UNIT 13: AMINES
IUPAC NOMENCLATURE
1.
Gabriel phthalimide synthesis
Gabriel
synthesis is used for the preparation of primary amines. Phthalimide on
treatment with ethanolic potassium hydroxide forms potassium salt of
phthalimide which on heating with alkyl halide followed by alkaline hydrolysis
produces the corresponding primary amine. Aromatic primary amines cannot be
prepared by this method because aryl halides do not undergo nucleophilic
substitution with the anion formed by phthalimide.
2.
Hoffmann bromamide degradation reaction
Hoffmann developed a method for preparation of primary
amines by treating an amide with bromine in an aqueous or ethanolic solution of
sodium hydroxide. The amine so formed contains one carbon less than that
present in the amide.
3.
Carbylamine reaction
Aliphatic and aromatic primary amines on heating with
chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines
which are foul smelling substances. Secondary and tertiary amines do not show
this reaction.
This reaction is known as carbylamine reaction or
isocyanide test and is used as a test for primary amines.
4.
Hinsberg Test:
Benzenesulphonyl chloride (C6H5SO2Cl), which is also
known as Hinsberg‘sreagent, reacts with primary and secondary amines to form
sulphonamides.
(a)
The reaction of benzenesulphonyl chloride
with primary amine yields N-ethylbenzenesulphonyl amide.
The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the
presence of strong electron withdrawing sulphonyl group. Hence, it is soluble
in alkali.
(b)
In
the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed.
Since N, N-diethylbenzene sulphonamide does not
contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence
insoluble in alkali.
(c)
Tertiary amines do not react with benzenesulphonyl chloride. This property of
amines reacting with benzenesulphonyl chloride in a different manner is used
for the distinction of primary, secondary and tertiary amines and also for the
separation of a mixture of amines.
5.
Sandmeyer Reaction
The Cl–, Br–and
CN–nucleophiles can easily be introduced in the benzene ring of
diazonium salts in the presence of Cu(I) ion.
6.
Gatterman Reaction
Chlorine or bromine can be introduced in the benzene
ring by treating the diazonium salt solution with corresponding halogen acid in
the presence of copper powder.
7.
Coupling reactions
The
azo products obtained have an extended conjugate system having both the
aromatic rings joined through the –N=N–bond. These compounds are often coloured
and are used as dyes. Benzene diazonium chloride reacts with phenol in which
the phenol molecule at its para position is coupled with the diazonium salt to
form p-hydroxyazobenzene. This type of reaction is known as coupling reaction.
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DISTINCTION
BETWEEN PAIRS OF COMPOUNDS
Give one chemical
test to distinguish between the following pairs of compounds.
(i)
Methylamine and dimethylamine (ii) Secondary
and tertiary amines
(v)
Aniline and N-methylaniline.
ANS.
(i) Methylamine and dimethylamine can be distinguished by the carbylamine test.
Carbylamine test: Aliphatic and aromatic primary amines on heating with
chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or
carbylamines.
Methylamine (being an aliphatic primary amine) gives a
positive carbylamine test, but dimethylamine does not.
(ii)
Secondary and tertiary amines can be distinguished by allowing them to react
with Hinsberg‘sreagent (benzenesulphonyl chloride, C6H5SO2Cl).
Secondary amines react with Hinsberg‘sreagent to form a product that is insoluble
in an alkali. For example, N, N−diethylamine reacts with Hinsberg‘s reagent to
form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary
amines, however, do not react with Hinsberg‘sreagent.
(iv)
Aniline and benzylamine can be distinguished by their reactions with the help
of nitrous acid, which is prepared in situ from a mineral acid and sodium
nitrite.
Benzylamine reacts with
nitrous acid to form unstable diazonium salt, which in turn gives alcohol with
the evolution of nitrogen gas.
On the other hand, aniline reacts with HNO2
at a low temperature to form stable diazonium salt. Thus, nitrogen gas is not
evolved.
(v)
Aniline and N-methylaniline can be distinguished using the Carbylamine test.
Primary amines, on heating with chloroform and ethanolic potassium hydroxide,
form foul-smelling isocyanides or carbylamines. Aniline, being an aromatic
primary amine, gives positive carbylamine test. However, N-methylaniline, being
a secondary amine does not.
REASONING QUESTIONS
Q1. Account for
the following:
(i)
pKb of aniline is more than that of
methylamine.
(ii)
Ethylamine is soluble in water whereas
aniline is not.
(iii)
Methylamine in water reacts with ferric
chloride to precipitate hydrated ferric oxide.
