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Published by NOR IRWAN SHAH MOHAMED NOOR, 2021-04-26 21:26:13

TMK CHAPTER 18 AMINES

TMK CHAPTER 18 AMINES

TUTORIAL 18 CHEMISTRY SK025
CHAPTER 18: AMINES
18.1: Introduction & 18.2: Nomenclature
NH2
1. Classify & name 1-butanamine
(a) 1 amine N-isopropyl-2-propanamine
(b) 2 amine 2-phenyl-1-propanamine
(c) 1 amine 2-chloroaniline @ o-chloroaniline
(d) 1 amine N-ethyl-N-propylaniline
(e) 3 amine

2. Structural formulae (b)
(a) CH3CH(NH2)CH3 HOOC

(c) (d) CH3CH2CH2CH2N(CH3)2

CH3 CH2 CH2 CH2 NCH2 CH3

18.3 : Physical Properties

1. Increasing order of boiling point
(a) CH3CH2CH2CH3 < CH3CH2CH2NH2< CH3CH2CH2OH
 All the molecules have a similar molecular mass.
 Alkane/butane is not capable of forming hydrogen bonding between molecules, it
only has weak van der Waals forces. Hence, the boiling point of CH3CH2CH2CH3
is the lowest.
 Amines and alcohols form hydrogen bonds between their molecules.
 The electronegativity of O is higher than N. Hence, the hydrogen bonding between
alcohol molecules are stronger than that between amine molecules. Therefore,
CH3CH2CH2OH has the highest boiling point.

(b) chloroethane< 1-propanamine < 1-propanol <ethanoic acid
 Chloroethane has no hydrogen bonding between molecules. It only forms weak
van der Waals forces. Hence, the boiling point of chloroethane is the lowest.
 1-propanamine, 1-propanol and ethanoic acid have hydrogen bonds between
their molecules
 The electronegativity of O is higher than N. Hence, the hydrogen bonding for 1-
propanol is stronger than 1-propanamine.
 Ethanoic acid has the highest boiling point because the molecules are arranged
as closely packed dimers resulting in a stronger hydrogen bond than in 1-
propanol.

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CHAPTER 18: AMINES

(c) N,N-dimethylmethanamine<N-methylethanamine<propanamine
 N,N-dimethylmethanamine only can form weak Van der Waals forces between
molecules.
 Both N-methylethanamine and propanamine can form hydrogen bond between their
molecules but propanamine can form more hydrogen bond.
 Therefore energy needed to overcome the forces N,N-dimethylmethanamine< N-
methylethanamine<propanamine.

2. Ascending order of basicity
basicity of aniline < ammonia <methanamine
 Methanamine is a stronger base than ammonia because alkyl group is an electron
releasing group which increases the electron density at nitrogen atom. Therefore, it
increases the tendency of nitrogen to donate electron and makes it more basic.
 Aniline is a weaker base than ammonia because the lone-pair electrons on the nitrogen
atom are delocalised on the benzene ring.

3. Compound that is more soluble in water
a) 1-butanamine
Explain:
 1-butanamine is a primary amine and has low steric effect, therefore can easily
form hydrogen bond with water.
 While N-ethyl-1-ethanamine is secondary amine and has higher steric effect,
therefore difficult to form hydrogen bond with water.

b) CH3CH2NHCH3
Explain:
 CH3CH2NHCH3 is secondary amine and (CH3)3N is tertiary amine.
 (CH3)3N has higher steric effect compared to CH3CH2NHCH3.
 (CH3)3N difficult to soluble in water.

