Baixe Carey - Organic Chemistry - sgchapt20 e outras Manuais, Projetos, Pesquisas em PDF para Química, somente na Docsity! CHAPTER 20 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION SOLUTIONS TO TEXT PROBLEMS 536 Back| 20.1 Forward o o I (b) | Carboxylic acid anhydrides bear two acyl groups on oxygen, as in RCOCR. They are named as derivatives of carboxylic acids. f Et CH;CH,CHCOH | CH,CH;CHCOCCHCH,CH, CoHs CH CoHs 2-Phenylbutanoie acid 2-Phenylbutanoic anhydride (c) | Butyl 2-phenyIbutanoate is the butyl ester of 2-phenyIbutanoic acid. i CH,CH,CHCOCH.CH,CH,CH, CoHs Butyl 2-phenylbutanoate (d) In 2-phenylbutyl butanoate the 2-phenylIbuty1 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, CH 2-Phenylbutyl butanoate MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website Back| Forward CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 537 o Il (e) | The ending -amide reveals this to be a compound of the type RCNH,. | CHLCHCHENH, CoHs 2-Phenylbutanamide (f) This compound differs from 2-phenyIbutanamide in part (e) only in that it bears an ethyl sub- stituent on nitrogen. l CHyCH.CHONHCH.CH; GH, 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,. CH.CH,CHC=N CoHs 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. O e Se nf cHs 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. | | CHCCI + CH,COH CH,COCCH, + HCI Benzoyl Benzoic acid Benzoic anhydride Hydrogen chloride chloride (c) | Acyl chlorides react with alcohols to form esters. j i CHSCI + CH,CHOH C;H,COCH.CH, + HCI Benzoyl Ethanol Ethyl benzoate Hydrogen chloride chloride The organic product is the ethyl ester of benzoic acid, ethyl benzoate. MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 540 Back| CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION (f) Water attacks the carbonyl group of benzoyl chloride to form the tetrahedral intermediate. ? 0H CHCCI + HO —— caca oH Benzoyl Water Tetrahedral chloride intermediate Dissociation of the tetrahedral intermediate occurs by loss of chloride and the proton on oxygen. H To o la Í coçit CH,COH + HCl 0H Tetrahedral Benzoicacid | Hydrogen intermediate chloride 20.5 One equivalent of benzoy1 chloride reacts rapidly with water to yield benzoic acid. í i CH,CCI + H,O C;H,COH + HCI Benzoyl Water Benzoicacid Hydrogen chloride chloride The benzoic acid produced in this step reacts with the remaining benzoyl chloride to give benzoic anhydride. o oo | | CHSCCI + CH,COH CH,COCCH, + HCI Benzoyl Benzoic acid Benzoic anhydride | Hydrogen chloride chloride 20.6 Acetic anhydride serves as a source of acetyl cation. 0 | CHCOSCCH, —— BrécH, — :0=CCH, Acetyl cation 20.7 (b) Acyl transfer from an acid anhydride to ammonia yields an amide. ls 1] | cH,COCCH, + 2NH, CH,CNH, + CH,CO” NH, Acetic anhydride Ammonia Acetamide Ammonium acetate The organic products are acetamide and ammonium acetate. Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website Back| Forward CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 541 (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.. o 9 | CN(CH,), O + 2CHJNH O — CX HN(CH), co” o | 0 Phthalic anhydride Dimethylamine Product is an amine salt and contains an amide function. 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. 