Class XII (Chemistry) Unit 13 Amines

Class XII Unit 13
AMINES
By
Arunesh Gupta
PGT (Chemistry)
KV Barrackpore (AFS)
1

Some natural amino acids
Natural anaesthetic drugs
Anaesthetic drug.
In nature, amines occur among proteins, vitamins, alkaloids, hormones etc. Adrenaline &
ephedrine are used to increase blood pressure. Benadryl is used in dentistry. Dyes are
synthesized from diazonium compounds
ARUNESH GUPTA PGT (CHEMISTRY)
2

Structures of some N containing organic compounds are addictive
analgesics.
ARUNESH GUPTA PGT (CHEMISTRY) 3

Amines:
Amines constitute an important class of organic compounds derived by
replacing one or more hydrogen atoms of NH3 molecule by alkyl/aryl group(s).
4

AMINES: When one hydrogen atom is replaced by –R or –Ar, gives rise to a
primary amine or 1⁰ amine.
The structure of secondary amines or 2⁰ amine is given by R2NH or R-NH-R′
whereas tertiary amines or 3⁰ amines are characterized by R-NR′R′′ or R2N-R′ or
R3N.
If the aryl or alkyl group of tertiary and secondary amines are same they are called
simple amines,(R3N:) but if they are attached to different groups they are called
mixed amines (RNHR’).
Amines have one unshared electron pair on the nitrogen atom, therefore, they
behave as Lewis bases. Amines are sp3 hybridised.
Structure: The nitrogen atom in amine is sp3-hybridised.
5

In the IUPAC system, the amines are regarded as alkanamines. R – N<
The three hybrid orbitals are involved in bond formation and one hybrid
atomic orbital contains the lone pair of electrons, giving the pyramidal
geometry of amines.

Preparation of amines:
1) Reduction of nitro compounds: Reducing agents are H2 + Pd (or Pt or Ni) in
ethanol, Sn (or Fe or Zn) + HCl etc.

2) Ammonolysis of alkyl halides to form 1°, 2°, 3° & 4° amines:
The free amine can be obtained from ammonium salt by reaction with
a strong base. R - NH3
+X- + NaOH  R-NH2 + H2O + Na+X-
Ammonolysis gives a mixture of primary (1°), secondary (2°), tertiary
(3°) & quaternary (4°)amines. But with excess ammonia, R-X mostly
form R – NH2.
R-X
𝐍𝐇𝟑 (𝐞𝐱𝐜𝐞𝐬𝐬)
R-NH2

3. Reduction of nitriles (R – CN) using LiAlH4 or (H2 + Ni ) or (Na/Hg +
C2H5OH) to form 1° or primary amine.
4) Reduction of amides: by LiAlH4 to form amine
R-OH
𝐏𝐂𝐥𝟓
R-Cl
𝐀𝐥𝐜𝐨𝐡𝐨𝐥𝐢𝐜 𝐊𝐂𝐍
R-CN
𝐇𝟐+𝐍𝐢 𝐜𝐚𝐭𝐚𝐥𝐲𝐬𝐭
R-CH2-NH2 (1° amine with 1 C atom more)
R-CH2OH
𝐎 𝐊𝐌𝐧𝐎𝟒| 𝐇+
R-COOH
𝐍𝐇𝟑,∆
R-CO-NH2
𝐒𝐧+𝐇𝐂𝐥
RCH2-NH2 (1° amine with no. of same C atom

*Schmidt Reaction:
R – COOH + N3H (hydrazoic acid)
𝐂𝐨𝐧𝐜.𝐇𝟐𝐒𝐎𝟒,∆
R-NH2 + H2O + N2
(No. of C atom decreases by one C atom ⇒ stepping down)
Gabriel phthalimide synthesis:
This method is used to prepare pure 1°aliphatic amine.( 1° aromatic amine cannot be
prepared. C6H5-X can not form phenyl carbocation due to+ R effect of –X group forming
partial double bond character in C – X bond.
11

Hoffmann bromamide degradation reaction:
Amides are reacted by Br2 in aqueous or ethanolic solution of NaOH to form
primary amines.
Mechanism:

Commercial preparation of aniline
*** R-CHO + H2N-R’
𝐇+
RCH=N-R’
𝐇𝟐+𝐍𝐢
R-CH2-NH-R’ (2° amine)

