Derivatives of Carboxylic Acid carboxylate acid chloride nitrile

Derivatives of Carboxylic Acid carboxylate acid chloride nitrile acid anhydride ester amide

Nomenclature of Acid Halides • IUPAC: alkanoic acid alkanoyl halide • Common: alkanic acid alkanyl halide I: 3 -aminopropanoyl chloride I: 4 -nitropentanoyl chloride c: g-nitrovaleryl chloride c: b-aminopropionyl chloride I: hexanedioyl chloride c: adipoyl chloride Rings: (IUPAC only): ringcarbonyl halide I: benzenecarbonyl bromide I: 3 -cylcopentenecarbonyl chloride c: benzoyl bromide 2

Nomenclature of Acid Anhydrides • Acid anhydrides are prepared by dehydrating carboxylic acids acetic anhydride ethanoic acid I: benzenecarboxylic anhydride c: benzoic andhydride I: cis-butenedioic anhydride c: maleic anhydride I: butanedioic acid c: succinic acid I: butanedioic anhydride c: succinic anhydride Some unsymmetrical anhydrides I: benzoic methanoic anhydride c: benzoic formic anhydride I: ethanoic methanoic anhydride c: acetic formic anhydride 3

Nomenclature of Esters • Esters occur when carboxylic acids react with alcohols I: methyl ethanoate c: methyl acetate I: phenyl methanoate c: phenyl formate I: t-butyl benzenecarboxylate c: t-butyl benzoate I: isobutyl cyclobutanecarboxylate c: none I: cyclobutyl 2 -methylpropanoate c: cyclobutyl a-methylpropionate I: dimethyl ethanedioate c: dimethyl oxalate 4

Nomenclature of Cyclic Esters, “Lactones” Cyclic esters, “lactones”, form when an open chain hydroxyacid reacts intramolecularly. 5 to 7 membered rings are most stable. I: 4 -hydroxybutanoic acid c: g-hydroxybutyric acid I: 4 -hydroxybutanoic acid lactone c: g-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. I: 5 -hydroxypentanoic acid lactone I: 4 -hydroxypentanoic acid lactone c: d-valerolactone c: g-valerolactone I: 3 -hydroxypentanoic acid lactone c: b-valerolactone I: 6 -hydroxy-3 -methylhexanoic acid lactone c: b-methyl-e-caprolactone 5

Nomenclature of Amides 1° amide 2° amide N-substituted amide 3° amide N, N-disubstituted amide § 1° amides: ‘alkanoic acid’ + amide ‘’alkanamide’ § a ring is named ‘ringcarboxamide’ I: butanamide c: butyramide I: 3 -chlorocyclopentanecarboxamide c: none I: p-nitrobenzenecarboxamide c: p-nitrobenzamide § 2° and 3° amides are N-substituted amides I: N, 2 -dimethylpropanamide c: N, a-dimethylpropionamide I: N-phenylethanamide c: N-phenylacetamide c: acetanilide I: N-ethyl-N-methylcyclobutanecarboxamide c: none 6

Nomenclature of Cyclic Amides, “Lactams” Cyclic amides, “lactams”, form when an open chain aminoacid reacts intramolecularly. 5 to 7 membered rings are most stable. I: 4 -aminobutanoic acid c: g-aminobutyric acid I: 4 -aminobutanoic acid lactam c: g-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. I: 3 -amino-2 -bromopropanoic acid lactam c: a-bromo-b-propionolactam I: 5 -aminohexanoic acid lactam c: d-caprolactam I: 4 -amino-3 -methylbutanoic acid lactam c: b-methyl-g-butyrolactam 7

Nomenclature of Nitriles are produced when 1° amides are dehydrated with reagents like POCl 3 • • • IUPAC: alkane + nitrile ‘alkanenitrile’ IUPAC rings: ‘ringcarbonitrile’ Common: alkanic acid + ‘onitrile’ ‘alkanonitrile’ I: ethanenitrile c: acetonitrile I: 4 -iodobutanenitrile c: g-iodobutyronitrile I: 3 -methoxycyclohexanecarbonitrile c: none I: p-thiobenzenecarbonitrile c: p-mercaptobenzonitrile I: 2 -cyanocyclopentanecarboxylic acid c: none 8