0 9 Lo. CO” Na O + 2N0H — Cr + HO CO” Nat O | 0 Phthalic anhydride Sodium Sodium phthalate Water hydroxide 20.8 (b) The tetrahedral intermediate is formed by nucleophilic addition of ammonia to one of the car- bonyl groups of acetic anhydride. (qo HO O cH dodeu, cutoben, Cm, te, Tetrahedral intermediate Dissociation of the tetrahedral intermediate occurs by loss of acetate as the leaving group. HNE O va , ? CHE TOCCH; —— CHCNH, + HNTOCCH, NH, Ammonia + tetrahedral Acetamide Ammonium acetate intermediate (c) - Dimethylamine is the nucleophile; it adds to one of the two equivalent carbonyl groups of phthalic anhydride. c HQ NH; No o + HN(CH), —— o O Phthalic Dimethylamine Tetrahedral anhydride intermediate MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 542 Back| CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION A second molecule of dimethy lamine abstracts a proton from the tetrahedral intermediate. (CHNH Cy t NCH;), bxca), 3 Deus o —— co” HN(CH,, O I o Tetrahedral intermediate + second Product of reaction molecule of dimethylamine (d) - Hydroxide acts as a nucleophile to form the tetrahedral intermediate and as a base to facilitate its dissociation. Formation of tetrahedral intermediate: co o oH DI Ú o o o Phthalic anhydride Som HQ oH o + HO —— o + “OH o Tetrahedral intermediate Dissociation of tetrahedral intermediate: 5h ARO o OQ on | A CoH — (O eo Co” Õ | o In base, the remaining carboxylic acid group is deprotonated. Lo —. lo CO—H “OH co OP Oem a a O o Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website Back| CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 545 20.13 Because ester hydrolysis in base proceeds by acyl-oxygen cleavage, the !ºO label becomes incorporated into acetate ion (O = 0). | N 9, CH;CH.CH.CH,CH,Q7CCH; + 38H —— CHCH;CH;CHCH.OH + CH, o Pentyl acetate Hydroxide 1-Pentanol Acetate ion ion 20.14 Step 1: Nucleophilic addition of hydroxide ion to the carbonyl group Lo CHC—OCH,CH, :0H Hydroxide Ethyl benzoate Anionic form of ion tetrahedral intermediate Step 2: Proton transfer from water to give neutral form of tetrahedral intermediate :0H Lo " N CHC—OCH,CH, + HÉ-OH CHC—CHCH, + “0H :0H :0H Anionic form of Water Tetrahedral Hydroxide tetrahedral intermediate intermediate ion Step 3: Dissociation of tetrahedral intermediate HO: / Cj es n / HO + Cos -BCHCH; HOM + CH CH,CH, :0H QH Hydroxide Tetrahedral intermediate Water Benzoic acid Ethoxide ion ion Step 4: Proton transfer from benzoic acid + HOH Benzoic acid Hydroxide Benzoate ion Water ion 20.15 The starting material is a lactone, a cyclic ester. The ester function is converted to an amide by nucleophilic acyl substitution. cH, | CHNH, + P CHNHÉCH,CH.CHCH, y 0H Methylamine 4-Pentanolide 4-Hydroxy-N-methylpentanamide Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 546 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 20.16 -Methanol is the nucleophile that adds to the carbonyl group of the thioester. » com CH,CSCH,CH,OCAH, + CHÓH —— CH COSCHCHOCH, — CH,COCH, + HSCH,CH,OC,H, OCH, -2-Phenoxyethyl Methanol Tetrahedral intermediate Methyl acetate | 2-Phenoxyethanethiol ethanethiolate 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. oo o cu tolca, + 2ºCHNH, — cm Anta, + cado CHNH, Acetic Methylamine N-Methylacetamide Methylammonium anhydride acetate (c) The acyl group is HC—. Because the problem specifies that the acyl transfer agent is a methyl ester, methyl formate is one of the starting materials. o o | HCN(CH;) + CH;OH |, HCOCH, + HN(CH;), Methyl Dimethylamine N.N-Dimethylformamide Methyl formate alcohol 20.18 Phthalic anhydride reacts with excess ammonia to give the ammonium salt of a compound known as phthalamic acid. o o I CNH, O + 2NH, CX . Co” NH, I o Phthalic Ammonia Ammonium phthalamate anhydride (CsHoN,05) Phthalimide is formed when ammonium phthalamate is heated. o I o CNH, — NH + NH, + HO O NH, I o Ammonium phthalamate Phthalimide Ammonia — Water 20.19 Step 1: Protonation of the carbonyl oxygen = + Os: H oH H PT TN ia/ /º / CHE + HO: CHE + 305 NHC,H, H Ne, H Acetanilide Hydronium ion Protonated form Water of amide Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 