Chem 226 — Problem Set #12 —
“Fundamentals of Organic Chemistry,” 4th edition, John McMurry.
Chapter 12
1.
Classify each of the following compounds as either a primary, secondary, or tertiary amine, or as a
quaternary ammonium salt.
(a) (CH3)2CHNH2
(c)
(b) (CH3CH2)2NH
CH3
(d)
N
CH2 N(CH3) 3 I
CH3
(a) primary: the nitrogen is joined to one carbon, (b) secondary: the nitrogen is joined to two carbons,
(c) tertiary: the nitrogen is joined to three carbons, (d) quaternary ammonium salt: the nitrogen is
joined to four carbons and bears a positive charge.
3.
Name the following compounds by IUPAC rules.
(a) CH3NHCH2CH3
(d)
(b)
(c)
N CH3
N
CH3
(e)
N
CH2 CH2 CH3
CH3
H2NCH2 CH2 CHNH2
(a) 1-(N-methylamino)ethane or N-methylethylamine, (b) tricyclohexylamine,
(c) N-methylpyrrole, (d) (N-methyl-N-propylamino)cyclohexane or N-methyl-Npropylcyclohexylamine, (e) 1,3-diaminobutane or 1,3-butanediamine
It must be admitted that, with three competing but similar nomenclature systems in existence, naming
amines can be a confusing undertaking.
5.
Which would you expect to be a stronger base, aniline or p-methylaniline? Explain.
Let’s compare these two compounds by having them function as bases in a reaction where water is the
acid.
NH2
H3C
+ H2 O
NH2 + H2 O
The electron donating methyl
group may destabilize this
molecule (compared to the
one above) by increasing
the electron density in the
aromatic ring. This effect, if
present, would be small since
there are no formal charges.
NH3 + OH
H3C
NH3 + OH
The electron donating methyl
group will stabilize this
molecule (compared to the
one above) by increasing
the electron density at the
positively charged nitrogen.
This effect would be substantial
owing to the positive charge on
the nitrogen.
The bottom line here is that the p-methylaniline is more basic than the aniline because the methyl group
stabilizes the p-methylanilium cation.
6.
Predict the product of the following raction.
CH3
N
+
H
HBr
CH3
N H
H
+
Br
This is a typical acid-base reaction for an amine, where the amine functions as base and thereby
accepts a proton from an acid.
7.
Which compound in each of the following pairs is more basic?
(a) CH3CH2NH2 or CH3CH2CONH2
(b) NaOH or
(c) CH3NHCH3
(d) CH3OCH3 or (CH3)3 N
or CH3NHC6 H5
C6H5NH2
The more basic compound is boxed in each case.
(a) Amides are not basic at the
nitrogen because, if the nitrogen
O
O
were to become protonated, the
CH3CH2 C NH2
CH3CH2 C NH2
resonance stabilization of the amide
structure would be sacrificed.
Resonance stabilization of amide. Were the nitrogen
to be protonated this resonance stabilization would
Another way of looking at this is that
disappear.
the unshared pair of electrons on the
nitrogen in amides is not really on the
nitrogen, exclusively, but is shared
between the oxygen and nitrogen and is, therefore, less available for donation (bases are electron-pair
donors).
(b) Hydroxide ion is a strong base. Perhaps that’s all that needs to be said, but we might note that in
picking up a proton the negatively charged hydroxide is joining to a positively charged proton; in the
case of the amine a neutral molecule is picking up a proton with the result that a positive charge resides
on the nitrogen.
(c) In dimethylamine the unshared pair of electrons is pretty much localized on the nitrogen (and more
available) while in N-methylaniline it is somewhat delocalized into the aromatic ring.
(d) Each of these compounds can be protonated, in the first case on the oxygen and in the second on
the nitrogen. This leads to a positively charged oxygen and a positively charged nitrogen; since oxygen
is more electronegative, this situation is a higher energy one for the oxygen.
9.
Propose structures for amides that might be precursors of the following amines.
(a) propylamine, (b) dipropylamine, (c) benzylamine
If we are going to reduce an amide to get an amine, we should be aware of the following:
O
R1
LiAlH4
C N
R1
CH2 N
R2
R2
R1 , R2 may by R or H
O
(a) CH3CH2 CNH2
CH3CH2CH2NH2
O
(b) CH3CH2 CNHCH2CH2CH3
CH3CH2CH2NHCH2CH2CH3
O
(c) C 6H5CNH2
11.
C6H5CH2NH2
How could you synthesize the following amines from benzene? More than one step is required in each
case.
(a) m-aminobenzoic acid, (b) 2,4,6-tribromoaniline
(a)
1) Mg, ether
Br2
Br
Fe
COOH
2) CO 2
+
3) H3O
O 2N
H2 N
HNO 3
H2SO4
Sn
COOH
COOH
HCl
The benzoic acid could also be made by Friedel-Crafts alkylation of benzene, followed by oxidation of
the alkyl side chain with KMnO4.
In this synthesis we have used the carboxyl group as the meta-directing group. It might appear that, as
an alternative, one could nitrate first and then brominate, having the nitro group direct the incoming
bromine to the meta position. That would work. But, converting the m-bromonitrobenzene to mnitrobenzoic acid would not work since a nitro group cannot be present when a Grignard reagent is
formed.
