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Polymers · TOPIC LESSON

Polymerization and Polymer Composition

Read the repeating connectivity, identify how chains grow, and connect molecular architecture to bulk properties.

JEE Main · JEE Advanced · NEET UG47 min study guide10 indexed sections

BY THE END, YOU CAN

  • Distinguish monomer, repeat unit, degree of polymerization and chain-end groups.
  • Explain chain-growth and step-growth without making them synonyms for addition and condensation.
  • Reconstruct common vinyl, diene, polyester and polyamide structures and monomers.
  • Compare copolymer sequences, tacticity, branching and cross-linking.
  • Calculate composition, average molar masses and ideal step-growth conversion requirements.
  • Connect polymer properties to applications with temperature and formulation caveats.

BEFORE YOU START

  • Alkene addition, radicals and nucleophilic acyl substitution.
  • Functional groups, stereochemistry and intermolecular forces.
  • Mole fractions, mass fractions and weighted averages.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. A repeat unit is a connectivity pattern, not an isolated monomer

A polymer contains macromolecules built from many linked units. The monomer is a starting molecule; the constitutional repeat unit is the structural segment repeated along the polymer chain. For ethene addition polymerization, CH₂=CH₂ becomes a chain segment –CH₂–CH₂–. The original pi bond is consumed, while the two carbon atoms become part of the saturated backbone.

Brackets around a repeat structure must cut actual backbone bonds, with connections continuing through both bracket ends. A formula such as[–CH₂–CH(Cl)–]ₙ represents poly(vinyl chloride), not a chain retaining an alkene C=C in every repeat. Substituents remain on the correct backbone carbon; chlorine is not removed during ordinary vinyl chloride polymerization.

The degree of polymerization counts repeat units according to the stated convention. A polyester repeat may contain residues from two monomer molecules, so “one repeat” is not always “one original molecule”. When converting chain molar mass to a repeat count, use the repeat mass after any eliminated small molecules and state whether end-group mass is neglected.

Real polymer samples usually contain chains of different lengths. A single value n in a sketch often represents an average or an idealized chain, not proof that every molecule has identical molar mass. Chain ends contribute little to a very long-chain mass but can dominate the chemistry of functional end groups.

ORIGINAL WORKED EXAMPLE

Recover a vinyl monomer

A polymer segment is –CH₂–CH(CN)–CH₂–CH(CN)–. What monomer gives this connectivity by ordinary addition polymerization?

  1. Identify the two-carbon backbone repeat –CH₂–CH(CN)–.
  2. Restore a double bond between those two backbone carbons without moving the CN substituent.
  3. The monomer is CH₂=CH–CN.
  4. The nitrile carbon is a substituent carbon, not one of the two vinyl backbone carbons.

Answer Acrylonitrile, CH₂=CHCN.

02 / UNDERSTAND & APPLY

2. Chain-growth transfers an active centre along a growing chain

In chain-growth polymerization, an active chain end reacts with monomer and regenerates an active end after each propagation step. In a radical example, an initiating radical adds to a vinyl double bond, creating a new carbon-centred radical. That radical adds to another monomer, extending the chain while leaving the radical at the new terminus.

Initiation creates active centres, propagation extends chains, and termination removes their ability to continue by that mechanism. Two radicals can combine to form a covalent bond, or disproportionation can transfer H so that one chain gains a saturated end and another an unsaturated end. Chain transfer moves activity to another molecule and can alter chain length without consuming all monomer.

Use single-headed fishhook arrows for individual-electron motion in radical mechanisms. A propagation step redistributes the two pi electrons and the radical electron to form a new sigma bond and a new radical centre. Ordinary double-headed curved arrows instead represent electron pairs and belong to polar mechanisms.

Chain-growth need not be radical. Cationic polymerization uses a cationic active end, anionic polymerization an anionic one, and coordination polymerization can insert monomer into a metal–carbon bond. Monomer substituents and the catalyst or initiator determine which pathway is viable. Polyisobutylene is associated with 2-methylpropene, not ordinary propene; confusing those names produces a wrong repeat structure.

High-molar-mass chains can appear early in a chain-growth process even while much monomer remains. This differs from ideal step-growth, where the average chain length rises sharply only as functional-group conversion becomes very high. These are mechanistic trends, not a guarantee of identical distributions for all industrial processes.

