CHAPTER 20 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION SOLUTIONS TO TEXT PROBLEMS 20.1 (b) Carboxylic acid anhydrides bear two acyl groups on oxygen, as in . They are named as derivatives of carboxylic acids. (c) Butyl 2-phenylbutanoate is the butyl ester of 2-phenylbutanoic acid. (d) In 2-phenylbutyl butanoate the 2-phenylbutyl group is an alkyl group bonded to oxygen of the ester. It is not involved in the acyl group of the molecule. CH 3 CH 2 CH 2 COCH 2 CHCH 2 CH 3 O C 6 H 5 2-Phenylbutyl butanoate CH 3 CH 2 CHCOCH 2 CH 2 CH 2 CH 3 O C 6 H 5 Butyl 2-phenylbutanoate CH 3 CH 2 CHCOCCHCH 2 CH 3 O O C 6 H 5 C 6 H 5 CH 3 CH 2 CHCOH O C 6 H 5 2-Phenylbutanoic acid 2-Phenylbutanoic anhydride RCOCR O O 536 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 537 (e) The ending -amide reveals this to be a compound of the type . ( f ) This compound differs from 2-phenylbutanamide in part (e) only in that it bears an ethyl sub- stituent on nitrogen. (g) The -nitrile ending signifies a compound of the type RC>N containing the same number of carbons as the alkane RCH 3 . 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. Rotation about the carbon–nitrogen bond is required to average the environments of the two methyl groups, but this rotation is relatively slow in amides as the result of the double-bond character im- parted to the carbon–nitrogen bond, as shown by these two resonance structures. 20.3 (b) Benzoyl chloride reacts with benzoic acid to give benzoic anhydride. (c) Acyl chlorides react with alcohols to form esters. The organic product is the ethyl ester of benzoic acid, ethyl benzoate. H11001C 6 H 5 CCl O Benzoyl chloride CH 3 CH 2 OH Ethanol H11001C 6 H 5 COCH 2 CH 3 O Ethyl benzoate HCl Hydrogen chloride H11001C 6 H 5 CCl O Benzoyl chloride C 6 H 5 COH O Benzoic acid H11001C 6 H 5 COCC 6 H 5 O O Benzoic anhydride HCl Hydrogen chloride C H O N CH 3 CH 3 H11001 H11002 C H O N CH 3 CH 3 2-Phenylbutanenitrile CH 3 CH 2 CHC N C 6 H 5 CH 3 CH 2 CHCNHCH 2 CH 3 O C 6 H 5 N-Ethyl-2-phenylbutanamide CH 3 CH 2 CHCNH 2 O C 6 H 5 2-Phenylbutanamide RCNH 2 O Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 538 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION (d) Acyl transfer from benzoyl chloride to the nitrogen of methylamine yields the amide N-methyl- benzamide. (e) In analogy with part (d), an amide is formed. In this case the product has two methyl groups on nitrogen. ( f ) Acyl chlorides undergo hydrolysis on reaction with water. The product is a carboxylic acid. 20.4 (b) Nucleophilic addition of benzoic acid to benzoyl chloride gives the tetrahedral intermediate shown. Dissociation of the tetrahedral intermediate occurs by loss of chloride and of the proton on the oxygen. (c) Ethanol is the nucleophile that adds to the carbonyl group of benzoyl chloride to form the tetrahedral intermediate. H11001C 6 H 5 CCl O Benzoyl chloride CH 3 CH 2 OH Ethanol Tetrahedral intermediate C 6 H 5 COCH 2 CH 3 OH Cl H11001C 6 H 5 COCC 6 H 5 O O Benzoic anhydride HCl Hydrogen chloride Tetrahedral intermediate C 6 H 5 COCC 6 H 5 O H Cl O H11001C 6 H 5 CCl O Benzoyl chloride C 6 H 5 COH O Benzoic acid C 6 H 5 COCC 6 H 5 HO Cl O Tetrahedral intermediate H11001C 6 H 5 CCl O Benzoyl chloride H 2 O Water H11001C 6 H 5 COH O Benzoic acid HCl Hydrogen chloride H11001C 6 H 5 CCl O Benzoyl chloride 2(CH 3 ) 2 NH Dimethylamine H11001C 6 H 5 CN(CH 3 ) 2 O N,N-Dimethylbenzamide (CH 3 ) 2 NH 2 Cl H11002 H11001 H11001C 6 H 5 CCl O Benzoyl chloride 2CH 3 NH 2 Methylamine H11001C 6 H 5 CNHCH 3 O N-Methylbenzamide H11001 CH 3 NH 3 Cl H11002 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website In analogy with parts (a) and (b) of this problem, a proton is lost from the hydroxyl group along with chloride to restore the carbon–oxygen double bond. (d) The tetrahedral intermediate formed from benzoyl chloride and methylamine has a carbon– nitrogen bond. The dissociation of the tetrahedral intermediate may be shown as More realistically, it is a second methylamine molecule that abstracts a proton from oxygen. (e) The intermediates in the reaction of benzoyl chloride with dimethylamine are similar to those in part (d). The methyl substituents on nitrogen are not directly involved in the reaction. Then H11001C 6 H 5 CN(CH 3 ) 2 O N,N-Dimethylbenzamide Dimethylammonium chloride (CH 3 ) 2 NH 2 Cl H11002 H11001 C 6 H 5 CN(CH 3 ) 2 O H Cl (CH 3 ) 2 NH H11001C 6 H 5 CCl O Benzoyl chloride (CH 3 ) 2 NH Dimethylamine Tetrahedral intermediate C 6 H 5 CN(CH 3 ) 2 OH Cl H11001C 6 H 5 CNHCH 3 O N-Methylbenzamide Methylammonium chloride H11001 CH 3 NH 3 Cl H11002 C 6 H 5 CNHCH 3 O H Cl CH 3 NH 2 H11001C 6 H 5 CNHCH 3 O N-Methylbenzamide HCl Hydrogen chloride Tetrahedral intermediate C 6 H 5 CNHCH 3 O H Cl H11001C 6 H 5 CCl O Benzoyl chloride CH 3 NH 2 Methylamine Tetrahedral intermediate C 6 H 5 CNHCH 3 OH Cl H11001C 6 H 5 COCH 2 CH 3 O Ethyl benzoate HCl Hydrogen chloride Tetrahedral intermediate C 6 H 5 COCH 2 CH 3 O H Cl CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 539 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) Water attacks the carbonyl group of benzoyl chloride to form the tetrahedral intermediate. Dissociation of the tetrahedral intermediate occurs by loss of chloride and the proton on oxygen. 20.5 One equivalent of benzoyl chloride reacts rapidly with water to yield benzoic acid. The benzoic acid produced in this step reacts with the remaining benzoyl chloride to give benzoic anhydride. 20.6 Acetic anhydride serves as a source of acetyl cation. 20.7 (b) Acyl transfer from an acid anhydride to ammonia yields an amide. The organic products are acetamide and ammonium acetate. H11001 Acetic anhydride CH 3 COCCH 3 O O 2NH 3 Ammonia H11001CH 3 CNH 2 O Acetamide H11001 CH 3 CO H11002 NH 4 O Ammonium acetate Acetyl cation CH 3 CO O CCH 3 O H11001 O CCH 3 H11001 O CCH 3 H11001 HCl Hydrogen chloride C 6 H 5 COCC 6 H 5 O O Benzoic anhydride H11001C 6 H 5 CCl O Benzoyl chloride C 6 H 5 COH O Benzoic acid H11001 H 2 O Water C 6 H 5 CCl O Benzoyl chloride H11001 HCl Hydrogen chloride C 6 H 5 COH O Benzoic acid H11001 HCl Hydrogen chloride C 6 H 5 COH O Benzoic acidTetrahedral intermediate C 6 H 5 CCl O H OH H11001C 6 H 5 CCl O Benzoyl chloride H 2 O Water Tetrahedral intermediate C 6 H 5 CCl OH OH 540 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The reaction of phthalic anhydride with dimethylamine is analogous to that of part (b). The organic products are an amide and the carboxylate salt of an amine. 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. 