Download Lecture - Ch 24

Chapter 24
Amines and
Heterocycles
Suggested Problems – 124,30-33,47-49,53-64,66
CHE2202, Chapter 24
Learn, 1
Naming Amines
• Alkyl-substituted (alkylamines) or arylsubstituted (arylamines)
• Classified as primary (RNH2), secondary
(R2NH), and tertiary (R3N)
– Depends on number of organic substituents
attached to nitrogen
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Naming Amines
• Quaternary ammonium salts: Compounds
that carry a positively charged nitrogen atom
with four attached groups
• Simple amines are named by adding the suffix
-amine to the name of the alkyl substituent
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Naming Amines
• The suffix -amine can be used in place of the
final -e in the name of the parent compound
• Amines with more than one functional group are
named by considering the –NH2 as an amino
substituent
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Naming Amines
• Symmetrical secondary and tertiary amines
are named by adding the prefix di- or tri- to the
alkyl group
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Naming Amines
• Unsymmetrically substituted secondary and
tertiary amines
– Named as N-substituted primary amines
– Largest alkyl group is the parent name, and other
alkyl groups are considered N-substituents
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Naming Amines
• Heterocyclic amines: Compound in which
the nitrogen atom occurs as part of a ring
– Each ring system has its own parent name
– Nitrogen atom is always numbered as position 1
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Worked Example
• Name the following compounds:
– a) CH3NHCH2CH3
– b)
• Solution:
– a) N-Methylethylamine
– b) N-Ethyl-N-methylcyclohexylamine
CHE2202, Chapter 24
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Structure and Properties of
Amines
• Bonding in alkylamines is similar to that in
ammonia
– N is sp3-hybridized
– C–N–C bond angles are close to 109° tetrahedral
value
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Structure and Properties of
Amines
• An amine with three different substituents on
nitrogen is chiral
– The two enantiomeric forms rapidly interconvert by
a pyramidal inversion at room temperature
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Structure and Properties of
Amines
• Amines with fewer than five carbon atoms are
water-soluble
• Primary and secondary amines form hydrogen
bonds, increasing their boiling points
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Basicity of Amines
• Lone pair of electrons on nitrogen makes
amines basic and nucleophilic
– React with acids to form acid-base salts and they
react with electrophiles
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Basicity of Amines
• Amines are stronger bases than alcohols,
ethers, or water
• Amines establish an equilibrium with water in
which water acts as an acid and transfers a
proton to the amine
• Basicity constant Kb is used to measure the
base strength of an amine
• High pKa → weaker acid and stronger
conjugate base
CHE2202, Chapter 24
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Basicity of Amines
RNH 2 + H 2O
RNH3+ + OH -
[RNH3+ ][OH - ]
Kb =
[RNH 2 ]
pK b = -logK b
• Basicity of an amine (RNH2) can be measured
by looking at the acidity of the corresponding
ammonium ion (RNH3+)
RNH3 + H 2O
RNH 2 + H3O +
[RNH 2 ][H3O+ ]
Ka =
[RNH3 ]
CHE2202, Chapter 24
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Basicity of Amines
 [RNH 2 ][H3O + ]   [RNH3+ ][OH - ] 
Ka × Kb = 


+
[RNH
]
[RNH
]
3
2



= [H 3O + ][OH - ] = K w = 1.00 × 10-14
Kw
Kw
Ka =
and K b =
Kb
Ka
pK a + pK b = 14
• Weaker base - Smaller pKa for ammonium ion
• Stronger base - Larger pKa for ammonium ion
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Basicity of Some Common
Amines
CHE2202, Chapter 24
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Basicity of Amines
• Amides (RCONH2) are nonbasic, in contrast
with amines
• Amides are stabilized by delocalization of the
nitrogen lone-pair electrons
• Amides are more stable than amines
– Stability is lost when protonated
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Basicity of Amines
• Primary and secondary amines can act as
very weak acids
• N-H proton can be removed by a
sufficiently strong base (eg. LDA)
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Worked Example
• Which compound in the following pair is
more basic
– CH3NHCH3 or pyridine
• Solution:
– CH3NHCH3 is more basic than pyridine
• pKa of CH3NHCH3 is 10.73
• pKa of pyridine is 5.25
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Basicity of Arylamines
• Arylamines are less basic than alkylamines
– The N lone-pair electrons in arylamines are
delocalized by interaction with the aromatic ring’s
 electron system
– Are less able to accept H+
• Energy difference between protonated and
nonprotonated forms is higher for arylamines
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Electrostatic Potential Maps
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Basicity of Arylamines
• Substituted arylamines can be either more
basic or less basic than aniline
• Electron-donating substituents increase the
basicity of the corresponding arylamine
• Electron-withdrawing substituents decrease
arylamine basicity
CHE2202, Chapter 24
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Worked Example
• Rank the following compounds in order of
ascending basicity
– p-nitroaniline, p-aminobenzaldehyde, pbromoaniline
• Solution:
Least basic
Most basic
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Biological Amines and the
Henderson-Hasselbalch Equation
• Henderson-Hasselbalch Equation:
 A - 
pH = pK a + log
 HA 
so
 A - 
log
= pH - pK a
 HA 
• To reflect structures at physiological pH:
– Cellular amines are written in their protonated form
– Amino acids in their ammonium carboxylate form
CHE2202, Chapter 24
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Worked Example
• Calculate the percentages of neutral and
protonated forms present in a solution of
0.0010 M pyrimidine at pH = 7.3
– The pKa of pyrimidinium ion is 1.3
• Solution:
 [RNH 2 ] 
log 
= pH - pK = 7.3 -1.3 = 6.0
+ 
 [RNH3 ] 
[RNH 2 ]
6
=
antilog(6.0)
=
10
[RNH3+ ]
[RNH 2 ] = 106 [RNH 3+ ]
– At pH = 7.3, virtually 100% of the pyrimidine
molecules are in the neutral form
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Synthesis of Amines
• Reduction of nitriles, amides, and nitro
compounds
– Amines can be prepared by reduction of nitriles
and amides with LiAlH4
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Synthesis of Amines
– Arylamines are prepared from nitration of an
aromatic compound and reduction of the nitro group
– Reduction by catalytic hydrogenation over platinum
is suitable if no other groups can be reduced
– Iron, zinc, tin, and tin(II) chloride are effective in
acidic solution
CHE2202, Chapter 24
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Worked Example
• Propose structures for either a nitrile or an
amide that might be a precursor of Nethylaniline
• Solution:
– The compound can be synthesized only by amide
reduction
• Amide reduction can be used to synthesize most
amines, but nitrile reduction can be used to
synthesize only primary amines
CHE2202, Chapter 24
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SN2 Reactions of Alkyl Halides
• Simplest method of alkylamine synthesis is by
SN2 alkylation of ammonia or an alkylamine
with an alkyl halide
– Ammonia and other amines are good nucleophiles
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SN2 Reactions of Alkyl Halides
• Primary, secondary, and tertiary amines all
have similar reactivity
– Initially formed monoalkylated substance
undergoes further reaction to yield a mixture of
products
• Secondary and tertiary amines undergo
further alkylation
CHE2202, Chapter 24
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SN2 Reactions of Alkyl Halides
• Azide ion, N3, displaces a halide ion from a
primary or secondary alkyl halide to give an
alkyl azide
• Alkyl azides are not nucleophilic
• Reduction gives only the primary amine
CHE2202, Chapter 24
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SN2 Reactions of Alkyl Halides
• Gabriel amine synthesis: A phthalimide
alkylation for preparing a primary amine from
an alkyl halide
• N-H in imides (–CONHCO–) can be removed
by KOH followed by alkylation and hydrolysis
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Worked Example
• Show two methods for the synthesis of
dopamine, a neurotransmitter involved in
regulation of the central nervous system
– Use any alkyl halide needed
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Worked Example
– Upper reaction is azide synthesis
– Lower reaction is Gabriel synthesis
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Reductive Amination of
Aldehydes and Ketones
• Reductive amination: Treatment of an
aldehyde or ketone with ammonia or an amine
in the presence of a reducing agent
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Mechanism
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Reductive Amination
• Ammonia, primary amines, and secondary
amines through reductive amination reaction
yield primary, secondary, and tertiary amines,
respectively
CHE2202, Chapter 24
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Worked Example
• How might the following amine be prepared
using a reductive amination reaction?
• Solution:
– Amine Precursor -
– Carbonyl Precursor - CH3CHO
CHE2202, Chapter 24
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Hofmann and Curtius
Rearrangements
• Carboxylic acid derivatives can be converted
into primary amines with loss of one carbon
atom by both the Hofmann rearrangement
and the Curtius rearrangement
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Mechanism – Hofmann
Rearrangement
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Mechanism – Hofmann
Rearrangement
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Curtius Rearrangement
• Heating an acyl azide prepared from an acid
chloride
• Migration of –R from C=O to the neighboring
nitrogen with simultaneous loss of a leaving
group
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Worked Example
• How is the following amine prepared using
Curtius rearrangements on a carboxylic acid
derivative?