(iv)
Although amino group is o–and p–directing
in aromatic electrophilic
substitution
reactions, aniline on nitration gives a substantial amount ofm-nitroaniline.
(v)
Aniline does not undergo Friedel-Crafts
reaction.
(vi) Diazonium
salts of aromatic amines are more stable than those of aliphatic amines.
Ans.
(i) pKb of aniline is more than that of methylamine:
Aniline
undergoes resonance and as a result, the electrons on the N-atom are
delocalized over the benzene ring. Therefore, the electrons on the N-atom are
less available to donate.
On the other hand, in case of methylamine (due to the
+I effect of methyl group), the electron density on the N-atom is increased. As
a result, aniline is less basic than methylamine. Thus, pKb of aniline is more
than that of methylamine.
(ii) Ethylamine is
soluble in water whereas aniline is not:
Ethylamine when added to water forms intermolecular
H−bonds with water. Hence, it is soluble in water.
But aniline does not undergo H−bonding with water to a
very large extent due to the presence of a large hydrophobic −C6H5 group. Hence,
aniline is insoluble in water.
(iii) Methylamine in water reacts with ferric chloride
to precipitate hydrated ferric oxide:
Due to the +I effect of −CH3 group, methylamine is
more basic than water. Therefore, in water, methylamine produces OH−ions
by accepting H+ ions from water.
Ferric chloride
(FeCl3) dissociates in water to form Fe3+ and Cl−ions.
Then, OH−ion
reacts with Fe3+ ion to form a precipitate of hydrated ferric oxide.
(v) Although
amino group is o, p− directing in aromatic electrophilic substitution
reactions, aniline on nitration gives a substantial amount of m-nitroaniline:
For this reason,
aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does
not undergo Friedel-Crafts reaction:
Due to the positive charge on the N-atom,
electrophilic substitution in the benzene ring is deactivated. Hence, aniline
does not undergo the Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more
stable than those of aliphatic amines: The diazonium ion undergoes resonance as
shown below:
This resonance accounts for the stability of the
diazonium ion. Hence, diazonium salts of aromatic amines are more stable than
those of aliphatic amines.
Q2. Why aromatic primary amines cannot be prepared by
Gabriel phthalimide synthesis?
Ans.
Gabriel phthalimide synthesis is used for the preparation of aliphatic primary
amines. It involves nucleophilic substitution (SN2) of alkyl halides
by the anion formed by the phthalimide. But aryl halides do not undergo
nucleophilic substitution with the anion formed by the phthalimide.Hence, aromatic primary amines cannot be prepared by this
process.
Q3.Give
possible explanation for each of the following:
(i)
Why are amines less acidic than alcohols
of comparable molecular masses?
(ii)
Why do primary amines have higher boiling
point than tertiary amines?
(iii)
Why are aliphatic amines stronger bases
than aromatic amines?
Ans.
(i) Amines undergo protonation to give amide ion.
Similarly, alcohol
loses a proton to give alkoxide ion.
In
an amide ion, the negative charge is on the N-atom whereas in alkoxide ion, the
negative charge is on the O-atom. Since O is more electronegative than N, O can
accommodate the negative charge more easily than N. As a result, the amide ion
is less stable than the alkoxide ion. Hence, amines are less acidic than
alcohols of comparable molecular masses.
(ii) In a molecule of tertiary amine, there are no
H−atoms whereas in primary amines, two hydrogen atoms are present. Due to the
presence of H−atoms, primary amines undergo extensive intermolecular H−bonding.
As a result, extra energy is required to separate the
molecules of primary amines. Hence, primary amines have higher boiling points
than tertiary amines.
(iii)
Due to the −R effect of the benzene ring, the electrons on the N- atom are less
available in case of aromatic amines. Therefore, the electrons on the N-atom in
aromatic amines cannot be donated easily. This explains why aliphatic amines
are stronger bases than aromatic amines.
SOLVED QUESTIONS
Short
Questions
Q1. Give the IUPAC name of the compound and classify
into primary, secondary or tertiary amines.
Q2. Give the IUPAC name of the compound and classify
into primary, secondary or tertiary amines.
2-Methylpropan-2-amine (10 amine)
Q5. Give the IUPAC name of the compound and classify
into primary, secondary or tertiary amines.
Q6. Write short notes on
diazotization
Aromatic primary amines react with nitrous acid
(prepared in situ from NaNO2 and a mineral acid such as HCl) at low
temperatures (273-278 K) to form diazonium salts. This conversion of aromatic
primary amines into diazonium salts is known as diazotization.
