Question 1

Intermolecular forces in nylon-6,6: the principal interchain bonding in linear polyamide 66 is predominantly which type?

Accepted Answer

Introduction / Context: Polyamide properties (strength, crystallinity, melting point) stem from strong interchain interactions. In nylons, amide groups create directional hydrogen bonds between chains. Concept / Approach: Unlike covalent crosslinks in thermosets, nylons are thermoplastics whose chains interact via hydrogen bonds (–C=O···H–N–), enhancing order and thermal resistance. Ionic bonding is not the main mechanism here. Final Answer: Hydrogen bonding

Question 2

Contact lens material identification: “Perspex” is the commercial name for which polymer used historically in optical applications?

Accepted Answer

Introduction / Context: Perspex, Lucite, and Plexiglass are trade names for PMMA, valued for optical clarity and biocompatibility in many lens and medical applications. Correct name recognition avoids material confusion with PS or PC. Final Answer: Polymethyl methacrylate (PMMA)

Question 3

Material class of PVC: Polyvinyl chloride is best classified as which type of material?

Accepted Answer

Introduction / Context: PVC is a vinyl thermoplastic used across pipes, cables, and films. It softens on heating and can be remelted and remoulded (within stabiliser limits), which defines thermoplastics. Concept / Approach: Thermosets cure irreversibly; PVC does not form permanent covalent crosslinks in standard grades (though crosslinkable PVC exists for specialty uses). Therefore, PVC is categorised as a thermoplastic. Final Answer: Thermoplastic

Question 4

In polymer science, which among the following is classified as an inorganic polymer (i.e., a macromolecule with a backbone containing non-carbon atoms such as Si–O)?

Accepted Answer

Introduction / Context: Classifying polymers as organic or inorganic depends on the composition of the main chain (backbone). Most commercial plastics are organic with carbon–carbon or carbon–heteroatom backbones, whereas inorganic polymers include repeating units with non-carbon atoms such as silicon and oxygen in the chain. Given Data / Assumptions: The options list common commercial polymers. We identify which one has a non-carbon backbone. Terminology: PTFE = polytetrafluoroethylene; PMMA = polymethyl methacrylate. Concept / Approach: Silicones (polysiloxanes) possess –Si–O–Si–O– repeating units and are therefore classified as inorganic or organo-inorganic polymers. PTFE, PMMA, and phenolic resins have carbon-based backbones and are grouped as organic polymers despite containing heteroatoms in side groups or crosslinks. Step-by-Step Solution: Check each option’s backbone: PTFE → –C–C–; PMMA → –C–C–; Bakelite → crosslinked carbon framework; Silicones → –Si–O–.Select the polymer with a non-carbon backbone: silicones.Confirm others remain organic. Verification / Alternative check: Materials texts universally describe polysiloxanes as inorganic/organosilicon polymers due to the siloxane linkage in the chain. Why Other Options Are Wrong: Teflon (PTFE): fully organic –C–C– backbone with fluorine substituents.Perspex (PMMA): methacrylate-based carbon chain; organic.Bakelite: phenol–formaldehyde thermoset; carbon framework. Common Pitfalls: Confusing the presence of heteroatoms in side groups with heteroatoms in the main chain. Only the backbone composition determines this classification. Final Answer: Silicones (polysiloxanes)

Question 5

Polymers can exist in multiple physical states under service or processing: which option correctly includes all the typical states?

