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Answer the following · Q2

Q.Write reactions of formation of :
a. Nylon 6 b. Terylene

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1 (a). Nylon 6 forms when epsilon-caprolactam, the cyclic 7-membered lactam monomer, is heated with water at high temperature; the ring opens and undergoes ring-opening polymerization to give the polyamide -[NH-(CH2)5-CO]n-.

Step 2 (b). Terylene forms when n molecules of ethylene glycol (HO-CH2-CH2-OH) condense with n molecules of terephthalic acid (HOOC-C6H4-COOH) at high temperature (about 533 K) in presence of a catalyst, eliminating n molecules of water and forming n ester linkages, giving the polyester -[O-CH2-CH2-O-CO-C6H4-CO]n-.

✓Final answer

a. Nylon 6: n [caprolactam] --[H2O, heat, ring-opening]--> -[NH-(CH2)5-CO]n-

b. Terylene: n HO-CH2-CH2-OH + n HOOC-C6H4-COOH --[catalyst, 533K, -nH2O]--> -[O-CH2-CH2-O-CO-C6H4-CO]n-

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