21 More About Amines • Heterocyclic Compounds

More About Amines • 21 Heterocyclic Compounds

CH3CH2NH2 CH3CH2NH ethylamine CH2CH3 diethylamine

mines are compounds in which one or more of the hydrogens of ammonia (NH3) A have been replaced by an alkyl group. Amines are among some of the most abundant compounds in the biological world. We will ap- CH3CH2NCH2CH3 preciate their importance in Chapter 23, when we NH look at amino acids and proteins; in Chapter 24, CH2CH3 when we study how enzymes catalyze chemical reac- triethylamine piperidine tions; in Chapter 25, when we investigate the ways in which coenzymes—compounds derived from vitamins—help enzymes catalyze chemical reactions; in Chapter 27, when we study nucleic acids (DNA and RNA); and in Chapter 30, when we take a look at how drugs are discovered and designed. Amines are also exceedingly important compounds to organic chemists, far too important to leave until the end of a course in organic chemistry. We have, therefore, al- ready studied many aspects of amines and their chemistry. For example, we have seen that the nitrogen in amines is sp3 hybridized and the lone pair resides in an empty sp3 orbital (Section 2.8). We also have examined the physical properties of amines—their hydrogen bonding properties, boiling points, and solubilities (Section 2.9). In Section 2.7, we learned how amines are named. Most important, we have seen that the lone-pair electrons of the nitrogen atom cause amines to react as bases, sharing their lone pair with a proton, and as nucleophiles, sharing their lone pair with an atom other than a proton.

an amine is a base: + + + − R NH2 H Br H NH3 Br an amine is a nucleophile: + + + − R NH2 CH3 Br R NH2 CH3 Br

In this chapter, we will revisit some of these topics and look at some aspects of amines and their chemistry that we have not discussed previously. 883 884 CHAPTER 21 More About Amines •Heterocyclic Compounds

Some amines are heterocyclic compounds (or heterocycles)—cyclic compounds in which one or more of the atoms of the ring are heteroatoms. A heteroatom is an atom other than carbon. The name comes from the Greek word heteros, which means “different.” A variety of atoms, such as N, O, S, Se, P, Si, B, and As, can be incorporated into ring structures. Heterocycles are an extraordinarily important class of compounds, making up more than half of all known organic compounds. Almost all the compounds we know as drugs, most vitamins, and many other natural products are heterocycles. In this chapter, we will consider the most prevalent heterocyclic compounds—the ones that contain the heteroatoms N, O, and S. A natural product is a compound synthesized by a plant or an animal. Alkaloids are natural products that contain one or more nitrogen heteroatoms and are found in the leaves, bark, roots, or seeds of plants. Examples include caffeine (found in tea leaves, coffee beans, and cola nuts) and nicotine (found in tobacco leaves). Morphine is an alkaloid obtained from opium, the juice derived from a species of poppy. Mor- phine is 50 times stronger than aspirin as an analgesic, but it is addictive and suppress- es respiration. Heroin is a synthetic compound that is made by acetylating morphine (Section 30.3).

O CH3 CH3 O H3C N N CH2CH2NH2 N HO N O N N Cl N N CH3 N CH3 C6H5 H  caffeine nicotine Valium serotonin

HO CH3C O

O O O NCH3 NCH3 HO H CH3C O H H H morphine heroin

Two other heterocycles are Valium®, a synthetic tranquilizer, and serotonin, a neuro- transmitter. Serotonin is responsible for, among other things, the feeling of having had enough to eat. When food is ingested, brain neurons are signaled to release serotonin. A once widely used diet drug (actually a combination of two drugs, fenfluramine and phentermine), popularly known as fen/phen, works by causing brain neurons to release extra serotonin (Chapter 16, p. 622). After finding that many of those who took fenfluramine had abnormal echocardiograms due to heart valve problems, the Food and Drug Administration asked the manufacturer of these diet drugs to withdraw the products. There is some evidence that faulty metabolism of serotonin plays a role in bipolar affective disorder.

21.1 More About Nomenclature

In Section 2.7, we saw that amines are classified as primary, secondary, or tertiary, depending on whether one, two, or three hydrogens of ammonia, respectively, have been replaced by an alkyl group. We also saw that amines have both common and systematic names. Common names are obtained by citing the names of the alkyl subsitutents Section 21.2 Amine Inversion 885

(in alphabetical order) that have replaced the hydrogens of ammonia. Systematic names employ “amine” as a functional group suffix.

CH3

CH3CH2CH2CH2CH2NH2 CH3CH2CH2CH2NHCH2CH3 CH3CH2CH2NCH2CH3 a primary amine a secondary amine a tertiary amine common name: pentylamine butylethylamine ethylmethylpropylamine systematic name: 1-pentanamine N-ethyl-1-butanamine N-ethyl-N-methyl-1-propanamine A saturated cyclic amine—a cyclic amine without any double bonds—can be named as a cycloalkane, using the prefix “aza” to denote the nitrogen atom. There are, however, other acceptable names. Some of the more commonly used names are shown here. Notice that heterocyclic rings are numbered so that the heteroatom has the lowest possible number.

CH3

H N NH N N N CH3 H H CH2CH3 azacyclopropane azacyclobutane 3-methylazacyclopentane 2-methylazacyclohexane N-ethylazacyclopentane aziridine azetidine 3-methylpyrrolidine 2-methylpiperidine N-ethylpyrrolidine

Heterocycles with oxygen and sulfur heteroatoms are named similarly. The prefix for oxygen is “oxa” and that for sulfur is “thia.”

O S O O oxacyclopropane thiacyclopropane oxacyclobutane oxacyclopentane oxirane thiirane oxetane ethylene oxide tetrahydrofuran

OO tetrahydropyran 1,4-dioxane

PROBLEM 1◆ Name the following compounds:

H CH CH3 N 3 a.CH3 c. e. HN CH CH3 O 3

CH2CH3 S O b. d. f. CH N 3 H CH2CH3

21.2 Amine Inversion

The lone-pair electrons on nitrogen allow an amine to turn “inside out” rapidly at room temperature. This is called amine inversion. One way to picture amine inversion is to compare it to an umbrella that turns inside out in a windstorm. 886 CHAPTER 21 More About Amines •Heterocyclic Compounds

amine inversion

p orbital

3 1 R ‡ R 3 2 3 R2 sp sp R N N R1 N R3 3 R2 sp R1 R2

transition state The lone pair is required for inversion: Quaternary ammonium ions—ions with four bonds to nitrogen and hence no lone pair—do not invert. Notice that amine inversion takes place through a transition state in which the sp3 nitrogen becomes an sp2 nitrogen. The three groups bonded to the sp2 nitrogen are coplanar in the transition state with bond angles of 120°, and the lone pair is in a p orbital. The “inverted” and “non-inverted” amine molecules are enantiomers, but they cannot be separated because amine inversion is rapid. The energy required for amine inversion is approximately 6 kcal>mol (or 25 kJ>mol ), about twice the amount of energy required for rotation about a carbon–carbon single bond, but still low enough to allow the enantiomers to interconvert rapidly at room temperature.

21.3 More About the Acid–Base Properties of Amines

Amines are the most common organic bases. We have seen that ammonium ions have pKa values of about 11 (Section 1.17) and that anilinium ions have pKa values of about 5 (Sections 7.10 and 16.5). The greater acidity of anilinium ions compared with ammonium ions is due to the greater stability of their conjugate bases as a result of electron delocalization. Amines have very high pKa values. For example, the pKa of methylamine is 40.

CH2CH3 + + + + + CH3CH2CH2NH3 CH3NH2 CH3CH2NH NH3 CH3 NH3 CH3NH2

CH3 CH2CH3 pKa = 10.8 pKa = 10.9 pKa = 11.1 pKa = 4.58 pKa = 5.07 pKa = 40

Saturated heterocycles containing five or more atoms have physical and chemical properties typical of acyclic compounds that contain the same heteroatom. For example, pyrrolidine, piperidine, and morpholine are typical secondary amines, and N-methylpyrrolidine and quinuclidine are typical tertiary amines. The conjugate acids 3-D Molecules: of these amines have pKa values expected for ammonium ions. We have seen that the Aziridinium ion; Pyrrolidine; basicity of amines allows them to be easily separated from other organic compounds Piperidine; Morpholine (Chapter 1, Problems 70 and 71). O

+++ + + N NN N N HH HH HH H CH3 H the ammonium ions of:pyrrolidine piperidine morpholine N-methylpyrrolidine quinuclidine pKa = 11.27 pKa = 11.12 pKa = 9.28 pKa = 10.32 pKa = 11.38

PROBLEM 2◆

Why is the pKa of the conjugate acid of morpholine significantly lower than the pKa of the conjugate acid of piperidine? Section 21.4 Reactions of Amines 887

PROBLEM 3◆ a. Draw the structure of 3-quinuclidinone.

b. What is the approximate pKa of its conjugate acid? c. Which has a lower pKa , the conjugate acid of 3-bromoquinuclidine or the conjugate acid of 3-chloroquinuclidine?