Nomenclature Practice Exercise I: cyclobutanecarbonitrile c: none I: bromomethyl ethanoate c: bromomethyl acetate I: sodium ethanoate c: sodium acetate I: 3 -bromo-N-methylpentanamide c: b-bromo-N-methylvaleramide I: pentanedioic anhydride c: glutaric anhydride I: 3 -oxobutanoyl chloride c: b-oxobutyryl chloride I: 2 -ethyl-5 -hydroxypentanoic acid lactone c: a-ethyl-d-valerolactone I: 6 -amino-6 -chlorohexanoic acid lactam c: e-chloro-e-caprolactam 9

Relative Reactivity of Carbonyl Carbons § Nucleophiles (electron donors), like OH-, bond with the sp 2 hybridized carbonyl carbon. § The order of reactivity is shown. acid chloride acid anhydride aldehyde most reactive ketone ester carboxylic acid amide nitrile carboxylate 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 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 (p. Kb of – 40 and -21). R and H are not leaving groups, so the carbonyl group becomes an alkoxide as the sp 2 C becomes a tetrahedral sp 3 C. tetrahedral alkoxide with sp 3 carbon. • • A second addition of a nucleophile cannot occur since alkoxides are not nucleophilic. The reaction is usually completed by protonation of the alkoxide with H 3 O+ forming an alcohol. This later reaction is simply an acid/base reaction. The characteristic reaction of aldehydes and ketones is thus ‘nucleophilic addition’. 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 bond breaks and the pair of electrons are stabilized on the electronegative O atom. A tetrahedral alkoxide is temporarily formed. Chlorine is a fair leaving group. sp 2 carbonyl reforms sp 2 carbonyl C • • alkoxide C js sp 3 In carboxylic acid derivates, one of the groups that was bonded to the carbonyl C is a leaving group. When this group leaves, the sp 3 tetrahedral alkoxide reverts back to an sp 2 C=O group. Thus substitution occurs instead of addition. In many cases, the substitution product contains a carbonyl that can react again. 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). Draw the mechanism. acyl group • • 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 p. Kb, and describe it as a strong or weak base. acid derivative leaving group p. Kb strength as base +21 non basic +9 weak base -2 strong base -21 v. strong base 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) hydrolysis alcoholysis aminolysis Grignard reduction hydride reduction 14

Nucleophilic Acyl Substitution of Carboxylic Acids • Nucleophilic acyl substitution converts carboxylic acids into carboxylic acid derivatives, i. e. , acid chlorides, anhydrides, esters and amides. SOCl 2 acid chloride acid anhydride D -H 2 O NH 3, D, -H 2 O amide ROH H+ ester 15

Conversion of Carboxylic Acids to Acid Halides • The S atom in SOCl 2 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 SN 2 mechanism yielding an acid chloride. • Use curved arrows to draw the initial steps of the mechanism shown below. 1 2 1 3 • • 3 2 PBr 3 will substitute Br for OH converting a carboxylic acid to an acid bromide Draw and name the products of the following reactions. 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. ethanoic acid • • acetic anhydride ethanoic anhydride Cyclic anhydrides with 5 or 6 -membered rings are prepared by dehydration of diacids. I: butanedioic acid I: butanedioic anhydride c: succinic acid c: succinic anhydride Draw a reaction showing the preparation of cyclohexanecarboxylic anhydride. 17

Conversion of Carboxylic Acids to Esters Two methods are used: SN 2 reaction of a carboxylate and Fischer Esterification 1. SN 2 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 (p. Kb = 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. E 2 sodium propionoate I: 5 -bromopentanoic acid c: d-bromovaleric acid isobutylene I: sodium 5 -bromopentanoate c: sodium d-bromovalerate I: 5 -hydroxypentanoic acid lactone c: d-valerolactone 18

Conversion of Carboxylic Acids to Esters SN 2 no reaction SN 2 E 2 (SN 2) no reaction SN 2 E 2 SN 1 E 2 v. gd. moderate 4. 7 E 2 SN 1, E 1 fair weak 9 6. 0 / 7. 0 SN 2 SN 2 SN 2 E 2 SN 1 E 2 E 2 SN 1 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 2 SO 4 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 H 2 O formed. The reaction is reversible. A large excess of H 2 O 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. • 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. I: propanedioic acid c: malonic acid I: diethyl propanedioate c: diethyl malonate cyclopentylmethyl benzoate • Draw the reagents that will react to produce the following ester. Why will an SN 2 reaction of a carboxylate and an alkyl halide not work here? I: isopropyl 2 -methylpropanoate c: isopropyl isobutyrate • Isopropyl bromide is a 2° alkyl halide and would undergo an E 2 rather than SN 2 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, (RNH 3+) • 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. 23