547 Step 2: Nucleophilic addition of water Ho g OH :0H Pt CHE CH,C—NHC,H, H NHC,H, dr PN HH Water Protonated form Oxonium ion of amide Step 3: Deprotonation of oxonium ion to give neutral form of tetrahedral intermediate :0H :0H H H o Z “ as CHC—NHC,H, + :0$ cHC—NHCH, + H—O | SH SH “09 :0H PN 9 H H Oxonium ion Water Tetrahedral Hydronium ion intermediate Step 4: Protonation of amino group of tetrahedral intermediate :0H H “qn j H L ns + Z CHC—NHCH; + H2OE — CHE—NCHs + 108 :9H H :9H H H Tetrahedral Hydronium ion N-Protonated formof Water intermediate tetrahedral intermediate Step 5: Dissociation of N-protonated form of tetrahedral intermediate com H 2H cH,CLNCH, cnc7o + HNC,H, :0H H OH N-Protonated form of Protonated form Aniline tetrahedral intermediate of acetic acid Step 6: Proton-transfer processes em 4 + OH + ARCH, + HNCHs H Hydronium ion Aniline Anilinium ion 02H H o H 4 / 7 af CHE +iof CHE + HO 0H H OH H Protonated form Water Acetic acid Hydronium ion of acetic acid Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 550 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Steps 4and 5: Proton transfers to give an amide :0H H :05-H H O: H =) / +n06 — Reú DoZ reZ, + H—6C + H GNH, H NH, H Iminoacid Hydronium Conjugate acid Water Amide Hydronium ion of amide ion 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 phenyImagnesium bromide, followed by hydrolysis of the imine. NH o CHCHCSN + CHM pegado Iondenca) SO cudcnca 5CH; + CHMgBr 6HsCCH,CH, hear 6HSCCH,CH, Propanenitrile Phenylmagnesium Imine Ethyl phenyl ketone bromide (not isolated) 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. cl i ra m-Chlorobenzoyl bromide (b) -Trifluoroacetic anhydride is the anhydride of trifluoroacetic acid. Notice that it contains six fluorines. Il CF,COCCF, Trifluoroacetic anhydride (c) This compound is the cyclic anhydride of cis-1,2-cyclopropanedicarboxylic acid. Ha /S AH o Ago cis-1,2-Cyclopropanedicarboxylic cis-1,2-Cyclopropanedicarboxylic acid anhydride (d) | Ethyl cycloheptanecarboxylate is the ethyl ester of cycloheptanecarboxylic acid. o | (tona, Ethyl cycloheptanecarboxylate Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 551 (e) 00) (8) dy) 20.26 (a) (b) 1-Phenylethyl acetate is the ester of 1-phenylethanol and acetic acid. i conto 5 cH, 1-Phenylethyl acetate 2-Phenylethyl acetate is the ester of 2-phenylethanol and acetic acid. í encon (5 2-Phenylethyl acetate The parent compound in this case is benzamide. p-Ethylbenzamide has an ethyl substituent at the ring position para to the carbonyl group. i CH,CH, CNH, p-Ethylbenzamide The parent compound is benzamide. In N-ethylbenzamide the ethy1 substituent is bonded to nitrogen. l CNHCH,CH, N-Ethylbenzamide 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. 650 4 3 Zu CH,CH,CH,CH,CHC=N cH, 2-Methylhexanenitrile This compound, with a bromine substituent attached to its carbonyl group, is named as an acyl bromide. It is 3-chlorobutanoyl bromide. | CHCHCH.CBr a 3-Chlorobutanoyl bromide 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. i encon 5 Benzyl acetate Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 552 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION (c) The group attached to oxygen is methyl; this compound is the methyl ester of phenylacetic acid. f cmocen=( 5 Methyl phenylacetate o (d) This compound contains the functional group Lodo 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. tl CICH,CH,COCCH,CH,CI 3-Chloropropanoic anhydride (e) This compound is a cyclic anhydride, whose parent acid is 3,3-dimethylpentanedioic acid. o HC HC o 3.3-Dimethylpentanedioie anhydride (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. CHÇHCH,CHC=N