Lecture 4
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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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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.

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.


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  • In butadiene:


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X= halogens,PVC

CH3 ,Propylene

CN ,acrylonitrile

Phenyl gp(bz ring) ,styrene


Polymerization mechanisms

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.


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b.Let us consider a condensation reaction of

monofunctional groups, like classical esterification:


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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


Condensation polymerization

Condensation Polymerization

  • Examples:

    • Polyesters

    • Nylons

    • Polycarbonates


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  • 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


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Condensation Polymerization — Esterification

  • Polyester from a di-alcohol and a di-acid

    • Example (Callister eq. 16.8):


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  • Another polyester: PETE

    • terephthalic acid+ ethylene glycol

      n.HOOC-C6H4-COOH + n.HO-CH2CH2-OH 

      -(-OOC-C6H4-COO-CH2CH2-)n + 2nH2O


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  • 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


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  • Nylon 6,6— 2 monomers: 6-carbon diamine + 6-carbon diacid

    hexamethylene diamine + adipic acid


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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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