Question 1

Fibre origin: which of the following is a natural fibre/material based on a naturally occurring polymer?

Accepted Answer

Introduction / Context: Distinguishing natural from synthetic fibres is foundational in materials selection. Natural fibres derive from polymers produced by biological systems, while synthetic fibres come from man-made polymerisation routes. Given Data / Assumptions: Cellulose is the structural polysaccharide found in plant cell walls. Dacron is the trade name for PET (polyethylene terephthalate) fibre. Nylon-6 is a synthetic polyamide (polycaprolactam). Concept / Approach: Cellulose is a natural polymer built from glucose units via beta-1,4-glycosidic bonds, forming fibres such as cotton and flax. Dacron and nylon-6 are fully synthetic polymer fibres produced through controlled industrial polymerisation and spinning processes. Therefore, the natural material among the options is cellulose. Step-by-Step Solution: Classify cellulose → natural polysaccharide; source of natural fibres (cotton, linen).Classify Dacron → PET (synthetic polyester).Classify nylon-6 → synthetic polyamide (PA6).Select cellulose as the natural entry. Verification / Alternative check: Textile references list cellulose-based fibres as “natural fibres,” while PET and nylon are “man-made” or “synthetic” fibres, sometimes “regenerated” for viscose/rayon but still derived through processing. Why Other Options Are Wrong: Dacron: synthetic polyester.Nylon-6: synthetic polyamide.None of these: incorrect because cellulose qualifies. Common Pitfalls: Confusing “cellulose acetate” (regenerated/derivative) with natural cellulose; forgetting that trade names often obscure chemistry. Final Answer: Cellulose

Question 2

Catalyst system: the low-pressure Ziegler process for polyethylene employs which catalyst combination?

Accepted Answer

Introduction / Context: Ziegler–Natta polymerisation revolutionised polyolefins by enabling low-pressure synthesis of linear, high-crystallinity polymers. Recognising the hallmark catalyst system helps connect process conditions with product properties (e.g., HDPE vs LDPE). Given Data / Assumptions: We refer to classic heterogeneous Ziegler–Natta catalysis for polyethylene. Transition-metal halides and organo-aluminium compounds are typical components. Low-pressure operation is implied. Concept / Approach: The archetypal system uses a titanium halide (commonly TiCl4 supported on MgCl2 in modern variants) combined with an organo-aluminium cocatalyst such as triethylaluminium (AlEt3). The pair activates insertion polymerisation of ethylene with controlled stereochemistry and suppressed branching, yielding HDPE. Nickel, vanadium pentoxide, and iron alone are not the characteristic Ziegler polyethylene catalyst pair in the low-pressure route posed here. Step-by-Step Solution: Recall Ziegler–Natta pair → TiCl4 + AlEt3 (or related Al-alkyl).Associate with low-pressure PE → HDPE via insertion mechanism.Eliminate single metals/oxides that are not the classic combination. Verification / Alternative check: Polymerisation texts and industrial practice cite TiCl4/Al-alkyl systems (and later MgCl2 supports) as standard for Z-N catalysis of ethylene/propylene. Why Other Options Are Wrong: Ni, V2O5, Fe: not the canonical low-pressure Ziegler PE catalyst pair indicated in the question. Common Pitfalls: Confusing Phillips chromium catalysts (Cr/SiO2) with Ziegler–Natta; the question specifies the Ziegler route with organo-aluminium cocatalyst. Final Answer: Aluminium triethyl combined with titanium tetrachloride

Question 3

Polymer additives: organophosphate esters such as tricresyl phosphate and tributyl phosphate are primarily used in plastics as what type of additive?