Accepted Answer

Introduction / Context: Unlike simple molecules, polymers display a spectrum of physical states depending on temperature, crystallinity, and molecular architecture. Recognizing these states is essential for processing and application design. Given Data / Assumptions: Three states are listed: melt/rubbery, glassy, and semicrystalline. Thermal transitions such as glass-transition temperature (Tg) and melting temperature (Tm) govern these states. Concept / Approach: At temperatures above Tm (for crystallizable polymers) or above Tg (for amorphous polymers), chains have sufficient mobility for melt or rubber-like behavior. Below Tg, amorphous regions are glassy and brittle. Many engineering polymers are semicrystalline, containing both crystalline lamellae and amorphous regions; their macroscopic state reflects both morphology and temperature. Step-by-Step Solution: Identify each state and when it appears (relative to Tg and Tm).Note that polymers may pass through glassy → rubbery → melt with increasing temperature.Conclude that all listed states are possible. Verification / Alternative check: DSC and DMA measurements routinely evidence glass transition and melting, correlating with glassy, rubbery, and melt states. Why Other Options Are Wrong: Any single state alone is incomplete; polymers exhibit all depending on conditions. Common Pitfalls: Assuming a polymer has only one fixed state; processing windows prove otherwise. Final Answer: All of the above: (a), (b) and (c)

Question 6

Hot-emulsion SBR (styrene–butadiene rubber) production: typical reactor conditions are approximately which temperature and gauge pressure?

Accepted Answer

Introduction / Context: SBR is commonly manufactured via free-radical emulsion polymerisation. “Hot SBR” denotes operation around 50 °C (versus “cold SBR” near 5 °C). Recognising typical operating windows helps avoid confusing SBR processes with high-temperature bulk polymerisations. Given Data / Assumptions: Hot SBR temperature is near 50 °C. Pressure must maintain monomer in emulsion and control volatility; moderate gauge pressures are typical. Concept / Approach: Emulsion polymerisation uses water, surfactant, initiator, monomers (butadiene and styrene), and shortstops. Temperatures around 50 °C accelerate kinetics relative to cold SBR. Pressures of a few kgf/cm2 are sufficient for butadiene containment and reactor operation in stirred vessels. Step-by-Step Solution: Select options consistent with emulsion conditions (not bulk high-T).50 °C aligns with “hot SBR”.3–4 kgf/cm2 matches typical moderate pressures. Verification / Alternative check: Process descriptions in polymer reaction engineering texts list ~50 °C and a few bar for hot SBR reactors. Why Other Options Are Wrong: 250 °C: far above emulsion SBR temperatures, risking degradation.1 kgf/cm2 may be too low to manage butadiene vapor pressure and throughput. Common Pitfalls: Confusing SBR with high-pressure polyethylene processes which do run at much higher T and P. Final Answer: 50 °C and 3–4 kgf/cm2

Question 7

Polycaprolactam (nylon-6) via ring-opening/condensation of caprolactam at 240–280 °C typically reaches what overall monomer conversion under steady industrial operation?

Accepted Answer

Introduction / Context: Nylon-6 is produced by hydrolytic ring-opening polymerisation of caprolactam. Knowledge of typical conversion informs downstream finishing, devolatilisation, and monomer recovery (e.g., lactam extraction and recycle). Given Data / Assumptions: Process temperature: 240–280 °C. Industrial reactors often operate with water as a chain-transfer/initiating species. We seek approximate conversion in primary polymerisation before finishing. Concept / Approach: Primary polymerisation yields high but not complete conversion; typical values are around 85–95%. Residual monomer is removed in finishing to meet product specs. Hence “about 90%” is a representative figure for the main reaction stage. Step-by-Step Solution: Match operating range to known conversion band (high but not near 100%).Select the closest listed value: about 90%.Note that final product may have lower residual monomer after post-treatment. Verification / Alternative check: Process monographs show significant monomer removal steps (extraction/stripping), evidencing incomplete in-reactor conversion. Why Other Options Are Wrong: 50% or 75%: too low for practical productivity.99%: unusually high for the main reactor without extensive finishing. Common Pitfalls: Assuming condensation polymerisations reach full conversion in one pass; equilibrium and kinetics limit this. Final Answer: About 90%

Question 8

Applications of polyurethane (PU): identify the option that is NOT a typical end use for polyurethane materials.