Chemistry of Acid Halides • In the same way that acid chlorides are produced by reacting a carboxylic acid with thionyl chloride (SOCl 2), acid bromides are produced by reacting a carboxylic acid with phosphorus tribromide (PBr 3). 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. 24

Chemistry of Acid Halides • Draw a reaction showing how propylbenzene can be produced by a Friedel Crafts acylation reaction. 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, -NH 2 to produce an amide. Hydride reduction produces a 1 alcohol, and Grignard reaction produces a 3 alcohol. 25

Hydrolysis: Conversion of Acid Halides into Acids • Acid chlorides react via nucleophilic attack by H 2 O producing carboxylic acids and 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. • 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. I: ethanoyl chloride c: acetyl chloride I: isopropyl ethanoate c: isopropyl acetate • 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 Na. OH 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). 27

Practice on Synthesis of Esters • Write equations showing all the ways that benzyl benzoate can be produced. Consider Fischer esterification, SN 2 reaction of a carboxylate with an alkyl halide, and alcoholysis of an acid chloride. • Answer the same question as above but for t-butyl butanoate 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 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 Na. OH (provided it is not soluble in the organic layer). Using Na. OH in an aminolysis reaction is referred to as the Schotten-Baumann reaction. I: N, N-dimethylbenzenecarboxamide c: N, N-dimethylbenzamide • Write equations showing how the following products can be made from an acid chloride. N-methylacetamide propanamide 29

Reduction of Acid Chlorides to Alcohols with Hydride • Acid chlorides are reduced by Li. Al. H 4 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 Li. Al. H 4 in a second step to yield a 1 alcohol. • Draw the reaction and name the product when 2, 2 -dimethylpropanoyl chloride is reduced with Li. Al. H 4 I: 2, 2 -dimethyl-1 -propanol c: neopentyl alcohol 30

Reduction of Acid Chlorides to Aldehydes with Hydride • The aldehyde cannot be isolated if Li. Al. H 4 (and Na. BH 4) 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. q q DIBAH is weaker than Li. Al. H 4. DIBAH is neutral; Li. Al. H 4 is ionic. DIBAH is similar to Al. H 3 but is hindered by its bulky isobutyl groups. q Only one mole of H: - is released per mole of DIBAH. 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 Li. Al. H 4 reduction. The 1 st equivalent of Grignard reagent adds to the acid chloride, loss of Cl- from the tetrahedral intermediate yields a ketone, and a 2 nd equivalent of Grignard immediately adds to the ketone to produce an alcohol. I: 2 -phenyl-2 -propanol • The ketone intermediate can’t be isolated with Grignard reaction but can be with Gilman reagent (diorganocopper), R 2 Cu. Li. 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) 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 I: 1, 1 -dicyclopentylethanol I: 1 -phenyl-1 -propanone c: ethyl phenyl ketone I: 2, 2 -dimethylpropanoyl chloride I: 2, 2 -dimethyl-1 -propanol I: cyclohexanecarbonyl chloride I: cyclohexanecarbaldehyde 33

Preparations of Acid Anhydrides Preparation of Acid Anhydrides: Dehydration of carboxylic acids as previously discussed is difficult and therefore limited to a few cases. acetic anhydride 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. • Draw all sets of reactants that will produce the anhydride shown with an acid chloride. 34

Reactions of Acid Anhydrides 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 Li. Al. H 4 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. 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. Na. OH) • 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 (Li. Al. H 4 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 2 -(methoxycarbonyl)benzoic acid • Draw acetominophen; formed when p-hydroxyaniline reacts with acetic anhydride 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 Name pineapple methyl butanoate 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 Structure 37