21.4 Reactions of Amines

The lone pair on the nitrogen of an amine causes it to be nucleophilic as well as basic. We have seen amines act as nucleophiles in a number of different kinds of reactions: in nucleophilic substitution reactions—reactions that alkylate the amine (Section 10.4)— such as

+ ++ CH3CH2BrCH3NH2 CH3CH2NH2CH3 CH3CH2NHCH3 HBr methylamineBr– ethylmethylamine in nucleophilic acyl substitution reactions—reactions that acylate the amine (Sections 17.8, 17.9, and 17.10)—for example,

O O + + + – C 2 CH3NH2 C CH3NH3Cl CH CH Cl CH CH NHCH 3 2 methylamine 3 2 3 an amide O O O C C ++C O CH O CH CH N 3 3 3 – N N OCCH3 H + HH piperidine an amide in nucleophilic addition–elimination reactions—the reactions of aldehydes and ketones with primary amines to form imines and with secondary amines to form enamines (Section 18.6)—such as

catalytic + + H + O H2NCH2 NCH2 H2O

benzylamine an imine

catalytic + ++H ONHN H2O

pyrrolidine an enamine and in conjugate addition reactions (Section 18.13)—for instance,

CH3 O CH3 O + CH3CCHCHCH 3NH CH3C CH2CH

CH3 NCH3

CH3 888 CHAPTER 21 More About Amines • Heterocyclic Compounds

We have seen that primary arylamines react with nitrous acid to form stable arenediazonium salts (Section 16.12). Arenediazonium salts are useful to synthetic chemists because the diazonium group can be replaced by a wide variety of nucleophiles. This reaction allows a wider variety of substituted benzenes to be prepared than can be prepared solely from electrophilic aromatic substitution reactions.

+ − HCl − Nu ++– NH2 N NCl Nu N2 Cl NaNO2 an arenediazonium salt

Amines are much less reactive than other compounds with electron-withdrawing groups bonded to sp3 hybridized carbons, such as alkyl halides, alcohols, and ethers. The relative reactivities of an alkyl fluoride (the least reactive of the alkyl halides), an alcohol, an ether, and an amine can be appreciated by comparing the pKa values of the conjugate acids of their leaving groups, recalling that the stronger the base, the weaker is its conjugate acid and the poorer the base is as a leaving group. The leaving group - of an amine ( NH2) is such a strong base that amines cannot undergo the substitution and elimination reactions that alkyl halides undergo.

relative reactivities

most >>∼ least reactive RCH2F RCH2OH RCH2OCH3 RCH2NH2 reactive

strongest acid, HF H2O RCH2OH NH3 weakest acid, weakest conjugate strongest conjugate pKa = 3.2 pKa = 15.7 pKa = 15.5 pKa = 36 base base

Protonation of the amino group makes it a weaker base and therefore a better leaving group, but it still is not nearly as good a leaving group as a protonated alcohol. Re- call that protonated ethanol is more than 13 pKa units more acidic than protonated ethylamine.

+ + CH3CH2OH2 CH3CH2NH3 pKa =− 2.4 pKa = 11.2

So, unlike the leaving group of a protonated alcohol, the leaving group of a protonated amine cannot dissociate to form a carbocation or be replaced by a halide ion. Proto- nated amino groups also cannot be displaced by strongly basic nucleophiles such as HO- because the base would react immediately with the acidic hydrogen, and protonation would convert it into a poor nucleophile.

+ ++– CH3CH2NH3 HO CH3CH2NH2 H2O

PROBLEM 4 Why is it that a halide ion such as Br- can react with a protonated primary alcohol, but cannot react with a protonated primary amine?

PROBLEM 5 Give the product of each of the following reactions:

O catalytic + + H a. CCH3 CH3CH2CH2NH2 Section 21.5 Reactions of Quaternary Ammonium Hydroxides 889

O + b. CH3CCl 2 N H

1. HCl, NaNO2 c. NH2 2. H2O, Cu2O, Cu(NO3)2

O catalytic + + H d. CCH3 CH3CH2NHCH2CH3

21.5 Reactions of Quaternary Ammonium Hydroxides

The leaving group of a quaternary ammonium ion has about the same leaving tendency as a protonated amino group, but it does not have an acidic hydrogen that would protonate a basic reactant. A quaternary ammonium ion, therefore, can undergo a reaction with a strong base. The reaction of a quaternary ammonium ion with hydroxide ion is known as a Hofmann elimination reaction. The leaving group in a Hofmann elimination reaction is a tertiary amine. Because a tertiary amine is only a moderately good leaving group, the reaction requires heat.

CH3 CH3 + ∆ CH3CH2CH2NCH3 CH3CH CH2 ++NCH3 H2O − CH3 HO CH3

A Hofmann elimination reaction is an E2 reaction. Recall that an E2 reaction is a August Wilhelm von Hofmann concerted, one-step reaction—the proton and the tertiary amine are removed in the (1818–1892) was born in Germany. He first studied law and then same step (Section 11.1). Very little substitution product is formed. changed to chemistry. He founded the German Chemical Society. Hofmann mechanism of the Hofmann elimination taught at the Royal College of Chem- istry in London for 20 years and then CH3 CH3 + returned to Germany to teach at the CH3CH CH2 NCH3 CH3CH CH2 ++NCH3 H2O University of Berlin. He was one of HCH CH the founders of the German dye in- 3 3 dustry. Married four times—he was − HO left a widower three times—he had 11 children.

PROBLEM 6 What is the difference between the reaction that occurs when isopropyltrimethylammoni- um hydroxide is heated and the reaction that occurs when 2-bromopropane is treated with hydroxide ion?

The carbon to which the tertiary amine is attached is designated as the a -carbon, so the adjacent carbon, from which the proton is removed, is called the b -carbon. (Recall that E2 reactions are also called b -elimination reactions, since elimination is initiated by removing a proton from the b -carbon; Section 11.1.) If the quaternary ammonium b ion has more than one -carbon, the major alkene product is the one obtained by re- In a Hofmann elimination reaction, the moving a proton from the b -carbon bonded to the greater number of hydrogens. In the hydrogen is removed from the B -carbon following reaction, the major alkene product is obtained by removing a hydrogen from bonded to the most hydrogens. 890 CHAPTER 21 More About Amines • Heterocyclic Compounds

the b -carbon bonded to three hydrogens, and the minor alkene product results from removing a hydrogen from the b -carbon bonded to two hydrogens.

-carbon -carbon

∆ + ++ CH3CHCH2CH2CH3 CH2 CHCH2CH2CH3 CH3CH CHCH2CH3 CH3NCH3 H2O 1-pentene 2-pentene CH CH3NCH+ 3 major product minor product 3 − trimethylamine CH3 HO

In the next reaction, the major alkene product comes from removing a hydrogen from the b -carbon bonded to two hydrogens, because the other b -carbon is bonded to only one hydrogen.

-carbon -carbon CH3 CH3 ∆ + + CH3CHCH2NCH+ 2CH2CH3 CH3CHCH2N CH2 CHCH3 H2O − propene CH3 CH3 HO CH3 CH3 isobutyldimethylamine

PROBLEM 7◆ What are the minor products in the preceding Hofmann elimination reaction?

We saw that in an E2 reaction of an alkyl chloride, alkyl bromide, or alkyl iodide, a hydrogen is removed from the b -carbon bonded to the fewest hydrogens (Zaitsev’s rule; Section 11.2). Now we see that in an E2 reaction of a quaternary ammonium ion, the hydrogen is removed from the b -carbon bonded to the most hydrogens (anti- Zaitsev elimination). Why do alkyl halides follow Zaitsev’s rule, while quaternary amines violate the rule? When hydroxide ion starts to remove a proton from the alkyl bromide, the bromide ion immediately begins to depart and a transition state with an alkene-like structure results. The proton is removed from the b -carbon bonded to the fewest hydrogens in order to achieve the most stable alkene-like transition state.

Zaitsev elimination anti-Zaitsev elimination alkene-like transition state carbanion-like transition state

δ− δ− δ− δ− OH OH OH OH H H H H CH CH CHCH CH CH CCH CH CHCH CH CH CH CHCHCH CH 3 3 3 2 2 δ− 2 2 2 3 3 δ− 2 3 δ− δ− + + Br Br N(CH3)3 N(CH3)3 more stable less stable more stable less stable

When, however, hydroxide ion starts to remove a proton from a quaternary ammonium ion, the leaving group does not immediately begin to leave, because a tertiary amine is not as good a leaving group as Cl-, Br-, or I-. As a result, a partial negative charge builds up on the carbon from which the proton is being removed. This gives the transition state a carbanion-like structure rather than an alkene-like structure. By removing a proton from the b -carbon bonded to the most hydrogens, the most stable carbanion- like transition state is achieved. (Recall from Section 11.2 that primary carbanions are more stable than secondary carbanions, which are more stable than tertiary carbanions.) Steric factors in the Hofmann reaction also favor anti-Zaitsev elimination. Section 21.5 Reactions of Quaternary Ammonium Hydroxides 891

Because the Hofmann elimination reaction occurs in an anti-Zaitsev manner, anti- Zaitsev elimination is also referred to as Hofmann elimination. We have previously seen anti-Zaitsev elimination in the E2 reactions of alkyl fluorides as a result of fluoride ion being a poorer leaving group than chloride, bromide, or iodide ions. As in a Hofmann elimination reaction, the poor leaving group results in a carbanion-like transition state rather than an alkene-like transition state (Section 11.2).