Physical properties:
1) The lower aliphatic amines are gaseous in nature. They have a fishy
smell.
2) Primary amines with three or four carbon atoms are liquids at room
temperature due to inter molecular hydrogen bonds whereas higher
ones are solids.
3) Aniline and other arylamines are generally colourless. However, they
get coloured when we store them in open due to atmospheric
oxidation.
⇒The electron density on aniline increases due to + R effect of -NH2
group and due to high electron density aniline gets readily oxidized in
air to give colour products.
4) Lower aliphatic amines can form hydrogen bonds with water
molecules. Therefore, such amines are soluble in water.

16

Basic Nature of Amines:
Amines are Lewis bases due to the presence of lone pair of electrons on
the nitrogen atom. In aqueous solution, it forms hydroxyl ions.
R – NH2 + H2O ⇌ R - NH3
+ + OH-
According to law of chemical eqution : Dissociation constant of base (Kb)
Kb =
𝐑𝐍𝐇𝟑
+
𝐎𝐇−
𝐑𝐍𝐇𝟐
and pKb = - log Kb
⇒ Higher value of Kb ⇒Lower value of pKb ⇒ more basic

Basic property of amines:
Amines reacts with mineral acids like HCl, H2SO4 etc to form water soluble
salts
Try:
Aniline is less soluble in water but completely soluble in dilute HCl. Why?
18

BASIC STRENGTH :
(C2H5)2N-H > (C2H5)3N > (CH3)2N-H > C2H5NH2 > CH3NH2 > (CH3)3N > C6H5-CH2NH2
> C6H5-N(CH3)2 >C6H5-NHCH3 > C6H5-NH2 (decreasing order)
Qu. Alkyl amine is more basic than ammonia. Why?
Ans: because the electron donating effect of the alkyl group increases the
electron density on the nitrogen and thus making alkyl amine more basic.
Example: CH3-NH2 is more basic than NH3. or (Kb)methyl amine > (Kb)ammonia
And this is a general trend, the more alkyl groups (more +I effect) on the
nitrogen, the stronger the basicity of the amine.
The order of basicity of amines in gaseous phase follows:
Tertiary amine > Secondary amine > primary amine > ammonia.
20

Basic strength of aliphatic amines in aqueous medium is controlled by (i) Inductive
effect (ii) Steric effect (iii) Hydration effect.
(i) Inductive effect: +I effect is maximum in 3° amine due to the presence of three –R
groups than that of Type equation here.1° amine. So electron cloud density is
maximum on N atom in 3° amine.
(ii) ⇒ Basic strength: 1° < 2° < 3° amine.
(ii) Steric effect: The steric repulsion is maximum in 3° amine, so the three –R
groups will repel the incoming acid maximum for 3° amine than 1° amine. Basic
strength: 1° > 2° > 3° amine.

(iii) Hydration effect: amine forms maximum hydrated conjugate acid due to the
presence of three –H atoms. Stability of hydrated conjugate acid:
RNH3
+(H2O)3 > R2NH2
+(H2O)2 > R3NH+(H2O).
So Hydration effect⇒ basic strength: 1° > 2° > 3° amine.
Considering all the above three factors together, we conclude the basic
strength:
(a) (CH3)2NH > CH3NH2 > (CH3)3N > NH3
(b) (C2H5)2NH > (C2H5)3N > C2H5-NH2 > NH3
22

Basic strength of aniline: Aniline is less basic than aliphatic amine because –NH2
group shows +R effect & due to resonance, the electron density on N atom
decreases & basic strength decreases.
In case of substituted aniline,
electron releasing groups like
–OCH3, -CH3 increases the
basic strength whereas the
electron withdrawing groups
like –NO2, -SO3H, -COOH, -X
decreases the basic strength.

Chemical reactions:
1) Alkylation: Alkylation of 1° amine forms 2° amine, 3° amine & 4° quaternary
ammonium salt.