03 / UNDERSTAND & APPLY

3. Common addition polymers preserve their substituents

Ethene gives polyethylene,[–CH₂–CH₂–]ₙ. Propene gives polypropylene,[–CH₂–CH(CH₃)–]ₙ. Vinyl chloride gives PVC,[–CH₂–CH(Cl)–]ₙ. Styrene gives polystyrene,[–CH₂–CH(C₆H₅)–]ₙ. Tetrafluoroethene gives PTFE,[–CF₂–CF₂–]ₙ. Each formula retains every monomer atom in the ideal addition repeat, apart from chain-end details.

Methyl methacrylate CH₂=C(CH₃)CO₂CH₃ gives PMMA with both CH₃ and CO₂CH₃ attached to the same substituted backbone carbon. Its ester group is pendant, rather than a linkage in the main backbone. That distinction matters in predicting hydrolysis: changing a pendant ester does not automatically cleave every carbon–carbon backbone bond.

A conjugated diene can undergo 1,4 polymerization, leaving one C=C per repeat. For isoprene CH₂=C(CH₃)–CH=CH₂, the common 1,4 repeat is –CH₂–C(CH₃)=CH–CH₂–. Natural rubber is predominantly cis-1,4-polyisoprene; trans-1,4-polyisoprene, associated with gutta-percha, has different packing and properties. Other microstructures such as 1,2 addition are possible with different conditions.

Neoprene is based on chloroprene, CH₂=C(Cl)–CH=CH₂. Styrene–butadiene rubber contains residues of styrene and butadiene; nitrile rubber contains butadiene and acrylonitrile residues. A copolymer name identifies its monomer sources but does not by itself specify their exact mole fractions or sequence distribution.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Monomer to repeat-unit connectivityPropene gives a polypropylene backbone CH2-CH with pendant methyl on the CH. Brackets cross continuing bonds rather than enclosing disconnected monomers. Butadiene 1,4-addition gives CH2-CH=CH-CH2 with one remaining double bond per repeat.Vinyl monomer → chain repeatPropene retains its methyl substituentCH₂ = CH — CH₃chain growthCH₂CHCH₃nPolypropylene: [—CH₂—CH(CH₃)—]ₙBoundary bonds continue to adjacent repeats.The double bond becomes a backbone single bond.A diene retains one C=C after 1,4-additionCH₂=CH—CH=CH₂[—CH₂—CH=CH—CH₂—]ₙOne repeat: C₄H₆; no small molecule is lost.Cis/trans geometry requires a separate drawing.
Connectivity schematics. Chain ends are omitted; n counts repeats. The butadiene formula does not specify cis/trans microstructure, and the polypropylene projection does not specify tacticity.

ORIGINAL WORKED EXAMPLE

Check the repeat formula of polypropylene

Find the ideal polypropylene repeat formula and repeat molar mass using C=12.0 and H=1.0.

  1. The repeat is –CH₂–CH(CH₃)–.
  2. Its carbon count is three, including the methyl substituent.
  3. Its hydrogen count is 2+1+3=6.
  4. The repeat mass is 3×12+6×1=42.0 g mol⁻¹.

Answer C₃H₆ and 42.0 g mol⁻¹ per repeat.

04 / UNDERSTAND & APPLY

4. Step-growth can join monomers, oligomers and longer chains

In step-growth polymerization, functional groups on molecules of many sizes can react with one another. A dimer can react with a monomer, another dimer or a longer chain, provided compatible groups remain. Many small species coexist early, and long chains require a high extent of functional-group reaction and appropriate stoichiometric balance.

A diacid and a diol can form a polyester by repeated ester formation, while a diacid and a diamine can form a polyamide. With carboxylic acids as starting groups, water is eliminated per newly formed ester or amide linkage in the ideal condensation balance. Using acid chlorides changes the small-molecule byproduct to HCl; the repeat connectivity can be the same.

Chain-growth/step-growth classify how chains grow. Addition/condensation describe a different aspect of the chemistry, especially whether small molecules are eliminated in building the polymer. Many familiar examples align chain-growth with addition and step-growth with condensation, but the labels are not universally interchangeable.