20.8 (b) The tetrahedral intermediate is formed by nucleophilic addition of ammonia to one of the car- bonyl groups of acetic anhydride. Dissociation of the tetrahedral intermediate occurs by loss of acetate as the leaving group. (c) Dimethylamine is the nucleophile; it adds to one of the two equivalent carbonyl groups of phthalic anhydride. Phthalic anhydride Dimethylamine H11001 Tetrahedral intermediate HN(CH 3 ) 2 O HO N(CH 3 ) 2 O O O O Acetamide Ammonium acetateAmmonia H11001 tetrahedral intermediate CH 3 CNH 2 O H11001 H 4 N H11002 OCCH 3 O H11001 O H CH 3 C OCCH 3 O NH 2 H 3 N Tetrahedral intermediate NH 3 O CH 3 COCCH 3 O CH 3 COCCH 3 NH 2 OHO H11001 2NaOH Sodium hydroxide Phthalic anhydride O H11001 H 2 O WaterSodium phthalate CO H11002 Na H11001 CO H11002 Na H11001 O O O O H11001 2(CH 3 ) 2 NH DimethylaminePhthalic anhydride O H 2 N(CH 3 ) 2 H11001 Product is an amine salt and contains an amide function. CO H11002 CN(CH 3 ) 2 O O O O CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 541 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A second molecule of dimethylamine abstracts a proton from the tetrahedral intermediate. (d) Hydroxide acts as a nucleophile to form the tetrahedral intermediate and as a base to facilitate its dissociation. Formation of tetrahedral intermediate: Dissociation of tetrahedral intermediate: In base, the remaining carboxylic acid group is deprotonated. CO CO H11002 O O H11001 H 2 O H CO H11002 CO H11002 O O OH H11002 O OH O O H COH CO H11002 O O HO H11002 H11001 H 2 O Tetrahedral intermediate H11001 O HO OH O O H11002 O OH O H 2 O H11001 H11002 OH Phthalic anhydride H11001O O O O H11002 O OH H11002 OH O Product of reactionTetrahedral intermediate H11001 second molecule of dimethylamine O N(CH 3 ) 2 O O H (CH 3 ) 2 NH CN(CH 3 ) 2 CO H11002 O O H 2 N(CH 3 ) 2 H11001 542 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 20.9 The starting material contains three acetate ester functions. All three undergo hydrolysis in aqueous sulfuric acid. The product is 1,2,5-pentanetriol. Also formed in the hydrolysis of the starting triacetate are three molecules of acetic acid. 20.10 Step 1: Protonation of the carbonyl oxygen Step 2: Nucleophilic addition of water Step 3: Deprotonation of oxonium ion to give neutral form of tetrahedral intermediate Step 4: Protonation of ethoxy oxygen H11001 Oxonium ion C 6 H 5 C OCH 2 CH 3 OH HO H H11001 O H H Water H11001C 6 H 5 C OCH 2 CH 3 OH HO Tetrahedral intermediate O H H H H11001 Hydronium ion H11001 O H H H11001C 6 H 5 C OCH 2 CH 3 OH O H11001 HH C 6 H 5 C OCH 2 CH 3 OH HO Tetrahedral intermediate O H H H H11001 Hydronium ion Protonated form of ester Oxonium ionWater O H H C 6 H 5 C OH H11001 O OCH 2 CH 3 HH C 6 H 5 C OCH 2 CH 3 OH H11001 Ethyl benzoate Hydronium ion C 6 H 5 C OCH 2 CH 3 Protonated form of ester H11001H11001 H H OH H11001 Water H H OC 6 H 5 C OCH 2 CH 3 OH H11001 O H11001 3CH 3 COH O H 2 O H H11001 O O CH 3 COCH 2 CHCH 2 CH 2 CH 2 OCCH 3 OCCH 3 O HOCH 2 CHCH 2 CH 2 CH 2 OH OH 1,2,5-Pentanetriol (C 5 H 12 O 3 ) Acetic acid CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 543 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Step 5: Dissociation of protonated form of tetrahedral intermediate This step yields ethyl alcohol and the protonated form of benzoic acid. Step 6: Deprotonation of protonated form of benzoic acid 20.11 To determine which oxygen of 4-butanolide becomes labeled with 18 O, trace the path of 18 O-labeled water as it undergoes nucleophilic addition to the carbonyl group to form the tetrahedral intermediate. The tetrahedral intermediate can revert to unlabeled 4-butanolide by loss of 18 O-labeled water. Alternatively it can lose ordinary water to give 18 O-labeled lactone. The carbonyl oxygen is the one that is isotopically labeled in the 18 O-enriched 4-butanolide. 20.12 On the basis of trimyristin’s molecular formula C 45 H 86 O 6 and of the fact that its hydrolysis gives only glycerol and tetradecanoic acid CH 3 (CH 2 ) 12 CO 2 H, it must have the structure shown. CH 3 (CH 2 ) 12 CO OC(CH 2 ) 12 CH 3 OC(CH 2 ) 12 CH 3 O O O Trimyristin (C 45 H 86 O 6 ) H11001 H H11001 OH O OH Tetrahedral intermediate O O 18 O-labeled 4-butanolide H 2 O Water H11001 4-Butanolide O O 18 O-labeled water H 2 O H H11001 Tetrahedral intermediate O OH OH (O H11005 18 O) Benzoic acid C 6 H 5 C OH O Water O H H Hydronium ion H11001 OH H H H11001H11001 Protonated form of benzoic acid H11001 C 6 H 5 C OH OH Oxonium ion C 6 H 5 C OCH 2 CH 3 OH OH H H11001 Protonated form of benzoic acid H11001 C 6 H 5 C OH OH Ethyl alcohol HOCH 2 CH 3 H11001 544 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 20.13 Because ester hydrolysis in base proceeds by acyl–oxygen cleavage, the 18 O label becomes incorporated into acetate ion . 20.14 Step 1: Nucleophilic addition of hydroxide ion to the carbonyl group Step 2: Proton transfer from water to give neutral form of tetrahedral intermediate Step 3: Dissociation of tetrahedral intermediate Step 4: Proton transfer from benzoic acid 20.15 The starting material is a lactone, a cyclic ester. The ester function is converted to an amide by nucleophilic acyl substitution. Methylamine 4-Pentanolide 4-Hydroxy-N-methylpentanamide H11001CH 3 NH 2 O CH 3 O CH 3 NHCCH 2 CH 2 CHCH 3 OH O Benzoic acid Hydroxide ion O C 6 H 5 C OH H11001 WaterBenzoate ion H11002 O C 6 H 5 C O H11001 HOHHO H11002 H11001H11001H11001 Tetrahedral intermediate C 6 H 5 C OCH 2 CH 3 OH OH C 6 H 5 C O OH Benzoic acid H11002 OCH 2 CH 3 Ethoxide ion HO H11002 Hydroxide ion HOH Water H11001H11001 C 6 H 5 C OCH 2 CH 3 OH OH Tetrahedral intermediate H11002 C 6 H 5 C OCH 2 CH 3 O OH Anionic form of tetrahedral intermediate OH H11002 Hydroxide ion Water OHH H11001 C 6 H 5 C H11002 OCH 2 CH 3 O OH Anionic form of tetrahedral intermediate C 6 H 5 C O OCH 2 CH 3 Ethyl benzoate HO H11002 Hydroxide ion H11001 Pentyl acetate Hydroxide ion CH 3 CH 2 CH 2 CH 2 CH 2 OH 1-Pentanol CCH 3 O H11002 O Acetate ion H11001 H11002 OHCH 3 CH 2 CH 2 CH 2 CH 2 O CCH 3 O (O H11005 18 O) CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 545 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 20.16 Methanol is the nucleophile that adds to the carbonyl group of the thioester. 20.17 (b) Acetic anhydride is the anhydride that must be used; it transfers an acetyl group to suitable nucleophiles. The nucleophile in this case is methylamine. (c) The acyl group is . Because the problem specifies that the acyl transfer agent is a methyl ester, methyl formate is one of the starting materials. 20.18 Phthalic anhydride reacts with excess ammonia to give the ammonium salt of a compound known as phthalamic acid. Phthalimide is formed when ammonium phthalamate is heated. 20.19 Step 1: Protonation of the carbonyl oxygen H11001H11001CH 3 C O NHC 6 H 5 Acetanilide HO H H H11001 Hydronium ion CH 3 C NC 6 H 5 H OH H11001 Protonated form of amide O H H Water heat H11001H11001 CNH 2 O O CO H11002 NH 4 Ammonium phthalamate O O Phthalimide NH 3 Ammonia H 2 O Water H11001 NH Phthalic anhydride Ammonia H11001 Ammonium phthalamate (C 8 H 10 N 2 O 3 ) 2NH 3 O O CNH 2 CO H11002 NH 4 H11001 O O O Dimethylamine N,N-Dimethylformamide HN(CH 3 ) 2 H11001H11001 Methyl formate HCOCH 3 O HCN(CH 3 ) 2 O Methyl alcohol CH 3 OH HC O Methylamine N-Methylacetamide 2CH 3 NH 2 H11001H11001 Acetic anhydride CH 3 COCCH 3 O O CH 3 CNHCH 3 O Methylammonium acetate CH 3 CO H11002 CH 3 NH 3 O H11001 H11001H11001 Methanol CH 3 OH S-2-Phenoxyethyl ethanethiolate CH 3 CSCH 2 CH 2 OC 6 H 5 O Tetrahedral intermediate CH 3 C SCH 2 CH 2 OC 6 H 5 OCH 3 OH 2-Phenoxyethanethiol HSCH 2 CH 2 OC 6 H 5 Methyl acetate CH 3 COCH 3 O 546 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Step 2: Nucleophilic addition of water Step 3: Deprotonation of oxonium ion to give neutral form of tetrahedral intermediate Step 4: Protonation of amino group of tetrahedral intermediate Step 5: Dissociation of N-protonated form of tetrahedral intermediate Step 6: Proton-transfer processes H11001H11001HO H H H11001 Hydronium ion O H H WaterProtonated form of acetic acid OH CH 3 C OH H11001 O CH 3 C OH Acetic acid H11001H11001H 2 NC 6 H 5 Aniline H 3 NC 6 H 5 Anilinium ionHydronium ion O H11001 H H H Water O H H H11001 N-Protonated form of tetrahedral intermediate CH 3 C OH NC 6 H 5 OH H H H11001 H11001 OH CH 3 C OH H11001 Protonated form of acetic acid H 2 NC 6 H 5 Aniline Tetrahedral intermediate CH 3 C OH NHC 6 H 5 OH H11001H11001 O H H WaterN-Protonated form of tetrahedral intermediate CH 3 C OH NC 6 H 5 OH H H H11001 HO H H H11001 Hydronium ion Tetrahedral intermediate CH 3 C OH NHC 6 H 5 OH Oxonium ion CH 3 C OH NHC 6 H 5 H11001 O HH H11001H11001O H H Water HO H H H11001 Hydronium ion H11001O H H Water Protonated form of amide OH CH 3 C NHC 6 H 5 Oxonium ion CH 3 C OH NHC 6 H 5 O H11001 HH H11001 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 547 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 20.20 Step 1: Nucleophilic addition of hydroxide ion to the carbonyl group Step 2: Proton transfer to give neutral form of tetrahedral intermediate Step 3: Proton transfer from water to nitrogen of tetrahedral intermediate Step 4: Dissociation of N-protonated form of tetrahedral intermediate Step 5: Irreversible formation of formate ion 20.21 A synthetic scheme becomes apparent when we recognize that a primary amine may be obtained by Hofmann rearrangement of the primary amide having one more carbon in its acyl group. This amide may, in turn, be prepared from the corresponding carboxylic acid. CH 3 CH 2 CH 2 NH 2 CH 3 CH 2 CH 2 CO 2 HCH 3 CH 2 CH 2 CNH 2 O H11001H11001 Water HOHOH H11002 Hydroxide ion Formic acid O HC OH Formate ion O HC O H11002 H11001H11001H11001 HC NH(CH 3 ) 2 O H OH H11001 N-Protonated form of tetrahedral intermediate Dimethylamine HN(CH 3 ) 2 HO H11002 Hydroxide ion H 2 O Water O HC OH Formic acid H11001H11001HC N(CH 3 ) 2 OH OH Tetrahedral intermediate OH H11002 Hydroxide ion Water OHH N-Protonated form of tetrahedral intermediate HC NH(CH 3 ) 2 OH OH H11001 H11001H11001HC N(CH 3 ) 2 OH OH Tetrahedral intermediate HC N(CH 3 ) 2 O H11002 OH Anionic form of tetrahedral intermediate OH H11002 Hydroxide ion Water OHH H11001 HC N(CH 3 ) 2 O OH H11002 Anionic form of tetrahedral intermediate HO H11002 Hydroxide ion HCN(CH 3 ) 2 O N,N-Dimethylformamide 548 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The desired reaction scheme is therefore 20.22 (a) Ethanenitrile has the same number of carbon atoms as ethyl alcohol. This suggests a reaction scheme proceeding via an amide. The necessary amide is prepared from ethanol. (b) Propanenitrile may be prepared from ethyl alcohol by way of a nucleophilic substitution reac- tion of the corresponding bromide. 20.23 Step 1: Protonation of the nitrile Step 2: Nucleophilic addition of water Step 3: Deprotonation of imino acid H11001H11001 Water OH 2 Hydronium ion H 3 O H11001 NH RC HO H H11001 Protonated form of imino acid NH RC OH Imino acid H11001 RC NH Protonated form of nitrile H11001 H 2 O Water NH RC OH 2 H11001 Protonated form of imino acid H11001 H 2 O WaterHydronium ion HO H H H11001 RC NH Protonated form of nitrile H11001 Nitrile RC N CH 3 CH 2 OH Ethyl alcohol CH 3 CH 2 Br Ethyl bromide CH 3 CH 2 CN Propanenitrile NaCN PBr 3 or HBr CH 3 CH 2 OH Ethyl alcohol CH 3 COH O Acetic acid CH 3 CNH 2 O Acetamide Na 2 Cr 2 O 7 , H 2 O H 2 SO 4 , heat 1. SOCl 2 2. NH 3 CH 3 CH 2 OH Ethyl alcohol CH 3 CNH 2 O Acetamide Ethanenitrile CH 3 CN P 4 O 10 CH 3 CH 2 CH 2 CO 2 H Butanoic acid CH 3 CH 2 CH 2 CNH 2 O Butanamide CH 3 CH 2 CH 2 NH 2 1-Propanamine 1. SOCl 2 2. NH 3 Br 2 H 2 O, NaOH CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 549 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Steps 4 and 5: Proton transfers to give an amide 20.24 Ketones may be prepared by the reaction of nitriles with Grignard reagents. Nucleophilic addition of a Grignard reagent to a nitrile produces an imine. The imine is not normally isolated, however, but is hydrolyzed to the corresponding ketone. Ethyl phenyl ketone may be prepared by the reaction of propanenitrile with a phenyl Grignard reagent such as phenylmagnesium bromide, followed by hydrolysis of the imine. 20.25 (a) The halogen that is attached to the carbonyl group is identified in the name as a separate word following the name of the acyl group. (b)Trifluoroacetic anhydride is the anhydride of trifluoroacetic acid. Notice that it contains six fluorines. (c) This compound is the cyclic anhydride of cis-1,2-cyclopropanedicarboxylic acid. (d) Ethyl cycloheptanecarboxylate is the ethyl ester of cycloheptanecarboxylic acid. COCH 2 CH 3 O Ethyl cycloheptanecarboxylate HH HO 2 C CO 2 H cis-1,2-Cyclopropanedicarboxylic acid HH O OO cis-1,2-Cyclopropanedicarboxylic anhydride Trifluoroacetic anhydride CF 3 COCCF 3 O O CBr Cl O m-Chlorobenzoyl bromide Propanenitrile Phenylmagnesium bromide C 6 H 5 MgBr Ethyl phenyl ketone O C 6 H 5 CCH 2 CH 3 H11001CH 3 CH 2 CN Imine (not isolated) NH C 6 H 5 CCH 2 CH 3 Diethyl ether H 2 O, H H11001 heat H11001H11001 OH NH RC Imino acid Hydronium ion HO H H H H11001 Hydronium ion HO H H H11001 H11001 O NH 2 RC AmideWater O H Conjugate acid of amide HO NH 2 RC H11001 550 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (e) 1-Phenylethyl acetate is the ester of 1-phenylethanol and acetic acid. ( f ) 2-Phenylethyl acetate is the ester of 2-phenylethanol and acetic acid. (g) The parent compound in this case is benzamide. p-Ethylbenzamide has an ethyl substituent at the ring position para to the carbonyl group. (h) The parent compound is benzamide. In N-ethylbenzamide the ethyl substituent is bonded to nitrogen. (i) Nitriles are named by adding the suffix -nitrile to the name of the alkane having the same number of carbons. Numbering begins at the nitrile carbon. 20.26 (a) This compound, with a bromine substituent attached to its carbonyl group, is named as an acyl bromide. It is 3-chlorobutanoyl bromide. (b) The group attached to oxygen, in this case benzyl, is identified first in the name of the ester. This compound is the benzyl ester of acetic acid. CH 3 COCH 2 O Benzyl acetate CH 3 CHCH 2 CBr Cl O 3-Chlorobutanoyl bromide CH 3 CH 2 CH 2 CH 2 CHC N CH 3 2-Methylhexanenitrile 56431 CNHCH 2 CH 3 O N-Ethylbenzamide CH 3 CH 2 CNH 2 O p-Ethylbenzamide CH 3 COCH 2 CH 2 O 2-Phenylethyl acetate CH 3 COCH CH 3 O 1-Phenylethyl acetate CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 551 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The group attached to oxygen is methyl; this compound is the methyl ester of phenylacetic acid. (d) This compound contains the functional group and thus is an anhydride of a car- boxylic acid. We name the acid, in this case 3-chloropropanoic acid, drop the acid part of the name, and replace it by anhydride. (e) This compound is a cyclic anhydride, whose parent acid is 3,3-dimethylpentanedioic acid. ( f ) Nitriles are named by adding -nitrile to the name of the alkane having the same number of car- bons. Remember to count the carbon of the C>N group. (g) This compound is an amide. We name the corresponding acid and then replace the -oic acid suffix by -amide. (h) This compound is the N-methyl derivative of 4-methylpentanamide. CH 3 CHCH 2 CH 2 CNHCH 3 CH 3 O N-Methyl-4-methylpentanamide CH 3 CHCH 2 CH 2 CNH 2 CH 3 O 4-Methylpentanamide CH 3 CHCH 2 CH 2 CN CH 3 4-Methylpentanenitrile H 3 C O O O H 3 C 3,3-Dimethylpentanedioic anhydride 3-Chloropropanoic anhydride ClCH 2 CH 2 COCCH 2 CH 2 Cl O O COC O O CH 3 OCCH 2 O Methyl phenylacetate 552 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (i) The amide nitrogen bears two methyl groups. We designate this as an N,N-dimethyl amide. 