Chemistry 213
Clark College
Homework 4: Amines SOLUTION
1. Predict the products, give the reagents or give the reactant required to complete the following
reactions.
NH2
F
1) NaNO2, HCl,
0°C
a)
2) HBF4
NH2
b)
H3C
c)
OH
1) NaNO2, HCl,
0°C
2) H3O+
H3C
CH3I
large excess
NH2
N
I
O
N K+
1)
Br
NH2
O
2) N2H4, !
d)
Or 1) NaN3 2) Reduction
(no over-alkylation!)
O
1)
e)
Br
N K+
O
2) N2H4, !
NH2
2. Provide an acceptable name for the following amines.
O
N H OH
Cyclohexyldimethylammonium
hydroxide
Homework 4
NH2
1-phenyl-2-propanamine
or 1-phenyl-2-aminopropane
Spring 2008
N
H
4-methyl-5-(methylamino)heptan-2-one
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Chemistry 213
Clark College
3. Pyridoxamine is a form of vitamin B6 that contains two nitrogen atoms. Which nitrogen is the
stronger base? Explain your reasoning behind your choice.
To think about the basicity of a group, we should consider the stability of the conjugate acid, much like
we should consider the stability of the conjugate base when determining the acidity of a proton.
We can write an acid-base reaction for the amine group like this:
H2N
H3N
HO
OH
HO
+ H2O
OH
N
+
OH
N
The amine is sp3-hybridized, so the lone pair is far away from the nucleus and can seek an electrophile.
One the ammonium ion is formed, there are no neighboring groups that can donate electron density to
help stabilize the positive charge on nitrogen. Both factors help promote the basicity of the free amine.
We can write an acid-base reaction for the pyridinium group like this:
H2N
H2N
HO
OH
HO
+ H2O
OH
N
+
OH
N
H
2
The nitrogen is sp -hybridized, so the lone pair is held more closely to the nucleus and is less available
to seek an electrophile. Also, the pyridinium cation can be resonance-stabilized, but doing so will break
aromaticity, so this does not help stabilze the conjugate acid. These factors show that the pyridine
nitrogen is a less-effective base than the free amine.
4. Convert the following molecules into benzylamine (PhCH2NH2). Use a different method for each
molecule.
OH
PBr3
TEA
Br
NH2
1) NaN3
2) H2, Pt
O
Cl 1)
N K+
NH2
O
2) N2H4, !
O
H
H
Homework 4
NH3
H+
NH
Spring 2008
H2
Ni
NH2
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Chemistry 213
Clark College
5. Synthesis.
NO2
HNO3
H2SO4
O
NO2
1) NaNO2, HCl,
0°C
2) CH3OH, H+
Cl
Cl2
FeCl3
NO2
HNO3
H2SO4
NO2
OCH3
NO2
NH2
H2
Ni
Cl
AlCl3
HNO3
H2SO4
O
Technically, the ketone
should be protected...
O
NH2
1) NaNO2, HCl,
0°C
2) H3PO2
NO2 O
1) NaNH2
2)
NO2 O
1) BH3
2) CH3COOH
HBr
ROOR
Br
O
1)
O
N
H
Ph
Homework 4
Br
Spring 2008
Ph
H
H+
N K+ 2) N2H4
!
O
NH2
Page 3 of 6
Chemistry 213
Clark College
6. Provide a synthesis for albuterol from the given starting material.
OH
O
OH
N
H
O
OH
Albuterol (a bronchodialator)
O
O
H
O
O
Cl
H
AlCl3
O
NaCN
HCN
O
NC
O
O
OH
H2
Ni
O
O
N
H
H+
Br
O
a poor SN2 rxn
OH
O
H2N
OH
OH
OH
N
H
OH
An alternate route (I pondered some more yesterday) using a coupling agent and an epoxide, avoiding
the poor SN2 reaction and taking fewer steps!
O
O
Cl2
O
AlCl3
1)
O
Cl
2
O
CuLi
2) H2O
O
mCPBA
OH
N
H
Homework 4
OH
1)
O
NH2
O
2) H+
O
OH
Spring 2008
Page 4 of 6
Chemistry 213
Clark College
7. Draw stepwise mechanism for the following reactions.
a)
Br
CH3CH2NH2
NaOH (dil.)
Br
N
CH2CH3
NH2
Br
Br
Br
Br
N
N
H
H H
OH
N
N
H
CH2CH3
CH2CH3
OH
H
O
b)
NH2
1) H+
2) NaBH4
3) H2O
N
H+
O
O
H+
OH H
N H
H
N H
OH H
N
NH2
H
H B H
H
H
N
Homework 4
H
O
H
N
N
H
N
OH2H
N
H
H
N
Spring 2008
Page 5 of 6
Chemistry 213
Clark College
Extra Credit (5 points)
Provide the mechanism for the following reaction.
O
OH
H
HN
OH
H
N
dil. acid.
O
OH H
OH
H
HN
H
O
OH
OH
H N
HN
OH
OH
N
OH2
N
OH H
OH
N
Resonance-delocalized C+
OH
H
OH
N
N
Resonance-delocalized C+
Homework 4
Spring 2008
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