Download 21 More About Amines • Heterocyclic Compounds

21
More About Amines •
Heterocyclic Compounds
CH3CH2NH
CH3CH2NH2
ethylamine
CH2CH3
diethylamine
A
mines are compounds in which one or
more of the hydrogens of ammonia (NH 3)
have been replaced by an alkyl group.
Amines are among some of the most abundant
compounds in the biological world. We will apCH3CH2NCH2CH3
preciate their importance in Chapter 23, when we
look at amino acids and proteins; in Chapter 24,
CH2CH3
when we study how enzymes catalyze chemical reactriethylamine
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, already studied many aspects of amines and their chemistry. For example, we have seen
that the nitrogen in amines is sp 3 hybridized and the lone pair resides in an empty sp 3
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.
NH
piperidine
an amine is a base:
R
+
NH2
+
H
Br
H
NH3
CH3
Br
R
NH2
+
Br−
an amine is a nucleophile:
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.
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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. Morphine is 50 times stronger than aspirin as an analgesic, but it is addictive and suppresses respiration. Heroin is a synthetic compound that is made by acetylating morphine
(Section 30.3).
O
H3C
O
N
N
CH3
CH3
N
N
O
CH2CH2NH2
HO
N
N
CH3
N
caffeine
N
Cl
CH3
N
H
C6H5
Valium
nicotine
serotonin
O
HO
CH3C
O
HO
O
H
H
NCH3
CH3C
O
O
H
O
NCH3
H
morphine
heroin
Two other heterocycles are Valium®, a synthetic tranquilizer, and serotonin, a neurotransmitter. 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
(in alphabetical order) that have replaced the hydrogens of ammonia. Systematic
names employ “amine” as a functional group suffix.
CH3
CH3CH2CH2CH2CH2NH2
CH3CH2CH2CH2NHCH2CH3
a primary amine
pentylamine
1-pentanamine
a secondary amine
butylethylamine
N-ethyl-1-butanamine
common name:
systematic name:
CH3CH2CH2NCH2CH3
a tertiary amine
ethylmethylpropylamine
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
azacyclopropane
aziridine
N
H
azacyclobutane
azetidine
N
H
3-methylazacyclopentane
3-methylpyrrolidine
N
CH3
2-methylazacyclohexane
2-methylpiperidine
CH2CH3
N-ethylazacyclopentane
N-ethylpyrrolidine
Heterocycles with oxygen and sulfur heteroatoms are named similarly. The prefix for
oxygen is “oxa” and that for sulfur is “thia.”
O
O
S
O
oxacyclopropane
oxirane
ethylene oxide
thiacyclopropane oxacyclobutane oxacyclopentane
thiirane
oxetane
tetrahydrofuran
O
O
O
tetrahydropyran
1,4-dioxane
PROBLEM 1 ◆
Name the following compounds:
H
N
a.
CH3
CH3
CH3
c.
e.
HN
CH3
O
CH3
CH2CH3
S
b.
d.
N
H
f.
CH3
O
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.
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amine inversion
p orbital
sp2
sp3
N
R1
R3
R1
R3
R3
R1
‡
N
N
R2
sp3
R2
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 sp 3
nitrogen becomes an sp 2 nitrogen. The three groups bonded to the sp 2 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.
+
CH3CH2CH2NH3
CH2CH3
+
CH3CH2NH
+
CH3NH2
CH3
pKa = 10.8
+
CH3
NH3
CH3NH2
CH2CH3
pKa = 10.9
3-D Molecules:
Aziridinium ion; Pyrrolidine;
Piperidine; Morpholine
+
NH3
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
of these amines have pKa values expected for ammonium ions. We have seen that the
basicity of amines allows them to be easily separated from other organic compounds
(Chapter 1, Problems 70 and 71).
O
+
+
N
H
H
H
+
+
N
N
N
H
H
H
H
CH3
N
+
H
the ammonium ions of:
pyrrolidine
pKa = 11.27
piperidine
pKa = 11.12
morpholine
pKa = 9.28
N-methylpyrrolidine
pKa = 10.32
quinuclidine
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
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
+
CH3CH2Br
+
CH3NH2
CH3CH2NHCH3 + HBr
CH3CH2NH2CH3
–
methylamine
ethylmethylamine
Br
in nucleophilic acyl substitution reactions—reactions that acylate the amine
(Sections 17.8, 17.9, and 17.10)—for example,
O
CH3CH2
C
O
CH3
C
O
+
Cl
2 CH3NH2
methylamine
CH3CH2
+
+
NHCH3
CH3NH3Cl–
an amide
O
O
C
O
C
CH3
+
CH3
N
H
C
N
+
O
N
H
piperidine
–
+
OCCH3
H
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
O
+
catalytic
H+
H2NCH2
benzylamine
O
+
an imine
catalytic
H+
HN
+
NCH2
+
N
pyrrolidine
H2O
an enamine
and in conjugate addition reactions (Section 18.13)—for instance,
CH3
CH3C
CH3
O
CHCH
+
CH3NH
CH3
CH3C
O
CH2CH
NCH3
CH3
H2O
Reactions of Amines
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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
NaNO2
NH2
N
N
Cl−
Nu−
Nu
+
N2
+
Cl–
an arenediazonium salt
Amines are much less reactive than other compounds with electron-withdrawing
groups bonded to sp 3 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 (-NH 2) is such a strong base that amines cannot undergo the substitution
and elimination reactions that alkyl halides undergo.
relative reactivities
most
reactive
strongest acid,
weakest conjugate
base
RCH2F
>
RCH2OH
HF
pKa = 3.2
∼
RCH2OCH3
H2O
>
RCH2OH
pKa = 15.7
least
reactive
RCH2NH2
NH3
pKa = 15.5
weakest acid,
strongest conjugate
base
pKa = 36
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. Recall that protonated ethanol is more than 13 pKa units more acidic than protonated
ethylamine.