20.27 (a) Acetyl chloride acts as an acyl transfer agent to the aromatic ring of bromobenzene. The reac- tion is a Friedel–Crafts acylation. Bromine is an ortho, para-directing substituent. (b) Acyl chlorides react with thiols to give thioesters. (c) Sodium propanoate acts as a nucleophile toward propanoyl chloride. The product is propanoic anhydride. (d) Acyl chlorides convert alcohols to esters. (e) Acyl chlorides react with ammonia to yield amides. H11001Cl CCl O p-Chlorobenzoyl chloride Cl CNH 2 O p-Chlorobenzamide NH 3 Ammonia H11001CH 3 CH 2 CH 2 CCl O Butanoyl chloride C 6 H 5 CH 2 OH Benzyl alcohol CH 3 CH 2 CH 2 COCH 2 C 6 H 5 O Benzyl butanoate Propanoic anhydride CH 3 CH 2 COCCH 2 CH 3 O O H11001 Propanoyl chloride CH 3 CH 2 CCl O Propanoate anion CH 3 CH 2 CO O H11002 Acetyl chloride CH 3 CCl O S-Butyl ethanethioate CH 3 CSCH 2 CH 2 CH 2 CH 3 O 1-Butanethiol CH 3 CH 2 CH 2 CH 2 SHH11001 AlCl 3 H11001H11001 Br CCH 3 O p-Bromoacetophenone Br CCH 3 O o-Bromoacetophenone Br Bromobenzene Acetyl chloride CH 3 CCl O CH 3 CHCH 2 CH 2 CN(CH 3 ) 2 CH 3 N,N-Dimethyl-4-methylpentanamide O CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 553 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) The starting material is a cyclic anhydride. Acid anhydrides react with water to yield two car- boxylic acid functions; when the anhydride is cyclic, a dicarboxylic acid results. (g) In dilute sodium hydroxide the anhydride is converted to the disodium salt of the diacid. (h) One of the carbonyl groups of the cyclic anhydride is converted to an amide function on reac- tion with ammonia. The other, the one that would become a carboxylic acid group, is con- verted to an ammonium carboxylate salt. (i) Acid anhydrides are used as acylating agents in Friedel–Crafts reactions. ( j) The reactant is maleic anhydride; it is a good dienophile in Diels–Alder reactions. (k) Acid anhydrides react with alcohols to give an ester and a carboxylic acid. O O Acetic anhydride CH 3 COCCH 3 H11001H11001 3-Pentanol CH 3 CH 2 CHCH 2 CH 3 OH Acetic acid CH 3 CO 2 H 1-Ethylpropyl acetate O CH 3 CH 2 CHCH 2 CH 3 OCCH 3 H11001 1,3-Pentadiene CH 3 Maleic anhydride 3-Methylcyclohexene-4,5- dicarboxylic anhydride O CH 3 O O O O O H11001 Succinic anhydride O OO AlCl 3 3-Benzoylpropanoic acid OO CCH 2 CH 2 COH Benzene H11001 Ammonia 2NH 3 Succinic anhydride O OO Ammonium succinamate OO NH 4 H11002 OCCH 2 CH 2 CNH 2 H11001H 2 O H11001 Sodium hydroxide 2NaOH Succinic anhydride O OO Sodium succinate OO Na H11001 H11002 OCCH 2 CH 2 CO H11002 Na H11001 H 2 O H11001 Water H 2 O Succinic anhydride O OO Succinic acid OO HOCCH 2 CH 2 COH 554 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (l ) The starting material is a cyclic ester, a lactone. Esters undergo saponification in aqueous base to give an alcohol and a carboxylate salt. (m) Ammonia reacts with esters to give an amide and an alcohol. (n) Lithium aluminum hydride reduces esters to two alcohols; the one derived from the acyl group is a primary alcohol. Reduction of a cyclic ester gives a diol. (o) Grignard reagents react with esters to give tertiary alcohols. (p) In this reaction methylamine acts as a nucleophile toward the carbonyl group of the ester. The product is an amide. (q) The starting material is a lactam, a cyclic amide. Amides are hydrolyzed in base to amines and carboxylate salts. H11001 Sodium hydroxide NaOH Sodium 4-(methylamino)butanoate O CH 3 NHCH 2 CH 2 CH 2 CO H11002 Na H11001 N-Methylpyrrolidone N CH 3 O H 2 O H11001H11001 Ethyl phenylacetate O C 6 H 5 CH 2 COCH 2 CH 3 Ethyl alcohol CH 3 CH 2 OH N-Methylphenylacetamide O C 6 H 5 CH 2 CNHCH 3 Methylamine CH 3 NH 2 4-Butanolide O O 1. 2CH 3 MgBr 2. H 3 O H11001 4-Methyl-1,4-pentanediol HOCH 2 CH 2 CH 2 CCH 3 OH CH 3 1,4-Butanediol HOCH 2 CH 2 CH 2 CH 2 OH 4-Butanolide O O 1. LiAlH 4 2. H 2 O H11001 Ammonia NH 3 4-Hydroxybutanamide O HOCH 2 CH 2 CH 2 CNH 2 4-Butanolide O O H 2 O H11001 Sodium hydroxide NaOH Sodium 4-hydroxybutanoate O HOCH 2 CH 2 CH 2 CO H11002 Na H11001 4-Butanolide O O H 2 O CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 555 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (r) In acid solution amides yield carboxylic acids and ammonium salts. (s) The starting material is a cyclic imide. Both its amide bonds are cleaved by nucleophilic attack by hydroxide ion. (t) In acid the imide undergoes cleavage to give a dicarboxylic acid and the conjugate acid of methylamine. (u) Acetanilide is hydrolyzed in acid to acetic acid and the conjugate acid of aniline. (v) This is another example of amide hydrolysis. (w) One way to prepare nitriles is by dehydration of amides. H11001 H 2 O Cyclopentanecarboxamide O CNH 2 Cyclopentyl cyanide CN P 4 O 10 H11001H11001 H11001 Water H 2 O Sulfuric acid H 2 SO 4 N-Methylbenzamide O C 6 H 5 CNHCH 3 Benzoic acid O C 6 H 5 COH Methylammonium hydrogen sulfate CH 3 NH 3 HSO 4 H11002 H11001 H11001H11001 H11001 Water H 2 O Hydrogen chloride HCl Acetanilide O C 6 H 5 NHCCH 3 Acetic acid O CH 3 COH Anilinium chloride C 6 H 5 NH 3 Cl H11002 H11001 H11001H11001 H11001 Water 2H 2 O Hydrogen chloride HCl Methylammonium chloride CH 3 NH 3 Cl H11002 N-Methylsuccinimide N CH 3 OO Succinic acid OO HOCCH 2 CH 2 COH H11001 H11001H11001 Sodium hydroxide 2NaOH Methylamine CH 3 NH 2 N-Methylsuccinimide N CH 3 OO Disodium succinate OO H11002 OCCH 2 CH 2 CO H11002 Na H11001 Na H11001 H 2 O H11001 Hydronium ion H 3 O H11001 N-Methylpyrrolidone N CH 3 O 4-(Methylammonio)butanoic acid OH H CH 3 NCH 2 CH 2 CH 2 COH H11001 556 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (x) Nitriles are hydrolyzed to carboxylic acids in acidic media. (y) Nitriles are hydrolyzed in aqueous base to salts of carboxylic acids. (z) Grignard reagents react with nitriles to yield ketones after addition of aqueous acid. (aa) Amides undergo the Hofmann rearrangement on reaction with bromine and base. A methyl carbamate is the product isolated when the reaction is carried out in methanol. (bb) Saponification of the carbamate in part (aa) gives the corresponding amine. 