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 C 6 H 5 CO 2 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. p-Methylbenzoic acid or 4-methylbenzoic acid (p-toluic acid) H 3 CCO 2 H HO 2 CCO 2 H Ethanedioic acid (oxalic acid) CC H CO 2 H H H 3 C (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- onance 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 K a of 10 H1100226 , 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, pK a H11005 36. The position of equilibrium lies to the right. H11001CH 3 CO 2 H Acetic acid (stronger acid) H 2 N H11002 Amide ion (stronger base) H11001CH 2 CO 2 H11002 Acetate ion (weaker base) NH 3 Ammonia (weaker acid) H11001CH 3 CO 2 H Acetic acid (weaker acid) NO 3 H11002 Nitrate ion (weaker base) H11001CH 3 CO 2 H11002 Acetate ion (stronger base) HNO 3 Nitric acid (stronger acid) H11001CH 3 CO 2 H11002 Acetate ion (weaker base) Acetylene (weaker acid) HC CHH11001CH 3 CO 2 H Acetic acid (stronger acid) Acetylide ion (stronger base) HC H11002 C H11001CH 3 CO 2 H Acetic acid (weaker acid) Br H11002 Bromide ion (weaker base) H11001CH 3 CO 2 H11002 Acetate ion (stronger base) HBr Hydrogen bromide (stronger acid) Acetic acid (stronger acid) tert-Butoxide (stronger base) tert-Butyl alcohol (weaker acid) CH 3 CO 2 H (CH 3 ) 3 CO H11002 Acetate ion (weaker base) CH 3 CO 2 H11002 (CH 3 ) 3 COHH11001H11001 Delocalization of negative charge into carbonyl group is not possible in peroxyacetate ion. CH 3 CO O O H11002 __ __ __ 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 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, RCH 2 CO 2 H, 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>CCO 2 H; C-2 is sp-hybridized and therefore rather electron-withdrawing and acid-strengthening. This is borne out by its measured ionization constant K a , which is 1.4 H11003 10 H110022 (pK a 1.8). 19.6 For carbonic acid, the “true K 1 ” is given by True K 1 H11005H5007 [H H11001 [H ][ 2 H C C O O 3 ] 3 H11002 ] H5007 The “observed K” is given by the expression 4.3 H11003 10 H110027 H11005 [H H11001 ][HCO 3 H11002 ] H5007H5007 [CO 2 ] Butanoic acid K a 1.5 H11003 10 H110025 (pK a 4.7) CH 3 CH 2 CH 2 CO 2 H Methanesulfonylacetic acid K a 4.3 H11003 10 H110023 (pK a 2.4) CH 3 SCH 2 CO 2 H O O CH 3 S O O Pyruvic acid K a 5.1 H11003 10 H110024 (pK a 3.3) CH 3 CCO 2 H O Acrylic acid K a 5.5 H11003 10 H110025 (pK a 4.3) H 2 C CHCO 2 H Propanoic acid K a 1.3 H11003 10 H110025 (pK a 4.9) CH 3 CH 2 CO 2 H Lactic acid K a 1.4 H11003 10 H110024 (pK a 3.8) CH 3 CHCO 2 H OH Hydroxyl group stabilizes negative charge by attracting electrons. CH 3 CH C O OH O H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website which can be rearranged to [H H11001 ][HCO 3 H11002 ] H11005 (4.3 H11003 10 H110027 )[CO 2 ] and therefore True K 1 H11005H5007 (4.3 H11003 [H 1 2 C 0 H11002 O 7 3 ) ] [CO 2 ] H5007 H11005 H11005 1.4 H11003 10 H110024 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 bicar- bonate solution because the equilibrium that leads to carbon dioxide formation is favorable, not be- cause 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 S N 2 reaction with cyanide ion is limited to primary and secondary alkyl halides. 19.8 Incorporation of 18 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 18 O-enriched water (the 18 O 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 18 O. Any one of these three hydroxyl groups may be lost in the dehydration step; when the hydroxyl group that is lost is unlabeled, an 18 O label is retained in the benzoic acid. H11001 H H11001 Tetrahedral intermediate C 6 H 5 COH OH OH H11001 H 2 O H11001 C 6 H 5 C OH OH 18 O-enriched benzoic acid H11001C 6 H 5 C O OH C 6 H 5 C O OH H11002H H11001 H11001 H H11001 Benzoic acid C 6 H 5 COH O C 6 H 5 COH OH H11001 H11001 C 6 H 5 COH OH HH O Tetrahedral intermediate C 6 H 5 COH OH OH H 2 O O tert-Butylmagnesium chloride (CH 3 ) 3 CMgCl 2,2-Dimethylpropanoic acid (61–70%) (CH 3 ) 3 CCO 2 H tert-Butyl chloride (CH 3 ) 3 CCl Mg diethyl ether 1. CO 2 2. H 3 O H11001 2-Chloroethanol HOCH 2 CH 2 Cl 2-Cyanoethanol HOCH 2 CH 2 CN NaCN H 2 O 3-Hydroxypropanoic acid HOCH 2 CH 2 CO 2 H H 3 O H11001 heat (4.3 H11003 10 H110027 )(99.7) H5007H5007 0.3 CARBOXYLIC ACIDS 505 __ __ __ 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 H9251-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 decar- boxylation in the manner shown. heat H11001CH 3 (CH 2 ) 6 CH CO H O CO HO 2-Heptylmalonic acid CO 2 Carbon dioxide CH 3 (CH 2 ) 6 CH C OH OH CH 3 (CH 2 ) 6 CH 2 COH O Nonanoic acid Br 2 , PCl 3 NaI acetone Octadecanoic acid CH 3 (CH 2 ) 15 CH 2 CO 2 H 2-Bromooctadecanoic acid CH 3 (CH 2 ) 15 CHCO 2 H Br 2-Iodooctadecanoic acid CH 3 (CH 2 ) 15 CHCO 2 H I CH 2 O O O OH O CH 2 H 2 C CH CH 2 OH CH 2 HOCH 2 HO 2 C H 2 C CO 2 H 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 (c) The phenyl and methyl substituents attached to C-2 of malonic acid play no role in the decar- boxylation process. 19.12 (b) The thermal decarboxylation of H9252-keto acids resembles that of substituted malonic acids. The structure of 2,2-dimethylacetoacetic acid and the equation representing its decarboxylation were given in the text. The overall process involves the bonding changes shown. 19.13 (a) Lactic acid (2-hydroxypropanoic acid) is a three-carbon carboxylic acid that bears a hydroxyl group at C-2. (b) The parent name ethanoic acid tells us that the chain that includes the carboxylic acid func- tion contains only two carbons. A hydroxyl group and a phenyl substituent are present at C-2. (c) The parent alkane is tetradecane, which has an unbranched chain of 14 carbons. The termi- nal methyl group is transformed to a carboxyl function in tetradecanoic acid. Tetradecanoic acid (myristic acid) CH 3 (CH 2 ) 12 COH O 2-Hydroxy-2-phenylethanoic acid (mandelic acid) CHCO 2 H OH 2-Hydroxypropanoic acid CH 3 CHCO 2 H OH 321 Enol form of 3-methyl- 2-butanone C CH 3 OH CH 3 C CH 3 3-Methyl-2-butanone CH 3 CCH(CH 3 ) 2 O H11002CO 2 2,2-Dimethylacetoacetic acid O H C H 3 CCH 3 O C O CH 3 C C CH 3 O C OH OO CH heat CO 2 H11001 H 3 C CC OH OH CHCOH O CH 3 2-Phenylpropanoic acid Carbon dioxide CARBOXYLIC ACIDS 507 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (d) Undecane is the unbranched alkane with 11 carbon atoms, undecanoic acid is the correspond- ing carboxylic acid, and undecenoic acid is an 11-carbon carboxylic acid that contains a double bond. Because the carbon chain is numbered beginning with the carboxyl group, 