Preparation of Esters 1. SN 2 reaction of a carboxylate anion with a methyl or 1 alkyl halide 2. Fischer esterification of a carboxylic acid + alcohol + acid catalyst 3. Acid chlorides react with alcohols in basic media 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. 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 (Na. OH, KOH, etc. ) makes soap. A soap is long hydrocarbon chain with an ionic end group. I: sodium dodecanoate bar soap c: sodium laurate • 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. c: potassium laurate liquid soap • The leaving group, methoxide (OCH 3 -), like all alkoxides, is a strong base (p. Kb = -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. • 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 • 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 the nucleophile is an alcohol rather than water. A dry acid catalyst must be used, e. g. , HCl(g) or H 2 SO 4. 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 dicyclobutyl 1, 4 -benzenedicarboxylate dicyclobutyl terephthalate diethyl 1, 4 -benzenedicarboxylate diethyl terephthalate 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. I: 2 -methylpropanamide c: a-methylpropionamide • Write an equation showing how the following amide can be prepared from an ester. • Note that the amide intermediate must deprotonate to form a stable, neutral amide. Thus the amine must have at least one H. NH 3, 1° and 2° amines will work but not 3°. 43

Hydride Reduction of Esters • Esters are easily reduced with Li. Al. H 4 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. • Draw and name the products. I: 4 -hydroxybutanoic acid lactone c: g-butyrolactone I: 1, 4 -butanediol c: none • The hydride intermediate can be isolated if DIBAH is used as a reducing agent instead of Li. Al. H 4. 1 equivalent of DIBAH is used at very low temp. (-78 C). I: 4 -hydroxypentanoic acid lactone I: 4 -hydroxypentanal c: g-hydroxyvaleraldehyde c: g-valerolactone 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. triphenylmethoxide methyl benzoate benzophenone triphenylmethanol I: 4 -hydroxybutanoic acid lactone 4 -methyl-1, 4 -pentanediol c: g-butyrolactone 45

Practice with Esters • What ester and Grignards will combine to produce the following 2 -phenyl-2 -propanol 1, 1 -diphenylethanol 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. • 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, NH 2‑, (a very reactive base – a difficult step requiring reflux) 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). N-methylcyclohexanecarboxamide cyclohexanecarboxylic acid 5 -aminopentanoic acid lactam d-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 H 2 SO 4 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. N, N-dimethylcyclopentanecarboxamide 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 Li. Al. H 4. The product is an amine rather than an alcohol. The amide carbonyl group is converted to a methylene group (-C=O -CH 2). 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. N, N-dimethylcyclopentanecarboxamide • Write equations showing how the above transformation can be carried out. benzoyl chloride 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 (p. Ka = 17) to be abstracted by Grignards. • Write equations showing how the following transformation can be carried out. 51

Chemistry of Nitriles • The carbon atom in the nitrile group is electrophilic because it is bonded to an electronegative N atom and a 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 SN 2 reaction of cyanide ion (CN-) with methyl halides or a 1 alkyl halide. 2º alkyl halides also work but some E 2 product also forms. 3º alkyl halides will result in mostly an alkene (E 2) product instead of a nitrile. (p. Kb of CN- = 4. 7) bromoethane ethyl bromide propanenitrile 2. Another method of preparing nitriles is by dehydration of a 1 amide using any suitable dehydrating agent such as SOCl 2, POCl 3, P 2 O 5, or acetic anhydride. Initially, SOCl 2 reacts with the amide oxygen atom and elimination follows. This method is not limited by steric hindrance. 52

Reactions of Nitriles • Like carbonyl groups, the nitrile group is strongly polarized and the nitrile C is electrophilic. Nucleophiles thus attack yielding an sp 2 hybridized 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. 53

Hydrolysis of Nitriles into Carboxylic Acids • Nitriles are hydrolyzed in either acidic or basic aqueous solution to yield carboxylic acids plus ammonia or an amine. • 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. 54

Hydrolysis of Nitriles into Carboxylate Salts • 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. • Show the following transformation can be carried out without using a Grignard. 55

Reduction of Nitriles 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 Li. Al. H 4 gives 1 amines. Li. Al. H 4 is a very good nucleophile and can break 2 bonds forming a dianion. • If less powerful DIBAH is used, only 1 equivalent of hydride can add. Subsequent addition of HOH yields the aldehyde. 2 -methylbenzaldehyde 56

Reduction of Nitriles with Grignards • 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 Li. Al. H 4 and so can only add once – a dianion is not formed. 1 -phenyl-1 -propanone ethyl phenyl ketone 57

Multistep Synthesis Problems • Write equations to show the following transformations can be carried out. 58