Carboxylic acid ppt

Carboxylic acid ppt - Derivatives of Carboxylic Acid : O: R...

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Unformatted text preview: Derivatives of Carboxylic Acid : O: R C R .. C.l : . : O: C R R C .. O .. C .. _ O: .. carboxylate acid chloride :O: :O: R .. OH .. N: C nitrile : O: C acid anhydride R :O: R C :O: .. O .. ester R R C amide .. H N H 1 Nomenclature of Acid Halides • • IUPAC: alkanoic acid → alkanoyl halide Common: alkanic acid → alkanyl halide NH2 O NO2 CH2 CH2 C Cl Cl C CH2 4 C Cl O C I: 4-nitropentanoyl chloride c: γ -nitrovaleryl chloride c: β-aminopropionyl chloride O R .. C.l : . H3C CH CH2 CH2 C Cl I: 3-aminopropanoyl chloride O : O: I: hexanedioyl chloride c: adipoyl chloride Rings: (IUPAC only): ringcarbonyl halide C Cl O I: 3-cylcopentenecarbonyl chloride O C Br I: benzenecarbonyl bromide c: benzoyl bromide 2 :O: Nomenclature of Acid Anhydrides • R C .. O .. : O: R C Acid anhydrides are prepared by dehydrating carboxylic acids O O CH3 C OH + O ∆ H O C CH3 acetic acid CH3 - H 2O O O COC HO C I: benzenecarboxylic anhydride c: benzoic andhydride I: cis-butenedioic anhydride c: maleic anhydride OC O CO C O C CH3 acetic anhydride ethanoic anhydride ethanoic acid O O ∆ O C OH I: butanedioic acid c: succinic acid - H2O OC O CO I: butanedioic anhydride c: succinic anhydride Some unsymmetrical anhydrides O O O COCH I: benzoic methanoic anhydride c: benzoic formic anhydride CH3 C O O C H I: ethanoic methanoic anhydride c: acetic formic anhydride 3 Nomenclature of Esters • : O: Esters occur when carboxylic acids react with alcohols I: alkanoate c: alkanate O alkyl CH3 O C CH3 I: methyl ethanoate c: methyl acetate O C O CH(CH3)2 R C .. OH .. + H carboxylic ac i d .. O .. R al c oh ol - :O: H2O H+ R C .. O .. R es t er O O C O C(CH3)3 OCH I: phenyl methanoate c: phenyl formate I: t-butyl benzenecarboxylate c: t-butyl benzoate O O C CH(CH3)2 I: isobutyl cyclobutanecarboxylate c: none I: cyclobutyl 2-methylpropanoate c: cyclobutyl α-methylpropionate OO CH3 O C C O CH3 I: dimethyl ethanedioate c: dimethyl oxalate 4 Nomenclature of Cyclic Esters, “Lactones” O C O OH CH2 O CH2 CH2 H H+ - H2O I: 4-hydroxybutanoic acid c: γ -hydroxybutyric acid O C C CH2 CH2 O CH2 O Cyclic esters, “lactones”, form when an open chain hydroxyacid reacts intramolecularly. 5 to 7membered rings are most stable. I: 4-hydroxybutanoic acid lactone c: γ -butyrolactone ‘lactone’ is added to the end of the IUPAC acid name. ‘olactone’ replaces the ‘ic acid’ of the common name and ‘hydroxy’ is dropped but its locant must be included. O I: 5-hydroxypentanoic acid lactone O O c: δ-valerolactone I: 4-hydroxypentanoic acid lactone O c: γ -valerolactone CH3 I: 3-hydroxypentanoic acid lactone c: β-valerolactone O O H3C O I: 6-hydroxy-3-methylhexanoic acid lactone c: β-methyl-ε-caprolactone O 5 Nomenclature of Amides 1° amide O RCN O O 3° amide R RCN N,N-disubstituted amide R H 2° amide RCN R N-substituted amide H H 1° amides: ‘alkanoic acid’ + amide → ‘’alkanamide’ a ring is named ‘ringcarboxamide’ O CH3CH2CH2 C N Cl H CN H I: butanamide c: butyramide O O NO2 H CN H H H I: 3-chlorocyclopentanecarboxamide I: p-nitrobenzenecarboxamide c: p-nitrobenzamide c: none 2° and 3° amides are N-substituted amides I: N,2-dimethylpropanamide c: N,α-dimethylpropionamide CH3 O CH3CH CN CH3 H I: N-phenylethanamide c: N-phenylacetamide c: acetanilide O CN CH3 CH2CH3 I: N-ethyl-N-methylcyclobutanecarboxamide c: none O H3C C N H 6 Nomenclature of Cyclic Amides, “Lactams” O C O OH CH2 H N CH2 H CH2 ∆ - H2O I: 4-aminobutanoic acid c: γ -aminobutyric acid O C CH2 CH2 NH NH CH2 Cyclic amides, “lactams”, form when an open chain aminoacid reacts intramolecularly. 5 to 7membered rings are most stable. I: 4-aminobutanoic acid lactam c: γ -butyrolactam ‘lactam’ is added to the end of the IUPAC acid name. ‘olactam’ replaces the ‘ic acid’ of the common name and ‘amino’ is dropped but its locant must be included. O NH CH3 O I: 3-amino-2-bromopropanoic acid lactam c: α-bromo-β-propionolactam I: 5-aminohexanoic acid lactam c: δ-caprolactam Br NH O H3C NH I: 4-amino-3-methylbutanoic acid lactam c: β-methyl-γ -butyrolactam 7 Nomenclature of Nitriles O RCN • • • H H POCl3 RCN - H2O Nitriles are produced when 1° amides are dehydrated with reagents like POCl3 IUPAC: alkane + nitrile → ‘alkanenitrile’ IUPAC rings: ‘ringcarbonitrile’ Common: alkanic acid + ‘onitrile’ → ‘alkanonitrile’ N C CH2CH2CH2 CH3 C N I: 4-iodobutanenitrile c: γ -iodobutyronitrile I: ethanenitrile c: acetonitrile CH3O CN I HS CN I: p-thiobenzenecarbonitrile c: p-mercaptobenzonitrile COOH CN I: 3-methoxycyclohexanecarbonitrile c: none I: 2-cyanocyclopentanecarboxylic acid c: none 8 Nomenclature Practice Exercise O CN CH3 C O CH2Br I: cyclobutanecarbonitrile c: none O O O O CH3 C O- Na+ I: bromomethyl ethanoate c: bromomethyl acetate I: sodium ethanoate c: sodium acetate O O I: 3-bromo-N-methylpentanamide c: β-bromo-N-methylvaleramide CH3 C CH2 C Cl I: pentanedioic anhydride c: glutaric anhydride NHCH3 I: 3-oxobutanoyl chloride c: β-oxobutyryl chloride O Br O O O NH Cl I: 2-ethyl-5-hydroxypentanoic acid lactone c: α-ethyl-δ-valerolactone I: 6-amino-6-chlorohexanoic acid lactam c: ε-chloro-ε-caprolactam 9 Relative Reactivity of Carbonyl Carbons Nucleophiles (electron donors), like OH-, bond with the sp2 hybridized carbonyl carbon. The order of reactivity is shown. :O: R C .. O .. acid chloride : O: R C R C : O: R R C C R .. OH .. N: R C acid anhydride aldehyde : O: : O: ketone ester R C :O: R C :O: R C :O: nitrile carboxylate most reactive : O: carboxylic acid amide .. C.l : . R C H .. O .. .. R H N H .. _ O: .. least reactive 10 Nucleophilic Addition to Aldehydes and Ketones • • Recall that electron donors (Nu: -’s) add to the electrophilic carbonyl C in aldehydes and ketones. The C=O π bond breaks and the pair of electrons are stabilized on the electronegative O atom. R (alkyl groups) and hydrogens (H) bonded to the C=O carbon remain in place. R- and H- are too reactive (pKb of – 40 and -21). R and H are not leaving groups, so the carbonyl group becomes an alkoxide as the sp2 C becomes a tetrahedral sp3 C. - O RCH • • O RCH CH3 MgBr tetrahedral alkoxide with sp3 carbon. CH3 A second addition of a nucleophile cannot occur since alkoxides are not nucleophilic. The reaction is usually completed by protonation of the alkoxide with H3O+ forming an alcohol. This later reaction is simply an acid/base reaction. The characteristic reaction of aldehydes and ketones is thus ‘nucleophilic addition’. O - RCH CH3 H + H OH O+H RCH CH3 + H OH 11 Nucleophilic Acyl Substitution in Acid Derivatives • • Carboxylic acid derivatives commonly undergo nucleophilic substitution at the carbonyl carbon rather than addition. The first step of the mechanism is the same. The C=O π bond breaks and the pair of electrons are stabilized on the electronegative O atom. A tetrahedral alkoxide is temporarily formed. - O O R C Cl R C Cl sp2 carbonyl C • • O CH3 MgBr R C CH3 + CH3 Cl Chlorine is a fair leaving group. sp2 carbonyl reforms alkoxide C js sp3 In carboxylic acid derivates, one of the groups that was bonded to the carbonyl C is a leaving group. When this group leaves, the sp3 tetrahedral alkoxide reverts back to an sp2 C=O group. Thus substitution occurs instead of addition. In many cases, the substitution product contains a carbonyl that can react again. O O - R C CH3 R C CH3 CH3 MgBr CH3 H H O+H OH R C CH3 + H O H CH3 Note that because the C=O group reforms, the nucleophile can react a second time. 12 Nucleophilic Acyl Substitution in Acid Derivatives • In carboxylic acid derivatives, the acyl group (RCO) is bonded to a leaving group (-Y). O RC RCY acyl group • • - O Nu:- O O RCY R C Nu + Y: - Draw the mechanism. Nu The leaving group (-Y) becomes a base (Y:-) . The acid derivative is reactive If the base formed is weak (unreactive). Weak bases are formed from good leaving groups. For the carboxylic acid derivatives shown, circle the leaving group. Then draw the structure of the base formed, give its pKb, and describe it as a strong or weak base. acid derivative leaving group pKb strength as base Cl +21 non basic +9 weak base OR -2 strong base NH2 -21 v. strong base O R C Cl O O RCOCR O RCOR O R C NH2 - O OCR - 13 Nucleophilic Acyl Substitution in Acid Derivatives We will study the reaction of only a few nucleophiles with various carboxylic acid derivatives and we will see that the same kinds of reactions occur repeatedly. •Hydrolysis: Reaction with water to produce a carboxylic acid •Alcoholysis: Reaction with an alcohol to produce an ester •Aminolysis: Reaction with ammonia or an amine to produce an amide •Grignard Reaction: Reaction with an organometallic to produce a ketone or alcohol •Reduction: Reaction with a hydride reducing agent to produce an aldehyde or alcohol Draw the structures of the expected products of these nucleophilic substitution reactions, then circle the group that has replaced the leaving group (-Y) O RCY+ O RCY+ O RCY+ O RCY+ O RCY+ H OH H OR H NH2 O hydrolysis alcoholysis RCOH O RCOR O aminolysis R C NH2 R MgX Grignard reduction LiAlH3 H hydride reduction OH RCR R OH RCH H 14 Nucleophilic Acyl Substitution of Carboxylic Acids O O RCOH • RCY Nucleophilic acyl substitution converts carboxylic acids into carboxylic acid derivatives, i.e., acid chlorides, anhydrides, esters and amides. : O: R C .. C.l : . SOCl2 R acid chloride :O: R C .. O .. : O: : O: C acid anhydride R ∆ -H2O C .. OH .. :O: NH3, ∆ , -H2O R C amide ROH H+ :O: R C .. O .. .. H N H R ester 15 Conversion of Carboxylic Acids to Acid Halides • The S atom in SOCl2 is a very strong electrophile. S is electron deficient because it is bonded to 3 electronegative atoms (Cl and O). Cl is a leaving group. • The hydroxyl O atom in a carboxylic acid has non bonded pairs of electrons, making it a nucleophile. This O atom bonds with S (replacing a Cl) and forming a chlorosulfite intermediate. The chlorosulfite group is a very good leaving group. It is easily displaced by a Cl- ion via an SN2 mechanism yielding an acid chloride. • Use curved arrows to draw the initial steps of the mechanism shown below. 1 :O: R C .. OH .. 