10-undecenoic acid has its double bond at the opposite end of the chain from the carboxyl group. (e) Mevalonic acid has a five-carbon chain with hydroxyl groups at C-3 and C-5, along with a methyl group at C-3. ( f ) The constitution represented by the systematic name 2-methyl-2-butenoic acid gives rise to two stereoisomers. Tiglic acid is the E isomer, and the Z isomer is known as angelic acid. The higher ranked sub- stituents, methyl and carboxyl, are placed on opposite sides of the double bond in tiglic acid and on the same side in angelic acid. (g) Butanedioic acid is a four-carbon chain in which both terminal carbons are carboxylic acid groups. Malic acid has a hydroxyl group at C-2. (h) Each of the carbon atoms of propane bears a carboxyl group as a substituent in 1,2,3-propane- tricarboxylic acid. In citric acid C-2 also bears a hydroxyl group. HO 2 CCH 2 CCH 2 CO 2 H OH CO 2 H 2-Hydroxy-1,2,3-propanetricarboxylic acid (citric acid) HO 2 CCHCH 2 CO 2 H OH 2-Hydroxybutanedioic acid (malic acid) CC H H 3 C CO 2 H CH 3 (E)-2-Methyl-2-butenoic acid (tiglic acid) CC H H 3 C CH 3 CO 2 H (Z)-2-Methyl-2-butenoic acid (angelic acid) CH 3 CH CCO 2 H CH 3 2-Methyl-2-butenoic acid HOCH 2 CH 2 CCH 2 CO 2 H OH CH 3 3,5-Dihydroxy-3-methylpentanoic acid (mevalonic acid) 10-Undecenoic acid (undecylenic acid) H 2 C CH(CH 2 ) 8 CO 2 H 508 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (i) There is an aryl substituent at C-2 of propanoic acid in ibuprofen. This aryl substituent is a benzene ring bearing an isobutyl group at the para position. ( j) Benzenecarboxylic acid is the systematic name for benzoic acid. Salicylic acid is a derivative of benzoic acid bearing a hydroxyl group at the position ortho to the carboxyl. 19.14 (a) The carboxylic acid contains a linear chain of eight carbon atoms. The parent alkane is octane, and so the systematic name of CH 3 (CH 2 ) 6 CO 2 H is octanoic acid. (b) The compound shown is the potassium salt of octanoic acid. It is potassium octanoate. (c) The presence of a double bond in CH 2 ?CH(CH 2 ) 5 CO 2 H is indicated by the ending -enoic acid. Numbering of the chain begins with the carboxylic acid, and so the double bond is be- tween C-7 and C-8. The compound is 7-octenoic acid. (d) Stereochemistry is systematically described by the E–Z notation. Here, the double bond be- tween C-6 and C-7 in octenoic acid has the Z configuration; the higher ranked substituents are on the same side. (e) A dicarboxylic acid is named as a dioic acid. The carboxyl functions are the terminal carbons of an eight-carbon chain; HO 2 C(CH 2 ) 6 CO 2 H is octanedioic acid. It is not necessary to iden- tify the carboxylic acid locations by number because they can only be at the ends of the chain when the -dioic acid name is used. ( f ) Pick the longest continuous chain that includes both carboxyl groups and name the compound as a -dioic acid. This chain contains only three carbons and bears a pentyl group as a sub- stituent at C-2. It is not necessary to specify the position of the pentyl group, because it can only be attached to C-2. Malonic acid is an acceptable synonym for propanedioic acid; this compound may also be named pentylmalonic acid. Pentylpropanedioic acid CH 3 (CH 2 ) 4 CHCO 2 H CO 2 H 21 3 CC H H 3 C H (CH 2 ) 4 CO 2 H (Z)-6-Octenoic acid o-Hydroxybenzenecarboxylic acid (salicylic acid) OH CO 2 H 2-( p-Isobutylphenyl)- propanoic acid CH 3 CHCO 2 H CH 2 CH(CH 3 ) 2 CARBOXYLIC ACIDS 509 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (g) A carboxylic acid function is attached as a substituent on a seven-membered ring. The com- pound is cycloheptanecarboxylic acid. (h) The aromatic ring is named as a substituent attached to the eight-carbon carboxylic acid. Numbering of the chain begins with the carboxyl group. 19.15 (a) Carboxylic acids are the most acidic class of organic compounds containing only the elements C, H, and O. The order of decreasing acidity is K a pK a Acetic acid CH 3 CO 2 H 1.8 H11003 10 H110025 4.7 Ethanol CH 3 CH 2 OH 10 H1100216 16 Ethane CH 3 CH 3 H1101510 H1100246 H1101546 (b) Here again, the carboxylic acid is the strongest acid and the hydrocarbon the weakest: K a pK a Benzoic acid C 6 H 5 CO 2 H 6.7 H11003 10 H110025 4.2 Benzyl alcohol C 6 H 5 CH 2 OH 10 H1100216 –10 H1100218 16–18 Benzene C 6 H 6 H1101510 H1100243 H1101543 (c) Propanedioic acid is a stronger acid than propanoic acid because the electron-withdrawing effect of one carboxyl group enhances the ionization of the other. Propanedial is a 1,3-dicar- bonyl compound that yields a stabilized enolate; it is more acidic than 1,3-propanediol. K a pK a Propanedioic acid HO 2 CCH 2 CO 2 H 1.4 H11003 10 H110023 2.9 Propanoic acid CH 3 CH 2 CO 2 H 1.3 H11003 10 H110025 4.9 Propanedial O?CHCH 2 CH?O H1101510 H110029 H110159 1,3-Propanediol HOCH 2 CH 2 CH 2 OH H1101510 H1100216 H1101516 (d)Trifluoromethanesulfonic acid is by far the strongest acid in the group. It is structurally related to sulfuric acid, but its three fluorine substituents make it much stronger. Fluorine substituents 6-Phenyloctanoic acid CH(CH 2 ) 4 CO 2 H CH 2 CH 3 CO 2 H 510 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website increase the acidity of carboxylic acids and alcohols relative to their nonfluorinated analogs, but not enough to make fluorinated alcohols as acidic as carboxylic acids. K a pK a Trifluoromethanesulfonic acid CF 3 SO 2 OH 10 6 H110026 Trifluoroacetic acid CF 3 CO 2 H 5.9 H11003 10 H110021 0.2 Acetic acid CH 3 CO 2 H 1.8 H11003 10 H110025 4.7 2,2,2-Trifluoroethanol CF 3 CH 2 OH 4.2 H11003 10 H1100213 12.4 Ethanol CH 3 CH 2 OH H1101510 H1100216 H1101516 (e) The order of decreasing acidity is carboxylic acid H11022 H9252-diketone H11022 ketone H11022 hydrocarbon. K a pK a Cyclopentanecarboxylic acid 1 H11003 10 H110025 5.0 2,4-Pentanedione 10 H110029 9 Cyclopentanone 10 H1100220 20 Cyclopentene 10 H1100245 45 19.16 (a) A trifluoromethyl group is strongly electron-withdrawing and acid-strengthening. Its ability to attract electrons from the carboxylate ion decreases as its distance down the chain increases. 