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 RCOCR. They are named as derivatives of carboxylic acid CH CH.CHCOH CH CH..CH C6Hs C6H 2-Phenylbutanoic acid 2-Phenylbutanoic anhydride (c) Butyl 2-phenylbutanoate is the butyl ester of 2-phenylbutanoic acid CHCH, CHCOCH, CH, CH,CH (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, CH,CH,COCH,CHCH, CH 2-Phenylbutyl butanoate 536 Back Forward Main Menu TOC Study Guide Toc Student OLC MHHE Website
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. CH3CH2CH2COCH2CHCH2CH3 O C6H5 2-Phenylbutyl butanoate CH3CH2CHCOCH2CH2CH2CH3 O C6H5 Butyl 2-phenylbutanoate CH3CH2CHCOCCHCH2CH3 O O C6H5 C6H5 CH3CH2CHCOH O C6H5 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 RCNH, CH3,CHCNH CAH 2-Phenylbutanamide (f) This compound differs from 2-phenylbutanamide in part(e)only in that it bears an ethyl sub- sutuent on nitrogen CHa CH, CHCNHCH, CH CH N-Ethyl-2-Phenylbutanamide (g) The -nitrile ending signifies a compound of the type RC=N containing the same number of carbons as the alkane rch CH2 CH. CHO≡N CH 2-Phenylbutanenitrile 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. CH CH3 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 C6H- CCI C6HSCOH CBHSCOCC H5 HCI Benzoic acid Benzoic anhydride Hydrogen (c) Acyl chlorides react with alcohols to form esters C6H CCI CH3 CH,OH- C6H_ COCH, CH3 HCI Benzoyl Ethanol chloride The organic product is the ethyl ester of benzoic acid, ethyl benzoate. 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 substituent on nitrogen. (g) The -nitrile ending signifies a compound of the type RC>N containing the same number of carbons as the alkane RCH3. 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 imparted 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. C6H5CCl O Benzoyl chloride CH3CH2OH Ethanol C6H5COCH2CH3 O Ethyl benzoate HCl Hydrogen chloride C6H5CCl O Benzoyl chloride C6H5COH O Benzoic acid C6H5COCC6H5 O O Benzoic anhydride HCl Hydrogen chloride C H O N CH3 CH3 C H O N CH3 CH3 2-Phenylbutanenitrile CH3CH2CHC N C6H5 CH3CH2CHCNHCH2CH3 O C6H5 N-Ethyl-2-phenylbutanamide CH3CH2CHCNH2 O C6H5 2-Phenylbutanamide RCNH2 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 C6HsCCI 2CH,NH C6HSCNHCH3 CHaNH3 CI Benzoyl Methy lamine N-Methy benzamide chloride (e) In analogy with part(d), an amide is formed. In this case the product has two methyl groups CI t 2(CH)2NH H3)2+(CH3)2NH2 (f) Acyl chlorides undergo hydrolysis on reaction with water. The product is a carboxylic acid CsH +H2O—CH5COH+HCl Hyd chloride 20.4(b) Nucleophilic addition of benzoic acid to benzoyl chloride gives the tetrahedral intermediate own C6HsCCI CH COH C6H-COCC6H5 Benzoic acid Dissociation of the tetrahedral intermediate occurs by loss of chloride and of the proton on the oxygen. C6H-COCC6H5 C6H COCC6H5 HCI Tetrahedral intermediate Benzoic anhydrid chloride (c) Ethanol is the nucleophile that adds to the carbonyl group of benzoyl chloride to form the tetrahedral intermediate C6HSCCI CH, CH,OH C6H, CH3 Ethanol 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-methylbenzamide. (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. C6H5CCl O Benzoyl chloride CH3CH2OH Ethanol Tetrahedral intermediate C6H5COCH2CH3 OH Cl C6H5COCC6H5 O O Benzoic anhydride HCl Hydrogen chloride Tetrahedral intermediate C6H5COCC6H5 O H Cl O C6H5CCl O Benzoyl chloride C6H5COH O Benzoic acid C6H5COCC6H5 HO Cl O Tetrahedral intermediate C6H5CCl O Benzoyl chloride H2O Water C6H5COH O Benzoic acid HCl Hydrogen chloride C6H5CCl O Benzoyl chloride 2(CH3)2NH Dimethylamine C6H5CN(CH3)2 O N,N-Dimethylbenzamide (CH3)2NH2 Cl C6H5CCl O Benzoyl chloride 2CH3NH2 Methylamine C6H5CNHCH3 O N-Methylbenzamide CH3NH3 