CONDENSATION

POLYMERIZATION
(POLYCONDENSATION)

Step-Growth polymerization
 Also known as step-growth polymerization
 The chain growth occurs in a slow, stepwise
manner
 2 monomer molecules react to form a dimer.
 the dimer can react with another monomer to
form a trimer
dimer + monomer  trimer
dimer + dimer  tetramer
trimer + tetramer 



polymer
 MW increases slowly over a period of time
 Produce a small molecule as a by-product (H2O,
HCl, CH3OH, etc)
 Monomer molecules consisting of at least 2
functional groups
 two types of linear condensation polymerization
(polycondensation)

1) Type 1 (A – B)

nA–R-B A–R–B
n

2) Type 2 (A – A & B – B)

nA – R – A + n B – R’ – B A – R – AB – R’ – B
n
 To obtain high-molecular weight polymer:
1. the amount of functional groups are equal at the
start of the reaction.
2. the monomers are pure.
3. one of the monomers does not get lost in
preference to the other.
 POLYESTERS

 Classified as polyesters because the repeating units

contain ester groups (–COO–)
 General reaction:

Type 1 (A – B)

O O
n HO – R – C – O – R* –O–R–C– + R*– OH
n
Example: O

HO (CH2)5 C OH
hydroxycaproic acid

(CH2)5 C O
n

Poly(6-hexanoate)
 Type 2 (A – A & B – B)

O O
n R*– O – C – R – C – O – R* + n HO – R’ – OH

O O
C – R – C – O – R’ – O + R*OH
n
Example: reaction between terephthalic acid and ethylene glycol

O O

HO C C OH + HO CH2CH2 OH

O O

HO C C O CH2CH2 OH + H 2O
O O

C C O CH2CH2 O + H2O
n

Poly(ethylene terephthalate), PET

Example: reaction between dimethyl terephthalic and ethylene glycol

O O

n H3C O C C O CH3 + n HO CH2CH2 OH

O O

C C O CH2CH2 O + CH3OH
n
 POLYAMIDES (nylon)
 Classified as polyamides because the repeating units
contain amide groups (–CO–NH–)
 Formed by reaction between amine and acid groups
 3 principal methods for the condensation polymerization
of polyamides:
1) reaction between a dicarboxylic acid and diamine

O O H H
n HO – C – R – C – OH + n H – N – R’ – N – H

O O H H
C – R – C – N – R’ – N + H2O
n
Example: reaction between adipic acid and hexamethylene diamine

O O H H
n HO – C – (CH2)4 – C – OH + n H – N – (CH2)6 – N – H

O O H H
C – (CH2)4 – C – N – (CH2)6 – N + H2 O
n
nylon 6,6
2) dehydration – condensation of an amino acid

Example:
H O
H2N – (CH2)5 – COOH – N – (CH2)5 – C –n + H2O

Aminocaproic acid Nylon 6

3) The reaction between a diacid chloride and diamine

 Also known as interfacial polymerization

 An interfacial polymerization takes place when the two
monomer are present in two immiscible solvents.
Reaction occurs at the interface between the two liquids
 O O H H
n Cl – C – R – C – Cl + n H – N – R’ – N – H

O O H H
C – R – C – N – R’ – N + HCl
n
Example:
O O

Cl C C Cl + H2N – CH2CH2 – NH2

Terephthaloyl chloride Ethylene diamine

O O
H H
C C N CH2CH2 N + HCl
n

Poly( ethylene terephthalamide)

Example:
O O
Cl C C Cl + H2N NH2

Terephthaloyl chloride p-phenylenediamine

H H O O
N N C C
n
Kevlar

Poly(p-phenyleneterephthalamide)
Exercise: Draw the chemical structure of the monomer(s) for the
following polymers

O O
1)
O CH2CH2CH2 O C C
n
O H
2)
C N
n

3) nylon 10, 6
 O O
4)
HN NHC C
n

nomex
Poly(m-phenyleneisophthalate)
Carother’s Theory
for linear condensation polymerization (polycondensation)
Number Average Degree of Polymerization (DP)

Type 1 (A – B)

 DP is defined as the average no. of repeating unit per

polymer molecules.

DP = total no. of monomer

total no. of polymer
No
DP = 1 DP = [A]o = [B]o 2
N [A] [B]
Where,
[A]o & [B]o = concentration of functional group A/B at
t=0
[A] & [B] = concentration of functional group A/B at
t=t
Conversion Factor (P)

 Also known as ‘extent of reaction’.

 P is the fraction of functional groups that have reacted.

[A] = [A]o (1 – P) 3

 From equation 1 , 2 & 3

* 1 Carother’s
DP = 4 Theory
1–P
 DP as a function of conversion, P (Plot DP versus P)
DP

P
0 1.0

 High MW is only achieved at very high degree of conversion

 At 90% conversion (P=0.90), DP is only 10
 At 95% conversion, P = ____ , DP = ____
 At 98% conversion, DP is still only 50. Thus, very high DP
are generally not obtainable for linear polymers by step
polymerization
 Type 2 (A – A & B – B)

DP = No 1+r
DP = 5
N 1 + r – 2rP

Where,
r = NA
NB

Theoretical Limit

1+r as P  1
DP = 6
1–r (complete reaction)
 Number Average Molecular Weight, Mn

* Mn = DP Mo

* Mn = Mo
7
1–P

Mo = average molecular weight of the structural units ?