3,3,3-Trifluoropropanoic acid is a stronger acid than 4,4,4-trifluorobutanoic acid. (b) The carbon that bears the carboxyl group in 2-butynoic acid is sp-hybridized and is, therefore, more electron-withdrawing than the sp 3 -hybridized H9251 carbon of butanoic acid. The anion of 2- butynoic acid is therefore stabilized better than the anion of butanoic acid, and 2-butynoic acid is a stronger acid. (c) Cyclohexanecarboxylic acid is a typical aliphatic carboxylic acid and is expected to be sim- ilar to acetic acid in acidity. The greater electronegativity of the sp 2 -hybridized carbon CH 3 C CCO 2 H 2-Butynoic acid K a 2.5 H11003 10 H110023 (pK a 2.6) CH 3 CH 2 CH 2 CO 2 H Butanoic acid K a 1.5 H11003 10 H110025 (pK a 4.8) CF 3 CH 2 CO 2 H 3,3,3-Trifluoropropanoic acid (pK a 3.0) K a 9.6 H11003 10 H110024 CF 3 CH 2 CH 2 CO 2 H 4,4,4-Trifluorobutanoic acid (pK a 4.2) K a 6.9 H11003 10 H110025 O CH 3 CCH 2 CCH 3 O O CO 2 H CARBOXYLIC ACIDS 511 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website attached to the carboxyl group in benzoic acid stabilizes benzoate anion better than the cor- responding sp 3 -hybridized carbon stabilizes cyclohexanecarboxylate. Benzoic acid is a stronger acid. (d) Its five fluorine substituents make the pentafluorophenyl group more electron-withdrawing than an unsubstituted phenyl group. Thus, pentafluorobenzoic acid is a stronger acid than benzoic acid. (e) The pentafluorophenyl substituent is electron-withdrawing and increases the acidity of a car- boxyl group to which it is attached. Its electron-withdrawing effect decreases with distance. Pentafluorobenzoic acid is a stronger acid than p-(pentafluorophenyl)benzoic acid. ( f ) The oxygen of the ring exercises an acidifying effect on the carboxyl group. This effect is largest when the oxygen is attached directly to the carbon that bears the carboxyl group. Furan-2-carboxylic acid is thus a stronger acid than furan-3-carboxylic acid. (g) Furan-2-carboxylic acid has an oxygen attached to the carbon that bears the carboxyl group, whereas pyrrole-2-carboxylic acid has a nitrogen in that position. Oxygen is more Furan-2-carboxylic acid (pK a 3.2) K a 6.9 H11003 10 H110024 CO 2 H O Furan-3-carboxylic acid (pK a 3.9) K a 1.1 H11003 10 H110024 CO 2 H O Pentafluorobenzoic acid K a 4.1 H11003 10 H110024 (pK a 3.4) p-(Pentafluorophenyl)benzoic acid (K a not measured in water; comparable with benzoic acid in acidity) CO 2 H F F F FF CO 2 H F F F FF Pentafluorobenzoic acid K a 4.1 H11003 10 H110024 (pK a 3.4) Benzoic acid K a 6.7 H11003 10 H110025 (pK a 4.2) CO 2 HCO 2 H F F FF F Benzoic acid K a 6.7 H11003 10 H110025 (pK a 4.2) CO 2 H Cyclohexanecarboxylic acid K a 1.2 H11003 10 H110025 (pK a 4.9) CO 2 H 512 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website electronegative than nitrogen and so stabilizes the carboxylate anion better. Furan-2- carboxylic acid is a stronger acid than pyrrole-2-carboxylic acid. 19.17 (a) The conversion of 1-butanol to butanoic acid is simply the oxidation of a primary alcohol to a carboxylic acid. Chromic acid is a suitable oxidizing agent. (b) Aldehydes may be oxidized to carboxylic acids by any of the oxidizing agents that convert primary alcohols to carboxylic acids. (c) The starting material has the same number of carbon atoms as does butanoic acid, and so all that is required is a series of functional group transformations. Carboxylic acids may be ob- tained by oxidation of the corresponding primary alcohol. The alcohol is available from the designated starting material, 1-butene. Hydroboration–oxidation of 1-butene yields 1-butanol, which can then be oxidized to butanoic acid as in part (a). (d) Converting 1-propanol to butanoic acid requires the carbon chain to be extended by one atom. Both methods for achieving this conversion, carboxylation of a Grignard reagent and forma- tion and hydrolysis of a nitrile, begin with alkyl halides. Alkyl halides in turn are prepared from alcohols. CH 3 CH 2 CH 2 CO 2 HCH 3 CH 2 CH 2 MgBr CH 3 CH 2 CH 2 BrCH 3 CH 2 CH 2 OH CH 3 CH 2 CH 2 CNor CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 CO 2 HCH 3 CH 2 CH CH 3 1. B 2 H 6 2. H 2 O 2 , HO H11002 H 2 CrO 4 1-Butene 1-Butanol Butanoic acid CH 3 CH 2 CH 2 CO 2 HCH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH CH 2 K 2 Cr 2 O 7 H 2 SO 4 , H 2 O Butanal CH 3 CH 2 CH 2 CH O Butanoic acid CH 3 CH 2 CH 2 COH O H 2 CrO 4 CH 3 CH 2 CH 2 CH 2 OH 1-Butanol CH 3 CH 2 CH 2 CO 2 H Butanoic acid Furan-2-carboxylic acid (pK a 3.2) K a 6.9 H11003 10 H110024 CO 2 H O CO 2 H N H Pyrrole-3-carboxylic acid (pK a 4.4) K a 3.5 H11003 10 H110025 CARBOXYLIC ACIDS 513 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Either of the two following procedures is satisfactory: (e) Dehydration of 2-propanol to propene followed by free-radical addition of hydrogen bromide affords 1-bromopropane. Once 1-bromopropane has been prepared it is converted to butanoic acid as in part (d). ( f ) The carbon skeleton of butanoic acid may be assembled by an aldol condensation of acetaldehyde. Oxidation of the aldehyde followed by hydrogenation of the double bond yields butanoic acid. (g) Ethylmalonic acid belongs to the class of substituted malonic acids that undergo ready ther- mal decarboxylation. Decarboxylation yields butanoic acid. 19.18 (a) The Friedel–Crafts alkylation of benzene by methyl chloride can be used to prepare 14 C-labeled toluene (C* H11005 14 C). Once prepared, toluene could be oxidized to benzoic acid. Benzene Toluene CH 3 Benzoic acid CO 2 H Methyl chloride CH 3 ClH11001 AlCl 3 K 2 Cr 2 O 7 , H 2 SO 4 H 2 O, heat ** * heat CH 3 CH 2 CH 2 CO 2 H Butanoic acid H11001 CO 2 Carbon dioxide CO 2 H CH 3 CH 2 CHCO 2 H Ethylmalonic acid H 2 CrO 4 H 2 Pt CH 3 CH CHCH O 2-Butenal CH 3 CH CHCO 2 H 2-Butenoic acid CH 3 CH 2 CH 2 CO 2 H Butanoic acid KOH, ethanol heat 2CH 3 CH O Acetaldehyde O CH 3 CH CHCH 2-Butenal CH 3 CH 2 CH 2 CO 2 HCH 3 CH CHCH O 2CH 3 CH O HBr peroxides H 2 SO 4 heat CH 3 CHCH 3 OH 2-Propanol CH 3 CH CH 2 Propene CH 3 CH 2 CH 2 Br 1-Bromopropane PBr 3 or HBr KCN DMSO H 2 O, HCl heat CH 3 CH 2 CH 2 Br 1-Bromopropane CH 3 CH 2 CH 2 CN Butanenitrile CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CH 2 CH 2 CO 2 H Butanoic acid CH 3 CH 2 CH 2 MgBr PBr 3 or HBr Mg diethyl ether 1. CO 2 2. H 3 O H11001 CH 3 CH 2 CH 2 Br 1-Bromopropane CH 3 CH 2 CH 2 OH 1-Propanol CH 3 CH 2 CH 2 CO 2 H Butanoic acid 514 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Formaldehyde can serve as a one-carbon source if it is attacked by the Grignard reagent de- rived from bromobenzene. This sequence yields 14 C-labeled benzyl alcohol, which can be oxidized to 14 C-labeled benzoic acid. (c) A direct route to 14 C-labeled benzoic acid utilizes a Grignard synthesis employing 14 C-labeled carbon dioxide. 