For example, on treatment with NaNO2 and
HCl at 273−278 K, aniline produces benzenediazonium chloride, with NaCl and H2O
as by-products.
Q7. Write short
notes on ammonolysis
When
an alkyl or benzyl halide is allowed to react with an ethanolic solution of
ammonia, it undergoes nucleophilic substitution reaction in which the halogen
atom is replaced by an amino (−NH2) group. This process of cleavage
of the carbon-halogen bond is known as ammonolysis.
Though primary amine is produced as the major product,
this process produces a mixture of primary, secondary and tertiary amines, and
also a quaternary ammonium salt
Q8.
Write short notes on acetylation.
Acetylation (or ethanoylation) is the process of
introducing an acetyl group into a molecule. Aliphatic and aromatic primary and
secondary amines undergo acetylation reaction by nucleophilic substitution when
treated with acid chlorides, anhydrides or esters. This reaction involves the
replacement of the hydrogen atom of −NH2 or > NH group by the
acetyl group, which in turn leads to the production of amides. To shift the
equilibrium to the right hand side, the HCl formed during the reaction is
removed as soon as it is formed. This reaction is carried out in the presence
of a base (such as pyridine) which is stronger than the amine.
pyridine
C2
H5NH2 +CH3COCl --------->C2H5NHCOCH3+
HCl
Q9.Why
are amines basic in character?
ANS. Like ammonia, the
nitrogen atom in amines RNH2 is trivalent and bears an unshared pair
of electrons. Thus it acts like a Lewis base and donates the pair of electrons
to electron-deficient species which further increases due to +I effect of alkyl
radical.
Q10.
Arrange the following in decreasing order of their basic strength:
C6H5NH2,
C2H5NH2, (C2H5)2NH,
NH3
The
decreasing order of basic strength of the above amines and ammonia follows the
following order:
(C2H5)2NH
> C2H5NH2
> NH3 > C6H5NH2
SOLVED EXAMPLES (2 Marks)
Q1.
Write chemical equations for the following reactions:
(i)
Reaction of ethanolic NH3 with C2H5Cl.
(ii) Ammonolysis
of benzyl chloride and reaction of amine so formed with two moles of CH3Cl
Q2. Write chemical
equations for the following conversions:
(i)
CH3
–CH2 –Cl into CH3 –CH2 –CH2
–NH2
(ii)
C6H5–CH2 –Cl into C6H5 –CH2 –CH2 –NH2
Q3.Write
structures and IUPAC names of
(i)
The amide which gives propanamine by
Hoffmann bromamide reaction.
(ii)
the amine produced by the Hoffmann
degradation of benzamide.
(iii)
Benzamide
is an aromatic amide containing seven carbon atoms. Hence, the amine formed
from benzamide is aromatic primary amine containing six carbon atoms.
(Aniline or benzenamine)
ANS.
Q5. Write the reactions of (i) aromatic and (ii)
aliphatic primary amines with nitrous acid.
(iii) Aliphatic
primary amines react with nitrous acid (prepared in situ from NaNO2 and a
mineral acid such as HCl) to form unstable aliphatic diazonium salts, which
further produce alcohol and HCl with the evolution of N2 gas.
Q6. How will you
convert:
(i)
Ethanoic acid into methanamine
(ii) Hexanenitrile
into 1-aminopentane
ANS.
(i)
(ii)
Q7. How will you
convert:
(i)
Methanol to ethanoic acid
(ii)
Ethanamine into methanamine
Q8. How will you
convert
(i ) Ethanoic acid
into propanoic acid
(ii)
Methanamine into ethanamine
Q9. How will you
convert
(i)
Nitromethane into dimethylamine
(ii)
Propanoic acid into ethanoic acid?
Q10. An aromatic compound ‗A‘on treatment with aqueous
ammonia and heating forms compound ‗B‘which on heating with Br2 and KOH forms a
compound ‗C‘of molecular formula C6H7N. Write the structures and IUPAC names of
compounds A, B and C.
Ans.