3 • • + S Cl 3 2 :O : 1 Cl 2 thionyl chloride :O : - HCl R C .. O .. :O : S :O: R Cl C Cl + SO2 + HCl + H chlorosulfite intermediate PBr3 will substitute Br for OH converting a carboxylic acid to an acid bromide Draw and name the products of the following reactions. SOCl2 O H3C C OH O H3C C O H O HCl + SO2 + H3C SOCl2 C Cl O HCl + SO2 + H3C C Cl I: p-methylbenzenecarbonyl chloride c: p-methylbenzoyl chloride I: ethanoyl chloride c: acetyl chloride 16 Conversion of Carboxylic Acids to Acid Anhydrides • High temperature dehydration of carboxylic acids results in two molecules of the acid combining and eliminating one molecule of water. O O CH3 C OH H O C CH3 + O ∆ - H 2O acetic acid CH3 C O C CH3 acetic anhydride ethanoic anhydride ethanoic acid • Cyclic anhydrides with 5 or 6-membered rings are prepared by dehydration of diacids. O O C 2HC OH ∆ 2HC 2HC OH - H2O 2HC C C O C O O I: butanedioic acid I: butanedioic anhydride c: succinic acid • O c: succinic anhydride Draw a reaction showing the preparation of cyclohexanecarboxylic anhydride. O O ∆ COC - H2O O 2 COH 17 Conversion of Carboxylic Acids to Esters Two methods are used: SN2 reaction of a carboxylate and Fischer Esterification 1. SN2 reaction of a carboxylate with a methyl halide or 1° alkyl halide is straightforward. 2° and 3° alkyl halides are not used because carboxylate is only a fair nucleophile and is basic enough (pKb = 9) that elimination of HX from the alkyl halide will compete with substitution. The carboxylate will be protonated and the alkyl halide eliminates HX becoming an alkene. : O: CH3CH2 C .. O .. + Na OH H _ propanoic acid : O: - C CH3CH2 H2O _ .. O: .. : O: I CH3 CH3CH2 C CH3 _ .. O: .. Na + + H3C O H NaOH CH2Br - H2O I: 5-bromopentanoic acid c: δ-bromovaleric acid O E2 CH3CH2C OH I CH3 sodium propionoate O C O Na+ OCH2 Br O NaI + CH2 C CH3 + NaI CH3 CH3 isobutylene O O - NaBr O .. SN2 acid/base : O: .. C + CH3CH2 Na+ O CH2 I: sodium 5-bromopentanoate c: sodium δ-bromovalerate I: 5-hydroxypentanoic acid lactone c: δ-valerolactone 18 Conversion of Carboxylic Acids to Esters - - - alkyl halide (substrate) good Nu strong base e.g., bromide e.g., ethoxide strong bulky base e.g., t-butoxide Br C2H5O (CH3)3CO Me SN 2 SN 2 SN2 no reaction 1° SN 2 SN 2 E2 (SN2) no reaction 2° SN 2 E2 3° SN 1 E2 alkyl halide (substrate) - v. gd. ……….. Numoderate.base ………. e.g., cyanide CN- pkb = …4.7 …… - v. gd. ……….. Numoderate base ……….. good Nu - good Nu nonbasic - very poor Nu nonbasic e.g., acetic acid CH3COOH SN1, E1 E2 E2 SN1, E1 - fair ……….. Nu weak ……….. base e.g., alkyl sulfide e.g., carboxylate 6.0 … pkb = ……/ alcohol (substrate) HI HBr HCl RCOO 9 pkb = ……… RS-, also HS- - Me SN2 7.0 SN2 SN2 Me SN 2 1° SN 2 SN 2 SN2 1° SN 2 2° SN 2 SN 2 E2 2° SN 1 3° E2 E2 E2 3° SN1 19 Conversion of Carboxylic Acids to Esters 2. Fischer Esterification: (RCOOH → RCOOR) Esters are produced from carboxylic acids by nucleophilic acyl substitution by a methyl or 1º alcohol. Heating the acid and alcohol in the presence of a small quantity of acid catalyst (H 2SO4 or HCl (g)) causes ester formation (esterification) along with dehydration. The equilibrium constant is not large (Keq ~ 1) but high yields can be obtained by adding a large excess of one of the reactants and removing the H2O formed. The reaction is reversible. A large excess of H2O favors the reverse reaction. Bulky (sterically hindered) reagents reduce yields. Since alcohols are weak nucleophiles, acid catalyst is used to protonate the carbonyl oxygen which makes the carbonyl C a better electrophile for nucleophilic attack by ROH. Proton transfer from the alcohol to the hydroxyl creates a better leaving group (HOH). Learn the mechanism since it is common to other reactions. : O: R .. C H : O+ H+ HSO4- .. OH R C : O: .. OH .. R R' ' • .. O .. H C + H H H .. OH .. : O: R C + R' .. O .. OH .. H proton transfer : O: R R' C O .. : : O: HSO4- .. +OH H R .. C OR' .. + H2O + H2 SO4 The net effect of Fischer esterification is substitution of the –OH group of a carboxylic acid with the –OR group of a methyl or 1° alcohol. 20 Conversion of Carboxylic Acids to Esters • Draw and name the products of the following reactions. O O HO CCH2C OH 2 CH3CH2 OH H+ I: propanedioic acid c: malonic acid O C OH • CH2Br CH3CH2O CCH2C OCH2CH3 I: diethyl propanedioate c: diethyl malonate C OCH2 cyclopentylmethyl benzoate Draw the reagents that will react to produce the following ester. O H3C CH3 HC C O CH H3C CH3 I: isopropyl 2-methylpropanoate c: isopropyl isobutyrate • O O 1. NaOH 2. O H+ CH3 HOCH CH3 + O H3C HC C OH H3C Why will an SN2 reaction of a carboxylate and an alkyl halide not work here? Isopropyl bromide is a 2° alkyl halide and would undergo an E2 rather than SN2 reaction. Draw the complete mechanism for Fischer esterification of benzoic acid with methanol. 