19.19 (a) An acid–base reaction takes place when pentanoic acid is combined with sodium hydroxide. (b) Carboxylic acids react with sodium bicarbonate to give carbonic acid, which dissociates to carbon dioxide and water, so that the actual reaction that takes place is (c) Thionyl chloride is a reagent that converts carboxylic acids to the corresponding acyl chlorides. (d) Phosphorus tribromide is used to convert carboxylic acids to their acyl bromides. Pentanoic acid 3CH 3 CH 2 CH 2 CH 2 CO 2 H Phosphorus tribromide PBr 3 Phosphorus acid H 3 PO 3 H11001H11001 Pentanoyl bromide 3CH 3 CH 2 CH 2 CH 2 CBr O Pentanoic acid CH 3 CH 2 CH 2 CH 2 CO 2 H Thionyl chloride SOCl 2 Sulfur dioxide SO 2 H11001H11001 Hydrogen chloride HClH11001 Pentanoyl chloride CH 3 CH 2 CH 2 CH 2 CCl O Pentanoic acid CH 3 CH 2 CH 2 CH 2 CO 2 H Sodium pentanoate CH 3 CH 2 CH 2 CH 2 CO 2 Na Sodium bicarbonate NaHCO 3 Carbon dioxide CO 2 H11001H11001 Water H 2 OH11001 Pentanoic acid CH 3 CH 2 CH 2 CH 2 CO 2 H Sodium pentanoate CH 3 CH 2 CH 2 CH 2 CO 2 Na Sodium hydroxide NaOH Water H 2 OH11001H11001 Benzene Bromobenzene Br Phenylmagnesium bromide MgBr Br 2 FeBr 3 Mg diethyl ether 2. H 3 O H11001 1. CO 2 Benzoic acid CO 2 H * * K 2 Cr 2 O 7 , H 2 SO 4 H 2 O Benzyl alcohol CH 2 OH * Benzoic acid CO 2 H * Benzene Bromobenzene Br Phenylmagnesium bromide MgBr Br 2 FeBr 3 Mg diethyl ether 2. H 3 O H11001 1. HCH O Benzyl alcohol CH 2 OH * * CARBOXYLIC ACIDS 515 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) Carboxylic acids react with alcohols in the presence of acid catalysts to give esters. ( f ) Chlorine is introduced at the H9251-carbon atom of a carboxylic acid. The reaction is catalyzed by a small amount of phosphorus or a phosphorus trihalide and is called the Hell–Volhard– Zelinsky reaction. The H9251-halo substituent is derived from the halogen used, not from the phosphorus trihalide. (g) In the case, bromine is introduced at the H9251 carbon. (h) H9251-Halo carboxylic acids are reactive substrates in nucleophilic substitution. Iodide acts as a nucleophile to displace bromide from 2-bromopentanoic acid. (i) Aqueous ammonia converts H9251-halo acids to H9251-amino acids. ( j ) Lithium aluminum hydride is a powerful reducing agent and reduces carboxylic acids to pri- mary alcohols. (k) Phenylmagnesium bromide acts as a base to abstract the carboxylic acid proton. CH 3 CH 2 CH 2 CH 2 CO 2 MgBr Bromomagnesium pentanoate H11001CH 3 CH 2 CH 2 CH 2 CO 2 H Pentanoic acid C 6 H 5 MgBr Phenylmagnesium bromide H11001 C 6 H 6 Benzene CH 3 CH 2 CH 2 CH 2 CH 2 OH 1. LiAlH 4 2. H 2 O 1-Pentanol CH 3 CH 2 CH 2 CH 2 CO 2 H Pentanoic acid H11001H11001CH 3 CH 2 CH 2 CHCO 2 H Br 2-Bromopentanoic acid 2NH 3 Ammonia CH 3 CH 2 CH 2 CHCO 2 H NH 2 2-Aminopentanoic acid NH 4 Br Ammonium bromide H11001H11001 acetone CH 3 CH 2 CH 2 CHCO 2 H Br 2-Bromopentanoic acid NaI Sodium iodide CH 3 CH 2 CH 2 CHCO 2 H I 2-Iodopentanoic acid NaBr Sodium bromide Pentanoic acid CH 3 CH 2 CH 2 CH 2 CO 2 H Bromine Br 2 Hydrogen bromide HBrH11001H11001 2-Bromopentanoic acid CH 3 CH 2 CH 2 CHCO 2 H Br PCl 3 (catalyst) Pentanoic acid CH 3 CH 2 CH 2 CH 2 CO 2 H Chlorine Cl 2 Hydrogen chloride HClH11001H11001 2-Chloropentanoic acid CH 3 CH 2 CH 2 CHCO 2 H Cl PBr 3 (catalyst) Pentanoic acid CH 3 CH 2 CH 2 CH 2 CO 2 H Benzyl alcohol C 6 H 5 CH 2 OH Water H 2 OH11001H11001 Benzyl pentanoate CH 3 CH 2 CH 2 CH 2 COCH 2 C 6 H 5 O H 2 SO 4 516 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Grignard reagents are not compatible with carboxylic acids; proton transfer converts the Grignard reagent to the corresponding hydrocarbon. 19.20 (a) Conversion of butanoic acid to 1-butanol is a reduction and requires lithium aluminum hy- dride as the reducing agent. (b) Carboxylic acids cannot be reduced directly to aldehydes. The following two-step procedure may be used: (c) Remember that alkyl halides are usually prepared from alcohols. 1-Butanol is therefore needed in order to prepare 1-chlorobutane. (d) Carboxylic acids are converted to their corresponding acyl chlorides with thionyl chloride. (e) Aromatic ketones are frequently prepared by Friedel–Crafts acylation of the appropriate acyl chloride and benzene. Butanoyl chloride, prepared in part (d), can be used to acylate benzene in a Friedel–Crafts reaction. ( f ) The preparation of 4-octanone using compounds derived from butanoic acid may be seen by using disconnections in a retrosynthetic analysis. CH 3 CH 2 CH 2 CCH 2 CH 2 CH 2 CH 3 O CH 3 CH 2 CH 2 CH CH 2 CH 2 CH 2 CH 3 OH CH 3 CH 2 CH 2 CH 2 MgBrH11001CH 3 CH 2 CH 2 CH O AlCl 3 Butanoyl chloride CH 3 CH 2 CH 2 CCl O H11001 Benzene Phenyl propyl ketone CCH 2 CH 2 CH 3 O SOCl 2 Butanoic acid CH 3 CH 2 CH 2 COH O Butanoyl chloride CH 3 CH 2 CH 2 CCl O SOCl 2 CH 3 CH 2 CH 2 CH 2 OH 1-Butanol [from part (a)] CH 3 CH 2 CH 2 CH 2 Cl 1-Chlorobutane 1. LiAlH 4 2. H 2 O PCC CH 2 Cl 2 CH 3 CH 2 CH 2 CH 2 OH 1-ButanolButanoic acid CH 3 CH 2 CH 2 COH O Butanal CH 3 CH 2 CH 2 CH O 1. LiAlH 4 2. H 2 O CH 3 CH 2 CH 2 CH 2 OH 1-ButanolButanoic acid CH 3 CH 2 CH 2 COH O CARBOXYLIC ACIDS 517 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The reaction scheme which may be used is (g) Carboxylic acids are halogenated at their H9251-carbon atom by the Hell–Volhard–Zelinsky reaction. A catalytic amount of PCl 3 may be used in place of phosphorus in the reaction. (h) Dehydrohalogenation of 2-bromobutanoic acid gives 2-butenoic acid. 19.21 (a) The compound to be prepared is glycine, an H9251-amino acid. The amino functional group can be introduced by a nucleophilic substitution reaction on an H9251-halo acid, which is available by way of the Hell–Volhard–Zelinsky reaction. (b) Phenoxyacetic acid is used as a fungicide. It can be prepared by a nucleophilic substitution using sodium phenoxide and bromoacetic acid. BrCH 2 CO 2 H Bromoacetic acid H11001 NaClC 6 H 5 OCH 2 CO 2 H Phenoxyacetic acid 1. C 6 H 5 ONa 2. H H11001 H 2 NCH 2 CO 2 H Aminoacetic acid (glycine) NH 3 , H 2 O BrCH 2 CO 2 H Bromoacetic acid CH 3 CO 2 H Acetic acid Br 2 P 1. KOC(CH 3 ) 3 DMSO 2. H H11001 CH 3 CH 2 CHCO 2 H Br 2-Bromobutanoic acid CH 3 CH CHCO 2 H 2-Butenoic acid Br 2 P CH 3 CH 2 CH 2 CO 2 H Butanoic acid CH 3 CH 2 CHCO 2 H Br 2-Bromobutanoic acid 1. diethyl ether 2. H 3 O H11001 Butylmagnesium chloride CH 3 CH 2 CH 2 CH 2 MgCl H11001 Butanal [from part (b)] CH 3 CH 2 CH 2 CH O 4-Octanol CH 3 CH 2 CH 2 CHCH 2 CH 2 CH 2 CH 3 OH 4-Octanone CH 3 CH 2 CH 2 CCH 2 CH 2 CH 2 CH 3 O H 2 CrO 4 CH 3 CH 2 CH 2 CH 2 Cl 1-Chlorobutane [from part (c)] CH 3 CH 2 CH 2 CH 2 MgCl