PROBLEM 8◆ Give the major products of each of the following reactions: + H3C N(CH3)3 CH3 + ∆ − ∆ a.CH3CH2CH2NCH3 c. HO − CH3 HO

CH3 H3C ∆ ∆ b. d. + − + − NHO NHO

H3CCH3 H3C CH3

For a quaternary ammonium ion to undergo an elimination reaction, the counterion must be hydroxide ion because a strong base is needed to start the reaction by removing a proton from a b -carbon. Since halide ions are weak bases, quaternary ammonium halides cannot undergo a Hofmann elimination reaction. However, a quaternary ammonium halide can be converted into a quaternary ammonium hydroxide by treating it with silver oxide and water. The silver halide precipitates, and the halide ion is replaced by hydroxide ion. The compound can now undergo an elimination reaction.

R R + + 2 R N R + Ag2O + H2O 2 R N R + 2 AgI − − R I R HO

The reaction of an amine with sufficient methyl iodide to convert the amine into a quaternary ammonium iodide is called exhaustive methylation. (See Chapter 10, Problem 8.) The reaction is carried out in a basic solution of potassium carbonate, so the amines will be predominantly in their basic forms.

exhaustive methylation

CH3 + K2CO3 CH3CH2CH2NH2 + CH3I CH3CH2CH2NCH3 excess − CH3 I

The Hofmann elimination reaction was used by early organic chemists as the last step of a process known as a Hofmann degradation—a method used to identify amines. In a Hofmann degradation, an amine is exhaustively methylated with methyl iodide, treated with silver oxide to convert the quaternary ammonium iodide to a quaternary ammonium hydroxide, and then heated to allow it to undergo a Hofmann elimination. Once the alkene is identified, working backwards gives the structure of the amine. 892 CHAPTER 21 More About Amines • Heterocyclic Compounds

A USEFUL BAD-TASTING added to toxic substances to keep them from being ingested COMPOUND accidentally. ® CH3 Several practical uses have been found for Bitrex , O CH2CH3 + a quaternary ammonium salt, because it is one of the most NHCCH N CH bitter-tasting substances known and is nontoxic. Bitrex® is put 2 2 on bait to encourage deer to look elsewhere for their food, it is CH2CH3 O CH3 put on the backs of animals to keep them from biting one − another, it is put on children’s’ fingers to persuade them to CO stop sucking their thumbs or biting their fingernails, and it is  Bitrex

PROBLEM 9 Identify the amine in each case. a. 4-Methyl-2-pentene is obtained from the Hofmann degradation of a primary amine. b. 2-Methyl-1-3-butadiene is obtained from two successive Hofmann degradations of a secondary amine.

PROBLEM 10 SOLVED Describe a synthesis for each of the following compounds, using the given starting material and any necessary reagents:

a. CH3CH2CH2CH2NH2 CH3CH2CH CH2

b. CH3CH2CH2CHCH3 CH3CH2CH2CH CH2 Br

c. CH CH CH CH N 2 2 H

SOLUTION TO 10a Although an amine cannot undergo an elimination reaction, a quaternary ammonium hydroxide can. The amine, therefore, must first be converted into a quaternary ammonium hydroxide. Reaction with excess methyl iodide converts the amine into a quaternary ammonium iodide, and treatment with aqueous silver oxide forms the quaternary ammonium hydroxide. Heat is required for the elimination reaction.

CH3I + + ∆ excess Ag2O + CH3CH2CH2CH2NH2 CH3CH2CH2CH2N(CH3)3 CH3CH2CH2CH2N(CH3)3 CH3CH2CH CH2 H2O K CO H2O 2 3 − − I HO

21.6 Phase Transfer Catalysis

A problem organic chemists face in the laboratory is finding a solvent that will dis- solve all the reactants needed for a given reaction. For example, if we want cyanide ion to react with 1-bromohexane, we encounter a problem: Sodium cyanide is an ionic compound that is soluble only in water, whereas the alkyl halide is insoluble in water. Therefore, if we mix an aqueous solution of sodium cyanide with a solution of 1-bromohexane in a nonpolar solvent, there will be two distinct phases—an aqueous phase and an organic phase—because the solutions are immiscible. How, then, can sodium cyanide react with the alkyl halide? Section 21.6 Phase Transfer Catalysis 893

− ? − CH3CH2CH2CH2CH2CH2Br + CN CH3CH2CH2CH2CH2CH2CN+ Br 1-bromohexane

The two compounds will be able to react with each other if a catalytic amount of a phase transfer catalyst is added to the reaction mixture.

phase transfer catalyst

+ − − R4N HSO4 − CH3CH2CH2CH2CH2CH2Br + CN CH3CH2CH2CH2CH2CH2CN+ Br

Quaternary ammonium salts are the most common phase transfer catalysts. How- ever, we saw in Section 12.9 that crown ethers can also be used as phase transfer catalysts.

phase transfer catalysts

CH2CH2CH2CH3 CH3 CH2CH3 + + + CH3CH2CH2CH2NCH2CH2CH2CH3 CH3(CH2)14CH2NCH3 CH2NCH2CH3 − − − HSO4 CH2CH2CH2CH3 HSO4 CH3 HSO4 CH2CH3 tetrabutylammonium hexadecyltrimethylammonium benzyltriethylammonium hydrogen sulfate hydrogen sulfate hydrogen sulfate

How does the addition of a phase transfer catalyst allow the reaction of cyanide ion with 1-bromohexane to take place? Because of its nonpolar alkyl groups, the quaternary ammonium salt is soluble in nonpolar solvents, but because of its charge, it is also soluble in water. This means that the quaternary ammonium salt can act as a mediator between the two immiscible phases. When a phase transfer catalyst such as tetrabutylammonium hydrogen sulfate passes into the nonpolar, organic phase, it must carry a counterion with it to balance its positive charge. The counterion can be either its orig- inal counterion (hydrogen sulfate) or another ion that is present in the solution (in the reaction under discussion, it will be cyanide ion). Because there is more cyanide ion than hydrogen sulfate ion in the aqueous phase, cyanide ion will more often be the ac- companying ion. Once in the organic phase, cyanide ion can react with the alkyl halide. (When hydrogen sulfate is transported into the oganic phase, it is unreactive because it is both a weak base and a poor nucleophile.) The quaternary ammonium ion will pass back into the aqueous phase carrying with it either hydrogen sulfate or bromide ion as a counterion. The reaction continues with the phase transfer catalyst shut- tling back and forth between the two phases. Phase transfer catalysis has been successfully used in a wide variety of organic reactions.

+ − + organic R N CN++RBr RCN R N Br− phase 4 4 starting target material compound ++ + − − − aqueous R4N C N Na HSO4 R4N Br phase 894 CHAPTER 21 More About Amines • Heterocyclic Compounds

21.7 Oxidation of Amines; The Cope Elimination Reaction

Amines are easily oxidized, sometimes just by being exposed to air. Amines, therefore, are stored as salts (e.g., as amine hydrochlorides), and drugs that contain amino groups are often sold as salts. Primary amines are oxidized to hydroxlyamines, which in turn are oxidized to nitroso compounds, which are oxidized to nitro compounds. Hydrogen peroxide, peroxy- acids, and other common oxidizing agents are used to oxidize amines. The oxidation reactions generally take place by mechanisms that involve radicals, so they are not well characterized.

O oxidation oxidation oxidation + R NH2 R NH OH R NO R N O− a primary amine a hydroxylamine a nitroso compound a nitro compound

Secondary amines are oxidized to secondary hydroxylamines, and tertiary amines are oxidized to amine oxides.

R R R +++− R NH HO OHR NH HO R N H2O OH OH a secondary amine a secondary hydroxylamine

R R R +++− RRN HO OHRR+N HO RR+N H2O − OH O a tertiary amine an tertiary amine oxide

Amine oxides undergo a reaction similar to the Hofmann elimination reaction, Arthur C. Cope (1909–1966) was called a Cope elimination reaction. In a Cope elimination reaction, a tertiary amine born in Indiana. He received a Ph.D. oxide rather than a quaternary ammonium ion undergoes elimination. The Cope elim- from the University of Wisconsin and ination reaction occurs under milder conditions than does a Hofmann elimination was a professor of chemistry at Bryn reaction. Mawr College, Columbia University, and MIT. CH3 CH3 ∆ + CH3CH2CH2NCH+ 3 CH3CH CH2 NCH3 − O OH a tertiary amine oxide a hydroxylamine

A strong base is not needed for a Cope elimination because the amine oxide acts as its own base. The Cope elimination, therefore, is an intramolecular E2 reaction and involves syn elimination.

mechanism of the Cope elimination reaction

CH3 + CH3 CH N ∆ 2 CH + − 3 CH3CH CH2 N CH3 CH3CH O OH H Section 21.8 Synthesis of Amines 895

The major product of the Cope elimination, like that of the Hofmann elimination, is In a Cope elimination, the hydrogen is the one obtained by removing a hydrogen from the b -carbon bonded to the greater removed from the B -carbon bonded to number of hydrogens. the most hydrogens.

CH3 CH3 + ∆ CH3CH2NCH2CH2CH3 CH2 CH2 + NCH2CH2CH3 − O OH

PROBLEM 11◆ Does the Cope elimination have an alkene-like transition state or a carbanion-like transition state?