2) Acylation: Aliphatic & aromatic 1° amine & 2° amine react with acid
chlorides, acid anhydrides & esters by SN reaction in presence of strong
base, called acylation, forming amides.
25

3) Carbylamine reaction or isocyanide reaction:
Aliphatic & aromatic primary amines on heating with chloroform & alcoholic
KOH forms isocyanides or carbylamines which are foul smelling substances.
Secondary or tertiary amines do not give this reaction.
This reaction is used to test for primary amines.
(Ph- or) R-NH2 + CHCl3 + 3KOH
𝚫
(Ph- or) R – NC + 3KCl + 3H2O. (Ph- C6H5)
Try: Give a chemical test to identify
(i) C2H5NH2 & CH3NHCH3 & (ii) C6H5NH2 & C6H5H(CH3)2
Mechanism:
26

Reaction of primary aliphatic amine with HNO2 forms unstable
aliphatic diazonium salts which liberate nitrogen gas and alcohols.
27

(ii) Secondary amines (both aliphatic and aromatic) react with
nitrous acid to form nitrosoamines which separate as yellow oily
liquids.
28

(iii) Tertiary aliphatic amines (3°) dissolve in a cold solution of nitrous
acid to form salts which decompose on warming to give nitrosoamine
and alcohol. For example,
(iii) Tertiary aromatic amines (3°) react with nitrous acid to give a
coloured nitroso derivative. This reaction is called nitrosation and as
a result, a hydrogen atom in the para position gets replaced by a
nitroso (-NO) group. For example,
29

Reaction of aromatic amines with nitrous acid (HNO2)
30

Hinsberg’s Test:
Test to distinguish between of primary, secondary and tertiary
amines.
An amine is shaken with Hinsberg’s reagent (benzene sulphonyl
chloride) in the presence of excess of aqueous KOH solution.
A primary amine forms a precipitate of N – alkyl benzene
sulphonamide which dissolves in aqueous KOH solution to form
water soluble potassium salt and upon acidification with dilute HCI
regenerates the insoluble sulphonamide.
31

A secondary amine forms a precipitate of N, N – dialkylbenzene suiphonamide
which remains insoluble in aqueous KOH and even after acidification with dilute HCl.
A tertiary amine does not react with benzene suiphonyl chloride and remains
insoluble in aqueous KOH.
However, on acidification with dilute HCI it gives a clear solution due to the formation
of the ammonium salt.
32

Hinsberg’s test for 1°, 2, 3° amines
33

Electrophilic substitution reaction of aniline.
In aniline, -NH2 group shows +R effect & due to resonance, it increases
electron density at ortho & para position as below & allows electrophile to
attack at o- & p- positions.
(i) Bromination: With bromine water, aniline forms
white precipitate of 2, 4,6-tribromoaniline.

(ii) Aniline to 4-bromo aniline: To form mono substituted products, -NH2 group
should be protected through acylation. –NH2 group in Aniline is a strong ortho, -
para activator. Aniline and ethanoyl chloride react to give N-phenylethanamide
which is a ortho-para moderate activator due to resonance. This product is an
amide compound -NH-CO-CH3 group results a steric impediment around that
group. Therefore, substitutions to ortho positions are more difficult than para places.
Therefore major product is para substituted compound, 4-bromoaniline.
35

(iii) Convert: (a) Aniline to 4-nitroaniline:
Convert (b) aniline into phenol

(iv) Convert: p-nitro aniline to 1,2,3-tribromoaniline

(v) Convert: p-nitroaniline into 2,4,6-tribromoaniline.
38

Nitration of aniline
It forms 51% p-nitro aniline & 47% m-nitro aniline & small amount of o-nitro
aniline on reaction with conc. HNO3 & conc. H2SO4.because, conc HNO3
protonated aniline to form C6H5-NH3
+ (anilinium ion) which is electron defficient
& deactivates the benzene ring at o- & p- positions & electrophile NO2
+ attacks
at meta position to give m-nitro aniline.
39

Sulphonation of aniline:
Aniline being a base reacts with cons. H2SO4 to form salt of hydrogen
sulphate which on heating rearrange to form p-amino benzene
sulphonic acid (or sulphanilic acid). It further rearranges internally by
proton transfer to form a zwitter ion or dipolar ion. It has high melting
point & is soluble in hot water.
40

