Level 4 — ApplicationPolymers

Polymers

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Subject: Chemistry | Chapter: Polymers Time Limit: 60 minutes | Total Marks: 50

Instructions: Attempt all questions. Show reasoning for full credit. Use $...$ notation where appropriate.

Q1. [12 marks] A polymer sample is fractionated and found to contain three groups of chains with the following data:

Fraction	Number of molecules $N_i$	Molar mass $M_i$ (g mol⁻¹)
A	$2.0\times10^{4}$	$10{,}000$
B	$3.0\times10^{4}$	$20{,}000$
C	$1.0\times10^{4}$	$50{,}000$

(a) Calculate the number-average molar mass $\bar{M}_n$ . (3) (b) Calculate the weight-average molar mass $\bar{M}_w$ . (4) (c) Compute the polydispersity index (PDI) and state what a PDI approaching 1 implies about the sample. (3) (d) One of these three fractions was produced by a living anionic polymerization while another came from free-radical polymerization. Predict which mechanism gives the narrower molar-mass distribution and justify. (2)

Q2. [10 marks] An aerospace engineer must select a polymer for each of the following unseen applications. For each, name ONE suitable polymer from the chapter and give the specific structural/chemical reason it fits:

(a) The outer flexible skin of a firefighter's re-entry suit that must resist flame without melting. (2) (b) The rigid matrix binding carbon fibres in a wing spar (a thermosetting composite). (2) (c) A lightweight tensile layer in a bullet-resistant vest requiring exceptional tensile strength along the chain. (2) (d) Explain, in molecular terms, why Kevlar and Nomex — both aromatic polyamides made from the same two monomer types — differ so much in stiffness. (4)

Q3. [10 marks] Consider the monomer 2-chloro-1,3-butadiene ( $\text{CH}_2=\text{CCl–CH=CH}_2$ ).

(a) Classify the polymer formed as addition or condensation, and explain why. (2) (b) Draw/describe the repeating unit of the polymer (neoprene). (3) (c) A student claims this monomer could also undergo free-radical polymerization. Write the three principal steps of a free-radical chain mechanism (generic), naming each. (3) (d) Neoprene retains a C=C double bond in its backbone. State one consequence of this for the material's properties. (2)

Q4. [10 marks] Nylon-6,6 and nylon-6 are both polyamides but are made differently.

(a) Nylon-6,6 is made from a diamine and a diacid. Identify both monomers by name. (2) (b) Nylon-6 is made from a single monomer. Name it and classify the polymerization type (chain vs step growth) — justify your classification given only one monomer is used. (3) (c) A newer biodegradable polymer, nylon-2-nylon-6, is designed to break down in the environment. Explain the structural feature that makes it biodegradable while ordinary nylon-6,6 is not. (3) (d) Predict whether a stoichiometric imbalance (excess diamine) in nylon-6,6 synthesis raises or lowers the achievable $\bar{M}_n$ , and explain using step-growth principles. (2)

Q5. [8 marks] PHBV (poly-3-hydroxybutyrate-co-3-hydroxyvalerate) is a biodegradable copolymer.

(a) State what "co" signifies and name the two monomer units. (2) (b) A 100 g sample of PHBV contains hydroxybutyrate (HB, repeat-unit mass $86$ g mol⁻¹) and hydroxyvalerate (HV, repeat-unit mass $100$ g mol⁻¹) in a 3:1 mole ratio of HB:HV. Calculate the mass percentage of HV units in the copolymer. (4) (c) Give one advantage of increasing HV content in PHBV for practical use. (2)

Answer keyMark scheme & solutions

Q1 (12 marks)

(a) $\bar{M}_n$ — number-average weights each chain equally. $\bar{M}_n=\frac{\sum N_iM_i}{\sum N_i}$ $\sum N_iM_i = (2.0\times10^4)(10^4)+(3.0\times10^4)(2\times10^4)+(1.0\times10^4)(5\times10^4)$ $= 2.0\times10^8 + 6.0\times10^8 + 5.0\times10^8 = 1.30\times10^9$ $\sum N_i = 6.0\times10^4$ $\bar{M}_n=\frac{1.30\times10^9}{6.0\times10^4}=21{,}667\text{ g mol}^{-1}$ (1 for formula, 1 for sums, 1 for answer)

(b) $\bar{M}_w$ — weights by mass. $\bar{M}_w=\frac{\sum N_iM_i^2}{\sum N_iM_i}$ $\sum N_iM_i^2 = (2.0\times10^4)(10^4)^2+(3.0\times10^4)(2\times10^4)^2+(1.0\times10^4)(5\times10^4)^2$ $= 2.0\times10^{12}+1.2\times10^{13}+2.5\times10^{13}=3.9\times10^{13}$ $\bar{M}_w=\frac{3.9\times10^{13}}{1.30\times10^9}=30{,}000\text{ g mol}^{-1}$ (1 formula, 2 sum of $N_iM_i^2$ , 1 answer)

