) — 714 [lactic acid] 10 10 0.04 At pH = 2.5, 25 times as much lactic acid is present as lactate ion. Scanned with | CamScanner’: 526 CHAPTER 19: Carboxylic Acids 19.3. (6) Propanoic acid is similar to acetic acid in its acidity. A hydroxyl group at C-2 is electron-withdrawing and stabilizes the carboxylate ion of lactic acid by a combination of inductive and field effects. 10) o Tou Hydroxyl group stabilizes negative charge by attracting electrons. Lactic acid is more acidic than propanoic acid. The measured pX,,’s are or A, OH Lactic acid Propanoic acid PK, 38 pKn49 (c) A carbonyl group is more strongly electron-withdrawing than a carbon-carbon double bond. Pyruvic acid is a stronger acid than acrylic acid. 10) 0 Spon A oO Pyruvie acid Acrylic acid pKn33 p= 43 (d) Viewing the two compounds as substituted derivatives of acetic acid, RCH,CO,H, we judge CH;SO,— to be strongly electron-withdrawing and acid-strengthening, whereas an ethyl group has only a small effect. 00 0 te) WwW | 2 OH OH Methanesulfonylacetic acid Butanoie acid pKy=24 pkye47 19.4 The compound can only be a carboxylic acid; no other class containing only carbon, hydrogen, and oxygen is more acidic. A reasonable choice is HC=CCO,H; C-2 is sp-hybridized and therefore electron-withdrawing and acid-strengthening. This is borne out by its measured pK, of 1.8. 195 Recall from Chapter | (text Section 1.14) that an acid-base equilibrium favors formation of the weaker acid and base. Also remember that the weaker acid forms the stronger conjugate base, and vice versa. (b) The acid-base reaction between acetic acid and tert-butoxide ion is represented by the equation | Scanned with | @camScanner’ 528 CHAPTER 19: Carboxylic Acids H’*][HCO; 43x 107= COs) [CO>] which can be rearranged to [H"][HCO}] = (4.3 x 107)[CO,] and therefore (43x 1077)[CO3] True K, [HsC03] (43x107)(997) 03 = 14x 107 Thus, when corrected for the small degree to which carbon dioxide is hydrated, it can be seen that carbonic acid is actually a stronger acid than acetic acid. Carboxylic acids dissolve in sodium bicarbonate solution because the equilibrium that leads to carbon dioxide formation is favorable, not because carboxylic acids are stronger acids than carbonic acid. 19.7. The order of steps is important. A Grignard reagent cannot be prepared from the starting diol because of the presence of the hydroxyl groups. The starting diol must first be converted to its dimethyl ether. This can be carried out with a Williamson ether synthesis: OH OCH; CHI a, Br Bil0; Br Ou OCH; 2-Bromo-1,3-dimethoxybenzene Formation of the Grignard reagent, followed by reaction with carbon dioxide and acidification gives the desired product. OCH; OCH, OCH, Mg, ether 1.00, — —= Br MgBr 2.140 COW OCH; OCH; OCH; 2,6-Dimethoxybenzoic acid 19.8 (6) 2-Chloroethanol has been converted to 3-hydroxypropanoic acid by way of the corresponding nitrile. NaCN 0° HOLA WOLA ; HO. ara HO Sy heat Ou 2-Chlorocthanol 2-Cyanoethanol 3-Hydroxypropanoic acid The presence of the hydroxy! group in 2-chloroethanol precludes the preparation of a Grignard reagent from this material, and so any attempt at the preparation of 3-hydroxypropanoic acid via the Grignard reagent of 2-chloroethanol is certain to fail. Scanned with | @camScanner’ CHAPTER 19: Carboxylic Acids 529 (c). Grignard reagents can be prepared from tertiary halides and react in the expected manner with carbon dioxide. The procedure shown is entirely satisfactory. Mg cl 1.C0, o Mgt = diethyl ether 2. HyO* 7 7 OH tert-Butyl chloride tert-Butylmagnesium 2,2-Dimethylpropanoic chloride acid (61-70%) Preparation by way of the nitrile will not be feasible. Rather than react with sodium cyanide by substitution, tert-butyl chloride will undergo elimination exclusively. The S,2 reaction with cyanide ion is limited to primary and secondary alkyl halides. 