Accepted Answer

Introduction / Context: Small-molecule additives can dramatically alter the flexibility, processability, and impact resistance of polymer matrices. Recognising common families—phthalates, adipates, and phosphates—helps you predict where they are used and why. Given Data / Assumptions: Examples listed: tricresyl phosphate (TCP), tributyl phosphate (TBP), tetrabutyl phosphate, triphenyl phosphate (TPP). We consider general PVC and engineering plastic formulations. The question seeks the functional class of these additives. Concept / Approach: Organophosphate esters act as plasticisers, reducing intermolecular forces between polymer chains and increasing free volume. This lowers glass transition temperature, enhances flexibility, and can impart flame-retardant tendencies for aromatic phosphates. They are not “hardeners” (curing agents), nor specific anti-shrinkage agents; transparency can be affected but is not their defining role. Step-by-Step Solution: Identify chemical family → phosphate esters.Map to function → plasticisation (Tg reduction, flexibility increase).Eliminate alternative roles that do not match the primary industrial use. Verification / Alternative check: Formulation guides list phosphate esters as plasticisers and secondary flame retardants in PVC, cellulose esters, and some engineering resins. Why Other Options Are Wrong: Hardeners: typically amines/anhydrides for epoxies, peroxides for unsaturated polyesters—different chemistry.Anti-shrinkage agents: niche additives; phosphates are not primarily used for this.Transparency improvers: not a standard classification; plasticisers may influence clarity indirectly. Common Pitfalls: Assuming all plasticisers are phthalates; phosphates are widely used, especially where flame performance matters. Final Answer: Plasticisers

Question 4

Polypropylene versus polyethylene (polythene): why is polypropylene generally preferred in many applications? Select the most accurate reason based on mechanical properties.

Accepted Answer

Introduction / Context: In polymer selection for consumer goods, packaging, automotive trims, and houseware, polypropylene (PP) and polyethylene (PE, often called polythene) are common choices. This question asks why PP is often preferred, focusing on mechanical property advantages relevant to real-world use. Given Data / Assumptions: Comparison at room temperature between commodity grades (e.g., PP homopolymer and LDPE/HDPE). Interest is in hardness and tensile strength for stiffness and load-bearing capacity. Flammability of both materials is comparable without special additives. Concept / Approach: Polypropylene typically exhibits higher hardness and tensile strength than many polyethylene grades, especially LDPE. This translates to better rigidity (higher flexural modulus) and improved creep resistance at ambient temperatures, which are desirable for living hinges, caps, appliance parts, and reusable containers. Step-by-Step Solution: Identify property targets: hardness and strength drive stiffness and durability.Compare typical data: PP has higher modulus and tensile strength than LDPE and is comparable to or stiffer than many HDPE grades.Evaluate flammability: both are combustible; “non-inflammable” is incorrect without flame retardants.Hence the best answer combines (b) and (c). Verification / Alternative check: Datasheets show PP tensile strength commonly higher than LDPE, with hardness and softening point (Vicat) also higher, supporting improved rigidity and heat resistance. Why Other Options Are Wrong: Non-inflammable: incorrect without FR additives; both PP and PE burn.Only hardness or only strength misses the broader advantage.“Better solvent for plasticisers” is irrelevant and not a benefit. Common Pitfalls: Assuming one material is inherently flameproof; overlooking grade-to-grade variations (copolymers, fillers, and orientation can change properties). Final Answer: Both (b) and (c)

Question 5

Alkyd resins: identify the polymer class most accurately describing these coating binders.