CHE2202, Chapter 24
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Worked Example
• Solution:
– The precursor is an acid chloride, which is
treated with NaN3, then with H2O and heat
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Reactions of Amines
• Primary and secondary amines can also be
acylated by nucleophilic acyl substitution
reaction
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Hofmann Elimination
• Converts amines into alkenes
• NH2 is very a poor leaving group; it is
converted to an alkylammonium ion, which is a
good leaving group
• An amine is completely methylated by reaction
with an excess amount of iodomethane to
produce the corresponding quaternary
ammonium salt
CHE2202, Chapter 24
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Hofmann Elimination
• Silver oxide is used for the elimination step
• Exchanges hydroxide ion for iodide ion in the
quaternary ammonium salt, thus providing the
base necessary to cause elimination
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Orientation in Hofmann
Elimination
• Major product is the less highly substituted
alkene
– Non-Zaitsev result is probably steric
– Due to the large size of the trialkylamine leaving
group
• The base must abstract a hydrogen from the
most sterically accessible
– Least hindered position
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Orientation in Hofmann
Elimination
CHE2202, Chapter 24
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Worked Example
• What products would you expect from
Hofmann elimination of the following
amine?
– If more than one product is formed, indicate
which is major
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Worked Example
• Solution:
– The first pair of products results from
elimination of a primary hydrogen and are the
major products
– The second pair of products results from
elimination of a secondary hydrogen CHE2202, Chapter 24
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Reactions of Arylamines
• Electrophilic aromatic substitution
– Amino substituents are strongly activating and
ortho- and para-directing groups in electrophilic
aromatic substitution reactions
– Reactions are controlled by conversion to amide
CHE2202, Chapter 24
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Reactions of Arylamines
– The amino group forms acid-base complex with
the AlCl3 catalyst preventing further reaction
– Therefore, we use the corresponding amide
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Reactions of Arylamines
– Modulating the reactivity of an amino-substituted
benzene allows many kinds of electrophilic
aromatic substitutions to be carried out
– Sulfa drugs were among the first pharmaceutical
agents to be used clinically against bacterial
infection
CHE2202, Chapter 24
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Worked Example
• Propose syntheses of m-chloroaniline from
benzene
• Solution:
– Chlorination occurs before reduction so that
chlorine can be introduced in the m-position
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Reactions of Arylamines
• Diazonium salts: The Sandmeyer reaction
– Primary arylamines react with HNO2 yielding
stable arenediazonium salts
• Corresponding alkanediazonium can not be isolated
– Arenediazonium salts are useful because the
diazonio group can be replaced by a nucleophile
in a substitution reaction
CHE2202, Chapter 24
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Reactions of Arylamines
– Sandmeyer reaction: Reaction of an
arenediazonium salt with the corresponding
copper(I) halide to yield aryl chlorides and
bromides
• Aryl iodides can be prepared by direct
reaction with NaI
CHE2202, Chapter 24
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Reactions of Arylamines
• An arenediazonium salt and CuCN yield the
nitrile, ArCN, which can be hydrolyzed to other
functional groups
• The diazonio group can be replaced by –OH
to yield a phenol and by –H to yield an arene
CHE2202, Chapter 24
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Reactions of Arylamines
• Reaction of arenediazonium salt with
copper(I) oxide in an aqueous solution of
copper(II) nitrate yield phenols
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Reduction to a Hydrocarbon
• By treatment of a diazonium salt with
hypophosphorous acid, H3PO2
• Diazonium replacement takes place through
radical pathways
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Worked Example
• How is p-bromobenzoic acid prepared from
benzene using a diazonium replacement
reaction
• Solution:
– Use of the diazonium replacement reaction that
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substitutes bromine for a nitro group
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Diazonium Coupling Reactions
• Arenediazonium salts undergo a coupling
reaction with activated aromatic rings to yield
azo compounds, Ar–N=N–Ar
• Are typical electrophilic aromatic substitutions
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Diazonium Coupling Reactions
• The electrophilic diazonium ion reacts with the
electron-rich ring of a phenol or arylamine
• Usually occurs at the para position but goes
ortho if para is blocked
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Diazonium Coupling Reactions
• Azo-coupled products have extended 
conjugation that lead to low energy electronic
transitions that occur in visible light
CHE2202, Chapter 24
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Worked Example
• Propose a synthesis of p(dimethylamino)azobenzene with benzene as
organic starting material
• Solution:
CHE2202, Chapter 24
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Heterocyclic Amines
• A cyclic organic compound that contains
atoms of two or more elements in its ring