PROBLEM 12◆ Give the products that would be obtained by treating the following tertiary amines with hydrogen peroxide followed by heat:

CH3 CH3

a.CH3NCH2CH2CH3 c. CH3CH2NCH2CHCH3

CH3

CH3NCH2CH2CH3

b. d. N CH3

CH3

21.8 Synthesis of Amines

Because ammonia and amines are good nucleophiles, they readily undergo SN2 reactions with alkyl halides. (X denotes a halogen.)

+ + RCH2 X RCH2 X NH3 RCH2 NH3 RCH2 NH2 RCH2 NH2 RCH2 NH + HX a primary amine RCH2 RCH2 + HX a secondary amine

RCH2 X

RCH2 RCH2 RCH2 RCH2 X + RCH2 N CH2R RCH2 N RCH2 NH

RCH2 RCH2 RCH2 a quaternary a tertiary amine ammonium salt + HX

Although these SN2 reactions can be used to synthesize amines, the yields are poor because it is difficult to stop the reaction after a single alkylation since ammonia and primary, secondary, and tertiary amines have similar reactivities. A much better way to prepare a primary amine is by means of a Gabriel synthesis (Section 17.17). This reaction involves alkylating phthalimide and then hydrolyzing the N-substituted phthalimide. 896 CHAPTER 21 More About Amines • Heterocyclic Compounds

Gabriel synthesis O O − + COOH + 1. HO H3O + NH NCHCH CH3CH2NH3 2. CH CH Br 2 3 ∆ 3 2 COOH a protonated − primary amine O O + Br phthalimide

- Primary amines also can be prepared in good yields if azide ion ( N3) is used as the nucleophile in an SN2 reaction. The product of the reaction is an alkyl azide, which can be reduced to a primary amine. (See Chapter 10, Problem 9.)

− N + H CH CH CH CH Br 3 CH CH CH CH NNN 2 CH CH CH CH NH 3 2 2 2 3 2 2 2 Pd/C 3 2 2 2 2 butyl bromide butyl azide butylamine

Other reduction reactions also result in the formation of primary amines. For example, the catalytic reduction of a nitrile forms a primary amine. (Recall that a nitrile can be obtained from the reaction of cyanide ion with an alkyl halide.)

NaC N H CH CH CH CH Br CH CH CH CH CN 2 CH CH CH CH NH 3 2 2 2 HCl 3 2 2 2 Pd/C 3 2 2 2 2 butyl bromide pentanenitrile pentylamine

Amines are obtained from the reduction of amides with LiAlH4 (Sections 18.5 and 20.1). This method can be used to synthesize primary, secondary, and tertiary amines. The class of amine obtained depends on the number of substituents on the nitrogen atom of the amide.

1. LiAlH4 C RCH2NH2 2. H O RNH2 2 a primary amine

1. LiAlH4 C RCH2NHCH3 2. H O R NHCH3 2 a secondary amine

1. LiAlH4 C RCH2NCH3 2. H O R NCH3 2 CH3 CH3 a tertiary amine

A primary amine can be obtained from the reaction of an aldehyde or a ketone with excess ammonia in the presence of H2 and Raney nickel. Because the imine does not have a substituent other than a hydrogen bonded to the nitrogen, it is relatively unstable, so the amine is obtained by adding H2 to the C “ N bond as it is formed. This is called reductive amination.

CH3CH2 CH3CH2 CH3CH2 H C ONH+ CNH 2 CHNH 3 Pd/C 2 excess CH3CH2 CH3CH2 CH3CH2 unstable

Secondary and tertiary amines can be prepared from imines and enamines by reducing the imine or enamine. Sodium triacetoxyborohydride is a commonly used reducing agent for this reaction. Section 21.9 Aromatic Five-Membered-Ring Heterocycles 897

O catalytic + + H NaBH(OCCH3)3 CHOCHCH CH3CH2NH2 NCH2CH3 2NHCH2CH3

an imine

O catalytic + CH3 CH3 + H NaBH(OCCH3)3 ONCH3NH N CH3 CH3 CH3 an enamine

A primary amine is obtained from the reduction of a nitroalkane, and an arylamine is obtained from the reduction of nitrobenzene.

+ Pd/C CH3CH2NO2 H2 CH3CH2NH2 nitroethane ethylamine

+ Pd/C NO2 H2 NH2

nitrobenzene aniline

PROBLEM 13◆ Excess ammonia must be used when a primary amine is synthesized by reductive amination. What product will be obtained if the reaction is carried out with an excess of the car- bonyl compound instead?

21.9 Aromatic Five-Membered-Ring Heterocycles Pyrrole, Furan, and Thiophene Pyrrole, furan, and thiophene are five-membered-ring heterocycles. Each has three 3-D Molecules: pairs of delocalized p electrons: Two of the pairs are shown as p bonds, and one pair Pyrrole; is shown as a lone pair on the heteroatom. Furan and thiophene have a second lone pair Furan; that is not part of the p cloud. These electrons are in an sp2 orbital perpendicular to the Thiophene p orbitals. Pyrrole, furan, and thiophene are aromatic because they are cyclic and planar, every carbon in the ring has a p orbital, and the p cloud contains three pairs of p electrons (Sections 15.1 and 15.3).

N OS H pyrrole furan thiophene

these electrons these are part of electrons the cloud are part of the cloud these N H O electrons are in an sp2 orbital perpendicular to the p orbitals

orbital structure of pyrrole orbital structure of furan 898 CHAPTER 21 More About Amines • Heterocyclic Compounds

Pyrrole is an extremely weak base because the electrons shown as a lone pair are part of the p cloud. Therefore, when pyrrole is protonated, its aromaticity is destroyed. Consequently, the conjugate acid of pyrrole is a very strong acid 1pKa =-3.82; that is, it has a strong tendency to lose a proton. The resonance contributors of pyrrole show that nitrogen donates the electrons de- picted as a lone pair into the five-membered ring.

− −

− − N +N + N +N +N HH HH H resonance contributors of pyrrole

δ− δ− δ− δ− δ+N H resonance hybrid

Pyrrolidine—a saturated five-membered-ring heterocyclic amine—has a dipole moment of 1.57 D because the nitrogen atom is electron withdrawing. Pyrrole—an unsat- urated five-membered-ring heterocyclic amine—has a slightly larger dipole moment (1.80 D), but as we see from the electrostatic potential maps, the two dipole moments are in opposite directions. (The red areas are on opposite sides of the two molecules.) Apparently, the ability of pyrrole’s nitrogen to donate electrons into the ring by resonance more than makes up for its inductive electron withdrawal (Section 16.3).

N N H H = 1.57 D = 1.80 D

pyrrolidine pyrrole In Section 7.6, we saw that the more stable and more nearly equivalent the resonance contributors, the greater is the resonance energy. The resonance energies of pyrrole, furan, and thiophene are not as great as the resonance energies of benzene and the cyclopentadienyl anion, compounds for which the resonance contributors are all equivalent. Thiophene, with the least electronegative heteroatom, has the greatest resonance energy of these five-membered heterocycles; and furan, with the most electronegative heteroatom, has the smallest resonance energy.

relative resonance energies of some aromatic compounds

>>>> − SNO H

Pyrrole, furan, and thiophene undergo Because pyrrole, furan, and thiophene are aromatic, they undergo electrophilic aro- electrophilic substitution preferentially matic substitution reactions. at C-2. mechanism for electrophilic aromatic substitution

+ + slow fast + + Y H + H NNN Y H H Y H Section 21.9 Aromatic Five-Membered-Ring Heterocycles 899

+ + Br2 HBr O O Br 2-bromofuran

+ (CH3CO)2O + HNO3 H2O N CH3 O2N N CH3 H H 2-methyl-5-nitropyrrole

Substitution occurs preferentially at C-2 because the intermediate obtained by attaching a substituent at this position is more stable than the intermediate obtained by attaching a substituent at C-3 (Figure 21.1). Both intermediates have a relatively stable resonance contributor in which all the atoms (except H) have complete octets. The Pyrrole, furan, and thiophene are more intermediate resulting from C-2 substitution of pyrrole has two additional resonance reactive than benzene toward elec- contributors, each with a positive charge on a secondary allylic carbon. The intermedi- trophilic aromatic substitution. ate resulting from C-3 substitution, however, has only one additional resonance contributor, which has a positive charge on a secondary carbon. This resonance contributor is further destabilized by being adjacent to an electron-withdrawing nitrogen atom, so its predicted stability is less than that of a resonance contributor with a positive charge on a secondary allylic carbon. If both positions adjacent to the heteroatom are occupied, electrophilic substitution will take place at C-3.

+ + Br2 HBr pyrrole H3C O CH3 H3C O CH3 3-bromo-2,5-dimethylfuran

Pyrrole, furan, and thiophene are all more reactive than benzene toward electrophilic substitution because they are better able to stabilize the positive charge on the carbocation intermediate, since the lone pair on the hetereoatom can donate electrons into the ring by resonance (Figure 21.1).

relative reactivity toward electrophilic aromatic substitution furan

> > > N O S H pyrrole furan thiophene benzene

Furan is not as reactive as pyrrole in electrophilic aromatic substitution reactions. The oxygen of furan is more electronegative than the nitrogen of pyrrole, so the oxy- thiophene gen is not as effective as nitrogen in stabilizing the carbocation. Thiophene is less reactive than furan toward electrophilic substitution because sulfur’s p electrons are in a 3p orbital, which overlaps less effectively than the 2p orbital of nitrogen or oxygen with the 2p orbital of carbon. The electrostatic potential maps illustrate the different electron densities of the three rings.

