CHAPTER 19 CARB○ XYLIC AC|DS SOLUTIONS TO TEXT 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. H3C、 COH (E)2-Butenoic acid (crotonic acid) (c) Oxalic acid is a dicarboxylic acid that contains two carbons. It is ethanedioic acid HO, CCO,H Ethanedioic acid (oxalic acid (d) The name given to C HSCO, 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. H,C -COH p-Methylbenzoic acid or 502 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CHAPTER 19 CARBOXYLIC ACIDS SOLUTIONS TO TEXT 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. (c) Oxalic acid is a dicarboxylic acid that contains two carbons. It is ethanedioic acid. (d) The name given to C6H5CO2H is benzoic acid. Because it has a methyl group at the para position, the compound shown is p-methylbenzoic acid, or 4-methylbenzoic acid. p-Methylbenzoic acid or 4-methylbenzoic acid (p-toluic acid) H3C CO2H HO2CCO2H Ethanedioic acid (oxalic acid) C C H CO2H H H 3C (E)-2-Butenoic acid (crotonic acid) 502 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CARBOXYLIC ACIDS 503 19.2 Ionization of peroxy acids such as peroxyacetic acid yields an anion that cannot be stabilized by res not possible is sanborn group is 19.3 Recall from Chapter 4(text Section 4.6) 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 equatio CHiCO,H +(CH3)3CO CH3 CO, +(CH3)COH Acetic acid tert-Butoxide Butyl alcohol (stronger acid) (stronger base) (weaker base) Icohols are weaker acids than carboxylic acids; the equilibrium lies to the right. (c) Bromide ion is the conjugate base of hydrogen bromide, a strong acid CH CO,H+ CH3CO2 HBI Acetic acid Bromide ion Acetate ion Hydrogen (weaker base) In this case, the position of equilibrium favors the starting materials, because acetic acid is a weaker acid than hydrogen bromide (d) Acetylide ion is a rather strong base, and acetylene, with a k of 10, is a much weaker acid than acetic acid. The position of equilibrium favors the formation of products CH2COH+HC≡C CH2CO3+HC≡CH (stronger acid) (stronger base) eaker base) (weaker acid (e) Nitrate ion is a very weak base; it is the conjugate base of the strong acid nitric acid. The position of equilibrium lies to the left CH3CO,H HNO Acetic acid Nitrate ion Acetate ion Nitric acid (weaker acid weaker base) (f) Amide ion is a very strong base; it is the conjugate base of ammonia, pKa= 36. The position of equilibrium lies to the right. CH,CO,H H,N CHCO Amide ion Ammonia (stronger base) (weaker base (weaker acid Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
CARBOXYLIC ACIDS 503 19.2 Ionization of peroxy acids such as peroxyacetic acid yields an anion that cannot be stabilized by resonance in the same way that acetate can. 19.3 Recall from Chapter 4 (text Section 4.6) 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 Alcohols are weaker acids than carboxylic acids; the equilibrium lies to the right. (c) Bromide ion is the conjugate base of hydrogen bromide, a strong acid. In this case, the position of equilibrium favors the starting materials, because acetic acid is a weaker acid than hydrogen bromide. (d) Acetylide ion is a rather strong base, and acetylene, with a Ka of 1026, is a much weaker acid than acetic acid. The position of equilibrium favors the formation of products. (e) Nitrate ion is a very weak base; it is the conjugate base of the strong acid nitric acid. The position of equilibrium lies to the left. ( f ) Amide ion is a very strong base; it is the conjugate base of ammonia, pKa 36. The position of equilibrium lies to the right. CH3CO2H Acetic acid (stronger acid) H2N Amide ion (stronger base) CH2CO2 Acetate ion (weaker base) NH3 Ammonia (weaker acid) CH3CO2H Acetic acid (weaker acid) NO3 Nitrate ion (weaker base) CH3CO2 Acetate ion (stronger base) HNO3 Nitric acid (stronger acid) CH3CO2 Acetate ion (weaker base) Acetylene (weaker acid) CH3CO2H HC CH Acetic acid (stronger acid) Acetylide ion (stronger base) HC C CH3CO2H Acetic acid (weaker acid) Br Bromide ion (weaker base) CH3CO2 Acetate ion (stronger base) HBr Hydrogen bromide (stronger acid) Acetic acid (stronger acid) tert-Butoxide (stronger base) tert-Butyl alcohol (weaker acid) CH3CO2H (CH3)3CO Acetate ion (weaker base) CH3CO2 (CH3) 3COH Delocalization of negative charge into carbonyl group is not possible in peroxyacetate ion. CH3CO O O __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