Accepted Answer

Introduction / Context: Alkyd resins are ubiquitous in paints, varnishes, and enamels. Recognising their chemical family helps predict curing behavior, compatibility, and performance in coatings technology. Given Data / Assumptions: Alkyds are produced from polyhydric alcohols and polybasic acids/anhydrides. They may be oil-modified (fatty acid chains attached). Used predominantly as film-forming binders. Concept / Approach: Chemically, alkyds are polyesters formed via condensation of polyols (e.g., glycerol, pentaerythritol) with dibasic acids/anhydrides (e.g., phthalic anhydride). Oil or fatty acid modification tailors flexibility, drying, and gloss. They are not polyamides or polyolefins, and their formation is by condensation (step-growth), not chain-growth addition. Step-by-Step Solution: Check synthesis route: condensation of polyol + dibasic acid → polyester.Relate properties: good film formation, oxidative crosslinking when oil-modified.Choose “Polyester.” Verification / Alternative check: Standard coating texts classify alkyds as oil-modified polyesters; drying occurs via oxidative crosslinking of unsaturated fatty acid side chains. Why Other Options Are Wrong: Polyamide: e.g., nylons; different chemistry.Polyolefin: chain-growth polymers like PE/PP.Addition polymer: incorrect mechanism for alkyd formation.Acrylic copolymer: distinct family (e.g., PMMA, styrene–acrylic). Common Pitfalls: Confusing curing mechanism (oxidative drying) with polymerisation route; assuming “alkyd” implies amide linkages due to sound-alike with “amide.” Final Answer: Polyester

Question 6

Which of the following is manufactured by addition (not condensation) polymerisation?

Accepted Answer

Introduction / Context: Polymerisation mechanisms fall into two broad families: chain-growth addition and step-growth condensation. Identifying which mechanism produces a given commercial polymer is a common exam theme because it governs structure, by-products, and processing. Given Data / Assumptions: Addition polymerisation: monomers add across double bonds; no small-molecule by-product. Condensation polymerisation: functional groups react, eliminating small molecules (e.g., water, methanol). Candidates include PTFE, PET, Nylon, phenolic, and melamine resins. Concept / Approach: PTFE forms by free-radical addition polymerisation of tetrafluoroethylene. In contrast, PET (Terylene) is a polyester from terephthalic acid/EG via condensation, nylons are polyamides from diamine/diacid (or lactam) via condensation, and phenolic/melamine resins are classic step-growth condensation networks. Step-by-Step Solution: Match PTFE to chain-growth of CF2=CF2 (no by-product) → addition.Identify PET/nylon/phenolic/melamine as step-growth with by-product formation.Select PTFE as the only addition example. Verification / Alternative check: Industrial PTFE is produced via suspension/emulsion polymerisation under free-radical initiators, confirming the addition route. Why Other Options Are Wrong: PET: condensation esterification reaction.Nylon: amide formation (condensation).Bakelite and melamine–formaldehyde: step-growth condensations forming thermoset networks. Common Pitfalls: Assuming “no by-product” equals thermoplastic only; forgetting that some condensation polymers can still be thermoplastics (e.g., PET, nylon) before crosslinking. Final Answer: Teflon (polytetrafluoroethylene, PTFE)

Question 7

Glyptal resins used in paints and lacquers are chemically which type of polymer?

Accepted Answer

Introduction / Context: “Glyptal” historically refers to polyesters produced from glycerol and phthalic anhydride, widely used as binders in coatings and electrical varnishes. Knowing the polymer type clarifies curing behavior and solvent resistance. Given Data / Assumptions: Monomers: glycerol (triol) and phthalic anhydride (dicarboxylic acid anhydride). Application: paints, varnishes, lacquers. Concept / Approach: Condensation of polyols with dibasic acids yields polyester chains. These can be further modified or crosslinked (e.g., with drying oils) for film formation. The backbone contains ester linkages, distinguishing it from amide, styrenic, or nitrile polymers. Step-by-Step Solution: Identify monomer functionality (triol + diacid anhydride).Condensation gives ester linkages → polyester.Therefore, choose “Polyester.” Verification / Alternative check: Coating literature classifies glyptal/alkyds within the polyester family, with oil length determining flexibility and drying behavior. Why Other Options Are Wrong: Polyamide: requires amide linkages (nylons).Polystyrene and polyacrylonitrile: chain-growth vinyl polymers; different chemistry and uses.Phenolic: phenol–formaldehyde networks, not glyptal. Common Pitfalls: Equating alkyds with “acrylics” due to coating context; confusing brand names with polymer families. Final Answer: Polyester

Question 8

Emulsion polymerisation of vinyl chloride for PVC: what is the typical reactor temperature maintained (°C) during production?