+ > Figure 21.1 2-position YY+ Y Structures of the intermediates that N N N can be formed from the reaction of HHH + H H H an electrophile with pyrrole at C-2 + Y+ Y Y and C-3. N H H H 3-position + N + N H H 900 CHAPTER 21 More About Amines • Heterocyclic Compounds

The relative reactivities of the five-membered-ring heterocycles are reflected in the Lewis acid required to catalyze a Friedel–Crafts acylation reaction (Section 15.13). Benzene requires AlCl3 , a relatively strong Lewis acid. Thiophene is more reactive than benzene, so it can undergo a Friedel–Crafts reaction using SnCl4 , a weaker Lewis acid. An even weaker Lewis acid,BF3 , can be used when the substrate is furan. Pyr- role is so reactive that an anhydride is used instead of a more reactive acyl chloride, and no catalyst is necessary.

O O CCH3 + 1. AlCl3 + CH3CCl HCl 2. H2O phenylethanone

O + 1. SnCl4 + CH3CCl HCl S 2. H2O S CCH3 O 2-acetylthiophene

O + 1. BF3 + CH3CCl HCl O 2. H2O O CCH3 O 2-acetylfuran

O O O + + CH3COCCH3 CH3COH N N CCH3 H H O 2-acetylpyrrole

The resonance hybrid of pyrrole indicates that there is a partial positive charge on the nitrogen. Therefore, pyrrole is protonated on C-2 rather than on nitrogen. Remem- ber, a proton is an electrophile and, like other electrophiles, attaches to the C-2 position of pyrrole.

+ + H H N + N H H H

pKa =−3.8

Pyrrole is unstable in strongly acidic solutions because once protonated, it can readily polymerize.

H H H polymer N +N +N N H H H H H H ' Pyrrole is more acidic 1pKa = 172 than the analogous saturated amine ' 2 1pKa = 362, because the nitrogen in pyrrole is sp hybridized and is, therefore, more electronegative than the sp3 nitrogen of a saturated amine (Table 21.1). Pyrrole’s acidity also is increased as a result of its conjugate base being stabilized by electron Section 21.9 Aromatic Five-Membered-Ring Heterocycles 901

TABLE 21.1 The pKa Values of Several Nitrogen Heterocycles H H + H N HN N

+N H +N +N N N N + N + H H H H H H =− =− = = = = pKa 3.8 pKa 2.4 pKa 1.0 pKa 2.5 pKa 4.85 pKa 5.16 Tutorial: Basic sites of nitrogen heterocycles HN NH NNH + N N N N + + H H HHHH pKa = 6.8pKa = 8.0pKa = 11.1 pKa = 14.4 pKa =∼17 pKa =∼36

delocalization. (Recall that the more stable the base, the stronger is its conjugate acid; Section 1.17).

+ + + H + H N N N N H − H − = pKa ~17 pKa = ~36

PROBLEM 14

When pyrrole is added to a dilute solution of D2SO4 in D2O, 2-deuteriopyrrole is formed. Propose a mechanism to account for the formation of this compound.

PROBLEM 15 Use resonance contributors to explain why pyrrole is protonated on C-2 rather than on nitrogen.

PROBLEM 16 ' Explain why pyrrole 1pKa 172 is less acidic than cyclopentadiene 1pKa = 152, even though nitrogen is considerably more electronegative than carbon.

Indole, Benzofuran, and Benzothiophene Indole, benzofuran, and benzothiophene contain a five-membered aromatic ring fused to a benzene ring. The rings are numbered in a way that gives the heteroatom the lowest possible number. Indole, benzofuran, and benzothiophene are aromatic because they are cyclic and planar, every carbon in the ring has a p orbital, and the p cloud of 3-D Molecules: each compound contains five pairs of p electrons (Section 15.1). Notice that the elec- Indole; trons shown as a lone pair on the indole nitrogen are part of the p cloud; therefore, the Benzofuran; conjugate acid of indole, like the conjugate acid of pyrrole, is a strong acid Benzothiophene 1pKa =-2.42. In other words, indole is an extremely weak base.

4 3 5 2 6 N1 OS 7 H indole benzofuran benzothiophene 902 CHAPTER 21 More About Amines • Heterocyclic Compounds

21.10 Aromatic Six-Membered-Ring Heterocycles Pyridine When one of the carbons of a benzene ring is replaced by a nitrogen, the resulting compound is called pyridine.

these electrons are 2 3-D Molecule: N in an sp orbital Pyridine perpendicular 4 to the p orbitals 5 3

6 2 N 1 pyridine orbital structure of pyridine

The pyridinium ion is a stronger acid than a typical ammonium ion because the acidic hydrogen of a pyridinium ion is attached to an sp2 hybridized nitrogen, which is more electronegative than an sp3 hybridized nitrogen (Section 6.9).

+ + H sp2 N + N H pyridine pyridinium ion pKa = 5.16

+ Tutorial: + H Lone-pair electrons on 3 sp N+ N nitrogen heterocycles H H H piperidine piperidinium ion pKa = 11.12

Pyridine is a tertiary amine, so it undergoes reactions characteristic of tertiary amines. For example, pyridine undergoes SN2 reactions with alkyl halides (Section 10.4), and it reacts with hydrogen peroxide to form an N-oxide (Section 21.7).

+ CH3 I − N +N I

CH3 N-methylpyridinium iodide

+ + HO OH H2O N +N +N − − OH+ HO O pKa = 0.79 pyridine-N-oxide

PROBLEM 17 SOLVED Will an amide be formed from the reaction of an acyl chloride with an aqueous solution of pyridine? Explain your answer. Section 21.10 Aromatic Six-Membered-Ring Heterocycles 903

SOLUTION An amide will not be formed because the positively charged nitrogen causes pyridine to be an excellent leaving group. Therefore, the final product of the reaction will be a carboxylic acid. (If the final pH of the solution is greater than the pKa of the carboxylic acid, the carboxylic acid will be predominantly in its basic form.) O O O + H O − RCCl + RC N 2 RCO + N N

Pyridine is aromatic. Like benzene, it has two uncharged resonance contributors. Because of the electron-withdrawing nitrogen, it also has three charged resonance contributors that benzene does not have.

+ + N N N N N −−− benzene resonance contributors of pyridine

The dipole moment of pyridine is 1.57 D. As the resonance contributors and the electrostatic potential map indicate, the electron-withdrawing nitrogen is the negative end of the dipole.

N = 1.57 D pyridine

Because it is aromatic, pyridine (like benzene) undergoes electrophilic aromatic substitution reactions (≠B is any base in the solution).

mechanism for electrophilic aromatic substitution B + H Y + slow fast + + Y Y + BH NNN

Electrophilic aromatic substitution of pyridine takes place at C-3 because the most sta- Pyridine undergoes electrophilic ble intermediate is obtained by placing an electrophilic substituent at that position aromatic substitution at C-3. (Figure 21.2). When the substituent is placed at C-2 or C-4, one of the resulting resonance contributors is particularly unstable because its nitrogen atom has an incomplete octet and a positive charge. The electron-withdrawing nitrogen atom makes the intermediate obtained from electrophilic aromatic substitution of pyridine less stable than the carbocation intermediate obtained from electrophilic aromatic substitution of benzene. Pyridine, therefore, is less reactive than benzene. Indeed, it is even less reactive than nitrobenzene. (Recall from Section 16.3 that an electron-withdrawing nitro group strongly deactivates a benzene ring toward electrophilic aromatic substitution.)

relative reactivity toward electrophilic aromatic substitution

NO2 NO2

>>>

N NO2 904 CHAPTER 21 More About Amines • Heterocyclic Compounds

Figure 21.2 N + + Structures of the intermediates that 2-position + Y Y Y can be formed from the reaction of N N N an electrophile with pyridine. H H H least stable + Y Y Y + Y+ 3-position H H H N + + N N N

Y H Y H Y H + + 4-position + N N N least stable

Pyridine, therefore, undergoes electrophilic aromatic substitution reactions only under vigorous conditions, and the yields of these reactions are often quite low. If the nitrogen becomes protonated under the reaction conditions, the reactivity is further decreased because a positively charged nitrogen is more electron withdrawing than a neutral nitrogen.

Br FeBr + Br 3 + HBr 2 300 °C N N 3-bromopyridine 30%

SO3H + 230 °C + H2SO4 H2O N N pyridine-3-sulfonic acid 71%

NO2 H SO + HNO 2 4 + H O 3 300 °C 2 N N 3-nitropyridine 22% We have seen that highly deactivated benzene rings do not undergo Friedel–Crafts alkylation or acylation reactions. Therefore, pyridine, whose reactivity is similar to that of a highly deactivated benzene, does not undergo these reactions either.

