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

Level 1 — Recognition

(Can you just see what the thing is?)

Recall Solution L1.1

The single test: does the monomer have a $C = C$ double bond, and is any small molecule thrown out?

(a) $C H_{2} = C H C l$ has a $C = C$ , nothing expelled → addition.
(b) diamine + diacid join by losing $H_{2} O$ → condensation.
(c) $C F_{2} = C F_{2}$ has a $C = C$ , nothing expelled → addition.
(d) amino acids link releasing $H_{2} O$ (peptide bonds) → condensation.

What we did & why: we only had to look for the double bond and ask "is anything lost?" — the two-question filter that separates the two whole families. See Condensation polymers — nylon, polyester for family (b) and (d).

Recall Solution L1.2

The trick to un-make a repeat unit: put the double bond back between the two backbone carbons.

(a) polyethene → ethene $C H_{2} = C H_{2}$
(b) PVC → chloroethene (vinyl chloride) $C H_{2} = C H C l$
(c) PTFE → tetrafluoroethene $C F_{2} = C F_{2}$
(d) polystyrene → styrene (phenylethene) $C H_{2} = C H (C_{6} H_{5})$

Why: each polymer's backbone is two carbons per repeat; restoring the $π$ bond between them rebuilds the alkene monomer.

Level 2 — Application

(Do the standard move cleanly.)

Recall Solution L2.1

Step 1 (what): identify the side group $X$ . Propene is $C H_{2} = C H (C H_{3})$ , so $X = C H_{3}$ . Step 2 (why): in addition polymerisation the $π$ bond opens; the two carbons that were double-bonded now bond to the neighbours on each side. Nothing is added or removed. $n C H_{2} = C H (C H_{3}) ⟶ - [C H_{2} - C H (C H_{3})]_{n} -$ Step 3 (check): backbone now has only single bonds; the $- []_{n} -$ open bonds show it continues forever; atom count per unit ( $C_{3} H_{6}$ ) equals the monomer ( $C_{3} H_{6}$ ). ✔

Recall Solution L2.2

What: insert the double bond between the two backbone carbons of one repeat unit. $- C H_{2} - C H (C N) - ⟹ C H_{2} = C H (C N)$ Why: un-polymerising is literally re-forming the alkene $π$ bond that opened during growth.

Monomer: acrylonitrile, $C H_{2} = C H (C N)$ .
Polymer: polyacrylonitrile (PAN) — the acrylic-fibre and carbon-fibre precursor (see Carbon fibre from PAN).

Recall Solution L2.3

Why this works: because addition polymers lose no atoms, the repeat-unit mass equals the monomer mass, and the whole chain is just $n$ copies. So $n = \frac{M _{polymer}}{M _{repeat unit}} = \frac{125 000}{62.5} = 2000.$ Check the repeat mass: $C_{2} H_{3} C l = 2 (12) + 3 (1) + 35.5 = 62.5$ . ✔ So $n = 2000$ units.

Level 3 — Analysis

(Take it apart; reason about why the property arises.)

Recall Solution L3.1

What differs: not the chemistry — the geometry.

LDPE chains are branched; branches act like stubby arms that stop chains lying close together.
HDPE chains are linear/unbranched; they lie side-by-side and pack tightly (more crystalline).

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

Why packing changes properties: closer packing means more chain sits inside a given volume → higher density. Closer packing also means the London (dispersion) forces between chains line up over longer stretches → stronger total attraction → more energy to pull chains apart → higher melting point and greater stiffness. Conclusion: same monomer, different branching → different packing → different density, stiffness, melting point. (More in HDPE vs LDPE — structure and density.)

Recall Solution L3.2

Bond argument (reactivity): every backbone carbon is wrapped in fluorine atoms bonded by very strong, short C–F bonds. Reagents cannot easily break these, and the fluorine "sheath" physically shields the carbon backbone. Result: chemically inert. Surface argument (non-stick): fluorinated surfaces have very weak intermolecular forces with other molecules — foods and sauces cannot grip the surface. Low surface attraction = low friction = non-stick. Why both matter: inertness keeps it from reacting with hot food/acids; low surface forces keep food from sticking. Two separate reasons, one material.

Level 4 — Synthesis

(Combine mechanism, structure and the whole family.)

Recall Solution L4.1

(a) Initiation — the weak O–O bond snaps to give two radicals (a dot = one unpaired electron): $R - O - O - R ⟶ 2 R - O ∙$ Why: a radical is electron-hungry and the loosely held $π$ electrons of $C = C$ are easy prey.

(b) Propagation — the radical grabs one carbon; one $π$ electron makes the new bond, the other is left as a radical on the far carbon, which then eats the next monomer: $R O ∙ + C H_{2} = C H_{2} ⟶ R O - C H_{2} - C H_{2} ∙$ $R O - C H_{2} - C H_{2} ∙ + C H_{2} = C H_{2} ⟶ R O - C H_{2} - C H_{2} - C H_{2} - C H_{2} ∙$ Why: the radical just walks to the new chain end and keeps going — a self-perpetuating chain-growth.

(c) Termination — two radical chain-ends meet and pair their electrons (combination), so both chains stop growing.

(d) Overall (ignoring chain ends): $n C H_{2} = C H_{2} ⟶ - [C H_{2} - C H_{2}]_{n} -$ Full mechanism reasoning: Free-radical mechanism.

Recall Solution L4.2

Choice: PMMA (acrylic / Perspex). Monomer: methyl methacrylate; repeat unit $- C H_{2} - C (C H_{3}) (C O O C H_{3}) -$ . Why PMMA: it is transparent and shatter-resistant, the two required properties — the classic glass substitute. Against alternatives:

PS (polystyrene) can be clear but is brittle — it shatters, failing the safety spec.
PTFE is opaque/white and waxy — fails transparency outright. Conclusion: only PMMA meets both the optical and the mechanical requirement.

Level 5 — Mastery

(Multi-step, exam-boss.)

Recall Solution L5.1

(a) Propene repeat unit $- C H_{2} - C H (C H_{3}) -$ is $C_{3} H_{6} = 3 (12) + 6 (1) = 42$ . (b) $n = \frac{210 000}{42} = 5000$ monomer units. (c) Same $n = 5000$ for PVC, repeat mass $62.5$ : $M_{PVC} = 5000 \times 62.5 = 312 500.$ (d) Why: addition polymerisation loses no atoms — the $π$ bond merely opens to form single bonds. So each repeat unit contains exactly the same atoms as its monomer, hence the same mass. (Contrast: condensation would lose $H_{2} O$ each join, so repeat-unit mass would be less than the sum of monomer masses.)

Recall Solution L5.2

Step 1 (what) — mass %: $C = 38.4%$ , $H = 4.8%$ , so $C l = 100 - 38.4 - 4.8 = 56.8%$ . Step 2 (why moles) — divide by atomic masses to turn mass ratios into atom ratios: $C : \frac{38.4}{12} = 3.2, H : \frac{4.8}{1} = 4.8, C l : \frac{56.8}{35.5} = 1.6.$ Step 3 — divide by the smallest ( $1.6$ ): $C : \frac{3.2}{1.6} = 2, H : \frac{4.8}{1.6} = 3, C l : \frac{1.6}{1.6} = 1.$ Empirical formula $= C_{2} H_{3} C l$ . (b) That is the repeat unit $- C H_{2} - C H C l -$ → PVC; monomer chloroethene $C H_{2} = C H C l$ . (c) Repeat mass $= 2 (12) + 3 (1) + 35.5 = 62.5$ . ✔ Why this route: percentage → moles → simplest ratio is the universal "empirical formula from combustion/analysis" method; matching it to a $C_{2} H_{3} X$ vinyl repeat unit pins the polymer.

Recall Solution L5.3

(a) $X = C N$ ; open the double bond: $n C H_{2} = C H (C N) ⟶ - [C H_{2} - C H (C N)]_{n} -$ (b) Warm, wool-like acrylic fibre (sweaters, blankets). (c) Heating PAN fibre drives off other atoms and leaves an ordered, aligned carbon backbone, so PAN is the precursor from which strong carbon fibre is made — full story in Carbon fibre from PAN.

Recall One-line self-test

Filter for addition vs condensation? ::: Monomer has $C = C$ and nothing is expelled → addition; a small molecule ( $H_{2} O$ / $H C l$ ) leaves → condensation. Degree of polymerisation formula? ::: $n = M_{polymer} / M_{repeat unit}$ , valid because addition loses no atoms.