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In Polymer Chemistry by Seymour-Carraher:

Saturated linear polymers such as HDPE are resistant to degradation, but slow degradation will occur in presence of $\ce{O2}$, UV, heat. Since tertiary carbon atoms are more readily attacked, PP is less resistant to degradation than HDPE.

It means a tertiary carbon atom makes polymer less resistant to photo-oxidation and photo-degradation. What makes tertiary carbon atom more likely to be attacked?

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  • $\begingroup$ In Polymer Chemistry by Seymour-Carraher: "Saturated linear polymers such as HDPE are resistant to degradation, but slow deg will occur in presence of O2, UV, heat. Since tertiary carbon atoms are more readily attacked, PP is less resistant to deg than HDPE." It means tertiary carbon atom makes polymer less resistant to photo-oxidation & photo-degradation. What makes tertiary carbon atom more likely to be attacked? $\endgroup$ – Masterbatch Oct 22 '15 at 8:06
  • $\begingroup$ You can always edit your question, to clarify it or add more information. I have hence inserted your comment. $\endgroup$ – Martin - マーチン Oct 22 '15 at 8:16
  • $\begingroup$ This question is not specific to just polymers and could be answered by considering in general, why are tertiary carbon atoms more likely to be oxidised. $\endgroup$ – Beerhunter Oct 22 '15 at 8:39
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This is admittedly a very brief and somewhat tempting answer and it might not be complete.

Photo-oxidation and -degradation occurs mainly via a radical mechanism. This is outlined on Wikipedia in more detail. In principle the initial step produces bond cleavage via UV radiation and with it radicals. Let $\ce{[P]_{$n$}}$ be a polymer, then $\ce{.[P]_{$n'$}}$ and $\ce{.[P]_{$n''$}}$ are decomposition products of this polymer, both of radical nature. $$\ce{P ->[h\nu] .[P]_{$n'$} + .[P]_{$n''$}}$$

In a saturated non chained polymer a homolytical bond cleavage would lead two primary radicals, i.e. $\ce{R-CH2.}$, which are very unstable. In a chained polymer, like PP, if the bond break occurs at the tertiary carbon, there would be a secondary radical $\ce{R-\overset{\bullet}{C}H-CH3}$ and another primary radical. Homolytical bond cleavage of a hydrogen carbon bond can even lead to a tertiary radical, $\ce{R-\overset{\bullet}{C}(CH3)2}$. All in all this can lead to a higher concentration of radicals, and therefore more access points for oxygen to attack, hence propagating the chain. One of the driving forces is the gain in entropy of this process.

$$\begin{align} \ce{.[P]_{$n''$} + O2 &-> [P]_{$n''$}OO.}\\ \ce{[P]_{$n$} + [P]_{$n''$}OO. &-> [P]_{$n''$}OOH + .[P]_{$n$}}\\ \ce{[P]_{$n''$}OOH &->[h\nu] [P]_{$n''$}O. + .OH}\\ \ce{[P]_{$n$} + [P]_{$n''$}O. &-> [P]_{$n''$}OH + .[P]_{$n$}}\\ \ce{[P]_{$n$} + .OH &-> H2O + .[P]_{$n$}}\\ \text{etc.}&\phantom{\ce{->}~~}\text{pp.} \end{align}$$

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Tertiary carbons require a low amount of energy to scission. The energy contained in UV rad up to 370 nm could cause chain scission in tertiary carbons. Other polymers have lower activation maxima ( like 340 nm for HDPE) or 315 nm for Nylon. PP at 370 is weaker. Use Planks formula to see energy contained in 370 nm UV and compare to the bond formation ( or scissioning ) energy.

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