+ AlCl3 CH3CH2Cl no electrophilic aromatic substitution reaction N

PROBLEM 18 Give the product of the following reaction: O + CH3OH CH3CCl N

Since pyridine is less reactive than benzene toward electrophilic aromatic substitution, it is not surprising that pyridine is more reactive than benzene toward nucleophili- c aromatic substitution. The electron-withdrawing nitrogen atom that destabilizes the intermediate in electrophilic aromatic substitution stabilizes it in nucleophilic aromatic substitution. Section 21.10 Aromatic Six-Membered-Ring Heterocycles 905

mechanism for nucleophilic aromatic substitution

− slow fast − + Y Y + Z N Z − N Z N Y

Nucleophilic aromatic substitution of pyridine takes place at C-2 and C-4, because at- Pyridine undergoes nucleophilic tack at these positions leads to the most stable intermediate. Only when nucleophilic aromatic substitution at C-2 and C-4. attack occurs at these positions is a resonance contributor obtained that has the greatest electron density on nitrogen, the most electronegative of the ring atoms (Figure 21.3). − − > Figure 21.3 2-position Structures of the intermediates that Y Y Y can be formed from the reaction of − N N N H H H a nucleophile with pyridine. most stable − Y Y Y + Y− 3-position H H H N − − N N N

Y H Y H Y H − − 4-position

N −N N most stable If the leaving groups at C-2 and C-4 are different, the incoming nucleophile will preferentially substitute for the weaker base (the better leaving group).

Br NH2 ∆ Pyridine is less reactive than benzene to- + − + − NH2 Br ward electrophilic aromatic substitution and more reactive than benzene toward N OCH3 N OCH3 nucleophilic aromatic substitution.

CH3 CH3 ∆ + − + − CH3O Cl

N Cl N OCH3

PROBLEM 19 Compare the mechanisms of the following reactions:

Cl NH2

+ − + − NH2 Cl N N

Cl NH2

+ − + − NH2 Cl

PROBLEM 20 a. Propose a mechanism for the following reaction:

KOH/H2O ∆ N N O H b. What other product is formed? 906 CHAPTER 21 More About Amines • Heterocyclic Compounds

Substituted pyridines undergo many of the side-chain reactions that substituted benzenes undergo, such as bromination and oxidation.

NBS ∆/peroxide N CH2CH3 N CHCH3 Br

CH3 COOH Na2Cr2O7 H2SO4 ∆ N N

When 2- or 4-aminopyridine is diazotized,a-pyridone or g-pyridone is formed. Apparently, the diazonium salt reacts immediately with water to form a hydroxypyridine (Section 16.10). The product of the reaction is a pyridone because the keto form of a hydroxypyridine is more stable than the enol form. (The mechanism for the con- version of a primary amino group into a diazonium group is shown in Section 16.12).

NaNO2, HCl H2O ° + 0 C − N NH N N N Cl N OH N O 2 H 2-aminopyridine 2-hydroxypyridine enol form -pyridone keto form + − NH2 N N Cl OH O

NaNO2, HCl H2O 0 °C N N N N 4-aminopyridine 4-hydroxypyridine H enol form -pyridone keto form

The electron-withdrawing nitrogen causes the a-hydrogens of alkyl groups attached to the 2- and 4-positions of the pyridine ring to have about the same acidity as the a-hydrogens of ketones (Section 19.1).

− − CH3 CH2 CH2 CH2 CH2 CH2

HO− − −

N N N N − N N

Consequently, the a-hydrogens of alkyl substituents can be removed by base, and the resulting carbanions can react as nucleophiles.

− CH3 CH2 O CH CH CH HO− + H O an aldol condensation 2 N N N CH CH CH 3 − 3 3 NH CH Br − 2 3 + Br an SN2 reaction N CH3 N CH− 2 N CH2CH3 Section 21.11 Biologically Important Heterocycles 907

PROBLEM 21◆ Rank the following compounds in order of decreasing ease of removing a proton from a methyl group:

CH3 CH3 CH3

N N + N − I CH2CH3

Quinoline and Isoquinoline Quinoline and isoquinoline are known as benzopyridines because they have both a benzene ring and a pyridine ring. Like benzene and pyridine, they are aromatic com- 3-D Molecules: pounds. The pKa values of their conjugate acids are similar to the pKa of the conjugate acid of pyridine. (In order for the carbons in quinoline and isoquinoline to have the Quinoline; Isoquinoline same numbers, the nitrogen in isoquinoline is assigned the 2-position, not the lowest possible number.)

54 6 3 + + H 7 2 N+ N H 8 1 pKa = 4.85 quinoline

5 4 6 3 + H+ NH 7 N 2 + 8 1 pK = 5.14 a isoquinoline

21.11 Biologically Important Heterocycles

Proteins are naturally occurring polymers of a-amino acids (Chapter 23). Three of the 20 most common naturally occurring amino acids contain heterocyclic rings: Proline contains a pyrrolidine ring, tryptophan contains an indole ring, and histidine contains an imidazole ring.

− − CH2CHCOO CH2CHCOO + NH3 + − NH3 N COO NNH + N HH H proline tryptophan histidine

Imidazole Imidazole, the heterocyclic ring of histidine, is the first heterocyclic compound we have encountered that has two heteroatoms. Imidazole is an aromatic compound because it is cyclic and planar, every carbon in the ring has a p orbital, and the p cloud Tutorial: contains three pairs of p electrons (Section 15.1). The electrons drawn as lone-pair Recognizing common electrons on N-1 (see p. 896) are part of the p cloud because they are in a p orbital, heterocyclic rings in complex whereas the lone-pair electrons on N-3 are not part of the p cloud because they are in molecules an sp2 orbital, perpendicular to the p orbitals. 908 CHAPTER 21 More About Amines • Heterocyclic Compounds

these electrons these electrons are part of are in an sp2 orbital the cloud perpendicular to the p orbitals N N H

orbital structure of imidazole The resonance energy of imidazole is 14 kcal>mol (59 kJ>mol), significantly less than the resonance energy of benzene (36 kcal>mol or 151 kJ>mol ).

− − 45

NNH NNH NNH NNH −NNH 3 1 + + + + 2 − resonance contributors of imidazole

Imidazole can be protonated because the lone-pair electrons in the sp2 orbital are not part of the p cloud. Since the conjugate acid of imidazole has a pKa of 6.8, imidazole exists in both the protonated and unprotonated forms at physiological pH 17.32. This is one of the reasons that histidine, the imidazole-containing amino acid, is an important catalytic component of many enzymes (Section 24.9).

+ + HN+ NH NNHH

pKa = 6.8

Neutral imidazole is a stronger acid 1pK = 14.42 than neutral pyrrole ' a 1pKa 172 because of the second ring nitrogen.

− + NNH NN+ H

pKa = 14.4

Notice that both protonated imidazole and the imidazole anion have two equivalent resonance contributors. This means that the two nitrogens become equivalent when imidazole is either protonated or deprotonated.

protonated imidazole imidazole anion

−− HN+ NH HN NH+ NN NN

δ+ δ+ δ− δ− HN NH N N

resonance hybrid resonance hybrid

PROBLEM 22◆ Give the major product of the following reaction: FeBr + 3 N NCH3 Br2

PROBLEM 23◆ List imidazole, pyrrole, and benzene in order of decreasing reactivity toward electrophilic aromatic substitution. Section 21.11 Biologically Important Heterocycles 909

PROBLEM 24◆ Imidazole boils at 257 °C, whereas N-methylimidazole boils at 199 °C. Explain this difference in boiling points.

PROBLEM 25◆ What percent of imidazole will be protonated at physiological pH 17.32?

Purine and Pyrimidine Nucleic acids (DNA and RNA) contain substituted purines and substituted pyrimidines (Section 27.1); DNA contains A, G, C, and T, and RNA contains A, G, C, and U. (Why DNA contains T instead of U is explained in Section 27.14.) Unsubsti- tuted purine and pyrimidine are not found in nature. Notice that hydroxypurines and hydroxypyrimidines are more stable in the keto form. We will see that the preference for the keto form is crucial for proper base pairing in DNA (Section 27.7).

6 7 4 5 N 1 3 5 N 8 N 2 26 N 49N N 3 H 1 purine pyrimidine

NH2 O NH2 O O N N CH3 N HN N HN HN

N N H N N N O N O N O N H 2 H H H H adenine guanine cytosine uracil thymine Porphyrin Substituted porphyrins are important naturally occurring heterocyclic compounds. A porphyrin ring system consists of four pyrrole rings joined by one-carbon bridges. Heme, which is found in hemoglobin and myoglobin, contains an iron ion (Fe2+) ligated by the four nitrogens of a porphyrin ring system. Ligation is the sharing of nonbonding electrons with a metal ion. The porphyrin ring system of heme is known as protoporphyrin IX; the ring system plus the iron atom is called iron protoporphyrin IX.

