Chemistry Reference
In-Depth Information
1
=
2
p
lim
C
1
2
ð½
H
2
O
=
C
0
K
Þ
(12-7)
2 in this example, the
C
arothers equation (Eq. 7-19)
shows the number average degree of polymerization
X
n
to be
Note also that since
f
av
5
1
=
2
C
0
K
1
X
n
5
(12-8)
p
5
1
½
H
2
O
2
It is evident from
Eqs. (12-7)
and
(12-8)
that a high conversion and high
molecular weight require low concentrations of the condensation product, water.
The lower the value of
K
, the more essential it is that the water concentration be
reduced. [The same considerations apply of course when an equilibrium exists
between a polymer and any condensation product such as in reaction (b) of
Fig. 7-2.]
In step-growth polymerizations with unfavorable values of
K
, it is therefore
standard practice to operate at high temperatures and reduced pressures to remove
the condensation products. This is typical of the manufacture of linear polyesters
where the final stages of the polymerization are at pressures near 1 mm Hg and
temperatures near 280
C. Alkyds (Section 7.4.2) are branched polyesters pro-
duced by esterification reactions of mixtures of polyhydric alcohols and acids
with varying functionalities. They are used primarily in surface coatings. Alkyd
syntheses are completed at temperatures near 240
C. It is not necessary to reduce
the pressure to pull residual water out of the reaction mixture, because the final
products are relatively low-molecular-weight fluids that are diluted with organic
solvents before further use. In one process variation, a small amount of a solvent
like xylene is added to the reactants to facilitate water removal by azeotropic dis-
tillation. Xylene residues in the final product are of no significance in this
instance.
The equilibrium is much more favorable to polymer formation in the produc-
tion of nylons than polyesters. This can be explained as being due to the greater
stability of amide as compared to ester linkages. Resonance structures can be
written for both groups as
HO
H
O
(12-9)
RN
C
′
R
NC
R
′
O
(12-10)
RO
C
′
R
OC
R
′
O
Oxygen is more electronegative than nitrogen and is less able to support a pos-
itive charge and so the amide linkage is more resistant to hydrolysis. As a result,
the equilibrium in reaction (10-11),