Polyurethane formation from a diol and a diisocyanate is a useful counterexample: stepwise addition forms urethane links without necessarily eliminating a small molecule in the ideal chain-forming reaction. Ring-opening polymerization provides further cases where the simple school-level addition/condensation division is not a complete mechanism classification.

Monomer functionality controls architecture. A purely difunctional ideal set favours linear chains; monofunctional molecules can cap chains, and functionality greater than two can create branching and eventually a network. Side reactions, unequal reactivity and cyclic oligomer formation can make actual products more complicated than the simplest model.

05 / UNDERSTAND & APPLY

5. Reconstruct monomers by cutting the correct linkage

PET is formed from ethane-1,2-diol and terephthalic acid or suitable derivatives. Its repeat can be written –O–CH₂CH₂–O–C(=O)–C₆H₄–C(=O)–, with the aromatic carbonyl substituents para to one another. Hydrolysis of the main-chain ester links separates glycol-derived and terephthalate-derived fragments.

Nylon 6,6 is based on hexane-1,6-diamine and hexanedioic acid, also called adipic acid. The repeat is –NH–(CH₂)₆–NH–C(=O)–(CH₂)₄–C(=O)–. Each monomer has six carbons, but the diacid’s six include its two carbonyl carbons, leaving only four CH₂ groups between them. Writing six CH₂ groups in the acid residue would give the wrong polymer.

Nylon 6 has repeat –NH–(CH₂)₅–C(=O)– and can be related to caprolactam ring opening or 6-aminohexanoic-acid connectivity. The six-carbon count again includes the carbonyl carbon. The final repeat does not have the same composition as a nylon 6,6 repeat simply because both names contain 6.

Polyamides can form interchain hydrogen bonds involving N–H donors and carbonyl oxygen acceptors, contributing to strength and fibre formation. Polyesters have polar ester groups but lack the corresponding amide N–H donors. Aromatic rings and chain regularity can increase rigidity; the exact material properties also depend on processing, crystallinity and molecular mass.

To infer monomers from a condensation polymer, identify the main-chain ester or amide connections and conceptually reverse their formation. Restore OH on the appropriate acyl fragment and H on the alcohol or amine fragment. Do not break arbitrary C–C bonds simply to make the name fit a remembered monomer pair.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Connectivity of PET and nylon-6,6 repeat units and monomersPET repeat contains O-CH2-CH2-O-C(=O)-para-phenylene-C(=O), leading back to ethane-1,2-diol and terephthalic acid. Nylon-6,6 repeat contains NH-(CH2)6-NH-C(=O)-(CH2)4-C(=O), leading back to hexane-1,6-diamine and hexanedioic acid.Cut a linkage to recover its monomersPET: an ester joins alcohol and acid residues[—O—CH₂—CH₂—O—C(=O)—p-C₆H₄—C(=O)—]ₙp-C₆H₄ means para-disubstituted phenylene.hydrolysisHO—CH₂—CH₂—OH+HOOC—p-C₆H₄—COOHEthane-1,2-diol + terephthalic acidRepeat mass: 62 + 166 − 2(18) = 192Nylon-6,6: amide links retain N in chain[—NH—(CH₂)₆—NH—C(=O)—(CH₂)₄—C(=O)—]ₙH₂N—(CH₂)₆—NH₂+HOOC—(CH₂)₄—COOHHexane-1,6-diamine + hexanedioic acidTwo amide links per ideal bulk repeat.Finite-chain end groups alter exact water counts.
Condensed structural formulas show every ester/amide linkage and distinguish the alcohol or amine residue from the acid residue. Hydrolysis products are shown in neutral parent form; their actual ionization depends on workup pH. Bulk repeat bookkeeping omits finite-chain end corrections.

ORIGINAL WORKED EXAMPLE

Repeat-mass accounting for PET

Use ethylene glycol 62.0 g mol⁻¹ and terephthalic acid 166.0 g mol⁻¹. Find the PET repeat molar mass in the long-chain condensation model.

  1. One repeat contains residues from one diol and one diacid.
  2. Its two ester connections correspond to loss of two waters per repeat in the long-chain interior accounting.
  3. Add monomer masses:62.0+166.0=228.0.
  4. Subtract 2×18.0=36.0.

Answer 192.0 g mol⁻¹ per PET repeat; exact finite-chain end groups require a separate correction.