21 Conversion of Carboxylic Acids to Amides • Amides are difficult to prepare by direct reaction of carboxylic acids with amines (RNH 2) because amines are bases that convert carboxylic acids to non electrophilic carboxylate anions and themselves are protonated to non nucleophilic amine cations, (RNH3+) : O: CH3 C .. O .. H + .. CH3NH2 : O: H3C C .. O: .. _ + NH3CH3 : O: ∆ - H2O CH3 C .. NH2 • High temperatures are required to dehydrate these quaternary amine salts and form amides. This is a useful industrial method but poor laboratory method. • In the lab amides are often prepared from acid chloride after converting the carboxylic acid to the acid chloride. • Explain why methylamine is a Bronsted base. Proton (H+) acceptor • Explain why methylamine is a Lewis base. Electron pair donor • Explain why methylamine is not an Arrhenius base Has no OH- group 22 Synthesis Problems Involving Carboxylic Acids • Write equations showing how the following transformations can be carried out. Form a carboxylic acid at some point in each question. H3C CH3 ∆ KMnO4 O ? Cl O O C C Cl O HO C SOCl2 C OH MgBr 2. Mg ether Br ? KMnO4 ∆ O C O CH3 C O CH3 O O C OH C OH O O 1. CO2 C OH ∆ - H2O H3O+ O ? O COC NaCN CN H2O ∆ O ∆ - H2O C OH H+ 2 CH3OH ∆ H+ 23 Chemistry of Acid Halides • In the same way that acid chlorides are produced by reacting a carboxylic acid with thionyl chloride (SOCl2), acid bromides are produced by reacting a carboxylic acid with phosphorus tribromide (PBr3). : O: R C : O: R C .. OH .. OH O (or PCl3) + SOCl2 R C + Cl SO2 + HCl O + R PBr3 C PBr2OH + Br Reactions of Acid Halides: Acid halides are among the most reactive of the carboxylic acid derivatives and are readily converted to other compounds. Recall that acid chlorides add to aromatic rings via electrophilic aromatic substitution (EAS) reactions called Friedel-Crafts Acylation with the aid of Friedel-Crafts catalysts. O R C Cl ...... AlCl3 R + .. O C .. R R + CO + AlCl4- acyl cation = acylium ion O + + O C .. C H + O R C - Cl R + HCl 24 Chemistry of Acid Halides • Draw a reaction showing how propylbenzene can be produced by a Friedel Crafts acylation reaction. O O Cl CCH2CH3 AlCl3 H2/Pt CCH2CH3 CH2CH2CH3 I: 1-phenyl-1-propanone c: ethyl phenyl ketone • Most acid halide reactions occur by a nucleophilic acyl substitution mechanism. The halogen can be replaced by -OH to produce an acid, -OR to produce an ester, -NH2 to produce an amide. Hydride reduction produces a 1° alcohol, and Grignard reaction produces a 3° alcohol. : O: R C acid .. OH : O: R H2O C ester .. OR .. O R R':- MgBr+ C Cl : O: R R [H] C R .. C NH2 amide C [H] R' ketone R':- MgBr+ : O: NH3 ROH O H aldehyde RCH2OH 1º alcohol OH RC R' R' 3º alcohol 25 Hydrolysis: Conversion of Acid Halides into Acids • Acid chlorides react via nucleophilic attack by H2O producing carboxylic acids and HCl. _ .. : O: : O: R C Cl R .. .. H2O H _ C O .. Cl - Cl : O: - Cl R : O: +H O: C + R + H H C .. OH HCl • Tertiary amines, such as pyridine, are sometimes used to scavenge the HCl byproduct and drive the reaction forward. 3º amines will not compete with water as a nucleophile because their reaction with acid halide stops at the intermediate stage (there is no leaving group). Eventually, water will displace the amine from the tetrahedral intermediate, regenerating the 3º amine and forming the carboxylic acid. .. _ .. _ : O: : O: R C Cl + : O: .. R R'3N C : O: Cl R C + NR'3 R'3N + .. R'3N N: pyridine + HCl : O: + R'3NH Cl- + R C .. H2O .. : O: .. OH R R C + NR'3 C +O H .. H .. .. O + H R'3N H • Draw the mechanism of the reaction of cyclopentanecarbonyl chloride with water. 26 Alcoholysis: Conversion of Acid Halides into Esters • Acid chlorides react with alcohols producing esters and byproduct HCl by the same mechanism as hydrolysis above. • Draw and name the products of the following reaction. O H3C C Cl + I: ethanoyl chloride c: acetyl chloride HO CH CH3 CH3 O H3C C O CH CH3 CH3 I: isopropyl ethanoate c: isopropyl acetate + HCl • Draw the mechanism of the reaction of benzoyl chloride and ethanol. • Once again, 3º amines such as pyridine may be used to scavenge the HCl byproduct or for water insoluble acid halides, aqueous NaOH can be used to scavenge HCl since it will not enter the organic layer and attack the electrophile (thus it cannot compete with the alcohol as the nucleophile). NaOH, H2O O R C Cl HCl CH2Cl2 ROH 27 Practice on Synthesis of Esters • Write equations showing all the ways that benzyl benzoate can be produced. Consider Fischer esterification, SN2 reaction of a carboxylate with an alkyl halide, and alcoholysis of an acid chloride. O C O CH2 H2SO4 O CH2OH C OH + O C O-Na+ CH2Br + N O CH2OH C Cl + • Answer the same question as above but for t-butyl butanoate O CH3 CH3CH2CH2C O C CH3 N O CH3 CH3CH2CH2C Cl + HO C CH3 CH3 CH3 This is the only method that will work. • Explain why the other methods will fail. 28 Aminolysis: Conversion of Acid Halides into Amides • Acid chlorides react rapidly with ammonia or 1° or 2° but not 3° amines producing amides. Since HCl is formed during the reaction, 2 equivalents of the amine are used. 1 equivalent is used for formation of the amide and a second equivalent to react with the liberated HCl, forming an ammonium chloride salt. Alternately, the second equivalent of amine can be replaced by a 3º amine or an inexpensive base such as NaOH (provided it is not soluble in the organic layer). Using NaOH in an aminolysis reaction is referred to as the Schotten-Baumann reaction. :O: C Cl + benzoyl chloride O .. 