Uses of Amines:
(i) Lower members of aliphatic amines like ethanamine,
diethylamine, triethylamine etc are used in synthesis of organic
compounds in laboratory & industry.
(ii) (ii) Quaternary ammonium salts of long chain tertiary amines are
used as detergents.
(iii)(iii) Aniline & other aromatic amines are used o manufacture of
dyes & drugs.
(iv)Diazonium salts of aniline are used to synthesize azodyes etc.
41

42

Qu.1(a) Classify the following amines as primary, secondary and tertiary:
Ans: (i) 1° (ii) -3° (iii) 1° (iv) 2°
Qu.(b) Out of CH3—NH2 and (CH3)3N, which one has higher boiling point?
Answer: CH3—NH2 has higher boiling point than (CH3)3N due to inter
molecular H bonding.
Qu.(c) Write IUPAC name of the following compound: (CH3CH2)2NCH3 .
Answer: N-Ethyl-N-methylethanamine
Qu.2) Write the structure of prop-2-en-l-amine.
Answer: H2C=CH—H2C—NH2
43

13.2. Write the structures of different isomeric amines corresponding to the molecular formula,
C4H11N.
(i) Write the IUPAC names of all the isomers
(ii) What type of isomerism is exhibited by different types of amines?
Ans: Eight isomeric amines are possible.
Isomerism exhibited by different amines
Chain isomers: (i) and (ii) ; (iii) and (iv) ; (i) and (iv)
Position isomers: (ii) and (iii) ; (ii) and (iv)
Metamers: (v) and (vi) ; (vii) and (viii)
Functional isomers  All the three types of amines.
44

3. How will you convert: (i) Benzene into aniline
(ii) Benzene into N,N-dimethylaniline (iii) Cl-(CH2)4-Cl into Hexane -1,6- diamine
45

Qu.4) Arrange the following in increasing order of their basic strength :
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2, (C2H5)2NH
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6HsNH2, C6H5CH2NH2
Ans. The Kb values are :
46

Qu.5) Complete the following acid-base reactions and name the products:
(i) CH3CH2CH2NH2+HCl ——–> ? M (ii) (C2H5)3 N+HCl ——–> ?
Ans.
47

Qu.6) Write structures of different isomers corresponding to the
molecular formula, C3H9N. Write IUPAC names of the isomers which will
liberated N2 gas on treatment with nitrous acid.
Ans: Four structural isomers are possible. These are:

Qu.7) Convert: 3-Methylaniline into 3-nitrotoluene.
Qu.8) Give one chemical test to distinguish between the following
pairs of compounds:
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
49

50

Qu.8) Give one chemical test to distinguish between the following pairs of
compounds: (iv) Aniline and benzylamine
(v) Aniline and N-Methylaniline.
51

Qu.9) Arrange the following:
(i) In decreasing order of pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH and
C6H5NH2
Answer: (i) Due to delocalization of lone pair of electrons of the N-atom over the
benzene ring,C6H5NH2 and C6H5NHCH3 are far less basic than C2H5NH2 and (C2H,)2NH.
Due to +I-effect of the -CH3 group, C6H5NHCH3 is little more basic that C6H5NH2. Among
C2H5NH2 and (C2H5)2NH, (C2H5)2NH is more basic than C2H5NH2 due to greater+I-effect
of two -C2H5 groups. Therefore correct order of decreasing pKb values is:
(ii) In increasing order of basic strength:C6H5NH2, C6H5N(CH3)2, (C2H5)2 NH and
CH3NH2.
Answer: (ii) Among CH3NH2 and (C2H5)2NH, mainly due to the greater +I effect of the two
- C2H5 groups over one -CH3 group, (C2H5)2NH is more basic than CH3NH2. In both
C6H5NH2 and C6H5N(CH3)2 lone pair of electrons present on N-atom is delocalized over
the benzene ring but C6H5N(CH3)2 is more basic due to +I effect of two - CH3 groups.
52