06 / UNDERSTAND & APPLY

6. Sequence, tacticity and branching are different structural questions

A homopolymer derives from one monomer species; a copolymer contains more than one. An alternating copolymer has an ideal ABAB sequence, a block copolymer contains longer A and B runs, and a graft copolymer has branches of one composition attached to a backbone of another. A statistical sequence is governed by incorporation probabilities and can depart from a perfectly random simple model.

Monomer feed fraction need not equal polymer composition. Different monomers can react at different rates with different chain ends, and the remaining feed can change during conversion. Unless the problem supplies equal reactivities or direct incorporation data, do not infer an exact copolymer formula from the initial feed ratio alone.

Tacticity describes the relative configuration of successive stereogenic backbone sites in polymers such as polypropylene. Isotactic, syndiotactic and atactic describe regular same-relative, regular alternating-relative and irregular stereochemical sequences in the conventional extended-chain depiction. These are configurational patterns, not merely arbitrary rotations of a flexible chain drawn on paper.

Branching concerns connectivity: side chains emerge from the main backbone. Cross-linking connects different chains into a network. Neither is the same as placing all methyl substituents on the same side in a tacticity drawing. A polypropylene methyl pendant group is part of its repeat structure, not by itself evidence of the long-chain branching meant in polyethylene architecture comparisons.

Coordination catalysts can control chain insertion and stereochemistry, allowing highly regular polymers. The details depend on the catalyst system, not on the generic phrase metal catalyst. Regularity often aids close packing and crystallinity, which in turn affects stiffness, melting behaviour and permeability.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Copolymer sequence and crosslinked network topologyAlternating ABABAB, block AAABBB and an irregular ABB AAB sequence illustrate residue order. Below, three polymer chain paths are joined by two sparse covalent interchain crosslinks. An irregular finite sequence alone does not establish statistical randomness.Sequence is different from architectureA and B label whole repeat residuesAlternatingA—B—A—B—A—BBlockA—A—A—B—B—BOne irregular sequenceA—B—B—A—A—BA random sequence is a statistical description.Sparse interchain cross-links make a networkGreen lines: flexible polymer-chain pathsRed connectors: interchain covalent cross-linksLinks limit permanent chain sliding after stretching.Line shapes are schematics, not molecular conformations.
Topology model, not an atom-resolved structure. The cross-link connectors do not specify the chemical identity or length of a sulfur bridge. Cross-link density and chain mobility jointly affect elastic response.

07 / UNDERSTAND & APPLY

7. Explain uses through packing, mobility and network structure

Thermoplastics soften on heating and can often be reshaped because their chains are not permanently connected into an extensive covalent network. Thermosets form cross-linked networks and do not simply remelt into freely flowing chains; sufficient heating can degrade them instead. These categories describe material behaviour, not a claim that every repeat unit contains a particular functional group.

Less-branched polyethylene can pack more efficiently and generally has higher crystallinity and density than highly branched polyethylene. HDPE is commonly associated with more rigid containers and piping applications; LDPE with flexible films and related products. Exact grades and processing histories matter, so a trade name or density label is not a complete performance specification.

PVC can be rigid or flexible depending on formulation, especially plasticizers that increase segmental mobility. It is used in pipes, coatings and insulation in suitable formulations. Polystyrene is commonly rigid and can be foamed for lightweight packaging. PMMA is transparent and used where optical clarity and stiffness are useful; PTFE has low surface adhesion and strong chemical resistance for suitable sealing and coating applications.

Fibres such as nylon and polyester benefit from chain alignment, intermolecular attraction and sufficient crystallinity. Drawing can orient chains and increase tensile strength along a direction, so a polymer’s properties are not determined by repeat formula alone. Directional mechanical behaviour is particularly important in oriented fibres.

The glass-transition temperature Tg marks a change in segmental mobility of amorphous regions; it is not the same as the melting temperature Tm of crystalline regions. A semicrystalline polymer can show both. Below Tg, an amorphous region is typically glassy; above Tg it becomes more mobile, though cross-links or crystalline domains can still prevent free flow.

Biodegradability depends on chemical linkages, morphology, environment and available organisms or enzymes. The presence of ester groups can support hydrolytic degradation but does not guarantee rapid disappearance in every environment. Natural origin, recyclability and biodegradability are separate claims. PHBV-type polyesters and biodegradable copolyamides are standard examples, but their actual breakdown conditions must be specified.