(b)
1. Sn
HCl
HNO 3
H2SO 4
NO2
NH2
2. NH3
Br2
Br
Br
NH2
Br
15.
Propose a synthesis of p-(dimethylamino)azobenzene from benzene.
1. Sn
HCl
HNO 3
H2SO 4
NO2
2. NH3
CH3Br
N(CH3)2 +
(CH3)2N
N N
NH2
NaNO2
HCl
N2 Cl
23.
How can you explain the fact that trimethylamine (bp 3oC) has a lower boiling point than dimethylamine
(bp 7oC)?
Since trimethylamine does not have a hydrogen attached to a nitrogen, it is not an associated liquid.
Dimethylamine has one hydrogen attached to the nitrogen, so it is an associated liquid. So, even though
the trimethylamine is more massive the dimethylamine has the higher boiling point owing to hydrogen
bonding.
27.
Show the products of the following reactions.
(a) CH3CH2 CH2 NH2 + CH3Br
CH3CH2CH2 NHCH3 + CH3CH2CH2N(CH3)2 + CH3 CH2 CH2N(CH3 )3 Br
(b)
NH2 + HBr
(c) CH3CH2 CONH2 + LiAlH4
(d)
NH3 Br
CH3 CH2CH2NH2
C N + LiAlH4
CH2 NH2
(a) It is very difficult to control the reaction between ammonia (or an amine) and an alkyl halide to give
only one class of amine as product. The exception to this is that if you want the quaternary ammonium
salt using an excess of the alkyl halide will give you mostly that product.
28.
How might you prepare the following amines from ammonia and any alkyl halides needed?
(a) hexylamine, (b) benzylamine, (c) tetramethylammonium iodide,
(d) N-methylcyclohexylamine
It is very difficult to control the reaction between ammonia (or an amine) and an alkyl halide to give only
one class of amine as product. The exception to this is that if you want the quaternary ammonium salt
using an excess of the alkyl halide will give you mostly that product. On the other hand it is easy to
write them down, so that is what we shall do.
CH3(CH2)5 Br + NH3
CH3(CH2 )5NH2
C6H5CH2Br
+ NH3
C 6H5CH2 NH2
CH3I
+ NH3
(CH3)4N + INH2
Br + NH3
29.
CH3
CH3I
NH
How might you prepare the following amines from 1-bromobutane?
(a) butylamine, (b) dibutylamine, (c) pentylamine
(a) CH3CH2CH2 CH2 Br
(b) CH3CH2CH2 CH2 NH2
(c) CH3CH2CH2 CH2Br
NH3
CH3CH2CH2CH2NH2
CH3 CH2CH2CH2Br
KCN
CH3CH2CH2CH2 CN
LiAlH4
CH3CH2CH2 CH2CH2NH2
30.
How might you prepare each of the amines in question #29 from 1-butanol?
We could convert the 1-butanol to 1-bromobutane using PBr3 or HBr and then use the reactions in
question #29. The following are additional possiblilities for (a) and (b).
(a) CH3CH2CH2 CH2 OH
CH3CH2CH2 COCl
CrO3
CH3CH2CH2COOH
H3O+
NH3
CH3CH2CH2 CONH2
SOCl2
1. LiAlH4
2. H2O
CH3CH2CH2 CH2NH2
from (a)
(b) CH3CH2CH2 CH2NH2
CH3 CH2CH2COCl
base
from (a)
CH3 CH2 CH2CH2NHCOCH2CH2CH3
1. LiAlH4
2. H2 O
CH3CH2CH2 CH2 NHCH2CH2CH2CH3
31.
How would you prepare benzylamine, C6H5CH2NH2, from each of the following starting materials?
(a) benzamide, (b) benzoic acid, (c) nitrobenzene, (d) chlorobenzene
(a) C 6H5CONH2
(b) C 6H5COOH
LiAlH4
SOCl2
C6H5CH2NH2
C6H5COCl
NH3
C6H5CONH2
then (a)
LiAlH4
(c) C 6H5NO2
C 6H5CN
(d) C 6H5Cl
LiAlH4
Mg
dry ether
C6H5NH2
NaNO2
HCl
+
-
C6 H5N 2 Cl
CuCN
C 6H5 CH2 NH2
C 6H5 MgCl
1) CO2
+
2) H3O
C6 H5COOH
then (b)
32.
Write equations for the reaction of p-bromobenzenediazonium bisulfate with the following reagents.
(a) H3O+, (b) HBr, CuBr, (c) H3PO2, (d) KCN, CuCN
H3O+
Br
N N HSO4
Br
OH
CuBr/HBr Br
Br
H3PO 2
Br
H
CuCN/KCN
Br
33.
CN
Show how you might prepare benzoic acid from aniline. A diazonio replacement reaction is needed.
NaNO2
NH2
H2SO 4
N N HSO4
H2SO 4
H2O heat
CuCN/KCN
COOH
CN
37.
Which compound is more basic, triethylamine or aniline? Does the following reaction proceed as
written?
+
-
(CH3 CH2)3NH Cl
weaker acid
+
NH2
weaker base
N+ H3 Cl- + (CH3CH2 )3N
stronger base
stronger acid
Aliphatic amines are ordinarily more basic than aromatic ones. Since the weaker acid/weaker base pair
will predominate in an acid-base equilibrium, the reaction would proceed to only a small extent in the
direction shown.