1. Poly(vinyl acetate) of number-average molecular weight 100,000 is hydrolyzed to
poly(vinyl alcohol). Oxidation of the latter with periodic acid to cleave 1,2-diol
linkages yields a poly(vinyl alcohol) with X n = 200. Calculate the percentages of head-
to-tail and head-to-head linkages in the poly(vinyl acetate).
Solution:
Repeating Unit: = 86.09 g/mole
100,000
PVA initial X n =
86.09
n oxidation generates n + 1 fragments
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Thus Number of oxidation = −1
200
= 4.81
In other hand, the initial X n drops to 200 with addition of acid. On average, the present
molecule is cleaved to form 5.81 smaller molecules, so it is cleaved at 4.81 diol linkages.
The percent head-to-head: = 4.81 100% = 0.41%
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poly(vinyl alcohol). Oxidation of the latter with periodic acid to cleave 1,2-diol
linkages yields a poly(vinyl alcohol) with X n = 200. Calculate the percentages of head-
to-tail and head-to-head linkages in the poly(vinyl acetate).
Solution:
Repeating Unit: = 86.09 g/mole
100,000
PVA initial X n =
86.09
n oxidation generates n + 1 fragments
1162
Thus Number of oxidation = −1
200
= 4.81
In other hand, the initial X n drops to 200 with addition of acid. On average, the present
molecule is cleaved to form 5.81 smaller molecules, so it is cleaved at 4.81 diol linkages.
The percent head-to-head: = 4.81 100% = 0.41%
1162