Qu.9) (iii) In increasing order of basic strength:
(а) Aniline, p-nitro aniline and p-toluidine (b)C6H5NH2, C6H5NHCH3, C6H5CH2NH2
(iv) In decreasing order of basic strength in gas phase:
C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3
Answer: (iii) (a) The presence of electron donating -CH3 group increases while the
presence of electron withdrawing -NO2 group decreases the basic strength of amines.
Answer:(b) In C6H5NH2 and C6H5NHCH3, N is directly attached to the benzene ring. As a
result, the lone pair of electrons on the N-atom is delocalised over the benzene ring.
Therefore, both C6H5NH2 and C6H5NHCH3 are weaker base in comparison to
C6H5CH2NH2. Among C6H5NH2 and C6H5NHCH3, due to +1 effect of-CH3 group
C6H5NHCH3 is more basic.
Answer: (iv) In gas phase or in non-aqueous solvents such as chlorobenzene etc, the
solvation effects i. e., the stabilization of the conjugate acid due to H-bonding are
absent. Therefore, basic strength depends only upon the +I-effect of the alkyl groups.
The +I-effect increases with increase in number of alkyl groups.Thus correct order of
decreasing basic strength in gas phase is,
53

Qu. (v) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2
Answer: (v) Since the electronegativity of O is higher than that of N,
therefore, alcohols form stronger H-bonds than amines. Also, the extent of H-
bonding depends upon the number of H-atoms on the N-atom, thus the extent
of H-bonding is greater in primary amine than secondary amine.
(vi) In increasing order of solubility in water: C6H5NH2,(C2H5)2NH,C2H5NH2
Answer: (vi) Solubility decreases with increase in molecular mass of
amines due to increase in the size of the hydrophobic hydrocarbon part
and with decrease is the number of H-atoms on the N-atom which
undergo H-bonding.
54

Qu.10) How will you convert:
(i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid. (iv) Ethanamine into methanamine.

Qu.10) How will you convert: (v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid?

Qu.11) Convert: (i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol

Qu.11) Convert: (iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromofluorobenzene
58

Qu.11) Convert: (v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-Chloroaniline (vii) Aniline to p-bromoaniIine

Qu.11) Convert: (viii) Benzamide to toluene
(ix) Aniline to benzyl alcohol.
60

Qu.12) Give the structures of A, B and C in the following reaction:
61

Answer 12)
62

Qu.13) 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.
Answer: It is given that ‘B’ upon heating with Br2 and KOH forms a
compound ‘C’. The compound ‘B’ is expected to be an acid amide. Since ‘B’
has been formed upon heating compound ‘A’ with aqueous ammonia, the
compound ‘A’ is an aromatic acid. It is benzoic acid.
The reactions involved are given as follows:-
63

Qu. 14) Why cannot aromatic primary amines be prepared by Gabriel
phthalimide synthesis?
Ans. The success of Gabriel phthalimide reaction depends upon the
nucleophilic attack by the phthalimide anion on the organic halogen
compound.
Since aryl halides do not undergo nucleophilic substitution reactions easily
due to partial double bond character in C – X bond (+R effect), therefore,
arylamines, i.e., aromatic, primary amines cannot be prepared by Gabriel
phthalimide reaction.
64

Qu.15) Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
Ans: (i) Loss of proton from an amine gives an amide ion while loss of a proton
from alcohol give an alkoxide ion.
R—NH2—>R—NH– + H+ and R—O —H—>R— O– + H+ .
Since O is more electronegative than N, so it will attract positive species
more strongly in comparison to N. Thus, RO- is more stable than RNH- as
conjugate base. Thus, alcohols are more acidic than amines. Conversely,
amines are less acidic than alcohols.
(ii) Why do primary amines have higher boiling point than tertiary amines?
(ii) Due to the presence of two H-atoms on N-atom of primary amines, they
undergo extensive intermolecular H-bonding while tertiary amines due to
the absence of H-atom on the N-atom do not undergo H-bonding. As a
result, primary amines have higher boiling points than tertiary amines of
comparable molecular mass.
65

(iii) Why are aliphatic amines stronger bases than aromatic amines?
(iii) Aromatic amines are far less basic than ammonia and aliphatic
amines because of following reasons:
(a) Due to resonance in aniline (+R effect of –NH2 group) and other
aromatic amines, the lone pair of electrons on the nitrogen atom
gets delocalised over the benzene ring and thus it is less easily
available for protonation. Therefore, aromatic amines are weaker
bases than ammonia and aliphatic amines.
(b) Aromatic amines are more stable than corresponding protonated
ion (say anilinium ion); Hence, they has very less tendency to
combine with a proton to form corresponding protonated ion, and
thus they are less basic.
66

THANK YOU

Class XII (Chemistry) Unit 13 Amines

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