Butylmagnesium chloride Mg 518 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Cyanide ion is a good nucleophile and will displace bromide from bromoacetic acid. (d) Cyanoacetic acid, prepared as in part (c), serves as a convenient precursor to malonic acid. Hydrolysis of the nitrile substituent converts it to a carboxyl group. (e) Iodoacetic acid is not prepared directly from acetic acid but is derived by nucleophilic substi- tution of iodide in bromoacetic acid. ( f ) Two transformations need to be accomplished, H9251 bromination and esterification. The correct sequence is bromination followed by esterification. Reversing the order of steps is not appropriate. It must be the carboxylic acid that is subjected to halogenation because the Hell–Volhard–Zelinsky reaction is a reaction of carboxylic acids, not esters. (g) The compound shown is an ylide. It can be prepared from ethyl bromoacetate as shown The first step is a nucleophilic substitution of bromide by triphenylphosphine. Treatment of the derived triphenylphosphonium salt with base removes the relatively acidic H9251 proton, form- ing the ylide. (For a review of ylide formation, refer to Section 17.12.) (h) Reaction of the ylide formed in part (g) with benzaldehyde gives the desired alkene by a Wittig reaction. H11001 Ylide from part (g) (C 6 H 5 ) 3 P CHCO 2 CH 2 CH 3 H11001 Triphenylphosphine oxide (C 6 H 5 ) 3 PO H11001 H11002 H11001 Ethyl cinnamate C 6 H 5 CH CHCO 2 CH 2 CH 3 Benzaldehyde C 6 H 5 CH O H11002 H11001BrCH 2 CO 2 CH 2 CH 3 Ethyl bromoacetate (C 6 H 5 ) 3 P Triphenyl- phosphine (C 6 H 5 ) 3 PCH 2 CO 2 CH 2 CH 3 H11001 NaOCH 2 CH 3 Ylide (C 6 H 5 ) 3 P CHCO 2 CH 2 CH 3 H11001 H11002 Br H11002 BrCH 2 CO 2 CH 2 CH 3 Ethyl bromoacetate CH 3 CH 2 OH H H11001 BrCH 2 CO 2 H Bromoacetic acid CH 3 CO 2 H Acetic acid Br 2 P NaI acetone Iodoacetic acid ICH 2 CO 2 H Bromoacetic acid BrCH 2 CO 2 H H 2 O, H H11001 heat Malonic acid HO 2 CCH 2 CO 2 H Cyanoacetic acid NCCH 2 CO 2 H BrCH 2 CO 2 H Bromoacetic acid [from part (a)] NaCN Na 2 CO 3 , H 2 O H H11001 Sodium cyanoacetate NCCH 2 CO 2 Na Cyanoacetic acid NCCH 2 CO 2 H CARBOXYLIC ACIDS 519 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 19.22 (a) Carboxylic acids are converted to ethyl esters when they are allowed to stand in ethanol in the presence of an acid catalyst. (b) Lithium aluminum hydride, LiAlH 4 , reduces carboxylic acids to primary alcohols. When LiAlD 4 is used, deuterium is transferred to the carbonyl carbon. Notice that deuterium is bonded only to carbon. The hydroxyl proton is derived from water, not from the reducing agent. (c) In the presence of a catalytic amount of phosphorus, bromine reacts with carboxylic acids to yield the corresponding H9251-bromo derivative. (d) Alkyl fluorides are not readily converted to Grignard reagents, and so it is the bromine sub- stituent that is attacked by magnesium. (e) Cyano substituents are hydrolyzed to carboxyl groups in the presence of acid catalysts. H 2 O, acetic acid H 2 SO 4 , heat CH 2 CO 2 H Cl m-Chlorophenylacetic acid (61%) CH 2 CN Cl m-Chlorobenzyl cyanide CF 3 MgBr Mg diethyl ether CF 3 Br m-Bromo(trifluoromethyl)- benzene 1. CO 2 2. H 3 O H11001 CF 3 CO 2 H m-(Trifluoromethyl)- benzoic acid Cyclohexanecarboxylic acid CO 2 H Br 2 P 1-Bromocyclohexanecarboxylic acid (96%) CO 2 H Br Cyclopropanecarboxylic acid COH O 1-Cyclopropyl-1,1- dideuteriomethanol (75%) COH D D 1. LiAlD 4 , diethyl ether 2. H 2 O H11001 Ethanol CH 3 CH 2 OH H11001 Water H 2 O (E)-2-Methyl-2-butenoic acid C H 3 C CH 3 CO 2 HH C Ethyl (E)-2-methyl-2-butenoate (74–80%) O C H 3 C CH 3 COCH 2 CH 3 H C H 2 SO 4 520 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) The carboxylic acid function plays no part in this reaction; free-radical addition of hydrogen bromide to the carbon–carbon double bond occurs. Recall that hydrogen bromide adds to alkenes in the presence of peroxides with a regioselec- tivity opposite to that of Markovnikov’s rule. 19.23 (a) The desired product and the starting material have the same carbon skeleton, and so all that is required is a series of functional group transformations. Recall that, as seen in Problem 19.17, a carboxylic acid may be prepared by oxidation of the corresponding primary alcohol. The needed alcohol is available from the appropriate alkene. (b) The target molecule contains one more carbon than the starting material, and so a carbon– carbon bond-forming step is indicated. Two approaches are reasonable; one proceeds by way of nitrile formation and hydrolysis, the other by carboxylation of a Grignard reagent. In either case the key intermediate is 1-bromo-2-methylpropane. The desired alkyl bromide may be prepared by free-radical addition of hydrogen bromide to 2-methylpropene. Another route to the alkyl bromide utilizes the alcohol prepared in part (a). Conversion of the alkyl bromide to the desired acid is then carried out as follows: 1-Bromo-2-methylpropane (CH 3 ) 2 CHCH 2 Br (CH 3 ) 2 CHCH 2 CN 3-Methylbutanoic acid (CH 3 ) 2 CHCH 2 CO 2 H H 2 O, H H11001 heat KCN (CH 3 ) 2 CHCH 2 MgBr 3-Methylbutanoic acid (CH 3 ) 2 CHCH 2 CO 2 H Mg diethyl ether 1. CO 2 2. H 3 O H11001 2-Methyl-1-propanol (CH 3 ) 2 CHCH 2 OH 1-Bromo-2- methylpropane (CH 3 ) 2 CHCH 2 Br PBr 3 or HBr tert-Butyl alcohol (CH 3 ) 3 COH 2-Methylpropene CH 2 (CH 3 ) 2 C H H11001 heat 1-Bromo-2- methylpropane (CH 3 ) 2 CHCH 2 Br HBr peroxides 3-Methylbutanoic acid (CH 3 ) 2 CHCH 2 CO 2 H 1-Bromo-2- methylpropane (CH 3 ) 2 CHCH 2 Br tert-Butyl alcohol (CH 3 ) 3 COH 2-Methylpropene CH 2 (CH 3 ) 2 C H H11001 heat 1. B 2 H 6 2. H 2 O 2 , HO H11002 (CH 3 ) 2 CHCH 2 OH 2-Methyl-1-propanol (CH 3 ) 2 CHCO 2 H 2-Methylpropanoic acid K 2 Cr 2 O 7 H 2 SO 4 , H 2 O 10-Undecenoic acid CH(CH 2 ) 8 CO 2 HH 2 C BrCH 2 CH 2 (CH 2 ) 8 CO 2 H 11-Bromoundecanoic acid (66–70%) HBr benzoyl peroxide CARBOXYLIC ACIDS 521 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Examining the target molecule reveals that it contains two more carbon atoms than the indi- cated starting material, suggesting use of ethylene oxide in a two-carbon chain-extension process. This suggests the following sequence of steps: (d) This synthesis requires extending a carbon chain by two carbon atoms. One way to form dicarboxylic acids is by hydrolysis of dinitriles. This suggests the following sequence of steps: (e) The desired alcohol cannot be prepared directly from the nitrile. It is available, however, by lithium aluminum hydride reduction of the carboxylic acid