20.28 (a) Acetyl chloride is prepared by reaction of acetic acid with thionyl chloride. The first task then is to prepare acetic acid by oxidation of ethanol. (b) Acetic acid and acetyl chloride, available from part (a), can be combined to form acetic anhydride. Acetic acid O CH 3 COH H11001H11001 Acetic anhydride OO CH 3 COCCH 3 Hydrogen chloride HCl Acetyl chloride O CH 3 CCl Acetic acid O CH 3 COH Acetyl chloride O CH 3 CCl Ethanol CH 3 CH 2 OH SOCl 2 K 2 Cr 2 O 7 , H 2 SO 4 H 2 O KOH H 2 O NHCOCH 3 H 3 C CH 3 O H 3 C CH 3 NH 2 C H 3 C CH 3 NH 2 O NaOCH 3 CH 3 OH H11001 Br 2 NHCOCH 3 H 3 C CH 3 O 2-Butanone O CH 3 CH 2 CCH 3 Propanenitrile CH 3 CH 2 CN 1. CH 3 MgBr 2. H 3 O H11001 NaOH, H 2 O heat H11001CH 3 OCN p-Methoxybenzonitrile CH 3 OCO H11002 Na H11001 O Sodium p-methoxybenzoate NH 3 Ammonia 3-Methylbutanoic acid O (CH 3 ) 2 CHCH 2 COH 3-Methylbutanenitrile (CH 3 ) 2 CHCH 2 CN HCl, H 2 O heat CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 557 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Ethanol can be converted to ethyl acetate by reaction with acetic acid, acetyl chloride, or acetic anhydride from parts (a) and (b). or or (d) Ethyl bromoacetate is the ethyl ester of bromoacetic acid; thus the first task is to prepare the acid. We use the acetic acid prepared in part (a), converting it to bromoacetic acid by the Hell–Volhard–Zelinsky reaction. Alternatively, bromoacetic acid could be converted to the corresponding acyl chloride, then treated with ethanol. It would be incorrect to try to brominate ethyl acetate; the Hell– Volhard–Zelinsky method requires an acid as starting material, not an ester. (e) The alcohol BrCH 2 CH 2 OH, needed in order to prepare 2-bromoethyl acetate, is prepared from ethanol by way of ethylene. Then ( f ) Ethyl cyanoacetate may be prepared from the ethyl bromoacetate obtained in part (d). The bromide may be displaced by cyanide in a nucleophilic substitution reaction. Ethyl bromoacetate BrCH 2 COCH 2 CH 3 O Ethyl cyanoacetate CCH 2 COCH 2 CH 3 N O NaCN S N 2 BrCH 2 CH 2 OH 2-Bromoethanol 2-Bromoethyl acetate CH 3 COCH 2 CH 2 Br O or CH 3 CCl O CH 3 COCCH 3 O O CH 3 CH 2 OH Ethanol BrCH 2 CH 2 OH 2-Bromoethanol Br 2 H 2 O H 2 SO 4 heat Ethylene CH 2 CH 2 CH 3 CO 2 H Acetic acid BrCH 2 CO 2 H Bromoacetic acid BrCH 2 COCH 2 CH 3 O Ethyl bromoacetate CH 3 CH 2 OH H H11001 Br 2 P H11001CH 3 CH 2 OH Ethanol Acetic anhydride O O CH 3 COCCH 3 CH 3 COCH 2 CH 3 O Ethyl acetate pyridine H11001CH 3 CH 2 OH Ethanol Acetyl chloride O CH 3 CCl CH 3 COCH 2 CH 3 O Ethyl acetate pyridine H11001CH 3 CH 2 OH Ethanol Acetic acid O CH 3 COH H11001CH 3 COCH 2 CH 3 O Ethyl acetate Water H 2 O H H11001 558 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (g) Reaction of the acetyl chloride prepared in part (a) or the acetic anhydride from part (b) with ammonia gives acetamide. (h) Methylamine may be prepared from acetamide by a Hofmann rearrangement. (i) The desired hydroxy acid is available from hydrolysis of the corresponding cyanohydrin, which may be prepared by reaction of the appropriate aldehyde with cyanide ion. In this synthesis the cyanohydrin is prepared from ethanol by way of acetaldehyde. 20.29 (a) Benzoyl chloride is made from benzoic acid. Oxidize toluene to benzoic acid, and then treat with thionyl chloride. (b) Benzoyl chloride and benzoic acid, both prepared from toluene in part (a), react with each other to give benzoic anhydride. Benzoic acid C 6 H 5 COH O Benzoyl chloride C 6 H 5 CCl O H11001 Benzoic anhydride C 6 H 5 COCC 6 H 5 O O Benzoic acidToluene C 6 H 5 CH 3 K 2 Cr 2 O 7 , H 2 SO 4 H 2 O, heat C 6 H 5 COH O Benzoyl chloride C 6 H 5 CCl O SOCl 2 H 2 O, H H11001 , heat 2-Hydroxypropanoic acid 2-Hydroxypropanenitrile OH CH 3 CHC N 1. HO H11002 , H 2 O, heat 2. H H11001 or CH 3 CHCOH OH O CH 3 CH 2 OH O CH 3 CH AcetaldehydeEthanol PCC CH 2 Cl 2 KCN H H11001 2-Hydroxypropanenitrile OH CH 3 CHC N CH 3 CHCOH OH O CH 3 CH O OH CH 3 CHC N Br 2 , HO H11002 , H 2 O O CH 3 CNH 2 CH 3 NH 2 MethylamineAcetamide [prepared as in part (g)] or Acetyl chloride CH 3 CCl O Acetic anhydride CH 3 COCCH 3 O O Acetamide CH 3 CNH 2 O NH 3 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 559 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Benzoic acid, benzoyl chloride, and benzoic anhydride have been prepared in parts (a) and (b) of this problem. Any of them could be converted to benzyl benzoate on reaction with benzyl alcohol. Thus the synthesis of benzyl benzoate requires the preparation of benzyl alcohol from toluene. This is effected by a nucleophilic substitution reaction of benzyl bromide, in turn pre- pared by halogenation of toluene. Alternatively, recall that primary alcohols may be obtained by reduction of the corresponding carboxylic acid. Then (d) Benzamide is prepared by reaction of ammonia with either benzoyl chloride from part (a) or benzoic anhydride from part (b). (e) Benzonitrile may be prepared by dehydration of benzamide. ( f ) Benzyl cyanide is the product of nucleophilic substitution by cyanide ion on benzyl bromide or benzyl chloride. The benzyl halides are prepared by free-radical halogenation of the toluene side chain. or Toluene C 6 H 5 CH 3 Benzyl bromide C 6 H 5 CH 2 Br NBS or Br 2 , light NaCN Benzyl cyanide C 6 H 5 CH 2 CN Toluene C 6 H 5 CH 3 Benzyl chloride C 6 H 5 CH 2 Cl Cl 2 light or heat NaCN Benzyl cyanide C 6 H 5 CH 2 CN Benzamide O C 6 H 5 CNH 2 P 4 O 10 heat Benzonitrile C 6 H 5 CN Benzoyl chloride O C 6 H 5 CCl Benzoic anhydride O O C 6 H 5 COCC 6 H 5 Benzamide O C 6 H 5 CNH 2 or NH 3 Benzoyl chloride O C 6 H 5 CCl Benzyl benzoate O C 6 H 5 COCH 2 C 6 H 5 Benzyl alcohol C 6 H 5 CH 2 OHH11001 pyridine Benzoic acid O C 6 H 5 COH Benzyl alcohol C 6 H 5 CH 2 OH 1. LiAlH 4 2. H 2 O Toluene C 6 H 5 CH 3 N-bromosuccinimide (NBS) or Br 2 , light Benzyl bromide C 6 H 5 CH 2 Br Benzyl alcohol C 6 H 5 CH 2 OH H 2 O HO H11002 560 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (g) Hydrolysis of benzyl cyanide yields phenylacetic acid. Alternatively, the Grignard reagent derived from benzyl bromide may be carboxylated. (h) The first goal is to synthesize p-nitrobenzoic acid because this may be readily converted to the desired acyl chloride. First convert toluene to p-nitrotoluene; then oxidize. Nitration must pre- cede oxidation of the side chain in order to achieve the desired para orientation. Treatment of p-nitrobenzoic acid with thionyl chloride yields p-nitrobenzoyl chloride. (i) In order to achieve the correct orientation in m-nitrobenzoyl chloride, oxidation of the methyl group must precede nitration. Once m-nitrobenzoic acid has been prepared, it may be converted to the corresponding acyl chloride. SOCl 2 m-Nitrobenzoic acid COH O 2 N O m-Nitrobenzoyl chloride CCl O 2 N O CH 3 CO 2 H HNO 3 H 2 SO 4 K 2 Cr 2 O 7 , H 2 SO 4 H 2 O, heat Toluene Benzoic acid m-Nitrobenzoic acid CO 2 H O 2 N COH O O 2 N SOCl 2 p-Nitrobenzoic acid CCl O O 2 N p-Nitrobenzoyl chloride HNO 3 H 2 SO 4 K 2 Cr 2 O 7 , H 2 SO 4 H 2 O, heat CH 3 Toluene CH 3 O 2 N p-Nitrotoluene (separate from ortho isomer) p-Nitrobenzoic acid COH O O 2 N Benzyl bromide C 6 H 5 CH 2 Br Benzylmagnesium bromide C 6 H 5 CH 2 MgBr Phenylacetic acid C 6 H 5 CH 2 COH O Mg diethyl ether 1. CO 2 2. H 3 O H11001 H 2 O, H H11001 , heat or 1. NaOH, heat 2. H H11001 Benzyl cyanide C 6 H 5 CH 2 CN Phenylacetic acid C 6 H 5 CH 2 COH O CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 561 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( j) A Hofmann rearrangement of benzamide affords aniline. 