+
CH3CH2OH2
pKa = − 2.4
+
CH3CH2NH3
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. Protonated 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
a.
CCH3
+
CH3CH2CH2NH2
catalytic
H+
Section 21.5
Reactions of Quaternary Ammonium Hydroxides
889
O
b. CH3CCl + 2
N
H
1. HCl, NaNO2
2. H2O, Cu2O, Cu(NO3)2
NH2
c.
O
CCH3
d.
+
CH3CH2NHCH2CH3
catalytic
H+
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
∆
+
CH3CH2CH2NCH3
CH3
CH3CH
CH2 + NCH3 + H2O
CH3 HO−
CH3
A Hofmann elimination reaction is an E2 reaction. Recall that an E2 reaction is a
concerted, one-step reaction—the proton and the tertiary amine are removed in the
same step (Section 11.1). Very little substitution product is formed.
mechanism of the Hofmann elimination
CH3
CH3CH
H
HO
CH2
+
NCH3
CH3
CH3
CH3CH
CH2 + NCH3 + H2O
CH3
−
August Wilhelm von Hofmann
(1818–1892) was born in Germany.
He first studied law and then
changed to chemistry. He founded the
German Chemical Society. Hofmann
taught at the Royal College of Chemistry in London for 20 years and then
returned to Germany to teach at the
University of Berlin. He was one of
the founders of the German dye industry. Married four times—he was
left a widower three times—he had
11 children.
PROBLEM 6
What is the difference between the reaction that occurs when isopropyltrimethylammonium 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
ion has more than one b -carbon, the major alkene product is the one obtained by removing a proton from the b -carbon bonded to the greater number of hydrogens. In the
following reaction, the major alkene product is obtained by removing a hydrogen from
In a Hofmann elimination reaction, the
hydrogen is removed from the B -carbon
bonded to the most hydrogens.
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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
CHCH2CH3
+ CH3NCH3 + H2O
2-pentene
minor product
1-pentene
major product
CH3NCH3
+
CHCH2CH2CH3 + CH3CH
CH2
CH3
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
CH3
-carbon
CH3
∆
CH3CHCH2NCH2CH2CH3
+
CH3
CH3CHCH2N
−
CH3 HO
CH3
+ CH2
CH3
CHCH3 + H2O
propene
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 (antiZaitsev 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
alkene-like transition state
anti-Zaitsev elimination
carbanion-like transition state
δ−
δ−
δ−
δ−
H
H
H
H
CH2
CH2CHCH2CH2CH3
OH
CH3CH
OH
CHCH3
δ−
Br
more stable
CH3CH2C
δ−
Br
less stable
OH
δ−
+
OH
N(CH3)3
more stable
CH3CHCHCH2CH3
+
δ−
N(CH3)3
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 carbanionlike 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
Because the Hofmann elimination reaction occurs in an anti-Zaitsev manner, antiZaitsev 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
CH3
∆
+
a. CH3CH2CH2NCH3
N(CH3)3
HO−
c.
∆
CH3 HO−
CH3
H3C
∆
b.
+
N
H3C
∆
d.
HO−
CH3
+
HO−
CH3
N
H3C
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
2R
+
N
R
R + Ag2O + H2O
2R
R I−
+
N
R + 2 AgI
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
CH3CH2CH2NH2 + CH3I
excess
K2CO3
+
CH3CH2CH2NCH3
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.
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CHAPTER 21
More About Amines • Heterocyclic Compounds
added to toxic substances to keep them from being ingested
accidentally.
CH3
CH2CH3
O
A USEFUL BAD-TASTING
COMPOUND
Several practical uses have been found for Bitrex®,
a quaternary ammonium salt, because it is one of the most
bitter-tasting substances known and is nontoxic. Bitrex® is put
on bait to encourage deer to look elsewhere for their food, it is
put on the backs of animals to keep them from biting one
another, it is put on children’s’ fingers to persuade them to
stop sucking their thumbs or biting their fingernails, and it is
NHCCH2
N
+
CH2
CH2CH3
CH3
O
CO−
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
b. CH3CH2CH2CHCH3
CH3CH2CH
CH3CH2CH2CH
CH2
CH2
Br
c.
CH2
N
H
CH
CH
CH2
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.
CH3CH2CH2CH2NH2
CH3I
excess
K2CO3
+
CH3CH2CH2CH2N(CH3)3
I
−
Ag2O
H2O
+
CH3CH2CH2CH2N(CH3)3
∆
CH3CH2CH
CH2 + H2O
−
HO
21.6 Phase Transfer Catalysis
A problem organic chemists face in the laboratory is finding a solvent that will dissolve 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
−
CH3CH2CH2CH2CH2CH2Br +
C
N
?