19.9 Incorporation of '$ into benzoic acid proceeds by a mechanism analogous to that of esterification. The nucleophile that adds to the protonated form of benzoic acid is '$0-enriched water (the "$0 atom is represented by the shaded letter © in the following equations). G: Git yy. :0H :0H Of Ott Ofe-Ofs 20H :0H to :0H wey Q Benzoic acid Tetrahedral intermediate The three hydroxyl groups of the tetrahedral intermediate are equivalent except that one of them is labeled with '$0. Any one of these three hydroxyl groups may be lost in the dehydration step; when the hydroxyl group that is lost is unlabeled, an '*O label is retained in the benzoic acid. ‘OH ot aw 6: :0H CY + Hs C\ + HO — ( + (\X 10H :0H 10H Q: Tetrahedral '$Q-enriched benzoic acid intermediate 19.10 (b) The 16-membered ring of 15-pentadecanolide is formed from 15-hydroxypentadecanoic acid. => O.; mes Disconnect this bond. ——~” on O 15-Pentadecanotide [S-Hydroxypentadecanoic acid Scanned with | @camScanner’ 554 CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution SOLUTIONS TO TEXT PROBLEMS In Chapter Problems Wei ae 20.1 (6) Acid anhydrides bear two acyl groups on oxygen, as in RCOCR . They are named as derivatives of carboxylic acids. " Oo oO az ‘OH o 0 ~ 2-Phenybutanoic acid 2-Phenybutanoic anyhydride (c) Butyl 2-phenylbutanoate is the butyl ester of 2-phenylbutanoic acid. pheny oO oO™ Butyl 2-phenylbutancate (d) In 2-phenylbuty! butanoate, the 2-phenylbutyl group is an alkyl group bonded to oxygen of the ester. It is not part of the acyl group of the molecule. oO ~~ 2-Phenylbuty! butanoate oO ti (e) The ending -amide reveals this to be a compound of the type RCNH,. Oo NH, 2-Phenylbutanamide (f) This compound differs from 2-phenylbutanamide in part (e) only in that it bears an ethyl substituent on nitrogen. Scanned with CamScanner | CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution 555 fe) ao H N-Ethyl-2-phenylbutanamide (g) The -nitrile ending signifies a compound of the type RC=N containing the same number of carbons as the alkane RCH,. Alternatively, the compound may be named as an alkyl cyanide. ZN Cc? 2-Phenylbutanenitrile or I-phenylpropyl cyanide 20.2 The methyl groups in N,N-dimethylformamide are nonequivalent; one is cis to oxygen, the other is trans. The two methyl groups have different chemical shifts. The third signal is due to the carbonyl carbon. 6313 26 TA. ke —_- oie i ) D CH. CH. 516 4 > 3 5364 Rotation about the carbon-nitrogen bond is required to average the environments of the two methyl groups, but this rotation is slow enough in amides that the two distinct methy! groups can be detected. This is the result of the double-bond character imparted to the carbon-nitrogen bond, as shown by the resonance contributor on the right. 20.3 Electron release from the p-methoxy group stabilizes the acylium ion, which lowers the activation energy for its formation, and increases the reaction rate. Oy js ae: <= c 7 i H,C H;C 20.4 (b) Benzoyl chloride reacts with benzoic acid to give benzoic anhydride. 10) te) fe) oO Oo--ov-oru:s i Benzoic acid Benzoic anhydride Hydrogen Benzoyl chloride oh Scanned with | CamScanner’: CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution 557 ° N(CH ‘0 + 2(CH,),NH (CHD? 1 NCHy)> 0 fo) Phthalic anhydride Dimethylamine Product is an amine salt and contains an amide functional group In this case, both the amide function and the ammonium carboxylate salt are incorporated into the same molecule. (d) The disodium salt of phthalic acid is the product of hydrolysis of phthalic acid in excess sodium hydroxide. fe) oO ‘O” Na’ 0 + 2NaOH ve + HO O° Na 9 ° Phthalic anhydride Sodium hydroxide Sodium phthalate Water 20.6 The starting material contains three acetate ester functions, All three undergo hydrolysis in aqueous sulfuric acid. ° ° 5 Nr = Hoon + 3 vou ty YT OH ° 1,2,$-Pentanetriol Acetic acid The product is 1,2,5-pentanetriol. Also formed in the hydrolysis of the starting triacetate are three molecules of acetic acid. 20.7 Step 1: Protonation of the carbonyl oxygen + ioe, :0-H Or — OL - : 30) Hydronium Protonated form of ester Water ion Ethyl benzoate Step 2: Nucleophilic addition of water Scanned with | CamScanner’: 558 CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Protonated form of ester Water Oxonium ion Step 3: Deprotonation of oxonium ion to give neutral form of tetrahedral intermediate Oxonium ion Water Tetrahedral Hydronium ion intermediate Step 4: Protonation of ethoxy oxygen + H-O \ wae H Tetrahedral Hydronium ion Oxonium ion Water intermediate Step 5: Dissociation of protonated form of tetrahedral intermediate This step yields ethyl alcohol and the protonated form of benzoic acid. :0-H —* Cp + H OQ - :0-H Oxonium ion Protonated form Ethyl alcohol of benzoic acid Step 6: Deprotonation of protonated form of benzoic acid Al s0-H*—\_ HH oO: H OWT — Of —H H : \ Protonated form Water of benzoic acid Protonated form Hydronium ion of benzoic acid Scanned with | CamScanner’: 560 CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acy! Substitution Ethyl benzoate Hydroxide Anionic form of ion tetrahedral intermediate Step 2: Proton transfer from water to give neutral form of tetrahedral intermediate + 36H Anionic form of Water Tetrahedral Hydroxide tetrahedral intermediate intermediate ion Step 3: Dissociation of tetrahedral intermediate 6 + OH = O-H ey soe ee Tetrahedral Hydroxide Benzoic acid Water Ethoxide ion intermediate ion Step 4: Proton transfer processes 6 6 Benzoic acid Hydroxide Benzoie acid Water ion — —o-n + Ethoxide ion Water Ethanol Hydroxide jon 20.12. (b) Stabilization of the ester carbonyl group is greater in the ester of cyclohexanol than in the ester of phenol, because of resonance involving the aromatic ring and oxygen in the latter. No such resonance is possible for the ester of cyclohexanol. Hydrolysis of the less stable ester of phenol occurs more rapidly. Scanned with | @camScanner’ CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acy! Substitution 561 6 6 : 2 og (c) The electron-withdrawing trifluoromethyl group destabilizes the ester carbonyl in the second compound by an inductive effect, so it hydrolyzes more rapidly. (d) The large tert-butyl group destabilizes due to steric hindrance in the tetrahedral intermediate in saponification of the second compound, resulting in a larger energy of activation and slower hydrolysis, (Note that this applies to saponification, and not to hydrolysis in acid.) 20.13 The starting material is a lactone, a cyclic ester. The ester function is converted to an amide by nucleophilic acyl substitution. CH; CH; ve > Methylamine _4-Pentanolide CH; CH, Ou 20: 0: OH HjC-NH HyC-NH 4-Hydroxy-N-methylpentanamide 20.14 Nucleophilic attack by the amine on the ester carbony] is the rate determining step that forms a tetrahedral intermediate. Go: slow a sg OTR os sa OCH; ° OCH 20.15 (6) Recall that the two identical groups bonded to the hydroxyl-bearing carbon of the alcohol arose from the Grignard reagent. That leads to the following retrosynthetic analysis: (C4tt:c—<] = [>—cor + 2CyHsMeX OH fe) Thus, the two pheny] substituents arise by addition of a phenyl Grignard reagent to an ester of cyclopropanecarboxylic acid. Scanned with | CamScanner’: CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution 563 6: H,C., | &) N’ H : Ao. =-z: A resonance structure of B can be drawn with a ring analogous to benzene: HC An N DMO HONG: H 20.20 (b) Acetic anhydride is the acid anhydride that must be used; it transfers an acetyl group to suitable nucleophiles. The nucleophile in this case is methylamine. fe) ie) 0 fe) Ak gf SCH a * Ag CH,NH, Acetic anhydride Methylamine N-Methylacetamide Methylammonium acetate oO I * (c) The acyl group is HC—. Because the problem specifies that the acyl transfer agent is a methyl ester, methyl formate is one of the starting materials. 