(c) PDI $\text{PDI}=\frac{\bar{M}_w}{\bar{M}_n}=\frac{30000}{21667}=1.385$ PDI → 1 means all chains have nearly identical length (monodisperse / uniform). (2 for value, 1 for interpretation)

(d) Living anionic polymerization gives the narrower distribution (PDI near 1) because chains initiate together and grow simultaneously with no termination/chain transfer, so all chains reach similar length. Free-radical polymerization has random termination and transfer → broad distribution. (2)

Q2 (10 marks)

(a) Nomex — meta-linked aromatic polyamide; the meta orientation and strong H-bonded amide network give excellent flame and heat resistance without melting. (1 name + 1 reason)

(b) Epoxy resin — thermosetting; crosslinks (cures) into a rigid 3-D network that binds/wets carbon fibres in CFRP. (2)

(c) Kevlar — para-aramid with fully extended, ordered chains giving very high tensile strength; used in body armour. (2)

(d) Kevlar is para-linked (1,4): chains are straight/rod-like, pack into highly ordered, parallel, strongly H-bonded sheets → high crystallinity and stiffness. Nomex is meta-linked (1,3): the kinked chain geometry prevents such regular alignment, lowering crystallinity and stiffness but retaining thermal stability. (2 for para vs meta, 2 for consequence on packing/stiffness)

Q3 (10 marks)

(a) Addition polymer — the diene adds to itself across double bonds with no small molecule eliminated; atom economy 100%. (2)

(b) Repeating unit of neoprene (1,4-addition): $-(\text{CH}_2-\text{CCl}=\text{CH}-\text{CH}_2)-$ One double bond retained per unit; chlorine on carbon 2. (3)

(c) Free-radical mechanism steps:

Initiation — initiator decomposes to radicals which add to monomer forming a chain-carrying radical.
Propagation — radical adds monomer repeatedly, growing the chain.
Termination — two radicals combine (coupling) or disproportionate, ending growth. (1 each)

(d) The residual C=C allows cross-linking / vulcanisation (giving elasticity/resilience) but also makes it susceptible to oxidative/ozone degradation at those double bonds. (Either consequence accepted.) (2)

Q4 (10 marks)

(a) Hexamethylenediamine (1,6-diaminohexane) + adipic acid (hexanedioic acid). (1 each)

(b) Caprolactam. Type: step-growth (condensation) polymer — although it is one monomer, it is bifunctional (has both amine and acid character within the ring); ring-opening polymerization builds amide links stepwise, no double-bond chain propagation. (1 name + 2 justification)

(c) Nylon-2-nylon-6 contains amide linkages derived from an α-amino acid (glycine, the "nylon-2" unit); these peptide-like linkages are recognised and cleaved by microbial/enzymatic action, making it biodegradable. Ordinary nylon-6,6 has no such α-amino-acid units and its regular H-bonded structure resists enzymatic attack. (3)

(d) Excess diamine (stoichiometric imbalance) lowers $\bar{M}_n$ . In step-growth, high molar mass needs near-perfect 1:1 stoichiometry and high conversion (Carothers); an excess of one monomer caps chain ends with unreactive amine groups, limiting further growth. (2)

Q5 (8 marks)

(a) "co" = copolymer (two different repeat units in one chain). Units: 3-hydroxybutyrate and 3-hydroxyvalerate. (2)

(b) Take 3 mol HB : 1 mol HV. Mass HB $=3\times86=258$ g; Mass HV $=1\times100=100$ g. Total $=358$ g. $\%\text{HV mass}=\frac{100}{358}\times100=27.9\%$ (1 masses HB, 1 mass HV, 1 total, 1 answer)

(c) Higher HV content lowers melting point / increases flexibility (toughness) and improves processability, and can tune biodegradation rate. (2)

[
  {"claim":"Mn of Q1 = 21666.67","code":"N=[2e4,3e4,1e4]; M=[1e4,2e4,5e4]; num=sum(n*m for n,m in zip(N,M)); den=sum(N); Mn=num/den; result = abs(Mn-21666.6667)<1"},
  {"claim":"Mw of Q1 = 30000","code":"N=[2e4,3e4,1e4]; M=[1e4,2e4,5e4]; num=sum(n*m*m for n,m in zip(N,M)); den=sum(n*m for n,m in zip(N,M)); Mw=num/den; result = abs(Mw-30000)<1"},
  {"claim":"PDI approx 1.3846","code":"N=[2e4,3e4,1e4]; M=[1e4,2e4,5e4]; Mn=sum(n*m for n,m in zip(N,M))/sum(N); Mw=sum(n*m*m for n,m in zip(N,M))/sum(n*m for n,m in zip(N,M)); result = abs(Mw/Mn-1.3846)<0.01"},
  {"claim":"Q5 mass % HV = 27.93","code":"mHB=3*86; mHV=1*100; pct=mHV/(mHB+mHV)*100; result = abs(pct-27.93)<0.1"}
]