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04-BS-12 · May 2017

Question 13 of 13: Identifying the Four Monomers of a Poly(ester amide)

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

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National Exam 04-BS-12, Organic Chemistry — May 2017. 3 hours, closed-book examination (no textbook aid beyond one double-sided aid sheet); a Casio or Sharp approved calculator is permitted. The paper prints thirteen questions using plain "Question N:" numbering; all thirteen are answered in full below.

Reference texts: McMurry, Organic Chemistry, 9th ed. (functional-group reactivity, amide/β-lactam resonance, stereochemistry and specific rotation, SN1/SN2 and epoxide-opening regiochemistry, cyclohexane conformational analysis, IR/NMR spectroscopy, electrophilic aromatic substitution and multi-step synthesis design, polymer/step-growth chemistry); Atkins, Physical Chemistry, 11th ed. (entropy of intramolecular vs. intermolecular reactions).

Question 13: Identifying the Four Monomers of a Poly(ester amide)

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.

The drawn segment is one repeat unit, running (left to right) succinyl–NH–Leu–O–CH2CH2–O–Leu–NH–succinyl–NH–(CH2)4–CH(CO2CH2Ph)–NH–, with the wavy bonds joining it to the next unit's succinyl carbonyl. Tracing every C(=O)–X linkage bond-by-bond (classifying each as an amide, N–C(=O), or an ester, O–C(=O)) resolves the four building blocks, in the ratio succinic acid : leucine : ethylene glycol : lysine benzyl ester = 2 : 2 : 1 : 1:

succinic acid
leucine
ethylene glycol
lysine benzyl ester (lysine "in reverse")
  1. Succinic acid, HOOC–CH2CH2–COOH. Both ends form amide bonds to the neighbouring amino-acid nitrogens — a simple diacid spacer, not itself an amino acid.
  2. Leucine (an amino acid). Its own α-amine forms the amide bond to succinic acid; its own carboxyl is esterified to the incoming ethylene glycol — both of leucine's functional groups are consumed in the backbone (one as an amide, one as an ester), which is exactly what "ester amide" copolymer means at this residue.
  3. Ethylene glycol, HO–CH2CH2–OH. Both hydroxyls form ester bonds — one to leucine's carboxyl, the other to the next residue's carboxyl — a simple diol spacer.
  4. Lysine, used "in reverse," capped as its benzyl ester. Lysine has two amines (α and ε) and one carboxyl. Tracing the chain carefully (rather than assuming the first amide bond terminates the residue) shows both of lysine's amines form backbone amide bonds, each to a succinyl carbonyl — the ε-amino group to the succinyl unit on its left in the drawing, and the α-amino group (at the wavy bond) to the succinyl unit that starts the next repeat — while its α-carboxyl is capped as a pendant benzyl ester (–C(=O)OCH2C6H5), not consumed in the backbone at all. This capped carboxyl is what makes lysine a "derivative" rather than the free amino acid itself.
MonomerRoleAmino acid?
Succinic aciddiacid spacer (2× amide)no
Leucineamide (to succinic acid) + ester (to ethylene glycol)yes
Ethylene glycoldiol spacer (2× ester)no
Lysine benzyl esterboth amines form backbone amides; carboxyl capped as pendant esteryes (derivative)

The two naturally-occurring amino acids the question asks for are therefore leucine and lysine.

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