2 NH(CH3)2 C demethyl amine H + O .. NH(CH3)2 Cl - C N CH3 N CH3 CH3 CH3 + + NH2( CH3)2 Cl- I: N,N-dimethylbenzenecarboxamide c: N,N-dimethylbenzamide • Write equations showing how the following products can be made from an acid chloride. N-methylacetamide O N CH3 C NHCH3 propanamide O CH3CH2C NH2 O CH3 C Cl N + NH2CH3 O CH3CH2C Cl + NH3 29 Reduction of Acid Chlorides to Alcohols with Hydride • Acid chlorides are reduced by LiAlH4 to produce 1° alcohols. The alcohols can of course be produced by reduction of the carboxylic acid directly. • The mechanism is typical nucleophilic acyl substitution in which a hydride (H:-) attacks the carbonyl C, yielding a tetrahedral intermediate, which expels Cl -. The result is substitution of -Cl by -H to yield an aldehyde, which is then immediately reduced by LiAlH4 in a second step to yield a 1° alcohol. Cl O R C Cl 1. LiAlH4 - O O RCH RCH LiAlH4 H 2. H3O+ OH R C H + H2O H • Draw the reaction and name the product when 2,2-dimethylpropanoyl chloride is reduced with LiAlH4 CH3 O H3C C CH3 C Cl excess 1. LiAlH4 2. H3O + CH3 H3C C CH2OH I: 2,2-dimethyl-1-propanol c: neopentyl alcohol CH3 30 Reduction of Acid Chlorides to Aldehydes with Hydride • The aldehyde cannot be isolated if LiAlH4 (and NaBH4) are used. Both are too strongly nucleophilic. • However, the reaction will stop at the aldehyde if exactly 1 equivalent of a weaker hydride is used, i.e., diisobutylaluminum hydride (DIBAH) at a low temperature (-78°C). • Under these conditions, even nitro groups are not reduced. CH3 H CH3 CH3 CH3CHCH2 Al CH2CHCH3 CH3CHCH2 + Al CH3 CH2CHCH3 + H: _ H Al diisobutyl aluminum hydride (DIBAH) H DIBAH is weaker than LiAlH4. DIBAH is neutral; LiAlH4 is ionic. aluminum hydride DIBAH is similar to AlH3 but is hindered by its bulky isobutyl groups. H Only one mole of H:- is released per mole of DIBAH. O NO2 C Cl - 78°C 1 equiv. 1. DIBAH 2. H3O+ O NO2 CH p-nitrobenzaldehyde 31 Reduction of Acid Chlorides to Alcohols with Grignards • Grignard reagents react with acid chlorides producing 3° alcohols in which 2 alkyl group substituents are the same. The mechanism is the same as with LiAlH4 reduction. The 1st equivalent of Grignard reagent adds to the acid chloride, loss of Cl - from the tetrahedral intermediate yields a ketone, and a 2nd equivalent of Grignard immediately adds to the ketone to produce an alcohol. O O C Cl O CH3 MgBr C CH3 CH3 MgBr - C CH3 CH3 H3O+ OH C CH3 CH3 I: 2-phenyl-2-propanol • The ketone intermediate can’t be isolated with Grignard reaction but can be with Gilman reagent (diorganocopper), R2CuLi. Only 1 equivalent of Gilman is used at -78°C to prevent reaction with the ketone product. Recall the preparation of ketones (Ch. 19). This reagent does not react other carbonyl compounds (although it does replace halogens in alkyl halides near 0°C) 1 equiv. O 1. Li(CH3)2Cu H3C - 78°C CH C Cl H3C 2. H3O+ H3C H3C O CH C CH3 I: 3-methyl-2-butanone c: isopropyl methyl ketone 32 Practice Questions for Acid Chloride Reductions • Draw the reagents that can be used to prepare the following products from an acid chloride by reduction with hydrides, Grignards and Gilman reagent. Draw all possible combinations. I: ethanoyl chloride excess OH 1. O MgBr I: 1,1-dicyclopentylethanol CH3 C 2. O C CH2CH3 H3C C O 1. CH 2. C Cl H3O+ DIBAH 1 equiv. - 78°C 2CuLi O 1 equiv. - 78°C CH3CH2 C Cl + H3O I: 1-phenyl-1-propanone c: ethyl phenyl ketone CH3 O H3C C H3O+ or 2. excess 1. LiAlH4 2. 1. O H3O+ CH2 CH3 H3O+ CH3CH2 2CuLi 1. 1 equiv. - 78°C 2. CH3 OH H3C C Cl C Cl I: 2,2-dimethylpropanoyl chloride I: 2,2-dimethyl-1-propanol CH3 O C Cl I: cyclohexanecarbonyl chloride I: cyclohexanecarbaldehyde 33 :O: Preparations of Acid Anhydrides R C .. O .. : O: C R Preparation of Acid Anhydrides: Dehydration of carboxylic acids as previously discussed is difficult and therefore limited to a few cases. O O O O ∆ CH3 C OH + H O C CH3 CH3 C O C CH3 - H 2O acetic anhydride acetic acid A more versatile method is by nucleophilic acyl substitution of an acid chloride with a carboxylate anion. Both symmetrical and unsymmetrical anhydrides can be prepared this way. : O: .. O: .. HC _ : O: Na + sodium formate + Cl C : O: ether 25ºC CH3 SN2 HC .. O .. : O: C CH3 + NaCl acetic formic anhydride acetyl chloride • Draw all sets of reactants that will produce the anhydride shown with an acid chloride. O O O CH3 C O C O CH3 C O- Na+ + O O CH3 C Cl Cl C + Na+ -O C 34 :O: Reactions of Acid Anhydrides R .. O .. C : O: C R The chemistry of acid anhydrides is similar to that of acid chlorides except that anhydrides react more slowly. Acid anhydrides react with HOH to form acids, with ROH to form esters, with amines to form amides, with LiAlH4 to form 1° alcohols and with Grignards to form 3° alcohols. Note that ½ of the anhydride is wasted so that acid chlorides are more often used to acylate compounds. Acetic anhydride is one exception in that it is a very common acetylating agent. O O : O: 2R C acid H2O R .. C O C R - R': MgBr OH [H] : O: ester + acid R C .. HO .. O R' .. R :O: C : O: R R C 2R .. C NH2 amide C [H] R' ketone R':- MgBr + : O: NH3 R'OH O + H aldehyde 2 RCH2OH 1º alcohol OH RC R' R' 3º alcohol Write the mechanism for the following reactions and name all products: •aniline with acetic anhydride (2 moles aniline are needed or use 1 mole + aq. NaOH) •cyclopentanol with acetic formic anhydride (the formic carbonyl is more reactive). •methyl magnesium bromide with acetic propanoic anhydride (Grignards are not nucleophilic enough to react with carboxylate by products) •lithium aluminum hydride with acetic formic anhydride (LiAlH4 is so powerful a nucleophile that it will reduce even carboxylates). 