Accepted Answer

Introduction / Context: Polyurethane chemistry produces a family of materials from soft foams to tough elastomers and high-solids coatings. Recognising where PU dominates—and where other polymers are preferred—helps in materials selection. Given Data / Assumptions: PU forms via isocyanate + polyol reactions. Common uses include flexible/rigid foams, coatings, adhesives, and elastomeric parts. Beverage bottles are typically PET or sometimes HDPE, not PU. Concept / Approach: Processability and regulatory/food-contact requirements make PET the bottle material of choice (clarity, barrier, strength). PUs excel in cushioning, protective coatings, footwear midsoles, and bonding applications but are not blown into beverage bottles. Step-by-Step Solution: List core PU applications: foams, coatings, adhesives.Compare to bottle production (stretch-blow moulded PET).Select “bottles for beverages” as the exception. Verification / Alternative check: Packaging industry standards specify PET for most beverage bottles due to clarity and carbonated-drink pressure resistance. Why Other Options Are Wrong: (a), (b), (c): all are mainstream PU uses. Common Pitfalls: Confusing “polyurethane elastomer parts” with blow-moulded packaging; very different processing routes. Final Answer: Bottles for beverages

Question 9

Identify the principal chemical component of cotton fibre used in textiles and medical gauze.

Accepted Answer

Introduction / Context: Natural fibres vary in composition. Cotton, derived from the seed hair of the cotton plant, is prized for its cellulosic content which governs dyeing, moisture absorption, and mechanical properties. Given Data / Assumptions: We consider purified fibre (lint) used in textiles. Minor constituents (waxes, pectins) are disregarded. Concept / Approach: Cotton is composed predominantly of cellulose, a linear polysaccharide of anhydro-glucose units. Lignin is abundant in wood and bast fibres, not in mature cotton lint. Starch and gelatine are unrelated (storage polysaccharide and protein, respectively) and not principal components of cotton fibre. Step-by-Step Solution: Recall cotton → nearly pure cellulose after scouring/bleaching.Eliminate lignin (woody tissues), starch (seed/endosperm), gelatine (animal protein).Select cellulose. Verification / Alternative check: Textile science references list cotton as ~90%+ cellulose after processing. Why Other Options Are Wrong: They are not the dominant chemical component of cotton fibre. Common Pitfalls: Confusing lignified bast fibres (e.g., jute) with cotton. Final Answer: Cellulose

Question 10

Low-pressure Ziegler process for polyethylene: select the single correct statement about operating/features.

Accepted Answer

Introduction / Context: The Ziegler (and later Ziegler–Natta/Phillips) processes revolutionized polyethylene by enabling polymerisation at relatively low pressures with specific catalysts, yielding high-density grades. Given Data / Assumptions: We compare pressure, yield, density of product, and catalyst requirements. Only one statement should be correct. Concept / Approach: Low-pressure processes use specialized catalysts to obtain HDPE with high conversion. They do use catalysts; they do not target very low-density PE (that is characteristic of high-pressure free-radical LDPE). Operating pressures vary (often a few to a few tens of bar), so asserting a single fixed value like 30 kgf/cm2 as “the” standard is too specific for a universal statement. Step-by-Step Solution: Eliminate (c): product is HDPE, not very low density.Eliminate (d): catalysts are essential (Ziegler/Phillips).Treat (a) as over-specific; conditions vary by technology and reactor.Select (b) as the broadly correct and general statement. Verification / Alternative check: Process summaries show high single-pass conversions and high overall ethylene utilisation in Ziegler-based HDPE plants. Why Other Options Are Wrong: (a) Not universally true; pressure is not a single fixed value.(c) Misstates product density class.(d) Incorrect—catalyst is fundamental. Common Pitfalls: Conflating high-pressure LDPE (radical) with low-pressure catalytic HDPE. Final Answer: Achieves a high ethylene conversion/yield of roughly 95–98%.

Question 11

Viscose rayon classification: complete the definition—viscose rayon is best described as __________.