Accepted Answer

Introduction / Context: Poly(vinyl chloride) (PVC) can be produced by suspension or emulsion polymerisation. Reactor temperature control directly affects polymerisation rate, particle size, molecular weight, and safety. This question targets the typical temperature used in emulsion processes. Given Data / Assumptions: Water-based emulsion polymerisation with surfactants and free-radical initiators. Safety constraints due to vinyl chloride monomer volatility. Desire for controlled molecular weight distribution. Concept / Approach: Emulsion polymerisation of VCM commonly operates in the moderate temperature range around 40–60 °C; a typical nominal setpoint is roughly 50 °C. This balances initiator decomposition, conversion rate, and heat removal while avoiding excessive chain transfer and runaway risk. Step-by-Step Solution: Identify feasible thermal window for initiators used in emulsion PVC.Select a mid-range value (≈50 °C) consistent with practice.Rule out extreme temperatures that would be unsafe or chemically inappropriate. Verification / Alternative check: Process descriptions and safety guidelines specify polymerisation typically in the 40–60 °C range, with jacketed vessels and staged initiator feeds. Why Other Options Are Wrong: −20 °C: impractically low for initiator kinetics.250–500 °C: decomposition and runaway; not applicable to aqueous emulsion systems.95 °C: high for standard PVC emulsion processes and risks undesirable side reactions. Common Pitfalls: Confusing suspension versus emulsion conditions; assuming higher temperature always improves throughput without considering quality and safety. Final Answer: 50 °C

Question 9

Polymer structure/mechanism check: pick the wrong statement about polymer formation and properties.

Accepted Answer

Introduction / Context: Understanding how polymer architecture forms is vital to predicting behavior in processing and service. This question asks you to identify the incorrect statement, focusing on polymerisation mechanisms and structural effects. Given Data / Assumptions: Bifunctional monomers produce linear chains via step-growth or chain-growth depending on functionality. Crosslinking creates network structures with very different properties. Polyester formation route is tested. Concept / Approach: Polyesters are produced by condensation (step-growth) between diols and diacids/anhydrides, eliminating small molecules (e.g., water). They are not made by an addition mechanism. The other statements broadly align with polymer science: linear chains from bifunctional monomers are typically thermoplastic; crosslinking leads to elastomeric/thermoset behavior; branching and crosslinking reduce solubility and increase softening or glass transition. Step-by-Step Solution: Evaluate each statement against standard polymer chemistry.Identify the mechanism for polyester formation: condensation, not addition.Therefore option (d) is the wrong statement. Verification / Alternative check: Textbook step-growth kinetics and Carothers’ theory describe polyesters arising from diol–diacid reactions producing ester linkages plus by-products. Why Other Options Are Wrong: (a), (b), (c) are consistent generalisations about structure–property relations. Common Pitfalls: Equating “addition” with any polymerisation; forgetting that many thermoplastics (PET, nylons) are condensation polymers. Final Answer: Dibasic acids react with dihydric alcohols to give polyesters using an addition polymerisation reaction.

Question 10

Which product is NOT made by addition polymerisation?

Accepted Answer

Introduction / Context: Commodity vinyl and olefin polymers are typically formed via addition mechanisms, while many engineering polymers form via condensation. Correctly classifying the mechanism helps anticipate by-products and processing methods. Given Data / Assumptions: LDPE, PVC, PS, PP are vinyl/olefinic classes. Nylon-66 is a polyamide produced from a diamine and diacid. Concept / Approach: LDPE, PVC, PS, and PP are classic chain-growth addition polymers derived from unsaturated monomers. Nylon-66 (polyhexamethylene adipamide) is a step-growth condensation polymer produced by reacting hexamethylenediamine with adipic acid to form amide links and release small molecules (e.g., water). Step-by-Step Solution: Identify vinyl/olefin monomers → addition route.Recognise nylon-66 chemistry → condensation route.Select nylon-66 as the exception. Verification / Alternative check: Industrial processes: Ziegler–Natta/metallocene or radical routes for addition polymers; nylon salt polymerisation (condensation) for nylon-66. Why Other Options Are Wrong: LDPE, PVC, PS, PP are all produced by addition polymerisation. Common Pitfalls: Assuming “engineering polymer” implies addition; confusing nylon-6 (ring-opening of caprolactam) with addition (it is also step-growth via ring opening, effectively condensation-like). Final Answer: Polyhexamethylene adipamide (nylon-66)

Question 11

Why is neoprene relatively non-inflammable compared to many hydrocarbon rubbers?