504 CARBOXYLIC ACIDS 19.4(b) 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 nd field effects CHCH—C Lactic acid is more acidic than propanoic acid. The measured ionization constants are CHCHCOH CH..COH OH (pKa3.8) (pK4.9) (c) A carbonyl group is more strongly electron-withdrawing than a carbon-carbon double bond Pyruvic acid is a stronger acid than acrylic acid. CH,CCO,,HH,C=CHCO,H Acrylic acid K5.1×10-4 K5.5×10-5 (pKa3.3) (pK24.3) (d) Viewing the two compounds as substituted derivatives of acetic acid, RCH, CO, H, we judge CHS to be strongly electron-withdrawing and acid-strengthening, whereas an ethyl group has only a small effect CH3SCH, CO,H CHaCH,CH,CO, H Methanesulfonylacetic acid Butanoic acid pk24) (pK24.7 19.5 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 rather electron-withdrawing and acid-strengthening. This is borne out by its measured ionization constant Ka, which is 1.4X10(pK. 1.8) 19.6 For carbonic acid, the" true Ki" is given by [HJH True K H,CO3J The"observed K"is given by the expression 4.3×10-7 [HIIHCO3 J CO, Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
__ __ __ 504 CARBOXYLIC ACIDS 19.4 (b) Propanoic acid is similar to acetic acid in its acidity. A hydroxyl group at C-2 is electronwithdrawing and stabilizes the carboxylate ion of lactic acid by a combination of inductive and field effects. Lactic acid is more acidic than propanoic acid. The measured ionization constants are (c) A carbonyl group is more strongly electron-withdrawing than a carbon–carbon double bond. Pyruvic acid is a stronger acid than acrylic acid. (d) Viewing the two compounds as substituted derivatives of acetic acid, RCH2CO2H, we judge to be strongly electron-withdrawing and acid-strengthening, whereas an ethyl group has only a small effect. 19.5 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>CCO2H; C-2 is sp-hybridized and therefore rather electron-withdrawing and acid-strengthening. This is borne out by its measured ionization constant Ka, which is 1.4 102 (pKa 1.8). 19.6 For carbonic acid, the “true K1” is given by True K1 [H [H ][ 2 H C C O O 3] 3 ] The “observed K” is given by the expression 4.3 107 [H][HCO3 ] [CO2] Butanoic acid Ka 1.5 105 (pKa 4.7) CH3CH2CH2CO2H Methanesulfonylacetic acid Ka 4.3 103 (pKa 2.4) CH3SCH2CO2H O O CH3S O O Pyruvic acid Ka 5.1 104 (pKa 3.3) CH3CCO2H O Acrylic acid Ka 5.5 105 (pKa 4.3) H2C CHCO2H Propanoic acid Ka 1.3 105 (pKa 4.9) CH3CH2CO2H Lactic acid Ka 1.4 104 (pKa 3.8) CH3CHCO2H OH Hydroxyl group stabilizes negative charge by attracting electrons. CH3CH C O OH O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CARBOXYLIC ACIDS 505 which can be rearranged to H']HCO3]=(43×10-CO2l and therefore Tuek,(4.3×10-)CO2 H,CO3I 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. Carboxy lic acids dissolve in sodium bicar bonate solution because the equilibrium that leads to carbon dioxide formation is favorable, not be ause carboxylic acids are stronger acids than carbonic acid 19.7 (b) 2-Chloroethanol has been converted to 3-hydroxypropanoic acid by way of the corresponding HOCHCHCI HOCH.. CN HOCH..H The presence of the hydroxyl group in 2-chloroethanol precludes the preparation of a Grignard agent 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 (c) Grignard reagents can be prepared from tertiary halides and react in the expected manner with carbon dioxide. The procedure shown is entirely satisfactory. 1. CO (CH3)cCI (CH,) CMgCl 2.H2O (CH3)CCO,H r-Butyl chloride terT-Butylmagnesium 2, 2-Dimethylpropanoic 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 SN2 reaction limited to pri 19.8 Incorporation of O into benzoic acid proceeds by a mechanism analogous to that of esterification The nucleophile that adds to the protonated form of benzoic acid is O-enriched water(the 0 atom is represented by the shaded letter e in the following equations). OH oH CHOH H CsH-COH C6HSCOH C6HSCOH Benzoic acid Tetrahedral intermediate The three hydroxyl groups of the tetrahedral intermediate are equivalent except that one of them is labeled withO. Any one of these three hydroxyl groups may be lost in the dehydration step; when the hydroxyl group that is lost is