0 fo) nvg-ctl + HN(CH); ——+ we cH, + CH,OH Nw ) CH; Methyl formate Dimethylamine N.N-Dimethylformamide Methyl alcohol 20.21 The reaction that occurs between an amine and a carboxylic acid is proton transfer from the acid to the amine (acid-base). —— wg + NR; ——+ The carboxylate carbonyl is highly stabilized by electron delocalization and does not undergo nucleophilic acyl substitution and an ammonium ion is not a nucleophile; the nitrogen atom has no lone electron pairs. 20.22 Step 1: Protonation of the carbonyl oxygen Scanned with | CamScanner’: 564 CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution ay +H 6: aH te) H 77 / re -O: ——= +. CoH 10: yc Ces + H-O: eon Hs + 20: | H | H H H Acetanilide Hydronium ion Protonated form Water of amide Step 2: Nucleophilic addition of water +67 H 107 H an SL “| Cols 803) + + CoH; === H;C 7 H,C7f ~N \ “ hotel n HOH Water Protonated form Oxonium ion of amide Step 3: Deprotonation of oxonium ion to give neutral form of tetrahedral intermediate Scanned with | CamScanner’: 566 CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution Step 5: Irreversible formation of formate ion + YO-H — Formic acid Hydroxide ion Formic acid Water 20.24 Both parts of this problem deal with hydrolysis reactions. Penicillin G has two amide bonds capable of being cleaved by hydrolysis. Bond a is cleaved in part (a) and bond b is cleaved in part (6). (a) The molecular formulas of penicillin G (CjgH,gN 04S) and the product of B-lactamase-catalyzed hydrolysis (Cy gHygN.O.S) differ by H,O. Therefore, all the atoms in penicillin G are retained on hydrolysis which is consistent only with cleaving the f-lactam ring. Therefore, the product is OH : 7~On a (6) The product of penicillin acyl transferase-catalyzed hydrolysis has a molecular formula (CgH,yN 03S) with eight fewer carbons than penicillin G. Therefore, the acyl group attached to the nitrogen of the side chain is cleaved from the molecule. The product is H INF og x co YY ‘OH Kan 20.25 (a) Ethanenitrile has the same number of carbon atoms as ethyl alcohol. This suggests a reaction scheme proceeding via an amide, “~ 9 oF a ou Ok HyC-C=N Nil, Ethyl alcohol Acetamide Ethanenitrile The necessary amide is prepared from ethanol. Scanned with | @camScanner’ CHAPTER 20: Carboxylic Acid Derivatives: Nucleophilic Acyl Substitution 567 Na;Cr0;, H;0 ° 1. Soc}, ° —~oH H,SO,, heat vn 2.NHy = 7] Ethyl alcohol Acetic acid Acetamide (b) Propanenitrile may be prepared from ethyl alcohol by way of a nucleophilic substitution reaction of the corresponding bromide. PBr; NaCN o™N Heit. a OH orBr sy Con Ethyl alcohol Ethyl bromide Propanenitrile 20.26 Step 1: Protonation of the nitrile aren DEE + cH ata = R—=N-H +: i Nitrile Hydronium ion Protonated form Water of nitrile Step 2: Nucleophilic addition of water N-H — r< :0-H / H Water Protonated form Protonated form of nitrile imino acid Step 3: Deprotonation of imino acid N-H H N-u JH r<, + rg — nr + HO: :0- Su 7 4} 5 H H H Protonated form Water Imino acid Hydronium ion imino acid Steps 4 and 5: Proton transfers to give an amide in TN AH CN H aL ft + We — R + 2 — { + H-O: ‘O: H Conjugate acid Water Amide Hydronium ion Imino acid Hydronium ion of amide Scanned with | CamScanner’: 