obtained by hydrolysis of the nitrile. heat H 2 O, H H11001 2. H 2 O 1. LiAlH 4 3-Phenylbutanenitrile CH 3 CHCH 2 CN C 6 H 5 3-Phenyl-1-butanol CH 3 CHCH 2 CH 2 OH C 6 H 5 3-Phenylbutanoic acid CH 3 CHCH 2 COH C 6 H 5 O 1. LiAlH 4 2. H 2 O HO 2 C(CH 2 ) 3 CO 2 H Pentanedioic acid HOCH 2 (CH 2 ) 3 CH 2 OH 1,5-Pentanediol BrCH 2 (CH 2 ) 3 CH 2 Br 1,5-Dibromopentane NCCH 2 (CH 2 ) 3 CH 2 CN 1,5-Dicyanopentane HO 2 CCH 2 (CH 2 ) 3 CH 2 CO 2 H H11013 HO 2 C(CH 2 ) 5 CO 2 H Heptanedioic acid KCN HBr or PBr 3 H 2 O, H H11001 heat HO 2 C(CH 2 ) 5 CO 2 H NC(CH 2 ) 5 CN Br(CH 2 ) 5 Br HBr Mg (CH 3 ) 3 COH tert-Butyl alcohol (CH 3 ) 3 CBr 2-Bromo-2- methylpropane (CH 3 ) 3 CMgBr tert-Butylmagnesium bromide (CH 3 ) 3 CCH 2 CH 2 OH 3,3-Dimethyl-1- butanol (CH 3 ) 3 CCH 2 CO 2 H 3,3-Dimethylbutanoic acid K 2 Cr 2 O 7 , H H11001 H 2 O 2. H 3 O H11001 H 2 CCH 2 O 1. (CH 3 ) 3 CCH 2 CO 2 HCH 2 CH 2 OH(CH 3 ) 3 C (CH 3 ) 3 CMgX H11001 H 2 CCH 2 O 522 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) In spite of the structural similarity between the starting material and the desired product, a one-step transformation cannot be achieved. Instead, recall that H9251-bromo acids are prepared from carboxylic acids by the Hell– Volhard–Zelinsky reaction: The problem now simplifies to one of preparing cyclopentanecarboxylic acid from cyclopentyl bromide. Two routes are possible: The Grignard route is better; it is a “one-pot” transformation. Converting the secondary bro- mide to a nitrile will be accompanied by elimination, and the procedure requires two separate operations. (g) In this case the halogen substituent is present at the H9252 carbon rather than the H9251 carbon atom of the carboxylic acid. The starting material, a H9252-chloro unsaturated acid, can lead to the desired carbon skeleton by a Diels–Alder reaction. The required trans stereochemistry is a consequence of the stereospecificity of the Diels–Alder reaction. H11001 1,3-Butadiene CO 2 H Cl H H C C (E)-3-Chloropropenoic acid CO 2 H Cl trans-2-Chloro-4- cyclohexenecarboxylic acid CO 2 H Cl H11001 CO 2 H Cl H H C C CO 2 H Cl Br Cyclopentyl bromide MgBr Cyclopentylmagnesium bromide CO 2 H Cyclopentanecarboxylic acid Mg diethyl ether 1. CO 2 2. H 3 O H11001 Br Cyclopentyl bromide CN Cyclopentyl cyanide CO 2 H Cyclopentanecarboxylic acid KCN H 2 O, H H11001 heat CO 2 H Br 1-Bromocyclopentane- carboxylic acid CO 2 H H Cyclopentane- carboxylic acid Br 2 P Br Cyclopentyl bromide CO 2 H Br 1-Bromocyclopentane- carboxylic acid CARBOXYLIC ACIDS 523 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Hydrogenation of the double bond of the Diels–Alder adduct gives the required product. (h) The target molecule is related to the starting material by the retrosynthesis The necessary bromine substituent can be introduced by electrophilic substitution in the acti- vated aromatic ring of m-xylene. The aryl bromide cannot be converted to a carboxylic acid by way of the corresponding nitrile, because aryl bromides are not reactive toward nucleophilic substitution. The Grignard route is necessary. (i) The relationship of the target molecule to the starting material requires that there be two synthetic operations: oxidation of the methyl group and nitration of the ring. The orientation of the nitro group requires that nitration must follow oxidation of the p-Chlorotoluene CH 3 Cl 4-Chloro-3-nitrobenzoic acid CO 2 H NO 2 Cl 2,4-Dimethylbenzoic acid 1. Mg, diethyl ether 2. CO 2 3. H 3 O H11001 CO 2 H CH 3 CH 3 1-Bromo-2,4- dimethylbenzene Br CH 3 CH 3 m-Xylene Br CH 3 CH 3 1-Bromo-2,4- dimethylbenzene CH 3 CH 3 Br 2 acetic acid (or Br 2 , FeBr 3 ) 2,4-Dimethylbenzoic acid CO 2 H CH 3 CH 3 Br CH 3 CH 3 m-Xylene CH 3 CH 3 H 2 Pt CO 2 H Cl trans-2-Chloro-4- cyclohexenecarboxylic acid CO 2 H Cl trans-2-Chlorocyclo- hexanecarboxylic acid 524 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website methyl group of the starting material Nitration of p-chlorobenzoic acid gives the desired product, because the directing effects of the chlorine (ortho, para) and the carboxyl (meta) groups reinforce each other. ( j) The desired synthetic route becomes apparent when it is recognized that the Z alkene stereoisomer may be obtained from an alkyne, which, in turn, is available by carboxylation of the anion derived from the starting material. The desired reaction sequence is Hydrogenation of the carbon–carbon triple bond of 2-butynoic acid over the Lindlar catalyst converts this compound to the Z isomer of 2-butenoic acid. 19.24 (a) Only the cis stereoisomer of 4-hydroxycyclohexanecarboxylic acid is capable of forming a lactone, as can be seen in the following drawings or with a molecular model. The most stable conformation of the starting hydroxy acid is a chair conformation; however, in the lactone, the cyclohexane ring adopts a boat conformation. cis-4-Hydroxycyclohexane carboxylic acid (boat conformation) C O OH OH Lactone C O O cis-4-Hydroxycyclohexane carboxylic acid (chair conformation) HOC OH O H 2 , Lindlar Pd CH 3 C CCO 2 H 2-Butynoic acid (Z)-2-Butenoic acid CC H 3 C HH CO 2 H NaNH 2 NH 3 1. CO 2 2. H 3 O H11001 CH 3 CCH Propyne CH 3 C CCO 2 H 2-Butynoic acid CH 3 C CNa Propynylsodium CH 3 C CCO 2 HCC H 3 C HH CO 2 H CO 2 H11001CH 3 CC H11002 HNO 3 H 2 SO 4 CO 2 H Cl p-Chlorobenzoic acid CO 2 H NO 2 Cl 4-Chloro-3- nitrobenzoic acid K 2 Cr 2 O 7 , H 2 SO 4 H 2 O CH 3 Cl p-Chlorotoluene CO 2 H Cl p-Chlorobenzoic acid CARBOXYLIC ACIDS 525 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) As in part (a), lactone formation is possible only when the hydroxyl and carboxyl groups are cis. Although the most stable conformation of cis-3-hydroxycyclohexanecarboxylic acid has both substituents equatorial and is unable to close to a lactone, the diaxial orientation is accessible and is capable of lactone formation. Neither conformation of trans-3-hydroxycyclohexanecarboxylic acid has the substituents close enough to each other to form an unstrained lactone. 