cH, 4-MethyIpentanenitrile (g) This compound is an amide. We name the corresponding acid and then replace the -oic acid suffix by -amide. | CHSÇHCHCH,CNH, cH, 4-Methylpentanamide (h) This compound is the N-methyl derivative of 4-methy Ipentanamide. I CHyQHCHSCH,CNHCH; cH, N-Methyl-4-methylpentanamide Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| Forward CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 555 (1) The starting material is a cyclic ester, a lactone. Esters undergo saponification in aqueous base to give an alcohol and a carboxy late salt. o | (Ao + mou É HOCH,CH,CH,CO” Na* o 4-Butanolide Sodium Sodium 4-hydroxybutanoate hydroxide (m) -Ammonia reacts with esters to give an amide and an alcohol. HO (o + NH, —> HOCH,CH,CH.CNH, O 4-Butanolide Ammonia 4-Hydroxybutanamide (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. LLIAIH, /=0 =mo > HOCH,CH,CH.CH,0H 4-Butanolide 1. 4-Butanediol (o) Grignard reagents react with esters to give tertiary alcohols. oH o. 1.2CH;MgBr I O ZH0 HOCHCH.CH.SCH, CH, 4-Butanolide 4-Methyl-1.4-pentanediol (p) In this reaction methylamine acts as a nucleophile toward the carbonyl group of the ester. The product is an amide. “ I I CHNH, + CH,CHCOCH,CH, — — CH,CH,CNHCH, + CH,CH,0H Methylamine Ethyl phenylacetate N-Methylphenylacetamide Ethyl alcohol (9) The starting material is a lactam, a cyclic amide. Amides are hydrolyzed in base to amines and carboxylate salts. HO | Lo + Na0H —> CH,NHCH,CH,CH,CO” Na* N cH, N-Methylpyrrolidone Sodium Sodium 4-(methylamino)butanoate hydroxide MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 556 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION (1) | In acid solution amides yield carboxylic acids and ammonium salts. | I Lo + Hot CH,NCH.CH,CH,COH N | | H cH, N-Methylpyrrolidone Hydronium 4-(Methylammonio)butanoic ion acid (s) | The starting material is a cyclic imide. Both its amide bonds are cleaved by nucleophilic attack by hydroxide ion. o no cd Lo, + 2NaOH Na” “OCCH,CH,CO Na” + CH;NH, 0Avº=o 22 2 cH, N-Methylsuccinimide Sodium Disodium succinate Methylamine hydroxide (1) In acid the imide undergoes cleavage to give a dicarboxylic acid and the conjugate acid of methylamine. I | too o o + 2H0 + HCl HOCCH,CH,COH + CH;NH, CI N' CH, N-Methylsuccinimide Water Hydrogen Succinic acid Methylammonium chloride chloride (4) -Acetanilide is hydrolyzed in acid to acetic acid and the conjugate acid of aniline. I I CH;NHCCH, + HO + HCl C;H;NH, CIT + CHCOH Acetanilide Water Hydrogen Anilinium Acetic acid chloride chloride (v) This is another example of amide hydrolysis. | | C,H,CNHCH, + HO + HSO, CHCOH + CHNH, HSO, N-Methylbenzamide Water Sulfuric acid Benzoic Methylammonium acid hydrogen sulfate (w) One way to prepare nitriles is by dehydration of amides. o l PO CNH, — o, N + HO Cyclopentanecarboxamide Cyclopentyl cyanide Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website Back| CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 557 (x) — Nitriles are hydrolyzed to carboxylic acids in acidic media. HCL,H,O | (CH)CHCH,C=N > (CH,,CHCH,COH 3-Methylbutanenitrile 3-Methylbutanoic acid (») Nitriles are hydrolyzed in aqueous base to salts of carboxylic acids. NaOH, H,O 7 cH,O C=N o CHO CO Na” + NH, p-Methoxybenzonitrile Sodium p-methoxybenzoate Ammonia (2) Grignard reagents react with nitriles to yield ketones after addition of aqueous acid. o 1. CH,MgBr I CHCHC=N ão + CH;CHCCH, Propanenitrile 2-Butanone (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. HC. CH, HC. CH, —NOcH, + B, ão” :OH NHÇOCH, hu, (bb) Saponification of the carbamate in part (aa) gives the corresponding amine. HC. CH, HC, CH, E —ae— E NHGOCH, 