The Notes for
Polymer and Coatings Science-
Chapter Three

Crosslinking an epoxide
In order to convert the resins into a crosslinked structure, it is necessary to add a curing agent. Most of the curing agents in common use can be classified into three groups:
  1. amines
  2. polyfunctional amines
  3. acid anhydrides
Chemistry of epoxies- Nucleophilic Displacement
Formation of a quaternary base:



Have John check reaction on the bottom of 125.

The negative charged monovalent oxygen can attack the primary epoxy carbon as shown below:



You can see where the extra lone pair of electrons (blue) on the oxygen went into forming a new bond. It should make sense that the extra lone pair of electrons on the oxygen with the green negative charge should be available to attack another epoxide.

The rate of this reaction is increased by the presence of phenols.

 Reactive amines- 1,2 (polyfunctional amines):
Both aliphatic and aromatic compounds having at least three active hydrogens present in 1 and/or 2 amine groups are widely used as curing agents for epoxy resins.

After the reaction takes place the final product is a highly crosslinked material. The reaction proceeds very fast.

Primary and secondary amines are also capable of reacting. It should make sense that primary amines would terminate that strand of a growing chain, and that secondary amines would merely increase the distance between crosslinks.

 Why do epoxies bond to almost everything? Epoxies provide good adhesion because there is a good chain to chain interaction (this makes for good cohesive strength) and the N and O have electron pairs that donate to the metals. A compound which contains N-H's, O-H's, or -COOH's will stick to a surface.

KINETICS OF A POLYESTER CONDENSATION

THIS IS THE LAST THING IN THE NOTES


Construction has just begun...
There may be errors in the math below...
Checks for continuity have not yet been performed...
Proceed only with a high level of critical scrutiny...
Thank you.

 Actually, read everything with critically scrutiny. The polymer building is constructed with the intent of creating suggestions of what ideas connect together. If you read a famous authors passage on something, and you tie it into something the author has placed in another chapter because of something you read here, then the Polymer Building has achieved the builder's objective.



There are going to be several assumptions here. We'll list them up front.
Assumptions simplify the math:
  1. Assume an equal amount of carboxylic acid and diol
  2. Assume 100% efficient reaction (i.e., no loss of either monomer to heat induced degradation or evaporation, or any thing of the sort.)
  3. No acid catalyst is added.
There is a lie built into the first expression you see below. The math is carried out correctly, so the results obtained are valid, but since one of the premises is false, the results are not truth. Here is the starting equation. 
                k
                 1
    aA + bB   ------>    cC   +  dD          (1)
              <------
                k
                 2 
-COOH   -OH           polymer  water
You are given this equation, it isn't something you have to derive

remember "products minus reactants" when you construct a differential equation. 
    d[A]        c   d          a   b
    ---- = k [C] [D]   -  k [A] [B]           (2)
     dt     2              1
This is something you covered in Freshman Chemistry. You should be able to make one of these for any reaction you are given.

but to follow tradition, we must make some algebraic changes: This is arbitrary. By putting the negative sign in front of the d[A]/dt, a disappearance reaction will have positive k value. 
   -d[A]        a   b            c   d
    ---- = k [A] [B]    -   k [C] [D]     (3)
     dt     1                2
Because initial concentration of diacid equals the initial concentration of diol, and a 100% efficient reaction is assumed, and we can say that the concentration of the diacid is always equal to the concentration of the diol.

This gives [A] = [B].
This simplifies things.

Assume the reaction is first order with respect to [A] and [B] so that a = b = 1.
This makes it simpler.

Assume the back reaction which corresponds to k(2) is negligible such that we can write: 
   -d[A]        1   1               1   1      2    -d[A]        2
   ----- = k [A] [B]   and since [A] [B]  = [A] ,   ----- = k [A]
     dt     1                                         dt     1

           (4)                       (5)                  (6)
This makes it easier

Do some "preintegration" algebra to separate the [A] terms from t and get: 
   -d[A]   
   ----- = k dt      we will integrate from start time, t(0), to final time,
       2    1        t(f), where [A(0)] corresponds to t(0) and 
    [A]                          [A(f)] corresponds to t(f).