Accepted Answer

Introduction / Context: Neoprene (polychloroprene) is a synthetic rubber valued for oil resistance, weathering, and relatively low flammability. Understanding the chemical reason for its flame resistance is helpful in materials selection for hoses, belts, and gaskets in demanding environments. Given Data / Assumptions: Neoprene is derived from chloroprene monomer (2-chloro-1,3-butadiene). Comparison is with hydrocarbon rubbers lacking halogens. No special flame retardants are assumed beyond inherent chemistry. Concept / Approach: Halogens (e.g., chlorine) in polymer chains inhibit flame propagation by releasing HCl and interfering with radical chain reactions in the flame zone. This lowers heat release and slows combustion compared to purely hydrocarbon backbones. Step-by-Step Solution: Identify neoprene structure: polymer of a chlorinated diene.Relate halogen content to flame inhibition mechanisms.Conclude that chlorine presence reduces flammability. Verification / Alternative check: Halogenated polymers like PVC and neoprene exhibit higher limiting oxygen index (LOI) values than comparable hydrocarbon polymers, corroborating reduced flammability. Why Other Options Are Wrong: Cross-link density affects softening and char but is not the primary reason for inherent flame resistance.Linear structure does not inherently reduce flammability.Absence of chlorine is the opposite of the correct reasoning.Fillers may influence fire behavior but are not the intrinsic cause. Common Pitfalls: Confusing vulcanisation effects with chemistry of the backbone; assuming any cross-linked rubber is non-flammable. Final Answer: Because chlorine atoms in the monomer reduce flammability

Question 12

Branched-chain versus linear polymers: which property is typically lower for branched materials?

Accepted Answer

Introduction / Context: Polymer architecture strongly influences packing efficiency, crystallinity, and properties. Branching disrupts chain alignment, which in turn affects density, mechanical strength, and thermal transitions. Given Data / Assumptions: Comparison at similar molecular weights and without heavy fillers. Examples: LDPE (branched) vs HDPE (linear). Concept / Approach: Branching reduces the ability of chains to pack into crystals, lowering crystallinity. Lower crystallinity typically leads to reductions in density, melting point, and tensile strength compared to linear analogs. Among the listed properties, “density” is a hallmark difference (e.g., LDPE ≈ 0.91–0.93 g/cm^3 vs HDPE ≈ 0.94–0.97 g/cm^3). Step-by-Step Solution: Relate branching → reduced packing → reduced crystallinity.Reduced crystallinity → lower density and often lower Tm and tensile strength.Select the most definitive property difference: density. Verification / Alternative check: Data sheets for PE grades show clear trends: linear (HDPE) has higher density than branched (LDPE). Similar trends appear in other polymer families when branching increases. Why Other Options Are Wrong: Melting point and tensile strength also tend to be lower, but “density” is the most directly and consistently reduced.“None of these” is incorrect since density is affected. Common Pitfalls: Attributing differences solely to molecular weight rather than architecture; ignoring branching type (short- vs long-chain branching). Final Answer: Density

Question 13

Flexible foams for mattresses are commonly made from which polymer family?