18.4 : Preparation & 18.5: Chemical Properties

1. Product

(a) A: CH3CH2NH2 excess

(b) B: (CH3)3CCH2CH2NH2

(c) C: i) LiAlH4, ether ii) H2O

(d) D:

NO2

(e) E: NO2

Br2, in NaOH(aq)

2. (a) Synthesis

OH PCl3 @ PCl5 Cl KCN @ NaCN CN
CH3 CHCH3 CH3 CHCH3 CH3 CHCH3

i)LiAlH 4,ether
ii)H 2O

CH2 NH2
CH3 CHCH3

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CHAPTER 18: AMINES

(b) CH3 CH3
Br NH2
CH3 excess NH3

HBr, H2O2

3. Reaction equation
(a)

+ + +CH CH NH NaNO2, HCl H C CH CH CH OH CH3 CH2 Cl N2

3 2 2 < 5 °C 2 2 32

(b) (CH ) NH NaNO2, HCl (CH ) NNO

3 2 < 5 °C 32

CH3 CH3 CH3

NaNO2, HCl + - + -

CH3 CH2N CH3 Cl + CH Cl
H
(c) CH3 CH2N CH3 CH3CH2N 3
NO
< 5 °C

(d) NaNO2, HCl +

NH2 NN
below 5 °C

above 5 °C

OH + N2(g)

4. Chemical test

(a)

Test/Reagent Br2(aq)

Compound aniline benzamide

NH2 O
C

NH2

Observation white precipitate is formed white precipitate does not form
Equation
Br NH2
NH2 3Br2(aq) + 3HBr

Br Br

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CHAPTER 18: AMINES

(b)

Test/Reagent nitrous acid, NaNO2/HCl

Compound Ethanamine (1°) N,N-diethylethanamine (3°)

CH3CH2NH2 CH3CH2N CH2CH3
CH2CH3

Observation gas bubbles are released clear solution is formed

Equation NaNO ,HCl
2 H2C=CH2 + CH3CH2OH + CH3CH2Cl + N2 (g)
CH3CH2NH2 below 5 °C

NOCl- CH2CH3

NaNO ,HCl CH+ + CH + -

CH3CH2 N CH 2CH 3 below 5 2 3CH2 N CH2CH3 HCl
3CH2 N
°C CH2CH3
CH2CH3
CH2CH3

5. Structure
C: H ratio suggests the presence of benzene ring.
Hinsberg’s test shows that the compound is 1 amine.
Reaction with nitrous acid shows that it is primary aromatic amine:

NH2 NH2 @ H3C NH2
CH3
@

H3C

Equation SO2Cl SO2 NH
H3C
NH2 +
CH3 precipitate

HCl KOH

K-+
SO2N

H3C
clear solution

NaNO2,HCl +
NH2
NN
5C

CH3 CH3

o

above 5 C

OH + N2(g)

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CHAPTER 18: AMINES

CH3

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CHAPTER 18: AMINES

MEKA 18

1. Classify & name
a) 2-methyl-2-butanamine (1°)
b) N-ethylaniline (2°)
c) N-ethyl-N-isopropyl-1-propanamine (3°)

2. Increasing order of boiling point
propanone< 2-propanamine < 2-propanol <ethanoic acid.

Reason:
 Ethanoic acid, 2-propanol and 2-propanamine can form hydrogen bond between their

molecules while propanone only can form weak Van der Waals force between their
molecules.
 Ethanoic acid forms closely packed dimers resulting in a stronger hydrogen bond
than 2 propanol and 2-ethanamine.
 Since, nitrogen in 2-propanamine is less electronegative compare to oxygen,
therefore, the strength of hydrogen bond between 2-propanol molecule > 2-
propanamine.

3. Structure CH3 NH2 CH3
a) methanamine H3C N CH3

NH2 trimethylamine

aniline

b) Decreasing order of basicity
trimethylamine > methanamine > aniline

Reason:
Basicity is the tendency of a compound to donate electrons. The easier for amines to
donate electrons, the more basic it is.

 Trimethylamine is more basic than methanamine because:
- Trimethylamine has more alkyl groups than methanamine which act as the

electron releasing group.
- Three alkyl groups increase the electron density at nitrogen atom in

trimethylamine.
- So, tendency of nitrogen to donate electron is higher in trimethylamine.

 Methanamine is more basic than aniline because:
- Methanamine has alkyl group but aniline does not.
- Alkyl group in methanamine increases the electron density at nitrogen atom.
- This increase the tendency of nitrogen to donate electron.

 Aniline is the least basic because:

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CHAPTER 18: AMINES
- The lone pair electrons of nitrogen atom are delocalized into the benzene ring.