09 / UNDERSTAND & APPLY

9. Count molecules for Mn and weight molecules for Mw

If Ni molecules have molar mass Mi, the number-average molar mass is Mn=ΣNiMi/ΣNi. It weights each molecule equally. The mass-average, commonly called weight-average, molar mass is Mw=ΣNiMi²/ΣNiMi. It gives greater weight to heavier molecules. Their ratio Đ=Mw/Mn is the molar-mass dispersity and is at least one for a positive distribution.

A sample can therefore have a small number of very long chains that raise Mw substantially more than Mn. End-group analysis and ideal colligative measurements are linked to molecule number and hence Mn, while other techniques weight the distribution differently. A quoted average molecular mass is incomplete unless the average type is identified.

For a sufficiently long homopolymer with negligible end-group mass, number-average degree of polymerization is approximately Mn divided by repeat molar mass. In a copolymer, the average repeat mass must be calculated from incorporated monomer fractions under the specified model. Mole fractions and mass fractions are not interchangeable when the repeat masses differ.

For ideal linear AB step-growth with equal group reactivity, no rings and no branching, start with N₀ monomer molecules. If fraction p of the initial A groups reacts intermolecularly, pN₀ bonds form and the number of molecules falls to N₀(1−p). Thus Xn=N₀/N=1/(1−p). The derivation explains why 99% conversion gives only 100 average units. Stoichiometric imbalance, capping and intramolecular cyclization invalidate that simple version.

Mn=ΣNiMi/ΣNi; Mw=ΣNiMi²/ΣNiMi; Đ=Mw/Mn

Number and mass averages refer to the same specified molecular population.

Xn=1/(1−p)

Ideal linear AB step-growth, equal reactivity, no cyclization/branching/capping.

ORIGINAL WORKED EXAMPLE

Two molecular populations

A sample has equal numbers of chains with molar masses 10,000 and 30,000 g mol⁻¹. Find Mn, Mw and Đ.

  1. Mn=(10,000+30,000)/2=20,000 g mol⁻¹.
  2. Mw=(10,000²+30,000²)/(10,000+30,000).
  3. The numerator is 1.00×10⁹ and denominator 40,000, giving 25,000 g mol⁻¹.
  4. Đ=25,000/20,000=1.25.

Answer Mn=20,000; Mw=25,000 g mol⁻¹; Đ=1.25.

ORIGINAL WORKED EXAMPLE

Conversion close to one still matters

For the stated ideal AB step-growth model, compare p=0.980 and p=0.995.

  1. At 0.980, Xn=1/0.020=50.0.
  2. At 0.995, Xn=1/0.005=200.
  3. A1.5 percentage-point increase in conversion raises the average chain length fourfold.
  4. The formula requires all the stated ideal assumptions.

Answer Xn=50.0 and 200 respectively.

10 / UNDERSTAND & APPLY

10. Composition examples and solved practice

Keep three counts separate: monomer molecules supplied, residues incorporated and polymer molecules produced. In a condensation problem, also count eliminated small molecules and terminal groups. In a copolymer problem, use the incorporated composition supplied rather than silently substituting the feed composition.

ORIGINAL WORKED EXAMPLE

Copolymer mole fraction versus mass fraction

An ideal incorporated copolymer contains 60.0 mol% ethene-derived units of mass 28.0 and 40.0 mol% propene-derived units of mass 42.0 g mol⁻¹. Find average unit mass and mass fraction of propene-derived units.

  1. On a 1.000 mol unit basis, ethene-derived mass is 0.600×28.0=16.8 g.
  2. Propene-derived mass is 0.400×42.0=16.8 g.
  3. Total mass is 33.6 g, so average unit mass is 33.6 g mol⁻¹.
  4. The propene-derived mass fraction is 16.8/33.6=0.500.

Answer Average unit mass 33.6 g mol⁻¹; propene-derived mass fraction 50.0%.

ORIGINAL WORKED EXAMPLE

Exact finite-chain condensation

An ideal linear molecule forms from 20 AB monomers, each 150 g mol⁻¹, with one water eliminated per intermolecular link and no cyclization. Find its molar mass.