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. KCr,0,, H5SO, | soct, I cHcHOH = cHCOH —> — caca Ethanol Acetic acid Acetyl chloride (b) -Acetic acid and acetyl chloride, available from part (a), can be combined to form acetic anhydride. o | I ai cH,COH + CHCCI | — CH,COCCH, + HCI Aceticacid | Acetyl chloride Acetic anhydride | Hydrogen chloride Forward] MainMenu| TOC] StudyGuideTOC| StudentOLC| | MHHE Website 560 Back| CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION (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. N-bromosuccinimide (NBS) HO CoHsCH, cr Bra light CHsCHBr o” CoHsCH,OH Toluene Benzyl bromide Benzyl alcohol Alternatively, recall that primary alcohols may be obtained by reduction of the corresponding carboxylic acid. o I LLAIH, C;H.COH o C;H;CH,0H Benzoic acid Benzyl alcohol Then I pyeidine I cHCCI + CHCHOH FT, coACoCcH,CH, Benzoyl Benzyl alcohol Benzyl benzoate chloride (d) | Benzamide is prepared by reaction of ammonia with either benzoyl chloride from part (a) or benzoic anhydride from part (b). o oo I | NH, | CHCCI o CH COCCGH ——> CH,CNH, Benzoyl Benzoic anhydride Benzamide chloride (e) | Benzonitrile may be prepared by dehydration of benzamide. 1 PO = CHCNH, — 52 CHC=N t Benzamide Benzonitrile (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. Ch NaCN CHCH, —mo CHCHCI —ÉE CHCHC=N heat Toluene Benzyl chloride Benzyl cyanide or NBS CHCH, — CHCHBr —£N CH,CH;C=N light Toluene Benzyl bromide Benzyl cyanide MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website Back| Forward CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 561 (g) Hydrolysis of benzyl cyanide yields pheny lacetic acid. HO, Hº, heat | CHCHCEN > CH.CH,CoH 1. NaOH, heat 2.H* Benzyl cyanide Phenylacetic acid Alternatively, the Grignard reagent derived from benzyl bromide may be carboxylated. Mg 1.Co, | CHSCHBr quem” C/HCHMgBr ão CHsCH;COH Benzyl bromide Benzylmagnesium Phenylacetic acid bromide (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. HNO, K€60, H;SO, CH; —5o— O cH os Toluene p-Nitrotoluene p-Nitrobenzoic acid (separate from ortho isomer) o | CoH Treatment of p-nitrobenzoic acid with thionyl chloride yields p-nitrobenzoyl chloride. o o | SOCL, | | o; CoH —+> ON Col p-Nitrobenzoic acid p-Nitrobenzoyl chloride (i) | In order to achieve the correct orientation in m-nitrobenzoyl1 chloride, oxidation of the methyl group must precede nitration. CH, BEOrHsO, coH HNO, com 3 HO, heat 2 HO, 2 Toluene Benzoic acid m-Nitrobenzoic acid Once m-nitrobenzoic acid has been prepared, it may be converted to the corresponding acyl chloride. Ê o | soci, COH ————— m-Nitrobenzoic acid m-Nitrobenzoyl chloride MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 562 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION (7) | A Hofmamn rearrangement of benzamide affords aniline. O | HO. CNH, + Bh —9— NH, Benzamide Bromine Aniline [prepared as in part (d)] o 20.30 The problem specifies that cncuecuca, is to be prepared from '*O-labeled ethyl alcohol (9 ="0). o cucadear + CH,CH,0H cncudocuca, Propanoyl Ethyl alcohol Ethyl propanoate chloride Thus, we need to prepare 'S0-labeled ethyl alcohol from the other designated starting materials, acetaldehyde and !$O-enriched water. First, replace the oxygen of acetaldehyde with SO by the hydration-dehy dration equilibrium in the presence of !