35 Practice Questions for Acid Anhydrides • Show the product of methanol reacting with phthalic anhydride O O C C OCH3 2-(methoxycarbonyl)benzoic acid O + CH3OH C C OH O O • Draw acetominophen; formed when p-hydroxyaniline reacts with acetic anhydride O HO O O NH2 + CH3 C O C CH3 HO O N C CH3 + CH3 C OH H N-(4-hydroxyphenyl)acetamide 36 Chemistry of Esters • Esters are among the most widespread of all naturally occurring compounds. Most have pleasant odors and are responsible for the fragrance of fruits and flowers. Write chemical formulas for the following esters Flavor pineapple methyl butanoate Structure Name bananas isopentyl acetate apple isopentyl pentanoate rum isobutyl propanoate oil of wintergreen methyl salicylate [methyl 2-hydroxybenzoate) nail polish remover ethyl acetate new car smell (plasticizer for PVC) dibutyl phthalate O CH3CH2CH2C OCH3 CH3 O CH3C OCH2CH2CHCH3 O CH3 CH3(CH2)3COCH2CH2CHCH3 O CH3 CH2CH2C OCH2CHCH3 O C O CH3 OH O O CH3C OCH2CH3 C O (CH2)3CH3 C O (CH2)3CH3 O 37 Preparation of Esters 1. SN2 reaction of a carboxylate anion with a methyl or 1° alkyl halide : O: R C .. _ O: .. : O: Na + R' + Br R SN2 C .. O R' .. 2. Fischer esterification of a carboxylic acid + alcohol + acid catalyst : O: R C : O: + .. OH + R' OH H R C .. O R' .. 3. Acid chlorides react with alcohols in basic media : O: O R C Cl + R' OH R C .. O R' .. + HCl 38 Reactions of Esters • Esters react like acid halides and anhydrides but are less reactive toward nucleophiles because the carbonyl C is less electrophilic. Both acyclic esters and cyclic esters (lactones) react similarly. Esters are hydrolyzed by HOH to carboxylic acids, react with amines to amides, are reduced by hydrides to aldehydes, then to 1°alcohols, and react with Grignards to 3° alcohols. : O: R .. C H2O : O: R C acid NH3 + .. H OH : O: R C R':- MgBr+ O R' .. .. NH2 amide O R [H] C - R': MgBr : O: R ketone R' C H aldehyde [H] RCH2OH + R'OH + OH RC R' R' 3º alcohol 1º alcohols 39 Base Hydrolysis of Esters • Esters are hydrolyzed (broken down by water) to carboxylic acids or carboxylates by heating in acidic or basic media, respectively. • Base-promoted ester hydrolysis is called saponification (Latin ‘soap-making’). Boiling animal fat (which contains ester groups) in an aqueous solution of a strong base (NaOH, KOH, etc.) makes soap. A soap is long hydrocarbon chain with an ionic end group. O bar soap I: sodium dodecanoate C c: sodium laurate O- Na+ • The mechanism of base hydrolysis is nucleophilic acyl substitution in which OH - adds to the ester carbonyl group producing a tetrahedral intermediate. The carbonyl group reforms as the alkoxide ion leaves, yielding a carboxylate. O - - c: potassium laurate OCH3 CH3(CH2)10C O CH3 O CH3(CH2)10C O H O CH3(CH2)10C O liquid soap - K+ + CH3OH K+OH- • The leaving group, methoxide (OCH3-), like all alkoxides, is a strong base (pKb = -2). It will deprotonate the carboxylic acid intermediate converting it to a carboxylate. The alkoxide, when neutralized, becomes an alcohol. 40 Acid Hydrolysis of Esters • Acidic hydrolysis of an ester yields a carboxylic acid (and an alcohol). The mechanism of acidic ester hydrolysis is the reverse of Fischer esterification. The ester is protonated by acid then attacked by the nucleophile HOH. Transfer of a proton and elimination of ROH yields the carboxylic acid. The reaction is not favorable. It requires at least 30 minutes of refluxing. • Draw the complete mechanism of acid hydrolysis of methyl cyclopentanecarboxylate. O .. H+ OH O H H2O C OCH3 + H O H + + H2O OH C OCH3 C OCH3 C OCH3 .. OH H O C OH + H3O+ + CH3OH • Acid hydrolysis of an ester can be reversed by adding excess alcohol. The reverse reaction is called Fischer Esterification. Explain why base hydrolysis of an ester is not reversible. 41 Alcoholysis of Esters :O: R C .. O .. R + H ester .. O .. :O: R' H alcohol + R C .. O .. R' H + ester .. O .. R alcohol • Nucleophilic acyl substitution of an ester with an alcohol produces a different ester. The mechanism is the same as acid hydrolysis of esters except that that the nucleophile is an alcohol rather than water. A dry acid catalyst must be used, e.g., HCl(g) or H2SO4. If water is present, it will compete with the alcohol as the nucleophile producing some carboxylic acid in place of the ester product. • The process is also called Ester Exchange or Transesterification O OC H2SO4 O dicyclobutyl 1,4-benzenedicarboxylate dicyclobutyl terephthalate CO 2 CH3CH2OH O diethyl 1,4-benzenedicarboxylate diethyl terephthalate CH3CH2O C O C OCH2CH3 + 2 OH cyclobutanol 42 Aminolysis of Esters • Amines can react with esters via nucleophilic acyl substitution yielding amides but the reaction is difficult, requiring a long reflux period. Aminolysis of acid chlorides is preferred. • Draw the mechanism aminolysis of methyl isobutyroxide with ammonia. H3C H3C - O CH C OCH3 - OCH3 NH3 ∆ H3C O H CH C N H + H3C H H3C O CH C N H H3C H + CH3OH I: 2-methylpropanamide c: α-methylpropionamide • Write an equation showing how the following amide can be prepared from an ester. O (CH3)2N C O C N(CH3)2 O CH3O C O C OCH3 + 2 NH(CH3)2 • Note that the amide intermediate must deprotonate to form a stable, neutral amide. Thus the amine must have at least one H. NH3, 1° and 2° amines will work but not 3°. 