Accepted Answer

Introduction / Context: Mattress cores and cushioning rely on flexible foams with tunable resilience, compression set, and airflow. The dominant commercial class for flexible furniture foams is polyurethane (PU) due to versatile chemistry and scalable foaming processes. Given Data / Assumptions: Mass-produced flexible slabs and molded cushions. Room-temperature comfort and durability are prime concerns. Concept / Approach: Flexible PU foams are formed by reacting polyols with isocyanates (e.g., TDI/MDI) while generating gas (often CO2 from water–isocyanate reaction) to create open-cell structures. Formulators tune hardness (ILD), resilience, and breathability by selecting polyol functionality, isocyanate index, and additives. Step-by-Step Solution: Identify application: mattresses and upholstery require flexible, open-cell foam.Match polymer family: polyurethanes deliver the needed property window.Select “Polyurethanes.” Verification / Alternative check: Industry statistics show PUs dominate flexible foam markets; alternatives like latex foam (NR/SBR) exist but are a smaller share. PVC and PS typically form rigid foams; silicones are specialty elastomers; polyamides are not used for flexible mattress cores. Why Other Options Are Wrong: PVC, PS: commonly used for rigid foams or solid parts.Silicone rubber: flexible but expensive and not standard for mattresses.Polyamides: engineering solids, not foams for bedding. Common Pitfalls: Confusing latex (rubber) mattresses with polyurethane market prevalence; assuming any foamed plastic suits cushioning. Final Answer: Polyurethanes

Question 14

Poly(vinyl chloride) (PVC) processing: why are stabilisers added during manufacture and compounding?

Accepted Answer

Introduction / Context: PVC is a versatile thermoplastic used in pipes, cables, profiles, and films. However, the polymer's chlorine-bearing backbone is susceptible to dehydrochlorination when exposed to heat and shear in processing. Stabilisers are therefore an essential additive family in PVC compounding. Given Data / Assumptions: PVC is processed at elevated temperatures where HCl elimination can start. Thermal degradation leads to discoloration, embrittlement, and viscosity changes. Commercial stabilisers include Ca–Zn, organotin, and mixed metal systems. Concept / Approach: Dehydrochlorination forms conjugated polyenes along the backbone, causing rapid autocatalytic degradation (HCl catalysis). Stabilisers scavenge HCl, replace labile chlorine sites, and inhibit chain unzipping, thereby preserving molecular weight and mechanical properties. Step-by-Step Solution: Identify the main degradation pathway: heat-induced HCl loss from PVC.Recognise stabiliser action: HCl scavenging and substitution at labile sites.Conclude the primary purpose: prevent thermal degradation. Verification / Alternative check: Processing windows for PVC widen markedly when appropriate stabiliser packages are used; colour hold and torque stability tests demonstrate effectiveness. Why Other Options Are Wrong: Impact and elasticity depend on plasticisers and impact modifiers, not primarily stabilisers.Reducing melt viscosity or glass transition is the role of plasticisers and processing aids.Crystallinity changes are not the main stabiliser function. Common Pitfalls: Confusing stabilisers with plasticisers or lubricants; assuming a single additive can address all property targets. Final Answer: To prevent thermal degradation during processing and service

Question 15

Identify the fibre family: which of the following fibres are polyamides?

Accepted Answer

Introduction / Context: Classifying fibres by polymer chemistry helps predict dyeing behaviour, thermal properties, and end-use. Polyamide fibres are known generically as nylon and are used in textiles, cords, and engineering applications. Given Data / Assumptions: Dacron is polyester (polyethylene terephthalate). Rayon is regenerated cellulose. Orion (more commonly Orlon) is acrylic (polyacrylonitrile-based). Nylon is the generic name for polyamides. Concept / Approach: Polyamides contain amide linkages in the backbone, conferring strong hydrogen bonding, good abrasion resistance, and toughness. Nylon fibres (e.g., nylon-6, nylon-66) are the canonical polyamide fibres. Step-by-Step Solution: Map brand/common names to polymer families.Recognise nylon as polyamide by definition.Select “Nylon.” Verification / Alternative check: Textile standards list nylon under polyamide; Dacron under polyester; rayon under cellulosics; Orlon under acrylics. Why Other Options Are Wrong: Dacron: polyester, not polyamide.Rayon: regenerated cellulose.Orion/Orlon: acrylic.Kevlar is an aramid (aromatic polyamide) fibre but is not the generic “nylon” used in common apparel context; the question targets the standard textbook match, nylon. Common Pitfalls: Confusing trade names with polymer families; overlooking that aramids are a specialised subset of polyamides. Final Answer: Nylon

Question 16

Phenolic vs other thermosets: formaldehyde condenses with which substrate to produce a non-phenolic resin?