unlabeled, anO label is retained in the benzoic acid. CHOH + H C6H5CY hO C HsC Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
which can be rearranged to [H][HCO3 ] (4.3 107 )[CO2] and therefore True K1 (4.3 [H 1 2C 0 O 7 3 ) ] [CO2 ] 1.4 104 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 (b) 2-Chloroethanol has been converted to 3-hydroxypropanoic acid by way of the corresponding nitrile. The presence of the hydroxyl 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. (c) Grignard reagents can be prepared from tertiary halides and react in the expected manner with carbon dioxide. The procedure shown is entirely satisfactory. 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 SN2 reaction with cyanide ion is limited to primary and secondary alkyl halides. 19.8 Incorporation of 18O into benzoic acid proceeds by a mechanism analogous to that of esterification. The nucleophile that adds to the protonated form of benzoic acid is 18O-enriched water (the 18O atom is represented by the shaded letter in the following equations). The three hydroxyl groups of the tetrahedral intermediate are equivalent except that one of them is labeled with 18O. Any one of these three hydroxyl groups may be lost in the dehydration step; when the hydroxyl group that is lost is unlabeled, an 18O label is retained in the benzoic acid. H Tetrahedral intermediate C6H5COH OH OH H2O C6H5C OH OH 18O-enriched benzoic acid C6H5C O OH C6H5C O OH H H Benzoic acid C6H5COH O C6H5COH OH C6H5COH OH H H O Tetrahedral intermediate C6H5COH OH OH H2O O tert-Butylmagnesium chloride (CH3)3CMgCl 2,2-Dimethylpropanoic acid (61–70%) (CH3)3CCO2H tert-Butyl chloride (CH3)3CCl Mg diethyl ether 1. CO2 2. H3O 2-Chloroethanol HOCH2CH2Cl 2-Cyanoethanol HOCH2CH2CN NaCN H2O 3-Hydroxypropanoic acid HOCH2CH2CO2H H3O heat (4.3 107 )(99.7) 0.3 CARBOXYLIC ACIDS 505 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
506 CARBOXYLIC ACIDS 19.9(b) The 16-membered ring of 15-pentadecanolide is formed from 15-hydroxypentadecanoic Disconnect this bond COH 5-Pentadecanolide 15-Hydroxypentadecanoic (c) Vernolepin has two lactone rings, which can be related to two hydroxy acid combinations CHECH OH HOCH CH=CH H H, H,C OH COH Be sure to keep the relative stereochemistry unchanged. Remember, the carbon-oxygen bond of an alcohol remains intact when the alcohol reacts with a carboxylic acid to give an 19.10 Alkyl chlorides and bromides undergo nucleophilic substitution when treated with sodium iodide in acetone(Section8. 1). A reasonable approach is to brominate octadecanoic acid at its a-carbon atom, then replace the bromine substituent with iodine by nucleophilic substitution CHa(CH,)15CH,CO,H CH3(CH)15 CHCO acetone CH3(CH) CHCO2H 2- Bromooctadecanoic acid 2-lodooctadecanoic acid 19.11 (b) The starting material is a derivative of malonic acid. It undergoes efficient thermal decar- boxylation in the manner shown HO H CH,(CH2)CHO CH3(CH,)CH c O, CH3(CH2)CH, COH Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
19.9 (b) The 16-membered ring of 15-pentadecanolide is formed from 15-hydroxypentadecanoic acid. (c) Vernolepin has two lactone rings, which can be related to two hydroxy acid combinations. Be sure to keep the relative stereochemistry unchanged. Remember, the carbon–oxygen bond of an alcohol remains intact when the alcohol reacts with a carboxylic acid to give an ester. 19.10 Alkyl chlorides and bromides undergo nucleophilic substitution when treated with sodium iodide in acetone (Section 8.1). A reasonable approach is to brominate octadecanoic acid at its -carbon atom, then replace the bromine substituent with iodine by nucleophilic substitution. 19.11 (b) The starting material is a derivative of malonic acid. It undergoes efficient thermal decarboxylation in the manner shown. heat CH3(CH2)6CH C O H O C O HO 2-Heptylmalonic acid CO2 Carbon dioxide CH3(CH2)6CH C OH OH CH3(CH2)6CH2COH O Nonanoic acid Br2, PCl3 NaI acetone Octadecanoic acid CH3(CH2)15CH2CO2H 2-Bromooctadecanoic acid CH3(CH2)15CHCO2H Br 2-Iodooctadecanoic acid CH3(CH2)15CHCO2H I CH2 O O O OH O CH2 H2C CH CH2 OH CH2 HOCH2 HO2C H2C CO2H CH OH Disconnect this bond. O OH COH O O 15-Pentadecanolide 15-Hydroxypentadecanoic acid 506 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website