43 Hydride Reduction of Esters • Esters are easily reduced with LiAlH4 to yield 1° alcohols. The mechanism is similar to that of acid chloride reduction. A hydride ion first adds to the carbonyl carbon temporarily forming a tetrahedral alkoxide intermediate. Loss of the –OR group reforms the carbonyl creating an aldehyde and an OR - ion. Further addition of H: - to aldehyde gives the 1° alcohol. Draw the mechanism and show all products. O LiAlH4 R C O R' LiAlH4 O - - OH H3O+ RCH RCH H RCH - O H OR' excess 1. LiAlH4 • Draw and name the products. O I: 4-hydroxybutanoic acid lactone O c: γ -butyrolactone 2. + R' O H CH2OH OH I: 1,4-butanediol c: none H3O+ • The hydride intermediate can be isolated if DIBAH is used as a reducing agent instead of LiAlH4. 1 equivalent of DIBAH is used at very low temp. (-78 °C). O I: 4-hydroxypentanoic acid lactone c: γ -valerolactone O O 1. CH3 2. D I BAH 1 equiv. - 78°C H3O+ I: 4-hydroxypentanal H c: γ -hydroxyvaleraldehyde OH CH3 44 Grignard Reduction of Esters • Esters and lactones react with 2 equivalents of Grignard reagent to yield 3 ° alcohols in which the 2 substituents are identical. The reaction occurs by the usual nucleophilic substitution mechanism to give an intermediate ketone, which reacts further with the Grignard to yield a 3° alcohol. MgBr O C O C H3 methyl benzoate MgBr O triphenylmethoxide O C - - C H OCH3 H3O+ O benzophenone C + CH3OH triphenylmethanol O O O CH3 MgBr O I: 4-hydroxybutanoic acid lactone CH3 MgBr CH3 O - - CH3 CH3 O - OH + H3O CH3 CH3 OH 4-methyl-1,4-pentanediol c: γ -butyrolactone 45 Practice with Esters • What ester and Grignards will combine to produce the following 2-phenyl-2-propanol CH3 CO+ OR C OH CH3 1. 2 CH3MgBr 2. H3O+ 1,1-diphenylethanol OH C CH3 O RO C CH3 + 1. 2 CH3MgBr 2. H3O+ 46 Chemistry of Amides • Amides are usually prepared by reaction of an acid chloride with an amine. Ammonia, monosubstituted and disubstituted amines (but not trisubstituted amines) all react. O .. NH3 O R C R NH2 C NR3 Cl NH2R 1º amine 1º amide NHR2 2º amine C NHR 2º amide 3º amine O O R no reaction R C NR2 3º amide • Amides are much less reactive than acid chlorides, acid anhydrides, or esters. Amides undergo hydrolysis to yield a carboxylic acids plus an amine on heating in either aqueous acid or aqueous base. • Basic hydrolysis occurs by nucleophilic addition of OH- to the amide carbonyl, followed by elimination of the amide ion, NH2‑,(a very reactive base – a difficult step requiring reflux) O - - NH2 C NH2 aq Na+ O O H- O COH CO - Na+ + NH3 I: sodium cyclohexanecarboxamide 47 Hydrolysis of Amides • Acidic hydrolysis occurs by nucleophilic addition of HOH to the protonated amide, followed by loss of a neutral amine (after a proton transfer to nitrogen). O CN H3O+ CH3 H H2O O H CH 3 CN + H O H CH 3 + CNH O H CH 3 CN + H O H OH H N-methylcyclohexanecarboxamide H H3O+ + H2O NH3CH3 O NH2CH3 + COH cyclohexanecarboxylic acid O O ∆ NaOH - NH2 O NH H2O H3O+ ∆ O OH + NH3 5-aminopentanoic acid lactam δ-valerolactam 48 Alcoholysis of Amides (to Esters) • Alcoholysis of amides occurs by the same acid catalyzed mechanism as acid hydrolysis except that the amido group of the amide is replaced with by an alcohol rather than water. Dry acid, e.g., HCl(g) or H2SO4 must be used otherwise water would compete with the alcohol as the nucleophile producing some carboxylic acid product in place of an ester. • The reaction will require a long reflux period because amides are weak electrophiles and alcohols are weak nucleophiles. O C N(CH3)2 OH CH3CHCH2CH3 H2SO4 N,N-dimethylcyclopentanecarboxamide O C + + NH2(CH3)2 O CH3CHCH2CH3 sec-butyl cyclopentanecarboxylate •Write a mechanism for this reaction. Refer to acid hydrolysis mechanism if necessary. 49 Hydride Reduction of Amides • Amides are reduced by LiAlH4. The product is an amine rather than an alcohol. The amide carbonyl group is converted to a methylene group (-C=O → -CH2). This is unusual. It occurs only with amides and nitriles. Initial hydride attack on the amide carbonyl eliminates the oxygen. A second hydride ion is added to yield the amine. The reaction works with lactams as well as acyclic amides. H -H O - Li+ C N CH3 CH3 - O Al AlH3 O C N CH3 C N CH3 H AlH3 H H H CH3 CH3 AlH2O- N,N-dimethylcyclopentanecarboxamide H + C N CH3 H O C Cl CH3 O ? C NHCH3 •Write equations showing how the above transformation can be carried out. O NH2CH3 C Cl benzoyl chloride O C NHCH3 N 1. LiAlH4 + 2. H3O H C NHCH3 H N-methylbenzamide 50 Grignard Reduction of Amides • Grignards deprotonate 1º and 2º amides and are not reactive enough to add to the imide ion product. N-H protons are acidic enough (pKa = 17) to be abstracted by Grignards. : O: R C .. _ : O: + .. :CH3 MgBr H N R _ CH 4 R 2º amide C .. N R :CH3 +MgBr imide anion • Write equations showing how the following transformation can be carried out. Br Mg, ether O 1. CO2 MgBr C OH 2. H3O+ ? CH2 N(CH3)2 SOCl2 OH C N CH2CH3 H2O ∆ H+ or OH- O NH(CH3)2 O C N(CH3)2 C Cl ? 