Accepted Answer

Introduction / Context: Condensation of formaldehyde with various nucleophilic substrates yields different thermoset families. Phenolic resins result from phenol-formaldehyde reactions, while melamine-formaldehyde resins constitute a separate class with distinct performance. Given Data / Assumptions: Phenolic substrates include phenol, resorcinol, and p-substituted phenols like p-cresol. Melamine is a triazine amine, not a phenolic compound. Formaldehyde acts as a crosslinker via methylol formation. Concept / Approach: Phenolic resins require phenolic hydroxyl groups. Amino resins (e.g., melamine–formaldehyde) form by condensation at amino groups, producing methylene and methylene-ether linkages typical of MF resins, known for hardness and heat resistance. Step-by-Step Solution: List phenolic candidates: phenol, resorcinol, para-cresol.Identify melamine as non-phenolic (amino) substrate.Therefore, melamine with formaldehyde does not produce a phenolic resin. Verification / Alternative check: Commercial tabletop laminates and dinnerware often use melamine–formaldehyde, not phenolics, highlighting the different resin family. Why Other Options Are Wrong: Phenol/resorcinol/p-cresol all yield phenolic-type networks with formaldehyde. Common Pitfalls: Assuming all formaldehyde condensations are “phenolic”; overlooking amino resins as a major thermoset class. Final Answer: Melamine

Question 17

Comparing nylons: relative to nylon-66, nylon-6 is generally… (choose the correct statement).

Accepted Answer

Introduction / Context: Nylon-6 and nylon-66 are two widely used polyamides with similar structures and overlapping applications. Subtle differences in melting point, abrasion resistance, and hardness influence material selection. Given Data / Assumptions: Nylon-66 typically melts around 255–265 °C; nylon-6 around 215–225 °C. Nylon-66 often shows slightly higher stiffness, hardness, and abrasion resistance. Nylon-6 is noted for better dye uptake and impact at low temperatures. Concept / Approach: The higher melting point and crystallinity of nylon-66 stem from its even-numbered monomers (HMDA + adipic acid) enabling efficient hydrogen bonding and packing. Nylon-6 (from caprolactam) has lower Tm and generally slightly lower hardness and abrasion resistance relative to nylon-66. Step-by-Step Solution: Check melting point claim: nylon-6 does not exceed nylon-66 → (c) false.Check hardness/abrasion: nylon-66 commonly superior → (a) and (b) false.Thus, the correct choice among the given is “None of these.” Verification / Alternative check: Datasheets and handbooks consistently report nylon-66 higher Tm and often higher wear resistance; nylon-6 better dyeability. Why Other Options Are Wrong: (a), (b), (c) contradict typical property data.Option (e) is true in many cases, but it is not among the choices for the main selection—hence (d) fits the question’s intent. Common Pitfalls: Overgeneralising across grades filled with glass/minerals; mixing copolyamides with homopolymers. Final Answer: None of these

Question 18

Identify the copolymer among the following polymer examples.