19.25 (a) The most stable conformation of formic acid is the one that has both hydrogens anti. A plausible explanation is that the syn conformation is destabilized by lone-pair repulsions. (b) A dipole moment of zero can mean that the molecule has a center of symmetry. One structure that satisfies this requirement is characterized by intramolecular hydrogen bonding between the two carboxyl groups and an anti relationship between the two carbonyls. CC O H O O H O Intramolecular hydrogen bond Intramolecular hydrogen bond Anti O H C H O Syn C H O O H Repulsion between lone pairs C H O H O Syn: less stable conformation of formic acid C H O O H Anti: more stable conformation of formic acid trans-3-Hydroxycyclohexanecarboxylic acid: lactone formation impossible CO 2 H OH CO 2 H HO cis-3-Hydroxycyclo- hexanecarboxylic acid CO 2 HHO Lactone O O CO 2 H HO 526 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Another possibility is the following structure; it also has a center of symmetry and an anti re- lationship between the two carbonyls. Other centrosymmetric structures can be drawn; these have the two hydrogen atoms out of the plane of the carboxyl groups, however, and are less likely to occur, in view of the known pla- narity of carboxyl groups. Structures in which the carbonyl groups are syn to each other do not have a center of symmetry. (c) The anion formed on dissociation of o-hydroxybenzoic acid can be stabilized by an intra- molecular hydrogen bond. (d) Ascorbic acid is relatively acidic because ionization of its enolic hydroxyl at C-3 gives an anion that is stabilized by resonance in much the same way as a carboxylate ion; the negative charge is shared by two oxygens. 19.26 Dicarboxylic acids in which both carboxyl groups are attached to the same carbon undergo ready thermal decarboxylation to produce the enol form of an acid. This enol yields a mixture of cis- and trans-3-chlorocyclobutanecarboxylic acid. The two products are stereoisomers. H11001 C OH OH Cl CO 2 H H H Cl HO 2 C H H Cl cis-3-Chlorocyclobutane- carboxylic acid trans-3-Chlorocyclobutane- carboxylic acid H11001 CO 2 C OH OH Cl heat C C Cl O O O OH H Compound A Acidic proton in ascorbic acid OHHO O O ONaHCO 3 O OH O H11002 OHO O H11002 HOCH 2 OHH HOCH 2 OHH HOCH 2 OHH O O C O H O H11002 o-Hydroxybenzoate ion (stabilized by hydrogen bonding) O OCH 3 o-Methoxybenzoate ion (hydrogen bonding is not possible) C O H11002 C H C H O O O O CARBOXYLIC ACIDS 527 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 19.27 Examination of the molecular formula C 14 H 26 O 2 reveals that the compound has an index of hydrogen deficiency of 2. Because we are told that the compound is a carboxylic acid, one of these elements of unsaturation must be a carbon–oxygen double bond. The other must be a carbon–carbon double bond because the compound undergoes cleavage on ozonolysis. Examining the products of ozonol- ysis serves to locate the position of the double bond. The starting acid must be 5-tetradecenoic acid. The stereochemistry of the double bond is not re- vealed by these experiments. 19.28 Hydrogenation of the starting material is expected to result in reduction of the ketone carbonyl while leaving the carboxyl group unaffected. Because the isolated product lacks a carboxyl group, how- ever, that group must react in some way. The most reasonable reaction is intramolecular esterifica- tion to form a H9253-lactone. 19.29 Compound A is a cyclic acetal and undergoes hydrolysis in aqueous acid to produce acetaldehyde, along with a dihydroxy carboxylic acid. The dihydroxy acid that is formed in this step cyclizes to the H9254-lactone mevalonolactone. 19.30 Compound A is a H9254-lactone. To determine its precursor, disconnect the ester linkage to a hydroxy acid. Compound A O HH CH 3 O HH CH 3 CO 2 H OH 3,5-Dihydroxy-3- methylpentanoic acid CH 3 CH 2 CO 2 H HO OH HO HO CH 2 O CH 3 OH Mevalonolactone O CH 3 OH O H11013 Compound A 3,5-Dihydroxy-3-methylpentanoic acid Acetaldehyde H 3 O H11001 O O CH 3 CH 3 CH 2 CO 2 H CH 3 CH 2 CO 2 H HO OH O CH 3 CHH11001 Levulinic acid 4-Hydroxypentanoic acid (not isolated) 4-Pentanolide (C 5 H 8 O 2 ) H 2 , Ni O OH 3 C CH 3 CHCH 2 CH 2 COH OH O CH 3 CCH 2 CH 2 COH OO CH 3 (CH 2 ) 7 CH HC(CH 2 ) 3 CO 2 H Cleavage by ozone occurs here. Nonanal 5-Oxopentanoic acid CH 3 (CH 2 ) 7 CH CH(CH 2 ) 3 CO 2 H 1. O 3 2. H 2 O, Zn H11001 O O 528 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The precursor has the same carbon skeleton as the designated starting material. All that is necessary is to hydrogenate the double bond of the alkynoic acid to the cis alkene. This can be done by using the Lindlar catalyst. Cyclization of the hydroxy acid to the lactone is spontaneous. 19.31 Hydration of the double bond can occur in two different directions: (a) The achiral isomer is citric acid. (b) The other isomer, isocitric acid, has two stereogenic centers (marked with an asterisk*). Iso- citric acid has the constitution With two stereogenic centers, there are 2 2 , or four, stereoisomers represented by this constitu- tion. The one that is actually formed in this enzyme-catalyzed reaction is the 2R,3S isomer. 19.32 Carboxylic acid protons give signals in the range H9254 10–12 ppm. A signal in this region suggests the presence of a carboxyl group but tells little about its environment. Thus, in assigning structures to compounds A, B, and C, the most useful data are the chemical shifts of the protons other than the carboxyl protons. Compare the three structures: The proton that is diagnostic of structure in formic acid is bonded to a carbonyl group; it is an aldehyde proton. Typical chemical shifts of aldehyde protons are 8–10 ppm, and therefore formic acid is compound C. CH O Compound C H9254 8.0 ppm OH H9254 11.4 ppm CC HO 2 C H CO 2 H H HO 2 CCH 2 CO 2 HHCOH O Malonic acidMaleic acidFormic acid HO 2 CCHCHCH 2 CO 2 H OH CO 2 H * * Isocitric acid HO 2 CCH 2 CCH 2 CO 2 H OH CO 2 H Citric acid has no stereogenic centers. CC H HO 2 C CH 2 CO 2 H CO 2 H H 2 O HO 2 CCH 2 CCH 2 CO 2 H OH CO 2 H H11001 HO 2 CCHCHCH 2 CO 2 H OH CO 2 H 5-Hydroxy-2-hexynoic acid CH 3 CHCH 2 C CCO 2 H OH H 2 Lindlar Pd (Not isolated) Compound A CC H 2 C H CO 2 H H CH 3 CHOH CH 3 O O CARBOXYLIC ACIDS 529 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The critical signal in maleic acid is that of the vinyl protons, which normally is found in the range H9254 5–7 ppm. Maleic acid is compound B. Compound A is malonic acid. Here we have a methylene group bearing two carbonyl substituents. These methylene protons are more shielded than the aldehyde proton of formic acid or the vinyl pro- tons of maleic acid. 19.33 Compounds A and B both exhibit 1 H NMR absorptions in the region H9254 11–12 ppm characteristic of carboxylic acids. The formula C 4 H 8 O 3 suggests an index of hydrogen deficiency of 1, accounted for by the carbonyl of the carboxyl group. Compound A has the triplet–quartet splitting indicative of an ethyl group, and compound B has two triplets, suggesting —CH 2 CH 2 —. 