20.30 The problem specifies that is to be prepared from 18 O-labeled ethyl alcohol . Thus, we need to prepare 18 O-labeled ethyl alcohol from the other designated starting materials, acetaldehyde and 18 O-enriched water. First, replace the oxygen of acetaldehyde with 18 O by the hydration–dehydration equilibrium in the presence of 18 O-enriched water. Once 18 O-enriched acetaldehyde has been obtained, it can be reduced to 18 O-enriched ethanol. 20.31 (a) The rate-determining step in basic ester hydrolysis is nucleophilic addition of hydroxide ion to the carbonyl group. The intermediate formed in this step is negatively charged. The electron-withdrawing effect of a CF 3 group stabilizes the intermediate formed in the rate- determining step of ethyl trifluoroacetate saponification. Because the intermediate is more stable, it is formed faster than the one from ethyl acetate. H11001 CF 3 COCH 2 CH 3 OH O H11002 CF 3 COCH 2 CH 3 O Ethyl trifluoroacetate HO H11002 Hydroxide ion Rate-determining intermediate H11001CH 3 COCH 2 CH 3 O Ethyl acetate HO H11002 Hydroxide ion CH 3 COCH 2 CH 3 OH O H11002 Rate-determining intermediate CH 3 CH 2 OH NaBH 4 , CH 3 OH or 1. LiAlH 4 2. H 2 O CH 3 CH O H11001 H11001CH 3 CH O Acetaldehyde H 2 O 18 O-enriched water CH 3 CH O 18 O-enriched acetaldehyde H 2 O Water CH 3 CH OH OH Hydrate of acetaldehyde Propanoyl chloride O CH 3 CH 2 CCl H11001 Ethyl propanoate O CH 3 CH 2 COCH 2 CH 3 Ethyl alcohol CH 3 CH 2 OHO O (O H11005 18 O) O CH 3 CH 2 COCH 2 CH 3 O CNH 2 NH 2 Benzamide [prepared as in part (d)] O H11001 Br 2 Bromine Aniline HO H11002 H 2 O 562 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Crowding is increased as the transition state for nucleophilic addition to the carbonyl group is approached. The carbonyl carbon undergoes a change in hybridization from sp 2 to sp 3 . The tert-butyl group of ethyl 2,2-dimethylpropanoate causes more crowding than the methyl group of ethyl acetate; the rate-determining intermediate is less stable and is formed more slowly. (c) We see here another example of a steric effect of a tert-butyl group. The intermediate formed when hydroxide ion adds to the carbonyl group of tert-butyl acetate is more crowded and less stable than the corresponding intermediate formed from methyl acetate. (d) Here, as in part (a), we have an electron-withdrawing substituent increasing the rate of ester saponification. It does so by stabilizing the negatively charged intermediate formed in the rate-determining step. (e) Addition of hydroxide to 4-butanolide introduces torsional strain in the intermediate because of eclipsed bonds. The corresponding intermediate from 5-butanolide is more stable because the bonds are staggered in a six-membered ring. Eclipsed bonds H H OH O O H11002 Less stable; formed more slowly Staggered bonds H H OH O H11002 O More stable; formed faster Rate-determining intermediate from methyl m-nitrobenzoate COCH 3 O H11002 OH N H11001 more stable than Rate-determining intermediate from methyl benzoate COCH 3 O H11002 OH O H11002 O HO H11002 H11001CH 3 COCH 3 O Methyl acetate Hydroxide ion CH 3 COCH 3 HO O H11002 Rate-determining intermediate; less crowded than intermediate from tert-butyl acetate H11001CH 3 COCCH 3 CH 3 O CH 3 tert-Butyl acetate CH 3 COCCH 3 CH 3 O H11002 HO CH 3 Rate-determining intermediate; crowded HO H11002 Hydroxide ion H11001 H11002 OH Hydroxide ion Ethyl 2,2-dimethylpropanoate CH 3 C COCH 2 CH 3 CH 3 O CH 3 sp 2 Rate-determining intermediate; crowded CH 3 C CH 3 CH 3 COCH 2 CH 3 O H11002 OH sp 3 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 563 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) Steric crowding increases more when hydroxide adds to the axial carbonyl group. 20.32 Compound A is the p-toluenesulfonate ester (tosylate) of trans-4-tert-butylcyclohexanol. The oxygen atom of the alcohol attacks the sulfur of p-toluenesulfonyl chloride, and so the reaction proceeds with retention of configuration. The second step is a nucleophilic substitution in which benzoate ion displaces p-toluenesulfonate with inversion of configuration. Saponification of cis-4-tert-butylcyclohexyl benzoate in step 3 proceeds with acyl–oxygen cleav- age to give cis-4-tert-butylcyclohexanol. 20.33 Reaction of ethyl trifluoroacetate with ammonia yields the corresponding amide, compound A. Com- pound A undergoes dehydration on heating with P 4 O 10 to give trifluoroacetonitrile, compound B. Grignard reagents react with nitriles to form ketones. tert-Butyl trifluoromethyl ketone is formed from trifluoroacetonitrile by treatment with tert-butylmagnesium chloride followed by aqueous hydrolysis. (CH 3 ) 3 CMgCl tert-Butylmagnesium chloride CF 3 CC(CH 3 ) 3 O tert-Butyl trifluoromethyl ketone H11001NCF 3 C Compound B 1. diethyl ether 2. H 3 O H11001 NH 3 P 4 O 10 heat CF 3 COCH 2 CH 3 O Ethyl trifluoroacetate CF 3 CNH 2 O Trifluoroacetamide (compound A) CF 3 CN Trifluoroacetonitrile (compound B) H11001 Benzoate ion CO O H11002 trans-4-tert-Butylcyclohexyl p-toluenesulfonate (compound A) CH 3 OS O H O cis-4-tert-Butylcyclohexyl benzoate (compound B) H OC O H11001 trans-4-tert-Butylcyclohexanol OH p-Toluenesulfonyl chloride CH 3 SCl O O trans-4-tert-Butylcyclohexyl p-toluenesulfonate (compound A) CH 3 OS O O Cis diastereomer: greater increase in crowding when carbon changes from sp 2 to sp 3 ; formed more slowly OCH 2 CH 3 OHO H11002 C Trans diastereomer: smaller increase in crowding when carbon changes from sp 2 to sp 3 ; formed more rapidly OH O H11002 C OCH 2 CH 3 564 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 20.34 The first step is acid hydrolysis of an acetal protecting group. Step 1: All three alcohol functions are converted to bromide by reaction with hydrogen bromide in step 2. Step 2: Reaction with ethanol in the presence of an acid catalyst converts the carboxylic acid to its ethyl ester in step 3. Step 3: The problem hint points out that zinc converts vicinal dibromides to alkenes. Of the three bromine substituents in compound D, two of them are vicinal. Step 4 is a dehalogenation reaction. Step 4: Step 5 is a nucleophilic substitution of the S N 2 type. Acetate ion is the nucleophile and displaces bromide from the primary carbon. Step 5: Compound E NaOCCH 3 CH 3 CO 2 H CH 3 CH 2 OC(CH 2 ) 5 CH CH(CH 2 ) 7 CH 2 OCCH 3 O O Compound F (C 20 H 36 O 4 ) O Compound D Zn ethanol CH 3 CH 2 OC(CH 2 ) 5 CH CH(CH 2 ) 7 CH 2 Br O Compound E (C 18 H 33 BrO 2 ) Compound C ethanol H 2 SO 4 CH 3 CH 2 OC(CH 2 ) 5 CH CH(CH 2 ) 7 CH 2 Br O BrBr Compound D (C 18 H 33 Br 3 O 2 ) Compound B HBr HOC(CH 2 ) 5 CH CH(CH 2 ) 7 CH 2 Br O Br Br Compound C (C 16 H 29 Br 3 O 2 ) Compound A H 2 O, H H11001 heat HOC(CH 2 ) 5 CH CH(CH 2 ) 7 CH 2 OH O HO OH Compound B (C 16 H 32 O 5 ) CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 565 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Step 6 is ester saponification. It yields a 16-carbon chain having a carboxylic acid function at one end and an alcohol at the other. Step 6: In step 7, compound G cyclizes to ambrettolide on heating. Step 7: 20.35 (a) This step requires the oxidation of a primary alcohol to an aldehyde. As reported in the litera- ture, pyridinium dichromate in dichloromethane was used to give the desired aldehyde in 84% yield. (b) Conversion of to @CH?CH 2 is a typical case in which a Wittig reaction is appropriate. (c) Lithium aluminum hydride was used to reduce the ester to a primary alcohol in 81% yield. (d) The desired sex pheromone is the acetate ester of compound D. Compound D was treated with acetic anhydride to give the acetate ester in 99% yield. Acetyl chloride could have been used in this step instead of acetic anhydride. 