NivaarExam PrepOfficial exam papers ↗

04-BS-12 · December 2019

Question 13 of 13: Step-Growth Polymers — Polyesters and Polyamides

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam 04-BS-12, Organic Chemistry — December 2019. 3 hours, closed-book examination (no calculator; one hand-written aid sheet permitted); NOTES on page 1 state that TEN (10) questions constitute a complete exam paper and only the first 10 as they appear in the answer book are marked, but this sitting prints 13 numbered questions — every question and sub-part below is answered in full.

Reference texts: McMurry, Organic Chemistry, 9th ed. (functional groups, stereochemistry, SN1/SN2 mechanisms and stereochemistry, steroid/bile-acid amphiphilicity, named-drug synthesis design, reaction-energy diagrams, polymer chemistry, arene-oxide metabolism, and mass-spectral/IR/NMR structure elucidation). Every molecular formula, exact mass, and stereochemical (R/S, cis/trans, meso/chiral) assignment below.

Question 13: Step-Growth Polymers — Polyesters and Polyamides (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

1,4-cyclohexanediol
succinic acid
terephthaloyl chloride
1,4-diaminobutane

a) Diol + diacid → polyester. A diol condensing with a diacid is the standard step-growth recipe for a polyester: each esterification loses one H2O, and repeating the condensation at both ends of every monomer builds an alternating –O–(1,4- cyclohexylene)–O–C(=O)–CH2CH2–C(=O)– chain: poly (1,4-cyclohexylene succinate).

b) Diacid chloride + diamine → polyamide. An acid chloride condensing with a diamine is the standard step-growth recipe for a polyamide (nylon-type): each amide bond forms with loss of HCl, giving an alternating –NH–(CH2)4–NH– C(=O)–C6H4–C(=O)– chain: poly(tetramethylene terephthalamide).

Reverse direction — identifying monomers from a repeat unit.

a) The repeat unit –O–CH2CH2–O–C(=O)– CH=CH–C(=O)– (the alkene drawn trans) is exactly an ethylene-glycol/unsaturated-diacid polyester — the monomers are ethylene glycol and fumaric acid (the trans isomer of butenedioic acid, the standard unsaturated diacid used industrially to make unsaturated polyester resins).

ethylene glycol
fumaric acid

b) The repeat unit shows an ortho-disubstituted benzene ring (both ester carbonyls emerging from adjacent ring carbons) linked through –O–CH2CH2 –O– bridges — the monomers are phthalic acid (benzene-1,2-dicarboxylic acid, the ortho isomer) and ethylene glycol.

phthalic acid
ethylene glycol
Back to the paper →