*O-enriched water. Q 0H o | | I CHCH + HO === CHCH === CACH+H oH Acetaldehyde !8O-enriched Hydrate of 8O-entiched Water water acetaldehyde acetaldehyde Once “0O-enriched acetaldehyde has been obtained, it can be reduced to “O-enriched ethanol. a | NaBH,. CH,0H cHCH — > CH.CH,0H LLIAIH, 2.H,0 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. j 7 CH.COCH,CH, + HO — CHCOCHCH, oH Ethyl acetate Hydroxide Rate-determining ion intermediate The electron-withdrawing effect of a CF; group stabilizes the intermediate formed in the rate- determining step of ethyl trifluoroacetate saponification. i Z CFCOCH,CH, + HO” CRÇOCH:CH, OH Eshyl trifluoroacetate Hydroxide Rate-determining ion intermediate Because the intermediate is more stable, it is formed faster than the one from ethyl acetate. Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 565 20.34 The first step is acid hydrolysis of an acetal protecting group. Step 1: H,O.H | o Compound A — HOC(CH;)ÇH GH(CHoyCHOH HO oH Compound B (CoH09) All three alcohol functions are converted to bromide by reaction with hydrogen bromide in step 2. Step 2: | CompoundB — HE, HOCCH;) GH —CH(CH,)CH:Br Br Br Compound C (CuHosBrO) Reaction with ethanol in the presence of an acid catalyst converts the carboxylic acid to its ethyl ester in step 3. Step 3: ethanol | Compound C ELSO, CHCHOACH OH H(CH) CH Br SO, 2 25 2 2 Br Br Compound D (CosHssBr;O,) 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: Zn ethanol Í Compound D CH,CH,OC(CH,).CH=CH(CH,),CH,Br Compound E (CisHssBrO) Step 5 is a nucleophilic substitution of the Sy2 type. Acetate ion is the nucleophile and displaces bromide from the primary carbon. Step 5: 7 o o NaoCCH, |, | CompondE con” CH,CH,OC(CH,),CH=CH(CH,),CH,OCCH, Compound F (CoHO,) Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 566 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 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: o 1. KOH, ethanol | CompoundF HOC(CH,).CH=CH(CH,),CH,0H Compound G (CroHO) In step 7, compound G cyclizes to ambrettolide on heating. “s heat “ HO,C “HO 0750 Compound G Ambrettolide 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. o HOCH,CH=CH(CH,) CO;CH, —per HÓCH=CH(CH,,CO.CH, Compound A o Compound B o (b) Conversion of — to —CH=CH, is a typical case in which a Wittig reaction is appropriate. | CH HCCH=CH(CH;),CO,CH, CEO, | c=CHCH=CH(CH;,CO;CH, Compound B Compound C (observed yield, 53%) (c) Lithium aluminum hydride was used to reduce the ester to a primary alcohol in 81% yield. LLIAIH, H;C=CHCH=CH(CH;,CO,CH, So H,C=CHCH=CH(CH,),CH,0H Compound C Compound D (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. o | cutotem, H,C=CHCH=CH(CH,),CH,OCCH, H,C=CHCH=CH(CH,),CH,0H 2 2 2 pyridine Compound D (E)-9,11-Dodecadien-1-yl acetate 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 Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 567 both Grignard functions of the reagent attack the lactone. The second attack is intramolecu- lar, giving rise to the cyclopentanol ring of the product. + aa: ,CH,CH,CH,MgBr —— LI emma CNO odio) 2MeBr 4-Butanolide 143-HydroxypropyD- cyclopentanol (88%) (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). Thiolactone Tetrahedral Lactam intermediate 20.37 (a) Acylchlorides react with alcohols to form esters. 0 o 0 | I pyridine I cH,o CCI + Sm o “a 0H $ p-Methoxybenzoyl Benzoin C=o chloride OCH, Benzoin p-methoxybenzoate (compound A; 95%) MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website 570 CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION Dehydration of the cyanohydrin followed by hydrolysis of the nitrile group and esterification of the resulting carboxylic acid yields methyl methacrylate. oH I H,So, HO, H' CH,0H, H* CHE, —a + HC=CCN —— H=(COH —— H=(CO,CH; CN CH, CH, CH, Acetone Methyl methacrylate cyanohydrin 20.42 Thecompound contains nitrogen and exhibits a prominent peak in the infrared spectrum at 2270 cm”!; itis likely to be a nitrile. Its molecular weight of 83 is consistent with the molecular formula CJHN. The presence of four signals in the ô 10 to 30-ppm region of the !C NMR spectrum suggests an unbranched carbon skeleton. This is confirmed by the presence of two triplets in the 'H NMR spec- trum atô 1.0 ppm (CH, coupled with adjacent CH,) and at ô 2.3 ppm (CH,CN coupled with adjacent CH,). The compound is pentanenitrile. CH,CH,CH,CH,C=N Pentanenitrile 20.43 The compound has the characteristic triplet-quartet pattern of an ethyl group in its !