Accepted Answer

Introduction / Context: Man-made cellulosic fibres are derived from natural cellulose that is chemically processed and then regenerated into filaments. Viscose rayon is the archetype of this family. Given Data / Assumptions: The original options missed the exact wording; applying the Recovery-First Policy, we correct the set to include the proper term. We distinguish between nitrate, acetate, and viscose processes. Concept / Approach: Viscose rayon is produced by xanthating cellulose to form cellulose xanthate, dissolving it in alkali, and extruding into an acid bath to regenerate cellulose filaments. This outcome is regenerated cellulose, not nitrate (celluloid) nor acetate (cellulose acetate). Step-by-Step Solution: Identify viscose process → cellulose derivatisation and regeneration.Match to “regenerated cellulose”.Eliminate nitrate and acetate variants. Verification / Alternative check: Textile chemistry references classify viscose rayon squarely as regenerated cellulose fibre. Why Other Options Are Wrong: Cellulose nitrate/acetate belong to different fibre families with distinct processing.“None” is invalid because the correct descriptor exists. Common Pitfalls: Confusing “rayon” generically with acetate; acetate is a separate man-made cellulosic. Final Answer: Regenerated cellulose

Question 12

Typical molecular-weight scale for polymers: select the most appropriate order-of-magnitude range for number-average molecular weights observed in commercial polymers.

Accepted Answer

Introduction / Context: Polymeric materials derive their properties from very high molecular weights compared to small molecules. Understanding the typical range helps interpret viscosity, mechanical strength, and processing behavior. Given Data / Assumptions: We refer to number-average (Mn) or weight-average (Mw) molecular weights for common commercial polymers. Extremes may exist, but typical commercial ranges are needed. Concept / Approach: Commodity and engineering polymers generally span from a few hundred (oligomers/prepolymers) up to several million g/mol. Values above 10^7 g/mol are uncommon in standard processing due to extreme viscosity and limited melt flow. Therefore, 10^2–10^7 captures the practical landscape. Step-by-Step Solution: Compare each range with known polymer data (e.g., PE, PS, PVC, PET).Select the broad but realistic 10^2–10^7 range.Exclude unrealistically high bands for routine processing. Verification / Alternative check: Gel-permeation chromatography data for commercial resins commonly report Mw in the 10^4–10^6 range. Why Other Options Are Wrong: 10–10^3 is too low for typical high-molecular-weight resins.10^7–10^9 and higher are extraordinary and not standard for melt processing. Common Pitfalls: Confusing degree of polymerisation (DP) with molecular weight units; DP times repeat-unit mass gives MW. Final Answer: 10^2 to 10^7

Question 13

Role of diphenylamine in rubber compounding: what is the primary purpose of adding diphenylamine to rubber formulations?

Accepted Answer

Introduction / Context: Antioxidants are critical additives in rubber to inhibit degradation that leads to cracking, hardening, or loss of elasticity. Aromatic amines, including diphenylamine and its derivatives, are classic antidegradants in many diene rubbers. Given Data / Assumptions: We consider the function of diphenylamine within typical rubber recipes. We separate antioxidant roles from curing chemistry. Concept / Approach: Diphenylamine acts as a stabiliser that scavenges radicals and inhibits oxidative chain reactions, thereby protecting rubber during service and processing. Vulcanisation is driven by sulphur/accelerator systems, not by diphenylamine. Flame retardancy requires other additives (e.g., halogenated systems, ATH), and thermosetting arises from crosslinking via curing agents, not antioxidants. Step-by-Step Solution: Identify diphenylamine class → antioxidant/antidegradant.Associate with protection against oxygen, heat, and ozone effects.Exclude vulcanisation, flame retardancy, and curing roles. Verification / Alternative check: Compounding guides list aromatic amines/phenolics as antioxidants improving heat-aging resistance. Why Other Options Are Wrong: (a) Vulcanisation requires sulphur/accelerators.(c) Non-inflammability needs dedicated FR systems.(d) Antioxidants do not crosslink rubber. Common Pitfalls: Assuming any additive that improves durability must be a curing agent; functions are distinct. Final Answer: To protect against oxidative deterioration during air/heat exposure

Question 14

Buna-S rubber nomenclature: in industrial and polymer science terminology, Buna-S is more commonly referred to as which material?

