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04-BS-12 · December 2017

Question 13 of 13: Poly(ester amide) Monomer Identification

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

Notes on this paper

National Exam 04-BS-12, Organic Chemistry — December 2017. 3 hours, closed-book examination (no calculator required); 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. (acid–base strength of drugs, pharmacokinetics/lipophilicity, β-lactam reactivity, SN2 stereochemistry, Williamson-ether-type syntheses, alkyne alkylation, IR/NMR structure elucidation, radical vs. ionic HBr addition, electrophilic aromatic substitution & synthesis design, acid strength/resonance & induction, polymer/monomer identification). Every molecular formula, mass balance, and stereochemical (R/S) assignment below.

Question 13: Poly(ester amide) Monomer Identification (20 marks)

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.

Approach. Trace the backbone bond-by-bond, classifying every carbonyl-heteroatom linkage as an amide (C(=O)–N) or an ester (C(=O)–O), and identify each stretch of backbone between amide/ester junctions as a distinct monomer residue.

  1. Diacid residue –C(=O)CH2CH2C(=O)– forms TWO amide bonds ⇒ succinic acid, HOOC–CH2CH2–COOH. Both ends of this four-carbon diacyl fragment terminate in –NH– on the backbone, so this monomer contributes no ester linkage at all — it is a simple saturated diacid, the shortest possible one that still separates its two carbonyls by a flexible –CH2CH2– spacer.
  2. Monomer 1: succinic acid (diacid)
    Monomer 2: leucine (amino acid)
  3. –NH–CH(CH2CH(CH3)2)–C(=O)–O–, an isobutyl-bearing α-amino-acid residue whose amine forms an amide (with succinic acid) and whose carboxyl forms an ester (with the diol) ⇒ leucine, (CH3)2CHCH2CH(NH2)COOH. The –CH2CH(CH3)2 side chain hanging off the backbone α-carbon is leucine's diagnostic isobutyl group; using leucine's own –COOH as an ester partner (rather than a second amide) is exactly what turns this from a simple polyamide (nylon-type) into a poly(ester amide).
  4. –O–CH2CH2–O– bridges two ester carbonyls ⇒ ethylene glycol, HOCH2CH2OH. A simple diol, the shortest one available, supplying the second oxygen needed to complete the leucine ester on one side and a second amino-acid ester on the other.
  5. The final backbone stretch — a long –NH–(CH2)4–CH(–NH–)– chain with a pendant benzyl ester (–C(=O)–O–CH2C6H5) hanging off its own α-carbon — is lysine, used "in reverse". Lysine, H2N–(CH2)4–CH(NH2)–COOH, is the one natural amino acid with two amine groups (α and ε). Here both amines are used as backbone amide partners (the ε-NH2 condenses with the second succinic-acid diacyl unit; the α-NH2 continues the backbone onward), while lysine's own α-carboxyl is capped as a benzyl ester (a protecting/pendant group, not a backbone linkage) — this pendant hydrophobic benzyl group is exactly the kind of tunable side chain that lets formulation chemists adjust the coating's hydrophobicity/degradation rate.
  6. Monomer 3: ethylene glycol (diol)
    Monomer 4: lysine (amino acid, used "in reverse")
#MonomerRole in backbone
1Succinic aciddiacid — two amide linkages
2Leucine (amino acid)amine→amide; carboxyl→ester
3Ethylene glycoldiol — two ester linkages
4Lysine (amino acid, "in reverse")both amines→amide (backbone); carboxyl→pendant benzyl ester

The two amino acids are leucine and lysine.

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