Cl Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SU BSTITUTION 539 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 CsHSCOCH,CH 6H5 COCH, CH3 HCI Tetrahedral intermediate Ethyl benzoate Hydrogen (d) The tetrahedral intermediate formed from benzoyl chloride and methylamine has a carbon- C6HSCCI CH,NH2 C6H-CNHCH3 Benzoyl Methylamine Tetrahedral intermediate The dissociation of the tetrahedral intermediate may be shown as CBHSCNHCH C6HSCNHCH3 HCl Tetrahedral intermediate More realistically, it is a second methylamine molecule that abstracts a proton from oxygen CH3NH C6HSCNHCH C6HSCNHCH3 CH3NH, CI N-Methylbenzamide Methylammonium chloride (e) The intermediates in the reaction of benzoyl chloride with dimethy lamine are similar to those n part(d). The methyl substituents on nitrogen are not directly involved in the reaction. C6H- CCI (CH,)2NH C6H5CN(CH3) 2 Tetrahedral intermediate Then (CH3)2NH C6HSCN(CH3)2 C6HSCN(CH3) (CH3)2NH, CI N,N-Dimethylbenzamide Dimethylammonium chloride 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 C6H5CN(CH3)2 O N,N-Dimethylbenzamide Dimethylammonium chloride (CH3)2NH2 Cl C6H5CN(CH3)2 O H Cl (CH3)2NH C6H5CCl O Benzoyl chloride (CH3)2NH Dimethylamine Tetrahedral intermediate C6H5CN(CH3)2 OH Cl C6H5CNHCH3 O N-Methylbenzamide Methylammonium chloride CH3NH3 Cl C6H5CNHCH3 O H Cl CH3NH2 C6H5CNHCH3 O N-Methylbenzamide HCl Hydrogen chloride Tetrahedral intermediate C6H5CNHCH3 O H Cl C6H5CCl O Benzoyl chloride CH3NH2 Methylamine Tetrahedral intermediate C6H5CNHCH3 OH Cl C6H5COCH2CH3 O Ethyl benzoate HCl Hydrogen chloride Tetrahedral intermediate C6H5COCH2CH3 O H Cl CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 539 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
540 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION (f) Water attacks the carbonyl group of benzoyl chloride to form the tetrahedral intermediate CsH CCI H,O CH-CCI tetrahedral oride Dissociation of the tetrahedral intermediate occurs by loss of chloride and the proton on C6HSCOH HCI Benzoic acid 20.5 One equivalent of benzoyl chloride reacts rapidly with water to yield benzoic acid C6HSCCI H,O C6HSCOH HCI Hydrogen chloride The benzoic acid produced in this step reacts with the remaining benzoyl chloride to give benzoic anhydride C6HSCCI CBH_ COH CBH_COCC H5 HCl enz Benzoic acid Benzoic anhydride Hydrogen 20.6 Acetic anhydride serves as a source of acetyl cation CH, CO-CCH3 CCH O≡CCH Acetyl catio 20.7(b) Acyl transfer from an acid anhydride to ammonia yields an amide. 00 CH, COCCH3 2NH, CHaCNH,+ CH3 CO NH4 Acetic anhydride ammonia etamide Ammonium acetate The organic products are acetamide and ammonium acetate 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. Acetic anhydride CH3COCCH3 O O 2NH3 Ammonia CH3CNH2 O Acetamide CH3CO NH4 O Ammonium acetate Acetyl cation CH3CO O CCH3 O O CCH3 O CCH3 HCl Hydrogen chloride C6H5COCC6H5 O O Benzoic anhydride C6H5CCl O Benzoyl chloride C6H5COH O Benzoic acid H2O Water C6H5CCl O Benzoyl chloride HCl Hydrogen chloride C6H5COH O Benzoic acid HCl Hydrogen chloride C6H5COH O Tetrahedral Benzoic acid intermediate C6H5C Cl O H OH C6H5CCl O Benzoyl chloride H2O Water Tetrahedral intermediate C6H5CCl OH OH 540 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SU BSTITUTION 541 (c) The reaction of phthalic anhydride with dimethylamine is analogous to that of part(b). The mide and the carboxylate salt of CN(CH3) 0+2(CH3)2NH H,N(CH Phthalic anhydride Dimethylamine amide fu In this case both the amide function and the ammonium carboxylate salt are incorporated into e same molecule (d) The disodium salt of phthalic acid is the product of hydrolysis of phthalic acid in excess sodium hydroxide