19.34 (a) The formula of compound A (C 3 H 5 ClO 2 ) has an index of hydrogen deficiency of 1—the carboxyl group. Only two structures are possible: Compound A is determined to be 3-chloropropanoic acid on the basis of its 1 H NMR spec- trum, which shows two triplets at H9254 2.9 and H9254 3.8 ppm. Compound A cannot be 2-chloropropanoic acid, because that compound’s 1 H NMR spectrum would show a three-proton doublet for the methyl group and a one-proton quartet for the methine proton. (b) The formula of compound B (C 9 H 9 NO 4 ) corresponds to an index of hydrogen deficiency of 6. The presence of an aromatic ring, as evidenced by the 1 H NMR absorptions at H9254 7.5 and C H9254 11.9 ppmTriplet H9254 2.9 ppm Compound A CH 2 ClCH 2 O OH Triplet H9254 3.8 ppm 3-Chloropropanoic acid ClCH 2 CH 2 C O O OH 2-Chloropropanoic acid and CH 3 CHC OH Cl CH 3 CH 2 OCH 2 COH 4.1 ppm (s) 11.1 ppm (s)3.6 ppm (q) 1.3 ppm (t) Compound A CH 3 OCH 2 CH 2 COH 2.6 ppm (t) 11.3 ppm (s)3.7 ppm (t) 3.4 ppm (s) Compound B OO HO 2 CCH 2 CO 2 H H9254 12.1 ppmH9254 3.2 ppm Compound A CC H HO 2 C CO 2 H H H9254 6.3 ppm H9254 12.4 ppm Compound B 530 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 8.2 ppm, accounts for four of the unsaturations. The appearance of the aromatic protons as a pair of doublets with a total area of 4 suggests a para-disubstituted ring. That compound B is a carboxylic acid is evidenced by the singlet (area H11005 1) at H9254 12.1 ppm. The remaining 1 H NMR signals—a quartet at H9254 3.9 ppm (1H) and a doublet at H9254 1.6 ppm (3H)—suggest the fragment CH—CH 3 . All that remains of the molecular formula is —NO 2 . Combining this information identifies compound B as 2-(4-nitrophenyl)propanoic acid. SELF-TEST PART A A-1. Provide an acceptable IUPAC name for each of the following: A-2. Both of the following compounds may be converted into 4-phenylbutanoic acid by one or more reaction steps. Give the reagents and conditions necessary to carry out these con- versions. A-3. The species whose structure is shown is an intermediate in an esterification reaction. Write the complete, balanced equation for this process. C 6 H 5 CH 2 COCH 2 CH 3 OH OH C 6 H 5 CH 2 CH 2 CH(CO 2 H) 2 C 6 H 5 CH 2 CH 2 CH 2 Br (Two methods) CH 3 CHCHCO 2 H(c) Br CH 2 CH 3 CO 2 H(b) C 6 H 5 CHCHCH 2 CH 2 CO 2 H(a) CH 3 CH 3 O 2 N CHCO 2 H CH 3 2-(4-Nitrophenyl)propanoic acid (compound B) XY CARBOXYLIC ACIDS 531 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-4. Give the correct structures for compounds A through C in the following reactions: A-5. Give the missing reagent(s) and the missing compound in each of the following: A-6. Identify the carboxylic acid (C 4 H 7 BrO 2 ) having the 1 H NMR spectrum consisting of H9254 1.1 ppm, 3H (triplet) H9254 2.0 ppm, 2H (pentet) H9254 4.2 ppm, 1H (triplet) H9254 12.1 ppm, 1H (singlet) A-7. Draw the structure of the tetrahedral intermediate in the esterification of formic acid with 1-butanol. A-8. Write a mechanism for the esterification reaction shown. PART B B-1. Which of the following is a correct IUPAC name for the compound shown? (a) 1,1,3-Triethylhexanoic acid (b) 2,2,4-Triethylhexanoic acid (c) 3,5-Diethyl-3-heptylcarboxylic acid (d) 3,5,5-Triethyl-6-hexanoic acid (CH 3 CH 2 ) 2 CCH 2 CH(CH 2 CH 3 ) 2 CO 2 H H11001 H H11001 CH 3 OHCH 3 COH O H11001CH 3 CO 2 CH 3 H 2 O (d)? ?NaCN HCl CH 3 CH 2 CHCOH O OH CH 3 CH 2 CH O (c)? ? NaSCH 3 CH 3 CH 2 CHCOH O SCH 3 CH 3 CH 2 CH 2 COH O (b ? PCC CH 2 Cl 2 CH 3 CH 2 CH 2 CH O CH 3 CH 2 CH 2 COH O Br CO 2 H(a)? ? 1. CO 2 2. H 3 O H11001 CC 6 H 5 CCH(CH 3 ) 2 CO 2 O H11001(b) heat (CH 3 ) 2 CHCH 2 CH 2 CO 2 HAB(a) KI acetone Br 2 P 532 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-2. Rank the following substances in order of decreasing acid strength (strongest A weakest): (a)4H11022 2 H11022 1 H11022 3(c)3H11022 1 H11022 2 H11022 4 (b)1H11022 2 H11022 4 H11022 d)2H11022 4 H11022 1 H11022 3 B-3. Which of the following compounds will undergo decarboxylation on heating? (a) 2 and 3 (c) 3 only (b) 3 and 4 (d) 1 and 4 B-4. Which of the following is least likely to form a lactone? (a) (c) (b)(d) B-5. Compare the two methods shown for the preparation of carboxylic acids: Method 1: Method 2: Which one of the following statements correctly describes this conversion? (a) Both method 1 and method 2 are appropriate for carrying out this conversion. (b) Neither method 1 nor method 2 is appropriate for carrying out this conversion. (c) Method 1 will work well, but method 2 is not appropriate. (d) Method 2 will work well, but method 1 is not appropriate. Br CO 2 H RBr RCN RCO 2 H NaCN H 2 O, HCl heat RBr RMgBr RCO 2 H Mg diethyl ether 1. CO 2 2. H 3 O H11001 OH CO 2 H OH CH 2 CO 2 H OH CO 2 H CH 3 CHCH 2 CH 2 CO 2 H OH 12 34 O CO 2 CH 3 CO 2 H CO 2 H CO 2 H O O O CH 3 CH 2 CH 2 CO 2 H 1 CH 3 CH 2 CH 2 CH 2 OH 32 CHCO 2 HCH 3 CH 4 CCO 2 HCH 3 C CARBOXYLIC ACIDS 533 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-6. Which one of the following is not an intermediate in the generally accepted mechanism for the reaction shown? B-7. Identify compound C in the following sequence: B-8. What is the final product (B) of this sequence? (e) None of these CH 2 CO 2 H (d) CH 3 CO 2 H (b) CO 2 H (c) CH 3 CO 2 H H11001 para (a) Br 2 light CH 3 AB 1. KCN 2. H 3 O H11001 , heat (CH 3 ) 2 CHCH C OH OH (b) (CH 3 ) 2 CHCH 2 CH O (c) (CH 3 ) 2 CHCH 2 COH O (d) (CH 3 ) 2 CHCCH 2 OH O (e) (CH 3 ) 2 CHCCH 3 O (a) (CH 3 ) 2 CHCH 2 C compound A compound B compound CN HCl, H 2 O heat PCC CH 2 Cl 2 1. LiAlH 4 2. H 2 O CF 3 COH OH H11001 (a) CF 3 C OCH(CH 3 ) 2 OH OH H H11001 (b) CF 3 C OCH(CH 3 ) 2 OH OH (c) CF 3 C OCH(CH 3 ) 2 O OH H11002 (d) CF 3 C OCH(CH 3 ) 2 OH OH 2 H11001 (e) H11001CF 3 COH O CH 3 CHCH 3 OH H 2 OH11001CF 3 COCHCH 3 CH 3 O H 2 SO 4 534 CARBOXYLIC ACIDS __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-9. Which one of the following undergoes decarboxylation (loses carbon dioxide) most readily on being heated? (a)(d) (b)(e) (c) B-10. Which of the compounds in the previous problem yields a H9254-lactone on being reduced with sodium borohydride? B-11. What is compound Z? (a)(d) (b e) (c) CH 3 CH 2 CHC N OCH 2 CH 3 CH 3 CH 2 CH 2 COCH 2 CH 3 O CH 3 CH 2 CH 2 CH(OCH 2 CH 3 ) 2 CH 3 CH 2 CH 2 CH NOCH 2 CH 3 CH 3 CH CHCOH O CH 3 CH 2 CH 2 Br X NaCN Y H 3 O H11001 heat Z CH 3 CH 2 OH H H11001 O OHO O OHO O O OH HO O OHO O O OH CARBOXYLIC ACIDS 535 __ __ __ Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website