20.36 (a) The reaction given in the problem is between a lactone (cyclic ester) and a difunctional Grignard reagent. Esters usually react with 2 moles of a Grignard reagent; in this instance Compound D CHCHH 2 C CH(CH 2 ) 7 CH 2 OH (E)-9,11-Dodecadien-1-yl acetate CHCHH 2 C O CH(CH 2 ) 7 CH 2 OCCH 3 pyridine CH 3 COCCH 3 O O Compound C CHCHH 2 C CH(CH 2 ) 7 CO 2 CH 3 1. LiAlH 4 2. H 2 O Compound D CHCHH 2 C CH(CH 2 ) 7 CH 2 OH Compound B CH(CH 2 ) 7 CO 2 CH 3 HCCH O Compound C (observed yield, 53%) CHCHH 2 C CH(CH 2 ) 7 CO 2 CH 3 (C 6 H 5 ) 3 PCH 2 H11001 H11002 CH O Compound A CH(CH 2 ) 7 CO 2 CH 3 HOCH 2 CH Compound B CH(CH 2 ) 7 CO 2 CH 3 HCCH O PDC CH 2 Cl 2 heat Compound G HO 2 C HO Ambrettolide OO Compound F HOC(CH 2 ) 5 CH CH(CH 2 ) 7 CH 2 OH O Compound G (C 16 H 30 O 3 ) 1. KOH, ethanol 2. H H11001 566 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website both Grignard functions of the reagent attack the lactone. The second attack is intramolecu- lar, giving rise to the cyclopentanol ring of the product. (b) An intramolecular acyl transfer process takes place in this reaction. The amine group in the thiolactone starting material replaces sulfur on the acyl group to form a lactam (cyclic amide). 20.37 (a) Acyl chlorides react with alcohols to form esters. H11001 pyridine CCH O OH Benzoin CH 3 O CCl O p-Methoxybenzoyl chloride CCH O O C OCH 3 O Benzoin p-methoxybenzoate (compound A; 95%) O N H HS Lactam S OH NH Tetrahedral intermediate S NH 2 O Thiolactone H11001 4-Butanolide 1-(3-Hydroxypropyl)- cyclopentanol (88%) O O H11001 BrMg CH 2 CH 2 CH 2 CH 2 MgBr O CH 2 MgBr H11002 O O CH 2 MgBr H11001 OBrMg H11001 H11002 OBrMg H11001 H11002 O MgBr H11001 H11002 HO OH H 3 O H11001 H11002 H11002 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 567 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Of the two carbonyl groups in the starting material, the ketone carbonyl is more reactive than the ester. (The ester carbonyl is stabilized by electron release from oxygen.) Compound B has the molecular formula C 6 H 10 O 2 . The initial product forms a cyclic ester (lactone), with elimination of ethoxide ion. (c) Only carboxyl groups that are ortho to each other on a benzene ring are capable of forming a cyclic anhydride. (d) The primary amine can react with both acyl chloride groups of the starting material to give compound D. 20.38 Compound A is an ester but has within it an amine function. Acyl transfer from oxygen to nitrogen converts the ester to a more stable amide. The tetrahedral intermediate is the key intermediate in the reaction. CH 2 CH 2 N O Ar H R OC Compound A (Ar H11005 p-nitrophenyl) CH 2 CH 2 N O Ar R OH C Tetrahedral intermediate Compound B (Ar H11005 p-nitrophenyl) ArCNCH 2 CH 2 OH R O H11001 CH 3 CH 2 CH 2 CH 2 NH 2 S Br Br ClC O O CCl Compound D S Br Br OO N CH 2 CH 2 CH 2 CH 3 heat COH HOC O COH O O O O O H 2 OH11001 HOC O Compound C CH 3 C O O OCH 2 CH 3 C CH 2 CH 2 CH 3 H11002 H11001 CH 3 CH 2 O H11002 O O H 3 C H 3 C Compound B OO CH 3 CCH 2 CH 2 COCH 2 CH 3 OCH 3 OMgI CH 3 CCH 2 CH 2 COCH 2 CH 3 CH 3 MgI (1 eq) 568 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 20.39 (a) The rearrangement in this problem is an acyl transfer from nitrogen to oxygen. This rearrangement takes place in the indicated direction because it is carried out in acid solution. The amino group is protonated in acid and is no longer nucleophilic. (b) The trans stereoisomer of compound A does not undergo rearrangement because when the oxygen and nitrogen atoms on the five-membered ring are trans, the necessary tetrahedral intermediate cannot form. 20.40 The ester functions of a polymer such as poly(vinyl acetate) are just like ester functions of simple molecules; they can be cleaved by hydrolysis under either acidic or basic conditions. To prepare poly(vinyl alcohol), therefore, polymerize vinyl acetate to poly(vinyl acetate), and then cleave the ester groups by hydrolysis. 20.41 (a) Each propagation step involves addition of the free-radical species to the H9252-carbon of a mole- cule of methyl methacrylate. (b) The correct carbon skeleton can be constructed by treating acetone with sodium cyanide in the presence of H 2 SO 4 to give acetone cyanohydrin. H11001 H 2 SO 4 CH 3 CCH 3 O Acetone HCN Hydrogen cyanide CH 3 CCH 3 OH CN Acetone cyanohydrin ROCH 2 CH 2 C CO 2 CH 3 CH 3 C CO 2 CH 3 CH 3 CH 2 C CO 2 CH 3 CH 3 ROCH 2 CH 2 C CO 2 CH 3 CH 3 C CO 2 CH 3 CH 3 H11001 H 2 C CCOCH 3 O CH 3 ROCH 2 C CO 2 CH 3 CH 3 ROCH 2 CH 2 C CO 2 CH 3 CH 3 C CO 2 CH 3 CH 3 H11001 H 2 C CCOCH 3 O CH 3 H 2 C CHOCCH 3 O Vinyl acetate H 2 O H H11001 or HO H11002 Poly(vinyl acetate) CH 2 CHCH 2 CH CH 3 CO OO OCCH 3 n Poly(vinyl alcohol) CH 2 CHCH 2 CH HO OH n H H11001 H H11001 Tetrahedral intermediate Ar C NH HO O Compound A (Ar H11005 p-nitrophenyl) O Ar C NHHO Compound B (Ar H11005 p-nitrophenyl) Ar C NH 3 O O H11001 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 569 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Dehydration of the cyanohydrin followed by hydrolysis of the nitrile group and esterification of the resulting carboxylic acid yields methyl methacrylate. 20.42 The compound contains nitrogen and exhibits a prominent peak in the infrared spectrum at 2270 cm H110021 ; it is likely to be a nitrile. Its molecular weight of 83 is consistent with the molecular formula C 5 H 9 N. The presence of four signals in the H9254 10 to 30-ppm region of the 13 C NMR spectrum suggests an unbranched carbon skeleton. This is confirmed by the presence of two triplets in the 1 H NMR spec- trum at H9254 1.0 ppm (CH 3 coupled with adjacent CH 2 ) and at H9254 2.3 ppm (CH 2 CN coupled with adjacent CH 2 ). The compound is pentanenitrile. 20.43 The compound has the characteristic triplet–quartet pattern of an ethyl group in its 1 H NMR spec- trum. Because these signals correspond to 10 protons, there must be two equivalent ethyl groups in the molecule. The methylene quartet appears at relatively low field (H9254 4.1 ppm), which is consistent with ethyl groups bonded to oxygen, as in @OCH 2 CH 3 . There is a peak at 1730 cm H110021 in the infrared spectrum, suggesting that these ethoxy groups reside in ester functions. The molecular formula C 8 H 14 O 4 reveals that if two ester groups are present, there can be no rings or double bonds. The re- maining four hydrogens are equivalent in the 1 H NMR spectrum, and so two equivalent CH 2 groups are present. The compound is the diethyl ester of succinic acid. 20.44 Compound A (C 4 H 6 O 2 ) has an index of hydrogen deficiency of 2. With two oxygen atoms and a peak in the infrared at 1760 cm H110021 , it is likely that one of the elements of unsaturation is the carbon–oxygen double bond of an ester. The 1 H NMR spectrum contains a three-proton singlet at H9254 2.1 ppm, which is consistent with a unit. It is likely that compound A is an acetate ester. The 13 C NMR spectrum reveals that the four carbon atoms of the molecule are contained in one each of the fragments CH 3 , CH 2 , and CH, along with the carbonyl carbon. In addition to the two car- bons of the acetate group, the remaining two carbons are the CH 2 and CH carbons of a vinyl group, CH?CH 2 . Compound A is vinyl acetate. Each vinyl proton is coupled to two other vinyl protons; each appears as a doublet of doublets in the 1 H NMR spectrum. 