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CHAPTER 18: AMINES

- Electron density at the nitrogen atom decreases
- This decreases the tendency of nitrogen to donate the electrons.
- So, it is the least basic among three compounds given.

4. Chemical test Nitrous acid test NHCH3 NH2
Test/ Reagent Reagent:NaNO2, HCl
Temperature: below 5°C
compound
CH2NH2

Observation 1° aliphatic amine 2° aromatic amine 1° aromatic amine
Formation of gas Yellow oily liquid Clear solution formed. No
bubbles below 5°C. formed. gas bubbles released below
5°C.
CH2NH2
NaNO2, HCl CH2 N + - +
below 5 °C
NCl H 2C
-N2
-
+
Cl

H2C Cl H2C OH

+

NHCH3 O N N CH3

NaNO2, HCl
below 5 °C

NH2 +-
N N Cl
NaNO2, HCl

below 5 °C

5. Complete equation

M: N:

O O
C
C CH CH CH
Cl
O 2 3

CH3

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CHAPTER 18: AMINES

P: Q:

O H
C N CH2CH3 C N CH CH

H 23

HH

Reagent R: NaNO2, HCl, < 5oC

6. reagent B: KMnO4, H+, heat
a) A: i. LiAlH4 ii. H2O

b) C: NH3, ∆ D: Br2, NaOH(aq) E:i. LiAlH4, ether ii. H2O

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CHAPTER 18: AMINES

KUMBE 18 Boiling point
1. a) N,N-dimethylmethanamine is a 3o amine. It cannot form hydrogen bond between
molecules. It can only form Van der Waals force between molecules. 1-propanamine
b) can form hydrogen bonds between molecules. Hydrogen bond is stronger compare to
Van der Waals forces.
c)
Solubility
Methylamine has smaller alkyl group compare to trimethylamine.
Methylamine has smaller hydrophobic area compare to trimethylamine.
Thus, methylamine easy to form hydrogen bond with water molecules compare to
trimethylamine.

Basicity NH2
NH2

< NH3 <

Aniline is the least basic because the lone pair of nitrogen atom is delocalised on N
atom and benzene ring thus making it less available to donate lone pair than ammonia.
Alkyl group in cyclohexanamine is electron donating group that increase the electron
density of N. Thus making cyclohexanamine easy to donate lone pair compare to
ammonia.

2. a) Higher melting point
Y.
Because Y forms hydrogen bonds between molecules. X does not.

b) Chemical test Observation / Equation
Test / Reagents
O
O2N
CH3 H2N C OH

Reaction with No precipitate formed. Precipitate formed.
bromine water, Br2
(aq) O Br
Br2 (aq) O

H2N C OH H2N C OH

Br

c) Synthesis 

i. Fe, HCl 

O2N CH3 ii. NaOH H2N CH3

X KMnO4,3H O+ 

H2N OH

SK
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CHEMISTRY SK025
CHAPTER 18: AMINES

Y

SK
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CHAPTER 18: AMINES

3. Structural formulae
X is a secondary aromatic amine because it gives oily yellow liquid when reacts with nitrous
acid and form N,N,4-trimethylaniline when treated with chloromethane.
Structure X:

H

H3C N CH3

4. Structural formulae
P: CH3CH2CN
Q: CH3CH2CH2NH2
R: CH3CH2CH2Cl + CH3CH2CH2OH + CH3CH CH2

Chemical equation:

CH CH CN i. LiAlH4,ether CH CH CH NH

32 ii. H2O 322 2

NaNO2,
HCl, below

5o C

+ +CH3CH2CH2Cl +CH3CH CH2 N2(g)
CH3CH2 CH2OH

5. Synthesis NO2 NH2
a) i. Fe, HCl

conc. HNO3, conc. H2SO4 ii. NaOH

55 °C

O
H3C C Cl

H
N CH3

C
O

b) CH3 KMnO , H O+ O
C
CH Br

3 43 OH

FeBr3 



CH3NH2 

O
C CH3

N
H

SK
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