CO +h,o cO Sodium Sodium phthalate Water 20.8(b) The tetrahedral intermediate is formed by nucleophilic addition of ammonia to one of the car- bonyl groups of acetic anhydride CH3COCCH CH COCCH Tetrahedral Dissociation of the tetrahedral intermediate occurs by loss of acetate as the leaving group HN: CHCTOCCH3 CH,, HN OCCH NE Ammonia tetrahedral Ammonium acetate (c) Dimethylamine is the nucleophile: it adds to one of the two equivalent carbonyl groups of phthalic anhydride HQ N(CH3)2 Phthalic Dimethylamine Tetrahedral anhydride 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 carbonyl 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 Tetrahedral intermediate HN(CH3)2 O HO N(CH3)2 O O O O Ammonia tetrahedral Acetamide Ammonium acetate intermediate CH3CNH2 O H4N OCCH3 O O H CH3C OCCH3 O NH2 H3N Tetrahedral intermediate NH3 O CH3COCCH3 O CH3COCCH3 NH2 HO O 2NaOH Sodium hydroxide Phthalic anhydride O H2O Sodium phthalate Water CO Na CO Na O O O O 2(CH3)2NH Phthalic anhydride Dimethylamine O H2N(CH3)2 Product is an amine salt and contains an amide function. CO CN(CH3)2 O O O O CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 541 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
542 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION A second molecule of dimethylamine abstracts a proton from the tetrahedral intermediate L)NH N(CH3)2 CO H,N(CH3)2 Tetrahedral mediate sec Product of reaction imethy lamin (d) Hydroxide acts as a nucleophile to form the tetrahedral intermediate and as a base to facilitate Formation of tetrahedral intermediate: +m一 thalia anhydride OH HO O Dissociation of tetrahedral intermediate oH COH +h. In base, the remaining carboxylic acid group is deprotonated. Co-H OH CO 十 CO 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 O O H2O H CO CO O O OH O OH O O H COH CO O O HO H2O Tetrahedral intermediate O HO OH O O O OH O H2O OH Phthalic anhydride O O O O O OH OH O Tetrahedral intermediate second Product of reaction molecule of dimethylamine O N(CH3)2 O O H (CH3)2NH CN(CH3)2 CO O O H2N(CH3)2 542 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SU BSTITUTION 543 20.9 The starting material contains three acetate ester functions. All three undergo hydrolysis in aqueous sulfuric acid H, O CH, COCH, CHCH,CH,CH,OCCH3 HOCH CHCH.CH CHOH 3CH-COI OCC OH The product is 1, 2, 5-pentanetriol. Also formed in the hydrolysis of the starting triacetate are three 20.10 Step 1: Protonation of the carbonyl oxygen O OH ChC OCHCH OCH2CH3 Ethyl benzoate Protonated form of ester Water Step 2: Nucleophilic addition of water H CHICA E C HSC--OCH,CH H, CH3 Water Protonated form of ester Step 3: Deprotonation of oxonium ion to give neutral form of tetrahedral intermediat C6HsC-QCH2CH,+jo: C6H5C-OCH2CH3 H- H edral Hydronium Step 4: Protonation of ethoxy oxygen H H CsH-C-OCH, CH2 + H C6HSC-OCH,CH3+: O Tetrahedra Oxonium ion 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 Oxonium ion C6H5C OCH2CH3 OH HO H O H H Water C6H5C OCH2CH3 OH HO Tetrahedral intermediate O H H H Hydronium ion O H H C6H5C OCH2CH3 OH O H H C6H5C OCH2CH3 OH HO Tetrahedral intermediate O H H H Hydronium ion Water Protonated form of ester Oxonium ion O H H C6H5C OH O OCH2CH3 H H C6H5C OCH2CH3 OH Ethyl benzoate Hydronium ion C6H5C OCH2CH3 Protonated form of ester H H H O Water H H C6H5C O OCH2CH3 OH O 3CH3COH O H2O H O O CH3COCH2CHCH2CH2CH2OCCH3 OCCH3 O HOCH2CHCH2CH2CH2OH OH 1,2,5-Pentanetriol (C5H12O3) Acetic acid CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 543 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