Accepted Answer

Introduction / Context: A copolymer consists of two or more different monomer species incorporated in the same polymer chain. Recognising copolymers helps anticipate property tuning via composition and sequence control. Given Data / Assumptions: Nylon-66 forms from hexamethylene diamine and adipic acid. Polypropylene, PVC, PTFE, and PMMA listed here are homopolymers of a single repeating monomer unit. Concept / Approach: Nylon-66 is an alternating condensation copolymer because its repeat unit derives from two distinct monomers (a diamine and a diacid). In contrast, polypropylene (from propylene), PVC (from vinyl chloride), PTFE (from tetrafluoroethylene), and PMMA (from methyl methacrylate) are homopolymers in their standard commercial forms. Step-by-Step Solution: Define copolymer vs homopolymer.Match each example to its monomer set.Select nylon-66 as the copolymer in the list. Verification / Alternative check: Polymer naming conventions for nylons reflect the carbon counts of the two monomers (6,6), indicating two-monomer chemistry. Why Other Options Are Wrong: PP, PVC, PTFE, PMMA are classically single-monomer polymers in their basic grades. Common Pitfalls: Assuming “condensation polymer” implies homopolymer; overlooking that many step-growth polymers are copolymers by nature. Final Answer: Nylon-66

Question 19

Anti-tacking performance: which resin class generally exhibits the highest resistance to tackiness (non-blocking surfaces)?

Accepted Answer

Introduction / Context: “Anti-tacking” refers to resistance to surface stickiness and blocking, important for coatings, laminates, and moulded parts requiring hard, mar-resistant finishes. Resin chemistry and crosslink density drive surface hardness and tack behaviour. Given Data / Assumptions: Thermoset networks tend to have higher surface hardness and lower tack than thermoplastic or lightly crosslinked systems. MF, phenolic, epoxy, and alkyd are common coating/moulding resins. Concept / Approach: Melamine–formaldehyde (MF) resins form highly crosslinked, hard, glassy networks with excellent mar and block resistance; they are widely used in decorative laminates and baked enamels. Phenolics are also hard but often darker and not targeted for clear, glossy anti-block surfaces. Epoxies are tough but may remain more tacky before full cure or under certain cure schedules. Alkyds are thermosettable via oxidation but are generally lower in hardness and more prone to blocking without MF crosslinkers. Step-by-Step Solution: Compare resin families by cured surface hardness and blocking resistance.MF typically provides the hardest, most non-tacky surfaces.Select melamine–formaldehyde as the best anti-tacking performer. Verification / Alternative check: Coating formulations often use melamine crosslinkers with alkyds or acrylics specifically to improve hardness and block resistance, confirming MF's superior anti-tack attributes. Why Other Options Are Wrong: Phenolic: hard but less used for clear anti-block surfaces.Epoxy: excellent adhesion and toughness; anti-tack not inherently superior without additives.Alkyd: can block without sufficient crosslinking; MF is often added to improve it. Common Pitfalls: Confusing “hardness” with “brittleness”; ignoring cure schedule and catalysts that influence tack. Final Answer: Melamine–formaldehyde resin

Question 20

Carbohydrate building blocks: β-D-glucose is the repeating monomer in which natural polymer?

Accepted Answer

Introduction / Context: Polysaccharides are formed by linking monosaccharides in specific configurations. The stereochemistry (alpha vs beta) of glycosidic bonds dictates structure and properties of the polymer. Given Data / Assumptions: Cellulose consists of β-D-glucose units. Starch and glycogen comprise α-D-glucose units. Proteins are polymers of amino acids, not sugars. Concept / Approach: Cellulose features β(1→4) glycosidic linkages between glucose units, forming linear chains that hydrogen-bond into strong microfibrils. Starch uses α-linkages (amylose: α(1→4), amylopectin: α(1→4) with α(1→6) branches), leading to different digestibility and crystallinity. Step-by-Step Solution: Identify the stereochemistry: β-glucose implies cellulose.Exclude α-linked polysaccharides (starch, glycogen).Exclude proteins (amino acid polymers). Verification / Alternative check: Cellulose's β-linkages render it indigestible to humans without cellulase enzymes, unlike starch. Why Other Options Are Wrong: Starch/glycogen: α-glucose polymers.Protein: peptide backbone, not carbohydrate. Common Pitfalls: Confusing α and β linkages; assuming all glucose polymers behave similarly. Final Answer: Cellulose

Polymer Technology Questions