Phase Transfer Catalysis
N + Br−
CH3CH2CH2CH2CH2CH2C
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
+
CH3CH2CH2CH2CH2CH2Br +
−
C
N
−
R4N HSO4
CH3CH2CH2CH2CH2CH2C
N + Br−
Quaternary ammonium salts are the most common phase transfer catalysts. However, 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
HSO4− CH2CH2CH2CH3
CH2NCH2CH3
HSO4− CH3
tetrabutylammonium
hydrogen sulfate
HSO4− CH2CH3
hexadecyltrimethylammonium
hydrogen sulfate
benzyltriethylammonium
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 original 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 accompanying 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 shuttling back and forth between the two phases. Phase transfer catalysis has been
successfully used in a wide variety of organic reactions.
+
−
organic
phase
R 4N C
aqueous
phase
R 4N C
+
−
N + R Br
starting
material
+
N Na HSO4−
R
C
+
N + R 4N Br −
target
compound
+
R 4N Br−
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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, peroxyacids, 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.
R
oxidation
NH2
R
NH
oxidation
OH
R
N
oxidation
O
R
+
O
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
NH
+
HO
OH
R
R
−
+
NH
HO
R
N
OH
a secondary hydroxylamine
R
R
N
R
+
HO
OH
R
+N
R
R
+
HO−
R
N
+
R
+
H2O
O−
OH
a tertiary amine
Arthur C. Cope (1909–1966) was
born in Indiana. He received a Ph.D.
from the University of Wisconsin and
was a professor of chemistry at Bryn
Mawr College, Columbia University,
and MIT.
H2O
OH
a secondary amine
R
+
an tertiary amine oxide
Amine oxides undergo a reaction similar to the Hofmann elimination reaction,
called a Cope elimination reaction. In a Cope elimination reaction, a tertiary amine
oxide rather than a quaternary ammonium ion undergoes elimination. The Cope elimination reaction occurs under milder conditions than does a Hofmann elimination
reaction.
CH3
CH3CH2CH2NCH
3
+
∆
CH3
CH3CH
CH2 +
−
O
NCH3
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
CH2
+
N
−
O
CH3CH
H
CH3
∆
CH3
CH3CH
CH2 + N
OH
CH3
Section 21.8
The major product of the Cope elimination, like that of the Hofmann elimination, is
the one obtained by removing a hydrogen from the b -carbon bonded to the greater
number of hydrogens.
CH3
+
CH3CH2NCH2CH2CH3
∆
Synthesis of Amines
895
In a Cope elimination, the hydrogen is
removed from the B -carbon bonded to
the most hydrogens.
CH3
CH2
CH2 + NCH2CH2CH3
−
OH
O
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.)
NH3
RCH2
X
RCH2
+
NH3
RCH2
NH2
RCH2
X
+
RCH2
a primary amine
NH2
RCH2
RCH2
+ HX
RCH2
RCH2
N
RCH2
CH2R
+ HX
a secondary amine
RCH2
RCH2
NH
RCH2
X
RCH2
N
RCH2
a quaternary
ammonium salt
a tertiary amine
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.
RCH2
RCH2
RCH2
+ HX
X
+
NH
RCH2
896
CHAPTER 21
More About Amines • Heterocyclic Compounds
Gabriel synthesis
O
NH
O
1. HO−
2. CH3CH2Br
N
O
COOH
H3O+
∆
CH2CH3
+
+ CH3CH2NH3
COOH
a protonated
primary amine
+ Br−
O
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.)
−
N3
CH3CH2CH2CH2Br
CH3CH2CH2CH2N
butyl bromide
+
N
N
butyl azide
H2
CH3CH2CH2CH2NH2
Pd/C
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
CH3CH2CH2CH2Br
N
CH3CH2CH2CH2C
HCl
butyl bromide
N
pentanenitrile
H2
CH3CH2CH2CH2NH2
Pd/C
pentylamine
Amines are obtained from the reduction of amides with LiAlH 4 (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.
O
1. LiAlH4
C
R
NH2
RCH2NH2
2. H2O
a primary amine
O
1. LiAlH4
C
R
NHCH3
RCH2NHCH3
2. H2O
a secondary amine
O
1. LiAlH4
C
R
NCH3
RCH2NCH3
2. H2O
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 H 2 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 H 2 to the C “ N bond as it is formed. This is
called reductive amination.
CH3CH2
C
CH3CH2
CH3CH2
O + NH3
excess
C
NH
CH3CH2
H2
Pd/C
CH3CH2
CHNH2
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
O
CH
catalytic
H+
O + CH3CH2NH2
CH
NCH2CH3
NaBH(OCCH3)3
CH2NHCH2CH3
an imine
O
O + CH3NH
catalytic
H+
CH3
N
CH3
NaBH(OCCH3)3
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
+ H2
CH3CH2NO2
CH3CH2NH2
nitroethane
ethylamine
NO2
Pd/C
+ 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 carbonyl 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
pairs of delocalized p electrons: Two of the pairs are shown as p bonds, and one pair
is shown as a lone pair on the heteroatom. Furan and thiophene have a second lone pair
that is not part of the p cloud. These electrons are in an sp 2 orbital perpendicular to the
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
H
O
S
pyrrole
furan
thiophene
these
electrons
are part of
the cloud
these
electrons
are part of
the cloud
N
orbital structure of pyrrole
3-D Molecules:
Pyrrole;
Furan;
Thiophene
H
O
orbital structure of furan
these
electrons are
in an sp2 orbital
perpendicular
to the p orbitals
897
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 depicted as a lone pair into the five-membered ring.