− − OOCCH2CH2 CH2CH2COO H H C C C C 3 C C C CCH3 NHN C NNC HC FeII CH NH N C N NC H2C C C CCH3 H C C a porphyrin ring system C C C H CH3 HC CH2 iron protoporphyrin IX heme

Hemoglobin is responsible for transporting oxygen to cells and carbon dioxide away from cells, whereas myoglobin is responsible for storing oxygen in cells. Hemo- globin has four polypeptide chains and four heme groups; myoglobin has one polypeptide chain and one heme group. The iron atoms in hemoglobin and myoglobin, 3-D Molecule: Heme in addition to being ligated to the four nitrogens of the porphyrin ring, are also ligated to a histidine of the protein component (globin), and the sixth ligand is oxygen or carbon dioxide. Carbon monoxide is about the same size and shape as O2 , but CO binds 910 CHAPTER 21 More About Amines • Heterocyclic Compounds

2+ more tightly than O2 to Fe . Consequently, breathing carbon monoxide can be fatal because it prevents the transport of oxygen in the bloodstream. The extensive conjugated system of porphyrin gives blood its characteristic red color. Its high molar absorptivity (about 160,000) allows concentrations as low as 1 * 10-8 M to be detected by UV spectroscopy (Section 8.10). The biosynthesis of porphyrin involves the formation of porphobilinogen from two molecules of d -aminolevulinic acid. The precise mechanism for this biosynthesis is as yet unknown. A possible mechanism starts with the formation of an imine between the enzyme that catalyzes the reaction and one of the molecules of d -aminolevulinic acid. An aldol-type condensation occurs between the imine and a free molecule of d - aminolevulinic acid. Nucleophilic attack by the amino group on the imine closes the ring. The enzyme is then eliminated, and removal of a proton creates the aromatic ring. a mechanism for the biosynthesis of porphyrin − − − − OOCCH2 OOCCH2 OOCCH2 CH2COO − CH2 CH2 HC CH2 base E NH2 O C EEN C N C C O enzyme CH2 CH2 CH2 CH2 + NH2 NH2 H2O NH2 NH2 -aminolevulinic acid

− − − − − − OOCCH2 CH2CH2COO OOCCH2 CH2CH2COO OOCCH2 CH2CH2COO E +NH2 C C H C C OH + N H E NH C CH2 E N CCH2 H2NCH2 H + −H+ H2NCH2 NH2 H2NCH2 NH2 + H2O

− − − − OOCCH2 CH2CH2COO OOCCH2 CH2CH2COO E NH2

N+ N H2NCH2 H H H2NCH2 H porphobilinogen

Four porphobilinogen molecules react to form porphyrin.

H + + + + NH3 N N N + N H3NCH2 H H3N CH2 H H3NCH2 H CH2 H

repeat three more times using an N intramolecular reaction for H + the third repetition N N NH HN H3NCH2 H CH2 H N H subsequent oxidation increases the unsaturation

porphyrin Section 21.11 Biologically Important Heterocycles 911

The ring system in chlorophyll a, the substance responsible for the green color of plants, is similar to porphyrin but contains a cyclopentanone ring, and one of its pyrrole 3-D Molecule: Chlorophyll a rings is partially reduced. The metal atom in chlorophyll a is magnesium (Mg2+)

CH2

HC CH3

H C CH CH 3 NN 2 3 MgII

H3C N N CH3 H H O C O O OCH3 O

chlorophyll a

Vitamin B12 also has a ring system similar to porphyrin, but one of the methine bridges is missing. The ring system of vitamin B12 is known as a corrin ring system. 3+ The metal atom in vitamin B12 is cobalt (Co ). The chemistry of vitamin B12 is discussed in Section 25.7.

NH2 O H H3C O H3C NH2 H N H 2 O N H H2N CN H C III CH O 3 N Co N 3 H3C N CH H 3 H 3-D Molecule: H2N Vitamin B12 CH3 O H CH3 O N CH3 HN NH2 H O N OH CH3 H3C O O H P O −O O CH2OH vitamin B12

PROBLEM 26◆ Is porphyrin aromatic?

PROBLEM 27 Show how the last two porphobilinogen molecules are incorporated into the porphyrin ring. 912 CHAPTER 21 More About Amines • Heterocyclic Compounds

PORPHYRIN, BILIRUBIN, AND JAUNDICE reduced to bilirubin, a yellow compound. If more bilirubin is formed than can be excreted by the liver, bilirubin accumulates The average human turns over about 6 g of hemo- in the blood. When the concentration of bilirubin in the blood globin each day. The protein portion (globin) and the iron are reaches a certain level, bilirubin diffuses into the tissues, caus- reutilized, but the porphyrin ring is broken down. First it is re- ing them to become yellow. This condition is known as duced to biliverdin, a green compound, which is subsequently jaundice.

Summary

Amines are compounds in which one or more of the hydro- Quaternary ammonium salts are the most common phase gens of ammonia have been replaced by R groups. Amines are transfer catalysts. classified as primary, secondary, or tertiary, depending on Primary amines can be synthesized by means of a Gabriel whether one, two, or three hydrogens of ammonia have been synthesis, by reduction of an alkyl azide or a nitrile, by replaced. Amines undergo amine inversion through a transi- reductive amination, and by reduction of an amide. tion state in which the sp3 nitrogen becomes an sp2 nitrogen. Pyrrole, furan, and thiophene are aromatic compounds Some amines are heterocyclic compounds—cyclic that undergo electrophilic aromatic substitution reactions compounds in which one or more of the atoms of the ring is preferentially at C-2. These compounds are more reactive an atom other than carbon. Heterocyclic rings are numbered than benzene toward electrophiles. When pyrrole is protonat- so that the heteroatom has the lowest possible number. A ed, its aromaticity is destroyed. Pyrrole polymerizes in strong- natural product is a compound synthesized by a plant or ly acidic solutions. Indole, benzofuran, and benzothiophene an animal. Alkaloids are natural products containing one or are aromatic compounds that contain a five-membered aro- more nitrogen heteroatoms and are found in the leaves, matic ring fused to a benzene ring. bark, roots, or seeds of plants. Replacing one of benzene’s carbons with a nitrogen forms Because of the lone pair on the nitrogen, amines are both pyridine, an aromatic compound that undergoes electrophilic bases and nucleophiles. Amines react as nucleophiles in nu- aromatic substitution reactions at C-3 and nucleophilic aro- cleophilic substitution reactions, in nucleophilic acyl sub- matic substitution reactions at C-2 and C-4. Pyridine is less stitution reactions, in nucleophilic addition–elimination reactive than benzene toward electrophilic aromatic substitu- reactions, and in conjugate addition reactions. tion and more reactive toward nucleophilic aromatic sub- Amines cannot undergo the substitution and elimination stitution. Quinoline and isoquinoline are aromatic com- reactions that alkyl halides undergo, because the leaving pounds with both a benzene ring and a pyridine ring. groups of amines are too basic. Protonated amines also can- Imidazole is the heterocyclic ring of the amino acid his- not undergo the reactions that protonated alcohols and pro- tidine. The conjugate acid of imidazole has a pKa of 6.8, tonated ethers undergo. Amines are easily oxidized. allowing it to exist in both the protonated and unprotonated Saturated heterocycles containing five or more atoms have forms at physiological pH 1pH = 7.32. Nucleic acids physical and chemical properties typical of acyclic com- (DNA and RNA) contain substituted purines and substitut- pounds that contain the same heteroatom. ed pyrimidines. Hydroxypurines and hydroxypyrimidines Quaternary ammonium hydroxides and amine oxides are more stable in the keto form. A porphyrin ring system undergo E2 elimination reactions known as Hofmann elim- consists of four pyrrole rings joined by one-carbon bridges; ination reactions and Cope elimination reactions, respec- in hemoglobin and myoglobin, the four nitrogen atoms are tively. In both reactions, the proton from the b -carbon ligated to Fe2+. The metal atom in chlorophyll a is Mg2+ 2+ bonded to the greater number of hydrogens is removed. and the metal atom in vitamin B12 is Co .

Summary of Reactions

1. Reaction of amines as nucleophiles (Section 21.4). a. In alkylation reactions: R R R + + + ′ RBr ′ – ′ RBr – ′ RBr ′ R NH2 R NH2 Br R NH R NH Br R N R N R R R R R R Br– + HBr + HBr Summary of Reactions 913

b. In acylation reactions: O O ++′ ′ + – C 2 R NH2 C R NH3 Cl R Cl R NHR′

c. In nucleophilic addition–elimination reactions: i Reaction of a primary amine with an aldehyde or ketone to form an imine: catalytic R + R ++H O RRNH2 NCH2 H2O R R ii Reaction of a secondary amine with an aldehyde or ketone to form an enamine: R catalytic RR + ++H ONRNH H2O R R RR d. In conjugate addition reactions: O O + ′ RCH CHCR R NH2 RCH CH2CR NHR′

2. Primary arylamines react with nitrous acid to form stable arenediazonium salts (Section 21.4). + HCl – NH2 NNCl NaNO2

3. Oxidation of amines: Primary amines are oxidized to nitro compounds, secondary amines to hydroxylamines, and tertiary amines to amine oxides (Section 21.7). O oxidation oxidation oxidation R NH2 R NH OH R NO R N − O R R oxidation + R NH R NH2O OH a secondary amine a secondary hydroxylamine

R R oxidation + RRN RR+NH2O − O a tertiary amine a tertiary amine oxide

4. Elimination reactions of quaternary ammonium hydroxides or tertiary amine oxides (Sections 21.5 and 21.7).

CH3 CH3 + ∆ + + RCH2CH2NCH3 Hofmann RCH CH2 NCH3 H2O − elimination HO CH3 CH3

CH3 CH3 CH3 + ∆ H2O2 + RCH2CH2NCH3 RCH2CH2NCH3 Cope RCH CH2 NCH3 − O elimination OH in both eliminations, the proton is removed from the -carbon bonded to the most hydrogens 914 CHAPTER 21 More About Amines • Heterocyclic Compounds