     (7)
Ideally, with regard to the subscripts, there should be a consistency such that every variablue would have either a 0 (zero) or an f subscript to emphasize whether the variable is concerned with the start of the reaction or the final time (some point in time into the reaction.) Unfortunately, this convention isn't followed. :(

Any world wide web traveling second grader can integrate (7) and get what is shown below, but a graduate students years after calculus may need to look up the Mean Value Theorem. 
         1          1
- - { (------) - (------) } = k  {t(f) - t(0)}          (8)
       [A(f)]     [A(0)]       1
which is 
       1          1
  { (------) - (------) } = k  {t(f) - t(0)}           (9)
     [A(f)]     [A(0)]       1
Define time 't' to be the length of time from t(0) to t(f): 
t = t(f) - t(0), and therefore, k  {t(f) - t(0)} =  k  t    (10)
                                 1                   1
Solve for 1/[A(t)]: 
  1        1
------ = ------  + k  t       (11)
[A(f)]   [A(0)]     1



[A(0)] is, as before, the concentration of A at t(0), and
[A(f)] is the concentration of A at t(f).
Time f doesn't have to be the point when the reaction is complete. We could set f to be any time we want. We might do the calculation to see the extent ten minutes into the reaction, for which t=10 minutes would be time final. 

                           [A(0)] - [A(f)]
Now, define p to be    p = ---------------         (12)
                               [A(0)]


The variable p is the fraction of molecules reacted at t(f). Remember that the concentration is decreasing as the reactants are used up. If the original concentration is 100, and at t(f),the concentration is 80, then we have (100-80)/100 = 20/100 = 0.2, which indicates 20%.

Note: this p will tie into the Carothers Equation (but that's getting ahead of ourselves...)

Define Xn(f) to be the extent of polymerization at t(f). It should make sense that the extent of polymerization at time zero, Xn(0), equals zero. If the original concentration is 100, and at t(f), the concentration is 2, then Xn(f) is 50. This will tie into degree of polymerization which is discussed later on.

In the notes Xn(f) is Xn with a bar over the X.

         [A(0)]                                       [A(0)] - [A(f)]
 Xn(f) = ------  (13)  which we will compare to   p = --------------- (12)
         [A(f)]                                            [A(0)]
Now, on comparing Xn and p, we can change p to 
        [A(f)]         1                                    1
p = 1 - ------ = 1 - ----- (14)   which leads to  1 - p = ----- (15)
        [A(0)]       Xn(f)                                Xn(f)
We moved p to the right side and moved (1/Xn) to the left. We then solve for Xn(f). 
             1
  Xn(f)  = -----  (16)
           1 - p


Now, we go back to 
  1        1
------ = ------  + k  t   (11)  
[A(f)]   [A(0)]     1
if we multiply this by [A(0)] we get 
 [A(0)]     [A(0)]
(------) = (------)  +  [A(0)]k  t    (17)
 [A(f)]     [A(0)]             1
which simplifies to 
Xn(f) = 1 + [A(0)]k  t    (18)
                   1
We recall 
          1 
Xn(f) = -----  (16)
        1 - p
which we use along with (18) to make 
  1
----- = 1 + k  [A(0)]t   (19)
1 - p        1
Degree of polymerization: Traditionally, polymer chemists write the following: 

__     1
DP = -----
     1 - p
or
[Odian-56] 
  
                                                                              0
                       __       1                                       [COOH]      1
[Rodriguez-67] shows   x   =  -----  ;  [Allcocke 2nd-268] shows   DP = ------  = -----
                        n     1 - p                                     [COOH]    1 - p
is the extent of reaction (Odian-77.)

Now, we can tie this into something we considered a while back, the need for stoichiometric balance.

There is more math to come. Some equations are shown below which need to be worked into the mathematical workup:













P(A) and P(b) are extent of reaction calculations for reacts A and B.

 0
P   is the amount of A at the start of the reaction.
 A

 t
P   is the amount of A at time t.
 A

              0         t 
likewise for P     and P  .
              B         B
As to the meaning of "amount", you can do concentration, moles or molecules, because whatever units you use, you have them in both the numerator and the denominator, and they cancel out to a unitless quantity.

Since A and B react together, if we start with equal amounts, we always have A=B, and we can say P(A) = P(B). 

                            0
                           N
                            A
We define capital gamma = -----
                            0
                           N 
                            B
X(n) can be expressed in terms of capital gamma, as is shown below:





If we shoot for [A(0)] = [B(0)] but have a mismatch of 1% such that capital gamma = 0.99, then Xn = 200, and for a polymer with a repeat functionality of molecular weight 100, we would limited to a polymer with a 20,000 molecular weight.

THE END



Last Update- July 5, 1995- wld