Chemical evidence for chain folding in polyethylene crystals is obtaine fragment

Question

Chemical evidence for chain folding in polyethylene crystals is obtaine fragments are separated chromatographically and their molecular back, then reentering and recrossing the crystal, and so o d by etching polymer the chain at the fold surface. The resulting chain weights determined by ain is pictured as crossing through the crystal, emerging and folding n. According to this idea, the shortest the chromatograms should equal the crystal thickness in length. The second h measures the length of the loop crystals with fuming nitric acid, which cleaves chain showing up in shortest chain exceeds twice this value by some amount, whic made by the chain outside the crystal. Molecular weights for the two shortest chairs in an experiment of this sort were 1260 and 2530. Since the cleaved chain end in nit observed ro and car boxyl groups, 60 should be subtracted from each of these molecular weights to give the ylene chain weight. Calculate the degree of polymerization of each molecule and the Å. What polyeth th (use the length of the unit cell along the chain axis, 2.53 Å, as the distance per repeat unit). Compare the latter with the crystal thickness determined by x ray diffraction, 105 does the ratio of chain lengths for the first and second peaks suggest about the tightness of folding?

Explanation / Answer

degree of the polymerization can be calculated using the formula

degree of the polymerization = molecular weight of (C2H4)n/C2H4

degree of the polymerization of 1st chain = (1260-60)/28 = 42.85

degree of the polymerization of 2nd chain = (2530-60)/28 = 88.21

Chain length of the first chain = 42.85 * 2.53Å =108.41Å (corresponds to the thickness), which is nearly the same as determined by the X-Ray diffraction technique

Chain length of the second chain = 88.21 * 2.53Å = 223.17Å (corresponds to the length of the loop)

Ratio of the chain length is approximately 1:2, which is suggestive of the quite tight folding)

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