5. Synthesis of amines (Section 21.8). a. Gabriel synthesis of primary amines:

O O − + COOH 1. HO H O + NH NR 3 + R NH 2. R Br ∆ 3 COOH O O + Br–

b. Reduction of an alkyl azide or a nitrile:

− + N3 – H2 RBr RNNN RNH2 Pd/C H RCN 2 RCHNH Pd/C 2 2

c. Reduction of a nitroalkane or nitrobenzene:

+ Pd/C CH3CH2CH2NO2 H2 CH3CH2CH2NH2

+ Pd/C NO2 H2 NH2

d. Aldehydes and ketones react (1) with excess ammonia plus H2>metal catalyst to form primary amines, (2) with a primary amine followed by reduction with sodium triacetoxyborohydride to form secondary amines, and (3) with a secondary amine followed by reduction with sodium triacetoxyborohydride to form tertiary amines:

R R excess NH3 + C O CH NH2 H2O H2, Pd/C R R O R catalytic R R H+ + NaBH(OCCH3)3 C O CH3NH2 C NCH3 CH NHCH3 R R R

O catalytic CH3 CH3 H+ + NaBH(OCCH3)3 O CH3NH N N CH CH CH3 3 3

6. Electrophilic aromatic substitution reactions. a. Pyrrole, furan, and thiophene (Section 21.9):

+ (CH3CO)2O + HNO3 H2O N N NO2 H H

+ + Br2 HBr OOBr

O 1. SnCl + CH CCl 4 + HCl 3 2. H O S 2 S CCH3 O Problems 915

b. Pyridine (Section 21.10): Br + FeBr3 + Br2 300 °C HBr N N Br Br

+ FeBr3 PCl3 Br2 90 °C +N +N N − − O O 7. Nucleophilic aromatic substitution reactions of pyridine (Section 21.10). ∆ + − + − NH2 Cl NNCl NH2

Br NH2

∆ ++− − NH2 Br N N

Cl NH2

N ∆ N + − + − NH2 Cl N N

AU: These terms are bold in text. Add to list? Key Terms Or make lightface in text? alkaloid (p. 884) Hofmann elimination reaction (p. 889) purine (p. 909) amine inversion (p. 885) imidazole (p. 907) pyridine (p. 902) amines (p. 883) iron protoporphyrin IX (p. 909) pyrimidine (p. 909) Cope elimination reaction (p. 894) ligation (p. 909) pyrrole (p. 897) exhaustive methylation (p. 891) natural product (p. 884) quaternary ammonium ion (p. 889) furan (p. 897) phase transfer catalysis (p. 893) reductive amination (p. 896) heteroatom (p. 884) phase transfer catalyst (p. 893) thiophene (p. 897) heterocycle (p. 884) porphyrin ring system (p. 909) heterocyclic compound (p. 884) protoporphyrin IX (p. 909)

Problems

28. Name the following compounds: CH Cl CH NH 3 3 a. b. c. d.

N CH3 N CH CH N CH3 H H 3 2 H 29. Give the product of each of the following reactions:

O − 1. C N C6H5Li a.C c.CH3CH2CH2CH2Br e. + 2. H2/Pd Cl N Cl N H O NH2 + − + b.HO d.+ CH CCl f. CH3CH2CH2Br CH 3 N Br O 3 916 CHAPTER 21 More About Amines • Heterocyclic Compounds

− + 1. NH 1. Mg/Et O + 2 2 g.C6H5NN h. i. 2. CH3CH2CH2Br S Br 2. CO2 N N CH 3. H+ H 3

30. List the following compounds in order of decreasing acidity: H N+ N

+N +N N N N N +N H H H H H HH HH

31. Which of the following compounds is easier to decarboxylate?

N COH N CH2COH O O

32. Rank the following compounds in order of decreasing reactivity in an electrophilic aromatic substitution reaction:

OCH3 NHCH3 SCH3

33. One of the following compounds undergoes electrophilic aromatic substitution predominantly at C-3, and one undergoes electrophilic aromatic substitution predominantly at C-4. Which is which?

N CCH2CH3 N NHCH2CH3 O

34. Benzene undergoes electrophilic aromatic substitution reactions with aziridines in the presence of a Lewis acid such as AlCl3 . a. What are the major and minor products of the following reaction?

CH3 + N AlCl3

CH3

b. Would you expect epoxides to undergo similar reactions? 35. A Hofmann degradation of a primary amine forms an alkene that gives butanal and 2-methylpropanal upon ozonolysis and work- up under reducing conditions. Identify the amine. 36. The dipole moments of furan and tetrahydrofuran are in the same direction. One compound has a dipole moment of 0.70 D, and the other has a dipole moment of 1.73 D. Which is which? 37. Show how the vitamin niacin can be synthesized from nicotine. O COH N

N CH3 N nicotine niacin

38. The chemical shifts of the C-2 hydrogen in the 1H NMR spectra of pyrrole, pyridine, and pyrrolidine are d2.82, d6.42, and d8.50, Match each chemical shift with its heterocycle. 39. Explain why protonation of aniline has a dramatic effect on the compound’s UV spectrum, whereas protonation of pyridine has only a small effect on that compound’s UV spectrum. Problems 917 ' 40. Explain why pyrrole 1pKa 172 is a much stronger acid than ammonia 1pKa = 362.

+ − + + + H NH3 NH2 H = N −N pKa 36 H pKa =∼17

41. Propose a mechanism for the following reaction:

+ + trace H 2 H2CO N N CH2 N H H H

42. Quinolines are commonly synthesized by a method known as the Skraup synthesis, which involves the reaction of aniline with glycerol under acidic conditions. Nitrobenzene is added to the reaction mixture to serve as an oxidizing agent. The first step in the synthesis is the dehydration of glycerol to propenal.

H2SO4 + CH2 CH CH2 ∆ CH2 CHCH O 2 H2O OH OH OH glycerol

a. What product would be obtained if para-ethylaniline were used instead of aniline? b. What product would be obtained if 3-hexen-2-one were used instead of glycerol? c. What starting materials are needed for the synthesis of 2,7-diethyl-3-methylquinoline? 43. Propose a mechanism for each of the following reactions:

OO + H CH3OH a.CH3CCH2CH2CCH3 b. + Br H C H2O 2 3 O CH3 ∆ O CH3OO OCH3

44. Give the major product of each of the following reactions: O CH3NCH3 CH3 CCH3 − + + 1. HO a.HNO3 d.CH3I g. 2. H2C O O + N + N 3. H

CH3

+ + + b.Br2 e.PCl5 h. CH3CH2MgBr S NO2 N O N H H

CH3 H3C CH3 + ∆ N 1. H2O2 c.CH CHCH NCH CH f. i. N CH2CH3 3 2 2 3 2. ∆ − CH3 CH3 HO 1. H2O2 CH3 2. ∆

45. When piperidine undergoes the indicated series of reactions, 1,4-pentadiene is obtained as the product. When the four different methyl-substituted piperidines undergo the same series of reactions, each forms a different diene: 1,5-hexadiene, 1,4-pentadiene, 2-methyl-1,4-pentadiene, and 3-methyl-1,4-pentadiene. Which methyl-substituted piperidine yields which diene?

1. excess CH3I/K2CO3 1. excess CH3I/K2CO3 2. Ag2O, H2O 2. Ag2O, H2O 3. ∆ CH3NCH2CH2CH2CH CH2 3. ∆ CH2 CHCH2CH CH2 N H CH3 piperidine 918 CHAPTER 21 More About Amines • Heterocyclic Compounds

46. a. Draw resonance contributors to show why pyridine-N-oxide is more reactive than pyridine toward electrophilic aromatic substitution b. At what position does pyridine-N-oxide undergo electrophilic aromatic substitution?

47. Propose a mechanism for the following reaction: OO O + + − CH3COCCH3 CH3CO + N CH3 N CH2OCCH3 − O O

48. Explain why the aziridinium ion has a considerably lower pKa (8.0) than that of a typical secondary ammonium ion (10.0). (Hint: Recall that the larger the bond angle, the greater the s character, and the greater the s character, the more electronegative the atom.)

H + H H N N + + H

aziridinium ion pKa = 8.04

49. Pyrrole reacts with excess para-(N,N-dimethylamino)benzaldehyde to form a highly colored compound. Draw the structure of the colored compound.

50. 2-Phenylindole is prepared from the reaction of acetophenone and phenylhydrazine, a method known as the Fischer indole synthesis. Propose a mechanism for this reaction. (Hint: The reactive species is the enamine tautomer of the phenylhydrazone.) O + + H + + CCH3 NHNH2 ∆ NH3 H2O N H

51. What starting materials are required for the synthesis of the following compounds, using the Fischer indole synthesis? (Hint: See Problem 50.)

CH2CH3

a.CH2CH3 b. c. N N N H H H

52. Organic chemists work with tetraphenylporphyrins rather than porphyrins because tetraphenylporphyrins are much more resistant to air oxidation. Tetraphenylporphyrin can be prepared by the reaction of benzaldehyde with pyrrole. Propose a mechanism for the formation of the ring system shown here:

HC O N BF3 H oxidation + NH HN tetraphenylporphyrin N H H N

53. Propose a mechanism different from the one shown in Section 21.11 for the biosynthesis of porphobilinogen.