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 (6 4.1 ppm), which is consistent with ethyl groups bonded to oxygen, as in —OCH,CH,. There is a peak at 1730 cm”! in the infrared spectrum, suggesting that these ethoxy groups reside in ester functions. The molecular formula C,H,0, 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 !H NMR spectrum, and so two equivalent CH, groups are present. The compound is the diethyl ester of succinic acid. Q | | CH,CH,OCCH,CH,COCH,CH, Diethyl succinate 20.44 Compound A (C,H,O;) has an index of hydrogen deficiency of 2. With two oxygen atoms and a peak in the infrared at 1760 cm”!, it is likely that one of the elements of unsaturation is the carbon-oxygen double bond of an ester. The !H NMR spectrum contains a three-proton singlet at 8 2.1 ppm, which is consistent with a CH,C unit. It is likely that compound A is an acetate ester. | o The PC NMR spectrum reveals that the four carbon atoms of the molecule are contained in one each of the fragments CH,, CH,, 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, and CH carbons of a vinyl group, CH=CH,. Compound A is vinyl acetate. 820.2 ppm o / 896.8 ppm SNocu la 8 167.6ppm 2 ô 141.8 ppm Each vinyl proton is coupled to two other vinyl protons; each appears as a doublet of doublets in the '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. Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 571 SELF-TEST PART A A-1. Give a correct IUPAC name for each of the following acid derivatives: | (a) CH;CH,CH,OCCH,CH,CH, | (b) C;HCNHCH, i (e) (CH;),CHCH,CH,CCI A-2. Provide the correct structure of (a) -Benzoic anhydride (b) | N(1-Methylpropylacetamide (c) | Phenyl benzoate A-3. What reagents are needed to carry out each of the following conversions? | (a) CHCH,CO,H C;H;CH,CCI 7 | (b) (CH):CCNH, (CH);CNH, | (e) (CH), CHCH,NH, C;H,CNHCH,CH(CH;), + CH;0H A-4. Write the structure of the product of each of the following reactions: 1. NaOH, H,O (a) Cyclohexyl acetate ZH - ? (two products) (b) Cyclopentanol + benzoyl chloride ide, o Heat) (c) O + CHCHOH ——> 2 o (d) Ethyl propanoate + dimethylamine —— ? (two products) 0 ] HO, H;SO, (e) HC CNHCH, e ? (two products) Back) Forward] MainMenu] TOC] StudyGuide TOC] —StudentOLC| | MHHE Website 572 Back| CARBOXYLIC ACID DERIVATIVES: NUCLEOPHILIC ACYL SUBSTITUTION 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. | C,H,COCH,CH,NHCH, — — CH,NCH,CH,0H CAHC=0 A-6. Give the correct structures, clearly showing stereochemistry, of each compound, A through D, in the following sequence of reactions: soc, NH; 25 A B Br, NaoH POr HO heat c D (CHsN) (CgHN) 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 ne Soma, from ne Some A-9. Outline a synthesis of benzyl benzoate using toluene as the source of all the carbon atoms. o I COCH, Benzyl benzoate A-10. The infrared spectrum of a compound (C;HçCINO) has an intense peak at 1680 cm”!. Its 'H NMR spectrum consists of a doublet (3H, ô 1.5 ppm), a quartet (1H, 6 4.1 ppm), and a broad singlet (2H, ô 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? oH cH-é-on d (a) Benzoic acid and HCI (b) -Benzoyl chloride and H,;O (c) Both (a) and (b) equally likely (d) Neither (a) nor (b) MainMenu) TOC] StudyGuideTOC| Student OLC| | MHHE Website