20.45 Solutions to molecular modeling exercises are not provided in this Study Guide and Solutions Man- ual. You should use Learning By Modeling for this exercise. CH 3 C OCH CH 2 O H9254 96.8 ppm H9254 141.8 ppm H9254 167.6 ppm H9254 20.2 ppm CH 3 C O CH 3 CH 2 OCCH 2 CH 2 COCH 2 CH 3 O O Diethyl succinate CH 3 CH 2 CH 2 CH 2 C N Pentanenitrile H 2 O, H H11001 heat CH 3 OH, H H11001 H 2 SO 4 heat CH 3 CCH 3 OH CN Acetone cyanohydrin H 2 C CCN CH 3 H 2 C CCO 2 H CH 3 H 2 C CCO 2 CH 3 CH 3 Methyl methacrylate 570 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SELF-TEST PART A A-1. Give a correct IUPAC name for each of the following acid derivatives: (a) (b) (c) A-2. Provide the correct structure of (a) Benzoic anhydride (b) N-(1-Methylpropyl)acetamide (c) Phenyl benzoate A-3. What reagents are needed to carry out each of the following conversions? (a) (b) (c) A-4. Write the structure of the product of each of the following reactions: (a) (b) (c) (d) (e) ? (two products)CNHCH 3 H 2 O, H 2 SO 4 heat H 3 C C O ? (two products)Ethyl propanoate H11001 dimethylamine H H11001 (cat) H11001 CH 3 CH 2 OH ? O O O Cyclopentanol benzoyl chloride ?H11001 pyridine Cyclohexyl acetate ? (two products) 1. NaOH, H 2 O 2. H H11001 (CH 3 ) 2 CHCH 2 NH 2 C 6 H 5 CNHCH 2 CH(CH 3 ) 2 ? O H11001 CH 3 OH (CH 3 ) 3 CCNH 2 (CH 3 ) 3 CNH 2 ? O C 6 H 5 CH 2 CO 2 HC 6 H 5 CH 2 CCl ? O (CH 3 ) 2 CHCH 2 CH 2 CCl O C 6 H 5 CNHCH 3 O CH 3 CH 2 CH 2 OCCH 2 CH 2 CH 3 O CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 571 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-5. The following reaction occurs when the reactant is allowed to stand in pentane. Write the structure of the key intermediate in this process. A-6. Give the correct structures, clearly showing stereochemistry, of each compound, A through D, in the following sequence of reactions: A-7. Write the structure of the neutral form of the tetrahedral intermediate in the (a) Acid-catalyzed hydrolysis of methyl acetate (b) Reaction of ammonia with acetic anhydride A-8. Write the steps necessary to prepare from . A-9. Outline a synthesis of benzyl benzoate using toluene as the source of all the carbon atoms. A-10. The infrared spectrum of a compound (C 3 H 6 ClNO) has an intense peak at 1680 cm H110021 . Its 1 H NMR spectrum consists of a doublet (3H, H9254 1.5 ppm), a quartet (1H, H9254 4.1 ppm), and a broad singlet (2H, H9254 6.5 ppm). What is the structure of the compound? How would you pre- pare it from propanoic acid? PART B B-1. What are the products of the most favorable mode of decomposition of the intermediate species shown? (a) Benzoic acid and HCl (b) Benzoyl chloride and H 2 O (c) Both (a) and (b) equally likely (d) Neither (a) nor (b) C 6 H 5 C Cl OH OH COCH 2 O Benzyl benzoate BrH 3 CNH 2 H 3 C C (C 7 H 15 N) D (C 8 H 13 N) CO 2 H AB CH 3 SOCl 2 NH 3 Br 2 NaOH H 2 O P 4 O 10 heat C 6 H 5 COCH 2 CH 2 NHCH 3 O CH 3 NCH 2 CH 2 OH C 6 H 5 CO 572 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-2. What is the correct IUPAC name for the compound shown? (a) 3-Bromopropyl 2-chloro-4-butylbutanoate (b) 2-Chloro-4-phenylbutyl 3-bromopropanoate (c) 3-Chloro-1-phenylbutyl 1-bromopropanoate (d) 3-Chloro-1-phenylbutyl 3-bromopropanoate (e) 7-Bromo-3-chloro-1-phenylbutyl propanoate B-3. Rank the following in order of increasing reactivity (least A most) toward acid hydrolysis: (a)1H11021 2 H11021 3(c)1H11021 3 H11021 2 (b)3H11021 1 H11021 2(d)2H11021 1 H11021 3 B-4. The structure of N-propylacetamide is B-5. Choose the response that matches the correct functional group classification with the fol- lowing group of structural formulas. (a) Anhydride Lactam Lactone (b) Lactam Imide Lactone (c) Imide Lactone Anhydride (d) Imide Lactam Lactone B-6. Choose the best sequence of reactions for the transformation given. Semicolons indicate separate reaction steps to be used in the order shown. (a)H 3 O H11001 ; SOCl 2 ; CH 3 NH 2 (b)HO H11002 /H 2 O; PBr 3 ; Mg; CO 2 ; H 3 O H11001 ; SOCl 2 ; CH 3 NH 2 (c) LiAlH 4 ; H 2 O; HBr; Mg; CO 2 ; H 3 O H11001 ; SOCl 2 ; CH 3 NH 2 (d) None of these would yield the desired product. H 3 CCO 2 CH 3 H 3 CCH 2 CNHCH 3 O ? OO N H O N H O O (d)CH 3 CH NCH 2 CH 2 CH 3 (b)CH 3 CN(CH 2 CH 2 CH 3 ) 2 O (c)CH 3 CNHCH 2 CH 2 CH 3 O (a)CH 3 CH 2 CNHCH 3 O CH 3 COCH 2 CH 3 O 1 CH 3 CCl O 2 CH 3 CNHCH 3 O 3 CH 2 CH 2 CHCH 2 OCCH 2 CH 2 Br Cl O CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 573 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-7. A key step in the hydrolysis of acetamide in aqueous acid proceeds by nucleophilic addi- tion of (a)(d) (b e) (c) B-8. Which reaction is not possible for acetic anhydride? (a) (b) (c) (d ) B-9. All but one of the following compounds react with aniline to give acetanilide. Which one does not? (a)(b)(c)(d) (e) B-10. Identify product Z in the following reaction sequence: CHCH 2 BrH 2 C NaCN YZ 1. C 6 H 5 MgBr, 2. H 3 O H11001 diethyl ether OCCH 3 O CH 3 O O H 3 CO O O H 3 C CH 3 CH 3 CH O CH 3 CCl O NH 2 NHCCH 3 O Aniline Acetanilide H11001 NaCl CH 3 CCl H11001 CH 3 CO 2 H11002 Na H11001 OO (CH 3 C) 2 O H11001 C 6 H 6 CH 3 CC 6 H 5 H11001 CH 3 CO 2 H O AlCl 3 O (CH 3 C) 2 O H11001 CH 3 CH 2 OH CH 3 CO 2 CH 2 CH 3 H11001 CH 3 CO 2 H O (CH 3 C) 2 O H11001 2HN(CH 3 ) 2 CH 3 CN(CH 3 ) 2 H11001 CH 3 CO 2 H11002 H 2 N(CH 3 ) 2 H11001 OO (CH 3 C) 2 O HO H11002 to CH 3 CNH 2 O HO H11002 to CH 3 CNH 2 H11001 OH H 3 O H11001 to CH 3 CNH 2 H11001 OH H 2 O to CH 3 CNH 2 H11001 OH H 3 O H11001 to CH 3 CNH 2 O 574 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (a)(d) (b e) (c) B-11. Which of the following best describes the nucleophilic addition step in the acid-catalyzed hydrolysis of acetonitrile (CH 3 CN)? B-12. Saponification (basic hydrolysis) of will yield: [ H11005 mass-18 isotope of oxygen] B-13. An unknown compound, C 9 H 10 O 2 , did not dissolve in aqueous NaOH. The infrared spectrum exhibited strong absorption at 1730 cm H110021 . The 1 H NMR spectrum had signals at H9254 7.2 ppm (multiplet), 4.1 ppm (quartet), and 1.3 ppm (triplet). Which of the following is most likely the unknown? COH CH 2 CH 3 O (a) COCH 2 CH 3 O (b) CH 2 OCCH 3 O (c) CH 2 COCH 3 O (d) OCH 2 CCH 3 O (e) C 6 H 5 CO H11002 (c) H11001 O HOCH 3 C 6 H 5 CO H11002 (e) OCH 3 OH C 6 H 5 CO H11002 (b) H11001 O HOCH 3 C 6 H 5 CO H11002 (d) H11001 O HOCH 3 C 6 H 5 CO H11002 (a) H11001 O HOCH 3 OC 6 H 5 COCH 3 O H11001 (d)H 3 CCNH O HH (b)H 3 CCN O HH (e)H 3 CCN O HH (c)H 3 CCNH O H11002 H11001 H (a)H 3 CCN O H H11002 CHCH 2 CNHC 6 H 5 H 2 C O CHCH 2 CHC 6 H 5 H 2 C NH 2 CHCH 2 CHC 6 H 5 H 2 C OH CHCH 2 NHCC 6 H 5 H 2 C O CHCH 2 CC 6 H 5 H 2 C O CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 575 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website