544 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUB Step 5: Dissociation of protonated form of tetrahedral intermediat This step yields ethyl alcohol and the protonated form of benzoic acid C6HSC-OCH,CH3 CHSC HOCH2 CH3 OH :OH H Protonated form Ethyl alcohol Step 6: Deprotonation of protonated form of benzoic acid H O H CaH CN CsSC H Benzoic acid Hydronium benzoic acid 20.11 To determine which oxygen of 4-butanolide becomes labeled withO, trace the path ofO-labeled water(0=O)as it undergoes nucleophilic addition to the carbonyl group to form the tetrahedral H,e sO-labeled The tetrahedral intermediate can revert to unlabeled 4-butanolide by loss of O-labeled water Alternatively it can lose ordinary water to give O-labeled lactone 8o-jJabeled Water The carbonyl oxygen is the one that is isotopically labeled in theO-enriched 4-butanolide. 20.12 On the basis of trimyristin's molecular formula CasHs,Os and of the fact that its hydrolysis gives only glycerol and tetradecanoic acid CH(CH)12 CO, H, it must have the structure shown CHa(CH2)n2CO OC(CH,)I2CH OC(CH)12CH3 (CasH86O6) 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 18O, trace the path of 18O-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 18O-labeled water. Alternatively it can lose ordinary water to give 18O-labeled lactone. The carbonyl oxygen is the one that is isotopically labeled in the 18O-enriched 4-butanolide. 20.12 On the basis of trimyristin’s molecular formula C45H86O6 and of the fact that its hydrolysis gives only glycerol and tetradecanoic acid CH3(CH2)12CO2H, it must have the structure shown. CH3(CH2)12CO OC(CH2)12CH3 OC(CH2)12CH3 O O O Trimyristin (C45H86O6) H OH O OH Tetrahedral intermediate O O 18O-labeled 4-butanolide H2O Water 4-Butanolide O O 18O-labeled water H2O H Tetrahedral intermediate O OH OH (O 18O) Benzoic acid C6H5C OH O Water O H H Hydronium ion H O H H Protonated form of benzoic acid C6H5C OH O H Oxonium ion C6H5C OCH2CH3 OH OH H Protonated form of benzoic acid C6H5C OH OH Ethyl alcohol HOCH2CH3 544 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 545 20.13 Because ester hydrolysis in base proceeds by acyl-oxygen cleavage, the O label becomes ncorporated into acetate ion(e=O) CH3CHLCH CH2CH,OCCH3 .H CHa CH,CH,CH,CH,OH CCH3 Pentyl acetate Hydroxide l-Pentanol Acetate 20.14 Step 1: Nucleophilic addition of hydroxide ion to the carbonyl group C6HSC--OC OCH.C Step 2: Proton transfer from water to give neutral form of tetrahedral intermediate OH chCOCHCH +Hcoh CHSC-OCH, CH,+ Anionic form of Water Tetrahedral Hydroxide Step 3: Dissociation of tetrahedral intermediate H-0: HO:+ CH.C-OCH CH,- HOH OCHCH Hydroxide Tetrahedral intermediate Benzoic acid Ethoxide i Step 4: Proton transfer from benzoic acid :OH—C6 HO Benzoic acid Hydroxide Benzoate ion Water 20.15 The starting material is a lactone, a cyclic ester. The ester function is converted to an amide by nucleophilic acyl substitution CHNH+ CH3NHCCH CH,CHCH3 OH Methylamine 4-Pentanolide 4-Hydroxy-N-methylpentanamide 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 18O 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 CH3NH2 O CH3 O CH3NHCCH2CH2CHCH3 OH O Benzoic acid Hydroxide ion O C6H5C O H Benzoate ion Water O C6H5C O HO HOH Tetrahedral intermediate C6H5C OCH2CH3 H O OH C6H5C O OH Benzoic acid OCH2CH3 Ethoxide ion HO Hydroxide ion HOH Water C6H5C OCH2CH3 OH OH Tetrahedral intermediate C6H5C OCH2CH3 O OH Anionic form of tetrahedral intermediate OH Hydroxide ion Water H OH C6H5C OCH2CH3 O OH Anionic form of tetrahedral intermediate C6H5C O OCH2CH3 Ethyl benzoate HO Hydroxide ion Pentyl acetate Hydroxide ion CH3CH2CH2CH2CH2OH 1-Pentanol CCH3 O O Acetate ion CH3CH2CH2CH2CH2O CCH3 OH O (O 18O) CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 545 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website