Accepted Answer

Introduction / Context: Buna-S is a historically important synthetic rubber that helped transform tire and general rubber industries. The question tests your knowledge of trade names and corresponding chemical identities in elastomers. Recognising these equivalences is crucial for materials selection, processing parameters, and understanding performance trade-offs in real engineering applications. Given Data / Assumptions: Buna-S is the legacy/trade description. Options include different polymers with distinct chemistries and properties. We must map Buna-S to its standard chemical name. Concept / Approach: “Buna” historically refers to butadiene-based rubbers; the letters after the hyphen indicate comonomers. Buna-S specifically denotes a copolymer synthesised from butadiene (Bu) and styrene (St), hence SBR: styrene–butadiene rubber. SBR is produced predominantly by emulsion polymerisation (hot or cold processes) and is widely used in passenger car tire treads, footwear, belts, and many moulded goods owing to balanced abrasion resistance and cost. Step-by-Step Solution: Identify the historical naming scheme: Buna-S → butadiene + styrene.Match to the modern, generic name: styrene–butadiene rubber (SBR).Exclude other candidates that do not share the same monomer set. Verification / Alternative check: Materials handbooks consistently equate Buna-S with SBR. Product datasheets for tires and industrial rubbers also use SBR as the generic name. Why Other Options Are Wrong: Polyurethane (PU): formed from diisocyanates and polyols; not butadiene/styrene based.Teflon (PTFE): fluoropolymer derived from tetrafluoroethylene; very different family.Bakelite: thermosetting phenolic resin; not an elastomer. Common Pitfalls: Confusing Buna-S (SBR) with Buna-N (NBR, nitrile rubber from butadiene + acrylonitrile). The suffix letter(s) matter. Final Answer: SBR (styrene–butadiene rubber)

Question 15

Thermoplastic resins: choose the statement that best characterises thermoplastics compared with thermosetting polymers.

Accepted Answer

Introduction / Context: Thermoplastics and thermosets are the two broad classes of polymeric materials encountered in engineering. Distinguishing their behaviours, especially under heat and mechanical stress, is foundational for process selection (injection, extrusion, thermoforming) and end-use performance (toughness, impact, recyclability). Given Data / Assumptions: We compare general properties, not edge-case formulations. Thermoplastics soften on heating and harden on cooling reversibly. Thermosets form crosslinked networks that do not melt. Concept / Approach: Thermoplastics possess linear or branched chains with limited crosslinking, enabling chain mobility when heated. This makes them reprocessable and, in many cases, tougher and less brittle than highly crosslinked thermosets. Thermosets, in contrast, cure irreversibly to rigid networks that resist creep at elevated temperatures but tend to be more brittle and cannot be remelted or readily reclaimed by melt processing. Step-by-Step Solution: Assess option (a): incorrect; thermoplastics soften when hot and harden when cooled—statement is internally inconsistent.Assess option (b): incorrect; many thermoplastics can be reprocessed/recycled via melt routes.Assess option (c): describes thermosets, not thermoplastics.Option (d) aligns with typical behaviour—thermoplastics are usually less brittle than thermosets. Verification / Alternative check: Impact tests (e.g., Izod/Charpy) and DMA measurements generally show higher toughness and ductility in many thermoplastics versus thermosets. Why Other Options Are Wrong: (a) Misstates temperature behaviour.(b) Ignores recyclability of thermoplastics.(c) Confuses with thermoset definition. Common Pitfalls: Assuming all plastics behave identically upon heating. Crosslink density dictates melt/flow capability. Final Answer: They are generally less brittle than thermosetting resins.

Question 16

Alkyd resins are produced by reacting a polybasic acid (e.g., phthalic anhydride) with a polyhydric alcohol (e.g., glycerol). Identify the correct product class.