It is given that compound ‗C‘having the molecular formula, C6H7N is formed by
heating compound ‗B‘with Br2 and KOH. This is a Hoffmann bromamide degradation
reaction. Therefore, compound ‗B‘is an amide and compound ‗C‘is an amine. The
only amine having the molecular formula, C6H7N is aniline, (C6H5NH2).The given
reactions can be explained with the help of the following equations:
Long
Answers Questions
Q1. Arrange the
following:
(i)
In decreasing order of the pKb values:
C2H5 NH2, C6H5NHCH3, (C2H5)2 NH and C6H5NH2
C6H5NH2,
C6H5N(CH3)2 , (C2H5)2 NH and CH3NH2
(iii) In
increasing order of basic strength:
Aniline,
p-nitroaniline and p-toluidine
ANS. (i) The order of
increasing basicity of the given compounds is as follows:
C6H5NH2 < C6H5NHCH3
< C2H5NH2 < (C2H5)2NH
We know that the higher the basic strength, the lower
is the pKb values.
C6H5NH2 > C6H5NHCH3
> C2H5NH2 > (C2H5)2NH
(ii)The increasing order of the basic strengths of the
given compounds is as follows:
C6H5NH2 < C6H5N(CH3)2
< CH3NH2 < (C2H5)2NH
(ii) The
increasing order of the basic strengths of the given compounds is :
p-Nitroaniline <
Aniline < p-Toluidine
Q2.
Arrange the following
(i)
In decreasing order of basic strength in
gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
(ii)
In increasing order of boiling point:
C2H5OH,
(CH3)2NH, C2H5NH2
(iii) In increasing order
of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2.
Ans. (i) The given
compounds can be arranged in the decreasing order of their basic strengths in
the gas phase as follows:
(C2H5)3N
> (C2H5)2NH > C2H5NH2 > NH3
(ii) The
given compounds can be arranged in the increasing order of their boiling points
as follows:
(CH3)2NH < C2H5NH2 < C2H5OH
(iii)
The more extensive the H−bonding, the
higher is the solubility. C2H5NH2
contains
two
H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2 undergoes
more extensive H−bonding than (C2H5)2NH. Hence, the solubility in water of
C2H5NH2 is more than that of (C2H5)2NH.
Q3. Accomplish the
following conversions:
(i)
Nitrobenzene to benzoic acid
(ii)
Benzene to m-bromophenol
(iii)
Benzoic acid to aniline
(iii)
Q4.
Accomplish the following conversions:
(i)
Aniline to 2,4,6-tribromofluorobenzene
(ii)
Benzyl chloride to 2-phenylethanamine
(iii)
Chlorobenzene to p-chloroaniline
(i)
Aniline to p-bromoaniline
(ii)
Benzamide to toluene
(iii) Aniline
to benzyl alcohol.
(iv) ANS.
Long
Answers Questions
Q1. Give the
structures of A, B and C in the following reactions:
ANS. (i)
(iii)
(iv)
Q2. Complete the
following reactions:
(v)
ANS.
(i)
Assignments
1.
Write IUPAC Name of C6H5N(CH3)3Br
?
2. Which
reaction is used for preparation of pure aliphatic & aromatic primary
amine?
3. Name
one reagent used to distinguish between primary, secondary & tertiary
amines?
4.
What is the directive influence of amino
group in arylamines?
5.
Why are benzene diazonium salts soluble in
water?
6.
Which is more basic: CH3NH2&
(CH3)3N?
7.
Which is more basic, aniline or ammonia?
8.
Write the IUPAC name of C6H5NHCH3?
9.
Mention two uses of sulphanilic acid?
Level
2
1.
Write the use of quaternary ammonium
salts?
2.
What product is formed when aniline is
first diazotized and then treated with Phenol in alkaline medium?
3.
How is phenyl hydrazine prepared from
aniline?
4.
What is the IUPAC name of a tertiary amine
containing one methyl, one ethyl and one n-propyl group?
5.
Explain why silver chloride is soluble in
aqueous solution of methylamine?
6.
Write the IUPAC name of C6H5N+(CH3)3Br-?
7.
Primary amines have higher boiling points
than tertiary amines why?
8.
Why is it necessary to maintain the
temperature between 273 K & 278 K during diazotization?
9.
Arrange the following in order of
decreasing basic strength: Ethyl amine, Ammonia, Triethylamine?
10.
Why aniline is acetylated first to prepare
mono bromo derivative?
LEVEL
3
1. Arrange the following in decreasing order of
their basic strength.
C6H5NH2,
C2H5NH2, (C2H5)2NH,
NH3 2. Write chemical equation for the conversion
CH3-CH2-Cl into CH3–CH2-CH2-NH2
3.
Write the equation involved in
Carbylamines reactions?
4.
How will you distinguish the following
pairs?
(i)
Methanamine and N-methyl methanamine
(ii)
Aniline and ethylamine
5. Write chemical reactions involved in following name
reactions.