2. H3O+ 1. LiAlH4 CH2OH 1. LiAlH4 O C OH 2. H3O+ 51 δ− N: + Chemistry of Nitriles δ C R • The carbon atom in the nitrile group is electrophilic because it is bonded to an electronegative N atom and a π bond in the nitrile is easily broken, i.e., as if it were providing a leaving group. Preparation of Nitrile: 1. Nitriles are easily prepared by SN2 reaction of cyanide ion (CN-) with methyl halides or a 1° alkyl halide. 2º alkyl halides also work but some E2 product also forms. 3º alkyl halides will result in mostly an alkene (E2) product instead of a nitrile. (pKb of CN - = 4.7) bromoethane ethyl bromide SN2 CH3CH2 Br Na+ CN- propanenitrile CH3CH2 C N 2. Another method of preparing nitriles is by dehydration of a 1 ° amide using any suitable dehydrating agent such as SOCl2, POCl3, P2O5, or acetic anhydride. Initially, SOCl2 reacts with the amide oxygen atom and elimination follows. This method is not limited by steric hindrance. O C NH2 SOCl2 , benzene C N + SO2 + 2 HCl 80ºC SOCl2 Recall: R R Cl OH POCl3 in pyr idine OH PBr3 R _ H2O Br 52 + Reactions of Nitriles δ C R δ− N: • Like carbonyl groups, the nitrile group is strongly polarized and the nitrile C is electrophilic. Nucleophiles thus attack yielding an sp2 hybridized imine anion. _ .. : O: : O: sp 2 C C E R Nu Nu:- + products sp 3 δ+ C − δ N: C R - sp + E Nu: .. _ N: products Nu sp 2 imine anion • Nitriles are hydrolyzed by HOH to amides and subsequently to carboxylic acids, reduced by hydrides to amines or aldehydes, and by Grignards to ketones. H2O O R C 1º amide R NH2 _ POCl3 N: C R C R ketone [H] [H] H2O O RMgX O R CH2 1º amine NH2 R C H aldehyde 53 δ− N: + Hydrolysis of Nitriles into Carboxylic Acids δ C R • Nitriles are hydrolyzed in either acidic or basic aqueous solution to yield carboxylic acids plus ammonia or an amine. RCN O H2O ∆ H+ or OH- + NH3 RCOH • In acid media, protonation of N produces a cation that reacts with water to give an imidic acid (an enol of an amide). Keto-enol isomerization of the imidic acid gives an amide. The amide is then hydrolyzed to a carboxylic acid and ammonium ion. It is possible to stop the reaction at the amide stage by using only 1 mole of HOH per mole of nitrile. Excess HOH forces carboxylic acid formation. H3O+ R C N: R C + N .. .. R H C + N :O .. R .. C OH acid H3O+ : O: R C amide .. NH2 N +O H .. H H .. .. H H : O: C R H H2O .. R N C H : O: hydroxyimine H 54 + Hydrolysis of Nitriles into Carboxylate Salts R δ C δ− N: • In basic media, hydrolysis of a nitrile to a carboxylic acid is driven to completion by the reaction of the carboxylic acid with base. The mechanism involves nucleophilic attack by hydroxide ion on the electrophilic C producing a hydroxy imine, which rapidly isomerizes to an amide. Further hydrolysis yields the carboxylate salt. : O: - OH R C N: R C :O .. H H H .. _ N: _ OH- R C : O: .. H N R C hydroxy imine : O: NH3 + H : O: .. _ + O : Na .. R C .. NH2 amide ∆ OHH2O • Show how the following transformation can be carried out without using a Grignard. Br ? O C OH ∆ H3O+ Na+CNSN2 CN ∆ H3O+ O C NH2 55 + Reduction of Nitriles R δ C δ− N: Alcoholysis of Nitriles doesn’t work. Alcohols are weak nucleophiles and nitriles are weak electrophiles Aminolysis of Nitriles doesn’t work. Amines are weak nucleophiles and nitriles are weak electrophiles. Reduction with Hydrides: Reduction of nitriles with 2 equivalents of LiAlH4 gives 1° amines. LiAlH4 is a very good nucleophile and can break 2 π bonds forming a dianion. H R C N: 1. H:- R C H .. _ N: R - H: C .. _ 2 N: H 2. H2O .. H dianion H2O .. R C NH2 H + 2 OH- • If less powerful DIBAH is used, only 1 equivalent of hydride can add. Subsequent addition of HOH yields the aldehyde. C N 1. LiAlH4 2. CH3 o-methylbenzonitrile H2O 1 equiv. 1. DIBAH , toluene , -78ºC 2. H2O or H3O+ CH2NH2 CH3 O CH 2-methylbenzaldehyde CH3 56 + Reduction of Nitriles with Grignards R δ C δ− N: • Grignards add to nitriles giving intermediate imine anions which when hydrolyzed yield ketones. The mechanism is similar to hydride reduction except that the attacking nucleophile is a carbanion (R-). Grignards are not as strongly nucleophilic as LiAlH4 and so can only add once – a dianion is not formed. 1. R C - R: MgBr R + N: R C .. _ N: 2. R R + H3O .. - .. R H N C imine anion :O .. NH4 + NH3 + R .. C R R C H H R : O: H3O + N : O: C N .. +O H H R .. H H H H C N 1. CH3CH3 MgBr 2. benzonitrile + H3O O C CH2CH3 1-phenyl-1-propanone ethyl phenyl ketone 57 Multistep Synthesis Problems • Write equations to show how the following transformations can be carried out. Br CH2NH2 ? CN Na+ CN- 1. LiAlH4 2. H3O+ O Br CH ? 1. 2. O DIBAH CN 1 equiv. - 78°C H3O+ from above C ? 1. MgBr ether 2. Br + H3O Br ? C 1. 2. O H2O ∆ MgBr ether H3O+ C OH H+ from above 2 equiv. HO CN SOCl2 O C Cl 58 ...
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This note was uploaded on 01/21/2012 for the course ECON 2a taught by Professor Xi during the Spring '11 term at Cornell College.

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Carboxylic acid ppt - Derivatives of Carboxylic Acid : O: R...

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