Accepted Answer

Introduction / Context: Polycondensation reactions between polybasic acids and polyhydric alcohols yield a family of coating resins whose performance can be tuned by oil modification and monomer selection. Recognising this chemistry is essential in paints, varnishes, and coil-coating industries. Given Data / Assumptions: Polybasic acid example: phthalic anhydride. Polyhydric alcohol example: glycerol (triol). Resulting resin class is tailored for coatings and varnishes. Concept / Approach: Alkyds are oil-modified polyesters formed by condensation of polyacids (or anhydrides) with polyols, often incorporating fatty acids or oils to enhance flexibility, flow, and drying characteristics. While unsaturated/saturated polyesters are also formed via acid-alcohol condensation, the term “alkyd” specifically denotes the oil-modified polyester systems widely used in solvent-borne and water-reducible coatings. Step-by-Step Solution: Map the generic reaction: polyacid + polyol → polyester backbone.Recognise the coatings context and oil modification → alkyd resin.Exclude other classes that do not match this exact industrial nomenclature. Verification / Alternative check: Coatings textbooks list alkyds as phthalic anhydride–glycerol based resins, frequently modified with drying oils for oxidative cure. Why Other Options Are Wrong: Unsaturated polyester: used for FRP with styrene crosslinking; different usage focus.Saturated polyester: film/engineering plastics; not termed “alkyd”.Amino resins: urea/melamine–formaldehyde, distinct chemistry. Common Pitfalls: Equating any polyester from acid-alcohol condensation with “alkyd”. Oil modification and coatings context define alkyds. Final Answer: Alkyd resins

Question 17

Free-radical addition polymerisation is typically initiated by substances that decompose to form radicals. Which reagent class commonly serves as the initiator?

Accepted Answer

Introduction / Context: Initiation of free-radical polymerisation requires a clean, controllable source of radicals. Selecting the correct initiator system is central to controlling molecular weight, rate, and polymer microstructure in industrial processes like PE (high-pressure), PS, PMMA, and many acrylics. Given Data / Assumptions: We focus on addition polymerisation via radical mechanism. Common initiator families include peroxides and azo compounds. Strong mineral acids are not radical sources. Concept / Approach: Organic peroxides (e.g., benzoyl peroxide, dicumyl peroxide) thermally decompose to generate radicals capable of adding to vinyl monomers and propagating chains. Acids like sulphuric or hydrofluoric serve as catalysts for ionic mechanisms (e.g., esterification, alkylation) but are not radical initiators for vinyl polymerisation. Step-by-Step Solution: Identify the mechanism: free-radical → need radical donor.Peroxides homolyse the O–O bond to form radicals.Exclude mineral acids; they promote ionic rather than radical pathways. Verification / Alternative check: Polymer reaction engineering references list peroxides and azo initiators (e.g., AIBN) as standard radical sources. Why Other Options Are Wrong: Sulphuric acid and hydrofluoric acid are not radical initiators for vinyl addition; they catalyse other reactions.“None” is invalid since a correct class exists. Common Pitfalls: Confusing initiators (radical) with catalysts for ionic or condensation polymerisations. Final Answer: Organic peroxides

Question 18

Crystallisation of celluloid (a nitrate–camphor plastic) is undesirable; which additive is traditionally used to prevent crystallisation and maintain flexibility?