  1. A connected linear chain of 20 units has 19 linking bonds.
  2. Starting monomer mass per mole of such chains is 20×150=3000 g mol⁻¹.
  3. Eliminated water mass is 19×18=342 g mol⁻¹.
  4. Subtract to get 2658 g mol⁻¹; subtracting 20 waters would ignore the end-group correction.

Answer 2658 g mol⁻¹.

ORIGINAL PRACTICE · 1

What monomer gives[–CF₂–CF₂–]ₙ?

Reveal answer & reasoning

Tetrafluoroethene, CF₂=CF₂.

Restore the vinyl double bond between the two backbone carbons; all four fluorines stay attached.

ORIGINAL PRACTICE · 2

Does ordinary PVC addition polymerization remove HCl from every repeat?

Reveal answer & reasoning

No.

Vinyl chloride retains its atoms in the addition repeat –CH₂–CH(Cl)–. HCl elimination is a different degradation or reaction process.

ORIGINAL PRACTICE · 3

Name the two monomer types required for a simple polyester condensation.

Reveal answer & reasoning

A diol and a dicarboxylic acid or suitable diacyl derivative.

Their two functional groups allow repeated ester-link formation. A hydroxy acid can provide both functions in one molecule.

ORIGINAL PRACTICE · 4

Why does the adipic-acid residue of nylon 6,6 have only four CH₂ groups between carbonyls?

Reveal answer & reasoning

Its two carbonyl carbons complete the six-carbon count.

HOOC–(CH₂)₄–COOH has six carbons total. Six CH₂ groups plus two carbonyls would be an eight-carbon diacid.

ORIGINAL PRACTICE · 5

Is polyurethane formation necessarily a condensation with water loss?

Reveal answer & reasoning

No.

Ideal diol–diisocyanate addition is step-growth but can form urethane links without small-molecule elimination.

ORIGINAL PRACTICE · 6

For Mn=84,000 g mol⁻¹ and polypropylene repeat 42.0 g mol⁻¹, estimate Xn while neglecting ends.

Reveal answer & reasoning

2000.

Divide 84,000 by 42.0. This is a number-average repeat count under the stated long-chain approximation.

ORIGINAL PRACTICE · 7

At p=0.990 in the ideal AB model, what is Xn?

Reveal answer & reasoning

100.

Xn=1/(1−0.990)=1/0.010. Ninety-nine percent functional-group reaction does not mean infinite chains.

ORIGINAL PRACTICE · 8

Can a nonuniform molecular population have Mw smaller than Mn?

Reveal answer & reasoning

Not under the stated positive-mass definitions.

Mw/Mn is at least one; equal values require a uniform molar mass. A smaller reported ratio signals inconsistent data or definitions.

ORIGINAL PRACTICE · 9

Does a block copolymer have to be 50:50 in A and B?

Reveal answer & reasoning

No.

Block describes sequence architecture, not an obligatory composition. Block lengths determine the incorporated proportion.

ORIGINAL PRACTICE · 10

Why can cross-linked rubber recover rather than flow permanently after modest stretching?

Reveal answer & reasoning

Cross-links constrain chain sliding while flexible segments change conformation.

Elastic recovery arises from network-chain conformational behaviour within its usable strain and temperature range, not unlimited reversible extension.

ORIGINAL PRACTICE · 11

Does Tg equal the crystalline melting point Tm?

Reveal answer & reasoning

No.

Tg concerns segmental mobility in amorphous regions; Tm concerns melting crystalline order. A semicrystalline polymer may show both.

ORIGINAL PRACTICE · 12

What remains unsaturated in the common 1,4-polyisoprene repeat?

Reveal answer & reasoning

One C=C bond per isoprene-derived repeat.

The repeat –CH₂–C(CH₃)=CH–CH₂– retains an internal double bond, allowing cis/trans microstructure and further chemistry.

KEEP THE BIG IDEAS

Quick revision

  • Repeat units must preserve correct backbone and substituent connectivity.
  • Growth mechanism and small-molecule elimination classify different aspects.
  • Polyester/polyamide repeats often combine two monomer residues.
  • Architecture, tacticity and processing influence properties beyond composition.
  • Mn and Mw weight a distribution differently; dispersity measures its spread.
  • High ideal step-growth chain length demands conversion extremely close to one.
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