• Most heterocyclic compounds possess same
chemistry as their open-chain counterparts
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Heterocyclic Amines
• Pyrrole
– Simplest five-membered unsaturated
heterocyclic amine
• Undergoes electrophilic substitution reactions
• Chemical properties are not consistent with the
structural features of amine or a conjugated diene
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Heterocyclic Amines
– Ring is reactive toward electrophiles
– Electrophilic substitutions occur at C2
• Reaction leads to intermediate cation having 3
resonance forms
– Reaction at C3 gives only 2 resonance forms
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Heterocyclic Amines
• Imidazole
– Common five-membered heterocyclic amine
– Constituent of the amino acid histidine
• One of two nitrogens is basic
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Worked Example
• Draw an orbital picture of thiazole
– Assume that both the nitrogen and sulphur atoms
are sp2-hybridized
– Show the orbitals that the lone pairs occupy
• Solution:
– Contains six  electrons
– Each carbon contributes one electron, nitrogen
contributes one electron, and sulfur contributes
two electrons to the ring  system
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Worked Example
– Sulfur and nitrogen have lone electron pairs in
sp2 orbitals that lie in the plane of the ring
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Heterocyclic Amines
• Pyridine
– Nitrogen-containing heterocyclic analog of
benzene
• The sp2-hybridized nitrogen atom is less basic than
the sp3-hybridized nitrogen in an alkylamine
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Heterocyclic Amines
– Undergoes electrophilic aromatic substitution
reactions with difficulty
• Acid-base complexation between the basic ring’s
nitrogen atom and the incoming electrophile
• High dipole moment pulls density out of ring
– Six-membered diamine pyrimidine is found
commonly in biological molecules
CHE2202, Chapter 24
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Worked Example
• Electrophilic aromatic substitution reactions of
pyridine normally occur at C3
– Draw the carbocation intermediates resulting from
reaction of an electrophile at C2, C3, and C4
– Explain the observed result
• Solution:
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Worked Example
– Reaction at C3 is favored over reaction at C2 or
C4
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Worked Example
– Positive charge of the cationic intermediate of
reaction at C3 is delocalized over three carbon
atoms rather than over two and the
electronegative pyridine nitrogen as occurs in
reaction at C2 or C4
CHE2202, Chapter 24
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Heterocyclic Amines
• Polycyclic heterocycles
– Quinoline, isoquinoline, and indole contain
both a benzene ring and a heterocyclic
aromatic ring
– Purine contains two heterocyclic rings joined
together
– Quinoline and isoquinoline are less reactive
towards electrophilic substitution
– Indole undergoes electrophilic substitution
more easily than benzene
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Heterocyclic Amines
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Heterocyclic Amines
– Purine has three basic, pyridine-like nitrogen
atoms
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Worked Example
• Which nitrogen atom in the hallucinogenic
indole alkaloid N,N-dimethyltryptamine is
more basic?
– Explain
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Worked Example
• Solution:
– Side chain nitrogen atom of N,Ndimethyltryptamine is more basic than the ring
nitrogen atom
– Aromatic nitrogen electron lone pair is part of
the ring  electron system
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Spectroscopy of Amines
• Infrared spectroscopy
– 1° and 2° amine are identified by characteristic N–
H stretching absorptions at 3300 to 3500 cm1
– Amine absorption bands are sharper and less
intense than hydroxyl bands
– N–H bend (scissor) is noticed above 1600 cm–1
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Nuclear Magnetic Resonance
Spectroscopy
• N–H hydrogens appear as broad signals without
clear-cut coupling to neighboring C–H hydrogens
• In D2O exchange of N–D for N–H occurs, and the
N–H signal disappears
• Hydrogens on the carbon next to nitrogen are
deshielded due to the electron-withdrawing effect
of nitrogen
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Nuclear Magnetic Resonance
Spectroscopy
• Hydrogens on C next to N absorb further
downfield than alkane hydrogens
• N-CH3 gives a sharp three-H singlet at 2.2
to 2.6 
CHE2202, Chapter 24
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Worked Example
• Compound A, C6H12O, has an IR absorption at
1715 cm-1 and gives compound B, C6H15N,
when treated with ammonia and NaBH4
– What are the structures of A and B?
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Worked Example
• Solution:
– It is inferred that B is a primary amine, from the
spectrum
– 1H NMR spectrum shows a 9-proton singlet, a
one-proton quartet, and a 3-proton doublet
– Absorption due to the amine protons is not
visible
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Mass Spectrometry
• Compound with odd number of nitrogen
atoms has odd-numbered molecular weight
– Presence of N can be detected observing the
spectrum
• Alkylamines cleave at the C–C bond
nearest the nitrogen to yield an alkyl radical
and a nitrogen-containing cation
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Mass Spectrum of Nethylpropylamine
• The two main modes of a cleavage give
fragment ions at m/z = 58 and m/z = 72
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