Accepted Answer

Introduction / Context: Celluloid, one of the earliest commercial plastics, is based on cellulose nitrate plasticised with camphor. Control of crystallisation and brittleness was a key historical challenge; plasticisers disrupt chain packing, improving flexibility and processability. Given Data / Assumptions: Celluloid = nitrocellulose + plasticiser. We seek the classic additive that prevents crystallisation and embrittlement. Options include common additives and components. Concept / Approach: Camphor acts as a plasticiser: it interposes between polymer chains, reduces intermolecular attractions, and inhibits ordered packing, thereby limiting crystallisation. This maintains clarity and flexibility. Glycerol plasticises some hydrophilic polymers but is not the traditional agent in celluloid. Nitrocellulose is the polymer itself, not an additive. “None” is invalid because the well-known plasticiser is camphor. Step-by-Step Solution: Recall composition of celluloid: nitrocellulose + camphor.Recognise camphor’s role in suppressing crystallisation and brittleness.Eliminate distractors that are either base polymer or unrelated plasticisers. Verification / Alternative check: Historical manufacturing notes and museum conservation literature cite camphor as the principal plasticiser in celluloid objects. Why Other Options Are Wrong: Glycerol: used with PVA or gelatin systems, not classic for celluloid.Nitrocellulose: not an additive; it is the base polymer.None: incorrect because camphor is correct. Common Pitfalls: Confusing moisture plasticisation (e.g., glycerol in hydrophilic polymers) with organic plasticisers used in early plastics. Final Answer: Camphor

Question 19

Identify the monomer used to produce neoprene (polychloroprene) synthetic rubber in commercial practice.

Accepted Answer

Introduction / Context: Neoprene is the trade name for polychloroprene, a versatile synthetic rubber noted for oil resistance, weathering resistance, and good mechanical strength. Knowing its monomer origin clarifies performance attributes and processing behaviour. Given Data / Assumptions: The polymer family is polychloroprene. We pick the correct monomer from common diene candidates. Concept / Approach: Polychloroprene is formed by polymerising chloroprene (2-chloro-1,3-butadiene). Although acetylene chemistry historically featured in early monomer syntheses (e.g., for chloroprene manufacture), acetylene is not the monomer polymerised to form neoprene. Isoprene polymerises to natural rubber-like polyisoprene, not neoprene. Step-by-Step Solution: Match trade name to chemical family: neoprene → polychloroprene.Identify the exact monomer: chloroprene.Exclude other dienes that yield different rubbers. Verification / Alternative check: Rubber technology texts list chloroprene emulsion polymerisation as the standard route to neoprene. Why Other Options Are Wrong: Acetylene: feedstock/intermediate for various chemicals, not the final monomer here.Isoprene: monomer for polyisoprene (natural/synthetic), not neoprene.None: incorrect; a specific correct monomer exists. Common Pitfalls: Confusing raw feedstocks (used upstream) with the actual polymerised monomer. Final Answer: Chloroprene (2-chloro-1,3-butadiene)

Question 20

Linear polymers typically exhibit which classification in terms of thermal behaviour and processing?

Accepted Answer

Introduction / Context: The architecture of polymer chains—linear, branched, or crosslinked—governs thermal response and processing. Recognising the link between structure and behaviour helps select moulding, extrusion, or curing pathways. Given Data / Assumptions: Linear polymers possess chains without extensive crosslinking. We seek the general classification that matches most linear materials. Concept / Approach: Linear (and lightly branched) polymers soften and flow upon heating due to increased chain mobility, enabling melt processing; they are categorised as thermoplastics. Thermosets, by contrast, contain dense covalent crosslinks that do not melt on heating but instead char or degrade. Elastomers are lightly crosslinked to exhibit rubbery elasticity at service temperature; however, not all linear polymers are elastomeric. Step-by-Step Solution: Map structure → behaviour: linear chains → melt reprocessability.Select “thermoplastic” as the most representative class.Exclude thermoset and elastomer categories which require crosslinking for their defining properties. Verification / Alternative check: Commodity resins such as PE, PP, PS, and PET are largely linear/branched and are melt-processable thermoplastics. Why Other Options Are Wrong: Thermosetting: needs crosslinked network, not typical of linear chains.Elastomeric: requires light crosslinking and glass transition below service temperature—structure/function differs.“Extremely high softening point only”: not a classification and not universally true. Common Pitfalls: Assuming “linear” implies “rubbery.” Rubberiness depends on Tg and crosslink density, not linearity alone. Final Answer: Thermoplastic

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