Chemistry 367L/392N

Macromolecular ChemistryMacromolecular Chemistry

Lecture 13 Lecture 13

CH

R

H2C + CO CH2 CH

R

C

O

Chemistry 367L/392N

Free Radical CopolymerizationFree Radical Copolymerization

~

~

~

~

~

~

~

~

Assume chain end concentrations are constant at “steady state”

Free Radical Co-polymerization

Chemistry 367L/392N

If r1 > 1

means M

If r1 >1 this means that ~M1• adds M1 more readily than M2. If r1 is zero then M1 does not undergo homopolymerization!!

This important relationship can be expressed in This important relationship can be expressed in

terms of mole fractions rather than concentrationsterms of mole fractions rather than concentrations

Chemistry 367L/392N

Copolymerization Equation

This gives the Instantaneous mole fraction of M1 in the copolymer

Chemistry 367L/392N

Finemann-Ross Equation

Finemann – Ross Equation

Chemistry 367L/392N

Chemistry 367L/392N

Let’s actually calculate r1 and r2 from some data

We run the copolymerization for a very short time starting with different weight ratios of 2 monomers.

The assumption is that f does not change in this

experiment

We isolate the small amount of polymer generated

and determine the ratio of monomers incorporated….by nmr, for example

We apply the Fineman-Ross analysis and extract

the reactivity ratios.

ChrisChris

Chemistry 367L/392N

Chemistry 367L/392N

r1 > 1, r2 < 1 or r1 < 1, r2 > 1

10

5

0.5

0.1

If r1 > 1 (k11 > k12) and r2 < 1 (k22 < k21) then M1 is more active than M2 whether either ~M1

• and ~M2•. F1 is

always larger than f1 and the curve is always above the diagonal.

The situation is similar for r1 < 1, r2 > 1 but the curve is below the diagonal.

Chemistry 367L/392N

as r1 x r2 = 1 (r2=1/r1), it becomes an “ideal copolymerization”

][M

][Mr

]d[M

]d[M

2

11

2

1=

In this case ~M1• and ~M2 • have the

same preference for adding either of the two monomers.

Ideal copolymerization

F

f

Chemistry 367L/392N

Condition under which F, the mole fraction of monomer in the polymer and f, the mole fraction in the feed

remain constant through the whole polymerization….that

is, these ratios are not a function conversion

Chemistry 367L/392N

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

A = styrene

B = butadiene

ra = 0.75, rb = 1.3

F

a (

po

lym

er

co

mp

ositio

n)

fa (monomer composition)

Ideal copolymer

r1=r2= 1.3

Chemistry 367L/392N

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

A = styrene

B = maleic anhydride

ra = 0.05, rb = 0.0

F

a (

po

lym

er

co

mp

ositio

n)

fa (monomer composition)

Alternating Copolymer

r1=0.05, r

2=0.0

Chemistry 367L/392N

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

A = vinyl acetate

B = styrene

ra = 0.01, rb = 50

F

a (

co

mp

ositio

n o

f p

oly

me

r)

fa (Composition of monomer)

Rich in one monomer

r1=0.01, r2= 50

Chemistry 367L/392N

0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

1.0

A = Acrylonitrile

B = Butadiene

ra = 0.046, rb = 0.36

F

a (

po

lym

er

co

mp

ositio

n)

fa (monomer composition)

Quite typical copolymerisation

Azeotropic Composition

r1=0.046, r2= 0.36

Chemistry 367L/392N

F

f

Chemistry 367L/392N

Poly(styrene-alt-maleic anhydride)

OOO

+δ

δ

OOO+

OOO

[ ]n

Charge transfer complex

Chemistry 367L/392N

Charge transfer complex polymerization(alternating copolymer).

H2C CH

R

+ SO2 CH2 CH

R

SO2

CH

R

H2C + CO CH2 CH

R

C

O

Chemistry 367L/392N

rr1 1 > 1, r> 1, r2 2 < 1 or r< 1 or r1 1 < 1, r< 1, r2 2 > 1> 1

10

5

0.5

0.1

If r1 > 1 (k11 > k12) and r2 < 1 (k22 < k21) then M1 is more active than M2 whether either ~M1

* and ~M2*. F1 is

always larger than f1 and the curve is always above the diagonal.

The situation is similar for r1 < 1, r2 > 1 but the curve is below the diagonal.

Chemistry 367L/392N

What is constant here???

This is a bit discouraging!

What is constant here???What is constant here???

This is a bit discouraging!This is a bit discouraging!��

Chemistry 367L/392N

Estimating Reactivity Ratios Estimating Reactivity Ratios

measurement of r1 and r2 is possible by applying the Fineman Rose equation but it is a bit tedious!

Chemistry 367L/392N

Chemistry 367L/392N

Q and eQ and e

� Generalizations:– This is a purely empirical relationship

– Q and e come from measurements of r1 and r2

– Ideal condition is same Q and e values

– Proceeds poorly if Q1 and Q2 are very different

– Tends toward alternating if Q’s are the same and

e’s arelarge but of opposite sign.

Chemistry 367L/392N

Chemistry 367L/392N

I+IV or II+III →

alternating

I+III, II+IV and same →Statistical Copolymers

Χ

Χ I+II or III+IV → Poor Copolymerization

Chemistry 367L/392N

Chemistry 367L/392N

Conclusions-radical copolymers

� Co-polymers are important

– Properties depend on composition

� In general, these materials are

heterogeneous

� Mayo equation allows calculation of composition knowing r1 and r2

� Finemann Ross approach allows determination of r1 and r2

� Alfrey – Price allows estimate of r1 and r2

Chemistry 367L/392N

We discussed A B type step growth polymers -ABABABA-

If is the number of A molecules at the beginning of the

polymerization and is the number if B molecules,

We define r, the stoichiometric imbalance as

What about step growth copolymers??

0

AN0

BN

0

0

B

A

N

Nr =

If p is the conversion (as in Carothers equation) then

fraction of B at p is p or pr and the number of unreacted groups NA and NB is …

0

AN0

BN

0)1( AA NpN −=r

NprNprNB A

B

00 )1()1( −=−=

Chemistry 367L/392N

)(2

1BA NNN +=

At this point, the number of A and B end groups is NA + NB , but 2 ends per chains…the number

N of chains in the jar is….

Which is….

−+−=

r

NprNpN A

A

00 )1()1(

2

1

Or…..

−+= p

r

NN A 2

11

2

0

Now..the total number of repeat units, Nr is ….

)(2

1 00

BAr NNN += One repeat unit formed per reaction

Chemistry 367L/392N

Remember….0

0

B

A

N

Nr = So….

+=

+=

r

rN

r

NNN A

o

Ao

Ar

1

22

1 0

DP is the number of monomer units divided by number of chains

−+

+

==

pr

N

r

rN

N

NDP

A

A

r

21

12

1

20

0

Which luckily reduces to…pr

rDP

21

1

−+

+=

SupplementaryHomework…

Show Algebra

To here!

Chemistry 367L/392N

rpr

rDP

21

1

−+

+=

Note that if there is no stoichiometric imbalance, r =1 and we get…..

This is nice!!!

pDP

−=

1

1 The Carothers Equation!!

When A is totally consumed (p =1) then….

r

rDP

−

+=

1

1