(i) Hoffmann Bromoamide reaction.
(ii)
Diazotisation reaction.
Basic character of
amines in aqueous and in gaseous state, pka and pkb
values
1
MARK QUESTIONS
Q1. Arrange the following in decreasing order of their
basic strength: C6H5NH2, C2H5NH2,
(C2H5)2NH, NH3
Q2. Arrange the following in decreasing order of the pKb
values: C2H5NH2, C6H5NHCH3,
(C2H5)2NH and C6H5NH2
Q3. pKb
of aniline is more than that of methylamine. Why?
Q4. Ethylamine is soluble in water whereas aniline is
not. Give reason. Q5. Methylamine in water reacts with ferric chloride to
precipitate hydrated ferric oxide. Why?
Q6. Although amino group is o–and p–directing
in aromatic electrophilic substitution reactions, aniline on nitration gives a
substantial amount of m-nitroaniline. Give reason.
Q7. Aniline does
not undergo Friedel-Crafts reaction. Why?
Q8. Diazonium salts of aromatic amines are more stable
than those of aliphatic amines. Why?
Q9.
Gabriel phthalimide synthesis is preferred for synthesising primary amines. Give
reason
Q10. Why cannot aromatic primary amines be prepared by
Gabriel phthalimide synthesis?
Q11. Why do primary amines have higher boiling point
than tertiary amines?
Q12. Why are aliphatic amines stronger bases than
aromatic amines? Q13. Direct nitration of aniline is not carried out. Give
reason.
Q14. The presence of base is needed in the ammonolysis
of alkyl halides. Why?
2
MARKS QUESTIONS
Q1. Write
structures and IUPAC names of
(i) the
amide which gives propanamine by Hoffmann bromamide reaction.
(ii) the
amine produced by the Hoffmann degradation of benzamide. Q2. Give one chemical
test to distinguish between the following pairs of compounds.
(i)
Methylamine and dimethylamine (ii)
Ethylamine and aniline
Q3. Write short
notes on the following:
(i) Carbylamine
reaction (ii) Diazotisation
Q4. Explain the
following with the help of an example.
(i) Hofmann’s bromamide reaction (i
Q5. Explain the
following with the help of an example.
(i)
Ammonolysis (ii) Gabriel phthalimide synthesis
Q6. How can you convert an amide into an amine having
one carbon less than the starting compound? Name the reaction.
Q7. Give a
chemical test to distinguish between:
(a)
C6H5NH2&
CH3NH2
(b)
CH3NHCH3& (CH3)3N
Q8. Give the IUPAC
names of:
(a)
(CH3)2CHNH2
(b)
(CH3CH2)2NCH3
Q9. Write the structures
of:
(a)
3-Bromobenzenamine
(b)
3-Chlorobutanamide
Long
Answer Questions
Q1. How will you
convert?
(i) Benzene into aniline (ii) Benzene into N,
N-dimethylaniline (iii) Aniline to Sulphanilic acid
Q3.
How will you carry out the following conversions (Write Chemical equations and
reaction conditions):
(a) Aniline to
Phenol
(b)Acetamide to
Ethylamine
(c) Aniline to p-nitro
aniline
VALUE BASED QUESTIONS:
1. Sushil’s friend want to play Holi with
synthetic colours, eggs, muddy water etc. Sunil persuades his friends to play
Holi with natural colours. He reminds them that last time one of their friends
had developed skin allergy after playing Holi with synthetic colours. It took
him a long time to recover Sushil’s friends agreed and prepared natural colours
using leaves and flowers
i)
Mention the values shown by Sushi
ii)
Write the name of two pigments present in natural colours
iii)
Write the names and reaction of preparation of two azo dyes
2. A mother brought her 2
year old child to a clinic with a complaint that the child would not stop
crying and was profusely vomiting. The doctor noticed the child shirt collar
colour was faded.
a) Why dyes are coloured? Give 2 reasons?
b) What is coupling reaction? Give
equation
c) What are the values involved in this?
3.
Ram & Shyam are good friends, from two different localities. Ram used to go
to cake point which
used
a colour light as insect attractants, but Shyam used to go to a way side XYZ
bakery. Shyam was
affected
with poor health and ulcer in the stomach and admitted in the hospital. While
Ram stayed
healthy.
The Doctor diagnosed and found Shyam’s poor health due to poor quality of the
food.
a)
How do you distinguish primary and secondary amine?
b)
Which organic compound is used as insect attractants?
c) What is the
lesson learnt by Shyam from Ram?
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