Question

Predict which polymer in each pair has a higher melting temperature through the total strength of secondary bonding (use pictures).

a. PP or PAN

b.PET or PBT

c. PVdF or PTFE

d. Nylon 6 or nylon 12

Answer 1

Answer 2

Pairwise Comparison:

1. PP (Polypropylene) vs. PAN (Polyacrylonitrile)

• Higher Tm: PAN

• Reason: PAN has strong dipole-dipole interactions (due to polar –C≡N groups) vs. PP’s weak van der Waals forces.

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2. PET (Polyethylene terephthalate) vs. PBT (Polybutylene terephthalate)

• Higher Tm: PET

• Reason: PET’s shorter alkyl chain (ethylene) increases packing efficiency and dipole interactions vs. PBT’s longer butylene chain.

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3. PVdF (Polyvinylidene fluoride) vs. PTFE (Polytetrafluoroethylene)

• Higher Tm: PVdF

• Reason: PVdF has stronger dipole-dipole bonds (–CH₂–CF₂–) vs. PTFE’s symmetrical, nonpolar –CF₂– units.

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4. Nylon 6 vs. Nylon 12

• Higher Tm: Nylon 6

• Reason: Nylon 6 has a higher density of hydrogen bonds (shorter hydrocarbon chain) vs. Nylon 12’s longer, flexible chain.

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 Symmetry: Asymmetry (e.g., PVdF) often enhances polarity vs. PTFE’s symmetry

Answer:

The polymer with higher melting temperature (Tm) in each pair is determined by stronger secondary bonding (e.g., dipole-dipole, hydrogen bonding, van der Waals forces). Here’s the prediction: