Addition polymers — polyethene, PVC, PTFE (Teflon), polypropylene, polystyrene, PMMA, polyacrylonitrile: Worked examples

Link back to the parent: the Hinglish parent and prerequisites Alkenes and the C=C double bond, Free-radical mechanism, Intermolecular forces and polymer properties, HDPE vs LDPE — structure and density, Condensation polymers — nylon, polyester, Carbon fibre from PAN.

Three symbols we will reuse — earn them now

Before any working, define every letter this page leans on, so nothing appears "out of nowhere".

The scenario matrix

Every arrow-question in an exam lands in exactly one of these cells. We will hit all of them.

#	Case class	The question type	Worked in
A	Forward	monomer → polymer (draw/write)	Ex 1
B	Reverse	polymer repeat unit → monomer	Ex 2
C	Multi-substituent monomer	monomer with two side groups (PMMA)	Ex 3
D	Degenerate side group	$X = H$ (symmetric monomer, PE/PTFE) — no stereocentre	Ex 4
E	Numeric / mole reasoning	find $n$ from molar masses	Ex 5
F	Reverse numeric	find molar mass from $n$ ; find $%$ by mass	Ex 6
G	Classification twist	addition vs condensation — spot the impostor	Ex 7
H	Word / real-world	choose the right plastic for a job	Ex 8
I	Exam trap + tacticity	correct repeat unit; isotactic/syndiotactic/atactic	Ex 9

Case A — Forward: monomer → polymer

Worked example Example 1 (cell A) — Polymerise chloroethene

Statement: Write the equation for polymerising vinyl chloride $C H_{2} = C H C l$ .

Forecast: Guess before reading: will the product still contain a $C = C$ ? How many bonds does the "Cl carbon" end up with?

Identify the double bond and the side group. The monomer is $C H_{2} = C H C l$ ; here the side group is $X = Cl$ (using the placeholder $X$ we just defined). Why this step? Every addition monomer is a vinyl type $C H_{2} = C H X$ . Naming $X$ tells us what dangles off the finished chain — see the red group in the figure below.
Open the $π$ bond. The double bond becomes a single bond; each carbon frees one electron to bond to a neighbouring unit. Why this step? The whole trick of addition polymerisation (Free-radical mechanism) is that the loosely-held $π$ electrons of the $C=C$ reach out to grab the next monomer. Look at how the double line becomes a single line in the figure.
Write the repeat unit with open bonds and subscript $n$ . $n C H_{2} = C H C l ⟶ - [C H_{2} - C H C l]_{n} -$ Why this step? The brackets + $n$ + dangling bonds say "this pattern repeats thousands of times" — it is a macromolecule, not a small molecule.

Verify: Count atoms in one repeat unit: $2 C, 3 H, 1 Cl$ — identical to the monomer. Nothing expelled ⇒ genuine addition. ✔

Figure — Addition polymers — polyethene, PVC, PTFE (Teflon), polypropylene, polystyrene, PMMA, polyacrylonitrile

Case B — Reverse: polymer → monomer

Worked example Example 2 (cell B) — Recover the PTFE monomer

Statement: A polymer has repeat unit $- C F_{2} - C F_{2} -$ . Find its monomer.

Forecast: Guess: where does the double bond go back — between the two carbons shown, or somewhere else?

Locate the two backbone carbons of one repeat unit. They are the two $C F_{2}$ groups. Why this step? Polymerisation joined monomers by opening a $C = C$ inside each monomer, so to reverse it we re-form that bond between the very same two carbons.
Re-draw the double bond between them and cap the open ends. $C F_{2} = C F_{2} (tetrafluoroethene)$ Why this step? The dangling bonds in the polymer were single bonds to neighbours; giving each carbon its "half" back re-creates the one $C = C$ .

Verify: Monomer has $2 C, 4 F$ ; repeat unit has $2 C, 4 F$ . Exactly one $C = C$ restored, no leftover fragments. ✔

Case C — Monomer with two side groups

Worked example Example 3 (cell C) — PMMA has

two substituents Statement: Methyl methacrylate is $C H_{2} = C (C H_{3}) (C O O C H_{3})$ . Write its repeat unit.

Forecast: Guess: one backbone carbon carries two groups here, not one — will the $C H_{2}$ carbon change?

Spot the asymmetry. Left carbon $= C H_{2}$ (two H's); right carbon carries both $C H_{3}$ and $C O O C H_{3}$ . Why this step? The general form $C H_{2} = C H X$ had one side group; PMMA is $C H_{2} = C X Y$ with two blanks filled ( $X = CH_{3}$ , $Y = COOCH_{3}$ ). The $C H_{2}$ end is unchanged; all the "personality" sits on the crowded carbon (red in figure).
Open the double bond, keep both groups on the same carbon. $n C H_{2} = C (C H_{3}) (C O O C H_{3}) ⟶ - [C H_{2} - C (C H_{3}) (C O O C H_{3})]_{n} -$ Why this step? Nothing is lost; both bulky groups stay put, which is exactly why PMMA is rigid and transparent (Intermolecular forces and polymer properties).

Verify: Repeat-unit atoms $=$ monomer atoms; backbone now all single bonds. ✔

Case D — Degenerate (symmetric) monomer

Worked example Example 4 (cell D) — When

$X = H$ : polyethene Statement: Polymerise ethene $C H_{2} = C H_{2}$ . Why is there no "side-group carbon" here?

Forecast: Guess: both carbons are identical — does the repeat unit simplify?

Recognise the degenerate case. Here $X = H$ , so both carbons are $C H_{2}$ — the monomer is symmetric. Why this step? This is the zero/degenerate input of our matrix: the two ends are indistinguishable, so no carbon becomes a special "substituted" carbon.
Open the bond; both units are $C H_{2}$ . $n C H_{2} = C H_{2} ⟶ - [C H_{2} - C H_{2}]_{n} -$ Why this step? Same rule as always, but the product looks the simplest possible — a plain hydrocarbon backbone.
Note the geometry branch. With no bulky group, chains can be linear (HDPE) or branched (LDPE) — see HDPE vs LDPE — structure and density. Symmetry is why PE has this density split while bulkier monomers do not, and why PE has no tacticity (there is no side group to point different ways — see Ex 9).

Verify: $2 C, 4 H$ in, $2 C, 4 H$ out. ✔

Case E — Numeric: find $n$

Case F — Reverse numeric & percentage by mass

Worked example Example 6 (cell F) — Molar mass from

$n$ , and $%$ chlorine Statement: (a) A PVC chain has $n = 1600$ ; find $M_{polymer}$ . (b) What percentage of PVC's mass is chlorine?

Forecast: Guess: is chlorine more or less than half the mass, given $C l = 35.5$ is heavy?

Multiply for total mass. Using $M_{unit} = 62.5 g mol^{- 1}$ from Ex 5: $M_{polymer} = n M_{unit} = 1600 \times 62.5 = 100000 g mol^{- 1}$ Why this step? Reverse of Ex 5 — bricks $\times$ brick mass $=$ total.
Chlorine fraction in one repeat unit. $% Cl = \frac{35.5}{62.5} \times 100 = 56.8%$ Why this step? $%$ by mass compares one component's mass to the whole unit's mass; it is the same for one brick as for the whole chain, so we can use one unit.

Verify: (a) $100000/62.5 = 1600$ ✔. (b) $35.5/62.5 = 0.568$ , and $0.568 \times 100 = 56.8$ ✔. Chlorine is indeed over half — matches the heavy $C l$ atom. ✔

Case G — Classification impostor

Case H — Word / real-world problem

Worked example Example 8 (cell H) — Pick the plastic

Statement: You need a coating for a frying pan: it must be non-stick, chemically inert, and survive heat. Which addition polymer, and why?

Forecast: Guess: which side group $X$ makes bonds that almost nothing can break?

Match the property to the side group. Non-stick + inert + heat-stable points to fluorine everywhere: PTFE (Teflon), monomer $C F_{2} = C F_{2}$ . Why this step? The C–F bond is short and very strong, and fluorine's tight electron cloud gives an extremely low-friction, unreactive surface (Intermolecular forces and polymer properties).
Rule out others. PE melts too easily; PVC can degrade/release HCl on heating; PS is brittle. None matches all three needs. Why this step? Choosing a plastic = matching side-group chemistry to the job — the core skill of this topic.

Verify: PTFE's inertness and non-stick nature are exactly why real frying pans use it. ✔ (Related: PAN → Carbon fibre from PAN shows a different "job-to-polymer" match.)

Case I — Exam trap: the correct repeat unit, and tacticity

Worked example Example 9 (cell I) — Draw the repeat unit

properly, then classify its tacticity Statement: A student draws polystyrene as $C H_{2} = C H (C_{6} H_{5})$ with an $n$ outside. (a) List the mistakes and give the correct repeat unit. (b) Explain what tacticity means for this polymer.

Forecast: Guess: how many separate errors can hide in one wrong structure — and once fixed, does the side group $C_{6} H_{5}$ have a "direction"?

Find the leftover double bond. The backbone must be all single bonds after polymerising. Why this step? The $π$ bond opened to form the chain — see Alkenes and the C=C double bond. A repeat unit still showing $C = C$ is the single most common exam error.
Add the open continuation bonds and brackets. The unit must show dangling bonds so it links to neighbours. Why this step? Without them the drawing reads as a small molecule, not a macromolecule.
Present the correct repeat unit (this is a structure, not an arrow-transformation): $- [C H_{2} - C H (C_{6} H_{5})]_{n} -$ Why this step? Single-bond backbone, open bonds, subscript $n$ — the three marks examiners look for. Note there is no " $n C H_{2} = \dots \to$ " here; the question asked only for the repeat unit itself.
Classify the tacticity. In an asymmetric monomer like styrene, each side group ( $C_{6} H_{5}$ ) can point to the same side of the backbone or alternate:
- isotactic — all side groups on the same side (regular → packs well, more crystalline);
- syndiotactic — side groups alternate side-to-side (also regular, ordered);
- atactic — side groups point randomly (irregular → cannot pack, amorphous; ordinary polystyrene is atactic). Why this step? Tacticity only exists when there is a side group with two possible sides — so PE (Ex 4, $X = H$ ) has none, but PS, PP and PMMA all do. It changes crystallinity and hence stiffness/melting point (Intermolecular forces and polymer properties).

Verify: Backbone single bonds ✔, open bonds ✔, subscript $n$ ✔, atom count preserved ( $8 C, 8 H$ per unit) ✔; three tacticity classes named ✔.

Recall Quick self-test across the matrix

Give the monomer of $- C H_{2} - C H (C N) -$ ::: Acrylonitrile, $C H_{2} = C H (C N)$ (cell B, PAN). Repeat-unit mass of PP $- C H_{2} - C H (C H_{3}) -$ ? ::: $3 (12) + 6 (1) = 42 g mol^{- 1}$ (cell E). Number of monomers if PP chain has $M = 84000 g mol^{- 1}$ ? ::: $84000/42 = 2000$ (cell E). Why does polyester not belong here? ::: It is condensation — loses $H_{2} O$ , no $C = C$ (cell G). Which tacticity has all side groups on the same side? ::: Isotactic (cell I). Why has polyethene no tacticity? ::: Its side group is just H, with no distinct "side" to point to (cell D).