−
−
−
+N
N
H
+N
H
+N
H
−
+N
H
H
resonance contributors of pyrrole
δ−
δ−
δ−
δ+N
δ−
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 unsaturated 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
H
N
H
= 1.57 D
= 1.80 D
pyrrole
pyrrolidine
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
>
>
Pyrrole, furan, and thiophene undergo
electrophilic substitution preferentially
at C-2.
>
>
S
−
N
H
O
Because pyrrole, furan, and thiophene are aromatic, they undergo electrophilic aromatic substitution reactions.
mechanism for electrophilic aromatic substitution
+
+
N
H
+ Y
slow
H
N
H
Y
fast
N
H
Y
+ H+
Section 21.9
O
+ Br2
899
+ HBr
Br
O
Aromatic Five-Membered-Ring Heterocycles
2-bromofuran
N
H
CH3
+ HNO3
(CH3CO)2O
O2N
+ H2O
CH3
N
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
intermediate resulting from C-2 substitution of pyrrole has two additional resonance
contributors, each with a positive charge on a secondary allylic carbon. The intermediate 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.
Pyrrole, furan, and thiophene are more
reactive than benzene toward electrophilic aromatic substitution.
Br
H3C
O
CH3
+ Br2
H3C
CH3
O
+ HBr
pyrrole
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
H
O
S
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 oxygen 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.
+
Y
2-position
N
H
N
H
+
Y+
Y
+
N
H
H
Y
Y
H
3-position
N
H
H
+
H
+N
H
Y
+N
H
H
thiophene
> Figure 21.1
Structures of the intermediates that
can be formed from the reaction of
an electrophile with pyrrole at C-2
and C-3.
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 AlCl 3 , a relatively strong Lewis acid. Thiophene is more reactive
than benzene, so it can undergo a Friedel–Crafts reaction using SnCl 4 , a weaker Lewis
acid. An even weaker Lewis acid, BF3 , can be used when the substrate is furan. Pyrrole 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
2. H2O
+ CH3CCl
+ HCl
phenylethanone
O
S
+ CH3CCl
1. SnCl4
2. H2O
CCH3
S
+ HCl
O
2-acetylthiophene
O
O
+ CH3CCl
1. BF3
2. H2O
CCH3
O
+ HCl
O
2-acetylfuran
O O
N
H
O
+ CH3COCCH3
CCH3
N
H
+ CH3COH
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. Remember, a proton is an electrophile and, like other electrophiles, attaches to the C-2 position of pyrrole.
N
H
+ H+
H
+N
H
H
pKa = −3.8
Pyrrole is unstable in strongly acidic solutions because once protonated, it can readily
polymerize.
H
H
N
H
+N
H
H
+N
H
H
N
H
polymer
H
Pyrrole is more acidic 1pKa = ' 172 than the analogous saturated amine
1pKa = ' 362, because the nitrogen in pyrrole is sp 2 hybridized and is, therefore,
more electronegative than the sp 3 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
TABLE 21.1 The pKa Values of Several Nitrogen Heterocycles
H
H
+N
H
+N
pKa = 1.0
+
H
pKa = 6.8
pKa = 2.5
N
N
N
N
H
N
H
pKa = −2.4
NH
N
HN
+N
H
pKa = −3.8
+
+
N
H
HN
H
NH
+
H
H
pKa = 8.0
H
pKa = 11.1
pKa = 14.4
N+
H
pKa = 4.85
N+
H
pKa = 5.16
N
H
N
H
pKa = ∼17
pKa = ∼36
Tutorial:
Basic sites of nitrogen
heterocycles
delocalization. (Recall that the more stable the base, the stronger is its conjugate acid;
Section 1.17).
+ H+
N
N
H
N
H
−
pKa = ~17
N
+ H+
−
pKa = ~36
PROBLEM 14
When pyrrole is added to a dilute solution of D 2SO 4 in D 2O, 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
each compound contains five pairs of p electrons (Section 15.1). Notice that the electrons shown as a lone pair on the indole nitrogen are part of the p cloud; therefore, the
conjugate acid of indole, like the conjugate acid of pyrrole, is a strong acid
1pKa = -2.42. In other words, indole is an extremely weak base.
4
3
7
N1
H
5
2
6
indole
O
benzofuran
S
benzothiophene
3-D Molecules:
Indole;
Benzofuran;
Benzothiophene
901
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.
3-D Molecule:
Pyridine
these
electrons are
in an sp2 orbital
perpendicular
to the p orbitals
N
4
5
6
3
N
2
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 sp 2 hybridized nitrogen, which is
more electronegative than an sp 3 hybridized nitrogen (Section 6.9).