Mo = average MW of structural units Mo = MW of repeating units
Mo = MW of repeating unit / 2

type 1

type 2
 The repeating unit of a chain derived from A – A / B – B
type monomers consists of 2 structural units.

 While the repeating unit in the case of type A – B

monomer, is the same as the structural unit
 For condensation polymers (A–B type monomers)
and vinyl polymers
Mo = MW of repeating unit

 For condensation polymers: A–A/B–B type monomers

Mo = average MW of structural units

* Mo = MW of repeating unit
2
 Weight Average Molecular Weight, Mw

*
Mw = Mo (1 + P) 8
1–P

Weight Average Degree of Polymerization, DPw = 1 + P

1–P
 Polydispersity Index, PDI

PDI = Mw
Mn

* PDI = 1 + P 9
Example:

A hydroxyl acid, HO(CH2)5COOH is polymerized and it is

found that the product has a Mn of 20 000 g/mol.
Calculate:
i. the extent of reaction, P
ii. the average degree of polymerization, DP
iii. Mw of the polymer
iv. polydispersity index
Solution:
O
HO(CH2)5COOH O(CH2)5C
n

Mo : C = 6 x 12 = 72
H = 10 x 1 = 10
O = 2 x 16 = 32
114 g/mol
i. the extent of reaction, P

Use equation 7

Mn = Mo
1–P

P = 1 – Mo = 1 – 114 g/mol
Mn 20 000 g/mol

= 0.9943
ii. the degree of polymerization, DP

1 1
DP = = = 175
1–P 1 – 0.9943

OR
Mn = DP Mo
DP = Mn = 20 000g/mol = 175
Mo 114 g/mol
iii. Mw of the polymer

Use equation 8

Mw = Mo (1 + P) = 1 + 0.994 (114 g/mol)

1–P 1 – 0.994
= 39 900 g/mol
iv. polydispersity index

PDI = Mw = 39 900 g/mol = 1.99

Mn 20 000 g/mol

OR

PDI = 1 + P = 1 + 0.99 = 1.99

 Exercise:

Nylon 6,6 of Mn 20 000 g/mol are synthesised through

polycondensation polymerisation of adipic acid and
hexamethylene amine.

Using the Carother’s Theory, calculate:

i. the extent of reaction.
ii. Mw of the Nylon 6,6.
iii. Polydispersity index of Nylon 6,6.
Rates of Polycondensation
Kinetics of polyesterification
 Reaction between two functional groups (COOH & OH)
 Consider:

A + B AB

Where,A = functional group of COOH

B = functional group of OH
– d[A] = k [A] [B]
dt
 Case 1: No acidic catalyst added

 If no catalyst is added, carboxylic acid groups (A) must

function as the acid catalyst.

– d[A] = k [A]2 [B]

dt
 Assume that [A] = [B] ( to obtain high MW polymer)

– d[A] = k [A]3
dt
 After integration,

1 1 2kt 10
[A]2 [Ao] 2

 Substitution of equation 3 into 10

* 1
2 [A]o2 kt + 1
(1 – P)2 11

remember : DP = 1/1–P

* DP2 2 [A]o2 kt + 1
12
DP2 2 [Ao] 2 kt + 1

DP2 versus t

DP2

m = 2 k [Ao] 2

t
 Case 2: Acid-catalyzed polyesterification

– d[A] = k [A] [B]

dt
 Assume that [A] = [B]
– d[A] = k [A]2
dt
 After integration,

1 1 kt 13
[A] [Ao]
 Substitution of equation 3 into 13

* 1
[Ao] kt + 1
1–P 14

* DP [Ao] kt + 1 15
DP [A]o kt + 1

DP versus t
DP

m = k [A]o

t
 Exercise:
Adipic acid (hexanedioic acid) and ethylene glycol were
polymerized at 109 oC using H2SO4 as a catalyst. The extent
of the reaction, P at various times are as follows:

t / min 20 40 80 120

P 0.800 0.909 0.949 0.960

At what time will the Mn reach 10 000 g/mol?

 Solution:

Use equation 15 because it is an acid-catalyzed polyesterification

DP [Ao] kt + 1

t / min 20 40 80 120

P 0.800 0.909 0.949 0.960

DP

DP = 1/(1-P)
 1 [COOH]o kt + 1  Plot graph 1/1-P versus t
1–P

[COOH]o k

30

25 y = 0.1716x + 4.857

20
1/1-P

15

10

0
0 20 40 60 80 100 120 140
t (min)
 Mn = DP Mo

DP = Mn = 10 000 g/mol = 116

Mo 86 g/mol
??
From the graph, [COOH]oK = 0.17 min-1
DP [COOH]okt + 1

116 = (0.17 min-1) t + 1

t = ….. min
 O O
HO – C – (CH2)4 – C – OH + HO CH2CH2 OH

Adipic acid Ethylene glycol

O O
C – (CH2)4 – C – O – (CH2)2 – O + H2 O
n

C: 8(12) = 96; O: 4(16) = 64; H: 12(1) = 12

172 g/mol

A–A/B–B type monomers

Mo = MW of repeating unit Mo = 172 = 86 g/mol

2 2