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Lecture # 4

Lecture # 4. MONOMERS: BUILDING BLOCKS FOR POLYMER MANUFACTURE : The paraffins ((like methane ) are unable to polymerize due to lack of a double bond-hence they are not considered to be monomers.

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Lecture # 4

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  1. Lecture # 4 MONOMERS: BUILDING BLOCKS FOR POLYMER MANUFACTURE: The paraffins ((like methane )are unable to polymerize due to lack of a double bond-hence they are not considered to be monomers. Ethylene, on the other hand, represents the simplest and most commonly used monomer, essentially due to the existence of an aliphatic double bond that starts the polymeric chain.

  2. In butadiene:

  3. X= halogens,PVC CH3 ,Propylene CN ,acrylonitrile Phenyl gp(bz ring) ,styrene

  4. Polymerization Mechanisms • 1- Polymerization by condensation: a.is the formation of polyamide (in our case Nylon 6-6) by reacting a di-acid with a diamine.

  5. b.Let us consider a condensation reaction of monofunctional groups, like classical esterification:

  6. Condensation Polymerization Characteristics • Gives off a small molecule (often H2O) as a byproduct • Stepwise polymerization • long reaction times • Bifunctional monomers linear polymers • Trifunctional monomers cross-linked (network) polymers

  7. Condensation Polymerization • Examples: • Polyesters • Nylons • Polycarbonates

  8. Polyesters from ahydroxy-acid • Acid and base functionality on one monomer: e.g., n .HO-CH2CH2CH2CH2COOH -(-CH2CH2CH2CH2COO-)n- + nH2O • General reaction: n . HO-R-COOH  -(-R-COO-)n- + nH2O

  9. Condensation Polymerization — Esterification • Polyester from a di-alcohol and a di-acid • Example (Callister eq. 16.8):

  10. Another polyester: PETE • terephthalic acid+ ethylene glycol n.HOOC-C6H4-COOH + n.HO-CH2CH2-OH  -(-OOC-C6H4-COO-CH2CH2-)n + 2nH2O

  11. Nylon 6 frompolymerization of an amino acid • acid and base groups on one monomer H2N-(CH2)5-COOH n H2N-(CH2)5-COOH  (-(CH2)5-CONH)n + nH2O • 6-carbon monomer  6-carbon mer

  12. Nylon 6,6— 2 monomers: 6-carbon diamine + 6-carbon diacid hexamethylene diamine + adipic acid

  13. Problem — Nylon 6,6 Calculate the masses of hexamethylenediamine and adipic acid needed to produce 1000 kg of nylon 6,6. • Solution: • The chemical reaction is: nH2N-(CH2)6-NH2 + nHOOC-(CH2)4-COOH -(-HNOC-(CH2)4-CONH-(CH2)6-)n+ 2nH2O • The masses are in proportion to molecular weights. Per merof nylon 6,6: C6H16N2+ C6H10O4C12H22O2N2 + 2H2O 72+16+28 + 72+10+64 144+22+32+28 + 2x18 116 + 146 226 + 36 116x103kg146x103kg103 kg 226 226 513 kg HMDA + 646 kg adipic acid Note: 159 kg of H2O byproduct

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