+ H+
N
N+
H
sp2
pyridine
pyridinium ion
pKa = 5.16
+ H+
Tutorial:
Lone-pair electrons on
nitrogen heterocycles
sp3
H
N
+
N
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
+ H2O
OH
+N
N
OH + HO−
pKa = 0.79
PROBLEM 17
+N
O−
pyridine-N-oxide
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
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
RCCl +
+
RC
H2O
N
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
pyridine
= 1.57 D
Because it is aromatic, pyridine (like benzene) undergoes electrophilic aromatic
substitution reactions (≠B is any base in the solution).
mechanism for electrophilic aromatic substitution
+
slow
+ Y+
N
B
H
Y
Y
fast
+
N
+
BH
N
Electrophilic aromatic substitution of pyridine takes place at C-3 because the most stable intermediate is obtained by placing an electrophilic substituent at that position
(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
Pyridine undergoes electrophilic
aromatic substitution at C-3.
903
904
CHAPTER 21
More About Amines • Heterocyclic Compounds
Figure 21.2 N
+
2-position
Structures of the intermediates that
can be formed from the reaction of
an electrophile with pyridine.
Y
+
N
+
N
H
Y
N
H
Y
H
least stable
+
+
Y+
Y
3-position
Y
H
N
H
+
N
H
Y
Y
H
N
N
H
Y
H
Y
+
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
FeBr3
300 °C
+ Br2
+ HBr
N
N
3-bromopyridine
30%
+ H2SO4
SO3H
230 °C
+ H2O
N
N
pyridine-3-sulfonic acid
71%
+ HNO3
NO2
H2SO4
300 °C
+ H2O
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.
+ CH3CH2Cl
AlCl3
no electrophilic aromatic substitution reaction
N
PROBLEM 18
Give the product of the following reaction:
O
+ CH3CCl
CH3OH
N
Since pyridine is less reactive than benzene toward electrophilic aromatic substitution, it is not surprising that pyridine is more reactive than benzene toward nucleophilic 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
Y−
+
N
slow
fast
Z
−N
+ Z−
Y
Z
N
Y
Nucleophilic aromatic substitution of pyridine takes place at C-2 and C-4, because attack at these positions leads to the most stable intermediate. Only when nucleophilic
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).
−
2-position
−
Y
−N
Y
N
H
N
H
Y
H
Pyridine undergoes nucleophilic
aromatic substitution at C-2 and C-4.
> Figure 21.3
Structures of the intermediates that
can be formed from the reaction of
a nucleophile with pyridine.
most stable
−
+
Y−
Y
3-position
Y
H
N
−
N
H
Y
Y
H
N
N
H
Y
−
H
Y
−
4-position
H
−
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
+
N
∆
−
NH2
OCH3
OCH3
N
CH3
CH3
∆
+ CH3O−
N
+ Br−
Cl
+ Cl−
OCH3
N
PROBLEM 19
Compare the mechanisms of the following reactions:
Cl
NH2
+
−
+ Cl−
NH2
N
N
Cl
NH2
+
−
+ Cl−
NH2
PROBLEM 20
a. Propose a mechanism for the following reaction:
N
b. What other product is formed?
KOH/H2O
∆
N
H
O
Pyridine is less reactive than benzene toward electrophilic aromatic substitution
and more reactive than benzene toward
nucleophilic aromatic substitution.
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.
N
NBS
∆/peroxide
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 conversion of a primary amino group into a diazonium group is shown in Section 16.12).
N
NH2
NaNO2, HCl
0 °C
H2O
N
+
N
N Cl−
N
2-aminopyridine
+
NH2
N Cl−
N
NaNO2, HCl
0 °C
N
N
H
OH
2-hydroxypyridine
enol form
O
-pyridone
keto form
OH
O
N
N
H
H2O
N
4-hydroxypyridine
enol form
4-aminopyridine
-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
N
−
CH2
−
HO−
N
CH2
CH2
−
N
N−
N
N
Consequently, the a-hydrogens of alkyl substituents can be removed by base, and the
resulting carbanions can react as nucleophiles.
−
O
CH2
CH3
CH
CH
CH
HO−
+ H2O
an aldol condensation
N
N
CH3
N
CH3
N
CH3
−
NH2
N
CH
− 2
CH3Br
an SN2 reaction
CH3
+ Br−
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 compounds. 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
same numbers, the nitrogen in isoquinoline is assigned the 2-position, not the lowest
possible number.)
5
N+
H
4
6
3
7
2
N
8
pKa = 4.85
3-D Molecules:
Quinoline; Isoquinoline
+ H+
1
quinoline
5
NH
+
4
6
3
7
N2
8
pKa = 5.14
+ H+
1
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−
+
COO−
N
+
H
H
proline
CH2CHCOO−
NH3
N
H
tryptophan
N
NH
+ NH
3
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
contains three pairs of p electrons (Section 15.1). The electrons drawn as lone-pair
electrons on N-1 (see p. 896) are part of the p cloud because they are in a p orbital,
whereas the lone-pair electrons on N-3 are not part of the p cloud because they are in
an sp 2 orbital, perpendicular to the p orbitals.
Tutorial:
Recognizing common
heterocyclic rings in complex
molecules
908
CHAPTER 21
More About Amines • Heterocyclic Compounds
these
electrons
are part of
the cloud
these electrons
are in an sp2 orbital
perpendicular
to the p orbitals
N
H
N
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).
4
3
N
2
5
−
NH
N
1
−
NH
N
+
NH
−
N
+
NH
−N
+
NH
+
resonance contributors of imidazole
Imidazole can be protonated because the lone-pair electrons in the sp 2 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
+
N
NH + H+
pKa = 6.8
Neutral imidazole is a stronger acid 1pKa = 14.42 than neutral pyrrole
1pKa ' 172 because of the second ring nitrogen.
N
NH
N
N
−
+ 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
HN
+
NH
δ+
HN
HN
imidazole anion
NH
+
N
N
−
δ+
−
δ−
N
NH
resonance hybrid
N
N
δ−
N
resonance hybrid
PROBLEM 22 ◆
Give the major product of the following reaction:
N
NCH3 + Br2
FeBr3
PROBLEM 23 ◆
List imidazole, pyrrole, and benzene in order of decreasing reactivity toward electrophilic
aromatic substitution.
Section 21.11
Biologically Important Heterocycles
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.) Unsubstituted 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).
7
6
4
N
5
1N
3N
8
2
N
2
N9
H
4
3
pyrimidine
NH2
N
NH2
O
N
N
HN
N
H
H2N
adenine
6
N
1
purine
N
5
N
O
N
N
H
O
guanine
O
HN
N
H
O
cytosine
CH3
HN
N
H
O
uracil
N
H
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 (Fe 2+) 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.
−OOCCH CH
2
2
H3C
N
C
HN
C
N
N
a porphyrin ring system
C
H2C
C
H
C
N
C
N
C
FeII
HC
NH
C
C
C
C
CH2CH2COO−
H
C
C
C
N
CH3
CH
C
C
C
H
CH3
C
CH3
C
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. Hemoglobin has four polypeptide chains and four heme groups; myoglobin has one
polypeptide chain and one heme group. The iron atoms in hemoglobin and myoglobin,
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
3-D Molecule:
Heme
909
910
CHAPTER 21
More About Amines • Heterocyclic Compounds
more tightly than O2 to Fe 2+. 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 daminolevulinic 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
E
−OOCCH
2
−OOCCH
2
CH2
CH2
NH2
O
enzyme
E
C
−OOCCH
2
−
HC
base
C
N
CH2
CH2
NH2
NH2
CH2COO−
+
E
N
H2O
CH2
C
C
O
CH2
CH2
NH2
NH2
-aminolevulinic acid
−OOCCH
2
E
−OOCCH
2
CH2CH2COO−
+NH2
C
N
H2NCH2 H
E
C
NH
+
H2NCH2
C
−OOCCH
2
H+
CH2
E
CH2CH2COO−
H
C
C
N
C
CH2
H2NCH2
OH
NH2
H2O
−OOCCH
2
CH2CH2COO−
−H+
NH2
+
E
−OOCCH
2
CH2CH2COO−
CH2CH2COO−
NH2
+
H2NCH2 N
H
H2NCH2 N
H
H
porphobilinogen
Four porphobilinogen molecules react to form porphyrin.
+
H3NCH2 N
H
NH
N
H
N
H
HN
+
H3N
CH2 N
H
repeat three more
times using an
intramolecular reaction for
the third repetition
subsequent oxidation
increases the unsaturation
porphyrin
H
N
H3NCH2 H + CH2
N
H
+
H3NCH2 N
H
N
H
+
CH2
+
NH3
Section 21.11
Biologically Important Heterocycles
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
rings is partially reduced. The metal atom in chlorophyll a is magnesium (Mg 2+)
3-D Molecule:
Chlorophyll a
CH2
HC
CH3
H3C
N
CH2CH3
N
MgII
H3C
N
N
CH3
H
H
O
O
C
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.
The metal atom in vitamin B12 is cobalt (Co 3+). The chemistry of vitamin B12 is discussed in Section 25.7.
NH2
O
O
NH2
H
H2N
CN
H2N
O
H
H3C
H3C
H3C
N
N
H3C
H
CoIII
H
O
O
CH3 NH
2
N
OH
−O
CH3
N
HN
P
CH3
N
CH3
CH3
H
H3C
O
O
H
H
H2N
O
N
H
O
O
CH3
O
CH2OH
vitamin B12
PROBLEM 26 ◆
Is porphyrin aromatic?
PROBLEM 27
Show how the last two porphobilinogen molecules are incorporated into the porphyrin
ring.
3-D Molecule:
Vitamin B12
911
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
in the blood. When the concentration of bilirubin in the blood
reaches a certain level, bilirubin diffuses into the tissues, causing them to become yellow. This condition is known as
jaundice.
The average human turns over about 6 g of hemoglobin each day. The protein portion (globin) and the iron are
reutilized, but the porphyrin ring is broken down. First it is reduced to biliverdin, a green compound, which is subsequently
Summary
Amines are compounds in which one or more of the hydrogens of ammonia have been replaced by R groups. Amines are
classified as primary, secondary, or tertiary, depending on
whether one, two, or three hydrogens of ammonia have been
replaced. Amines undergo amine inversion through a transition state in which the sp 3 nitrogen becomes an sp 2 nitrogen.
Some amines are heterocyclic compounds—cyclic
compounds in which one or more of the atoms of the ring is
an atom other than carbon. Heterocyclic rings are numbered
so that the heteroatom has the lowest possible number. A
natural product is a compound synthesized by a plant or
an animal. Alkaloids are natural products containing one or
more nitrogen heteroatoms and are found in the leaves,
bark, roots, or seeds of plants.
Because of the lone pair on the nitrogen, amines are both
bases and nucleophiles. Amines react as nucleophiles in nucleophilic substitution reactions, in nucleophilic acyl substitution reactions, in nucleophilic addition–elimination
reactions, and in conjugate addition reactions.
Amines cannot undergo the substitution and elimination
reactions that alkyl halides undergo, because the leaving
groups of amines are too basic. Protonated amines also cannot undergo the reactions that protonated alcohols and protonated ethers undergo. Amines are easily oxidized.
Saturated heterocycles containing five or more atoms have
physical and chemical properties typical of acyclic compounds that contain the same heteroatom.
Quaternary ammonium hydroxides and amine oxides
undergo E2 elimination reactions known as Hofmann elimination reactions and Cope elimination reactions, respectively. In both reactions, the proton from the b -carbon
bonded to the greater number of hydrogens is removed.
Quaternary ammonium salts are the most common phase
transfer catalysts.
Primary amines can be synthesized by means of a Gabriel
synthesis, by reduction of an alkyl azide or a nitrile, by
reductive amination, and by reduction of an amide.
Pyrrole, furan, and thiophene are aromatic compounds
that undergo electrophilic aromatic substitution reactions
preferentially at C-2. These compounds are more reactive
than benzene toward electrophiles. When pyrrole is protonated, its aromaticity is destroyed. Pyrrole polymerizes in strongly acidic solutions. Indole, benzofuran, and benzothiophene
are aromatic compounds that contain a five-membered aromatic ring fused to a benzene ring.
Replacing one of benzene’s carbons with a nitrogen forms
pyridine, an aromatic compound that undergoes electrophilic
aromatic substitution reactions at C-3 and nucleophilic aromatic substitution reactions at C-2 and C-4. Pyridine is less
reactive than benzene toward electrophilic aromatic substitution and more reactive toward nucleophilic aromatic substitution. Quinoline and isoquinoline are aromatic compounds with both a benzene ring and a pyridine ring.
Imidazole is the heterocyclic ring of the amino acid histidine. The conjugate acid of imidazole has a pKa of 6.8,
allowing it to exist in both the protonated and unprotonated
forms at physiological pH 1pH = 7.32. Nucleic acids
(DNA and RNA) contain substituted purines and substituted pyrimidines. Hydroxypurines and hydroxypyrimidines
are more stable in the keto form. A porphyrin ring system
consists of four pyrrole rings joined by one-carbon bridges;
in hemoglobin and myoglobin, the four nitrogen atoms are
ligated to Fe 2+. The metal atom in chlorophyll a is Mg 2+
and the metal atom in vitamin B12 is Co 2+.
Summary of Reactions
1. Reaction of amines as nucleophiles (Section 21.4).
a. In alkylation reactions:
R′
NH2
R′
+
–
NH2 Br
R
R′
NH
R
+ HBr
RBr
R
+
NH
R
R
R
R
RBr
–
Br
R′
N
R
+ HBr
RBr
R′
+
N
R
R
Br–
Summary of Reactions
913
b. In acylation reactions:
O
O
C
R
+
Cl
2 R′NH2
C
R
R′NH3+ Cl–
+
NHR′
c. In nucleophilic addition–elimination reactions:
i Reaction of a primary amine with an aldehyde or ketone to form an imine:
catalytic
H+
R
+
O
R
NH2
R
NCH2
R
+
R
H2O
R
ii Reaction of a secondary amine with an aldehyde or ketone to form an enamine:
catalytic
H+
R
+
O
R
NH
R
R
R
R
R
+
N
H2O
R
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
NaNO2
NH2
N
Cl–
N
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
R
NH2
oxidation
R
NH
OH
oxidation
R
N
oxidation
O
R
N
O−
R
R
R
oxidation
NH
R
+
N
H2O
OH
a secondary amine
a secondary hydroxylamine
R
R
R
N
R
oxidation
R
N
R
+
+
H2O
−
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
+
RCH2CH2NCH3
−
HO
CH3
∆
Hofmann
elimination
CH2 + NCH3 + H2O
RCH
CH3
CH3
CH3
RCH2CH2NCH3
CH3
H2O2
+
RCH2CH2NCH3
−
O
∆
Cope
elimination
in both eliminations, the proton is removed from the
CH3
RCH
CH2 + NCH3
OH
-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
−
NH
COOH
+
1. HO
2. R Br
N
O
H3O
∆
R
O
+
R
+
NH3
COOH
+ Br–
b. Reduction of an alkyl azide or a nitrile:
−
R
Br
R
C
N3
R
H2
N
Pd/C
+
H2
N–
N
N
R
CH2NH2
R
Pd/C
NH2
c. Reduction of a nitroalkane or nitrobenzene:
CH3CH2CH2NO2
+ H2
Pd/C
NO2
+ H2
Pd/C
CH3CH2CH2NH2
NH2
d. Aldehydes and ketones react (1) with excess ammonia plus H 2>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
C
excess NH3
O
CH
H2, Pd/C
R
O
+
H2O
R
R
C
NH2 +
CH3NH2
O
R
catalytic
H+
C
R
R
NaBH(OCCH3)3
NCH3
R
O
+
catalytic
H+
CH3NH
O
CH3
CH3
NaBH(OCCH3)3
N
CH3
a. Pyrrole, furan, and thiophene (Section 21.9):
(CH3CO)2O
NO2
N
H
+ Br2
O
Br
O
+ H2O
+ HBr
O
S
+ CH3CCl
1. SnCl4
2. H2O
S
N
CH3
6. Electrophilic aromatic substitution reactions.
N
H
NHCH3
R
CH3
+ HNO3
CH
CCH3
O
+ HCl
Problems
b. Pyridine (Section 21.10):
Br
FeBr3
300 °C
+ Br2
N
+ HBr
N
Br
FeBr3
90 °C
+ Br2
+N
Br
PCl3
+N
O−
N
O−
7. Nucleophilic aromatic substitution reactions of pyridine (Section 21.10).
−
+
NH2
∆
+ Cl−
Cl
N
N
NH2
Br
NH2
−
+
NH2
∆
+ Br−
N
N
Cl
N
NH2
+
−
NH2
N
∆
+ Cl−
N
N
AU: These terms are bold in text. Add to list?
Or make lightface in text?
Key Terms
alkaloid (p. 884)
amine inversion (p. 885)
amines (p. 883)
Cope elimination reaction (p. 894)
exhaustive methylation (p. 891)
furan (p. 897)
heteroatom (p. 884)
heterocycle (p. 884)
heterocyclic compound (p. 884)
purine (p. 909)
pyridine (p. 902)
pyrimidine (p. 909)
pyrrole (p. 897)
quaternary ammonium ion (p. 889)
reductive amination (p. 896)
thiophene (p. 897)
Hofmann elimination reaction (p. 889)
imidazole (p. 907)
iron protoporphyrin IX (p. 909)
ligation (p. 909)
natural product (p. 884)
phase transfer catalysis (p. 893)
phase transfer catalyst (p. 893)
porphyrin ring system (p. 909)
protoporphyrin IX (p. 909)
Problems
28. Name the following compounds:
a.
Cl
CH3
NH
b.
N
H
CH3
CH3
c.
d.
CH3
CH3CH2 N
H
N
H
29. Give the product of each of the following reactions:
O
a.
−
C
+
Cl
1. C N
2. H2/Pd
c. CH3CH2CH2CH2Br
C6H5Li
e.
N
Cl
N
H
O
−
+ HO
b.
N
Br
d.
O
CH3
+ CH3CCl
NH2
f.
+
CH3CH2CH2Br
915
916
CHAPTER 21
More About Amines • Heterocyclic Compounds
−
+
+ C6H5N
g.
h.
N
N
H
N
CH3
1. NH2
2. CH3CH2CH2Br
i.
S
Br
1. Mg/Et2O
2. CO2
3. H+
30. List the following compounds in order of decreasing acidity:
H
N+
+N
+N
H
H
N
H
N
H
H
N
N
H
N
H
+N
H
H
31. Which of the following compounds is easier to decarboxylate?
or
N
N
COH
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
N
CCH2CH3
NHCH2CH3
O
34. Benzene undergoes electrophilic aromatic substitution reactions with aziridines in the presence of a Lewis acid such as AlCl 3 .
a. What are the major and minor products of the following reaction?
CH3
+
AlCl3
N
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 workup 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
nicotine
N
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+
N
H
−
NH3
pKa = 36
N
−
NH2 + H+
pKa = ∼17
41. Propose a mechanism for the following reaction:
2
N
H
+ H2C
trace H+
O
CH2
N
H
N
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.
CH2
OH
CH
CH2
OH
OH
H2SO4
∆
CH2
CH
CH
O + 2 H2O
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:
a.
H3C
H+
H2O
∆
CH3
O
O
O
CH3CCH2CH2CCH3
b.
O
+ Br2
CH3OH
CH3O
O
OCH3
44. Give the major product of each of the following reactions:
O
CH3
CH3NCH3
CCH3
+ HNO3
a.
O
+ CH3I
d.
1. HO−
2. H2C
3. H+
g.
+N
N
O
CH3
b.
NO2
S
+ Br2
c. CH3CHCH2NCH2CH3
CH3
∆
f.
HO−
N
H
CH3
H3C
+
+ CH3CH2MgBr
h.
O
N
H
CH3
CH3
+ PCl5
e.
N
i.
1. H2O2
2. ∆
CH2CH3
N
1. H2O2
2. ∆
CH3
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?
N
H
piperidine
1. excess CH3I/K2CO3
2. Ag2O, H2O
3. ∆
CH3NCH2CH2CH2CH
CH3
1. excess CH3I/K2CO3
2. Ag2O, H2O
CH2 3. ∆
CH2
CHCH2CH
CH2
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:
O O
O
+ CH3CO−
+ CH3COCCH3
+N
N
CH3
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
CCH3 +
NHNH2
H+
∆
+ 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.
b.
CH2CH3
N
H
c.
N
H
N
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
BF3
+
N
H
N
H
NH
H
N
HN
oxidation
tetraphenylporphyrin
53. Propose a mechanism different from the one shown in Section 21.11 for the biosynthesis of porphobilinogen.